THE CALCULUS FOR BEGINNERS CAMBRIDGE UNIVERSITY PRESS ilonUon FETTER LANE, E.G. C. F. CLAY, MANAGER Binburgb: io, PRINCES STREET Berlin: A. ASHER AND CO. ILfipMs: F. A. BROCKHAUS fhto Boris: G. P. PUTNAM'S SONS Bombay nnt) Calcutta: MACMILLAN AND Co., LTD. Toronto: J. M. DENT AND SONS, LTD. THE MARUZEN-KABUSHIKI-KAISHA All rights reserved THE CALCULUS FOR BEGINNERS v J/W^MERCER, M.A. Head of the Mathematical Department, Royal Naval College, Dartmouth Author of Trigonometry for beginners Cambridge : at the University Press 1914 First Edition, 1910. Reprinted, 1912, 1914. PREFACE IN writing this short course, I have been guided by my conviction that it is much more important for the be- ginner to understand clearly what the processes of the Calculus mean, and what it can do for him, than to acquire facility in performing its operations or a wide acquaintance with them. I had much rather that a boy, confronted with a problem, should, after analysing it, be able to say " If I could differentiate (or integrate) this function of x, I could solve the problem " than that he should be able to perform the operation without seeing its bearing on the problem. The book is intended primarily for those who are or will be interested in the applications of the Calculus to Physics and Engineering. I believe that boys, especially on the Science side of a School, could profitably begin the study of the subject at a much earlier age than is usual. I believe, too, that even those who are to be Mathematicians would do well to go through some such course as is here indicated before pro- ceeding to the more formal treatises, for comparatively young students gain by approaching a subject inductively before studying it deductively. Even if they are possessed of mathematical ability, it is good that they should have a thorough comprehension of the goal in particular cases VI PREFACE before undertaking general proofs, and should be reminded by a preliminary course depending on more laborious methods, of the advantage of replacing " mere counting " by mathematical processes. As I have already indicated, no great prominence is given to methods of differentiating and integrating complicated expressions, in fact, x n is the only function whose differential coefficient is required in the first 250 pages. Later, sin a;, cosic, etc., e? and log a; are dealt with, and the Engineer needs little beyond these. Before reaching " ~ ," I have devoted many pages some may think too many to the ideas of rate of change and of the limiting value of the ratio of two continually diminish- ing quantities. Unless the ground is cleared in this way, a boy is apt to begin and continue indefinitely his study of the subject in a state of mental fog. I do not believe that anything is gained by hurrying over the early stages. On the contrary, I believe that the ultimate result of reducing the time spent on them is a very great loss of power. When cc n has been differentiated, the result is applied to problems of maxima and minima, geometrical properties of curves, approximations, etc. Then the inverse problem "Given -~, find y" is discussed, and this leads up to the CLOD Integral Calculus and its application to Areas, Volumes, Work, etc. These should not be looked upon as so many detached problems each requiring its own particular rule, but if the meaning of integration is properly understood, they will all be seen to depend on one general principle. This will explain why I have given so much space to the discussion of a few typical problems ab initio. After this, rules are investigated for differentiating PREFACE Vll sin a;, cosx, tana?, etc., e*, \og e x, products, quotients, etc., and more problems are dealt with of the same nature as before but requiring a knowledge of these rules. Chapters XII. XIV. form a sort of appendix with short notices of the application of the Calculus to the Solution of Equations, Methods of Integration and Polar Co-ordinates. No great amount of previous mathematical knowledge is assumed. A boy is supposed to know his Elementary Algebra and Trigonometry and to have some slight acquaint- ance with the co-ordinate geometry of the straight line. He should be able to write down the equation of a line through a given point with a given gradient and should know the relation between the gradients of perpendicular lines. He is also expected to have had some practice in drawing graphs from their equations and to know what these graphs mean. Many of the illustrations given are arithmetical, algebraical and geometrical, but in addition to these many will be found drawn from the scientific knowledge usually at the command of boys of this age. Several of the examples are intended to lead up to the ideas afterwards presented in the text, and in the more difficult ones, especially in some which will be found in the miscellaneous sets, the boy of mathematical ability will find plenty of scope for his skill in the manipulation of symbols. It would probably be well to omit the section on Points of Inflexion (pp. 127 135) on a first reading It is found there, because there did not seem to be a more convenient place for it. I take this opportunity of thanking Professor Hobson of Cambridge for some notes on the earlier chapters, which he sent to me after seeing the manuscript, but this must not be taken to mean that anything I have written has his sanction. Vlll PREFACE I am very grateful also to Mr A. W. Siddons of Harrow, for some criticisms and suggestions which he made after wading through most of the original manuscript. The subject has been taught here during the last few years to boys of 16 on the lines of this book and I am much indebted to my colleagues who have helped in various ways during the passage of the book through the press. Thanks are also due to Professor Lamb of Manchester for permission to reprint two tables from his "Infinitesimal Calculus," to Professor Ewing, Director of Naval Education, for permission to use questions set to Naval Officers, and to the Controller of His Majesty's Stationery Office for permission to include some questions set at Examinations by the Board of Education and the Civil Service Com- missioners. J. W. MERCER. EOYAL NAVAL COLLEGE, DARTMOUTH, June 1910. CONTENTS CHAPTER I FUNDAMENTAL NOTIONS SECTIONS PAGES 1 37 Uniform speed, average speed, speed at an instant . 1 46 Rate of increase. Function. Uniform gradient, average gradient, gradient at a point. Tangent at a point of a curve. Rate of increase as gradient. Test of approximate equality. Ratio of continually diminishing quantities. Ess. I. VII. DIFFERENTIATION FROM FIRST PRINCIPLES 38 71 Notation. Meaning of A, etc 47 74 Meaning of -=- . Its connection with -r- . dt A* dy f . -f- as a rate of increase. dx (%/U -f as ratio of time rates of increase of x and y. dx y -f- as gradient (HP X CONTENTS SECTIONS PAGES Formal definition of -/- and summary of process of cLx finding it. Speed and acceleration as gradients. Differentiation of # 3 . Geometrical illustration. Sign of g. Exs. VIII. XIII. CHAPTER III DIFFERENTIATION OF X n 7295 Notation 75101 Differentation of x n (i) with, (ii) without Binomial Theorem. Differentiation of kx n , a? n +c. Graphical and kinematical illustrations. Use of -j in drawing a curve from its equation. Applications to Geometry. Some properties of the parabola. The functional notation. Higher differential coefficients. Derived curves. Relation between the signs of ~ , ~^ and tho shape of a curve. Exs. XIV. XXII. Miscellaneous Examples on Chapter III. . 102105 CHAPTER IV MAXIMA AND MINIMA 96117 Turning points of a curve .... 106135 Tests for discriminating between maximum and minimum. Points of inflexion. Exs. XXIII. -XXV. CONTENTS XI CHAPTER V SMALL ERRORS AND APPROXIMATIONS SECTIONS PAGES 118127 The relation Ay = -j- . tuc approximately . . 136144 Relative error. The relation f(x + h)=f(x) + hf'(x} approxi- mately. Exs. XXVI. XXIX. CHAPTER VI THE INVERSE OPERATION 128133 Given ^, find y ...... 145152 CLX - = gives a family of curves. / Exs. XXX. XXXIV. Miscellaneous Examples on Chapters T. VI. . 153 159 CHAPTER VII INTEGRAL CALCULUS. AREAS OF PLANE CURVES. MEAN ORDINATE 134 167 Areas by summation 160 204 Meaning of integration. n Meaning of / f(x) . dx. J a, Definite and indefinite integrals. Approximate rules for Area : Trapezoidal and Simpson's rules. Sign of Area. Approximation to value of definite integral. Xll CONTENTS SECTIONS PAGES Area represented by difference of ordinates of integral curve. The relation f(x + A) =/(*) + hf(x) + } ^f"(x) A approximately. Mean ordinate. Exs. XXXV. XLVII. FURTHER APPLICATIONS OF THE INTEGRAL CALCULUS 168 202 Distance from speed-time formula . . . 205 253 Speed-time and space- time graphs. Work done in stretching string. Aspects of integration. Resultant thrust on immersed area. Work done by expanding gas. Volume of solids of revolution (frustum of cone, sphere). Moment of inertia of rectangle. Moment of inertia of circular disc. Radius of gyration. Theorems of perpendicular and parallel axes. Centre of gravity (parabola, paraboloid of revolution, quadrant of circle). Centre of pressure. Mean values. Mean speed with respect to (i) time, (ii) distance. Exs. XLVIII. LVI. Miscellaneous Examples on Chapters VII. and VIII. . 254260 CHAPTER IX DIFFERENTIATION OF TRIGONOMETRICAL RATIOS 203-213 Exs. LVII. LIX 261277 CONTENTS Xlll CHAPTER X HARDER DIFFERENTIATION SECTIONS PAGES 214 244 Differentiation of a product of two or more functions 278310 Differentiation of a quotient. Differentiation of a function of a function. -r- when y and x are given as functions of a third variable. Integration sometimes simplified by the use of a third variable. Differentiation of inverse functions. Differentiation of inverse trigonometrical func- tions. Differentiation of implicit functions. Exs. LX. LXXIV. CHAPTER XI DIFFERENTIATION OF n x 245267 Ify=* ^- = N.y 311344 Approximate values of N for different values of n. Definition of e. Relation between the logarithms of the same number to different bases. Graphic treatment. Differentiation of \og e x, log a #. Approximation to value of e. Hyperbolic functions. Logarithmic differentiation. Compound Interest Law. / j\n e as Lt ( 1 + - ) . -*oo \ */ Exs. LXXV. LXXXIII. Miscellaneous Examples on Chapters IX. XI. 345361 xiv CONTENTS CHAPTER XII APPROXIMATE SOLUTION OF EQUATIONS SECTIONS PAGES 268277 Graphical solution 362374 Successive approximation by calculus. Exs. LXXXIV. CHAPTER XIII METHODS OP INTEGRATION 278282 List of standard forms 375384 r j r The formula I $ (u) . -^ . dx = I <f> (u) , du. Integration by substitution. 1 Integration by parts. Exs. LXXXV. LXXXVIL Miscellaneous Exercises on Chapter XIII. . 385 387 CHAPTER XIV POLAR CO-ORDINATES 283289 Meaning of polar co-ordinates .... 388394 Angle between tangent to a curve and the radius vector. Areas. Exs. LXXXVIII., LXXXIX. ANSWERS 395432 Table of exponential and hyperbolic functions of numbers from to 2'5 at intervals of '1. Table of logarithms to base e 433 INDEX . . 435440 CHAPTER I FUNDAMENTAL NOTIONS Speed 1. Uniform, speed. To measure the speed of a moving body, we must know the distance it travels in some definite time. The speed is said to be uniform if equal distances are travelled in equal times, however small those equal times may be. Thus if a train has a uniform speed of 60 miles an hour, it will travel 30 miles in every half hour, 10 miles in every 10 mins., 1 mile in every min., 88 ft. in every sec., 8 '8 ft. in every ! sec., and so on, and we could state the speed as 60 miles/hour, or 1 mile/minute, or 88 ft. /sec., etc. We can deduce any one of these expressions for the speed from any one of the relations between distance and time : e.g. 60 miles an hour is 60 x 5280 ft. in 60 x 60 sees. 60 x 5280 ,, or ~n R?T ft - m 1 sec> 60 x 60 or 88 ft. /sec., 8 *8 8*8 ft. in -1 sec. is-^p ft. in 1 sec. or 88 ft./sec., and generally if a body moving with uniform speed travels s ft. o in an interval of t sees., - will have the same value whatever be I M. C. I 2 CALCULUS FOR BEGINNERS [CH. I the duration of the interval and at whatever stage of the journey c it be taken, and in this case - ft./sec. is said to be the speed of the body at every instant. 2. Average speed. Suppose a train travels 60 miles in a certain hour, but that we do not know that its speed has been uniform during the hour, we can say that its average speed during the hour is 88 ft./sec., meaning that if it kept up a uniform speed of 88 ft. /sec. for an hour, i.e. travelled 88 feet in every separate second, the distance covered in the hour would be 60 miles, the same as the distance actually covered. Actually it might cover 70 ft. in one second, 90 ft. in another, and so on, provided that the total distance was 60 miles. If a body travels 3 ft. in j sec., we say that its average speed during this quarter of a second is 12 ft./sec., meaning that if a body had a uniform speed of 12 ft./sec., it would in sec. travel 3 ft, the same as the distance actually travelled. The distance travelled in any other quarter of a second would not as a rule be 3 feet, unless the motion were uniform ; suppose for example that in the next quarter of a second the distance travelled were 5 feet, then the average speed during this next quarter of a second would be 20 ft./sec. and during the complete half second 16 ft./sec. Generally, if a body travels s ft. in t sees., its average speed during this interval of t sees, is - ft./sec., and as a rule the value of - will depend on the duration of the t interval and on the stage of the journey at which it is taken. 1-3] SPEED AT AN INSTANT -3 EXERCISES. I. 1. A train leaves Bristol at 4.0 p.m. and arrives at Exeter, 75J miles distant, at 5.55 p.m. Find its average speed (i) in miles per hour, (ii) in feet per secohd. 2. A train travels 40 miles in the first hour, 35 in the second, 45 in the third, 50 in the fourth and 40 in the fifth. Find in miles per hour its average speed for the whole journey. 3. A stone falls 1 ft. in - sec. Find its average speed during this interval (i) in miles per hour, (ii) in yards per minute. 4. A body travels -0035 ins. in -02 of a second. Find in ft./sec. its average speed during this interval. 6. A body travels -00063 ft. in -001 of a second. Find in miles/hour its average speed during this interval. 3. Speed at an instant. Suppose we are studying the motion of some moving body and that we are able to measure the dis- tances it travels in certain small intervals of time after passing some fixed mark. In our first experiment we will suppose that we get the distance travelled in one second. The next experiment we will suppose to be made with improved apparatus which will enable us to find the distance travelled in half a second, and so on, each experiment being a more refined one than its pre- decessor. Tabulate the results of the experiments as follows : Time in sees. Distance in ins. (1) 1 48 1 (2) 2 24 1 (3) 4 12 1 / A\ G * 8 (5) 1 16 3 1-2 4 CALCULUS FOR BEGINNERS [CH. I In this case we should probably conclude that the body was moving with a uniform speed of 48 ins./sec. If however the results were (1) 1 48 ( 2 ) 5 21 (3) I 9-75 (4) I 4-69 (5) 1 2-30 it is obvious that the speed is not uniform. What shall we say then is the speed of the body as it passes the fixed point? The average speeds during the intervals used in the different experiments are as shewn in the following table : Interval in sees. Average speed in ins.jsec, (1) 48 (2) \ 42 (3) \ 39 (4) I 37-52 (5) 1 36-80 These results are all different. Now suppose we took the result of the first experiment and assumed that the body moved for the whole second with a uniform speed of 48 ins./sec. and from this calculated how far it would go in ^ sec., we should get 3 ins., which is by no means the same as 2-30 ins., the actual distance travelled. If, however, we used the result of the second experiment, and assumed that the body moved for half a second 3] SPEED AT AN INSTANT 5 with a uniform speed of 42 ins./sec., and calculated from this the distance it would go in y^ sec., we should get 2-625 ins., and in the same way if we assumed uniform speeds of 39 and 37'52 ins./sec., we should get 2-44 and 2-34 ins. respectively. These distances 3, 2'625, 2-44, 2-34 ins. approximate more and more to the true distance 2 -3 ins. If we tried to deduce the distance travelled in ^ sec. we should get a better approxi- Oa mation by using the average speed 36-8 ins. /sec. than by using any of the others, but the result 1*15 ins. would not be correct. Still, if the apparatus at our disposal did not permit us to measure the distance travelled in any shorter period of time than y^ sec., we should content ourselves by saying that the speed of the body as it passed the given point was approximately 36-8 ins./sec. If we could improve our apparatus and measure the distance travelled in ^TWC sec., say -361 ins., we should say that the speed of the body as it passed the given point was still more nearly 36 ! ins./sec. If from this we calculated the distance travelled in any shorter interval, say sec., our result -036 1 ins. would be nearer the truth than a result obtained by using any of the other average speeds. But however refined our measurements, we could never by a calculation of this kind obtain the speed of the body as it passed the given point. We calculate each time the average speed of the body during a short interval following the instant in question and the effect of improving our apparatus is that we are able to shorten the interval for which the average speed is calculated. We cannot say that the body moves for any interval of time, however short, with uniform speed, but the shorter the interval of time for which we make this assumption the smaller is the error made in the calculated distance for a still shorter interval. CALCULUS FOR BEGINNERS [CH. I EXERCISES. U. 1. Three men A, B, C begin to observe the motion of a train just as it reaches a quarter-mile post. A finds that in the next second it goes 44 ft., B that it takes 4| sees, for the whole train (210 ft. long) to pass him, C that it takes 26 sees, to reach the next quarter-mile post. Find in ft./sec. what each would obtain for the average speed of the train. What is the best available result for the speed of the train as it passes the first post ? If each assumes that the speed he obtains is uniform over the time cousidered, and uses it to calculate the distance travelled by the train in the next half second after leaving the post, what would their results be, and which would be nearest to the truth? [Nearest tenth of a foot.] 2. A train leaves the terminus A and passes through the stations B, C, D, etc. at the times indicated in the table. The second column gives distances from A. [80 chains = 1 mile.] A D C D E F G H I Find the average speed between the stations in chains (nearest chain) per minute. What would you guess to be the approximate speed as the train passed through (i) F, (ii) H ? n. ch. 00 Time oj passing 5.20 1 43 5.24 2 41 5.26 3 4 4 81J 04 5.27J 5.284 G 8 34* 5.34 8 12 5.35 Fig- 1. 4] SPEED AT AN INSTANT 7 3. A point P starts from A (Fig. 1) and moves round a circle of radius 100 cms. with uniform speed, making a complete revolution in 1 min. PM is perpendicular to OA. What is OM when L AOP is (i) 60, (ii) 55, (iii) 65, (iv) 57, (v) 63, (vi) 59, (vii) 61? What is the average speed (cms. /sec.) of M as L AOP increases (i) from 55 to 60, (ii) from 60 to 65, (iii) from 57 to 60, (iv) from 60 to 63, (v) from 59 to 60, (vi) from 60 to 61? What is approximately the speed of M when / AOP = 60? 4. A point P starts from A and moves round a circle of radius 100 cms. with uniform speed, making a complete revolution in 1 min.' OP meets the tangent at A in T. What is AT when L AOP is (i) 60, (ii) 55, (iii) 65, (iv) 57, (v) 63, (vi) 59, (vii) 61? What is the average speed (cms./sec.) of T as L AOP increases (i) from 55 to 60, (ii) from 60 to 65, (iii) from 57 to 60, (iv) from 60 to 63, (v) from 59 to 60, (vi) from 60 to 61 ? What is approximately the speed of T when L AOP = 60? 4. Now suppose the distance travelled in a given time is not determined by experiment but given by an algebraical formula, say s = 5 + 3t + 2t3, s ft. being the distance from some fixed point at the end of t sees, measured from some fixed instant, and suppose we want the speed at the end of 4 sees. If in the formula we put t = 4, we get s = 145 ; i.e. the distance from the fixed point at the end of 4 sees, from the fixed instant is 145 ft. Similarly the distance from the fixed point at the end of 5 sees, from the fixed instant is 270 ft. Reckoning from the end of the fourth second, i.e. from the instant at which we are to find the speed, we see that the distance travelled in the next second is 125 ft. and the average speed during this second is 125 ft. /sec. Now calculate the distance travelled in 4^- sees, from the beginning. It is 200f ft., and the distance described in the -~ sec. after the end of the fourth second is 55f ft. .'. the average speed during this ^ sec. is 55^-=-- or 111^ ft. /sec. Similarly find the distances and average speeds for -2, !, '01, 8 CALCULUS FOR BEGINNERS [CH. I 001, -0001 of a second after the end of the fourth second. The results should be as tabulated. Time in sees. Distance in ft. Average speed in ft. I sec. 1 1-25 125 5 55-75 111-5 2 20-776 103-88 1 10-142 10142 01 0-992402 99-2402 001 0-099024002 99-024002 0001 0-009900240002 99-00240002 This table could be continued to any extent, and it appears that the average speed as we diminish the interval of time over which it is calculated is always greater than 99 ft./sec., ap- proaches more and more to 99 ft./sec., and apparently could be made as near to 99 ft./sec. as we like if we extend the table far enough, i.e. if we make the interval of time small enough. To make quite sure of this let us consider th'e distance travelled in (4 + h) sees., where h can eventually be as small a fraction as we please. The distance is 5 + 3 (4 + h) + 2 (4 + h) 3 ft. or 145 + 99A + 24& 2 + 2A ft, i.e. the distance travelled in the interval h sees, after the end of the fourth second is (99A + 24A 2 + 2A 3 ) ft. .'. the average speed during this interval is 99A+24/i 2 +2A 3 r or (99 + 24A + 2A 2 ) ft. /sec. [Notice that this result includes all the results in the table, e.g. if we put h= -1 we get 99 + 2-4 + -02 or 101-42.] As h is made smaller and smaller (247i + 2/i 2 ) becomes smaller and smaller and can be made as small as we please if h be made small enough. In other words the average speed during the interval h can be made as near to 99 ft./sec. as we please if we take the interval short enough. 4] SPEED AT AN INSTANT 9 This, so to speak, ideal speed of 99 ft./sec. is called the speed of the body at the end of 4 sees. As this is an extremely important idea, it may be well to summarise the steps by which we arrived at our result. We found the distance travelled by the body in a short interval immediately following the end of the fourth second and calculated the average speed during this interval. We then did the same thing with a shorter interval, then with a shorter, and so on, and found that our average speeds continually approached 99 ft./sec. and that if we took our interval short enough, we could make the average speed come as near to 99 ft./sec. as we wished. This limiting value of the average speed which can be approached as nearly as we please by taking the interval short enough, is called the speed of the body at the end of 4 sees. EXERCISES. III. 1. If = 5 + 3f + 2t 3 , find the average speed during the interval A sees, which immediately precedes the end of the fourth second. Shew that it is less than 99 ft./sec., continually approaches 99 ft./sec. as h is made smaller and can be brought as near to 99 ft./sec. as we please if we make h small enough. Also find the average speed during the interval of h seconds which extends from ^ sees, before to - sees, after the end of the fourth second, and -j a shew that this average speed also can be brought as near to 99 ft./sec. as we please if we make h small enough. 2. If s = 5t + 3t 2 find the distances travelled in 3, 4, 3'5, 3-2, 3-1, 3 - 01, 3'001 sees, from the beginning. Hence find the average speeds during the intervals 1, -5, -2, -1, -01, -001 sees, immediately following the end of the third second. Also find the distance travelled in (3 + h) sees, and the average speed during the interval h sees, after the end of the third second. What is the speed at the end of the third second ? 10 CALCULUS FOR BEGINNERS [CH. I Find also the average speed during (i) the interval of h seconds immediately preceding the end of the third second, (ii) during the interval of h seconds which extends from the end of ( 3 - - j sees, to the end of 3 + o ) secs -> (iii) during the interval between the end of (3-j) sees, and ( the end of (3 + ) sees., and shew in each case that by making the interval short enough we can bring the average speed as near as we please to 23 ft./sec. 3. In Qu. 2 find (i) the average speed during the fourth second, (ii) the speeds at the end of 3, 4, 3 sees., (iii) the mean (half the sum) of the speeds at the end of 3 and 4 sees. 4. If s S + 9t 3 , find (i) the average speed during the fourth second, (ii) the speed at the end of 3J sees., (iii) the mean of the speeds at the end of 3 and 4 sees. 6. If s = t*-'6t + 5, find the distance gone in 4 sees, and in (4 + /<) sees. Hence get the speed at the end of the fourth second. 5. From Exs. 1 and 2 we see that the speed at a given instant may be deduced from the average speed during a short interval, (i) immediately following, or (ii) immediately preceding, or (iii) including, the instant in question and may be defined as that speed which any of these average speeds can be made to approach as nearly as we please if the interval be made short enough. See 28. Rate of Increase. 6. The question of speed might be presented somewhat differently, as shewn in the following examples. Suppose a train to travel uniformly between two stations distant 15 and 17 miles from the terminus, and to pass those stations at 5.20 and 5.23. Let s feet be the distance of the train from some fixed point of the path, say the terminus, at the end of t sees, from some fixed instant, say 5 o'clock. 5-8 J RATE OF INCREASE 11 Then as t increases from 20 x 60 to 23 x 60 a 15 x 5280 to 17 x 5280, i.e. as t increases by 3 x 60, s increases by 2 x 5280, 1 2x528Q or 582 3x60 *' or 5S is the rate of increase of s per unit increase of t. [With our earlier terminology we should say the speed was 58| ft/sec.] , Increase in a Notice that this 58| is = . Increase in t 7. It will avoid confusion if we remember in cases like this where we are dealing with quantities of two different kinds, viz. : distance and time, that a and t are numbers, s being the number of feet in the distance and t the number of seconds in the time. Increase in s When we say T : - = 58|, we are stating, not that Increase in t 58| is the result of dividing 2 x 5280 feet by 3 x 60 seconds, but that it is the result of dividing 2 x 5280 by 3 x 60; 2 x 5280 being the number of feet and 3 x 60 the number of seconds. The statement in 2 that if a body travels s feet in t seconds o its average speed during the interval is - ft. /sec. is equivalent to this : If the number of feet in the distance travelled be divided by the number of seconds in the time taken, the resulting number is the number of ft./sec. in the average speed. , Increase in s . 8. If the motion is not unitorm. i.e. it the ratio = - : - is Increase in t not constant, as in 4, we might state our results as follows: When I is increased by -5, a is increased by 55'75. Increase in a 55*75 -,-,-, Kr , The ratio = -. = = = 111-50, Increase in t 'O 12 CALCULUS FOR BEGINNERS [CH. I and we say that 111-50 is the average rate of increase of s per unit increase of t between t = 4 and t = 4-5; for if the body were moving uniformly in such a way that an increase of -5 in t produced an increase of 55-75 in s, then 1 would produce 111-50 in s. Similarly 101-42 is the average rate of increase of s per unit increase of t between t= 4 and t = 4-1, for if s were increased by 10-142 in every -1 sec. s would be increased by 101 '42 in 1 sec. and so on. 99 + 24A + 2h 2 is the average rate of increase of s per unit increase of t between t= 4 and t = 4 + h and 99 is said to be the rate of increase of s per unit increase of t when t = 4. Or we might say 111-50 is the average rate of increase of s with respect to t between t = 4 and t = 4-5, meaning that as t increases from 4 to 4*5, s increases 111-50 times as much, and 99 is the rate of increase of s with respect to t when t = 4, this 99 being the number towards which our succession of average rates continually tends. 9. Suppose the formula connecting s and t to be s = 20t - 2t 2 , and to fix our ideas we will suppose that s is measured to the right. If < = 4, 8 = 4:8: if t = 4: + h, s = 48 + 4A-2A 2 and proceeding as before we find that the speed at the end of 4 seconds is 4 ft./sec. to the right. If <=5, s = 50: if t = 5+h, * = 50-2A 2 . If we go through the usual steps we say that in the interval of A seconds which follows the end of the 5th second, the body moves - 2k" 2 ft. to the right and its average speed in the interval is 2h ft./sec. The significance of the negative sign is easily seen. At the end of 5 seconds, the distance of the body to the right of the 8-10] RATE OF INCREASE 13 starting point is 50 feet: at the end of (5 + h) seconds it is (50 2h?) ft., i.e. in the interval of h sees, the body has moved 2& 2 ft. to the left and the average speed is 2h ft./sec. to the left. If t 5 h, s = 50 2/i 2 and in the interval h sees, before the end of the 5th second, the body has moved 2A 2 ft. to the right and its average speed is 2h ft./sec. to the right. This average speed can be made as small as we like if h is made small enough, i.e. if the interval be made short enough. During any interval, however short, immediately before the end of the 5th second the body is moving to the right and during any interval however small immediately after the end of the 5th second, it is moving to the left and in each case the average speed during the interval can be made as small as we please if we make the interval short enough. We say then, that at the end of 5 seconds the speed of the body is zero and that the direction of motion changes at the end of 5 seconds. 10. If t = 7, s = 42 : if t = 7 + h, s = 42 - 8h - 2h\ Proceeding as before we say that in the interval of h seconds following the end of the 7th second the body moves 8A 2h? ft. to the right and its average speed in that direction is ( 8 2/t) ft./sec. and the speed at the end of 7 sees, is - 8 ft./sec. The negative sign is again easily explained. At the end of 7 sees, the body is 42 ft. to the right of the starting point. At the end of (7 + h) sees, it is 42 8h 2A 2 ft. to the right, i.e. in the interval of h sees, it has moved (8h + 2h 2 ) ft. to the left, the average speed in this direction being (8 + 2h) ft./sec. and the speed at the end of 7 sees, being 8 ft./sec. to the left. A negative speed is thus seen to indicate that the body is moving in the opposite direction to that in which s is measured. Or we might say As t increases from 7 to 7 +/*, s changes from 42 to 42 8h 2h\ 14. CALCULUS FOR BEGINNERS [CH. I Or, as t increases by h, s increases by Sh - 1h* or decreases by Sh + 2h?, and the average rate of increase of s per unit increase of t is - 8 - 2h, or the average rate of decrease of s per unit increase of t is 8 + 2h. These two statements mean the same thing; in fact, to say that the rate of increase of s with respect to t is negative simply means that s is decreasing as t is increasing. 11. Function. When two quantities are connected in such a way that to each value of the first there corresponds a definite value of the second, the second quantity is said to be a function of the first, e.g. in the case considered in 4, s is a function of t the time being given we can calculate the distance of the body from the fixed point. The relation between s and t is shewn in the form of an equation s = 5 + St + 2t 3 , [In 3 the relation was exhibited by means of a table of pairs of corresponding values.] and we have seen how to find (i) The average rate of increase of s per unit increase of t or the average rate of increase of s with respect to t between two r~ Ti . Increase in s ~] specified values of t. It is .= : . Increase in t (ii) The rate of increase of s per unit increase of I or the rate of increase of s with respect to t for a specified value of t. rr , . , , Increase in s lit is the value which the ratio =^- = can be made to Increase in t approach as nearly as we please if the increase in t be made small enough.] 10-13] RATE OF INCREASE 15 12. The area of a square is a function of the length of the side. If A sq. ins. is the area of a square whose side is x ins. A = x 2 . Now suppose we have a square plate whose side is 1" and that it is expanding under the action of heat so that it remains a square. The original area is 1 sq. in., i.e. the original value of A= 1. If the side be increased to 1*1" the area increases to 1*21 ins., i.e. if x be increased by !, A is increased by -21, Increase in A -21 T-- - = =2-1 (see 7), Increase in a; '1 and this is the average rate of increase of A per unit increase of x between x = 1 and x= I'l. . Similarly if the side be increased from 1" to 1-001" the area is increased from 1 sq. in. to 1 '002001 sq. ins., and the average rate of increase of A per unit increase of x between x = 1 and x =1-001 is 2-001. If the increase in the side be made smaller and smaller this average rate of increase can be brought as near to 2 as we please, for if x is increased from 1 to 1 + h, A is increased from 1 to 1 + 2h + h? Increase in A 2h + h? . and : = j - 2 + h, increase in a? h which can be made as near to 2 as we please if h is made small enough. And we say that when x = 1, the rate of increase of A per unit increase of x, or the rate of increase of A with respect to x, is 2. 13. If the plate be supposed to be cooling and contracting we shall find that if x is diminished by '1, A is diminished by -19, or we might say 16 CALCULUS FOR BEGINNERS [CH. I if x is increased by *1, A is increased by '19, so that or we might say Decrease in A '19 Decrease in a? '1 Increase in A '19 = 1-9. and Increase in a? ! Similarly if x changes from 1 to 1 - h A 1 to 1 - 2h + . Decrease in A 2A A 2 or Decrease in a; h Increase in A 2k + h? = 2-* Increase in a? h and as before, when x = 1, the rate of increase of A with respect to x is 2. 14. It thus appears that if the rate of increase of one quantity with respect to another is positive, the two quantities are increasing or decreasing together, but if the rate is negative, one is increasing while the other is decreasing. 15. The pressure of a given mass of gas at constant tempera- ture is a function of the volume. Let V c. ins. denote the volume of a given mass of gas and p Ibs. per sq. in. its pressure, and suppose the following table given P V 30 40 40 30 48 25 50 24 i.e. As V decreases from 40 to 24, p increases from 30 to 50, Increase of p 20 5 Decrease of V 16 ~ 4* 13-16] RATE OF INCREASE 17 or - is the average rate of increase of p per unit decrease of V between V = 40 and V = 24. Similarly g v is the average rate of increase of p per unit decrease of V 5 o or = is the average rate of increase of p with respect to V D between V = 30 and V = 25. 16. Now suppose the relation between p and V given by the formula pM = 1 200. We can by calculation form the following table : V 40 45 42 41 40-5 40 + 7* 1200 p 30 26-67 28-57 29-27 29-63 40 + /i and the average rates of increase of p with respect to V between (i) V = 40 and V = 45 (ii) V = 40 and V = 42 (iii) V =-. 40 and V = 41 (iv) V = 40 and V = 40-5 J will be found to be respectively '67, *71, *73, '74 and what we call the rate of increase of p with respect to V when V = 40 is the quantity towards which this succession of numbers continually tends as the increase in V is made smaller. To make quite certain what this limit is, find the average rate of increase of p with respect to V between V = 40 and 30 V = 40 + h. It is JJT j- ; and by making h small enough we 30 can make this as near or -75 as we like. 40 Thus -75 is the rate at which p is increasing with respect to v when v = 40. M. c. 2 18 CALCULUS FOR BEGINNERS [CH. 1 EXERCISES. IV. 1. In the following table A sq. ins. is the area of a segment of height h ins. in a circle of diameter 1000 ins. h 100 101 102 103 104 A 40875 41477 42081 42687 43296 Find the average rate of increase of A with respect to h between the following values of h : (i) 100 and 104; (ii) 100 and 103; (iii) 100 and 102; (iv) 100 and 101. 2. Use tables to find the average rate of increase of sin with respect to 6 (0 being the number of radians in the angle) (i) between = -5061 and 0=-5236, (ii) between 0='5236 and 6 = -5411. With 4-figure tables what is the best value you can get tor the rate of in- crease of sin0 with respect to when 6 = -z1 o 3. A square is expanding in such a way that it remains a square. Find the rate of increase of the number of sq. ins. in the area per unit increase of the number of sq. ins. in the side when the side of the square is (i) 3", (ii) a". 4. If a cubical block expands so as to remain cubical, find the rate of increase of the number of cubic inches in the volume with respect to the number of inches in the edge when the edge ia (i) 1"; (ii)5"; (iii) a". 6. If a circle expands so as to remain circular, and if the radius, circumference and area are respectively r ins., C ins. and A sq. ins., find the rates of increase of (i) A with respect to r, (ii) C r, (iii) A C. when r=3. 6. The radius of a spherical bubble is increasing at the rate of 1 cm. per second. At a certain instant its radius is 4 cms. What will its radius be I, -5, -2, !, -01, h sees, after this? In each case find the surface and volume. 17] GRADIENT 19 Find during each interval (ij the average rate of increase of the number of cubic cms. in the volume with respect to the number of cms. in the radius ; (ii) the average rate of increase of the number of sq. cms. in the surface with respect to the number of cms. in the radius ; (iii) the average increase per sec. of the volume ; (iv) ,, ,, ,, surface. Also find at the end of 4 sees. (v) the rate of increase of the number of c.c. in the volume per unit increase of the number of cms. in the radius ; (vi) the rate of increase of the number of sq. cms. in the surface per unit increase of the number of cms. in the radius ; (vii) the rate of increase of the volume per sec. ; (viii) ,, surface . 7. The speed of a body (v ft. /sec.) at the end of t sees, is given by the formula v=3 + 4t 2 . Find the average rate of increase of v with respect to t between (i) t=2andt=3; (ii) f=2 and t = 2 + fc, and deduce the rate of increase of v with respect to t when t = 2. [Note. The first result is 20 and this means that in 1 sec. the speed has increased 20 ft. /sec. or that the average acceleration is 20 ft./sec 2 . The last result gives the acceleration at the end of 2 sees.] Gradient. 17. Uniform gradient. Fig. 2 shows a portion of a straight line Kl_ P, Q are any two points on the line. PM, Q.N are the ordinates of P, Q and PR is parallel to OX. The A PQR has always the same shape wherever we take RQ P and Q. In particular the ratio is constant. PR This ratio is called the gradient of the line. It is the tangent of the angle RPQ or the tangent of the angle which KL makes with OX. 22 20 CALCULUS FOR BEGINNERS [CH. 1 18. We may suppose that we pass from P to Q by two steps, (i) a horizontal step PR, (ii) a vertical step RQ. The gradient is Fig. 2. the ratio of the vertical step to the horizontal step, and is the same whatever the lengih of the horizontal step and from whatever point we start. Following the usual convention we call a horizontal step positive if it is to the right and a vertical step positive if it is upwards; so that the gradient is positive if both steps are positive or both negative, and the gradient is negative if the steps are of opposite signs. e.g. in Fig. 2 Horizontal step P^ = + 10, Vertical step RjQj = + 5, >> P-s^2 = ~^ : > ;> n R a Q 3 = 2, and the gradient of the line is + jr . 18, 19] UNIFORM GRADIENT 21 In Fig. 3 Horizontal step P^ = + 10, Vertical step R J Q 1 = - 15, f22 = ~ 4, II 11 RjQa = + 6, 3 and the gradient of the line is ^ . Fig. 3. 19. Expressing this in the language of coordinates we maj say that if (a^) (x.gj^) be the coordinates of any two points on the line, the gradient is this being a fraction which has the #2 a a same value whatever the coordinates of the points chosen. e.g. in Fig. 2 coordinates of P l are (10, 10) and of QI (20, 15), 15-10 1 and gradient = ^- lo =2 ; coordinates of P 2 are ( 2, 4) and of Q 2 ( 6, 2), 2-4 1 and gradient __=-. 22 CALCULUS FOR BEGINNERS [CH. 1 20. Another way of looking at the question is as follows : In Fig. 2 if we take a horizontal step + 1 [P 3 R 3 ], the vertical step is +^r [R 3 Q 3 ] ; so that we may say that the gradient gives 2i the vertical ascent per unit length of horizontal advance, or the increase of y per unit increase of x. 21. Example. Shew that with our definition the gradient of y = mx + n is m. Let (#1^1) (#22/2) be any two points on the line, .'. 2/ 2 = mx 2 + n > and 2/1= I m&\ + n ; ' 2/2 - /i = * 0*2 - i), 3/2 &! or m is the gradient of the line. 22. Average gradient. If P, Q bo two points on a path RQ which is not straight as in Fig. 4, is called the average PR gradient of the path between P and Q; e.g. the average /> gradient between P and Q is ^. This is the gradient of the line y PQ, and we may say that if we move from P along the path so as to make a horizontal step PR, the vertical step RQ is the same as if we had moved along the line PQ which has a uniform gradient <r . y The average gradient will depend on the length of the horizontal step and on the point from which we start, e.g. the 1 fi average gradient between P and Qj is y^r and between P 2 and Qj i y .. . 9-5 20-23] AVERAGE GRADIENT 23 Generally if (x^) (#2^2) ^ e the coordinates of two points P, Q on a curve the average gradient between P and Q is and this will be different for different positions of P and Q, i.e. for different values of the coordinates. o/r 4 - rb- trr -5 Fig. 4. Gradient at a point of a curve. 23. Figure 5 shews a portion of the curve P is the point (1, 1), Qj is (3, 9). Let PQj be joined and produced indefinitely both ways. Draw the ordinates PM, Q 1 N l ,and draw PF^ parallel to OX, 24 CALCULUS FOR BEGINNERS Y [CH. I 7 Fig. 5. 4 X 23] GRADIENT AT A POINT 25 Then as x increases from 1 to 3, y 1 to 9 > RiQ, Increase in y 8 i.e. - = f -. - = - = 4 = average rate of increase of PRj Increase in a; 2 y per unit increase in x between x= 1 and x = 3. This is tan RjPQj or tan XK X P or the tangent of the angle which PQ a makes with OX and is called the gradient of the chord PQj. [.Notice. If PR 2 = 1 and RaSj be drawn parallel to OY meeting PQj in Su then R 2 Sj = 4.] Now suppose the line PQj to rotate about P until it cuts the curve in Q 2 corresponding to x = 2. We shall get in an exactly similar way tan R 2 PQ 2 or tan XK 2 P or the gradient of PG^ or the average rate of increase of y per unit increase of x between x = 1 and x = 2 is j or 3. Let the line continue to rotate and cut the curve successively in Q 3 , 04, Q 5 , Qg corresponding to a3 = l'4, 1'2, I'l, I'Ol, and we get Gradient of PQ 3 or average rate of increase of y per unit increase of x between x 1 and x = 1*4 is 2-4. Gradient of PQ 4 or average rate of increase of y per unit increase of x between a; 1 and x= 1*2 is 2-2. Gradient of PQ 5 or average rate of increase of y per unit increase of x between x = 1 and x = 1 ! is 2-1. Gradient of PG^ or average rate of increase of y per unit increase of x between x 1 and x = 1-01 is 2-01. If we continue this process making the increase in x and therefore at the same time the increase in y smaller and smaller or bringing the point Q nearer and nearer to P, the gradient of the rotating line comes gradually nearer to 2, is always greater than 2, but can be made as near to 2 as we please by bringing Q near enough to P. 26 CALCULUS FOR BEGINNERS [CH. I To make quite sure of this let the line cut the curve in Q for which x = 1 + h and /. y = 1 + 2A + A 9 (Fig. 6). Increase in y 2h + h 2 Then gradient of PQ = ^- - = 5- = 2 + h, Increase in a; n and by making h small enough we can make this as near 2 as we like. 2h-h 2 Fig. 6. If we continue the rotation, the line will cut the curve again on the other side of P, say in a point for which x = 1 h and (Fig. 7). Then gradient of PQ = Decrease in y 2A - h? = 2-A, Decrease in a; A i.e. the gradient of a chord joining P to a point Q for which x < 1 is less than 2, but can be made as near to 2 as we like, by bringing Q near enough to P. Thus if a chord through P cut the curve in a point to the right of P, however near to P, its gradient > 2 ; if the chord cut the curve in a poiut to the left of P, its gradient < 2 but in either case the gradient can be made as near to 2 as we please by bringing the second point near enough to P. The gradient of the curve at P is said to be 2. It is the same thing that we have previously called the rate of increase of y per unit increase of x t when x=l. 23-25] GRADIENT AT A POINT 27 24. The line through P (PT) whose gradient is 2 is called the tangent at P. This line does not cut the curve again in a point near P, for if it did its gradient would not be 2. In fact PT is not one of the series of chords PQ 15 PQ 2 , PQ 3 , etc. for we have seen that none of these chords has a gradient 2, but by bringing Q near enough to P we can bring PQ as near to coincidence with PT as we like, i.e. we can make the angle TPQ as small as we like. This is expressed shortly by saying that the tangent PT is the limiting form of the chord PQ when Q is brought indefinitely near to P. If PT (Fig. 8) is the tangent at P and PR = 1 then if RT be drawn parallel to OY RT = 2. Fig. 8. 25. In these figures we have taken the same scale along both axes and what we have called the gradient of a line is the tangent of the angle which it makes with OX. 28 CALCULUS FOB BEGINNERS [CH. I . Actual length of R,Q, e.g. the gradient of PQ, is .- 2__ _t _ l or tan PKjX. Actual length of PRj If the scales are not the same the gradient, being Increase in y . Number of y-units in Increase in x Number of a>units in PRj ' and this is not the tangent of PK X X. e.g. if y = y? be drawn with the scales shewn in Fig. 9, the R O gradient may be written provided it is understood that RjQj PRj is to be measured by means of the scale marked on OY, and PRj by means of the scale marked on OX. Thus gradient = - = 4 as before but now 4 is not tan RjPQj in Fig. 9. It is the tangent of the angle that we should get in place of RjPQj if we altered the scales so that the oMinit and the y-unit were of the same length, i.e. it is tan R 1 PQ 1 in Fig. 5. 25] GRADIENT 29 We may look upon Fig. 9 as a deformed copy of Fig. 5 and consider either that horizontal distances are correctly represented and vertical distances shortened in the ratio 1 to 2 or that vertical distances are correctly represented and horizontal distances lengthened in the ratio 2 to 1. The tangent of the actual angle F^PQi that we see in Fig. 9 is not 4 but 2. EXERCISES. V. 1. If y=x*, find y when x=3, 3-5, 3-1, 3-01, 3 + fe, 2-9, 3-fc. Hence get the gradients of the chords joining the point on the curve y=x% at which a;=3 to the points at which a;=3-5, 3-1, 3-01, 3 + ft, 2-9, 3- A. What is the gradient of the curve at the point (3, 9) ? Draw the graph of y=x% between x = I and x = 4 [taking " as the unit for both x and y~\. Through (3, 9) draw the line whose gradient is 6. It should be a tangent to the curve. What angle does it make with OX ? 2. What is the gradient of the chord joining the points where a:=3-/j and x=3 + h. Shew that by making h smaller and smaller we can bring this chord as near as we please to coincidence with the tangent at (3, 9). 3. Draw y=x z between x = l and x = 4 taking 1" as the unit along OX and -2" as the unit along OY. As in Qu. 1 draw through (3, 9) the line whose gradient is 6. It should be a tangent. What angle does it make with OX ? 4. Find the gradient of the chord joining the point on the curve y=x s at which a; =2 to the point at which x = 2 + h. Hence get the gradient at the point (2, 8). 5. Find the gradient of the chord joining the points where x= 2 - h and x = 2 + h on the curve y = x 3 . Shew that by making h smaller and smaller we can bring the chord as near as we please to coincidence with the tangent at (2, 8). Similarly for the chord joining the points where x=2-m and 6. Find the gradient of y=x* at the points where x = Q, 1, 3^4. 7. In the curve y=x 3 you have found the values of y at the points where x = 0, 1, 2, 3, 4 and also the gradients at these points. Make use 30 CALCULUS FOR BEGINNERS [CH. I of these ten facts to draw the curve from x=0 to x=4. Take 1" as the a:-unit and -1" as the y-unit. What angle does the tangent at (4, 64) make with OX ? 8. Find the gradient of the curve y = at the points where 9. Find the gradient of y=4x + 3 at the points where x=0, 1, 2. 10. If y = x 3 (4 - x) find the values of y when x = I and when x = 1 + 7. Use your result to find ' ' the gradient of the curve y = x 3 (4 - x) at the point where x=l," explaining carefully what the phrase means. Similarly find the gradient where x=3. Using these results and the values of y when x =0, 1, 2, 3, 4 draw the graph between x = Q and a; = 4. 26. In curves with which we shall deal, the limiting position of the chord PQ when Q is brought indefinitely near to P will be the same on whichever side of P we take Q, and will be the Fig. 10. same as the limiting position of the chord RS (P lying between R and S) when R and S are each brought indefinitely near to P. [See Figs. 10, 11.] 26] TANGENT 31 Fig. 11. If a curve had a sharp point at P as in Fig. ] 2, it would not be true that the limiting position of the chord PQ is the same on whichever side of P we take Q. In fact there are two tangents at P, viz., PT 1} PT a , one to each branch. Fig. 12. 32 CALCULUS FOB BEGINNERS [CH. I 125 4; Fig. 13. 26, 27] RATE OF INCREASE AS GRADIENT 33 27. If one quantity be a function of another, the relation may be shewn analytically by an equation or geometrically by a graph. e.g. in 4 a is a function of t, this fact being expressed in the form of an equation s = 5 + 3t + 2t 3 . This relation may be exhibited graphically. In Fig. 13, the portion of the graph between t = 3 and t = 5 is shewn, P, Q 1? GLj, Q 3 being the points on the graph corresponding to* = 4, 6, 4-5, 4-2. The number of units of length in F^Qj is the increase in s as t increases from 4 to 5 [or it is the number of feet travelled between the ends of the 4th and 5th seconds]. The increase in t is the number of units of length in PR lf Increase in s Thus = ; or what we have called the average rate Increase in t of increase of s with respect to t between t = 4 and t = 5 is the gradient of the chord PQ^ Similarly the gradients of PQ 2 and PQ 3 give the average rates of increase of s with respect to t between t = 4 and t = 4'5 and between t = 4 and t = 4*2. The limit to which this average rate of increase of with respect to t tends as the increase of t is diminished more and more, or what we have called the rate of increase of s with respect to t when t = 4 is the limit to which the gradient of the chord PQ tends as Q is brought nearer and nearer to P, i.e. the gradient of the curve at P. In this particular case, then, when the space-time graph is drawn, the gradient of the curve at any point gives the speed at the instant corresponding to that point. Similarly if the speed-time graph is drawn, the gradient of the curve at any point gives the acceleration at the instant corresponding to that point. M. C. 3 34 CALCULUS FOR BEGINNERS [CH. I Similarly, if we draw the graph pM = 1200 the gradient of PQ (3'33\ J is the average rate of increase of p with 29 27 43 Fig. 14. respect to V between V = 40 and V = 45. The gradient of the curve at P is the rate of increase of p with respect to v when v = 40. 27, 28] APPARENT EXCEPTION 35 28. There are cases in which the statement in 5 appears not to be true. e.g., suppose a marble falls vertically from a height of 36 feet. It strikes the ground with a speed of 48 ft./sec., and will rebound with reduced speed, say 40 ft./sec. Fig. 15 is the space-time graph, the abscissae representing the number of seconds from the instant when it begins to fall and the ordinates the height above the ground in feet. Fig. 15. The speed 48 ft. /sec. is the limit of the average speed taken over a short interval preceding the instant when it strikes the ground, and the speed 40 ft./sec. is the limit of the average speed taken over a short interval following the instant when it leaves the ground. 32 36 CALCULUS FOR BEGINNERS [CH. I If we take these two instants to be the same, we seem to have two different gradients at A, or in other words to have two different speeds at the instant of impact according as we consider the limit of the average speed over a short interval preceding or over a short interval following the instant. Fig. 16. 28, 29] RATE OF INCREASE AS GRADIENT 37 As a matter of fact the two instants are not the same and the marble is in contact with the ground for some short time during which there is compression followed by recovery. Suppose for instance that the time of contact is T ^ 7 sec. Fig. 16 shews roughly what the true space-time graph is like. It is a portion of the graph, the horizontal and vertical scales being each magnified 100 times as compared with Fig. 15. A corresponds to the instant when the marble strikes the ground and B to the instant when it leaves the ground. It will now be seen that the statement in 5 still applies to the speeds at the instants corresponding to A and B. EXERCISES. VI. 1. Draw pretty accurately between x =3 and a: =5 the graph of y=log 10 ar. Draw by eye the tangent at the point corresponding to ar=4.- What is the rate of increase of Iog 10 a; per unit increase in x when x=4 ? 2. A body is moving in a straight line and its distances (s ft.) from a fixed point in the line at the end of (t sees.) are given by the following table t 1 2 3 4 5 it 7 22 45 76 115' Draw as accurate a graph as possible and from it get the speed at the end of 3 sees. 3. Get the results of Exs. TV. 5, p. 18, graphically. 29. The processes by which we arrived at our results in the preceding illustrations were in principle the same. In finding the speed of a moving body at a given instant, we calculated the average speed during a short interval of time starting from the instant in question. We then found that as the interval over which the average speed was calculated was made shorter and shorter, the average speed continually approached some definite speed from which we could make it 38 CALCULUS FOB BEGINNERS [CH. I differ by as little as we pleased by making the interval short enough. This speed we called the speed of the body at the given instant. In finding the gradient of a curve at a given point we calculated the gradient of a line passing through the given point and cutting the curve again in a neighbouring point. We then found that as the neighbouring point was moved nearer and nearer to the given point, this gradient continually approached some definite value from which we could make it differ by as little as we pleased by making the points close enough together. This value we called the gradient of the curve at the given point. In finding the rate at which y is changing with respect to x for a given, value of x, we made a small change in x, calculated the corresponding change in y and found the ratio of the change in y to the change in x. We then found that as the change in x was made smaller and smaller, this ratio continually approached some definite value from which we could make it differ by as little as we pleased by making the change in x small enough. This value we called the rate at which y was changing with respect to x for the given value of x. 30. Our final result was in each case obtained by finding the ratio of two numbers each of which was supposed to become less than any assigned number however small. Now it might at first sight appear that if two quantities are constantly decreasing they must become more and more nearly equal. This is because we have a wrong idea of the test of approximate equality ; we think the two quantities must approach to equality because their difference becomes smaller and smaller. In estimating whether two quantities are nearly equal or not, we must look not merely at the difference between them, but at the ratio which this difference bears to the quantities compared. For example, two lengths of 3 and 4 inches are not so nearly equal as two lengths of 100 and 101 inches, for though the actual difference is the same in both cases, viz. : 1 inch, this difference 29, 30] TEST OF APPROXIMATE EQUALITY 39 is -^ of the smaller length in the first case, but only ^7 of the smaller length in the second case. Lengths of 1000 and 1003 inches are still nearer to equality, for though the actual difference is 3" and therefore greater than the difference in either of the other two cases, this difference is only 1fif)ft of the smaller length, and 3 J_ 1000 < 100* The point is that there are different standards of size in the three cases. The same difference of length is of less importance when we are dealing with 1000 inches than when we are dealing with only 3 or 4. An error of 1 foot made in measuring the length of a room would be considerable, but the same error made in finding the height of a mountain would be of very little account. A loss of 200 would be serious for a man with an income of 300, but of little moment to a man with an income of 100,000. An error of 10 million miles in the Sun's distance (about 90 millions of miles) would be considerable, but the same error would be inappreciable in the distance of Sirius (more than 500,000 times that of the Sun). A carpenter would neglect an error of ^ of an inch, in fact he would not detect it, but in cases where it was necessary to work to -. nnnn of an inch, it would be a gross error. We have no right to say that all microscopic creatures are of the same length, because they are all so small as to be invisible to the naked eye. In comparing two such creatures, we must, so to speak, think microscopically, i.e. use a microscopic standard of length. It is just as true that a creature ^7; of a millimetre 40 CALCULUS FOE BEGINNERS [CH. I in length is twice as long as a creature r-^r of a millimetre in length as that a man 6 ft. in height is twice as high as a child of 3 ft. in height To say that the difference between two quantities is great or small is meaningless. The terms are purely relative and both may be used of the same quantity in different cases. The important thing is the ratio of the difference to the quantities compared. The test of approximate equality might be put in another way : 4 -l-3333 101 -1-01 1003 _1. OS 3~ 33 -' TOO" 01 ' 1000- 3 ' and we say that the last pair of numbers are more nearly equal than either of the other pairs, because their ratio is nearer to 1 than eillier of the other ratios. EXERCISES. VII. Which of the following pairs are most and which least nearly equal ? 1. (a) 7" and 8" ; (b) 9 ft. Sins, and 11 ft. ; (c) -0054" and -0062". a. (a) 93,000,000 miles and 92,900,000 miles. (b) 9-30 inches and 9'29 inches. (c) 93,000,000 miles and 93,100,000 miles. (d) 930 yds. and 931 yds. 3. (a) x and y ; (b) Wx and W*y ; (c) JL and JL ; where y = l-Qlx. 31. We shall now give a few geometrical illustrations to shew that if x and y are two quantities which continually decrease towards zero and become zero (or vanish) together, the ratio of a: to y may (1) increase continually, i.e. it may be greater after a decrease than it was before, still greater after a further decrease and so on, becoming greater than any number that can 30-32] RATIO OF CONTINUALLY DIMINISHING QUANTITIES 41 be named, however large, if the decrease in x and y be carried far enough ; or (2) decrease continually, i.e. it may be less after a decrease than it was before, still less after a further decrease and so on, becoming less than any fraction that can be named, however small, if the decrease in x and y be carried far enough ; or (3) increase or decrease towards a limit, i.e. con- tinually approach some value which it never reaches, but to which it can be brought as near as we like if the decrease of x and y be carried far enough ; or (4) always remain the same. 32. (1) Fig. 17 shews a circle centre O. OA, OB are two radii at right angles. P is a point which is supposed to move along the Q circumference nearer and nearer to A. PM is perpendicular to OA and AP is produced to meet OB produced in Q. As P approaches A, PM and MA both diminish without limit, i.e. assign any length however small, we can find a position of P so that PM or MA shall be less than that length. If we sup- pose P to be at A, there is no triangle PMA at all, or as we may put it, PM and MA vanish together, but we have no right to say that PM and MA are all the time tending to become equal. For every position of P which is distinct from A, PMA is similar to PM QO A QOA, i.e. = . MA OA But as P approaches A, Q recedes further and further from O, i.e. QO increases without limit, 42 CALCULUS FOR BEGINNERS [cil. I QO PM and OA remains the same, .'. increases without limit, i.e. - OA MA increases without limit, i.e. state any number however great, PM may be made to contain MA more than that number of times before P reaches A. Suppose the radius of the circle to be 1" and call L AOP 6- Then PM = sin 6 ins. and MA = (1 - cos 0} ins. We can form this table : 20 10 5 Q1O -1 O " * PM 3420 1736 0872 0436 -01745 MA PM MA 0603 5-67 0152 11-43 0038 -00095 -00015 22-90 45-83 114-6 6 3<X i<y 4' 2' PM 0087265 0029089 0011636 0005818 MA PM MA 0000381 229 0000042 688 0000007 1719 0000002 3438 etc. sin TN.B. Since , 7 .=cot ~ the numbers in the last line .can L 1 - cos 6 2 be written down from a table of cotangents.] Suppose our eyes incapable of seeing any length less than th of an inch ; then when L AOP is about 2, MA is 1000 invisible but PM is more than = of an inch. zo . Again, when the angle is about 4', PM is just visible, but we should have to use a microscope magnifying about 1700 times to make MA visible. 33. (2) If in the same figure we consider , the ratio can be made less than any fraction we like to assign, however small, before P reaches A. 32 -34] RATIO OF CONTINUALLY DIMINISHING QUANTITIES 4* 34. (3) In Fig. 18, OA is a fixed radius of a circle, centre O. P, Q. are two points on the cir- cumference, so that Q is the raid- point of the arc AP; PM, QN are perpendicular to OA. As P approaches A, Q ap- proaches A, and AM, AN both diminish without limit. If we suppose P to have arrived at A, Q will also be at A, and AM, AN will have ceased to exist. But we must not say that AM, AN are all the time tending to equality. Call L AOQ 0, then L AOP = 20. If the radius of the circle be r, AN=r(l-cos0) Fig. 18. AM=r(l-cos20) = 2rsin 8 & AM AN Now as 6 diminishes cos ^ increases and can be made as near m to 1 as we please if & be made small enough. Thus is a continually increasing ratio whose limit is 4, AM i.e. is always less than 4, but state a number whose defect from 4 is as little as we please, we can place P so that shall be equal to this number. AN 10 5 3-9924 1 3-9997 30' 3-99992 10' 3-999992 3-999999... 44 CALCULUS FOR BEGINNERS [CH. I 35. (4) In Fig. 19, let XOY be a triangle right-angled at X, such that OX = 2XY, and let a point P move along YO towards O. Draw PM perpen- dicular to OX. Kg. 19. Then as P approaches O, PM and OM both diminish without limit. If we suppose P to have reached O the A PMO has ceased to exist. Bub for every position of P distinct from O, however near to it OM OX ~MP ~ XY ~ "' OM i.e. is a constant ratio, however small OM and MP are. MP 36. Of course, it may happen, that, if two quantities continually approach zero and vanish together, their ratio continually approaches the limit 1, but it must be borne in mind that this is an exceptional case. The following illustration is an important example of this case. In Fig. 20, if the point P move along the circumference of a circle towards A, the arc PA and the chord PA diminish together and can each be made less than any assigned length, however small, 35, 36] RATIO OF CONTINUALLY DIMINISHING QUANTITIES 45 if P be brought near enough to A ; but we have no right to con- 3iT*C AP elude from this that the ratio -: , continually approaches 1. chord AP Fig. 20. As a matter of fact it does, as may be seen from the table below. LAOP in degrees 10 9 8 7 6 3 Arc^P 1745329 1570796 1396263 1221730 1047198 Chord AP 1743114 1569182 1395130 1220970 1046720 AioAP 1-0013 6 1-00103 4 1-0008 3 1-0006 2 1-0005 1 Chora AP LAOP in degrees Arc^P 0872665 0698132 0523599 0349066 0174533 Chord AP 0872388 0697990 0523538 0349048 0174530 AxoAP 1-0003 1-0002 1-0001 1-00005 1-00002 Chord AP 46 CALCULUS FOR BEGINNERS [CH. I 37. Another difficulty which often occurs is this : How can a quantity continually decrease without eventually disappearing altogether ? In Fig. 21, OAB is a quadrant of a circle, centre O. BX is the tangent at B. B T T' T" T'" M M'M" A Fig. 21. Suppose T to travel along *BX, at any rate whatever, say 1000 miles an hour, and in every position of T, join OT cutting the circumference in P. It is obvious that as T travels along BX, P moves nearer to A, but P will never reach A, however far T be supposed to travel along BX. The lengths PM, MA considered in 32 can be supposed to go on decreasing for ever without actually disappearing and it has PM been shewn that during the motion of P towards A, will MA increase without limit. CHAPTEE II DIFFERENTIATION FROM FIRST PRINCIPLES 38. WE shall now go through some of the investigations of Chapter I in a slightly different manner, using the notation of the subject. 39. We have s = 5 + 3t -f 2 3 , giving the distance s feet travelled in t seconds. If we take a slightly greater value of t, say t + At, we shall obtain a slightly different value of s, say s + As. [A is merely shorthand for " an ' increment ' or ' increase ' in the value of t," A being the Greek equivalent of capital D. A by itself means nothing, and A2 must be considered as one symbol; in fact, Ai takes the place of A in 4.] There will be the same relation between (s + As) and (t + A<) as between s and t ; i.e. s + As = 5 + 3 (t + A<) + 2 (t + A<) 3 = 5 + 3* + 2^ + (3 + 6 2 ) A< + Qt (A) 3 -f 2 (A*) 8 , .*. As = (3 + W) A* + Qt (A) 2 + 2 (A<) 3 . This gives the extra distance travelled in the extra interval of time A. 48 CALCULUS FOB BEGINNERS [CH. II [e.g. If we want the distance travelled in the half-second following the end of the 3rd second, we put t=3, At=-. A /. =? = (3 + 6* 2 ) + 6* . A* + 2 Since As feet is the distance described in the interval A sees As .". gives the average speed during this interval in ft./sec. Av As A is made smaller and smaller Qt . A + 2 (A) a becomes smaller and smaller and can be made as small as we please if A is made small enough. i.e. can be made as near to (3 + 6< 2 ) as we please if Ai be A made small enough. As The limit to which continually approaches and from which we can make it differ by as little as we please by taking A small enough is called -y- . dt /Jo (7$ Thus -5- = 3 + 6^ and -3- ft. /sec. is the speed of the body at the end of t seconds. \V. definition of speed at an instant 4.] 40. Notice that if s = 5 + 3t + 2 As (i) = (3 + 6< 2 ) + 6* . A< + 2 (A*) 2 is an accurate statement whatever the value of A, e.g. take =4, A = ^ and we get 99 + 12+^ or 111| ft./sec. as the average speed during the interval of second immediately after the end of the 4th second. 39, 40] DIFFERENTIATION FROM FIRST PRINCIPLES 49 As (ii) = 3 + 6^ 2 is approximately true if A be small and t\t becomes more and more nearly true as A is made smaller, but there is no value of A2 however small for which the statement is accurately true. As [See table in 4 where the third column gives for LA 6 different values of AZ when t = 4 and in this case 3 + Qt" 2 = 99.] In other words, the average speed during a small interval A following the end of t seconds is approximately (3 + 6^ 2 ) ft./sec. and this is more nearly true, the shorter we make the interval A& e.g. if we want approximately the distance described my^th of a second after the end of the 3rd second we have As = (3 + 6 2 ) A approximately, meaning that if A be small the distance described in A2 seconds immediately following the end of t seconds is nearly (3 + 6 2 ) At feet. Put t = 3 and A=y^ and we have approximately as the 57 distance required in our special case, yr-- or -57 ft. [Shew that the distance is really -571802 ft.] This gives an error of about '3 / o . If we used the formula to give the distance in ycnnr sec. we should get '057 ft. instead of -057018002 ft., an error of about 03 / an d so on. ds (iii) -j- = 3 + 6f 2 is accurately true. It is in fact a convenient of* As way of saying that the unattainable limit which is trying to reach is 3 -t- 6^, or - r stands for the ideal value which -~ is at t\t striving to attain. M. c. 4 50 CALCULUS FOR BEGINNERS [CH. II 41. ds and dt have no separate meanings, dt does not stand for a very small increment in t, nor ds for a small increment iii s, for we have just seen that however small such increments are made, their ratio is never 3 + 6Z 2 . ds -TT then must be treated as a whole, and simply means dt the limit to which is constantly tending as At and with At As it As becomes smaller and smaller. 7 can be brought as At ds near as we please to -? if we make At small enough, dt ds does not mean the result of dividing ds by dt as dt means the result of dividing As by At. The fractional At ds form -T7 is retained merely to remind us of the fraction dt of which it is the limiting value. 42. It is a common error to say that -r- is the value of - dt A< when As and A are each zero. If this statement be examined, it will be seen to have no meaning. As Let us go back to the stage at which was obtained. ttv We had As = (3 + 6* 2 ) A + 6t . (A*) 2 + 2 . (A<) 3 . We then divided both sides of the equation by A. Now if A = we have no right to do this. [It is certainly true that 8x0 = 3x0 but we cannot divide both sides by and deduce 8 = 3. Similarly if a;xO = 7/xOwe cannot conclude that x =y, though if x x T^= y x y^ it is true that x = y.] We should in fact be trying to estimate the speed of the body from the fact that it had travelled feet in sees. ds 41-44] MEANING OF -7- 51 43. We have found that the speed at the end of t seconds is (3-J-6* 2 ) ft/sec. The result obtained in 4 is merely a particular case of this, for when = 4, 3 +6f 2 = 99 and our formula enables us to write down the speed at any instant, e.g. to get the speed at the end of 10 seconds, put t- 10 and we get 603 ft. /sec. EXERCISES. VIII. 1. Obtain a formula giving the speed at the end of t seconds of a body whose distance (s feet) from a fixed point at the end of t seconds is given by (i) 8 = 16(2, (ii) 8 = lOOt -16*2, (iii) s=5 + t 3 , (iv) s = 5t + 9, and in each case find the speeds at the end of 3, 5, 17 seconds. 2. If 8 = 16^2, what is the speed at the end of 2 seconds? Assuming this speed to remain constant for the next tenth of a second, how far would the body move, and what percentage error would be made in taking this as the true distance ? Do the same for a hundredth of a second. 3. If the speed of a body increase by 12 ft./sec. in 3 seconds, what is the average acceleration during the 3 seconds ? 4. If the speed at the end of 4 seconds is 78 ft./sec. and at the end of 9 seconds 146 ft./sec., what is the average acceleration between the end of the 4th and the end of the 9th second ? [v. Ex. 7, p. 19.] 6. If the speed at the end of t seconds is v ft./sec. and at the end of (t + At) seconds, (v + Av) ft./sec. , what is the average acceleration during the interval At? By what symbol should the acceleration at the end of t seconds be denoted ? 6. In Ex. 1, find the accelerations at the end of 3, 5, 17 seconds. 7. With the formula 8 = 5 + t 3 what is approximately the distance described in of a second immediately following the end of the 3rd second? DU What is approximately the increase of speed during this interval ? ds 44. DEFINITION, -rr is called the differential coefficient dt of s with respect to t, and "to differentiate a with respect to t" means to find -7- . at 42 52 CALCULUS FOR BEGINNERS [CH. II ds 45- Instead of saying that if s = 3t + 2t 3 then -3-= 3 + G^ 2 we 0V might say dt = 3 + Gf. This form of stating the result will sometimes be found convenient. 46. Consider a square plate whose side is x ins. and let its area be y sq. ins. so that y = a?. Suppose the side increased to (x+ Aa:) ins., and as a result let the area be increased to (y + Ay) sq. ins. .'. y + Ay = (x + Aa;) 2 , .'. Ay = 2xAa + (Aa;) 2 . i.e. the increase in area due to an increase Aa; in the side is 2.rAa; + (Aa;) 2 sq. ins. [In Fig. 22 each of the shaded rectangles has area x&x and the square in the top corner has area (Aa;) 2 .] Increase in number of sq. ins. in area Aw ^ . _ n - = _ 2. Increase in number of ins. in side Ax dii and as before we deduce ?= 2x. ax _ 2. = 2x + Aa; Ax >*) 2 x x Fig. 22 45-47] -r- AS A RATE OF INCREASE 53 CLOS 47. As before notice that if y = a? AV (i) = 2o? + Ax is an accurate statement whatever the Ax- value of Ax, e.g. if x = 3, and Ax = -2, Ay * = 6-2, i.e. Increase in y = 6*2 times increase in x = 6-2x -2 = 1-24. i.e. if the side be originally 3" and increase by -2" the area increases by 1-24 sq. in. A?/ (ii) -~- = 2x is approximately true and becomes more nearly 1.+OC true as Ax is made smaller. e.g. if x = 3, Ay = 6Ax nearly, i.e. if a square have a side of 3 ins. and a small increase be made in the side, the increase in the number of sq. ins. in the area will be nearly 6 times the increase in the number of ins. in the side. Suppose for example the side increases from 3 to 3-02 ins. the area increases approximately by *12 sq. ins. Actually the increase is "1204 sq. in. [-0004 sq. in. is the area of the small square in Fig. 22.] fjft (iii) -~ = 2x is accurately true, and is simply a convenient tUB way of saying that the approximate statement in (ii), Increase in y ^ = 2x, Increase in x can be brought as near to the truth as we please by making the increase in x small enough. Taking again the special case when x = 3, ~ = 6, this might be stated thus : ux When x = 3, the rate at which y is increasing compared with x is 6, or the x-rate of increase of y is 6, or the rate of increase of y per unit increase of x is 6. 54 CALCULUS FOR BEGINNERS [CH. II 48. If we introduce the idea of time into the question and suppose the side of the square to be steadily growing, then in the same time in which Aic ins. is added to the side, Ay sq. ins. Aw is added to the area and ~-=6'2 tells us that y increases 6'2 Aa? times as much as x increases in the same time. If we call the time in which the change takes place A< sees, we have i.e. average rate of increase of y during this interval = 6'2 times the average rate of increase of x during the interval. In the general case At/ Ay If A, and therefore at the same time Arc and Ay, be made smaller and smaller, -~ and can be made as near as we please A A dx , dy , Ay dy to -T- and -?- respectively, as near as we please to -?- , and dt dt Ao? dx 2o? + Aa; as near as we please to 2x. dx l But -37 and -=- are what we call the rates at which y and x at at respectively are increasing at the end of t sees. di/ dij flit* Thus -~- = 2x being equivalent to -j- = 2oj . -j- may be looked CtOC ut Cut upon as telling us that y is increasing 2x times as fast as x. di/ In the special case when x = 3, -f-=, and this tells us that Cww at the particular instant when the side, of the square is 3 inches y is increasing 6 times as fast as x, or the number of square inches in the area is increasing 6 times as fast as the number of inches in the side. 48-50] DIFFERENTIATION FROM FIRST PRINCIPLES 55 49. Example. The side of a square increases uniformly at the rate of ! inch per second. At what rate is the area growing when the side of the square is 6 inches ? If y sq. ins. is the area of a square of side x ins. --x ' dx ** Ul/ and when x = 6, -^=12. i.e. at the instant when the side of the square is 6 ins. y is increasing 12 times as fast as a?, or the number of square inches in the area is increasing 12 times as fast as the number of inches in the side. But x is increasing '1 per sec. /. y ,, 12 x ! or 1-2 per sec. i.e. the area is increasing at the rate of 1*2 sq. in. per sec. 50. Suppose the pressure (p Ibs./sq. in.) and the volume (V cub. ins.) of a given mass of gas are connected by the relation joV-1200 or j = ^p. (1) If we take a slightly different value of V, V + AV we shall get a slightly different value of p, p + Ap such that 1200 1200 1200 1200AV A <v\ (2) P V + AV V (V + AV)V' Ap 1200 AV (V + AV)V' As AV is made smaller and smaller this comes nearer and . 1200 1200 ... nearer to and can be made as near to ^ as we like it AV be made small enough. dv - . dp_ 1200 \/ ' "XT, vl" ' 56 CALCULUS FOR BEGINNERS [CH. II 51. The meaning of the negative sign is simply this : As V increases p decreases, so that AV and &p have opposite signs, therefore ^ is negative. We might read the final result either The rate of increase of p per unit increase of V = 5-, or the rate of decrease of p per unit increase of V = -f ^ , or p is decreasing 5 times as fast as V is increasing. 52. As before, notice, if pV = 1200 (i) ^ = AvW * s an accurate statement whatever the values of V and AV. e.g. if V = 40, and AV = 5 Ajp_ 1200 _2 AV~ 40 x 45~~3' 2 Le. decrease in p = ^ (increase in V) = 3. 9 .... Ap 1200 . (n) -^- = is approximately true and becomes more nearly true the smaller we make AV. e.g. if V = 20, &p = - 3 . AV nearly. i.e. if the volume be 20 cub. ins. and a small increase be made in the volume, the decrease in the number of Ibs./sq. in. in the pressure will be nearly 3 times the increase in the number of cub. ins. in the volume. Suppose for example the volume increases from 20 to 20'1 cub. ft., the pressure decreases approximately 4 3 Ibs./sq. in., i.t). from 60 to 59-7 Ibs./sq. in. [Actually the decrease is '2985.] 51-53] DIFFERENTIATION FROM FIRST PRINCIPLES 57 .... dp 1200. (111) -j-- -- -j- is accurately true and is merely a convenient A/? way of saying that ~ can be brought as near as we please to if AV be made small enouh. ,o dv is a formula giving the rate at which p is changing with respect to V for any assigned value of V. i.e. when the volume is 10 cubic feet, . increasing 1 . (decreasing p is . . } 12 times as fast as V is < . decreasing J (.i (.increasing EXERCISES. IX. 1. A eq. ins. is the area of a circle of radius r ins. What is the relation between A and r ? If r be increased to r + Ar, find the corresponding increase AA in A, and dA prove as above that -r-=2?rr. This tells us that approximately AA = 2n-r . Ar. Interpret this as a rule for the approximate area of a thin circular ring, and apply it to find approximately the area of a circular ring whose inner radius is 3 ft. and thickness ! inch. 2. Shew that the percentage error made in taking 2irr. Ar for the area 50. Ar of the ring is less than - . What is approximately your percentage error in Ex. 1 ? 3. The radius of a circle is increasing at the rate of 1 mm. per second. At what rate is the area increasing when the radius is (i) 1 mm., (ii) 1 cm., (iii) 20 cms. ? 4. What will be the approximate increase in area during the next tenth of a second after the radius reaches 20 cms. ? 58 CALCULUS FOR BEGINNERS [CH. II 6. The side of a square is measured and found to be 8 inches. If an error of '01 inch is made in measuring the side, find approximately the error in the calculated area. 6. The area of a circle grows at the rate of 2 sq. his. per second. At what rate is the radius growing (i) when the radius is 5 inches, (ii) when the area is 20 sq. ins. ? 7. If pv=80, find the rate of increase of p with respect to v when t;=(i) 10, (ii) 80. 8. OX, OY are two lines at right angles. A and B are points in OX, OY respectively such that the area of the triangle OAB is always 10 square inches. If A moves along OX at a constant speed of O'l inch per second, what is the speed of B (i) when OA = 8", (ii) when OB = S"? 54. Let P be a point whose coordinates are (x, y) on the curve y = a? and let Q be a point on the curve near to P. Its coordinates will differ slightly from those of P and may be denoted by (a; + Aa;, y + Ay). (Fig. 23.) Fig. 23. 54, 55] -j- AS GRADIENT 59 Then PR = Aa;, RQ = Ay. .*. tan RPQ or tan XKP1 Aw r r, I" = T [ V - n te On Sca H 25 ]' or gradient or PQ ) Aa: L Now y + Ay = (a; + Ace) 2 = a; 2 + 2a;Aa; + (Ax) 2 , .'. Ay = 2x . Ax + (Ax)*, .'. t^=2a; + Aa;; Ace i.e. gradient of PQ= 2a; + Aa;. Now suppose Q to move along the curve towards P, so that Aa; and Ay become smaller and smaller, then by bringing Q near Aw enough to P, we can make - as near to 2x as we please. Ace Av The limit which ^ continually approaches and from which it can be made to differ by as little as we please if dv we make Ax small enough is called -^ . (iij Thus in this case ~= 2cc, and this is called the gradient of CL3C the curve at the point P. 55. Notice, \iy = a? A?/ (i) - = 2o3 + Aa; is an accurate statement whatever the value Ace of Ace, e.g. take x = 1, Ace= -2, and we have 2-2 as the gradient of the chord joining the point for which x = I to the point for which 1 ?/ (ii) ~ = 2cc is approximately true if Ace is small and becomes AdB more and more nearly true as Ace is made smaller, but there is no value of Aa; however small for which = 2a;, in other words if Ace 60 CALCULUS FOR BEGINNERS [CH. II we draw a chord through P cutting the curve in a point Q as near to P as we please, the gradient of this chord will never be exactly 2x, but can be made as near to 2x as we please if we bring Q /\ 1 1 near enough to P. For instance if x = 3, ^ = 6 nearly, i.e. the Ax gradient of a chord joining (3, 9) to a neighbouring point on the curve is nearly 6 and becomes nearer to 6 the closer this neighbouring point is to (3, 9). (iii) -~ = 2o; is accurately true and is merely a convenient dx A?/ way of saying that the unattainable limit which 2 is trying to reach is 2x. In fact 2x is the gradient of a line through P towards which the chord PQ is continually tending as Q approaches P, but with which it can never be made to coincide however near Q is to P. This line towards which PQ continually tends is called the tangent at P. dv Thus -^ gives the gradient of the tangent to the curve at P, or as it is sometimes called, the gradient of the curve at P. dv 56. The result ~ = 2x may be looked upon as a QiZ formula giving the gradient at any point of the curve. Thus, if 05 = 3, -^=6, i.e. the gradient of the curve at the point (3, 9) is 6. 57. Notice that although ~ is not the gradient of any ctx chord through P however near the other end of the chord may be to p, yet it is the gradient of an actual line through P, namely the line which we call the tangent to the curve at P. 58. We can now see better the significance of the approxi- mation in (ii). 55-59] dy -f- AS GRADIENT dx 61 Let P be the point (3, 9) and let Q be a point on the curve near to P. (Fig. 24.) Let PT the tangent at P cut QR in Q'. Then the statement is that the gradient of PQ is nearly 6, in RO RQ' other words is nearly 6 ; but since - = 6, this is equivalent RP RP to saying that RQ is nearly the same as RQ'. e.g. suppose PR = '02 then Q is the point (3-02, 9-1204). /. RQ= -1204 and RQ' = 6 x-02 = -12. Fig. 24. The error in taking RQ = RQ' is about ^ / . o If PR = -001 then Q is the point (3-001, 9-006001). .'. RQ = -006001 and RQ' = 6 x -001 = -006. The error is about -^ / o . 59. A caution similar to that issued in 41 applies here also, dy and dx have no separate meanings : they do not stand for small changes in length, for we have seen that however small the increments in x and y are supposed to be, their ratio is never exactly 2x. "We must not say that is the value of ^ when Aa; and dx AOJ Ay Ay are each zero. The whole process by which we obtained ~ 62 CALCULUS FOR BEGINNERS [CH. II becomes unintelligible if we imagine affairs pushed so to speak to the limit. The whole argument depends on the existence of a triangle PQR, which even though every side is out of the range of visibility can be supposed magnified so as to become visible, e.g. [with unit 1" along each axis] if x = 1 and Ao;= -0001 then Ay '00020001 and if we used a microscope magnifying 1000 times the triangle PQR would appear as a triangle in which PR = !" and RQ= -20001". But if we push Q to absolute coinci- dence with P, no amount of magnifying will separate the points P, Q, R. 60. If the equation of a curve is given as y = a?, we may look upon this as a formula for finding the value of y corre- sponding to any assigned value of x, or for finding the ordinate corresponding to any given abscissa. In other words it is a formula by means of which we may obtain the positions of any number of points on the curve. dij -j- = 2x is a formula giving the gradient corresponding to any given abscissa. e.g. if x = 5, y= 25 and ^ = 10. ax i.e. (5, 25) is the point on the curve whose abscissa is 5 and the gradient at this point is 10. 61. In drawing the curve y = y? it is a great help to make use of both formulae. Suppose we draw it between a? = 3 and x 3, we have the following table x -3 -2 -1 1 2 3 y 9 4 10149 g -6 -4 -2 2 4 6 Plot the points (-3,9) (-2,4) (-1,1) (0,0) (1,1) (2,4) (3,9). 59-61] dy - AS GRADIENT dx 63 Through these points draw lines whose gradients are 6, -4, 2, 0, 2, 4, 6 respectively. These lines then are to be tangents to the curve (Fig. 25). If we try to draw a curve passing through these 7 points and touching these 7 lines, we shall find that the run of the curve is pretty well determined. 64 CALCULUS FOR BEGINNERS [CH. II EXERCISES. X. 1. Use the result just obtained to get the gradient of y = x 2 at the points where ar=-3, -2, -1, 0, 1, 2, 3. Draw the graph j/=x 2 , and through the points where x=-3, -2, -1, 0, 1, 2, 3 draw lines with gradients just found. They should be tangents to your 1" curve. Take 1" as the a;-unit and - as the y-unit. m 2. Find the gradient of j/ = x 3 at the points where x=-B, -1, 0, 5. Shew that the gradient of t/=x 3 is never negative. Draw a figure. 3. Find the gradient of y=x*-4x at the points where x = 0, 1, ^4, 2. Draw a figure. 4. If y=x 2 , what is -p when x = 7? Suppose we move along the curve (tvG from the point where a; = 7 to the point where x = 7'03, what is the gradient of the chord joining these points ? What would be the percentage error if we took the value of -^ just obtained as the gradient ? else 6. Shew that the percentage error when 2x is taken instead of 2x + Ax or 2x . Ax instead of 2x . Ax + (Ax) 2 is less than - and if x 2 + 2x Ax be /Ax\ 2 taken instead of (x + Ax) 2 the percentage error < 100 ( I . Use these results to find approximately the p.c. error (i) if 2-01 2 be taken as 4-04 ; (ii) if7'03 2 49-42; (iii) if the gradient of the chord joining the point where x = 5 to the point where x = 5'02 be taken as 10; (iv) if RQ in Fig. 24 be taken as equal to RQ', P being the point (10, 100) and PR = -5. 6. Find (i) l*O (/ \*V 62. The process by which we obtained -j- in 39 ; -j- in cCt else (I'D 46 and 54, -j- in 50 should be carefully studied, as it is the same in all cases. 62-64] DIFFERENTIATION FROM FIRST PRINCIPLES 65 Taking the first case : We had s given as a function of t [s = 5 + 3t + 2t s ]. (1) We gave t a small increment (Atf) and found the corresponding increment in s As (2) We found the ratio . A^ As (3) We found the limit to which tended as A< was made A smaller and smaller, and from which it could be made to differ by as little as we please by taking A< small enough and this limit ds 63. Generally if a quantity y be a function of another quantity x, a small change, Ax, in x, will produce in y a corresponding small change which we call Ay, and in Av the cases with which we have to deal tends continually Ax to some limit from which it can be made to differ by as little as we please by making Ax small enough. This dv limit is denoted by -~- , and in the Differential Calculus the dZ ul/ fundamental problem is to determine - for different forms of ax the relation connecting y and x. dy 64. -p gives the rate of increase of y per unit in- crease of x or the rate of increase of y with respect to x for any value of x. It tells us that y is increasing so many times as fast as x. In the special case when y and x are replaced by s and t with ds the usual meanings, -y- gives the speed at a given instant. Ct'U dv Similarly -j- gives the acceleration at a given instant. at M. a 5 66 CALCULUS FOR BEGINNERS [CH. II If y and x are the coordinates of a point on a curve whose equation is the given equation connecting y and x, -=- gives the cLx gradient of the curve at any point. 65. If we draw the graph of [Fig. 26 shews it drawn between t = and t = 3.J 70 60 50 40 30 s 20 Fig. 26. 64-66] SPEED AND ACCELERATION AS GRADIENTS 67 ds the gradient at any point corresponding to time t will be -j- citi or 3 + 6* J . ds But -r gives the speed at the end of time t. Cut Thus we see that if the space-time graph be drawn the speed at any instant is given by the gradient of the graph at the point corresponding to that instant. (See 27.) e.g. in the figure Q is the point corresponding to t = 2, s = 27. The gradient of the tangent at Q is 27. The speed at the end of 2 seconds is 27 ft./sec. 66. Similarly if we draw the speed-time graph the gradient at any point is [shew that this is 12<] and -=- dt dt gives the acceleration in ft./sec. a V 40 Fig. 27. 2 3 t 52 68 CALCULUS FOR BEGINNERS [CH. II Thus if the speed-time graph be drawn the acceleration at any instant is given by the gradient of the graph at the point corresponding to that instant. e.g. Q' is the point corresponding to t= 2, v= 27. The gradient of the tangent is 24. The acceleration at the end of 2 sees, is 24 ft./sec. 2 67. Generally if y is given as a function of x [or y=f(x), f(x) being algebraic shorthand for a function of x\ and if the graph of y=f(x) be drawn, the rate at which y is increasing compared with x for any value of x is the gradient of the graph at the point corresponding to this value of a?. Thus if we could draw accurately the graph of y f (x) and also draw accurately the tangent at a given point we should be cty able to read off the value of - for the value of x corresponding MX to that point. EXERCISES. XL [To be done as accurately as possible by drawing tangents ; v. 67.] 1. Draw the graph of y = */25 - x 2 or a: 2 + j/ 2 = 25. It is a circle centre at the origin, radius 5. What is ^ when x = 1, 3. - 4 ? ax 3. Draw the graph of j/=- between x = and a; =5. What is ^ when x=l, 2 ? dx 6?77 68. Example. Find from first principles -j- when y = x*. CX/3C Let x receive a small increment Aa; and let the resulting increment in y be Ay. 06-68] DIFFERENTIATION FROM FIRST PRINCIPLES 69 Then y + Ay = (x + Aa?) 3 = a? + 3a; 2 . Aa: + 3a; (Ai) 2 + (Aa:) 3 and */ <&> .'. Ay = 3 2 . Ax + 3a; (Aa;) a + (Aa:) 3 . .'. ^ = 3a; 2 + 3 . A + (A) 2 . A Now if Aaj be made smaller and smaller, each of the last two terms on the right can be made as small as we please by taking Aa; small enough. A?/ .'. - can be made as near to 3a? as we please by making Aa? small enough. EXERCISES. XIL 1. Find -p if y = k . x 3 where k is any constant. dx Make use of the result of Ex. 1 to solve the remaining questions. 2. If a body is travelling ia a straight line and its distance from a fixed point in the line at the end of t seconds is given by s = 4t 3 , where s is the number of feet, find the speed at the end of 3 seconds. 3. If the graph y = 5x? be drawn, what is the gradient at the point (2, 40) and what is the equation of the tangent at that point ? 4. If the edge of a cube be increasing at the rate of '001 of an inch per second, find at what rate the volume is increasing when the edge is 2 feet. 6. If the radius of a sphere be increasing at the rate of 3 inches per second, at what rate is the volume increasing when the radius is 3 feet ? 6. If the edge of a cube be measured and found to be 8 inches and if an error of 5^ inch has been made, what is the approximate error in the U calculated volume? 7. If an error of ! / is made in measuring the radius of a sphere, find approximately the perceutage error in the calculated volume. 70 CALCULUS FOR BEGINNERS [CH. II 69. The result obtained in 68 that if y a? Ay = 3ar>Aa; + 3x (A*) 2 + (Ax) 3 may be geometrically illustrated as follows. The edge of a cube is x ins. and is increased by Ax ins. On inspection of Fig. 28 the volume added will be seen to be made Fig. 28. up of 3 slabs each'of volume x 2 Ax, 3 bars each of volume x (Asc) 8 and a cubical block of volume (Ax) 3 . /\/y* The ratio of a bar to a slab is . and the ratio of the small x A 3? block to a bar is . x G9-71] DIFFERENTIATION FROM FIRST PRINCIPLES 7l When we say that the increase in volume is approximately 3o; 2 Aa; we omit the bars and the small block. Suppose for example the edge of the original cube is 10 ins. and that it is increased by Ao; 1 1 in. [i.e. x= 10, Aa; = -1, so that = ^. .1 a; 100 J = 300 x -1 = 30, (Az) 3 =-001. So that we take 30 cub. ins. as the increase in volume instead of 30-301. 70. Another illustration of the fact that if y = a?, then dy -3a* "1 - C/iA/ ax If x = 3, y = 27. Take an increased value of x and find the increased value of y as shewn in the table : x + Ax y + Aj, Ax Ay Ay Ax 3-5 42-875 5 15-875 31-750 3-2 32-768 2 5-768 28-84 3-1 29-791 1 2-791 27-91 3-01 27-270901 01 270901 27-0901 3-001 27-027009001 -001 -27009001 27*009001 Thus as A* is made smaller - continually approaches 3x* and can be made as near to it as we like if we take Aa; small enough. 71. Sign of ~ . In 8 50 -r? turned out to be negative dx cN because p decreased when V increased. Generally : Suppose y = any function of x. CALCULUS FOR BEGINNERS [CH. II Ay If - is positive A.r and Ay have the same sign ; i.e. an increase in x produces an increase in y and a decrease in x produces a decrease in y. Suppose the graph of y =f(x) be drawn, P being the point (x, y) and Q (a; + Ace, y + Ay). Then if Ax and Ay are both positive P and Q will be placed as in (1), if both negative as in (2). (Fig. 29.) Ay- Ax-f- rig. 29. In each case the chord PQ has a positive gradient. But if A* be + and Ay , P and Q will be placed as in (3), if Aa; and Ay +, as in (4). In each case the chord PQ has a negative gradient. dy So, if is positive for a given value of x, x and y are 71] dy SIGN OF -f- das 73 dv both increasing or both decreasing, but if - is negative, u,Jx x is increasing and y decreasing or vice versa. di/ In the graph if -~ is positive at the point (x, y) the gradient (KB of the tangent is positive and the curve in the neighbourhood of ftV P is shaped like (5) or (6), but if -~ is negative, the shape is like (7) or (8). (Fig. 30.) Fig. 30. 74 CALCULUS FOR BEGINNERS [CH. II EXERCISES. XIII. 1. Find from first principles -^- when y = - . Illustrate your result by making a table like that in 70. [Use a table of reciprocals.] 2. Find from first principles ~ when y = x + - . What is -~ when x=l and when x = 2 ? dx Draw the graph of y=x + - between x = - and x=3, and illustrate your X Ji answer by reference to this graph. 3. What is the equation of the tangent at the point (2, 2J) on the curve y = x + - ? x ds 4. Find from first principles when s= 3 + 2i + 1 2 . (tt If a body move in a straight line, so that its distance from a fixed point at the end of t sees, is given by s = 3 + 2t + t 2 , find its speed at the end of (i) 3 seconds, (ii) 10 seconds. 6. Find from first principles -^- : CiX (i) ify=p, (ii) ify=2*3+3. Find the equations of the tangent and normal to each of the curves 7 y = -j and j/=2x 3 + 3 at the point corresponding to x = l. 6. Find from first principles -^ if y = *Jx~. ["Use JxTh- ^J^*) = L Jx + h+Jx v/ar + A+x J 7. Find from first principles -p when j/ = fc (a constant). ~ 8. From first principles, shew that if y =#' then -^- = -x dx o r n . , , . Ay 2x + Ax "I [Put y*=x> and get J - %2+% Ay + (Aj/)2 .J 9. From first principles, shew that if a; 2 +u 2 = a 2 , then ^= dx T . Aw 2z + Aa; ~1 _ . Get -= - - - -. Interpret this result g ;ometrically. CHAPTER III DIFFERENTIATION OF a* (ll/ 72. IN future instead of saying that " -2- is the quantity which ! continually approaches as Ace continually approaches zero, and from which we can make it differ by as little as we please by making Aas small enough " we say that " -^- is the limit of - (or Lt. - j when Ax is indefinitely diminished," but it must be constantly borne in mind that this is merely an abbreviated form of the longer statement. 73. A sign which has been recently introduced into mathematical text-books enables us to compress this phrase still further. The sign is -* . z * a means that z continually approaches a and can be brought as near to a as we please. Thus in 68 we have -^= 3x 2 + 3x. Ax + (Ax) 2 and we may say Ax Ay that as Ax -^ 0, - -*- Sx 2 , meaning that as Ax approaches zero, - continually approaches Sx 2 and can be made as near to 3x 2 as* 76 CALCULUS FOR BEGINNERS [CH. Ill we please if Aa; be brought near enough to zero. The statements in inverted commas in 72 may be conveniently written rf y_ Lt *y ~T~ *** A - Aa;-0 Ax A?/ so that when we have found that as Ax - 0, ~ -*- Bar 2 we may say ^=3ar>. J oa? 74. If y = x, find S. Let as receive a small increment Aa; and let the corresponding increment in y be Ay. Then y + A^ = (03 + '&x) n , and y = a;", .'. Ay = (a; + Aa?) n - aj n . [Now it is proved in algebra that if h < x (x + h) n = x n + nx n ~ l h + n ( n ~ l ) x n-w +><> for all values of n : also that starting from any term and taking any number of terms however large, the sum never exceeds a finite limit.] .'. A?/= nx n ~ l (Aa;) + . ^af~ 2 (Aa;) 2 + terms containing 2i higher powers of (Aa;) = nx n ~ l Aa? + (Ax) 2 x L, where L is a finite quantity. .*. -^ = nx"- 1 + L . Aa;. Aa; AV .*. as Aas -*- 0, -^ - naf-*. Aa; or = nx n ~ a . dx 73-76] DIFFERENTIATION OF x n 77 75. The proof of the Binomial Theorem for indices Avhich are not positive integers is a very difficult piece of mathematics, which will probably never be met with by many readers of this = x n , = which does not depend on this theorem, such a proof is subjoined. book. For those who prefer a proof that if y = x n , = nx 76. We shall first establish the following theorem : For all values of n, positive and negative, integral and z n 1 fractional,' as z approaches nearer and nearer to 1, z 1 approaches nearer and nearer to n and can be made as near to n as we like if z be brought near enough to 1, or shortly : t> n 1 T z L lit. - - ~n. 2^1 s-1 (1) Let n be a positive integer. Then by actual division z n -l - - = z n - l + z n -* + z n - 3 +...+ z + 1. z 1 There are n terms on the right, each of which can be made as near to 1 as we please if z is brought near enough to 1. z 1 or 1-P-l -331 = 331 wvi -3-0301, 1-1-1 ~ -1 ' 1-OP-l -030301 1-01 - 1 -01 etc., and in this case n = 3. tn (2) Let n be a positive fraction, say n = - where p and q are positive integers. 78 CALCULUS FOR BEGINNERS [CH. Ill p Also let zP = u, .'. 3 = u q and z = u p . Z-\ W-1 ~ W-l As a -*- 1, so does w. Since ^ and g- are positive integers, , W P -1 , . . as -*!, -- ?--*-j9 and a w -l p .. Lib. - r =- or n. 1-1- 1 1-0741-1 -0741 e -s- TTT r = 1-1-1 ^r - ' 741j 1 -01 s - 1 1-00749-1 _ -00749 _ 1-01-1" 1-01-1 -01 ' and in this case = ='75. 4 (3) Let i be negative = - m where m is positive. 1-1 " " 2-1 ~ a-1 "T^T """*' a-1 ' z m 1 Since m is positive, Lt. -- = m, and Lt. z m = 1. -*! z ~ 1 s-fl z n 1 .*. Lt. - 7- = m or w. *i -! M**-l -9535-1 ag - : -- " 465 99504- 1_ 1-01-1 -01 and in this case n = - = - -5. 76, 77] DIFFERENTIATION OF X n 79 77. Now with the notation of 74, Ay _ (x + Aa;) re - x n Aa; (x + Aa;) - x / ( x x + Ate ( x ~n 1 Z 1 n ) H ) X 4- where z stands for x Now as Ace - 0, or z -> 1 and the limit of ~, is x 2-1 therefore n. . -f. % ._! Ijt. = nx , -5 = nx" dx This is a most important formula and includes all the results we have obtained hitherto. e.g. H,.<* = W. EXERCISES. XIV. 1. Write down ~ when dx dy dx = sf>, x 20 , x, a;- 3 , ^ar, ., x 1 -*, x-*, x, 5, 2. In the above piece of work ( 74 or 77) what difference would it make if y = kx n where k is a constant, i.e. some number independent of x? 3. What difference would it make if y = x n + c where c is a constant? 80 CALCULUS FOR BEGINNERS [CH. Ill 78. The results of the last two examples are very important, (i) Taking y = kx n we should have y + Ay k (x + &x) n . .'. Ay = [(x + Ax') rt - x n ]. i.e. the value of Ay in this case is k times its value in the case when y = x n . The rest of the work is the same to the end, except that this factor k remains, and we get eventually ^ = knx"- 1 . dx (ii) Taking y = x n + e, we should have y + Ay =- (03 + Aa;) n + c. .*. Ay = (x + Ace) 71 x" exactly as when y x n . dy .'. in this case ~ = nx"" 1 . dx EXERCISES. XV. en j/=3x 5 , 2x, -, ax x I. Write down ^ when j/=3x 5 , 2x, -, 5Jx, 2a;2 + 3, - -8, - 1 2. Write down if =-t 2 + 5. di o a. Write down - if pv* >4 = 500. dv 4. Write down if R= k (1 + a0) where k and a are constants. 79. Graphical illustration of the result of 78 (i). Draw on the same sheet, using the same scales y = a; 2 ...(l)andy=3 2 ...(2). [Fig. 31.] In (1)^ = 2*. In (2)^ = to ^ ' dx ^ ' dx 78/79] DIFFERENTIATION OF kx n 81 So that at the point (2, 4) on (1), the gradient of the curve is 4 and at the point (2, 12) on (2), the gradient of the curve is 12. i.e. the gradient of (2) at any point is 3 times the gradient of (1) at the point with the same abscissa, i.e. at the point vertically under it. II. 0. 82 CALCULUS FOR BEGINNERS [CH. Ill EXERCISES. XVI. 1. Shew that the tangents at (2, 4) and (2, 12) to the curves y=x z and y = 3x 2 meet OX in the same point. 2. Shew that if a tangent be drawn to y = x 2 at the point (c, c 2 ) and a tangent to y kx 2 at the point (c, kc 2 ) the gradient of the second tangent is k times that of the first and that the two tangents meet OX in the same point. 3. Make and prove similar statements about y x 3 and y = kx* and generally about y = x n and y = kx n . 80. Kinematical illustration. Suppose that two bodies are moving in the same straight line and that the distance of the first from a fixed point in the line at the end of a given time is determined by the equation s = fi, and that the distance of the second body from the same point is given by s = 3t 2 ; then the speed of the first at any instant is 3 times that of the second at the same instant ; in fact, in any interval of time however small, the second body moves 3 times as far as the first. 81. Graphical illustration of the result of 78 (ii). Draw on the same sheet, using the same scales [Fig. 32] 2/=ic 2 ...(l) and y = x*+2. ..(2), then in each case -^- = '2x, dx so that at the point (2, 4) on (1) the gradient is 4 and at the point (2, 6) on (2) the gradient is 4. i.e. the gradient at two corresponding points is the same. This is obvious from the fact that (2) may be obtained by sliding (1) bodily parallel to OY through a distance 2. 82. Kinematical illustration. Suppose that two bodies are moving in the same straight line and that their distances from a fixed point in the line at the end of a given time are 80-83] DIFFERENTIATION OF X n + C given by the equations s = t 2 and s = t 2 + 2 respectively, then their speeds are the same at every instant, for the second body is always the same distance, 2 feet, ahead of the first. 12 10 QY 2 3 Fig. 32. 83. We found in 39 that if s = 5 + 3t + 2 s , ds dt fjo fiQ /7Q Now if s = 5,^ = 0; if s = St, 33 = 8; if s=W, ^= (tt Ctt (Aft ds d(5) d(3t) d(2f) so that in this case -^ -- ~-i + \ ' + \ ' . dt dt dt dt 6-2 84 CALCULUS FOR BEGINNERS [CH. Ill Generally if y = u + v where u and v stand for two functions of x whose differential coefficients we know, dy du dv dx dx dx ' for if x receive a small increment ACT, there will be a resulting small increment in u which we may call Aw, and a small increment in v which we may call A-y, and Ay, the increment in y, is equal to the sum of these. In fact y + Ay = u + AM + v + Aw, so that Ay = AM + Aw. A'// A?t i\v ^ y. _j_ ^ Ao; A# Aa: ' If Aos be made smaller and smaller -- and - can be made A Aa; to differ by as little as we please from -=- and -7- respectively. ax dx Ay du dv . . . - can be made to diner from -y- + -j- by as little as we Ax- ax ax ' please; ie. dy _ du dv dx dx dx ' dy du dv dw Similarly if y = u + v + w+ ... , -j- = -, l--y- + -5- + ... . dx dx dx dx EXERCISES. XVII. 1. Write down ^ (i) if j/=3 + 2a; + 4x2, (ii) if y = 5-- + \. uX ' X X 2. Writedown^ if *= 5(3- Jt2 + i. (it '2 3. Draw on the same sheet the graphs of y = 3, j/ = 2a;, y = 4a; 2 , and j/ = 3 + 2x + 4a; 2 , and verify that at any point [say where x=l] the gradient of the fourth is equal to the sum of the gradients of the other three. 4. If y = ox 2 + bx + c, shew that f - \ = 4aj/ + fc 2 - 4ac. 83, 84] DIFFERENTIATION OF X n 85 84. Example. Find (i) the gradients of the tangents to the curve y = 2a; 2 -3o;-l at the points where x = 1, 0, 1, 2 ; (ii) at what point the tangent is parallel to OX. Also find y when x = 1, 0, 1, 2 and make use of all this information to draw the graph of y = 2X 2 3x 1 between x = 1 and x = 2. y = 2x* 3x 1. .'. we X ~ l 1 2 y 4 -1 -2 1 dy dx -7 -3 1 5 Also the tangent is parallel to OX when the gradient is 0, 3 i.e. when x = -r, and for this value of x, y - 2^. Plot the points (- 1, 4) (0, - 1) (1, - 2) (2, 1) (| , - 2^ and through these points draw lines whose gradients are -7, 3, 1, 5, respectively. Then our curve must go through all these points and touch all these lines. [Fig. 33.J EXERCISES. XVIII. 1. If v=x3-3x, what is ? Find the values of y and - corre- dx Q-* spending to x -2, -1, 0, 1, 2, and draw the curve y = x 3 -3x between x= -2 and x = 2. 3. If i/=a;*-4a;3-f-4x2-3 what is -j-1 Find the values of y and ^| when x= -1, 0, 1, 2, 3, and draw the curve y = x* - 4z 3 + 4*2 - 3 between x= - 1 and x = 3. 86 CALCULUS FOR BEGINNERS [CH. Ill 3. If y = 2a: 3 -9*2 + 12* -3 find ^. Also find the values of y and -^- when x = 0, 1, 2, 3. From these data, draw the graph of y = 2x 3 -$x- + l'2x-3 between a;=0 and x=3. 4. Find the points on y =x* - 3x 2 + 2x at which the gradient is zero and draw the graph between x=0 and x=3. Fig. 33. 85, 86] APPLICATION TO GEOMETRY 87 85. In the following examples we make use of the following theorem : If a line whose gradient is m be drawn through the point (h, k") its equation is y k = m (x - h). That this is so is easily seen, for (1) The equation represents a straight line since it is of the first degree in x and y. (2) It is satisfied when x = h and y = k and therefore passes through (h, k). (3) Its gradient is m since it may be written y mx + (k mh). e.g. the line through (3, 5) whose gradient is 4 is y - 5 = 4 (x - 3) or y = 4x - 7. 86. Ex. 1. Find the equations of the tangent and normal (perpendicular to the tangent) at the point for which x = 3 on the curve y ~ 2x + 3x 3 . We have ^ = 2 + 9a; 2 . ax Whena;=3,y = 87 and ^ = 83. ax .'. the tangent is the line through (3, 87) whose gradient is 83. .'. its equation is y 87 = 83 (x 3) or y = 83x-162. The normal is perpendicular to the tangent. .'. its gradient is ^ , oo and its equation y - 87 = ^ (x - 3) or x+83y = 7224. 88 CALCULUS FOR BEGINNERS [CH. Ill or 2 Ex. 2. In the curve y=~, P is the point where x = h. Find the equations of the tangent and normal at P. 3C fLfJ 3f* Wehavey = , -'.^=77-. 4a ax 2a W , dy h .. when a: = ft, y=-r and -~- = -. la dx 2a . h ls ~ (h^\ h, J whose gradient ., ,. n? h . .. . . its equation is y - = (x - h), or 2a or The normal is the line through (h, - J whose gradient 2a '. its equation is y = =- (x - Ji), ^rr. 3. From the results of Ex. 2 we can deduce some a; 2 interesting properties of the curve y = . (Fig. 34.) 4:C& First we notice that the curve passes through the origin, and is symmetrical about OY, i.e. the same value of y is obtained by putting x = h as by putting x h. Also y is always positive. / 7t 2 \ Let P be the point f 7t, -: ) , and let the tangent and normal at P meet OY in T and G and let PN be perpendicular to OY. 86] APPLICATION TO GEOMETRY h* 80 In (1) put x = 0, .*. y 1U> f A 2 \ A a h* i.e. co-ordinates of T are ( 0, -r- ) or OT = -: , but ON =7-. \ 4a/ 4a 4a /. OT = ON. Fig. 34. A In (2) put rr-0. .'. y = +2a, i.e. the co-ordinates of G 4# are ( 0, /- + 2a] or OG = -j- + 2a, but ON =- . \ 4a / 4a 4 a .'. NG = 2 a, i.e. NG is the same for all positions of P. 90 CALCULUS FOR BEGINNERS [CH. Ill Again if S is the point (0, a), h? ST - SO + OT = a + 7- , 4a h? and SG =OG - OS = a + -j . 4a /. ST = SG, ie. S is the centre of the semi-circle on TG as diameter. But this semi-circle passes through P, since TPG is a right angle. .'. ST = SG = SP. EXERCISES XIX. 1. Taking the unit as -^ along each axis and a = l, what is the m equation of the curve in Exs. 2 and 3, 86? Draw the curve from x= - 4 to x= + 4. Let P be the point at which x=3a. Take OT = ON and verify that TP is the tangent at P. Take NG = 2a and verify that GP is perpendicular to TP. 2. If P be any point on j/= , S the point (0, a), KK' the line y= - a, PM perpendicular to KK', prove SP=PM. [This curve is called a parabola, S is its focus and KK' its directrix.] 9. In your figure to Question 1 shew the focus and directrix, and verify SP=PM. x z 4. If the tangent at any point P on y= meets the directrix in Z, Qct shew that PSZ is a right angle. 6. Find the equations of the tangent and normal at the poiut (1, 1) to the curve y = x 3 . If the tangent meet OX and OY in T and ( respectively and the normal meet OX and OY in G and g respectively, find the co- ordinates of T, t, G, g. 6. Find the equations of the tangent and normal at the point where x = 2 on the curve t/ = 3a; 4 : also at the point where x=3 on the curve 86, 87] THE FUNCTIONAL NOTATION 91 / h s \ x 3 7. P is the point f h, -| 1 on the curve y = -%. PN and Pn are perpen- dicular to OX, OY. The tangent at P meets OX, OY in T, t and the normal meets OX, OY in G, g. Prove OT = ON, Ot = 2On, ON. NG=3.O7i 2 , Ore.n</ = o 8 8. Find the equations of the tangent and normal at each of the points on the curve y 2 -=x 3 where x = 4. Draw a figure. 8. Find the equations of the tangent and normal to the curve at the point where x= 3. 10. The equation of a curve is x + y = a%x3. Find the gradient of the tangent at the point where x l&a. Find where this tangent meets the line x+y 0. 11. Find the equations of the tangent and normal at the point where .T = 2 on the curve y = 3x + a; 2 . If the point be P and if the tangent and normal meet the y-axis in T and G respectively find the area of ATPG. The functional notation. 87. The fact that y is a function of x is sometimes expressed in the form y=f(x), e.g. if y y?~ Ix + 8, f(x) is x- 7x + 8. /"(2) means the result of substituting 2 for x in this expression, i.e. /(2) = 2 2 -7x2+8-- 2. Similarly /(0)=8, /(-3) = 38, f(2p) = 4p 2 - 14jt> + 8, f(z> + 2) = (z 2 + 2) 2 - 7 (s 2 + 2) + 8 = z 4 - 3z 2 - 2, etc. Other notations sometimes used are F (x), < (x), etc. It is very important to bear in mind thaty^a?) does not stand for the product of x and some quantity f, but is really an instruction written in shorthand to perform a certain operation or series of operations on x. Thus in the above example where f(x) stands for .r 2 7x + 8, the symbol/ placed before a number means square the number, subtract 7 times the number and add 8. 92 CALCULUS FOR BEGINNERS [CH. Ill EXERCISES. XX. 2. If 0(x) = So? -2*2 + , find 0(6), <f>(y + l). cc 3. Itf(x)=x + sinx [x being in radians] find/f |Y/(1),/(0). 88. If y=f(x), the differential coefficient of y with respect to x may be written -^ or 7 or /' (a;). rfcc ox Thus if ^Sa^ f (x) is itself a function of or, and we may therefore evaluate such expressions as /' (3), /')),/'(* + 3), etc. e.g. /'(3) = 9x3 2 -18x3+7 = 34. With our usual notation we should say that this was the d'U value of -r- when x = 3. ax Thus we may say that the gradient at any point of the curve y =f(x) is /' (a:), or that the gradient at the point for which x = a isf (a). Similarly if s=f(t), ds -j- may be written y (<), and we may say that /' (t) gives the speed at the end of time t. 88-90] HIGHER DIFFERENTIAL COEFFICIENTS 93 Higher differential coefficients. 89. If y =/() then or/' (x) is also a function of x and may therefore itself be differentiated. The result of this differentiation may be written dx df'(x) dx dx d?y These are abbreviated into - and/" (x) respectively. e.g. if y =/() = 2x 4 -3x 3 + 2x, 90. Just as -p or/' (x) tells us the rate of increase of y orf(x) CvX d*y with respect to x, so -7^ or /" (x) tells us the rate of increase of DNv -, or/' (x) with respect to x. dx Considered in relation to the graph y =/(*), -~- or/' (05) gives cLx d?u the gradient at any point on the graph ; -~ or/" (a;) gives the doc? rate at which the gradient is increasing with respect to x. If in the last illustration we put x = 2, we get /(2) = 2x2 4 -3x2 s + 2x2 = 12. /' (2) = 8 x 2 3 - 9 x 2 2 + 2 = 30. /"(2)=24x2 2 -18 x2 = 60. CALCULUS FOR BEGINNERS [CH. Ill i.e. at the point (2, 12) on the curve y = 2o^ - Sx 3 + 2x, y is increasing 30 times as fast as x, or the gradient of the curve is 30, and the gradient is increasing 60 times as fast as x. (Fig. 35.) 30 '"- ztjZ^^d: 20 ;;;;li;ii;lil nf---T : 10 1 i iiiiiiii o:;; l f~-2~^-.^ 60!; jijjijijj ; ;;! 40::: 20;:; :;|i ! ill!!!!!!';; jiii jiji; - 3^ 100 :: :::::::::::::::::! JCTjto'^^i HJ:;;; | i:;;;;;! O 1 i 4 2 3 igs. 35, 36, 37. Or we may say that at the point (2, 30) on the curve = Bar* - which is called the first derived curve of y = 2x* - 3x? + 2x, the gradient is 60. (Fig. 36.) 90-93] DERIVED CURVES 95 91. If we draw the graphs y =/(), y=f'(*\ y =f" (*)> the number of units in the ordinate of y =/' (x) corresponding to any value of x gives the gradient at the corresponding point on y =f(x), and the number of units in the ordinate of y =f" (x) gives the gradient at the corresponding point on y =f (x}. e.g., looking at Figs. 35, 36, 37 which give the graphs of y =f(x), y =f (x), y f" (x) for the particular case when f(x)= 2x 4 -3x s + 2x, P, P', P" are the points corresponding to x 2. The ordinate of P" is 60 units and 60 is the gradient at P'. The ordinate of P' is 30 imits and 30 is the gradient at P. ds 92. If s=f(t), -j- or f (t) (sometimes written s) gives the ut speed at end of time t. cti} If we call this v we have seen [v. Ex. (5) p. 51] that -y- gives ttt the acceleration at end of time t. dv . cPs . <Ps Thus -j-^ or f" (t) [sometimes written s] gives the acceleration ut at end of time t. The graphs of f(t), f (t), f" (t) are respectively the space- time, speed-time and acceleration-time graphs. 93. We get interesting special cases when f(x) is of the 1st or 2nd degree. e.g. if /(a) = 2o: 2 -3a:-t-U / = 4*-3> and f" (x) - 4. J 96 CALCULUS FOR BEGINNERS [CH. Ill The graphs of y=f(x), y=f'(x), y-f"(x) are shewn in Fig. 38. y f ( x ) is a straight line, and the gradient at every point is the same. y =f" (a;) is a straight line, the ordinate at every point being the same, i.e. it is a straight line parallel to OX. 1 2 Fig. 38. In the general case, if f( x } =a y ? + bx + c, where a, b, c are constant, f (x) = 2ax + b t and f" (x) = 2a> 93] DERIVED CURVES 97 If s =f(t) = at 2 + bt + c, v =/' (t) = 2at + b, and a =/" (t) = 2a. i.e. if s be a quadratic function of t, the acceleration is constant. In this case the speed-time graph is a straight line and the acceleration-time graph a straight line parallel to OX. If f(x) = 4:X + 3, /'(a) = 4, /"(*)= a The graphs are shewn in Fig. 39. y =f(x) is a straight line, y =f (<) is a straight line parallel to OX, y ~f" (x) is OX. Generally if f(x) = ax + b t where a and b are constant, / (x) = a, /*(*)-<>. If s =f(t) = at + b, v=f'(t) = a, a=f"(t) = 0. i.e. if s be a linear function of t the speed is constant and the acceleration zero. EXERCISES. XXI. 1 . If / (*) = 2z + - , find /' (x) and /" (x) . X Alsofind/(2),/'(2),/"(2). Draw the graphs y=f(x), y = f (x), y=f"(x) between z=-l and *= + !. Make a statement like that in 91, taking x- ^ . 2. If = 3-4 + 2t 2 + t 3 find the speed and acceleration at the end of 3 seconds. Draw the space-time, speed-time and acceleration-time graphs. M. C. 1 98 CALCULUS *OR BEGINNERS [CH. Ill fjm 94. We have seen (71) that if -^- is positive, y is increasing dy . as x increases, but that if -f- is negative y is decreasing as x increases. Q. .. . ... dx) . dy . . Similarly if ^ or - - is positive, - is increasing as x increases, but if ~ is negative, ~- is decreasing as a? increases. 94] dy d?v SIGNS OF -/ AND -~ dx da? 99 Now if between P and Q a curve has a shape as in Fig. 40 (a) it is clear that as x increases from OM to ON, y increases and also Ul/ the gradient or -~ increases. [The positive directions of the x CvCC and y axes are supposed to be as usual 'right' and 'up' respectively.] (a) (b) (c) (d) Fig. 40. i.e. for all points between P and Q dy . . (Fy . -?- is + and ~ is + . dx dy? 72 100 CALCULUS FOR BEGINNERS [CH. Ill Shew similarly that corresponding to a shape like Fig. 40 (6) dy . , o? 2 v . -f- is + and -r^ is . ax ax- For Fig. 40 (c) ^ is - and ^ is + . dx dx* For Fig. 40 (d) ^ is - and ^ is - . dx dx- 95. e.g. consider the curve y = 2x* 9x 2 + l'2x 3. We have ^-. = Qx 2 - 1 8* + 1 2 ax = 6 (ar-1) (x- 2), and ^=12a;-18. do: 2 For all values of x between 2 and 3, - is + and -^ is + . dx dx 2 .'. between x = 2 and x = 3 the curve is of the form shewn in Fig. 40 (a). For all values of x between and 1, -^ is + and -. is , dx dx 2 .'. between x = and x = 1 the curve is of the form shewn in Fig. 40 (6). For all values of x between 1A and 2, -~ is and ~ is + . dx dx 2 .'. between xl^ and x= 2 the curve is of the form shewn in Fig. 40 (c). For all values of x between 1 and 1 1, -^- is and -~ z is . CvSO CL3& .'. between x= 1 and x=\^ the curve is of the form shewn in Fig. 40 (d). [See Fig. 41.] 04, 95] du d 2 y SIGNS OF -f- AND -~, doc dx* Fig. 41. 101 EXERCISES. XXII. What is the form of the following curves ? : (i) between x = 2 and x=3, (ii) x = and x = l. : (i) a;=l and x = 2, (ii) x= -1 and x= -2, (iii) a; = 3anda; = 4. (i) when a; is positive, (ii) ,, negative. a. = 102 CALCULUS FOR BEGINNERS [CH. Ill MISCELLANEOUS EXAMPLES ON CHAPTER III. 1. Find from first principles (i) -^-ify = x^. (ii) -p if pv z =k (a constant). a. Find the equations of the tangent and normal to the curve y = - at each of the points (2, ^ ) , (5, = j and in each case find the area of the triangle formed by the tangent and the axes of co-ordinates. 3. Do the same for the point ( c, - ) . State a geometrical property of \ c J the tangent at any point of the curve. 4. P is the point (3, 9) on the curve y=x 2 . Find the equations of the tangent and normal at P, and the co-ordinates of the points T, G where they meet the y-axis. If PN is perpendicular to the ?/-axis, find the lengths of NT and NG. 6. A body moves in a straight line in such a way that its distance (s ft.) from a fixed point in the line at the end of t seconds is given by Find its distance from the fixed point, its speed and acceleration at the end of 5, 6, 7 seconds. 6. The same if =3 + t-5t 2 + 7t 3 . 7. In the curve j/ = 2a;+3rr 2 , find the gradient at the point P where x = 2. Find also the gradient of the chord joining P to Q, Q being the point where x =2 -01. The tangent at P and the line through P parallel to OX meet the ordinate of Q in T and R respectively. Find the percentage error made in using RT instead of RQ. Explain the connection between the statements RT = RQ approximately and Ay = ~.Ax approximately. ax 8. Find the equation of the tangent to the curve y = 3x* - 4x 8 - 12x 2 + 5, at the point where x 3. Also find the co-ordinates of the points at which the tangent is parallel to the x-axis. 95] MISCELLANEOUS EXAMPLES 103 9. P is the point on y = cx n (c a constant) at which x = h. The tangent and normal at P meet the x-axis in T, G, and the y-axis in t, g. PN, Pn are perpendicular to OX, OY. Find .. OT ., Ot .. ON.NG . On. ?i0 * (m) --' (lv) ON oS* -5S 10. A trough is in the shape of a right prism with its ends equilateral triangles. The length of the trough is 10 feet. It contains water, which leaks at the rate of 1 cubic foot per minute. Find in ins. /sec. the rate at which the level is sinking when the depth of water is (i) 3 inches, (ii) 1 foot. 11. A current C of electricity is changing according to the law C where t is seconds. The voltage V is such that V-RC+L.*?. where R = 0-5, and L = 0'01. Find V when t=2. 12. A man runs a given total distance in such a way that the time, t sees. , he takes to run any part, s feet, of the distance is given by the equation t=as n , a and n being constants. Compare his average speed over the whole distance with his speed at the end. 13. A body of constant mass m is moving with variable speed v. JLf If K is its kinetic energy and M its momentum shew that = M. State what your units are. 14. If pV=k (a constant) shew that Tn = r> andifpV n =&, ^ = ~ n U' 16. The coefficient of cubical expansion of a substance at temperature 6 is the rate of increase of volume per unit increase of temperature. The volume (V c.c.) of a gram of water at 9 C. is given by where a=8'38x!0- 9 . rfV Find and hence get the coefficient of cubical expansion of water at da C. and at 20 C. 104 CALCULUS FOB BEGINNERS [CH. Ill 16. The specific heat of a substance at temperature 6 is the rate of increase of Q per unit increase in 0, where Q is the number of heat units required to raise the temperature of 1 gm. from some standard temperature to0. The total heat required to raise the temperature of 1 gm. of water from to C. is given by Q= e + 2 x 10-6 02 + 3 x io-7 03. Find the specific heat of water at 80 C. 17. For diamond the formula is Q= -09470 + -0004970 2 - -0000001203. Find the specific heat of diamond at 80 C. 18. If y = 2x*-2x3-x2 + l find y when x=-l, - | , 0, 1, 2. Also find -T- when x= - 1, 2 and find what values of x make -;- = 0. dx dx Using all this information draw the graph of between x=-l and a; = 2. 19. In a certain case of straight-line motion, the number of feet in the distance from a fixed point, at a given instant t seconds after the start, is given by the formula Calculate (i) The average speed between the end of the 2nd and the end of the 4th second. (ii) The mean of the speeds at the instants 2 and 4 seconds after the start. (iii) The speed 3 seconds after the start. 2O. Find the equations of the tangent and normal to y = x 3 at the point (2, 8). If P is the point (2, 8) and if the tangent and normal meet the x-axis in T and G and PN is perpendicular to OX, find the lengths NT, NG. (h 2 \ x 2 h, 1 on the parabola y= ; PT is the tangent at P. PM is perpendicular to the line y = -a. S is the point (0, a). Prove SM is perpendicular to PT and meets it on the rr-axis. 22. A vessel containing water is in the form of an inverted hollow cone with vertical angle 90. If the depth of water be x feet, what is the volume of water ? If water flows in at the rate of 1 cubic foot per minute, at what rate is the level of water rising when the depth is 2 feet ? 21. 95] MISCELLANEOUS EXAMPLES 105 23. Find the co-ordinates of the point of intersection of the tangents to y = So; 2 + 5x - 7 at the points where x = 2 and a; = 5. Verify that this point and the mid-point of the chord joining the two points of contact lie on a line parallel to OY. 24. The shape of the vertical section of a hill is given (approximately) by the graph of y = -05x 2 + -25x from x = Q to x=6 and by the graph of 15 froma? = 6 to x = 12. Calculate the ordinates when a; = l, 2, 3, ... 12 and the gradient at each of the corresponding points. Shew that when x = 6 the gradient is the same for each part of the hill. Draw the figure taking - as the unit each way. m 25. Find the equation of that tangent to the parabola y = Bx 2 which is parallel to 'Ax - 2y = 7. 26. If s = 3t - 7t 2 + 16t 3 , s being the number of feet and t the number of seconds, find the formula for the acceleration at the end of t seconds. If 8 Ibs. is the mass of the moving body, what is the force acting on it at the end of 4 seconds ? 27. A body is moving in a straight line and its distance (.s feet) from a fixed point at the end of t seconds is given by the following table : t 1 2 3 4 5 6 s 7 22 45 76 115 162 Find graphically the speed at the end of 5 seconds. 28. A wheel in t seconds rotates through (5t + 4( 3 ) radians from some standard position. Find its angular velocity and acceleration after 5 seconds. 29. The side walls of a truck are vertical and its section by a plane parallel to the side walls is a trapezium, upper side 14 feet, lower side 10 feet, depth 3 feet. The width of the truck is 4 feet. It is originally full of sand (sp. gr. 1-5) which leaks out in such a way that the depth of sand diminishes at a constant rate of 1 inch per minute. Find the mass of sand in the truck after t seconds. If the truck is made to move in a straight line with a uniform speed of 15 miles an hour, what is the momentum of the sand in the truck at the end of t seconds ? What is the resultant force on the truck at the end of t seconds ? [Kemember force is given by rate of change of momentum.] Find the force at the end of (i) 5 minutes, (ii) 20 minutes. 30. If di/ find graphically the value of -2- when x = CHAPTER IV MAXIMA AND MINIMA 96. PLOT roughly the curve y = 2ac? - $a? + 12x - 3 between x = and x = 3. You will get a curve as in Fig. 42. s : - Fig. 42. d/u . Between A and B, -^ is positive, CHB B and C negative, C and D ,, positive again, 96, 97] MAXIMA AND MINIMA 107 or in other words as we advance from left to right, y increases up to B and then begins to decrease ; as we come to C, y goes on decreasing and then begins to increase. (71.) Points like B and C are called turning points and at B y is said to have a maximum value, and at C a minimum value. Notice, that although we call the value of y at B a maximum, yet this is not by any means the greatest value y can have. A maximum point is one where the ordinate (representing the function) stops increasing and begins to decrease; in other words the value of y at a maximum point is greater than the value of y at neighbouring points on either side, and similarly at a minimum point the value of y is less than the value of y at neighbouring points on either side. 97. Now suppose we wish to find the exact position of these turning points B and C. At a turning point the tangent to the curve is parallel to OX, i.e. the gradient of the curve is zero, or . ax dv .'. to find the turning points, we get -~ and find the dv values of x which make ^ = 0. dx e.g. in the case above :. -f- = when x = I or 2. dx When x = 1, y - 2 ; when x = 2, y = 1. .'. (1, 2) and (2, 1) are the turning points. 108 CALCULUS FOR BEGINNERS [CH. IV 98. In order to find whether a turning point corresponds to a maximum or minimum we may either draw a rough graph or proceed as follows : when x 1, y = 2 ; when x= 1 + h, y = 2(l + A) 3 - 9 (1 + A) 2 + 12 (1 + A) - 3 = 2 - 3A 2 + 2A 3 . Now if h is a small fraction, 27i 3 is small compared with 3A 2 , and the sign of 3A 2 + 2A 3 is that of the first term, and is there- fore negative whether h is positive or negative. Thus when x=\ t y = 2 ; but if x is slightly greater or less than 1, y < 2. .'. x = 1 corresponds to a maximum value of y. Similarly, shew that when x ~ 2, y = 1 ; but if x is slightly greater or less than 2, y> 1. .'. x = 2 corresponds to a minimum value of y. 99. There is an exceptional case which we shall consider later. Suppose y = So/ - IQx 3 + SOar* - 24a + 5. ax = 12 (-!)( -!)(*- 2), and -?- = when as = 1 or 2. aa5 Ifoj = l, y = -2. If x= 1 + A, y = - 2 - 47i 3 + 3A 4 . If A is a small positive fraction, y < 2. If A is a small negative fraction, y > 2. Thus when x = 1, y 2. If x is slightly greater than 1, y < 2. If as is slightly less than 1, y> 2. So that here there is neither maximum nor minimum. The graph is shewn in the first part of Fig. 56, p. 130, and the point P is called a point of inflexion. 98-100] MAXIMA AND MINIMA 109 100. We may state the results of 97 : "The values of x which make f (x) a maximum or minimum are roots of f'(x) = 0." In the case considered / (x) = 2a? - 9x 2 + I2x - 3, Fig. 43. 110 CALCULUS FOR BEGINNERS [CH. IV If we plot the curves y =f(x) and y =/' (x) taking the same x-scale each time, the points where y =f (x) cuts the cc-axis will correspond to turning points on y =f(x). (Fig. 43.) Another Test for Maxima and Minima. 101. It is evident from Fig. 44 that if we advance from dv left to right, i.e. in the positive x-direction, f'(x) or -p changes from + to as we pass through a maximum point and from to + as we pass through a minimum point. (See 71.) orffx) a maximum a mnimum or f(x) Fig. 44. Consider the example in 96. / (x) = Sec 8 -9^+ 12* -3, = 6 (x -!)(- 2). The values of x, which make/' (a;) = 0, are 1 and 2. If x is slightly less than \,f'(x) is + for x- 1 is -| and x 2 is / ' If x is slightly more than 1, /' (x) is - for x - 1 is 05- 'I is +] '. - 2 is -/ ' 100-102] MAXIMA AND MINIMA 111 This may be abbreviated conveniently : If =!-, /=(-)(-) If =!+, /'(*) = (+)(-) .'. as we pass from left to right through the point (1, 2) /' (x) changes from + to . .'. (1, 2) is a maximum point. Similarly shew that (2, 1) is a minimum point. 102. If we apply this test to the case in 99, where f(x) = 3x* - 16x* + 30a; a - 24a + 5, we have f'(x) = l2(x-l)(x-l)(x-2). The values of x which make /' (a;) = are 1 and 2. If x=\- /'() = (-)(-)(-)=-) If x=I + , / = ( + )( + )(-) = -/' i.e. f(x) does not change sign "as we pass through the point (1, 2), and this point is neither a maximum nor a minimum. EXERCISES. XXIII. 1. Find the turning points of the curve y = x 3 - Bx and plot enough of the curve to shew these points. 2. Shew that y=x 3 + 3x has no turning points. Plot it between x= - 2 and x = + 2. Also plot y =/' (x). 3. Find the turning points of the curve 4. Find the turning points of the curve and plot between x=-2 and x=+2. Also plot y =f (x) between the same limits. 5. Shew that y=x*+2x z + l has only one turning point and plot between x = - 2 and x = + 2. Plot y =/' (x). 9. Find the turning points on Plot between x= - 2 and x= +2. Plot y=f (x). 7. Shew that y=x + - has two turning points and that the maximum SB point is lower than the minimum. 112 CALCULUS FOR BEGINNERS [CH. IV 8. Find the turning point of y = 2z 2 -3a; + 5. Is it a maximum or a minimum point? 9. Shew that y = a.r 2 + bx + c has always one turning point and that it is a maximum or minimum according as a is negative or positive. 10. Shew that y = ax z + bx* + ex + d has two turning points or none. 103. Examples. (1) The stiffness of a rectangular beam varies as bcP where b is the breadth and d the depth. Find the breadth and depth of the stiffest beam the cross section of which has a perimeter of 4 feet. Let the depth be x feet. .'. the breadth is (2 -a;) feet, and the beam will be stiffest when (2 a;) 3? is greatest. Call this y. y = (2-x)x* = 2** -x\ :. ^ = 6** -4**. dx For a maximum or minimum value of y --. dx 3 .'. x = or = . A x = obviously corresponds to a zero value of y. *jc = - ives a maximum. * It is clear that there must be a maximum between x = and x = 2, for y = when x = or 2 and is positive for intermediate values of x, or we may say ft. dx~ When *=|-, when = + , .*. x = g gives a maximum value of y. 103J MAXIMA AND MINIMA 113 Thus the stiffest beam is 1| ft. deep and ft. broad. Fig. 45 shews the graph of y = (2 x)x 3 between x and 1 1-5 Fig. 45. (2) The volume of an open cylindrical canister is to be 100 cubic inches. Find the most economical dimensions, i.e. the dimensions which give the least surface area. Let the radius of base be x ins. and the height h ins. Then surface area = (TTO? + 2-n-xti) sq. ins. We want the least value of this. At present the surface involves two variables x and h but we are iven 7,_ 100 Surface area =7rjc 2 + 2x . 200 100 M. C. 114 CALCULUS FOR BEGINNERS Calling this y, we have [CH. IV dx . a? dy For a maximum or minimum value of y, -f = 0. <ttB and _ Sr ioo TT 100 To6 3J / o 1 1 j 100 TT 3 /lQO = - x -- -= ./ - = 3-17, v 10QS V * 350 300 250 20( 150 100 50 1 2345 Fig. 46. The corresponding value of y is 30 \/!OTT = 94*6. [Shew that this is a minimum value.] .*. The canister has the radius of the base and the height each 3'17 ft. and 94*6 sq. ins. of material are required to make it. 103] MAXIMA AND MINIMA [That x - h can be seen as follows: X J- 100 f " x=Jl -\ Fig. 46 shews the graph of y=ira 115 200 x (3) Find two factors of 24 whose sum is a minimum. 24 Let the factors be x and and call their sum y. x y Fig. 47. 82 116 CALCULUS FOR BEGINNERS [OH. IV 24 Thus y x + , oc __ * dx~ ar" /. ^ = o whenar> = 24. ax is a maximum or minimum when = + 4-899... If a: = 4'899, sum of factors = 9'798...-| .'. Sum is a minimum If x 5, other factor is 4 '8, sum = 9'8j when x = \/24. Similarly shew that sum is a maximum when Thus the required factors are 4 '899, 4 '899. Fig. 47 shews the graph of 24 104. The following example illustrates a useful device. A ship B is 75 miles due east of a ship A. B sails west at 9 knots and A south at 12 knots. Find when their distance apart is least. Find also what the east distance is. P, Q are positions after x hrs. [Fig. 48.] AP = 75 - 9cc = 3 (25 - 3x) miles, AQ = 12o; =3(4tt), /. PQ 2 = 9 [(25 - 3ic) 2 + (4o;) 2 ] = 9 [625 -150Z+250 2 ] 103-105] MAXIMA AND MINIMA 11 Put = x* .*. y is a minimum when x= 3. [Why minimum?] .'. least distance is after 3 hrs, A r 75-9x - > p I2x Fig. 48. When <r = 3, PQ 2 = 225xl6, .'. PQ=60. Instead of finding when PQ. is a minimum we found when PQ 2 was a minimum. If we had expressed PQ in terms of a?, we should have 15^25 6a; + y? and we do not yet know how to differentiate this. 105. Notice that we have to express the quantity, of which we wish to find the maximum or minimum value, as a function of one unknown. In some cases it is first of all necessary to express it in terms of two or more unknowns, and then by means of given conditions to express all these unknowns in terms of one of them. 118 CALCULFS FOR BEGINNERS [CH. IV Thus in the second example the surface was written first as a function of x and h, then by means of the given relation, ?rx 2 A= 100, we were able to express h in terms of x, and so get the surface as a function of x only. 106. A common error in questions of maximum and minimum values arises from a confusion between variables and constants. Take, for instance, the second example of 103. If y sq. ins. is the surface area y = irx* + 2-n-xh. The error consists in saying that -- = 2-n-x + 2irh. ax This would only be right if h were constant, i.e. if A remained the same when x changed, which is obviously not the case, for since the volume remains the same, h must increase when x decreases and decrease when x increases. In fact h is a function of x, being connected with x by the equation 100 or h = y, 7TC 3 and this value must be substituted for h before we proceed to get dy dx' EXERCISES. XXIV. 1. Find the number such that when it is added to its reciprocal the sum is a minimum. 2. The sum of the square and the reciprocal of a number is to be a minimum. Find the number. Draw a graph shewing how this sum changes with the number (positive numbers only). 3. Divide a number a into two parts so that three times tbe square of one part plus four times the square of the other shall be a minimum. 105, 106] EXERCISES 119 4. Find the dimensions of the rectangle of greatest area with a perimeter of 10 feet. Draw a graph shewing how the area changes with the length of the rectangle. 6. Prove that the greatest rectangle that can be inscribed in a given circle is a square. 6. A rectangle PQRS is inscribed in a triangle so that P lies in AB, Q in AC and R, S in BC. If BC = a, QR=a; and the altitude of the triangle = h, shew that PQ = a ( 1 - - J . Shew that the rectangle of maximum area has its height one-half that of the triangle. 7. Find the dimensions of the greatest cylinder that can be inscribed in a cone of radius r and height h, the planes of the bases of the cone and cylinder being the same. 8. The surface of a hollow cylinder without top is 100 sq. ins. Find the maximum volume. 9. The volume of a solid cylinder is V cubic ins. Find its dimensions if the total surface is a minimum. 10. A sheet of tin is 6 feet by 4 feet. A square is cut out of each corner, and a tank made by bending up the projecting portions. Find the side of the squares cut out so that the volume of the tank may be a maximum. 11. The number of tons of coal consumed per hour by a certain ship is 0'3 + 0-OOlV 3 where V knots is the speed. For a voyage of 1000 miles at V knots, find the total consumption of coal. Find for what speed the coal consumption is least. Do the same for 100 miles. 12. A running track is in the form of a rectangle ABCD with semi- circles on AB, CD. If the perimeter is a quarter of a mile find the maximum area. 13. A window is in the form of a rectangle surmounted by a semi-circle. If the perimeter is 30 feet, find the dimensions so that the greatest possible amount of light may be admitted. 14. The strength (i.e. resistance to breaking) of a rectangular beam varies as bd z where b is breadth and d depth. Find the breadth and depth of the strongest beam that can be cut from a circular log of diameter 2 feet. 120 CALCULUS FOR BEGINNERS [CH. IV 15. A shot is projected in vacuo with velocity ft./sec. in a direction making an angle a with the horizontal. Its height above the ground at the end of t seconds is tu ein o - ^ gt 2 . Find the greatest height and the time of i reaching it. 16. A shot is projected with velocity u ft./sec. in a direction making an angle a with the horizontal. An inclined plane making an angle /3 with the horizontal passes through the point of projection. The distance of the shot from this plane at the end of t sees, is tu sin (a - ft) - - g cos . t 2 . 2i Find its greatest distance from the plane and the time of reaching it. 17. By the G.P.O. regulations the combined length and girth of a parcel must not exceed 6 feet. Find the volume of the greatest parcel that can be sent (i) in the shape of a rectangular solid with square ends, (ii) in the shape of a cylinder, (iii) in the shape of a cone. 18. An open tank whose base is a square has to contain 1000 cubic feet of water. Find the least cost of lining it with lead at 4d. per sq. ft. 19. A man has 108 square feet of iron which he is going to make into a rectangular tank, lidless, and such that the depth is to be double the breadth. What will be the length, breadth and depth of the tank con- structed under these conditions which will hold the greatest volume of water? 20. Find the circular cylinder of largest volume which can be cut from a sphere of radius 6 inches, the plane ends being perpendicular to the axis. 21. Find the volume of the greatest cylinder that can be inscribed in a sphere of radius a. What is the ratio of the height of the cylinder to the diameter of the base? 22. Find the angle of the circular sector of greatest area that can be made having a given perimeter a. 23. A dynamo is in two parts whose weights are x and y. The cost of the machine is (r/ + 4a:). The usefulness of the machine is proportional to (a; 2 + 3xy). Find the values of x and y for maximum usefulness in a dynamo worth 10. 106] EXERCISES 121 24. The cost per hour of driving a steamer through water varies as the cube of the speed relative to the water. Shew that the most economical speed 3 against a current of V miles/hour is ^ V. [If C is cost per mile, find when 31 ^ is a maximum.] C 25. Given 200 sq. feet of canvas, find the greatest conical tent that can be made out of it. [Get V 2 in terms of r and I (slant height).] 26. Given a circular sheet of paper. Find the angle of the sector which must be cut out so that the remainder may be folded to give a conical vessel of maximum volume. 27. Find the least area of canvas that can be used to construct a conical tent whose cubical capacity is 800 cubic feet. 28. Find the volume of the greatest right cone that can be described by the revolution about a side of a right-angled triangle of hypotenuse 2 feet. 29. There are n voltaic cells each of E.M.F. E volts and internal resistance r ohms; x cells are arranged in series and - rows in parallel. The current C amperes that the battery will send through an external resistance R is given by Given n=20, R = -25, r = -2, find how many cells should be arranged in series to give the maximum current. 30. A battery, internal resistance r ohms and E.M.F. E, sends a current through an external resistance R. The power W given to the external circuit is given by RF2 W ~(R + r)2* Given E and r find what the external resistance must be so that the power may be the greatest possible. 31. The power W given to an external circuit by a generator of internal resistance r ohms and E.M.F. E when the current is C amperes is given by W=CE-C2r. If E=60, r=-3, find C so that W may be a maximum. 122 CALCULUS FOR BEGINNERS [CH. IV 33. The annual cost of giving a certain amount of electric light to a certain town, the voltage being V and the candle-power of each lamp C, is found to be A=a + y for electric energy, o _ 7 nV , , and B = ^ -i -- - for lamp renewals. C T The following figures are known when C is 10. V 100 200 A 1500 1200 B 300 500 Find a, b, m, n. If C=20 what value of V will give the minimum total cost? 33. Shew that the Mechanical Power of a motor is greatest when the efficiency is 50 / . [Mec. Power = e.Ca; EfBciency=pj E-e=Ca. Ra. E and Ra are con- stants.] 34. A given weight W hangs from a point B of a straight horizontal lever ABC which can turn about A. The lever weighs w Ibs. per foot length. If the length of AB is given, say a feet, find the length of lever for which the requisite vertical supporting force at C will be a minimum. 35. A piece of wire of length 6 feet is to be cut into six portions, two equal of one length and four equal of another. The two former are each bent into the form of a circle and these are held in parallel planes and fastened together by the four remaining pieces which are perpendicular to the planes of the circles. The whole thus forms a model of a cylinder. Calculate the lengths into which the wire must be divided so as to produce a cylinder of maximum volume. 36. In measuring electric current by a tangent galvanometer the per- centage error due to a given small error in the reading is proportional to tan x + cot x. Find the value of x for which this is a minimum. [Put 37. A log is in the form of a frustum of a cone, 20 feet long, the diameters of the ends being 2 feet and 1 foot. Find the length of the greatest beam of square section that can be cut from it. Do the same if the length is I feet and the diameters of the ends a and b feet. What is the relation between a and b if the length of the greatest beam is the same as the length of the log ? 107] EXERCISES 123 38. Find the length of the shortest line which will divide a given triangle into two parts of equal area. 39. An electric current flows round a coil of radius a. A small magnet is placed with its axis on the line perpendicular to the plane of the coil through its centre. If x is the distance of the magnet from the plane of the coil, the force exerted on it by the current is proportional to Find x so that the force may be a minimum. [Put a; 2 + a 2 =j/.] 4O. A tank standing on the ground is kept full of water to a depth a ft. Water issues from a small aperture at a depth ft ft. below the surface with velocity *J%gh ft. /sec. Find ft in order that the water may strike the ground at the greatest possible distance from the tank. Another way of distinguishing between Maximum and Minimum. 107. If f(x) is a function of x which is increasing as x increases, or decreasing a? x decreases, f'(x)is positive [71], but if f(x) is decreasing as x increases, or increasing as x decreases, f'(x) is negative. [Fig. 49.] increases f (x) increasing as X increases f (x) decreasing as X f (x) + Fig. 49. 124 CALCULUS FOR BEGINNERS [CH. IV Now as we pass through a maximum point /' (x) is decreasing as x increases. In Fig. 50 (a) the gradient of the tangent (1) is positive; that of (2) is positive and less than that of (1) and so on, that of (4) is zero, that of (5) is negative, that of (6) is negative and numerically greater than that of (5) and so on, so that all the way f (x) is algebraically decreasing. Similarly from Fig. 50 (6) we see that as we pass through a minimum pointy (x) increases as x increases, for we start with a large negative gradient \j and gradually increase up to a large positive gradient s** 2 3 Ax* Fig. 50. But by the above if f (x) increases as x increases, f" (x) is positive, and \if (x) decreases as x increases/" (x) is negative. /. If f(x) is a maximum, /' (x) - and f " (x) is negative. minimum, /' (x) = and f" (x) is positive. 107-109] MAXIMA AND MINIMA 125 We see then that where the curve is concave down- wards -r-^ is , but where it is concave upwards -~ is +. ax* dx 2 (See 94.) Fig. 51. [Positive directions of axes as usual; v. p. 99.] Sign of l!l or f ( x ) p ig. 51. 108. Consider the example in 96. /() = 2or J -9x 2 i 12a; - 3, = 6 (a; 2 -3a;--2), /"(*)= 6 (2* -3). The values of x which make/' (x) are 1 and 2. If x =l,f"(x) is-. If * = 2,/'is+. .'. x = 1 gives a maximum value of f(x) \ and x 2 gives a minimum value of f(x) ) ' [See Fig. 41, p. 101.] 109. If we apply this test to the case in 99, where f( x ) 3* 4 16& 8 + 30ic 2 - 24a? -f 5, The values of x which make/' (a;) = are 1 and 2. If x 2,/" (x) is +, .'. f(x) is a minimum, but if a;= I,/" (x) = 0, and the test fails. 126 CALCULUS FOB BEGINNERS [CH. IV Fig. 52. 110. In the case of 108, putting we have and /' (x) = 6 (a? - 3x + 2), If we draw the graphs of y =/(#), y =/' (x), and y =f" (x) (Fig. 53) we get: When = 1 or 2, f(x) is maximum or minimum, i,e. y =f (x) cuts the %-axis. 110, 111] MAXIMA AND MINIMA When x 1|, 127 and f (x) is a minimum, /' = 0, y =f" (x) cuts the a;-axis. Kg. 53. Points of inflexion. 111. A point like P where the gradient is a maximum or minimum is called a point of inflexion. 128 CALCULUS FOlt BEGINNERS [CH. IV The portion BPC of the curve is shewn magnified in Fig. 54. Between B and P the gradient diminishes (algebraically) steadily from to - 1 *5. la 1-8 IE Y-L ^ w '@L 1-2 ^ a Fig. 54. Fig. 55. Between P and C it increases (algebraically) from 1 *5 to 0. The gradient is a minimum at P and the curve crosses the tangent there. If we suppose a point to advance along the curve from B to C, and a line to be drawn from any point K. parallel to the tangent at each position, this line swings in a clockwise direction until the moving point reaches P, and then back again in a counter- clockwise direction. (Fig. 55.) 111-113] POINTS OF INFLEXION 129 The tangent at a point like P is sometimes called a stationary tangent. Notice (v. Fig. 41, p. 101) that from A to P the curve is concave downwards and -~ is : from P to D the curve is concave dy? j fy upwards and ~ 2 is +. At P where we change from one kind of bending to the other 112. If we call a point where a curve crosses the -axis a zero point we see (1) To a turning point on y = f (x) corresponds a zero point on y = f ' (x). (2) To a point of inflexion on y = f(x) corresponds a turning point on y = f ' (x) and a zero point on y = f"(x). 113. In the case of 109 : f(x) = 3x 4 - IGo 3 + SOar* - 24* + 5, /' () = 12 (y? - 4ic 2 + 5x - 2) = 12 (x - 1) 3 (x - 2), f"(x) = 12 (3x*-8x+ 5) = 12 (x-l) (Sec -5). When x= 2, f( x ) i & a minimum, /=<>, f"(x)is+. When x = 1, f(x) is neither a maximum nor a minimum [i.e. there is a point of inflexion], /' (x) = and is a maximum, /"<)= a There is also another point of inflexion on y=f(x) corre- 5 spending to x = 5 . ^ M. C. 9 130 CALCULUS FOR BEGINNERS This makes f" (x) = 0, f (x) a minimum. [CH. IV 114. Fig. 56 shews the gra y=. y=- y=* 3hs of flrb f (T^ r<). -,fc'! 3 ^3QX?-i& : 3!I.' \\ : jljjj jjjjj 1; ;; i;i;:;;;:;;;];;i;;;;J!;;;; '' / \ ' ' ' ; ;;| - ;; - fr ;rL BfV p ] :;-l;|;;;;;;!:i||;;;: I ii 21%-* i| jijiiii Notice when x= 1, f (x) - 0, but y=f (x) does not cross the ce-axis, and there i 1 ? not a turning point on y=f(x) corresponding 113-116] POINTS OF INFLEXION 131 to this value of x. f" (%) = Q, and y=f" (x) crosses the a>axis, arid we have a turning point on y =/' (x) and a point of inflexion on y =/(*). 115. We have seen that if a be a value of x which makes f (x) = Q, it may happen that /() is not a maximum or mini- mum value of /(a:), and that usually if/"(a) = 0, it will not, because then as a rule f (x) does not change sign as x passes through the value a. It may happen, however, that even if/" (a) = O,/' (a;) changes sign as x passes through the value a, in which case we get a turning point corresponding to x = a. Similarly if b be a value of x which makes/" (x) = Q, it may happen that/' (6) is not a max. or a min. value of /'(a?) and that usually if /"' (6) = 0, it will not, because then, as a rule /" (x) does not change sign as x passes through the value b. 116. Summary. If a is a value of x which makes f'(x) = and if /" (a) ^fe 0, there is a turning point corresponding to x a and the value of f(x) is a max. or iniii. according as/" (a;) is or +. If/' (a) = and also/" (a) = we cannot tell without further investigation whether there is a turning point or not. If /' (x) changes sign as x passes through the value a, there will be a turning point and f(x) will be a max. or min. according as /' (x) changes from + to or from to +. If b be a value of x which makes /" (a?) = and if /'" (b) J= there is a point of inflexion corresponding to x = b. If /" (b) = and also /'" (b) = we cannot tell without further investigation whether there is a point of inflexion or not. If /" (x) changes sign as x passes through the value 6, there is a point of inflexion. e.g. if y=x*(x+\], we have f(x) = x 5 + x 4 , f (x) 92 132 CALCULUS FOR BEGINNERS [CH. IV Now/' (a;) = 0, when x = or '8. When as = - -8, /" (x) is - . .'. x = '8 gives a maximum value of f(x). When x = 0, /" (a;) = 0. But f'(x) .'. !* = <)*, /'(*) = (- .*. x = gives a minimum value of /(a:). Again /" (a;) = 4x 2 (5* + 3). .'. /" (a;) = when x = or - -6, /'" (a) = 60* 2 + 24* = llx (5x + 2). When x = - -6, /'" (x) = (-) (-) = +. .'. there is a point of inflexion when x = - -6 and the gradient is a minimum. [Of course we could have seen this without using f" (x), for if =--6-,/"(*) = <+)(-) = -J if = - -6 +,/"(*) = (+)(+) = 4, .'. /' (a;) is a minimum and there is a point of inflexion.] When aj = 0,/"'() = 0. In this case, if x = , f" (x) = (+) (+) = + ; if x = +,/"(*) = (+)(+) = +. .*. there is not a point of inflexion. [See Fig. 57.] 117. If we go on and calculate /"" (a;), etc. we get /"" (a?) = 1 20x + 24 = + 24 when x = 0, and generally, it will be found that if f (a:) = when x = c, and if the first of the derived functions of x which does not vanish when x = c is an even one, i.e. if it is one of the set/" (a;), / IT (x) etc., x = c gives a turning point, but if the first one which does uot vanish when x = c is an odd one, i.e. if it is one of the set /'" (*),/ v (x) etc., x= c does not give a turning point. For instance in the above/"" (a;) was the first derived function which did not vanish when x = 0. For a proof, the reader must refer to more advanced text-books. 116, 117] MAXIMA AND MINIMA 133 134 CALCULUS FOR BEGINNERS [CH. IV EXERCISES. XXV. 1. Find the maximum and minimum values of 2a; 3 -3x 2 -36a; + 10. Also find the point of inflexion on and shew that it is the mid-point of the line joining the turning points. 2. Find the maximum and minimum values of 3. In the curve y=x*-2x 3 , shew that there is a minimum point where 3 x - and a point of inflexion where x = 0. m Find the co-ordinates of the other point of inflexion. Draw the graph between x= - 1 and x =3. a; 2 5 4. Find a minimum value of -^- + 2 + -5 and shew that there is no o x maximum. 6. Shew that there are two turning points on y=x 6 (x-2). How many points of inflexion are there? Draw a sketch between 0;= -1 and x=2. 6. If -^- (x + 1) 3 (x - 2)* shew that x= - 1 gives a minimum value of y, fljp but that x = 2 gives neither a max. nor a rnin. 7. If /(a;)=a;*-8x3 + 24a: 2 -32a; shew that/' (2)=0 and/" (2)=0. Is the point corresponding to x=2 a turning point or a point of inflexion? Draw a rough sketch of y = f(x) from x= - 1 to x 5. 8. In the curve find the turning points and the point of inflexion and shew that the point of inflexion is midway between the turning points. If the point of inflexion be taken as origin, the axes being parallel to the original axes, shew that the equation of the curve is y=2x*-24:x, and that the curve is symmetrical in opposite quadrants. 117] EXERCISES 135 . Shew that y = ax 3 + bx* f ex + d has always one point of inflexion and find its co-ordinates. Shew also that if the curve has two turning points, the point of inflexion is midway between them. Shew that the equation of the curve referred to parallel axes through the point of inflexion is and that the curve is symmetrical in opposite quadrants. 10. Shew that y = ax* + bx 3 + ex 2 + dx + e has two points of inflexion or none according as 36 2 - 8ac is + or . 11. Draw on a large scale the graph of y = 2x 4 - 3x? + 2 x between a; = and 2 = 1. CHAPTER V SMALL ERRORS AND APPROXIMATIONS 118. WE have already had a few examples of the application of the Differential Calculus to this kind of problem : y being a given function of x, what will be the approximate change in y, due to a small change in x 1 A?y We make use of the fact that is approximately equal to JT-, the approximation being better as Aa; is diminished. The CM? approximation may be put in the form dv Ay = . Ax approximately. e.g. if A sq. ins. is the area of a circle of radius r ins. A = 71^. Now suppose a small error, which we may call Ar, has been made in measuring r, there will be a corresponding small error in A, which we may call AA, and we have seen that AA dA -= approximately, Ar dr J i.e. AA = 2-irr . Ar approximately. Suppose the radius was measured as 10 inches and that there was an error of *1 inch. Then we have r = 10, Ar = !. .'. AA = 27T x 10 x -1 approximately = 2ir approximately, ie. the error in the area is approximately 6*28 sq. ins. [Shew that actually the error is 6'31 sq. ins.] 118-120] SMALL ERRORS AND APPROXIMATIONS 137 If the error in the radius was -01" we should get for the error in the area -6284 sq. ins. instead of '6287. 119. The result AA = 2irr. Ar approximately, admits of a simple geometrical interpretation which is important. It tells us that if the radius (r) of a circle be increased or decreased by a small amount (A?-) the increase or decrease in area is approximately 2rrr . Ar or Area of thin circular ring is approximately circum- ference of either boundary x breadth. e.g. Suppose we have a ring with inner radius 20" and breadth -06", the area is approximately 2ir x 20 x -06 = 2 Air= 7'54 sq. ins. approximately. [Shew that area is actually 2'4036n-= 7 '55 sq. ins.] 120. We might get this result from first principles thus. Area of ring, inner and outer radii a and b, = Tr(& 2 -a 3 ) = IT (b + a) (b a) = ( 2ir . ) t> where t is the breadth = Circumference of concentric circle lying midway between the two circumferences x breadth. [Fig. 58.] Fig. 58. 138 CALCULUS FOE BEGINNERS [CH. V As the breadth is made less, the circumference of this intermediate circle comes nearer to the circumference of either boundary and can be brought as near to it as we please by making the breadth small en'ough. EXERCISES. XXVI. 1. The radius (r) of a sphere is increased by a small length Ar. Shew that the increase in volume is approximately 4ar 2 . Ar. State this as a formula for the approximate volume of a thin spherical shell. 2. The radius of a sphere Is 6". Find approximately the diminution in volume if the radius is diminished by ^ ". 3. Find approximately the volume of a thin spherical shell, internal radius 1 foot, thickness -2". [Shew tbat your result is less than 2/ in error.] 4. A stone is thrown at an angle of 40 to the horizontal with a velocity of 80 ft./sec. Find the range R = and the approximate I Jf _J increase in the range if the velocity is changed to 82 ft./sec., the angle of projection remaining unchanged. 6. A cylindrical well is said to be 25 feet deep and 6 feet in diameter. Find the error in the calculated volume if there is an error of (i) 1" in the diameter, (ii) 3" in the depth. 6. The radius of the base of a cone is r and its vertical angle 2a. Find the approximate increase in volume due to a small increase Ar in the radius, the vertical angle remaining constant. Hence shew that the volume of a conical shell, internal radius r, thickness t, t being small, is approximately rrlt where I is the slant height. 7. The radius of a sphere is found by measurement to be 18-5", with a possible error of !". Find the consequent errors possible in (i) the surface area, (ii) the volume, as calculated from this measurement. 8. R = R (l + at + & 2 ) is a formula for the electrical resistance of a metal, R being the resistance at C. and t C. the temperature. Find the rate of increase of R per unit increase of t. If R =l-6, a = -00388, b= -000000587, find the approximate change in R when the temperature rises from 100 to 101. 120-122] RELATIVE ERROR 139 121. Since A = nr 2 and A A = 2irr . Ar approximately, AA Ar 7T =3 'T' AA or the ratio of the error in the area to the original area, is called the relative error in the area. Thus the relative error in the area is approximately twice the relative error in the radius. If, for example, the radius increases by 1 / o , the area will increase approximately by 2 / o , Ar 1 AA 2 for T = loo' " "A" = 100 a pp roximatel y- EXERCISES. XXVII. 1. If V c. ins. is the volume of a cube of edge x ins. prove AV Az -rj- = 3 . approximately. V X Hence find approximately the percentage increase in volume due to an increase of *5 / in the edge. 2. Prove that the relative increase in the volume of a sphere is approximately 3 times the relative increase in the radius. 3. If pv= k, prove = approximately. If the pressure increase 1/ , what is the approximate change in the volume ? 4. Ifpv l *=k, prove = - 1'4 approximately. 6. If y = x n , prove =n . approximately. " 122. Elasticity of volume. Suppose v c. ins. to be the volume of unit mass of a fluid and p Ibs./sq. in. the pressure, p being some given function of v. If p be increased to p + AJL>, v will become v + Av (Aw being, 140 CALCULUS FOR BEGINNERS [CH. V of course, negative, since an increase in pressure will produce a decrease in volume). Aw will be the diminution of volume. Aw (the ratio of this diminution to the original volume) is called the volume strain. The ratio of A/> (the increase of pressure required to produce this) to the volume strain is v . and the limit of this when Av Aw is indefinitely diminished, i.e. V ==-, is called the elasticity of volume or the bulk modulus of the fluid. 6. If pv- constant, prove that the elasticity of volume is p. 7. If j>v n = constant, prove that the elasticity of volume is np. 8. If p be given as a function of v, say p=f(v) and the graph p=f(v) be drawn, the v-axis being horizontal, shew that if P be a point on the graph corresponding to any given value of v and MQ be drawn through the foot of the ordinate of P parallel to the tangent at P, meeting the p-axis in Q, then OQ will give the elasticity of volume for this value of v. [Fig. 59.] Fig. 59. 9. A formula giving the deflection (D) of a beam in terms of the length (I), the load (W), the Moment of Inertia of the Cross-Section (I) and WJ 3 1 Young's modulus (E) is D = x . 122, 123] SMALL ERRORS AND APPROXIMATIONS 141 Shew that if a small error be made in I, the resulting percentage error in D is approximately 3 times the percentage error in I. 10. The velocity of discharge (v ft. /sec.) of water through a long pipe is LJ given by u 1>87 = --jrp-j-7-xd 1 -* where H ft. ~ head of water, d ft. = diameter of 'UUU4L pipe, L ft. = length of pipe. Find the velocity of discharge when the head is 35 feet, the length of pipe 1 mile and its diameter 18 inches. If a small error Ad is made in estimating the diameter, prove that the error (Av) in the velocity = - x - x Ad approximately. If the error in the diameter is ", find Av. 11. If t sees, is the time of oscillation of a pendulum of length I feet i. [0=32-2.] Shew that the relative error in t is approximately half that in I. Calculate t when Z=4 and find approximately the error in t corresponding to an error of ~ " in the length. m 12. If pv l '% = k and the pressure increase 2/ , what is approximately the percentage change in volume ? 123. = -r- approximately, Ax ax i.e. if x is increased by Ace, y is increased by Ace x -J?- approximately. Suppose y =/(), then -j- =f'(x), and the above statement is ctx equivalent to this : If x is increased by a small quantity A, then f(x) is increased by hf'(x) approximately. \h taking the place of Ax and/'(x) J.1 Or dx f (x + h) = f (x) + hf (x) approximately. 14-2 CALCULUS FOR BEGINNERS [CH. V 124. Let P be the point (a;, y) on the curve y =f(x) and let Q be a neighbouring point whose abscissa ON is x + h. [Fig. 60.] M N Fig. 60. Then NQ =f(x + h), MP =f(x). :. if PR is parallel to OX, RQ =f(x + h) -f(x). If PT is the tangent at P, the gradient of PT is/' (x) and RT = hf'(x), so that in taking f(x + h) -/(*) as equal to hf (x) we take PQ as coincident with PT, in other words we consider the portion PQ of the curve as straight. 125. Ex. Find the value of 32* - 4X 3 + 6a: 2 - 7x + 8 when x = 3-01. Calling the expression y (or) we have = 662, /'(3)=1136. .'./(8-01) = 662 + 1136 x -01 approximately = 673 '36 approximately. 126. Notice that if /' (a;) = 0, i.e. if the tangent at P is parallel to OX, this approximation will not shew any difference between f(x) and f(x + h). See Exs. 5 and 6 below. 124-127] SMALL ERRORS AND APPROXIMATIONS 143 EXERCISES. XXVIII. 1. Shew that if f(x}=ax + b where a and b are constants the statement f(x + h)=f(x)+hf'(x) is accurately true for all values of h. 2. Find approximately the value of 5x 2 + 4x + 9 when x =2-0087. 3. Find approximately the value of Gx 3 - 7* 2 + 2x - 9 when x = 3-002. 4 4. Find approximately the value of 7a; 2 -3x + - when x=2-01. 5. Find approximately the value of 2z 3 - 9* 2 + 12s- 3 when x= (i) 2-005, (ii) 1-995. In each case find also the accurate value. 6. Find approximately the value of 3a;*-16aj3 + 30a;2-24a; + 5 when x = (i) 2-01, (ii) 1-01. In each case find the accurate value. 127. The approximation f (x + h) =f (x) 4- hf'(x) is, as we have seen, equivalent to the statement that the gradient of PQ [Fig. 61] is that of the tangent at P. If we suppose P and Q to be such that f (x) and f" (x) do not change sign between P and Q, so that between these points the curve has one of the shapes shewn on p. 99, it is clear that the gradient of PQ is between the gradients of the tangents at P and Q and equal to the gradient at some intermediate point. Thus the quantity [represented by RQ] added to f(x) to make f(x + h} is between hf (x) [RT] and hf'(x + h) [RUj and is actually hf'(x+Qh) where 6 is some proper fraction, so that (x -f 6h} is between x and (x + h). e.g. in 125 /' (3) = 1136, /' (3-01) = 1151-68. So that /(3-01)-/( 3 ) is between -01 x/'(3) and -01 x/'(3'01), i.e. between 11 '36 and 11 '52. Actually it is H'43826..., and as a matter of fact /(3-005) = 1143-8200, so that in this case f(x + h) -f(x) = hf (x + ^h). [Correct to 3 decimal places.] The more rapidly f (x) is changing, the less reliable will f(x) + hf'(x) be as an approximation tof(x + h). 144 CALCULUS FOR BEGINNERS [CH. V Fig. 61. EXERCISES. XXIX. 1 . If / (x) = ax* + Ix + c, shew that / (x + h) =f (x) + hf ( x -f U). In particular, if/(a;) = 7a; 2 -lla;+15, shew that/(3'4)=/(3) + -4/' (3-2). a. If / (x) = x 3 - 6x - 5, shew that / (1-4) -/(I) =-0-656, 4/' (1-4) =-0-048, 4/' (1-205) =-0-65757. 3. If/(a;)=4-4a+16a;-7, shew that ft /(3-2)-/ (3) = 3-9856, (ii) / (4) -/ (3) = 43, 2/'(3)=3-2, /'(3) = 16, 2/' (3-2) = 4-8384, /' (4) = 80, 2/' (3-102) = 3-9852. /' (3-539) = 43-003. CHAPTER VI THE INVERSE OPERATION 128. GIVEN , find y. CwB di/ We know that if y = c w , ~f- = a; TO ~ 1 . cKE ttW Can we say that if -~ nx n-1 then y = x n t Not unless we are CwB certain that x n is the only function whose differential coefficient is nx n -\ As a matter of fact we have seen that if y = x n + any d'u constant whatever, - = nx n ~\ .". if we are given that Cnv -j- = wa3 n ~ 1 , we can only conclude that y = JB W + some constant, and if no further information is given " some constant " may be replaced by "any constant whatever." As a rule sufficient information is given to enable us to fix the constant for the particular case in question. In practice -f- will not present itself in such a convenient fUD form as rac"" 1 , e.g. we might have the value of -~ given not as Gfos 5x* but as ar 4 or 3x*. To deal with such cases we have only to U'l/ remember that if y = k . x n where k is a constant, -^- = k. nx n ~\ o o Thus 3x 4 = -.5oj 4 and therefore arises from differentiating -= , X s . O M. C. 10 146 CALCULUS FOR BEGINNERS [CH. VI Remembering what was said about the addition of an arbitrary constant, we see that if -^ = 3x* then y = -=x 5 + c where c is some dx o constant. Similarly if -^ = Sjx, i.e. 5o?, we know that x* must come from differentiating x% + ', i.e. a;^ : but if y = x*. -^=H# dx 2 EXERCISES. XXX. 1. Write down the values of y corresponding to the following values dm (i) So; 2 . (ii) S* 3 . (iii) -^. (iv) 7. (v) 5x~%, ( v i) - x n . (vii) ^. (viii)^. (ix) i. (x) |*i (xi) 2*3 + 3*. (xii) | 2 + 4. (xiii) 7* 2 + 3* + -5 . (xiv) X** a. What is /(*) if /'(*) = (i) A (ii) 8a?-* + l. (iii) 2 d 2 ^ 129. If -T^ is given there will be two arbitrary constants QBP in the value of y, as the following example will shew : Given ^ = 2* + 3, find y. Since -^~ means -=- where z stands for ~- , we have da? dx dx .'. z = x? + 3x + a where a is any constant. 128-130] THE INVERSE OPERATION 147 (*?/ i.e. ~ = a? + 3x + a. ax x 3 3x 2 " y=3-+ 2- +ax+b where b is any constant. EXERCISES. XXXI. 1. Find the values of y corresponding to the following values of 3-^ dx^ (i) 3x2. (ii) |. (iii) 7. (iv) ^ + 5. (v) ^. (vi) V*. 2 . If /" (x) = 3x - 8, find / (x). 3. If /"(*) = 5, find f(x). 4. Iff" (x) = 2x 3 -3x + l, find/{a;). 130. The following examples will shew how to fix the value of the arbitrary constant in special cases. CtAJ Ex. 1. Given that -^- = 3x, and that t/=5 when a? = 2, fiud flHD y in terms of x. 3 We have y = x z + c. c must be chosen so that this shall be satisfied when x = 2 and y = 5. 10-2 148 CALCULUS FOR BEGINNERS [CH. VI EXERCISES. XXXII. du 1. Given -r^ = a; 2 + l and y = 3 when a; = 3, express y in terms of a;. 2. Given =v + ^ and p = 5 when v = 2, express^ in terms of v. 3. Given -^ =N /a; and y = 7 when a; =4, find y in terms of x. dA 4. Given -=-=5y + 4 and A = 6 when y = l, express A in terms of y. 6. Given -p = (a; + l) (a; + 2) andy=12 when a; =3, express y in terms of a;. 6. Given -5- = (2t - 3) 2 ands = 52 when t = 5, express s in terms of t. 7. Given = 4irr 2 and V = 367r when r=3, find V in terms of r. 8. Given -J?- = and y = 8 when a; =27, find y in terms of x. ax *jx 9. The speed of a body at the end of t sees, is given by the formula v=u + at where u and a are constants. Find the relation between s and t, given s=0 when t=0. <Z 2 w 10. Given j-^=3x; also y = ll when a; =2, and y = 47 when a; =4, find y. ^-32- ~I n vii . cuovs u ^ v T i_i^u *L> v c^ii^ ~7"~ - dx 2 djj i:d y. 11. Given j-|=32; also y=0 when x = 0, and ^ = 5 when a; = 0, 131. jE'x. 2. The gradient of a curve at the point (x, y) is 2a; + 3, and the curve passes through (1, 2). Find its equation. We have ^ = 2x + 3. dx .'. y = a; 2 + 3a; + c [where c is some constant to be fixed]. 131, 132] THE INVERSE OPERATION This has to be satisfied by x= 1, y - 2. 149 .'. the equation of the curve is y = x 2 + 3x - 2. 132. If we had not been given the second piece of in- formation we should have had the equation of the curve y = a? + 3a; + e where c is any constant whatever. Fig. 02. 150 CALCULUS FOR BEGINNERS [CH. VI For different values of c this really represents a whole family of curves got by taking y = a? + 3* and sliding it through any distance parallel to Oy. is that particular member of the family which goes through (1, 2). (Fig. 62.) "We might say that -^- = 2x + 3 represents a family of curves, (tx viz. that obtained by drawing y = a? + 3x + c for different values of c. EXERCISES XXXIII. 1. Find the equation of a curve whose gradient at any point (x, y) is 3 - 4 and which passes through (2, 3). a. Find the equations of the curves with the same gradient-law as in Ex. 1 passing through (i) (0, 0), (ii) (-3, 1). 3. Draw the 3 curves of Exs. 1 and 2. *. The gradient of a certain curve at any point (x, y) is 2x + l, and the curve passes through the point (1, 5). Find the equation of the curve and the equation of the tangent at the point (1, 5). 6. The gradient of a certain curve at the point (x, y) is 2a; 2 + 3x - 7, and the curve passes through (1, 2). Find its equation and also the equation of the normal at the point (1, 2). 6. The gradient of a curve at the point (x, y) is 3x - 5. Shew that it has a minimum point and find its co-ordinates (i) if the curve passes through the origin, (ii) if the curve passes through ( - 3, 2). 7. Find the equation of a curve which passes through the point (3, 5) and whose gradient-law is -^- = 1 - 4r + 2 . dx In what point and at what angle does this curve cut the y-axis? 133. Ex. 3. A point is moving in a straight line so that its speed at the end of t sees, is (3t + 2^) ft/sec, and its distance from a fixed point in the line at the end of 2 seconds is 20 feet. Find its distance from the fixed point at the end of t sees. 132, 133] THE INVERSE OPERATION 151 If the required distance is s ft. we have at 3 [where c is a constant to be determined]. Now we are told that when t = 2, s = 20. -Eoj. 4. A point is moving in a straight line so that its acceleration at the end of t seconds is (3 + 4f) ft./sec. a At the end of 2 seconds its speed is 16 ft. per second and its distance from a fixed point in the line is 20^ feet. Find its distance from the fixed point at the end of t seconds. Let the required distance be s ft. and let its speed at the end of t sees, be v ft./sec. We have ^ = 3 + 4*. at :. v = a + 3 + 2 a , where a is some constant to be determined. Now when t = 2, v 16. .'. 16 = a + 6+8. .'. a = 2. /. v = 2 + 3< + 2< 2 . i.e. ^=2 + 3<+2<?. at 3 2 ;P + -~1>*, *t O where 6 is some constant to be determined. 152 CALCULUS FOR BEGINNERS [CH. VI Now when t = 2, s = 20 . /. 20 = b + 4 + 6 + 5J. .'. 6 = 5. EXERCISES. XXXIV. 1. The speed of a body moving in a straight line is given hy t> = 60-32f. Its distance from a fixed point O in the line at the end of 1 second is 44 feet, find its distance from O at the end of t seconds. 2. The acceleration of a body moving in a straight line is 32 ft. /sec. 2 Its speed at the end of 2 seconds is 154 ft. sec. and its distance from a fixed point O in the line is 234 feet. Find its speed and distance from O at the end of 3 seconds. 3. The acceleration of a body moving in a straight line at the end of t seconds is (2 + 3t) ft./sec. 2 Its distances from a fixed point O in the line at the end of 1 and 2 seconds are respectively 7 ft. and 15 ft. ; find s in terms of t and get the distance from O and the speed of the body (i) when t=0, (ii) when t=5. 4. A body moves in a straight line with constant acceleration a ft./sec. 2 Its initial speed is M ft./sec. If v ft./sec. is its speed at the end of t sees., and a ft. its distance from the initial position at the end of t sees., prove and s= 6. If J- = g^ express p as a function of v, given that p = 18 - 95 when 20. 6. If -~ = ax + b express y as a function of x, given the following pairs of corresponding values of x and y : a; 1 2 y 3 5 9. 133] MISCELLANEOUS EXAMPLES 153 MISCELLANEOUS EXAMPLES ON CHAPTERS I VI. A. 1. From first principles find -^ when dx Also write down the equation of the tangent to the curve represented by y = 3x 3 + 2x + 1 at the point where x= 1. 2. A body starts from rest and moves in a straight line in such a way that its speed after t seconds is (16t - 4t 2 ) ft. /sec. Find the distance described during the fourth second. 3. The strength of a rectangular beam varies as fed 2 where b is the breadth and d the depth. Find the depth and breadth of the strongest rectangular beam of perimeter 3 feet. 4. The radius of a spherical bubble grows at the rate of - 01 inch per second. At what rate is (i) the surface, (ii) the volume increasing when the radius is (a) 1 inch, (b) 1 foot? 6. The volume of a cylindrical tin canister closed at both ends is 300 c.c. Find the most economical dimensions. B. 1. (i) Find from first principles ~ when y=-* tlx x (ii) Find the equations of the tangent and normal to the curve xy = l at the point P, where x = 5. (iii) If the tangent meets the axis in T and t, shew that PT = P*. 2. Writedown (i) ~ when y = 5*Jx + /- . dx y x d\/ lex* (ii) if V = - (R -x), where R is constant. 3. Water is poured at the rate of 5 cubic feet per minute into a vessel in the shape of a hollow cone with a vertical angle of 90. When the vessel contains 12 c. ft., at what rate is the depth increasing ? 154 CALCULUS FOR BEGINNERS [CH. VI 4. The sum of the perimeters of two equal squares and a circle is 100 feet. When is the sum of the areas least, and when is it greatest ? Are these really maximum and minimum values ? 6. There is a certain curve such that the gradient at any point (x, y) is Bx. Find its equation, given that it passes through the origin. C. 1. If j/ = 3# 2 + 2, obtain from first principles the differential coefficient of y with respect to x, and shew with a diagram the geometrical meaning of your result. For what value of y is this differential coefficient equal to unity ? 2. If y = 2x 3 -27x*-132x + 2 find its maximum and minimum values. 3. A lamp is 60 feet above the ground. A stone is let drop from a point at the same level as the lamp and 20 feet away from it. Find the speed of the shadow of the stone on the ground (i) after 1 second, (ii) when it has fallen 30 feet. [0=32.] 4. Find the co-ordinates of the point of intersection of the tangents to j/ = 3a; 2 +5j;-7 at the points where x=2 and x=5, and the equation of the line joining this point of intersection to the mid-point of the chord joining the points of contact. 6. The graph of/' (x) is the parabola y + 2 = 3 (x- 1)2. What information does this give respecting .; (a;) ? What is the gradient at the point on y =f(x) where x=l 1 D. 1. A body is thrown into the air with a velocity of 100 ft. /sec. at a certain elevation. If the resistance of the air be neglected, the horizontal distance from the point of projection (a; feet) and the height above the ground (y feet) after t seconds are given by the equations x = 80t, y = 60t - 16t 2 . Find the horizontal and vertical velocities after 3 seconds, and the magni- tude and the direction of their resultant. By eliminating t between the 3 1 given equations, get the equation of the path in the form y = j x - -^ x-, and find the direction of the tangent to this path when x = 240. a. Find the maximum and minimum values of (*-!)(* -3). 133J MISCELLANEOUS EXAMPLES 155 3. Find the equations of the tangents to the curve at the points where the curve crosses the axes and also the co-ordinates of the point of inflexion. 4. The volume of a sphere is V cuhio inches and the surface S square inches. Shew that V = ^- . S i 6VT The surface of a sphere is given as 100 square inches, and from this the volume is calculated. If the surface is really 101 square inches, find the approximate error in the calculated volume. dy_ curve goes through the point (1, 1). Find its equation. 5. The gradient of a certain curve obeys the law -^=4a; 2 +3 and the clx E. 1. The surface of a sphere is given and the volume calculated from it. If there is an error of '4 / in the surface, find approximately the resulting percentage error in the calculated volume. 2. An open tank with a square base and vertical sides is to have a capacity of 4000 cubic feet. Find the dimensions so that the cost of lining it with lead may be a minimum. 3. The equation of the path of a projectile thrown in vacuo with a x 2 velocity of 64 ft./sec. at an angle of 45 to the horizon is y x- ^^ , the I2o origin being the point of projection, the x-axis horizontal, and the unit along each axis 1 foot. Find: (i) The co-ordinates of the highest point, (ii) The equation of the tangent at the point for which a; =32. (iii) Where the direction of motion makes an angle 30 with the horizon. 4. If y=2x*-2x3-x z + l, find y when x = -1, --,0,1,2. Also find -^ and the values of -j- when x - 1, 2. ax dx Also find what values of * make -j^=0. dx 156 CALCULUS FOR BEGINNERS [CH. VI Using all this information, draw the graph of y = 2x 4 - 2x 3 - x 2 + 1 between x = - 1 and x = 2. , 1" x-scale unit 2", y-scale unit ^ . 6. In a certain case of straight line motion, the velocity in feet per second, at a given instant t seconds after the start, is given by the formula Calculate : (i) The average velocity between the end of 2 seconds and the end of 4 seconds. (ii) The arithmetic mean of the velocities at the inatauts 2 and 4 seconds after the start. (iii) The velocity 3 seconds after the start. 1. If v be the volume of a given mass of water at C., and v its volume at 6 C., Hallstrom's formula is for values of between and 30, where a =-000057577, 6 = -0000075601, c= -00000003509. What is the temperature of maximum density ? 2. Draw the graph of y^=x(x-l) z between x=0 and x=S. Find the co-ordinates of the turning points and the equations of the tangents at (1, 0). 3. The normal at a point of y 3 =x meets OX in G. Prove that the minimum value of OG is -7698. 4. A body of mass 20 Ibs. is moving in a straight line so that s feet being its displacement from some standard position at the end of t seconds. Find its kinetic energy and momentum and the force acting on it at the end of 4 seconds. 6. Find approximately the value of 16 - 2-5x + 47x 2 , when x = '997. 133] MISCELLANEOUS EXAMPLES 157 G. 1. In a certain case of straight line motion the acceleration is proportional to the time which has elapsed since the moving body passed a fixed point A. Find the distance of the body from A after 10 seconds, given that its speed at A is 8 ft. /sec. and its distance from A after 1 second is 9 ft. 2. If y = ax 2 + bx where a and 6 are constants, prove rttfir .*%-* dx 2 dx 3. A piece of wire, a inches long, is cut into two parts, one of which is bent into the form of a square and the other into that of a circle. Find where it must be cut so that the sum of the areas enclosed may be a minimum and shew that the diameter of the circle is equal to the side of the square. 4. P is the point on the curve y=zP whose co-ordinates are (h, ft 3 ). The tangent and normal to the curve at P meet the axis of y in T and G respectively, and PN is the perpendicular from P on the y axis. Prove that OT=2ON, and that as P changes its position on the curve, NG varies inversely as N P. 6. The pressure and volume of a gas are connected by the relation pv 8 = constant. When the volume is 20 cubic feet the pressure is 30 Ibs. per sq. ft. Find the elasticity of volume when the volume is 40 cubic feet. H. 1. If/() = (* + !) (as -2)3, find /'(*),/"(*), and/'" (x). Find the values of x which make (i) / (*) = 0, (ii) /'(*) = 0, (iii) /"(*) = 0, (iv) /'" (x) = 0. Find the co-ordinates of the maximum and minimum points on y=f(x), y =/'(*), y =/"(*), and the points of inflexion on y=f(x) and y=f (x). Also calculate the values off(x),f (x\, f" (x), /'" (x), when x= - 1, 0, 1, 2, 3. Using the information you have now collected draw on squared paper the curves y =/(*), y =/'(*). y =/"(*). y =/'"(*) using the same x-scalo for all. 158 CALCULUS FOR BEGINNERS [CH. VI 2. A conical vessel has vertical angle 60. Water is flowing into it at the rate of 1 cubic foot per minute. Shew that when the water is 1 foot deep the level is rising at about -19 inch per second. 3. The tangent at P to the curve aj/ 2 =x 3 meets OX in T and OY in t. PN is the ordinate of P. Prove OT = i ON and Ot=| PN. O a 4. A solid sphere radius r floats just immersed in water. Find where the sphere must be cut by a horizontal plane, so that if the top part be removed the thrust on the plane surface of the remainder may be a maximum. 6. Draw the circle x z + y z -6x-8y + 21=0, and by drawing find the points on it at which -p = - . 1. A vertical line moves with uniform speed of 1" per second, so that its lower end always lies in the axis and its upper end on the parabola y=x%. [The unit being 1" each way.] At what rate is the length of the line growing when it is 10" from the origin? 3.2 3. P, Q are any two points on the parabola y=. V is the mid- ft0 point of the chord PQ and T is the point of intersection of the tangents at P and Q.. Shew that TV is parallel to OY. [Take co-ordinates of P (,,) a of a (,.?).] 3. ABCD is a rectangle. On AB, on the side remote from CD, an equilateral triangle ABE is drawn. If AB=a;" and BC=y" write down (i) the perimeter, (ii) the area of the figure EADCB. If the perimeter is 33 inches, what is the maximum area of the figure? 4. A man 6 ft. high walks at a uniform rate of 5 ft./sec. away from a lamp 10 ft. high. Find (i) the rate at which the length of his shadow grows, (ii) the rate at which the end of his shadow moves. 6. The formula for the velocity of sound is _ /Volume Elasticity Density [If the volume is in c.c., the pressure in dynes per sq. cm. and the density in gms. per c.c., this gives the velocity in cms. /sec.] 133] MISCELLANEOUS EXAMPLES 159 Find the velocity of sound in air, at C. , the height of the mercury barometer being 76 cms., given that density of air under these conditions is 001293 gms. and that the sp. gr. of mercury is 13-6. The pressure and volume of the air are supposed to be connected by the formula p V 1 '- = const. J. 1. A body moves in a straight line in such a way that its acceleration t seconds after it has passed a fixed point O on the line is (3t + 2t 3 ) ft. /sec. 2 Find its distance from O after 5 seconds, given that its speed when it passes O is 10 ft./sec. 2. A uniform sphere diameter 2 T 1 B inches is turned down uniformly to diameter 2-J? inches. Find approximately the percentage by which its weight is reduced. [Use no tables.] 3. Shew that the expression 1 + 12%-x 3 increases in value as x increases from -2 to +2 and after that diminishes as x increases. 4. In a certain curve the subtangent at any point is always equal to 3 times the abscissa of the point. Express this as a relation between the (x, y) and -~ of the point. [If the tangent at P meet OX in T and PN is the ordinate of P, TN is the subtangent.] Shew that y = 2x* fulfils the above condition. 6. A steamer has to go 20 miles up a river which is flowing at 3 miles an hour. The resistance of the water is proportional to the square of the relative velocity and the coal consumed in a given time is proportional to the product of this resistance into the distance the steamer has moved relative to water. If the cost of the coal is to be made a minimum, find the speeds of the steamer relative to shore and relative to water and the time taken on the journey. CHAPTER VII INTEGRAL CALCULUS. AREAS UNDER PLANE CURVES. MEAN ORDINATE 134. Integration. Suppose we want the area of the figure bounded by HK part of the curve y=f(x), the a?-axis, and the ordinates AH, BK corresponding to x = a, x = b. Fig. 63. Divide AB into any number of equal parts and complete the rectangles as shewn in Fig. 63. 134, 135] AREA BY SUMMATION 161 We get an approximation to the area we require by taking it as equal to the sum of these rectangles. This is too small by the sum of the blank portions HOP, PDQ, etc. If AB were divided into twice a.s many parts we should get a better approximation to the area, the sum of the 10 rectangles in Fig. 64 being nearer to the area sought than the sum of the 5 rectangles in Fig. 63 [the sum of the two blank portions between H and P < the portion HOP in the previous figure]. A L M A L M Fig. 65. Fig. 66. By dividing AB into more and more parts we obtain a series of closer and closer approximations to the area sought. 135. If we complete rectangles as in Ficr. 65, we get an approximation to the area required by taking the sum of these external rectangles. This is too large by the sum of the black portions HC'P etc. If AB were divided into twice as many parts we should get a better approximation, the sum of the 10 rectangles in Fig. 66 being nearer to the area sought than the sum of the 5 rectangles in Ficr. 65. M. a 11 162 CALCULUS FOR BEGINNERS [CH. VII [The sum of the two small pieces between H and P in Fig. 66 < the small piece HC'P in Fig. 65.] By dividing AB into more and more parts we obtain a series of closer and closer approximations to the area sought 136. Combining Figs. 63 and 65 into one (Fig. 67) we see that the difference between the sum of the outside and the sum of the inside rectangles is the sum of the rectangles CC', DD', EE', FF', = rectangle G F'. If we increase the number of parts into which AB is divided the area of GF' diminishes, for its height remains constant and its breadth diminishes, and by making the number of divisions large enough, we can make the area of GF' as small as we please. i.e. by taking a sufficiently large number of divisions, we can make the difference between the sum of the outside rectangles and the sum of the inside rectangles as small as we please ; and since the area sought lies between these two sums, it is the limit towards which the sum of outside or inside rectangles tends as the number of divisions is indefinitely increased, and the breadth of each division consequently indetinitely diminished. The determination of this limit is called Integration. 135-137] INTEGRATION 163 137. Note. If the shape of the curve is like that shewn in Figs. 68 and 69, the sum of the first set of rectangles is too large and the sum of the second set too small, otherwise the argument is the same and the area sought is the limit towards which the sum of either set tends as the number of divisions is indefinitely increased. \ A B Fig. 68; A B Fig. 69. If the shape is as in Figs. 70 and 71, we cannot say with certainty of the sum of either set that it is too great or too small, C' Fig. 71. 112 164 CALCULUS FOR BEGINNERS [CH. VII for in each case some of the strips are greater and some less than the corresponding portions of the required area. By dividing the area as in Fig. 72, we can apply the theorem to each part separately. Y W Op 45 10 14j 5 X Pig. 72. Pig. 73. 138. To make the principle clear we will take an easy special case. HK is a portion of the line y = 3. (Fig. 73.) AH, BK are the ordinates corresponding to x = 4 and x = 14. Divide AB into 10 equal parts and complete the rectangles as shewn. The different ordinates are the values of y corresponding to x = 4, 5, ... 14. 137, 138] AREA BY SUMMATION 165 Now y = 3ox Hence the ordinates are 12, 15, 18 ... 42. The width of each rectangle = 1. .*. sum of inside rectangles = 12 + 15 + 18 + ... + 39 (10 terms) Sum of outside rectangles = 15 + 18 + ... + 42(10 terms) The difference between these is 30, the area of the rectangle GK. [Height 30 = BK-AH, width 1.] The area ABKH lies between 255 and 285. Now suppose AB divided into 100 equal parts. The different ordinates will be the values of y corresponding toa? = 4, 4-1, 4-2, ... 13-9, 14. .'. the ordinates are 12, 12-3, 12-6, ...417, 42. The width of each rectangle is !. .'. sum of inside rectangles = -1 [12 + 12-3 + 12-6 + ... + 41-7] 100 terms = .! x 100 x 53-7 = 268 . 5> 2 and sum of outside rectangles = -l[12-3+12-6+...+42] = . lx 100^54-3 = 2?1 . 5 The difference between these is 3, the area of a rectangle, height 30 and width !. The area ABKH lies between 268-5 and 271-5. 166 CALCULUS FOR BEGINNERS [CH. VII Similarly shew that if AB be divided into 1000 equal parts the area ABKH lies between 269-85 and 270-15. We have two sets of numbers: 255, 268-5, 269-85,... 285, 271-5, 270-15,... apparently both continually approaching a limit 270. To make quite sure of this, suppose AB divided into n equal parts each = h, so that nh = 10. The ordinates are the values of y corresponding to a? = 4, 4+A, 4 + 2A, ..., 14 -A, 14. Hence the ordinates are 12, 12 + 3A, 12 + GA, ...,42-SA, 42. The width of each rectangle is h. .*. sum of inside rectangles = A[12 + (12 + 3A) + (12 + 6A) + ... + (42 - 37*)] (n terms) and sum of outside rectangles = h [(12 + 3A) + (12 + 6A) + ... + 42] . m The difference is 30/t the area of a rectangle height 30, width A. The area ABKH lies between 270 - 15h and 270 + 15h. Now, as the number of strips is indefinitely increased, i.e. as A -- 0, each of the areas (1) and (2) can be made as near as we please to 270. .'. the area ABKH which is the limit of either sum is 270. Since the limit of the sum of inside rectangles when A -* is the same as the limit of the sum of outside rectangles, we need only calculate one of the sums. 138, 139] UNITS 167 We may say either sum of inside rectangles = 270 15 A, .'. area reqd. = Limit of this when h -- 0, = 270 ; or sum of outside rectangles = 270 + 157*, .'. area reqd. = Limit of this when h -*- 0, = 270. 139. Note on units. In Fig. 73 the a-scale and the y-scale are the same, and the unit of area is the small square which is shaded, whose side is the unit of length. The area ABKH is 270 times this unit. /. if the unit be -1" each way, the unit of area is -01 sq. in. and area ABKH = 270 x -01 = 2'7 sq. ins. If however the scales are different as in Fig. 74 the unit of area is no longer a square but a rectangle whose sides are equal respectively to the oj-unit and the y-unit. The area ABKH is 270 times this rectangle. Y 40 30 20 10 2 3 4 5 6 7 8 9 10 11 12 13 14 Fig. 74. If a-unit = -3" and y-unit = -05", unit of area = '015 sq. ins., and area ABKH = 270 x -015 = 4-05 sq. ins. X 168 CALCULUS FOR BEGINNERS [CH. VII EXERCISES. XXXV. 1. Find by the method of 138 the area bounded by If the unit is fo" each way what is the area in sq. ins.? What is it if the ar-unit is J" and the y-unit T V I 2. Find the area bounded by y = 3x + 2, o;=0, y = B, y = 7. [Divide into strips parallel to Ox.] What is the area in sq. ins. if the ar-unit is 3" and the y-unit "? 3. P is the point (10,100) on the curve y = x z . PN is the ordinate of P and O is the origin. Find the area OPN bounded by ON, NP and the curve. (i) Divide ON into 10 equal parts, (ii) ,, ,,100 (iii) ,, ,, n ,, each h, and in every case find the sum of the inside and outside rectangles, using the formula .. b (iv) Find the limit of each of the last results as h -* 0. 4. Find the area OPN bounded by y^x 3 , the ar-axis and the ordinate x=3. Divide ON into n equal parts and use the formula 140. We shall now solve the problem of 138 by adopting an entirely different point of view. P is any point (a;, y) on the line y = 3x. (Fig. 75.) An ordinate is supposed to advance from the position AH corresponding to x = 4, to the position BK corresponding to #= 14, its length continually changing in such a way that its upper end 140J INTEGRATION 169 is always in the line HK. When the ordinate reaches MP, it has swept out a certain area which is obviously a function of x, i.e. it is determined when x is known and changes as a? changes. Call this area A. 7 A MN Fig. 75. "We will consider the change in the area due to a short advance of the ordinate. Let Q be a point (a; + Ao;, y + Ay) on the line and complete the rectangles as in the figure. Then the area PMNQ is the increment in the area A, due to an. increment AOJ in a;, and may therefore be denoted by AA. 170 CALCULUS FOR BEGINNERS [CH. Vil Now PMNQ lies between PMNR and SMNQ. i.e. AA lies between y . Ace and (y + Ay) . Ace. .*. lies between y and (y + Ay). As Arc, and with it Ay -* 0, (y + Ay) -* y. dA - = y = 3x. .". A is a function of x which when differentiated gives 3x. 3 where c is some constant to be determined. Now since A stands for the area traced out by an ordinate starting from AH, A = when the ordiuate coincides with AH, i.e. when x = 4. = -24. We want the value of A corresponding to x = 14. = 294-24 = 270. 3 Notice 294 is the value of x 2 when x= 14 3 and 24 is the value of -=a? when x = 4, and the area sought is the difference of these. 140, 141] INTEGRATION 171 EXERCISES. XXXVI. To be solved by the method of 140. 1. Find the area bounded by y = 3x, y = 0, a? =3, x = 9. 3. Find the area bounded by y = 3x, y = 0, x = 0, x=9. 3. Find the area bounded by y = Bx, y = 0, x = 0, x=3. 4. Find the area bounded by t/ = - x + 2, y = 0, x = 5, a:=10. IB 5. Find the area bounded by y = 5x + 7, J/ = 0, x = Q, x=5. 6. Find the area bounded by y = Bx + 2, aj = 0, y = 3, j/ = 7. 7. Find the area bounded by y = 4x- 3, a:=0, y=0, y = ll. 141. Example. Find the area OBK bounded by y = 0, o?=10 and part of y = x*. As before let M P be the ordinate of the point P (x, y) and PMNQ a strip of width A. [Fig. 76.] MN , X Fig. 76. Let A be the area OMP, then AA lies between ar'Aa; and x + Aa;' 2 Aa;. 172 CALCULUS FOR BEGINNERS [CH. VII This time our moving ordinate which is supposed to trace out the area starts from O where x 0. .*. when x = 0, A = 0. r 3 ' A-- ~3' 10* ,\ area OBK = -^- = 333^. EXERCISES. XXXVIL 1. Find the area bounded by y=x z , y = 0, x = B, a: = 14. 2. Find the area in the first quadrant bounded by y = x 2 , x-0, y = l, y = 4. 3. Find the area bounded by y=x 5 , y=0, x = 2, x=5. 4. Find the area bounded by y = 2x 2 + Bx + 1, y = 0, x = 3, x = 7. 5. Find the area bounded by y=x 3 + x, y = 0, x=3, a: = 11. 6. Find the area bounded by y = x 3 , x=Q, y = S, j/ = 27. 7. Find the area bounded by y = x 3 , y = Q, x=a, x = b. 142. Notice that in leading up to the equation //A ~ = 3x in 140 ax we could have supposed the moving ordinate to start from any position. The fact that it was supposed to start from the position AH was used in the determination of the constant c in 3 A = H a; 2 + c. 4i 3 The equation A - ^ a? 24 a is a formula for the area traced out by an ordinate which starts from AH and moves into any other position. 141, 142] INTEGRATION 173 Suppose now that the moving ordinate started from A'H' (a; = 2). [Fig. 11.} I OA'A B X Fig. 77. We should get in the same way as a formula for the area traced out by an ordinate which starts from A'H' and moves into any other position. Thus area A'H'HA = . 4 2 - 6, area A'H'KB = - . 14 2 -6. 2 - - 4 a 2 ' as before, the constant 6 disappearing in the subtraction. Therefore if we are finding the area between two given ordinates x - 4, a; =14 we can deduce at once from dx 174 CALCULUS FOR BEGINNERS [CH. VII that area between x= 4 and x= 14 rs i 14 or as it may be written ^ or ; and the same result would be obtained if we wrote for 3 3 where c is any constant whatever. 3 In this particular case, A = ^ ic 2 is a formula for the area traced 2i out by an ordinate which starts from the origin, so that when we T3 ~l 14 T3 I 14 = UarM e|-<#-6 L^ 1 4 I ^ _J 4 say we are merely saying that AHKB = OKB - OHA = A'H'KB - A'H'HA. 143. In Ex. xxxvn (4), the result is 309- 34|, 309^- and 34| 2X 3 3JC 2 being the values of ~- + -^- + x when x = 7 and x =3 respec- O L tively. 2a; 3 3JC 2 j- + -^r- + a; gives the area traced out by an ordinate starting o a from OL 309^ is area OLKB ; 34^ is area OLHA. [Fig. 78.] We miht have said 295| being area A'H'KB and 21^- the area A'H'HA, where A'H' is the ordinate x = 2, the constant 13 being chosen so that -5- + -jf + a; 13^ = when x = 2. o 2 142-144] INTEGRATION 175 HO 120 100 80 60 40 20 01234S678 Fig. 78. & a 4 b* , a 4 144. In Ex. xxxvn (7) the result is j-j. -j and 4 bein<* the values of ^- when x = b and * = respectively. We e 4 fa; 4 "! 6 might write the result A = T meaning f A 1 f QtA Value of - when x = b - Value of ^ when a - aj and the work might be stated shortly thus : ^ = .' being a function [regardless of arbitrary constant] which when differentiated gives a?. 176 CALCULUS FOR BEGINNERS [Ctf. VTT 145. Notice in Ex. (4), instead of finding the values of when x= 7 and when x = 3 we might more conveniently say = 274f. 146. Generally. If we want the area bounded by y =f(x} [any function of x], y = 0, x = a, x = b, we have, making the same construction as in 140 : Area of strip PMNQ lies between yAo; and (y + Ay) A#. [Fig. 79.] A MN. B X Fig. 79. i.e. AA is between yAa; and (y + Ay) Ase. .'. is between y and y + Ay ; but as A, and with it Ay, - 0, y + Ay -- y, 145-148] INTEGRATION 177 .'. A is some function of x which when differentiated gives /(#). Suppose <f> (x) to be such a function, .'. A = d> (x) + c. -L*^ Now A = when x = a, .'. c =-$(), .'. A = < (a;) < (a), .*. area between AH and BK = <f> (b) - (a) = \<f> (x)~| 147. Notice that PMNR is the area which would be traced out by the moving ordinate between M and N if it kept through- out the length which it has at M. SMNQ is the area which would be traced out if it kept throughout the length which it has at N. d\ Notice also that so far as finding --is concerned, it is suffi- ax cient to say AA =f(x) . Ao: approximately, AA .*. =f(x) approximately, it being understood that the statement AA -f(x) . Aos approximately means that AA is betweeny(a;) . Aa; and/(o; + Ace) . AiK. EXERCISES. XXXVIII. Find formulae for the area traced out by an ordinate of y = 5x s + 2 which starts from the position given by (i) # = 0, (ii) x=3, (Hi) x= -2 and moves into any other position. Deduce from each of these formulae the area traced out by an ordinate which moves from x=5 to =10. 148. Returning to the investigation in 138, p. 166, each term in the series (1) such as (12 + 6h)h is the area of a rect- angular strip. (12 + Qh) is the value of y or 3x when x = 4 + 2h. M. C. 12 178 CALCULUS FOR BEGINNERS [CH. VII The series is 12/i + (12 + 3A) h + (12 + 6A) A + ... + (42 - 3A) h ; [12, 12 + 3A, 12 + 6A, ...(42-3A), being the values of y or 3x when x = 4, 4 + A, 4 + 2A, ... (14 A)], or more shortly x=U-h 2 3x . h, a=4 meaning "Take values of x, starting at 4 and increasing by h at a time up to (14 A) ; for each value of x, find the corre- sponding value of 3x', multiply each of these values by A, and add all the results." The actual area sought is the limit of this sum when A~*-0. X=li z=14-fc 2 3x . h would mean the same as 2 3a:.A with the z=4 z=4 addition of an extra strip 42h (42 being the value of 3x when 03= 14) and this can be made as small as we please by taking A small enough. x=i4 x=n-h i.e. the limits when A -* of 2 3a? . A and 2 3x . h are x=4 x=4 the same. a =14 .*. we may say that the area ABKH is Lt 2 3.rA. h-*~o 1=4 149. Generally if the curve is y=f(x), adopting the usual notation as in 146, the area of the rectangular strip PMNR is 2/A.r, and the sum of the areas of all such strips between =6 AH and BK may be denoted by 2 yAo?, meaning : take values of x=a x starting at a and increasing Aa; at a time up to b ; for each value of x get the corresponding value of y from y=f(x); multiply each value of y by Arc and add all the results. The area of the figure bounded by HA, AB, BK and the curve is the limit to which this sum tends as Ace -*- and this is 148-151] INTEGRATION 179 rb r r denoted by / ydx I being simply a lengthened form of the J a LJ letter si. rb x=b 150. Thus I ydx is defined to be Lt 2 yAx, and .'a Ax-*-Ox=a we have seen that in order to get this limit, we must find a function of x which when differentiated gives y, and subtract the value of this function when x = a from its value when x = b. where = f(x). dx /i4 e.g. j[ (8)dte that is to say instead of finding the area by a process of direct summation, we find the rate at which the area increases with respect to x and from this deduce the area as a function of x. In other words instead of finding A from A= ( f(x)dx= Lt 2 f(x) Aaj, a summation which in most cases it would be impossible or at any rate inconvenient to effect, we find A from and we say that rb -]ft () Ja where </> (x) is a function of x such that d(fr(x) ,. . TT : t/ 151. Notice that just as dy and dx in the expression -j- have no separate meanings, but the form ~ is preserved to 122 180 CALCULUS FOR BEGINNERS [CH. VII remind us of of which -^- is the limit, so dx in the expression *Au? CISC /14 [14 I (3#) dx has no separate meaning, but the form I (So;) dx is x=u preserved to remind us of 2 (3a;) Ax of which it is the limit. x=4 x=14 Az represents a definite length and 2 (3x) Ase represents the sum x=4 of a finite number of rectangles. L 14 (3x) eta stands for the limit of this sum as Ax -* 0. 4 fit 152. An expression like I (3a;) dx is called a definite J4 integral, 4 and 14 being called the limits of the integral. It is read "Integral of (3a;) dx between 4 and 14." We have seen that in order to find the value of this we must first discover a function of x which when differentiated gives (80). This function is written I (3a;) dx and is read " Integral (3) dx." Thus /1 J An expression like I (3x) dx is called an indefinite integral. y = l(3x)dx, are merely different forms of the same statement. When the 151, 152] DEFINITE AND INDEFINITE INTEGRAL 181 value of an indefinite integral has been found, it should always be checked by differentiation. // 1 9 \ ^<y4 9 o 9 I / o -K \ r "^ $ ** Check : d^ /3jB_ 4 _ 2 f 2 \_3 23 ^ / 1\ dx\ 4 3 x J 4' 3'2'' \ x~) i 2 = 3x 3 -x +- 2 . The first step in the evaluation of a definite integral is the determination of the indefinite integral*. EXERCISES. XXXIX. Write down the values of the following indefinite integrals and in each case check by differentiation ; 1. 2x*dx. 2. . dx. J . I (ax 2 + bx + c)dx. 4. / - -? .- J J ^v* 5. (5t + 6t 2 -t)d. 6. . \ldx. 8. f Ux*~ dx. *' * Notice that the indefinite integral always contains an arbitrary constant, but in the subtraction which forms part of the process of evaluating the definite integral, this constant disappears and may therefore be omitted, so that, instead of saying _ we may say / , 182 CALCULUS FOR BEGINNERS [CH. VII Hence write down the values of the following definite integrals : 10. n h 11. /'('^ + ?) fc 12. ri J-i a 13. P X JjS^/x 14. 16. /8 Jo r \ Idx. J-3 15. 17. ./ 30 v1 ' 4 (Y3x3-5x2 + 2x- J 3\ f 2 / l \ i \ _LW o*^/ 18. ^ 1 Find the values of the following: z=7 19. 2 2x 2 . Ax (i) when Ax = l, (ii) when Ax= -05. X=5 20. 2 7. Ax (i) when Ax=-l, (ii) when Ax = '01. z=-3 Z=2 / 1 \ 21. 2 I x + -^ Ax when Ax ='2. =i V 2 / Compare your results with those of Exs. 10, 16, 18. 153. The process of finding an area may now be written down as follows : (taking again the example of 140) Area of strip = (3aj) Ax app. {z=u .*. Sum of strips = 2 (3a;) Aw app. and area required is the x=i limit of this sum when Ax -*- 0. ! /M .*. Area required = I (3x) dx Ji After a little practice the part in brackets {} may be omitted. 153] INTEGRATION 183 EXERCISES. XL. 1. P is the point (2, 4) on the parabola y=x i . PM and PN are perp. to OX and OY. Find (i) area OMP, (ii) area ON P. Verify that their sum is the area of the rectangle OMPN. 2. The same where P is (2, 8) on y = x 3 . f h z \ x 2 3. P is the point ( h, J on the parabola y . PM and PN are perpendicular to OX and OY. Find separately areas OMP, ONP and shew that they are respectively one-third and two-thirds of the rect. OMPN. / / t 3\ x 3 4. P is the point I h, -5 ) on the curve y = -5 , PM, PN are perp. to OX, V a* I a 2 OY. Prove that OMP, ONP are respectively one-quarter and three-quarters of the rect. OMPN. / h n \ x n 5. P is the point I h, - r I on the curve y=-=, . PM, PN are per- \ a" 1 / a* pendicular to OX, OY. Prove OMP= -^ rect. OMPN. n + 1 6. Find the area bounded by = 0, x=-2, x = 7. 7. Find the area bounded by c, y=0, x=-p, x=p. 8. The co-ordinates of H, K, L are respectively (-a, h) (a, k) (0, I). [Fig. 80.] A curve whose equation is yp + qx + rx 2 passes through H, L, K. Find p, q, r in terms of a, h, k, I. 9. In Qu. 8, if HA, KB are the ordinates of H, K, prove that the area 184 CALCULUS FOB BEGINNERS [CH. VII 154. The result of Exs. XL. (9) leads to a very important rule called Simpson's Rule for finding approximately the area of any figure. A B X Fig. 80. Suppose we want the area bounded by the curved line PQRSTUV, the ordinates AP, GV, and AG perp. to AP and GV. [Fig. 81.] Q R A B C D E F G Fig. 81. We can get a first approximation as follows : Divide AG into any number of equal parts and draw ordinates through the points of division as in the figure. Join PQ, QR etc. 154, 155] APPROXIMATE RULES FOR AREA 185 Then if AP = A n BQ = A 2 , e *c. and AB = BC =etc. =a we have Area of trapezium ABQP = - (Aj + A 3 ). Area of trapezium BCRQ= - (/* 2 + A 8 ) and so on. Area of trapezium FGVU = ^ (A 6 + hj). .'. by addition, Area of figure is approximately d or generally, for any number of equidistant ordinates Area is approximately ^. Distance between consecutive ordinates x {sum of extreme ordinates + twice sum of intermediate ordinates}. This is the Trapezoidal rule and its accuracy is obviously increased by taking more and more strips. Notice that this result is obtained by substituting for the curved boundary, a boundary composed of straight lines through pairs of consecutive points. In other words the portion of the boundary between two consecutive points is replaced by a line whose equation is of the first degree or of the form y =p + qx. 155. Now if we suppose the portion of the boundary passing through 3 consecutive points to be replaced by a curve whose equation is of the second degree of the form y = p + qx + rx*, we shall get a closer approximation. Considering the portion PQR. Take B as origin and axes along BC and BQ. Then co-ordinates of P, Q, R are respectively ( a, Aj) (0, A 2 ) (a, A 8 ) 186 CALCULUS FOR BEGINNERS [CH. VII and if y p + qx + ne 2 passes through these three points, we have A! = p qa> + ra? whence 3 = p + qa + ra h 3 fir, Now area ACRP bounded by PA, AC, CR and y =p + qx + rx z / of ( p + qx + rar) dx - px + J-a L 2 3 J-a 2ra 8 ~3~* Substituting the values of p and r, this becomes 2/4 2 a + (^i - 2^ + ^-3) = o (^i + 4A 2 + 7< 3 ), and this is taken as being approximately the area of the portion PACR of the given figure. If there is an even number of strips as in the figure, we shall get in exactly the same way Area RCET = ^ (h 3 + 4A 4 + h 6 ) approx. ci and Area TEGV - ~(h 6 + 4ft 6 + h,) 3 Adding we get Area of figure = ^ {/t x + h 7 + 4 (h 2 + h 4 + h 6 ) + 2 (h 3 + /i & )} approx. and similarly for any even number of strips. We thus have Simpson's rule : Divide the area into an even number of strips of equal width by an odd number of ordinates ; the area is approximately 7j . Width of a strip x {sum of extreme ordinates + twice sum 3 of other odd ordinates -t- 4 times sum of even ordinates}. 155-157] SIGN OF AREA 187 EXERCISES. XLI. 1. Find by integration the area bounded by y=0, x=2, #=12 and the curve y = x 3 . 3. Find the approximate area in (1) by using 3 ordinatea and applying (i) the trapezoidal rule, (ii) Simpson's rule. 3. Do the same as in (2) using 11 ordinates. 4. Do the same as in (1), (2) and (3) for the area bounded by y = 0, x = 2, x = 7 and y = x*. 156. Note on sign of area. If we advance from left to right, i.e. if Ao; is positive, y&x representing the area of our typical strip is positive if y is positive, i.e. if the strip lies above the ic-axis and negative if the strip lies below the a:-axis [Fig. 82.] I + y - y + y - LX + Ao: + AX Ao: y kx \ + y&x y*x y&x + * ~~ 1 J } I Fig. 82. If we advance from right to left, i.e. if Aw is negative, y&x is positive if the strip lies below and negative if the strip lies above the x-axis. a? 157. Thus if the curve be y = - 2 and P, Q be respectively the a f of points ( a, - a), (a, a) [Fig. 83], / z dx will give the area PMO J-a a and will be negative since we are advancing from left to right 188 CALCULUS FOR BEGINNERS [CH. VII and the area is below the axis, i.e. during the whole motion of the ordinate from MP to O, Ax is + and y is . /; a Fig. 83. z dx will give the area OMP and will be positive since we are advancing from right to left and the area is below the axis, i.e. Ace is and y is . f a y? [ y? Similarly I , dx will be positive and I dx will be h ' Ja o? negative. / y? ~2 dx we shall get zero, which simply means -a a that the moving ordinate in passing from MP to NQ sweeps out two numerically equal areas of opposite signs. If we want the actual area shaded, we must find one of the portions PMO, ONQ separately. .'. Shaded area = -- . 157, 158] SIGN OF AREA 158. As another example, consider the curve 189 between x= - 2 and x= + 5. [Fig. 84.] The area bounded by the curve, the a:-axis, and the extreme ordinates is seen from the figure to consist of 3 parts, two below the oj-axis and one above. Fig. 84. 190 CALCULUS FOR BEGINNERS [CH. VII r5 r~ %y3 y^~\ > Now I (4 + 3o: - or) dx -\ 4x+ - j-a 4 * J -s 2 . 21-. 133 The co-ordinates of the points R and S are (1, 0) and (4, 0). ~ f 1 -2 i.e. actual area of portion PMR = 2|-. This is area RAS. r5 i.e. actual area of portion SNQ = 2|-. i.e. the definite integral gives the algebraic sum of the 3 areas PMR, RAS, SNQ, or the excess of the actual area PMR over the sum of the actual areas of RAS and SNQ. EXERCISES. XLII. f2 1. Shew that I (l-x)dx=0. Jo Draw a figure to explain the result. /I (x 3 -3x*+2x)dx=0. Draw a figure to explain the result. It the curve y = x 3 - Sx 2 + 2x meet the x-axis in O, B, D (in order from left to right), and the ordinates at O and D meet the line y = - ? in E, F, what is the area of the figure bounded by the curve OE, EF, FD? 158J INTEGRATION 191 /+! 3. Find I (x 2 -x 3 )dx and interpret the result. 4. Find the area bounded by y =0^-6x2 + 9*4- 5 the x-axis and the maximum and minimum ordinates. 6. Draw the curve y*=x(x-l) 2 between x=0 and x=2 and find the area of the loop. 6. P, Q, R are the points on y = x 3 at which z=0, 2, 4. Find a, b, c BO that the parabola may pass through P, Q, R. Draw the two curves on as large a scale as possible and shew that the two closed portions contained between the curves are equal in area. Also find the area of each portion. 7. P, Q, R are the points on y = x 3 at which x h-k, h, h + k, Find a, b, c so that the parabola may pass through P, Q, R. Shew that the area bounded by y = 0, x = h-k, x=h + k and y=x 3 is the same as that bounded by j/ = 0, x h-k, x=-h + k and this parabola. 8. If the area bounded by y = x 3 , y = and any two ordinates be found by Simpson's Rule, what does (7) tell you about the result? 9. What is the area bounded by y=p + qx + rx* + sx3, y=0, x=-k, x-+k? Notice that your result is independent of 3 and s. 10. If P, Q, R be the points (-a, ftj) (i\ fc 2 ) (a, fcs), and if the curve y=p + qx + rx z + sx s passes through P, Q, R, shew that p, r have the same values as in 155 but that q, s are indeterminate. Find the area bounded by this cubic, t/ = 0, and the ordinates at P and R. Shew that it is the same as that obtained in 155. 1 1 . Draw roughly y - (.T + 1 ) (2 - x) . Find the area of the part above the x-axis. 192 CALCULUS FOR BEGINNERS [CH. VII 12. Trace the curve and find the areas of the two portions bounded by the curve and the x-axis. 13. Find the area common to the two curves y z =iax and a: 2 =4ay. 14. Find the area bounded by y%=x 3 and x=2. 15. P, Q are the points on the parabola y = 5 + 3x - 2z 2 where x= and x1^. Find the area bounded by the chord PQ and the portion of the curve above it. If the tangent to the curve which is parallel to PQ meet the ordinates of P, Q in M, N, shew that the area just obtained is two-thirds of the parallelogram PQNM. 16. P, R, Q are three points on y = ax* + bx + c corresponding to x=-h, x = 0, x = h. The tangent at R meets the ordinates at P, Q in M, N. Shew that this tangent is parallel to PQ and that area PRQ bounded by the curve and the chord PQ is two-thirds of the parallelogram PQNM. 159. The fact that any definite integral may be interpreted as an area can be used to find sometimes accurately, sometimes approximately the value of a given definite integral. ra . _ Ex. Find I *Ja?-x>dx. Jo At present we do not know how to find the indefinite integral V a 2 x 1 dx. But if we draw the graph of y = \/a 2 or 2 between x = and x = a we get a quadrant of a circle, centre at the origin, radius a. [Fig. 85.] The given definite integral is the area of this. 159, 160] APPROXIMATE INTEGRATION 103 Even if we did not recognise the curve as part of a circle we could find several points on it, i.e. calculate the lengths of a number of ordinates and apply Simpson's Kule to find the approximate area. Fig. 85. EXERCISES. XLIII. 1. Shew from a figure that = 2 | J - 3. Find approximately by Simpson's rule [use 11 ordinates] f 3 1 3. Similarly find I y J 2 L + 160. If we draw and -dx. x dx and __1 1 = 4^ + [3 1 I , y 2 l-a; dx. (which may be called the integral curve of t/ = ^ + 2 j between the same values of x (say 2 and 12), the number of units of length in the difference of the extreme ordinates in the second graph is M. 0, 13 194 CALCULUS FOR BEGINNERS [CH. VII equal to the number of units of area in the figure bounded by the curve, the ar-axis and the extreme ordinates in the first graph. [Fig. 86.] S:: GO 40 20 **- -20 -40 Fig. 86. Or No. of units of length in gk (or g'k') (55) = No. of units of area in AHKB and generally if the second graph be y=f(x) and the first y /' (x) a corresponding result will be true. 160] 195 Notice that it does not matter which curve of the family we use. e.g. in the figure hk is y - - y? + 2x (c = 0), h'k'isy = jx> + 2x-'20 (c = -20). The ordinate of ti = (the ordinate of h) - 20 and the ordinate of k' - (the ordinate of k) - 20. So that difference between ordinates of h' and k' difference between ordinates of h and k. 132 196 CALCULUS FOR BEGINNERS [CH. VII 161. As another example take the graphs of y = 6a; 2 -18aj+ 12 [/'(*)] and y = 2a 8 -9a? + 12+c [/()]. [Fig. 87.] No. of units of area in STLP = No. of units of length in Ip ts, i.e. in up or in lp'ts, i.e. in u'p'l2^. No. of units of area in PLQ = No. of units of length in mq lp, i.e. in vq or in mq' lp\ i.e. in v'q' = l. No. of units traced out as ordinate moves from L to R = No. of units of length in nr Ip or in nr - lp' = 0. i.e. area PLQ = area QRK. No. of units traced out as ordinate moves from Q to R = No. of units of length in nr mq or in nr mq' = vq or v q' = 1. i.e. area QRK = 1 unit. 162. We saw on p. 142 that if h is small to a first approximation. We can now obtain a closer approximation. As a first example suppose so that /'(*) = 4* + 3. In this case y =f (x) is a straight line. [Fig. 88.] Now, No. of units of length in nq mp = No. of units of area in PMNQ. The gradient of PQ is/" (x). .'. RQL=hf"(x). .'. No. of units of area in PMNQ 161, 162] INTEGRAL CURVES 197 /. f(x + h) -f(x) = hf (x) + \ Vf" (x) or This is accurately true in this case and in every case where f(x) is of the form aa? + bx + c, since in all such cases y =f (a;) is a straight line. y=f(x) f(x+h) f(x) rn n y=f(x) ?B R hf"(x) If' x M N Fig. 88. 198 CALCULUS FOR BEGINNERS [CH. VII EXERCISES. XLIV. Verify f(x + h) =f (x) + hf (x) + i fc/' (*), (i) when / (x) = (ii) (iii) 163. If y =f (x) is not a straight line, we have as before f(x + h) -f(x) = No. of units of area in PMNQ = No. of units of area in PMNT approximately [Fig. 89] i.e. f(x + h) =f(x) + hf (x) + - tff" (x) approximately. 2 Notice that in this case we take the arc of y =f (x) as being approximately a straight line, whereas in obtaining the first approximation we took the arc of y =f(x) as being approxi- mately straight e.g. taking the example on p. 143. Find approximately the value of 3af-4a? + 6x*-7x+ 8 when a: = 3-01. /' (*) = 15x 4 -l2x*+ 12* - 7, /"(*) = 60^-240; +12. = 662, /'(3) = 1136, /"(3)=1560, /. /(3-01) =/(S) + -01 /' (3) + f" (3) app. = 662 + 1 136 x -01 + x -0001 2 = 673-4380 app. [Actually /(3-01) - 673-438264503.] MEAN ORDINATE 199 163, 164] 164. Mean or average ordinate. Suppose AB to be a portion of the curve y f(x\ A, B being the points corresponding to x = a and x = b. [Fig. 90.] 5 I M Fig. 89. Fig. 90. The mean ordinate of the curve between x - a and x = b is the height of the rectangle with base MN equal in area to the figure AMNB. It is in fact that ordinate of constant length which in moving from M to N traces out the same area that the variable ordinate of the curve does in moving from M to N. [Compare idea of average speed p. 2.] 200 CALCULUS FOR BEGINNERS [CH. VII 165. Suppose for example, we want the mean ordinate of the curve y = x 2 between x = \ and x = 5. [Fig. 91.] and T 2 3 4 5 Fig. 91. [* > 7 f^l 5 124 AreaAMNB=/ x*dx= -^ = ~3~ .*. Mean ordinate = -5 - = = lOi. 6 x 4 6 If CD be the line y = 10, Area CMND = area AMNB. EXERCISES. XLV. 1. Find the mean ordinate of the curve y x 2 between x=4 and a; = 10. Find also the arithmetic mean of the two ordinates a;=4, a; = 10 and the midway ordinate corresponding to a; = 7. a. Find the mean ordinate of the line j/ = 3x + 5 between x = 3 and Find also the arithmetic mean of the two ordinates x = 3 and x = 15 and the midway ordinate. 165, 166] MEAN ORDINATE 201 3. Prove by calculus and geometrically that the mean ordinate of y=mx + n between x = a and x=b, the arithmetic mean of the ordinates x = a, x = b and the midway ordinate are all equal. 4. Find the mean ordinate of y=x 3 (i) between x = 3 and x=7, (ii) between x = a and x=b. 6. Find the mean ordinate of y=x z -4x + 3 between x=l and x=5. 6. Find the mean ordinate of y = x% between x=0 and x=+3, also between x= - 3 and x= +3 and explain why these results are the same. 166. If we divide the area into a number of strips (say 20) of equal width (-2), and complete the inside rectangles in the usual way with an extra one on the right [Fig. 92], the area of "1 2 3 4 5 Fig. 92. the figure bounded by AM, MN', N'B' and the zig-zag boundary AB' is and MM' = 4-2. So that the mean height of this figure is 2 fl 2 + 1 '2 2 + . . . + 5 2 ] I 2 + 1 -2 2 + . . . + 5 3 4-2 21 = the Arithmetic mean of the 21 ordinates from AM to BN inclusive. 202 CALCULUS FOR BEGINNERS [CH. VII [This is 10^ = 10-47.] If we take 100 strips the Arithmetic mean of the 101 ordinates is 1043. The more strips we take the nearer does the Arithmetic mean approach to 10^. This is only what might be expected, for we have shewn that by increasing the number of strips the area AMN'B' can be brought as near to the area bounded by AM, MN, NB and the curve BA as we please. The value 10^ may thus be called the mean or average value of a; 2 between x = 1 and x = 5, and it is the limit to which the Arithmetic mean of a number of equidistant ordinates the first of which is AM and the last BN, approaches as the number of ordinates is indefinitely increased. EXERCISES. XLVI. 1. Find the mean value of x 2 + 3x between x=2 and z=4. Find also the Arithmetic mean of 11 values of x 2 + Bx corresponding to x = 2, 2-2, 2-4, 2-6, ...4. 2. Find the mean value of , between x = 1 and a; =4. x e 3. Use Simpson's rule to find approximately the mean value of - between x = l and x = 2. (11 ordinates.) Also use a table of reciprocals to find the Arithmetic mean of 11 values of - corresponding to o; = l, !!, 1-2, ... 2. *C 4. Find the mean value of z 3 + 2x z + 1 between x = 2 and x = 7. 167. If part of the curve y =J (x) is below the a-axis the corresponding portion of area must be reckoned negative. Thus to get the mean ordinate of y = y? between x = 2 and a? = 3. 166, 167] We have MEAN ORDINATE 208 65 and this gives actual area ONB - actual area OMA. 65 13 . . Mean ordmate = = = - = 3 i . 4x5 4 If CD be y = -^, Area CMND = actual area ONB -actual area OMA [Fig. 93] or = algebraic sum of areas ONB, OMA. Fig. 93. If we take 26 equidistant ordinates distance apart -2 the Arithmetic mean will be (_ 2) 3 + (- 1 -8) 3 + (- 1 -6) 3 +...(- -2) 3 + O 3 + (-2) 3 + (-4) 3 + . . . + 3 8 ~2Q _ 2-2 3 + 2-4 3 +...3 8 _ 91 _ ~~26~ ~26~ * and as before the greater the number of ordinates we take the nearer will the Arithmetic mean be to 3f . 204 CALCULUS FOR BEGINNERS [CH. VII EXERCISES. XLVII. I. Find the mean ordinate of y = 2x + 6 between x= - 11 and x=5. Draw a figure to explain the result. 3. Find the mean ordinate of y=a? (i) between and 3, (ii) between - 3 and 3. Draw a figure. 3. Find the mean value of 4 + 3x - x z between x = - 2 and x = 3. Find also the Arithmetic mean of 11 equidistant ordinates. 4. Find the mean height with respect to x of that portion of y = 4 + 3a;-x 2 which lies above the z-axis. 6. Find the mean value with respect to the abscissa of the square of the ordinate of a semicircle of radius a. 6. Find the mean value of 2x 3 - 3a; a - 36z -f 30 between x= -3 and CHAPTER VIII FURTHER APPLICATIONS OF THE INTEGRAL CALCULUS 168. So far we have only considered the Integral Calculus in its application to the areas of plane curves. We shall now shew how similar methods may be applied to other problems. Ex. 1. The speed of a body in ft. /sec. at the end of t seconds is given by V = 3t2. Find the distance travelled in 12 seconds from rest. Suppose the whole time 12 sees, to be divided into n equal intervals each h sees., so that nh=l'2, and suppose the speed to remain constant during each interval and equal to the speed at the beginning of the interval. The speeds at the beginning of successive intervals are 0, 3A 2 , 3 (2hy, ... 3 (n^lhf, and the total distance described on our assumption would be h [0 + 3# + 3 . (2A) 2 + ... + 3 (n^\hy\ 01J 2 2 + 3 2 +... (n-1) terms] =-(-l) n(2-l) = 1728 - 216h + 6h2 (= aj. 206 CALCULUS FOR BEGINNERS [CH. VIII Now suppose the speed to remain constant during each inter- val and equal to the speed at the end of the interval. The total distance would be h [3/t 2 + 3 . (2&) 2 + . . . + 3 (w/i) 2 ] = 3h 3 [P + 2 2 + 3 2 + ... n terms] o 73 = -- . n (n + 1) (2n + 1) = 1728 + 216h + 6h2 (= S 2 ). Now the actual distance (S) lies between Sj and S. 2 . But by diminishing h indefinitely we can make Sj and S 2 each as near to 1728 as we like. .'. S is the limit of either Sj or S 2 as h-*-Q, and S= 1728. Notice that it is unnecessary to find both S, and 83. We can say either S 1 = 1728 - 216A+ 6A 2 , .'. S=1728, or S=1728 + 216A + 6/t 2 , .'. S=1728. EXERCISES. XLVm. If v=5 + 7t find by this method the distance travelled, (i) in the first 5 seconds, (ii) in the 10th second. Also in each case find what the distance would be if we supposed the time to be divided up into intervals of -01 sec. and the speed to remain constant during each interval and equal to the speed at the beginning of the interval. Find by how much per cent, the distance calculated on this assumption differs from the actual distance. 169. The method employed in 168 corresponds exactly to that employed in the determination of an area in 138. There we had a continually changing ordinate (y) whose length was given as a function of the abscissa (a;). Here we have a continually changing speed (v) whose magnitude is given as a function of the time (t). There we got an approximation to the area traced out 168-170] DISTANCE FROM SPEED-TIME FORMULA 207 by the moving ordinate by dividing the distance travelled by it into small pieces and supposing the ordinate to remain the same length as it passed from end to end of each small piece : we then added the areas of the thin rectangles so obtained and finally found the limit of the sum of these areas as the number of them was indefinitely increased and the breadth of each consequently indefinitely diminished. Here we get an approximation to the distance travelled by the body by dividing the whole time into small intervals and supposing the speed to remain constant during each interval ; we then add the small distances so obtained and finally find the limit of the sum of these distances, as the number of them is indefinitely increased and the length of each interval consequently indefinitely diminished. 170. We shall now approach the problem from a different point of view corresponding to that adopted in 140, in finding the area traced out by a moving ordinate. Let s ft. be the distance described in t sees, and s + As ft. be the distance described in t + A sees. The speed at the beginning of the interval A is 3t* f t./sec. and at the end 3 (t + A) 2 . /. As lies between 3< 2 . A< and 3 (t + A*) 2 A, As .-. lies between 3i! 2 and 3 (t + A) a . L\t But as At ^ 0, (t + A) 2 ^ < 2 . /. s = t 3 + c where c is some constant. ds * [So far as finding -T- is concerned we might simply say CLt As = 3t 2 . A app. /. ?=8A v. 147.] at 208 CALCULUS FOR BEGINNERS [CH. VIII Since we are finding the distance the body has travelled from its position when t = 0, we have s = when t 0, .'. = + c or c = 0. .'. Distance in 12 sees. = 12 3 = 1728 ft. Notice that this is EXERCISES. XLIX. 1. Find by this method the distance travelled in the 5th second when r=3t 2 [ft. -sec. units]. 2. Find the distance travelled between the ends of the 3rd and 10th seconds when w=2t 3 + 3t + 5. 171. Generally if v =f(t) and we want the distance between the end of a sees, and the end of b sees., we say As lies between f (t) . A< andy( + A2) . A and as before we deduce Let < (t) be a function of t which when differentiated gives /<*) .*. s = < (t) + c. Now s = when t = a. i.e. * = < (t} <f> (a), and the distance required = <j> (b) </> (a) = (t) \ . 172. Each term in the first series in 168 is the distance travelled in a time h, the speed being supposed constant through- out the interval. The series is 170-174] DISTANCE FROM SPEED-TIME FORMULA 209 (0, 3A 2 , 3(2/t) 2 ... being the values of v or 3t 2 when = 0, h, 2h } ***j /> t=12-h =12-fc or shortly 2 3 2 . h or 2 v . A. t=o t=o t=12-A As in 148 we may shew that the limits of 2 3 3 . A and <=o e=i2 2 3Z 2 . h when A -*- are the same. t=o .'. we may say that the actual distance required is the limit =12 of 2 3Z 2 . h when h is indefinitely diminished. <=o 173. Generally if v =f(t}, the distance described in a small interval A following the end of t seconds is approximately v. A and the sum of all such distances between the end of a seconds and the end of b seconds may be denoted approximately by <=b 2 v&t. t=a The actual distance is the limit of this when A -*- and this is denoted by / vdt or [ f(t}. dt, Ja. J a and we have seen that this is <j>(b) <f> (a) or < (t) \ e.g. In our problem ( 168, 170) distance - I W . dt = [f] - 1728. Jo L Jo 174. This problem can be reduced to the problem of finding an area as follows : The relation v - 3 2 can be exhibited in the form of a graph. [Fig. 94.] M. C. 14 210 CALCULUS FOR BEGINNERS [CH. VIII Distances along the horizontal axis represent seconds of time and distances along the vertical axis represent speed in ft./sec. OM represents t seconds, i.e. OM contains t horizontal units of length. MP represents a speed of 3 2 ft./sec., i.e. MP contains 3< J vertical units of length. ON represents (t+ A) seconds and NQ 3 (t + A) 8 ft./sec. The number of feet described in the interval A on our first hypothesis is v A, and this is the number of units of area in the rectangle PMNR. The number of feet described in 12 seconds will be the number of units of area in all strips like PMNR between O and the ordinate corresponding to t = l2. t=12 In other words 2 vA< may be looked upon either as the t=o number of feet described in all the intervals or as the number of 174, 175] DISTANCE FROM SPEED-TIME GRAPH 211 units of area in all the strips and the limit of this when A< - f 12 (which we call I vdt) is either the number of feet described in JO 12 seconds or the number of units of area in the curvilinear figure OAB where AB is the ordinate 03= 12. General statement for speed-time graph. 175. Generally. If the speed be given as a function of the time, say v=f(t), and the speed-time graph be drawn, the 400 300 V 200 2000 1500 S 1000 5 2 4 6 8 10 12 Fig. 95. 142 212 CALCULUS FOR BEGINNERS [CH. VIII number of units of area bounded by the curve, v = Q, t = a, t b, gives the number of units of length in the distance described between the end of a seconds and the end of b seconds. If the speed-time and space-time graphs be drawn, the number of units of length in the difference between two ordinates of the space-time graph is equal to the number of units of area in the corresponding portion of the speed-time graph. e.g. number of units of area in OPM is number of units of length in mp or m'p. [Fig. 95.] i.e. the number of units of area lying between the speed-time curve, the time-axis and the ordinates t = a, t = b is the number of units of length in the distance travelled between the end of a seconds and the end of b seconds. Another example. 176. If w=112-32, rt>A = ["ll2-16< 1 T= 112x3-16 x 21=0. The meaning of this is not hard to find. From t = 2 to t = 3 J, v is positive. From t = 3 to t = 5, v is negative. t=5 So that of the terms which go to make up 2 v&t, some are t=2 positive and some negative and our result tells us that the sums of positive and negative terms are numerically equal. As a matter of fact our formula gives the speed at any time of a body projected vertically with speed 112 ft./sec. From t = 2 to = 3J the body ascends and from t = 3^ to t = 5 it descends an equal distance, so that its distance from the starting point, or what we may call the effective distance described, is zero. If we want the actual distance described we must integrate from 2 to 3. 175-177] SPEED-TIME AND SPACE-TIME GRAPHS 213 If we draw the graph v=112-32< we get a straight line cutting y - where x = 3. From x = 2 to x = 3| the area is above the a;-axis and from x = 3 to x = 5 below and the two portions of area are numerically equal but of opposite sign. [Fig. 96.] 100 50 -50 200 150 I '100 50 123456 t Fig. 96. The corresponding ordinates in the space-time graph are equal. 177. Ex. 2. Find the work done in stretching an elastic string from a length of 18 to a length of 25 inches, given that the natural length is 8 inches and that it has a length of 14 inches when sustaining a pull of 3 Ibs. wt. 214 CALCULUS FOR BEGINNERS [CH. VIII By Hooke's law if the pull be T Ibs. wt. and the stretch s inches, T = ms (where m is a constant). Now when T = 3, s = 6, .. T = S. Divide the total extension of 7 inches into n equal parts, each h inches, so that nh=7 and suppose that as the string was stretched from 18 to (18 + A) inches the pull remained the same as it was at 18 inches and that as it was stretched from (18 + h) to (18 + 2A) the pull remained the same as at (18 + h), and so on. The pulls corresponding to lengths 18, 18+ A, 18 + 2A, ...25 -A, h 17-h are 5, 5 + ^, o + n, ... ^ , and the work done on our assumption would be /_ h\ 17 /i~l . h 5 + 1 5 + 5 J + (5 + h) + ... - in. Ibs. wt. L V z ' ^ J _nh 27 -h 27 1 , , 27 , 7, /189 7h\ . = - r . nh --.nh.h = -r- . 7 -rh[ 7- I in. Ibs. wt. = W,. 4 4 4 \ 4 4 / If we supposed that as the string was stretched from 18 to (18 + A) inches the pull was the same as at 18 + h, and so on, we should get as the work done 189 7A\ . ] in. IDS. wt. = Vj. Now the actual work done lies between Wj and W 2 but each of 189 these can be made as near j- as we like if h be made small enough. .'. Actual work (W in. Ibs. wt.) is the limit of either W x or W, when h - and is 47 j in. Ibs. wt. 177-179] WORK DONE IN STRETCHING STRING 215 Notice that we need not calculate both Wj and W 2 . We can say .,, u , 189 7A . 189 either Wj - j- , . . W = , 189 77* 189 w 2 = - T - + T , .. w=- r . 178. The method employed here corresponds to that employed in 138 and 168. We shall now solve the problem by a method corresponding to that employed in 140 and 170. 179. Let W in. Ibs. wt. be the work done in stretching the string from 18 ins. to x ins. and (W + AW) in. Ibs. wt. the work done in stretching it from 18 ins. to (x + Ace) ins. ic 8 The pull corresponding to length x ins. is Ibs. wt. 2 The pull corresponding to length (x + Ax) ins. is C+ Ace Ibs. wt. .". the work done in stretching from x ins. to (x 4- Aa;) ins. lies x-8 aj + Aoi-8 . between - Aaj and = Aa; in. Ibs. wt. '-i 2i 05-8 A . x + Ace 8 A i.e. AW lies between Ace and = Ace in. Ibs. wt. 2i '2i AW ,. cc-8 , 05 + Aa;-8 . .*. lies between ^ and in. IDS. wt. Aa; 2 2 dw x-8 _ x . % " dS-~2~-2~ a? :. W = -r - 4ce + c. 4 dW * So far as finding -r is concerned it is sufficient to ay ax Q AW = ^ Aa; approximately. 2 AW x-8 ,\ = - approximately. 216 CALCULUS FOR BEGINNERS [CH. VIII Now when x = 18, W = 0. 18 2 .'. = 4. 18 + c, x 2 /. W=^-4x-9. 625 .*. work done in stretching to 25 inches= j- = 56^-9 = 47 in. Ibs. wt. yS, Notice that 564- and 9 are the values of -7 4a3 when x = 25 4 and 18 respectively, and r-y.1 -i 25 work done = - 4cc . Jtt 180. Each term in the series in 177 is the work done in stretching the string through a small distance A, the pull being supposed to remain constant throughout this small stretch. The series is h 5 + ( 5 + - ) + (5 + K) + . . . in. Ibs. wt. L V */ J (5, 5 + ^ , 5 + h ... being the values of T or ^ , when x = 1 8. 2i 2i 18 + A, 18 + 2^...). x=25-h x8 =25-A Or shortly 2 h . ^ or 2 hT. z=18 x=18 =25-fc As in 148 we may shew that the limits of 2 AT and x=18 *=25 2 AT when h -*- are the same. =18 .'. we may say that the actual distance required is the limit a;=25 of 2 AT when A -*- 0. z=18 179-182] WORK 217 181. Generally if T=f(x) gives the pull in terms of the length, the work done in stretching from length x to length (a; + Ace) is approximately T . Aa; and the sum of all such works done as the length increases from a to b may be denoted approximately by x=b 2 T . Aa x=a The actual work done is the limit of this when Ax -+ and this is denoted by rb /& I T dx or ; f (x) . dx, J a J a and we have seen that this is <f> (b) <j> (a) or </> (x)\ where dx ~' e.g. In our problem rx 8 Vx* ~\ w _ . dx = \ 4* = 471 in. Ibs. wt. 2 L 4 JM 182. This can also be reduced to finding the area of a curve as follows. Tdx where T is a given function of x, ^- x ~ 8 in this case ^ . m x 8 This relation T= ^ can be exhibited in the form of a graph. [Fig. 97.] Distances along the horizontal axis represent length of string in ins., and distances along the vertical axis represent pull of string in Ibs. wt. In this case the graph is a straight line. OM represents x ins., i.e. OM contains x horizontal units, MP represents a pull of T ( = ^ J Ibs. wt.,i.e. MP contains T ( = ^ J vertical units. 218 CALCULUS FOR BEGINNERS [CH. VIII ON represents (x + Ace) ins. and NQ a pull of x + Ace i T + AT = Ibs. wt. -2 -4* 20 B 10 15 A MN 25 Fig. 97. The work done on our first hypothesis as the string is stretched from length x to length x + Ao; is T . Ao; in. Ibs. wt. and this is the number of units of area in the rectangle PMNR. The work done in stretching the string from 18 to 25 ins. will be the number of units of area in all the strips like PMNR between A and B. z=25 In other words S T . Aa; may be looked upon either as the 2=18 number of in. Ibs. wt. of work done on our first assumption or as the number of units of area in all strips like PMNR, and the (26 limit to which this tends as Aa;--0 which we call I T dx is J18 either the number of in. Ibs. wt. of work done or the number of units of area in the figure ABKH. 182, 183] ASPECTS OF INTEGRATION 219 183. The following statements should now be intelligible. (1) Integration is a process of summation. If f(x) be any function of x, the values which f(x) takes as x increases from a to b by equal increments h are /(a), f(a + h),f(a + 2h).... The limit of the sum h ./(a) + hf(a + h) + ... hf(b) when h -*-0 x=b or as it may be briefly written the limit of S f(x) . Aa; when x=a rt> Ace -* is written I /(a?) . dx and the finding of this limit is Ja called integration. (2) Integration is a process of anti-differentiation. To find the limit of the sum just mentioned it is necessary to discover a function of x, say <f> (x), such that ^ ' =f(x), i.e. to dx find a function (f> (x) which when differentiated gives /(). This process of finding < () which may be called a process of anti-differentiation is also called integration. If (3) Any definite integral may be interpreted as an area. f(x) . dx is the area bounded by the curve y =f(x), the -axis and the ordinates x = a and x = 6. Therefore, even if we cannot find the value of the definite integral by the ordinary methods, that is to say if we cannot dis- cover a function < (x) such that *V =f( x }i we can nn d its CtOb value approximately by drawing the curve y=f(tt) between x = a 220 CALCULUS FOR BEGINNERS [CH. VIII and x = b and applying some one of the approximate methods previously mentioned for finding the area. 184. In any problem requiring the Integral Calculus for its solution it is advisable to keep in mind what we should do if we wanted an approximate solution. Ex. 1 . Find the area bounded by y = cc 3 , y = 0, x = 3, x = 1. We have to find the area traced out by an ordinate which moves from the position AH to the position BK, its length changing continually in accordance with the law y = a?. [Fig. 98.] To get an approximate area we should divide AB into a num- ber of equal parts like MN, and suppose the ordinate to keep the same length MP as it moved from M to N. The sum of all such rectangles as MR would be an approxima- tion to the area required, and we have seen that the actual area required is the limit of the sum of such rectangles when their width is indefinitely diminished. We have seen that so far as the value of this limit is concerned it is immaterial whether we suppose the ordinate in its passage 183, 184] EXAMPLES OF INTEGRATION 221 from M to N to preserve its initial length M P or its final length NQ, or of course any intermediate length. The work may be stated thus : Suppose the area divided into a number of strips by lines parallel to OY. Let PMNQ be a typical strip, MP being the ordinate corre- sponding to abscissa x. .'. MP = x 3 and area PMNQ = x=1 approximately and sum of strips = 5 a? 3 Ace approximately. z=3 x s dx Ex. 2. The speed (v ft./sec.) of a body at the end of t seconds is given by v = 6t+17t 2 . Find the distance travelled between the end of the. 2nd and the end of the 5th second. Here we have to find the distance travelled in a certain time by a body whose speed continually changes in accordance with the law v = 6t + 17t 2 . To get an approximate result we should divide the whole time into a number of small intervals and suppose the speed to keep throughout each interval the value which it has at the beginning of the interval. We should then calculate the distance travelled in each interval on this assumption and the sum of all the distances would be an approximation to the required distance. The actual distance is the limit of this sum when the duration of each interval is indefinitely diminished. We have seen that so far as the value of this limit is concerned it is immaterial whether we suppose the speed throughout an interval to remain 222 CALCULUS FOR BEGINNERS [OH. VIII the same as it was at the beginning or at the end of the interval, or indeed at any intermediate instant. The work may be stated thus : Suppose the whole time divided into a number of small intervals. The interval A following the end of t seconds may be taken as the typical interval. The speed at the beginning of this interval is (6J+17* 2 ) ft/sec. .*. Distance travelled in this interval = (6f+ 17 2 ) A feet approximately and the total distance is 2 (Qt+ 17< 2 ) A approxi- mately. /B Actually, distance is / (6< + Iltydt Ja = 63 + 663 = 726 feet. Ex. 3. A tank 6 feet deep is filled with water. Find the magnitude of the resultant thrust on an end which is 4 feet wide. The pressure at any point is proportional to the depth and is therefore continually changing as we pass downwards from top to bottom of the tank. We should get an approximation to the required thrust by dividing the area into a number of thin strips by horizontal lines, and supposing the pressure to remain constant over each strip. If PQNM is such a strip we suppose the pressure not to change as we pass from the level PM to the level QN but to keep throughout the value that it has at the level PM. [Fig. 99.] 184] EXAMPLES OF INTEGRATION 223 We should then calculate the thrust on the atrip PQN M on this assumption, and the sum of the thrusts on all the strips would be an approximation to the required result. 4' X 1 Fig. 99. The actual resultant thrust is the limit of this sum when the width of each strip is indefinitely diminished. The limit would be the same if we supposed the pressure at every point of PQNM to be the same as it is at the level QN or at any level intermediate between PM and QN. The work may be stated thus : Suppose the end divided into thin strips by horizontal lines. PQNM is such a strip, the depth of P being x feet and PQ = Ax feet. The pressure at the level PM = wt. of x cubic feet of water per sq. foot= 62'5a; Ibs. wt./sq. ft. The area of the strip = 4Aa? sq. ft. .*. Thrust on strip = 62'5a? x 4Aa; Ibs wt. approximately = 250ajAo3 Ibs. wt. approximately. =6 .'. Total thrust on end = 2 SSOceAa? Ibs. wt. approximately. *=o 224 CALCULUS FOR BEGINNERS [CH. VIII Actually, thrust = = 250 ra: = 4500 Ibs. wt. Ex. 4. Find the work done by a gas in expanding from 2 to 3 cubic feet, the pressure and volume being connected by the law pv = constant, given that the pressure is 2 160 Ibs. wt./sq. ft. when the volume is - cubic foot. 4 Let p Ibs. wt./sq. ft. be the pressure] v cub. ft. the volume / * .'. pv = 21 60x^ = 540. Suppose the gas contained in a cylinder closed at one end and fitted with a piston free to slide along the cylinder. Call the area of the piston a sq. ft. [Fig. 100.] R 5 P/Q v Ay a Fig. 100. When the volume of the gas increases from v to v + &v the piston moves feet and the force on the piston at the beginning Ctr of this small movement is . a Ibs. wt. v 184] FURTHER APPLICATIONS 225 .*. Work done by gas when volume increases from v to v + Av is 5-10 Aw,, ., . a . ft.-lbs. wt. approximately = . Aw ft.-lbs. wt. approximately, and the total work done in expanding from 2 to 3 cubic feet is = 3 540 2 - - Av app., =2 r540 . dv. v At present we do not know how to find the value of the indefinite integral I - dv, but we can obtain an approximate value for the definite integral by looking upon it as an area and applying Simpson's rule. [Fig. 101.] Taking 11 ordinates, we have Extreme Even Odd 270 257-14 245-45 180 234-78 225 216 207-69 200 192-86 186-21 450 1094 ; 13 871-00 4376-52 1742-00 450 6568-52 Area = ^ x 6568-5 U = 219 nearly. .'. Work done = 219 ft.-lbs. wt. M. 0. 15 226 CALCULUS FOR BEGINNERS [CH. VIII [We shall see later, Exs. LXXIX. 6, that this is correct to three figures.] 300 200 100 2-2 2-4- 2-6 Fig. 101. 2-8 Ex. 5. If in the last example the pressure and volume wero connected by a law of the form pv n = K (constant), say pv 1 ' 13 = K, wo should get in exactly the same way "['*>* Work done where Now 2160 540 - ._ 13 (2' 13 3' 1 = 5 1 JJL e -13 18^~ = ^{7631 --7239} = 163 ft.-lbs. wt. FURTHER APPLICATIONS 227 EXERCISES. L. 1. The speed of a body (v ft./sec.) at the end of * sees, from a fixed instant is given by [where u and a are constants]. Shew that the distance travelled in these t seconds ia ut+ iat2. Shew also that u is the speed at the fixed instant and that the accelera- tion is constant and equal to a. 2. If v = 3t + 7t 2 , find the space passed over between the beginning of the 3rd and the end of the 5th second. 3. If a = 3< + 7< 2 [ ft./sec. 2 is acceleration], find the increase of speed between the end of the 4th and the end of the 8th second. Also, if the speed when t = Q is 20 ft./sec., find the speed at the end of the 4th second. Find a formula for the speed at the end of t seconds and the distance passed over between the end of the 4th and the end of the 8th second. 4. Find the work done in the expansion of a quantity of steam from 2 cubic feet at 4000 Ibs. per sq. ft. pressure to 8 cubic feet. The steam expands so as to satisfy the law pv " 9 = constant. 5. The pressure (p Ibs. w f .. per sq. ft.) and the volume (v cubic feet) of a gas are connected by the law pv*'' l = constant. When the volume is 40 cubic feet the pressure is 100 Ibs. wt. per sq. ft. Find the work done in compressing the gas from 40 to 35 cubic feet. 6. Find the resultant thrust on a rectangular area 6 ft. by 4 ft. im- mersed in water with its long sides vertical and the upper side 3 feet below the surface. 7. Find the resultant thrust on a triangular plate immersed in water with its plane vertical, its vertex A in the surface, and its base BC horizontal. BC=4 ft. and the perpendicular from A to BC = 3 ft. 8. Same with BC in the surface. 9. Find the work done in stretching an elastic string from length a ft. to length b ft. given that the natural length is I ft. and the pull when it is stretched to twice its natural length is k Ibs. wt. 152 228 CALCULUS FOR BEGINNERS [CH. VIII 10. A ship of 600 tons displacement goes a distance of 900 feet after steam has been shut off. Supposing that the resistance of the water varies as the square of the distance which it has to go before coming to rest and that the initial resistance is 15 Ibs. wt. per ton, find the work done before the ship comes to rest. This will he the initial Kinetic Energy of the ship. Hence find the initial speed in knots. 11. A body of mass 5 Ibs. moves in a straight line under the action of a force F Ibs. wt. which obeys the law Given that the body starts from rest, i.e. v=0 when t = 0, find (i) the speed at the end of 5 seconds, (ii) the distance travelled in 5 seconds. 12. A body of mass m Ibs. moves in a straight line so that its accelera- tion (a ft./sec. 2 ) is given by a= - w 2 x, where w is a constant and x ft. is the distance from a fixed point O in the line. Find the work done against the force as the body moves from O to a distance r. If the speed at distance r is zero, what is the speed at O ? 13. Supposing the force of the earth's attraction at points above the surface to vary inversely as the square of the distance from the earth's centre, find the work done in moving a body of mass m Ibs. from the earth's surface to a height of 4000 miles. [Take earth's radius =4000 miles.] 14. Find the weight of a rod AB 10 feet long, cross-section 1 sq. in., whose density changes from point to point in such a way that it is propor- tional to the distance from one end A, given that the density at B is 1 Ib. per cubic inch. Volumes of solids of revolution. 185. If a plane area be made to rotate about any line in its plane a solid of revolution is generated. Familiar instances of such solids are the cylinder, cone and sphere. A cylinder is generated by the rotation of a rectangle about one side. 185, 186] VOLUMES OF REVOLUTION 229 A cone is generated by the rotation of a right-angled triangle about one of the sides containing the right angle. A sphere is generated by the rotation of a semi-circle about the diameter. 186. To find the volume of a solid of revolution. oc Ex. 1. OK is the line y=^'> AH > BK are tne ordinates jc = 2, a x = 9. The area AHKB (Fig. 102) is rotated through a complete revolution about OX generating a frustum of a cone (Fig. 103). Required the volume of this solid. H 6" A Pig. 102. M N Fig. 102 a. Suppose the area AHKB divided into strips by lines parallel to Qy and the inner rectangles completed as in Fig. 102 a. If the area composed of these rectangles rotate about OX we shall get a solid as in Fig. 103 a consisting of a number of cylinders, and just as the sum of the rectangles in Fig. 102 a can be made as 230 CALCULUS FOR BEGINNERS [CH. VIII near as we like to the area AH KB in Fig. 102, if the strips are made thin enough, so the sum of the cylinders in Fig. 103 a can be made as near as we like to the volume required (Fig. 103). Fig. 103. Fig. 103 a. 186] VOLUMES OF REVOLUTION 231 Now if OM=#and = y = - and the volume of m the typical slice in Fig. 103 a formed by the revolution of PMNR y? = TT. M P 2 . M N = TT . y 2 . Ao; = TT . . Aa;. 4 x=9 yS .*. Sum of slices in Fig. 103 a = S IT Aa; approximately, *=2 4 and actual volume of solid in Fig. 103 721 Ex. 2. To find volume of sphere, radius a. The sphere may be considered as generated by the rotation of the area ABA' about OX. [Fig. 104.] Suppose the area split into strips by lines parallel to OY. PMNQ is such a strip. 232 CALCULUS FOR BEGINNERS [CH. VIII When the rotation about A'A takes place this strip will generate a slice of the sphere. Volume of slice = TT . PM 2 . MN approximately = Try- Ace approximately = TT (or x") Aa; approximately. (v. 147.) /a TT (a 2 a; 2 ) d -a 2a 3 EXERCISES. LL 1. Find the volume of a cone radius of base 6", height 10". 2. Find the volume of a cone radius of base r", height h". 3. In a sphere of radius 6" find the volume of a cap of height 2". 4. In a sphere of radius r" find the volume of a cap of height h". 6. In a sphere of radius 10" find the volume of a slice contained between two parallel planes at distances 2" and 5" from the centre (i) on the same side, (ii) on opposite sides of the centre. 6. Find the volumes of the solids formed by the revolution of ^+T = 1 > (i) about OX, (ii) about OY. y 4 [These solids are called respectively oblate and prolate spheroids.] 7. Find the volume of that portion of the first solid in Qu. 6 which is cut off by planes perpendicular to the axis of rotation through the points (-1,0), (2,0). 8. Find the volume of the solid formed by the rotation about OY of that x z portion of the parabola y = which lies between the origin and y=k. Shew that it is half the volume of the circumscribed cylinder. This solid is called a paraboloid of revolution. 186-188] MOMENTS OF INERTIA 233 9. Trace roughly the curve 36?/ 2 = x (4 - a:) 2 . Find the area of the loop and the volume obtained by revolving the loop about OX. 10. Find the equation of the parabolic arc with axis parallel to axis of y which passes through the points ( -2, 3), (0, 5), (2, 3). Find the volume formed by rotating the area bounded by this arc and the extreme ordinates about OX. 11. Find the volume of the solid formed by the rotation (i) about OX, (ii) about OY of that part of xy = 20 which lies between x = '2 and x=5. 12. Shew that the volume formed by the rotation about OX of that portion of y =f (x) [where / (x) denotes any function of x] which lies between [b x=a and x = b is I wy 2 .dx. Ja Moments of inertia. 187. Definition. If a particle of mass m be placed at a distance r from a fixed axis, its moment of inertia about that axis is mr 2 . If a number of particles m^, m a , m^ ... be placed at distances r n r 2> r s ..."from a fixed axis, the moment of inertia of all these masses about the fixed axis is m^Ti + m. 2 r + m 3 r^ + ... or shortly 2 mr 3 . If instead of a series of disconnected particles we have a continuous body, we split it up into strips or slices as when we were finding areas or volumes, find the sum of the moments of inertia of all the pieces and the limit to which this sum tends as the pieces are indefinitely diminished. 188. Ex. Find the moment of inertia about a short side of a thin plate of length 3 feet and breadth 2 feet, the mass being 20 Ibs. XX' is the axis about which the M.I. is required. [Fig. 105.] PMNQ is a thin strip. AP = o?feet. PQ=Ao?. .'. area PMNQ-2Acc square feet. 20, 20 .'. mass PMNQ = --(2A)--~- A# Ibs., 234 CALCULUS FOR BEGINNERS [CH. VIII arid the distance of any point in the strip from XX' lies between x and x + Ao:. X pa x' MN Fig. 105. 20 .'. The M.I. of the strip = -^- Ace . as 2 approximately. (v. 146.) .*. M. I. of rectangle 20-, r20,- s 20x27 ., . /20 = Jo T 189. Radius of gyration. If the M. i. of a body of mass M about an axis be M& 3 , k is called the radius of gyration about this axis. In this example since the mass is 20 Ibs., k* = 3, Le. radius of gyration = \/3 feet. 190. Sometimes a more convenient method of splitting into elements than by lines parallel to an axis may be found. e.g. Find the moment of inertia of a circular disc radius a about an axis through the centre perpendicular to its plane. We naturally seek for a set of points at the same distance from the axis and this suggests splitting the disc into thin con- centric rings. Such a ring is shewn in figure 106. Let OP = x, PQ = Ao;. 188-191] MOMENTS OF INERTIA 235 Then area of ring= 2irx&x approximately, and if m be mass of unit area Mass of ring = 'Zirxm&.x approximately, M. i. of ring = ^TTxrn &x x a; 2 approximately, .'. M.I. f of disc = / Jo Fig. 106. Now if M be mass of disc, M = rnira*. Ma 2 /. M. I. Of dlSC = -~- . /a 2 at 10n = V 2=J1' Radius of gyrat 191. Suppose we try to find the M.I. of a circular disc radius a about a diameter. Take the diameter as the y-axis. 236 CALCULUS FOR BEGINNERS [CH. VIII Here points equidistant from the axis lie on a parallel line and this suggests division into strips by lines parallel to OY. [Fig. 107.] Y Fig. 107. Fig. 108. Let PpqQ. be such a strip, and let OM=a?, MN=Aaj, = y(Wa 2 -x*). Mass of strip = 2y&xxm approximately = 2m >Ja? y? . A. M. I. of strip = 2m \fa- tf&xxa? approximately, /a _ M. I. of disc = / 2mx 2 Va 2 - a?dx. J -a Here we have a function which at present we do not know how to interate. and 192. We can avoid the difficulty by making use of the following important theorem. If OX, OY are two perpendicular axes in the plane of a lamina and OZ is a third axis perpendicular to the plane, and if the moments of inertia about OX, OY, OZ are respectively l z , !, I,, then Suppose a mass m situated at P in the plane XOY. [Fig. 108.] 191-193] MOMENTS OF INERTIA 237 Draw PM, PN perpendicular to OX, OY and join OP. Then OP is perpendicular to OZ. M. i. of mass about OX - m . PM a . M.I. of mass about OY = m . PN a . M. I. of mass about OZ = m . PO 3 . Bub PO 2 = PM 2 + PN 2 . .'. M.I. of mass about OZ =sum of M. i.'s about OX and OY. This is true for any number of masses in the plane XOY whence the theorem follows. 193. To apply this to the present case. Take an axis OZ perpendicular to the plane of the disc. We ft /I 2 have just found that l a = - t ~ ( 190), / t I -I- I But by symmetry \ x = \ y . _ Ma 2 a; ~ y ^ ^M [N.B. This shews that mira? . a 9 f J-a ~5 - ij er - x*dx = or I y? \/a a x*dx= -5- . J-a EXERCISES. LI!/ 1. Find the moment of inertia of a uniform rod 10 feet long, mass 20 Ibs., about an axis perpendicular to its length (a) through one end, (b) through a point 3 ieet from one end. 2. Find the M.I. about AB of a rectangular plate ABCD in which AB=a ft., BC = & ft., the mass being M Ibs. 238 CALCULUS FOR BEGINNERS [CH. VIII 3. Find the si. i. of the same plate about a line parallel to AB and DC and midway between them. 4. Find the M. i. of the same plate about a line perpendicular to its plane (i) through A, (ii) through the centre. 5. Find the M. i. of a sphere, radius a, mass M, about a diameter. 6. Find the radius of gyration about OY (i) of a lamina in the form of x 2 part of the parabola y = bounded by the curve and y = k, (ii) of the solid formed by the revolution of this area about OY. 7. Find the M. i. of a rod AB length a ft. about a line through O per- pendicular to the plane OAB. O is on right bisector of A B at distance p. 8. Find the radius of gyration about the axis of (i) a hollow cone, (ii) a solid cone (radius r, height h). 9. Find the moment of inertia of a triangular lamina ABC of mass m Ibs. about the side BC, given BC = 2 ft. and the perpendicular from A to BC=3 ft. [First find the length of a line parallel to BC at a distance x ft. from it and terminated by the other sides.] 10. Find the radius of gyration about each side of a triangular lamina, mass M, sides 4, 5, 6 feet. 11. Find the M. i. of a solid cylinder about the axis of figure. 12. Find the M.I. of a hollow cylinder, internal and external radii R and r, about the axis of figure. 13. A grindstone, radius 1 foot, mass 100 Ibs., is rotating about its axis at 100 revolutions per minute. Find its kinetic energy. 14. Find the moment of inertia about the axis of revolution of the x 2 t/ 3 solid formed by the revolution of -2 + r2 = l about (i) the x-axis, (ii) the y-axis. 15. Find the moment of inertia about its axis of a frustum of a cone, the radii of the ends being 3 ft. and 4 ft. and the thickness 5 feet. Centre of gravity. 194. If we have a number of particles of masses m n m z , m^ &c. situated in a plane at points whose co-ordinates referred to some fixed axes in the plane are (a^t/j), (xg/J), (x 3 y 3 ) &c., and if x and y 194] CENTilE OF GRAVITY 239 are the co-ordinates of the centre of gravity of the masses [Fig. 109], then and ^ + m. A x z + or 2m y=^ [If we call m^ the moment of the mass m^ about OY, these results merely express the fact: 8a 7a 6a 5a 4a 3a 2a m, Fig. 109. a 2a 3a Fig. 110. Moment about OY (or OX) of sum of masses supposed collected at centre of gravity = sum of moments about OY (or OX) of the separate masses, for this is equivalent to 240 CALCULUS FOR BEGINNERS [CH. VIII Now suppose we are dealing with a continuous body instead of a set of disconnected particles, we proceed as when we required the area or volume, that is to say, suppose the body split up into suitable strips or slices. 195. The method will be best seen by examples. x 2 Ex. 1. HK is a portion of the curve y = ; AH and BK are the ordinates x = a, x = 3a. [Fig. 110.] Find the centre of gravity of the area ABKH. Suppose the area divided into strips like PMNQ by lines parallel to OY. x* Let OM = x, MN = A*, MP = y = . Let m be the mass of unit area. Area of strip = y&x approximately, (v. 146.) Mass of strip = my&x approximately. Moment of mass about OY = myAa; x x approximately. Moment of mass about OX = my Ace x ^ approximately. Sum of masses of strips = 2 my&x approximately. x a x=3a Sum of moments about OY = 2 mxy&x approximately. z=3a ? ,2 Sum of moments about OX = 2 m^ Ax approximately. x=a " rSa f3<L /yi2 O .*. Mass of ABKH = I mydx = \ m dx = -^-ma 3 . Jo, Jet /**** /"* ftlSu Moment of mass about OY= I mxydx= I dx=2Qma?. Ja Ja a fZa y2 Moment of mass about OX = / m ^ dx Ja " =/' Ja mx* . 121 194, 195] CENTRE OF GRAVITY '241 20m 3 30 = T3 a ~26 T""* 121 . "5-"" 363 y ~~^~ -I30 a a;' Ex. 2. HK is a portion of the curve y = ; CH and DK are the lines y = a, y = 9a. [Fig. 111.] Find the centre of gravity of the solid formed by a complete revolution of the area CHKD about OY. lltiiizf-i~ 3?^rES3| m 4-lji -! ! i i i i ^ :: ill jiii 1 6a N : Mg 0> a Fig. 1 2a 3a 11. M. C. 16 242 CALCULUS FOR BEGINNERS [CH. VIII Divide the area into strips like PMNQ by lines parallel to OX. Let OM = y, MN = Ay and MP = :(= Jay}. When the area is rotated about OY this strip will generate a slice of the solid which will be approximately a cylinder, radius of base MP, height MN. Volume of slice = TT . MP 2 . MN = ira; 2 Ay approximately = Tray . Ay approximately. . . If m be mass of unit volume, Mass of slice = miray . Ay approximately. Moment of slice about OX = miray&y x y approximately y 2 Ay approximately. mira x 80a 2 , ,-, I" 90 rwiTray 2 ! 90 .*. Mass of solid = J miray ay = = Ja * Ja r 90 PmTray 3 ] 90 nura x 728a 8 Moment about OX = I nnray^dy = 5-^- = - - Ja L 3 J a 728 = 57- 728 91 and by symmetry 5 = 0. / 91 \ . . Co-ordinates of centre of gravity are 1 0, TF ) . 196. Suppose we try to find the centre of gravity of the area of a quadrant of a circle, say OAB, radius a. [Fig. 112.] Area of strip PMNQ = yAa; or va 2 & 2 Aa: approximately. Mass of strip - m va 2 y? Ax approximately. Moment of strip about OY = in va 2 a; 2 A;r x x approximately. 195, 196] .'. Mass of OAB Moment about OY CENTRE OF GRAVITY 243 Fig. 112. Here we have apparently two integrals neither of which we can evaluate. The value of the fir?t is however easily obtained, for it is ^ of mass of whole disc = -. mira?. 4 [~N.B. This shews that I* N/a 2 ~^o? dx = ~ .1 To avoid the difficulty of evaluating the second integral con- sider the moment of the strip about OX. It is i/ my&x x ^ approximately 2i in , A = -jr y* Aa; approximately in = (a?~ a; 2 ) Aa? approximately. 16-2 244 CALCULUS FOR BEGINNERS [CH. VIII f a m ,*. Moment of whole area about OX =1 -^ (a z x 3 ) dx and it is obvious from symmetry that the moment about OY is the same. [This tells us that x va a y? dx s a 3 .! o J ma" 3 " 4a __2 ~~ O_ ftnra OTT EXERCISES. LIII. 1. OP is the line y = kx. P is the point on it whose abscissa is a. Find by integration the co-ordinates of the o. G. of the triangle OMP where M P is the ordinate of P. 2. Find the distance from the base BC of the c. o. of a triangle ABC whose altitude is h. x 2 3. Find the c. o. of that portion of the parabola y= cut off by the line y = k. 4. Find the c. G. of a cone. [Take your origin at the vertex and suppose the cone generated by the revolution of part of the line y = kx about the axis of*.] 6. Find the c.o. of the area bounded by the x-axis, the ordinates a; = 3, x= 10 and the curve y = 3a; 2 - 2x - 1. 6. Find the c. G. of the trapezium whose corners are (a, 0), (-o, 0), (P, 9), (~P, q)- 7. Find the c. G. of a semicircle, radius a. 8. Find the c. G. of a hemisphere, radius a. 9. The lengths of the parallel sides of a trapezium are a, b and the distance between them is h. Shew that the c. G. lies on the line joining the mid-points of the parallel sides and that its distance from the side a is ^ . - r- . [First find the 3 a + o length of a line parallel to the side a distance x from it.] What does your result give you if o=0 ? 196, 197] CENTRE OF GRAVITY 245 1O. Shew that the co-ordinates of the o. o. of a uniform lamina bounded by y = 0, 3 = a, x=b and the curve y=f(x) are given by If the lamina is uniform this result is independent of the density and the point is sometimes called the Centre of area of the figure bounded by y = 0, x = a, x=b and the curve y =f (x). 11. A lamina, area A sq. ft., is totally immersed in water with its plane vertical and its o. a. at a depth h ft. below the surface. Shew that the resultant thrust on it = to Aft Ibs. wt. where w Ibs. is the weight of a cubic foot of water. 197. (i) The moment of inertia of a lamina about any axis in its plane is equal to the moment of inertia about a parallel axis through the centre of gravity + the moment of inertia about the first axis of the whole mass supposed concentrated at the centre of gravity. Let particles of masses m^, m^ <fec. lie in the plane of the paper at points Pj, P s , &c., and let G be their o. Q. H B Fig. 113. Suppose we want the M.I. of these masses about AB and let HG be a parallel axis through G. 246 CALCULUS FOR BEGINNERS [CH. VIII Let Xi, x 2 , &c. be the distances of the masses from HG, and let h be the distance between the two axes. M. i. of mass m l about AB = m l (x^ Kf = and similarly for each of the other masses. .". M. i. of all the masses about AB = 2 (my?) 2h (%mx) and M.I. GH=2 l (mx?). Also since G is the c. a. of the masses ^7 r = .*. M. I. about AB = M. I. about GH + h?. (Sm) = M.I. about GH + M.I. about AB of the whole mass (2m) supposed placed at G. Hence the theorem. (ii) Now let O be any point in the plane of the lamina, G its C. G. ; OX, OY any two axes at right angles through O; GX', GY' two parallel axes through G. K X Fig. 114. Then, if \ x be the M.I. about OX, \ x ' about GX', &c., we have I. = I.' + Mb 2 , \ y = \y + Ma 2 . i.e. 197, 198] THEOREM OF PARALLEL AXES 247 where \ a and I,' are M.I. about axes through O and G perpendicular to the plane of the lamina. (iii) Now let AB be parallel to the plane of the lamina, GH a parallel axis through G. Fig. 115. Then P^ 3 X* + h? 2ha^ cos 0, where 6 is the angle between the plane ABHG and the plane of the lamina. . . 2 (mPQ 2 ) = 2 (ma: 2 ) + /* 2 2 (m) - 2 h cos 02, (mx), and as before 2 (mx) = 0. So that M.I. of masses about AB = M. I. about GH + M.J. about AB of 2 (m) placed at G. 198. Hence the moment of inertia of a plane lamina about any axis in its plane, parallel to its plane, or perpendicular to its plane = the moment of inertia about a parallel axis through the centre of gravity + the moment of inertia about the original axis of the whole mass supposed concentrated at the centre of gravity. 248 CALCULUS FOR BEGINNERS [CH. VIII Ex. The M. I. of a circular disc, mass M, radius a, about an axis through the centre O, perpendicular to its plane, is Ma 2 2 ' If we want the M. I. about an axis through A a point on the circumference perpendicular to the plane, the theorem tells us that it is Ma 2 -s-+ Ma* a (Ma 2 being the M.I. about A of the whole mass supposed concen- trated at the centre of gravity O). EXERCISES. LIV. 1. Find the radius of gyration of a circular lamina about a tangent. 2. Find the M.I. of a circular lamina, radius a, about a line perpendicular to its plane and meeting it in a point distance c from the centre. 3. Find the M.I. of a circular lamina, radius a, about a line in its plane distance c from the centre. 4. Find the M.I. of a thin circular hoop, radiug a, about an axis perpendicular to its plane through a point in the rim. 6. Find the H.I. of a circular lamina about a line parallel to its plane at a distance c from the centre. 6. Find the M.I. of a solid cylinder about a line through the centre perpendicular to the axis of figure. 7. Find the radius of gyration of a solid cone about a line through the vertex perpendicular to the axis. Centre of pressure. 199. In Ex. 3, 184 we found the resultant thrust on a rectangular area immersed in water. Suppose we wish to find through what point of the plate this resultant pressure acts. [This point is called the centre of pressure.] Our approximate solution would be the finding of the line of action of the resultant of the thrusts on the strips into which the rectangle is supposed to be divided. 198, 199] CENTRE OF PRESSURE 249 The principle of which we make use is that if we take moments about any line, the moment of the resultant of any number of forces = the sum of the moments of the separate forces. Now it is obvious that the centre of pressure lies on the vertical line of symmetry. It therefore remains to find its distance from AB. Take moments about AB. The thrust on our typical strip PQNM = 250* A,w Ibs. wt. approximately and may be taken to act at the centre of the rect- angle PQNM. .\ Moment of this thrust about AB = 250* Aa; x x Ibs. ft. approximately and the sum of the moments of all the thrusts i-e F250 ~1 6 = I 250^. dx = ~y? = 250 x 72 Ibs. ft. / L ^ -I o The resultant thrust we have seen to be f J T250 ~l 6 250xdx= \..a?\ = 250 x 18 Ibs. wt. L ^ Jo If X ft. be the distance below AB of the centre of pressure the moment of this resultant thrust = 250 x 18 x X Ibs. ft. .-. 250 x 18xX = 250x72. . . X = 4. .. Centre of pressure is 4 ft. below the surface. EXERCISES. LV. 1. Find the centre of pressure of a rectangle a feet by b feet, placed in water with its plane vertical and one of the a feet sides (i) in the surface, (ii) c feet below the surface. 2. Find the centre of pressure of a triangle, base a feet, height h feet, placed in water with its plane vertical and (i) its vertex, (ii) its base in the surface, (iii) its vertex c feet and its base (c + h) feet below the surface, (iv) its base c feet and its vertex (c + h) feet below the surface. 3. The parallel sides of a trapezium are a, b and the distance between them is h. It is placed in water with its plane vertical ; find the centre of pressure if the side a is (i) in the surface, (ii) c ft. below the surface. 250 CALCULUS FOR BEGINNERS [CH. VIII Mean values. 200. We have already ( 1 64) defined the mean value of f(x) between x = a and x = b as being the height of a rectangle whose base is (b a) and whose area is equal to that bounded by the curve y =f(x), the sc-axis and the ordinates x = a and x = b. This mean value is Ja f(x) . dx b-a and we might take this as the definition of the mean value. It will be instructive to obtain this result by a method which does not involve the idea of area. In ordinary arithmetic the mean or average of a set of quantities is obtained by dividing the sum of the quantities by their number [i.e. it is the Arithmetic mean of the quantities]. Thus if 5 men have heights 5 ft. 8 ins., 5 ft. 6 ins., 6 ft., 6 ft. 2 ins. and 5 ft. 2 ins. the average height is b' 8" + 5' 6" + 6' + 6' 2" + 5' 2" 28' 6" p D 0-2- . 5 o The combined height of 5 men each of height 5 ft. 8| ins. is the same as that of the given 5 men. Now suppose we have a function of x, /(a?), and suppose x to pass from the value a to the value b by n equal steps e^ch = A so that nh = b a. Successive values of x are a, a + h, a+2h, ... (b-h), b, and the corresponding values of f(x) are /(a), /(a + A), /(a + 2A), . . ./(b - A), /(&), (n + 1) in number. The Arithmetic mean of these values is /() +f(a + h) +f(a + 2A) + ... +f(b-h) +f(b) n+l ' 200, 201] MEAN VALUES 251 hf(a) + hf(a + h) + ... + hf(b- h) + hf(b) (b-a) + h ' since nh b a. We define the mean value of f(x) between a and b as the limit of this when h -*- 0, i.e. when the number of steps from a to b is indefinitely increased. We have seen that the numerator can be written "?/<) **, x=o where Aw takes the place of h and the limit of this when AOJ - is rt> f(x) dx. Ja Also the limit of the denominator is b a. .'. The mean value of f(x) between a and b is L b f(x) . dx b a 201. In questions of mean value it is important to make clear what the independent variable is. The following example will illustrate what is meant. A stone falls to the ground from the top of a tower 400 feet high. Find its mean speed. [g= 32.] We may think of the speed as a function of either (i) the time, or (ii) the distance fallen. If we take (i) we suppose the whole time of fall divided into n equal parts, the speeds at the ends of these equal intervals of time calculated, and the Arithmetic mean of them obtained. Our mean speed is the limiting value of this Arithmetic mean when n is indefinitely increased. It is easily found that 5 sees, is the time of falling and that 7> = 32< is the formula which expresses v as a function of . 252 CALCULUS FOR BEGINNERS [CH. VIII .'. The mean speed on the first assumption is Cvdt /* Jo _ Jo 32*.<ft o _ o Qr . f. I 5 5 -- = - 5 - = 8O it. /sec. If we take (ii) we suppose the whole distance, 400 feet, divided into n equal parts, the speeds at the points of division calculated and the Arithmetic mean obtained. Our mean speed is the limiting value of this Arithmetic mean when n is indefinitely increased. The formula connecting v and s is v* = 64s. .'. The mean speed on the second assumption is 202. Ex. 1. Find the average speed between the end of the 2nd and the end of the 5th second of a body whose speed changes according to the law v = Qt + 17 &. i.e. find the mean value of Qt + 17 2 between t = 2 and t = 5. This is - _. - = = 242 ft./sec. Z o /6 Since I (6< + 17 2 ) rf< or 726 ft, is the distance travelled in the J2 3 seconds between the end of the 2nd and the end of the 5th second, this speed of 242 ft./sec. is the uniform speed with which the body would have to move to cover the same distance in the same time (v. definition of average speed in Chap. L). Ex. 2. The pressure and volume of a gas are connected by the law pv = 540, p being in Ibs. wt./sq. ft. and v in cub. ft. Find the mean pressure as the gas expands from 2 to 3 cub. ft. i.e. find the mean value of - between v = 2 and v = 3. v 201, 202] MEAN VALUES 253 / 3 540 I . dv This is h " = 219 Ibs.wt. /sq.ft. [v. Ex. 4, 184.] o 2i This is the uniform pressure which would have to act on the piston to do the same amount of work that is actually done. EXERCISES. LVL 1. Find the mean sectional area of a sphere supposed cut by a series of equidistant parallel planes. Interpret the result geometrically. 2. Find the mean sectional area of a cone supposed cut by a series of planes parallel to the base. [Radius of base r, height ft.] 3. If v = u + at where u and a are constants, shew that the average speed between the end of ti seconds and the end of t% seconds is equal to half the sum of the speeds at these instants. 4. If v=u + at + bt* where u, a, b are constants, find the average speed between the end of ^ seconds and the end of t% seconds. Also find half the sum of the speeds at these instants. O. The pressure and volume of a gas are connected by the law pv l - 2 = constant. If the pressure is 400 Ibs. per sq. ft. when the volume is 3 cub. ft., find the mean pressure as the gas expands from 2 to 4 cub. ft. 6. Find the mean gradient (i) with respect to x, (ii) with respect to y, oty = x z , between the points (2, 4), (5, 25). 7. A stone falls freely from rest; shew that the mean speed with respect to the time is half the final speed, but the mean speed with respect to the distance is two-thirds of the final speed. 8. A stone falls freely from rest. Shew that the mean kinetic energy with respect to the time is two-thirds of the mean kinetic energy with respect to the distance fallen. 254 CALCULUS FOB BEGINNERS [CH. Vlll MISCELLANEOUS EXAMPLES ON CHAPTERS VII. AND VIIL 1. Draw roughly the curve y = 6 - 4x - x z cutting the x-axis in A, B and the y-axis in C. Find the areas AOC, BOG. 2. The parabola j/ 2 = 6x forms a paraboloid of revolution by revolving about the x-axis. Find the volume of a segment of this paraboloid of length 10 units, and shew that it is half the volume of the circumscribing cylinder. 3. Find the mean value of 10x 2 - X s between x =3 and x= 5. Explain your result graphically. 4. A point starts with initial velocity u feet per second, and moves along a straight path with a constant acceleration a feet per second per second. Prove by integration that its displacement, a, at any time t sees. after the start is given by / i \ feet. 6. ABC is a triangle having a right angle at C ; BC is horizontal, and A, which is vertically above BC, is in the surface of a liquid ; find the centre of pressure. If the triangle were turned about BC until A were vertically below C, find where the centre of pressure would now be. B. 1. Write down the values of J (2x 2 - 3. Sketch the curve t/=x(5-x) 2 between x= - 1 and x=6. Find the area bounded by the x-axis and that portion of the curve which lies between x=0 and x = 5. Find also the maximum ordinate and the area of the circumscribed rectangle. 3. The equation of an ellipse is x 2 + 4j/ 2 = 4. Shew that the volume generated by the rotation of this ellipse about the x-axis is half the volume generated by its rotation about the t/-axis. 202J MISCELLANEOUS EXAMPLES 255 4. Find the abscissae of the point in which the curve y= cuts the x-axis. Calculate the maximum and minimum ordinates and the gradient at each of the points where the curve cuts the x-axis. Sketch the curve roughly between the extreme points in which it cuts the -axis. Find the areas of the two portions in the 2nd and 4th quadrants. 6. Find the Moment of Inertia of a circular disc, radius a, about a line parallel to its plane at a distance c from its centre. Hence find the Moment of Inertia of a solid cone about a line through its C.Q. perpendicular to the axis. 0. 1. Consider only the part of the curve y = 4x-x s , for which neither x nor y is negative. Make a rough sketch of this part. Find the equations to the tangents at the points where it cuts the axis of x. Find the length of its greatest ordinate. Find the area between the curve and the axis of x. 2. A solid is formed by the revolution about the s-axis of the curve y2 = a + bx + ex 2 + dx 3 . Find volume of solid between x=-h and x=h and prove that this volume is given by Simpson's rule where Aj and A 3 are the areas of the end sections, A2 of the middle section (corresponding to a:=0). 3. Find the moment of inertia of a uniform thin triangular lamina ABC about the side BC. 4. A quantity of steam expands so as to follow the law pv' 6 = constant. Find the work done in expanding from 1 to 4 cub. ft., given that when the volume is 2 cub. ft. the pressure is 50 Ibs. per sq. ft. Also find the mean pressure during this expansion. 5. A square side a is placed in water with a corner in the surface and the diagonal through that corner vertical. Find the depth of the centre of pressure. 256 CALCULUS FOR BEGINNERS [CH. VIII (V J o D. dx 1. Find 2. Draw roughly y=(a; + 2) (x-1) (x-4). If the points in which this cuts the a;-axis be, in order from the left, A, B, C, find the areas bounded by the x-axis and (i) that portion of the curve between A and B, (ii) that portion between B and C. 3. Two parallel planes cut a sphere in circles whose diameters are 12 inches and 10^/3 inches respectively, the perpendicular distance between the planes being 13 inches. Find the radius of the sphere and shew that the planes cut off spherical segments whose volumes are in the ratio 112 : 625. 4. An isosceles triangle 10 metres in altitude is immersed in water with its base in the surface and vertex pointing downwards. Find the depth of its centre of pressure supposing the pressure at the surface to be that due to a column of water 10 m. high. 6. Find the moment of inertia about its axis of a grindstone 120 cm. in diameter, 20 cm. thick, whose density is 2*14 gms. per c.c. Also find its kinetic energy at 15 revolutions per minute. E. fs 1. Shew that (x 2 - 4x + 3) dx = 0. Jo Draw a figure to explain the result. 2. A rectangle 5 ft. by 3 ft. revolves about a straight line parallel to the longer side and 3 ft. from the nearest one. Find the volume generated and the radius of gyration of the solid so formed about the axis of revolution. 3. Indicate in a diagram the curves 7/a;i-2=2 and j/x 1>5 =2. At what angle do these curves intersect each other ? Find the area of the closed figure bounded by these curves and the ordinate x = 2. Find the co-ordinates of the centre of gravity of this area. 202] MISCELLANEOUS EXAMPLES 257 4. Assuming that above the surface the force of the earth's attraction varies inversely as the square of the distance from the centre and below the surface directly as the distance, find the work done in moving a mass of m Ibs. from a feet below the surface to the surface, calling the earth's radius R feet ; also the work done in moving the mass from the surface to a feet above. Find the ratio of the work done in moving a mass from 1000 miles below the surface to the surface, to that done in moving it from the surface to 1000 miles above, taking the earth's radius as 4000 miles. 6. A zone of a sphere has thickness t. A cylinder has the same thickness and the area of its base equal to the Arithmetic mean of the areas of the ends of the zone. Shew that volume of the zone = the volume of cylinder + the volume of sphere, diameter t. F. 1. Find the values of (i) I3x%.dx, (ii) 2. A curve has to be drawn through the points (0, 0), (1, 4), (2, 1), (3, 5). Assuming the equation to be of the form y ax 3 + bx" + ex + d deter- mine a, b, c, d and find by integration the area of the curve between the extreme ordinates and the axis of x. The area comes out to be that of a rectangle on the same base with height 2. Draw a figure shewing the curve and this rectangle, and shew that the result can be obtained from symmetry. 3. There is a curve y = ax n . If y = 2'34 when x=2 and y= 20-61 when #=5 find a and n. Let the curve rotate about the s-axis forming a surface of revolution. Find the volume between the sections at x = 2 and x = 5. 4. AB is a rod I ft. long and O a point in the rod a ft. from A. Find M. i. of the rod about an axis through O perpendicular to rod. Find O so that this M.I. is a minimum. 5. A zone of a sphere has thickness t, A cylinder has the same thickness and the diameter of its base equal to the diameter of the section midway between the ends of the zone. Shew that the volume of the zone = the volume of the cylinder -the volume of a hemisphere, diameter . M. C. 17 258 CALCULUS FOR BEGINNERS [CH. VIII fc G. 1. Trace roughly y = x x 3 from x = to x = l. Find the area between it and the x-axis. If your x-unit is 5" and your y-unit 10" what is the actual area in square inches ? What is the mean height of this portion of the curve ? Find also the area bounded by the curve, the x-axis and the maximum ordinate between x = and x=l. 2. A hemisphere 1 foot in radius has to be divided into two equal parts by a plane parallel to the base. Prove that if the distance of the plane from the centre is x feet, then x is a root of the equation x 3 -3x + l = 0. 3. Prove that the area contained between the parabola y s =4ax, the axis of y and the line through the point (h, k) on the parabola parallel to the x-axis is - hk. o Prove that the radius of gyration about the x-axis of a lamina shaped /3~ / 3 3 \ like this area is k ./ = and that the co-ordinates of the c.o. are ( ^ h, - k \ . 4. A torque T Ib. ft. acts on a shaft and is proportional to the angle turned through (T = fc0). Shew that the work done in turning the shaft through an angle a radians is - Tjtt ft. Ibs. where Tj is the value of the torque when the shaft has 2i turned through an angle a. 6. A cigar-shaped strut, 40 ft. long, is of diameter ( 10 - -^- n ) inches at a section x feet from the middle. Find its total volume. 1. Trace the curve 5y = (x + 3) (x- 1) (x-3) from x= -4 to x=4. If it cuts the x-axis in A, B, C reading from the left and the y-axis in D, find the areas of the figures bounded by the curve and (i) OA, OD, (ii) OB, OD, (iii) BC. Also find the mean value of y between x= - 4 and z=4. 202] MISCELLANEOUS EXAMPLES 259 2. Find (i) The area comprised between the a;-axis and that portion of y=6x-&x-12 which lies below it. (ii) The c.o. of the area. (iii) The volume generated by the revolution of this area about OX. (iv) The C.G. of this volume. 3. Find the equation of a parabola, whose axis is along the axis of y, and which passes through the three points ( - 3, 2), (0, 2), (3, 2). Use your result to find the volume of a cask, height 6 feet, equal circular ends of radii 2 feet, radius of the middle section 2 feet, and such that the curve of a vertical section is a parabola. 4. Find the radius of gyration of a uniform solid hemisphere about a diameter of the base. 6. A body mass 10 Ibs. moves in a straight line in such a way that its acceleration at any instant varies as its distance from a fixed point O in the line and is directed towards O. At a distance of 1 foot from O the acceleration is 64 ft. /sec. 2 Find the work done in bringing the body to O from a distance of 6 feet. L 1. Shew that the value of I ft. (5 + mx) dx i" -a is the same for all values of m. Interpret this result geometrically. Do the same for 2. The bounding radii of a sector of a circle of radius a include an acute angle 0. If this sector revolves about one of its bounding radii, shew that the volume of the spherical sector thus generated is 8 17-a 260 CALCULUS FOR BEGINN 7 ERS [CH. VIII 3. The section of a girder is of the form shewn in the accompanying figure (Fig. 116). It is 3" high, the greatest breadth is 2" and the breadth of the straight part in the middle which is 1" high is 1". Fig. 116. The section may thus be regarded as made up of a rectangle 3" x 1" with 4 semicircles 1" in diameter. Find the radius of gyration of the section about the axis AB through the centre perpendicular to the greatest length. 4. Shew that I x t jd 2 -x 2 dx=0. J 6. Find the average length of the ordinates to the parabola y*=8x erected at equidistant intervals from x Q to x=6. J. 1. Find the volume formed by the revolution about OX of that portion of the curve x + y = 2x*, for which x and y are both positive. 3. A hemispherical bowl radius 9" is being filled with water at the rate of half a cubic foot per minute. Find the rate at which the depth is increasing when the water is 6 inches deep. 3. A solid iron cylinder, diameter 6 inches, length 2 feet, rotates about an axis along a diameter of one end. If it performs 100 revolutions per minute, find its K. E. , given that a cubic inch of iron weighs '28 Ibs. 4. Find the area enclosed between y z =x 5 and x 2 =y s , and the volume generated by the revolution of this area about the x-axis. 6. If the radius of the earth is r feet and a body of mass m Ibs. fall from a height a feet to the earth, find the work done on it by the earth's attraction and shew that if a be very large compared with r the work done is approxi- mately mr ft. -Ibs. wt. CHAPTER IX DIFFERENTIATION OF TRIGONOMETRICAL RATIOS 203. WE have so far confined ourselves to problems depending on the differentiation and integration of x n . We shall now investigate rules for dealing with other functions. Differential coefficient of sin x. 204. Let y - sin x, where x is the number of radians in the angle. Then with the usual notation y + Ay = sin (x + Aa;). .". Ay = sin (x + Aa;) - sin x = 2 cos ( x + -- ) sin -^- . \ * / J / AccX . Ao; 2 cos ( x + -. ] sm ~^- V 2 / 2 = Aa; Aic . Aa; 262 CALCULUS FOR BEGINNERS [CH. IX Now, as Aa; -* 0, cos (a; + -r- ) -* cos x and - -*- 1. \ L J Aa; T (v-73.) /. The limit of when Aa; -*- is cos x. Ax i.e. dy dx = COS X. Illustrations of this result. 205. (1) X Angle in = No. of y degrees radians in / = sin x 40 -6981317 -6427876 x + Ax y + Ay Ax Ay Ay Ax 41 7155850 6560590 0174533 0132714 760 40 30' 7068583 6494480 0087266 0066604 763 40 10' 7010406 6450132 0029089 0022256 7651 40 6' 6995861 6439011 0014544 0011135 7656 40 1' 6984226 6430104 0002909 0002228 7659 And cos 40 ='7660. Thus we see that as the increase in x is made smaller, increase in sin x -*-. : approaches the value cos x. What we proved in increase in x 204 is that it can be brought as near to cos a? as we like if the increase in x is made small enough. 206. (2) Draw the graph of y = sin x (where x is C.M.). At a few points, say x Q, x=-7, x=l, a=l'3 draw lines whose gradient is cos x and verify that they are tangents to the graph. [Fig. 117.] 204-206] DIFFERENTIATION OF SIN X 263 Angle x in deg. &c. 1 5 44' 2 11 28' 3 17 11' 4 22 55' 5 28 39' 6 34 23' 7 40 06' 8 45 50' 9 51 34' 1-0 57 18' 1-1 G3 02' 1-2 68 45' 1-3 74 29' 1-4 80 13' i : ^A llill - y <ty _ . -j- = COSX 1 100 199 295 389 479 565 644 -765 717 783 842 -540 891 932 964 -268 985 -T - 'I' y/X^- - - 11 ii| ; iffc>F- iiiijiiiiiiiHIil^M '-//" ;S:|:?::j:j!:: - - ' ' ^- -p- / k 4 r o . .4 c e* - i j r .0 .0 i.n 4.1 1.0 i.o <.ii Fig. 117. 264 CALCULUS FOR BEGINNERS [CH. IX Geometrical proof. 207. P, Q are points on a circle radius r such that AOP x radians, POQ = Ax radians. [Fig. 118.] OS bisects L POQ. Fig. 118. MP y = sin x = , r A y + Ay = sm x + Aa: = , r RQ and Now PQ r . Ay _ RQ _ RQ PQ ' Ate ~ ^ ~ PQ PQ L. PQR = z. ACS = X + PQ Aw T 207, 208] DIFFERENTIATION OF SIN X 265 [The arms of L PQR are perpendicular to those of L. AOS.] Ay / AoA PQ . . -r^ = cos (x + -^- } . -=- . A /y X V / V * / PQ / At\ PQ As Aa; -* 0, cos ( x + -^- ) -*- cos x and -^ - 1. V 2* / PQ . dy = ( , " da; It is very important to remember that x is the number of radians in the angle. 208. Ex. 1. To find the gradient of the curve y = sina: at the point where x = *5. We have dy -~ = cos x. ax .'. Gradient where x= '5 is cos *5, i.e. cos(*5 radians) = cos 28 39' = -8776. To find the equation of the tangent at this point we have when x = -5 y = sin -5 = sin 28 39' = -4795. /. We want the equation of the line through (-5, -4795) whose gradient is (-8776). This is y - -4795 = -8776 (* - -5), or y--8776x=-0407. Ex. 2. If y = sina;, dy -f- = cos x. ax .'. Ay = cos x . Aa; approximately. 266 CALCULUS FOR BEGINNERS L CH - Ix e.g. if x = [c. M. of 30], sin x = '5, cos x = '8660 and Ay = '8660 x Ax approximately. Suppose Ace = c. M. of 1 = -0175, then Ay = -8660 x -0175 approximately = '0152 approximately, i.e. if the angle increases from 30 to 31 the increase in the sine is approximately -0152, or sin 31 = '5152 approximately. [Actually sin 31" = *5150 correct to 4 figures.] This might be stated thus : f(x + h) =f(x) + hf (x) approximately. Here f (x) = sin x\ .*. f (x) = cos x) ' Also x = ?, h = -0175. /. sin 31 = sin 30 + -0175 x cos * D = -5000 + -0175 x -8660, &c. Ex. 3. Find maximum and minimum values of 4 cos x + 3 sin x. Let y = 4 cos as + 3 sin x. .; ^=-4sino;+3cosa;. [Ex. 1, Exs. LVII.] ax dy ^ If y is a maximum or minimum, -j- = 0. .'. tan x = -. . 4 i.e. a: = c.M. of 36 52' -143 OS' 216 52' -323 08' &c. &c. 208, 209] DIFFERENTIATION OF SIN X 267 Take for example C = C.M. of (- 143 08'), y = 4 cos (- 143 08') + 3 sin (- 143 08') = - 4 cos 36 52' - 3 sin 36 52' 4 ?_ If # = C.M. of (-142) y = - 4 cos 38" - 3 sin 38 = -3-1520-1-8471 = -4-9991. If a = c.M. of (-144) y = - 4 cos 36 - 3 sin 36 = -3-2360-1-7634 = -4-9994. .*. - 5 corresponding to x = C.M. of ( 143 08') is a minimum value. Or we might say, since ~ = 4 sin x + 3 cos x % ax .'. -. = 4 cos x 3 sin x. dx* and when x c.M. of 143 08', cos x and sin x are both negative. .'. ~. is positive, ClZF and a; = C.M. of 143 08' gives a minimum value of y. 209. {This particular kind of problem is more easily solved by ordinary Trigonometry, for 4 cos x + 3 sin x = 5 sin (x + a), 4 where a is the acute angle whose tangent is 5, i.e. 53 8', and 268 CALCULUS FOR BEGINNERS [CH. IX 5 sin (x + a) always lies between - 5 and + 5, having the value - 5 when sin (a; + a) = 1, i.e. x + 53 8' = 270, 630, ..., -90, -450, ..., and having the value + 5 when sin (a; + a) - + 1 , i.e. x +53 8' = 90, 450, ..., -270, -630.} EXERCISES. LVII. dy 1. Prove that if y=cosx, = - sin x. dx Illustrate as in 205, 206. 2. Get - =cos x by taking it as sin/; AaA _._ f Aa; Lt ^ 3. Find the gradients of the curve y = 2 sin x + 3 cos x n o / o\ Q at the points where x = 0, 1, 2, - , , tan' 1 - , i&n~ l { - - } , ir-tan" 1 ^. O O O \ &j t Draw the graph of y = 2 sin x + 3 cos* from x=-v to X=T, and check your results. Find the equations of the tangents to y = 2 sin x + '6 cos x at the points where x=0, 1, -^, tan" 1 ! -5), \ / dy 4. If y-sinnx where n is a constant, prove =ncosnx, and if dy y = cosnz, =-nsinnx. dx dy A. Write down -~ for the following values of y : dx (i) cos 5x, (ii) 3 sin 2x, (Hi) j sin 4x - - cos Bx, (iv) osinnx + 6 cosna:, where o, 6, n are constants. 6. The radius of a circle starts in the position OA and rotates with uniform angular velocity a> radians/sec. OP is its position at the end of t seconds. OB is the radius perpendicular to OA and PM, PN are perpen- dicular to OA, OB. If the radius is a, what are OM (x) and ON (y)? 209] DIFFERENTIATION OF SIN # AND COS X 269 What is the speed of M at the end of t seconds ? What is the acceleration of M at the end of t seconds ? What is the speed of N at the end of t seconds? What is the acceleration of N at the end of t seconds ? In each case shew that the acceleration varies as the distance from O [you should get M's acceleration = - <a 2 x], and that M oscillates backwards and forwards from A to A' and back, performing a complete oscillation in O seconds. M is said to have Simple Harmonic Motion. w 7. If a; increases at a uniform rate, shew that the rate of increase of sin x decreases as x increases from to - , i.e. that sin x increases faster l when x is near zero and more slowly when x is near . 31 [Look at the difference columns in the table of Natural Sines.] 8. Given sin 60 ='8660 and cos 60 = -5000, find approximately sin 61, sin 60 30', cos 61, cos 60 30'. 0. Find -r^, (i) if y = sin a;, (ii) if y = cos x. In each case shew that -~^ + y = 0. 1O. Find -j^, (i) if y = acosnx, (ii) if y = a sin nx. d 2 y In each case shew that ~ > + n 2 y = 0. ax- il. Find maximum and minimum values of (i) 2 sin x + 3 cos x, (ii) 4 cos x - 5 sin x. What is (i) I sin a;, da;, (ii) \ co&x.dxt 13. Find the area bounded by the a;-axis and that portion of y = s'm x which lies between x = and x = rr. Draw a figure and be quite clear what the answer means, e.g. if 1" repre- sents ^ along OX and 1" represents 1 along OY, what is the area in square 2 inches ? ri fi Find (i) / sin x . dx, (ii) I cos x . dx, Jo Jo 12. 14 f2 f2 (iii) I sin x . dx, (iv) I cos x . dx. Jo Jo 270 CALCULUS FOR BEGINNERS [CH. IX /2ir r2ir sin x . dx, (ii) I cos x . dx, o jo Sir Sir /2 /" 2 sin x . dx, (iv) / cos x . dx. 2 2 Draw figures to explain the results. 16. What is (i) I sin nx . dx, (ii) Icosnx.dx? [v. Ex. 4.] 17. Prove and draw figures to illustrate the following results : fir [2 [Sir (i) I sin x . dx = 2 I sin x . dx = I sin x . dx Jo Jo Jo /"~2~ = 21 sin x . dx, Jo cos a; . o f2 /" (ii) I sin x .dx I Jo Jo /n fZir rSn- cos a; . dx = 0= / cos x. dx =/ cosx.dx, o Jo Jo (iv) |' r sinx.dx = 2 | sin 2x . dx = 3 I siia.3x.dx. Jo Jo Jo 18. Write down (i) / sin 2x . dx, (ii) I cos 2x . dx, (iii) I (3 sin 2x + 4 cos 3x) dx, (iv) I (1 + cos 2x) dx, /a /"" cos 2x . dx, (vi) I cos 2x . dx, ~2 (vii) I sin x cos x . dx [remember sin 2x = 2 sin x cos x], (viii) I sin 2 x . dx [cos 2x = 1 - 2 sin' 2 x], (ix) Fcoa 2 x . dx [cos 2x = 2 cos 2 x - 1], 209] SIN X AND COS X 271 /IT |'T sin 2 x . dx, (xi) I cos 2 x . dx. o Jo What is the sum of (viii) and (ix) ? How could you foretell this result without finding the separate integrals ? 19. Find the C.G. of the area bounded by the x-axis and that portion of y = sinx that lies between a; = and x = ir. [If you try to find x by the ordinary method, you will meet with an inte- gral which you do not know how to evaluate. How can you evade the difficulty?] [v 20. What is the value of I x sin x . dx ? Jo 21. Find (i) I sin Bx cos Ix . dx, (v) I sin px cos qx . dx, (ii) I cos Bx sin Ix . dx, (vi) I cospx cos qx . dx, (in) I cos Bx cos 2x . dx, (vii) I sin px sin qx . dx. I (iv) I sin Bx sin 2x . dx, 22. In Ex. 2, 208, what would you get if you used the approximation 23. The area of a triangle is calculated from the statement that two sides are 2 feet and 3 feet and the included angle 40. If the sides are measured correctly, but an error of 10' is made in measuring the angle, find approximately in square inches the error in the area. Use no tables. You are given sin 40 = -6428 ; cos 40= '7660. 24. Find the area bounded by the curve y = B sin x + 2 cos x, the x-axis and the ordinates x = and x = ir. 25. Find the mean value of sin x between x =- and x=- . 4 o Check approximately by calculating ^ [sin 45 + sin 46 + . . . + sin 60J. 26. The distance ( feet) of a particle moving in a straight line, from a fixed point in the line, at the end of t seconds is given by s = acos t + b sin t. Shew that the acceleration is - s ft./sec> Find the speed and acceleration at the end of one second, if a = 4, 6 = 3. 27. Find the mean value of the ordinate of a semicircle when the ordinates are erected at equal intervals (i) along the arc, (ii) along the diameter. Draw figures shewing 15 ordinates in each case. 272 CALCULUS FOR BEGINNERS [CH. J] Differential coefficient of tanx. 210. With same notation as in 201 y tau a?, y + Ay = tan (x + Ax). .". Ay = Ian (a; + Ax) tan x sin (x + Ax) sin x cos (a; + Ax) cos x _ sin Aa; cos (a; + Ax) cos x ' . Ay sin Aa; 1 1 A* Aa; * cos (x + Aa;) ' cos x ' . sin Ax 1 1 Vti hen Aa; -- 0, 1 and - -. -- Ax cos (a; + Ax) cos x dv 1 /. ~ = ., = sec 2 x (= 1 + tan 3 x). dx cos^x 211. Illustration of this result. Angle in a; y degrees =C.H. of angle =tanx 30 -5235988 -5773503 y-l-Ay Aa; Ay 31 5410521 6008606 30 30' 5323254 5890450 30 12' 5270894 5820139 30 6' 5253441 5796797 30 1' 5238897 5777382 Fill up the blank columns and shew that the last column comes nearer and nearer to sec 2 30. 210-212] TAN a; 273 Geometrical proof. 212. As in 207, L AOP = x radians, L. AOQ = x + Ax radians. OP, OQ meet the tangent at A in p, q. [Fig. 119.] Fig. 119. qr, QR are perpendicular to OP. V = tan x = . r y + Ay = tan (as + A) = and sr. c. 274 CALCULUS FOR BEGINNERS |_CH. IX Ay pq . sec x QR QR 'p^ Now = = = sec (x + A.-C). QR OQ OA .Ay . , QR . . - = sec x . sec (a; + Aa:) . -^^ . Aa; PQ QR As Arc -*- 0, sec (x + Aa;) -*- sec x and -^ -*- 1. PQ .. dx EXERCISES. LVIII. 1. Draw y=tan or from x = to x = l-2 and verify that tbe gradient is sec 2 a; at a few points. 3. Draw a sketch of y = tan x from x = - 2ir to x = + 2?r. How can you see that is positive for all values of * ? dx dy 3. (i) If y = cotx, prove =-cosec 2 x. dx dy (n) If y-8ecx, prove see x tans. dx dy (iii) If y = cosec x, prove = - cosec x cot x. dx 4. If y = t&nnx, prove -^- = n sec 2 nx, and find -p- (i) if y = cotna;, (ii) if ;, (iii) if y = cosec na. 6. Write down : (i) I sec 2 x . dx, (ii) I cosec 2 x . dx, (iii) I secxta.nx.dx, (iv) I cosec x cot a; . dx, (v) I (1+ tan 2 x) dx, (vi) | (l + cot 2 x) dx, 212, 213] TAN x 275 . ... [ sino; . ... f cosx (vii) I y- dx, (vni) / . dx, ' J cos 2 x ' J sin 2 a; (ix) I sec 2 3x . dar, (x) \ia.vtx.dx, (xi) I tan 2 3 . dx. 6. If i/ = sec x + tan x t shew that . y dx 1. Knowing tan 45= 1, and see 45= /s /2, find approximately tan 45-1. 8. AOB is a quadrant of a circle, centre O. BX is the tangent at B. A point T travels along BX and in every position of T, TO is joined cutting the circumference in P. If / BOP = 0, shew that the speed of P = speed of T x cos 2 0. If the speed of T be uniform 5000 ft. /sec. find the speed of P when 6 = (i) 45, (ii) 60, (iii) 80, (iv) 89, (v) 89$, (vi) 89|. (v. 37.) Also if OB be 10 ft. and OP have a uniform angular velocity of 1 radian per minute, find in ft./sec. the speed of T for the same values of 6. 9. An electric current is measured by a tangent galvanometer, the current being proportional to the tangent of the deflection. If the deflection is read as 45 and an error of 1% is made in reading it, shew that the percentage error in the current is approximately -5 . a 10. If x is the deflection in a tangent galvanometer and a given small error is made in reading the deflection, shew that the percentage error in the current is proportional to (taux + cotx). 213. Ex. A man in a boat 6 miles from shore wishes to reach a village 14 miles distant from the point of the shore nearest to him. He can walk 4 miles an hour and row 3 miles an hour. Where should he land in order to reach the village in the least possible time ? What will this least time be ? [The shore is supposed straight.] 182 276 CALCULUS FOR BEGINNERS [CIL IX B is the boat. V the village. NV the shore. BN perpen- dicular to NV. P any point in NV. Call the angle NBP 0. [Fig. 120.] V B 6m N Fig. 120. Then BP = 6 sec 6, and PV = 14-Gtan0. .'. Total time taken if he lands at P is /6sec0 14-6 tan0 Call this t, so that 3 7 t = 2 sec 6 - tan B + . . dt_ _ " d6~ For a minimum value of t t dt dO = 0. (3 \ 2 tau B - H sec \ = 0. .'. 2 sin 6 cos 6 2 cos = 0, since sec ^ 0. 213] DIFFERENTIATION OF TRIGONOMETRICAL RATIOS 277 3 .*. sin = -r. 4 .*. sec = j= , v7 and tan = ^= . \/7 I Q /. NP = 6tan0 = ~ =6-801, and ~V7 2x/7 2- 2- [Shew that this is a minimum value.] .'. He must land 6-804 miles from N and his time will be 4-823 hours. EXERCISES. LIX. 1. A wall 6 feet high runs parallel to and 5 feet from another wall. Find the length of the shortest ladder that will reach from the ground to the second wall over the first. 2. An elastic string has one end fixed at A and the other B moves along OY perpendicular to OA. If OA=2 feet, find at what rate the other end is moving when / OAB = 60, supposing the angle OAB to be increasing at a uniform rate of 1 per second. 3. With the same string as in Ex. 2, find the rate at which ^OAB is increasing when L OAB = 60, supposing B to be moving at 1 inch per second. 4. A man 6 feet high walks at 6 feet per second away from a lamp-post 10 feet high. Find the rate at which his shadow is increasing. If 6 be the angle made with the ground by the line joining the top of his head to the top of the lamp- post when he is x feet from the post, prove =4 cot &. Hence find the rate at which 9 is decreasing when he is 8 feet from the lamp. CHAPTER X PRODUCT. QUOTIENT. FUNCTION OF FUNCTION. INVERSE FUNCTION. IMPLICIT FUNCTIONS Differential coefficient of a product. 214. SUPPOSE y = a? sin x. Here y is the product of two functions, viz. a? and sin x, each of which we can differentiate separately. y + Ay = (x + Aa;) 2 sin (a; + Aa;). .'. Ay = or 1 {sin (x + Ace) - sin a;} + 2x Aa: . sin (a; + Aa;) + (Ate) 2 sin (a; + Aa;). . Ay sin (a; + Aa;) - sin x . . ~ = a,- 2 . a '- + 2aj sm (a; + Aa;) H- Aa; . sin (a; + Aa?). sin(a; + Aa;)-sinaj Now as Aa; - 0, * *- cos x, sin (a; + Aa;) - sin a?, OdB and Aa?-*-0. .*. -jr- = x* cos x+2x sin a;. cc 215. We might have arranged this as follows, so as to shew more clearly the significance of the different components of the result. 214-216] DIFFERENTIATION OF PRODUCT 279 Ay = (x + Ax) 2 sin (x + Ax) - (a; + A*) 2 sin x + (x + Aw) 8 sin x - or 2 sin x, where the term (x + Aa;) 2 sin x has been added and subtracted as a link between (x + Aa;) 2 sin (x -f Ax) and a 2 sin x. . Ay . sin (x + Ax) sin x (x + Ax) 2 - a? .'. = (a; 4 Aa;) 1 - - + sin a ^ ^ Ax Ax Ax = A (x + Ax) 2 + B . sin x, say. is Ax--0, Also (x + Ax) 2 x' J . d since . d.x? . Now, as Ax -^ 0, A -- ; . i.e. cosx. and B -*- . , i.e. 2x. dx dx dv .'. -^- = x 2 cos x + 2x sin x, c?(ar s sinx) rfsinx rfx a or * ; = x 3 . ; H sin x . -^ . dx dx dx EXERCISES. LX. 1. Go through the work again, using as a link a; 2 sin (a; + Ax). a. Find in the same way -p , (i) when y = (2x + 3) cos a;, (ii) when y = x 3 ia,nx, 216. In the general case let y = uv, where u and v are functions of x. If x receive a small increment Aa 1 , this will produce small increments AM, Aw in u and v and these will produce a small increment Ay in y. Thus y + ky = (u + AM) (v + Aw) = MV + u . A + w . AM + A?t . Aw. .*. Ay = (M + AM) Av + v A?*. Aw \ Av Aw .*. -r^- = (M + AM) . -r + v . . Ace Ao: AJC 280 CALCULUS FOR BEGINNERS [C1I. X Av dv AM du As Ace -* 0, (u + A?n -- u, --- - -=- and --- - -y- . Aw cte Aaj do . dy_ dy du " dx ~~ dx dx ' 217. Notice that this result may be stated as follows : (i) Differentiate uv as if u were constant. This gives u - ^ ' dx (ii) Differentiate uv as if v were constant. This gives v . -y- . Qu6 (iii) Add these together. If we divide both sides by y or uv, we get 1 dy _ 1 du 1 dv y' dx u' dx v' dx' 218. Ex. 1. If Then by the formula which agrees with what we get -by writing y = x 13 and differentiating in the ordinary way. Ex. 2. If y = cos x . tan x, dij ~ = cos x . sec 2 x 4- tan x ( sin x) dx 1 sin 2 x cos a; cos x cos 2 a; = COS X, cos as which agrees with what we get by writing T sin of! y = sin x\ = cos x x . L cos xj 216-219] DIFFERENTIATION OF PRODUCT 281 Ex. 3. If y sin 2 x = sin x . sin x, dy -f- sin x . cos x + sin a; . cos x ax = 2 sin x cos x. EXERCISES. LXL Find ~- for the following values of y : I. x sin x. 2. a -2 tanar. 3. (3x 2 + 2.r + l) (2.r + 3), (i) by treating as a product, (ii) by multiplying out. 4. 2 sin x cos x. [Cf. Exs. LVII. 4.] 3 __A 5. x 2 " x a; % (i) by treating as a product, (ii) by writing as one term. 6. tan x . cosec x, (i) by treating as a product, (ii) by writing it sec x. 7. 4s 6 sin or. 8. 5x 2 cos3#. 9. cos 2 x. 1O. (i) cos 2 5x, (ii) sin 2 5x. What is the sum of these two results ? How can you see that this must be so without actually finding them ? sin x II. Jxcosx, 12. jj- x* Product of more than two functions. 219. Suppose y = uvw, where u, v, w are all functions of x. uv is a function of x. Call it z. .'. y = zw. dy dw dz :. By our rule -~ = z . -7- + w . -r- : aa3 ax ax but since z = uv, dz dv du (!.< dx dx ' dii dw f dv du\ . . -f- = uv . -= h w I u -=- + v -=-] dx ax \ ax dx/ du dv dw 282 CALCULUS FOB BEGINNERS [CH. X Ex. Prove similarly that if y is the product of four functions, say uvtvz, then dy du dv dw dz 3^. = VWZ -= + WZU -3- + ZUV -3 + UVW -3 . dx dx dx dx dx 220. If we divide both sides of the result in 219 by y or uvw t we get 1 dy 1 du 1 dv 1 dw ? i i y' dx u' dx v' dx w' dx' and similarly if y uvwz, I dy 1 du I dv 1 div 1 dz y' dx u' dx v' dx w' dx z' dx' [Another way of proving these results will be seen later, 263.] 1 dy _ 1 2 3 4 y' dx x+l 2x + 3 3a; + 5 4x + 7' 2 3 4 "I j. _i_ 4. I EXERCISES. LXII. (i) by treating y as the product of z 2 , (x + 1) , (x + 3), (ii) by treating y as the product of a; 2 and a; 2 (iii) by taking y = x 4 + 4s 3 + 3x 2 . (i) by treating y as the product of (x + 1), (x + 2), -, 2 (ii) by taking y = x + 3 + -. X (y d. If y = :r 2 sins cos x, find . 219-221] DIFFERENTIATION OF QUOTIENT 283 4. If y = 6. If u is any function of x, find ... d 8 .... du 3 .... du* . du ' (11) ' (U1) m terms of " and ' Differential coefficient of a quotient. sin x 221. Suppose y = j sn sin (x + Ax) sin x a; 2 sin (a; + Ax) (a; + A.*) 2 sin x a? (x + A*) 2 _ ar'lsin (a;+ Ax) -sina:}-sina; {(a; + Acc) 2 -x 2 } a 2 (a; + A*) 2 ~"' where a? sin x has been added and subtracted as a link between a? sin (a; + Aa;) and (a; + Ax) 2 sin x. . Ay Aa; 2 - B sin x sin (a? + Aa;) sin x where A stands tor , Aa; (x + Ax) 2 - a? and B stands for J - '- - . Ax Now the limits of A and B as Ax-*-0 are cosx and 2x respectively [the differential coefficients of sin x and x 2 ], and the limit of (a; + Ax) 2 is a?. dy x 2 cos x 2x sin x x cos x 2 sin x " dx x* x 8 rf(sinx) d(a?) ar. ^-j ^-sinx. V~^ dx dx 284 CALCULUS FOR BEGINNERS [CU. X 222. Generally, if y = -, u + Ait u + AM v + Aw v (v + Aw) AM At? . Ay Ax Ax Ax v (v + Av) ' and as Ax -- the limits of , , (v + Av) are -= , -7- and r Ax Aa; ax ax respectively. du dv . dy _ dx ~ dx " dx~ v a ""' Ex. 1. Take y = ^ -= , /r 1 ?/ ^ / >*" v vR'** 4.1-' v Qo'2 Lvf/ JtO A M OiX' T*O A i/^O _48x 9 _16 ~ 9^" ~ 3 ^ ' 4 which agrees with what we get by taking y = -x*. o sinx x.2. y = dy cos x x cos x sin x ( sin x) 1 v ' _ ' __ ^^^^^^ COf* '* -j - s - s set - X, ax cos x cos x which agrees with what we have previously determined as. the differential coefficient of tan x. 222, 223] DIFFERENTIATION OF FUNCTION OF FUNCTION 285 EXERCISES. LXIII. Find -/- for the following values of y: x + 1 x x + 2 ' .T-i + 1 ' sin x r in 3. (i) quotient, (11) sin ex-. x L tfj 4. -: [(i) quotient, (ii) x cosec xl sm x 6. 1 sin x 1 1 -f sin x ' ' cos x " O. - , where u is any function of x. Function of a function. 223. The only functions of x which we have yet dealt with are powers and roots of x, sin x, cos x, &c. and products of these, and we have shewn how to find the rate at which any one of these changes with respect to x. For instance we have d sin x which we may read : The differential coefficient of the sine of any angle with respect to the angle itself = the cosine of the angle, or, the rate of change of the sine of an angle per unit increase in the angle = cosine of the angle, it being understood that by " angle " we mean " number of radians in the angle." We may write the above d . sin (angle) d . angle = cosine (angle). 286 Thus CALCULUS FOR BEGINNEIiS d . sin 4:X - = cos 4or, [CH. X = cos &c. Similarly, from the result d - dx c . sn a; -; - : - a . sin x Write down d.(y? 1. 7. EXERCISES. LXIV. d.st? ' d . tan 4x d . cos 2 (4 x + 7) d . cos (4as + 7) ' d.afi d.x*' d.sec 6 * d . sec x ' 3. d . d. sin x 8. 1O. d.(2sinx + 3 cos xfi d . (2 sin x + 3 cos z) " 224. Now suppose we have a function y = sin3, and we want -j- . We do not get cos 3a;, for the fact corresponding to d sin x L d sin 3x d sin 3x - =cosa is not ; - =cos3a;, but .. . = cos 3x, i.e. dx dx d(3x) sin 3cc is increasing cos 3x times as fast as 3x, but dx 223-225] DIFFERENTIATION OF FUNCTION OF FUNCTION 287 i.e. 3x is increasing 3 times as fast as x. .'. combining these two statements sin 3o5 is increasing 3 cos 3x times as fast as x, d . sin 3x or = = 3 cos 3x dx td . sin 3o; _ d . sin 3a; d . 3afl dx d,3x dx J ' 225. As another example take y = v/tana;. If y = *Jx, we know that -~- -^ , but we must not say that if y = \/tan x, ~- = . , for the statement corresponding to : 1 cwtan x 1 = T= is not but 7 T^. AO XHJU ' = . dx Zjx dx 2 N/ tan x' c?vtan x 1 tana? 2\/tana;' i.e. vtana; is increasing . times as fast as tana?. 2 v tan x dtanx Now -j = seo j a;, i.e. tan x is increasing sec 2 a; times as fast as x. i- . . . sec 2 aj . . v tan a; is increasing . times as tast as a;, 2v tan x d \/tanx sec 2 x or c c/vtan a; c? \/tana: </ tan a;~| cfe . 288 CALCULUS FOR BEGINNERS [CH. X EXERCISES. LXV. By similar reasoning find -^- for the following values of y: 1. cos 5x. 2. tan 3.T. 3. sin 3 x [i.e. (sin a:) 3 ]. 4. sin(z 2 + 3). 6. sin *fx. G. 7. (3x 2 + 5x + l). 8. 226. We might have presented the argument of 224 somewhat differently. y = sin M, where u = 3x. Now suppose y to be any function of u, and u any function of x, If x receive a small increment AOJ, this will produce a small increment AM in M, and this increment in u will produce a small increment Ay in y. Ay Ay AM however small Aa;, AM, Ay are. . Ay Ay AM c/y ty du But asArc^O =*, - , - -+--, ^- , -7- respectively. Aas AM Aa; ax du ax 5z=^ dx ~ du ' dx ' But in this case -^- = cos M and -=- = 3. an ax .'. = 3 cos M cw; = 3 cos 3a?. 2.26, 227] FUNCTION OF A FUNCTION 289 Ex. 1. y = su\ 3 x. Put y v? where u = sin x. dy du -j- = ow 2 , -i- = cos x. du dx .'. ~ 3 ax = 3 sin 2 x cos x. This might be written d (sin 3 x) d (sin 8 x) d (sin a;) dx d (sin a;) fo; 3 sin 2 x x cos x. Ex. 2. y = A/3aT+l. Put = iju where u = 3x + 4. nnn o- -i i / <V CW v 227. Similarly -^ = -/. . . ^a;. 1. Suppose y = \/sin (403 2 + 6x). We may say y \/w where u = sin v, where v = 4* 2 + 6a5. dy \ du dv ^- du -j- dv = cos v, -y = dx 6. . - = dx = x cos v x (80; + 6) w 3 + 3) cos (4o; 2 + 6x) \/sin (4a; 2 + 6.7;) M. 0. 19 290 CALCULUS FOR BEGINNERS [CH. X dy d vsin (4x 2 + 6r) d&in (4ce 2 + 603) for ^ = L dx d sin (4x 2 + 6*) d(*ia? + 6x) . dx = . x cos (4a; 2 + Qx] x (&e + 6). 2 Vsin (4aj + 6a;) Ex. 2. y = sin 4 (3x + 4). Put y = u* where u - sin (3# + 4). Put u = sin v, v = 3x + 4. dy , du do ~ = 4w , -7 = cos v. - 3. du dv dx . dy -f~ 4w x cos wxo - 12 sin 3 (Sx + 4) cos (3.x + 4), tfX "T" T: j Cv Sill ( tjQG ~r 4: ) tt ( OiK *T~ rr ) = 4 sin 3 (3* + 4) x cos (3x + 4) x 3. EXERCISES. LXVI. Find -J- for the following values of y : 1. v'sin x. 2. sin /. 3. sin 4 x. 4. sin 4a;. 6. sin (x 4 ). 6. fcin" x. 1. sin nx. 8. rasinx. 9. cos n x. 1O. cosnx. 11. sin(4r + 5). 12. (3 sin x + 4 cos x) 2 . 13. tan 2 x. 14. sec 2 x. [Why are the results of (13) and (14) the same?] 15. (a sin x + b cos x) n . 16. tan n x. 17. sec n ar. 18. *Jt&nx. 19. tan ,Jx. 2O. 21. tx* + 3z + 1. 22. 227] 23. 29. 32. 35. (6* + 1)20. 1 24. 27. 3O. 33. 36. (a* + b) n . cos (sin x). sin x. sin 8 (4* + 5). 25. 28. 31. 34. 37. (3* 2 + 2* + 1) 10 ZJax+b. (* 2 + l)6' sin (cos *). 1 jSx* + 2x + 1. sin (ax + 6). sin" (ax + b). njain. (ax + b). 38. tan" (ax + 6). 4O. ^/sin (2* 2 + 5* + 6). 42. (a sin 8 x + b cos 3 *) 5 . 44. cos^Sx. 46. sin (a + to"). 48. sin m (a; n ). 6O. 4 cos 3 a; - 3 cos *. [In Exs. 51 60, u denotes any function of x.] 51. sinu. 62. sin 3 w. 53. u 2 . 39. sec" (a* + 6). 41. (a sin 3 as + 6 cos 3 *). 43. *Js sin 2 x - 4 cos 2 *. 46. tan s (x 2 ). 47. */5 tan 2x + 3 sec2 2ar. 49. 3 sin x - 4 sin 3 x. 64. M n . 56. 66. -. tt 67. 68. cos n M. 59. sin ^i. 6O. *Ja? + u?. Write down the values of / ydx corresponding to the following values of y : 61. sin 2 x cos *. 62. sin" x cos x. 63. cos n x sin a;. 64. sinna;. 66. cos nx. 66. sec 2 nx. 67. tan 2 nx. 68. sin (ax + b). 69. cos (a* + 6). 7O. (ax+b)*. 71. *(* 2 + 4)6. 72. sin *. 73. sec 5 x . tan x. 74. 2x+B 75. 2* + 3 192 292 CALCULUS FOR BEGINNERS [CH. X 228. By using the methods of 218, 219, 222, 226 we may differentiate more complicated expressions. Ex. 1. If We have y = Ju, where u = du = 2 1-cc , du 2 and = dx (1 + a;) 2 1 [ 222] dx Ex. 2. If * 4 , , > find method, y = , where u = x Jx* 4, v = v a; 2 1. v du dv dy _ dx dx dx v* du To get - r we have u - x x vo; 2 4. ax du .*. -rr- var* 4 + x x a05 4 -4 + 2a; 2 -4 4 v a; 2 4 and x. *Jx? - 4. a; 2 -! /a.2_4 2 no? method, y = an*, where u = ^/ ^-. 3x ~~ X -T- ox 228, 229] MISCELLANEOUS EXPRESSIONS 293 x 2 4 Now u = Jv, where v = -. r 1 (>-4J* (>-!)*" jEfo 3. y = sin 3 2#.cos 4 5x, i.e. y = uv, where u = sin 8 2x and v = cos 4 5x. ~r~ == V j W -= , dx dx dx du d.s\n 3 2 dx d sin 2x d.2x ' dx = 3 sin 2 2x . cos 2x . 2 = 6 sin 2 2x cos 2. 0B Similarly = 20 cos 3 5x . sin 5a;. CttiC du .*. -j- = 6 sin 2 2o; cos 2x cos 4 5ic 20 sin 3 2x cos 3 5as sin 5x dx = 2 sin 2 2x cos 3 5a? (3 cos 2x cos 5cc 10 sin 2a; sin 5cc) = sin 2 2cc cos 3 5x (13 cos 7x- - 7 cos 3x). 229. Notice that we could differentiate quotients without the special rule. Take as an example the one first worked out sing y= -^- We could put this in the form y x~ z x sin x, and treat it as a product. 294 CALCULUS FOK BEGINNERS [CH. X dy .*. -f = ar a x cos x + (- 2ar 3 ) sin x _ cos x 2 sin x x cos x 2 sin x 7-3 ~ * tiff n R ' n * Ex. 2. v = , cos a:' \'Q could put this in the form y = sin x x (cos a;)" 1 = u x v, where u = sin x and = (cos a;)" 1 . We know j- = cos x. dx dv d (cos a;)" 1 d (cos a;) and - 1 -= A- x 1- ' dx d (cos x) dx = ( 5) (-sin x) \ cos 2 x) v COS 2 X ' dy dv du MOW ^=14 + v ax ax ax sin a; 1 = sin x x H x cos x cos 2 x cos a; = tan 2 x + 1 = sec 2 x. EXERCISES. LXVIL Find -- for the following values of y : /l-x* i-z s /I -sin 2 x l - /v/1 5' 8 - xJa 2 -**. 3. A /, r-s-- V U-* 2 V 1 + sin 2 a: 230] MISCELLANEOUS EXPRESSIONS 4. x^cos^x. 6. 6. f.r 2 + x + 1)4 / 2 . T s _ 3^2 + 4)2. 7. 295 8. 1O. sin 3 a; . sin 3s. 0. 12. 2 cos 3 5a; - 3 sin 4 2z. 14. sin n xcosnx. 15. 16. sin" 1 w cos" v. 11. x (x z + 4) ^z 2 - 4. 13. sin" x . sin nar. ^", where M and v are functions of x. . / \/ 17 smJ5a: ., 5 + cos 5o; 18. 19. ao . /. \/ ^ 230. Sometimes y and x are each given as the function ot a third variable. e.g. the co-ordinates (measured horizontally and vertically) at the end of t seconds, of a body projected with a velocity whose horizontal and vertical components are 30 and 40 ft./sec., are .-. = 30, = 40-32*. at at i.e. x is increasing 30 times as fast as t, and y is increasing (40 _ 32<) times as fast as t. 40-32* . . . y is increasing ^ times as fast as x, dy _ 40 - 30 dy Idx ~dtdi' From the given equations we might eliminate t. The first o-ives t q^: , and substituting in the second, we get 296 CALCULUS FOR BEGINNERS [CH. X . dy _ 4 32a? "dx = 3~900 _ 4 32 x 3Qt ~3 900 40-32* = ~r as before, uU but in many cases the elimination of t would be inconvenient if not practically impossible. The general case. 231. Suppose x and y are given in terms of a third variable t. So that *=/(<) and y = <(<) Suppose u receives a small increment A. This will produce small increments A and Ay in x and y. Now Ay _ Ay /A* A^~ Sr/A*' however small A, Aa:, Ay are, but as A2, and with it Ay and Ay Ay Aa? dy dy dx Ace, --0, - . - . ^ -gr i -5 , - rr respectively. ' Ao; ' A< ' A< dx' dt' dt y _ y /^ ' dx" dt/dt ' EXERCISES. LXVIIL 1. If =3t + 4t 2 anda; = 2 + l, find - in two ways. dx 2. If z = 4 + 3t + 2j3and?/ = 7 + t 3 , find -. (US 3. If a; = acos5 + tsin0 and y=a8m0-bcos$, find 2, 230, 231] x AND y IN TERMS OF THIRD VARIABLE 297 x <i 4. Shew that the co-ordinates of any point of the parabola y may be written in the form x = am, y = am 2 . Find -/-, (i) from-j-^ and -= , (ii) fromw = . dx dm dm x ' a * and shew that the results agree. 6. Shew that whatever 6 is, x=a cos 0, y = a sin 9 satisfy the eqno.tion Prove -/ = - cot = - - . dx y Interpret this result geometrically. Shew that the equation of the tangent at the point a cos o, a sin a is x cos a + y sin a = a. 6. Shew that whatever is, x = a cos 0, y = b sin satisfy the equation dy b Wx Prove 3^= cot = 5- . dx a a?y Hence shew that the equation of the tangent at the point (a cos a, b sin a) is x y . - cos a + f- sin a=l. a o 7. The co-ordinates of a point on a cycloid are given by x = a (9 - sin 0), y=o(l-cos0). Prove -j^ = cotjj. dx 2 8. The co-ordinates of a point on an epicycloid are given by x = (a + b) cos 0-6 cos J-T ' y = (a + 6) sin b sin dy , a + 26 Prove -~ = tan 0. da; 26 298 CALCULUS FOR BEGINNERS [CH. X 9. A body is projected from O with velocity ft./sec. in a direction making an angle a with the horizontal. Taking as axes the horizontal and vertical through O in the plane of projection, the co-ordinates of the body at the end of t seconds are ( tu cos a, tu sin a - 5 gt z j . Prove that at the end of t seconds the direction of motion makes an angle tan" 1 ( ) with \ u cos a / the horizontal. Hence find the time of reaching the greatest height. 232. The same problem might be presented in another form, c^sin x Ex. Fin'd We have da? i.e. find -=-- where u = sin x and v = x*. dv du dv du /dv dx/ dx cosx Illustration of the meaning of this result. Angle 50 X = C.M. 8726646 x+ Ax 51 8901179 50 30' 8813913 50 10' 8755735 50 05' 8741191 and Find: a m. \ a? sinx 664572 7660444 (x + Ax) 3 sin (x + Ax) 705251 7771460 684709 7716246 671242 7679110 667901 7669785 cos a; -2( B13. 3x2 EXERCISES. LXIX. d (tan x) A (?in x) 0111016 040679 0055802 020137 0018666 006670 0009341 003329 = -2729 = -2771 = -2798 = -2806 3. 5. d (sin 2x) d(sinx) d sin x d cos x ' 232, 233] x AND y IN TERMS OF THIRD VARIABLE 299 233. The device of expressing x and y in terms of some third variable is sometimes useful in integration. Take the example in 191 which we failed to do directly because we could not evaluate a? \/a 2 - # 2 dx. -a If however we work in terms of ( AOP) we have, x = a cos 6. .". Aa; = - a sin . A0 app. and y = a sin 6. [Aa; and A0 are of opposite signs.] .'. Area of strip PpqQ. = 2 2 sin 2 6 . A0 app. and M. I. = 2ma 4 sin 2 6 cos 2 6 A0 app. /. M. I. of disc = I 9 2ma 4 sin 2 6 cos 2 6dO Jo ma 4 [* . n . 7 . ma 4 /"",, = I sm 2 20 . dO = - \ (1 -cos 4(9) dB * JQ * Jo EXERCISES. LXX. 1. Find the ares of a quadrant of yfi yZ [Put x=acos 0, y = 5sin 8.] 3. The co-ordinates of a point on a cycloid are given by x=a (0 - sin 6), y = a (1 - cos 8), where 6 is the angle turned through by the rolling circle. Find the area between the ar-axis and the portion of the curve traced out in one complete revolution of the rolling circle. 300 CALCULUS FOR BEGIN NEKS [CH. X x 2 w 2 3. Find the M.I. of the ellipse -5 + TO = ! a 2 6 2 (i) about the 2 -axis, (ii) ahout the y-axis, (iii) about a line through the origin perpendicular to the plane of the ellipse. 4. Find the resultant thrust on a circle, radius a ft., just immersed in water with its plane vertical. Also find the centre of pressure. 5. Find the c.o. of a quadrant of an ellipse. Inverse functions. 234. If y is a function of x, then a? is a function of y ; for instance if y = x + 3, x = ^-r . and generally if y =f(x), x is some function of y, say x = <j> (y). In the above example f(x) is 4ce + 3 and < (y) is . Notice that /{< (x)} =f ~ = * . + 3 = x, (v. 87.) _ Q and <j> {/(x)} = < (4* + 3) = = x, or the two operations denoted by y and < neutralise one another. f(x) and < (a;) are called inverse functions. Sometimes < (x) is written y*" 1 (a:). e.g. sin x and sin" 1 x are inverse functions sin (sin" 1 x) = x and sin" 1 (sin x) = x. EXERCISES. LXXI. 1. If y= j , find x in terms of y. a. Verify /{/-I (*)}=*, and -=x. 234-237] INVERSE FUNCTIONS 301 235. If y=x>, x = Jy. Here y is said to be a single-valued function of x, i.e. given any value of x, there is only one corresponding value of y, but x is a double-valued function of y, i.e. given any value of y, there are two corresponding values of x. e.g. if x = 3, y = 9 ; but if y = 9, x = 3 or - 3. In this case/(cc) is y?, and < (y) or/- 1 (y) is >Jy and one of the values of \f(x)} is x. Differentiation of inverse functions. 236. If dtl From the first of these relations we get -~ = 4, and from the ctx , dx 1 1 second -7- = - = . dy 4 dy dx EXERCISES. LXXII. Verify that ~ x -. =1 in each of the following cases : 1. 3z + 47/ = 7. 2. y = x s . 3. .r?/ = *. 4. ysl. 43? i 237. We shall now prove the general theorem dy dx a x _ = 1. dx dy j\'l ' Z\T* The relation x - = 1 is always true however small Aa; and Ace Ay Ay are. But by diminishing Ace and Ay indefinitely we can bring Aw , A* dy , dx and as near as we like to - and respectively. Ao; Ay dx dy , dy dx _. ~j~ x ~j~ := i ax ay This formula is of great use in differentiating inverse functions, for it often happens that -5- is easier to get than ~ . 302 CALCULUS FOR BEGINNERS [CH. X 238. So long as y is a single-valued function of x, and x of y, this formula presents no difficulty. Suppose y = y?, so that x = vy. mu dy dx 1 Then -f- = 2x and -r = -=. dx dy 2 V y ' the sign being the same as in the line above. .. r dx i.e. if x = +*Jy. -;- = dy .. i dx ,f .._v?. 5 . , , . dx 1 so that in each case -r- = TT- , ay 2 dy dx and thus -T^.^-S:!. c??/ Suppose for instance we want the value of -^ when x = 3 [and therefore y = 9]. We have ^ = 2a;, dx dy so that when x = 3, ~ =Q. dx d'U Similarly, when a; = 3, -^- Q. CHP Suppose however we take x = + vy, then when y = 9, a; = + 3, dx 1 and i- = + <&c 1 so that when y = 9. -=- - 7; , dy 6 1 ^ corresponding to the case when x = 3 and ^ corresponding to the case when x 3 238, 239] INVERSE FUNCTIONS 303 .". If in each case we are dealing with the same values of both x and y t dy dx x = 1. dx dy 239. If we draw the graph of y = x z , then x = 3 gives y = 9 and determines one point P x [Fig. 121], ^ gives tan XTjP,, dx . gives tan Y^P Jt Fig. 121. 304 CALCULUS FOR BEGINNERS [CH. X and tan XT^ . tan ^t 1 P 1 = 1 since the angles are complementary : but if y = 9, x = 3, and two points P x and P 2 are determined: dx gives tan XTjPj or tan XT 3 P 2 according as we take x = 3 or 3 and dx -j- gives tan Y^Pj or tan YtjP 2 (negative angle). ay Now tan XTjPj x tan Y^Pj = tan XT a P 2 x tan Y 2 P 2 = but tan XTj?! x tan Y< 2 P 2 = 1 and tan XT,P, x tan =:!} i.e. ~ * T- = 1 if we deal with the same point throughout dx dy (either Pj or P,), but if we take ? with reference to the co-ordi- CBB (iy nates of P, and with reference to those of P, we do not get 1. dy The inverse trigonometrical ratios. dii 240. Given y = sin" 1 x, to get ~ we have CL3C x sin y. /. -j- = cos y = +\/l - a?. dy The ambiguity of sign is accounted for as follows : If we take y = sin" 1 x to mean that y is some angle whose sine is x, then y is a many-valued function of x : for any value of x between 1 and I, say -866, there is an infinite number of values of y, viz. : T 27r T-JT Sir 4ir 5ir lOir UTT 3' T' ~3~' "3 '"' ~T' ~T' ~~3~' ~~3~"" /"/ay and for some of these ~- is +, for others (see Fig. 122). ' 239, 240] SIN" 1 X 305 If we agree to understand by sin l x the angle between - 7T and + -jr whose sine is x, then a glance at the figure shews that. z -~ is always positive. [Fig. 122.] Fig. 122. If y = sin" 1 x, we take dy 1 M. 0. 20 306 CALCULUS FOR BEGINNERS [CH. X x y = sin" 1 a; = o. M. of acute angle whose sine is '5000. 5000 5235988 x + Ax y + Ay As Ay Ay As 5878 6283 0878 1047 1-19 + 5592 5934 0592 0698 1-18- 5299 5585 0299 0349 1-17 5150381 5410.521 0150381 0174533 1-16 5075384 5323254 0075384 0087266 1-157 1 1 /4 -A/ -./!; J33 = 1'166. V/S 241. Similarly taking cos" 1 x to mean the angle between and IT whose cosine is x, we can prove that if y = cos" 1 x dy _ 1 dx = ~ Notice that since whatever x is, . , TT sin" 1 x + cos' 1 x = , d.sin^x d.cos^x _ ' '-dF- "dF- 242. If y = tan-ix, x = tan y. dx .*. -f- = sec 3 y = 1 + tan 2 y - 1 + a?. dy - 2l- JL ' dx l + x2' Usually tan" 1 x is taken to mean the angle -between -= and 2i + -- whose tangent is x, but the result just obtained shews that 2 even though we make no such restriction, but take tan" 1 x to mean any angle whose tangent is *, ~ is always the same. 240-242] TAN-IS 307 The graph shews that - is always positive. [Fig. 123.] Fig. 123. EXERCISES. LXXIII. dy 1 1. If y = cot" 1 x, prove -p = (b 3. Ify=sec~ 1 a?, prove ^- = ^ L 8. If y=cosec~ 1 *, prove -r-= - 202 308 CALCULUS FOR BEGINNERS [CH. X Implicit functions. 243. It often happens that the relation between x and y is in such a form that it is impossible or undesirable to find y in terms of x. e.g. we might have /-2 = ............... (1). 6R?y We can get -without finding y in terms of x. ClX We must remember that dy* = dy*^dy = 2 ^ dx dy'dx dx' Differentiate both sides of (1) with respect to x. ... ( x _ 6y + 4) + (4* + y + 1) = 0. . dy 4x + y + 1 " dx " x - 6y + 4 ' In this particular case we could have solved (1) as a quadratic in y. Rearranging we get -V - ( + 4) y - (2x 2 + a; - 2) = 0. (x + 4) N/(o;+' * (a; + 4) NSSar* + 2Qa;-8 6 Taking the positive sign we have 50x + 20 dy _ 1 ( fa ~ 6 I 25*+ 10 243, 244] IMPLICIT FUNCTIONS 309 Now 25^ + 20a;-8 = Qy-x- 4. ... ^=1/1 , 25a;+10l dx 6 1 6y - as 4J 4o3 + y + 1 6y x 4 ' which agrees with the previous result. Similarly when we take the negative sign. 244. Ex. Find the equation of the tangent at the point (2, 3) on the curve We have 4x- 4 (x^ + y) + Uy ^= 0. \ ttuC / W'iC 4a; /. Gradient at (2, 3) = fe 4x 14 8-12 2 .'. Equation of tangent is y 3 = y= (a; 2), 2x - 17y + 47 = 0. EXERCISES. LXXIV. 1. If x 2 + y 2 = a 2 , find -~ ; and find the equations of the tangents at the ctx 3 points where x=-a. fl'fj __ __ Also find ~ by taking y = Jo? - x 3 . ctx a. if^ + S-i, find ^. a a o* dx 3. If a;?/ = ft 2 , shew that -/ = - y - . Hence shew that if the tangent at ax x P meets the axes in Q and R, PO = PQ=PR. 310 CALCULUS FOR BEGINNERS [CH. X 4. If *V= **! shew that ^ = - ^ . ax ox Verify by writing y = . dy 5. Find -2 ^ ax 2 + 2^a;r/ + 6t/ 2 =l ; a, ft. b being constants. 6. Verify that (2, 1) is a point on the curve 3x 2 - 4xy + 2j/2 _ Q x + i y _ 3 = Q, and find the equation of the tangent to the curve at this point. 7. Find j- if 2cosx + 3cosy = 4. 8. If x 2 + j/ 2 - 3x + lOy - 15 = 0, find /- . Use this to find the equations of the tangent and normal at (4, -11) to the circle a; 2 + y 2 -3a; + 10j/-15 = 0. Verify that the normal passes through the centre. 9. Find i 10. If x m y n =a m+* shew that -= -^. ax nx 11. If Bin m x cos* y = const. dy m shew that ~- = cot x cot y. dx n CHAPTER XI DIFFERENTIATION OF * e* log,*. HYPERBOLIC FUNC- TIONS. LOGARITHMIC DIFFERENTIATION. COMPOUND INTEREST LAW. SOME DIFFERENTIAL EQUATIONS 245. HERE n is constant and the index x is variable. With the usual notation put y = w", then y + Ay = ri**** = n* . n** t .-. &y = n*(n* x -\\ Ay , n Aa! -l and T^ = w H - Ao, 1 Ax n* 3 * 1 Now Lt - - is obviously a function of n and is in- Ax ^o Aa; dependent of x. .*. we have = N . n* or Ny, where N is a constant, ie. it is ax the same for all powers of the same number n. 246. We can find an approximate value for this constant in particular cases, e.g. if n = 2, . Ax Ax By logarithms we get 2' 2 -l = -743, 2 2" 1 -! = -718, ^-* ^-~ 312 CALCULUS FOR BEGINNERS [CH. XI The last two were got by using 7-figure tables. With 4-figure 2' 01 1 tables the best we can do is to say that ^ = '7 approximately. With 7-figure tables it is impossible to distinguish between 2' 0001 -1 .'. we may say that if y = 2 Z , ^= -693 x 2* approximately. 247. It is important to realise that this number is '693 because we are dealing with powers of 2. Take the case when x = 5, y = 2 5 , if x = 5-1, y = 2" - 2 5 x 2' 1 , Increase in y _ 2 8 ' 1 - 2 8 _ p 1 Increase in a; '1 |_ ! Similarly if x = 5 -01, y = 2 8l " = 2 s x 2' 01 , Increase in y _ _ 6 |~2' 01 1~| and T : - * x rpj I , Increase in a; \_ '01 J and so on. Take the case when x = 7, y = 2 7 , if x = 7-1, 2 / = 2=2 7 x2- 1 , Increase in y 2 7 - 2 7 f2 " - 1~| and : - = - T = 2 7 x . Increase in x L * J Similarly if *=7-01, y = 2 7 ' 01 - 2 7 x 2 fll Increase in -)/ r2 01 -l~l and : - = 2 7 x , Increase in a: L '01 J and so on. The quantities in square brackets are the same whatever power of 2 we start from, and -693 is the limiting value to which 2' 1 - 1 2' 01 - 1 the series of fractions ^ , >Q . etc. continually approach as the index -*- 0. 24G-248] e 313 If we are dealing with powers of 3, MO takes the place of 693 [this being the limiting value to which the series of fractions 3'i - 1 3' 01 - 1 q > .Qi etc. continually approach as the index -* 0.] And similarly for powers of other numbers. EXERCISES. LXXV. As in 246 shew that 1. If y = 3 z , ^=1'10 x C* approximately. CLX 3. If y = 5 X , -/-- 1-61 x 5 X approximately. Q.3C 3 . If y = 2-5*, ^ = 0-92 x 2 -5* approximately. (UB 4. If y = 2-7*, ^' = 0-99 x 2-7* approximately. 5 . If y = 2-8*, - = 1-03 x 2'8* approximately. 248. The results of Exs. LXXV. 4, 5 suggest that there is a value of n between 2'7 and 2-8 such that if y = n y> , J^=] *n x . dtOts This value of n is called e. Its value can be calculated to any degree of accuracy required, by methods which have no place in this book. Correct to 9 decimal places it is 2-718281828. dv e is defined to be such that if y = 6*, then -~ = y. cue So far as numerical results are concerned, it will be suffi- cient for us to know that it is 2*7183 correct to 5 significant figures. EXERCISES. LXXVI. 2 -7183 -ooi_l Shew that approximately ^TT = 1, and hence that if y = 2-7183*, ^=2-7183 a: approximately. 00 314 CALCULUS FOR BEGINNERS [CH. XI 249. We can now express the constant N in 245 in terms of n and e. d.P First, since i = e , CHB d.e d.ef d.ax it follows that ; = -= - . = = a . d. ax a .ax ax Now by the definition of a logarithm n = d*', .'. if y = n-, dy i.e. -f- = iogt n x ,*. So that N is log e n. 250. We have thus two very important results: if y = e*, g = y, if y = n", where stands for the number 2718281828. ... 251. The results in Exs. LXXV. therefore give us the following approximate values for logarithms to the base e. No. log. to base e 2 -693 2-5 -92 2-7 -99 2-8 1-03 3 MO 5 1-61 249-252] e 315 252. Tables have been made giving logarithms of numbers to the base e [called natural or Hyperbolic, or Napierian loga- rithms, the last from the inventor John Napier of Merchiston 15501617]. If these are not available, the logarithm of any number to the base e can be obtained from the logarithm to the base 10 by using the formula Iog 10 n log e n=^ . log This is a particular case of a general theorem, viz. For let Iog 6 n = x and Iog 6 a = y, :. n=b x ...(l) and a = 6*.. .(2). . To get log a n we want n as a power of a. i (2) gives b = av y f l\a> x /. from (1) n= \av) = av, x log b n i.e. log a n = - ==--. y Iog 6 a The special case when n = b is important. This gives As we shall be mainly concerned with the bases e and 10 we take the particular results got by substituting e and 10 for a and 6. Thus and log, 10 = Iog 10 e[i.e. log,, 27183...] is -43429, and its reciprocal is 2-3026, each being correct to 5 significant figures. 316 CALCULUS FOR BEGINNERS [CH. XI We thus have log e n = Iog 10 n x 2-3026 and Iog 10 n = log, n x -43429, e.g. log, 2 = Iog 10 2 x 2-3026 = 30100x2-3026 = 6931. Graphic treatment. 253. Fig. 124 shews the graph of y = 2*. 70 60 50 30 20 10 Fig. 124. 252-254] GRAPHIC TREATMENT 317 Now if we keep the same axes and make a suitable change in the cc-scale, this can be made to stand for the graph of y = n x , where n is any number whatever. e.g. 4 = 2 2 , so that 4 X = 2*, 4 s = 2 6 , 4 1 ' 5 = 2 s etc., the index when a number is expressed as a power of 4 being half the index when it is expressed as a power of 2. Therefore if the a;-scale be as indicated in the second line, the graph is that of Similarly since 8 = 2 s and 8* = 2 s *, the graph will be that of y = 8 X if the jc-scale be as indicated in the third line. Any number can be expressed as a power of 2, for instance 3=10' 4771 and 2 = 10' S01 , 4771 /. 3 = 2' 3U1 = 2 1-58B . .', if the ce-scale be as indicated in the fourth line, the graph will be that of y = 3 X . And so on for any value of n. The oj-axis is in each case OX. 254. We shall now prove an important geometrical property of the curve y = n x [Fig. 1 25]. P is the point (a, n a ), Q is the point (a + h, n a+h ), RN _ NP PS~ SQ' . RN_ n a I " h ~n a+h n a w*-l* h 318 CALCULUS FOR BEGINNERS [CH. XI and is independent of a, i.e. of the position of P on the curve but depends only on the length of the horizontal step PS. R N M Fig. 125. EXERCISES. LXXVIL Verify this property on the graph of y = 2* (i) by joining the points for which x = 2 and a: =3 and also those for which a; =4 and x = 5. (ii) by joining the points for which x = and x = 2, also those for which x=2 and x=4t, also those for which x = 3 and x=5. 255. Now let A-*-0 and we get the property that in the curve y = n x the subtangent is the same at every point. i.e. if NP be the ordinate of P and if the tangent at P meet OX in T, then TN is the same for all positions of P on the curve. [Fig. 126.] Now if P be the point (x, y) and if TN = t, we have y = t x gradient of TP ^y^. w here t is constant ax t 254-257 GEAPHIC TREATMENT 319 256. There is one of the curves y = n x for which t=\ t in other words on some scale or other TN will be of unit length. Y1 Let this particular curve be called y e", Then as before e is defined by the property that 257. The numerical value of e can be found approximately from the graph as follows. Let A be the point on the graph the tangent at which goes through O. [This point can be found fairly accurately with a straight edge.] Then if AB be the ordinate of A and we choose our or-scale so that OB is unit length the graph will be y = d e , and moreover BA gives the value of y corresponding to # = 1, i.e. the number of y- units in BA is e. From Fig. 124 we can see that e is something between 2*5 and 3. From Fig. 127 which is part of Fig. 124 drawn to a larger scale we see that e is between 2 '7 and 2-75. 320 CALCULUS FOB BEGIMNERS [CH. XI 4 - 2 Scale for 2 X 1 4 e 257-259] LOG e o; 323 1O. Prove that y^e-' 3 * and y=e- 3x sin2x touch each other at the points corresponding to x= -, -^-, - , etc., and find the gradient where 11. Prove that the curve y=e" 3x sin 2x has turning points where a IT a a .'Sir a , etc., and points of inflexion where - ir Sir where a^tan" 1 . 13. On the same sheet plot between x=0 and x=2ir t (i) y = e~' 3x , (ii) .y = s'm2x t (iii) y = e-' 3z sin 2a;. [Fig. 129.] Differentiation of 258. If y = \og e x, x = e y , dx This is a most important result. We deduce that being the only power of x we have hitherto been unable to x integrate. 259. Ex, 1. If y = Iog 10 x, we use the fact that Iog 10 x = log e x x -43429, dy _ -43429 dx~ x 21 a 324 CALCULUS FOR BEGINNERS [CH. XI 4 p" \ 7 Z?j &7 -f^ ^ N<NJ y 259] LOG. x 325 Ex. 2. If y = log e sina;, dy d \og e sin x d sin x dx rfsina; dx 1 .cosx sin x cot a:. EXERCISES. LXXIX. 1. Illustrate the fact that if y = log,, x, -~ = -, by filling op the following table : x y=log.x 2 x+Ax y + Ay Ax Ay OdB 2-1 2-01 2-001 [If you have no table of logs to base e, yon must say log, 2 = Iog 10 2 x 2 -3026. ] Also make a similar table for x =6. Illustrate also by reference to the graph of y=log.x. [Pig. 130.] a. Find for the following values of y : dx (i) log. 3*, (ii) log,*', (iii) log. (3* + 5), (iv) log cos x, (v) log. tan x, \ (vi) log. (3x 2 + ox + 1), (vii)* log. sin 2x, (viii) log s x, (ix) log. tan |, (x) log. (sec x + tan x), (xi) log, tan ( ^ + - j , (xii) log.^Tl, (xiii) log.{ (xiv) log.(a; + Va; 2 + a 2 ). (**) log. (x + (xvi) log ( X 2 < a 2), (xvii) I (Z J? 3/ ~~ fit (xvjii) x log. x, (xix) x 2 log. x, (xx) log. (ax + b), (xxi) log. (ax 2 + fex + c), (xxii) log. u, where u is any function of x. 326 CALCULUS FOR BEGINNERS [CH. XI 3. Write down / ydx for the following values of y : (iv) tan x, (vii) M (xiii) (xvi) (rvii) x+~2 (v) cosec x, 1 ax + b ' V 1 (vi) sec x, (viii) log e x, (xv) [Get this equal to a; + 3 5. I Fig. 130. ft I /! nooi 4. Shew that / - dx = I - dx= -dx, J \x J 3 a; J so x /b 1 f kb 1 h -da;= / -dx=log,-. a * J ka 3 ' " 259 j LOG, ab 32? 6. Find the area bounded by (i) xy = l, y = 0, x = %, x = 9, (ii) xy = l, y = 0, x = 3, a: = 18. 6. Find the work done by a gas in expanding from 2 to 3 cubic feet, the pressure and volume being connected by the law pv = const., given that the pressure is 2160 Ibs. wt./sq. ft. when the volume is J c ft. [v. Ex. 4, 184]. 7. In the last example find the work required to compress the gas from (i) 30 cubic feet to 20 cubic feet, (ii) 3 cubic inches to 2 cubic inches. 8. Evaluate (o r dx f 2 dX J o >jx* + 9 ' IT fs (iv) / tan x dx, 4 IT /I cosec xdx. 4 ft. Find the c. G. of the area contained between xy = l, y = 0, x=l, x=4. 10. Find the C.G. of the area contained between x 2 j/ = l, y = 0, x = 1, x = 4. 11. Find the mean value of - between x = l and x = 2. [Cp. Exs. XLVI. 3.] 13. P, Q are two points on the same meridian whose latitudes are x, x + Ax (radians). Shew that arc PQ=fc. Ax nautical miles, where k= . 7T On a Mercator chart, taking as unit the distance representing a sea-mile at the equator, the distance corresponding to PQ will be k . Ax multiplied by some factor between sec a; and sec (x + Ax), and the whole distance from the equator to a point B, latitude X, will be represented on the Mercator chart by a distance / k sec x . dx. Jo Find the distance on a Mercator chart from the equator to a spot in lat. 45 and compare with what you find in the table of meridional parts. 13. For values of x > - 1 2y =x* - I2x + 30 log e (x + 2). Find the maximum and minimum values of y, and evaluate them numeric- ally. 328 CALCULUS FOR BEGINNERS [CH. XI 260. Another method of approximating to the value of e. If * = lo.*i and r* 1 I -cte = log e &- Jo, x Put 6 = e and a = 1 and we have e= Draw the graph of y . Then the area bounded by the 3? curve, the ovaxis and the ordinates x = 1 and a; = e is 1 unit of area, i.e. it is equal to the area of the square OPQR [Fig. 131]. 6 1-2 1-4 1-6 1-8 2 2-2 Fig. 131. 2-4 2-6 K2-8 3 X If we can place an ordinate KL so that the^area OKLR = area OPQR or area PKLS = area SQR the value of x given by OK is e. Now it is obvious from the figure that e lies between 2'6 and 3 [mere counting of whole square inches is sufficient to settle this]. 260 j APPROXIMATION TO VALUE OF 6 329 We shall get a very good approximation to the area OABR by drawing rectangles as in Figs. 68, 69, p. 163, each of width -01, and taking the arithmetic mean of the sums of inside and outside rectangles. [When x is positive, -/ is negative, and -^ positive, cfcc CLXJ so that the curve is like Fig. 40 (c), 94.] The sum of the outside rectangles is ni ri i i i 1 + 1-01 and the summation is easily effected by means of a table of reciprocals. Using 4-figure tables [the numbers to be added are very conveniently placed for addition] we get '01 x 95-8596. The sum of the inside rectangles < the sum of the outside rectangles by .'. Area OABR < -958596 and > -952442, a good approximation will be '955519 [as a matter of fact the first 5 figures are right], Now try OA'B'R where A'B' is x = 27. Sum of outside rectangles will be greater than in previous case by * b ? 37810 ' and is therefore -996406. The sum of the inside rectangles is less than this by 01 l- r '006296. .*. Area OA'B'R < '996406 and > -990110, a good approximation is 993258. 330 CALCULUS FOR BEGINNERS [CH. XI Now try OA"B"R where A"B" is x= 2-8. Area OA"B"R < 1 '032839 and > 1-026410, a good approximation is 1*029624. .'. e is between 2 '7 and 2*8. Try 2-75. Area < 1-014789 and > 1-008425, a good approximation is 1-011607. Try 2-72. Area < 1-003800 and > -997476, a good approximation is 1-000638 Try 2-71. Area < 1-000110 and > -993800, a good approximation is '996955. .'. e < 2-72 and > 2-71 and apparently nearer to 2 - 72. If we use a table of reciprocals which gives more figures and divide the area into strips '001 aj-units wide, we shall be able to shew that e> 2-718 and <2-719. Hyperbolic Functions. A* c ft"*** 261. DEFINITIONS. = is called the hyperbolic sine A gX 1 Q X of x (written sinh x) ; -^ is called the hyperbolic cosine px ^ ft""*" sinh iK of x (written cosh x) ; x - x or - ^ is called the hyper- bolic tangent of x (written tanhx); - r is cotho;, r is sech x % and . . is cosech x. sinh a; 260, 261] HYPERBOLIC FUNCTIONS 331 EXERCISES. LXXX. 1 . Prove cosh 2 x - sinh 2 x = 1 . [Just as the co-ordinates of a point on the circle x 2 +y 2 =a 2 mav t e taken to be (acos0, asinfl), so the co-ordinates of any point on the rectangular hyperbola x 2 -y* = a z may be taken to be (a cosh 0, a sinh 0).] 2. Shew that cosh x is never less than 1, by writing it 3. Shew that sinh ( - x) = - sinh x and cosh ( - x) = cosh x. 4. On the same sheet draw (i) y = e x , (ii) y=e~ x , (ui) y = sinhz, (iv) j/ = cosh.T between x =-2-5 and a; =2-5. 6. Shew that tanh x lies between - 1 and + 1. Write it 1 - Shew also that tanh ( - x) = - tanh x and draw the graph of y = tanh x between x=-2 - 5 and x = 2-5. 6. Shew that cosh 2* = cosh 2 x + sinh 2 x = 2 cosh 2 a; -1 = l + 2sinh 2 o; sinh 2x=2 sinh x cosh .c. dcoshx =inhx, 8. If y = A cosh x + B sinh x, shew that y-^ - n%y = 0. 9. What are (i) / sinh a; dx, (ii) Icoshxdx? 10. Shew that x sinh 2x "2 + ~4~ ' + 8i ^ X . [v.Ex.6.] - 332 CALCULUS FOR BEGINNERS [CH. XI 11. If y=sinli~ 1 x[i.e. a;=sinhw], prove = and if - a [Remember cosh y is necessarily positive.] 12. If y = sinh" 1 x, prove = x + *Jx* + 1 , i.e. sinh^x is the flame as log (a; + >Jx* + 1). Similarly shew that sinh" 1 - is the same as log. [Compare results of Exs. LXXIX. 2 (xiv) and Exs. LXXT. 11.] 13. If y=coli- 1 x(a;>l), prove -= and if y=s 14. If y^cosh" 1 ^, prove e'=a: N /a:2_i j j >e- cosh" 1 * is the same as log,, (a; V* 2 - 1 ) or Io 3 (* + v/* 2 - 1), since _ Similarly shew that cosh- 1 - is the same as log - . [Compare results of Exs. LXXIX. 2 (zv) and Ezs. LXXX. 13.] 16. If y= taut- 1 x(x 2 <l), prove d Z=_i_; and If y = tanli" 1 - (z 2 < o 2 ), prove = 2 ^ a . 16. If y=coth~ 1 x or tanh" 1 -(x 2 > Improve = -- 5 - ; and if y = coth" 1 - or tanh" 1 - (a; 2 > a 2 ), prove = -- 5 - - . 17. If j/ = tanh~* x (x 3 <l), prove e 2 "= 1 - , i.e. tanh" 1 * is the same as X 3C n lSe I - Similarly tanh" 1 - (a; 2 <a 8 ) is the same as ^ log -- . Also if y=coth~ 1 -(a; 8 >a 2 ), prove y = -log -- . [Compare results of Exs. LXXIX. 2 (xvi and xvii) and Exs. LXXX. 15, 16.] 261-263] LOGARITHMIC DIFFERENTIATION 333 262. Note. If y = sinh" 1 x, the quadratic for e v gives e* = x \/ar a + 1, but since e v is essentially positive, e v must be x+ *Jx* + 1 and there is no ambiguity. If y = cosh' 1 x,e v = x \Ac 2 1 and both values are admissible, i.e. for any value of x (> 1) there are two equal and opposite values of y such that cosh y = x. If we take the positive value, ~ will be , ; if we take the negative value, -~ will be dx Vcc 2 -! <*" . . These facts will bo readily seen from the graphs of vie 2 1 y = cosh" 1 x and y = sinh" 1 x. If we agree that cosh' 1 a; shall mean the positive value, i.e. if we take cosh" 1 a; as being log e (x+ va^ 1) we may say , d cosh" 1 x 1 1 4~ 00 / 1 + *C If v = tanh~ 1 a;(ar ! <l), 6^ = - -- and e = + v /- - , but as l-a; "VI- x before, e v must be positive, so that /I +x 6 =vr^ , I +x and = In this case therefore there is no ambiguity. Logarithmic Differentiation. 263. The process of differentiating a product is often simplified by taking logarithms before differentiating. , dy .. (a; + 3) 3 (2a;-l) Ex. I. Find g|if y = ^ (3 ^ 2)5 >. We have log y = 2 log (x + 3) + 3 log (2z - 1) - 5 log (3a: + 2), 1 dy _ 2 6 15 ' ~ 334 CALCULUS FOB BEGINNER'S [CH. XI dy _ / 2 6 15 \ * ~~ ~ (7a;-77)(a; (3x + 2) 6 Ex. 2. If y = mw3 where u, v, w, z are functions of x, log ?/ = log u + log v + log w + log z. I dy I du I dv 1 c?w 1 dz _ _ _ _ _ i _ _ i _ _ t _ _ y ' dx u ' dx v ' dx w' dx z ' dx ' dy _ __ du dv dw dz a (Cp. 219.) ~ = VWZ . -7=- + WZU . -rr- + ZUV . -r- + UVW . -j- . ax ax dx dx dx EXERCISES. LXXXI. Find ~ if da 1. y = a. y = 6. y=x t . 6. y=x x . 7. y = (sin a;) 00 '" 5 . The Compound Interest Law. 264. Suppose money put out to compound interest at 5 "/ and suppose in the first instance that interest is payable yearly. Then if x be the capital at the beginning of any year, the increase in capital during the year is '05x. If interest be payable half-yearly, the increase in capital during any half-year is -Q5x x - where x is the capital at the Z beginning of the half-year. 263-265] COMPOUND INTEREST LAW 335 If interest be payable every th of a year, the increase in n capital during any period is 'Q5x x where x is the capital at the beginning of the period. Now suppose the increase in capital to be continuous and not to increase suddenly at the end of stated periods. If x be the capital at the end of t years, the increase (As;) during the next interval A< is between 05x.A< and -05 (x + Ax) . A*, Ai . . - - is between -Q5x and -05 (a; + Ace) ; Cu *-* and generally if interest is at r/ = r?)f) = kx (where k = Int. on 1 for 1 yr. at r/ ), dx i.e. x is a function of t such that -y- is proportional to x. Civ i.e. a; is a function of the form Ae w . Now when t = 0, x = P [P being original capital]. .'. P = Ae = A. 1 .'. x = Pe** = Pe 100. 265. If y is a function of x such that dx y and x are said to be connected by a Compound Interest Law. Notice if x increases in A.P., y increases in G.P., for suppose x takes successively values a, a + 6, a +26, ..., 336 CALCULUS FOR BEGINNERS [CH. XI the corresponding values of y are and these form a G.P. with common ratio e**. Notice also that A is the value of y when x = 0. Examples of the Compound Interest Law. 266. (1) Newton's Law of Cooling. Under certain conditions the rate of fall of temperature of a cooling body is proportional to the excess of its temperature above that of surrounding bodies. i.e. if be this excess of temperature at the end of t seconds, A0 _ , n ("the temperature of surrounding bodies'] dt ~~ being supposed to remain constant where k is some constant. [The sign is to indicate that decreases as t increases.] From this we get = Ae" 6 *, where A is some constant. The constants A and k can be determined experimentally by making two observations. e.g. Suppose when t=W, 0=17"| and when t = 30, = 12J' then 17 = Ae~ l 12 = Ae- 30 *J 17 s A 2 - 12' .-. A = 20-2. And j2 = ^*> '* A = 20 10ge l2 = -017. /, $ = 20-2e--<>. 265, 266] COMPOUND INTEREST LAW * 337 (2) Atmospheric pressure. Let the pressure be p Ibs. per sq. ft. at a height h ft. above the earth's surface, and let (p + &p) be the pressure at height h + A/t [Ajo is negative if A/i is positive]. Also let wlbs. be the weight of a cubic foot of air at pressure p and (w + Aw) its weight at pressure (p + A/)). Then Ap is the weight of A h cubic feet of air whose average density is between w and w + Aw. i.e. Ap is between wA/i and (w + Aw) AA, dp = _ w Now if the temperature be supposed constant w varies as p by Boyle's Law, or w = w , where p Ibs./sq. ft. is the pressure and w Ibs./c. ft. the density at the surface. dh P(, and when A = 0, p=p , .'. A=JO O . _w^ ' P = Po e *> e.g. if the pressure at the earth's surface be 2100 Ibs./sq. foot and the density of air at the surface be '08 Ibs./cubic foot, and if the temperature be supposed the same from the surface to height A, p = 2100e--oooo4A. (3) A body is moving in a straight line and the retardation at any instant = kv ft./sec. a , where k is constant and v ft./sec. the speed at that instant. If v ft./sec. is the speed when < = find formulae for (i) the speed at the end of t sees., (ii) the distance travelled in t sees. M. o. 22 338 CALCULUS FOR BEGINNERS [CH. XI We have y- = kv. at where A is constant ; and v = v when t 0, .*. A = v :. v = v e- w , (i) k and s = when t 0, .". c = -y 267. We have found in 264 that the amount at the end of t years of P at Compound Interest, the increase in capital being supposed continuous, is Pe** where ^ = rKA- This was obtained by consideration of the increase of capital in a small time A, which led to the differential equation dx 7 We shall now consider the question from a different point of view. Suppose interest paid at intervals of -th of a year. Then if x be the capital at the beginning of any interval, the capital at / &\ the end will be x ( 1 + - ) . \ n) 266, 267] COMPOUND INTEREST LAW 339 Thus we have Original capital = P. . *. Capital at end of first interval = P{ 1 + - ) . \ nj ' 2nd - -( 1+ s)( I + s) -(it* ,, 3rd etc. t years The amount when the increase in capital is supposed continuous is the limit of this when x is increased indefinitely. (k\ nt 1 + - ) when n is increased indefinitely , ja, 1 > O Special cases. >fc=land=l, IA1+ *\ To illustrate the meaning of this last result calculate by loarithms (1 \50 / 1 NlOO / 1 \1000 1 + 50J ' ( l+ TOO/ ' ( l + TOOOJ ' You should get 2, 2-25, 2-37, 2-44, 2-594, 2-692, 2-705, 2-717. The theorem is that as n is increased we get nearer and nearer to e (2'7 182818...) and can get as near to e as we like by taking n large enough. 222 CALCULUS FOR BEGINNERS [CH. XI EXERCISES. LXXXII. 1. A body starts 2 feet from O and moves so that when it is x feet from O its speed ia 3a; feet/second. Find a formula giving its distance from O at the end of t seconds. Find the distance travelled in 3 seconds and the time taken to travel 100 feet. a. A rope passes round a drum, centre O. A and B being the points where the rope leaves the drum. P is a point on the drum such that L AOP = radians. The tension of the rope varies according to the law -r-=^iT where u. is the coefficient of friction. act If TO is the tension at A and TI at B, shew that TjsToe'*" where a is the angle AOB. If n='7, T =20 (Ibs.wt.), arcAB = lfoot and diameter of drum = 1 foot, find Tj. Also find at what point the tension is 40 Ibs. wt. 3. A cistern full of water has a leak in the bottom. Assuming that the rate at which the water escapes is proportional to the pressure, shew that the rate at which the height of water in the cistern diminishes is proportional to the height Ddx . -| ue. 3 *-J. If the height of the cistern is 10 feet, and the level falls 1 foot in the first minute, when will the cistern be half empty ? 4. Suppose a fly-wheel, moment of inertia I Ibs. ft. 2 , rotating in a fluid which produces a resisting torque cw Ib. ft., where u is the angular velocity; shew that the angular velocity at the end of time t is given by W=Wo t *i where k is, & constant, and UQ is the initial angular velocity, and find k in terms of c and I. If the wheel is initially rotating at 200 revolutions a minute, and after a minute at 100 revolutions a minute, after what time will the number of revolutions per minute be (i) 50, (ii) 25, (iii) 20 ? 267] 341 5. Mallock's formula for the retardation due to air resistance of a projectile is k(v-850), where t; ft./sec. is the speed and & a constant depending on the form and weight of the shell. If V stands for (w-850), dV shew that the acceleration is -r at Hence get a formula giving the speed at the end of t seconds. 6. A body is moving in a straight line and the retardation at any instant is 0'03 ft./sec. 2 , where v ft. /sec. is the speed at that instant. If the initial speed is 1-5 ft./sec., find the distance of the body at the end of 2 minutes from its position when t=0 and shew that it can never reach a point 50ft. from that position. 7 A rod hanging vertically carries a weight of 100 Ibs. The weight of the rod is 1-5 oz. per cubic inch. Find the law connecting the cross-section with the distance from the end if the tensile stress is everywhere 400 Ibs./sq. in. (Fig. 132.) \y sq. ins. is area of cross-section x ins. from lower end.] 400 342 CALCULUS FOR BEGINNERS [CH. XI EXERCISES. LXXXIII. 1. If y = A cos nx + B sin nx, where A and B are any constants whatever, shew that _ dx 2 8. If 7/ = Ae n *+ Be'** or if y = A cosh nx+ B sinhnx, shew that -*-. dx 2 3. Write down solutions (each involving two arbitrary constants) of , S + 4,-0. (u) g-4,=0. 4. If -5- + 4w = and if, when z=0, y = 5 and = 4, find in terms ax- ax of x. 6. If -=-^-4y = and if, when z=0, t/ = 5 and -^=4, find y in terms uX CtJC of x. 6. A body moves in a straight line and ft. is its distance from a fixed point O in the line at the end of t seconds. The acceleration is 9s ft./sec. 2 towards O. When t=0, the body is 15ft. to the right of O and when = ^,it is 20ft. to the right of O. Shew that s = 15cos3 + 20sin 3t 3 = 25 sin (3t + 0) , where 6= tan' 1 ^ . Hence shew that the body oscillates backwards and forwards between two points 25 feet on either side of O, and that the time of a complete oscillation is seconds. O What is the speed of the body (i) when t = 0, (ii) when it passes through O? 7. Ify=ze~ ix , where z is a function of x, shew that and hence that if z A cos 2x + B sin 2x, or y = e~ 2x (Acos2x + B sin 2x), 267] DIFFERENTIAL EQUATIONS 343 8. If y ze^ x , where z is & function of x, shew that and hence that if z = Ae& + Be ~ % x or y = Ae 31 + Be 2 *, d?y _ dy . cA- 5 + *y= Q ' 9. If y =ze~ Sx , where z is a function of x, shew that 3+Z+*- and hence that if = Az + B or y = e~** (Ax + B), 1O. If y = ze~P x , where z is a function of x and j>, q are constants, shew that If ? +2p ~ +qy = 0, what value should 2; have da; 2 do; (i) ifg>p 2 , (ii) if2<p 2 , (iii) ifg= 11. By the device indicated in Exs. 7 10 get solutions of -* +*-* - 1 2 . Find A so that if y = Ae 2 *, - 5y = Se 22 . d 2 ?/ 13. Find A so that if y = A sin 2x, ^ + 5y = 3 sin 2s. 14. Find A and B so that if y = A sin 6ar + B cos 6x, f| + 4 ^ das 2 dx 344 CALCULUS FOR BEGINNERS [CH. XI 1. If y = xe**, shew that ^ - 5 ^ + 6y = e 3 *. CuC* (IX !. Find A so that if y = Aa-2z, \ - 4y = 3e- x . 17. Shew that - | + 2p-p + g'?/:=r ) where p, q, r are constant?, is the uX (tX same as 5-5+20 -r- + g=0, where M is (11 ); and hence that the solutions ax 2 dx \ qj of ^ + 2p -j= + qy =r are got by adding the constant quantity - to those of dx GLX q 18. Write down solutions of 267] MISCELLANEOUS EXAMPLES 345 MISCELLANEOUS EXAMPLES ON CHAPTERS IX-XI. A. 1. A ladder 20 feet long has one end on the ground and the other in contact with a vertical wall. The lower end slips along the ground. Shew that when the foot of the ladder is 16 feet away from the wall the upper end is moving 1 times as fast as the foot. IT I. Evaluate I (7 cos0-2sm0 + l)d0, -}dx. 3. A point moves in a straight line according to the law -r- = Prove that the distance described from t=0 until it comes to rest is 6| feet. 4. A variable torque produced by a couple of moment M sin Ib. ft. acts on a shaft, M being constant and 9 the angle turned through by the shaft. Find the work done in turning the shaft from 0=0 to G=ir. ' 6. Find the mean value of tan x between TT 2ir x=7 and x= 7r . b 9 Check your result by finding the Arithmetic mean of tan 30, tan 31, ... tan 40. B. f?V 1. Find -j* in the following cases : x _ (i) y = sm 2 3a;, (ii) y = x a cos 4x, (iii) j^log^ (iv) y = -s*sinfx+V (v) x*-2xy -3y2=0. 346 CALCULUS FOR BEGINNERS [CH. XI L 2. Write down the values of (i) / e ' 3* dx, (ii) I (2 sin 3* + 4 cos 6x) dx, dx m I 4 J o 3. A rod 10 feet long elides with its ends A, B on two lines at right angles which intersect in O. If A has a uniform speed of 3 ft./sec., find the speed and acceleration of B when A is 8 ft. from O. 4. If y =xe~ mx (m positive), find the maximum value of y. 6. O is the centre of a circle, radius 2a. From a point P inside the circle, PB and PC are drawn to the circumference each of length a. If OP=x, shew that for the figure PBC (bounded by PB, PC and arc BC) to have maximum area 0. 1. If x find -^ and find the length of the tangent intercepted between the curve and dx the y-axis for the point where x=h. 9. Find the area bounded by the curve y = sin 2 , j/=0, ar=0, *=1, I (i) by integration, (ii) by Simpson's rule (11 ordinates). Draw a figure. 3. The height of a tower is calculated from the observation of the elevation 6 at a distance d ft. If d = 200 and = 35, find the consequent error in the calculated height due to (i) an error of 1 inch in measuring d, (ii) an error of 10' in measuring 0. 4. A chord divides a circle into two segments of heights hi, h^. If A is the area of either segment and the angle which its arc subtends at the centre, prove 267] MISCELLANEOUS EXAMPLES 347 6. In a simple horizontal engine a is the length of the crank and I the length of the connecting rod. Shew that when the crank makes an angle B with the line of dead centres the angular velocity of the connecting rod is a cos where u is the angular velocity of the crank. D. ... _,. dy , 1. (i) Find when y = lo x -j a > Find 3. Find the intersections of y=^2 with y=lt and find the equations of the normals at these points. 3 . Sketch roughly the curve y = 3 coax -2 sin x between # = 0andx=:-. 39 Find the area bounded by the axes and that part of the curve which lies between =0 and the point where the curve first cuts the x-axis. 4. The blade of a fan consists of a uniform circular disc, centre O, from which a small portion has been cut away by a chord AB equal to the radius r. Find the distance of the c.o. from AB. 6. The thickness x at the distance r from the centre of a disc of varying thickness is given by x=be~ ar . Find the volume of the disc, if its radius is R. What is it in the special case when R = 4, 6 =-5, a= '08? E. 1. Find f'(x) and f"(x) for the following values of / (x) : (i) xlog e x, (ii) e*sina;, (iii) - , (iv) sin**. 348 CALCULUS FOR BEGINNERS [CH. XI 2. Draw a triangle BAG, the angle A being about 120, and produce AC to D. Let AD represent the plan of a door hinged at A, which is being shut into a position along BA produced by a rod BC turned round the fixed point B by a spring and sliding on the door at C. If the angular velocity of the door is u, find the velocity with which the end C of the rod is sliding along the door at any instant. 3. If a;=acos 3 0, y = asin 3 0, express -j= and 3-^ as functions of 0. dx dx 2 4. c is the distance between the centres of two spheres, radii a and b. Find a point in the line joining the centres so that the total spherical surface visible from it shall be a maximum. [Surface of segment of sphere = 2wr h, where r is radius of sphere and / height of segment.] 6. The half water line section of a small boat is bounded by the curve t ranging from to 10. Determine the area of the water line section. P. 1. Differentiate with respect to x. = - % , cos 3 2x, * sin bx, log. tan ( 2x 4- -. } . x*-x+l \ 4/ If 8=be nt +ce~ nt , where b, c, n are constants, shew that the acceleration is proportional to the displacement. 3. In the catenary or y=ccosh -; shew that the length of the perpendicular let fall from N, the foot of the ordinate PN, upon the tangent at P is of constant length. Also if the normal at P meet the axis of x in G, shew that PG varies as PN. 267] MISCELLANEOUS EXAMPLES 349 3. A stick 3 feet long rotates about one end in a vertical plane with uniform angular velocity, making one complete revolution in each second. A light is 6 feet vertically above the centre of rotation and casts a shadow of the stick on the floor which is 6 feet below the centre. Find the speed of the end of the shadow when the stick makes an angle of (i) 30, (ii) 60, (iii) 90 with the downward vertical. 4. Shew that the radius of gyration of a triangular lamina about an axis through its o.o. perpendicular to its plane is 6 5. A torque acts on a shaft. When the shaft has turned through an angle the torque is G sin sin (0 - a) lb.-ft., where G and a are constant. Find the work done in a complete revolution. G. 1. Find -3^ for the following values of y : dx (i) 2 i ^ f (ii) (2-3x2)4, (iii) log. (2*-**), (iv) tan4x, (v) xe" 3 *, (vi) sin 2. A lighted candle is raised above a horizontal table. Find the height for maximum illumination of the surface at a given point, whose distance from the point vertically under the candle is a feet. [If A is the flame, P the point, 6 the angle between AP and the normal to the table at P, the intensity of illumination varies directly as cos 6 and inversely as AP 2 .] 3. The roof of a house is inclined at an angle 60 to the horizontal and the height of the wall is h feet. A ladder whose length is I feet rests with one end on the ground at a distance y feet from the foot of the wall and the other end on the roof at a distance x feet from the top of the wall. If the inclination of the ladder to the ground is , prove -^ = - - - ^- tan 6. dx & 350 CALCULUS FOR BEGINNERS [CH. XI 4. Evaluate the following integrals : (i) J ***<?*, (ii) P -L-d-r, (iii) j\os(l-x)dx, 6 (iv) I 1 x (1 - a?)* dx, (v) /V-**d* f (vi) 6. A thin uniform rod OA, 6 feet long, swings in a vertical plane about a horizontal axis through one end O. Express the velocity of a point in the rod distant r feet from O at the instant when the rod makes an angle & with the vertical, in terms of . Hence write down an approximate expression at for the kinetic energy of a small portion, length Ar, of the rod, and fiud by integration the total kinetic energy of the rod at this instant. If the rod falls from the horizontal position, shew that when OA is vertical, the velocity of A is 24 feet/sec, nearly. H. 1. Find ~- for the following values of y : (1) -- (2) r (3) (2x- 3) (3x2-^ + 1) (4x3 -2x2 -3), (4) tan 5*. (5) ^ (6) sec 2 3x, (7) * (a 2 + x 2 ) Ja^Hfi. 2. Find the maximum and minimum values of sin x- sin 2x between x = and x=2ir. Sketch the curve y = sin x - sin 2x. 3. OA, AB are the crank and connecting rod of a steam engine, OA rotating about O and B sliding along a fixed line through O. If OA = a, AB = c, OB=r, / BOA = 0, dr ar sin 6 express cos0 in terms of r, a, c and shew that J7 .= -. dO r-a cos O A = 3^", A B = 12", / O A B = 90, angular velocity of O A = 100 revolutions per minute; find velocity of B. 2.67] MISCELLANEOUS EXAMPLES 351 4. The motion of the needle of a galvanometer is given by the equation 6 = &e~^ t sin3f, where is the angle in radians made by the needle with the zero position at the end of t seconds. Find the angular velocity of the needle at time t and shew that the extreme excursions to the right and left of the zero position occur at intervals of - seconds and that the angles corre- o IT spending to these extreme excursions form a a. P. of common ratio - e 6. irx 5. The equation of a curve is y = b sin 2 . Find the mean height of that portion for which x lies between and a. I. (2x ir\ T + 4 )' dy , d?y find - and -r4, dx dx 2 and shew that T^~ 6 ^ + 7ry = - dx 2 dx 9 J 2. A crank CP revolves uniformly n times a minute and the other end A of a rod PA moves to and fro in the straight line AC. If CP=aft., PA = 2aft., CA=a;ft., shew that (1) sin = 2 sin 0, (2) x = a cos 6 + 2a cos $, where is L ACP and <j> is / CAP. Hence get -^- and -=- in terms of and if>. dt dt What are the angular velocity of AP and the speed of A when n=40, o = l, 0=39? 3. On the same sheet trace the curves between a; = and x = l, using as large a scale as possible. Find the slope of each at the points where x = 0, !, '2, -4, -6, '8, 1. 352 CALCULUS FOR BEGINNERS [CH. XI 4. Find the work done in slowly compressing a gas from a volume of 10 c. ins. to 6 c. ins., the initial pressure being atmospheric and the temperature being kept constant. Take atmospheric pressure to be 14-75 Ibs. per sq. inch. [pv = const.] If the compression takes place under adiabatic conditions, i.e. so rapidly that no heat is lost, find the work done. [In this case pv l ' tl = constant.] 5. Draw a figure to shew that Hence prove - w w f 2 . , fa 1 [2 TT sm^xax= I cos^x.ax=^ l.ax = . JO JO 2jo 4 I. (i) If prove ^ = *. ax y Hence prove that if the normal at any point P on x 2 -t/ 2 =a 2 meet the x-axis in G, PO=PG. sin nx (ii) Shew that y= - 7*2-1 is a solution of -r-^ + V = sin nx. dx* a. Given log, 6 = 1-7918, find log.,6-1. 3. A chain hangs in the form given by the equation -\ v* c J or 7/ = ccosh-, the axis of y being vertical. Find the area of the figure enclosed between the curve and the chord which joins the lowest point of the curve to the point where x=2c. 4. Find the C.G. of a circular segment whose arc subtends an angle 2a at the centre of a circle of radius r. Find also the C.G. of a sector of angle 2a and of an arc of angle 2a. 5. The sides of a triangular lamina are 3, 4, 5 feet. Find the radius of gyration about the axis of revolution of the solid formed by the revolution of the triangle about the longest side. 267] MISCELLANEOUS EXAMPLES 353 E. 1. If p = ab m + ne where a, 6, m, n are constants [De la Boche's formula connecting vapour pressure and absolute temperature], shew that dp _ mp log b d0 2. If the equation of a curve is xy+y + x=Q, find the equations of the tangent and normal at the point where a; =4. 3. P moves in a straight line AB and O is a fixed point. Shew that the angular velocity of P about O at any instant is - , where r ft. is the length OP, v ft./sec. the velocity of P at the instant, and 6 the angle between PO and AB. 4. If a plane area [A sq. ft.] revolve about an axis in its plane which does not intersect it, the volume generated is Al c. ft., where I ft. is the length of the path traced by the o.o. of the area. [Theorem of Pappus, end of 3rd century A.D.] 6. Find the area included between the curve x 2 y = x 3 + a 3 , the x-axis and the ordinates x a, x = 2a. Find also the co-ordinates of the C.G. of this area. L. 1. If where A is a constant, prove 8. If (* - a)* is & factor of / (a;), shew that (x - a)"-* is a factor of /' (x). Hence given that 3x 3 - 1x* - 8x + 20 = has two equal roots, solve the equation. 3. OA, OB are the bounding radii of a quadrant of a circle, PQRS is a rectangle having one corner (P) in OA, one (Q.) in OB and two (R, S) in the arc AB. Find the maximum area of this rectangle. [Work in terms of /COR, C being the mid-point of arc AB.] M. C. 23 354 CALCULUS FOR BEGINNERS [CH XI 4. A point in a certain mechanism moves so that its position at the end of t seconds is given by irt 2-rrt x + 4 =3 cos + cos- , o o . irt J/ = sm T . Find the position, velocity and acceleration of the point when t = 2. [Distances are in feet.] 6. The x-axis is taken along the axis of the cylinder of an engine sc that the piston moves between x=0 and a; = 10 [unit 1 inch]. The thrust on the piston is 81 Ibs. wt. from x=Q to * = 6 and 1000.E- 1 ' 4 Ibs. wt. from x G to a; =10. Find the work done in one stroke in ft. Ibs. wt. Also draw the graph shewing the relation between the thrust and x, and get the work done by counting squares in the appropriate area. M. 1 . Find the differential coefficients with respect to x of 2. A point moves in a plane so that at end of t sees, its coordinates are x = t, j/ = sin2jrt, the unit being 1". Draw the path from t = to t = 2. Find the angle the direction of motion of the point makes with the x-axis at tune t = 0'2. Find area contained between x-axis and the path from t = to t=0-5. 3. A and B are two sources of heat, 20 feet apart. The intensity of A is twice that of B. Find the position between them where the combined heating effect is least. 4. If V at and C = 1008in600t, R being 2 and L being O'OOS, find V in the form 5. A point is moving over a straight line of length 2a with simple harmonic motion, period T seconds. Find its speed and acceleration at a time t seconds from rest. If V be the maximum value of v, shew that r Jc 267] MISCELLANEOUS EXAMPLES 355 N. 1. Find equations of tangent and normal to -^ + ^ = 1 at the point (h, k). Find the distances from the origin at which this tangent and normal cut the x-axis and the product of these distances. 2. A ladder has one end on the ground and the other projects over a vertical wall 10 feet high. The lower end slides along the ground with uniform speed 1 foot per second. Find the angular velocity of the ladder when its foot is 10 feet from the wall, the motion being supposed to take place in a plane perpendicular to the wall. 3. A straight road runs along the edge of a common and a person on the common, at a distance of 1 mile from the nearest point (A) of the road, wishes to go to a distant place (C) on the road in the least time possible. If his rates of walking on common and road are 4 and 5 miles an hour respectively, prove that he must strike the road at a point B, 1$ miles from A. IT n fi 4. Evaluate (i) cos (3 - 2x) dx, (ii) sin x. cos 2 x dx, Jo JO It TT /"2 /"30 /"'1047 (iii) ' sm*xGOs 2 xdx, (iv) sinx.dx, (v) xdx. Jo Jo Jo Account for the fact that the last two results are approximately equal. 6. CA is a radius of a circle, O its mid-point. OB perpendicular to CA meets the circumference in B. The head of a shell is formed by the Devolution of the arc AB about BO. Find its volume if CA = 2a. [If P be any point on the arc, work in terms of / ACP (=0).] O. 1. Find ^ if dx (i) (ii) (iii) ?/ = 3u 2 -7u + 2 where u=2x s + 3x + 2, (iv) sin x = log, y. . d(3w 2 -7w + ll) , .. dx .... dsinx Fmd W - - - - 232 356 CALCULUS FOR BEGINNERS [CH. XI 2. (i) If y=xlogx, shew that y is least when x = ~. 6 (ii) If y= j , shew that y is greatest when x=e. 3. (i) Shew that v is acceleration. at (ii) If v 2 =a + fc, prove that the acceleration is constant. (iii) If t> 2 =as 2 + 6s + c, prove that the acceleration varies as the distance from a certain fixed point in the line of motion. (iv) If s = ~vt, prove that the acceleration is constant. S3 4. A curve whose equation is xy = constant passes through the point (4, 5). Find the area of the figure bounded by the curve, the axis of x and the two ordinates x =4, x = 12. Find also the position of the centre of gravity of this area and the radius of gyration of the area about the x-axis. Find the mean value between 0=0 and 0=- of a (i) sin 6, (iv) (ii) cos 6, (v) cos 2 6. (iii) sin 6 cos 6, P. 1. Find -3^ when y is ax (i) x 2 sin x, tana: (ii) a;^ 2 *, (v) ^a; sin x. (iv) x , (vii) ^/sin >Jx. (iii) e * cos 2 6x, 'l + cosa; 3. A line is drawn through (1, 3) cutting the axes in A and B respectively, so that OA . OB is a minimum. Find the equation of the line. 3. A body mass m Ibs. is moving in a straight line. If at the end of t sees, its displacement from some standard position is < ft., its speed v ft. /sec. and its acceleration a ft. /sec. 2 , and if k foot-poundals be the kinetic energy, shew that dk , dk -j-=ma and = ma.r. d dt 267] MISCELLANEOUS EXAMPLES 357 4. If a piston-rod drives a crank of length r through a connecting-rod of length I, prove that its acceleration at the end of the stroke is u*r ( 1 - ) , where u is the uniform angular velocity of the crank. 6. A cart weighing 200 Ibs. containing 400 Ibs. of sand, ascends a straight hill rising 40 feet vertically altogether. The sand is assumed to be thrown out uniformly so that the cart reaches the top empty. Find by integration the whole work done against gravity in the ascent. Q. 1. Give equations of tangent to x 2 + 2y*=5 at each of the points where 2/=i. 2. AB is a straight line ; P, Q two given points on opposite sides of it, PM, QN are perpendicular to AB. PM =p ft., QN =q ft., MN =a ft. R is any point in AB such that MR=a; ft. If the speed of light on the P-side of AB is u ft. /sec. and on the Q-side v ft. /sec., find the time light would take to travel from P to Q by the path PRQ. If R' be the position of R for which this time is the least and if PR', QR' make angles 0, <f> with the normal to AB, prove sin u 3. P is any point (h, Jc) on the curve y = be a . PN is the ordinateof P, PT and PG the tangent and normal meeting OX in T and Q. ft 2 Prove NT constant and NG = . a 4. In a tangent galvanometer the current is proportional to the tangent of the angle of deflection (c = fctan &). Find the approximate error in the inferred value of the current if an error A0 is made in reading the deflection. Shew that the relative error in the current ( J is approximately ' and for a given error in the deflection is least when 0=45. 6. Prove that the area common to the two parabolas is ^(a + b)Jab, and find the o. o. of this area. 358 CALCULUS FOE BEGINNERS [CH. XI R. 1. The equation s=ae~ M sin-=- gives the displacement (s feet) of the end of a stiff spring from the position of equilibrium at the end of t seconds. If a = 10, T = '9, b ='75, find the speed and acceleration of the end of the spring at the end of 5 seconds. V 2 V - 12 2. Shew that rr~3 g ^ has a minimum value when V = 6 and that it has no maximum value. 3. In a simple horizontal engine the crank is r feet and the connectiug- rod I feet long. At any instant the crank and connecting-rod make angles 6 and <j> respectively with the line of dead centres. If the angular velocity of the crank be constant and equal to w, shew that the angular velocity of the connecting-rod is * .'l2i " RP, lcos<(> I where P is the end of the crank and R the point where the connecting-rod meets the perpendicular to the line of dead centres through the centre of the crank circle. a v fZ [4 4. Find (i) | sin 3x cos 2x dx, (ii) I cos 3x sin 2x dx, (iii) I cos Bx cos 2x dx, (iv) I sin 3* sin 2x dx ; Jo jo and shew that if m and n are whole numbers r sin mx cos nx dx = /O 2m /sii according as (m + n) and (m-n) are both even or both odd. - 2 2 Shew also that fir frr I sin mx sin nx dx = I cos mx cos nx dx = 0. Jo Jo 6. Draw roughly y = sin* from x = Q to x = -^ . Consider the area bounded by the curve, the x-axis and the ordinate 267] MISCELLANEOUS EXAMPLES 359 Divide the base into 10 equal parts. Shew that the area lies between 919 and 1-08. Divide the base into n equal parts each ~ or h. &n Shew that the area lies between ft . if . fif h\ h . IT . ITT h\ sm^sm^.-J and _sm 3 n ( - + ) . sin - am - Then shew that when h is indefinitely diminished the limit of each of these is 2 sin 2 T or 1. 4 IT f* Compare with what you get by finding I . sin xdx in the ordinary way. [Assume sin a + sin (a + /3) + sin (a + 2) + . . . to n terms ' J s. 1. (i) If shew that (ii) If q = shew that 5f-6^ at* at 2. OX, OY are two lines at right angles to each other. A, B are two points 5 feet apart in a rod which is constrained to move so that A is always in OX and B in OY. Supposing A to move along OX at a uniform rate of 1 inch per second, find the rate of motion of B along OY and the angular velocity of AB at the instant when OB is 3 ft. 3. The co-ordinates of any point on a cycloid being given in the form x=a (0-sin 0), y = a(i-coa 6), shew that the whole area bounded by one complete arch and the line on which the generating circle rolls is 3ira 2 . 360 CALCULUS FOR BEGINNERS [CH. XI 4. A submarine cable consists of a circular core surrounded by a con- centric circular covering. The speed of signalling through it varies as -^ x* where x is the ratio of the radius of the covering to that of the core. Find x so that the speed may be a maximum. 5. Find the centre of pressure of a circular area (radius r) with its plane vertical and its centre at a distance c below the surface (or). T. C /v*3 i y8 1. Find m I x * +a * dx (ii) 1 cot xdx, ir f* (iv) I tun 2 a; da;. () J 2* It [2 (Hi) I sin 2 2xdx, (v) [0 2. Water is poured at a constant rate into a conical glass, which is filled in 2 minutes, the height of the cone being 12 inches. At what rate in inches per minute is the surface of water rising (1) when the glass is filled to half its height, (2) when half the liquid has been poured in ? 3. The equation to the Folium of Descartes is Shew that any point given by 3am lies on the curve. _.. , du , dx dy Find -r^ and , and hence -/- . am am ax Find the equations to the tangents at the points where m= - and m = l. Draw the curve carefully for positive values of m taking a=l. [Unit 1".] Find -j- in the above otherwise. dx 267] MISCELLANEOUS EXAMPLES 361 4. If = A sin n + B cos n, where A, B, n are constants, find the maximum kinetic energy, the space-average and the time-average of the kinetic energy for one journey between the extreme positions and shew that the three results are proportional to 6, 4, 3. 6. The cross-section of an I girder has the following dimensions : Top Flange 6x2 ins. Bottom Flange 15 x 2 ins. Web 18 x 2 ins. Find the radius of gyration of this section about an axis in its plane through its C.G. perpendicular to the web. CHAPTER XII APPROXIMATE SOLUTION OF EQUATIONS 268. SUPPOSE we wish to solve x3 _ 6x 2 + 9x - 1 = 0. The ordinary graphical method is to draw the graph of and find where it cuts the a;-axis. The degree of accuracy attained will depend on the care expended in drawing the graph and on the size of the scales used. We have seen that a knowledge of the gradients at different points is a great help in drawing the graph. Put f (x) =a?-Ga? + 9x-l, then / / (*) = 3* a -12a; + 9=3(a;-l)(a:-3). Make a table of values : x 01234 /() -1 3 1-1 3 /'(*) 90-309 If we draw the graph (Fig. 133) the roots appear to lie between (i) and -2, (ii) 2-3 and 2-4, (iii) 3-5 and 3-6. 269. If we want better approximations, it is better to work at each root separately. 268, 269] EQUATIONS 3(53 Take the first root, that is the one lying between and -2. x -2 /(*) - 1 -568 /' (x) 9 6-72 Fig. 133. P, Q (Fig. 134) are the points (0, - 1) and (-2, -568). PS, QT are the lines through these points with gradients 9 and 6 '7 2, i.e. they are the tangents at P and Q. It is evident from the figure without actually drawing the curve that it cuts the axis very nearly where x = '12. We find /(-12) =- -004672. If we want something nearer, the figure suggests trying -13. /(13) = -070797. 364 CALCULUS FOB BEGINNERS .*. the root is between -12 and '13. x -12 -13 f(x) - -00467 -07080 /'() 7-60 7-49 P' is (-12, - -00467), Q' is (-13, -07080). [CH. XII Fig. 134. It will be found, even on a large scale drawing (Fig. 135), difficult to distinguish between P'Q' and the tangents at P' and Q'. 1205 seems the best approximation. As a matter of fact /(-1 205) = - -0009 nearly. 269, 270] EQUATIONS 365 270. In the example just worked out /(O) =- 1, /(I) = 3, i.e. the point on the graph of y =f(x) corresponding to x = is below, while that corresponding to x = 1 is above the aj-axis, therefore we might say that the graph of y =f(x) cuts the OS-axis at some intermediate point. Fig. 135. It may happen that y =f(x) cuts the cc-axis at more than one point between the points corresponding to je = and x = l, e.g. referring to Fig. 133, and there are 3 roots between and 4. and there is 1 root between 1 and 3. 366 CALCULUS FOR BEGINNERS [CH. XII Also if / (x) has the same sign for two different values of x, say a and b, it does not follow that there is no root between a and b ; for instance, taking the same figure, /(2) = l,/(4) = 3 and there are 2 roots between 2 and 4_ Iff (a) and/ (6) have opposite signs, there is an odd number of roots between a and b. If /(a) and/(6) have the same sign, there is no root or an even number of roots between a and b. 271. There is an exceptional case if f (x) becomes infinite between a and b. e.g. suppose /(#) Fig. 136. 270-273] EQUATIONS 367 Here /(- -5) = - 175, but there is no root between -5 and - 5. A glance at Fig. 136 will shew the reason of this. When x = 0, y is infinite. When x is very small and negative, y is very large and negative. When x is very small and positive, y is very large and positive. so that as x passes through the value 0, y suddenly changes from a very large negative value to a very large positive value. It thus appears that \i.f(a) andy*(6) have opposite signs we must take care that/(a;) does not become infinite for some value of x between a and b, otherwise our inference that there is a root between a and b may be wrong. 272. If f(a) and f(b) have opposite signs and if moreover f (x) and f" (x) do not change sign between x = a and x = b, then between x=a and x = b the graph of y=f(x) will have one of the shapes shewn in 94, and will cut the o;-axis only once between x = a and x = 6. 273. A method of continual approximation to the value of a root will now be considered which does not require the actual drawing of the graph. Taking the same equation as before, we have /' (x) = 3a; 2 - 1 2* + 9 > = 3 (x - 1 ) (x - 3), / = 6(*-2). Now/(2)=l,/(3) = -l. 368 CALCULUS FOR BEGINNERS [CH. XII Also from o? = 2 to x-3, f'(x) is negative and f"(x) is positive. .'. the curve y =/() between x = 2 and x 3 is like Fig. 137, and there is only one root between 2 and 3. M Fig. 137. If PT is the tangent at P and the chord PQ cuts the .r-axis in R, the point A where the curve cuts the aj-axis is between T and R. Now/' (2) =-3. .'. equation of PT is y 1 = 3 (x - 2). Putting y = 0, x = 2 + . [This is OT where O is the origin.] This value of x [2J] is a better approximation to the root than 2. 2 Again gradient of PQ is - y , and equation of PQ is Putting y = 0, x = 2 + J. [This is OR.] This value of x [2] is a better approximation to the root than 3. .'. we know that the root is between 2J and 2J and therefore between 2 -33 and 2-50. ["We must be careful here not to overstate the smaller or to understate the larger value, for instance we cannot safely say that the root is between 2-34 and 2-50.] 273, 274] EQUATIONS 369 To get a closer appi-oximation, use the points P', Q' corre- sponding to a; = 2-33 and as = 2'50 exactly as we have just used P and Q. /(2-33) = 0-045937, /(2'5) = - 0-375, /'(2-33) = -2-6733. Equation P'T' y - -045937 = - 2-6733 (x - 2-33). /. atT' * = 2-33 + -. = 2-33 + (-017+). Equation P'Q' y - -Q45937 = - ' 42Q ? 37 ( x - 2-33) = - 2-4761 (x- 2-33). . 04^007 /. at R' x= 2-33 + ^f. = 2-33 + (-019 -). .'. the root lies between 2-347 and 2-349. If we want a still closer approximation we use the points P", Q" corresponding to x= 2-347 and 05 = 2'349. /(2-S47) = 0-000781923, /(2-S49) =-0-004491451, /' (2-347) =-2-638773. Equation P"T" y - -000781923 = - 2-638773 (* - 2-347). Equation P"Q" y- -000781923 = - 2-636687 (x - 2-347). .*. at R" .'. the root lies between 2-3472963 and 2-3472966. 274. We might have " closed in " on the root more rapidly by drawing first the portion of the graph of -x z - 6x 2 M. c. 24 370 CALCULUS FOR BEGINNERS [CM. XII which lies between x = 2 and x = 3 from the data This would shew quite clearly that the root lies between 2-3 and 2-4. In working out the smallest root, we might proceed graphically as far as the stage reached in 269. Fig. 135 seems to shew that the root is between -1205 and '1207. Also/' (x) is positive and /" (x) negative so that the curve is like Fig. 40 (6), 94. Proceeding on the lines of 273, we find that the root lies between -12061474 and -12061477. 275. Sometimes it may not be so easy as in the example just considered to decide whether f (x) keeps the same sign between the values of x chosen, and it may be useful, to argue as follows: Between x = 2 and x = 3, f" (x) is clearly positive. i.e. f (x) continually increases j but/' (2) = 3 and /' (3) = 0. .'. /' (x) is negative all the way. 276. The rough figure (137) should be drawn, so that there shall be no doubt at which end of the arc the tangent ought to be drawn. The following example will shew what may happen if we neglect this precaution. Suppose f(x) = Saj 8 80; + 1 "j f'(x) = Qx*-8 \. f"(x) = lSx ) Here /(I) =-4, /(2) = 9, and between x = 1 and x = 2, /' (x) and /" (x) do not change sign ; they are both positive. .*. there is 1 root between 1 and 2. 274-277] EQUATIONS 371 Now if we start from P (the point on y =f(x) corresponding to x = 1), we have since y (1) = 1 : Equation of tangent PT is y + 4 = 1 (x 1). .". at T, x = 5. But 5 is not a better approximation to the root than 1. Since f (x) and f" (x) are both positive between x = 1 and x = 2 the curve is of the form shewn in Fig. 138 and a glance at the figure will shew that we should have worked from Q (corresponding to x = 2). Q Fig. 138. U is nearer to A than N is. T is not nearer to A than M is. Of course with a curve of this form it may happen that T comes nearer to A than M does, but on the other hand it may not, whereas U must be nearer to A than N is. What we ought to do in this case is to consider 2 as the first approximation to the root and get a better one by finding the co-ordinates of U. 277. As another example take the equation coso?=o; and let it be required to find the least positive root. An inspection of the tables shews that x is between [42] and [43] where [42] stands for the C.M. of 42". 24-2 372 CALCULUS FOB, BEGINNERS [CH. XII If f( x ) is a: cos x } f (x) = 1 + sin x I . f" (x) - cos x } "Working with 5-figure tables /[42] = -73304 - -74314 = - -01010 ( -00001) j /[43] = -75049 - -73135 = + -01914 (+ -00001)} ' Also between x = [42] and x = [43]/' (x) and/" (x) are both positive and the curve is like Fig. 139. /' [43] = 1-68200 (+ -000005). Q Pig. 139. Equation QU is y- -01914 = 1-68200 (a;- [43]). 01 Q14- * /. atu aj = Equation PQ is -- 01914 = 02924 _r 43 .-,v 02924. ol745 <*-[ 43 ]> l-6756(a;-[43 J). .'. atR * We must be careful to understate the value of the first of these fractions, and it will be safe to take it as -^ !^r . Similarly the value I*o82005 of the other fraction should be overstated. 277] EQUATIONS 373 .'. x lies between -73906 and 73912, or since -01137 and 01143 each differ from [39'j by less than [!'], we may say x is approximately the C.M. of 42 21'. As a matter of fact for this value of x f(x) = -73914 - -73904 = -0001. EXERCISES. LXXXIV. 1. Shew that a: 3 -a:- 1=0 has one and only one root between 1 and 2, and then shew by the method of 273 that this root lies between 1-1 and 1-6. 2. 4z 6 -24a; 3 + 2 = has a root near 3. Shew that 2-7 is a better approximation. 3. x* - 2x - 1 = has a root between - 1 and 0. Shew that - -3 is a better approximation than and ? than 1. 4. 3x* - 40z 3 + 1200z - 2900 = has a root between 3 and 4. Shew that it is between 3-3 and 3 -6. 6. x a -Bx 2 + 2x-5 = has a root near 2-9. Shew that it is between 2-9041 and 2'9042. 6. Shew that x s -2x-5 = Q has a root between 2 and 3, and find its value correct to 3 decimal places. 7. Find all three roots of 20x 3 -24x 2 +3=0 correct to 3 decimal places. 8. Shew that x 3 -- 5x + 2=0 has one and only one root between and 1, and find its value correct to 4 decimal places. 9. Shew that 3x 3 =x + l has only one real root and find its value correct to 5 decimal places. 10. Shew that x 3 - 30x 2 + 2600 = has a root between 12 and 13 and find its value correct to 2 decimal places. [First write for x, y + 10.] 11. Shew that if a floating sphere of radius r" and specific gravity *, sink in water to a depth x", then x 3 - If the sphere be of wood (sp. gr. -6) and its radius be 10", to what depth will it sink? [Correct to nearest hundredth of an inch.] 374 CALCULUS FOR BEGINNERS [CH. XII 12. x 3 - Wx^ + iOx - 35=0 has a root between 1 and 2. Find its value correct to 4 decimal places. 13. Find the real roots of x*-3x 5 = 0, each correct to 3 significant figures. 14. A plane is drawn parallel to the base of a hemisphere of 1 foot radius, dividing it into two parts of equal volume. Find its distance from the centre correct to the nearest thousandth of an inch. 15. x -sin x =-02 is satisfied if x is the C.M. of some angle near 30. Find the angle so that your error does not exceed 1 minute. 16. Solve approximately t&nx = 2x between x = and x = ^. Your 2 answer must not be more than -001 in error. 17. Solve tan x= - - x between and IT correct to 3 decimal places. 2 18. Shew from a graph that the line By = 2x + 1 cuts the curve y = tan x roughly where x = - . Find the value of x at the point of intersection correct to 3 decimal places. 19. Solve the equation log g a; = -a: correct to 3 decimal places. O SO. Sketch the shape of the curve y = x 2 sinx: suppose x to range from OtOTT. Determine approximately the maximum ordinate within this range. CHAPTER XIII SOME METHODS OF INTEGRATION Further examples in integration. 278. OUR success in evaluating integrals of the form ]f(x) dx has so far depended on our ability to recognise f(x) as the result of differentiating some standard function, e.g. we know that d (tan x) 7 = sec 2 x, dx and hence that f sec 2 x dx = tan x + c. It will be convenient at this stage to collect the most impor- tant standard results. They have all been worked out either in the text or in the exercises. It is understood that an arbitrary constant should be added to every function in the column headed z dx. When a more complicated expression is to be integrated, our object will be to reduce it to one or other of these standard forms. A few of the simpler methods of effecting such reduction are given in this chapter, and it will be seen that many functions yield to these methods of treatment, but no rule can be given for the integration of any given function. 370 CALCULUS FOR BEGINNERS [CH. XIII y dx 2 / zdx x X* nx*- 1 except when 11= -1 T n +l n+1 sinx cosx cosx sinx cosx -sinx sinx cosx tanx Bec 2 x sec 2 x tanx cotx - cosec 2 x cosec 2 x -cotx secx sec x tan x sec x tan x secx cosec x - cosec x cot x cosec x cot x cosec x Bin" 1 - a 1 1 , x sin- 1 - a V/a 2 -* 2 v/a 2 - x 2 tan" 1 x a 1 ta.n-1 X a a 2 + x 2 a 2 + x 2 a a e* * e* t* log,x 1 1 x log,* sinhx coshx coshx sinhx coshx sinhx sinhx coshx tauhx sech 2 x sech 2 x tanhx sinh" 1 - sinh- 1 - a 1 1 a or locr *&+* Vi 2 ^ x + Jx^ + a? be a J cosh" 1 - cosh" 1 - 1 ^2,1^ 1 a orlop a; + V a;2 - 2 ^-A ; Va % a2 x+>v/x 2 -a 2 a ( ->1 J ( 1 x tanh-i- \u y - tanh" 1 - a (x'ca 9 ) 1 a a a + x a 2 -x 2 1 o + x 2 ge a-xj Or 2a 10g a-x 278, 279] METHODS OF INTEGRATION 377 y dy dx Izdx X \ 1 coth" 1 - a a or tanh" 1 - V x a 2 1 1 1 , .a or tanh" 1 - a x x 2 - a 2 * x' 2 - a 2 ,11 **! (x3>a>) 1 ln x + a 2o &8 x-a log e g i n x loge cos x loge (sec x + tan x) 1 cotx -tanx cotx tanx loge s i Q x -logeCosx or loge secx i log, (sec x + tan x) or loge tan (T + O)| secx secx | or log e * an ( 7 + H ) log e (cosec x + cot x) ~j / - log e (cosec x + cot x) or - log tan - I cosec x cosec x J x Or loge * an 9 \ 279. Corresponding to the formula of differentiation df(u) df(u) du ., . du V \ f %f \ / Qp J. Inl \ we have the formula of integration for since , and du =f ( u ), ff ().*. dx =/() and /() = I/' (u) du, or where <^> (u) stands for/' (u). That is to say -v- . dx occurring under the integral sign may dx be replaced by du. 378 CALCULUS FOR BEGINNERS [CH. XIII Ex. 1. [sin 3 x cos x dx = I sin 3 x ^ . dx ~ I sin 3 x d (sin x) = Iu 3 du [where u is sin a;] u* sin* x 4 4 Everything here depends on noticing that the expression to be integrated (sin 3 x cos x) is the product of a function of sin x (viz. sin 3 x) and cos x, which is the differential coefficient of sin x. Generally if the expression, f(x), to be integrated can be written in the form <f> (u) . -r- we can say that dx \f(x).dx=\<t>(u).du, and it may happen that I < (w) . du is easier to get than ff(x).dx. Ex. 2. Find ( , 2a; + 3 dx. J >Ja? + 3x + 5 Here we notice that (2a? + 3) is ' ^- and thus dx (2x + 3)dx may be replaced by d (ar* + 3x + 5) and the integral may be written or I j- where u is (x 2 + 3x + 5), which is 2/Ju or 2\Ae a + 3x + 5. Ex. 3. lsiii s x.dx= Isin 4 a;. sino;c&c = f - (1 - cos 2 a) 2 d ( in 2x2^ / 2 1 1 = -J^- M >^ = -( M -3 W+ 5 W cos a where = cos x 2 1 = cos x + = cos 3 x - - cos 5 x. 3 5 279] METHODS OF INTEGRATION 379 dx Ex. 4. The expression under the >J 11 T2' and j. i du . . integral is - j - where u stands for 1 u-i u * i, i 6 *- 1 = fo sinh J T=- = rr smh" 1 j=- v** /ll 'v V 12 or EXERCISES. LXXXV.* Find I zdx for the following values of z : 1. sin x cos 8 x. 2. xsina; 2 . 3. 1 6- -s -^ =. 6. 1 gdn-'as 7. -sin (log e a;). 8. . O. l0 ' 9^^ 2<9 )' 4^9 (^> 9 )' "' 9+1F3- 13. cos 3 x. 14. tan 8 x sec 2 x. 15. tan 3 x. * In this and following Exs. results should be checked by differentiation. 380 CALCULUS FOE BEGINNERS [CH. XIII X6. *j2a + x. 17- sin 3 a; cos 2 ar. 18. _ - - R . [(i) x*-3x + 2pos. (ii) x* - 3x + 2 neg.] 2 -3x + 2 19 3x + l 20. tan(aa; + 6). 21. sina;cos2a;. 1 as. . 26. cos 5 a; Integration by Substitution. 280. This depends on the same principle as the last method, viz. that Ex. 1. Find \*Jd i -a?.dx. Put x = a sin 0; then v a 2 x 2 = a cos 0, and efcc so that I v 2 # 2 cfo; becomes I a cos . d (a sin 0), = I a cos . a cos c?0 x 280] INTEGRATION BY SUBSTITUTION 381 Ex. 2. Find \x /2x + 5 . dx. z 5 (i oc Put 2x + 5 = z, .*. x = , and = - , 'Jj dz 2t so that I x v 2x + 5 . dx = I Jz.-^ = | f(a* - 4 5 EXERCISES. LXXXVI. Find: 3 - 5. I x*Ja*-x*.dx. 6. I o?*jcfl-x*.dx. 1. I . a . da;. 8. I >/a a + o; 2 .da;[a;=osmh0]. y v x + ") /y . / Vx - a 2 rfz [a; = a cosh tfl. 1O. / / . , ^*- J J x*Ja?-x 2 11. I - -dx[x-2 = z]. 12. 13. \ ^dx. 14. /- J 2a-x j , 15. By means of the substitution tan x = sinh <j>, or seca; = cosh0, shew that Isec 3 xdx= /cosh 2 (f>.d<t>, and hence evaluate Isec 3 a;dx. (See Exs. LXXXVII. 15.) 382 CALCULUS FOB BEGINNERS [CH. XIII Integration by Parts. 281. This is an inversion of the rule for differentiating a product. d (x sin x) Ex. - ; -as sin x + xcosx (1). ax If one of the members of the right-hand side can be integrated so can the other. (1) may be written d (x sin x) d cos x -. = + X COS X. dx ax d (x sin x + cos x) /. x cos x = = , dx or I x cos x dx = x sin x + cos x. Similarly by differentiating x cos x get I x sin x . dx, and x log x get I log x . dx. 282. The work can be arranged more conveniently. ,. , d(uv) dv du We have - \ '- = u -=- + v -7- , dx dx dx .'. uv = \udv + Ivdu, or hi d v = u v I v du, and it may happen that Ivdu is easier to get than ludv. Ex. 1. \xco&xdx= I x d (sin a:) = x sin x l&inxdx = x sin x + cos x. 281, 282] INTEGRATION BY PABTS 383 We might have said / x cos xdx = lcosx.d(~} y? [a? = -^ cos x I d (cos x) A J Z = cos x + I sin x . dx, f 1$ but this leads to an integral I -~- sin xdx which is no easier to get than the original one / x cos x dx. Ex. 2. I log e xdx = x \og e x ~ I X( % (lge x ) f 1 , *:xlog e x- Jx.-ax = X loge X X. EXERCISES. LXXXVIL Find: 1. I x log e x dx. 6. I x sin xdx. 2. I xe x dx. 7. Ix sin mx cos nx dx. 3. I x z e x dx. 8. / a; 3 log e o;<ZaJ. I 4. I tan" 1 xdx. 0. x 3 6. I sin' 1 xdx 1O. I x sec* xdx. 384 CALCULUS FOE BEGINNERS [CH. XIII 11. Shew that / e? sin x dx = -e x cos x + I f* cos x dx, and I e?cosxdx=e x Binx- I e'sinxdx. Hence get I e x siuxdx = -e ae (sin-cosa:), and I e* cos x dx=^e" (sin* + cos x). 12. Similarly get I e^sin^x + ^dx and I e * cos (bx + c) dx. 13. Shew that I siu 6 xdx= - sin* cos a; + I 4 sin 3 x cos 2 * dx. Hence 5 I sin 6 x dx = - sin 4 x cos x + 4 I sin 3 x dx. Similarly shew n I ain n xdx= - sin"" 1 x cos x + (n - 1) I sin n ~ 2 a;. dx, and n / cos n x dx = cos* 1 " 1 x sin x + (n - 1) I cos n ~ 2 x . dx. 14. Shew that a it /sm n xdx= / si n J o ir r /"a /a sin 7 xdx and J sit J o ; o Hence find / sin 7 x dx and | sin 8 x dx. 16. Shew that sec 3 x dx = sec x tana;- / tan 2 # sec a- da?. Hence 2 / sec s xda;=secxtana;+ / secai.d*. I sec 3 x dx = sec x tan a?- / i ! / sec s xda;=secxtanx+ / -sec a;, i 282] EXERCISES IN INTEGRATION 385 MISCELLANEOUS EXERCISES ON CHAPTER XIII. Find I z dx for the following values of z : 7 1 m log e ax cos .T a ' 1 + sino;" C 9 ..... 10 a; 2 -5 2 v /a; 13. xaiuBx. 16. ar(3-2a:)i 14. x^sinBx. 15. (3 2s;)*. 22. sec a;. \ Evaluate the following definite integrals: 25. 28. 27. r ;;. 29. ein 6 xdx. 31. . 2 3O. I cos 8 x dx. 32 11. 0. J o 4 + 5 cos a; 25 386 CALCULUS FOR BEGINNERS [dl. XIII 83. O is the lowest point of the chain of a suspension bridge, P any other point on the chain whose co-ordinates are (x, y) referred to horizontal and vertical axes through O. If the resultant load on the portion OP of the chain is wx Ibs. (w being constant) and if T, T Ibs. wt. are the pulls a P and O respectively, shew that dy _ w Tx = T x ' and hence get the equation of the curve of the chain. / w* Also shew that T = T ^J 1 + ^- a a; 2 . 34. A body rr.nss m Ibs. starts with a speed of u ft. /sec. and moves horizontally, the resistance to motion being Kt> 2 poundals, where K is a constant, and v ft./sec. its speed at any instant. Shew that the speed at the end of t seconds is - ^ ft./sec. 35. Assume that the resistance to the movement of a ship through the water is of the form (a 2 + 6 2 t; 2 ) Ibs. wt., where v is the speed and a, b are constants, m Ibs. being the mass of the ship. If the engines are stopped, find a formula for the time in which the speed falls to one-half of its original amount u. 36. A body of mass m Ibs. falls from rest vertically in a medium in which the resistance is K 2 Ibs. wt. where K is constant and v ft./sec. is the speed. Shew that o wliere Vl= Hence get v in terms of t and t>i , and shew that r i 2 , / , gt\ = log ( cosh ) . 9 \ vij 87. A block slides in a straight line OA. The resistance to its movement when it is at a distance x ft. from O is R Ibs. wt. a 3 -f- a; 2 where OA is a ft. Find the work done in pushing the block from A to O. 282] EXERCISES IN INTEGRATION 387 38. A train is moving on the level, the power exerted by the engine being constant and the total resistance proportional to the speed. Shew that -^ = - - bv, where a and 6 are constants, and hence that a - 6u 2 = Ae~ 2bt , at v where A is a constant. If the constant power is 400 H.P. and if the resistance is 12 Ibs. wt. per ton at 40 miles an hour, and if the mass of the train be 200 tons, find the values of the constants a, b, A ; t being reckoned from the instant at which the speed is 40 miles an hour. Shew that the speed can never reach 50 miles an hour. Find the speed after half-a-minute and after half-an-hour. [0 = 82.] 25-2 CHAPTER XIV POLAR CO-ORDINATES 283. WE have hitherto considered the position of a point in a plane as determined by its distances from two fixed axes. These distances are the " Cartesian co-ordinates " of the point. There is another system in which the position of a point is determined by a distance and an angle. The position is obviously determined if we know the length, direction and sense of the line joining the point to some fixed point. For instance if A is a given point and we say that B is 100 yards N. 30 E. from A, B is determined. This is the principle of " Polar Co-ordinates." 284. O is a fixed point (fig. 140), OX a fixed line through p Fig. 140. 283-287] POLAR CO-OTCDINATES 389 O. The position of P is known if we are given the length OP and the angle, XOP, through which a line must swing to get from the position OX to the position OP. These are called the "polar co-ordinates " of P, and are denoted by (r, 0). O is called the pole and OX the "initial line." 285. There is an infinite number of ways of writing the co-ordinates of a point P, e.g. if XOP = 30 and OP is 3 inches, taking 1" as the unit of length we may give the co-ordinates of P as (3, 30), (3, 390), (3, - 330) etc. 286. If r is negative the radius bounding the angle must be produced backwards through O, e.g. if the point is ( 3, 210), first draw OP' so that XOP'= 210, then produce P'O backwards through O so that OP is 3". This brings us to the same point P as in 285. EXERCISES. LXXXVIII. 1. Taking 1" as the unit of length, plot the points (3, 40), (-3, 140), ( - 3, 220), (3, 320), ( - 3, - 50), (3, - 500), ( - 3, - 1000). 2. Draw the loci : (i).r=a, (ii) 0=40, (iii) r=acos0, (iv) rcos0 = a, (v) r=asin0, (vi) rsin 0=a. 287. Figure 141 shews part of the curve r=f(0), P is the point (r, 0) and Q the point (7- + Ar, 6 + A0), PS is perpendicular to OQ, and R is the point where a circle, centre O radius OP, cuts OGL .'. POQ = A0 and RQ = An QS QR + RS QR RS , , JMow cot PQR = = = + and by sufficiently SP SP SP SP J J RS diminishing A0 we can make as small as we like ( 32) ; also SP 390 CALCULUS FOR BEGINNERS [CH. XIV = - = ^ . -f. and by sufficiently diminishing A0 SP r sin Ad r. A<9 sin A0 1 dr we can brine this as near to - . -=7. as we please. r dO Fig. 141. .'. if < be the angle between OP and the tangent at P (fig. r r 'dr' 1 dr 142) cot = -. g or r du Ex. If Fig. 142. r = a (1 cos 0), dr 287, 288] POLAR CO-ORDINATES 891 1 dr sin 6 6 2' e.g. it 6 = 90, <f> = 45 ; if $ = 1 20, [See fig. 143.] = 60 ; it 6= 60, <= 30. These facts may be found useful in drawing the curve, just as the knowledge of the gradient was a help in drawing a curve whose equation was given in Cartesian co-ordinates. 288. Areas. Suppose we want the area bounded by HK, part of the curve r =/(#), and the radii OH, OK. 392 CALCULUS FOR BEGINNERS [CH. XIV Divide Z HOK into any number of equal parts and with O as centre describe circular arcs as shewn in fig. 144. Fig. 144. The area required lies between the sum of the internal sectors OHC, OPD, OQE, ... and the sum of the external sectors OPC', OQD', .... The difference between these sums is the sum of the areas CHC'P, DPD'Q ..., and this is the same as the area GK. By sufficiently diminishing the angle of the sectors or increasing the number of the sectors, we can make this area GK as small as we like, and thus the area required is the limit of the sum of either the internal or external sectors as the number of sectors is indefinitely increased or the angle of the sectors indefinitely diminished [cp. 136]. If P be any point (r, ff) on the arc HK and if Q be (r + Ar, 6 + A0), the area of the sector POD is ^ r 3 . A0, and area e=/s HOK = Lt 2 A0H.-0 6=a where a and ft are the angles XOH, XOK respectively. rft i - ^ J a * 288, 289] POLAR CO-ORDINATES 393 289. Ex. If the curve be r = a (1 - cos 6) [Fig. 143], area OAB 3 f*l = / - J TT 2 ( z Vi o /> 1 + cos 2#\ = / - ( 1 - 2 cos 4- J o?0 = -1505a 2 . EXERCISES. LXXXIX.* 1. Find values of r corresponding to 0=0, 20, 40, 60, 80, 90, ... 360 ; if r=a (1 -cos0). Also find the values of corresponding to = 0, 60, 90, 120, 150, 180. Draw the graph of r=a (1 - cos 6), using these values. (See Fig. 143.) 2. Find tan <j> if (i)r=asin0, (ii)r=acos0, (iii) r cos a, (iv)rsin0 = a. Interpret the results geometrically. 3. Draw graphs of (i) r = asin20, (ii) r=acos20, (iii) r=asinC0, (iv) r=acos30, (v) r sin 29= a, (vi) r sin 30 = a, ., a (vn) - = l + cos0, ' r .. a . 1 (vm) - = l + -cos0, (ix) - = 1 + 2 cos (x> -2= a 2 cos 20, (xi) r=3 + 2cos0. 4. Find the area enclosed by r = a cos 0. * Paper specially ruled for polar co-ordinates can be obtained. Lines are drawn through O (see Fig. 143) at intervals of 5, and concentric circles are described at small equal distances. 394 CALCULUS FOR BEGINNERS [CH. XIV 6. Find the area of one loop oir=a sin 20. 6. Find the area of a loop of r 2 =a 2 cos 20. 7. Draw the curve r=a0. (The spiral of Archimedes.) 8. In the curve r = ae ke , shew that <f> is constant. [This is called the equiangular or logarithmic spiral.] 9. Draw the curves (i) r=ae , (ii) r=ae . 10. Shew that r=6e e may be written r = c e+log < 6 and hence that r 6e e is r=e 6 turned round the pole through a certain angle. Generally shew that r=oe fctf , r=le* e represent the same spiral in different positions. ANSWERS EXERCISES I. p. 3. 1. (i) 39-39 m./hr. (ii) 57-77 ft./sec. 2. 42 m./hr. 3. (i) 2^ m./hr. (ii) 80 yds./min. 4. 0146 ft./sec. . -43 m./hr. EXERCISES II. p. 6. 1. 44 ft./sec.; 461 ft./sec.; 50-77 ft./sec. A's is best. 22ft.; 23Jft.; 25-4 ft. A's. 2. [All in chs./min.] 31, 39, 57, 52, 59, 58, 62, 62. (i) 59, (ii) 62. 3. 50, 57-36, 42-26, 54-46, 45 -40, 51-50, 48-48 cms. 8-832 -006, 9-288 -006, 8-92-01, 9'20-01, 9-00-03, 9-12-03 cms./sec. Between 8-97 and 9-12 cms./sec. 4. 173-21, 142-81, 214-45, 153'99, 196-26, 166-43, 180-40 cms. 36-480-012, 49'488-012, 38-44-02, 46'10-02, 40-68-06, 43-14-06 cms./sec. Between 41-4 and 42-3 cins./sec. EXERCISES III. p. 9. L. (99- 24A + 2A2) ft./sec., ( 99 + ^ ft./sec. 2. 42, 68, 54-25, 46-72, 44-33, 42-2303, 42-023003 ft. 26, 24-5, 23-6, 23-3, 23 03, 23-003 ft./sec. (42 + 23ft + 3fc2) ft. ; (23 + 3/i) ft./sec.; 23 ft./sec.; (23 - 3/z) ft./sec. ; 23 ft./sec. ; {23 + 3 ( - m)} ft. /sec. 3. (i) 26 ft./sec. (ii) 23, 29, 26 ft./sec. (iii) 26 ft./sec. 4. (i) 333 ft./sec. (ii) 330J ft./sec. (iii) 337 ft./sec. . 249 ft., (249 + 253/t + 96/i 2 + 16/i 3 + fc 4 ) ft., 253 ft./seo. 396 CALCULUS FOR BEGINNERS EXERCISES IV. p. 18. 1. (i) 605J. (ii) 604. (iii) 603. (iv) 602. 2. '87 ; '86 ; about -87. 3. 6; 2a. 6. r S V m 244 r 2 4. 3; 75; 3c 4 5 64*- 1007T 256 500 [2. { 4-5 Six 364-5 >. (i) 67T. 4-2 70-56T 296-352 (ii) 2*-. (iii) 3. 4-1 67-247T 275-684 3 ' 3 *" 4-01 64-3204jr 257-924804 3 *" 3 " 4+7t (64 + 32ft + 4 7(2) n 3 p 2 + 4871 + 47^ 3 17 201-76 196-84 3 192-4804 19 (1) 8 -, 3 -, 3 .., 3 . 3 *"' 3 (ii) 367T, 347T, 32-87T, 32-47T, 32-04ir, (32 + 4/z) x. 244 (iii) -5~T etc. c.c./sec. [see (i)]. (iv) 36ir etc. sq. cms./sec. [see (ii)]. o (v) 64*-. (vi) 327T. (vii) 64*- c.c./sec. (viii) 32?r sq. cms./sec. 7. (i) 20. (ii) 16 + 4/1. (iii) 16. EXERCISES V. p. 29. 1. 9, 12-25, 9-61, 9-0601, 9 + 6/i + ft 2 , 8-41, 9-6/ + / t 2 ; 6-5, 6-1, 6-01, 6 + h, 5-9, Q-h; 6; tan-i6=80 32'. 2. 6. 3. tan-il-2=5012'. 4. 12 + 6/j + 7i2. 12. 5. 12 + ft 2 ; 12 + 6(ji-m) + (jt2 +nm+m 2). e . 0,3,27,48. 7. tan- 1 4-8 =78 14'. a. 8,0. 9. 4; 4; 4. 10. 3; 3 + 8A + 6fc 2 -fc 4 ; 8; 0. 1. -109. EXERCISES VI. p. 37. 2. 27 ft./sec. EXERCISES VUL p. 5L (i) =32f. (ii) r = 100-32t. (iii) u = 3t 2 . (iv) v = 5. (i) 96, 160, 544. (ii) 4, - 60, - 444. (iii) 27, 75 867. (iv) 5, 5, 5 (ft./sec). ANSWERS 397 2. 64 ft./sec. ; 6-4 ft. ; 2-44 o/ . -64 ft. ; -249 o/ . 3. 4 ft./sec. 2 4. 13-6 ft./sec. 2 5. ^ ft./sec. 2 , ~. 6. (i) 32, 32, 32. (ii) - 32, - 32, - 32. (iii) 18, 30, 102. (iv) 0, 0, (ft./sec. 2 ). 7. -54ft.; -36 ft./sec. EXERCISES IX. p. 57. 1. 22-G3sq. ins. 2. -14%. 3. (i) 2ir, (ii) 207T, (iii) 400jr (sq. mm./sec.). 4. 407r sq. mm. 5. -16sq. ins. 6. (i) = '064 ins. /sec. (ii) ./_L= '126 ins./sec. 7. (i) --8. (ii) -Jr. 8. (i) ^= -03125 ins./sec. (ii) -32 ins./sec. o(J Oa EXEEOISES X. p. 64. !. -6, -4, -2, 0,2, 4, 6. 2. 27,3,0,75. 3. -4,0,12,28. 4. 14; 14-03; about -21 o/ . 5. (i) -0025. (ii) -00184. (iii) -2. (iv) 2-5. 9. (i) 3 -to. (ii)8. (iii) -| 3 . EXERCISES XL p. C3. EXERCISES XII. p. 69. 1. 3kx*. 2. 108 ft./sec. 3. 60; y = 60x-80. 4. 1-728 c. ins./sec. 8. 9ir = 28-27 c. ft./seo. 6. 9 '6 c. ins. 7. -3<Y . 398 CALCULUS FOR BEGINNERS EXERCISES XIIL p. 74. i 2. 1 0: -. a. 3x- x a ' x ' 4 4. 2 + 2t; 8 ft. /sec.; 22 ft./seo. /\ ^-^ /' *\ /* Q EXEKCISES XIV. p. 79. | x-t, - 3x-, j x~S, - i , l-8x-8, -A*-", 1, 0, 7 5 ^ ! -* 2 -* - fa7 '* ' ' X ' "" 2. - = kn**-i. 3. -= ax ax EXERCISES XV. p. 80. 3. gt. 3. --^5. . a&. EXERCISES XVII. p. 84. 1. (i) 2 + 8x, (ii) ^ 2 -l. 2. 15t 2 -t. ANSWERS 399 EXERCISES XVIII. p. 85. 1. 3x*-3; x -2 -1 12 y -2 2 0-22 ^ 9 0-3 09 dx 2. 4.T a: -1 1 23 y 6-3-2-36 $^ -24 24 dx 3. Gx 2 -18a; + 12 = 6(x-l) (x-2); x 0123 y -3 2 1 6 12 12 - - - -nr and > ' - or (1 '" 77 ' ~ ' 38 ^ and (-423> ' 385) - EXERCISES XIX. p. 90. 5. 3*-y-2 = 0, rr + 3y-4 = 0; T,o<(0, -2), G (4, 0), g 80 6. 9Gz-y-144=0, a; + 96y -4610 = 0; x S. 3x-y-4 = 0| 3a; + 7/-4=Oi a; + 3i/-28 = 0j" ' *-3y -28 = 0) " 9. ^ = 40x-57, 11. y = 7x-4, x + ly-12, -=- sq. units. EXERCISES XX. p. 92. 3. 5+1,1-8415,0. 400 CALCULUS FOR BEGINNERS EXERCISES XXI. p. 97. . 2-^; - 3 ; 4, If, *. 2. 35 ft. /sec.; 22 ft./sec X X EXERCISES XXII. p. 101. Letters refer to figures on p. 99. 1. (i) a. (ii) c. a. (i) b. (ii) a. (iii) d. 3. (i) b. (ii) a. MISCELLANEOUS EXAMPLES ON CHAPTER III. p. 102. .. 2 _i 2ft 2 1. (i) gX S. (h) __ = - J: . 2. 3. x + c z y-2c = 0\ Area of triangle formed by the tangent at 1 I ; 2. any point and the axes of co-ordinates is c ~ constant. " J ; (0, - 9), (0, 9) ; 18, s . ar + 6y-57 = 0j 2 6. t 5 6 7 6. t 5 6 7 s 207 301 413 s 758 1341 2166 v 85 103 121 v 476 697 960 a 18 18 18 a 200 242 284 7. 14, 14-03, -21 o/o . 8. y = 144x-400. (0, 5), (-1, 0), (2, -27). 9. (i) 1--. (ii) 1-n. (iii) n. (iv) -. 10 - (^ T = >0693 ins./sec. (ii) = -0173 ins./sec. 1UU 1UU 11. 2-65. 12. Average speed = n times speed at end. 15. (25-8)a, -6 -70ix 10- 6 ,2 -68x10-*. 16. 1-00896. ANSWERS 401 17. 0-172. 13. a; -1-^01 4 -12 36 da; IS. (i) 115 ft./scc. (ii) 123 ft./sec. (iii) 111 ft./sec. , _ 19r Ifil 2 3 ; ;96> " - or '955 ins./niin. 23. (Si, 40J). 24. x 1 2 3 4 5 6 7 8 9 10 11 12 t/ -30 -70 1-20 1-80 2-50 3-30 4'08 4-73 5-25 5-63 5-88 6'00 ^ -25 -35 -45 -55 -65 "75 '85 '72 -58 -45 -32 -18 -05 dx 25. 3z-2M= s . 26. 92-5 Ibs. wt. 27. 27 ft./sec. o 28. 305 radians/sec. 120 radians/sec. 2 (i) -4-Slbs. wt., (ii) -4-2 Ibs. \vt. 30. L EXERCISES XXIII. p. 111. /2 11\ / 3 83\ 1. (1, -2)min., (-1, 2) max. 3. (^, ^J min., (^ - -, -j J 4. ( - 1, 0) min., (0, 1) max., (1, 0) min. 6. (0, 1) min. 6. ( - 1, 0) min., (0, 5) max., (2, - 27) min. 7. ( - 1, - 2) max., (1, 2) min. 8. (-, 3|J minimum; M. c. 402 CALCULUS FOR BEGINNERS EXERCISES XXIV. p. 118. 1. 1. 3. 4/-5 = -7937. 3. a, |a. 4. 1\ ft. by 2Jft. 7. Radius ^ r , height -5. 8. -5- . -~- =108-6 c. ins. O O iJ ^y t>7T 9. 2r=ft=. / . 1O. 9-42 inches. 11. 4/150 = 5-31 knots. 12 = 3-183 acres. 13. Total height = width = -. =8-40 feet. 7T + 4 14. A = M55 ft., 2 ^=1-633 ft. 15. ~pft.j sees. 2^i cos ft 9 cos /3 17. (i) 2 c. ft. (ii) -=2-546 c. ft. (iii) = '849 c. ft. 7T OTT 18. 7. 18s. 9<L [Ht. = J side of base.] 10. 4-8 ft., 3 ft., 6 ft. 2O. rad.=2 N /6 ins., ht. = 4,/3 ins., vol. =96^^3=522 c. ins. a i. J_ (=-5774) of vol. of sphere; 1 : /2 or -7071 : 1. v^ 15 50 22. 2 radians =114 36'. 23. , - 25. r= 10 \32= 6-06, Z=10 y = 10' 5, volume = 330 c. ft. 26. 66-1. 27. 860-9 sq. ft. [A = tx/2]. 28. ^ = 3-224 c. ft 20. 5. 30. R=r. 31. 100. 32. a=900; 5 = 60,000; m= -404-8; n = 733'6; V=358. 34 . /'^L. 35. 4 of 6 ins. each and 2 of 2 feet each. 36. 45. 37. 13i feet; ___. . 6 = a . y\ a ffl 38. A/26C . sin 5- , where A is the least angle. 3O. - . ANSWERS 403 EXERCISES XXV. p. 134. 1. -71 (whena=3) min., 54 (when a:= -2) max.; (> ~ 8 i)- a. 12J ( when x = - J Min., 44 (when x = 4) Max. 3. (1, -1). 4. 2 + 2 ./I = 4-58 (when x =4/15). 5. (0, 0) max., f~y , - ? (y Yl or [1-71, - 7-25] min. 1 point of inflexion 7. A minimum point. 8. (3, -47) Min., ( - 1, 17) Max., (1, - 15) point of inflexion. EXERCISES XXVI. p. 138. a. 45-24 sq. ins. 3. 362 c. ins. 4. 197 ft., 9'8 ft 6. (i) 19-6 c. ft. (ii) 7-07 c. ft. 6. irr 2 coto.Ar. 7. (i) 46-5 sq. ins. (ii) 430 o. ins. 8. R (a + 2bt), -0064. EXERCISES XXVII. p. 139. I. 1-5 <V . 3. Decrease 1 %. 1O. 6-08 ft. /sec., -025. II. 2-214, -0115. 13. 1-1 % decrease. EXERCISES XXVIII. p. 143. a. 37-2088. 3. 96-244. 4. 24-24. 6. (i) 1. (ii) 1. 1-00007525, 1-00007475. 6. (i) -3. (ii) -2. - 2-99D39197, -2-00000397. 20 2 404 CALCULUS FOB BEGINNERS EXERCISES XXX. p. 146. [Add an arbitrary constant in each case.] 1. (i) a*. (ii)|**. (Hi) -\. (iv) 7*. (v) 10A ( (vii) -jp. (viii)jal () g*. (x) gg*^. (a) ^ + ^ (xii) --+4z. (xiii) o;3 + a; 2 -. (xiv) EXERCISES XXXI. p. 147. [o and 6 are arbitrary constants.] x* ..5 ... 7 1 4 s ( v ) ^z +ax + b - (vi) -** + *+ * 2. EXERCISES XXXI. p. 147. (iv) Q- + -x 2 + ax + 6. 4 5 (v) jg+ax + o. (vi) ^ x^ + ax + o. x^ 5 x^ x" - 4x 2 + ax + o. 3. - x 2 + ax + o. 4. ^ - -^ 2i & JLO -i EXERCISES XXXH. p. 148. ,. ,---i + J. a. , . , ,. V = r3. 8. y = , O j> 10. t/ = + 4x-l. 11. = ANSWERS 405 EXERCISES XXXHI. p. 150. ; 2 . a. (i) y = Zx-1x*. (ii) y = 28 + 3x - 2x. 4. y = 5 25\ .... /5 92\ 3, - T ). (11) (3, - T ). ' -*3; (0, -7); 8 8'. o EXERCISES XXXIV. p. 152. 1. (60-162)ft. 3. 186 ft./sec.; 404ft. 3. s=(3 + t 2 + t + 5. (i) 5 ft., 1 ft./sec. (ii) 97^ft-> 48|ft./seo. 6. = MISCELLANEOUS EXAMPLES ON CHAPTERS I-VI. p. 153. A. 1. ^=9x2 + 2. y = lla;-5. 3. 6 ft. 3. 1 ft., ft. 4. (i) (a) -08w or -2513 sq. ins./seo. (i) (6) -96^ or 3-016 sq. ins./sec. (u) (a) -04ir or -1257 c. ins./sec. (ii) (b) 5-16-rr or 18-096 c. ins./sec. 3 /]50 6. Height = diameter =2 * / = 7'26 cms. Surface = 248 sq. cms. V 7T B. (i) --2- (ii) 406 CALCULUS FOR BEGINNERS 4. Greatest when radius of circle = ft. = 15-9 ft. and side of square = 7T (not a true max.). Area = 796 sq. ft. Least when side of square = diameter of circle= -- - ft. = 8-97 ft. (true TT-fO min.). Area = - 5 = 224 sq.ft. 7T + O . =. 2. Max. 142 when x= -2 ; Min. - 2055 when a; =11. 3. (i) 150ft./sec. (ii) 58-4 ft./sec. 4. (3$, 400, s=3J. 5. f(x) = x*-3x 2 + x + C} -2. D. 1. 80 ft./sec., - 36 ft. /sec., 87'72 ft./seo. at angle 65 46' with downward vertical. a . Max. (16-90) when ,= at ... 160-56J28. ..... , ..-., Mm. - ^^ ( - 5-Oo) when x= (2-097). iif 3. /I 32\ \3' 27/ 4. -^- = 1-41 c. ins. 6. y = -x 3 + 3x--^-. V* E. 1. -6 o/ . 2. 20 ft. by 20 ft. by 10 ft. 3. (i) (64,32). (ii) *-2j/ + 16=0. (iii) [J64 (l=F JgV y] or (27-146, 21J) and (100-954, 21J). x -1 -i 1 2 -l). -12; 36; -i, 0,1. y 4 i o 13 6. (i) 46 ft./sec. (ii) 49 ft./sec. (iii) 45 ft./sec. ANSWERS 407 F. 1. 3-91C. 2. Q, -385^; y = x-l; y=-x + l. 4. 1-09 x 10< ft. ibs. wt. ; 3740 Ibs. ft. /sec. ; 60 Ibs. wt. 5. 18-18. G. 1. 1080 ft. 3. Lengths of pieces are - - , - - inches, or '56a, '44a H. . (x-2)2(4x + l); 6 (x -2) (2x- 1) ; 6(4x-5). (i) x=-lor2. (ii) x=-jor2. (iii) x - or 2. (iv) x = j. 1 2187X JMax. pt. on y =f (x) ^- , j . iMin. pt. on y=/' (a;) (2, 0). Min. pt. on y =/" (x) ^ , - ^J . . of infl. on y =f(x] (2, 0), Q , - g) . Pts x -1 0123 f(x) -8-2 4 /'(x) -27 4 5 13 /"(x) 54 12 -6 30 /'"(x) -54 -30 -6 18 42 4. J below centre. 5. (3-99, 2-26) ; (1*05, 4-46). o I. 1. 20ins./sec. 3. (i) (3x + 2y) ins. (ii) f X y + ^-- sq. ins. 4. (i) 7ift./sec. (ii) 12^ ft./sec. 5. 3 '32 x 10* cms./seo. 408 CALCULUS FOB BEGINNERS J. 1. 425ft. a. 4-5 %. 3. is + from x= - 2 to x=2 and - when x > 2. dx V_ dy dx 4. j- = 3z. 6. 1J m./hr., 4J m./hr., EXERCISES XXXV. p. 168. 1. 97J; -975 sq. ins.; 4-875 sq. ins. 2. 4; 6 sq. ins. 3. (i) 285, 385. (ii) 328'35, 338-35. (iii) |(200-30A+/i2), jj (200 + 307 + ft2). (iv) 333^. . 20*. [inside rects.{^| 2 , outside EXERCISES XXXVI. p. 171. 1. 108. a. 121J. 3. 13J. 4. 28f. 6. 97$. 6. 4. 7. 23 J. EXERCISES XXXVII. p. 172. 1. 905J. a. 4f. 3. 152J. 4. 274J. 6. 3696. 6. 48^. * &4 " 4 . EXERCISES XXXVIII. p. 177. (i) A = -x4 + 2x. (ii) A = |a:4 + 2a; (iii) A=|a; 4 +2ar-16. (iv) 11728|. ANSWERS . 409 EXERCISES XXXIX. p. 181. [In 1 9 an arbitrary constant should be added.] 2 . 2x __. 3. *3 + + c*. 4. .. 6 . -^. 7. 7* . 8. ^* 2 -~. 1O. 223J. 11. 4-032. 12. ^ + 2c. 13. L 14. 1164*. O J 15. -0695. 16. 56. 17. 614^ J. 18. 2. 19. (i) 278. (ii) 225-9875. 2O. (i) 56. (ii) 56. 21. 2-431. EXERCISES XL. p. 183. 1. 2|. (ii) 5|. 2. (i) 4. (ii) 12. 3. (i) g. (ii) ^. k-h 6. 211i. 7. -- + 2 C2 ,. 8. p = l, q = EXERCISES XLL p. 187. 1. 5180. 2. (i) 6055. (ii) 5180. 3. (i) 5215. (ii) 5180. 4. (1) 3355. (2) (i) 4046-41. (2) (ii) 3381-04. (3) (i) 3382-91. (3) (ii) 3355-04. EXERCISES XLIL p. 19tt 2. 1-4. 8. |. 4. 14. 6. A. 6. a = 0, b= -8, c=6. 4. 7. a = h (h?-k*), 6=A;2-3/i2, c=3A. Each area = O. 2pk + ~. 10. 1(^ + 4^+ AS)- " 12. 8j\, 1J. 13. <& 14. ' 2 = 4< 410 CALCULUS FOR BEGINNERS EXERCISES XLIII. p. 193. a. -69315. 3. -1419, --2027. EXERCISES XLV. p. 200. 1. 52, 58, 49. a. 32, 32, 32. 4 . (i ) Ho. (ii) ('^H''**). 6. 14. .. 3. EXERCISES XLVL p. 202. 1. 18J, 18-4. a. . 3. -69315, -6989. 4. 170H- EXERCISES XL VII. p. 204. 1. 0. 3. (i) 6. (ii) 0. 3. 3J, 2-75. 4. 4J. 6. |a*. 6. 11J. EXERCISES XL VIII. p. 206. (i) 112J, 112-325, '155%. (ii) 71*, 71-465, -05%. EXERCISES XLIX. p. 208. 1. 61 ft. 2. 5131 ft. EXERCISES L. p. 227. 3/2 7/8 3. 304 ft. 3. 1117i ft./sec. 193ift./sec. = ^1 + -11 + 20, 2544ft. 2t o 4. 1-19 x 10* ft. Ibs. wt. 6. 560 ft. Ibs. wt. 6. 9000 Ibs. wt. 7. 750 Ibs. wt. 8. 375 Ibs. wt. O. * ( b ~ ) f^ + g _ {] ft lbs wt . I \_ i J 10. 2,700,000 ft. Ibs. wt. 6-71 knots. 11. 192 ft./sec. 346| ft. 771/,)2y2 13. - - ft. lbs. wt., ur ft./sec. 13. 10,560,000m It. Ibs. wt. 9g 14. 60 lbs. ANSWERS EXERCISES LI. p. 232. 1. 1207r=376-99 c. ins. 3. irr 2 Ac. ins. 3. ^=67-02 c. ins. o o 4. ~(3r-fc). 6- (i) 261ir=820c. ins. (ii) ir= 2060 c. ins. o o qo_. 6. (i) 16?r = 50-27. (ii) 247r = 75-40. 7. =33-51. 8. ^. 0. 2 = 2-84 = l-862. 10. =5-x*. 11. (i) 120jr = 37G-99. (ii) 60ir=188'5. EXERCISES LII. p. 237. 1. (a) 666 Ib. ft. 2 (6) 246$ Ib. ft. 2. M ^. 3. M ~ . O XA r\ /i r-\ R /i iv/i 2 4. (i) M g . (n)M-jg . 6. -Ma 2 . 8. (i). Wr. 9. . 10. 2-025, 1-62, 1-35 ft. 11. M-. Ift. M. . 13. 85-7 ft. Ibs. wt. ^ A 90 14. (i) ^M6. (ii)|Ma 3 . 16. 6-34M. EXERCISES LIII. p. 244. .. (7-71,85.34). 8. . / 412 CALCULUS FOR BEGINNERS a 7. from centre, on a radius perpendicular to diameter. oTT 8. Q a from centre, on a radius perpendicular to base. 8 EXERCISES LIV. p. 248. 2. M 2 +c2. 8. . 2Ma 6. M-t-c'. 6. EXERCISES LV. p. 249. 1. On vertical line of symmetry (i) -5-, (ii) '-= . - below the top side. O tC T O 3 1 2. On the median drawn to the base a (i) T ft below, (ii) h ahove, . 2c + ft ft 3. On the line joining the mid-points of the parallel sides ... o4-37> ft . ft (a + 3fc) + e(2a + 4fc) j, I 1 ) ST . (") r-; STT nr- rrv below the side a. l 'a--2b 2 ' ft (a + 26) + c (3a + 36) 2 EXERCISES LVt p. 253. 2 1. - Tra 8 . [Area of base of cylinder, same height and volume.] 4. 5. 420 Ibs. wt./sq. ft. 6. (i) 7. (ii) 7f . ANSWERS 413 MISCELLANEOUS EXAMPLES ON CHAPTERS VII AND VIII. p. 254. A. I. 2 ^f^(38.41); 2 -^^- 2 (3-75). a. 300,. 3. 95*. o 6. -r of the way from A down the median. 5 of the way from A up the median, o B. 2 Q 3 2s 1 . =a?--x 32 500 /101*\ /Q016\ (18if); -^(92if). 3. -3-;^-. 4. -5, 0, 4; 42-04, -28-56; 45, -20, 36; 135&, 74|. 3. M - (where A is the altitude from A) = ^ s (s - a) (s - 6) (s - c). 4. 139 ft. Ibs. wt. ; 46-5 Ibs. wt./sq. ft. 6. D. I. 30; 4; 26. 2. (i) 20J. (ii) 20J. 3. 10 inches. 4. 3| m. 6. 8-713 x 108 gm. cm.2 ; 1-075 x 10 9 ergs. CALCULUS FOB BEGINNERS E. 2. 135*- = 424-1 c. ft. x /22 7 5=4'743 ft. 3. tan-i^- = 4ll'; 0-122; (1-61, 1-12). 4. ^ (2aR - a2) ft. Ibs. wt. ; ,^ ft. Ibs. wt. ; 35 : 32. (R + a) P. 1. (i) g*' + c. (ii) 37 J. 3. -4514, 2-374 ; 1160. 7 O1 73 I.-*, ?. 0;?4. 4. ~(P- G. 1. i, 12isq. ins.; , 2iins.; ^, 6H sq. ins. 6. 7r = H. 1. 7{, IS, H; . 2. (i)27. (ii).-y- (lii) 291-6T=916. (iv) |, '7 o.ft. 6. 360 ft. Ibs. wt. J. , 12 9. = 3-82ins./min. 3. 439 ft. Ibs. wt. 6. r+o aib3. w t. r ANSWERS 415 EXERCISES LVII. p. 268. 3. 2, -1-444, -3-560, 0-232, -3598, 0, V!3 = 3-606, -^13= -3-606. y = 1x-\, j/ + -144.r=4-748, y + 3-598* = 7'77. y = 3-606* + 3 -545, y = 3-606. 5. (i) -5 sin 5*. (ii) 6 cos 2*. (iii) cos 4a; + sin 3x. (iv) n(acosnx- & sin no;). 6. a cos art, a sin art, -aaisinart, -aw 2 cos art, aw cos art, -aw 2 sin art. 9. (i) -sinx. (ii) -cos a:. 1O. (i) -a>i 2 cosnx. (ii) -an 2 sinnx. 11. (i) ^13, -^13. (ii) ^41, -^41. 4 12. (i) -cosz + c. (ii) sinx + c. 13. 2, - = 1'273 sq. ina. 14. (i) 0-4598. (ii) 0-8415. (iii) 1. (iv) 1. 15. (i) 0. (ii) 0. (iii) 0. (iv) -2. 16. (i) --COS7UC + C. (ii) -sinnx + c. n n 18. [An arbitrary constant to be added in each case except those marked *.] 1 1 34 (i) --cos2a;. (ii) -siu2x. (iii) - ^ cos 2x + - sin 3ar. (iv) ar + -sin2x. (v) 0*. (vi) 0*. (vii) - Z cos2ar. 2i 4 .. x sin2 . x sin2a; . . IT* , ., T* ( T1U ) _ _ -^ . () - + __. ( X ) - . (xi) . 19. , or (1-5708, 0-3927). 2O. . 21. [Add an arbitrary constant.] ... l/cos5x \ .... l/cosSa; sin5x -5- cos(p-g)x p-q \ )' 1 fsin(p + q)x sin (y-g)A V P + q ~Y-q )' sin(p-g)x\ P-g /' 416 CALCULUS FOR BEGINNERS 22. -5150. 23. -96 sq. ina. 24. 6. 25. -791. 26. - 1-745 ft./sec., - 4-685 ft./sec. 27. (i) - = -6366r. (ii) ^=-7854r. 7T 4 EXERCISES LVIII. p. 274. 4. (i) -n cosec 2 n*. (ii) + n sec nx tan nx. (iii) - ;i cosec nx cot n.r. 6. An arbitrary constant to be added in every case. (i) tan*, (ii) -cot a;, (iii) sec a;, (iv) cosec*. (v) tan*. (vi) - cot x. (vii) sec x. (viii) - cosec a;. (ix) - tan 3*. (x) tan x - x. (xi) - tan 3* - x. o B. 2500, 1250, 150-75, 1-523, 0-381, 0-095 ft./sec. ; i, |, 5-53, 547, 2189, 8754 ft./sec. EXERCISES LIX. p. 277. BOA nv 96 1. 15-535 ft. 2. -1396 ft./sec. 3. 35-S mins./sec. or radians/sec. 4. 9 ft./sec.; 17 11' per sec. or ^ radian/sec. EXERCISES LX. p. 279. 3. (i) -(2* + 3)sin* + 2cos*. (ii) * 3 sec 2 * + 3x 2 tan x. EXERCISES LXI. p. 281. 1. *cos* + sin*. a. * 2 sec 2 * + 2* tan *. 3. 7 4. 2 cos 2*. 5. T*~*T. e. sec* tan*. 7. 4* 6 cos* + 20* 4 sin*. 8. -15* 2 sin 3* +10* cos 3*. 9. -2 cos* sin*. 1O. (i) - 10 cos 5* sin 5*. (ii) 10 sin 5* cos 5*. cos x cos * 2 sin * ANSWERS 417 EXERCISES LXII. p. 282. 2 1. 4r 3 + 12.r 2 + 6. 2. 1--=. 3. x sin 2x + a: 2 cos 2ar. 4. 3 (4x EXERCISES LXIII. p. 285. 1 - a; 2 x cos x - sin x 2 ' "' Bin x- x cos a; sin 2 * (x 2 - 2 cos a; sin a; tj m _ O , ^ (1 + sina;) 2 ' ' cos 2 ar ' EXERCISES LXIV. p. 286. 4. sec 2 4x. 2 N /x 2 + a; + l' '. 2cos(4x + 7). 8. 5 ji IO. -(2 sinx + 3 cosx)*. EXERCISES LXV. p. 288. 1. -5sin5x. 2. 3sec 2 3x. 3. 3 sin 2 x cos a?. 1 4. 2xcos(x 2 + 3). 6. ^ y-cos^/x. 6. 14(2x + 3). 7. 4(6a; + 5)(3x 2 + 5x + l) 3 . 8. 21 sin 2 (7x + 5) cos (Ix + 5). EXERCISES LXVI. p. 290. lf C08X . a. j-coajx. 3. 4 sin 3 x cos x. 2^/sina: 2 v a; 4. 4cos4x. 6. 4x 3 cos(a; 4 ). 6. n sin"- 1 x cos x. M. C. 27 418 CALCULUS FOR BEGINNERS 7. ncosnx. 8. ncosa?. 9. -nco8 n ~ 1 xsinx. 1O. -nsinnar. 11. 4cos(4x + 5). 12. 2 (3 sin x + 4 cos x) (3 cos x -4 sin a;). 13. 2 tan x sec 2 . 14. 2 sec 2 a; tan x. 15. n (o sin x +b cos x)* 1 " 1 (a cos x- b sin z). sec 2 * 16. ni&n n ~ l x sec 2 *. 17. nsec B x tanx. 18. 2 A^ tan j; 1 la 8* + 3 21. 23. 120 (6x+ ip. 24. 26. - .-s -rj. 27. na(ax + b)*~ 1 . (x 2 +l) 8 28. r a(ax + ft) . 29. - sin x. cos (cos x). 3O. - cos x. sin (sin x). 3 a cos (ax + b) 34. acos(ax + &). 35. v ' . 36. 12sin 2 (4a; + 5)cos(4:c + 5). 37. wasin"- 1 (aa; + 6)cos (ax + b). 38. na tan n-1 (ax + b) sec 2 (aa: + 6). SO. na sec" (ax + b) tan (ax + b). (ix + 5) cos (2z 2 + 5x + 6) 4O. s =-^. 41. 3 sin x cos x (a sm x - b cos a;). 42. 15 sin as cos a; (a sin x - b cos a;) (a sin 3 x + b cos 3 a;) 4 . 7 sin x cos a; 43. 44. - lo cos* 3a; sm 3x. ^3 sin 2 x - 4 cos 2 x 45. 6a;tan 2 (x2) S ec 2 (x 2 ). 46. n6x n -icos (a + 6x). 5 sec 2 2x + 6 sec 2 2x tan 2.T 47. - . . 48. mnx n ~ l sin" 1 " 1 (.c n ) cos (x n ). 49. 3 cos x - 12 sin 2 x cos x = 3 (4 cos 3 x - 3 cos &) = 3 cos 3x. 50. -12cos 2 xsinx + 3sinx = 3(4sin 3 x-3sinx)= -3sin3x. 51. cosw.^. 52. 3 sin 2 u. cos w-^*. 53. 2u.^. ax da; dx ANSWERS 419 64. *u-l 65. *-..- 56. . ax 2*Ju dx it 2 da; cos u du du 67. . . 68. - n cos"" 1 w . sin 3- . 2^/sinu dx * 59. CO.,/.. 60. -,==.. V" V **" 2 o An arbitrary constant to be added from 61 75. 61. 5 sin 3 *. 62. - rain** 1 *. 63. -- rcos w+1 z. n + 1 n + 1 64. --cosnx. 66. -siuna;. 66. -tan war. n n n 67. -t&nnx-x. 68. --cos (ax + 6). 69. -sin (ax + 6). n a a ^ 70. 73. ^ec^x. 74. --. 76. EXERCISES LXVII. p. 294. 2z a 2 - 2x 2 2 sin x 1. ; r. 2. . 3. - 4. 2xcosx(cosx-xsinx). 6. (2x + 1) 6 (3x + 4) 4 (7 6 7 . -^ 8 . ( a 2_a;2)t 2*Jx 9. (ax + 6)"*-i (ex -f d)"- 1 {(w + n) acx + mad + nbc}. 10. 8 sin 2 x sin 4x. 11. 4 (^ ~ x2 ~ 4 ) . - 15 cos 2 5x sin 5x - 12 sin 3 2x cos 2x 12. - - . 13. ^2 cos 3 5x - 3 sin* 2as 272 420 CALCULUS FOR BEGINNERS ,,,_i _i f du dv\ 14. n sin"" 1 x cos (n + 1) x. 15. u m l v n 1 1 mv +nu ] . V (IX CLxJ 16. sin" 1 " 1 u cos"" 1 u ( wi cos M cos v- n sin w sin v 1 . V &3/ (tXJ 5 ^/sin 5x (15 cos 5x + cos IPs + 2) __!L__ f* j. /; (du du\ r ^~ M ^j v 2 I EXERCISES LXVIH. p. 296. t 2 _ cos 9 + b sin g ' ~ usina 4. 2?w. 9. - seconds. EXERCISES LXIX. p. 298. 5 2cos2j; 4-3 1. 5x8. 2. sec 3 x. 3. - . 4. ; -s. 6. -cotz. 3 coso; bt + 2 EXERCISES LXX. p. 299. M& Ma 2 M - ; 8 - -"' -~ 4. 7ra 3 io [to = weight of 1 c. ft. of water] ; z ft. below centre. 6. EXERCISES LXXI. p. 300. ANSWERS 421 EXERCISES LXXIV. p. 309. 3 o. EXERCISES LXXVIII. p. 321. 1 . (i) W. (ii) I (i) Se 3 *. (ii) 2ff.*. (iii) 2e 2 *+3. (iv) ae ***. (v) (vi) e x (x + 'i). (vii) e z (x 2 + 2a;). ('iii) e* z (kx n + nx n ~* (ix) e* (sin x + cos x). (x) e 2z (2 sin 3x + 3 cos 3x). (xi) e aa; {a sin (6ar + c) + 6 cos (fix + c)}. (xii) e~ -3* { - -3 sin 2x + 2 cos 2o; } . (xiii) ,. du (xiv) e. . (xv) (xvi) e3*(3sin*5a; + 20sin35xcos5x). (xvii) 7 a: log e 7=7 :B xl-9459. 7. [Add an arbitrary constant.] (i) e*. (ii) \e**. (iii) le**. (iv) \e*\ (v) e tanz . O A (vi) -e*. (vii) y- s=7*x-5139. le ' 8. -e 2 = 47-209; 2-36 sq. ins. 1O. -0-237. EXERCISES LXXIX. p. 325. 13 3 2. (i) -. (ii) -. (iii) 5 =. (iv) -tanas, (v) 2cosec2x. X X OX "T* 6a; + 5 .. -910 F 1 H ( V1 ) 0-5 5 r ( V11 ) 2cot2x. (vui) = : I. 3z 2 + 5x + l v ' v ' x L log e 3j 422 CALCULUS FOR BEGINNERS (ix) cosecx. (x) secx. (xi) secx. (xii) ^ ' ^ 2ax + b .. 1 du (XX) - r . (XXl) - 5 - r - . (XXll) - . -j- . v ' z 2 6a; c dx 3. [An arbitrary constant should be added in each case.] (i) log e s. (ii) log e x or log e jx. (iii) log (2* + 3). (iv) - log e cos x or log e sec x. (v) log tan ^ (vi) log e tan + or log e (seca;+tana:). (vii) (viii) -log(ax + b). (ix) log e (x 2 + 4* - 7). (x) a & (xi) log.^ + ^^). (xii)lo g< . (xiii) a: (xiv) xlog e x-x. (xv) + 2x + log e x. (xvi) - + 3x-7log e (x + 2). (xvii) j + * 2 - a; + 5 log e (x -3). 6. (i) and (ii) log <J 6=l'792. 6. 219 ft. Ibs. wt. 7. 219ft. Ibs. wt. 8. (i) log. 3 =1-0986. (ii) \ log e ^= 0-1062. (iii) ~ log. 3 (iv) i log. 2 = 0-3466. (v) log. (^2 + 1) = -8813. IB 10 . , or (1-8484, 0-21875). 11. ^2=0-6931. 12. 3030. 13. Max. value (-ll + 301og < 3) = 10-98, when *=1. Min. value ( - 27 -f 30 log, 5) = 10-64, when x = 3. ANSWERS 423 EXERCISES LXXXI. p. 334. 3.34.1 (37s 4 + 81* 3 + 24*2 + x-3). (96*6 + 24x - 19* 2 + 32* + 12). /cos 2 a; \ 7. (Binx)c** ( sm x . log e sin x J . EXERCISES LXXXII. p. 340. 1. o;=2e3; 16206ft.; 1-304 sees. 2. 81-1 Ibs. wt. ; ^AOP=0644'. 3. After 6-575 minutes. 4. h=- j; (i) 2 mins., (ii) 3 mins., (iii) 3-32 mins. 6. v - 850 = (u - 850) e~ w . [ u ft. /sec. is speed when t = 0.] 6. 48-634 ft. [=50(1 --).] J -32_ 7. y = -e 12800 . [j/ sq. ins. is cross-section x ins. from the lower end.] EXERCISES LXXXIII. p. 342. 8. (i) y = A cos 2z + B sin 2x. (ii) y = Ae 2iE +Bc- 2a: or A' cosh 2a; + B' sinh 2o5. 7 3 4. ?/ = 5cos2a; + 26in2a;. 6. j/ = - 2 6. (i) 60 ft. /sec. (ii) 75 ft./seo. 424 CALCULUS FOR BEGINNERS * '* V 10. (i) (iii) Ax+B. 11. (i) y = ke-*+Be-* x . (ii) j/ = e-2*(Acosx+B sinx). (iii) y = e-2* (Ax + B). (iv) y = Ae** + Be*. (v) y = e Aeos*+Bsin. (vi) y = e* x (*x+ B). 76 3 12. -3. 13. 3. 14. --g, - 6l . 16. j. / /3 /3 \ 11 18. (i) y = Ae-*+B- 3a! + 2. (ii) y = e& ( A cos ^- x + B sin ^- a; J + y (iii) y = e MISCELLANEOUS EXAMPLES ON CHAPTERS IX XI. p. 344. A. 2. 4i- 2 + 7 = 5 ' 149 ; | + log, 2 = 2- 193. 4. 2Mft.lba. v^ 6. -7033; -7034. B 1. (i) 6 sin Bx cos 3x. (ii) -^x~^ (8x sin 4a; + cos4a;). (iii) -tan-. (iv) e 22 (ii) --cos3x + -sin6x + c. (iii) 1. 3. 4 ft./sec. ; 4 ft./sec. 2 ; both towards O if A is moving away from O. 4. when x = . me m 0. fl 2_ j-2 1 1. -at ; a. 2. ^; -500. 3. -.70"; -87 ft. ANSWERS 425 D. 2-2* 2 1 1. (l) - -r. (ll) = :. 2* (1 +*)*(!-*)* 2. (2, 1), (-1, 1); 2x-3j/-l=0, x-3y + 4=0. 3. ^13-2 = 1-606. 4. -8933r. 6. (l-c-"" 2 ); 14-2. E. 1. (i) log,,a; + l, -. (ii) e x (sino:+cosa;), 2e*coacB, x ... xcosx sinx , (iv) 4 sin 3 a; cos x, 4 sin 2 a; (3 cos 2 x - sin 2 x). 2. - 6 tan C . w (where & is CA and C the angle ACB). 4. 7 : <: from centre of first sphere, ; c from centre of second. a^ + 67 6. 2 + 2jr =63-496. P. 1. T-g ; TTg, - 6 cos 2 2x sin 2ar, e 01 (a sin fa; + 6 cos bx), 4 sec 4x. PN 2 2. c; PG= - . 3. 25-1 ft./sec., 24-1 ft./sec., 18-85 ft./seo. c 0. ?rG cos a ft. Ibs. G. (iv) 4sec 2 4a;. (v) e-3* (1 - 3*). (vi) a. 18 1 2. -^=. 4. (i) 22 (ii) 0-3054. (iii) 0-2456. (iv) 5. v/ ^ & w (v) 0-0429. (vi) 0-3466. 5. - 426 CALCULUS FOR BEGINNERS H. (iii) 144x6 _ 280s* + 168x 3 -120.r2 + 78o;- 15. (iv) 5 sec 2 5*. 1 /j4 _L /j2 T 2 _ 4.J-4 (v) - ^. (vi) 6sec23*tan3*. (vii) - - (1 - x 2 )? V a 2 - a; 2 a. -0-3690 and -1-7602 Minimum. 0-3690 and 1-7602 Maximum. 3. 38-2 ins./sec. 4. be'-* ( 3cos3- jjsin3n . 6. 5 \ J / ^ dd> irn cos Ar airn s < a. -j7 = -^ . ; 3-= - -57T- r^ > 1>714 radians/sec. ; dt 60 cos</> dt, 30 co -3-715 ft. /sec. 4. 6-28 ft. Ibs. wt., 6-98 ft. Ibs. wt. J. a. 1-8083. a. c2('l+Vl = c 2 x 1-1353. V V 2r sin 3 o 2r sin a * ? sin a , 4. ^-j : ; ; from centre. 3 (a - sin a cos a) 3a a K. 6. 2a 2 ; (l-513a, l'003a). L. 8. 2, 2, -|. a. 02(^/2 -l)=0-4142a. 4. ( 6, J ; ft. /sec. making 210 with OX. (3-96) ft./sec.2 making - 13 54' with OX. 6. 59-3 ft. Ibs. wt. ANSWERS 427 K/7 1. 6tan3xsec3x, x (1 + 2 log 4x), 2 (5*+3) (2x2 + 7)" (170x2+72x4- 175), 3x(a2+*2)i, a. 62 45'; - = 0-3183. 3. 11-16 ft. - from A. 8-85 ft. from B. 4. A = 100^13 = 360 -6. g = tan- 1 l-5=c.M. of 5619'. 6. aw sin wt, aw 2 COB w. N. . to to ^.fo/.^fl M. 2 . a *~ b 2 + 2~ ' 2 2 - IK 2 ~ z ' ~>T 2. ^Q radians/sec. 4. (i) --350. (ii) -. (iii) ^. (iv) -0055. (v) -0055. 6. 3-165a 3 . O. (iii) (12a^ + 18* + 5) (6*2 + 3). (iv) y cos a: or e" n!B . cos *. & ^T- ? - (vi ) 55-5- < vii ) 4. 20 log. 3 = 21-97 ; (7-28, 1-52) ; 1-84. /\ ^ /* \ ^ \ -* /% -^ 6 ' (l) T' (ll) T* (m) ' (W) 2' P. I. (i) x 2 cos x + 2x sin *. (ii) 2xe 2a! (a; + l). (iii) ^(aco&Zbx-ZbcoBbxsinbx). (iv) - ^ . . sin x + x cos x ... 1 . x .... cosjx 00 / -, (vi -o sm o- ( vu ) , -=' 2 ^/a; sm x 4 a. 3x + = 6. 6. 16000 ft. Ibs. wt. 428 CALCULUS FOR BEGINNERS Q- B. 1. - 1-482 ft./sec. ; 6-187 ft./sec. 2 4. (i) |. (ii) -0243. (iii) 0. (iv) 0. S. a. IJins./sec.; radians/sec. 4. e5 = l-649. oo r 2 6. 7- below centre. 4c T. x 3 a 3 1 . (i) -g- + log e a; + c. (ii) log a sin x + c or log e A sin at. (iii) J. (iv) 1-^ = 0-2146. (v)^ 3. 8ins./min.; 3*175 ins./min. 3am(2-7ft3) 3a(l-2m 3 ) m(2-m *' 6 . 7 . 82ins . EXERCISES LXXXIV. p. 373. e. 2-095. 7. -0-315, 0-446, 1-069. 8. 0-4142. . 0-85138. 1O. 12-03. 11. 11-34 ins. 12. 1-1839. 13. 1-80, -1-13. 14. 4-167 ins. 15. 28 23 / . 1. 1-166. 17. 2-289. 18. 0-657. 19. 4-536. SO. 3-945. ANSWERS 429 EXERCISES LXXXV. p. 379. [Add an arbitrary constant in each case.] 1. - 5 cos 8 x. 3. --cosz 2 . 3. log e (x 2 + 3x + 5). O a 10 2x + 3 6. 21og e (jc 2 + 3a;+5) + - 7T - r tan- 1 - Trr . 7. -cos (log, 8. e 8 -" 8 . 9. log. (1 + log.*). 1. , ,2x 1 . 10. 2a; 11. " T coth r- or log- . 13. 13 sin x - 5 sin 3 x. 14. 5 tan 6 ar. 16. s tan 2 x + log., cos x. O U & 2 . ,s cos 3 a; cos 5 a; 16. -(2a + a;)*. 17. a~ + 5 1 8. (i) - 2 coth-i (2x - 3) or log e ^| . a/ ~~ A 2 a; (ii) - 2 tanh- 1 (2x - 3) or log e = . 3 1 19. ^logj^ + lJ + tan-Jx. 3O. - - log e cos (era; + b). 2 1 / 31. --cos s a; + cosx. 33. --*/2-3a: 2 . as. _t a nh-i- + lo g (3-x2. 34. 38. 2 tanh- l Jx. 3. 30. log. 2.^* + --(log. a:) 2 . 430 CALCULUS FOR BEGINNERS EXERCISES LXXXVI. p. 381. 1 f jZ - 2^3 V a ax \. ' " 6 (* -3) ' 2ax ~ 1 ~ , , x ax 6. - 15 8. ^f x 1 14. sinh" 1 -. 15. - [log, (sec a; + tan x) + tan a; sec x] (i J EXERCISES LXXXVII. p. 383. . 2. e*z-l. 3. 1 . _ 4. a; tan- 1 x -jrlog^l-l- a; 2 ). . a;sin~ 1 a;+ Jl-x\ 6. - x cos a; + sin x. a; -^rn 8in(7K+n)a: ) + - ( x cos (m - ?t)ar -- sin (TO - n) x } I . m-n\ TO-n ' JJ B. - 13. -g p. e"* [a sin (fee + c)- 6 cos (for + c)]; -jp p . c * [6 sin (bx + c) + a cos (bx + c)]. 16 357T F6 . 4 . 2 7.5.3.1 iT\ 35' 256' l_7.5.8 ; 8.6.4.2'2_T ANSWERS 431 MISCELLANEOUS EXAMPLES ON CHAPTER XIII. p. 385. 3. ~- + 2x-2log e (x + l). 4. tan- 1 ^. 5. log e a.log e x + -(log e x)*. e. log, (1 + sin*). 3 . , , 1 , 8mh or log 8. ^V^+sM* 3 - 2 " 2 )- 1O. x-o 11. x^5( O JO 12. -sin 4 2ar. 13. -^x cos 3a; + 5 sin 3a5. o o J 122 8 14. ~-a; 2 cos3a; + -xsin3a5+2=cos3a;. 15. - (3- 16 ' -3-2x)(27 + 30,). 17. 18. l ogs seo + a 19. - tanh-i - + log e (3 - x^J. .. / a\ ~, < i + 241og,(x-2). 21. log, t=log e tan-. 22. - log,, tan I - I =log. tan ( -r + - } . 23. = sin~ 1 - . \4 2/ \4 a/ a 24. - = -tan- J -. 25. 1-568. 26. 0-430. 27. -0-148. a a a 28. |3. 29. 0-0266, 3O. 0-430 f^l . oo I 2ob I 81. 5tan-i = 0-2145. 82. log,2=0'2310. 33. y = -.x^. 432 CALCULUS FOR BEGINNERS 36. - - tan- 1 a " 2 . 36. v = vi tanh . 37. ^-^- ft. Ibs. wt. 38. If v ft./sec. is speed, and t sees, time, _110 9 A _198 ~~T~' '~3080' ' -"86"* 41-77 m./hr.; 49-9998 m./hr. EXERCISES LXXXIX. p. 393. 2. (i) tan0. (ii) -cote, (iii) cotfl. (iv) -tantf. ira a ira s a 3 1O. r=ae is the same as r=e * r=6e is the same as r=e Table of the Exponential and Hyperbolic Func- tions of Numbers from O to 2-5, at Intervals of !. X ex e~* cosh x sinh x tanh x 1-000 1-000 1-000 1 1-105 905 1-005 100 100 2 1-221 819 1-020 201 197 3 1-350 741 1-045 305 291 4 1-492 670 1-081 411 380 5 1-649 607 1-128 521 462 6 1-822 549 1-185 637 537 7 2-014 497 1-255 759 604 8 2-226 449 1-337 888 664 9 2-460 407 1-433 1-027 716 10 2-718 368 1-543 1-175 762 1-1 3-004 333 1-669 1-336 801 12 3-320 301 1-811 1-509 834 1-3 3-669 273 1-971 1-698 862 14 4-055 247 2-151 1-904 885 15 4-482 223 2-352 2-129 905 1-6 4-953 202 2-577 2-376 922 1-7 5-474 183 2-828 2-646 935 1-8 6-050 165 3-107 2-942 947 1-9 6-686 150 3-418 3-268 956 20 7-389 135 3-762 3-627 964 2-1 8-166 122 4-144 4-022 970 2-2 9-025 111 4-568 4-457 976 23 9-974 100 5-037 4-937 980 24 11-023 091 5-557 5-466 984 25 12-182 082 6-132 6-050 987 Table of Logarithms to Base e. o 1 2 3 4 5 6 7 8 9 I 095 182 262 336 405 470 531 588 642 2 693 742 788 833 875 916 956 993 1-030 1-065 3 1-099 1-131 1-163 1-194 1-224 1-253 1-281 1-308 1-335 1-361 4 1-386 1-411 1-435 1-459 1-482 1-504 1-526 1-548 1-569 1-589 5 1-609 1-629 1-649 1-668 1-686 1-705 1-723 1-740 1-758 1-775 6 1-792 1-808 1-825 1-841 1-856 1-872 1-887 1-902 1-917 1-932 7 1-946 1-960 1-974 1-988 2-001 2-015 2-028 2-041 2-054 2-067 8 2-079 2-092 2-104 2-116 2-128 2-140 2-152 2-163 2-175 2-186 9 2-197 2-208 2-219 2-230 2-241 2-251 2-262 2-272 2-282 2-293 log 10=2-303, log 102=4-605, log 103=6'908. M. C. INDEX The numbers refer to pages. Acceleration 19 ,, from gradient of speed-time graph 33, 67, 68 Approximate equality, test of 39 ,, solution of equations 362-373 Approximations 136-144 to f(x + h) 141, 196-198 Arbitrary constants 145-147, 173, 174, 181 Area of plane curve 160-192 by summation 161, 165, 166 from 5rW(*) 170, 177 OfX approximate rules for 184-186 sign of 187-190 ,, ; , from ordinates of integral curve 193-196 Area traced by moving ordinate 169-177 Atmospheric pressure as function of height 337 Average speed 2 ,, gradient 22, 23 rate of increase 12, 15, 16 ordinate 199-203 values 250-253 Centre of gravity 238-244 ,, ,, of plane area 240 ,, ,, ,, solid of revolution 241 ,, ,, ,, quadrant of circle 243, 244 ,, ,, ,, quadrant of ellipse 300 Centre of pressure 248, 249 ,, of circle 300 ,, ,, rectangle 249 ,, triangle 249 436 INDEX Centre of pressure of trapezium 249 Circular ring, thin 137 disc, moment of inertia of 235 Compound Interest Law 334-339 Cycloid 297 area of 299 Definite integral 179, 180 ,, approximation to value of 192, 193 represented by area 192, 219 Derived curves 94-97 Differential coefficient 51 ,, coefficients, higher 93 equations, Exs. on 342-344 Differentiation, meaning of 51 from first principles 47-71 of a* 69, 70 * 75-79 kx n 80-82 x* + c 80, 82, 83 sin x 261-265, sin- 1 x 304, 305, sinh x 331, sinh- 1 x 332, 333 ,, cos a; 268, cos- 1 * 306, cosh a; 331, cosh" 1 a; 332, 333 tana; 272-275, tan- 1 a; 306, tanha; 331, tanh -1 x 332 cot a: 274, cot" 1 a; 307, coth- 1 x 332 sec a; 274, sec" 1 a; 307 ,, ,, coseca: 274, cosec^a: 307 ,, sum 84 ,, product 278-282 ,, quotient 283, 284 ,, ,, function of function 285-290 ,, ,, inverse functions 301-307 ,, ,, implicit functions 308, 309 ,, when two variables are each given as functions of a third 295, 296 of rf 311, 314 ,, c* 313, 319 log,* 323; log a a; 323 list of standard results in 376, 377 Diminishing quantities, ratio of continually 40-45 dt , As . Jt and At 48 - 5 f and 55-57 dv Av INDEX 437 and 52-54, 58-65 fix Ax da -t- as speed 48 ^ ae gradient 58-61, 85, 86 dx dy ~- as rate of increase 52, 53 dx ~ as ratio of time-rate of increase of w to time-rate of increase of x 54 dx ^ , sign of 66, 71-73, 99 d3j dy and dx no meaning when separated 61, 62, 180 dy d z y Z. and , sign of, in relation to shape of curve 99 dx dx i At, meaning of 47 Distance from speed-time formula 205-210, 221 ,, graph 211-213 e, definition of 313 e, approximation to value of 319, 328-330 e as Lt fl + -V 339 e x , differentiation of 313, 319 Elasticity of volume 139 graphical representation of 140 Ellipse, moment of inertia of 300 centre of gravity of quadrant of 300 Equality, approximate 39 Equations, approximate solution of 362-373 Exs. on differential 342-344 Error, relative 139 Errors, approximation to value of small 136-144 Family of curves given by -^=/(x) 149 dx Frustum of cone, volume of 229, 230 Function 14 represented graphically 33 Functional notation 91, 92 438 INDEX Function of function differentiated 285-290 / (x + h) =/ (x) + hj ' (x) approximately 141 / (x + h) =f (x) + Jif (x) + ^ h*f" (x) approximately 196-198 Fundamental notions 1-46 Geometry, Applications of Differential Calculus to 87-90 Gradient 19-29 uniform 19-21 average 22-23 at a point 23-26, 38 ,, representing rate of increase 33, 34 of space-time graph represents speed 33 of speed- time graph represents acceleration G3 Graphical solurion of equations 362-366 Gyration, radius of 234 Harmonic motion, simple 269 Higher differential coefficients 93 Hyperbolic functions 330-333 logarithms 315 Implicit functions, differentiation of 308, 309 Inertia, moments of 233-237 , theorem of perpendicular axes 237 parallel axes 245-247 Inflexion, points of 127-133 Integral curve 193-196 definite 179, 180, 209 indefinite 180, 181 Integration 160-172 ,, meaning of 1GO-162 ,, as a process of summation 219 > anti-differentiation 219 of - 323 x list of standard results in 376, 377 some methods of 377-383 by change of variable 299 examples of 220-226 Inverse functions, differentiation of 301-307 operation, the fgiven ^ find y\ 145-152 INDEX 439 Limit, limiting value 9, 30, 38, 41, 48, 75, 162 Limits of definite integral 180 Logarithms, Napierian 314, 315 ,, of same number to different bases 315 log e x, differentiation of 323 Logarithmic differentiation 333 Maxima and minima 106-133 ,, tests for 108, 110, 123 ,, Exs. on 112-117 Mean ordinate 199-203 ,, values 250-253 Moments of inertia 233-237 of rectangle 233, 234 ,, circular disc 235, 236 ,, ellipse 300 Napierian logarithms 315 Natural 315 n x , differentiation of 311 ,, graph of 316 Negative speed 13 ,, rate of increase 16, 17 Newton's law of cooling 336 Pappus, theorem of 353 Parabola 88-90 Parallel axes, theorem of 245-247 Perpendicular axes, theorem of 236, 237 Points of inflexion 127-133 Polar co-ordinates 388-393 Product, differentiation of 278-282 Quadrant of circle, centre of gravity of 243 Quotient, differentiation of 283, 284 Eadius of gyration 234 Bate of increase- 10-17, 38 ,, ,, ,, meaning of negative 13, 16, 17 ., ,, ,, ,, ,, zero 13 ,, ,, represented by gradient 33, 34 Katio of continually diminishing quantities 41-46 44-0 INDEX Rectangle, moment of inertia of 233, 234 Revolution, solids of 228-232 sec a:, differentiation of 274 Shape of curve 99-101 Sign of area 187-190 Sign of ^ and ^ 56, 71-73, 99 ax dx* Simpson's rule 185, 186 sinar, differentiation of 261-265 Solids of revolution, volume of 228-232 Space-time graph 33, 35, 36, 66, 211, 213 Speed 1-10 ,, uniform 1 average 2 ,, at an instant 3-10, 37 given by gradient of space-time graph 33, 66, 67 Speed-time graph 33, 67, 211, 213 Sphere, volume of 231, 232 Square plate, expanding 15, 52, 53 Standard forms in differentiation and integration 376, 377 Stationary tangent 129 Tables 433 Tangent at a point of a curve 27 tanx, differentiation of 272, 273 Thrust on immersed area 223 Trapezoidal rule 184, 185 Trigonometrical ratios, differentiation of 261-277 ,, ,, inverse 304-307 Turning points 107-111, 129-132 Uniform gradient 20, 21 ,, speed 1 Volume of frustum of cone 229, 230 sphere 231, 232 Volumes of solids of revolution 228-232 Volume strain 140 Work done in stretching elastic string 213-218 ,, by expanding gas 224-226 Zero speed 13 CAMBRIDGE \ PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS. 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