Livro de cálculo - calculus made easy - silvanus thompson, Manuais, Projetos, Pesquisas de Química

Baixe Livro de cálculo - calculus made easy - silvanus thompson e outras Manuais, Projetos, Pesquisas em PDF para Química, somente na Docsity! The Project Gutenberg EBook of Calculus Made Easy, by Silvanus Thompson This eBook is for the use of anyone anywhere at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online at www.gutenberg.org Title: Calculus Made Easy Being a very-simplest introduction to those beautiful methods which are generally called by the terrifying names of the Differentia Author: Silvanus Thompson Release Date: October 9, 2012 [EBook #33283] Language: English Character set encoding: ISO-8859-1 *** START OF THIS PROJECT GUTENBERG EBOOK CALCULUS MADE EASY *** Produced by Andrew D. Hwang, Brenda Lewis and the Online Distributed Proofreading Team at http://www.pgdp.net (This file was produced from images generously made available by The Internet Archive/American Libraries.) transcriber’s note Minor presentational changes, and minor typographical and numerical corrections, have been made without comment. All textual changes are detailed in the LATEX source file. This PDF file is optimized for screen viewing, but may easily be recompiled for printing. Please see the preamble of the LATEX source file for instructions. CALCULUS MADE EASY: BEING A VERY-SIMPLEST INTRODUCTION TO THOSE BEAUTIFUL METHODS OF RECKONING WHICH ARE GENERALLY CALLED BY THE TERRIFYING NAMES OF THE DIFFERENTIAL CALCULUS AND THE INTEGRAL CALCULUS. BY F. R. S. SECOND EDITION, ENLARGED MACMILLAN AND CO., LIMITED ST. MARTIN’S STREET, LONDON  COPYRIGHT. First Edition 1910. Reprinted 1911 (twice), 1912, 1913. Second Edition 1914. What one fool can do, another can. (Ancient Simian Proverb.) CALCULUS MADE EASY viii Chapter Page XV. How to deal with Sines and Cosines . . . . . . . . . . . 162 XVI. Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 172 XVII. Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 XVIII. Integrating as the Reverse of Differentiating 189 XIX. On Finding Areas by Integrating . . . . . . . . . . . . . . 204 XX. Dodges, Pitfalls, and Triumphs . . . . . . . . . . . . . . . . 224 XXI. Finding some Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 232 Table of Standard Forms . . . . . . . . . . . . . . . . . . . . . . . . 249 Answers to Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . 252 PROLOGUE. Considering how many fools can calculate, it is surprising that it should be thought either a difficult or a tedious task for any other fool to learn how to master the same tricks. Some calculus-tricks are quite easy. Some are enormously difficult. The fools who write the textbooks of advanced mathematics—and they are mostly clever fools—seldom take the trouble to show you how easy the easy calculations are. On the contrary, they seem to desire to impress you with their tremendous cleverness by going about it in the most difficult way. Being myself a remarkably stupid fellow, I have had to unteach myself the difficulties, and now beg to present to my fellow fools the parts that are not hard. Master these thoroughly, and the rest will follow. What one fool can do, another can. CHAPTER I. TO DELIVER YOU FROM THE PRELIMINARY TERRORS. The preliminary terror, which chokes off most fifth-form boys from even attempting to learn how to calculate, can be abolished once for all by simply stating what is the meaning—in common-sense terms—of the two principal symbols that are used in calculating. These dreadful symbols are: (1) d which merely means “a little bit of.” Thus dx means a little bit of x; or du means a little bit of u. Or- dinary mathematicians think it more polite to say “an element of,” instead of “a little bit of.” Just as you please. But you will find that these little bits (or elements) may be considered to be indefinitely small. (2) ∫ which is merely a long S, and may be called (if you like) “the sum of.” Thus ∫ dx means the sum of all the little bits of x; or ∫ dt means the sum of all the little bits of t. Ordinary mathematicians call this symbol “the integral of.” Now any fool can see that if x is considered as made up of a lot of little bits, each of which is called dx, if you add them all up together you get the sum of all the dx’s, (which is the CALCULUS MADE EASY 4 much smaller by comparison is one second! Again, think of a farthing as compared with a sovereign: it is barely worth more than 1 1000 part. A farthing more or less is of precious little importance compared with a sovereign: it may certainly be regarded as a small quantity. But compare a farthing with £1000: relatively to this greater sum, the farthing is of no more importance than 1 1000 of a farthing would be to a sovereign. Even a golden sovereign is relatively a negligible quantity in the wealth of a millionaire. Now if we fix upon any numerical fraction as constituting the pro- portion which for any purpose we call relatively small, we can easily state other fractions of a higher degree of smallness. Thus if, for the purpose of time, 1 60 be called a small fraction, then 1 60 of 1 60 (being a small fraction of a small fraction) may be regarded as a small quantity of the second order of smallness.∗ Or, if for any purpose we were to take 1 per cent. (i.e. 1 100 ) as a small fraction, then 1 per cent. of 1 per cent. (i.e. 1 10,000 ) would be a small fraction of the second order of smallness; and 1 1,000,000 would be a small fraction of the third order of smallness, being 1 per cent. of 1 per cent. of 1 per cent. Lastly, suppose that for some very precise purpose we should regard 1 1,000,000 as “small.” Thus, if a first-rate chronometer is not to lose or gain more than half a minute in a year, it must keep time with an accuracy of 1 part in 1, 051, 200. Now if, for such a purpose, we ∗The mathematicians talk about the second order of “magnitude” (i.e. great- ness) when they really mean second order of smallness. This is very confusing to beginners. DIFFERENT DEGREES OF SMALLNESS 5 regard 1 1,000,000 (or one millionth) as a small quantity, then 1 1,000,000 of 1 1,000,000 , that is 1 1,000,000,000,000 (or one billionth) will be a small quantity of the second order of smallness, and may be utterly disregarded, by comparison. Then we see that the smaller a small quantity itself is, the more negligible does the corresponding small quantity of the second order become. Hence we know that in all cases we are justified in neglecting the small quantities of the second—or third (or higher)—orders, if only we take the small quantity of the first order small enough in itself. But, it must be remembered, that small quantities if they occur in our expressions as factors multiplied by some other factor, may become important if the other factor is itself large. Even a farthing becomes important if only it is multiplied by a few hundred. Now in the calculus we write dx for a little bit of x. These things such as dx, and du, and dy, are called “differentials,” the differential of x, or of u, or of y, as the case may be. [You read them as dee-eks, or dee-you, or dee-wy.] If dx be a small bit of x, and relatively small of itself, it does not follow that such quantities as x ·dx, or x2 dx, or ax dx are negligible. But dx× dx would be negligible, being a small quantity of the second order. A very simple example will serve as illustration. Let us think of x as a quantity that can grow by a small amount so as to become x+dx, where dx is the small increment added by growth. The square of this is x2 + 2x · dx + (dx)2. The second term is not negligible because it is a first-order quantity; while the third term is of the second order of smallness, being a bit of, a bit of x2. Thus if we CALCULUS MADE EASY 6 took dx to mean numerically, say, 1 60 of x, then the second term would be 2 60 of x2, whereas the third term would be 1 3600 of x2. This last term is clearly less important than the second. But if we go further and take dx to mean only 1 1000 of x, then the second term will be 2 1000 of x2, while the third term will be only 1 1,000,000 of x2. x x Fig. 1. Geometrically this may be depicted as follows: Draw a square (Fig. 1) the side of which we will take to represent x. Now suppose the square to grow by having a bit dx added to its size each way. The enlarged square is made up of the original square x2, the two rectangles at the top and on the right, each of which is of area x · dx (or together 2x · dx), and the little square at the top right-hand corner which is (dx)2. In Fig. 2 we have taken dx as quite a big fraction of x—about 1 5 . But suppose we had taken it only 1 100 —about the thickness of an inked line drawn with a fine pen. Then the little corner square will have an area of only 1 10,000 of x2, and be practically invisible. Clearly (dx)2 is negligible if only we consider the increment dx to be itself small enough. Let us consider a simile. CHAPTER III. ON RELATIVE GROWINGS. All through the calculus we are dealing with quantities that are grow- ing, and with rates of growth. We classify all quantities into two classes: constants and variables. Those which we regard as of fixed value, and call constants, we generally denote algebraically by letters from the be- ginning of the alphabet, such as a, b, or c; while those which we consider as capable of growing, or (as mathematicians say) of “varying,” we de- note by letters from the end of the alphabet, such as x, y, z, u, v, w, or sometimes t. Moreover, we are usually dealing with more than one variable at once, and thinking of the way in which one variable depends on the other: for instance, we think of the way in which the height reached by a projectile depends on the time of attaining that height. Or we are asked to consider a rectangle of given area, and to enquire how any increase in the length of it will compel a corresponding decrease in the breadth of it. Or we think of the way in which any variation in the slope of a ladder will cause the height that it reaches, to vary. Suppose we have got two such variables that depend one on the other. An alteration in one will bring about an alteration in the other, because of this dependence. Let us call one of the variables x, and the CALCULUS MADE EASY 10 other that depends on it y. Suppose we make x to vary, that is to say, we either alter it or imagine it to be altered, by adding to it a bit which we call dx. We are thus causing x to become x + dx. Then, because x has been altered, y will have altered also, and will have become y + dy. Here the bit dy may be in some cases positive, in others negative; and it won’t (except by a miracle) be the same size as dx. Take two examples. (1) Let x and y be respectively the base and the height of a right- angled triangle (Fig. 4), of which the slope of the other side is fixed x dx y y dy 30◦ Fig. 4. at 30◦. If we suppose this triangle to expand and yet keep its angles the same as at first, then, when the base grows so as to become x+dx, the height becomes y + dy. Here, increasing x results in an increase of y. The little triangle, the height of which is dy, and the base of which is dx, is similar to the original triangle; and it is obvious that the value of the ratio dy dx is the same as that of the ratio y x . As the angle is 30◦ it will be seen that here dy dx = 1 1.73 . ON RELATIVE GROWINGS 11 (2) Let x represent, in Fig. 5, the horizontal distance, from a wall, of the bottom end of a ladder, AB, of fixed length; and let y be the x y O A B Fig. 5. height it reaches up the wall. Now y clearly depends on x. It is easy to see that, if we pull the bottom end A a bit further from the wall, the top end B will come down a little lower. Let us state this in scientific language. If we increase x to x+dx, then y will become y−dy; that is, when x receives a positive increment, the increment which results to y is negative. Yes, but how much? Suppose the ladder was so long that when the bottom end A was 19 inches from the wall the top end B reached just 15 feet from the ground. Now, if you were to pull the bottom end out 1 inch more, how much would the top end come down? Put it all into inches: x = 19 inches, y = 180 inches. Now the increment of x which we call dx, is 1 inch: or x+ dx = 20 inches. CALCULUS MADE EASY 14 implicit function in x and y; it may be written y = x2 + 10 2 (explicit function of x) or x = √ 2y − 10 (explicit function of y). We see that an explicit function in x, y, z, etc., is simply something the value of which changes when x, y, z, etc., are changing, either one at the time or several together. Because of this, the value of the explicit function is called the dependent variable, as it depends on the value of the other variable quantities in the function; these other variables are called the independent variables because their value is not determined from the value assumed by the function. For example, if u = x2 sin θ, x and θ are the independent variables, and u is the dependent variable. Sometimes the exact relation between several quantities x, y, z ei- ther is not known or it is not convenient to state it; it is only known, or convenient to state, that there is some sort of relation between these variables, so that one cannot alter either x or y or z singly without affecting the other quantities; the existence of a function in x, y, z is then indicated by the notation F (x, y, z) (implicit function) or by x = F (y, z), y = F (x, z) or z = F (x, y) (explicit function). Sometimes the letter f or φ is used instead of F , so that y = F (x), y = f(x) and y = φ(x) all mean the same thing, namely, that the value of y depends on the value of x in some way which is not stated. We call the ratio dy dx “the differential coefficient of y with respect to x.” It is a solemn scientific name for this very simple thing. But we are not going to be frightened by solemn names, when the things themselves are so easy. Instead of being frightened we will simply pro- nounce a brief curse on the stupidity of giving long crack-jaw names; and, having relieved our minds, will go on to the simple thing itself, ON RELATIVE GROWINGS 15 namely the ratio dy dx . In ordinary algebra which you learned at school, you were always hunting after some unknown quantity which you called x or y; or some- times there were two unknown quantities to be hunted for simultane- ously. You have now to learn to go hunting in a new way; the fox being now neither x nor y. Instead of this you have to hunt for this curious cub called dy dx . The process of finding the value of dy dx is called “dif- ferentiating.” But, remember, what is wanted is the value of this ratio when both dy and dx are themselves indefinitely small. The true value of the differential coefficient is that to which it approximates in the limiting case when each of them is considered as infinitesimally minute. Let us now learn how to go in quest of dy dx . CALCULUS MADE EASY 16 NOTE TO CHAPTER III. How to read Differentials. It will never do to fall into the schoolboy error of thinking that dx means d times x, for d is not a factor—it means “an element of” or “a bit of” whatever follows. One reads dx thus: “dee-eks.” In case the reader has no one to guide him in such matters it may here be simply said that one reads differential coefficients in the follow- ing way. The differential coefficient dy dx is read “dee-wy by dee-eks,” or “dee-wy over dee-eks.” So also du dt is read “dee-you by dee-tee.” Second differential coefficients will be met with later on. They are like this: d2y dx2 ; which is read “dee-two-wy over dee-eks-squared,” and it means that the operation of differentiating y with respect to x has been (or has to be) performed twice over. Another way of indicating that a function has been differentiated is by putting an accent to the symbol of the function. Thus if y = F (x), which means that y is some unspecified function of x (see p. 13), we may write F ′(x) instead of d ( F (x) ) dx . Similarly, F ′′(x) will mean that the original function F (x) has been differentiated twice over with respect to x. SIMPLEST CASES 19 Numerical example. Suppose x = 100 and ∴ y = 10, 000. Then let x grow till it becomes 101 (that is, let dx = 1). Then the enlarged y will be 101 × 101 = 10, 201. But if we agree that we may ignore small quantities of the second order, 1 may be rejected as compared with 10, 000; so we may round off the enlarged y to 10, 200. y has grown from 10, 000 to 10, 200; the bit added on is dy, which is therefore 200. dy dx = 200 1 = 200. According to the algebra-working of the previous paragraph, we find dy dx = 2x. And so it is; for x = 100 and 2x = 200. But, you will say, we neglected a whole unit. Well, try again, making dx a still smaller bit. Try dx = 1 10 . Then x+ dx = 100.1, and (x+ dx)2 = 100.1× 100.1 = 10, 020.01. Now the last figure 1 is only one-millionth part of the 10, 000, and is utterly negligible; so we may take 10, 020 without the little decimal at the end. And this makes dy = 20; and dy dx = 20 0.1 = 200, which is still the same as 2x. Case 2. Try differentiating y = x3 in the same way. We let y grow to y + dy, while x grows to x+ dx. Then we have y + dy = (x+ dx)3. Doing the cubing we obtain y + dy = x3 + 3x2 · dx+ 3x(dx)2 + (dx)3. CALCULUS MADE EASY 20 Now we know that we may neglect small quantities of the second and third orders; since, when dy and dx are both made indefinitely small, (dx)2 and (dx)3 will become indefinitely smaller by comparison. So, regarding them as negligible, we have left: y + dy = x3 + 3x2 · dx. But y = x3; and, subtracting this, we have: dy = 3x2 · dx, and dy dx = 3x2. Case 3. Try differentiating y = x4. Starting as before by letting both y and x grow a bit, we have: y + dy = (x+ dx)4. Working out the raising to the fourth power, we get y + dy = x4 + 4x3 dx+ 6x2(dx)2 + 4x(dx)3 + (dx)4. Then striking out the terms containing all the higher powers of dx, as being negligible by comparison, we have y + dy = x4 + 4x3 dx. Subtracting the original y = x4, we have left dy = 4x3 dx, and dy dx = 4x3. SIMPLEST CASES 21 Now all these cases are quite easy. Let us collect the results to see if we can infer any general rule. Put them in two columns, the values of y in one and the corresponding values found for dy dx in the other: thus y dy dx x2 2x x3 3x2 x4 4x3 Just look at these results: the operation of differentiating appears to have had the effect of diminishing the power of x by 1 (for example in the last case reducing x4 to x3), and at the same time multiplying by a number (the same number in fact which originally appeared as the power). Now, when you have once seen this, you might easily conjecture how the others will run. You would expect that differentiating x5 would give 5x4, or differentiating x6 would give 6x5. If you hesitate, try one of these, and see whether the conjecture comes right. Try y = x5. Then y + dy = (x+ dx)5 = x5 + 5x4 dx+ 10x3(dx)2 + 10x2(dx)3 + 5x(dx)4 + (dx)5. Neglecting all the terms containing small quantities of the higher orders, we have left y + dy = x5 + 5x4 dx, CALCULUS MADE EASY 24 and dy dx = 1 2 x− 1 2 . Agreeing with the general rule. Summary. Let us see how far we have got. We have arrived at the following rule: To differentiate xn, multiply by the power and reduce the power by one, so giving us nxn−1 as the result. Exercises I. (See p. 252 for Answers.) Differentiate the following: (1) y = x13 (2) y = x− 3 2 (3) y = x2a (4) u = t2.4 (5) z = 3 √ u (6) y = 3 √ x−5 (7) u = 5 √ 1 x8 (8) y = 2xa (9) y = q √ x3 (10) y = n √ 1 xm You have now learned how to differentiate powers of x. How easy it is! CHAPTER V. NEXT STAGE. WHAT TO DO WITH CONSTANTS. In our equations we have regarded x as growing, and as a result of x being made to grow y also changed its value and grew. We usually think of x as a quantity that we can vary; and, regarding the variation of x as a sort of cause, we consider the resulting variation of y as an effect. In other words, we regard the value of y as depending on that of x. Both x and y are variables, but x is the one that we operate upon, and y is the “dependent variable.” In all the preceding chapter we have been trying to find out rules for the proportion which the dependent variation in y bears to the variation independently made in x. Our next step is to find out what effect on the process of differenti- ating is caused by the presence of constants, that is, of numbers which don’t change when x or y change their values. Added Constants. Let us begin with some simple case of an added constant, thus: Let y = x3 + 5. Just as before, let us suppose x to grow to x+dx and y to grow to y+dy. CALCULUS MADE EASY 26 Then: y + dy = (x+ dx)3 + 5 = x3 + 3x2 dx+ 3x(dx)2 + (dx)3 + 5. Neglecting the small quantities of higher orders, this becomes y + dy = x3 + 3x2 · dx+ 5. Subtract the original y = x3 + 5, and we have left: dy = 3x2 dx. dy dx = 3x2. So the 5 has quite disappeared. It added nothing to the growth of x, and does not enter into the differential coefficient. If we had put 7, or 700, or any other number, instead of 5, it would have disappeared. So if we take the letter a, or b, or c to represent any constant, it will simply disappear when we differentiate. If the additional constant had been of negative value, such as −5 or −b, it would equally have disappeared. Multiplied Constants. Take as a simple experiment this case: Let y = 7x2. Then on proceeding as before we get: y + dy = 7(x+ dx)2 = 7{x2 + 2x · dx+ (dx)2} = 7x2 + 14x · dx+ 7(dx)2. WHAT TO DO WITH CONSTANTS 29 algebraical expressions, and enable you to work out by yourself the examples given at the end of this chapter. (1) Differentiate y = x5 7 − 3 5 . 3 5 is an added constant and vanishes (see p. 25). We may then write at once dy dx = 1 7 × 5× x5−1, or dy dx = 5 7 x4. (2) Differentiate y = a √ x− 1 2 √ a. The term 1 2 √ a vanishes, being an added constant; and as a √ x, in the index form, is written ax 1 2 , we have dy dx = a× 1 2 × x 12−1 = a 2 × x− 12 , or dy dx = a 2 √ x . (3) If ay + bx = by − ax+ (x+ y) √ a2 − b2, find the differential coefficient of y with respect to x. As a rule an expression of this kind will need a little more knowledge than we have acquired so far; it is, however, always worth while to try whether the expression can be put in a simpler form. First we must try to bring it into the form y = some expression involving x only. The expression may be written (a− b)y + (a+ b)x = (x+ y) √ a2 − b2. CALCULUS MADE EASY 30 Squaring, we get (a− b)2y2 + (a+ b)2x2 + 2(a+ b)(a− b)xy = (x2 + y2 + 2xy)(a2 − b2), which simplifies to (a− b)2y2 + (a+ b)2x2 = x2(a2 − b2) + y2(a2 − b2); or [(a− b)2 − (a2 − b2)]y2 = [(a2 − b2)− (a+ b)2]x2, that is 2b(b− a)y2 = −2b(b+ a)x2; hence y = √ a+ b a− bx and dy dx = √ a+ b a− b. (4) The volume of a cylinder of radius r and height h is given by the formula V = πr2h. Find the rate of variation of volume with the radius when r = 5.5 in. and h = 20 in. If r = h, find the dimensions of the cylinder so that a change of 1 in. in radius causes a change of 400 cub. in. in the volume. The rate of variation of V with regard to r is dV dr = 2πrh. If r = 5.5 in. and h = 20 in. this becomes 690.8. It means that a change of radius of 1 inch will cause a change of volume of 690.8 cub. inch. This can be easily verified, for the volumes with r = 5 and r = 6 are 1570 cub. in. and 2260.8 cub. in. respectively, and 2260.8− 1570 = 690.8. Also, if r = h, dV dr = 2πr2 = 400 and r = h = √ 400 2π = 7.98 in. WHAT TO DO WITH CONSTANTS 31 (5) The reading θ of a Féry’s Radiation pyrometer is related to the Centigrade temperature t of the observed body by the relation θ θ1 = ( t t1 )4 , where θ1 is the reading corresponding to a known temperature t1 of the observed body. Compare the sensitiveness of the pyrometer at temperatures 800◦ C., 1000◦ C., 1200◦ C., given that it read 25 when the temperature was 1000◦ C. The sensitiveness is the rate of variation of the reading with the temperature, that is dθ dt . The formula may be written θ = θ1 t41 t4 = 25t4 10004 , and we have dθ dt = 100t3 10004 = t3 10, 000, 000, 000 . When t = 800, 1000 and 1200, we get dθ dt = 0.0512, 0.1 and 0.1728 respectively. The sensitiveness is approximately doubled from 800◦ to 1000◦, and becomes three-quarters as great again up to 1200◦. Exercises II. (See p. 252 for Answers.) Differentiate the following: (1) y = ax3 + 6. (2) y = 13x 3 2 − c. CHAPTER VI. SUMS, DIFFERENCES, PRODUCTS AND QUOTIENTS. We have learned how to differentiate simple algebraical functions such as x2 + c or ax4, and we have now to consider how to tackle the sum of two or more functions. For instance, let y = (x2 + c) + (ax4 + b); what will its dy dx be? How are we to go to work on this new job? The answer to this question is quite simple: just differentiate them, one after the other, thus: dy dx = 2x+ 4ax3. (Ans.) If you have any doubt whether this is right, try a more general case, working it by first principles. And this is the way. Let y = u+v, where u is any function of x, and v any other function of x. Then, letting x increase to x+ dx, y will increase to y + dy; and u will increase to u+ du; and v to v + dv. And we shall have: y + dy = u+ du+ v + dv. SUMS, DIFFERENCES, PRODUCTS 35 Subtracting the original y = u+ v, we get dy = du+ dv, and dividing through by dx, we get: dy dx = du dx + dv dx . This justifies the procedure. You differentiate each function sep- arately and add the results. So if now we take the example of the preceding paragraph, and put in the values of the two functions, we shall have, using the notation shown (p. 16), dy dx = d(x2 + c) dx + d(ax4 + b) dx = 2x + 4ax3, exactly as before. If there were three functions of x, which we may call u, v and w, so that y = u+ v + w; then dy dx = du dx + dv dx + dw dx . As for subtraction, it follows at once; for if the function v had itself had a negative sign, its differential coefficient would also be negative; so that by differentiating y = u− v, we should get dy dx = du dx − dv dx . CALCULUS MADE EASY 36 But when we come to do with Products, the thing is not quite so simple. Suppose we were asked to differentiate the expression y = (x2 + c)× (ax4 + b), what are we to do? The result will certainly not be 2x × 4ax3; for it is easy to see that neither c × ax4, nor x2 × b, would have been taken into that product. Now there are two ways in which we may go to work. First way. Do the multiplying first, and, having worked it out, then differentiate. Accordingly, we multiply together x2 + c and ax4 + b. This gives ax6 + acx4 + bx2 + bc. Now differentiate, and we get: dy dx = 6ax5 + 4acx3 + 2bx. Second way. Go back to first principles, and consider the equation y = u× v; where u is one function of x, and v is any other function of x. Then, if x grows to be x + dx; and y to y + dy; and u becomes u + du, and v becomes v + dv, we shall have: y + dy = (u+ du)× (v + dv) = u · v + u · dv + v · du+ du · dv. QUOTIENTS 39 Now perform the algebraic division, thus: v + dv u+ du u v + du v − u · dv v2 u+ u · dv v du− u · dv v du+ du · dv v − u · dv v − du · dv v − u · dv v − u · dv · dv v2 − du · dv v + u · dv · dv v2 . As both these remainders are small quantities of the second order, they may be neglected, and the division may stop here, since any further remainders would be of still smaller magnitudes. So we have got: y + dy = u v + du v − u · dv v2 ; which may be written = u v + v · du− u · dv v2 . CALCULUS MADE EASY 40 Now subtract the original y = u v , and we have left: dy = v · du− u · dv v2 ; whence dy dx = v du dx − u dv dx v2 . This gives us our instructions as to how to differentiate a quotient of two functions. Multiply the divisor function by the differential coef- ficient of the dividend function; then multiply the dividend function by the differential coefficient of the divisor function; and subtract. Lastly divide by the square of the divisor function. Going back to our example y = bx5 + c x2 + a , write bx5 + c = u; and x2 + a = v. Then dy dx = (x2 + a) d(bx5 + c) dx − (bx5 + c) d(x 2 + a) dx (x2 + a)2 = (x2 + a)(5bx4)− (bx5 + c)(2x) (x2 + a)2 , dy dx = 3bx6 + 5abx4 − 2cx (x2 + a)2 . (Answer.) The working out of quotients is often tedious, but there is nothing difficult about it. Some further examples fully worked out are given hereafter. DIFFERENTIATION 41 (1) Differentiate y = a b2 x3 − a 2 b x+ a2 b2 . Being a constant, a2 b2 vanishes, and we have dy dx = a b2 × 3× x3−1 − a 2 b × 1× x1−1. But x1−1 = x0 = 1; so we get: dy dx = 3a b2 x2 − a 2 b . (2) Differentiate y = 2a √ bx3 − 3b 3 √ a x − 2 √ ab. Putting x in the index form, we get y = 2a √ bx 3 2 − 3b 3√ax−1 − 2 √ ab. Now dy dx = 2a √ b× 3 2 × x 32−1 − 3b 3√a× (−1)× x−1−1; or, dy dx = 3a √ bx+ 3b 3 √ a x2 . (3) Differentiate z = 1.8 3 √ 1 θ2 − 4.4 5 √ θ − 27◦. This may be written: z = 1.8 θ− 2 3 − 4.4 θ− 15 − 27◦. The 27◦ vanishes, and we have dz dθ = 1.8×−2 3 × θ− 23−1 − 4.4× ( −1 5 ) θ− 1 5−1; or, dz dθ = −1.2 θ− 53 + 0.88 θ− 65 ; or, dz dθ = 0.88 5 √ θ6 − 1.2 3 √ θ5 . CALCULUS MADE EASY 44 In the indexed form, y = a+ x 1 2 a− x 12 . dy dx = (a− x 12 )(1 2 x− 1 2 )− (a+ x 12 )(−1 2 x− 1 2 ) (a− x 12 )2 = a− x 12 + a+ x 12 2(a− x 12 )2 x 12 ; hence dy dx = a (a−√x)2√x. (11) Differentiate θ = 1− a 3 √ t2 1 + a 2 √ t3 . Now θ = 1− at 23 1 + at 3 2 . dθ dt = (1 + at 3 2 )(−2 3 at− 1 3 )− (1− at 23 )× 3 2 at 1 2 (1 + at 3 2 )2 = 5a2 6 √ t7 − 4a 3 √ t − 9a 2 √ t 6(1 + a 2 √ t3)2 . (12) A reservoir of square cross-section has sides sloping at an angle of 45◦ with the vertical. The side of the bottom is 200 feet. Find an expression for the quantity pouring in or out when the depth of water varies by 1 foot; hence find, in gallons, the quantity withdrawn hourly when the depth is reduced from 14 to 10 feet in 24 hours. The volume of a frustum of pyramid of height H, and of bases A and a, is V = H 3 (A+ a+ √ Aa). It is easily seen that, the slope being 45◦, if the depth be h, the length of the side of the square surface of the water is 200 + 2h feet, so that the volume of water is h 3 [2002 + (200 + 2h)2 + 200(200 + 2h)] = 40, 000h+ 400h2 + 4h3 3 . DIFFERENTIATION 45 dV dh = 40, 000+800h+4h2 = cubic feet per foot of depth variation. The mean level from 14 to 10 feet is 12 feet, when h = 12, dV dh = 50, 176 cubic feet. Gallons per hour corresponding to a change of depth of 4 ft. in 24 hours = 4× 50, 176× 6.25 24 = 52, 267 gallons. (13) The absolute pressure, in atmospheres, P , of saturated steam at the temperature t◦ C. is given by Dulong as being P = ( 40 + t 140 )5 as long as t is above 80◦. Find the rate of variation of the pressure with the temperature at 100◦ C. Expand the numerator by the binomial theorem (see p. 137). P = 1 1405 (405 + 5× 404t+ 10× 403t2 + 10× 402t3 + 5× 40t4 + t5); hence dP dt = 1 537, 824× 105 (5× 404 + 20× 403t+ 30× 402t2 + 20× 40t3 + 5t4), when t = 100 this becomes 0.036 atmosphere per degree Centigrade change of temperature. Exercises III. (See the Answers on p. 253.) (1) Differentiate (a) u = 1 + x+ x2 1× 2 + x3 1× 2× 3 + · · · . (b) y = ax2 + bx+ c. (c) y = (x+ a)2. (d) y = (x+ a)3. CALCULUS MADE EASY 46 (2) If w = at− 1 2 bt2, find dw dt . (3) Find the differential coefficient of y = (x+ √ −1)× (x− √ −1). (4) Differentiate y = (197x− 34x2)× (7 + 22x− 83x3). (5) If x = (y + 3)× (y + 5), find dx dy . (6) Differentiate y = 1.3709x× (112.6 + 45.202x2). Find the differential coefficients of (7) y = 2x+ 3 3x+ 2 . (8) y = 1 + x+ 2x2 + 3x3 1 + x+ 2x2 . (9) y = ax+ b cx+ d . (10) y = xn + a x−n + b . (11) The temperature t of the filament of an incandescent electric lamp is connected to the current passing through the lamp by the re- lation C = a+ bt+ ct2. Find an expression giving the variation of the current corresponding to a variation of temperature. (12) The following formulae have been proposed to express the re- lation between the electric resistance R of a wire at the temperature SUCCESSIVE DIFFERENTIATION 49 The corresponding symbol for the differential coefficient is f ′(x), which is simpler to write than dy dx . This is called the “derived function” of x. Suppose we differentiate over again, we shall get the “second derived function” or second differential coefficient, which is denoted by f ′′(x); and so on. Now let us generalize. Let y = f(x) = xn. First differentiation, f ′(x) = nxn−1. Second differentiation, f ′′(x) = n(n− 1)xn−2. Third differentiation, f ′′′(x) = n(n− 1)(n− 2)xn−3. Fourth differentiation, f ′′′′(x) = n(n− 1)(n− 2)(n− 3)xn−4. etc., etc. But this is not the only way of indicating successive differentiations. For, if the original function be y = f(x); once differentiating gives dy dx = f ′(x); twice differentiating gives d ( dy dx ) dx = f ′′(x); and this is more conveniently written as d2y (dx)2 , or more usually d2y dx2 . Similarly, we may write as the result of thrice differentiating, d3y dx3 = f ′′′(x). CALCULUS MADE EASY 50 Examples. Now let us try y = f(x) = 7x4 + 3.5x3 − 1 2 x2 + x− 2. dy dx = f ′(x) = 28x3 + 10.5x2 − x+ 1, d2y dx2 = f ′′(x) = 84x2 + 21x− 1, d3y dx3 = f ′′′(x) = 168x+ 21, d4y dx4 = f ′′′′(x) = 168, d5y dx5 = f ′′′′′(x) = 0. In a similar manner if y = φ(x) = 3x(x2 − 4), φ′(x) = dy dx = 3 [ x× 2x+ (x2 − 4)× 1 ] = 3(3x2 − 4), φ′′(x) = d2y dx2 = 3× 6x = 18x, φ′′′(x) = d3y dx3 = 18, φ′′′′(x) = d4y dx4 = 0. Exercises IV. (See page 253 for Answers.) Find dy dx and d2y dx2 for the following expressions: (1) y = 17x+ 12x2. (2) y = x2 + a x+ a . (3) y = 1 + x 1 + x2 1× 2 + x3 1× 2× 3 + x4 1× 2× 3× 4. SUCCESSIVE DIFFERENTIATION 51 (4) Find the 2nd and 3rd derived functions in the Exercises III. (p. 45), No. 1 to No. 7, and in the Examples given (p. 40), No. 1 to No. 7. CALCULUS MADE EASY 54 time dt be called dy, the rate of spending it will be dy dt , or rather, should be written with a minus sign, as −dy dt , because dy is a decrement, not an increment. But money is not a good example for the calculus, because it generally comes and goes by jumps, not by a continuous flow—you may earn £200 a year, but it does not keep running in all day long in a thin stream; it comes in only weekly, or monthly, or quarterly, in lumps: and your expenditure also goes out in sudden payments. A more apt illustration of the idea of a rate is furnished by the speed of a moving body. From London (Euston station) to Liverpool is 200 miles. If a train leaves London at 7 o’clock, and reaches Liverpool at 11 o’clock, you know that, since it has travelled 200 miles in 4 hours, its average rate must have been 50 miles per hour; because 200 4 = 50 1 . Here you are really making a mental comparison between the distance passed over and the time taken to pass over it. You are dividing one by the other. If y is the whole distance, and t the whole time, clearly the average rate is y t . Now the speed was not actually constant all the way: at starting, and during the slowing up at the end of the journey, the speed was less. Probably at some part, when running downhill, the speed was over 60 miles an hour. If, during any particular element of time dt, the corresponding element of distance passed over was dy, then at that part of the journey the speed was dy dt . The rate at which one quantity (in the present instance, distance) is changing in relation to the other quantity (in this case, time) is properly expressed, then, by stating the differential coefficient of one with respect to the other. A velocity, scientifically expressed, is the rate at which a very small distance in any given direction is being passed over; and may therefore WHEN TIME VARIES 55 be written v = dy dt . But if the velocity v is not uniform, then it must be either increasing or else decreasing. The rate at which a velocity is increasing is called the acceleration. If a moving body is, at any particular instant, gaining an additional velocity dv in an element of time dt, then the acceleration a at that instant may be written a = dv dt ; but dv is itself d ( dy dt ) . Hence we may put a = d ( dy dt ) dt ; and this is usually written a = d2y dt2 ; or the acceleration is the second differential coefficient of the distance, with respect to time. Acceleration is expressed as a change of velocity in unit time, for instance, as being so many feet per second per second; the notation used being feet÷ second2. When a railway train has just begun to move, its velocity v is small; but it is rapidly gaining speed—it is being hurried up, or accelerated, by the effort of the engine. So its d2y dt2 is large. When it has got up its top speed it is no longer being accelerated, so that then d2y dt2 has fallen to zero. But when it nears its stopping place its speed begins to slow down; may, indeed, slow down very quickly if the brakes are put on, CALCULUS MADE EASY 56 and during this period of deceleration or slackening of pace, the value of dv dt , that is, of d2y dt2 will be negative. To accelerate a mass m requires the continuous application of force. The force necessary to accelerate a mass is proportional to the mass, and it is also proportional to the acceleration which is being imparted. Hence we may write for the force f , the expression f = ma; or f = m dv dt ; or f = m d2y dt2 . The product of a mass by the speed at which it is going is called its momentum, and is in symbols mv. If we differentiate momentum with respect to time we shall get d(mv) dt for the rate of change of momentum. But, since m is a constant quantity, this may be written m dv dt , which we see above is the same as f . That is to say, force may be expressed either as mass times acceleration, or as rate of change of momentum. Again, if a force is employed to move something (against an equal and opposite counter-force), it does work ; and the amount of work done is measured by the product of the force into the distance (in its own direction) through which its point of application moves forward. So if a force f moves forward through a length y, the work done (which we may call w) will be w = f × y; where we take f as a constant force. If the force varies at different parts of the range y, then we must find an expression for its value from WHEN TIME VARIES 59 and acceleration 5 seconds after the body began to move, and also find the corresponding values when the distance covered is 100 feet. Find also the average velocity during the first 10 seconds of its motion. (Suppose distances and motion to the right to be positive.) Now x = 0.2t2 + 10.4 v = ẋ = dx dt = 0.4t; and a = ẍ = d2x dt2 = 0.4 = constant. When t = 0, x = 10.4 and v = 0. The body started from a point 10.4 feet to the right of the point O; and the time was reckoned from the instant the body started. When t = 5, v = 0.4× 5 = 2 ft./sec.; a = 0.4 ft./sec2. When x = 100, 100 = 0.2t2 + 10.4, or t2 = 448, and t = 21.17 sec.; v = 0.4× 21.17 = 8.468 ft./sec. When t = 10, distance travelled = 0.2× 102 + 10.4− 10.4 = 20 ft. Average velocity = 20 10 = 2 ft./sec. (It is the same velocity as the velocity at the middle of the interval, t = 5; for, the acceleration being constant, the velocity has varied uniformly from zero when t = 0 to 4 ft./sec. when t = 10.) (2) In the above problem let us suppose x = 0.2t2 + 3t+ 10.4. v = ẋ = dx dt = 0.4t+ 3; a = ẍ = d2x dt2 = 0.4 = constant. When t = 0, x = 10.4 and v = 3 ft./sec, the time is reckoned from the instant at which the body passed a point 10.4 ft. from the point O, CALCULUS MADE EASY 60 its velocity being then already 3 ft./sec. To find the time elapsed since it began moving, let v = 0; then 0.4t + 3 = 0, t = − 3 .4 = −7.5 sec. The body began moving 7.5 sec. before time was begun to be observed; 5 seconds after this gives t = −2.5 and v = 0.4×−2.5 + 3 = 2 ft./sec. When x = 100 ft., 100 = 0.2t2 + 3t+ 10.4; or t2 + 15t− 448 = 0; hence t = 14.95 sec., v = 0.4× 14.95 + 3 = 8.98 ft./sec. To find the distance travelled during the 10 first seconds of the motion one must know how far the body was from the point O when it started. When t = −7.5, x = 0.2× (−7.5)2 − 3× 7.5 + 10.4 = −0.85 ft., that is 0.85 ft. to the left of the point O. Now, when t = 2.5, x = 0.2× 2.52 + 3× 2.5 + 10.4 = 19.15. So, in 10 seconds, the distance travelled was 19.15 + 0.85 = 20 ft., and the average velocity = 20 10 = 2 ft./sec. (3) Consider a similar problem when the distance is given by x = 0.2t2 − 3t + 10.4. Then v = 0.4t − 3, a = 0.4 = constant. When t = 0, x = 10.4 as before, and v = −3; so that the body was moving in the direction opposite to its motion in the previous cases. As the acceleration is positive, however, we see that this velocity will decrease WHEN TIME VARIES 61 as time goes on, until it becomes zero, when v = 0 or 0.4t − 3 = 0; or t = 7.5 sec. After this, the velocity becomes positive; and 5 seconds after the body started, t = 12.5, and v = 0.4× 12.5− 3 = 2 ft./sec. When x = 100, 100 = 0.2t2 − 3t+ 10.4, or t2 − 15t− 448 = 0, and t = 29.95; v = 0.4× 29.95− 3 = 8.98 ft./sec. When v is zero, x = 0.2× 7.52 − 3× 7.5 + 10.4 = −0.85, informing us that the body moves back to 0.85 ft. beyond the point O before it stops. Ten seconds later t = 17.5 and x = 0.2× 17.52 − 3× 17.5 + 10.4 = 19.15. The distance travelled = .85 + 19.15 = 20.0, and the average velocity is again 2 ft./sec. (4) Consider yet another problem of the same sort with x = 0.2t3− 3t2 + 10.4; v = 0.6t2 − 6t; a = 1.2t − 6. The acceleration is no more constant. When t = 0, x = 10.4, v = 0, a = −6. The body is at rest, but just ready to move with a negative acceleration, that is to gain a velocity towards the point O. (5) If we have x = 0.2t3−3t+10.4, then v = 0.6t2−3, and a = 1.2t. When t = 0, x = 10.4; v = −3; a = 0. The body is moving towards the point O with a velocity of 3 ft./sec., and just at that instant the velocity is uniform. CALCULUS MADE EASY 64 (4) If a body move according to the law s = 12− 4.5t+ 6.2t2, find its velocity when t = 4 seconds; s being in feet. (5) Find the acceleration of the body mentioned in the preceding example. Is the acceleration the same for all values of t? (6) The angle θ (in radians) turned through by a revolving wheel is connected with the time t (in seconds) that has elapsed since starting; by the law θ = 2.1− 3.2t+ 4.8t2. Find the angular velocity (in radians per second) of that wheel when 11 2 seconds have elapsed. Find also its angular acceleration. (7) A slider moves so that, during the first part of its motion, its distance s in inches from its starting point is given by the expression s = 6.8t3 − 10.8t; t being in seconds. Find the expression for the velocity and the acceleration at any time; and hence find the velocity and the acceleration after 3 seconds. (8) The motion of a rising balloon is such that its height h, in miles, is given at any instant by the expression h = 0.5 + 1 10 3 √ t− 125; t being in seconds. Find an expression for the velocity and the acceleration at any time. Draw curves to show the variation of height, velocity and acceleration during the first ten minutes of the ascent. WHEN TIME VARIES 65 (9) A stone is thrown downwards into water and its depth p in me- tres at any instant t seconds after reaching the surface of the water is given by the expression p = 4 4 + t2 + 0.8t− 1. Find an expression for the velocity and the acceleration at any time. Find the velocity and acceleration after 10 seconds. (10) A body moves in such a way that the spaces described in the time t from starting is given by s = tn, where n is a constant. Find the value of n when the velocity is doubled from the 5th to the 10th second; find it also when the velocity is numerically equal to the acceleration at the end of the 10th second. CHAPTER IX. INTRODUCING A USEFUL DODGE. Sometimes one is stumped by finding that the expression to be differ- entiated is too complicated to tackle directly. Thus, the equation y = (x2 + a2) 3 2 is awkward to a beginner. Now the dodge to turn the difficulty is this: Write some symbol, such as u, for the expression x2 + a2; then the equation becomes y = u 3 2 , which you can easily manage; for dy du = 3 2 u 1 2 . Then tackle the expression u = x2 + a2, and differentiate it with respect to x, du dx = 2x. INTRODUCING A USEFUL DODGE 69 Hence dy dx = − (1 + x) 1 2 2(1 + x) √ 1− x − (1− x) 12 2(1 + x) √ 1 + x = − 1 2 √ 1 + x √ 1− x − √ 1− x 2 √ (1 + x)3 ; or dy dx = − 1 (1 + x) √ 1− x2 . (6) Differentiate y = √ x3 1 + x2 . We may write this y = x 3 2 (1 + x2)− 1 2 ; dy dx = 3 2 x 1 2 (1 + x2)− 1 2 + x 3 2 × d [ (1 + x2)− 1 2 ] dx . Differentiating (1 + x2)− 1 2 , as shown in example (2) above, we get d [ (1 + x2)− 1 2 ] dx = − x√ (1 + x2)3 ; so that dy dx = 3 √ x 2 √ 1 + x2 − √ x5√ (1 + x2)3 = √ x(3 + x2) 2 √ (1 + x2)3 . (7) Differentiate y = (x+ √ x2 + x+ a)3. Let x+ √ x2 + x+ a = u. du dx = 1 + d [ (x2 + x+ a) 1 2 ] dx . y = u3; and dy du = 3u2 = 3 ( x+ √ x2 + x+ a )2 . CALCULUS MADE EASY 70 Now let (x2 + x+ a) 1 2 = v and (x2 + x+ a) = w. dw dx = 2x+ 1; v = w 1 2 ; dv dw = 1 2 w− 1 2 . dv dx = dv dw × dw dx = 1 2 (x2 + x+ a)− 1 2 (2x+ 1). Hence du dx = 1 + 2x+ 1 2 √ x2 + x+ a , dy dx = dy du × du dx = 3 ( x+ √ x2 + x+ a )2( 1 + 2x+ 1 2 √ x2 + x+ a ) . (8) Differentiate y = √ a2 + x2 a2 − x2 3 √ a2 − x2 a2 + x2 . We get y = (a2 + x2) 1 2 (a2 − x2) 13 (a2 − x2) 12 (a2 + x2) 13 = (a2 + x2) 1 6 (a2 − x2)− 16 . dy dx = (a2 + x2) 1 6 d [ (a2 − x2)− 16 ] dx + d [ (a2 + x2) 1 6 ] (a2 − x2) 16 dx . Let u = (a2 − x2)− 16 and v = (a2 − x2). u = v− 1 6 ; du dv = −1 6 v− 7 6 ; dv dx = −2x. du dx = du dv × dv dx = 1 3 x(a2 − x2)− 76 . Let w = (a2 + x2) 1 6 and z = (a2 + x2). w = z 1 6 ; dw dz = 1 6 z− 5 6 ; dz dx = 2x. dw dx = dw dz × dz dx = 1 3 x(a2 + x2)− 5 6 . INTRODUCING A USEFUL DODGE 71 Hence dy dx = (a2 + x2) 1 6 x 3(a2 − x2) 76 + x 3(a2 − x2) 16 (a2 + x2) 56 ; or dy dx = x 3 [ 6 √ a2 + x2 (a2 − x2)7 + 1 6 √ (a2 − x2)(a2 + x2)5] ] . (9) Differentiate yn with respect to y5. d(yn) d(y5) = nyn−1 5y5−1 = n 5 yn−5. (10) Find the first and second differential coefficients of y = x b √ (a− x)x. dy dx = x b d {[ (a− x)x ] 1 2 } dx + √ (a− x)x b . Let [ (a− x)x ] 1 2 = u and let (a− x)x = w; then u = w 12 . du dw = 1 2 w− 1 2 = 1 2w 1 2 = 1 2 √ (a− x)x . dw dx = a− 2x. du dw × dw dx = du dx = a− 2x 2 √ (a− x)x . Hence dy dx = x(a− 2x) 2b √ (a− x)x + √ (a− x)x b = x(3a− 4x) 2b √ (a− x)x . CALCULUS MADE EASY 74 Exercises VII. You can now successfully try the following. (See page 256 for Answers.) (1) If u = 1 2 x3; v = 3(u+ u2); and w = 1 v2 , find dw dx . (2) If y = 3x2 + √ 2; z = √ 1 + y; and v = 1√ 3 + 4z , find dv dx . (3) If y = x3√ 3 ; z = (1 + y)2; and u = 1√ 1 + z , find du dx . CHAPTER X. GEOMETRICAL MEANING OF DIFFERENTIATION. It is useful to consider what geometrical meaning can be given to the differential coefficient. In the first place, any function of x, such, for example, as x2, or √ x, or ax + b, can be plotted as a curve; and nowadays every schoolboy is familiar with the process of curve-plotting. dx dy P Q R y x dx O X Y Fig. 7. Let PQR, in Fig. 7, be a portion of a curve plotted with respect to the axes of coordinates OX and OY . Consider any point Q on this curve, where the abscissa of the point is x and its ordinate is y. Now observe how y changes when x is varied. If x is made to increase by CALCULUS MADE EASY 76 a small increment dx, to the right, it will be observed that y also (in this particular curve) increases by a small increment dy (because this particular curve happens to be an ascending curve). Then the ratio of dy to dx is a measure of the degree to which the curve is sloping up between the two points Q and T . As a matter of fact, it can be seen on the figure that the curve between Q and T has many different slopes, so that we cannot very well speak of the slope of the curve between Q and T . If, however, Q and T are so near each other that the small portion QT of the curve is practically straight, then it is true to say that the ratio dy dx is the slope of the curve along QT . The straight line QT produced on either side touches the curve along the portion QT only, and if this portion is indefinitely small, the straight line will touch the curve at practically one point only, and be therefore a tangent to the curve. This tangent to the curve has evidently the same slope as QT , so that dy dx is the slope of the tangent to the curve at the point Q for which the value of dy dx is found. We have seen that the short expression “the slope of a curve” has no precise meaning, because a curve has so many slopes—in fact, every small portion of a curve has a different slope. “The slope of a curve at a point” is, however, a perfectly defined thing; it is the slope of a very small portion of the curve situated just at that point; and we have seen that this is the same as “the slope of the tangent to the curve at that point.” Observe that dx is a short step to the right, and dy the correspond- ing short step upwards. These steps must be considered as short as MEANING OF DIFFERENTIATION 79 dx dy dx dy dx dy O X Y Fig. 12. dx dy dx dy dx dy O X Y Fig. 13. reached, as in Fig. 14. If a curve first descends, and then goes up again, as in Fig. 15, presenting a concavity upwards, then clearly dy dx will first be negative, with diminishing values as the curve flattens, then will be zero at the point where the bottom of the trough of the curve is reached; and from this point onward dy dx will have positive values that go on increasing. In such a case y is said to pass by a minimum. The minimum value of y is not necessarily the smallest value of y, it is that value of y corresponding to the bottom of the trough; for instance, in Fig. 28 (p. 99), the value of y corresponding to the bottom of the trough is 1, while y takes CALCULUS MADE EASY 80 O X Y Fig. 14. O X Y y min. Fig. 15. elsewhere values which are smaller than this. The characteristic of a minimum is that y must increase on either side of it. N.B.—For the particular value of x that makes y a minimum, the value of dy dx = 0. If a curve first ascends and then descends, the values of dy dx will be positive at first; then zero, as the summit is reached; then negative, as the curve slopes downwards, as in Fig. 16. In this case y is said to pass by a maximum, but the maximum value of y is not necessarily the greatest value of y. In Fig. 28, the maximum of y is 21 3 , but this is by no means the greatest value y can have at some other point of the curve. N.B.—For the particular value of x that makes y a maximum, the value of dy dx = 0. If a curve has the peculiar form of Fig. 17, the values of dy dx will always be positive; but there will be one particular place where the slope is least steep, where the value of dy dx will be a minimum; that is, less than it is at any other part of the curve. MEANING OF DIFFERENTIATION 81 O X Y y max. Fig. 16. O X Y Fig. 17. If a curve has the form of Fig. 18, the value of dy dx will be negative in the upper part, and positive in the lower part; while at the nose of the curve where it becomes actually perpendicular, the value of dy dx will be infinitely great. dx dx dy dy Q O X Y Fig. 18. Now that we understand that dy dx measures the steepness of a curve at any point, let us turn to some of the equations which we have already learned how to differentiate. CALCULUS MADE EASY 84 P Q R b O X Y Fig. 22. Let us illustrate this by working out a particular instance. Taking the equation y = 1 4 x2 + 3, and differentiating it, we get dy dx = 1 2 x. Now assign a few successive values, say from 0 to 5, to x; and calculate the corresponding values of y by the first equation; and of dy dx from the second equation. Tabulating results, we have: x 0 1 2 3 4 5 y 3 31 4 4 51 4 7 91 4 dy dx 0 1 2 1 11 2 2 21 2 Then plot them out in two curves, Figs. 23 and 24, in Fig. 23 plotting the values of y against those of x and in Fig. 24 those of dy dx against those of x. For any assigned value of x, the height of the ordinate in the second curve is proportional to the slope of the first curve. MEANING OF DIFFERENTIATION 85 −3 −2 −1 0 1 2 3 4 5 5 6 7 8 9 x y b 1 4x 2 y = 1 4 x 2 + 3 Fig. 23. −3 −2 −1 1 2 3 4 5 1 2 3 4 5 x dy dx 0 dy dx = 1 2 x Fig. 24. If a curve comes to a sudden cusp, as in Fig. 25, the slope at that point suddenly changes from a slope upward to a slope downward. In O X Y Fig. 25. that case dy dx will clearly undergo an abrupt change from a positive to a negative value. The following examples show further applications of the principles just explained. CALCULUS MADE EASY 86 (4) Find the slope of the tangent to the curve y = 1 2x + 3, at the point where x = −1. Find the angle which this tangent makes with the curve y = 2x2 + 2. The slope of the tangent is the slope of the curve at the point where they touch one another (see p. 76); that is, it is the dy dx of the curve for that point. Here dy dx = − 1 2x2 and for x = −1, dy dx = −1 2 , which is the slope of the tangent and of the curve at that point. The tangent, being a straight line, has for equation y = ax + b, and its slope is dy dx = a, hence a = −1 2 . Also if x = −1, y = 1 2(−1) + 3 = 2 1 2 ; and as the tangent passes by this point, the coordinates of the point must satisfy the equation of the tangent, namely y = −1 2 x+ b, so that 21 2 = −1 2 × (−1) + b and b = 2; the equation of the tangent is therefore y = −1 2 x+ 2. Now, when two curves meet, the intersection being a point com- mon to both curves, its coordinates must satisfy the equation of each one of the two curves; that is, it must be a solution of the system of simultaneous equations formed by coupling together the equations of the curves. Here the curves meet one another at points given by the solution of  y = 2x 2 + 2, y = −1 2 x+ 2 or 2x2 + 2 = −1 2 x+ 2;