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Calculus FOR Engineers % Tr.i.cJu o^^jr r.i^ LIBRARY OF THE UNIVERSITY OF CALIFORNIA. aass CALCULUS FOE ENGINEEES, QtambxiUQt: PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS. THE CALCULUS FOE ENGINEEES BY JOHN PERRY, M.E, RSc, F.R.S. WH. SCH., ASSOC. MEMB. INST. C.E., PROFESSOR OF MECHANICS AND MATHEMATICS IN THE ROYAL COLLEGE OF SCIENCE, LONDON: PRESIDENT OF "IBan^Xtsiil^^O^ OF ELECTRICAL ENGINEERS. FIFTH IMPRESSION. Honlron : EDWARD ARNOLD, 37, BEDFORD STREET. lAll Rights reserved,] Engirie- ring Library .V^^ Digitized by tine IntemVt Arciiive in 2008 witftfanding from IVIicrosoft Corporation littp://www.arcliive.org/details/calculusforenginOOperrricli PEBFACB. This book describes what has for many years been the most important part of the regular course in the Calculus for Mechanical and Electrical Engineering students at the Finsbury Technical College. It was supplemented by easy work involving Fourier, Spherical Harmonic, and Bessel Functions which I have been afraid to describe here because the book is already much larger than I thought it would become. The students in October knew only the most elementary mathematics, many of them did not know the Binomial Theorem, or the definition of the sine of an angle. In July they had not only done the work of this book, but their knowledge was of a practical kind, ready for use in any such engineering problems as I give here. One such student, Mr Norman Endacott, has corrected the manuscript and proofs. He has worked out many of the exercises in the third chapter twice over. I thank him here for the care he has taken, and I take leave also to say that a system which has, year by year, produced many men with his kind of knowledge of mathematics has a good deal to recommend it. I say this through no vanity but because I wish to encourage the earnest student. Besides I cannot claim more than a portion of the credit, for I do not think that there ever before was such a complete J 2/ 12b VI PREFACE. harmony in the working of all the departments of an educational institution in lectures and in tutorial, labora- tory, drawing office and other practical work as exists in the Finsbury Technical College, all tending to the same end ; to give an engineer such a perfect acquaintance with his mental tools that he actually uses these tools in his business. Professor Willis has been kind enough to read through the proofs and I therefore feel doubly sure that no important mistake has been made anywhere. An experienced friend thinks that I might with advantage have given many more illustrations of the use of squared paper just at the beginning. This is quite possible, but if a student follows my instructions he will furnish all this sort of illustration very much better for himself Again I might have inserted many easy illustrations of integration by numerical work such as the exercises on the Bull Engine and on Beams and Arches which are to be found in my book on Applied Mechanics. I can only say that I encourage students to find illustrations of this kind for themselves; and surely there must be some limit to spoon feeding. JOHN PERRY. EoYAL College of Science, London, 16th March, 1897. TABLE OF CONTENTS. PAGE Introductory Remarks ..,.,,.. 1 Chap. I. The Study of ^** . 6 Chap. II. The Compound Interest Law and the Harmonic Function 161 Chap. III. General Differentiation and Integration . 267 CALCULUS FOE ENGINEERS. INTRODUCTORY. 1. The Engineer has usually no time for a general mathe- matical training — mores the pity — and those young engineers who have had such a training do not always find their mathe- matics helpful in their profession. Such men will, I hope, find this book useful, if they can only get over the notion that because it is elementary, they know already all that it can teach. But I write more particularly for readers who have had very little mathematical training and who are willing to work very hard to find out how the calculus is applied in Engineer- ing problems. I assume that a good engineer needs to know only fundamental principles, but that he needs to know these very well indeed. 2. My reader is supposed to have an elementary know- ledge of Mechanics, and if he means to take up the Electrical problems he is supposed to have an elementary knowledge of Electrical matters. A common-sense knowledge of the few fundamental facts is what is required; this knowledge is seldom acquired by mere reading or listening to lectures; one needs to make simple experiments and to work easy numerical exercises. In Mechanics, I should like to think that the mechanical engineers who read this book know what is given in the elementary parts of my books on Applied Mechanics and the Steam and Gas Engine. That is, I assume that they know P 1 2 CALCULUS FOR ENGINEERS. the elementary facts about Bending Moment in beams, Work done by forces and the Efficiency of heat engines. Possibly the book may cause them to seek for such knowledge. I take almost all my examples from Engineering, and a man who works these easy examples will find that he knows most of what is called the theory of engineering. 3. I know men who have passed advanced examinations in Mathematics who are very shy, in practical work, of the common formulae used in Engineers' pocket-books. How- ever good a mathematician a student thinks himself to be, he ought to practise working out numerical values, to find for example the value of a^ by means of a table of logarithms, when a and h are any numbers whatsoever. Thus to find V'014, to find 2*36o~<''2«, &c., to take any formula from a pocket-book and use it. He must not only think he knows ; he must really do the numerical work. He must know that if a distance 2 4 5 4 has been measured and if one is not sure about the last figure, it is rather stupid in multiplying or dividing by this number to get out an answer with many significant figures, or to say that the indicated power of an engine is 324"65 Horse power, when the indicator may be in error 5 per cent, or more. He must know the quick way of finding 3'216 x 4571 to four significant figures without using logarithms. He ought to test the approximate rule (1 +a)"=l + ?ia, or (1 + a)" (1 + ^)'« = 1 + 7ia + m^, if a and /S are small, and see for himself when a = '01 or — -01, or yS = ± -025 and ti = 2 or ^ or — 1^, and m = 4 or 2 or — 2 or J or any other numbers, what errors are involved in the assumption. f As to Trigonometry, the definitions must be known. For example, Draw BA C an angle of, say, 35°. Take any point B and drop the perpendicular. Measure AB and ^0 and AC as accurately as possible. Is AC^ -{- BC^ = AB^ 1 Work BC AG this out numerically. Now -r^ = sin 35°, -rv^ = cos 35°, BC ^^ ^^ -j-^ = tan 35°. Try if the answers are those given in the tables. Learn how we calculate the other sides of the INTRODUCTORY. 3 triangle ABC when we know one side and one of the acute angles. Learn also that the sine of 130° is positive, and the cosine of 130° is negative. Also try with the book of tables if sin {A-\- B) — sin A , cos B + cos A . sin B, where A and B are any two angles you choose to take. There are three other rules like this. In like manner the four which we obtain by adding these formulae and subtract- ing them, of which this is one, 2 sin a . cos /S = sin (a + y8) + sin (a — /8) ; also cos 2^ = 1 - 2 sin^ A =2 cos^ ^ - 1. Before readers have gone far in this book I hope they will be induced to take up the useful (that is, the elementary and interesting) part of trigonometry, and prove all rules for themselves, if they haven't done so already. Calculate an angle of 1*6 degrees in radians (1 radian is equal to 57'296 degrees) ; see how much the sine and tangent of this angle differ from the angle itself Remember that when in mathematics we say sin x, x is supposed to be in radians. I do not expect a man to know much about advanced algebra, but he is supposed to be able to give the factors of a?^ + 7^ 4- 12 or of x"^ — o? for example ; to be able to simplify expressions. It is not a knowledge of permutations or com- binations or of the theory of equations, of Geometrical Conies or tangent planes to quadrics, that the Engineer wants. Happy is the Engineer who is also a mathematician, but it is given to only a few men to have the two so very different powers. A prolonged experience of workshops, engineers and students has convinced me that although a Civil Engineer for the purposes of surveying may need to understand the solution of triangles, this and many other parts of the Engineer's usual mathematical training are really useless to the mechanical or electrical engineer. This sounds un- orthodox, but I venture to emphasise it. The young engi- neer cannot be drilled too much in the mere simplification of algebraic and trigonometrical expressions, including ex- pressions involving J —1, and the best service done by 1—2 4 CALCULUS FOR ENGINEERS. elementary calculus work is in inducing students to again undergo this drilling. But the engineer needs no artificial mental gymnastics such as is furnished by Geometrical Conies, or the usual examination-paper puzzles, or by evasions of the Calculus through infinite worry with elementary Mathematics. The result of a false system of training is seen in this, that not one good engineer in a hundred believes in what is usually called theory. 4. I assume that every one of my readers is thoroughly well acquainted already with the fundamental notion of the Calculus, only he doesn't know it in the algebraic form. He has a perfect knowledge of a rate, but he has never been accustomed to write -^ ; he has a perfect knowledge of an area, but he has not yet learnt the symbol used by us, lf(x).dos. He has the idea, but he does not express his idea in this form. 1 assume that some of my readers have passed difficult examinations in the Calculus, that they can differentiate any function of x and integrate many ; that they know how 'to work all sorts of difficult exercises about Pedal Curves and Roulettes and Elliptic Integrals, and to them also I hope to be of use. Their difficulty is this, their mathematical know- ledge seems to be of no use to them in practical engineering problems. Give to their afs and y's a physical meaning, or use p's and vs instead, and what was the easiest book exercise becomes a difficult problem. I know such men who hurriedly skip in reading a book when they see a ~ y or a sign of integration. 5. When I started to write this book I thought to put the subject before my readers as I have been able, I think — I have been told — very successfully, to bring it before some classes of evening students; but much may be done in lectures which one is unable to do in a cold-blooded fashion sitting at a table. One misses the intelligent eyes of an audience, warning one that a little more explanation is needed INTRODUCTORY. 5 or that an important idea has already been grasped. An idea could be given in the mere drawing of a curve and illustrations chosen from objects around the lecture-room. Let the reader skip judiciously; let him work up no problem here in which he has no professional interest. The problems are many, and the best training comes from the careful study of only a few of them. The reader is expected to turn back often to read again the early parts. The book would be unwieldy if I included any but the more interesting and illustrative of engineering problems. I put off for a future occasion what would perhaps to many students be a more interesting part of my subject, namely, illustrations from Engineering (sometimes called Applied Physics) of the solution of Partial Differential Equations. Many people think the subject one which cannot be taught in an elementary fashion, but Lord Kelvin showed me long ago that there is no useful mathematical weapon which an engineer may not learn to use. A man learns to use the Calculus as he learns to use the chisel or the file on actual concrete bits of work, and it is on this idea that I act in teaching the use of the Calculus to Engineers. This book is not meant to supersede the more orthodox treatises, it is rather an introduction to them. In the first chapter of 160 pages, I do not attempt to differentiate or integrate any function of x, except x^. In the second chapter I deal with e^^, and sin (aw + c). The third chapter is more difficult. For the sake of the training in elementary Algebraic work, as much as for use in Engineering problems, I have included a set of exercises on general dinerentiation and integration. Parts in smaller type, and the notes, may be found too difficult by some students in a first reading of the book. An occasional exercise may need a little more knowledge than the student already possesses. His remedy is to skip. CHAPTER I. x». 6. Everybody has already the notions of Co-ordinate Geometry and uses squared paper. Squared paper may be bought at sevenpence a quire : people who arc ignorant of this fact and who pay sevenpence or fourteen pence a sheet for it must have too great an idea of its value to use it properly. When a merchant has in his office a sheet of squared paper with points lying in a curve which he adds to day by day, each point showing the price of iron, or copper, or cotton yarn or silk, at any date, he is using Co-ordinate Geometry. Now to what uses does he put such a curve ? 1. At any date he sees what the price was. 2. He sees by the slope of his curve the rate of increase or fall of the price. 3. If he plots other things on the same sheet of paper at the same dates he will note what effect their rise and fall have upon the price of his material, and this may enable him to pro- phesy and so make money. 4. Examination of his curve for the past will enable him to prophesy with more certainty than a man can do who has no records. Observe that any point represents two things; its horizontal distance from some standard line or axis is called one co-ordinate, we generally call it the x co-ordinate and it is measured horizontally to the right of the axis of y ; some people call it the abscissa ; this represents time in his case. The other co-ordinate (we usually call it the 3/ co-ordinate or tlie ordinate, simply), the vertical distance of the point above some standard line or axis; this represents his price. In the newspaper you will find curves showing how the thermometer and barometer are rising and falling. I once read a clever article upon the way in which the English population and wealth and taxes were increasing; the reasoning was very SQUARED PAPER. 7 difficult to follow. On taking the author s figures however and plotting them on squared paper, every result which he had laboured so much to bring out was plain upon the curves, so that a boy could understand them. Possibly this is the reason why sDme writers do not publish curves: if they did, there would be little need for writing. 7. A man making experiments is usually finding out how one thing which I shall call y depends upon some other thing which I shall call x. Thus the pressure p of saturated steam (water and steam present in a vessel but no air or other fluid) is always the same for the same temperature. A curve drawn on squared paper enables us for any given temperature to find the pressure or vice versa, but it shows the rate at which one increases relatively to the increase of the other and much else. I do not say that the curve is always better than the table of values for giving information ; some information is better given by the curve, some by the table. Observe that when we represent any quantity by the length of a line we represent it to some scale or other ; 1 inch represents 10 lbs. per square inch or 20 degrees centigrade or something else ; it is always to scale and according to a convention of some kind, for of course a distance 1 inch is a very different thing from 20 degrees centigrade. When one has two columns of observed numbers to plot on squared paper one does it, 1. To see if the points lie in any regular curve. If so, the simpler the curve the simpler is the law that we are likely to find. 2. To correct errors of observation. For if the points lie nearly in a simple regular curve, if we draw the curve that lies most evenly among the points, using thin battens of w^ood, say, then it may be taken as probable that if there were no errors of observation the points would lie exactly in such a curve. Note that when a point is — 5 feet to the right of a line, we mean that it is 5 feet to the left of the line. I have learnt by long ex- perience that it is worth while to spend a good deal of time subtracting from and multiplying one's quantities to fit the numbers of squares (so that the whole of a sheet of paper is needed for the points) before beginning to plot. Now let the reader buy some squared paper and without asking help from anyone let him plot the results of some 8 CALCULUS FOR ENGINEERS. observations. Let him take for example a Whitaker's Almanack and plot from it some sets of numbers ; the average temperature of every month last year; the National Debt since 1688; the present value of a lease at 4 per cent, for any number of years ; the capital invested in Railways since 1849 ; anything will do, but he had better take things in which he is interested. If he has made laboratory observa- tions he will have an absorbing interest in seeing what sort of law the squared paper gives him. 8. As the observations may be on pressure jj and tem- perature t, or p and volume v, or v and t, or Indicated Horse Power and Useful Horse Power of a steam or gas engine, or amperes and volts in electiicity, and we want to talk generally of any such pair of quantities, I shall use x and y instead of the p'q and v^ and ^ s and all sorts of letters. The short way of saying that there is some law connecting two variable quantities x and y is F (x, y) =0...(1), or in words "there is some equation connecting x and y." Any expression which contains x and y (it may contain many other letters and numbers also) is said to be a function of x and y and we use such symbols as F(x, y),f{x, y), Q(x, y) etc. to represent functions in general when we don't know what the expres- sions really are, and often when we do know, but want to write things shortly. Again we use F{x) or f{x) or any other convenient symbol to mean "any mathematical ex- pression containing x" and we say " let f(x) be any function of x!* Thus y =f(x) ... (2) stands for any equation which would enable us when given x to calculate y. X' ifi The law ^ + ^t^ = 1 comes under the form (1) given above, whereas if we calculate y in terms of x and get y = 4- 1 \/25 — x'^ we have the form (2). But in either case we have the same law connecting y and x. In pure mathe- matics X and y are actual distances ; in applied mathematics X and y stand for the quantities which we are comparing and which are represented to scale. 9. ' Graph ' Exercises. I. Draw the curve y = 2 + -^j-^x-. Take ^ = and we find 2/ =2; take x = l, then 7/ = 20333 ; 'graph' exercises. 9 take X = 2, then y = 2 + IS^o = 2-133; and so on. Now plot these values of x and y on your sheet of squared paper. The curve is a parabola. II. Draw the curve y = 1 — ^cc-\- -J^x^ which is also a parabola, in the same way, on the same sheet of paper. III. Draw the curve xy = 120. Now if x = l, y=120; if X = 2, y = 60 ; if x = S, y = 4<0; if a; = 4, y = 30 and so on : this curve is a rectangular hyperbola. IV. Draw yx'-'^' = 100 or y = 100^-i-«^ If the student cannot calculate y for any value of x, he does not know how to use logarithms and the sooner he does know how to use logarithms the better. V. Draw y = ax''^ where a is any convenient number. I advise the student to spend a lot of time in drawing members of this great family of useful curves. Let him try ?i = — 1 (he drew this in III. above), 7i=— 2, ?i=— 1-J, n=—^, ?i=— 0*1, n = 0, n = "I, ri = f , n = l, n=l\,n = '2 (this is No. I. above), ?i = 3, w = 4 &c. YI. Draw y = a sin (bx + c) taking any convenient numbers for a, h and c. Advice. As hx-{-c is in radians (one radian is 57'2958 degrees) and the books of tables usually give augles in degrees, choose numbers for h and c which will make the arithmetical work easy. Thus take 6 = 1-7- 114"6, take c the number of radians which correspond to say 30° ( this is ^ or '5236 Let a = 5 say. Now let x — 0, 10, 20, &;c., and calculate y. Thus when «?= 6, 3/ = 5 sin [yrxTp + "5236) ; but if the angle is converted into degrees we have 2/ = 5 sin (16 + 30 degrees) = b sin 33° = 2723. Having drawn the above curve, notice what change would occur if c were changed to or ^ or ^ or ^ . Again, if a were changed. More than a week may be spent on this curve, very profitably. 10 CALCULUS FOR ENGINEERS. VII. Draw y = ae***. Try 6 = 1 and a = 1 ; try other values of a and h ; take at least two cases of negative values for h. In the above work, get as little help from teachers as possible, but help from fellow students will be very useful especially if it leads to wrangling about the subject. The reason why I have dwelt upon the above seven cases is this: — Students learn usually to differentiate and integrate the most complicated expressions: but when the very simplest of these expressions comes before them in a real engineering problem they fight shy of it. Now it is very seldom that an engineer ever has to face a problem, even in the most intri- cate part of his theoretical work, which involves a knowledge of more functions than these three y = ax^y y = a sin {hx + c), y = ad^, but these three must be thoroughly well understood and the engineering student must look upon the study of them as his most important theoretical work. Attending to the above three kinds of expression is a student's real business. I see no reason, however, for his not having a little amusement also, so he may draw the curves X' 4- y^ = 25 (Circle), ^ + f^ = 1 (Ellipse), |g-^ = l (Hyperbola), and some others mentioned in Chapter, III., but from the engineer's point of view these curves are comparatively un- interesting. 10. Having studied y = e~^^ and y = 6 sin (ex + ^) a student will find that he can now easily understand one of the most important curves in engineering, viz : y = be~** sin (ex + g). He ought first to take such a curve as has already been studied by him, y = h sin (ex + g) ; plot on the same sheet of paper y = e~^^ ; and multiply together the ordinates of the two curves at many values of x to find the ordinate of the EQUATIONS TO LOCI. 11 new curve. The curve is evidently wavy, y reaching maximum and minimum values; y represents the displacement of a' pendulum bob or pointer of some measuring instrument whose motion is damped by fluid or other such friction, x being the time, and a student will understand the curve much better if he makes observations of such a motion, for example with a disc of lead immersed in oil vibrating so slowly under the action of torsional forces in a wire that many ob- servations of its angular position (using pointer and scale of degrees) which is called y, x being the time, may be made in one swing. The distance or angle from an extreme position on one side of the zero to the next extreme position on the other side is called the length of one swing. The Napierian logarithm of the ratio of the length of one swing to the next or one tenth of the logarithm of the ratio of the first swing to the eleventh is evidently a multiplied by half the periodic time, or it is a multiplied by the time occupied in one swing. This logarithmic decrement as it is called, is rather important in some kinds of measurement. 11. When by means of a drawing or a model we are able to find the path of any point and where it is in its path when we know the position of some other point, we are always able to get the same information algebraically. Example (1). A point jPand a straight line DD being given ; what is the path of a point P when it moves so that its distance from the point F is always in the same ratio to its distance from the straight line? Thus in the figure let FF=exPI> ... (I), where e is a constant. Draw FFX at right angles to DD. y If the distance FD is called a; and the perpendicular FG is y ; our problem is this ; — What is the equa- tion connecting x and y? Now all we have to do is to express (1) in terms of x and ^. Let FF be called a. Thus FF= \/FG^+FG^= Vy^ + (^'-a)2 so that, squaring (1) we have This is the answer. If e is 1 the curve is called a parabola. If e is greater than 1, the curve is called an Fig. 1. hyperbola. If e is less than I, the curve is called an ellipse. 12 CALCULUS FOR ENGINEERS. Example (2). The circle APQ rolls on the straight line OX, What is the Path of any point P on. the circumference? If whenP touched the line it was at 0, let OX and OF bo the axes, and let SP be X and PT be y. Let the radius of the circle be a. Let the angle PCQ be called <^. Draw CB, peq^endicular to PT. Observe that PB=a . sin PCB=a sin (</> - 90) = - a cos (/>, BC= a cos PCB = a sin (/>. Now the arc QP= a .cf) = OQ. Hence SiS x=OQ — BC, and /c have y = BT^-PB, x=axf)- a sin 0| y = a —acoscfij (3). If from (3) we eliminate we get one equation connecting x and y. I^ut it is better to retain </> and to have two equations because of the greater simplicity of calculation. In fact the two equations (3) may be called the equation to the curve. The curve is called the oycloid as all my readers know already. Example (3). A crank and connecting rod work a slider in a straight path. Where is the slider for any position of the crank ? Let the path be in the direction of the centre of the crank shaft. P.^- Fig. 3. If A is the end of the path, evidently ^0 is equal to l-\-r, r being length of crank. MECHANISMS. 13 It is well to remember in all such problems that if we project all the sides of a closed figure upon any two straight lines, we get two in- dependent equations. Projecting on the horizontal we see that .(1). s+l cos (f)+r cos 6=1 + r\ Projecting on the vertical ^ sin ^ = r sin ^ J If we eliminate from these equations we can calculate s for any value of 6. The student ought to do this for himself, but I am weak enough to do it here. We see that from the second equation of (1) COS0 so that the first becomes -V^-i sin2^, s=l\l-^l-^^^sm^0J+r{l-cosd)..* (2). Students ought to work a few exercises, such as; — 1. The ends A and B of a rod are guided by two straight slots OA and OB which are at right angles to one another ; find the equation to the path of any point P in the rod. 2. In Watt's parallel motion there is a point which moves nearly in a straight path. Find the equation to its complete path. In fig. 4 the Mean Position is shown as OABC. The best place for P is such that BP/PA = OAICB. Draw the links in any other Fig. 4. * Note that if as is usual, -j^isa. small fraction, then slnca ,Jl-a=l-\a ■when a is small, we can get an approximation to the value of s, which can be expressed in terms of d and 2d. This is of far more importance than it here seems to be. When the straight path of Q makes an angle a with the Hne joining its middle point and O, if a is not large, it is evident that s is much the same as before, only divided by cos a. When a is large, the algebraic expression for s is rather complicated, but good approximations may always be found which will save trouble in calculation. 14 CALCULUS FOR ENGINEERS. position. The complete path of P would be a figure of 8. 3. Find the equation to the path of a point in the middle of an ordinaiy con- necting rod. 4. A^ the end of a link, moves in a straight path COC\ being the middle of the path, with a simple haraaonic motion OA=asmpti where t is time ; the other end B moves in a straight path OBD which is in a direction at right angles to COC ; what is ^'s motion ? Show that it is approximately a simple harmonic motion of twice the frequency of ^. 5. In any slide valve gear, in which there are several links, &c. driven from a uniformly rotating crank ; note this fact, that the motion of any point of any link in any particular direction consists of a fundamental simple harmonic motion of the same frequency as the crank, together with an octave. The proper study of Link Motions and Radial valve gears from this point of view is' worth months of one's life, for this contains the secret of why one valve motion gives a better diagram than another. Consider for example the Hackworth gear with a curved and with a straight slot. What is the difference ? See Art. 122. 12. Plotted points l3ring in a straight line. Proofs will come later; at first the student ought to get well ac- quainted with the thing to be proved. I have known boys able to 'prove mathematical propositions who did not really know what they had proved till years afterwards. Take any expression like y = a + hx, where a and h are numbers. Thus let y = 2 + l^x. Now take ^7 = 0, a?=l, a; = 2, a? = 3, <Src. and in each case calculate the corresponding value of y. Plot the corresponding values of a; and y as the co-ordinates of points on squared paper. You will find that they lie exactly in a straight line. Now take say y = 2-\-Sx or 2 + ^a; or 2 — ^a; or 2 — Sx and you will find in every case a straight line. Men who think they know a little about this subject already will not care to take the trouble and if you do not find yourselves interested, I advise you not to take the trouble either; yet I know that it is worth your while to take the trouble. Just notice that in every case I have given you the same value of a and consequently all your lines have some one thing in common. What is it? Take this hint, a is the value of y when a? = 0. Again, try y=:2 + l^x, y = 1 +l^x, y = 0-i-lix, y = ^l + lix, 2/ = -2 + l^a?, and so see what it means when h is the same in every case. You will find that all the lines with the same b have the SLOPE OF A LINE. 16 same slope and indeed I am in the habit of calling h the slope of the line. If y = a 4- 6^, when oc = x-^, find y and call it y^y when a; = a?! + 1, find y and call it y^. It is easy to show that y.^—y^ = h. So that what I mean by the slope of a straight line is its rise for a horizontal distance 1. (Note that when we say that a road rises ^ or 1 in 20, we mean 1 foot rise for 20 feet along the sloping road. Thus ■jjj is the sine of the angle of inclination of the road to the horizontal; whereas our slope is measured in a different way). Our slope is evidently the tangent of the inclination of the line to the horizontal. Looking upon y as a quantity whose value depends upon that of x, observe that the rate of in- crease of y relatively to the increase of x is constant, being indeed h, the slope of the line. The symbol used for this rate is -p. Observe that it is one symbol; it does not mean /Y V 7/ -^ — ~ . Try to recollect the statement that if y= a + hx, a ^ X -~—h, and that if -,— = h, then it follows that y:=A-\-hx, dx dx where A is some constant or other. Any equation of the first degree connecting x and y such as Ax-\-By = G where A, B and C are constants, can G A be put into the shape y = ^ — ^x, so that it is the equation A to a straight line whose slope is — ^ and which passes C through the point whose x = 0, whose y = -n, called point ( 0, p ) . Thus 4^ + 2^/ = 5 passes through the point x = 0, y = 2^ and its slope is — 2. That is, y diminishes as x in- creases. You are expected to draw this line y = 2|^ — 2^7 and distinguish the difference between it and the line 3/ = 2 J + 2x. Note what is meant by positive and what by negative slope. Draw a few curves and judge approximately by eye of the slope at a number of places. 13. Problems on the straight line. 1. Given the slope of a straight line; if you are also 16 CALCULUS FOR ENGINEERS. told that it passes through the point whose x = 3, and whose y = 2, what is the equation to the line ? Let the slope be 0-35. The equation is y = a + 0'35.'^^^. where a is not known. But (3, 2) is a point on the line, so that 2 = a + 035 x 3, or a = 0"95 and hence the line is y = 0'95 + 35a?. 2. What is the slope of any line at right angles to y=za-\-bxl Let -45 be the given line, cutting OX in C. Then h = tan BOX. If DE is any line at right angles to the first, its slope is tan DEX or — tan DEC or - cot BGE or — ^ . So that y = A—j-a)is typical of all lines at right angles to y = a-{-h.v] A being any constant. 3. Where do the two straight lines Ax -\-By-\-C—0 and Mx + Ny -{■8 = meet ? Answer, In the point whose x and y satisfy both the equations. We have therefore to do what is done in Elementary Algebra, solve simultaneous equations. 4. When tan a and tan ^ are known, it is easy to find tan (a — /S), and hence when the straight lines y=a + hx and y = m-\-nx are given, it is easy to find the angle between them. 5. The line y = a-\-bx passes through the points a?=l, y = 2, and a? = 3, y = 1, find a and h. EMPIRICAL FORMULAE. 17 6. A line y = a-{-bx is at right angles to y = 2 + Sa) and passes through the point x = l, y = l. Find a and h. 14. Obtaining Empirical Formulae. When in the laboratory we have made measurements of two quantities which depend upon one another, we have a table showing corresponding values of the two, and we wish to see if there is a simple relation between them, we plot the values to convenient scales as the co-ordinates of points on squared paper. If some regular curve (a cm*ve without singular points as I shall afterwards call it) seems as if it might pass through all the points, save for possible errors of measurement, we try to obtain a formula y = f(x), which we may call the law or rule connecting the quantities called y and x. If the points appear as if they might lie on a straight line, a stretched thread may be used to help in finding its most probable position. There is a tedious algebraic method of finding the straight line which represents the positions of the points with least error, but for most engineering purposes the stretched string method is suffi- ciently accurate. If the ciu-ve seems to follow such a law as y~a-\-hx^, plot y and the square of the observed measurement, which we call x^ as the co-or- dinates of points, and see if they lie on a straight line. If the curve ax seems to follow such a law as y= j ■ (1), which is the same as ^.\.hy = a, divide each of the quantities which you call y by the corre- sponding quantity x ; call the ratio X. Now plot the values of X and of y on squared paper ; if a straight line passes through the plotted points, then we have such a law as X=A-{-By, or - — A-\-By^ or , so that (1) is true. ^ l-Bx' Usually we can apply the stretched thread method to find the probability of truth of any law containing only two constants. Thus, suppose measurements to be taken from the expansion part of a gas engine indicator diagram. It is im]3ortant for many purposes to obtain an empirical formula connecting p and v, the pressure and volume. I always find that the following rule holds with a fair amount of accuracy pv" = C where s and C are two constants. We do not much care to know C, but if there is such a rule, the value of s is very important*. To test if this rule holds, plot log jo and log v as the co- * There is no known physical reason for expecting such a rule to hold. At first I thought that perhaps most curves drawn at random approximately like hyperbolas would approximately submit to such a law as yx^= C, but I found that this was by no means the case. The following fact is. worth mentioning. When my students find, in carrying out the above rule that log p and log v do not lie in a straight line, I find that they have P. 2 18 CALCULUS FOR ENGINEERS. ordinates of points on squared pa^^er (common logarithms will do). If they lie approximately in a straight line, we see that \ogp+s\ogv=c a constant, and therefore the rule holds. When we wish to test with a formula containing three independent constants we can often i*educe it to such a shape as Av+Bw+Cz=l (2), where y, w, z contain x and y in some shajje. Thus to test if y=—- — , we have y + cxy —a-\-hx^ ov- + -xy — x=\. Here y itself is the old v, xy is the old w>, and x itself is the old z. If (2) holds, and if t?, \o and z were plotted as the three co-ordinatea of a ix)int in space, all the points ought to lie in a plane. By means of three sides of a wooden box and a number of beads on the ends of pointed wires this may be tried directly ; immersion in a tank of water to try whether one can get the beads to lie in the plane of the siuface of the water, being used to find the plane. I have also used a descrip- tive geometry method to find the plane, but there is no method yet used bv me which compares for simplicity with the stretched thread method in the other case. But no hard and fast rules can be given for trying all sorts of em- pirical formulae upon one's observed numbers. The student is warned that his formula is an empirical one, and that he must not deal with it as if he had discovered a natural law of infinite exactness. AVhen other formulae fail, we try y—a-\-hx-\-cx'*--\-^-\'^^") because we know that with sufficient terms this will satisfy any curve. When there are more than two constants, we often find them by a patient application of what is called the method of least squares. To test if the pressure and temperature of saturated steam follow the rule jo = a(^+/3)"... (3), where B is temperature, Centigrade, say, three constants have to be found. The only successful plan tried by me is one in which I guess at /3. I know that /3 is nearly 40. I ask one student to try ^ = 40, another to try i3 = 41, another ^=39 and so on; made a mistake in the amount of clearance. Too much clearance and too little clearance give results which depart in opposite ways from the straight hne. It is convenient in many calculations, if there is such an empirical formula, to use it. If not, one has to work with rules which instruct us to draw tangents to the curve. Now it is an excellent exercise to let a number of students trace the same curve with two points marked upon it and to let them all independently draw tangents at those points to their curve, and measure the angle between them. It is extraordinary what very different lines they will draw and what different angles they may obtain. Let them all measure by trial the radius of curvature of the curve at a point ; in this the discrepancies are greater than before. SLOPE OF A CURVE. 19 they are asked to liiid the rule (3) which most accurately represents p and 6 between, say jo = 71b. per sq. inch, and p = 150. He who gets a straight Hne lying most evenly (judging by the eye) among the points, when \ogp and log(^-f /3) are used as co-ordinates, has used the best value of 4. The method may be refined upon by ingenious students. (See end of Chap. I.) 15. We have now to remember that if y = a-^hx, then -^ — h, and if -^ = h, then y = A -{■ hx, Avhere A is some ax ax ^ constant. Let us prove this algebraically. \iy = a-\-hx. Take a particular value of x and calculate y. Now take a new value of x, call it ^ + hx, and calculate the new y, call it y -t- Sy, y ^^y — a-\-h{x-\- hx). Subtract y—a-\-hx and we get hy = hhx, or -^ = 6, ox and, however small hx or Zy may become their ratio is 6, we therefore say ~ = h. Fig. 6. 16. In the curve of fig. 6 there is positive slope {^ increases as x increases) in the parts AB, DF and HI and 2—2 20 CALCULUS FOR ENGINEERS. Fig. 7. negative slope {y diminishes as x increases) in the parts BD and FH. The slope is at 5 and F which are called points of maximum or points where i/ is a maximum; and it is also at D and H which are points of minimum. The point E is one in which the slope ceases to increase and begins to diminish : it is a point of inflexion. Notice that if we want to know the slope at the point P we first choose a point F which is near to P. (Imagine that in fig. 6 the little portion of the curve at P is magnified a thousand times.) Call PS=x, PQ=:y-NF=x+Bx,FL=ij+ht/, sotha,tPM=Bx,F3I==Si/. Now FM/PM or By/Bx is the average slope between P and F. It is tan FPM. Imagine the same sort of figure drawn but for a point F' nearer to P. Again, another, still nearer P. Ob- serve that the straight line FP or F'P or F"P gets gi-adually more and more nearly what we mean by the tangent to the curve at P. In every case hyjhx is the tangent of the angle which the line FP or F'P or F'P makes with the horizontal, and so we see that in the limit the slope of the line or dy/dx at P is the tangent of the angle which the tangent at P makes with the axis of X. If then, instead of judging roughly by the eye as we did just now in discussing fig. 6, we wish to measure very accurately the slope at the point P ; — Note that the slope is independent of where the axis of X is, so long as it is a horizontal line, and I take care in using my rule here given, to draw OX below the part of the curve where I am studying the slope. Draw a tangent PR to the curve, cutting OX in P. Then the slope is tan PRX. If drawn and lettered according to my instructions, observe that PRX is always an acute angle when the slope is positive and is always an obtuse angle when the slope is negative. Do not forget that the slope of the curve at any point means the rate of increase of y there with regard to a?, and WHAT IS SPEED? 21 that we may call it the slope of the curve or tan PRQ or by the symbol -^- or "the differential coefficient of y with regard to x," and all these mean the same thing. Every one knows what is meant when on going up a hill one says that the slope is changing, the slope is diminish- ing, the slope is increasing ; and in this knowledge he already possesses the fundamental idea of the calculus. 17. We all know what is meant when in a railway train we say ^'we are going at 30 miles per hour.^' Do we mean that we have gone 30 miles in the last hour or that we are really going 30 miles in the next hour ? Certainly not. We may have only left the terminus 10 minutes ago ; there may be an accident in the next second. What we mean is merely this, that the last distance of 3 miles was traversed in the tenth of an hour, or rather, the last distance of 0*0003 miles was traversed in 0*00001 hour. This is not exactly right ; it is not till we take still shorter and shorter distances and divide by the times occupied that we approach the true value of the speed. Thus it is known that a body falls freely vertically through the following distances in the following intervals of time after two seconds from rest, at London. That is between 2 seconds from rest and 2*1 or 2*01 or 2001, the distances fallen through are given. Each of these divided by the interval of time gives the average velocity during the interval. Intervals of time in seconds Distances in feet fallen through Average velocities We see that as the interval of time after 2 seconds is taken less and less, the average velocity during the interval approaches more and more the true value of the velocity at 2 seconds from rest which is exactly 64*4 feet per second. We may find the true velocity at any time when we know the law connecting s and t as follows. Let s = 16'lt2, the well known law for bodies falling freely at London. If t is given of any value we can calculate •1 •01 •001 6*601 •6456 •064416 6601 64*56 64416 22 CALCULUS FOR ENGINEERS. s. If t has a slightly greater value called t-\- St (here St is a symbol for a small portion of time, it is not S "Xt, but a very different thing), and if we call the calculated space s + Ss, then s -^ hs = U'l (t + Btf or l&l {f + 2t. St + (Sty]. Hence, subtracting, Ss = 16*1 {2t . St 4- (Sty], and this formula will enable us to calculate accurately the space Ss passed through between the time t and the time t+St. The average velocity during this interval of time is Ss -^ St or ^ = 32-2^+ 10- 1 8/. Please notice that this is absolutely correct ; there is no vagueness about it. Now I come to the important idea; as St gets smaller and smaller, ^ approaches more and more nearly 32*2^, the other term 16'1S^ becoming smaller and smaller, and hence we say that in the limit, Ss/St is truly 32-2^. The limiting value of ^ as St gets smaller and smaller is called , or the rate of change of s as t increases, or the differential coefficient of s with regard to t, or it is called the velocity at the time t Now surely there is no such great difficulty in catching the idea of a limiting value. Some people have the notion that we are stating something that is only approximately true ; it is often because their teacher will say such things as "reject 16'1S^ because it is small," or "let dt be an infinitely small amount of time" and they proceed to divide something by it, showing that although they may reach the age of Methuselah they will never have the common sense of an engineer. Another trouble is introduced by people saying " let Ss ds St — and kt or -y- is so and so." The true statement is, " as of at . . Ss St gets smaller and smaller without limit, ^ approaches more and more nearly the finite value 322^," and as I have already said, everybody uses the important idea of a limit every day of his life. SLOPE AND SPEED. 23 From the law connecting s and t, if we find -r- or the velocity, we are said to differentiate s with regard to the time t. When we are given ds/dt and we reverse the above process we are said to integrate. If I were lecturing I might dwell longer upon the correct- ness of the notion of a rate that one already has, and by making many sketch'es illustrate my meaning. But one may listen intently to a lecture which seems dull enough in a book. I will, therefore, make a virtue of necessity and say that my readers can illustrate my meaning perfectly well to themselves if they do a little thinking about it. After all my great aim is to make them less afraid than they used to be of such symbols as dy/dx and fy . dx. 18. Given s and t in any kind of motion, as a set of num- bers. How do we study the motion ? For example, imagine a Bradshaw's Railway Guide which not merely gave a few stations, but some hundred places between Euston and Rugby. The entries might be like this: s would be in miles, t in hours and minutes. .9 = would mean Euston. 10 o'clock 3 10. .10 5 10. .15 7 10.. 20 7 10. .23 9 10.. 28 12 10. .33 &c. One method is this : plot t (take times after 10 o'clock) horizontally and s vertically on a sheet of squared paper and draw a curve through the points. The slope of this curve at any place represents the velocity of the train to some scale which depends upon the scales for s and t. 24 CALCULUS FOR ENGINEERS. Observe places where the velocity is great or small. Between ^=10.20 and i = 10.23 observe that the velocity is 0. Indeed the train has probably stopped altogether. To be absolutely certain, it would be necessary to give s for every value of t, and not merely for a few values. A curve alone can show every value. I do not say that the table may not be more valuable than the curve for a great many purposes. If the train stopped at any place and travelled towards Euston again, we should have negative slope to our curve and negative velocity. Note that acceleration being rate of change of velocity with time, is indicated by the rate of change of the slope of the curve. Why not on the same sheet of paper draw a curve which shows at every instant the velocity of the train ? The slope of this new curve would evidently be the accelera- tion. I am glad to think that nobody has yet given a name to the rate of change of the acceleration. The symbols in use are s and t for space and time ; ds velocity i; or -^ , or Newton's symbol s ; acceleration -j- or -yz , or Newton's s. at dv Rate of change of acceleration would be -r- . Note that ^ is one symbol, it has nothing whatever to do d^ X s with such an algebraic expression as , — -— . The symbol is Gt X I supposed merely to indicate that we have differentiated s twice mth regard to the time. I have stated that the slope of a curve may be found by drawing a tangent to the curve, and hence it is easy to find the acceleration from the velocity curve. ACCELERATION. 26 19. Another way, better than by drawing tangents, is illustrated in this Table : -. acceleration t seconds s feet V feet per second in feet per second per second or or 8sJ8t 8vldt •06 •0880 14-74 •07 •2354 13-49 -125 •08 •3703 1222 -127 •09 •4925 10-95 -127 •10 •6020 9-66 -129 •11 •6986 8-35 -131 •12 •7821 704 -131 •13 •8525 In a new mechanism it was necessary for a certain purpose to know in every position of a point A what its acceleration was, and to do this I usually find its velocity first. A skeleton drawing was made and the positions of A marked at the intervals of time t from a time taken as 0. In the table I give at each instant the distance of A from a fixed point of measurement, and I call it s. If I gave the table for all the positions of A till it gets back again to its first position, it would be more instructive, but any student can make out such a table for himself for some particular mechanism. Thus for example, let s be the distance of a piston from the end of its stroke. Of course the all-accomplished mathematical engineer will scorn to take the trouble. He knows a graphical rule for doing this in the case of the piston of a steam engine. Yes, but does he know such a rule for every 26 CALCULUS FOR ENGINEERS. possible mechanism ? Would i^t be worth while to seek for such a graphical nile for every possible mechanism ? Here is the straightforward Engineers' common-sense way of finding the acceleration at any point of any mechanism, and although it has not yet been tried except by myself and my pupils, I venture to think that it will commend itself to practical men. For beginners it is invaluable. Now the mass of the body whose centre moves like the point A, being m (the weight of the body in pounds at London^ divided by 32*2)*, multiply the acceleration in feet per second per second which you find, by m, and you have the force which is acting on the body increasing the velocity. The force will be in pounds. * I have given elsewhere my reasons for using in books intended for engineers, the units of force employed by all practical engineers. I have used this system (which is, after all, a so-called absolute system, just as much as the c. a. s. system or the Foundal system of many text books) for twenty years, with students, and this is why their knowledge of mechanics is not a mere book knowledge, something apart from their practical work, but fitting their practical work as a hand does a glove. One might as well talk Choctaw in the shops as speak about what some people call the English system, as if a system can be English which speaks of so many poundals of force and so many foot-poundals of work. And yet these same philosophers are astonished that practical engineers should have a contempt for book theory. I venture to say that there is not one practical engineer in this country, who thinks in Poundals, although all the books have used these units for 30 years. In Practical Dynamics one second is the unit of time, one foot is the unit of space, one ijoiind (what is called the weight of 1 lb. in London) is the unit of force. To satisfy the College men who teach Engineers, I would say that "The unit of Mass is that mass on which the force of 1 lb. produces an acceleration of 1 ft. per sec. per see." We have no name for unit of mass, the Engineer never has to speak of the inertia of a body by itself. His instructions are "In all Dynamical calculations, divide the weight of a body in lbs. by 32'2 and you have its mass in Engineer's units — in those units which will give all your answers in the units in which an Engineer talks." If you do not use this system evei-y answer you get out will need to be divided or multiplied by something before it is the language of the practical man. Force in pounds is the space-roXe at which work in foot-pounds is done, it is also the f/we-rate at which momentum is produced or destroyed. Example 1. A Kaxnxner head of 2^ lbs. moving with a velocity of 40 ft. per sec. is stopped in '001 sec. What is the average force of the blow? Here the mass being 2^-^32•2, or -0776, the momentum (momentum is mass X velocity) destroyed is 3-104. Now force is momentum per sec. and hence the average force is 3•104-^•001 or 3104 lbs. Example 2. Water in a jet flows with the linear velocity of 20 ft. per sec. DIFFERENTIATION OF ax^. 27 20. We considered tlic case of falling bodies in which space and time are connected by the law 6' = iu^^? where cf the acceleration due to gravity is 32 "2 feet per second per second at London. But many other pairs of things are con- nected by similar laws and I will indicate them generally by y = ax2. Let a particular value of a be taken, say a = -^. Now take a) = 0, x = l, x==2, 00 = 3, &c. and in every case calculate y. Plot the corresponding points on squared paper. They lie on a parabolic curve. At any point on the curve, say where x = S, find the slope of the curve ( I call it ~j^] , do the same at a; = 4, ^ = 2, &c. Draw a new curve, now, with the same values of x but with ~- as the ordinate. This curve ax shows at a glance (by the height of its ordinate) what is the slope of the first curve. If you ink these curves, let the y curve be black and the -~ curve be red. Notice that the slope or -~ at any point, is 2a multiplied by the x of the point. We can investigate this algebraically. As before, for any value of x calculate y. Now take a greater value of x which I shall call x + Bx and calculate the new y, calling it y + 8y. We have then y -\- By = a (x -{- Bxf = a[x^^+2x.Bx + (8xy}. Subtracting ; By = a [2x . Bx + (Bx)-]. By Divide by Bx, -— = 2ax + a . Bx. (relatively to the vessel from which it flows), the jet being 0*1 sq. ft. in cross section ; what force acts upon the vessel ? Here we have 20 x -1 cu. ft. or 20 x -1 x 62-3 lbs. of water per sec. or a mass per second in Engineers' units of 20 x -1 x 62*3-1- 32-2. This mass is 3-^7, its momentum is 77 '4, and as this momentum is lost by the vessel every second, it is the force acting on the vessel. A student who thinks for himself will see tliat this force is the same whether a vessel is or is not in motion itself. 28 CALCULUS FOR ENGINEERS. Imagine Bw to get smaller and smaller without limit and use the symbol —^ for the limiting value of -A > ^^^ we have -p = 2aic, a fact which is known to us already from our CLX squared paper*. 21. Note that when we repeat the process of differentiation we state the result as -^ and tho answer is 2a. You must become familiar with these symbols. If y is a function of X, ,- is the rate of change of y with regard to a?; t4 is the rate of change of -^ with regard to x. Or, shortly; ^^ is the differential coefficient of ^ with regard to a? ; ;p is the differential coefficient of y with regard to x. Or, again ; integrate ~; and our answer is ~- ; inte- grate ~ and our answer is y. You will, I hope, get quite familiar with these symbols and ideas. I am only afraid that when we use other letters than x& and y^ you may lose your familiarity. * Symbolically. Let 2/=/(a;)..,(l), where /(a;) stands for any expression containing x. Take any value of x and calculate y. Now take a slightly greater value of x say x + 8x and calculate the new y ; call it y + 8y then y + dy^f{x + 8x) (2). Subtract (1) from (2) and divide by dx. Sy_ f{x + dx)-f{ x) 5^- 5i ^^^• ,^ , dw . f(x + 8x)-f(x) What we mean by ^~ is the limiting value of —— — ^-^-^ as dx is ax ox made smaller and smaller without limit. This is the exact definition of -^ . dx It is quite easy to remember and to write, and the most ignorant person may get full marks for an answer at an examination. It is easy to see that the differential coefficient of of {x) is a times the differential coefficient of / (x) and also that the differential coefficient of/ {x) + F.{x) is the sum of the two differential coefficients. MOTION. 29 The differential coefficient of where a, b and c are constants, is cix The integral of + b-\-kx with regard to w is A-\-bx + ^kaf, where A is any constant whatsoever. Similarly, the integral of 6 + A;^ with regard to z is A+bz-i- ^kz\ The integral oib-^kv with regard to v/i& A + bv-\- ^kv"^. It is quite easy to work out as an exercise that if y=aa^^ then ~-=Mx^, and again that if y = ax^^ then -p=4a^. All these are examples of the fact that if y = ax\ then -^—naod^'K In working out any of these examples we take it that —■ becomes -— OX OiX dii or that ^y=8xx-^ more and more nearly as dx gets smaller and smaller without limit. This is sometimes written y-\-by=y-{-bx . -f ,ov f{x+bx)^f{x) + bx'^.. (I). 22. Uniformly accelerated motion. dt d^s If acceleration, ^2~^ (!)• Integrate and we have ~z=b-\-at— velocity v. Observe that we have added a constant 6, because if we differentiate a constant the answer is 0. There must be some information given us which will enable us to find what the value of the constant b is. Let the information be v = Vo, when i = 0. Then b is evidently v^. ds So that velocity v— ^- =Vo + at (2). 30 CALCULUS FOR ENGINEERS. Again integrate and 5 = c + ?;„< + ^at-. Again you will notice that we add an unknown constant, when wc integrate. Some information must be given us to find the value of the constant c. Thus if s = So when ^ = 0, this 5o is the value of c, and so we have the most complete statement of the motion s = So + Vot + ^at^ (3). If (3) is differentiated, we obtain (2) and if (2) is differen- tiated we obtain (1). 23. We see here, then, that as soon as the student is able to differentiate and integrate he can work the fol- lowing kinds of problem. I. If s is given as any function of the time, differentiate and the velocity at any instant is found ; differentiate again and the acceleration is found. II. If the acceleration is given as some function of the time, integrate and we find the velocity; integrate again and we find the space passed through. Observe that s instead of being mere distance may be the angle described, the motion being angular or rotational. Better then call it 0. Then 6 or -^- is the angular velocity and S or J— is the angular acceleration. 24. Exercises on Motion with constant Accele- ration. 1. The acceleration due to gravity is doiunwards and is usually called g, g being 32*2 feet per second per second at London. If a body at time is thrown vertically upwards with a velocity of F© feet per second; where is it at the end of t seconds ? If s is measured upwards, the acceleration is — g and s^V^Jt — \gt^. (We assume that there is no resist- ance of the atmosphere and that the true acceleration is g downwards and constant.) Observe that v=Vo—gt and that v=0 when Vo—gt = or t = ~- . When this is the case find s. This eives the highest point and the time taken to reach it. When is s = again ? What is the velocity then ? KINETIC ENERGY. 31 2. The body of Exercise 1 has been given, in addition to its vertical velocity, a horizontal velocity Uo which keeps constant. If oo is the horizontal distance of it away from (I?X (17' the origin at the time ^, — = and -7- = Uq, x — ujt. If we call s by the new name ?/, we have at any time t, y = V,t - \gt\ iXj ^^ UQVy V of and if we eliminate t, we find y=—^x — ^g-- which is a Parabola. 3. If the body had been given a velocity V in the dii'ection a above the horizontal, we may use V sin a for Vq and V cos a for Uf^ in the above expressions, and from them we can make all sorts of useful calculations concerning pro- jectiles. Plot the curve when V= 1000 feet per second and a = 45°. Again plot with same V when a = 60°, and again when a = 30°. 25. Kinetic Energy. A small body of mass vi is at 5 = when ^ = and its velocity is Vq, and a force F acts upon it causing an acceleration F/ni. As in the last case at any future time v^Vf, + - t...(l), and s = -^Vot + I ~ t" (2), (2) maybe written s^^tilv^A — t\ and it is easy to see from (1) that this is s = \t{vQ-\-v), and that the average velocity in any interval is half the sum of the velocities at the beginning and end of the interval. Now the work done by the force F in the distance s is Fs. Calculating F from till (1), F={v — Vq)— and multiplying upon s we find that the t work is ^m (v^ — Vq^) which expresses the work stored up in a moving body in terms of its velocity. In fact the work done causes ^mvo^ to increase to imv^ and this is the reason why ^mv'^ is called the kinetic energy of a body. 32 CALCULUS FOR ENGINEERS. Otherwise. Let a small body of mass iii and velocity v pass through the very small space hs in the time ht gaining velocity hv and let a force F be acting upon it. Now ov F= m X acceleration or ^= 77i -^ and Bs=^v , Bt so that hv F,hs=^ nivBt -^ =1^1' vBv = BE, ot if BE stands for the increase in the kinetic energy of the body oE -K- = m . -y. But our equations are only entirely true when Bs, Bt, &c., are made smaller and smaller without limit : Hence as -Y- = mv, or in words, "the differential coefficient of av E with regard to v is mv" if we integrate with regard to v, E = \mv^ + c where c is some constant. Let E — when v==0 m that c = and we have E — |mt;l Practise differentiation and integration using other letters than X and y. In this case -^ stands for our old -^^ If we •^ dv dx had had —■ = mx it might have been seen more easily thaf y = Jmar* + c, but you must escape from .the swaddling bands of X and y. 26. Exercise. If x is the elon/^ation of a spring F when a force F is applied and if a; = - , a representing the stiffness of the spring ; F .Bx'is the work done in elongating the spring through the small distance Bx. If F is gradually increased from to i^ and the elongation from to x, what strain energy is stored in the spring ? The gain of energy from x to x-\-Bx is BE = F . Bx, or iilF rather ~=— = F = ax, hence E = ^ax^ + c. Now if ^ = when x=^0, we see that c = 0, so that the energy E stored is E=iax' = iFx..,{l). It is worth noting that when a mass M is vibrating at the end of a spiral spring ; when it is at the distance x from its ELEOTRTO CTT1CUI^,„„,^ 33 "... ^< ■ '"'' ■ ^". position of equilibrium, the potential energy is ^ax^ and the kinetic energy is Jilfy- or the ^/tot^l, energy is ^Mv^ + ^ax"^.... Note that when a forc0 F is required to pi-oduce an elongation or compression ^ in a rod, or a deflexion ^r in a beam, and ii F=^ ax where a is some constant, the energy stored up as strain energy or potential energy is ^ax^ or ^Fx. Also if a Torque T is required to produce a turning through the angle ^ in a shaft or spring or other structure, and if r = aO, the energy stored up as strain energy or potential energy is |a^ or ^T6. If T is in pound-feet and is in radians, the answer is in foot-pounds. Work done = Force x distance, or Torque x angle. 27. If the student knows anything about electricity let him translate into ordinary language the improved Ohm's law V = RC + L.dC/dt (1). Observe that if R (Ohms) and L (Henries) remain con- dC stant, if C and -^ are known to us, we know F, and if the law of F, a changing voltage, is known you may see that there must surely be some means of finding C the changing current. Think of L as the back electromotive force in volts when the current increases at the rate of 1 ampere per second. If the current in the primary of a transformer, and there- fore the induction in the iron, did not alter, there would be no electro-motive force in the secondary. In fact the E.M.F. in the secondary is, at any instant, the number of turns of the secondary multiplied by the rate at which the induction changes per second. Rate of increase of / per second is what we now call the differential coefficient of / with regard to time. Although L is constant only when there is no iron or else because the induction is small, the correct formula being V= BC-h /V ,- (2), it is found that, practically, (1) with L constant is of nearly universal application. See Art. 183. 28. If y = ax" and you wish to find -,- , I am afraid that 34 CALCULUS FOR ENGINEERS. I must assume that you know the Binomial Theorem which is : — (x + by = a?« + nbo)''-^ + ^ ^^^ " ^ 62a;«-2 + -^ ^ ^ h'x''-^ + &c. It is easy to show by multiplication that the Binomial Theorem is true when n = 2 or 3 or 4 or 5, but when n = ^ or J or any other fraction, and again, when n is negative, you had better perhaps have faith in my assertion that the Binomial Theorem can be proved. It is however well that you should see what it means by working out a few examples. Illustrate it with n = 2, then ?i = 3, 71 = 4, &c., and verify by multiplication. Again try n = — l, and if you want to see whether your series is correct, just recollect that (x + h)~^ is j and divide 1 by a? + 6 in the regular way by long division. Let us do with our new function of co as we did with aao^. Here y = aoo'\ y-\-hy = a{x + hxY =a[x^-\- n.hx . x^-^ Tl ill/ ~~ W H — ^-^ — - {hx)' x^~^ + terms involving higher powers of hx]. Now subtract and divide by hx and you will find !=•{ n . ^"-1 + ^^\ ^^ {hx) X''-' + &c. We see now that as 8x is made smaller and smaller, in the limit we have only the first term left, all the others having in them Bx or (8xy or higher powers of Bx, and they must all disappear in the limit, and hence, ^ = nax"-i. (See Notes p. 159.) Thus the differential coefficient of a^ is 6x^, of x^^ it is 2^x^^, and of x~^ it is — |^~i When we find the value of the differential coefficient of any given function we are said to differentiate it. When given -^ to find y we are said to integrate. The origin of /" . dec. 35 the words differential and integral need not be considered. They are now technical terms. Differentiate ax^ and we find nax'^~^. Integrate naaf^^ and we find ax^ -\- c. We always add a constant when we integrate. Sometimes we write these, ^- (ax'^) = nax'^~^ and dx^ ^ \nax'^~^ . dx = aaf\ Obsei've that we write / before and dx after a function when we wish to say that it is to be integrated with regard to X. Both the symbols are needed. At present you ought not to trouble your head as to why these particular sorts of symbol are used*. You will find presently that it is not difficult to learn how to differentiate any known mathematical function. You will learn the process easily ; but integration is a process of guessing, and however much practice we may have, ex- perience only guides us in a process of guessing. To some extent one may say that differentiation is like multiplica- tion or raising a number to the 5t]i power. Integration is like division, or extracting the 5th root. Happily for the engineer he only needs a very few integi-als and these are * When a great number of things have to be added together in an engineer's office — as when a clerk calculates the weight of each little bit of a casting and adds them all up, if the letter w indicates generally any of the little weights, we often use the symbol I,io to mean the sum of them all. When we indicate the sum of an infinite number of little quantities we replace the Greek letter s or S by the long English s or / . It will be seen presently that Integration may be regarded as finding a sum of this kind. Thus if y is the ordinate of a curve ; a strip of area is y . dx and jy . dx means the sum of all such strips, or the whole area. Again, if 8m stands for a small portion of the mass of a body and r is its distance from an axis, then r^ . 57/1 is called the moment of inertia of dm about the axis, and Sr^ . dm or / r^ . dm indicates the moment of inertia of the whole body about the axis. Or if 5 F is a small element of the volume of a body and m is its mass per unit volume, then ir^m . dV is the body's moment of inertia. 3—2 36 CALCULUS FOR ENGINEERS. well known. As for the rest, lie can keep a good long list of them ready to refer to, but he had better practise working them out for himself. Now one is not often asked to integrate n(Mf~^. It is too nicely arranged for one beforehand. One is usually asked to inteoi-ate 6a?"*...(l). I know that the answer is -,...(2). How do I prove this ? By differentiating (2) I obtain (1), there- fore I know that (2) is the integral of (1). Only I ought to add a constant in (2), any constant whatever, an arbitrary constant as it is called, because the differential coefficient of a constant is 0. Students ought to work out several examples, integrating, say, a*^, hsc^, hx^, ax~^, ci^, auo^. When one has a list of differential coefficients it is not wise to use them in the reversed way as if it were a list of integrals, for things are seldom given so nicely arranged. For instance J4>a^ .da; = a^. But one seldom is asked to integrate ^oc^, more likely it will be Sx^ or 5./"^, that is given. We now have a number of interesting results, but this last one includes the others. Thus if y = (f? or y — x^ or y = a?^ or 3/= a;"*, we only have examples of y — x*^, and it is good for the student to work them out as examples. Thus ax If 7i = 1 this becomes \x^ or 1. If ?i = it becomes ()x~'^ or 0. But we hardly need a new way of seeing that if y is a constant, its differential coefficient is 0. We know that if y = a-^hx + cx^ + ex^-\- &c. + (/x^\ f = + h + 2cx-^ Sex" + &c. + ng with this knowledge we have the means of working quite * I suppose a student to know that anj^thing to the power is unit}-. It is instructive to actually calculate by logarithms a high i-oot of any number to see how close to 1 the answer comes. A high root means a small power, the higher the root the more nearly does the power approach 0. Then ^ = + 6 + 2c^ + Sex" + &c. + ngx''-^ INTEGRATION OF *'»'. 37 half the problems supposed to be difficult, that come before the engineer. , The two important things to remember now, are : If y = ax'K then -~ = nax'^~^ ; and if -:,*- = bx^, then -^ ' dx ' dx ' y = i^''"+i + c, ^ m + 1 where c is some constant, or / bx'^dx = *•'«+! + c. m + 1 I must ask students to try to discover for themselves illustrations of the fact that if y = x'\ then -i^ = ??a'"-^ I do -^ dx not give here such illustrations as happened to suit myself; they suited me because they were my own discovery. I would suggest this, however : Take y = af\ Let x = l'()2, calculate y by logarithms. Now let X = 1*03 and calculate y. Now divide the increment of y by *01, which is the increment of x. Let the second x be 1"021, and repeat the process. Let the second x be 1*0201, and repeat the process. It will be found that ^^ is approaching the true value of ^^ which is 5 (l-02y. Do this again when y = x^"^ for example. A student need not think that he is likely to waste time if he works for weeks in manufacturing numerical and graphical illustrations for himself. Get really familiar with the simple idea that if V = j;" then -,- = nx'^-^ : ^ dx ' that \ax^ . dx — — — - ip*+i + constant : J 5 + 1 that \av^ . dv = \f^^ + constant. j sH-1 Practise this with s= '7 or "8 or I'l or —5 or —"8, and use other letters than x or v. / 38 CALCULUS FOR ENGINEERS. 29. Exercises. Find the following Integrals. The constants are not added. Ix". dx. Answer, \x^. \v^ . dv. Answer, ^if. /v~*. dv. The answer i.> - — v^~^- J 1-5 j\^V' .dv or jv^ , dv. Answer, ^v^, r^.dt. Answer, 2e^ j-dx or lx~^.dx. Here the rule fails to help us for of . . we get -zr which is x , and as we can always subtract an in- finite constant our answer is really indeterminate. In our work for some time to come we need this integral in only one case. Later, we shall prove that I— dx = log X, and 1 ^— dx = log (x + a), and if y = log a?, -— = - and l-dv = log v. If j9 = av^, then ~ = Sav\ If t; = mrK then ^ = _ ^^^rl at 30. If pv = Rt, where i^ is a constant. Work the following exercises. Find -— , if v is constant. Answer, — . Find ,- , if « is constant. Answer, — . at ^ p The student knows already that the three variables p, v and t are the pressure volume and absolute temperature of a ''dp gas. It is too long to Avrite -^ when v is constant." We ctt 1. PARTIAL DIFFERENTIATION. 39 shall use for this the synibol [-4i) , the brackets indicating that the variable not there mentioned, is constant. Find (;/ ) • Answer, As p=Rt . v~^ we have [-T-) = — Rtv~\ and this simplifies to —pv~^. Find f ^' ] . Answer, As v = Rt . p~^ we have (;t- ) = — Rtp~^, and this simplifies to — vp''\ Find (I) . Answer, As i = | . /, we have Q = ^ . Find the continued product of the second, fifth, and third of the above answers and meditate upon the fact that fdv\ fdt\ (dp\___ [dtj [dp) [dv)" Generally we may say that if u is a function of two variables x and y, or as we say 'dv> \dxj efficient of u with regard to x when y is considered to be constant. These are said to be partial differential coefiKcients. 31. Here is an excellent exercise for students : — Write out any function of x and y ; call it u. Find ( ^7- ) • Now differentiate this with regard to y, assuming that x is constant. The symbol for the result is dhi dy .dx' It will always be found that one gets the same answer if one differentiates in the other order, that is d^u d^u .(3). then we shall use the symbol ( t- ) to mean the differential co- dy . dx dx . dy 40 CALCULUS FOR ENGINEERS. Thus iry 11 = 0,^ + if+ a(i?ij + hx]f. Again, \-^\ = + 3?/= 4- ax- + ^hxy, and r/""V = ^ + ^ + 2aa; + 2bij, which is the same as before. A student ought not to get tired of doing this. Use other letters than w and y, and work many examples. The fact stated in (3) is of enormous importance in Thermodynamics and other applications of Mathematics to engineering. A proof of it will be given later. The student ought here to get familiar with the importance of what will then be proved. 32. One other thing may be mentioned. Suppose we have given us that ii is a function of x and y, and that (du ^ =aa? + hf-\-co(?y-\-gxy\ Then the integral of this is 11 = lax* + by^x + Jc^y + ^gxy -\-f( y), where f{y) is some arbitrary function of y. This is added because we always add a constant in integration, and as ?/ is regarded as a constant in finding (-7- J we add f(y), which may contain the constant y in all sorts of forms multiplied by constants. 33. To illustrate the fact, still unproved, that if y = log x, then ^- = - . A student ouo^ht to take such values of x as 3, dv X ^ 3001, 3*002, 3"003 &c., find y in every case, divide increments of y by the corresponding increments of x, and see if our rule holds good. STEAM AND POWER. 41 Note that when a mathematician writes logo) he always means the Napierian logarithm of x. 34. Example of / — =log « + constant. It is proved in Thermodynamics that if in a heat engine the work- ing stuff' receives heat // at temperature t, and if t^ is the temperature of the refrigerator, then the work done by a perfect heat engine would be ir.'-Jo,orH(l-f). If one pound of water at Iq is heated to ^j, and wc assume that the heat received per degree is constant, being 1400 foot-lbs. ; what is the work which a j^erfect heat engine would give out in equivalence for the totiil heat ? Heat energy is to be expressed in foot-pounds. To raise the temperature from t to t-\-dt the heat is 14005^ in foot-lb. This stands for JI in the above expression. Hence, for this heat we have the equivalent work 8 W= l-iOOdt ( 1 - -^ j , or, rather, "^^ =1400-1400^. at t Hence W= \AQOt - UOOIq log t + constant. Now W=0 when t=tQ, = 1400^0 - 1400^0 log ^0 + constant, therefore the constant is known. Using this value we find equivalent work for the heat given from t^ to ^i= 1400 (t^ - to) - 1400^0 ^^S j" • If now the pound of water at t^ receives the heat X^ foot-lb. (usually called Latent Heat) and is all converted into steam at the constant temperature t^ , the work which is thermodynamically equivalent to this is L^ll--]. We see then that the work which a perfect steam engine would give out as equivalent to the heat received, in raising the pound of water from t^ to t^ and then evaporating it, is 1400 {t, - 1,) - urn, loge ^^ + X, (^1 - ^f^ . Exercise. What work would a perfect steam engine perform per pound of steam at ^^ = 439 (or 102 lb. per sq. inch), or 165^0, if ^q = 374 or 100' C. Here L^ = 681 ,456 foot-pounds. The work is found to be 107,990 ft.-lb. per lb. of steam. Engineers usually wish to know how many pounds of steam are used per hour per Indicated Horse Power, w lb. per hour, means pr! ^^ ft.-lb. per minute. Putting this equal to 3.3(KM) we find 2V to be 18-35 lb. of steam per hour per Indicated Horse Power, as the requirement of a perfect steam engine working between the temperatm'es of 165° C and 100° C. 42 CALCULUS FOR ENGINEERS. 35. Exercises. It is proved iii Thermodynamics when ice and water or water and steam are together at the same temperature, if Si is the volume of unit mass of stuff in the higher state and Sq is the volume of unit mass of stuff in the lower state. Then where t is the absolute temperature, being 274 + 6" C, L being the latent heat in unit mass in foot-pounds. If we take L as the latent heat of 1 lb. of stuff, and Si and Sq are the volumes in cubic feet of 1 lb. of stuff, the formula is still correct, p being in lb. per sq. foot. I. In Ice-water, So = '01747, 8^ = '01602 at ^=274 (cor- responding to 0° C), p being 2116 lb. per sq. foot, and X = 79 X 1400. Hence ^ = - 278100. at And hence the temperature of melting ice is less as the pressure increases ; or pressure lowers the melting point of ice ; that is, induces towards melting the ice. Observe the quantitative meaning of -^ ; the melting point lowers at the rate of '001 of a degree for an increased pressure of 278 lb. per sq. foot or nearly 2 lb. per sq. inch. II. Water Steam. It seems almost impossible to measure accurately by experiment, Si the volume in cubic feet of one pound of steam at any temperature. So for water is known. Calculate s^ — Sq from the above formula, at a few temperatures having from Regnault's experiments the follow- ing table. I think that the figures explain themselves. e^c t absolute pressure in lb. per sq. inch P lb. per sq. foot Sp 5t assumed dp dt L in foot- pounds Sj-So 100 374 14-70 211G-4 81-5 105 379 17-r)3 2524 94 87-8 740,710 22-20 110 384 20-80 2994 LATENT HEAT. 43 It is here assumed that the vakie of dpjdt for IDS'" C. is half the sum of 81 "5 and 94. The more correct way of proceeding would be to plot a great number of values of hplht on squared paper and get dpjdt for 105° C. more accurately by means of a curve, f s, - So for 105° C. = 740710 -- (379 x 87*8) = 22-26. Now So = "OIG for cold water and it is not worth while making any correction for its warmth. Hence we may take s^ = 22*28 which is sufficiently nearly the correct answer for the present pui-pose. Example. Find Si for 275° F. from the following, L being t°F. 248° 4152 257° 266° I 275' 284° ; 293° 302° 4854 I 5652 6551 7563 8698 9966 Example. If the formula for steam pressure, p=a6^ where a and b are known numbers, and 6 is the temperature measured from a certain zero which is known, is found to be a useful but incorrect formula ibr re])reseiiting Reguault's experimental results ; deduce a fornuila for the volume s^ of one p<nnid of steam. We have also the well known formula for latent heat L—c-et^ where t is the absolute temperature and c and e are known numbers. Hence, as i? which is the same as dv • ' de ^ is haB^-\ s^-SQ=(c-et)^tbae^-\ After subjecting an empirical formula to mathematical operations it is wise to test the accuracy of the result on actual experimental numbers, as the formula represents facts only approximately, and the small and apparently insignificant terms in which it difl:ers from fact, may become greatly magnified in the mathematical operations. 36. Study of Curves. When the equation to a new curve is given, the practical man ought to rely first upon his power of plotting it upon squared paper. Very often, if we find ~ or the slope, everywhere, it gives us a good deal of information. If we are told that a.\, y^ is a point on a curve, and we are asked to find the equation to the tangent there, we have simply to find the straight line which lias the same slope as the curve there and which passes through x^, y^. The normal is the straight line which passes through x^, y^ and whose slope is minus the reciprocal of the slope of the curve there. See Art. 13. CALCULUS FOR ENGINEERS. P (fig. 8) is a point in a curve APB at which the tangent FS and the normal FQ are drawn. OX and Y are the axes. OR — x, RF — y, tan FSR = ;,- •, the distance SR is called the subtangent ; prove that it is equal to ?/ -i- -^ . The distance RQ is called the subnormal; it is evidently dii equal to y -^ . The length of the tangent FS will be found to be y\J\-\-{,\ , the length of the normal FQ is y W^^ ^£i • ^^^ Intercept OS is ^- - y '^ . Example 1. Find the length of the sub tangent and sub- normal of the Parabola y = mx'^, - dy dx 2nix. Hence Subtangent = mx- -=- 2?/iif or ^x. Subnormal = y x 2nix or 2m^x^. Example 2. Find the length of the subtangent of y = Wio;", dy_ dx mnx' Subtangent = nix'^ -i- vinx'^~^ — xju. CURVES. 45 Example 3. Find of Avhat curve the subnormal is constant in length, dy d^ 1 y-f- = a or -r = ~ n. ^ dx dy a-^ The integral of - ?/ Avith regard to y is a* = ~ 3/^ + a con- Cl Act stant h, and this is the equation to the curve, where h may have any value. It is evidently one of a family of parabolas. (See Art. 9 where xs and y's are merely interchanged.) Example 4. The point ^ = 4, ?/ = 3 is a point in the para- bola y = '^a^. Find the equation to the tangent there. The dy slope is -^ = 1- X f A'"^ or, as ;r = 4 there, the slope is f x ^ or -| . The tangent is then, y = m+ g^. To find m we have ^ = 3 when a? = 4 as this point is in the tangent, or 3 = 9/1 + | x 4, so that ??i is 1 J and the tangent is 3/ = li|- + %x. Example 5. The point ^ = 32, 1/ = 3 is evidently a point in the curve y = 2-{-^x-'. Find the equation to the normal there. dij The slope of the curve there is ^ = -^x' * = yj^ and the slope of the normal is minus the reciprocal of this or — 160. Hence the normal is y = m — \ Q>Ox. But it passes through the point x=S2, y = S and hence S = m— KiO x 32. Hence m = 5123 and the normal is y = 5123 — 160;r. Example 6. At what point in the curve y = aa?""" is there the slope 6? dy -f = - nax-'^-K ax The point is such that its x satisfies — nax~^^~'^ = h or, 1 — j-j . Knowing its x we know its y from the equation to the curve. It is easy to see and well to remem- ber that if ^1, 3/1 is a point in a straight line, and if the slope of the line is 6, then the equation to the line most quickly written is x — x^ 46 CALCULUS FOR ENGINEERS. Hence the equation to the tangent to a curve at the point a?i, 2/1 on the curve is •^^ y^ = the -^ at the point. X — Xx dx And the equation to the normal is x — Xi aij - = — the ^ at the point. Exercise 1. Find the tangent to the curve x^y^^ = a at the point x^, y^ on the curve. Answer, — X'\- - y=m-\- n. Exercise 2. Find the normal to the same curve. Answer, — (a? — a?,) m Xi (!/ - y.) = 0. Exercise 3. Find the tangent and normal to the parabola y" = ^ax at the point where x = a. Answer, y = x -\- a, y = Sa — x. Exercise 4. Find the tangent to the curve y =^ a -{-hx ■\- ca^ + ex^ at a point on the curve x^, y^. Answer, ^ ~^-^ = 6 + 2cXi -f ^ex^^ X — Xi 37. When y increases to a certain value and then diminishes, this is said to be a maximum value of y; when y diminishes to a certain value and then increases, this is said to be a minimum value of y. It is evident that for either case -^ = 0. See Art. 16 and fief. 6. dx ^ Example 1. Divide 12 into two parts such that the product is a maximum. The practical man tries and easily finds the answer. He tries in this sort of way. Let x be one part and 12— x the other. He tries x = 0, x = l, x = 2, &c., in every case finding the product. Thus a^|0|l|2i3 4 5 6 7 8 9 Product I i 11 i 20 27 32 35 36 35 32 27 MAXIMA AND MINIMA. 47 It seems as if ^' = 6, giving the product 36, were the correct answer. But if we want to be more exact, it is good to get a sheet of squared paper; call the product y and plot the corresponding values of x and y. The student ought to do this himself Now it is readily seen that where y has a maximum or a minimum value, in all cases the slope of a curve is 0. Find then the point or points where dy\dx is 0. Thus if a number a is divided into two parts, one of them X and the other a — x, the product is y = ^ (a — a;) or ax — x^, and £ =a — 2x. Find where this is 0. Evidently where 2x — a or x= \a. The practical man has no great difficulty in any of his problems in finding whether it is a maximum or a minimum which he has found. In this case, let a = 12. Then x = Q gives a product 36. Now if a? = 5 999, the other part is 6*001 and the product is 35-999999, so that a? = 6 gives a greater product than x = 5*999 or x= 6*001, and hence it is a maxi- mum and not a minimum value which we have found. This is the only method that the student will be given of dis- tinguishing a maximum from a minimum at so early a period of his work. Example 2. Divide a number a into two parts such that the sum of their squares is a minimum. If x is one part, a—x is the other. The question is then, if y = ar' + (a — xf, when is y a minimum ? 2/ = 2^-2 + a^ - 2ax, -^= 4^ — 2a, and this is when x — ^a. dx 2 Examjile 3. When is the sum of a number and its reciprocal a minimum ? Let x be the number and y =x + -. When is 2/ a minimum ? The differential coefficient of - or x~'^ being — x~^, we have dy -, 1 11..^. ^ . -IT" = 1 -, and this is when a? = 1. 48 CALCULUS FOR ENGINEEKS. The student ought to take numbers and a sheet of squared paper and try. Trying a;= 100, 10, 4 &c. we have 100 2/ 100-01 10 4 2 1 i i i 101 4-25 2-5 2 2-5 ^ 4i Now let him plot x and i/ and he will see that y is a minimum when a; = 1. Example 4. The strength of a rectangular beam of given length, loaded and supported in any 2:)articular way, is proportional to the breadth of the section multiplied by the square of the depth. If the diameter a is given of a cylindric tree, what is the strongest beam which may be cut from it ? Let X be its breadth. Then if you draw the rectangle inside the circle, you will see that the depth is Ja'^ — x-. Hence the strength is a maximum when y is a maximum if y = x((r--x-), or 3/ = a\x — x-\ j^ = a- — 3a", and this is when x—~=^, dx V3 In the same way find the siiffest beam which may be cut from the tree by making the breadth x the cube of the depth a maximum. This, however, mav wait till the student has read Chap. III. Example 5. Experiments on the explosion of mixtures (at atmospheric pressure) lead to a roughly correct rule p = 83-3-2a?,t where p is the highest pressure produced in the explosion, and X is the volume of air together with products of previous combustions, added to one cubic foot of coal gas before ex- plosion. Taking px as roughly proportional to the work done in a gas engine during explosion and expansion ; what value of X will make this a maximum ? FUEL ON A VOYAGE. 49 That is, when is 83^3 — S'2x^ a maximum ? Answer, When 83 — 6'4fX= 0, or x is about 13 cubic feet. I am afraid to make Mr Grover responsible for the above result which I have drawn from his experiments. His most interesting result was, that of the above 13 cubic feet it is very much better that only 9 or 10 should be air than that it should all be air. Example 6. Prove that ax — x^ is a maximum when x — \a. Example 7. Prove that x — x^ is a maximum when ^=:^V3. , Example 8. The volume of a circular cylindric cistern being given (no cover) when is its surface a minimum ? Let X be the radius and y the length ; the volume is irafy = a, say (1). The surface is ttx^ + 27rxi/ (2). When is this a minimum ? From (1), y is — ; using this in (2) we see that we TTX" must make , 2a . . TTX^ H a mmimum, X 27rx r=0 or x^ = - . a? TT „ iTX^y x^= — — or x=^y. TT The radius of the base is equal to the height of the cistern. Example 9. Let the cistern of Ex. 8 be closed top and bottom, find it of minimum surface and given volume. The surface is 2irx- + ^irxy, and proceeding as before we find that the diameter of the cistern is equal to its height. Example 10. If v is the velocity of water in a river and X is the velocity against stream of a steamer relatively to the luater, and if the fliel burnt per hour is a + bx^; find the p. 4 60 CALCULUS FOR ENGINEERS. velocity a? so as to make the consumption of fuel a minimum for a given distance m. The velocity of the ship relatively to the bank of the river is ac—v, the time of the passage is , and therefore the fuel burnt during the passage is OS —~ V m(a-\- ha?) x — v Observe that a -\- hx^ with proper values given to a and h may represent the total cost per hour of the steamer, in- cluding interest and depreciation on the cost of the vessel, besides wages and provisions. You cannot yet differentiate a quotient, so I will assume a? a — 0, and the question reduces to this: when is a x — v minimum ? Now this is the same question as : — when is X "^ V — r— a maximum ? or when is ar^ — vx~^ a maximum ? The a? differential coefficient is — 2ar^ + ^vx~\ Putting this equal to we find x = '^v, or that the speed of the ship relatively to the water is half as great again as that of the current. Notice here as in all other cases of maximum and mini- mum that the engineer ought not to be satisfied merely with such an answer. x='^v is undoubtedly the best velocity, it makes a?j{x — v) a minimum. But suppose one runs at less or more speed than this, does it make much difference ? Let V = 6, the best a? is 9, — x-=243 if «; = 9. .-r — 6 = 250 if a; = 10. = 256 if a;=8; and these figures tell us the nature of the extra expense in case the theoretically correct velocity is not adhered to*. * Assuming that you know the rule for the differentiation of a quotient — usually learnt at the very beginning of one's work in the Calculus, and without assuming a to be as above, we have (a;-i?)36a;*=a + 6x', 26x3_3tv«2_rt (1). CUllRENT FROM CELLS. 51 Example 11. The sum of the squares of two factors of a is a minimum, find them. If x is one of them, - is the X Qj^ (it/ ZiO' other, and y'=x^-{-— is to be a minimum, -^ = 2x , and x^ ax x^ this is when x^ = ct- ^r x — \'a. Example 12. To arrange n voltaic cells so as to obtain the maximum current through a resistance R. Let the E.M.F. of each cell be e and its internal resistance r. If the cells are arranged as x in series, n-jx in parallel, the E.M.F. of the battery is xe, and its internal resistance is - - . Hence the current G == xe-^ I \- R] a As the student cannot yet differentiate a quotient, we shall say that (7 is a maximum when its reciprocal is a ... fx'r . „\ xr R minimum, so we ask when is h R] -^^e or 1 — \ n J n X Given the values of a, 6 and v the proper value of x can be found by trial. Thus let the cost per day in pounds be 30 + ^V^^ so that a = 30, 6 = ^ and let r = 6. Find x from (1) which becomes .r3_ 9^2- 300 = (2). I find that a; = 11-3 is about the best answer. Tliis is a cubic equation and so has three roots. But the engineer needs only one root, he knows about how much it ought to be and he only wants it approximately. He solves any equation whatsoever in the following sort of way. Let x^-9x--S00 be called f{x). The question is, what value of x makes this 0? Try a;= 10, f{x) turns out to be - 200, X 10 j 8 12 11 11-3 1 /(^) -200 I -360 + 176 -57 -6 whereas we want it to be 0. Now I try a; = 8, this gives -360 which is further wrong. Now I try 12 and I get 176 so that x evidently lies between 10 and 12. Now I try 11 and find - 57. It is now worth while to use squared paper and plot the curve y=f{x) between a; = 10 and a;=12. One can find the true answer to any number of places of decimals by repeating this process. In the present case no great accm*acy is wanted and I take a: = 11*3 as the best answer. Note that the old answer obtained by assuming a = is only 9. A practical man will find much food for thought in thinking of these two answers. Note that the captain of a river steamer must always be making this sort of calculation although he may not put it down on paper. 4—2 52 CALCULUS FOR ENGINEERS. . . ?' jR minimum ? Its differential coefficient is and this is when R = — , which is the internal resistance of the battery. Hence we have the rule : Arrange the battery so that its internal resistance shall be as nearly as possible equal to the external resistance. Example 13. Voltaic cell of E.M.F. = e and internal re- sistance r; external resistance i?. The current is (7= ^z. The power given out is P = RC\ What value of R will make P a maximum ? P = R {r + Rf To make this suit such work as we have already done we may (?• + Ry say, what value of R will make ^^ — j~^ a minimum, or 7^ + 2Rr + jR^ D ~ <^r ')'^R~^ + 2r + P a minimum ? Putting its diiferential coefficient with regard to R equal to we have - i-^R-^ -t- 1 = so that R = r, or the external resistance ought to be equal to the internal resistance. Example 14. What is the volume of the greatest box which may be sent by Parcels post ? Let x be the length, y and z the breadth and thickness. The P. 0. regulation is that the length plus girth must not be greater than 6 feet. That is, we want v = xyz to be a maximum, subject to the condition that x-\-2{y + z) = Q. It is evident that y and z enter into our expressions in the same way, and hence y = z. So that a; + 4i/ = 6 and v = xy^ is to be a maximum. Here as a; = 6 — 4^ we have -y = (6 ■- 4^) y^ or 6^* — 4^' to be a maxi- fj'i) mum. Putting -T- = we have 12^ — 123/^ = 0. Rejecting v = for an obvious reason, y = 1, and hence our box is 2 feet long, 1 foot broad, 1 foot thick, containing 2 cubic feet. Find the volume of the greatest cylindric parcel which may be sent by Post. Length being I and diameter d,l + ird — Q IT . . 4 and -r Idr is to be a maximum. Answer, I = 2 feet, d = — feet, 4 TT Volume = 8 -^ TT or 2'55 cubic feet. WATER IN STEAM CYLINDER. 58 Example 15. Ayrton-Perry Spring. Prof. Ayrton and the present writer noticed that in a spiral spring fastened at one end, subjected to axial force F, the free end tended to rotate. Now it was easy to get the general formula for the elongation and rotation of a spring of given dimensions, and by nothing more than the above principle we found what these dimensions ought to be for the rotation to be great. Thus for example, the angle of the spiral being a the rotation was proportional to sin a. cos a. It at once followed that a ought to be 45°. Again, the wire being of elliptic section, x and y being the principal radii of the ellipse, we found that the rotation was proportional to To make this a maximum, the section (which is propor- tional to xy) being given. Let xy = 5, a constant, then the above expression becomes "-— ^ — r, and this is to be a maximum. s^ o sy^ Here we see that there is no true maximum. The larger we make y or the smaller we make y (for small values of y the rotation is negative but we did not care about the direc- tion of our rotation, that is, whether it was with or against the usual direction of winding up of the coils) the greater is the rotation. This is how we were led to make springs of thin strips of metal wound in spirals of 45°. The amount of rotation obtained for quite small forces and small axial elongations is quite extraordinary. The discovery of these very useful springs was complete as soon as we observed that any spring rotated when an axial force was applied. Students who are interested in the practical application of mathematics ought to refer to the complete calculations in our paper published in the Proceedings of the Royal Society of 1884. Example 16. From a Hypothetical Indicator Diagram the indicated work done per cubic foot of steam is w — 144^1 (1 -f- log r) — 144?"j[)3 — x. 54 CALCULUS FOR engineehs. where jh and jXi arc the initial and back pressures of the steam ; r is the ratio of cut off (that is, cut off is at -th of the stroke) and a? is a loss due to condensation in the cylinder, w depends upon r. 1st. If X were 0, what value of r would give most indicated work per cubic foot of steam ? We must make .— = 0, and we find — =^ — 144«.i = or ?' = — . If it is brake energy which is to be a maximum per cubic foot of steam, we must add to p^ a term represent- ing engine friction. 2nd. Mr Willans found by experiment in non-condensing engines that ?• = — ^— gave maximum indicated w. Now if we put in the above -,— = we have ^ — 144^^ — i =0. So that ~~ = ^^^ (p, + 10) - 144^3 or ^^^ = 1440. So that .r = 1440r + constant. Hence Mr Willans' practical rule leads us to the notion that the work lacking per cubic foot of steam is a linear function of 7\ This is given here merely as a pretty exercise in maxima and*minima. As to the practical engineering value of the result, much might be said for and against. It really is as if there was an extra back pressure of 10 lb. per sq. inch which represented the effect of condensation. Mr Willans found experimentally in a non-condensing engine that the missing water per Indicated Horee Power hour is a linear function of ?• using the same steam in the boiler, but this is not the same as our .^. We sometimes assume the ratio of condensed steam to indicated steam to be proportional to log r, but a linear function of r will agree just as well with such experimental results as exist. Example 17. The weight of gas which will flow per second through an orifice from a vessel where it is at ECONOMY IN CONDUCTORS. 55 pressure />o into another vessel where it is at the pressure jfj is proportional to «> V 1 — « ^ ; where a is pjpf^ and 7 is a known constant, when is this a maximum ? That is, when is ay — a y a. maximum ? See Art. 74, where this example is repeated. Differentiating with regard to a and equating to we find j^ In the case of air 7 = 1*41 and we find p = "527^0 > that is, there is a maximum quantity leaving a vessel per second when the outside pressure is a little greater than half the inside pressure. Example 18. Taking the waste going on in an electric conductor as consisting of (1) the ohmic loss ; the value of G^r watts, where r is the resistance in ohms of a mile of going and coming conductor and C is the current in amperes; (2) the loss due to interest and depreciation on the cost of the conductor. I have taken the price lists of manufacturers of cables, and contractors' prices for laying cables, and I find that in every case of similar cables, similarly laid, or suspended if overhead, the cost of a mile of conductor is practically proportional to the weight of copper in it, that is, inversely proportional to the resistance, plus a constant. The cost of it per year will depend upon the cost of copper per ton, multiplied by the number taken as representing rate per cent, per annum of interest and depreciation. We can state this loss per year or per second, in money per year or per second and the ohmic loss is in watts. We cannot add them together until we know the money value per year or per second of 1 watt. There are three things then that decide the value of the quantity which we call t^. I prefer to express the total waste going on in watts rather than in pounds sterling per annum and I find it to hey— C^r -\ f- 6, where h is some constant. The value of t may be taken as anything from 17 to 40 for the working of exercises, but 66 CALCULUS FOR ENGINEERS. students had better take figures of their own for the cost ot power, copper and interest*. For a given current (7, when is the y, the total waste, a minimum ? that is, what is the most economical conductor for a given current ? ^~^^~';^ ^^^ *^^^ ^^ ^ when r — -^. Thus if ^ = 40, r = ^. Now if a is the cross section of the conductor in square •04 inches, r = — nearly, so that (7= 1000a, or it is most econom- ical to provide one square inch of copper for every 1000 amperes of current. When w is a function of more than one independent * The weight of a mile of copper, a square inches in cross section, is to be figured out. Call it ma tons. If p is the price in pounds sterling of a ton of copper, the price of the cable may be taken as, nearly, pma + some constant. If R is the rate per cent, per annum of interest and depreciation, then the loss per annum due to cost of cable may be expressed in pounds as R -r^l'wa + some constant. If £1 per annum is the value of %v watts, (ob- 100 serve that this figure w must be evaluated with care. If the cable is to have a constaiat current for 24 hours a day, every day, w is easily evaluated), then the cost of the cable leads to a perpetual loss of zrj-r. wpma + some 100 constant. Now taking a= — , we see that our t^ is -^7=7^77 . ** r 2000 Men take the answer to this problem as if it gave them the most econom- ical current for any conductor under all circumstances. But although the above items of cost are most important, perhaps, in long cables, there are other Items of cost which are not here included. The cost of nerves and eyesight and comfort if a light blinks ; the cost in the armature of a dynamo of the valuable space in which the current has to be carried. If a man will only write down as a mathematical expression the total cost of any engineering contrivance as a function of the size of one or more variable parts, it is quite easy to find the best size or sizes; but it is not always easy to write down such a function. And yet this is the sort of problem that every clever engineer'is always working in his head; increasing something has bad and good effects; what one ought to do is a question in ' maxima and minima. Notice also this. Suppose we find a value of x which makes y a maxi- mum ; it may be, that quite different values of x from this, give values of y which are not very different from the maximum value. The good practical engineer will attend to matters of this kind and in such cases he will not insist too strongly upon the use of a particular value of x. ELECTRIC CONDUCTORS. 57 variable^ say x and y. Then | — - J = 0, y being, considered constant during the differentiation, and f — j = 0, x being considered constant during the differentiation, give two equa- tions which enable the values of x and y to be found which will make u a maximum or a minimum. Here, however, there is more to be said about whether it is a minimum or a maximum, or a maximum as to x and a minimum as to y, which one has found, and we cannot here enter into it. Sometimes in the above case although it is a function of X and y, there may be a law connecting x and y, and a little exercise of common sense will enable an engineer to deal with the case. All through our work, that is what is wanted, no mere following of custom; a man's own thought about his own problems will enable him to solve very difficult ones with very little mathematics. Thus for example, if we do not want to find the best conductor for a given current of Electricity: if it is the Power to be delivered at the distant place that is fixed. If the distance is n miles, and the conductors have a resistance of r ohms per mile (go and return), if \\ is the potential, given, at the Generating end, and C is the current. Then the potential at the receiving end being T' ; F^ - T^= Cnr. CV=PiH fixed, and the cost per mile is y = C^r + -...{l), where t^ is known. When is y a minimum? Here both C and r may vary, but not independently. V= V^ - Cnr and P=(7F, — (7%r ... (2). One simple plan is to state ?/ in terms of r CV —P alone or of C alone. Thus r from (2) is r= — ^ — (3). Substituting for this jn (1), we get CV,-P . f-a^n y=-ir-^cv;^p ^^^- Here everything is constant except (7, so we can find the value of C to make y a minimum, and when we know C we also know r from (3). At present the student is supposed to be able to difibrentiate only ^", so he need not proceed with the problem until he has worked a few exercises in Chap. III.* * To differentiate (4) is a very easy exercise in Chap. III. and leads to j>, = — i + - — - — -/xr^ — 5svo > and on putting this equal to we obtain d(J n {CVi~Py JT o the required value of C. It would not be of much use to proceed further 58 CALCULUS FOR ENGINEERS. In ray Cantor lectures on Hydraulic Machinery, I wrote out an expression for the total loss in pounds per annum in Hydraulic transmission of povirer by a piiwj. I gave it in terms of the maximum pressure, the power sent in, and the diameter d of the pipe. It was easy to choose d to make the total cost a minimum. If how- ever I had chosen jo, the pressure at the receiving end as fixed, and the power delivei-ed as fixed, and therefore Q, the cubic feet of water per second, and if I had added the cost of laying as proportional to the square of the diameter, I should have had an expression for the total cost like i(^ pqi when the values of a, h and c depend upon the cost of power, the interest on the cost of iron, &c. This is a minimum when its differen- tial coefficient with regard too? is zero or ^cld—hal(^d'^-\-Zh^^M~^^ and d can be obtained by trial. The letters 6 and c also involv^ the strength of the material, so that it wjxs possible to say whether wrought iron or ca»t iron was on the whole the cheaper. But even here a term is neglected, the cost of the Engine and Pumps. The following example comes in conveniently here, although it is not an example of Maximum or Minimum. An Electric Conductor gives out continuously a amperes of current in every mile of its length. Let x be the distance of any jKiint in miles from the end of the line remote from the generator, let C be the current there and V the voltage. Let r ohms per mile be the resistance of the conductor (that is, of one mile of going and one mile of coming conductor). The current given out in a distance hx is dC, or rather hx -j- , and the power is bx . V '-f- y «o that if P is the power per mile (observe the meaning of per), ''='-f :•••(')• Also if V is voltage at x and V+bVat x+8x] — As the i-esistance is i' . 8x the current is dV-rr . dx^ or rather, since these expressions are not correct until 8x is supposed smaller and smaller witliout limit, ^-l^ (^)- unless we had numerical values given us, 91 = 10 miles, P= 20000 watts, f^=H)00, find G and then r. Consult a Paper in the Joimial of the Institution of tJie Society of Tele- graph Engineers, p. 120, Vol. xv. 1886, if there is any further difficulty. It has not yet been sufficiently noted that if V^ and P and r are given, there is a limiting length of line n=JV/4rP, and when this is the case P is exactly equal to the ohmic loss in the con- ductor. ELECTRIC TRACTION. 59 As -T- = ^h C=t<.<,- if C is when .r = 0. Hence if r is constant (2) becomes dV rax = , so that T^ = Fq + i ^^-^-^ .(3), ^o being the voltage at the extremity of the line. . (1) becomes P^aV^+^ra'^x'^ (4). Taking ro = 200 volts, a = 25 amperes per mile, r=l ohm per mile, it is easy to see by a numerical example, how the power dispensed per mile, and the voltage, diminish as we go away from the generator. X '' P 200 5000 1 212-5 5312 2 250 6250 3 312-5 7812 4 400 10,000 If Fi is the voltage at the Dynamo and the line is n miles long T\=F^-|-iam2from(4). The power per mile at the extremity being l\ = aVQ, if we are given V^ and 1\ to find T'q, we shall find that n cannot be greater than l\-V2^o, and this gives the limiting length of the line. If we wish, as in Electric traction to get a nearer approach to uniform P, let us try C=ax-hx^ (5), where a. h, c are constants. 1 ^y 7^ - - - = ax - oyf\ r ax \=V,+\rax-^-^^x<^^ (6). dC As P=V ,- , or V {a - cbx^-'^), we can easily detennine the three constants a, b, c so that P shall be the same at any three points of thehne. Thus let r=l ohm, Fo=100 volts, and let P= 10000 watts, where .^;=0, x=l mile, .r=l| miles. We find by trial that (7=100.r-14-75.r2-iu and from this it is eas}'^ to calculate Cat any point of the line 60 CALCULUS FOR ENGINEERS. Example 19. A machine costs ax-\-hy/\t'& value to mc is proportional to xy, find the best values of x and y if the cost is fixed. Here xy is to be a maximum. Let c = cw; + 63/, so that V = J -~ j.^> ^^d ^y is r ^ — T ^^. This is a maximum Hence ax = hy = 0/2 makes xy a when j = 2 r X or ax — -^ . b 2 maximum. Examj)le 20. The electric time constant of a cylindric coil of wire is approximately II = mxyzl{ax +hy + cz), where x is the mean radius, y is the difference between the internal and external radii, z is the axial length and m, a, b, c are known constants. The volume of the coil is 27rxyz. Find the values of x, y, z to make u a maximum if the volume of the coil is fixed. Let then 27r . xyz = g ; when is — I a minimum ? That is, substituting for z, when yz xz xy 90 is ax + bi/ 4- ^ — = v, say, a minimum ? As x and y are p(ir- fectly independent we put f ^ J = and ( ;t~ ) = 0, or ft + o-.^^=o. ITT^/.^- Fig. 9. and + 6 - s^^^, = 0, 2'7rxy^ so that A'^y = ^^^ , a^r 627r and ax - or y — T, .1 - -7- a27r x^ = -^ or a'^27r ^ V a=27r W ¥2'ir' and 2^ 'lirxy or -? 7«¥ V C»27i^ HANGING CHAIN. 61 38. The chain of a suspension bridge supports a load by means of detached rods ; the loads are about equal and equally spaced. Suppose a chain to be really continuously loaded, the load being w per unit length horizontally. Any veiy flat uniform chain or telegraph wire is nearly in this condition. What is its shape ? Let be the lowest point. OX is tangential to the chain and horizontal at 0. OF is vertical. Let P be any point in the chain, its co-ordinates being x and y. Consider the equilibrium of the portion OP. OP is in equilibrium, under the action of Tq the horizontal tensile force at 0, Tthe inclined tangential force at P and wx the resultant load upon OP acting ver- -p. ^^ tically. We employ the laws of forces acting upon rigid bodies. A rigid body is a body which is acted on by forces and is no longer altering its shape. If we draw a triangle whose sides are parallel to these forces they are proportional to the forces, and if 6 is the inclination of T to the c. 5^ ^a Y \ I ^M \ , horizontal and T wx cos 0. •(1). wr tan^. .(2), but tan ^ is j^ , so that ^ = 7^ ;«;... (3); dx dx Tq Pig. 11. 1 w hence, integrating, y = ^7rr^ + constant. - -to Now we see that y is when x is 0, so that the constant is 0. Hence the equation to the curve is 1 ^ ^ (*). 62 CALCULUS FOR ENGINEERS. and it is a parabola. Now tan ^ is ^ o^, so that sec- 6 is 1-f^,^^ Andasr = rosec^, r=ToA/l+^U-^...(5). From this, all sorts of calculations may be made. Thus if I is the span and D the dip of a telegraph wire, if the whole curve be drawn it will be seen that we have only to put in (4) the information that when x—\l,y = D, and the greater tension elsewhere is easy to find. In the problem of the shape of any uniform chain, loaded only with its own weight, the integi'ation is not so easy. I give it in a note*. When it is so flat that we may take the * The integration in this note requires a knowledge of Chapter in. If the weight of the portion of chain OP, instead of being wx is %cs, where s is the length of the curve from to P, the curve y is called the Catenary, Equation (3) above becomes | = ^^.or,e«ingr,=.c.^V.i (X). If bs is the length of an elementary bit of chain, we. see that in the limit {bsY={8xf^{byf 60 that "i- = A / 3- + If and hence -^ = . — . This being integrated gives y + c=v^c2 + «-...( 2), "« 'slc' + s^ the constant added in integration being such that « = when y = 0. From (2) we find 6-2=^2^ 2j/c... (3), and using this in (1), we have dx _ c the integral of which is as when y=0, a?=0, if is the origin, no constant is to be added. Putting this in the exponential form c^lc _ y ^ ^. ^. JyT. ^ 2yc, transposing and squaring we find HEATING SURFACE OF BOILER a3 load on any piece of it as proportional to the horizontal pro- jection of it, we have the parabolic shape. | 39. Efficiency of Heating Surface of Boiler. If 1 lb. of gases in a boiler flue would give out the heat in cooling to the temperature of the water (0 maybe taken as proportional to the difference of temperature between gases and water, but this is not quite correct), we find from Peclet's experiments that the heat per hour that flows through a square foot of flue surface is, roughly, md\ Let ^ = ^i at the furnace end of a flue and 6 = 0^ at the chimney end. Let us study what occurs at a place in the flue. The gases having passed the area S in coming from the furnace to a certain place where the temperature is 6, pro- Or changing the origin to a point at the distance c below O, as at O in fig. 12 >Yhere SP is y' and RP is x, we have This is sometimes called Using (1) we find sometimes called y' = c cosh xjc. , I xlc -x!c\ s = c sinh xjc. .(4). Nv R A ly O » Fig. 12. Note that tables of the values of sinh u and cosh u have been publighed. Returning to the original figure, the tension at P being T, T W8 AB BG -p , and from (3), dy ■y+c, so that W8 S Hence T=zio{y + c) or T=wy' 64 CALCULUS FOR ENGINEERS. ceed further on to a place where S has become aS' 4- ^S and has become 6 + B0 (really SO is negative as will be seen). A steady state is maintained and during one hour the gase i lose the heat 7n0^ . BS through the area hS. If during the hour W lb. of gases lost at the place the amount of heat -W.Bd, then orrather ^ = -^-g. «• That is, integrating with regard to 0, where c is some constant. Putting in 6 = ^i the temperature at the furnace end when S=0, we hav©' ' Wl Wl 0= —-pf-^c orc = ^, 711 Ui m Oi • 80 that (2) becomes ^•=^^U^J (3). This shows how 6 diminishes as S increases from the furnace end, and it is worth a student's while to plot the curve connecting S and 6. If now S is the whole area of heating surface and ^ = ^2 at the smoke-box end, »-^a-i) <•> The heat which one pound of gases has at the furnace end is 6^, it gives up to the water the amount 61—62. Therefore the efficiency of the heating surface may be taken as . j^t_ ^i- 62 . . ^'^~dr ^^^' and it follows from (4) that 1 ^ 6,mS HEATING SURFACE. 65 Now if W is the weight of coals burnt j^er hour ; W=13W' if air is admitted just sufficient for complete combustion ; W = about 20 W in the case of ordinary forced draught ; W = about 26 W in the case of chimney draught. In these cases 0^ does not seem to alter inversely as W, as might at first siglit appear : but we do not know exactly how 0^ depends upon the amount of excess of air ad- mitted. We can only say that if W'-^S is the weight of coal per hour per square foot of heating surface and we call it tu, there seems to be some such law as B = z, , where a clepends 1 -h aw ^ upon the amount of air admitted. In practice it is found that a = 0*5 for chimney draught and 0'3 for forced draught, give fairly correct results. Also the numerator may be taken as greater than 1 when there are special means of heating the feed water. Instead of the law given above (the loss of heat by gases in a flue oc 6^^), if we take what is probably more likely, that the loss is proportional to 0, Then (1) above becomes de mO ^ ^' W or S= loe: ^ + constant (2). Let 0= 6j at furnace end or when S=0 so that our con- . W stant IS — log Oi and (2) becomes m S 'i'-^ai) »■ If S is the area of the whole flue and 6^ is the temperature at the smoke-box end, then ^ = Zlog^i (4), Sm ^ 66 CALCULUS FOR ENGINEERS. The efficiency E J-^^ (5) becomes E=^\-~^ = l-e ^^ (6). Or if w is the weight of fuel per square foot of heating surface as above (6) becomes j5'=l-e"a«' (7). 40. Work done by Expanding Fluid*. If p is the Eressure and v the volume at any instant, of a fluid which as already done work W in expanding, one good definition of pressure is p—'-r- •..(!)) or in words, pressure is the rate at which work is done per unit change of volume. Another way of putting this is: if the fluid expands through the volume hv there is an increment 8 IF of work done so that 'p .hv=hW^ or jo=-— , but this is only strictly tiiie when hv is made smaller and smaller without li«iit, and so (1) is absolutely true. Now if the fluid expands according to the law p%}'=c. a constant ... (2) : p = cv~'j and this is the differential coefficient dW of W with regai*d to v or, as we had better write it down, -j— = cv'. We therefore integrate it according to our rule and we have T^=Z7^i^-'-*-^+<^' ^^^' where C is some constant. To find C\ let us say that we shall only begin to count W from v=Vi. That is, W=0 when v^Vi, Then 0= -^ v.^-' + C, so that C= - ^ v.^-'. Insert this value of G in (3) and we have F=^^(^i--V-) (4), ■ which is the work done in expanding from v^ to r. Now if we want to know W when ^ = ^2) we have ^^'i2=rr-/^2'"'-^^'"'^ ^^^- * Observe that if for p and v we write y and x this work becomes very easy. WORK BY EXPANDING FLUID. 67 This answer may be put in other shapes. Thus from (2) we know that c^p^v^' or pav/, so that iir —P_^.u, 1-8. a formula much used in gas engine and steam engine calculations. There is one case in which this answer turns out to be useless ; try it when s — \. That is, find what work is done from Vy to Vg by ^ fluid expanding according to the law (it would be the isothermal law if the fluid were a gas) •pv—c. If you have noticed how it fails, go back to the statement dW dv (7). You will find that when you integrate x"^ with regard to m, the general answer has no meaning, cannot be evaluated, if m= - 1. But I have already said, and I mean to prove presently that the integral of x"^ is log^. So the integral of (7) is W=c\ogv-{-C. Proceeding as before we find that, in this particular case. ^1 (8). 41. Hypothetical Steam Engine Diagram. Let steam be admitted to a cylinder at the constant pressure jt?,. the volume increasing from to Vi in the cylinder. The work done is VjjOj. Let the steam expand to the volume Vg according to the law 'pijP^c. The work done is given by (6) or (8). Let the back pressure bejOg, then the work done in driving out the steam in the back stroke is p^^. We neglect cushioning in this hypothetical diagram. Let v^-~v^ be called r the ratio of cut-off. Then the nett work done altogether is Fig. 13. 5-^2 68 CALCULUS FOR ENGINEERS. If pe is the effective i)re.s.sure so that ^e^^ ^^ equal to the above iiett work Ipe is measured from actual indicator diagrarus, as the average pressure) ; putting it equal and dividing by v^ we have on simplifying ■M-) ■Vz- In the special case of s = l we find pe=Pi p^ in the same way. 42. Definite Integral. Definition. The symbol I /(a?) . dx tells us: — " Find the general integral of/ (a?); insert in it the value a for ic, insert in it the value 6 for a? ; subtract the latter from the former value*." This is said to be the (1), tells us to integrate u (which is a function of x and ?/), with regard to y, as if x were constant ; then insert F{x) for y and also f(x) for y and subtract. This result is to be integrated with regard to x, and in the answer a and b are in- serted for X and the results sub- tracted. I. If w = 1, dx . dy evidently means an element of area, a little rectangle. The result of the first process leaves Fig. 14. /: {F{x)-f{x)}dx (2) still to be done. Evidently we have found the area included between the curves y = F{x) and y~f{x) and two ordinates at a: = a and x = b. Beginners had better always use form (2) in finding areas, see fig. 14. II. If u is, saA', the weight of gold per unit area upon the above men- tioned area, then u . dx . dy is the weight upon the little elementary area dx .dy, and our integral means the weight of all the gold upon the area I have mentioned. When writers of books wish to indicate generally that they desire to integrate some jwoperty u (which at any place is a function of x, y, z), throughout some volume, they will write it with a triple integral, /// u .dx.dy. dz, and summation over a surface by I j V . dx . dy. AREAS OF CURVES. 69 integral of f{x) between the limits a and h. Observe now that any constant which may be found in the general integral simply disappears in the subtraction. Iq integrating between limits we shall find it convenient to work in the following fashion. Example, to find I x^ . dx. The general integral is ^xf^ J h and we write I x"^ .dx—\ ^a^ — ^d^ — ^bK Symbolically. If F{x) is the general integral of /{x) then rf(x) . dx = P F(x)] =F(a) - F(b), J b U J Note as evidently true from our definition, that I f(x),dx = - I f(x).dx, J b J a and also that T/w • ^^' = f Vw • ^^' + f "/(^) • ^^• 43. Area of a curve. some function of x and let PS be the curve. It is required to find the area MPQT. Nowiftheareai¥PQr be called A and OT=x, QT=y, OW = X -ir Sx, WR = y-\-8y, and the area MPRWheA + 8A then BA = siTesiTQIiW. Let y of the curve be known as / Fig TW , 15. Indeed some writers use V over an area, and jto . Ih dS to mean generally the summation of els to mean the summation of 2a along a line or what is often called "the line integral of ic. The line integral of the pull exerted on a tram car means the work done. The surface integral of the normal velocity of a fluid over an area is the total volume flowing per second. Engineers are continually finding line, surface and volume integrals in their practical work and there is nothing in these symbols which is not already perfectly well known to them. 70 CALCULUS FOR ENGINEERS. .(1). If the short distance QR were straight, Therefore ^ = y + i^y» as ha) gets smaller and smaller and in the limit -y— =?/ Hence A is such a function of x that y is its differential coefficient, or A is the integral of y. In fig. 16 CQD is the curve y=a-\- hx^ and EMGF is the curve showing so that A is the integral of y. In what sense does A represent the area of the curve CD ? The ordinate of the A curve, GT^ repre- sents to some scale or other, the area of the y curve MPQT from some standard Fig. 16. ordinate MP. The ordinate TQ represents to scale^ the slope of EF at G. Observe, however, that if we diminish or increase all the ordinates of the A curve by the same amount, we do not change its slope anywhere, and y, which is given us, only tells us the slope of A. Given the y curve we can therefore find any number of A curves ; we settle the one wanted when we state that we shall reckon area from a particular ordinate such as MP. Thus, in fig. 16 if the general integral of y is F(x)-{- c. If we use the value x = OM we have, area up to MP from some unknown standard ordinate =F(OM)-i-c. Taking x = ON, we have area up to NR from some unknown standard ordinate =F{ON)-\-c. And the area between MP and NR is simply the difference of these F(ON) — F{OM), the constant disappearing. Now the symbol I y .dx tells us to follow these instruc- J OM tions : — integrate y ; insert ON for x in the integral ; insert OM for ^ in it ; then subtract the latter. We see therefore AREA OF PARABOLA. 71 that the result of such an operation is the area of the curve between the ordinate at OM and the ordinate at ON. If y and a; represent any quantities whatsoever, and a curve be drawn with y as ordinate and x as abscissa, then the integral j ; y .dx is represented by the area of the curve, and we now know how to proceed when we desire to find the sum of all such terms rs y . 800 between the limits x=b and a)=a when Bw is supposed to get smaller and smaller without limit. Example. Find the area enclosed between the parabolic curve OA, the ordinate AB and the axis OB. Let the equation to the curve be y = ax^ (1), where PQ = y and OQ = x. Let QR = Bx. The area of the strip PQRS is more and more nearly ax^ . Sx, as Bx is made smaller and smaller; or rather the whole rOB area is I aa^ . dx, whic'h is ^0 Y M A X p c o| ' G I k B Fig. 17. OB -\ OSn Now what is a in terms o{ AB and x==OB. Hence by (1) AB = a.OB^, so that a = OB^ AB OB^' (2). When y^AB, Therefore the area = 5 =^, OJ^ ^AB . OB ; that is, frds of the area of the rectangle OMAB. Observe that the area of a very flat segment of a circle is like that of a parabola when OB is very small compared with BA, 72 CALCULUS FOR ENGINEERS. Exercise 1. Find the area between the curve y = mx " and the two ordinates at a? = a and x — h. The answer is 'h I irix"''^ . dx = 1-71 nt 1-n (^,i-»_ai-n). Observe (as in Art. 40) that this fails when n — 1; that is, in the rectangular hyperbola. In this case the answer is JaX m \ogx = m log - . The equation to any curve being y = a -\- hx -{- ca^ -{■ ex^ 4-/^, the area is A = ax f ^bx^ + ^cx^ 4- \ex* + i/ar*. Here the area up to an ordinate at x is really measured from the ordinate where x = 0, because A =0 Avhen x=0. We can at once find the area between any two given ordinates. Exercise 2. Find the area of the curve y — a^x between the ordinates at *• = a and x — ^. aTx^ .dx = a\ ix^'] = ^ (y8^ - a^). Exercise 3. Find the area of the curve yx" = a between the ordinates at a; = a and x — fi. Answer : a \ x~'- .dx = a — x~ = a(a-'-^-'). 44. Work done by Expanding Fluid. When we use definite integrals the work is somewhat shorter than it was in Art. 40. For i{p = cv~', the work done from volume v^ to volume V2 is /V2 Fv-i 1 "1 c cy~*. dv or c\ y^"* or _- — -(^'2^"'- V') The method fails when s = l and then the integral is c\ "logeV =cl0ge— . L»i J ^1 CENTKE OF GRAVITY. 73 45. Centre of Gravity, Only a few bodies have centres of gravity. We usually mean the centre of mass of a body or the centre of an area. Fig. 18. If each little portion of a mass be multiplied by its dis- tance from any plane, and the results added together, they are equal to the whole mass multi- plied by the distance of its centre, X, from the same plane. Expressed algebraically this is If each little portion of a plane area, as in fig. 18, be multiplied by its distance from any line in its plane and the results added together, they are equal to the whole area multiplied by the distance of its centre x from the same line. Expressed algebraically this is Saaf = xl^a. Example. Find the centre of mass of a right cone. It is evidently in the axis OB of the cone. Let the line OA rotate about OX, it will generate a cone. Consider the circular slice PQR of thickness hx. Let OQ = x, then PQ or AB y^OB'^- The mass of PR multiplied by the distance from to its centre is equal to the sum of the masses of all its parts each mul- tiplied by its distance from the plane YOY^. The volume of the slice PR being its area iry'^ multiplied by its thickness hx ; multiply this by rn the mass per unit volume and we have its mass miry- . hx. As the slice gets thinner and thinner, the distance of its centre from gets more and more nearly x. Hence we have to find the sum of all such terms as mirxy- . hx, and put it equal to the whole mass {^irm . ABr . OB) multiplied by x, the distance of Fig. 19. 74 CALCULUS FOR ENGINEERS. its centre of gravity from 0. Putting in the value of y^ in terms of x we have roB fAB\^ I rmr ( -^f^ ] ^'^ • dx equated to ^TrmAB^ . OB . x. rOB FOB -] Now af,dx== i^ = iOB\ and hence '"^'^ (Tfn] iOB^^^ir^ii . AB^.OB.x. Hence x = J OB. That is, the centre of mass is f of the way along the axis from the vertex towards the base. 46. It was assumed that students knew how to find the volume of a cone. We shall now prove the rule. The volume of the slice PR is it .y"^ ^ hx and the whole volume is £ V • d^ = \y (oi)' *■= • '^- = - i^) [p] or J of the volume of a cylinder on the same base AG and of the same height OB, If we had taken y — ax all the work would have looked simpler. Example. Find the volume and centre of mass of uniform material (of mass m per unit volume) bounded by a para- boloid of revolution. Let PQ = y, OQ=^x, QS=Bx. Let the equation to the curve OFA hey=aa^ (1). The volume of the slice PSR is Try^ . Bx; so that the whole rOB volume is 1 tt . a^x . dx or ^0 Y n ■^'^ A O^ [4 S B ^ Fig. 20. l-iraKOB^ (2). SURFACE OF REVOLUTION. "75 Now what is a ? When 7/ = AB, a)=: OB, so that from (1), AB = a . OB^ and a is ^-^ . Hence the volume is ^ir -7y^ OB'^ or ^TT.A&.OB (3). That is, half the area of the circle AC multiplied by the height OB. Hence the volume of the paraboloid is half the volume of a cylinder on the same base and of the same height. (The volumes of Cylinder, Paraboloid of revolution, and Cone of same bases and heights are as 1 : i : J.) Now as to the centre of mass of the Paraboloid. It is rOB evidently on the axis. We must find I jriir . y^x . da), or jmira) . a^js . da;, or 7/t7ra- Ix^ . dx, -OB or mTTcv^ • i^M ^^^^ *his is ^mira^ . 0B\ Inserting as before the value of «- or -jTW ^^ have the integral equal to IniTT.OB^.AB^. This is equal to the whole mass multiplied by the x of the centre of mass, x, or m^Tr . AB^ . OB . x, so that x = ^OB. The centre of mass of a paraboloid of revolution is |rds of the way along the axis towards the base from the vertex. Example. The curve y = ax^ revolves about the axis of X, find the volume enclosed by the surface of revolution between x = and x = h. The volume of any surface of revolution is obtained by integrating iry"^ . dx. Hence our answer is TT {'a'x-^ . dx = J^ Vx'^-^A = ^-^ 6^+\ J 2n 4- 1 Lo J 2/1 + 1 Find its centre of mass if in is its mass per unit volume. For any solid of revolution we integrate m . xiry^ . dx and 76 ^ CALCULUS FOR ENGINEERS. divide by the whole mass which is the integral of mir'fdx. If m is constant we have rb rb niTT I a;a"iv-"' . clx = niirtv^ I a;2"+i dx Jo Jo ~2n + 2Lo J~2n + 2^ ' and the whole mass is _ 6-"+\ so that Jj = _ ^ b. 2/1 +1 2?i -I- 2 Suppose m is not constant but follows the law 'ill = //i„ 4- caf. To find the mass and centre of mass of the above solid. Our first integral is TT |(?/toA' + caf+') ttU-'^ . dw, or (/ V [(vi^''+' + cx''''+'+^) clx, |_o 2n + 2 2/1 + 5 + 2 J ^ ass is a^TT I (?yio + co."*) x^ . c?a; aV \^, af''^' +-« ^ ^ ar^+'-^»] (2). L2m+1 2n + 5+l J ^'^ Substituting h for ^' iu both of these and dividing (1) by (2), we find x. An ingenious student can manufacture for himself many exercises of this kind which only involve the integration ofd?". An arc of the ellipse ~ + ^^ = l revolves about the axis of X, find the volume of the portion of the ellipsoid of revolu- tion between the two planes where x = and where x = c. Here y^ = — (a^ — x-). The integral of tti/^ is or or |[k-K]=-^;(a=c-|c'). LENGTHS OF CURVES '^ The volume of the whole ellipsoid is -t- iira; aud of ^ it IS -^ a^ ^^ o 77 ihere 47. Lengths of Curves. In fig. 21 the co-ordinates of P are x and y and of Q they are a; -\- Sx and 3/ + S^. If we call the length of the curve from some fixed place to P by the name s and the length PQ, 8s, then (8sy=(Bxy-\-(8i/y more and more nearly as 8x gets smaller, so that 8s Boo -^A-(I)■ or rather, in the limit Fig. 21. CD dx V'-©"- To find 5 then, we have only to integi'ate ^/'-(i)' It is unfortunate that we are only supposed to know as yet j x^ . dx, because this does not lend itself much to exer- cises on the lengths of curves. Example. Find the length of the curve y — a-\-hx (a straight line) between the limits x = Q and x = c. t^ ds [^ — ~ 5=1 j^.dx=\ ^/l+6^(^a; = x\/l+b' ^cVl+6^ 78 CALCULUS FOR ENGINEERS. Exercise. There is a curve whose slope is »Ja^x^—l, find an expression for its length. Answer: s= -«;'**+->/-. Other exercises on lengths of curves will be given later. 48. Areas of Surfaces of Revolution. When the curve APB revolving about the axis OX de- scribes a surface of revo- lution, we have seen that the volume between the ends AC A' and BDB' is the integral of iry^ with regard to x between the limits OG and OD. Again the elementary area of the surface is what is traced out by the ele- mentary length PQ or hs and is in the limit 2iry . ds. Hence we have to integrate ds dx Fig. 22. roD ^ I liTV .-T-.dx, and as the law of the curve is known, y . Joe dx 00 or can be expressed in terms of x. Example. The line y = a-\-bx revolves about the axis of X ; find the surface of the cone between the limits x=0 and x=c. -^ = 6, so that the area is 27r dx JO /.■^■-©■' = 2'7r*Jl+b^r{a-\-bx)dx = 2'ir's/l -^¥ I ax + ^boA = 27rVITF(ac + Jtc^. The problem of finding the area of a spherical surface is here given in small printing because the beginner is supposed to know only how to differentiate x'^ and this problem requires him to know that the differential coefficient of y^ AREAS OF SURFACES. 79 with regard to x is the differential coefficient with regard to y multiplied hy -~ , or 2y .-^ . As a matter of fact this is not a real difficulty to a thinking student. The student can however find the area in the following way. Let V be the 47r volume of the sphere of radius r, F= -^ r^, Art. 46. Let o V-{- SVhe the volume of a sphere of radius r*+ 8r, then which is only true when Sr is supposed to be smaller and smaller without limit. Now if S is the surface of the spherical shell of thickness Br, its volume is Bi^ . S. Hence Br .S = Br . ^irr^ and hence the area of a sphere is ^iri^. Example. Find the area of the surface of a sphere. That is, imagine the quadrant of a circle AB oi radius a, fig. 23, to A Fig. 23. revolve about OX and take double the area generated. We have as the area, 47r j y a / \ -\- {-^\ dx. In the circle x'^ +y^ = <^% or y = \/«^ - x-, X V' ..^..,.|=0.or|,= -^- 80 CALCULUS FOR ENGINEERS. Hence as l + m^^l+^ \dx') r y Guldinus's Theorems. PCD 49. If each elementary portion hs of the length of a curve be multiplied by x its distance from a plane (if the curve is all in one plane, x may be the distance to a line in the plane) and the sum be divided by the whole length of the curve, we get the x of the centre of the curve, or as it is sometimes called, the centre of gravity of the curve. Observe that the centre of gravity of an area is not necessarily the same as the centre of gravity of the curved boundary. Volume of a Ring^. BC, fig. 24, is any plane area ; if it revolves about an axis 00 lying in its own plane it will generate a ring. The volume of this ring is equal to the area of BG multiplied by the circumference of the circle passed through by the centre of area of BC. Imagine an exceed- Fig' 24. ingly small portion of the area a at a place P at the distance r from the axis, the volume of the elementary ring generated by this is a . 27rr and the volume of the whole ring is the sum of all such terms or F= 27r2ar. But ^ar = rA, if A is the whole area of BG. The student must put this in words for himself; r means the r of the centre of the area. Hence V= 2irf x A and this proves the proposition. II. Area of a Ring. The area of the ring surface is the length of the Perimeter or boundary of BG multiplied by the circumference of the circle passed through by the centre of gravity of the boundary. Imagine a very short length of the boundary, say hs^ at MOMENT OF INERTIA. 81 the distance r from the axis ; this generates a strip of area of the amount hs x 27rr. Hence the whole area is ^irZZs . r. But 2^5 . r = r X 6- if r is the distance of the centre of gi-avity of the boundary from the axis and s is the whole length of the boundary. Hence the whole area of the ring is 27rr X 8. Example. Find the area of an anchor ring whose sec- tion is a circle of radius a, the centre of this circle being at the distance JR, from the axis. Answer : — the perimeter of the section is 27ra and the circumference of the circle- described by its centre is 27rR, hence the area is 47r-ajR. Exercise. Find the volume and area of the rim of a fly- wheel, its mean radius being 10 feet, its section being a square whose side is 1'3 feet. Answer: Volume = (1-3)2 x 27r x 10 ; Area = 4 x 1-3 x 27r x 10. 50. If every little portion of a mass be. multiplied by the square of its distance from an axis, the sum is called the moment of inertia of the whole mass about the axis. It is easy to prove that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the centre of gravity together with the whole mass multiplied by the square of the distance between the two axes. Thus, let the plane of the paper be at right angles to the axes. Let there be a little mass m at '^P in the plane of the paper. Let be the axis through the centre of gravity and 0' be the other axi-s. We want the sum of all such terms as m . (O'Py. Now (O'py = (ovy + OP' +2. 00'. oq, where Q is the foot of a perpendicular from P upon 00\ the plane containing the two axes. Then calling Xm. (O'Py by the name /, calling Xm . OP^ by the name /q, the moment of inertia about the axis through the centre of gravity of the whole mass, then, / = (0'0)^Sm + /o + 2.00'.2:m.OQ. But tm.OQ means that each portion of mass m is multiplied by its distance from a plane at right angles to the paper through P. 6 S2 CALCULUS FOR ENGINEERS. the centre of gravity, and this must be by Art. 45. So that the proposition is proved. Or letting 2m be called M the whole mass I = Io + M.(0'0)2. Find the moment of inertia of a circular cylinder of length I about its axis. Let fig. 26 be a section, I the axis being 00. Con- — o sider an elementary ring shown in section at TQPR of inside radius r, its out- side radius being r + hr. Its sectional area is I .hr T R O— - Fig. 26. SO that the volume of the ring is 2irr . 1 . 8r and its mass is m^irrl . h\ Its moment of inertia about 00 is ^irml .t-^.dr and this must be integrated between the limits r — R the outside radius and ?• = to give the moment of inertia of the whole cylinder. The answer is /q = lirmlRK The MR^ whole mass M=ml7rR\ So that 7o = -«— • If we define the radius of gyration as Jc, which is such that Mk^ = Io, we have here k^ = ii2^ or A; = -p^ R. The moment of inertia about the axis NS is I=Io + M.R' = ^3IR% so that the radius of gyration about jsrs is R Vf . Moment of inertia of a circle about its centre. Fig. 27. Con- sider the ring of area between the Fig. 27. circles of radii r and r + Br, its area is 27rr . Br, more and more nearly as Br is smaller and smaller. Its moment of inertia is 27rr^ . dr and the integral of this between and R is ^ttR* where R is the radius of the circle. The square of the radius of R^ gyration is JttjR'^-t- the area =-^ MOMENTS OF INERTIA. 83 At any point in an area, fig. 28, draw two lines OX and OF at right angles to one another. Let an elementary area a be at a distance x from one of the lines and at a distance i/ from the other and at a distance r from 0. Observe that ax^ 4- ay- = «/'^ so that if the moments of inertia of the whole area about the two lines be added together the sum is the moment of inertia about the point 0. Hence the moment of inertia of a circle about a diameter is half the above, or JttjR*. The square of its radius of gyration is ^R-. The moment of inertia of an ellipse about a principal diameter AOA. Let OA = a, OB = b. .y^^ B y f \ > / / \ \ aI B Fig. 29. The moment of inertia of each strip of length BT is t times the monaent of inertia of each strip PQ of the circle, MT a because it is at the same distance from A OA and ytfi = y • MQ h This is a property of ellipse and circle well known to all engineers. But the moment of inertia of the circle of radius h about AOA is \it¥, so that the moment of inertia of the ellipse about AOA is \'Tr¥a. Similarly its moment of inertia about BOB is \'iTa%. The above is a mathematical device requiring thought, not practical enough perhaps for the engineer's every-day work; it is given because we have not yet reached the inte- 6—2 84 CALCULUS FOR ENGINEERS. gral which is needed in the straightforward working. The integral is evidently this. The area of the strip of length ST and breadth % is 2x . Sy, the equation to the ellipse beins: -„ + '7. = 1, so that ic=^T "Jh^ — y^ Then 2J if.2x.dy or 4^[ ^fs/b^ -if .dy=^ I. The student ought to return to this as an example in Chap. III. 51. Moment of Inertia of Rim of Fly-wheel. If the rim of a fly-wheel is like a hollow cylinder of breadth I, the inside and outside radii being R^ and R.,, the moment of inertia is ^irml I 1-^ .dr or ^irml — =i7rml(R/—Ri*). J jR, \_Ri ^ J The mass is 7r(jR2'-i^f)^//i'=^irsay,_so that I=\{R.^-\-R^^M. The radius of gyration is V^ (^R^ + Ri^). It is usual to calculate the moment of inertia of the rim of a fly-wheel as if all its mass resided at the mean radius of the rim or Oil? --^ — ^ . The moment of inertia calculated in this way is to the true moment of inertia as ^y^- — ^- . Thus if R.2 = R + a, Ri = R — a, the pretended I divided by the true / is 1 -^ (1 -f 0:2) ^^^ i^ ^ is small, this is 1 — -^^ nearly. If the whole mass of a fly-wheel, including arms and central boss, be M, there is usually no very great error in assuming that its moment of inertia is / = R'^M. 52. A rod so thin that its thickness may be neglected is of length I, its mass being m per unit length, what is its moment of inertia about an axis at right angles to it, through one end ? Let w be the distance of a point from one end. An elementary portion of length Sw of mass in . Boo has a moment of inertia x^ ,m .dx and the integral of this from ^ = to x = l \^ i^nP, which is the answer. As 7nl ^ .X £83s>Q is the whole mass, the ^U ' --1 ^^^^ square of the radius of Fig. 30. gyration is ^l^ /© the CENTRE OF GRAVITY. 85 moment of inertia about a parallel axis through the middle of the rod, at right angles to its length, is ^ml^ — tnl.liz) or ml^ (^ — {) or -^mlK So that the square of this radius of gyration is -^P. We shall now see what error is involved in neglecting the thickness of a cylindric rod. If 00 is an axis in the plane of the paper at right angles to the axis of a circular cylinder, through one end, and OP is <-» a; and R is the radius of the cylinder, its length being I ; if p is the mass of the cylin- der per unit volume ; the moment of inertia about 00 of the disc of radius It and thickness 8.v is irR^pbx . x*- + the moment of inertia of the disc about its own diameter. Now we saw that the radius of gyration of a circle about its diameter was — , and the radius of gyration of the disc is evidently the same. Hence its moment of inertia about its diameter is \ WirR^ . hx . p, or | irpTt^ . 8x. Hence the moment of inertia of the disc about is 7rR^p{xK8x+im.dx). If OAj the length of the rod, is I, we must integrate between and I, and so we find --<>- -A Fig. 31. '-nR^p(^+k mi\ The mass m per unit length is irR'^pj so that I—m 3 IP ?)■ This is the moment of inertia about an axis through the centre of gravity parallel to 00. 53. Example. Where is the Centre of Area of the parabolic segment shown in fig. 20 ? The whole area is lAGy<OB. The centre of area is evidently in the axis. 86 CALCULUS FOR ENGINEERS. The area of a strip PSR is 2i/ . Bx and we musb integrate 2xy . Bx. Now y — ax^ where a — AB ^ OBK rOB ^^ Hence 2 1 x vym^* . dx = ^AO . OB . x. The integral is 2 g^^ Ux^l or f -^j 0J5t, so that iAB . OB' = ^AB.OB.x or x = ^OB, Find the centre of area of the segment of the sym- metrical area bounded by + ?/ = ax'^ between x = b and x = c. i: We must divide the integral 2 1 x . ax"^ . dx by the area 2 I ax'' . dx. Or 2a r"-^"" X = C«+i _ hn+i ' n + 2' Many interesting cases may be taken. Observe that if the dimensions of the figure be given, as in fig. 20 : thus if AB and PQ and BQ are given, we may find the position of the centre of the area in terms of these magnitudes. 54. Moment of Inertia of a Rectangle. The moment of inertia of a rectangle about the line 00 ATTRACTION. 87 through its centre, parallel to one side. Consider the strip of area between OP = y and OQ = y + hj. Its area is b . By and its moment of inertia about 00 is b .y^ . By, so that the moment of inertia of the whole rectangle is bd' rhd rid-] b f.dyorb\if\ J -id L-idA "' li This is the moment of inertia which is so important in calculations on beams. Fig. 33. 55. Force of Gravity. A uniform spherical shell of attracting matter exercises no force upon a body inside it. On unit mass outside, it acts as if all its mass were gathered at its centre. The earth then exercises a force upon unit mass at any point P outside it which is inversely proportional to the square of r the distance of P from the centre. But if P is inside the earth, the attraction there upon unit mass is the mass of the sphere inside P divided by the square of r. 1. If the earth were homogeneous. If m is the mass per unit volume and R is the radius of the earth, the attraction on any outside point is — mR^ -r- r\ The attraction on any inside point is -^ mi^ -h t^ or — mr. The attraction then at the surface being called 1, at o any outside point it is R^ -r- r- and at any inside point it is r -f- R. Students ought to illustrate this by a diagram, 2. If m is greater towards the centre, say m — a — br, then as the area of a shell of radius r is 4<TTf\ its mass is 4mr^ . Ill . 5r, so that the whole mass of a sphere of radius r rr ^^ 47r r^ (a — br) dr, or -^ ar^ — irbr^. Hence on any Jo ^ 4i7r inside point the attraction is -;r- a?' — irbr^ and on any o outside point it is f ^ aR^ — irbR^j r^. is 88 CALCULUS FOR ENGINEERS. Dividing the whole mass of the earth by its volume -^ jK^ o we find its mean density to be a — ^hR, and the ratio of its mean density to the density at the surface is (4a-36i^)/(4a-46E). 56. Strength of thick Cylinders. The first part of the following is one way of putting the well known theory of what goes on in a thin cylindric shell of a boiler. It prevents trouble with + and — signs after- wards, to imagine the fluid pressure to be greater outside than inside and the material to be in compression. Consider the elementary thin cylinder of radius r and of thickness 8?\ Let the pressure inside be p and outside p + Bp and let the crushing stress at right angles to the radii in the material be q. Consider the portion of a ring PQSR which is of imit length at right angles to the paper. Radially we have p + 8p from outside acting on the area RS or (r + Br) BO if QOP = BO, because the arc RS is equal to radius multiplied by angle; and 2:> .r. BO from inside or {p + Bp){r-\-B7^)Bd-pr.Be the whole the radial force from the outside more and more nearly as B6 is smaller and smaller. This is balanced by two forces each q . Br inclined at the angle B6, and just as in page 165 if we draw a triangle, each of whose sides GA and ^J5 is pamllel to q . Br, the angle BAG being Bd, and BG representing the radial force, we see that this radial force is q . Br . Bd, and and Fig. 34. is on Fig. 35. this expression is more smaller and smaller. Hence more nearly true as BO is (p + Bp) (r + Br) B0-pr.Be = q.Br, B0, HYDRAULIC PRESS OR GUN. 89 or p .^7^ -\- r.Bp+ 8p.Sr = q.Sr, or rather p-\- r -¥- = <! (1), since the term Bp is in the limit. When material is subjected to crushing stresses p and q in two directions at right angles to one another in the plane of the paper, the dimensions at right angles to the paper elongate by an amount which is proportional to p + q. We must imagine the elongation to be independent of r if a plane cross section is to remain a plane cross section, and this reasonable assumption we make. Hence (1) has to be combined with p-\-q = 2A (2) where 2A is a constant. Substituting the value of q from (2) in (1) we have dp 2A 2p or -f- = ^ . a?' r r Now it will be found on trial that this is satisfied by P = ^+J (3). and hence from (2), q = A — -; (4). To find these constants A and B. In the case of a gun or hydraulic press, subjected to pressure po inside where r = ro and pressure outside where r = 7\. Inserting these values of p in (3) we have sothat ^=i^o.(i-l) = Po,;g^, 90 CALCULUS FOR ENGINEERS. The compressive stress — q may be called a tensile stress f, /-^V,'-V r^ ^^^. / is greatest at r = 7\ and is then fo = Po-^—', (6). This is the law of strength for a cylinder which is initially unstrained. Note that po can never be equal to the tensile strength of the material. We see from (5) that as r increases, / diminishes in proportion to the invei-se square of the radius, so that it is easy to show its value in a curve. Thus a student ought to take 7\ = 1% r^ = 0'8, p^ — 1500 lb. per sq. inch, and graph / from inside to outside. / will be in the same units as p. (5) may be taken as giving the tensile stress in a thick cylinder to resist bui^ting pressure if it is initially unstrained. If when p^ — there are already strains in the material, the strains produced by (5) are algebraically added to those already existing at any place. Hence in casting a hydraulic press we chill it internally, and in making a gun, we build it of tubes, each of which squeezes those inside it, and we try to produce such initial compressive strain at r = ro and such initial tensile strain at r = ri, that when the tensile strains due to Pq come on the material and the cylinder is about to burst there shall be much the same strain in the material from r^ to i\* * In the case of a eylindric body rotating with angular velocity a, if p is the mass per unit volume ; taking into account the centrifugal force on the element whose equilibrium is considered, above the equation (1) becomes p + r ^ - r^pa? = q and the solution of this is found iohep = A+ Br~^ + ^pa^r^ and by inserting the values of p for two values of 7- we find the constants A and B ; q is therefore known. If we take p~0 when r=rQ and also when This is greatest when r = r^. If the cylinder extends to its centre we must write out the condition that the displacement is where r=0, and it is necessary to write out the valuer GAS ENGINE. 91 TJim cylinder. Take ro = R and 7\=Il-\- 1, where t is very small compared with R, . 2Rr' + 2Rt + t' PoR /^ t t-\l(^ t\ Fig. 86. Now -jz and ^r-^^ and ^r^ become all smaller and smaller li ZH- ZK as t is thought to be smaller and smaller. We may take f=V3 . .. (7) (7) as a formula to be used ' t ' " when the shell is exceedingly ^j^ ^ thin and (8) as a closer ap- ^~^"^2 ^^^' proximation, which is the same as if we used the average radius in (7). In actual boiler and pipe work, there is so much uncertainty as to the proper value of/ for ultimate strength, that we may neglect the con-ection of the usual formula (7). 57. Gas Engine Indicator Diagram. It can be proved that when a perfect gas (whose law is pv = Rt for a pound of gas, R being a constant and equal to K—k the difference of the important specific heats; 7 is used to denote Kjk) changes in its volume and pressure in any of the strains, Eadial strain =pa - g/3 if a is the reciprocal of Young's Modulus and ^/a is Poisson's ratio, generally of the value 0-25. In this way we find the strains and stresses in a rotating solid cylinder, but on applying our results to the case of a thin disc we see that equation (2) above is not correct. That is, the solution is less and less correct as the disc is thinner. Dr Chree's more correct solution is not difficult. &2 CALCULUS FOR ENGINEERS. way, the rate of reception of heat by it per unit change of volume, which we call h (in work units) or -^— , is {jpv) + 'p (1), rf—ldv '-~i[t^A ^'^- Students ouorht to note that this -~ is a very different thing from [-i] , because we may give to it any value we please. We always assume Heat to be expressed in work units so as to avoid the unnecessary introduction of J^ for Joule's equiva- lent. Exercise 1. When gas expands according to the law ptf = c...{^) a constant, find li. Answer : // = ^^— :. p (4). y- 1 Evidently when 5 = 7, k = 0, and hence we have pyy = con- stant as the adiabatic law of expansion of a perfect gas. 7 is 1*41 for air and 1'37 for the stuff inside a gas or oil engine cylinder. When s= 1, so that the law of expansion is pv constant, we have the isothermal expansion of a gas, and we notice that here h=p, or the rate of reception of heat energy is equal to the rate of the doing of mechanical energy. Notice that in any case where the law of change is given by (3), h is exactly proportional to p. If .9 is greater than 7 the stuff is having heat withdrawn from it. If the equation (1) be integrated with regard to v we have J^oi = — 3T (piVi - p^Vo) + T^oi- • -(5). Here H^^ is the heat given to a pound of perfect gas between the states po, Vq, t^ and jOi, -yi, ti, and TFoi is the work done by it in expanding from the first to the second state. This expression may be put in other forms because we have the connection pv = Rt...(6). It is very useful in cal- culations upon gas engines. Thus, if the volume keeps ELASTICITY. 93 constant Wqi is and the change of pressure due to ignition and the gift of a known amount of heat may be found. If the pressure keeps constant, TFoi is p(vi — Vo) and the change of volume due to the reception of heat is easily found. /J JT rit Another useful expression is -p =1^~r +'P"'0) where k is a constant, being the specific heat at constant volume. Integrating this with regard to v we find Ho, = k(t,-to)+W,, (8). This gives us exactly the same answer as the last method, and may at once be derived from (5) by (6). In this form one sees that if no work is done, the heat given is k (ti — to) and also that if there is no change of temperature the heat given is equal to the work done. 58. Elasticity is defined as increase of stress -r increase of strain. Thus, Young's modulus of elasticity is tensile or compressive stress (or load per unit of cross section of a tie bar or strut) divided by the strain or fractional change of length. Modulus of rigidity or shearing elasticity is shear stress divided by shear strain. Volumetric elasticity e is fluid stress or increase of pressure divided by the fractional diminution of volume produced. Thus if fluid at p and v, changes to p-\-Bp, v-\-Sv: then the volumetric stress is Bp and the volumetric compressive strain is — Sv/v, so that by definition e = — Bp-. — , or e = — v J- . The •^ ^ V ov definition really assumes that the stress and strain are smaller and smaller without limit and hence e — — v -/-. . .(1). dv Now observe that this may have any value whatsoever. Thus the elasticity at constant pressure is 0. The elasticity at constant volume is — oo . To find the elasticity at con- stant temperature, we must find ( i^ ) • see Art. 30. As pv = Rt, p^ Rtv-\ Here Rt is to be constant, so that ^^- Rtv-^ and e=Rtv^^^p. ®- 94 CALCULUS FOR ENGINEERS. It is convenient to write this et and we see that Kt^ the elasticity at constant temperature, is p. This was the value of the elasticity taken by Newton ; by using it in his calcu- lation of the velocity of sound he obtained an answer which was very different from the experimentally determined velo- city of sound, because the temperature does not remain constant during quick changes of pressure. Exercise. Find the elasticity of a perfect gas when the gas follows the law pvy = c, some constant. This is the adiabatic law which we found Art. 57, the law connecting j9 and V when there is no time for the stuff to lose or gain heat by conduction, p = cv~y, so that -t- = — 7C^;~>'~^ and 6? = + vycv~y~^ or ycv~y, or yp. It is convenient to write this e^, and we see that in a perfect gas en = yet. When this value of the elasticity of air is taken in Newton's calculation, the answer agrees with the experimentally found velocity of sound. 59. Friction at a Plat Pivot. If we have a pivot of radius R carrying a load W and the load is uniformly distributed over the surface, the load per unit area is w=^ W-T-irR^ Let the angular velocity be a radians per second. On a ring of area between the radii r and r-\-8r the load is w27rr . 8r, and the friction is fjLiv27rr . Sr, where fjb is the coefficient of friction. The velocity is v = otr, so that the work wasted per second in overcoming friction at this elementary area is ^^irwoLr^ . Br, so that the total energy wasted per second is ^irwoLfi \ ?'^ . dr = ^irwafxR^ = |« ^ WR, Jo On a collar of internal radius Ri and external R^ we have 27rwafi r'.dr^ ^irwdfi (R^^ - R^% W^irw {R^^ - R^) and J Ri Jl3 _ J^ 3 hence, the energy wasted per second is '{afiW j^\ _ pV 60. Exercises in the Bending of Beams. When the Bending moment ilf at a section of a beam is known, we can calculate the curvature there, if the beam was CURVATURE. 95 straight when unloaded, or the change of curvature if the unloaded beam was originally curved. This is usually written 1 1 1 if - or = -^^ , where / is the moment of inertia of the cross section about a line through its centre of gravity, perpendicular to the plane of bending, and E is Young's modulus for the material. Thus, if the beam has a rectangular section of breadth h and depth d, then I=^hd^ (see Art. 54); if the beam is circular in section, I=-:R\ if J2 is the radius of tbe section (see Art. 50). If the beam is elliptic in section, I — T ^^^> if ^ aiid h are the radii of the section in and at 4 right angles to the plane of bending (see Art. 50). t Curvature. The curvature of a circle is the reciprocal of its radius, and of any curve it is the curvature of the circle which best agrees with the curve. The curvature of a curve is also " the angular change (in radians) of the direction of the curve per unit length." Now draw a very flat curve, with dii very little slope. Observe that the change in -^ in going from a point P to a point Q is almost exactly a change of angle change in -^ is really a change in the tangent of an angle, but when an angle is very small, the angle, its sine dii and its tangent are all equal . Hence, the increase in -^ from P to Q divided by the length of the curve PQ is the average curvature from P to Q, and as PQ is less and less we get more and more nearly the curvature at P. But the curve being very flat, the length of the arc PQ is really hx^ and the change in -/ divided by hx, as hx gets less and less, dv is the rate of change of -f- with regard to x, and the symbol ^2y ° ax ^2y for this is -—^ . Hence we may take -y^ as the curvature of dx^ '' ^ dx^ a curve at any place, when its slope is everywhere small. 96 CALCULUS FOR ENGINEERS. If the beam was not straight originally and if y' was its small deflection from straightness at any point, then ,^ was its original curvature. We may generalize the following work by using ^^ (y -y') instead of --j— everywhere. It is easy to show, that a beam of uniform strength, that is a beam in which the maximum stress / (if compressive ; positive, if tensile, negative), in every section is the same, has the same curvature every- where if its depth is constant. If d is the depth, the condition for constant strength is that -J '\d=±f a constant. But -y=^x curvature, hence curvature -E.d' Exercise. In a beam of constant strength if c^= , . d^V '2f Then ^-^ = -^ {a-\-hx). Integrating we find ET 1 TTI 2?' ^=<' + ^+i^'^^ ■^^.y = e^rCX-\-\ax''--V\.hx^, where e and c must be determined by some given condition. Thus if the beam is fixed at the end, where ^=0, and -^- = there, and CLX also y = there, then c = and e — 0. In a beam originally straight we know now that, if X is distance measured from any place along the beam to a section, and if y is the deflection of the beam at the section, and / is the moment of inertia of the section, then ^ = ^ a) dx2 EI ^ ^' where M is the bending moment at the section, and E is Young's modulus for the material. We give to -^ the sign which will make it positive if M is positive. If M would make a beam convex upwards and y is measured downwards then (1) is correct. Again, (1) would be right if M would make a beam concave up- wards and y is measured upwards. BENDING. 97 Example /. Uniform beam of length 1 fixed at one end, loaded with weight W at the other. Let x be the distance of a section from the fixed end of the beam. Then M= W{l — x), so that (1) becomes .(2). w <3L X— ■l-x— ^ Fig. 37. Integrating, we have, as E and / are constants, EI dy J . , From this we can calculate the slope everywhere. To find c, we must know the slope at some one place. Now we know that there is no slope at the fixed end, and 0, hence c = 0. Integrating again, EI To find G, we know that ^ = when x = 0, and hence (7=0, so that we have for the shape of the beam, that is, the equation giving us y for any point of the beam, hence -y^ = where x dx W la-) .(3). We usually want to know y when x = l, and this value of y is called D, the deflection of the beam, so that B Wl' SEI .(4). Example II. A beam of length I loaded with W at the middle and supported at the ends. Observe that if half of this beam in its loaded condition has a casting of P. 7 98 CALCULUS FOR ENGINEERS. cement made round it so that it is rigidly held ; the other half is simply a beam of length ^l, fixed at one end and W\ W Fig. 38. loaded at the other with ^W, and, according to the last example, its deflection is ZEI ^ 4:8. EI (5). The student ought to make a sketch to illustrate this method of solving the problem. Example III. Beam fixed at one end with load w per unit length spread over it uniformly. The load on the part FQ is wx PQ or w{l — x). P Q — X >* ^'1-X ■*■ Fig. 39. The resultant of the load acts at midway between P and Q, so, multiplying hy ^{1 — x), we find M at P, or M=\w{l-xy (6). Using this in (1), we have w dx^ or. Integrating, we have 2EJ w This gives us the slope everywhere ^0 ax BEAMS. 99 Now -r =0 where ^ = 0, because the beam is fixed ax there. Hence c = 0. Again integrating, 2EI w y = ^px" - ^laf + yijja;* + G, and as ?/ = where a; = 0, = 0, and hence the shape of the beam is y=^((ilV-ila^ + a,-) (7). 1/ is greatest at the end where a; — I, so that the deflection is ^=M&7^' "'^ = 8 EI <«>• if Tr='Z(;^, the whole load on the beam. Example IV. Beam of length I loaded uniformly with lu per unit length, supported at the ends. Each of the supporting forces is half the total load. The moment about P of ^zvl, p q at the distance PQ, is -------- -<,— X — *^-il-x-^ iwl against the hands of a > watch, and I call this i^i direction positive; the moment of the load Fig. 40. '^^ (2 ^ ~ ^) ^t the average distance ^PQ is therefore negative, and hence the bending moment at P is ^lul (hi — x) — ^w ( JZ — xy, or ^^Z- — ^wx'^ . . .(9), so that, from (1), EI ,^ = ^wP - \wx'^, y being the vertical height of the point P above the middle of the beam, see Art. 60. Integrating we have dy dx EI -^ = \wl~x — \wa? 4- c, a formula which enables us to find the slope everywhere. 7-2 100 CALCULUS FOR ENGINEERS. c is determined by our knowledge that ^ = where a; = 0, and hence c = 0. Integrating again, and C=0, because y = 0, where x = 0. Hence the shape of the beam is y = aaW'T^^^^^ " 2^)-"(l^X V is greatest where X = i?, and is what is usually called the deflection D of the beam, or ^= oQ±pf if" ^= ^^^ the total load. 61. Beams Fixed at the Ends. Torques applied at the ends of a beam to fix them (that is, to keep the end sections in vertical planes) are equal and opposite if the loading is symmetrical on the two sides of the centre of the beam. The torques being equal, the supporting forces are the same as before. Now if 7ti is the bending moment (positive if the beam tends to get concave upwards) which the loads and supporting forces would produce if the ends were not fixed, the bending moment is now ni — c because the end torques c are equal and opposite, and the supporting forces are unaltered by fixing. ^^"^ d^^=^r w- If the beam is uniform and we integrate, we find EI .-— = ini.dx — ex -\- const (2). Take x as measured from one end. We have the Uvo conditions : / = where x = 0, and ,- = where x = lAf I ax dx is the length of the beam. Hence if we subtract the value of (2) when ic = from what it is when x=l, we have n I ri 0=1 7n . dx — cl, or c = J j m . dx, that is, c is the average value of m all over the beam. The rule is then (for symmetric loads): — Draw the diagram of bending moment m as if the beam were merely suppoi^ted BEAMS FIXED AT THE ENDS. 101 at the ends. Find the average height of the diagram and loiuer the curved outline of the diagram by that amount. The resulting diagram, which will be negative at the ends, is the true diagram of bending moment. The beam is concave upwards where the bending moment is positive, and it is convex upwards where the bending moment is negative, and there are points of inflexion, or places of no curvature, where there is no bending moment. Example. Thus it is well known that if a beam of length I is supported at the ends and loaded in the middle with a load W, the bending moment is \Wl at the middle and is at the ends, the diagram being formed of two straight lines. The student is supposed to draw this diagram (see also Example II.). The average height of it is half the middle height or ^Wl, and this is c the torque which must be applied at each end to fix it if the ends are fixed. The whole diagram being lowered by this amount it is evident that the true bending moment of such a beam if its ends are fixed, is ^Wl at the middle, half-way to each end from the middle so that there are points of inflexion there, and — iTT^ at each end. A rectangular beam or a beam of rolled girder section, or any other section symmetrical above and below the neutral line, is equally ready to break at the ends or at the middle. Example. A uniform beam loaded uniformly with load w per unit length, supported at the ends ; the diagram for 711 is a parabola (see Example IV., where ilf = ^tul^ — ^wx^); the greatest value of m is at the middle and it is ^tul^ ; m is at the ends. Now the average value of m is | of its middle value (see Art. 43, area of a parabola). Hence c = -^wP. This average value of m is to be subtracted from every value and we have the value of the real bending moment every- where for a beam fixed at the ends. Hence in such a beam fixed at the ends the bending moment in the middle is i^^ivl-y at the ends — -^i^wP, and the diagram is parabolic, being in fact the diagram for a beam supported at the ends, lowered by the amount -^wP every- where. The points of inflexion are nearer the ends than in the last case. The beam is most likely to break at the ends. Students ought to make diagrams for various examples of 102 CALCULUS FOR ENGINEERS. symmetrical loading. Find m by the ordinary graphical method and lower the diagram by its average height. When the beam symmetrically loaded and fixed at the ends is not uniform in section^ the integral of (1) is ^t-i>-^lT (^). and as before this is between the limits and I, and hence to find c it is necessary to draw a diagram showing the value 111/ of -J everywhere and to find its area. Divide this by the area of a diagram which shows the value of y ever3rwhere, or the average height of the Mjl diagram is to be divided by the average height of the 1// diagram and we have c. Subtract this value of c from every value of m, and we have the true diagram of bending moment of the beam. Graphical exercises are much more varied and interest- ing than algebraic ones, as it is so easy, graphically, to draw diagrams of m when the loading is known. The solution just given is applicable to a beam of which the / of every cross section is settled beforehand in any arbitrary manner, so long as / and the loading are symmetrical on the two sides of the middle. Let us give to / such a value that the beam shall be of uniform strength every- M where ; that is, that -jz=fc or ft . . .(4), where z is the greatest distance of any point in the section from the neutral line on the compression or tension side and/, and/^ are the constant maximum stresses in compression or tension to which the material is subjected in every section. Taking /<. as numerically equal to ft and z = ^d, where d is the depth M of the beam, (4) becomes Trc?= ± 2/"... (5), the + sign being taken over parts of the beam where if is positive, the — sign where 31 is negative. As I -y c^^ = 0, or, using (5-), /iff^^^O (6), BEAMS FIXED AT THE ENDS. 103 the negative sign being taken from the ends of the beam to the points of inflexion, and the positive sign being taken between the two points of inflexion. We see then that to satisfy (6) we have only to solve the following problem. In the figure, EATUGQE is a diagram whose ordinates represent the values of -^ or the reciprocal of the depth of the beam which may be arbitrarily fixed, care being taken, however, that d is the same at points which are at the same distance from the centre. EFGE is a diagram of the values of m easily drawn when the loading is known. We are re- quired to find a point P, such that the area of EPTA = area Fig. 41. of POO'Tj where is in the middle of the beam. When found, this point P is a point of inflexion and PR is what we have called c. That is, m — PR is the real bending moment M at every place, or the diagram EFG must be lowered vertically till R is at P to obtain the diagram of M. Knowing M and d it is easy to find / through (5). It is evident that if such a beam of uniform strength is also of uniform depth, the points of inflexion are half- way between the middle and the fixed ends. Beams of uniform strength and depth are of the same curvature ever3rvvhere except that it suddenly changes sign at the points of inflexion. 61. In the most general way of loading, the bending moments required at the ends to fix them are different flrom one another, and if m^^ is the torque against the hands of a watch apphed at the end A, and ?«._, is the torque with the hands of a 104 CALCULUS FOR ENGINEERS. watch at the end /?, and if the bending moment in case the beam were merely supported is m : — Consider a weightless unloaded beam of the same length with the torques oui and wio applied to its ends; to keep it in equilibrium it is necessary to introciuce equal and opposite supporting forces P at the m, c m, ^A --T-- 4--— • X- — ^ B ' P > Fig. 42. ends as shown in the figure. Then Fl + m2 = m^f the forces &c. being as drawn in fig. 42, so that P=— i-^ — ^. If then these torques m.^ and wij are exerted they must be balanced by the forces P shown ; that is, at -B a downward force must be exerted ; this means that the beam at B tends to rise, and hence the ordinary supporting force at B must be diminished by amount P. At any place C the bending moment will be m (what it would be if the beam were merely supported at the ends) — m^-P. BC. . .(1). If one does not care to think much, it is sufficient to say : — The beam was in equilibrium being loaded and merely supported at the ends ; the bending moment at any place was m ; we have introduced now a new set of forces which balance, the bending moment at C due to these new forces is -(ma + P.iiC). So that the tnie bending moment at C is m - m^ - P. BC. 9)1 Suppose r/i2 = 0, then P= j-, and the bending moment at C is I m-^.BCovm-P.BC. 62. Beam fixed at the end A^ merely supported at B which is exactly on the same level as A. As Wg^^ ^"d letting BC^x, we have the very case just mentioned, and El'^^^^m-Px (2). We will first consider a uniform beam uniformly loaded as in Example IV., Art. 60. It will be found that when x is measured from the end of the beam, the bending moment in = \wlx -\wx'^-^ if the beam is merely supported at its ends and \o is the load per unit length. Hence (2) is EI^ = ^wlx-\wx''--Px (3), El-^=\wlx'^-\wx^-\Px^\c (4). BENDING. 105 We have also the condition that -i\=0 where ^=^...(5), for it is to be observed that we measure ,v from the unfixed end. Again integrating, £;ii/=^wla;^-^wa)*-^Pa^ + CA' (6). We need noi add a constant because y is when ^ is 0. We also have y=0 when x=l. Using this condition and also (5) we find 0=j\v>l^-^PP + c (7), 0=^^wl^-^PP+d (8), and these enable us to determine P and c. Divide (8) by I and subtract from (7) and we have (d=^wP - ^PP or P= Iwl, hence from (7 ), 0'=^wP — ^wP+CfC—- -^g wP. We have the true bending moment, ^wlx - J wx^ - \ ivlxj and (6) gives us the shape of the beam. 63. If the loading is of any kind whatsoever and if the section varies in any way a graphic metliod of inte- gration must be used in working the above example. Now if the value of an ordinate z which is a function of ^ be shown on a curve, we have no instrument which can be relied upon for showing in a new curve / z . dx, that is, the ordinate of the new curve representing the area of the z curve up to that value of .r from any fixed ordinate. I have sometimes used squared paper and counted the number of the squares. I have sometimes used a planimeter to find the areas up to certain values of .v, raised ordinates at those places representing the areas to scale, and drawn a curve by hand through the ten or twelve or more points so found. There are integrators to be bought ; I have not cared to use any of them, and perhaps it is hardly fair to say that I do not believe in the accuracy of such of them as I have seen. A cheap and accurate form of integrator would not only be very useful in the solution of graphical problems ; it would, if it were used, give great aid in enabling men to understand the calculus. Let us suppose that the student has some method of showing the value of I z. dx in a new curve ; the loading being of any kind what- y soever and / varying, since dx^ 106 CALCULUS FOn ENGINEERS, we have on integrating, "£-} i'^-'-^l~r-' w. We see that it ia necessary to make a diagram whose ordinate every- where is -j and we mast integrate it. Let I -^ o?^ be culled fi ; when w. x=lf IX becomes the whole area of the -j diagram and we will call this ^j. It is also necessary to make a diagram whose ordinate everywhere is -J and integrate it. Let J -.dxhQ called X, When x=l^ X becomes X the whole area of the j diagram and we will call this X^. Then as in (9), d * when x=lj o=f,,-p. x,+c ao). Integrating (9) again, we have Ey=\ii. dx - P \X . dx + ex + C. In this if we use y — O when x=0, we shall find C'=0, and again if y=0 when x=l, and if we use Mj and Xj as the total areas of the /x and X curves we have = Mi-P.Xi + e;* (11), from (10) and (11) i^ and c may be found, and of course P enables us to state the bending moment everywhere, -c is the slope when .'t- is 0. 64. Example. Beam of any changing section fixed * Without using the letters fi, X, yUj , X^ &c. the above investigation is : — e:i]--/:t--/:^^ <-'• Integrating again between the limits and I and recollecting that y is the same at both limits f4]=0=f7'"i..-p/"'f'^%.Z (11). a;=0 J J J- J J J- The integrations in (10) and (11) being performed, the unknowns P and c can be calculated ; the true bending moment everywhere is what we started with, m - Px, BENDING. 107 at the ends^ any kind of loading. Measuring x from one end where there is the fixing couple ?«2) M=m-m,-rj; (1), ^'g = ?-'?-''7 (^). ^j-,= I 7 • dx-m^ I ~r -P \ -y-+constant (3), Let fi = I — 'y-^ and fi^ the whole area of the -^ curve; (dx 1 Let F= / -y and Fj the whole area of the y curve; Let foe dx 1' X=l ' and Xi the whole area of the ~ curve; then = /Xi-m2Fi-PXi (4). Again integrating y=j/jL.dx — 111.2 1 Y ,dx — P \X .dx-\- const. Calling the integi^als from to I of the //,, F and X curves Mj, Yj, and X^, we have = Mi-7?i.,Yi-PXi* (5), and as rrio and P are easily found from (4) and (5), (1) is known. * We have used the Bymbols /a, A', Y, /j.^, X^, Y^, M, Y, Mi, Xi, Yi fearing that students are still a little unfamiliar with the symbols of the calculus ; perhaps it would have been better to put the investigation in its proper form and to ask the student to make himself familiar with the usual symbol instead of dragging in eleven fresh symbols. After (3) above, write as follows ; — U d}='=} 07"^- '"■'). T-n.-r w- Again integrating between Umits l-U=N"^a.-.J'l'^^-r['f'i^ (5, x=o J J J- J J ■'■ yoyo-' The integrations indicated in (4) and (5) being performed, the unknowns m^ and P can be calculated and used in (1). The student must settle for himself which is the better course to take; to use the formidable looking but really easily understood symbols of this note or to introduce the eleven letters whose meaning one is always forgetting. See also the previous note. 108 CALCULUS FOR ENGINEERS. 65. In Graphical work. Let A GB (fig. 43) represent m, the bending moment, if the beam were merely supported at the ends; let AD represent nv^ and let BE represent Wg. Join DE. Then the difference between the ordinates of Fig. 43. ACB and of ADEB represents the actual bending moment; that is the vertical ordinates of the space between the straight line DE and the curve AFCGB. It is negative from A to H and from / to B, and positive from H to I. F and G are points of inflexion. Useftil Analogies in Beam Problems. If io is the load per unit length on a beam and M is the bend- ing moment at a section (positive when it tends to make the beam convex up- wards*), a; being horizontal distance, to prove that dF = ^ ^^>- If at the section at P fig. 44, whose distance to the right of some origin is oc there is a bending moment M in- dicated by the two equal and p. ^, opposite arrow heads and a shearing force S as shown, being positive if the material to the right of the section is * This convention is necessary only in the following generalization. BENDING. 109 acted on by downward force, and if PQ is hx so that the load on this piece of beam between the sections at P and Q is w .hx\ if the bending moment on the Q section is M + hM and the shearing force S + hS, then the forces acting on this piece of beam are shown in the figure and from their equili- brium we know that hS — io.hx or — - = w (2), and taking moments about Q, 3I+S.Sx + iw (8xy=^M+ MI, or -tT- = o + 1^ . ox, ox ^ and in the limit as hx is made smaller and smaller ^ = S (3), and hence (1) is true. Now it is well known that in beams if y is the deflection d2y M d^"EI .(4). If we have a diagram which shows at every place the value of Wy called usually a diagram of loading, it is an exercise known to all students that we can draw at once by graphical statics a diagram showing the value of M at every place to scale ; that is we can solve (1) very easily graphi- cally*. We can see from (4) that if we get a diagram M showing yn- at every place, we can use exactly the same method (and we have exactly the same rule as to scale) to find the value of y ; that is, to draw the shape of the beam. Many of these exercises ought to be worked by all engineers. * We find M usually for a beam merely supported at the ends. Let it be AGB, fig. 43. If instead, there are bending moments at the ends we let AD and BE represent these and join DE. Then the algebraic sum of the ordinates of the two diagrams is the real diagram of bending moment. 110 CALCULUS FOR ENGINEERS. Example. In any beam whether supported at the ends or not : if -m; is constant, integrating (1) we find — - — h-\-iux and M = a-\-hx-\- ^tuaf (5). In any problem we have data to determine a and b. Take the case of a uniform beam uniformly loaded and merely supported at the ends. Measure y upwards from the middle and x from the middle. Then i/= where x = ^l and — ^l, and = tt — ^hl + ^wl-. I Hence 6 = 0, ti = — Itvl^ and (5) becomes M = - iwl" ■}- ^wx"" (6), which is exactly what we used in Example IV. (Art. 60) where we afterwards divided M by EI and integrated twice to find y. Let i be -^- or the slope of the beam. e,. dy . di M dM ^ dS dx ' dx EI dx dx we have a succession of curves which may be obtained from knowing the shape of the beam y by differentiation, or which may be obtained from knowing w, the loading of the beam, by integration. Knowing w there is an easy graphical rule for finding MjEI, knowing MjEI we have the same graphical rule for finding y. Some rules that are obviously true in the w to MjEI construction and need no mathematical proof, may at once be used without mathematical proof in applying the analogous rule from MjEI to y. Thus the area of the MjEI curve between the ordinates x^ and x^ is the increase of i from x^ to x.2y and tangents to the curve showing the shape of the beam at x^ and x^ meet at a point which is vertically in a line with the centre of gravity of the portion of area of the MjEI curve in question. Thus the whole area of the M/EI curve in a span IIJ is equal to the di/ increase in -~ from one end of the span to the other, and CONTINUOUS GIRDERS. Ill the tangents to the beam at its ends H, J meet in a point P which is in the same vertical as the centre of gravity of the whole M/EI curve. These two rules may be taken as the starting point for a complete treatment of the subject of beams by graphical methods. If the vertical from this centre of gravity is at the horizontal distance HG from H and GJ from «/, then P is higher than H by the amount HG x ij^, the symbol % being used to mean the slope at ^ ; J" is higher than P by the amount GJ x { at J. Hence J is higher than H by the amount HG.i^+GJ.ij, a relation which may be useful when conditions as to the relative heights of the supports are given, as in continuous beam problems. 67. Theorem of Three Moments. For some time, Railway Engineers, instead of using separate girders for the spans of a bridge, fastened together contiguous ends to prevent their tilting up and so made use of what are called continuous girders. It is easy to show that if we can be absolutely certain of the positions of the points of support, continuous girders are much cheaper than separate girders. Unfortunately a comparatively small settlement of one of the supports alters completely the condition of things. In many other parts of Applied Mechanics we have the same difficulty in deciding between cheapness with some uncertainty and a gi-eater expense with certainty. Thus there is much greater uncertainty as to the nature of the forces acting at riveted joints than at hinged joints and therefore a structure with hinged joints is pre- ferred to the other, although, if we could be absolutely certain of our conditions an equally strong riveted structure might be made which would be much cheaper. Students interested in the theory of continuous girders will do well to read a paper published in the Proceedings of the Royal Society/, 199, 1879, where they will find a graphical method of solving the most general problems.t I will take here as a good example of the use of the calculus, a uniform girder resting on supports at the same level, with a uniform 112 CALCULUS FOR ENGINEERS. load distribution on each span. Let ABC be the centre line of two spans, the girder originally straight, supported at A, B and G. The distance from J. to .S is ^i and from B to G 20£5S' -^^^^^^^^mxww^mm Fig. 45. is I2 and there are any kinds of loading in the two spans. Let A, B and G be the bending moments at A, B and G respectively, counted positive if the beam is concave upwards. At the section at P at the distance x from A let m be what the bending moment would have been if the girder on each span were quite separate from the rest. We have already seen that by introducing couples 77I2 and mi at A and B (tending to make the beam convex upwards at A and B) we made the bending moment at P really become what is given in Art. 61. Our m^ — — A^m^^ — B, and hence the bending moment at P is B-A_ d'j m-i- A -^ X EI ■(1), where m would be the bending moment if the beam were merely supported at the ends, and the supporting force at A is lessened by the amount ^ A-B .(2). I. Assume EI constant and integrate with regard to x and we have / B — A m .dx -\- Ax + ^x^ — J 1- Ci ^'■i (^^)- Using the sign jltn. dx . dx to mean the integration of the curve representing J7a.dx we have r r T) A \\m.dx.dx'^\Ax^+^x' — ^ + Ci^p -f e = ^Y . 3/ . . . ( 4). THEOREM OF THREE MOMENTS. 113 As 7/ is when ic=0 and it is evident that llm.dx.da; = when cc = 0, e is 0. Again y = when a;=lj^. Using the symbol //^ to indicate the sum jjm.dx.dic over the whole span, fi, + iAk'-{-ili'(B-A) + cA = (5). From (3) let us calculate the value of EI -—- at the point B, and let us use the letter a^ to mean the area of the m curve over the span, or I m . dx, so that EI ~ at 5 is a^-\-Ak + \k{B-A) + c, (6). But at any point Q of the second span, if we had let BQ = x we should have had the same equations as (1), (3) and (4) using the letters B for A and G for B and the constant Ca. Hence making this change in (3) and finding EI -^- at the point B where a: = 0, we have (6) equal to Co or '^ C2-Ci = ai + ^^i4-i^i(5-^) (7), and instead of (5) we have /i, + \BU 4- ^U {C-B)-¥ cA = (8). Subtracting (5) from (8) after dividing by Zj and l.j we have c,-c, = ^-f^ + iAh-iBk + lh{B-A)-ih{C-B)...(9). The equality of (7) and (9) is an equation connecting A, B and C, the bending moments at three consecutive supports. If we have any number of supports and at the end ones we have the bending moments because the girder is merely supported there, or if we have two conditions given which will enable us to find them in case the girder is fixed or partly fixed, note that by writing P. 8 114 CALCULUS FOR ENGINEERS. down (10) for every three consecutive supports we have a sufficient number of equations to determine all the bending moments at the supports. Example. Let the loads be w^ and w^ per unit length over two consecutive spans of lengths l^ and l^. Then m = \wlx — \wa^, jm.dx= \wla^ — \wa?, Hence rti = T^^'j ^^^ \\m.dx ,dx = -^wlaf^ — -^wa^. Hence /Aj = -^w^k\ fi^ = i^wj^i*. Hence ^ + tti — y^ becomes -^-fW.}.^ + r-^ Zi» — ^w^li^, or ^\ (ukI/ 4- w^li"), and hence the theorem becomes in this case Al, + 2B (I, + k) + Gl, + i (wj2^ + w^k') = (10). If the spans are similar and similarly loaded then A-h4^B-\-C+^wP=^0 (11). Case 1. A uniform and uniformly loaded beam rests on three equidistant supports. Here A = C = and B = — ^wl\ m = \w {Ix -- aP), and hence the bending moment at a point P distant x from A is \w {Ix - a^) + --^wl\ The supporting force at A is lessened from what it would be if the part of the beam AB were distinct by the amount shewn in (2), — ^ — or ^wl. It would have been ^l, so now it is really ^wl at each of the end supports, and as the total load is 2wl, there remains ^^-wl for the middle support. Case 2. A uniform and uniformly loaded beam rests on four equidistant supports, and the bending moments at these supports are A, B, G, D. Now A = D=0 and from symmetry B = G. Thus (11) gives us + 5J5+irf = or B=^G=^--^^wl\ SHEAR STRESS IN BEAMS. 115 If the span AB had been distinct, the first support would have had the load ^wl, it now has ^wl — -^^wl or -^qwI. The supporting force at D is also -^-^wl. The other two supports divide between them the remainder of the total load which is altogether ^wl and so each receives \^wl. The supporting forces are then -^^wl, \^wl, \^wl and -^-^wl. 68. Shear Stress in Beams. Let the distance measured from any section of a beam, say at 0, fig. 46, to the section at A be x, and let OB — x-\-hx. Let the bending moment at G'AG be M and at B'BD be M + hM, C D H E— IF C D — E Fig. 46. Fig. 47. Fig. 48. OAB (fig. 46) and A A (fig. 47) represent the neutral surface. We want to know the tangential or shear stress / at E on tlie plane GAC. Now it is known that this is the same as the tangential stress in the direction EF on the plane EF which is at right angles to the paper and parallel to the neutral surface at AB. Consider the equi- librium of the piece of beam ECDF, shown in fig. 47 as ECE, and shown magnified in fig. 48. We have indicated only the forces which are parallel to the neutral surface or at right angles to the sections. The total pushing forces on DF are greater than the total pushing forces on CE, the tangential forces mi EF making up for the difference. We have only to state this mathematically and we have solved our problem. At a place like ^ in the plane CAC at a distance y from the neutral surface the compressive stress is known to be M . , p = Y y^ ^^^ ^^ ^ ^® *^^ breadth of the section there, shown as IIH (fig. 47), the total pushing force on the area ECE is rAC \f Jlf rAC P= b'^y.dy or P^'y- hy.dy (1). J AE J- -^ J AE 8—2 116 CALCULUS FOR ENGINEERS. Observe that if b varies, we must know it as a function of y before we can integrate in (1). Suppose we call this total pushing force on EC by the name P, then the total push- ing force on DF will be P + 8^ . -r* . The tangential force on EF is fx area of EFovf.hx. EE, and hence f.Bx.EE = Bx.^~ovf= J:^ . i? (2). *^ dx -' EE dx ^ ' Example. Beam of uniform rectangular section^ of constant breadth 6 and constant depth d. Then p 12m p^ , i2iifr^ 1 F^^-^{\d^-AE% and hence /= ^ |(t^'3_^/;.)^^^ (8); so that /is known as soon as M is known. As to M, let us choose a case, say the case of a beam supported at the ends and loaded uniformly with iv lb. per unit length of the beam. We saw that in this case, x being distance from the middle M — \wl^ — \wx^. Hence - 7- = — wx, so that (3) is dx f=~{^l''-AE^)wx (4). If we like we may now use the letter y for the distance -4 -£^, and we see that at any point of this beam, x inches measured horizontally from the middle, and y inches above the neutral line the shear stress is /=-5«*-2'')'^ ^■'^- The — sign means that the material below EF acts on the material above EF in the opposite sense to that of the arrow heads shown at EF, fig. 48. SHEAR STRESS IN BEAMS. 117 Observe that where 2/ = the shear stress is greater than at any other point of the section, that is, at points in the neutral line. The shear stress is at C. Again, the end sections of the beam have greatest shear. A student has much food for thought in this result (5). It is interest- ing to find the directions and amounts of the principal stresses at every point of the beam, that is, the interfaces at right angles to one another at any point, across which there is only compression or only tension without tangential stress. We have been considering a rectangular section. The student ought to work exercises on other sections as soon as he is able to integrate hy with regard to 3/ in (1) where h is rAG any function of y. He will notice that I hy.dy is equal to J AE the area of EHGHE, fig. 47, multiplied by the distance of its centre of gravity from A A. Taking a flanged section the student will find that / is small in the flanges and gets greater in the web. Even in a rectangular section /' became rapidly smaller further out from the neutral line, but now to obtain it we must divide by the breadth of the section and this breadth is comparatively so great in the flanges that there is practically no shearing there, the shear being confined to the web ; whereas in the web itself / does not vary very much. The student already knows that it is our usual custom to calculate the areas of the flanges or top and bottom booms of a girder as if they merely resisted compressive and tensile forces, and the web or the diagonal bracing as if it merely resisted shearing. He will note that the shear in a section is great only where , or rather -^\-f] is great. But inasmuch as in Art. QQ we saw that -j- — S, the total shearing force at the section, there is nothing very extraordinary in finding that the actual shear stress anywhere in the section depends upon -r- . In a uniformly loaded beam -7- is greatest at the ends and gets less and less towards the middle and then changes sign, hence the bracing of a girder loaded mainly with its own weight is much slighter in the middle than at the ends. 118 CALCULUS FOR ENGINEERS. Deflection of Beams. If a bending moment M acts at a section of a beam, the part of length bx gets the strain-energy \ — '' , because M . dx/FI is the angular change (see Art. 26), and therefore the whole strain-energy in a beam due to bending moment is m-'^ <«\ If / is a shear stress, the shear strain-energy \}er unit volume is lf2/2iV...(7), and by adding we can therefore find its total amount for the whole beam. By equating tlic strain-energy to the loads multiplied by half the displacements produced by them we obtain interesting relations. Thus in the case of a beam of length /, of rectangular section, fixed at one end and loaded at the other with a load W; at the distance x from the end, M= Wx and the energy due to bending is 172^.2 dx=WH^lQEI (8). 1 p TF2; ^Ej^ I The above expression (5) gives for the shearing stress f=\^iid?-f)W (9). The shear strain-energy in the elementary volume h .bx . by is b.bx . by.f'^/2K Integrating this with regard to y from - ^c^ to -{-^d we find the energy in the slice between two sections to be SWH.bx/biYbd, so that the shear strain-energy in the beam is 3 WH/dNbd... (10). If now the load W produces the deflection z at the end of the beam the work done is ^ Wz,..{\l). Equating (11) to the sum of (8) and (10) we find .= i^ + iL ^^ (12) Note that the first part of this due to bending is the deflection as calculated in Art. 60, Example I. We believe that the other part due to shearing has never before been calculated. If the deflection due to bending is z^ and to shearing is z^^ zjz.^=iomy3Ed^. Taking A^= § ^ as being fairly correct, then zjs.^ = Al^j^d^. If a beam is 10 inches deep, when its length is 8-6 inches the deflections due to bending and shear are equal ; when its length is 86 inches, the deflection due to bending is 100 times that due to shear ; when its length is 0-79 inch, the deflection due to bending is only 1 /100th of that due to shear. Probably however our assumed laws of bending do not apply to so short a beam. BENDING OF STRIPS. 119 69. Springs which Bend. Let fig. 49 show the centre line of a sirring fixed at A, loaded at B with a small load W in the direction shown. To find the amount of yielding at B. The load and the deflec- tion are supposed to be very small. Consider the piece of spring bounded by cross sec- tions at P and Q. Let FQ = 8s, the length of the spring between B and P being called s. Fig. 49. The bending moment at P is W . PR or W.w if x is, the length of the perpendicular from P upon the direction of W. Let BR be called y. Consider first that part of the motion of B which is due to the change of shape of QP alone ; that is, imagine AQ to be perfectly rigid and PB a rigid pointer. The section at Q being fixed, the section at P gets an angular M change equal to Ss x the change of curvature there, or Bs -^rt- 8s Wx . . or — '^p^ ...(1), where E is Young's modulus and / is the moment of inertia of the cross section. The motion of B due to this is just the same as if PB were a straight pointer ; in fact the pointer PB gets this angular motion and the motion of B is this angle, multiplied by the straight distance PB or Ss. Wx EI PB .(2). Now how much of B's motion is in the direction of IF ? PR X It is its whole motion x -p^ or x p-^ and hence B'a motion in the direction of W is hs . Wx" EI .(3). 120 CALCULUS FOR ENGINEERS. Similarly B'a motion at right angles to the direction of W is ^-i^ (4). In the most general cases, it is easy to work out the integrals of (3) and (4) graphically. We usually divide the whole length of the spring from B to A into a large number of equal parts so as to have all the values of Bs the same, and then we may say (s being the s W whole length of the spring) that we have to multiply —w— upon the average values of -^ and -j- for each part. In a well made spring if b is the breadth of a strip at right angles to the paper and t its thickness so that / = -f^bt^ we usually have the spring equally ready to break everywhere or ~r42~~f> ''^ constant. When this is the case (3) and (4) become 2/.^^ ^ and ^-^'^^ y And if the strip is constant in thickness, varying in breadth in proportion to x, then If X and y are the x and y of the centre of gravity of the curve (see Art. 48) ^- is the total yielding parallel to W, li/t •^-^ is the total yielding at right angles to 'W. 70. Exercises. The curvature of a curve is When the equation to a curve is given it is easy to find STATIC PRESSURE OF FLUIDS. 121 -^ and v^and calculate - where r is the radius of curvature. ax da? r This is mere exercise work and it is not necessary to prove beforehand that the formula for the curvature is correct. 1. Find the curvature of the parabola y — ax^ at the point d? = 0, 3/ = 0. 2. The equation to the shape of a beam, loaded uni- w formly and supported at the ends is 3/ = aqtpt {^^^^^ ~ 2^'*), see Art. 60, where the origin is at the middle of the beam ; I is the whole length of the beam, lu is the load per unit length, E is Young's modulus for the material and / is the moment of inertia of the cross section. Take I = 200, w = 5, E = 29x 10«, / = 80, find the curvature where x = 0. Show that in this case f-^) may be neglected, in comparison with 1, and that really the curvature is represented by -vt,. Show that the bending moment of the above beam is ^^RFT ^^^ " ^'^•^' ^^^^ *^^^ ^^^^ ^^ greatest at the middle of the beam. 3. Find the curvature of the curve i/= a log x +bx-\-c at the point where x = Xi. 71. Force due to Pressure of Fluids. Exercise 1. Prove that if p, the pressure of a fluid, is constant, the resultant of all the pressure forces on the plane area A is Ap and acts through the centre of the area. -"t^j ; 2. The pressure in ''\ a liquid at the depth h \ being wh, where w is the ^^\^ weight of unit volume, what is the total force due to pressure on any im- mei*sed plane area ?- Let DE be the surface from which the depth h is measured and where the ^^S- ^0. 122 CALCULUS FOR ENGINEERS. pressure is 0. Let BG be an edge view of the area ; imagine its plane produced to cut the level surface of the liquid DE in D. Let the angle EDO be called a. Let the distance DP be called x and let DQ be called x + hx, and let the breadth of the area at right angles to the paper at P be called z. On the strip of area z . hx there is the pressure wh if h is PH the depth of P, and h^x sin a, so that the pressure force on the strip is wx . sin OL . z . hx, rue and the whole force is F=wsuyol\ x .z .dx (1 ). J 1)B Also if this resultant acts at a point in the area at a distance X from D, taking moments about P, rDC FX=tu sina x- .z.dx (2). J DB rDC Observe in (1) that I x ,z ,dx — Ax, J DB if A is the whole area and x is the distance of its centre of gravity from P. Hence, the average pressure over the area is the pressure at the centre of gravity of the area. CDC Observe in (2) that I x^z .dx — I the moment of inertia J DB of the area about P. Letting I — k^A , where k is called the radius of gyration of the area about P, we see that F=iu sin a . Ax, FX = w sin a . Ah^. Hence X = — . . .(3), the distance from D at which the x resultant force acts. Example, If DB = and the area is rectangular, of constant breadth h ; then CDC h I=h x^.dx = ^DC\ Jo «5 and ^ = 6 . PC so that k^ = ^DC\ Also x = J PC. Hence X = fPC, that is, the resultant force acts at | of the way CENTRIFUGAL FORCE IN FLUIDS. 123 down the rectangle from i) to (7 and the average pressure is the pressure at a point half way down. It is an easily remembered relation that we find in (8). For if we have a compoimd pendulum, whose radius of gyration is k and if x is the distance from the point of support to its centre of gravity and if X is the distance to its point of percussion, we have the very same equation (3). Again, if X is the length of the simple pendulum which oscillates in exactly the same time as the compound one, we have again this same relation (3). These are merely mathematical helps to the memory, for the three physical phenomena have no other relation to one another than a mathematical one. Whirling Fluid. 72. Suppose a mass of fluid to rotate like a rigid body about an axis with the angular velo- city of a radians per second. Let 00 be the axis. Let P be a particle weighing w lbs. Let OP o<> X. The centrifugal force in pounds of any mass is the mass multiplied by the square of its angular velocity, multi- plied by X. Here the mass w is - and the centrifugal force • "U) „ IS - o?x. Fig. 51. 9 Make PR represent this to scale and let PS represent w the weight, to the same scale, then the resultant force, repre- sented by PT, is easily found and the angle RPT which PT w makes with the horizontal. Thus tan RPT — w^— ot^x or 9 g — a^x, being independent oi w: we can therefore apply our results to heterogeneous fluid. Now if y is the distance of the point P above some datum level, and we imagine a curve drawn through P to which PT is (at P) tangential, and if at 124 CALCULUS FOR ENGINEERS. every point of the curve its direction (or the direction of its tangent) represents the direction of the resultant force; if such a curve were drawn its slope ~ is evidently — y- and its equation is y = — -2 log x + constant (1 ). The constant depends upon the datum level from which y is measured. This curve is called a line of force. Its direction at any place shows the direction of the total force there. We see that it is a logarithmic curve. Level Surfaces. If there is a curve to which PT is a Fig. 52. FLUID MOTION. 125 normal at the point P, it is evident that its slope is positive and in fact dx — —X, so that the curve is y — ^x^-\- constant (2), the constant depending upon the datum level from which y is measured. This is a parabola, and if it revolves about the axis we have a paraboloid of revolution. Any surface which is everywhere at right angles to the force at every point is called a level surface and we see that the level surfaces in this case are paraboloids of revolution. These level surfaces are sometimes called equi-potential surfaces. It is easy to prove that the pressure is constant everywhere in such a surface and that it is a surface of equal density, so that if mercury, oil, water and air are in a whirling vessel, their surfaces of separation are paraboloids of revolution. The student ought to draw one of the lines of force and cut out a template of it in thin zinc, 00 being another edge. By sliding along 00 he can draw many lines of Force. Now cut out a template for one of the parabolas and with it draw many level surfaces. The two sets of curves cut each other everywhere orthogonally. Fig. 52 shows the sort of result obtainable where aa^ hb, cc are the logarithmic lines of force and A A, BB, GO are the level paraboloidal surfaces. 73. Motion of Fluid. If AB is si stream tube, in the 126 CALCULUS FOR ENGINEERS. vertical plane of the paper, consider the mass of fluid between sections at P and Q of length Bs feet along the stream, and cross-section a square feet, where a and Bs are in the limit supposed to be infinitely small. Let the pressure at F be p lbs. per square foot, the velocity v feet per second, and let F be at the vertical height h feet above some datum level. At Q let these quantities be ^ + Sp, v + Bv and h + Bh. Let the fluid weigh w lbs. per cubic foot. Find the forces urging PQ along the stream, that is, forces parallel to the stream direction at FQ. pa acts on one end F in the direction of motion, and (p-\-Bp)a acts at Q retarding the motion. The weight of the portion between F and Q is a . Bs . w and, as if on an inclined plane, its retarding component is . , ^ height of plane ^ Bh weight X -, — ^Tu— r n — or a. Bs .w ^, ° length ot plane Bs Hence we have altogether, accelerating the motion from F towards Q, pa — (p + Bp) a — a . Bs . w . ^— . But the mass is — — , and -7- is its acceleration, and we have merely to put the force equal to — ^ — '- — . -=- . We have then, dividing by a, 5. ^ Bh Bs .w dv -bp-8s.w-^= -J- . ^ Bs g dt Now if Bt be the time taken by a particle in going from Bs F to Q, v — j- with greater and greater accuracy as Bs is . dv . shorter and shorter. Also, the acceleration , is more aud Bv . ^^ more nearly kt . (It is more important to think this matter out carefully than the student may at first suppose.) Hence if Bs is very small, Bs .-j- — -^.Bv — v ,Bv, so that at ot we have Bp + w. Bh-^-v .Bv = (1), FLOW OF FLUID. 127 or as we wish to accentuate the fact that this is more and more nearly true as Ss is smaller and smaller, we may write it as ^ + dh-\--.dv==0 (2)*, w g or integrating, ^ + 9- + I = constant (2). dp We leave the sign of integration on the -^ because w may vary. In a liquid where w is constant, /* + ^ + --= constant (3). 2g w ^ -^ 74. In a gas, we have lu ocpif the temperature could be kept constant, or we have the rule for adiabatic flow w x py, where 7 is the well-known ratio of the specific heats. In either of these cases it is easy to find I — and write out the law. This law is of universal use in all cases where viscosity may be neglected and is a great guide to the Hydi^aulic Engineer- Thus in the case of adiabatic flow w=cpy , the inteOTal of -^ is f dp 1 /" _i 1 V 1 1 we have h-{--;r--\ — ©"^constant (4). In a great many problems, changes of level are insignificant and we * After a little experience with quantities like 5p &c., knowing as we do that the equations are not true unless 5/), &c. are supposed to be smaller and smaller without limit and then we write their ratios as ^f , &c., we an get into the way of writing dp, &c. instead of 5p, &c. Again, if /(x) . dx-\-F{y) dy + <p [z) dz = (1), then 1/H 'dx+ I F {y) . dy + I <p{z) . d2 = a constant (2). There is no harm in getting accustomed to the integration of such an equation as (1), all across. 128 CALCULUS FOR ENGINEERS. often use i?^^.,^^*^ constant (4) for gases. Thas, if «« is the cs pressure and zvq the weight of a cubic foot of gas inside a vessel at places where there is no velocity and if, outside an orifice, the pressure isp; the constant in (4) is evidently + — jOq«, and hence, outside the C8 surface, v^=-^{pQ'-p') (5), and as c is Wq-^PqV it is easy to make all sorts of calculations on the quantity of gas flowing per second. Observe that if p is very little less than joq, if we use the approxi- mation (1 +«)" = 1 -\-naf when a is small, we find v'=^^{Po-p) (6), a simple rule which it is well to remember in fan and windmill problems. In a Thomson Water Turbine the velocity of the rim of the wheel is the velocity due to half the total available pressure ; so in an air turbine when there is no great difference of pressure, the velocity of the rim of the wheel is the velocity due to half the pressure difference. Thus if Pq of the supply is 7000 lbs. per square foot and if p of the exhaust is 6800 lbs. per square foot and if we take ?^'q=0*28 lb. per cubic foot, the velocity of the rim V is, since the difference of pressure is 200 lbs. per square foot, y % (100) = 151 feet per second. Returning to (5) ; neglecting friction, if there is an orifice of area A to which the flow is guided so that the streams of air are parallel, Q the volume flowing per second is Q=vA and if the pressure is p, the weight of stuff flowing per second is W==vAw, 1 1 or since w = cp^ ^ and iOQ=cp^ , If the student will now substitute the value of v from (5) and put a for p/pQ he will obtain W =^"'-N/r^^:(--^^^) ^^)- Problem. Find p the outside pressure so that for a given inside pressure there may be a maximum flow. FLOW OF GASES. 129 It is obvious that as p is diminished more and more, v the velocity mcreases more and more and so does Q. But a large Q does not necessarily mean a large quantity of gas. We want W^ to be large. When is W a maximum 1 That is, what value of a in (7) will make a^l-a-^ )ova^-a ^ a maximum 1 Differentiating with regard to a and equating to 2 , / -.v 1 ?„;— _(i+i)„9=o dividing by ay we find a = ( ^ ) JUL In the case of air 7/= 1-41 and we find p= •527po- That is, there is a maximum quantity leaving the vessel per second when the outside pressure is a little greater than half the inside pressure. » Problem. When p is indefinitely diminished what is v ? Answer -. v= j,./ -^ — . This is greater than the velocity of sound in the ratio a/ — — t- , being 2 -21 for air. That is, the limiting velocity in the case of air is 2413 feet per second x */ — - , where t is the absolute temperature inside the vessel and there is a vacuum outside. Students ought to work out as an example, the velocity of flow into the atmosphere. Returning to equations (2) and (4), we assumed h to be of little importance in many gaseous problems of the mechanical engineer. But there are many physical problems in which it is necessary to take account of changes in level. For example if (2) is integrated on the assumption of constant temperature and we assume v to keep constant, we find that p diminishes as h increases according to the compound interest law considered in Chap. 11. Again under the same condition as to V, but with the adiabatic law for w we find that p diminishes with h according to a law which may be stated as " the rate of diminution of temperature with A, is constant." These two propositions seem to belong more naturally to the subject matter of Chapter II. 130 CALCULUS FOR ENGINEERS. 75. A great number of interesting examples of the use of (2) might be given. It enables us to understand the flow of fluid from orifices, the action of jet pumps, the attraction of light bodies caused by vibrating tuning-forks, why some valves are actually sucked up more against their seats instead of being forced away by the issuing stream of fluid, and many other phenomena which are thought to be very curious. Example 1. Particles of water in a basin, flowing very slowly towards a hole in the centre, move in nearly circular paths so that the velocity v is inversely proportional to the distance from the centre^ Take v = - where a is some con- x stant and x is the radius or distance from the axis. Then (13) (Art. 73) becomes k^f-, + P = f;. 2gx- w Now at the surface of the water, p is constant, being the pressure of the atmosphere, so that, there and this gives us the shape of the curved surface. Assume c and a, any values, and it is easy to calculate h for any value of X and so plot the curve. This curve rotated about the axis gives the shape of* the surface which is a surface of revo- lution. Example 2. Water flowing spirally in a horizontal plane follows the law v = - if ^ is distance from a central point. w h^ Note that p=:G^—h- - . ^ g X'' The ingenious student ought to study how p and v vary at right angles to stream lines. He has ohly to consider the equilibrium of an elementary portion of fluid PQ, fig. 53, subjected to pressures, centrifugal force and its own weight in a direction normal to the stream. He will find that if ~- means the rate at which p FLOW OF LIQUID. 131 varies in a direction of the radius of curvature away from the centre of curvature and if a is the angle QPR, fig. 53, the stream being in the plane of the paper, which is vertical, dp wv^ . ,^, -T = lusma (1). ar g r If the stream lines are all in horizontal planes f = ^^' (2). ar g r ^ Example 3. Stream lines all circular and in horizontal planes in a liquid, so that h is constant. If -y = -, where 6 is a constant, r dp _ lu h- dr g ' T^' p:=— I , + constant (3). ^ ^ g r- We see therefore that the fall of pressure as we go out- ward is exactly the same as in the last example. Show that this law, V = hjr, must be true if there is no ' rotation ' (See Example 5). Example 4. Liquid rotates about an axis as if it were a rigid body, so that v = 6/-, then dr g ' This shows the law of increase of pressure in the wheel of a centrifugal pump when full, but when delivering no water. Exercise. The pressure at the inside of the wheel of a centrifugal pump is 2116 lbs. per sq. foot, the inside radius is 0*5 foot, the outside radius 1 foot. The angular velocity of the wheel is 6 = 30 radians per second ; draw a curve show- ing the law of p and r from inside to outside when very little water is being delivered. If the water leaves the wheel by a spiral path, the velocity everywhere outside being 9—2 132 CALCULUS FOR ENGINEERS. invei-sely proportional to r, draw also the curve showing the law of p in the whirlpool chamber outside. Example 5. The expression ti^ 1 which remains constant all along a stream line, may be called the total store of energy of 1 lb. of water in the stream if the motion is steady. ^^ dE 1 dv 1 dp dh , « ^. ,.,. Now -r- =~ V j--\ r-4-T- becomes irom equation (1), d?' g or w dr ar . r ~ g 2\r dr) This expression ^ \- + -r) is called the "average angular velocity" or "the rotation" or the 'spin' of the liquid. Hence dE 2v ^j- = — X rotation. dr g When liquid flows by gravity from a small orifice in a large vessel where, at a distance inside the orifice, the liquid may be supposed at rest, it is obvious the E is the same in d IR all stream lines, so that -T7 is 0, and there is no ' rotation ' anywhere. ^ If when water is flowing from an orifice in a vessel we can say that across some section of the stream the velocity is everywhere normal to the section and that the pressure is everywhere atmospheric, we can calculate the rate of flow. It is as well to say at once that we know of no natural foundation for these assumptions. However wrong the assumptions may be, there is no harm in using them in mere exercises on Integration. There being atmospheric pressure at the still water level, if v is the velocity at a point at the depth A, if a is an element of area of the section, Q= 2a 'J2gh the summation being effected over the whole section, Q being the volume flowing. Thus if the section is a vertical plane and if at the depth h it is of horizontal breadth z, through EXERCISES ON INTEGKATlON. ISS the area z . Sh water is flowing with the velocity '^2gh, so that ^2gh . z .Bhis the elementary volume flowing per second, and if hi and h^ are the depths of the highest and lowest points of the orifice, the total flow is Q = V2^ ( zh^ . dh. J hi Example 6. Rectangular section, horizontal breadth 6, Q = ^¥gh{\^.dh = |6\/2^ M 1 = lb \f2g (h.^ - h^^). Example 7. Triangular section, angle at depth h^, base horizontal of length b at depth h... Then within the limits of integration it will be found that z= j v (— /<i 4- h). fl2 — ill Hence Q = |^^ f(- hih^^ + A?) dh = ^-^-^f W-^hJi^ + W-)\ «2— "W h'i — hi L/t^ J If the ratio hJjhi be called r, it will be found that (^ = i^^f J6H-10ri+25l. When the student has practised integration in Chap. III., he may in the same way find the hypothetical flow through circular, elliptic and other sections. Keturning to the rectangular section, there is no case practically possible in which h^ is 0, but as this is a mere mathematical exercise let us assume hi = 0, and we have Q= |6 s^'lgh^. Now further assume that if there is a rect- angular sharp-edged notch through which water flows, its edge or sill being of breadth b and at the depth h^, the flow through it is in some occult way represented by the above answer, multiplied by a fraction called a coefficient of contraction, then Q = cb \l2gh^. Such is the so-called theory of the flow through a rectangular gauge notch. A true theory was based by Prof. James Thomson on his law of flow from similar orifices, one of the very few laws which the hydraulic engineer has to depend upon. We are sorry to think that nearly all the mathematics to be found in standard treatises on Hydraulics is of the above character, that is, it has only an occult connection with natural phenomena. 134 tJALClTLtJS FOR ENGINEERS. 76. Magnetic Field about a straight round wire. There are two great laws in Electrical Science. They concern the two circuits, the magnetic circuit and the electric circuit, which are always linked through one another. I. The line integral (called the Gaussage whatever the unit may be) of Magnetic Force round any closed curve, is equal to the current [multiplied by 47r if the current is in what is called absolute C.G.S. units (curious kind of absolute unit that needs a multiplier in the most important of all laws); multiplied by 47r/10 if the current is in commercial units called Amperes]. II. The line integral (called the Voltage whatever the unit may be) of Electromotive Force round any- closed curve is equal to the magnetic current (really, rate of change of induction) which is enclosed. [If the in- duction is in absolute C.G.S. units, we have absolute Voltage in C.G.S.; if the induction is in Webers the Voltage is in Volts. We are to remember that in a non-conducting medium the voltage in any circuit produces electric displacement, and the rate of change of this is cun-ent, and we deal with this exactly as we deal with currents in conducting material. When we deal with the phenomena in very small portions of space we speak of electric and magnetic currents per unit area, in which case the line integrals are called ' curls/ Leaving out the annoying 47r or 47r/10, we say, with Mr Heaviside, " The electric current is the curl of the magnetic force and the magnetic current is the negative curl of the electric force." When we write out these two statements in mathematical language, we have the two great Differential Equations of Electrical Analysis. The Electrical Engineer is continually using these two laws. Many examples will be given, later, of the use of the second law. We find it convenient to give here the following easy example of the first law. Field about a round wire. A straight round wire of or A amperes, so that G 10 on r °%10 MAGNETIC FIELD. loO If H is the magnetic force at a distance r from the centre of the wire, the Gaussage round the circle of radius r is Hx27rr, because H is evidently, from symmetry, the same all round. Hence, as Gaussage = 47rC, 2 Al Inside the wire, a circle of radius r encloses the total current — G, and hence H inside the wire at a distance r from the axis is 2rG r 2r A ^ ['' To ^ If BG is a cross section of the round wire of radius a, and if OD is any plane n through the axis of the f\ p Q p /^"^ wire, and \oJ y \0^ OP==r,OQ = ri-Br: ^ f". 54. then through the strip of area FQ, which is / centimetres long at right angles to the paper, and Sr wide, area I.Br, there is the induction If per sq. cm. [We take the perme- ability as 1. If /A is the magnetic permeability of the me- dium, the induction is yS = fiH per sq. cm.], or H .I.Br through the strip of area in question. If there are two parallel wires with opposite currents, and if OB is the plane through the axes of the two wires, the fields due to the two currents add themselves together. If 0' is the centre of the other wire, the total i^ at P is 2C ( ^p + yyp 77. Self-induction of two parallel wires. Let the radius of each wire be a, and the distance between their centres h, the length of each being I between two planes at right angles to both. The wires are supposed to be parts of two infinite wires, to get rid of difficulties in imagining the circuit com- pleted at the ends. The total induction from axis to axis is the sum of the [^ G dr two amounts, 4>l - - — from the outside of each wire to the 1 J a 136 CALCULUS FOR ENGINEERS. fa rC axis of the other and 41 \ —- dr from the axis of each wire Jo Cb' to its own surface. This is 21G j 2 log - + 1 1 , or - . j 2 log - + 1 [ in absolute units. Dividing by 10^ we have it in commercial units. This total field when the current is 1, is the self-induction L of the circuit (we imagine current to be uniformly distri- buted over the section of the wire), and y = ^ U^g~i + 1^ in c.G.s. units, in Henries per centimetre length of the two circuits. 78. Function of Two Independent Variables. Hitherto wo have been studying a function of one variable, which we have generally called x. In trying to under- stand Natural Phenomena we endeavour to make one thing only vary. Thus in observing the laws of gases, we measure the change of pressure, letting the volume only change, that is, keeping the temperature constant, and we find p cc - , Then we keep v constant and let the tempera- ture alter, and we find p cc t (where ^ = ^"^ C. + 274). After * Notice that one Henry is lO' absolute units of self-induction ; our commercial unit of Induction called the Weber is 10* absolute units of Induction. dA The Henry suits the law : \o\iB= RA+L — , The Weber suits Volts = J?^ + iV . ^ , where R is in ohms, A amperes, L Henries, N the number of turns in a circuit, I Weber's of Induction. In Elementary Work such as is dealt with in this book, I submit to the use of 47r and the difficulties introduced by the unscientific system now in use. In all my higher work with students, such as may be dealt with in a succeeding volume, I always use now the rational units of Heaviside and I feel sure that they must come into general use. TWO INDEPENDENT VARIABLES. 137 much trial we find, for one pound of a particular gas, the law pv = Rt to be very nearly true, R being a known constant. Now observe that any one of the three, p, v or t, is a function of the other two; and in fact any values what- soever may be given to two, and the other can then be found. Thus p = R^ (1), we can say that p is a, function of the two independent vari- ables t and V. If any particular values whatsoever of t and v be taken in (1) we may calculate p. Now take new values, say t + Bt and v + Sv, where St and 8v are perfectly independent of one another, then . s^ 7? ^ + ^^ J s> n t + Bt j^t p-\-6p = R ^ and 6/; = it s R - . V -\-ov ^ V ■\-ov V We see therefore that the change Sp can be calculated if the independent changes Et and Bv are known. When all the changes are considered to be smaller and smaller without limit, we have an easy way of expressing Bp in terms of Bt and Bv. It is '^-m^^-o (^>- This will be proved presently, but the student ought fii'st to get acquainted with it. Let him put it in words and compare his own words with these : " The whole change in ^j is made up of two parts, 1st the change which would occur inp if V did not alter, and 2nd the change in p if ^ did not alter." The first of these is St x the rate of increase of p with t when v is constant, or as we write it ( -^ ) Bt, and the second of these is Bv x the rate of increase of p with v if t is constant. This idea is constantly in use by every practical man. It is only the algebraic way of stating it that is unfamiliar, and a student who is anxious to understand the subject w^ill manufacture many familiar examples of it for himself. 138 CALCULUS FOR ENGINEERS. Thus when one pound of stuff which is defined by its p^ v and t^ changes in state, the change is completely defined by any two of the changes bp and hi\ or bv and bt, or bp and bt^ because we are supposed to know the characteristic of the stuff, that is, the law connecting p, v and t. Now the heat bH given to the stuff in any small change of state can be calculated from any two of bv, bt and bp, and all the answers ought to agree. As we wish to accentuate the fact that the changes are supposed to be exceedingly small we say dH=k . dt+ l.dv\ = K. dt-\-L.dp> (3), = r .dp+V.dv) where the coefficients k, I, A', A, P and V are 'all functions of the state of the stuff, that is of any two of v, t and p. Notice that k . dt is the heat required for a small change of state, defined by its change of temperature, if the volume is kept constant : hence k is called the specific heat at constant volume. In the same way K is called the specific heat at constant pressure. As for I and L perhaps they may be regarded as some kinds of latent hent, as the temperature is supposed to be constant. These coefficients are not usually constant, they depend upon the state of the body. The mathematical proof that if bH can be calcu- lated from bt and bv^ then dH=k . dt + l. dv, where k and I are some numl;ers which depend upon the state of the stuff, is this : — If bH can be calculated, then bH=L bt-^l.bv-\-a(bty + b{bv)^ + c{bt.bv)+e{btf + terms of the third and higher degrees in bt and bv, where k, I, a, b, c, e &c. are coefficients depending upon the state of the body. Dividing by either bt or bv all across, and assuming bt and bv to diminish without limit, the proposition is proved. Illustration. Take it that for one pound of Air, (1) is true and R is, say, 96, p being in lb. per sq. foot and v in cubic feet. As « = 96 - , Hence, from (2), hp = ^^^ .M — - .hv (4) Example. Let t = 300, p = 2000, t; = 1 44. If t becomes 301 and v becomes 14*5 it is easy to show that p will became 199283. But we want to find the change in pressure, using (2) or rather (4), ^ 96 , 2000 , ^^^„ ^i^= 14^ >< 1 - 14:4 >< •!=- 7-22 lb. per sq.ft., whereas the answer ought to be — 7*17. 'dp\ 96 fdp\ .dt)~ V ' \dvj~ V ' Bp = ^l,Bt-^.Sv.. ^ V V SPECIFIC HEATS. 139 Now try Bt = 1 and Bv = '01 and test the rule. Again, try ^^ = •01 and Bv = "001, or take any other very small changes. In this way the student will get to know for himself what the rule (1) really means. It is only true when the changes are supposed to be smaller and smaller without limit. Here is an exceedingly interesting exercise : — Suppose we put hp = in (2). We see then a connection between Bt and 8v when these changes occur at constant pressure. Divide one of them by the other ; we have ^ when p is constant, or rather fdp\ dv\ \dtj (dv\ \dtj ,^s \Tt)-'W{ '^''^• \dv) At first sight this minus sign will astonish the student and give him food for thought, and he will do well to manu- facture for himself illustrations of (5). Thus to illustrate it with pv = Rt Here 'dv\ _ R (dp\ _ R (dp\ _ Rt p /dv\ _ K /dp\ _ R \dtl ~ ~p ' \dt) ~ V ' Ivj t^ V and (5) states the truth that ^^_^^( _P The student cannot have better exercises than those which he will obtain by expressing hv in terms of ht and Sp, or ht in terms of Bp and hv for any substance, and illustrating his deductions by the stuff for which pv = Rt t 79. Further Illustrations. In (.3) we have the same answer whether we calculate from dt and dv, or from dt and dp, or from dp and c?y. Thus for example, Jt . dt+l .dv = K . de + L . dp (6). We saw that ^P={-j^) (^i+{-r-] dv, and hence substituting this for dp in (6) we have k.dt + Ldv = K.dC + Li'^^]dt-{-L{^^)dv: (1)"-^: 140 CALCULUS FOR ENGINEERS. This is true for any independent changes dt jind do\ let do=0^ and again let dt=0^ and we have ^-A-+^g) (7), ^At) («>• Again, in (6) substitute ^^=(j:)^^+(j~) ^^P^ ^^^^ we have k.dt + l(^dt-\-l(^\dp = K.dt + L.dp. Equating coefficients of dt and of dp as before we have ^•+'60=''^' '-'^ <©=^ (^°)- Again, putting h .dt-\-l .dv = P .dp+ V . doy and substituting we have /: . <lt + Ld,' = p(^£\ dt+p{^ dv+ V. dv, -d *=^(|) ("). al»o ;=/>(*)+ F (12). Again, putting K . dt + L . dp^ P . dp+ V . dv, and substituting we have K (j-j dp+K l-r-j dv + L . dp = P . dp+ V . dt\ ""■> ^(1)+^=^' c-^)' ^(1)='^ (»>• The relations (7), (8), (9), (lO), (ii), (12), (13) and (14) which are not really all iudepeiulcnt of one another (and indeed we may get others in the same way) are obtained merely mathematically and without assuming any laws of Thermodynamics. We have called THE TWO ELASTICITIES. 141 ZT, heat; t temperature &c., but we need not, unless we please, attach any physical meaning to th^ettei-s. The relations are true for any substance. Find what they become in the case of the stuff for which pv = Rt (the mathe- matical abstraction called a perfect gas). We know that -^ 1 = — , so that (7) becomes k=K-{-L— (7)*, (s) /dp\ ^_P ^^ ^j^3^^ (8) becomes l=-L^ (dv\_ R \dt)~ p (dv\ _ V \dp)~~p \dt)~ V (dp\_'P \dv) V V (8)*, SO that (9) becomes k + l — = K (9)* so that (10) becomes -Z- = X (10)*, so that (11) becomes ]c=P- (11)* so that (12) becomes l=-r(-+y (12)^ It is evident that these are not all independent ; thus using (10)* in (9)* we obtain (7)*. 80. Another Illustration. The Elasticity of our stuff is defined, see Art. 58, as dp dv Now if t is constant, we shall write this e^ = — v ( -, - j , or the elasticity when the temperature remains constant. If it is the adiahatic elasticity e„ which we require, we want to know the value of -^ when the stufl neither loses nor gains heat. In the last expression of (3) put dH=Oj and the ratio of our dp and our dv will then be just what is wanted or l-f) = — p> *^® ^ being affixed to indicate that ff is constant or that the stuff neither loses Y nor gains heat. Hence eij=v ^. Taking Y from (14) Art. 79 and P from (11), €n_ \dvj \dtj ei J 7T% 142 CALCULUS FOR ENGINEERS. but we have already seen as in (5) that* f ^^ ) -i- ( -^ ) = — ( -r-; ] and hence for any substance -5 = ^- (15). This ratio of the two specific Heats is usually denoted by the letter y. Note that neither of the two laws of Thermodynamics nor a Scale of temperature is referred to in this proof. 81. General Proof. If u is a function of a- and y, we may write the statement in the form u=f(x, y). Take particular values of X and y and calculate u. Now take the values x-\-bx and y + by, where bx and by are perfectly independent of one another, and calculate the new u^ call it u + bu. Now subtract and we can only indicate our result by bu=f{x+bx, y+by)-f{x, y). Adding And subtracting the same thing f{x^ y + by) we have bic=f{x+bx, y+by)-fix, y+by)+f{x, y+by)-f{x, y). This is the same as ^^_ /(^+&r,y+dy)-/(a;,y+&y ) g^, ^f{^,y+b y)-f{x,y) ^^^^^^^^ Now if bx and by be supposed to get smaller and smaller without limit, the coefficient of by or/(^+MzZ(^y) becomes ^^^^ or (f), by dy \dy) the X being constant. In fact this is our definition of a differential coefficient (see Art. 20, Note). Again, the coefficient of bx becomes the limiting value of — > ^/ J \ ^ if) ^ i^ecause by is evanescent. Writing then u instead of /(^, y) we have «»"=(i)<^-+(g)'»y (")• Thus if w = ax^ + 6y^ + cxy, du = i2ax + cy) dx-\-{2by + ex) dy. 82. Notice that although we may have dz=M. dx-\-JSr. dy (18), where M and iV are functions of x and y ; it does not follow that ^ is a function of x and y. For example, we had in (3) dll'=k.dt+l.dv^ where k and I are functions of t and v. Now H the total heat which has been given to a pound of stuff is not a function of v and t ; it is not a flinction of the state of the stuff. Stufi' ma^ COMPLETE DIFFERENTIAL. 143 receive enormous quantities of heat energy, being brought back to its original state again, and yet not giving out the same amounts of he^at as it received. The first law of Thermodynamics states however chat if dE = dH — p . dv, where p . dv is the mechanical work done, we can give to E the name Intrinsic Energy because it is something which is a function of the state of the stuff. It always comes back to the same value when the stuff returns to the same state. Our E is then some function of t and v, or of t and jo, or of p and v, but H is not ! The second law of Thermodynamics is this : — If dH be divided by t where t is ^°C. + 274, 6°Q. being measured on the perfect gas thermo- JTT meter, and if — be called c?0, then ^ is called the Entropy of the stuff, and </> is a function of the state of the stuff. 83. It is very important, if dz=M.dx-\-N .d)/ (18), where M and N are functions of x and y, to know when z is a function of X and y. If this is the case, then (18) is really that is, M is ( ;t- ) and iV is (-r-)^ ^"^ i^^"^« (^) = ( as^) (1^)' because it is known that -. — — ,- = -^ ^ . dy .dx dx . a// d2u d2u * Proof that - — —. = -— T= • dy . dx dx . dy We gave some illustrations of this in Art. 31, and if the student is not yet familiar with what is to be proved, he had better work more examples, or work the old ones over again. Let w=/(a?, y); (^) is the limiting value of f^'^ + ^'^^y)-f('^'y^ as 8x gets smaller and smaller. Now this is a function of y, so — ( ^ j or ^^ is, by our defini- tion of a differential coefiicient, the limiting value of 1 l f{x + 8x,y + dy)-f{x,y + Sy) f(x + 8x,y)-f{x,y) ] by \ 8x $x \ as 8y and 8x get smaller and smaller. 144 CALCULUS FOR ENGINEERS. Here we have an exceedingly important rule: — If dz=M.dx + N.dy (18), and if 2 is a function of x and v (another way of saying that z is a function of x and y is to say that dz=M ,da;-\-N .dy is a complete differentiaVjy then "£f (-)• (dM\ _ (o Working the reverse way, we find that ^ — -p is the limiting value of 1^ { f{x+Sx, y + S y )-f{x-\-dx, y ) f{x, y + Sy)-f{x ,y)\ 8x I Sy dy ) as Sy and 5x get smaller and smaller. Now it is obvious that these two are the same for all values of 5x and %, and we assume that they remain the same in the limit. * M .dx + N.dy (1), where M and N are functions of x and y, can always be multiplied by some fonction of x and y which will make it a complete differential. This multiplier is usually called an integrating factor. For, whatever functions of x and y, M and N may be, we can write dx-~N ^^'r and this means that there is some law connecting x and y. Call it ^,.„.c.t.en(f).(^-)| = ,3, and as -^ from (3) is the same as in (2) it follows that (— )"^(*7— ) = xr» and hence (—\= /xM, f — j = fiN, where /* is a function of x and y or else a constant. Multiplying (1) by /* we evidently get (^h-i^^y w. and this is a complete differential. It is easy to show that not only is there an integrating factor /x but that there are an infinite number of them. As containing one illustration of the importance of this proposition I will state the ^teps in the proof which we have of the 2nd law of Thermo- dirnamictt. 1. We have shown that for any substance, of which the state is defined by its t and v, dH=k.dt+l.dv (5), where k and I are functions of t and v. Observe that t may be measured on any curiously varying scale of tempera- ture whatsoever. We have just proved that there is some function /u. of t and THERMODYNAMICS. 145 84. The First Law of Thermodynamics is this : If dE= dH-p . dv, or dE=- k.dt + {l -p) dv, then dE is a complete differential that is, ^returns to its old value when V by which if we multiply (5) all across we obtain a complete differential ; indeed there are an infinite number of such functions. Then calling the result d0, d(f> = fi.dH=fik.dt + fxl.dv (6). Let us see if it is possible to find such a value of n that it is a function of t only. If so, as the differential coefficient of fik with regard to v {t being supposed constant) is equal to the differential coefficient of fil with regard to t {v being supposed constant), (dk\ _ dti (dX\ \di)t~ Tt'^^Utjv fdk\ _fdl\ or I — J =( ^1 +-. -^ » /* at But the first law of Thermodynamics (see Art. 84) gives us .(7). and hence /dk\ _ /dl\ _(dp\ \dv)t~\dt)v \dt) l/dp\ _ 1 djx T\dtJ "■ " |I ■ dt .(8), (9). This then is the condition that /j.. dH is & complete differential, fi being a function of temperature only. Obviously for any given substance (9) will give us a value of /x which will answer ; but what we really want to know is whether there is a value of /* which will be the same for all substances. 2. Here is the proof that there is such a value. I need not here give to students the usual and well- known proof that all rever- sible heat engines working between the temperatures t and t-5t have the same efficiency. Now let ABCD be a figure showing with in- finite magnification an ele- mentary Caru ot cycle. Stuff at A at the temperature t-8t\ A I shows the volume and AK the pressure. Let ^D be the isothermal for t - dt and BC the isothermal for t, AB and CD being adiabatics. Notice carefully that the distance AG or WB {JV is in DA produced to meet the ordinate at B) is {dpjdt) dt. Now the area of the parallelogram ABCD which represents the work done, is BW X XZ (if parallelograms on the same base and between the same parallels be drawn, this will become clear). Call XZ by the symbol 5v (the increase of volume in going along the isothermal from B to C), and we see that p \ R \ \ b % A^ L . w V V^ 1 Q A \ M e N ^^^8 A^ . •-^ t> O > : \ ( 2 Fig. 55 P. 10 146 CALCULUS FOR ENGINEERS. the nett work done in the Carnot cycle is (dpldt)5t. 5v. Now the Heat t and V return to their old values, (or another way of putting it is that dE for a complete cycle is 0). / We have seen that the differential coefficient of k with regard to v, t being constant, is equal to the differential coefficient of l—p with regard to ty v being constant, or (a-(a-(s « This statement, which is true for any kind of stuff, is itself sometimes called the first law of Thermodynamics. The Second Law of Thermodynamics is this ; -— or t taken in at the higher temperature is, from (3), Art. 78, equal to l.dv and hence — =j— - — = efficiency = -- ( - . j 8t... (10), and this is the same for all substances. As it is the same for all substances, let us try to find its value for any one substance. A famous experiment of Joule (two vessels, one with gas at high pressure, the other at low pressure witli stopcock between, immersed in a bath all at same temperature ; after equalization of pressure in the vessels, the temperature of the bath keeps its old value) showed that in gases, the intrinsic energy is very nearly constant at constant temperature, or what is the same thing, that / in gases is very nearly equal to p, and it is also well known that in gases at constant volume, p is & linear function of the temperature. Whether there really is an actual substance possible for which this is absolutely true, is a question which must now be left to the higher mathematicians, but we assume that there is such a substance and in it I \dty " p \dtj " + 274 ^ '* if is the Centigrade reading on the Air Thermometer. If then we take t = d + 2l4: as our scale of temperature and (11) as the universal value of - ( '--; I , then, from (9), -=--.-r-,or — = — ^or log t + log^= a constant, I \dt J ^ t fi at t /Jt. o or- or Ai=r-, where c is any constant. This being an integrating factor for (5), 1 we usually take unity as the value of c or ^=- as Camot's function. It is not probable that, even if there is one which is independent of j? or V, there really is so simple a multiplier as ^— o;^ (where is the Centigrade temperature on the air thermometer) or that there is such a substance as we have postulated above. Calling our divisor t the absolute temperature, we believe that for ordinary values of 0, t is ^ + 274, and the greater is, the more correctly is t represented by ^ + 274; but when is very small, in all probability the absolute temperature is a much more complicated function of 6. The great discoverers of the laws of Thermodynamics never spoke of - 274° C. as the absolute zero of temperature. PERFECT GAS. 147 d<f> = -. dt-\-- .dv ...{21), is a complete diftereutial, and hence the differential coefficient of - with regard to v, t being considered t constant, is equal to the differential coefficient of - with regard to t, V being constant t. 1 /^'\ ^\dt)~^ Hence or (ii-Q.-i » This statement, which is true for any kind of stuff, is itself some- times called the second law of Thermodynamics. Combining (20) and (22), we have for any stufl' (t)A (->- a most important law, T Applying these to the case of a perfect gas we find that (23) becomes - = -, or 1= — , or l=^p (24). Hence (20) is (-^ ) =0. It is not of much importance perhaps, practically, but a student ought to study this last statement as an exercise, k is, for any substance, a function of v and t, and here we are told that for a perfect gas, however k may behave as to temperature, it does not change with change of volume. Combining (24) with (9)* &c. (p. 141), already found, we have K-k—R, and as Regnault found that K is constant for air and other gases, k is also constant, so tliat y-i- y. We can now make exact calculations on the Thermodynamics of a perfect gas if we know K and R. l=p^ L= —V, P= , ]'= ^^-Aj-» where y=x- 85. The statements of (3) Art. 78 become for a pound of perfect gas dH = k.dt + p.dv >| = K . dt - V . dp V— 1 -Y— 1 dv. .(1). + The rule for finding the differential coefficient of a quotient is given in Art. 197. 10—2 148 CALCULUS FOR ENGINEERS. I often write this la«t in the shape d{pv)-{-p . dv (2), also d£J=k .dty or ^=X'^ + constant (3). It is easy to obtain from this other forms of E in terms of p and v. To the end of this article, I consider the stuff to be a perfect gas. Example 1. d(h = k . - +^ . dv. or as - = — . t t ' t V d<b = i- — + — . do. ^ t V Hence, integrating, <f)=k log t-\-R log V + constant, or (f> = log ^v" + constant (4). Again d<^ = --.dt-. dpy but - = — . t t P Hence d<f)= -dt dp. Integrating <f) =>: K log t-Ii log p + constant, or </> = log t^p " " + constant ... (5). Substituting for t its value -., we have (5) becoming (/) = log jo^y*'+ constant (6). The adiabatic law^ or constant, may be written down at once. Keducing from the above forms we find or t^-yp = constant, ■ or pv^ = constant. Students ni^,y manufacture other interesting exercises of this kind for themselves. Example 2. A pound of gas in the state p^^ v^^^ t^ receives the amount of heat ^^j^, what change of state occurs? We get our informa- tion from (1). I. Let the volume % keep constant. Then dH=k.dt from (I). The integral of this between ;„ and t^ is II^^^ = Jc{t^ — t^^ and we may calculate the rise of temperature to t^ . Or again, dH= - - dp. GAS ENGINE. 149 Hence, the integral, or H^^ = -^ {p^ - p^), and we may calculate the rise of pressure. II. Let Pq the pressure, keep constant. dH= K . dt, hence ^^i = ^^ (^i - ^o)- Again dH= -^ dv, hence H^^ = -^"^^ (v^ - v^). III. At constant temperature. dH=p.dv or Hq]^= \ p.dv=W, the work done by the gas in ex- J Vo panding. IV. Under any conditions of changing pressure and volume. Jf^^ = h (t^ — tf^ -\- work done. Also from (2), Hqi = r(i^i^i— iOo^*o) + work done. If ir=0, the work done=X*(^Q — ^i) We often write the last equation of (1) in the convenient shape ^^=7^f^^-W If in this we have no reception of heat, dff ^ ,. dp or _ = 0, then v^ + y.^ = 0, or -" + y — = or, integrating, log p-\-y log v = constant, or j02;'>'=: constant. This is the adiabatic law again. Example 3. In a well known gas or oil engine cycle of p .(7). Fig. 56. 150 CALCULUS FOR ENGINEEIIS. operations, a pouiul of gas at p.j,, v-,, t.,, iudicatcd by the point A is compressed adiabatically to B, wnere we have p^, v,, ^j. The work done upon the gas is evidently (from iv.) Jc{t^ — t^, being indeed the gain of intrinsic energy. Heat given at constant volume from B to C where we have Work done in adiabatic expansion CJ)=k(t^-t^). Nett work done = work in Ci>-work in AB= -^ = efficiency eJ'-^l' -^^^=1- ^^ (8). But we saw that along an adiabatic W~'^ is constant, and hence From this it follows that -t = t — \-] > ^"d ^^^^"^ of these t —t h h \^h) = * — ^ . Using this value in (8) we have efficiency = 1 - f^j (9), a formula which is useful in showing the gain of efficiency produced by diminishing the clearance v^. Students will find other good exercises in other cycles of gas engines. Change of State. 86. Instead of using equations (3) Art. V8, let us get out equations specially suited to change of state. Let us consider one pound of substance, m being vapour, 1 - m being liquid (or, if the change is from solid to liquid, m liquid, l-m solid), and let ^2 = cubic feet of one pound of vapour, *i = » )) of one pound of liquid, jt>= pressure, t temperature, p is a function of t only. Tf V is the volume of stuff in the mixed condition, v=ms2 + {l —m)sj = (*2 — «i) wi + Si, or v = mu-{-Xi (1), if we write u for ,% — Si. AVhen heat dJJ is given to the mixture, consider that t and m alter. In fact, take t and m as independent variables, noting that t and VI define the state. If 0-2 and o-^ be specific heats of vapour and li(iuitl, when in the saturated condition (for example, 0-2 is the heat given to one pound of vapour to raise it one degree, its pressure rising at LATENT HEAT. 151 the same time according to the proper law), then the 7>i lb. of vapour needs the heat ma^.dt^ and the 1— m of liquid needs the heat {\ — m)<j^.dt and also if dm of liquid becomes vapour, the heat L . dm is needed, if L is latent heat. Hence dII={(a-2-o-j)m + ai]dt+L.dm (2). If E is the Intrinsic Energy, the first law of Thermodynamics gives dE=^dlI-p,dv (3). Now if m and t define the state, v must be a function of m and t^ or Using this in (3) and (2) we find dE = |(cr2 - o-i) m + o-i -p {^j^ dt-\-\L-p {~y^ dm ... (4). Stating that this is a complete differential, or ^{(..,-,.,)m + .,-^(|)}=||i;-^(^)}. we have, noting from (1), that ( -i— ) = ?/, dL , dp fdv\ dp ,_. -^ + 0-1 — (To = -& . -1— I , or w -4- (5). dt^ ^ '" dt \dm)' dt ^ ' Now divide (2) by t and state that d(^ = — is a complete dif- ferential, * (6), dm \ ~^t ]~dt \i) dm dL L or -n-^f^\-^^2 = -7 ('^)- Hence, with (5) we have T = U -s^ (8), 87. To arrive at the fundamental Equation (8) more rapidly. In fig. 57 we have an elementary Carnot cycle for one pound * w7 ( T ) ~ g as will be seen later on when we have the rule for differentiating a quotient. But indeerl we may as well confess that to understand this article on change of state, students must be able to perform ' differentiation on a product or a quotient, 152 CALCULUS FOR ENGINEERS. of stuff. The co-ordinates of the point B are FB=s^ the volume, and BG the pressure jt? of 1 pound of liquid. At constant tem^^ratrire t, and Fig. 57. also constant pressure, the stuff expands until it is all vapour at FC^s^ ; CD is adiabatic expansion to the temperature t-bt at J). J) A is iso- thermal compression at t-bt and AB is the final adiabatic operation. The vertical height of the parallelogram is bt ~ , and its area, repre- dv senting the nett work, is bt . -£ (s.^- s^). The heat taken in, in the : dp operation BO is X, and the efficiency is bt -^ («2 ""*i) "^'^- ^^^^ ^^ i^ i'^ a Carnot cycle this is equal to — and so we obtain (8). t 88. The Entropy. From (6) we find o-^,-<ri = ^ we can write (2) as dt\t)' and or dH=tr^dt-^t.d(^^y\ Hence, the entropy d(l> = - — = ^^ dt + di-^\^ t t \ / jut .(9). <(- = - + c?^-f constant .(10). In the case of water, o-j is nearly constant, being Joule's equivalent. (We have already stated that all our heat is in work units), and = — +0-1 log f-j-f constant (11). Hence the adiabatic law for water-steam is TnT t -T- + o"! log £- = constant (12). It is an excellent exercise for students to take a numerical example. . EXAMPLES. TWO VARIABLES. 153 Let steam at 165°C. (or ^ = 439) expand adiabatically to 85°C. (ori = 359). Take a-i = 14(X) and L in work units, or take (t^ = \ and take L in heat imits. In any case, use a table of values of t and L. 1. At the higher ^2=4^9 1^^ '^2='"- (This is chosen at random.) Calculate ii\ at, say ^^ = 394, and also m^ at ^^=359. Perhaps we had better take L in heat units as the formula Z = 796 --695^ is easily remembered. Then (12) becomes r,r, (^ - -695) +log^^= m^ (^ - '695) +log^^ , logr+wi2(^-'695 ^ /2/ 695 h If we want m^ we vise t^ instead of t^. Having done this, find the coiTesponding values of v. Now try if there is any law like pi;* = constant, which may be approximately true as the adiabatic of this stuff. Rej^eat this, starting with m^=-% say, instead of '7. The t, <t> diagram method is better for bringing these matters most clearly before students, but one or two examples like the above ought to be worked. 89. When a complete differential dn, is zero, to solve the equation du = 0. We see that in the case, {x" - ^xy - t\f) dx + {if - ^xy - 'Ice") dy = 0, we have a complete differential, because -J- (xf^ — 4fxy — 2y^) = — 4;x — ^y, ^ if - 4^y - 2^0 = - 4y - 4a-, so that they are equal. Hence it is of the form Integrating a?^ — ^xy — 2?/^, since it is ( y j , Avith i-egard to 154 CALCULUS FOR ENGINEERS. X assuming y constant, and adding, instead of a constant, an arbitrary function of y, we get ^ as u = :^a^-2afy-2ifx + <i>{y). To find </) {y\ we know that \~j~]—y^~ ^^V — ^^' Hence -2a^- ^yx + ^ <f> {y) = y"" - ^.x-y - 2a^, ay Hence ^ </, (y) = y^ or </> {y) = \f. Hence xi — \a? — 2x^y — ^xy"^ -\- ^y^ =zc. We have therefore solved the given differential equation when we put this expression equal to an arbitrary constant. Solve in the same manner, 1 + ^ j (/^ - 2 ^ c?y = 0. Answer x^-y^= ex. ., , 2x.dx n Saf\ , holve f- ( -; —jdy=0. Answer x^ — y^ — cy^. J ^J J ' -Solve (3^- + 3y - ^ dx + {^x - 1 + 3/) dy = 0. Answer x^y"^ + ot^y^ + 4^^ ^ ^y"^ ^ ^^yz _ ^y%^ 90. In the general proof of (17) given in Art. 81, we assumed that X and y were perfectly independent. We may now if we please make them depend either upon one another or any third variable z. Thus if when any inde^Hindent quantity z becomes z + bz, x becomes x + bx and y becomes y+by, of course u becomes u + bu. Let (IG) Art. 81 be divided all across by bzy and let bz be diminished without limit, then (17) becomes du _ /o?w\ dx /du\ dy di ~ \dx) dz "^ XTy) ^ .(1). Thus let u = ax^-\-hy^+cxy^ and let x = ez^, y = yz'^. and consequently -T- = {2ax + cy) nez:^ ~^-\- {2by + ex) mgz^ ~ ^ EXAMPLES. 155 In this we may, if we please, substitute for x and y in terms of z, and so get our answer all in terms of z. This sort of example is rather interesting because it can be worked out in oiu* earlier way. In the expression for w, substitute for x and y in terms of z, and we find ii^aeh'^^+hgh^'^-'rcegz:^'^^^ and -^ = 2?iae%2n - 1 + 2 w6^%2m - 1 -I- (,i + m) cegz"" + "*-!. It will be found that this is exactly the same as what was obtained by the newer method. The student can easily manufacture examples of this kind for himself. For instance, let y=uv where u and v are functions of ^, then (1) tells us that dy dv . du a5 = "S+^d5E' a formula wliich is usually worked out in a very different fashion. See Art. 196. In (1) if y is really a constant, the formula becomes du _ du dx dz " dx ' Hz ' which again is a formula which is usually worked out in a very different fashion. See Art. 198. In (1) assume that z=x and that y is a function of x, then dx \dx) "^ \dy) dx ^'^^^ The student need not now be told that ;,- is a very different thing from (du\ \dx) • Example. Let u = ax^ + hy'^ + cxy^ and let y=gx'^. Hence (2) is, j- = {2ax + cy) + (26y + ex) mga/^-K More directly, substituting for y in u, we have u = ax^ + 6^2^2m ^ ^^^» + 1^ J- = 2ax + 2?»6^2^'2'« - 1 + (771 + 1 ) cgx^, and this will be found to be the same as the other answer. 156 CALCULUS FOR ENGINEERS. If M is a function of three independent variables it is easy to prove, as in Art. 81, that ,MJ'-\a.+m,,U^-)^, (3). 91. Example. When a mass m is vibrating with one degree of freedom under the control of a spring of stiffness a, so that if x is the displacement of the mass from its position of equiHbrium, then ax is the force with which the spring acts upon the mass ; we know that the potential energy is ^ ax^ (see Art. 26), and if v is the velocity of the mass at the same time t, the kinetic energy is ^tnv^, and we neglect the mass of the spring, then the total store of energy is When X is 0, r is at its greatest; when v is 0, x is at its greatest. 1. Suppose this store E to be constant and differentiate with regard to ^, then = mv^^ + ax--^^ (1)' . dx ... d^x « dv , or as V IS ~j- , writnig -y-^ for -j- we have ^1-^^-0 •••■ ^^)- which is (see Art. 119) tlie well known law of simple harmonic motion. 2. If the total store of energy is not constant but diminishes at a rate which is proportional to the square of the velocity, as in the case rf If of Fluid or Electromagnetic friction, that is, if -r = - Ev^ then (1) becomes - Fv^^mv y-4- ax—j , or (2) becomes d2x F dx a ^^ Compare (1) of Art. 142. 92. Similarly in a circuit with self induction L and resistance R, joining the coatings of a condenser of capacity K, if the current is C\ and if the quantity of electricity in the condenser at time t is K V so that dV C= -K --J-J ^LC'^ is called the kinetic energy of the system, and \KV^ is the potential energy, and the loss of energy by the system per second is RC'^. So that if E is the store of energy at any instant • E=^LC'' + \KV\ EXAMPLES. 157 or LC"^- V. C+EO^=0, at or L^-V+RC=0, at d2V . R dV^ 1 Tr /% aF + L^+EK'^=^ ^^^- Differentiating this all across and replacing K -r- with C we have a similar equation in C. Compare (4) of § 145. 93. A mass m moving with velocity v has kinetic energy \mv^. If this is its total store E^ If E diminishes at a rate proportional to the square of its velocity as in fluid friction at slow speeds, dE jy ., dv dv F ,^, or -j-= V (5). dt m ^ ' We have a similar equation for the dying out of current in an electric conductor, \LC'*- being its kinetic energy, and RC^- being the rate of loss of energy per second. 94. In (2) of Art. 90, assume that u is a constant and we find for example that if u=f{x, y) = G (dn^\_y)\ . ( df{x, y) \ di_ \ dx )^\ dy ) dx~ ' so that if /(^-j y)=o' or =0, we easily obtain g^. 1. Thus if x^^f = c, 2..+2y . ^^=0, or ^= -^'. 2. Alsoif^^+f-!-l=0,?|' + |^/ = 0,orf'=-^;i;, a^ W- ' a^ h^ dx ' dx a^ ij 3. Again if u = A a;"' + %", Hence i{ u = 0, or a constant, we have dy _ mAx'^~^ dx~ nBy'^~'^ 158 CALCULUS FOR ENGINEERS. 4. If u==Zi + 1:.. a,'" y- 2a; , 2y , 5. If ^-3 +3,3 _ 3«.^^ = i^ find ^ . Answer : ^, = ''''' — ^ . 6. If .rlogy— ylogA'=0, t/y^y /^log. t/y_y ^^logy-y\ .v — xj 95. Example. Find the equations to the tangent and normal to the ellipse ' ., + t^ = l, at the point A'l, yi on the curve. •^ 4. 2^ ^^-0 or at the tx)int ^^- -?^' ^i Hence the equation to the tangent is y-^yi^_62.ri a;—a;i ^ Vi ''' -^+-52-;^ + ^' and as x^ and y^ are the co-ordinates of a point in the curve, this is 1. Hence the tangent is — / + '^ = 1. a^ b^ The slope of the normal is v^ — , and hence the equation to the normal is ti^i = " ^J ^ x-Xi b^ x^ APPENDIX TO CHAP. I. Page 19. In an engineering investigation if one arrives at mathe- matical expressions which cannot really be thought about because they are too complicated, one can often get a simple empirical formula to replace them with small error within the limits between which they have to be used. Sometimes even such a simple expression as a-|-6.r, or x^ will replace a complicated portion of an expression with small error. Expertness in such substitution is easily attained, especially in calculations where some of the terms can be expressed numerically or when one makes numerical experiments. MAXIMA AND MINIMA. 159 Exercise 1. The following observed numbers are known to follow a law like y = a-\-bx, but there are errors of observation. Find by the use of squared paper the most probable values of a and h. X 2 3 4i 6 7 9 12 13 y 5-6 6-85 9-27 11-65 12-75 16-32 20-25 22-33 Ans. y = 2*5 + 1 'hx. Exercise 2. The following numbers are thought to follow a law like y=zax\{\-^sx). Find by plotting the values of y\x and y on squared paper that these follow a law ylx + sy=a and so find the most probable values of a and s. X •5 1 2 0-3 1-4 2-5 y •78 •97 r22 -55 1-1 1-24 Ans. y = 3^/(1 + 2.r). Exercise 3. If p is the pressiu-e in poimds per square inch and if v the volimie in cubic feet of 1 lb. of saturated steam, p 6-86 14-70 28-83 60-40 101-9 163-3 250-3 V 53-92 2636 14-00 6992 4-28 2-748 1-853 Plotting the common logarithms of p and v on squared paper test the truth oi pv^'^^^^Ti^. Exercise 4. The following are results of experiments each lasting for four hours ; / the indicated horse-power of an engine, transmitting B horse-power to Dynamo Machines which gave out E horse-power (Electrically), the weight of steam used per hour being TFlb., the weight of coal us^ per hour being G lb. (the regulation of the engine was by changing the pressure of the steam). Show that, approximately, fr=800+21/, i5=-95/-18, ^=-935-10, C=4-2/-62. 1 B ^ ^y C 190 163 143 4800 730 142 115 96 3770 544 108 86 69 3080 387 65 43 29 2155 218 19 1220 — Page 34. It has been suggested to me by many persons that I ought to have given a proof without assuming the Binomial Theorem, and then the Binomial becomes only an example of Taylor's. In spite 160 CALCULUS FOR ENGINEERS. of the eminence and experience of my critics, I believe that my method is the better — to tell a student that although I know he has not proved the Binomial, yet it is well to assume that he knows the theorem to be correct. The following seems to me the simplest proof which does not assume the Binomial. Let y—^^i x+bx=x\^ y + ^^=yi« (1) Supjjose n a positive integer ; then In the limit, when bx is made smaller and smaller, until ultimately Xi=x\ the left-hand side is -^ and the right-hand side is x^~'^+x'^~^ -f... to n terms : so that j-=7iar»~i. dx (2) Suppose n a positive fraction, and put n= - where I and m are positive integers. We have — — ^- = -^ = ~i,i — ;^ where a;'" =2, Xi — X X% — X Zt — * Xi=Zj^^j and so on. ^-linut of (h-^)ih'-'+^'h'~'+ +^'-') dx~ {zi-z){z{^-i + z.Zi''^-'^ + -|-2'»-i) (3) Suppose n any negative number = —m say, where vi is positive, then noticing that Xi~^-x~'^= *' we have ■x-'" 1 x\ Xy—X /It wi ^^"^ Now the limit of ^ =mx'^-'^ by cases (1) and (2) whether 111 Xi — x •' be integral or fractional. .'. -r-=— -^ ' mx'^ - 1 = — 771^- '»- 1 = MA*"" 1. Thus we have shown that -j- (^*) = 7w?»»-i, where ?i is any constant, positive or negative, integral or fractional. COMPOUND INTEREST LAW. 161 CHAPTER II. e* and sinx. 97. The Compound Interest Law. The solutions of an enormous number of engineering problems depend only upon our being able to differentiate x'\ I have given a few examples. Surely it is better to remember that the differential coefficient of x^^ is nx^~^, than to write hundreds of pages evading the necessity for this little bib of knowledge. We come now to a very different kind of function, e*, where it is a constant quantity e (e is the base of the Napierian system of logarithms and is 2*7 183) which is raised to a variable power. We calculate logarithms and exponential functions from series, and it is proved in Algebra that The continuous product 1 . 2 . 3 . 4 or 24 is denoted by .4 or sometimes by 4 ! Now if we differentiate e* term by term, we evidently obtain SO that the differential coefficient of ^ is itself e*. Simi- larly we can prove that the differential coefficient of e^^ is a^". This is the only function known to us whose rate of increase is proportional to itself; but there are a great many phenomena in nature which have this property. Lord Kelvin's way of putting it is that "they follow the compound interest law." P. 11 162 CALCULUS FOR ENGINEERS. dy Notice that if -i = ay (1), that is, the rate of increase of y is proportional to y itself, then y = be" (2), where h is any constant whatsoever ; h evidently represents the value of y when a; = 0. Here again, it will be well for a student to illustrate his proved rule by means of graphical and numerical illustrations. Draw the curve y = d^, and show that its slope is equal to its ordinate. Or take values of a?, say 2, 2-001, 2002, 2003, &c., and calculate the corresponding values of y using a table of Logarithms. (This is not a bad exercise in itself, for practical men are not always quick enough in their use of logarithms.) Now divide the increments of y by the corresponding incre- ments of X. An ingenious student will find other and probably more complex ways of getting familiar with the idea. However complex his method may be it will be valuable to him, so long as it is his own discovery, but let him beware of irritating other men by trying to teach them through his complex discoveries. 98. It will perhaps lighten our study if we work out a few examples of tbe Compound Interest Law. Our readers are either Electrical or Mechanical En- gineers. If Electrical they must also be Mechanical. The Mechanical Engineers who know nothing about electricity may skip the electrical problems, but they are advised to study them ; at the same time it is well to remember that one problem thoroughly studied is more instructive than thirty carelessly studied. Example 1. An electric condenser of constant capacity K, fig. 58, discharging through great resistance R. If v is the potential difference (at a particular instant) between the condenser coatings, mark one coating as v and the other as on your sketch, fig. 58. Draw an arrow-head representing the current G in the conductor ; then C =v-t-R. But q the quantity of electricity in the condenser is Kv LEAKAGE RESTSTANCE. 163 and the rate oi diminution oiq per second or — ^ or —K-r-. is the very same current. Hence ,^ -^ .dv di V ~R' dv di'- = — kr'- EX — _ _ ^ ^'' ~di~~KR'"' Fig. 58. That is, the rate of diminution of v per second, is propor- tional to V, and whether it is a diminution or an increase we call this the compound interest law. We guess therefore that we are dealing with the exponential function, and after a little experience we see that any such example as this is a case of (1), and hence by (2) v=^he~^' (8). It is because of this that we have the rule for finding the leakage resistance of a cable or condenser. For (log 6 -log 2;) = ^^. So that if Vi is the potential at time ti and if v^ is the potential at time fg KR{logb-\ogv,) = t^, KR (log h — log V2) = ^2. Subtracting, KR (log v^ — log Vo) — U — ti So that R = {U - t,)/K log ^ . It is hardly necessary to say that the Napierian logarithm of a number n, \ogn, is equal to the common logarithm logio?! multiplied by 2-3026. Such an example as this, studied carefully step by step by an engineer, is worth as much as the careless study of twenty such problems. Example 2. Newton's law of cooling. Imagine a body all at the temperature v (above the temperature of sur- 11-2 164 CALCULUS FOR ENGINEEKS. rounding bodies) to lose heat at a rate which is proportional to v. Thus let -T. = — «^j at where t is time. Then by (2) V = 6e-«« (4), or log h — log V = at. Thus let the temperature be Vi at the time t^ and v^ at the time t^, then log Vi — log I'g = a (^2 — ^i), so that a can be measured experimentally as being equal to Example 3. A rod (like a tapering winding rope or like a pump rod of iron, but it may be like a tie rod made of stone to carry the weight of a lamp in a church) tapers gradually because of its own weight, so that it may have everywhere in it exactly the same tensile stress / lbs. per square inch. If 3/ is the cross section at the distance X from its lower end, and if y + 5y is its cross section at the distance oc-{- Sx from its lower end, then /. By is evidently equal to the weight of the little portion between OS and w + 8x. This portion is of volume BiV x y, and if w is the weight per unit volume f.Bi/ = tv.y. B.€ or rather ;r; = rpl/- — Hence as before, y=ibe^ (5). If when a7 = 0, y = yo, the cross section just sufficient to support a weight W hung on at the bottom (evidently fy^zrzW), then yQ=b because e°=l. It is however unnecessary to say more than that (5) is the law according to which the rod tapers. Example 4. Compound Interest. £100 lent at 3 per cent, per annum becomes £103 at the end of a year. The interest during the second year being charged on the increased capital, the increase is greater the second year, and is greater and greater every year. Here the addition of interest due is made every twelve months ; it might be SLIPPING OF A BELT. 165 made every six or three months, or weekly or daily or every second. Nature's processes are, however, usually more continuous even than this. Let us imagine compound interest to be added on to the piincipal continually, and not by jerks every year, at the rate of r per cent, per annum. Let P be the principal at the end of t years. Then hP for the time ht is -— ^ P .ht or dP r 1^^ -zr- — -—■ P, and hence by (2) we have dt 100 ^ ^ ^ P^he^\ where 6 = Po the principal at the time t = 0. Example 5. Slipping of a Belt on a Pulley. When students make experiments on this slipping phenomenon, they ought to cause the pulley to be fixed so that they may see the slipping when it occurs. The pull on a belt at W is Ti, and this overcomes not only the pull To but also the friction between the belt and the pulley. Consider the tension I' in the belt at P, fig. 59, the angle QOP being 6] also the tension T+BT at >Sf, the angle QO>Sf being (9 + 8(9. Fig. 60 shows part of OPS greatly magnified, SO being very small. In calculating the force pres- sing the small portion of belt P>Si against the pulley rim, as we think of PS as a shorter and shorter length, we see that the resultant pressing force is T. hd*, so that /ju.T.BOis * When two equal forces T make a small angle dd with one another, find their equilibrant or resultant. The three forces are parallel to the sides of an isosceles triangle like fig. 01, where AB=CA represents 2', where Fig. 59. Fig. 61. 1G6 CALCULUS FOR ENGINEERS. the friction, if /x, is the coefficient of friction. It is this that BT is required to overcome. When fi.T .BO is exactly equalled by Fig. 60. BT sliding is about to begin. Then fi . T , B6 = BT or, ^ = fiT, the compound interest law. Hence T=^he^^. Insert now T^T, when ^ = 0, and T = T^ when e^QOW or 6, , and we have T, = h, T, = ToCf'K In calculating the horse-power H given by a belt to a pulley, we must remember that H={1\ - T,) F-=- 33000, if Ti and T^ are in pounds and V is the velocity of the belt in feet per minute. Again, whether a belt will or will not tear depends upon T^ ; from these considerations wo have the well-known rule for belting. Example 6. Atmospheric Pressure. At a place which is h feet above datum level, let the atmospheric pressure be p lbs. per sq. foot ; at h + Bh let the pressure be p-\-Bp {Bp is negative, as will be seen). The pressure at h is really greater than the pressure at h+Bh by the weight of air filling the volume Bh cubic feet. If w is the weight in lbs. of a cubic foot of air, —Bp = w. Bh. But lu = cp, where c is some constant if the temperature is constant. Hence — Bp= cp. Bh...(l), or, rather Jr = — cp. Hence, as before, we have the compound interest law; the rate of fall of pressure as we go up or the rate of increase of pressure as we come down being proportional to the pressure itself. Hence p = ae~'^^y where a is some constant. If ^; = po> when /i = 0, then a = j^oj so that the law is p=Poe-'^ (2). an As for c we easily find it to be -- , w^ being the weight Po of a cubic foot of air at the pressure po. If t is the constant (absolute) temperature, and Wq is now the weight of a cubic w 274 foot of air at 0° C. or 274° absolute, then c is — — . Po t BAG =56 and BC represents the equilibrant. Now it is evident that as 5^ is less and less, BG-^AB is more and more nearly 5^, so that the equilibrant is more and more nearly T . 50. ATMOSPHERIC PRESSURE. 167 If w follows the adiabatic law, so that pw-y is constant or w=cp^^y where 7 = 1*414 for air. Then (1) be- comes — Bp = cp^ly Bh or f~ = cBh or rather — I ^ = ch or 7 1 , 1 ^1 P ^ =oh+G. If ^ =po where A = 0, we can find G, and we have p y=po y— ch (3), 7 as the more usually correct law for pressure diminishing upwards in the atmosphere. Observe that when we have the adiabatic law pyy = 6, a constant, smd pv=Rt; it follows that the absolute temperature 1— - is proportional to p y , So that (3) becomes 1 7-1 li 7 So that the rate of diminution of temperature is constant per foot upwards in such a mass of gas. Compare Art. 74, (4), if v is 0. Example 7. Fly-wheel stopped by a Fluid Frictional Resistance. Let a be its velocity in radians per second, / its moment of inertia. Let the resistance to motion be a torque proportional to the velocity, say Fa, then Fa = — Ix angular acceleration (1), / J + i^a=0 (2), da F * = -/"• Here rate of diminution of angular velocity a, is proportional to a, so that we have the compound interest law or oL — a^e ^ (3), where a^ is the angular velocity at time 0. 168 CALCULUS FOR ENGINEERS. Compare this with the case of a fly-wheel stopped by solid fHction. Let a be the constant solid-frictional torque. (1) becomes a = — Ida/dt, or da/dt + a/I = 0, or a = — at/T -f a constant, or a = ao — a^// (4), where Oo is the angular velocity when t = 0. Returning to the case of fluid frictional resistance, if M is a varying driving torque applied to a fly-wheel, we have ^=^»+4; <«>• Notice the analogy here with the following electric circuit law. Example 8. Electric Conductor left to itself. Ohm's law is for constant currents and is V= RC, where R is the resistance of a circuit, C is the current flowing in it ; V the voltage. We usually have R in ohms, C in amperes, V in Volts. When the current is not constant, the law becomes V = RC + I.g (1). dC where -7- is the rate of increase of amperes per second, and L is called the self-induction of the circuit in Henries. It is evident that L is the voltage retarding the current when the current increases at the rate of one ampere per second. 1. IfF=Oin(l) dC__R^ dt~ L ' which is the compound interest law. -St Consequently C = CoC ^ (2). EXERCISES IN CURVES. 169 so that from (1) V==Ra-\-(Rh- Lgh) e-^K Now let R = Lg or ff—Y and we have V= Ra, so that the voltage may keep constant although the current alters. Putting in the values we have found, and using Vq for the constant voltage so that a= Fq -4- R, we find G^'^+be'L' (3). If we let C=0 when ^ = 0, then 6 = — ^, and hence we may write G= -^{l-e~L^) (4). The curve showing how G increases when a constant voltage is applied to a cii'cuit, ought to be plotted from ^ = for some particular case. Thus plot when Fo = 100, -R = l, L—01. What is the current finally reached ? 99. Easy Exercises in the Differentiation and Integration of e**^. 1. Using the formula of Art. 70, find the radius of curvature of the curve y = eF, where a; = 0. Answer : r = \/8. .( 2. A point ;ri, y^ is in the curve y = hef^, find the equation to the tangent through this point. Answer: y^Zyi — Vl^ X — Xi a Find the equation to the normal through this point. Answer : - — ^ = . X - Xy^ ?/i Find the length of the Subnormal. Answer : y -- or y^-ja. Find the length of the Subtangent. Answer \y-\--^- or a. 170 CALCULUS FOR ENGINEERS 3. Find the radius of curvature of the catenary y = - (e^'c + g-*/''), at any place. Answer : r — y^jc. At the vertex when a? = 0, r = c. 4. If y = Ae''^^ where i stands for V — 1. Show that if i behaves as an algebraic quantity so that i^ = — 1, i' = — i, {* = !, i^ = I, &c. then ^ = - a^y. 5. Find a so that y = Ael^ may be true when Show that there are two values of a and that y = Ae-"^ + 5e-3«. 6. Find the subtangent and subnormal to the /J Catenary y = ^ {^"^ + e"*'"), or, as it is sometimes written, y = c cosh xjc. Answer : X the subtangent is c coth - or c (e^^" + e~^l*')l{d^'*' — e~^l*^), c C 2ul7 c the subnormal is ^ sinh — or 2 (e^^^ — g-^^/c). 7. The distance PS, fig. 8, being called the length of the tangent, the length of the tangent of the above catenary is 2z cosh^ - / sinh - . 2 cj c The length of PQ may be called the length of the normal, rn and for the catenary it is c cosh^ - or y^/c. c 8. Find the length of an arc of the catenary c - — y— (e" -{-e « ). The rule is given in Art. 38. Fig. 62 shows the shape of the curve, being the origin, the distance A being c. The point P has for its co-ordinates x and y. CURVES. 171 Now -f-= ^(e'' —e ^). Squaring this and adding to 1 and extracting the square root gives us ^(e^ + e ^). The integral of this is ^ (e ^ — e ^ ) which is the length of the arc AP, as it is when x = 0. We may write it, s = c sinh w/c. \. R pA A O s Fig. 62. 9. Find the area of the catenary between OA and SP, fig. 62. Area =1 -^ (e^ ■{■ e *" ) dw^ OS OS c2 ros 1 _«~| Or the area up to any ordinate at x is & sinh x\c. or ^ (e '^ -e " ). The Catenary i/ = -{e^"^ + e~^"') revoh^es about the axis of x^ find the area of the hour-glass-shaped surface generated. See Art. 48. Area = - \ {e^ "^ -\- e~="''f . dx % ire- =-^(^ IC ^ ff-'lx IC' )+'ncx between the ordinates at x = x and .^'=0. It is curious that the forms of some volcanoes are as if their own sections obeyed the compound interest law like an inverted pump rod. The radii of the top and base of such a 172 CALCULUS FOR ENGINEERS. volcano being a and h respectively and the vertical height h, find the volume. See Art. 46. Taking the axis of the volcano as the axis of x, the curve ij = be^^ revolving round this axis will produce the outline of the mountain if c = -r log ^ . The volume is ttJ 6- . e"'"" . da)=^ ^- e-'^^ = 2^(e--l). IT Now a — be^^, so that our answer is ^ {a^ — 6-). 100. Harmonic Functions. Students ought to liavc already plotted sine curves like y = asm{bx + e)...(l)on squared paper and to have figured out for themselves the signification of a, b and e. It ought to be unnecessary here to speak of them. Draw the curve again. Why is it sometimes called a cosine curve ? [Suppose e to be ^ or 90°.] Note that however great a; may be, the sine of {bx-\-e) can never exceed 1 and never be less than — 1. The student knows of course that sin = 0, sin j (or 45°) = '707, sin ^ (or 90°) = 1, sin -J (or 135°) = '707, sin tt (or 180°) = 0, sin ^ (or 225°) = - '707, sin '-^ (or 270°) = - 1, sin ^~ (or 315°) = --707, sin 27r (or 360') = again and, thereafter, sin ^ = sin (^ — 27r). Even these numbers ought almost to be enough to let the wavy nature of the curve be seen. Now as a sine can never exceed 1, the greatest and least values of y are a and — a. Hence a is called the amplitude of the curve or of the function. When A" = 0, y — a sin e. This gives us the signification of e. Another way of putting this is to say that when ba; was = — e or A" = — J , y was 0. When x indicates time or when bx is the angle passed through by a crank or an eccentric, e gets several names ; Valve-motion engineers call SINE FUNCTIONS. 173 it the advance of the valve ; Electrical engineers call it the lead or (if it is negative) the lag. Observe that when bx — 27r we have everything exactly the same as when x was 0, so that we are in the habit of calling -J the periodic value of (i\ Besides the method given in Art. 9, I advise the student to draw the curve by the following method. A little know- ledge of elementary trigonometry will show that it must give exactly the same result. It is just what is done in drawing the elevation of a spiral line (as of a screw thread) in the drawing office. Draw a straight line OM. Describe Fig. 63. a circle about with a as radius. Set off the angle BOG equal to e. Divide the circumference of the circle into any mimber of equal parts numbering the points of division 0, 1, % 3, &c. We may call the points 16, 17, 18, &c., or 32, 33, 34, &c., when we have gone once, twice or more times round. Set off any equal distances from B towards M on the straight line, and number the points 0, 1, 2, &c. Now project vertically and horizontally and so get points on the curve. The distance BM represents to some scale or other the periodic value of x or 27r/6. If OG is imagined to be a crank rotating uniformly against the hands of a watch in the vertical plane of the paper, y in (1) means the distance of G above OM, hx means the angle that OG makes at any time with the position OM, and if x means time, then h is the angular velocity of the crank and 27r/6 means the time of one revolution of the crank 174 CALCULUS FOR ENGINEERS. or the periodic time of the motion, y is the displacement at any instant, from its mid position, of a slider worked vertically from G by an infinitely long connecting rod. A simple harmonic motion may be defined as one which is represented by s = a sin (bt + e), where s is the distance from a mid position, a is the amplitude, e the lead or lag or advance, and 6 is 27r/T or ^irf where T is the periodic time or /is the frequency. Or it may be defined as the motion of a point rotating uniformly in a circle, projected upon a diameter of the circle (much as we see the orbits of Jupiter's satellites, edge on to us), or the motion of a slider worked from a uniformly rotating crank-pin by means of an infinitely long connecting rod. And it will be seen later, that it is the sort of motion which a body gets when the force acting upon it is proportional to the distance of the body from a position of equilibrium, as in the up and down motion of a mass hanging at the end of a spring, or the bob of a pendulum when its swings are small. It is the simplest kind of vibrational motion of bodies. Many pairs of quantities are connected by such a sine law, as well as space and time, and we discuss simple harmonic motion less, I think, for its own sake, than because it is analogous to so many other phenomena. Now let it be well remembered although not yet proved that if dy y = a sin (bx + c) then -?i = ab cos (bx + c) and ly . dx = -- cos(bx + c). 101. When c = and t = 1 and a = 1 ; that is when y — ^mx (1), let us find the differential coefficient. As before, let a; be increased to x-\-hx and find y + By, y-^hy — sin {x + hx) (2). Subtract (1) from (2) and we find hy = sin {x + hx) — sin x, or 2 cos (x + \Bx) sin \hx. (See Art. 3.) Hence ^ = cos (a? + JS^) ^^^^ (3). a sin (Kv -{• c). 175 It is easy to see by drawing a small angle a and recollect- ing what sin a and a are, to find the value of sin a -r a as a gets smaller and smaller. Thus in the figure, let POA be the angle. The arc PA divided by OP is a, the angle in radians. And the perpendicular PB divided by OP is the pj g^ sine of the angle. Hence = -pr-r and it is evident that this is more and more a PA nearly 1 as a gets smaller and smaller. In fact we may take the ratios of a, sin a and tan a to one another to be 1, more and more nearly^ as a gets smaller and smaller. If we look upon ^Bx as a in the above expression, we see that in the limit, (3) becomes dy r • * -r^ = cos X if y = sin x.* * The proof of the more general case is of exactly the same kind. Here it is : i/ = actln(bx + c), y + Sy = a sin {b{x + Sx) + c}, Sy = 2a COB {bx+c + ^b, Sx) sin (^ 6 . dx). See Art. 3. Sx ^ ^ ^ ^b.Sx Now make 5a; smaller and smaller and we have — = ab cos (bx 4. c) and hence I a cos (bx + c) . dx = - sin (bx + c). Again, to take another case : — If ysa cos (bx + e), this is the same as y = asin( 6x+e + - J=a 8in(6aT + c), say. Hence -=^ = a6 cos (6a; + c) dx ^ = a&C08 ( 6a: + e + - j ^ = — ab sin (bx + e). Hence / a sin (bx + c) . dx =— - cos (bx + c)« 176 CALCULUS FOR ENGINEERS. And hence I cos A' dx sm a% 102. Now it is not enough to prove a thing like this, it must be known. Therefore the student ought to take a book of Mathematical Tables and illustrate it. It is unfortunate that such books are arranged either for the use of very ignorant or else for very learned persons and so it is not quite easy to convert radians into degrees or vice versa. Do not forget that in sin x or cos x we mean x to be in radians. Make out such a lit tie bit ot tab le as this, w hich is taken L at ranc iom. Angle in degrees X or angle in radians y = sina; 5y dy~$x Average '6y_ dx 40 •6981 •6427876 •0132714 •7583 41 •7156 •6560590 •0130716 •7512 •7547 42 •7330 •6691306 If it is remembered that Bi/ -^ Bx in each case is really the average value of -^^ for one degree or ^01745 of a radian, it will be seen why it is not exactly equal to the cosine of x. Has the student looked for himself, to see if "7547 is really nearly equal to cos 41° ? 103. It is easy to show in exactly the same way that if dy F y = cos X, ji- = - sin x and Isin x . dx = - cos x. The — sign is troublesome to remember. Here is an illustration : Angle in degrees X or angle in radians i/ = C08a; negative 8x Average 5y 5x 20 21 22 •3491 •3665 •3840 •9396926 •9335804 •9271839 •0061122 •0063965 -•3513 -•3666 - -3584 sin d. vn Notice that y diminishes as x increases. Notice that sin 21° or sin (-3665) = '3584. 104. Here is another illustration of the fact that the differential coefficient of sin x is cos x. Let AOP, fig. 65, be 6. Let AOQhe -h SO. Let PQ be a short arc drawn with as centre. Let OP = 0Q= 1. PR is perpendicular to QB. Fig, 65. Then AP ^y = sin e, BQ = sm(0 -^ Be) = i/ + By, QP = B0 and EQ = 8y. Now the length of the arc PQ becomes more and more neai-ly the length of a straight line between P and Q as B6 is made smaller and smaller. Thus Yyp 01^ ]Si is more and more nearly equal to cos PQR or cos 0. imit*' -J2J = cos d if y = sin 0. In the limit d9 Similarly if z = cos = OA, Bz = — BA = — RP and dz RP . >, d9=-QP = """^- Illustrations like these are however of most value when a student invents them for himself. Any way of making the fundamental ideas familiar to oneself is valuable. But it is a great mistake for the author of a book to give too many illustrations. He is apt to give prominence to those illustra- tions which he himself discovered and which were therefore invaluable in his own education. 105. Observe that ii ^=A sin ajc 4- B cos ax; <^% and ^=«V Compare this with the fact that if y = e«*, ;7-^=aV) ;7^=^V ^^ 12 p. 178 CALCULUS FOR ENGINEERS. the higher applications of Mathematics to Engineering this resemblance and difference between the two functions ^ and sin ax become im- portant. Note that if i stands for s] -\ so that i^= -\^ {*=1, &c. Then if y=e^ 1^~ ~^V> ;7:^=^"V j^^^ ^^ ^i*^ *^® s^^^® function. Comi)are Art. 99. 106. Exercise. Men >yho have proved Demoivre's theorem in Trigonometry (the proof is easy ; the proofs of all mathe- matical rules which are of use to the engineer are easy ; difficult proofs are only useful in academic exercise work) say that for all algebraic purposes, co^ax=\{e^^-\-e~^^)^ and sin CW7 = ^ (e^*** — e~^^). If this is so, prove our fundamental propositions. 107. Example. A plane electric circuit of area A sq. cm. closed on itself, can rotate with uniform angular velocity about an axis which is at right angles to the field, in a uni- form magnetic field H. H is supposed given in c.G.s. units; measuring the angle ^ as the angle passed through from the position when there is maximum induction HA through the circuit ; in the position 6, the induction through the circuit is evidently A.H. cos 6. If the angle 6 has been turned through in the time t with the angular velocity q radians per second, then 6 = qt. So that the induction /= AH cos qt. The rate of increase of this per second is —AqH ^in qt, and this is the electromotive force in each turn of wire. If there are n turns, the total voltage is — nAqH^m qt in C.G.S. units; if we want it in commercial units the voltage is - nAqH 10"8 sin qt volts, being a simple harmonic function of the time. Note that the term voltage is now being employed for the line integral of electromotive force even when the volt is not the unit used. Example. The coil of an alternator passes through a field such that the induction through the coil is /= ^0 + A^ sin {6 + fi) + Ar sin {rO + Cr), where 6 is the angle passed through by the coil. If q is the ILLUSTRATIONS. 179 relative angular velocity of the coil and field, 6 = qt. If there are n turns of wire on the coil, then the voltage is w -7- , or nq {Ai cos (qt + ej) + ^,.r cos (rqt + e^)]. So we see that irregularities of ?• times the frequency in the field are relatively multiplied or magnified in the electromotive force. 108. In Bifilar Suspension, if W is the weight of the suspended mass, a and b the distances between the threads below and above, h the vertical height of the threads; if the difference in vertical component of tension is n times the total weight W, and 6 is the angle turned through in azimuth, the momental resistance offered to further turning is {{l-n')Wjsmd (1). Note that to make the arrangement more ' sensitive ' it is only necessary to let more of the weight be carried by one of the threads than the other. The momental resistance offered to turning by a body which is 6 fi-om its position of equilibrium, is often propor- tional to sin 6. Thus if W is the weight of a compound pendulum and OG is the distance from the point of support to its centre of gravity, W . OG . sin 6 is the moment with which the body tends to return to its position of equilibrium. If M is the magnetic moment of a magnet displaced from equilibrium in a field of strength H, then HM sin 6 is the moment with which it tends to return to its position of equilibrium. A body constrained by the torsion of a wire or a strip has a return moment proportional to 0. When angular changes are small we often treat sin as if it were equal to 6. Sometimes a body may have various kinds of constraint at the same time. Thus the needle of a quadrant electrometer has bifilar suspension, and there is also an elec- trical constraint introduced by bad design and construction which may perhaps be like a6 + bO^. If the threads are stiff, their own torsional stiffness introduces a term proportional to 6 which we did not include in (1). Sometimes the constraint is introduced by connecting a little magnetic needle rigidly 12—2 180 CALCULUS FOR ENGINEERS. with the electrometer needle, and this introduces a term proportional to sin 6, In some instruments where the moving body is soft iron the constraint is nearly propor- tional to sin 26, Now if the resisting moment is M and a body is turned through the angle hO, the work done is M. Bd, Hence the work done in turning a body from the position 6^ to the position ^.^, where S.^ is greater than ^i, is / M .dd. . 6, Example. The momental resistance offered by a body to turning is asin^ where 6 is the angle turned through, what work is done in turning the body from 6^ to 6.J Answer, / a . sin 6 .d6 = — a (cos ^2 — cos d^=^a (cos 6^ — cos 6.). J e, Example. The resistance of a body to turning is partly a constant torque a due to friction, partly a term b6-{-c&-, partly a term esinO; what is the work done in turning from ^ = to any angle ? M the torque = a-{-bd + cd" + e sin =f(0) say, V the work done = aO + ibO' + icd'-\-e(l- cos 6) = F(e) say. This is called the potential energy of the body in the position 6. -^-j where / is the body's moment of inertia about its axis. When a body is at 6 its total energy E is \I i~\ + F {$). If the total energy remains constant and a body in the position 6 is moving in the direction in which 6 increases, and no force acts upon it except its constraint, it will con- tinue to move to the position 6^ such that So that when the form oi F {6) is known, 61 can be calculated, if we know the kinetic energy at 6. ILLUSTRATIONS. 181 Thus let M=e sin 0, so that F(0) = e(l- cos 0), then 1/ (jY + e (1 - cos (9) = e (1 - cos e,), from which ^i the extreme swing can be calculated. Exercise. Show that if the righting moment of a ship is proportional to sin 4)6 where is the heeling angle, and if a wind whose momental effect would maintain a steady inclination of llf degrees suddenly sends the ship from rest at ^ = and remains acting, and if we may neglect friction, the ship will heel beyond 33J degrees and will go right over. Discuss the effect of friction. A body is in the extreme position 0^, what will be its kinetic energy when passing through the position of equilibrium ? Answer, F (Oi). Thus let M^hO + ce^-^e sin 6. Calculate a, the angular velocity at = 0, if its extreme swing is 45°. Here 6, = ^ and F(e) = ^6^ -f ^c6' + e(l- cos 6), from which we may calculate a. Problem. Suppose we desire to have the potential energy following the law V=F(e)= a6^ + hO^ + ce'""^ + h sin 26, and we wish to know the necessary law of constraint, we see dV at once that as Jllf = -y^ , M = f al9'' -f 3&^ + mc€'''^ + 2/i cos W. Problem. A body in the position 6^, moving with the angular velocity a, in the direction of increasing 6, has a momental impulse m in the direction of increasing 6 suddenly given to it ; how far will it swing ? The moment of momentum was /a, it is now loi + m and if a' is its new angular velocity ot' = a + y . 182 CAI.CULUS FOR ENGINEERS. The body is then in the position 0^ with the kinetic energy |/ ( a + y ) and the potential energy F{Oq), and the sum of these equated to F{6^ enables 6i to be calculated. The student will easily see that the general equation of angular motion of the constrained body is fid) may include a term involving friction. 109. Every one of the following exercises must be worked carefully by students ; the answers are of great practical use but more particularly to Electrical Engineers. In working them out it is necessary to recollect the trigonometrical relations cos 2(9= 2 cos2^ - 1 = 1-2 sin2^, 2 sin ^ . cos </) = sin (^ + </>) + sin {6 - ^), 2 cos 6 . cos <fi = cos (^ + </)) + cos {$ — (j)), 2 sin ^ . sin <^ = cos (0 — (f)) — cos (6 + <^). Such exercises are not merely valuable in illustrating the calculus; they give an acquaintance with trigonometrical expressions which is of great genei-al importance to the engineer. The average value of /(.r) from ii' = ooi to .T = a?.^ is evidently the area I f(x) . dx divided by .r., — a'i. Every exercise from 6 to 20 and also 23 ought to be illustrated graphically by students. Good hand sketches of the curves whose ordinates are multiplied together and of the resulting curves will give sufficiently accurate illustrations. X sin 2ax sin- (Lxdx o r . , , cos (a-\-h)a; cos (a — h)x 3. sm ax , cos bx . dx = ~ ~ ^ rf- J 2(a + 6) 2(a-6) 2 4a X . sin 2ax 4>a EXERCISES. 183 f . . , , sm(a~b)x sin (a + h) x k sm ax . sin bx . d^• = —^ r-r ^, , . (— ; 2 (a — 6) 2 (« + b) 5. I cos a^ . cos bx.cix — — ^, ~ h sin {a + b) x sin (a — b) x 2(a + 6) "^ ~27a^;^6)~ -I' 6. The area of a sine curve for a whole period is 0. r2.1T r2jr "I I sinfl7.fZic=— cosiT = — (1 — 1) = 0. 7. Find the area of the positive part of a sine curve, that is I sin ^ . c^a; = — cos x = — (— 1 — 1) = 2. Since the length of base of this part of the curve is tt, 2 the average height of it is - . Its greatest height, or ampli- tude, is 1. 8. The area of y — a-\-b sin x from to 27r is 27ra and the average height of the curve is a. 9. Find the average value of sin- x from a; = to ^ = 27r. As cos 2^' = 1 - 2 sin^ x, sin^ ^ = | (1 - cos 2x). The integral of this is |.2J - J sin 2x, and putting in the limits, the area is (^27r-isin47r-04-isinO) = 7r. The average height is the area -j- 27r, and hence it is ^. 10. The average value of cos- x from ^; = to iz? = 27r is \. In the following exercises s and r are supposed to be whole numbers and unequal : 11. The average value of a niri^ {sqt ■\- e) from ^=0 to ^ = T is g , s being a whole number and q = ^irlT. T is the periodic time. 12. The average value of aQ,o^^isqt-\-e) from i = to 13. I cos sqt . sin sqt ,dt=^0. 184 CALCULUS FOR ENGINEERS. [^ 14. I sin sqt . sin rqt ,dt — 0. Jo rT 15. I cos sqt . cos rqt .dt — 0. Jo 16. I sin sqt . cos rqt .dt = 0. Jo 17. The average value of sin^ sqt from to ^T is ^. 18. The average value of cos^ sqt from to ^T is J. 19. I sm 55^ . sin rqt . a^ = 0. rhT 20. I cos 55^ . cos rqt ,dt = 0. 21. Find /sin a; . sin (x + e) . dx. Here, sin {x-\- e) = sin a? cos e -\- cos a; . sin e. Hence we must integrate sin'* x . cos e-^-smx . cos a; . sin e, a? sin 2a; /■ sin'' X .dx = (sin X . cos X .dx = \ (sin 2a? . (^a; = — J cos 2a;, and hence our integral is fx sin 2a;\ , ^ ( ^ J— J cos e — % cos 2a7 . sin e. 22. Prove that (sin qt . sin (^-^ + e) . f/^ /^ sin2gA 1 ^ , . 23. Prove that the average value of sin qt . sin {qt ± e) or of sin (qt + a) sin (5'^ + a ± e) for the whole periodic time T (if q = y) ^s i cos e. USE OF IM AGIN ARIES. 185 This becomes evident when we notice (calling qt-\- a = (p), sin (ft . sin ((f) ±e) = sin (ft (sin (j) cos e ± cos (f) . sin e) = sin^ <l> . cos e ± sin (f> . cos </> . sin e. Now the average value of sin- (/> for a whole period is |^, and the average value of sin <^ . cos <^ is 0. IT By making a = ^ in the above we see that the average value of cos qt . cos {qt ± e) is ^ cos e, or the averages value for a whole period of the product of two sine flinctions of the time, of the same period, each of amplitude 1, is half the cosine of the angular lag of either behind the other. 24. Referring to Art. 106, take cosa(9 = J(e'^ + e"'^), or take e^^ = cos ad -f i sin aO, e"^ = cos ad — i sin aO, and find L^ cos ad . dO. This becomes \j{e^''-''^'-Ve^'-''^')dd (b+ai)0 . 1 ib~ai)e) e + V . e y b — ai * (6 + ai 1 W f 1 aid 1 i^ "^r-; — ^-^ +1 — 6 ~ ai and on substituting the above values it becomes I e cos a^ . Similarly we have e cos a0.dd= - — ^ e^ (b cos a^ + a sin a^) . . .(1). fe^« sin adds = -^^-^ e^« (b sin a(9 - a cos a(9) (2). 186 CALCULTTS FOn ENGINEERS. 110. Notes on Harmonic Functions. In the fol- lowing collection of notes the student will find a certain amount of repetition of statements already made. 111. A function ^ = tt sin qt is analogous to the straight line motion of a slider driven from a crank of length a (rotating with the angular velocity q radians per second) by an infinitely long connecting rod. x is the distance of the slider from the middle of its path at the time t At the zero of time, ^ = and the crank is at right angles to its position of dead point, q = 27rf = -=- , if T is the periodic time, or if/ is the frequency or number of revolutions of the crank per second, taking 1 second as the unit of time. 112. A function x = a sin {qt -}- e) is j ust the same, except that the crank is the angle € radians (one radian is 57 '2957 degrees) in advance of the former position; that is, at time the slider is the distance a sin e past its mid-position*. * The student is here again referred to § 10, and it is assumed that he lias drawn a curve to represent x=6c""'sin {qt + i) (1). Imagine a crank to rotate uniformly with the angular velocity q, and to drive a slider, but imagine the crank to get shorter as time goes on, its length at any time being fl€~ . Another way of thinking of this motion is: — Imagine a point P to move with constant aiujnlar velocity round O, keeping in the equiangular spiral path APBCDEF; the motion in question is the motion of P projected upon the straight line MON and what we have called the logarithmic decrement is tt cot a if a is the angle of the spiral. SIMPLE HARMONIC MOTION. 187 113. A fiuictiou X = a sin (ryi + e) + o! sin {(jt + e') is the same as X = A sin (5'^ + ^); that is, the sum of two crank motions can be given by a single crank of proper length and proper advance. Show on a drawing the positions of the first two when ^ = 0, that is, set off YOF = e and OF = a, YOQ = e' and OQ = a/. Complete the parallelogram OPRQ and draw the diagonal OR, then the single crank OR = A, with angle of advance YOR = E, would give to a slider the sum of the motions which OP and OQ would separately give. The geometric proof of this is very easy. Imagine the slider to have a vertical motion. Draw OQ, OR and OP in their relative positions at any time, then project P, R and Q upon OX. The crank OP would cause the slider to be OP' above its mid-position at this instant, the crank OQ would cause the slider to be OQ' above its mid-po- ^^g- ^'^' sition,the crank OR would cause the slider to be OR' above its mid-position at the same instant; observe that OR' is always equal to the algebraic sum of OP' and OQ. We may put it thus: — "The s.H.M. which the crank OP would give, + the s.H.M. which OQ would give, is equal to the S.H.M. which OR would give." Similarly "the s.H.M. which OR would give, - the s.H.M. which OP would give, is equal to the s.H.M. which OQ would give." We sometimes say: — the crank OR is the sum of the two cranks OP and OQ. Cranks are added therefore and subtracted just like vectors. that is, the constant acute angle which OP everywhere makes with the curve, or Tfcot a = aT/2 and g = 27r/r, so that cot a = alq. If fig. 66 is to agree with fig. 67 in all respects NM being vertical and P is the position at time 0, then €= angle iVOP -7r/2. 188 CALCULUS FOR ENGINEERS. 114. These propositions are of great importance in dealing with valve motions and other mechanisms. They are of so much importance to electrical engineers, that many practical men say, "let the crank OP represent the current." They mean, " there is a current which alters with time according to the law (7= a sin (5^ 4- e), its magnitude is analogous to the displacement of a slider worked vertically by the crank OP whose length is a and whose angular velocity is q and OP is its position when t = 0." t 115. Inasmuch as the function w = a cos qt is just the same as asinf^^ + ^j, it represents the motion due to a crank of length a whose angle of advance is 90°. At any time t the velocity of a slider whose motion is {V = a sin (qt + e), due is v^aq cos (qt 4- f) = -^ or ^r = aq sin (,* + e + |). that is, it can be represented by the actual position at any instant of a slider worked by a crank of length representing aq, this new crank being 90° in advance of the old one. cl?x dv The acceleration or -^^ or -7- or v is shown at any instant by a crank of length aq^ placed 90° in advance of the v crank, or 180° in advance of the x crank, for Accel. = - aq^ sin {qt + e) = aq^ sin {qt-\-e + ir). The characteristic property of S.H. motion is that, numerically, the acceleration is q^ or 47r-/- times the displacement, /being the frequency. If anything follows the law a sin {qt + e), it is analogous to the motion of a slider, and we often say that it is repre- sented by the crank OP ; its rate of increase with time is analogous to the velocity of the slider, and we say that it is represented by a crank of length aq placed 90° in advance of the first. In fact, on a S.H. function, the operator djdt multiplies by q and gives an advance of a right angle. ELECTRIC CIRCUIT. 189 116. Sometimes instead of stating that a function is A sin (qt + e) we state that it is a sin qt + b cos qt. Evidently this is the same statement, if a? + h^ = A^ and if h tan e = -. a It is easy to prove this trigonometrically, and gra- phically in fig. 68. Let The crank OP is the sum of 08 and OQ, and tan e or tan FOP = -. a Fig. 68. 117. We have already in Art. 100 indicated an easy graphical method of drawing the curve x = asm (qt + e), where a; and t are the ordinate and abscissa. Much information is to be gained by drawing the two of the same periodic time, a; = a sin (qt + e) and x — a sin (qt + e), and adding their ordinates together. This will illustrate 113. 118. If the voltage in an Electric Circuit is V volts, the current G amperes, the resistance R ohms, the self-induction L Henries, then if t is time in seconds, V==RC+L dC dt' .(1). Now if = Co sin qt, T- = G^q cos qt, • so that Y= RGq sin qt + LG^q . cos qt, and by Art. 116 this is . V=Go^/W+LY^^^(qi + ^) (2); 190 CALCULUS FOR ENGINEERS. nR^ 4- L^q^ is called the impedance ; tan e = -^* = — — ^ , if/ is frequency ; 6 is lag of current behind voltage. Hence again if V = VoSin qt (3), then C = ;7=^Bin(qt-tan-^)l (4). Notice that if V is given as in (3) the complete answer for G includes an evanescent term due to the starting conditions see Arts. 98, 147, but (4) assumes that the simple harmonic V has been established for a long time. In practical electrical working, a small fraction of a second is long enough to destroy the evanescent term. 119. We may wiite the characteristic property of a simple harmonic motion as S+<»'*=o W- (compare Arts. 26 and 108) and if (1) is given us we know that it means x = a8mqt+hQ0>iqt or x =^ A ^m {qt + e) (2), where A and e, or a and h are any arbitrary constants. Example. A body whose weight is W lb. has a simple harmonic motion of amplitude a feet (that is, the stroke is 2a feet) and has a frequency / per second, what forces give to the mass this motion ? If X feet is the displacement of the body from mid- position at any instant, we may take the motion to be x= a sin qt or a sin 2irf. t, and the numerical value of the acceleration at any instant is ^ir^px and the force drawing the body to its mid -position is in pounds ^ir^f'^xW-^ 322, as mass in engineer's units is weight in pounds in London -r- 32 "2, and force is acceleration x mass. CONNECTING KOD. 191 120. If the connecting rod of a steam or gas engine were long enough, and we take W to be the weight of piston and rod, the above is nearly the force which must be exerted by the cross-head when the atmosphere is admitted to both sides of the piston. Observe that it is when a? is and is proportional to o), being greatest at the ends of the stroke. Make a diagram showing how much this force is at every point of the stroke, and carefully note that it is always act- ing towards the middle point. Now if the student has the indicator diagrams of an engine (both sides of piston), he can first draw a diagram show- ing at every point of the stroke the force of the steam on the piston, and he can combine this with the above diagram to show the actual force on the cross-head. Note that steam pressure is so much per square inch, whereas the other is the total force. If the student carries out this work by himself it is ten times better than having it explained. Since the acceleration is proportional to the square of the frequency, vibrations of engines are much more serious than they used to be, when speeds were slower. 121. As we have been considering the motion of the piston of a steam engine on the assumption that the connecting rod is infinitely long, we shall now study the effect of shortness of connecting rod. In Art. 11, we found 6" the distance of the piston from the end of its stroke when the crank made an angle 6 with its dead point. Now let x be the distance of the piston to the right of the middle of its stroke in fig. 3, so that our x is the old s minus r, where r is the length of the crank. Let the crank go round uniformly at q radians per second. Again, let t be the time since the crank was at right angles to its dead point position, so that 0— - = qt, and we find ^ or x z= - r cos -^ I \l - \/ 1 -^ sin2 oi , x= r iiin qt + I \i — \/ 1 — j~ cos^ qt [• 192 CALCULUS FOR ENGINEERS. Using the approximation that Vl — a = 1 — ^a if a is small enough we have X =^ r sm qt -\- ^. cos- qt But we know that 2 cos- qt-'l= cos 2qL (See Art. 109.) Hence x = r sin qt + ^i ^^os (2qt) - — (1). We see that there is a fundamental simple harmonic motion, and its octave of much smaller amplitude. Find -jr and also -f- . This latter is at dt^ acceleration = — rq- sin qt y cos '2.qt. It will be seen that the relative importance of the octave term is four times as great in the acceleration as it was in the actual motion. We may, if we please, write 6 again for 2^ + 2 and get d\ r-if' — - = r^' cos 6 + ~ cos 2(9. .2^2 r^q When ^ = 0, the acceleration is rif- H — ? . When B — 90^ the acceleration is — - V When = 180°, the acceleration is — rq^ + —j- , (q is 27rf, where / is the frequency or number of revolutions of the crank per second). If three points be plotted showing displacement x and acceleration at these places, it is not difficult by drawing a curve through the three points to get a sufficiently accurate idea of the whole diagram. Perhaps, as to a point near the middle, it might be better to notice that when the angle OPQ is 90°, as P is moving uniformly and the rate of change of the angle Q is zero, there is no acceleration of Q just then. This position of Q is easily found by construction. VALVE GEARS. 193 The most important things to recollect are (1) that accelerations, and therefore the forces necessary to cause motion, are four times as great if the frequency is doubled, and nine times as great if the frequency is trebled ; (2) that the relative importance of an overtone in the motion is greatly exaggerated in the acceleration. 122. Take any particular form of link motion or radial valve gear and show that the motion of the valve is always very nearly {t being time from beginning of piston stroke or qt being angle p?^sed through by crank from a dead point), x = a^ sin {qt + 61)4-02 sin {^qt + 63) (1). (There is a very simple method of obtaining the terms Oj and a^ by inspection of the gear.) When the overtone is neglected, a^ is the half travel of the valve and ei is the angle of advance. In a great number of radial valve gears we find that 6-2 = 90°. The best way of studying the effect produced by the octave or overtone is to draw the curve for each term of (1) on paper by the method of Fig. 63, and then to add the ordinates together. If we subtract the outside lap L from cc it is easy to see where the point of cut-off is, and how much earlier and quicker the cut-off is on account of this octave or kick in the motion of the valve. In an example take «i = 1, e^ = 40°, a^ = '2, e^ — 90"^. The practical engineer will notice that although the octave is good for one end of the cylinder it is not good for the other, so that it is not advisable to have it too great. We may utilize this fact in obtaining more admission in the up stroke of modern vertical engines ; we may cause it to correct the inequality due to shortness of connecting rod. Links and rods never give an important overtone of frequency 3 to 1. It is always 2 to 1 . In Sir F. Bramwell's gear the motion of the valve is, by the agency of spur wheels, caused to be a? = «! sin {qt -f Cj) + Oo sin {^qt -t- e,) (2). Draw a curve showing this motion when a, = 1-15 inch, e^ = 47", a._ = '435, e, = 62°. P. 13 194 CALCULUS FOR ENGINEERS. >A^ If the outside lap is 1 inch and there is no inside lap, find the positions of the main crank when cut-off, release, cushion- ing and admission occur. Show that this gear and any gear giving an overtone with an odd number of times the funda- mental frequency, acts in the same way on both ends of the cylinder. 123. If a? = Oi cos (^1^ + €i) -H 02 cos {q4 ■\- e^ (3), where q^^^irfi and q^—'^irf^, there being two frequencies; this is not equivalent to one S. H. motion. Suppose a^ to be the greater. The graphical method of study is best. We have two cranks of lengths a^ and tta rotating with different angular velocities, so that the effect is as if we had a crank A rotating with the average angular velocity of a^, but alternating between the lengths tti + tta and ai — tta ; always nearer a^s> position than Og's ; in fact, oscillating on the two sides of Oi's position. If q^ is nearly the same as q^ we have the interesting effect like beats in music*. Thus tones of pitches 100 and 101 produce 1 beat per second. The analogous beats are very visible on an in- candescent lamp when two alternating dynamo-electrical machines are about to be coupled up together. Again, tides of the sea^ except in long channels and bays, follow nearly the s. H. law ; a^ is produced by the moon and a^ by the sun if ai=2'la2, so that the height of a spring tide is to the height of a neap tide as 3*1 to 1*1. The times oi full are times of lunar full. The actual tide phase never differs more than 0*95 lunar hour from lunar tide ; 095 lunar hour = 098 solar. 124. A Periodic Function of the time is one which becomes the same in every particular (its actual value, its rate of increase, &c.) after a time T. This T is called the * Analytically. Take cos (^vf^t + eg) =co8 {^irf-J, - 2t (/^ -/2)f + e^}, therefore a; = r cos (2t/i« + 6) , where r^ = a-^ + a^ + 2a^a^ cos {27r (/^ -/g) t + Cj - Cg} , and the value of tan 6 is easily vrritten out. ROTATING MAONETTO FIELD. 195 periodic time and its reciprocal is called f the frequency. Algebraically the definition of a periodic function is where n is any positive or negative integer. 125. Fourier's Theorem can be proved to be true. It states that any periodic function whose coraplete period is T (and g is ^tt\T or 27r/) is really equivalent to the sum of a constant term and certain sine functions of the time fif) = ^0 + ^1 sin {qt + E^ + A^ sin {%qt + E^ + ^3 sin (3^'^ + £'3) 4- &:c (1). In the same way, the note of any organ pipe or fiddle string or other musical instrument consists of a fundamental tone and its overtones. (1) is really the same as /(^) = ^0 + «i sin 5^ + 61 cos qt + ag sin ^qt -f 6., cos Iqt + &c., if ar + &r = A^ and tan JS^^ = — , &c. t 126. A varying magnetic field in the direction x follows the law X = a sin qt where t is time. Another in the direction y, which is at right angles to x, follows the law Y—a cos qt. At any instant the resultant field is R = VZmT^ = a = a constant making with y the angle 6, where tan = F/X, or ^ = qt Hence the effect produced is that we have a con- stant field R rotating with angular velocity q. When the fields are X = Oi sin (qt + Cj) and F = aa sin (qt + e^), it is better to follow a graphical method of study. The resultant field is represented in amount and direction by the radius vector of an ellipse, describing equal areas in equal times. Let OX and OY, fig. 69, be the two directions mentioned. Let OAi in the direction OX = a^ . With 0-4 1 as radius describe 13—2 196 CALCULUS FOR ENGINEERS. a circle. Let YOO be the angle ei. Divide the circle into many equal parts starting at and naming the points of division 0, 1, 2, 3, &c. Draw lines from these points parallel to OY, Let 0A» in the direction OF be ag. Describe a circle with OA2 as radius. Set off the angle X'OO'as e^ and divide this circle at 0', 1, 2, 3, 4, &c., into the same number of parts as before. Let lines be drawn from these points parallel to OX, and where each meets the corresponding line from the other circle we have a point whose radius vector at any instant represents, in direction and magnitude, the resultant magnetic field. If OX and OY are not at right angles to one another, the above instructions have still to be followed. If we divide the circle OA2 into only half the number of parts of OJ-i we have the combination of X = Oj sin (qt + fi) and F= ttg sin (2qt 4- 62). If we wish to see the combination X = any periodic function and Y any other periodic function, let the curve SINE CURVE. 197 from Mo to N^ show F, M^N^ being the whole periodic time ; and let the curve from M^ to N^ show X, the vertical distance M^Ni Fig. 70. being the whole periodic time. If Pi and Pj are points on the two curves at identical times, let the horizontal line from P^ meet the vertical line from Pj in P. Then at that instant OP represents the resultant field in direction and magnitude. Carry out this construction carefully. It has a bearing on all sorts of problems besides problems on rotating mag- netic fields. 127. The area of a sine curve for a whole period or for any number of whole periods is zero. This will be evident if one draws the curve. By actual calculation; let s be an integer and g= -— -, •2. 1 r^ ' sin sqt.dt = cos sat because cos5 -^ Tor cos s2it = 1 and cos = 1. /; .-/cos.^P cos j 0, .J, Again, I ^0 cos sqt .dt = — 27r,,, ■^. 1 sm sqt .0 J 1 / . 27r ^ sin )=.. because sin s~^ T= sin s27r = and sin 0=0. 198 CALCULUS FOR ENGINEERS. 128. If the ordinates of two sine curves be multiplied together to obtain the ordinate of a new curve : the area of it is for any period which is a multiple of each of their periods. Thus if s and r are any integers T sin sqt . cos rqt . dt = (1), o sin sqt . sin rqt. dt = (2), /, I cos sqt . cos rqt . dt = . .(3). These ought to be tried carefully. '1st as Exercises in Integration. 2nd Graphically. The student cannot spend too much time on looking at these propositions from many points of view. He ought to see very clearly why the answers are 0. The functions in (1) and (2) and (3) really split up into single sine functions and the integral of each such function is 0. Thus 2 sin sqt . cos rqt = sin (s + r) ot 4- sin {s — r) qt, and by Art. 127, each of these has an area 0. The physical importance of the proposition is enormous. Now if 5 = r the statements (2) and (3) are untrue, but (1) continues true. For •(4), /•T rT I sin^ sqt . dt = / cos^ sqt . dt = ^T .'o Jo whereas (1) becomes the integral of ^ sin 2sqt which is 0. (4) ought to be worked at graphically as well as by mere inte- gration. Recollecting the trigonometrical fact that cos2^ = 2cos2^-l or 1 - 2 sin^ ^, and therefore that cos^ qt=^ cos 2qt + i, sin^ g'^ = J — J cos 2qt, the integration is easy and the student ought to use this method as well as the graphical method. 129. To illustrate the work graphically. Let 0C\ fig. 71, be T. Taking s = 2, the curve OPQRSC represents sin sqt ELECTRICAL ILLUSTRATION. 199 Its maximum and minimum heights are 1. Now note that sin^^^^ is always + and it is shown in OP'Q^R^Sfi. It fluctuates between and 1 and its average height is \ or Fig. 71. the area of the whole curve from to (7 is ^T. The fact that the average value of sinsqt. x sinrqt is 0, but that the average value of sin sqt x sin sqt is \, is one of the most important in practical engineering "work. 130. Illustration in Electricity. An electric dynamo- meter has two coils ; one fixed, through which, let us suppose, a current G flows ; the other moveable, with a current c. At any instant the resultant force or couple is proportional to Go and enables us to measure Gc. But if G and c vary rapidly we get the average value of Gc. Prof. Ayrton and the author have carried out the following beautifully illus- trative experiment. They sent a current through the fixed coil which was approximately, G=Gq sin 27rft. This was supplied by an alternating dynamo machine. Through the other coil they sent a current, c = Co sin 2'nf't whose frequency could be increased or diminished. It was very interesting to note (to the average practical engineer it was uncanny, unbelievable almost) that although great currents were passing through the two coils, there was no average force — in fact there was no reading as one calls it in the laboratory. Suppose / was 100 per second, /' was gradually increased from say 10 to 20, to 30 to 40 to 49. Possibly about 49 to 200 CALCULUS FOR ENGINEERS. 51 a vague and uncertain sort of action of one coil un the other became visible, a thing not to be measured, but asf increased the action ceased. No action whatever as f became 60, 70, 80, 90, 97, 98, 99, but as / approached 100 there was no doubt whatever of the large average force ; a reading could be taken and it represented according to the usual scale of the instrument ^CqCo; when/' increased beyond 100 the force suddenly ceased and remained steadily until /' became 200 when there was a small force to be measured ; again it ceased suddenly until /' became 300, and so on. We know that if C and c had been true sine functions there would have been absolutely no force except when the frequencies were exactly equal. In truth, however, the octaves and higher harmonics were present and so there were slight actions when /and/' were as 2 to 1 or 1 : 2 or 1 : 3, &c. This is an extremely important illustration for all electrical engineers who have to deal with alternating currents of electricity. 131. Exercise in Integration. C and c being alternating currents of electricity. When G = Co sin qt and c = cv sin(qt ± e) and these two currents flow through the two coils of an electro-dynamometer, the instrument records ^ CoCo . cos e as this is the average value of the product Cc. When G and c are the same, that is, when the same current G = G^ sin (qt + e) passes through both coils, the instrument records the average value of G^ . dt, or ^j G,'8m'{qt + e).(lt (1), which we know to be ^Gq^ The square root of any such reading is usually called the effective current, so that ~r^ Gq is what is known as the effective value of C^o sin qt V^ Effective current is defined as the square root of mean square of the current. Thus when an electrical engineer speaks of an alternating current of 100 amperes he means that the effective current is 100 amperes or that G= 141*4 sin {qt + a). Or the voltage 1000 means t; = 1414 sin (</^ + /3). EFFECTIVE CUKRENTS. 201 Exercise. What is the effective value of tto 4- ^1 sin {qt + ti) + A.2 sin {^qt + 63) + &;c. ? Notice that only the squares of terms have an average value, the integral of any other product being during a complete period. Answer: \/ a^ -^ ^{A-^ -{-A.^ -{■ kc). Observe the small importance of small overtones. If V = — (sin qt + J sin ^qt + i sin aqt &c.). (2), we shall see from Art. 135 that this is the Fourier expression for what is shown in the curve (fig. 72) the distance OM being called Vq and the distance OQ being the periodic time T, where 27r q= rp } s-i^d V is measured upwards from the line OQ. M Fig. 72. The effective « = ^^ Vl + * + ^V + ,V + &<=. (3). Again in fig. 73, where PM = Vq and OQ = T, IT' (sin qt — ^ sin Sqt + ^ sin 5qt — &c.) . . .(4). Fig. 73. The effective ?; = —-";. VI +-J^ + ^ + &c (5). Again note the small importance of everything except the fundamental term. 202 CALCULUS FOR ENGINEERS. Exercise. If C=Go-{-A^smqt-\- B^ cos qt 4- ^2 sin 2qt + B^ cos 2qt + <fec ((3), and if c = Co + «! sin qt 4- 6i cos g'< + aa sin 2^'^ 4- h^ cos 2^^ + &c. . . .(7). Average Cc = G^c^ + J (^itti 4- BJ)^ 4- ^./ta + BJ)^ + &c.). . .(8). It will be seen that there are no terms like A.]),^ or A.^^, 132. Let AB and BG be parts of an electric circuit. In yj^^ AB let the resistance be R ^- --^- -■-- „„ ->, ^j^^ j^^ there be no self-in- ^•vsAAAAAAAAA/Tfnrinnpnnr-*^ duction. In BG let the re- ^ ^ T,- ^ i ' ^ sistance be r and let there be ^'^' ^^- self-induction Z. If 0= Co sin ^i is the current passing. Let Vab &c. represent the voltage between the points A and B, &c. Let F^ mean the effective voltage between A and B. Vab = -^C'o sin qt, Vbc ■■= Go Vr« 4- ly sin ^^^ + tan-^ -^') , see Art. 118, Vac = Co \/(i24-r)2 4-?Y sin (^« 4- tan"^ ^^-) , F..= -^i2C„, F..= i^^^V^+>' F^o=-^(i2 + r)Ooyi4- ^'^' Observe that V^c is alTvays less than Vj^^ 4- Vbc^ or the effective voltage between A and C is always less than the sum of the effective voltages between A and B and between B and C. Thus take Co =141*4, jR=l, ?'=1, lq=l, and illustrate a fact that sometimes puzzles electrical engineers. 133. Rule for developing any arbitrary function in a Fourier Series. -|*The function may be represented as in fig. 75, P^ repre- sents the value of y at the time t which is represented by FOURIER ANALYSIS. 203 OE, 00 represents the whole periodic time T. At G the curve is about to repeat itself. (Instead of using the letter F ^^""^^^ / / |\ G ' E > C Fig. 75. t we may use x or any other. We have functions which are periodic with respect to space for example.) Assume that y can be developed as y^a^^-ai sin qt + hi cos qt -}- a.^ sin 2qt + h.^ cos 2qt + as sin 3^^ + 63 cos ^qt 4- &c. . . .(1), where 27r It is evident from the results given in Art. 127 that a^ is the average height of the curve, or the average value of y. This can be found as one finds the average height of an indicator diagram. Carry a planimeter point from to FPHGCO, and divide the whole area thus found by OC. If we have not drawn the curve ; if we have been given say 36 equidistant values of y, add up and divide by 36. The reason is this; the area of the whole curve, or the integral of y between the limits and T, is a^T, because the integi'al of any other term such as Wi sin qt or 63 cos Sqt is 0. In fact rT rT sin sqt .dt or I cos sqt . dt is 0, J .'0 if s is an integer. tti is twice the average height of the curve which results from multiplying the ordi nates y by the corresponding 204 CALCULUS FOR ENGINEERS. ordinate of sin^^; for, multiply (1) all across by sinqt, and integrate from to T, and we have by Art. 128 rT rr I y.sinqt.dt==0-\-aij ain' qt,dt -^-O -\-0 + &i}c,=^aiT...{l); Jo Jo dividing by T gives the average value, and twice this is evidently Oi. Similarly rT y.coaqt,dt-=^b/r (2). .0 In fact, by the principles of Ai't. 128, a.^ and bs are twice the average values of y sin sqt and y . cos sqt^ or 2 f^ ^^»=^t/ y. sin sqt, dt i .' (3). dt 134. In the Electrician newspaper of Feb. 5th, 1892, the author gave clear instructions for carrying out this process numerically Avhen 36 numbers are given as equidistant values of y. In the same paper of June 28th, 1895, the author de- scribed a graphical method of finding the coefficients. The graphical method is particularly recommended for de- veloping any arbitrary function. Students who refer to the original paper will notice that the abscissae are very quickly obtained and the curves drawn. In this particular case we consider the original curve showing y and time, to be wrapped round a circular cylinder whose circumference is the periodic time. The curve is pro- jected upon a diametral plane passing through ^ = 0. Twice the area of the projection divided by the circumference of the cylinder is di. Projected upon a plane at right angles to the first, we get bi in the same way. When the curve is wrapped round s times instead of once, and projected on the two diametral planes, twice the areas of each of the FOURIER ANALYSIS. 205 two projections divided by s times the circumference of the cylinder give cig and hg* Prof Hem'ici's Analyzers, described in the Proceedings of the Physical Society, give the coefficients rapidly and accurately. The method of Mr Wedmore, published in the Journal of the Institution of Electrical Engineers, March 1896, seems to me very rapid when a column of numbers is given as equidistant values of y, 135. When a periodic function is graphically represented by straight lines like fig. 72 or fig. 73 we may obtain the development by direct integration. Thus in fig. 76, the Electrician's Make and Break Curve: w P Q Fig. 76. y=OA, or 2^0 say, from t- to t = OP =^iT; y = from t = iT or OP, to t==T or OQ. 277 Evidently ao = Vo, 5' = -^; 2 ri^ / b M 2 fi^ 2 r*^ as = 7w I 2i'o . sin sqt . dt, h = jp\ 2vq cos sqt . dt. 4i;o T [■ hT 27r "^=-T-25;^Lr'- r^ , 4t;o T V^T 27r 1 * The method is based upon this, that a, = ^\y.m.nsqt.dt = -—^\y.d (cos sqt) = - — / 1/ • ^ (cos sqt). Drawing a complete curve of which y (at the time t) is the ordinate and COB sqt is the abscissa, we see that its area as taken by a Planimeter divided by sir gives a^ . This graphical method of working is made use of in developing arbitrary functions in series of other normal forms than sines and cosines, such as Spherical Zonal Harmonics and Bessels. By the above method ' ^'=hfy' d (sin sqt). 206 CALCULUS FOR ENGINEERS. 2^0 , ... 2vo aii= (cos SIT — cos 0) = / if « ii \- 2 if s ij if « is even\ IS odd / = — if .9 is odd, Sir 2v, ha=—^ (sin sir - sin 0) = 0. STT Hence the function shown in fig. 76 becomes y = Vf,+ -- (sin at + \ sin Zqt + J sin ^qt + &c.) (1). IT M Fig. 77. If the origin is half-way between and A (fig. 76), as in fig. 77, so that instead of what the electricians call a make and break we have Vq constant for half a period, then — Vq for the next half period, that is, reversals of y every half period, we merely subtract v^, then y = — ^ (sin qt-\-l sin ^qt + i sin oqt + &c.) . . .(2). Let the origin be half-way between and P, fig. 70; the ^ of (1) being put equal to a new t-\-\T, sin sqt where s is odd, becomes sin sq {t -f \T), sin s -^ (t-\-^T) or sin Uqti-s'^], where 5=1, 5, 9, 13 &c. this becomes cos sqt, „ 5 = 3, 7, 11, 15 &c. „ „ — cos sqty and consequently with the origin at a point half-Avay between and P, or y = vo + — (cos qt-^ cos Sqt + i cos Sqt - 1 cos Iqt + &c.). FOURIER ANALYSIS. 207 136. To represent a periodic function of x for all values of X it is necessary to have series of terms each of Tvhich is itself a periodic function. The Fourier series is the simplest of these. 137. If the values of y, a function of x, be given for all values of x between a; = and x — c\y can be expanded in a series of sines only or a series of cosines only. Here we regard the given part as only half of a complete periodic function and we are not concerned with what the series represents when x is less than or greater than c. In the previous case y was completely represented for all values of the variable. I. Assume y = ai sin qx + a^ sin 2qx + &c. where q = tt/c. Multiply by ^insqx and integrate between the limits and c. It will be found that all the terms disappear except as sin'^ sqx . dx which is Ja«c, so that a^ is twice the average value oiy,^ixisqx. Thus let y be a constant m, then /, 2 /•' C /a , 2m p m sm sqx .dx= cos sox esq Lo ^ _ = (cos 57r — 1 ) = — II s IS odd, csq^ ' SIT = if s is even. Hence m=^ — (sin qx-\-\ sin Zqx + \ sin hqx + &c.)*. II. Assume y = 6o + &i cos 5-5; + &2 cos 2^-^ + &c. Here 60 is evidently the mean value of y from a? = to a; = c. In the * Exercise. Develope y—mx from a;=0 to .r=c in a series of sines. mx — a^ sin qx + a^ sin ^qx + Ac, where g- = - , «8 IS ~ I mx . sm sg^a; . «« = -g^i" ^^^ ^^^ ~ *5^ • ^^^ -""/^ • For this integral refer to (70) page 365. Hence 2mc f IT 1 . 27r 1 . Stt „ \ '= 1 sin-a;--8in — a;+-sin — x- &c. j. TT \ c 2 c 6 c J 208 CALCULUS FOR ENGINEERS. same way as before we can prove that hg is twice the average vakie of t/ cos sqx*. 138. In Art. 118 we gave the equation for an electric circuit. The evanescent term comes in as before but we shall neglect it. Observe that if V is not a simple sine function of t, but a complicated periodic function, each term of it gives rise to a term in the current, of the same period. Thus if F=Fo + SF,8in(5^^ + 6'«)'^ (1), R \IB? + X V^2 V ^ R ] ^ ^ If Lq is very large compared with R we may take ^■^i-^Isq''''^^''^^^''^ •••^^^• Thus, taking the make and break curve for V, fig. 76, 4F F=F„+ — \Hmqt-\■\sm^t^&c.) (4), TT V 2VT ^=^- ^«^ (^os qt-hi cos 3^^ + ^ cos 5qt + &c.) . . .(5), T which is shown by the curve of fig. 73, being at j . 139. When electric power is supplied to a house or contrivance, the power in watts is the average value of CV where G is current in amperes and V the voltage. * Thus let y=ma? between x=0 and x = c. Evidently bQ=^mc, and we ^ , tmr 4wt / 1 „ 1m „ \ find y~~o ( cos9a; + -co8 3gx + 2^cos5gar + <S:c. J . There are many other normal forms in which an arbitrary function of X may be developed. Again, even of sines or cosines there are other forma than those given above. For example, if we wish generally to develope y a function of x between and c as y = Sa^ sin a„^x by the Fourier method, the essential principle of which is I sin a,j.T . sina^-r . rfa;=0, where m and n are different ; we must have o„ and o-, roots of — ^.— - =a. In the ordinary Fourier series s is oo . i — =: COS e — tan^ — v/r2 4- pQ"^ V rj ALTERNATING CURRENT POWER. 209 Let F= Fo sin qt and 0—0^ sin (g'^ — e). Then P = \GqVq cos ef, or half the product of the ampli- tudes multiplied by the cosine of the lag. When the power is measured by passing G through one coil of a dynamometer and allowing F to send a current c through the other coil, if this coil's resistance is r and self-induction I c= ,^' - sinfg^-tan-^^) (6). What is really measured therefore is the average value of Gc, or i Usually in these special instruments, large non-inductive resistances are included in the fine wire circuit and we may take it that Iq is so small in comparison with r that its square may be neglected. If so, then cos le — tan~^~) apparent power V rJ true power cos e Observe that tan~^ — is a very small angle, call it a, apparent power cos e cos a + sin e sin a -E-f = = cos a -H sm a . tan e. true power cos e Now cos a is practically 1, and sin a is small, and at first sight it might seem that we might take the answer as nearly 1. But if e is nearly 90° its tangent may be exceedingly large and the apparent power may be much greater than the true po'virer. It is seldom however that e approaches 90° unless in coils of great diameter with no iron present, and precautions taken to avoid eddy currents. Even when giving power to a choking coil or unloaded transformer, the effect of hysteresis is to cause e not to exceed 74°. 140. True Power Meter. Let EG and GD be coils wound together as the fixed part of a dynamometer, and let P. 14 210 CALCULUS FOR ENGINEERS. <c ^ <#^c _.^ DB be the moveable coil. The current C + o passes from E to G. Part of it c goes along the non-inductive resistance GF which. has a resistance R. The part G flows from G to D and D to B and through the house or contrivance. The instantaneous value of Rc.C is the instantaneous power. The coils EG and GD are care- fully adjusted so that when c = and the currents are continuous currents, there shall be no deflec- tion of the moveable coil DB. Hence the combined action of C + c in EG and of C in GD upon C in DB is force or torque proportional to cG, and hence the reading of the instrument is proportional to the power. With varying currents also there will be no deflection if there is no metal near capable of forming induced currents. Fig. 78. -rB o ^ o C— -0 141. The student ought to get accustomed to translating into ordinary language such a statement as (1) of Art. 119. Having done so, consider a mass of W lb.* hanging from a spring whose stiffness is such that a force of 1 lb. elongates it h feet. If there is vibration ; when W is at the level CO, fig. 79, a? feet below (we imagine it moving downwards) its position of equili- brium 00, the force urging it to the position of equilibrium is x-^h pounds, and as the W moving mass is — (neglect the mass of the spring itself or consider one-third of it as being added to the moving body), W . X — X the acceleration = r . 9 h xo The acceleration = -t4t • The acceleration is then pro- * The name W lb. is the weight of a certain quantity of stuff ; the inertia of it in Engineers' units is W-i-S2'2. Fig. 79. MECHANICAL VIBRATION. 211 portional to x, and our ^ stands for g-- in (1) of Art. 119, and (2) shows the law connecting a) and t Notice carefully that the + sign in (1) is correct. The body is moving downwards and x is increasing, so that doo/dt is positive. But -j— is negative, the body getting slower in its motion as a) increases. 142. Imagine the body to be retarded by a force which die is proportional to its velocity, or b -r. • Observe that this acts as J acts, that is upwards, towards the position of equi- librium. Hence we may write Wd^x dx X Jdi^^^di-^h'^^ (^>- We shall presently see what law now connects x and t in this damped vibration. 143. Suppose that in the last exercise, when the body is displaced x feet downwards, its point of support B is also y feet below its old position. The spring is really only elongated by the amount x—y, and the restoring force is —7-^ • Con- sequently (1) ought to be Wd'x dx x_y J dt'^^ drh'h ^^^' Now imagine that the motion y is given as a function of the time, and we are asked to find a; as a function of the time. y gives rise to what we call a forced vibration. If 2/ = we have the natural vibrations only. We give this, not for the purpose of solving it just now, although it is not difficult, but for the purpose of familiar- izing the student with differential equations and inducing him to translate them into ordinary language. 14—2 212 CALCULUS FOR ENGINEERS. 144. Notice that if the angular distance of a rigid body from its position of equilibrium is 0, if / is its moment of inertia about an axis through the centre of gravity, if H6 is the sum of the moments of the forces of control about the •in same axis, and if jP vr is the moment of frictional forces at which are proportional to velocity, 4'+^!+^^=^^' <'^)' if B' is the forced angular displacement of the case to which the springs or other controlling devices are attached. 145. The following is a specially good example. Referring back to Example 1 of Art. 98, we hud CR, the voltage in the circuit, connecting the coatings of the condenser. If we take into account self-induction L in this circuit, then the voltage V is ♦ ^C' + ii = '' w- We may even go further and say that if there is an alternator in the circuit, whose electromotive force is e at any instant (e, if a constant electromotive force would oppose G as shown in the figure) RG + L^ = v-e (5). But we saw that the current G =^ — K-jj (6). Using this value of G in (5) we get LK^ + RK^+v = e (7). Kow imagine that e is given as a function of the time and we are asked to find t; as a function of the time. e gives rise to what we call a forced vibratory current in the system. If e = we have the natural vibrations only of the system. Having v, (6) gives us (7. FORCED VIBRATIONS. 218 146. If (7) is compared with (2) or (3) we see at once the analogy between a vibrating mechanical system and an electrical one. They may be put W d^x ..dx X y ^f , . - ^S+^S+^=i'E^-*'^-i (^)- W The mass — corresponds with self-induction L. The friction per foot per second &, corresponds with the resistance R. The displacement x, corresponds with voltage v, or to be seemingly more accurate, v is Q the electric displacement divided by K. The want of stiffness of the spring h con-esponds with capacity of condenser K. The forced displacement y corresponds with the forced E.M.F. of an alternator.! 147. The complete solution of (8) or (9), that is, the expression of x or i; as a function of t, will be found to include: — (1) The solution if i/ or e were 0. This is the natural vibration of the system, which dies away at a rate which depends upon the mechanical friction in the one case and the electrical friction or resistance in the other case. We shall take up, later, the study of this vibra- tion. It ought to be evident without explanation, that if y or e is 0, we have a statement of what occurs when the system is left to itself (2) The solution which gives the forced vibrations only. The sum of these two is evidently the complete answer.! 148. Forced Vibration. As the Mechanical and Electrical cases are analogous, let us study that one about 214 CALCULUS FOn ENGINEERS. which it is most easy to make a mental picture, the mechanical case. We shall in the first place assume no friction and neglect the natural vibrations, which are however only negligible when there is some friction. Then (8) becomes dt?^wi.'—my <i*^>- Let y = a sin qt be a motion given to the point of support of the spiral spring which carries W ; y may be any compli- cated periodic function, we consider one term of it. We know tha t if y were 0, the natural vibration would be a; = 6 sin f i 4/ ^-jj + ^^^ ) , where h and m might have any values whatsoever. It is simpler to use n"^ for glWh as we have to extract its square root, n is 27r times the frequency of the natural vibrations of W. We had better write the equation as — -f i^^x = n^y = n-a sin qt (11). Now try if there is a solution, 00 = A r,in qt + B cos qt If so, since ,- = — Aq- sin qt — Bq^ cos qt ; equating the coefficients of sinqt and also those of cosqt, — Aq- -[■ n^A = n^a, so that A =— — ^ , and — Bq^ + ii^B = 0, so that B = unless n = q, n^ — q^ ^ * We see that we have the solution ^^n^Zrjf^'^'i^ (12)- This shows that there is a forced vibration of W which is synchronous with the motion of the point of support ; its amplitude being , times that of the point of support. Now take a few numbers to illustrate this answer. Let a = 1, let " be great or small. Thus ^ = -X means that the forced n ^ n ^^ frequency is one tenth of the natural frequency. FORCED VIBRATIONS. 215 ? Amplitude of j '/ Amplitude n W 's motion j n JF's motion •1 1 101 1 1 X •5 1^33:3 1^01 -50 •8 2^778 1-03 -16^4 •9 5^263 1^1 - 4-762 •95 1026 ; 15 - 0-8 •97 1692 20 - 0-333 •98 2525 50 - 0042 •99 50^25 100 - 0010 Note that when the forced frequency is a small fraction of the natural frequency, the forced vibration of TT is a faithful copy of the motion of the point of support B ; the spring and W move like a rigid body. When the forced is increased in frequency the motion of TT is a faithflil magni- fication of B's motion. As the forced gets nearly equal to the natural, the motion of W is an enormous magnification of B's motion. There is always some friction and hence the amplitude of the, vibration cannot become infinite. When the forced frequency is greater than the natural, W is always a half-period behind B, being at the top of its path when B is at the bottom. When the forced is many times the natural, the motion of W gets to be very small ; it is nearly at rest. Men who design Earthquake recorders try to find a steady point which does not move when everything else is moving. For up and down motion, observe that in the last case just mentioned, W is like a steady point. When the forced and natural frequencies are nearly equal, we have the state of things which gives rise to resonance in acoustic instruments ; which causes us to fear for suspen- sion bridges or rolling ships. We could easily give twenty interesting examples of important ways in which the above principle enters into engineering problems. The student may now work out the electrical analogue for himself and study Hertz' vibrations. 149. Steam engine Indicator vibration. The motion of the pencil is to faithfully record the force of 216 CALCULUS FOR ENGINEERS. the steam on the piston at every instant ; this means that the natural vibrations of the instrument shall be very quickly destroyed by friction. Any friction as of solids on solids will cause errors. Indeed it is easy to see that solid friction causes diagrams to be always larger than they ought to be. Practically we find that if the natural frequency of the instrument is about 20 times that of the engine, the diagram shows few ripples due to the natural vibrations of the indi- cator. If the natural frequency is only 10 times that of the engine, the diagram is so ' upset ' as to be useless. F The frequency of a mass — at the end of a spring whose if yieldingness is hy see Art. 141, is ^ ^gj Wh, neglecting friction. We shall consider friction in Art. 1 GO. What is the frequency of a mechanism like what we have in an indicator, controlled by a spring ? Answer : If at any point of the indicator w mechanism there is a mass - , and if the displacement of this point is .9, when the displacement of the end of the spring (really the piston, in any ordinary indicator) is 1 ; imagine w w that instead of — we have a mass s^ — at the end of 9 hXsHv' the spring. Thus the frequency is ^ a/ ^ To illustrate this, take the case shown in fig. 80 ; GAB is a massless lever, hinged at 0, with S- the weight W at B. The massless e spring is applied at A. I When A is displaced downwards from equilibrium through the dis- tance x, the extra pull in the spring €) X is -r . The angular displacement of ^ig- 80. the lever, clockwise, is j-j . Mo- UA ment of Inertia x angular acceleration, is numerically equal W to moment of force. The Moment of Inertia is — OB^, 9 VIBRATION INDICATOR. 217 The angular acceleration is yr-j , where x stands for -^ , so that -0i?^;^ + 7. 0^ = 0, g OA h ^+^.^.^ = 0. OR And yya ^^ what we called 5, so that s^W takes the place of our old W when W was hung directly fronri the spring. 150. Vibration Indicator. Fig. 81 shows an in- strument which has been used for indicating quick vertical vibration of the ground. A c o a o Fig. 81. The mass GPQ is supported at F by a knife edge, or by friction wheels. The centre of gravity G is in a horizontal line with P and Q. Let FG=a, GQ=:h, PQ = a+b=l The vertical spring AR and thread BQ support the body at Q. As a matter of fact AR is an Ayrton-Perry spring, which shows by the rotation of the pointer R, the relative motion of A and Q; let us neglect its inertia now, and consider that the pointer faithfully records relative motion of A and Q. It would shorten the work to only consider the forces at P and Q in excess of what they are when in equilibrium, but for clearness we shall take the total forces. When a body gets motion in any direction parallel to the plane of the paper, we get one equation by stating that the resultant force is equal (numerically) to the mass multiplied by the linear acceleration of the centre of gravity in the direction of the resultant force. We get another equation by stating 218 CALCULUS FOR ENGINEERS. that the resultant moment of force about an axis at right angles to the paper through the centre of gravity is equal to the angular acceleration, multi- plied by moment of inertia about this axis through the centre of gravity. I shall use x, x and ic to mean fj nf* Cm IT displacement, velocity and acceleration, or x, -j- and -v- . Let P and A get a displacement x^ downward. Let Q be displaced x downward. Let the pull in the spring be Q = Qq-\- c{x — Xi) where c is a known constant (c is the reciprocal of the h used in Art. 141). Let W be the weight of the body. Then if Pq and Qo be the upward forces at the points marked P and Q when in the position of equilibrium, Q„(a + Z>)= Wa and Po+Qo= W. J, bW ■ aW Hence ^o = ^^^, Qo = ^-7-^ (1), Q=Qo + c{x-x,). Now G is displaced downwards =- Xi H ? x, so that ^ a+6 a+b W-P-Q='^{b-£, + aB]^^ (2). The body has an angular displacement clockwise about its centre of mass, of the am< moment of inertia about G X -— X centre of mass, of the amount r« So that if / is its a-{-b --Qb + Pa = ^^^^^(x-x,) (3). W Hence (2) and (3) give us, if M stands for — , and if / = Mk^ where k is the radius of gyration about G, VIBRATION INDICATOR. 219 If ki is the radius of gyration about P, we find that (4) simplifies to I [^ if n stands for tta/ ir?= ^tt x natural frequency, and e^ stands for 1 — r-2 • Call X — Xihy the letter y because it is really y that an observer will note, if the framework and room and observer have the motion x^. Then 2^ y = x — XiOv x = y-\-Xx y + Xi-\- n^ (y + ^'i) = ^Xi 4- n^Xi . So that y + n^y = (^-l)xj_ (5), or y + w'2/ + p^i = ^ (6)- Let Xj^ = A sin qt. We are neglecting friction for ease in understanding our results, and yet we are assuming that there is enough friction to destroy the natural vibration of the body. We find that if we assume y = a sin qt, then _ al ([- . That is, the apparent motion y (and this is what the pointer of an Ayrton- Perry spring will show ; or a light mirror may be used to throw a spot of light upon a screen), is T-« -r^ — ^ times the actual motion of the framework and A-j* n^ — q^ room and observer. If q is large compared with n, for example if q is always more than five times n, we may take it that the apparent motion is y-^ times the real motion and is independent of frequency. Hence any periodic motion whatever (whose periodic time is less say than Jth of the periodic time of the apparatus) will be faithfUlly indi- cated. Note that if al ~ Tc^ so that Q is what is called the point of percussion, Q is a motionless or * steady ' point. But in practice, the instrument is very much like what is shown in 220 CALCULUS FOR ENGINEERS. the figure, and Q is by no means a steady point. Apparatus of the same kind may be used for East and West and also for North and South motions. 151. Any equation containing ■— or -^ or any other differential coefficients is said to be a " Difierential Equation/' It will be found that differential equations contain laws in their most general form. Thus if o) is linear space and t time, the statement -f^ = d^x . ^^ means that ,- , (the acceleration), does not alter. It is the most general expression of uniformly accelerated motion. When we integrate and get -7^= ci, we have introduced the more definite statement that the constant accelei-ation is known to be a. When we integrate again and get dx ^ = at + h, we are more definite still, for we say that h is the velocity when ^ = 0. When we integrate again and get X = ^af -i-bt-^-c, we state that x = c when ^ = 0. Later on, it will be better seen, that many of our great general laws are wrapped up in a simple looking expression in the shape of a differential equation, and it is of enormous importance that when the student sees such an equation he should translate it into ordinary language. 152. An equation like g-^S-«2-^l-^=^ «. if Py Q, R, S and X are functions of x only, or constants, is said to be a linear differential equation. Most of our work in mechanical and electrical engineering leads to linear equations in which P, Q, &c., are all constant with the exception of X. Thus note (8) and (9) of Art. 146. DIFFERENTIAL EQUATIONS. 221 Later, we shall see that in certain cases we can find the complete solution of (1) when X is ; that is, that the solu- tion found will include every possible answer. Now suppose this to be y =/(*'). We shall see that it will include four arbitrary constants, because ^^ is the highest differential coefficient in (1), and we shall prove that if, when X is not 0, we can guess at one solution, and we call it y= F (os), then j,=f{x) + F(x) is a solution of (1). We shall find in Chap. III. that this is the complete solution of (1). In the remainder of this chapter we shall only consider P, Q, &c., as constant ; let us say da^^^d^^^d^^^dx^^y-^ ^^^' where A, B, G, E are constants and Xis a function o{ a;. We often write (2) in the form 153. Taking the very simplest equation like (3). Let J-«y = o • W' it is obvious (see Art. 97) that y = Me^'' (5) is the solution, where M is any constant whatsoever. 154. Now taking ^-«'.y = (6), we see by actual trial that 2/ = if €«* + i\r6-«^ (7) is the solution, where M and N are any constants whatsoever. But if we take ^^n^y = (8), 222 CALCULUS FOR ENGINEERS. we see that as the a of (6) is like ni in (8) if i means V— 1, then y =: Me''^ -\- Ne-""^ (9) is the sohition of (8). If we try whether this is the case, by difFerentiat.Ton, assuming that i behaves like a real quantity and of course i^ = — 1, i^ = — i, i* = l, {^ = i, &c., we find that it is so. But what meaning are we to attach to such an answer as (9) ? By guessing and probably also through re- collection of curious analogies such as we describe in Art. 106, and by trial, we find that this is the complete solution also, y — Mi sin tix + N^ cos nx (10). As (10) and (9) are both complete solutions (Ai-t. 152) be- cause they both contain two arbitrary constants which may be unreal or not, we always consider an answer like (9) to be the same as (10), and the student will find it an excellent exercise to convert the form (10) into the form (9) by the exponential forms of sinew; and coscu?, Art. 106, recollecting that the arbitrary constants may be real or unreal. Besides, it is im- portant for the engineer to make a practical use of those quantities which the mathematicians have called unreal, 155. Going back now to the more general form (3) when X = O, we try if 2/ = Me'^ is a solution, and we see that it is so if , m^ + Am^ -\- Bni" -{■ Gm-\- E = ^ (1). This is usually called the auxiliary equation. Find the four roots of it, that is, the four values of m which satisfy it, and if these are called 7?ii, "nfi^y Wg, m^, we have as the complete solution of (3) when X = 0; M^, &c., being any arbitrary constants whatsoever. 156. Thus to solve if we assume y = e*"*, we find that m must satisfy ?M^ + oiit^ -f om^ — om — 6 = 0. DIFFERENTIAL EQUATIONS. - 223 By guessing we find that m = l is a root; dividing by m — 1 and again guessing, we find that m= — l is a root ; again dividing by m + 1 we are left with a quadratic expres- sion, and we soon see that m= — 2 and m = — 3 are the remaining roots. Hence is the complete solution, il/i, i/s) &c., being any constants whatsoever. 157. Now an equation like (1) may have an unreal root like 7n + 7ii, where t is written for J— 1, and if so, we know from algebra, that these unreal roots go in pairs ; when there is one like m + ni there is another like ni — ni. The corre- sponding answers for y are y — M € ('"~"^' ^ 4- N 6 <"*+*"^ * or e"*^ {i¥i€-^'^ + iVi€+"'*), and we see from (10) that this may be written y = e"** {31 sin nx + N cos nx], where M and iV" are any constants whatsoever. 158. Suppose that two roots m of the auxiliary equation, happen to be equal, there is no use in writing because this only amounts to (M^ + M,^) e*"^ or ife**^ where M is an arbitrary constant, whereas the general answer must have two arbitrary constants. In this case we adopt an artifice ; we assume that the two roots are m and 7n + h and we imagine h to get smaller and smaller without limit : = e"'^ (Ml -H if^e^^^), but by Art. 97, e^ = 1 + /^A- + ^ + j^3 4- &c., therefore y = e"^^ (m^ + M^ + MJiw + M^ ^ + ^c) - Now let i/2^ be called N and imagine h to get smaller 224 CALCULUS FOR ENGINEERS. and smaller, and M2 to get larger and larger, so that MJi may be of any required value we please, say N, and also as h gets smaller and smaller without limit we find If this reasoning does not satisfy the reader, he is to remember that we can test our answer and we always find it to be correct. 159. It is in this way that we are led to the fol- lowing general rule for the solving of a linear differential equation with constant coefficients. Let the equation be fl + A^^+Bf;^, + S^. +G^^ + % = 0...(1). Form the auxiliary equation m** + ^m^-i + ^m'*-^ + &c. + Gm + iT = 0. The complete value of y will be expressed by a series of terms : — For each real distinct value of m, call it a, , there will exist a term Mie*»'*'; for each pair of imaginary values a^ ± ySgi, a term 6»*^ (ilfa sin ^^ + iVs cos ^.^) ; each of the coefficients M^y M^, N2 being an arbitrary constant if the corresponding root occurs only once, but a polynomial of the r— 1th degree with arbitrary constant coefficients if the root occur r times. Ea^rcue. g+ 12^ + 66 f|+ 206 g aa:^ dx* dx^ do(? + 345^ + 234y = 0. Forming the auxiliary equation, I find by guessing and trying, that the five roots are - 3, - 3, - 2, - 2 + 3i, - 2 - 3i. Consequently the answer is y = (ifi + N^x) €-^ + M^e-"^ + €-^ (i/g sin Sx + N, cos 3^). NATURAL VIBRATIONS. 225 Exercise. 1. Integrate ^t^ - 4 ^ 4- 3y = 0. Answer : y= A^+ Bg^. % lutegrate g-loJ+34y = 0. Answer : y = e^ {A sin ^x + B cos 3^j. 3. Integrate g + 6| + 9, = 0. Answer : y = (^ + 5i») e~^. 4 Integrate g-l^g + esg- lo6|+ 169, = 0. Here vv' - 12m^ + 627?t2 - Ib^m + 169 = 0, and this will be found to be a perfect square. The roots of the auxiliary equation will be found to be 3 + 2i, 8 + 2'i, 3 - 1i, 3 - 2i Hence the solution is y = 6^* {(^1 + B^x) sin 2*- + (^2 + BrO)) cos 2^j. We shall now take an example which has an important physical meaning. Natural Vibrations. Example. 160. We had in Art. 146, a mechanical system vibrating with one degree of fi-eedom, and we saw that it was analogous with the surging going on in an Electric system consisting of a condenser, and a coil with resistance and self-induction. We neglected the friction in the mechanical, and the resistance in the electric problem. We shall now study their natural vibrations, and we choose the mechanical problem as before. If a weight of TTlb. hung at the end of a spring which elongates x feet for a force of x -^Ji lb., is resisted in its motion by friction equal to 6 x velocity, then wc had (8) of Art. 146, or IL^ h^ --0 d^x hq dx xq ^ ,^. -rfF + F-rf*+m = *^ (!>• i\ 15 ^26 CALCULUS FOR ENGINEERS. Let ^^ be called 2/ and let -^ = ?i= ; (1) becomes 5+2/.J + n'. = (2). Forming the auxiliary equation we find the roots to be We have different kinds of answers depending upon the values of / and n. We must be given sufficient information about the motion to be able to calculate the arbitrary con- stants. I will assume that when t ia the body is at w = and is moving with the velocity Vq. I. Let / be greater than n, and let the roots be — a and - 13. IL Let /be equal to n, the roots are —/and — / IIL Let /be less than n, and let the roots be — a ± bi. IV. Let/= 0, the roots are ± ni. Then according to our rule of Art. 159, In Case I, our answer is dos H ^ = 0, we can calculate A and B and so find x exactly in terms of ^ ; In Case II, our answer is In Case III, our answer is X = €-«* {A sin U + B cos U] ; In Case IV, our answer is X — A^mnt'\- B cos nt. 161. We had better take a numerical example and we assure the student that he need not grudge any time spent upon it and others like it. Let n = 3 and take various values of /. For the purpose of comparison we shall in all cases let a? = "*" dx when ^ = 0; and ^. = 20 feet per second, when i = 0. and if we are told that x = when < = and -y; = Vo when NATUKAIi VIBRATIONS. Case IV. Let/= 0, then a;== A sin nt + 5 cos nt, = AxO + Bxl, so that 5 = 0, da) 227 dt = nA cos ?«^ — iiB sin n^, 20 = SA, so that ^ = ^-. Plot therefore *•= 6*667 sin 3^. This is shown in curve 4, fig. 82. It is of course the ordinary curve of sines: undamped S.H. motion. Fig. 82. Case III. Lety = "S. The auxiliary equation gives 771 = - -3 ± \/-09 - 9 = - -3 ± 2-985t. Here a = '3 and 6 = 2-985 in x^e-^'^lAsinU-tBcosbt} (1). You may not be able to differentiate a product yet, although we gave the rule in Art. 90. We give many exercises in Chap. III. and we shall here assume that dx — = - ae-"* (A sin ht + B cos bt) + 6e-««(^cos6e-5sin60 (2). 15—2 228 CALCULUS FOR ENGINEERS. Put .£=0 when t = and -r: = 20 when t = 0. Then at B = from (1) and and hence os = 6-7€-*«* sin 2-985^.* This is shown in curve 3 of fig. 82. Notice that the period has altered because of friction. Case II. Let/= 8. The roots of the auxiliary equation are ?/i = — 3 and — 3, equal roots. Hence ^• = (^+^0^"'' (!)• Here again we have to differentiate a product and ~ = B€-''-S{A+Bt)e-'^' (2). Putting in a; = when ^ = and -17 = 20 when ^ = 0, ^ = from (1) and J5 = 20 from (2). Hence o) = 20^ . e-'K This is shown in curve 2 of fig. 82. Case I. Let f=o. The roots of the auxiliary equation are — 9 and -- 1, 0! = Ae-''^ + Be-\ ^=-9^6-««-l?6-'. at Putting in the initial conditions we have = A-hB, 20^-9A-B. Hence J. = - 21, 5 = 2^, «<'=2i (€-*-€-»«)• This is shown in curve 1 of fig. 82. Students ought to take these initial conditions OS =10 when t^O and -77 = when t = 0. dt VIBRATIONS. 229 This would represent the case of a body let go at time or, in the electrical case, a charged condenser begins to be discharged at time 0. Notice that if we differentiate (1), Art. 160, all across we have f using v for -j-) , d^v hq dv a ^ — ^ — P— ?) = dt^^ Wdt^ Wh We have therefore exactly the same law for velocity or acceleration that we have for x itself. Again, in the electrical case os K -r: represents current, if we differentiate all across we find exactly the same law for current as for voltage. Of course differences are produced in the solutions of the equations by the initial conditions. 162. When the right-hand side of such a linear differential equation as (2) Art. 152 is not zero and our solution will give the forced motion of a system as well as the natural vibrations, it is worth while to consider the problem from a point of view which will be illustrated in the following simple example. To solve (11) Art. 148, which is -J- -\- ii^x — n^a sin qt (1), the equation of motion of a system with one degree of freedom and without friction. Differentiate twice and we find d*w ^d^x , , . . Hence from (1), ^+(712 + 52)^+^^^ = (2). To solve (2), the auxiliary equation is m^ + (ii' + 50m2 + g'2|i2^0 (3), and we know that + ni are two roots and + qi are the other two roots. Hence we have the complete solution «? = ^ sin nt + B cos nt -^-G sinqt-^ D cos qt (4). 2S0 CALCULUS FOR ENGINEERS. © @ © Now it was by differentiating (1) that we introduced the possibility of having the two extra arbitrary constants C and D, and evidently by inserting (4) in the original equation, we shall find the proper values of C and D, as they are really not arbiti-ary. It will be noticed that by differentiating (1) and obtaining (2) we made the system more complex, gave it another degree of freedom, or rather we made it part of a larger system, a system whose natural vibrations are given in (4). When we let a mass vibrate at the end of a spring, it is to be remembered that the centre of gravity of the mass and the frame which supports it and the room, remain unaltered. Hence vibrations occur in the supporting frame, and there is friction tending to still the vibrations. If there is another mass also vibrating, this effect may be lessened. For example in fig. 83, if M vibrates at the end of the strip MA, clamped in the vice A, any motion of M to the right must be accompanied by motion of A and the support, to the left. But if we have two masses Mi and il/g (as in a tuning fork), moving in opposite directions at each instant there need be no motion of the supports, consequently the system MjMo vibrates as if there were less friction, and this principle is utilized in tuning forks. Should a motion be started, different from this, it will quickly become like this, as any part of the motion which necessitates a motion of the centre of gravity of the supports, is very quickly damped out of existence. The makers of steam engines and the persons who use them in cities where vibration of the ground is objected to, find it important to take matters like this into account. 163. If y is a known function of oc, we are instructed by Fig. 83. SYMBOLS OF OPERATION. 231 (3), Art. 152, to perform a complicated operation upon it. Sometimes we use such a symbol as {e'-\-Ae'-{-Bd^+ce+E)y = x, to mean exactly the same thing ; 6y meaning that we differ- entiate y with regard to x, 6^y meaning that we differentiate y twice, and so on. 6, 6^, &c., are symbols of operation easy enough to under- stand. We need hardly say that 6'^y does not mean that there is a quantity 6 which is squared and multiplied upon y: it is merely a convenient way of saying that y is to be differentiated twice. 66y would mean the same thing. On this same system, what does (6 -\-a)y mean ? It means ^+ay. What does (6'' + A0 + B)y mean? It means (Py dy ~T^ -}- -4 1^ -h By. (6-{-a)y instructs us to differentiate y and add a times y, for a is a mere multiplier although is not so, and yet, note that (6 -\-a)y= 6y + ay. In fact we find that 6 enters into these operational expressions as if it were an algebraic quantity, although it is not one. If u and V are functions of x we know that e {u + v)= 6u + Ov. This is what is called the distributive law. Again, if a is a constant, 6au = a6u, or the operation 6a is equivalent to the operation aO. This is called the com- mutative law. Again 6^6^ — 6^+^.^ ^\^\^ jg ^j^^ index law. When these three laws are satisfied we know that 6 will enter into ordinary algebraic expressions as if it were a quantity. 6 follows all these laws when combined with constants ; but note that if u and v are functions of x, vdu meaning v -j- , is a very different thing from 6 , uv. When we are confining our 232 CALCULUS FOR ENGINEERS. attention to linear operations we are not likely to make mistakes. Thus operate with 6-{-b upon (0 -f a) y. Now {e-\-a)y=-ey-\-ay or ^ + ay. Operating with 6 ■\-h means "diflFerentiate (this gives us /7/S"^^^) ^^^ ^^^ ^ times -^- + ay" Consequently it gives We see, therefore, that the double operation {e + h){d + a) gives the same result as [&'-\-{a-\-h)e^-ah]. In this and other ways it is easy to show that although 6 ia a symbol of operation and not a quantity, yet it enters into combinations as if it were an algebraic quantity, so long as all the quantities a, h, &c. are constants. Note also that is the same as {0 + h){d -\- a). The student ought to practise and see that this is so and get familiar with this way of writing. He will find that it saves an enormous amount of unnecessary trouble. Thus compare such expressions as (ae-hh)(ae-h^)y with jaa^-^ + (a^ +ah)e + h^} y, 164. Suppose that Dy is used as a symbol for some curious operation to be performed upon y, and we say that Dy = X ; does this not mean that if we only knew how to reverse the SYMBOLS OF OPERATION. 233 operation, and we indicate the reverse operation by D~^ or p. , then y = I)~^ X or yy ? We evidently mean that if we operate with D upon D~^X, we annul the effect of the D~^ operation. Now if -^ + ay = X , or l-j- -\- a] y — X , or {6 ■{■ a) y =X , let us indicate the reverse operation b}^ y-^^x-^^y^^^i^+^rx (1), X X ,^^ or , or ^ (2). ax Keeping to the last of these; at present ^ is a mere symbol for an inverse operation, but v-\-a y=-^a (^> submits to the usual rules of multiplication, because (3) is the same as (6 + a)y= X (4); and yet (4) is derived from (3) as if by the multiplication of both sides of the equation by {6 + a), Again,take^ + (a + 6)^+a62/ = Z (5), or {e^+{a-{-h)e + ab]y=^X (6), or (e + a)(d + h)y = X (7). Here the direct operation 6 + a performed upon (0 -{•h)y gives us X ; hence by the above definition (^+6)y=^„ (8). and repeating, we have 2/ ~(e-hh)\d-\-aj "^^^' 234 CALCULUS FOR ENGINEERS. But it is consistent with our way of writing inverse operations to write (6) as ^^e'-^(a+b)e+ab ^^^^' and so we see that there is nothing inconsistent in our treating the 6 -{-h and ^ + a of (9) as if ^ were an algebraic quantity. 165. We know now that the inverse operation {^+(a+6)(9 + a61-^ (1), may be effected in two steps ; first operate with (6 -\- h)~^ and then operate with (6 + a)~\ Here is a most interesting question. We know that if 6 were really an algebraic quantity, 1^1 1 ^ .(2). ^ + (a + 6) ^ + a6 b-a\d-\-a O-^b, And it is important to know if the operation irrs(5n"ff+b) ^^^' is exactly the inverse of ^ + (a + b) d + ab?...(4). Our only test is this ; it is so, if the direct operation (4) com- pletely annuls (3). Apply (3) to X and now apply (4) to the result; if we apply (4) to ^ X, we evidently obtain dX tf + ct 1 (d-{-b)X or -T- + 6X ; if we apply (4) to j^v X we evi- dently obtain (^ + a) X or -^ + aX, and 1 b — a We see therefore that (3) is the inverse of (4), and that we have the right to split up an inverse operation like the left-hand side of (2) into partial operations like the right-hand side of (2). We have already had a number of illustrations of this when the operand was 0. For it is obvious that if a^, oui, &c. are the roots of the auxiliary equation of Art. 159, it really means that e^'^A e>'-' + J5l9"-=^ + ka. ■\- GO ■\- H splits up into the factors {6 — ttj) {6 — a^), &c, f + 6X-(f + «X)^=X. SYMBOLS OF OPERATION. 235 Observe that if -J^ = X, or 6y = X, or y — -^, or y = 6~^X, the inverse operation ^^ simply means that X is to be inte- grated. Again, 0~^ means integrate twice, and so on*. * Suppose in our operations we ever meet with the symbols 6^ or d~^ or 6"^ &c., what interpretations are we to put upon them ? It is not very necessary to consider them now. Whatever interpretations we may put upon them must be consistent with everything we have ah"eady done. For example 6^ will be the same as 60^ and d~ -^ will be the same as d^ d~^ or 6~^6^. We have to recollect that all this work is integration and we use symbols to help us to find answers ; we are employing a scientific method of guessing, and our great test of the legitimacy of a method is to try if our answer is right; this can always be done. Most of the functions on which we shall be operating are either of the shape ^e"* or B sin bx or sums of such functions. Observe that ^^e«^=^a» if n is an integer either positive or negative. There is therefore, a likelihood that it will help in the solution of problems to assume that or that e^A6*^ = Aa^6^ (1). Again 0B sin bx = Bb cos bx = Bb sin. (bx + ^\ , e^B sin bx= - Bb^ sin 6a; = Bb^ sin (&a; + tt) , and ^ 0^B sin bx=Bb'' Bin (bx + n'^j (2). Evidently this is true when n is a positive or negative integer ; assume it true when n is a positive or negative fraction, so that 623 sin bx = Bb2 sin f bx + ^^ j .(3). There are certain other useful functions as well as e*"* and si7i bx such that we are able to give a meaning to the effect of operating with 6^ upon them. It will, for example, be found, if we pursue our subject, that we shall make use of a function which is for all negative values of x and which is a constant a for all positive values of x. It will be found that if this function is called /(.r) then 0if(x)=a4-x-^ (4), and the meaning of d^ or ^^ or ^~i or 9~^ &c. is easily obtained by differentiation or integration. The Mnemonic for this, we need not call it bn proof or reason, is d^x'^ =~=^~ x"^'^. Let w=i, m=0 and we have *^ \m-n ^ 286 CALCxn.us for engineers. 166. Electrical Problems. Circuit with resistance R and self-induction L, let -^ be indicated by 0, then V V=(R + Le)G or = R + Ld' In fact in all our algebraic work wo treat R + "LO as if it were a resistance. Condenser of capacity K farads. Let V volts, be voltage between coatings. Let G be current in amperes into the condenser, that is, the rate at which Q, its charge in coulombs, is increasing. Or ^ = 7^ = "7^ (-^^^ ^^' ^^ ^ ^ usually dV assumed to be constant, (7 = TT—r- . at The conductance of a condenser is KO, therefore Hence the current into a condenser is as if the condenser had a resistance — ^. Fig. 84. Circuit with resistance, self-induction and capacity, fig. 84. All problems are worked out as if we had a total resistance R + I-^ + xFZJ (!)• kS dhx^=-. — ^x 2. But |- i has no meaning. Give it a meaning by assuming |_IL2 ~ that whatjs true of integers, is true of all numbers, and use gamma function of I or ! ^ which is Jir instead of l~^' It is found that the solutions effected by means of this are correct. EASY RULE FOR ELECTRICAL PROBLEMS. 287 167. In any network of conductors we can say exactly what is the actual resistance (for steady currents) between any point A and another point B if we know all the resist- ances 7\, r^, &c. of all the branches. Now if each of these branches has self-induction ^i, &c. and capacity Ki, &c. what we have to do is to substitute r^ + l^d -f t^-t, instead of Ti in the mathematical expressions, and we have the resistance right for currents that are not steady. How are we to understand our results ? However com- plicated an operation we may be led to, when cleared of fractions, &c. it simplifies to this; that an operation like a-\-b0-\- cB' 4- dO' + ed'+fd' + &c a' + b'd + c'0'' + d'e' + e'e'~tf'e'-{-&}c. ^ ^' has to be performed upon some voltage which is a function of the time. On some functions of the time which we have studied we know the sort of answer which we shall obtain. Thus notice that if we perform (a -}- bd -\- &}c.) upon e** we obtain {a + b2 + col' + da^ + ea.^ +/a^ + &c.) e*^ (2). Consequently the complicated operation (1) comes to be a mere midtiplication by A and division by A', where A is the number a -f 6a + ca^ + &c. and A' is the number a + Vol + c V -|- &c. Again, if wc operate upon in sin {nt + e), observe that 6"- would give — mn^ sin {nt + e). and ^ „ „ +mn^mn{nt-\-e\ and so on ; whereas would give inn cos {nt + e). 6^ „ „ — mu^ cos {nt + e). 6^ „ „ + mn^ cos {nt + e). And hence the complicated operation (1) produces the same effect as -^ — ^^ , where p — a — cn^-hen* — &c., q=^b — dn- -^fn^ — &c. a = a' - c'n2 -\-e'n^ - &c., ^ = b' - dn" +fn' - &c. 238 CALCULUS FOR ENGINEERS. Observe Art. 118, that p + qO operating upon m sin (tit 4- e) multiplies the amplitude by Vp^ + q^ii^ and causes an advance of tan~^— . The student ought to try this again for himself, although he has already done it in another way. Show that (p + qO) sin nt = V^T^ sin (nt + tan-^ 2!!:^ . Similarly, the invei-se operation l/(a + ^6) divides the amplitude by Va" + yS^/i^ ^j^^j produces a lag of tan~^ — , and hence ^-j-|gmsm(7i«+e) a labour-saving rule of enormous importance. 168. In all this we are thinking only of the forced vibrations of a system. We have already noticed that when we have an equation like (1) or (2) Art. 152, the solution consists of two parts, say y —f(x)-{- F{x)\ where f{x) is the answer if X of (2) is 0, the natural action of the system left to itself, and F {x) is the forced action. If in (2) we indicate the operation { d' . d^ ^ d'' ^ d „\ , ^ then D (y) = X gives us Where I)-^{0) gives /(^) and B-^(X) gives F(x). Thus if^+ay^O, or fj- + a]y = 0, or (^ + a)y = 0, we know Art. 97, that y = ^.e-**^. O a so that if ^■-\-ciy'= X, the complete solution is Hence we see that x is not nothing^ but is Ae' ELECTRICAL EXAM^lS^^^V , ^^ 239 "We are now studying this latter paiX't^the forced part, only. In most practical engineering problems the exponential terms rapidly disappear. 169. Thus in an electric circuit where V=(R + Ld) C, if V = Vq sin qt, we have already found the forced value of G, ^_ Fosin^ R+Ld' and according to our new rule, or according to Art. 118, this becomes 0= . ZV s^^ f^^-tan-^^) ....(1). But besides this term we have one -R-^Ld'^'^TT^' and according to the above rule (Art. 168) this gives a term A,e ^' (2). Or we may get this term as in Art. 97, iJ(7+X^ = 0,or ^=-j;0. This is the compound interest law and give's us the answer (2), and the sum of (2) and (1) is the complete answer. If we know the value of G when ^ = 0, we can find the value of the constant Ai ; (2) is obviously an evanescent term. Thus again, suppose V to be constant = Vq, = R + ie- V It is evident that G = ~ is the forced current, for if we Y ^ operate on G=-~ with R + LOwe obtain Fo, and the e vanes- 240 CALCULUS FOR ENGINEERS. cent current is always the same with the same R and L whatever Fmay be, namely Ai€~l^, The complete answer is then ^t n G^A,e-L^+^^ (2). Let, for example, C = when t — Oy then V V and (2) becomes (7= — " (l - e"!') (3). The student ought to take Fo=100, jK=1, L=1 and show how C increases. We have had this law before. 170.. Example. A condenser of capacity K and a non-inductive resistance r in parallel; voltage V at their terminals, fig. 85. The two currents are c— V/r, C=KOV, and their sum is C+c=v(^^ + Ke\ or v(^-'~^^\ V so that the two in parallel act like a resistance .. ,T^n * o g^c> C K Fig. 85. UV^Vosinnt, r '' K C + c = — Vl + r'^K'^n'', sin ini + tan-^ rKn\ r c = — sin nt, C = V^n sin ( n^ 4- ^ ) • ELECTRICAL EXAMPLES. 241 171. A circuit with resistance, self-induction and capacity (fig. 86) has the alternating voltage V= Fosin z*^ established at its ends ; what is the current ? V Answer, G= -—, and by Art. 167 R + Ld + ^^ r- ^'^'^ ^^ V * l + RK.e + LK.e^ {l-LK7i') + RKe C = : Sin nt +K — tan ^ - — ^ rr » • ^{l-LKnJ + R^KHi-' V 2 l^-LKnV The earnest student will take numbers and find out by much numerical trial what this means. If he were only to work this one example, he would discover that he now has a weapon to solve a problem in a few lines which some writers solve in a great many pages, using the most involved mathe- matical expressions, very troublesome, if not impossible, to follow in their physical meaning. Here the physical mean- ing of every step will soon become easy to understand. ' Numerical Exercise. Take Fo = 1414 volts, K=l micro- farad or 10-«, R = 100 ohms and n = 1000, and we find the following effects produced by altering L. We give the following table and the curves in fig. 87 : Fig. 86. A BCD shows how the current increases slowly at first from A where Z = as X is increased, and then it increases more rapidly, reaching a maximum when Z= 1 Henry and diminish- ing again exactly in the way in which it increased. EFG shows the lead which at X = 1 changes rather rapidly to a lag. The maximum current (when LKn^=l) is the same as if we had no condenser and no self-induction, as if we had a mere non-inductive resistance R. It is interesting to note in the electric analogue of Art. 160 that this LKn^=l is the relation which would hold between L, K and n (neglecting the P. 16 242 CALCULUS FOR ENGINEERS. small resistance term) if the condenser were sending surging currents through the circuit R, L, connecting its two coatings. /y, ill Henries. 0-1 0-2 0-8 0-4 0-5 0-6 0-7 0-8 0-9 0-95 0-975 100 1025 Effective Lead of Henries. Effective ciuTent, in current, in current, in amperes. degrees. amperes. 0-995 84-28 1-05 8-944 1110 83-67 11 7071 1-240 82-87 1-2 4-472 1-414 81-87 1-3 3-162 1-644 80-53 1-4 2425 1 1-961 78-67 1-5 1-961 1 2-425 75-97 1-6 1-644 1 3-162 71-57 1-7 1-414 ! 4-472 63-43 1-8 1-240 ! 7071 450 1-9 1-110 8-944 26-57 20 0-995 9-701 1403 2-5 0-665 1000 30 0499 9-701 -14-03 Lead of current, in degrees. -26-57 -45-0 -63-43 -71-57 -75-97 -78-67 -80-53 -81-87 -82-87 -83-67 -84-28 -86-18 -87-13 10 T~ ~~~ — 1 -f ■"s. \ '»J \ s' A \ K \ „5 8 ^ B i ^ p / V \ iu 3 / f V \ v^ / ^^.^ V Q- '*" — -« -= -2 '4 100 o •6 -8 1-0 1-2 1-4 1-6 1-8 2-0 22 2-4 2-6 2-8 S'O 8ELF-1NDUCI10N IN HENRIES. 1003 Fig. 87. Experimenting with numbers as we have done in this example is much cheaper and much more conclusive in preliminary work on a new problem, than experimenting with alternators, coils and condensers. IDLE CURRENTS OF TRANSFORMERS. 243 172. Even if a transformer has its secondary open there is power being wasted in hysteresis and eddy currents, and the effect is not very different from what we should have if there was no such internal loss, but if there was a small load on. Assume, however, no load. Find the effect of a con- denser shunt in supplying the ''Idle Current." The current to an unloaded transformer, consists of the fundamental term of the same frequency as the primary voltage, and other terms of three and five times the frequency, manufactured by the iron in a curious way. With these "other terms" the con- denser has nothing to do ; it cannot disguise them in any way ; the total current always contains them. We shall not speak of them, as they may be imagined added on, and this saves trouble, for if the fundamental term only is considered we may imagine the permeability constant ; that is, that the primary circuit of an unloaded transformer has simply a constant self-induction. In fact between the ends of a coil (fig. 88) which has resistance R and self-induction L, place a condenser of capacity K. Let the voltage between the terminals, be F= Fo sin nt. Let G be the instantaneous current through the coil and let c be the current through the condenser, then + c is the current supplied to the system. Fo sin nt ^C*c V { p o C-.c'- Fig. 88. Now G: R+LO and c = Fo sin ??^ -7- -^^ , or c = Kn Fo cos nt, 0^o=[^^^Ke)v.sinnt _ 1 + RKd -H LKd' ^_ l-LK n' + RK.e y R + LO R + Ld by our rule of Art. 167. It is quite easy to write out by Art. 167 the full value of 16—2 244 CALCULUS FOR ENGINEERS. + c, but as we are not concerned now with the lag or lead, we shall only state the amplitude. It is evidently "V R' + JJn^ and the effective value of (7 + c (what an ammeter would give as the measure of the cuiTent), is this divided by \/2. Observe that C + c is least when (Note that if L is in Henries and n = 27r x frequency (so that in practice n = about GOO), K is in farads. Now even a condenser of J microfarad or J x 10"" farad costs a number of pounds sterling. We have known an unpractical man to suggest the practical use of a condenser that would have cost millions of pounds sterling.) When this is the case, the effective current C + c, is R/JR^ + L'^n- times the effective value of G. The student ought to take a numerical case. Thus in an actual Hedgehog Transformer we have found iJ = 24 ohms, L = 6'23 Henries 7i = 509, corresponding to a frequency o^ about 81*1 per second. The effective voltage, or Fo-rV2 is 2400 volts. In fig. 89 we show the effective current cal- culated for various values of K. The current curve ABCD is a hyperbola which is undistinguishable except just at the •9 2-8 i: h 1 ■""■ — -% ^^ ~ / r- vo '8 \ / •6 1 s. 1 / D s A i / \ ',* i/ \ % 4 r \ ll / i -6 s. 1 / B^ s. ; / J \ A ... ... —' — ._ ... -10 •* -5 -6 -7 -8 -9 10 M CAPACiTV in MirROfARAOS •2 1-3 1-4 \-h Fig. 89. vertex, from two straight lines. The total current is a minimum when K — Lj(R? -^ LSi^)^ in this case 618 micro- ELECTRICAL EXAMI»LES. 245 farad ; and the effect of the condenser has been to diminish the total cun-ent in the ratio of the resistance to the im- pedance. It is interesting on the curve to note how the great lag changes very suddenly into a great lead. 173. If currents are steady and if points A and £ are connected by parallel resistances 7\, 7\, r^y if V is the voltage between A and B, and if the three currents are Ci, C2, C3, and if the whole current is (7; then _F _7 _F In fact the three parallel conductors act like a conductance 1 1 IN - + -■ + - Also if G is known, then C 111, i\ r^ Vs Now let there be a self-induction I and a condenser of capacity k in each branch, and we have exactly the same instantaneous formulae if, for any value of r, we insert The algebraic expressions are unwieldy, and hence nu- merical examples ought to be taken up by students. 174. Two circuits in parallel. They have resistances Vi and ?-2 and self-inductions l^ and I2 . how does a total current ASL^i^i&SliSiSi C divide itself between them ? "^ISlSUUlSlSlMSi If the current were a con- tinuous current, Cj (fig. 90) in the branch 7\ would be " p- ' qq 246 CALCULUS FOR ENGINEERS. TT V • r.+ LO ,^ Jtieiice it IS now Ci = -. rz~n ^. If a= Oo sin nt, then by Art. 167 '^'"''"Vcn + ny + i/i + Lyrr sin «^ + tan^ — tan^ -^^ In the last case suppose that for some instrumental purpose we wish to use a branch part of C, but with a lead. We arrange that tan~^ tan~^ — ^^ shall be 7*2 ?•! + ?•, equal to the required lead, and we use the current in the branch r^ for our purpose. 175. Condenser annulling effects of self-induction. When the voltage between points A and B follows any law w^hatever^ and we wish the current flowing into Y A and out at B to be exactly -^ , whatever V may be, and when we have already between A and B a coil of resistance R and self-induction L, show how to arrange a condenser shunt to effect our object. Connect A and 5 by a circuit containing a resistance ?•, self-induction I and condenser of capacity K, as in fig. 91. The total current is evidently V V r 1 k '^ I ^ . M-hLd^ ^j. 1 ' r^ld + j^^ AJUMilfiJUMiU R L or, bringing all to a common Fig. 91. denominator and arranging terms, it is 1 + e (vK + RK) -^e^-{lK + LK) y R+e (RrK -\-L)+&' {RIK -I- LrK) + LIKO' Observe that as V may be any function whatsoever of the time Ave camiot simplify this operator as we did those of ELECTRICAL EXAMPLES. 247 Art. 167. Now we wish the effect of the operation to be the same as p F! Equating and clearing of fractions we see that R + e (BrK+ B?K) + &" (IilK+ RLK) must be identical with jR + (9 {RrK + X) + (9^ {RIK + LrK) + LIKO'. As V may be any function whatsoever of the time, the operations are not equivalent unless LIK = ; that is, l—O) so there must be no self-induction in the condenser circuit, RrK + R^K = RrK + L ; that is, i(^= 4 ; RIK + RLK = RIK + LrK ; that is, R = r; so the resistance in the condenser circuit must be equal to that in the other. In fact we must shunt the circuit R + L^ by a 1 Ii condenser circuit R + ==3 where K = =^ . 176. If in the last case F= Fq sin nt, the operator may be simplified into l-K(l-\-L)n'-^K(r+R)e R-K(Rl + rL) n^ + (9 {RrK + L - LlKn'} ' , . R-K(Rl-\- rL) n^ RrK + L- LlKn' ^^^'* l-K{l-\-L)n' ~ K(r + R) then although the adjustment alters when frequency alters, we have for a fixed value of n the current flowing in at A and out at B proportional to V and without any lag. If R — r Y the current is equal to ^ . 177. To explain why the effective voltage is some- times less between the mains at a place D, fig. 92, than at a place B further away from the generator. This is usually due to a distributed capacity (about J microfarad per mile is usual) in the mains. We may consider a dis- tributed capacity later ; at present assume one condenser of 248 CALCULUS FOR ENGINEERS. capacity K between the mains at B. Let the non-inductive resistance, say of lamps, between 000000/ ? *^® mains at B be r. Let the jn resistance and self-induction of Kfrnlr the mains between D and B be If R and L. Let v be the voltage — X ^^ ^» ^^^ ^ the current from D Fig. 92. to B. The current into the condenser is v -^ ifn^^ ^^^'• V The current through ?- is - , so that O=(if0 + i). (1). The cZro/) of voltage between D and 5 is {R + X^) or (jK + LO) (kB + ^) t;, or 1^ 4- (rK + ^) (9 + LKe\ V. Now if V = t;o sin nt, the fZrop is The voltage at D is the drop plus v, or 1 b A-f ir7 square of effective voltage at D square of effective voltage at B = {l + ^-LKn^\{RK + ^^'n\ and there are values of the constants for which this is less than 1. As a numerical example take r= 10, R = '\, K=lx 10-^ w = 1000, and let L change from to 05, "01, '02, '03 &c. TWO CIRCUITS. 249 The student will find no difficulty in considering this problem when r + ^^ is used instead of r in (1) ; that is, when not merely lamps are being fed beyond B, but also coils having self-induction. Most general case of Two Coils. 178. Let there be a coil, fig. 93, with electromotive force E, resistance i2, self-induction L, capacity K\ and another with e, r, RLC m ric I, h Let the mutual induction f § § I be m. Em § fe |o« Using then R for R -\- Ld ■\- -^^, I | I r for r -{- the equations are and r for r -1-/(9 4--^^^. ^^^- '^^• E = RG + m6c,] e = meC + rc j ^^' Notice hoAV important it is for a student not to trouble himself about the signs of and c &c. until he obtains his answers. From these we find Re — mOE ,^. '=w^.^^ ^^>' ^ rE-mOe ,„. ^^Rr-m^e' ^^^- We can now substitute for i2, r, E and e their values and obtain the currents. Observe that E may be a voltage established at the terminals of part of a circuit, and then R is only between these terminals. The following exercises are examples of this general case. There are a gi-eat. many other examples in which mutual induction comes in. 179. Let two circuits (fig. 94), with self-inductions, 250 CALCULUS FOR ENGINEERS. be in parallel^ with mutual induction m between them. umsmsmsu 1 r c Fig. 94. (1) At their terminals let v = Vo sin nt ; (2) of last exercise becomes c = fs iTTv. v- Or, chanerinof R into R-\-L6 and r into r 4- 16, R + {L-m)e ^""[Rr- (LI - m'') n^) + (Zr + IR) 6 ^* (2) How does a current A sin nt divide itself between two such circuits ? Since -p^ = ^ we can find at once -^ G r-mO G+c A ^ A R + {L-m)e and >. -— . Answer : c = Tr, x ^Vr r—7i — r^ operatmsf G + c {R + r)-{-{L+l-2m)0 ^ ^ on -4 sin nt. We think it is hardly necessary to work such examples out more fully for students, as, to complete the answers they have the rule in Art. 167. 180. In the above example, imagine each of the circuits to have also a mutual induction with the com- pound circuit. We shall use new letters as shown in fig. 95. ^ r ? c,lir, viznnnfg W ^ c+Co=C c,Lr„ Fig. 95. If V is the potential difference between the ends of the two circuits which are in parallel. Using r, /jl and m to stand for r 4- W, fiO and mO, V = ?-iCi + fiCi + nh (ci + Co). ROTATING FIELD. , 251 Hence the equations are V = (7\ + 71h) Ci + (/X + nil) Ca, V = (ft + m^) Ci + (r., + mg) c.,, ^. _ (r, + m,)-(/^ + mO ^ ' (ri + mi)(r2 + 7?i2)-(/A + mi)(/A + m.2) *"^ ^' with a similar expression for Cg. Also a total current C divides itself in the following way ^ 7-1+ ro — 2//- ^ '^' If we write these out in full, we have exceedingly pretty problems to study, and our study might perhaps be helped by taking numerical values for some of the quantities. If we care to introduce condensers, we need only write ^ + ^^ + r^ with proper affixes, instead of each ?•; fi becomes fiO and m becomes mO. To what extent may we make some of the ma negative? I have not considered this fully, but some student ought to try various values and afterwards verify his results with actual coils. Taking (2) without condensers _ f\+ (h+ fn-2 — 1^ — '^^h) p 181. Rotating Field. Current passes through a coil wound on a non-conducting bobbin ; the same current passes through a coil wound on a conducting bobbin. The coils are at right angles and have no mutual induction ; find the nature of the fields which are at right angles at the centre of the two bobbins. Let the numbers of turns be rii and 71.2 • Instead of a conducting bobbin imagine a coil closed on itself of resistance 7\, and ng turns and current c. For simplicity, suppose all three mean radii the same, and the coils 7i2 and % well intermingled. One field F^ is propor- tional to iiiG per square centimetre, call it iiiC The other F, is proportional, or let us say equal to,??2(7 + ^gC per square cm. Take the total induction / through each of the coils as 252 CALCULUS FOR ENGINEERS. proportional to the intensity of field at its centre, say b times. Then for the third coil, we have = nc + risdl or = rgC + bn-iO {n.C -\- n^c), bi^n-iOC so that — c = and hence F., = v..G rs-hbn,^e' \-^bn^^e n + bn^^e' If then G= Cq sin qty Fi = 71jGq sin qt, ■^■2= 2* = — , sin 7^-tan-i q): Art. 126, shows the nature of the rotating field. We can assure the student that he may obtain an excellent j-otating field in this way. It is evident that bn-i really means the self-induction of the bn^ . . third coil, and — - means its time constant. A coil of one turn, — that is, a conducting bobbin, will have a greater time constant than any coil of more than one turn wound in the same volume. It is evident that if the bobbin is made large enough in dimensions, we can for a given frequency have an almost uniform and uniformly rotating field by making n^^b— q=nj. This is one of a great number of examples which we might give to illustrate the usefulness of our sign of opera- tion 6. 182. In Art. 1 78 let E= V the primary voltage of a trans- former, the primary circuit having internal resistance B and self-induction L ; let the secondary have no independent E.M.F. in it; let its internal resistance be 7\ and self-induction I and let it have an outside non-inductive resistance p, of lamps. Let the voltage at the secondary terminals be i^ = cp. TRANSFORMERS. 253 Then in (1), (2) and (3) Art. 178, let F^V, e=0; instead of R use R-\-Ld. Instead of r use r + Id which is really i\-{- p + W, ....(1). (7 = Br + (Rl -trL)e + {LI - m"") ^ (r-^W)V (2). Rr + (Rl + rL)e+(Ll-m^)d' "" Note that the second equation of (1) Art. 178 is 0==meG + (r + W)c (2)* I. From (2)*, if C = Co6**, c = r + la — c 1) ma r + la If r is small compared with la — c u m I ' c 1 "o 1 Rss r V •o- I 1 F^ Fi^. 96. II. If C = Cq sin qt, again using (2)* Wr^-\'Pq'' V 2 rJ 1 Hence effective c effective G V^ + £' Except when the load on the secondary is less than it ever is usually in practice, r is insignificant compared with Iq (a practical example ought to be tried to test this) and we may take C=Gq sin qt. c= Gg-j- sin (qt — tt), or -c C m T .(3> 254 CALCULUS FOR ENGINEERS. It may become important in some application to re- member that the ratio of the instantaneous values of — c and G is that of r sin qt + Iq cos qt to mq cos qt, and this sometimes is oo . t Returning to (1). Let LI — m^ (this is the condition called no magnetic leakage) and let Rr be negligible. In any practical case, Ri' is found to be negligible even when r is so great as to be several times the resistance of only one lamp, t 'f'^- -^ = B^Z W' SO that — c is a faithful copy of F as a function of the time. C is so also. If N and n are the numbei*s of windings of the two coils on the same iron, m : L : l^Nn : N^ : 71^ (5), N so that — c = ^ (6); that is, the secondary current is the same as if the trans- formed voltage ( — 7^ ^ ) acted in the secondary circuit, but as if an extra resistance were introduced which I call the transformed primary resistance IR-^). If the volumes of the two coils were equal, and if the volumes of their insulations were equal, R ^^ would be equal to Vy the internal resistance of the secondary. Assume it so and then -^=2;tt7 ^'^^ DROP IN VOLTAGE. 255 JSf also /)c or v = — (8). P As Vi is usually small compared with p, and — * is called the drop in the secondary voltage due to load. ^2 IP As — = P, the power given to lamps ; - = -^ and the 2r fractional drop is — ^ P and is proportional to the Power, or to the number of lamps which are in circuit. 183. The above results may be obtained in another way. Let / be the induction, and let it be the same in both coils. Here again we assume no magnetic leakage, V^RC-^NOI (1), O^rc + ndl (2). Multiplying each equation by its N or n and dividing by its ^ or r and adding NV . (m n\ ^^ -^ = ^ + U+-)^^ (•^)' where A — NC + nc, and is called the current turns. Now when we know the nature of the magnetic circuit, that is, the nature of the iron and its section, a square centi- metres, and the average length X centimetres of the magnetic circuit, we know the relationship between A and /. I have gone carefully into this matter and find that whatever be the nature of the periodic law for A, so long as the frequency and sizes of iron &c. are what they usually are in practice, the term A is utterly insignificant in (3). Reject- ing it we find '^ SZhk '^^--^^^ '"' "^'"^ ""^^' •••^*^- 256 CALCULUS FOR ENGINEERS. Thus, in a certain 15 00- watt transformer, J? =27 ohms, N = 460 turns, internal part of r = '067 ohms, ii = 24 turns, effective V is 2000 volts or F= 2828 sin qt where q = 600 say, a= 360, \ = 31. When there is no load r = x ; on full load r = nearly 7 ohms. We have called R -^^ the transformed resistance of the . / 24 \2 primary. It is in this case 27 ( .^^1 or '073 ohms. If the primary and secondary volumes of copperthad been equal, no doubt this would have been more nearly identical with '067, the internal resistance of the secondary. ^r,.— or — '- is the fractional drop in / from what it is at no load. When at full load r* = 7 ohms the fractional drop is greatest, and it is only 1 per cent, in this case. Because of its smallness we took a fractional increase of the denominator as the same fractional diminution of the numerator of (4). Consider / at its greatest, that is, at no load ; -^V is the 2828 d integral of F or — -^^^ cos 600^. So that the amplitude . 2828 ^^^ ° ^^600x460 • Multiply this, the maximum value of / in Webers, by 10* to obtain c. a s. units, and divide by a = 360, and we find 2856 c. G. s. units of induction per sq. cm. in the iron, as the maximum in this transformer every cycle. 61 being ^ F / f 1 + vfa — ) > we have from (2) the same value of — re that we had before in (6) of Art. 183. 184. Returning to (7) of Art. 182. Let us suppose that there is magnetic leakage and that Vx is really r^ + VO. If one really goes into the matter it will be seen that this is what we mean by magnetic leakage. Then we must divide by p + 2n + 21' By instead of /> -f 2ri. In fact our old answer must be divided by 1+ 2r . MAGNETIC LEAKAGE. 257 or neglecting 2?'i as not very important in this connection ; 2^ our old answer must be divided by 1 -\ 6. This means . . P that the old amplitude of v must be divided by /- 1 H ^ or 1 H ^,— nearly, ^ P" if the leakage is small, and there is a lag produced of the amount tan-i — ^ . We must remember that a is 27r/* if f is the frequency. We saw that P, the power given to the lamps, is inversely proportional to p, so we see that the Aractional 2r P drop due to mere resistances is — \- , the fractional drop due to magnetic leakage is la^pF", and the lag due to magnetic leakage is an angle of afP radians where a is a constant which depends upon the amount of leakage, and / is the frequency. 185. Only one thing need now be commented upon in regard to Transformers. If Fis known, it has only to be inte- grated and divided by N to get I. Multiply by 10^ and divide by the cross-section of the iron in square centimetres, and we know how /3, the induction per sq. cm. in the iron, alters with the time. The experimentally obtained /3, H curve for the iron enables us to find for every value of ^ the corresponding value of H,w[\d H multiplied by the length of the magnetic cuTCUit in the iron gives the gaussage, or y^ x the ampere turns A. Hence the law of variation of A is known, and if there is no secondary current, we have the law of the primary current in an unloaded transformer or choking coil. This last statement is, however, inaccurate, as one never has a truly unloaded transformer, even when what is usually called the secondary, has an infinite resistance. 186. Sir W. Grove's Problem ; the effect of a condenser in the primary of an induction coil when using alternating currents. ABB, fig. 97, is the primary with electromotive force E— Ed sin nt, resistance R and self-induction L. BA is h P. IT 25S CALCULUS FOR ENGINEERS. condenser of capacity K, and ?' is a non-inductive resistance in parallel with the condenser. G qq^q'q^j^ -? the current in the primary, has an S JTI amplitude Co, say. ® Kli1j|r rpjjg condenser has the resist- *^" It is quite easy to write out the value of Cq when r and K have any finite values*. But for our problem we suppose r=0 or else r=oo . When r = 0, the resistance \^R-\-L6 and the current is EI{R + Z^), n^-—^ (1). When r=oc , the resistance is R + Le+-^^ or j^ , or ^^ j^ , by Art. 167, A rr.- £Jo'K'n^ Ej . \Kn J Now (2) is greatei' than (1) if 2KLn^ is greater than 1, so that the primary current is increased by a condenser of capacity greater than o;-— g. Again, there is a maximum current if iT = y— ^ ; in this case the condenser completely destroys the self-induction of the primary. * When both r and K have finite values, the parallel resistances between 2S and A, together form a resistance rl{l + rKd), and the whole resistance of the circuit for C is jR + L^ + - tptSO that 1 + rKd (1 + rKd)EQ sin lit {R + r- LrKii') + {RrK+L) d 02 = - " ~ (E + r - LrKn^f + (RrK-^LYii^ ' and the lag of C is easily written. ALTERNATORS IN SERIES. 259 187. Alternators in series. Let their e.:m.f. be ei and e.2 and let C be the current through both. The powers exerted are efi and 020. Now if ei=E sin (nt+a) and e.2=Esm (nt — a),t ei +e.j=2j&cos a . sin?^^. If I is the self-induction of each machine, r its internal resistance, and if 2R is the outside resistance and if Pi and P2 are the average powers developed in the two machines, ^ 2^cosa. sin?i^ Ecosa • / . , , In \ ^ = -zTTTi — ^ ^tt = ,-' sm nt — tan~^ v, 2;^ + 2r-f-2P V(P4-r)2 + ^'^'' V i^ + r/ = ilf cos a sin {nt — e) say, Pi = p/^ cos a . cos (a + e), Po = ^Jlfii' cos a . cos (a — e). Hence P2 is greater than Pi, and machine 2 is retarded whilst machine 1 is accelerated ; hence a increases until a = - , and when this is the case, cos a = 0, so that Pi = 0, P2 = and the machines neutralize each other, producing no current in the circuit. Alternators cannot therefore be used in series unless their shafts are fastened together. 188. As we very often have to deal with circuits in parallel we give the following general formula ; if the electro- motive forces ^1, ^2 and e^, fig. 98, are constant, v^e^ — c^Vi = e.2 - 0.^2 ^e-i-c^n ( 1 ), and Ci + C2 + C3 = (2). Given the values of e^, e._,, e^ and 7\, i\, r^ we easily find the currents, because , = (?i + ?. + ^V(i + i + l) (3). Fig. 98. 17—2 200 CALCULUS FOR ENGINEERS. Now if the es are not constant, we must use ?'i + liO, &c., instead of mere resistances. 189. Alternators in Parallel. Let two alternatoi-s, each of resistance r and self-induction I, and with electromotive forces, ei = E sin (iit + a), and e^ = E sin (nt — a), be coupled up in parallel to a non-inductive circuit of resistance R. What average electrical power will each of them create, and will they tend to synchronism ? If Ci and e.2 were con- stant or if I were 0, then v = ei — CiV = e., — cr = (ci + c^) R. And hence ^^ = 2r^H^+3-^-^ fcH^+S"^^ ^'~'2rR-¥r' Now alter r to ?• 4- 16, because the e's are alternating. The student will see that we may write 62 =61 (a — bO), e^ = 62 (a -H hO), where a^ -h 6W =1, a = cos 2a, 6/1 = sin 2a. Then I hi-^)-'Ki^') with a similar expression for Ca in terms of e, except that 6 is made negative. If we write out (1) by the rule of Art. 167, there is some such simplification as this : — Let tan <f) = — ^— - — =^ and tan ^^ = ^— ^ ^ 2Rr + r^ — Pn^ ^ R+r — aR tan ylr^ = t,-; d • ^ R + r-aR Then c, = if sin (ni -f a — <^ -}- i|ri), C2 = il/ sin (nt — a — <j) -{- yfr^, the angles <^, yjry, s^^ being all supposed to be between and + 90^ ALTERNATORS IN PARALLEL. 261 The average powers are P„_ = ME cos ((f>-^|r.,), where M'= ^ .^^.. l^^""^ , ^J' E^. Po COS (<l>-ylr^y It' 11= oo we see that , , In ^ . sin a , , tan 6 = — , tan -v/r, = , = — tan vr... ^ r ^ 1 — cos a ^ " Hence P,^ cos(^-^ ) P. COS(«/)+>^i) In this case it is obvious that Pi is greater than P2. P The author has not examined the general expression for p- with great care, himself, but men who have studied it say that it shows P, to be always greater than Po. Students would do well to take values for i\ I, R and a and try for themselves. If Pi is always greater, it means that the leading alternator has more work to do, and it will tend to go slower, and the lagging one tends to go more quickly, so that there is a tendency to synchronism and hence alternators will work in Parallel. 190. Struts. Consider a strut perfectly prismatic, of homogeneous material, its own weight neglected, the resultant force F at each end passing through the centre of each end. Let ACB, fig. 99, show the centre line of the bent strut. Let FQ = yhe the deflection at P where OQ = x. Let OA=OB = I. y is supposed everywhere small in comparison with the length 21 of the strut. Fy Fy is the bending moment at P, and ^ is the curva- ture there, if E is Young s modulus for the material and / is 262 CALCULUS FOR ENGINEERS. the least moment of inertia of the cross section everywhere, P, about a line through the centre of area of the sec- tion. Then as in Art. 60 the curvature bemg - t~* we have II--^^ (1) Now if the student tries he will find that, as in the many cases where we have had and again shall have this equation, (see Art. 119) 7/ = a cos X \lm <^) satisfies (1) whatever value a may have. When a: = we see that y — a, so that the meaning of a is known to us ; it is the deflection of the strut in the middle. The student is instructed to follow carefully 99. the next step in our argument. When A" = /, 2/ = 0. Hence FY a cos ^ a/ VTr = (3). * Notice that when we choose to call -~ the curvature of a curve, if the expression to which we put it equal is essentially positive, we must give such a sign to -r^ as will make it also positive. Now if the slope of the curve of fig. 99 be studied as we studied the curve of fig. 6, we shall find that -r-^ is negative from a; = to a: = O^ , and as y is positive so that ^ is positive, «re must use - —-^ on the right-hand side. It will be found that the comjilete (see Arts, loi and 159) solution of any such equation as (1) which may be written ia y = ^ ^^^ nx + B sin nx where A and B are arbitrary constants. A and B are chosen to suit the particular problem which is being solved. In the present case it is evident that, as ?/ = when x = l and also when x= - I, 0=iA cos 111 + B sin 111, = A cos 111 - B sin 7il, so that B ia 0. STRENGTH OF STRUTS. 263 Now how can this be true ? Either a = 0, or the cosine is 0. Hence, if hending occurs, so that a has some value, the cosine must he 0. Now if the cosine of an angle is the angle must be ^ or -^ or -^ , &c. It is easy to see why we confine our attention to » *. Hence the condition that bending occurs is is the load which will produce bending. This is called Euler's law of strength. The load given by (4) will produce either very little or very much bending equally well. It is very easy to extend the theory to struts fixed at both ends or fixed at one end and hinged at the other. For equilibrium under exceedingly great bending, the equation (1) is not correct, as -r~ is not equal to the curva- ture when the curvature is great, but for all engineering purposes it may be taken as correct. 191. We may take it that F given by (4), is the load which will break a strut if it breaks by bending. If / is the compressive stress which will produce rupture and A is the area of cross section, the load /J. will break the strut by direct crushing, and we must take the smaller of the two answers. In fact we see that /A is to be taken for short struts or for struts which are artificially*!" protected from bending, and (4) is to be taken for long struts. Now, even when great care is taken, we find that struts are neither quite straight nor homogeneous, nor is it easy to load them in the specified manner. Consequently when loaded, they deflect with even small loads^ and the}^ break with loads less than either /A or that given by (4). * This gives the least value of W. The meaning of the other cases is that y is assumed to be one or more times between x = and x = l, so that the strut has points of inflexion. t This casual remark contains the whole theory of struts such as are used in the Forth Bridge. 264 CALCULUS FOR ENGINEERS. Curiously enough, however, when struts of the same section but of different lengths are tested, their breaking loads follow, with a rough approximation to accuracy, some rule as to length. Let us assume that as F=fA for short struts, and what is given in (4) for long struts, then the formula F = -a .(5) may be taken to be true for struts of all lengths, because it is true both for short and for long ones. For if I is gieat we may neglect 1 in the denominator, and our (5) is really (4) ; again, when / is small, we may regard the denominator as only 1 and so we have W=fA. We get in this way an empirical formula which is found to be fairly right for all struts. To put it in its usual form, let I — Ak^, k being the least radius of gyration of the section about a line through its centre of gravity, then I± I- •(6), where a is ^//Ett^ or rather / and a are numbers best determined from actual experiments on struts. If F does not act truly at the centre of each end, but at the distance h from it, our end condition is that y = h when 00=^1. This will be found to explain why struts not perfectly ti-uly loaded, break with a load less than what is given in (4). Students who wish to pursue the subject are referred to pages 464 and 513 of the Engineer for 1886, where initial want of straightness of struts is also taken account of 192. Struts with Lateral Loads. We had better confine our attention to a strut with hinged ends. If the lateral loads are such that by themselves and the necessary lateral supporting forces, they produce a bending moment which we shall call ^ (.r), then (1) Art. 190 becomes Thus let a strut be uniformly loaded laterally, as by centrifugal force or its own weight, and then <j> {a:)~^ w' {l — x)^ if w' is the lateral load per unit length. STRUTS LATERALLY LOADED. 265 We find it slightly more convenient to take <f) (x) = ^Wl cos ^x where W is the total lateral load ; this is not a very different law. Hence dh/ F , Wl n ^ d^^+m^-^^W'''2i''=^ (1)- We find here that y="-r3 cos- r (2). Observe that when i^=0 this gives the shape of the beam. The deflexion in the middle is .. = -i?^ (3), and the greatest bending moment fi is M = ^yi+iTf/,or -i--^7(^^--) (^)- If TF=0 and if fx has any value whatever, the denominator of (4) must be 0. Putting it equal to 0, we have Euler's law for the strength of struts which are so long that they bend before breaking. If Euler's value of F be called U, or U=ETii^/4l^y (4) becomes >^=i^''^T&F • '^)- If Zc is the greatest distance of a point in the section from the neutral line on the compressive side, or if I-^Zc=^Z, the least strength modulus of the section, and A is the area of cross section, and if / is the maximum compressive stress to which any part of the strut is subjected, n F Z'^ A- Using this expression, if /3 stands for -r (that is Euler's Breaking load per square inch of section), and if w stands for -r (the true break- A ing load per inch of section), then (^-?)(^-f)=:^ («)• This formula is not difficult to remember. From it lo may be found. Example. Every point in an iron or steel coupling rod^ of length 26 inches, moves aljout a radius of r inches. Its section is rectangular, 266 CALCULUS FOR ENGINEERS. d inches in the plane of the motion and b at right angles to this. We may take W=lbdrn^'7-62940y in pounds, where 7i = number of revolu- tions per minute. Take it as a strut hinged at both ends, for both directions in which it may break. 1st. For bending in the direction in which there is no centrifugal force where/ is ^^^ 12' Euler's rule gives 48^2 .(7). Now we shall take this as the endlong load which will cause the strut to break in the other way of bending also, so as to have it equally ready to break both ways. 2nd. Bending in the direction in which bending is helped by centrifugal force. Our w of (6) is the above quantity of (7) divided by bdf or taking iS'=3xlO", b^ '?t'=6'17x^xlO«. Taking the proof stress / for the steel used, as 20000 lb. per sq. inch (remember to keep^ low, because of reversals of stress), and recollecting the fact that / in this other direction is r-s^ , we have (6) becoming b' 8-4 X 108 1 - 308 12 62 d^ )- nH^r-T-d (8). Thus for example, if 6= 1, Z= 30, r= 12, the following depths c? inches, ai'e right for the following speeds. It is well to assume d and calculate n from (8). <^ 1 I 1-5 2 2-5 3 4 6 0|205 277 327 368 437 545 Exercise. A round bar of steel, 1 inch in diameter, 8 feet long, or Z = 48 inches. Take i^=1500 lb. Show that an endlong load only sufficient of itself to produce a stress of 1910 lb. per sq. in., and a bending moment which by itself would only produce a stress of 816 lb. per sq. inch ; if both act together, produce a stress of 23190 lb. per sq. inch. For other interesting examples the student is refeiTed to The Philosophical Magazine for March, 1892. t CHAPTEE III ACADEMIC EXERCISES. 193. In Chapter I. we dealt only with the differentia- tion and integration of ic" and in Chapter II. with €«^ and sin aw, and unless one is really intending to make a rather complete study of the Calculus, nothing further is needed. Our knowledge of those three functions is sufficient for nearly every practical engineering purpose. It will be found, indeed, that many of the examples given in this chapter might have been given in Chapters I. and II. For the differen- tiation and integration of functions in general, we should have preferred to ask students to read the regular treatises, skipping difficult parts in a first reading and afterwards returning to these parts when there is the knowledge which it is necessary to have before one can understand them. If a student has no tutor to mark these difficult parts for him, he will find them out for himself by trial. By means of a few rules it is easy to become able to differentiate any algebraic function of cc, and in spite of our wish that students should read the regular treatises we are weak enough to give these rules here. They are mainly used to enable schoolboys to prepare for examinations and attain facility in differentiation. These boys so seldom learn more of this wonderful subject, and so rapidly lose the facility in question, because they never have learnt really what -^ means, that we are apt with beginners to discourage much practice in differentiation, and so err, possibly, as much as the older teachers, but in another way. If, however, a man sees clearly the object of his work, he ought to try to gain this facility in differentiation and to retain it. The knack is easily learnt, and in working the examples he will, at all 268 CALCULUS FOR ENGINEERS. events, become more expert in manipulating algebraic and trigonometric expressions, and such expertness is all-important to the practical man. In Chapters I. and II. we thought it very important that students should graph several illustrations of y = cwJ", y = ae^, y = a sin {hcc + c). So also they ought to graph any new function which comes before them. But we would again warn them that it is better to have graphed a few very thoroughly, than to have a hazy belief that one has graphed a great number. The engineer discovers himself and his own powers in the first problem of any kind that he is allowed to work out completely by himself. The nature of the problem does not matter; what does matter is the thoroughness with which he works it out. Graph y = tan ax. We assume that the student has already graphed y = «€** sin nx. 194. If y =f(x), so that when a particular value of on is chosen, y may be calculated ; let a new value of x be taken, x+Bx, this enables us to calculate the corresponding value ofy, or y + By^f(x-{-Bx). Now subtract and divide by Bx, and we find By_ f{x-^Bx)--f(x ) Bx~ Bx ' ^^^• We are here indicating, generally, what we must do with any function, and what we have already done with our famous three, and we see that our definition of dy/dx is, the limiting value reached by (1) as Bx is juade smaller and smaller without limit. 195. It is evident from this definition that the differential coefficient of af(x), is a multiplied by the differential co- efficient of fix), and it is easy to show that the differential coefficient of a sum of functions is equal to the sum of the differential coefficients of each. In some of the examples of Chapter I. we have assumed this without proof DIFFERENTIATION OF A PRODUCT. 269 We may put the proof in this form : — Let y=u + v + w, the sum of three given functions of x. Let X become os-{- hx,f\et u become ii + Su, v become v-{-Sv, and w become w -f ^iv. It results that if ij becomes y + 5y, then 8y = Bit -^ hv + Biu, - By Bit Bv Bw and 5^^ = ^ + ?- + F~ . ox ox ox ox , . ,, ,. . dy du dv dw and m the limit -r^ = ^- + t" + -? • ax ax ax ax 196. Differential Coefficient of a Product of two Func- tions. Let y = uv where u and v are functions of x. When x becomes x + Bx, let y-\-By = {ii + Bi()(v + Bv) — uv •{- ^l . Bv + v . Bu ■{■ Bu . Bv. Subtracting we find By = a . Bv -^^ v . Bu -{■ Bu . Bv> , By B^ Bu Bu ^ and f-^U'^-hv K- + ^'Bv. ox ox ox ox We now imagine Bx, and in consequence (for this is always assumed in our work) Bti^ Bv and By to get smaller du and smaller without limit. Consequently, whatever -r*- may be, -r- . Bv must in the limit become 0, and hence ax dy_ dv da dx dx dx' The student must translate this for himself into ordinary language. It is in the same way easy to show, by writyig uvto as uv X w, that if t/ = uvw then dy dw , du dv -r- = wv -7— + viv -,- + wit -J- " ax ax ax ax Illustrations. If y = 10a;'' then, directly, ^ = lOa^. But we may write it y = 5^' x 2x^. 270 CALCULUS FOR ENGINEERS. Our new rule gives ^ = 00^ (8^) + 20,-^ (15*'0 = ^O*-" + SOaf = 70af. The student ought to manufacture other examples for himself. 197. Differential Coefficient of a Quotient. Let ?/ = - when u and v are functions of x. Then y + oy= — —^ . Subtract and we find ^ __ w 4- Sw u _v .8u — to.^v "^ " v+Bv V v'^+v .Sv ' Bu Bv hi/ __ 8x Bx Bx" v'^ + v .Sv Letting Bx get smaller and smaller without limit, v . Bv becomes 0, and we have du dv dy dx dx dx v^ Here again the student must translate the rule into ordinary language, and he must get very well used indeed to du the idea that it is ^ -v" which comes first : — ax Denominator into differential coefficient of nume- rator^ minus numerator into differential coefficient of- denominator, divided by denominator squared. A few illustrations ought to be manufactured. Thus 24a;^ . ,i ^ . t dy .^ ^ y = -^— IS really Sa?^, and -j- — ¥)oi^. ox ax By our rule, ■£= ^^ ^—^ = 40^. d^^dy dz^ 271 dx dz ' dx' The student ouffht to work a few like y = -^r— - = 5x~^ or again y = — ^ = — 1^~^, and verify for himself. 198. If y is given as a function of z, and ^ is given as a function of x, then it is easy to express y as a function of oj. Thus if 2/ = 6 log (az^ 4- g) and ^ = c + c?^ + sin ex, then 2/ = 6 log {a (c-\- dx-\- sin eaj)^ + ^}. Now under such circumstances, that is, y =/{z) and 2 = F (x), if for X we take x + 5^, and so calculate z + S^, and with this same z -\-Bz we calculate y + %, then we can say that our 8y is in consequence of our Bx, and Bx Sz 8x ^ ^' This is evidently true because we have taken care that the two things written as Bz shall be the same thing. On this supposition, that the two things written as Sz remain the same however small they become, we see that the rule (1) is true even when Sx is made smaller and smaller without limit, and as we suppose that Bz also gets smaller and smaller without limit, dx"dz ' dx ^ ^* This is such an enormously important proposition that a student ought not to rest satisfied until he sees very clearly that it is the case. For we must observe that the symbol dz cannot stand by itself ; we know nothing of dz by itself; we only know of the complete symbols dy/'dz or dzjdx. We are very unwilling to plague a beginner, but it would be fatal to his progress to pass over this matter too easily. Therefore he ought to illustrate the law by a few examples. Thus let y = az^ and z=hx\ As -^ = Saz^, -y- = 2hx, we have ^ dz dx -r ' -T- = Qabz^x or Qab^x^. But by substitution, y = ab^af, and if we differentiate directly we get the same answer. A student ought to manufacture many examples for himsel£ .(3). 272 CALCULUS FOR ENGINEERS. An ingenious student might illustrate (2) by means of three curves, one connecting z and x, the other connecting z and y and a third produced by measurements from the other two, and by means of them show that for any value of x the slope of the y, X curve is equal to the product of the slopes of the other two. But in truth the method is too complex to be instructive. By an extension of our reasoning we see that dy _ dy dw du dv dx dw ' du ' dv ' dx 199. It is a much easier matter to prove that dv dx :t^x-- = i (4), dx dy ^ ^ by drawing a curve, because it is easy to see that -y- is the cotangent of the angle of which -~ is the tangent. Otherwise: — if by increasing i» by &c we obtain the increment By of y, and if we take this same By, so foimd, we ought to be able to find by calculation the very same Bx with which we started. Hence By Bx ^ _. On this proviso, however small Bx may become, (5) is true and therefore (4) is true. 200. To illustrate (2). If a gas engine indicator diagram is taken, it is easy to find from it by applying Art. 57, a diagram for h, the rate at which the stuff shows that it is receiving heat in foot-pounds per unit change of volume, on the assumption that it is a perfect gas receiving heat from some furnace. (In truth it is its own furnace; the heat comes from its own chemical energy.) Just as dW pressure is -j- , the rate at which work is done per unit change of volume : so ^ is -=- . Observe that h is in the dv same units as p, and to draw the curve for h it is not necessary to pay any attention to the scales for either p or v. They ILLUSTRATIONS. 273 may be measured as inches on the diagram. We know of no better exercise to bring home to a student the meaning of a differential coefficient, than to take the indicator diagram, enlarge it greatly, make out a table of many values of j> and V, and find approximately ^ for each value of v. This is better than by drawing tangents to the curve. Using rlTT these values, and having found the values of A or -p at every place, suppose we want to find the rate per second at which the stuff is receiving heat. If t represents time, -rr — —r- . -77 , and hence it is only necessary to multiply h , dv dv As -7T is represented by the velocity of the piston, and as the motion of the piston is, as a first approximation, simple harmonic, we describe a semicircle upon the distance on the diagram which represents the stroke, and the ordinates of the di) semicircle represent -j- . We have therefore to multiply every value of h by the corresponding ordinate of the semi- circle, and we obtain, to a scale easily determined, the diagram which shows at every instant -^ . Havins: seen that -^^ = -^ . -^ and that — = 1^-7-, we ° dx dv dx dx dy shall often treat dx or dy as if it were a real algebraic quantity, recollecting however that although dy or dx may appear by itself in an expression, it is usually only for facility in writing that it so appears ; thus the expression M.dx + N.dy = (1), may appear, where M and N are functions of x and y ; but this really stands for M+N-^ = (2). Again, if y = ax^, we may write dy = 2ax . dx (3), P. 18 274 CALCULUS FOR ENGINEERS. but this only stands for ~~ = 2ax (4). Our main reason for doing it is this, that if we wish to integrate (3) we have only to write in the symbol /, whereas, if we wish to integrate (4) we must describe the process in words, and yet the two processes are really the same. We have already used dx and dy in this way in Chap. I. Mere mathematical illustrations of Art. 198 may be manu- factured in plenty. But satisfying food for thought on the subject, is not so easy to find. The law is true; it is not difficult to prove it ; but the student needs to make the law part of his mental machinery, and this needs more than academic 'proof.' Let us now use these principles. 201. Let y = log x ; this statement is exactly the same as X = e'^ Hence -,- — e^ — x and -f- = - . We used the idea ay ax x that the integral of x~^ is logd?, in Chap. I., without proof. It is the exceptional case of the integration of «". 202. If the differential coefficient of sin a? is known to be cos X, find the differential coefficient of sin ax, y = sin ax = sin u\iu = ax, dy J du -Y- = cos u and -r = a, du dx xi X ^y dii du so that -!i r= -^ . — = cos uxa — a cos ax. dx du dx Find the differential coefficient of 3/ = cos ax, knowing that the differential coefficient of sin x is cos a?, y = cos ax = Bm.lax + -^\ — sin u say, where ;7- = «, dy dy du / 7r\ . ^ — ~l '~^ — ^^s u X a = a cos {ax-{--^j — — asinaa. FUNDAMENTAL CASES. 275 203. Let 2/ = log(x+a). Assume x-\-a= ti, or y = loer u, then -,- = 1 and -j^ = - , dy _dy dii _\ _ 1 rfa; dii ' dx u x+a' 204. y = tan x. Treat this as a quotient, y = , cos X dy __ cos a? . cos a; — sin a; (— sin x) _ 1 (ia? ~ cos'^ X cos** a^ ' The student ought to work this example in a direct manner also. 205. y = cot X. We now have choice of many methods. cos X Treat this as a quotient, y = -. , ^ ^ sm X dy _ sin x (— sin x) — cos x (cos x) _ 1 c^a; ~ sin^ x sin^ a? ' or we might have treated it in this way, y = u~^ if ii = taniJ7, ir'' X -T- = — i^~- X COS^'fl? 1 1 cosec^ ic. dx dx cos* fl? 11 1 tan^a; cos'^a? sin^a? 206. Let y = sin aar^, say y = sin t^, and it = aa-^. Then -r- = 2aa7, and -=^ = cos u, du so that ~ = cos ^6 X 2ax = 2aa7 cos ax^, dx Let 3/ = 6*""^*, say y = e", and w = a sin a?, so that dy „ c?u ^ = e", J- = a cos a?, so that ~- = e^a cos x, or a cos a? . e**'^"* 18—2 276 CALCULUS FOR ENGINEERS.- 207. y — sec x. We may either treat this as a quotient, or as follows ; y = (cos ijc)~^ = u~^ if u = cos x. du . dy dy du ,/ • x -Y- == — sin a?, j^ = -/ . J— = — u-^ (— sm a?) ax ax du ax ^ ' sin 33 sec X . tan x. cos'*;?? 208. In Art. 1 1 the equation to the cycloid was given in terms of an auxiliary angle <!>; x = a<t> — a sin (f), y = a — a cos (f). Find -— and ~ at any point. Here ^^^ J±^^^^ dx d(j) ' dx d(f> ' d(j) . ,,. ,. sin</> = a sm d>(a — a cos 0) = = ^~ . ^'^ ^^ l--cos<^ Also '^ = - (^] = — (^A X ^ dx^ dx \dx) d<t> \dx) dx (1 — cos <f)) cos <f) — sin d) (sin 6) . = ^^ ^^ — ^ JLX2 ^ -^ (a - a cos </)) (1 — cos <^)'* ^^ -1 ^_a '"a(l-cos</))2'~ 2/'*- 209. If a^ + 2/'*=a'* (1), If we want -,- in terms of x only we must find y from (1) and use it in (2). But for a great many purposes (2) is useful as it stands. x^ ifi In the same way, if — + ^ = 1, ?f 4.?^^=0 or ^--^^ a?'^ V" dx dx~ a'y' Agam, if --2-n=l» :r ="~2-- • FUNDAMENTAL CASES. 2V7 Also if u;^ 4- 2/^ = a^j If y = i^-^i sin 2a; + yg sin 4.x;, -j- = cos* a). If 2/ = i t^^^^ *' + t^^ ^' 1^>~ ^^^*' ^' Let // = V^2 + a^ = u^ if m = a-^ + ct^, ^— = 2a;, so that — - = 4?4-i X 2x or V 210. Let y = sin~^ x. In words, y is the angle whose sine is x. Hence x = sin y, dx dy cos y — 's/i— sin'-^ y = Vl — a;^. Hence dy dx ^/l-.a^' We have extracted a square root, and our answer may be + or — . We must give to -^ the sign of cos y. 211. Similarly if y = cos~^ x, dy 1 dx Vl-x^' 212. Let y — tan~^ x, so that x = tan y, ~ = —= 1 + tan^y = 1 +x\ dy cos^ y '^ 213. Similarly if 2/ = cot"^ x, then -f-= — ^ — -5 ■ 278 CALCULUS FOR ENGINEERS. 214. It will be seen that (2) and (4) of Arts. 198 and 199 give us power to differentiate any ordinary expression, and students ought to work many examples. They ought to verify the list of integrals given at the end of the book. A student ought to keep by him a very complete list of integrals. He cannot hope to remember them all. Some- times it is advisable to take logarithms of both sides before differentiating, as in the following case : y = x^. Here logy = ^ log ;r, 1 dy 1 , y ' dx X | = .^(l + log,.).t 215. In the following examples, letters like x, y, z^ v, w, 0, &c. are used for the variables ; letters like a, b, c, m, n, &c. are supposed constant. A student gets too familiar with x and y. Let him occasionally change x into ^ or ^ or v, and change ?/ also, before beginning to differentiate. Ho ought to test the answer of every integral by differentiation. List of Fundamental Cases. -7- x'' = nx''-\ L'"» . dx = *•»«+! ; dx J m 4- 1 ^(log^) = i, jl.dx^logx; d , . . [ , 1 . ^r-(sm mx)= m cos mx, j cos 'tnx .dx— — sm mx ; dx J m d . . . r 7 1 -J- ( cos mx)= — m sin mx, \ sm mx ,dx — cos nix ; dx^ ^ J m ' ., . a [ dx 1 (tan ax) = — ^ — , — - — = ~ tan ax ; cos^a./; j cos'' fr.r a dx d , . . a [ dx 1 ^ -^ (cot ax) = — -^— — , -^-T— - = cot ax \ dx^ ^vo^ax J^m^ax a ADVICE TO STUDENTS. 279 Sin" d , . , . 1 { dx -j- (sin ^x) = . , , = d .. , . 1 f dx 1 ^ , ^ Many integrals that at first sight look different are really those given above. Even the use of \/~ or ^~ instead of the numerical symbol of power or root, disguises a function to a beginner. Thus 1 . 1 _, 3 ■- IS -X*, ci\/x (I and its integral is 1 / x-^+' \ 3 3 or z:r-X\ a V- i + 1/ 2a 216. In some of the following integrals certain substitu- tions are suggested. The student must not be discouraged if he cannot see why these are suggested ; these suggestions are the outcome of, perhaps, weeks of mental effort by some dead and gone mathematician. Indeed, some of them are no better than this, that we are told the answer and are merely asked to test if it is right by differentiation. Just here, in learning the knack of differentiation and integi'ation, the student who has a tutor for a few lessons has a gi'eat advantage over a student who works by himself from a book. Nevertheless the hardworking student who has no tutorial help has some advantages ; what he learns he learns well and does not forget. The man who walks through England has some advantages over the man who only takes railway journeys. In learning to bicycle, I think that on the whole, it is better to be held on for the first few days ; learning the knack of differentiation and integration is not unlike learning to bicycle. Exercises and Examples. 1. i/ = xhgx, ^=l+%^'- 280 CALCULUS FOR ENGINEERS. -^ ' dx 2^Jx 3. y = losf (tan a?), -t-= - — ^r- - '^ ^^ ^' dx sin 2a; , 1 — tan X du ... . 4. y — , -T-— — (sin X 4- cos x\ •^ sec a; eta; ^ 1/1 \ ^y 1 o. y = log(log<.), ^ = ^1^. tto; — ^— — 6. a; = 6«< sin 6^, "^ = ^/«' + ^' • ^* «" (6^ + c), where tan c = - . a We here use the simplification of Art. 116. The student will note that by page 235, Q (standing for djdt), operating n times upon sin ht^ multiplies its amplitude by 6" and gives a lead of n right angles. He now sees that if operates n times upon e"* sin ht, it multiplies by {it? 4- 6^)"^^ and produces a lead nc. Thus ^1 = (a= + I/) €«« sin (6^ + 2c) ; and ^^ = (a-^ + 62)^€«' sin {ht + 3c). ^, fl-e dp 1 i^ = 2tan-yj-^^, J = -^7j^ 8. 2/ = log (6^ + 6-0, :£ c?i/ €^ — e" cZ^ e^ + e-^ ' 9- y = ^^^, t=t^/^. c?a; 10. y = aa^+hx + c, -^^ = 2aa; + &. 12. p^cv-''\ ^=-l-37ct;-'^' 13 EXERCISES. 281 dv 14. \av'''dv^--^v-''\ lav-'- 15. Haf + bt + c)dt=: ^at^ + ^bf + ct + g. 16. I V^ . (^« = \x^ . c?a; = §a-^ 17. /f i«/r3.<^e=JJ^ = -i. 19 ! dx 1 r dx 1 1 _, a; ^ "^' 2r^ V m 1 20. I v^oT+l; . dv. Here let a + 1; = 2/ so that dv = c?y, and we have jy^ . dy = ly^ = ^ (a + v)\ = f Sa f— + 3a^ I a^ /— , 4 — m 3 — m 2 — m 1— m and in this it is easy to substitute t i- a for y. 282 CALCULUS FOR ENGINEERS. 22. I '- -T . Let a + 6^ = y so that h .dx = dy, J (a + bx)^ 23 24 3. I -7^=r_ .dt=^-\/a^- 1\ evidently. . I . Let x — a — y,dx = dy 25. Since ~i-, = ^(-^ --^-l g^—d? zaxx — a x-^ a/ J a?-a? 2a ^ ^^ ^^ ^^ 2a ^^ + a ^. ., , r dx \ , x — a r fix 26. If x- + 2Ax + B has real factors, then - — ^— j ^ j x^-{-2Ax-\-B is of the form just given. But if there are no real factors, then the integral may r (2x ^^ ''"'^^'' j a? + 2Ax + A^ TB^rAi ^"^ if y = *' + ^ and a^ = B — A^ we have | ~ ^-„ which is - tan~* - . J y^ + a^ a a 27. I tan X. dx=— I dx. This is our first example J J cosx ^ of a great class of integrals, where the numerator of a fraction is seen to be the differential coefficient of the denominator. Lot ?/ := cos X, then dy = — sin x . dx, so that the above integral is — I -^ , or — log y, or — log (cos x). EXAMPLES. 283 28. Let/' (x) stand for the differential coefficient oif{x), and Ave are asked to find I — \r7-: — ■ Let f{x) = y, then f {oc) . dx = dy, so that the integral becomes /--=logy=log/(^). Hence, if the numerator of a fraction is seen to be the differ- ential coefficient of the denominator, the answer is log (denominator). _^ f X . dx If 2bx .dx 1 , , , „. 31. Reduce I ^ ^ — - to a simpler form. If the J Ct "T" OX "T" cx numerator were 2cx + b, the integral would come under our rule in Ex. 28. Now the numerator can be put in the shape n ,^ 7x ^^ -(2c^ + 6) + m- — , so we may write the integral as r 2cx + h J f nh\ C dx ^j a + bx + cx^^'^'''^V''~2c)j a + bx + cx' The latter integral is given in Example 26. na f x + b , . f2x .dx C b. dx = vUog(a2 + .7'^) + -tan-i-. „_ r sin X .dx 1 f—b sin x .dx 1 i . 7 v 33. — — T = -,- -— , = -rlog(a + £*cos*'). Ja + b cos i« b J CI + 6 cos a; 6 ° 284 CALCULUS Foil ENGINEERS. 34 14^ n_+jog^-log^^ J xAogx J xiogx _ /' (I + \ogx)dx Cdx J a; log a; J x = log (x log x) — log X = log X 4- log (log ;r) — log X = log (log a;). When expressions involve a"* and (a + 6^)**, try substi- tuting y = a + bx or y=^ - + b. X 35. Thus f .--^^-.- = -^-Jl_-. *17 f <^^' __Jl_lA] (f' + bx ^^'"^ J (a + 6^+1 " 2^;^ (a + ba^y- 2m - 1 2ma J (a-hbx-y and so we have a formula of reduction. C da J (a-hh 39 When expressions involve Va 4- bx try y^ = a-{- bx. rp,, C x.dx 2 (2a — 6^) / =— Thus -— ..^=_-A_^^ — ^s/a + bx. J Wa + bx ob^ 40. I - Vl 4- log ic . dx. Try ?/ = 1 -f logo;. Answer : f (1 + log a;)^. 41. I VT~^=^- Try€^ = i/. Answer: tan-^e*. 217. Integration by Parts. Since, if u and v are functions of x, d . ._ dv du dx^ ^^ dx dx' uv= j u.dv-\- I V . da, EXAMPLES. 286 or lu . dv = uv — I V . du (1). We may write (1) as ju '-r . dx=uv — Iv .-j-.doc. By means of this formula, the integral I u . dv may be made to depend upon j v .du. r rl 42. Thus to find p**. log a?, c?^. Let 2^ = log a? and -t-=^", so that V = r . Formula (1) gives us :r loe: x — \ ^ dx, a;«+i /, 1 \ or log X . 43. fx . 6" , dx. Let it — x;--j- = e^^, so that v==-€^"'; then formula (1) gives us \ x.e"^^ .dx = - .re"* I e"* . dr = - a?€"* ^e"* J a aj a a^ = - €«* (a? ) . a \ aJ 44. I e** . sin bx . dx. Call the answer A. Let u = sin 6^, v = - e^^, then formula (1) gives us ^ _ _ gaa; gj^ hx \ e^^ . COS bx ,dx = - e"^ sin 6^ 5. a a J a a But similarly 1 e^* . cos 6a; . dx, which we have called B, may be converted, if we take u = cos bx and v = - e"*; ^ _ _ guas cos hx-\- - I 6*** sin bx.dx — - e"* . cos bx-\-~A, a a J a a 286 CALCULUS FOR ENGINEERS. Hence A = - e"* sin h.v ( - e"^ cos hx-\- -A] ,so that a a \a a ) f . , , €"* (a sin bx—b cos bx) A = l e^"" sin bx .dx= - -^ ;— -y^ . €°^(a cos bx-\-b sin bx) €** COS bx . dx = — Similarly i^ = I a- + b"- 218. By means of Formulae of Reduction we reduce integrals by successive steps to forms which are known to us. They are always deduced by the method of integi'ation by parts. Thus I ar«e"* . dx = - A-«6«^ - - I x''-' . €«^ . dx. j a aj If then we have to integrate ar*e"*, we make it depend upon x^e"^', again using this formula of reduction we make ^a^ax depend upon ar^e"*, and so on, till we reduce to x^e"^ or t*"*, whose integral we know. Thus L-^e^ . (^ = o^-^t* - 3 U-^e* dx = x^e^-^\x'e'-2 jxe^ . dxl = (x^-Sx'-^Qx-(j)6\ Some General Exercises. 45. 2/ = « sin^ bx, -^^ab sin 2bx. 46. y=^b sin ax^, -~ = bnax^~^ cos ax^. ^ dx 47. y = (a + 6a;")»", -^ = nbx^-^m {a + 6a?»*)»*-^ 48. y = (a + 6^') e'^*, -^ = €'^(b + ac + bcx). EXERCISES. 287 49. y — a"^, -^=^a^ . log a. 50. y=\ogaiCy -r-=—\ . ^ ^ dx xioga -- a — tdv a dv 52. v=^^a^-t\ — = 53. u = v^ du Sv"" (l_V2)f' dv (1-V2)f _ ^/a + t dv _ Vg ( V^- Va) 57. 2/ = log (sin a?), -7^ = cot x. 58. „ = logya-^sJ=-^,. 1 /l — cost dy 1 61. ^ = tan"^ / , -TT = / . Jl+P-Jl-t" dt Ji-ti 62. a; = sec-i^, ^= "^ dt tjf-l' 63. 3/ = sin (log v)y-r = ■ cos (log v). 288 CALCULUS FOR ENGINEERS, ^- _l-\-x dy'^\ — 2x — ob^ 66. « = log (cot t;), -^ = - -^-s" . ■^ *=* ^ ^ dv sin 2v 67. 5=e'(l-f»X^=€*(l-3f''-^). 69. ^ = '^- /^-fd-r rf(9~ (€^-1)'^ ■ hr-. Ti- X Zl . Zl 4.1, 4. ^^^ COS ^ 71. If a? = tan 6 + sec ^, prove that -j^ = c?^ (1 - sin ey d^x 2 72. If a; = ^ log ^, prove that 3^ = g • 78. If 2/ = 6~^ cos a?, prove that -^^ + 4?/ = 0. .T>. T^ ^ 4.1,4. d*y 24 74. If2,= ^__, prove that ^ = ^-^-^. 75. L"»-^ (a 4- &a:") ^/« dx. (1) If ^/5' be a positive integer, expand, multiply, and integrate each term. (2) Assume a + hx^ — y^ ; and if this fails, (3) Assume aaj~" +h — y^: this also may fail. 76. \ af^ (a + x)^ . dx. Let a + x — y"^, then dx—2y. dy, and co^y^—a, so that we have 2 {(y*— 2ay^-h a^)y^ ,dy, or 2 I (3/^ — 2a3/* 4- ay) c?y, which is easy. EXERCISES. 289 — 2x~^ .dx = 2i/ .dy so that we have 78. [ — ^^ . Try a'^a?"* + 1 = ^^ ^nd we find 1 rc^^_ 1 _ 0) cPx 79. If a? = -4 sin nt +B cos w<, prove that -j-^ + ?i2a; = 0. 80. If 1* = xy, prove that ^^, = ^ ^ + ^-^^i • 81. Illustrate the fact that -^ — i- = ^ 7- (see Art. 83) ay . ax ax . ay ^ in the following cases : X w = tan~^ - , ^* = sin (aa;** + 63/**), w = sin {oi^y), u = a; sin y + y sin a?, « = ba^ log ai/, li = log (tan ^ ) , 82. y = e**^ sin"* 6a;, -^ = e*'* sin"*-^ 6a? (a sin hx + m6 cos hx). 83. ic = €-«'cos6^, ^ = (a« + 6=')~2 e""' cos(6^ -n^) where tan d = - . a ^1 A 1.2.3.4 84. 2/ = a;^loga?,^- ^— 19 290 CALCULUS FOR ENGINEERS. ,,.. d^y 2 COS 00 85. y = log(8m^),J=-^j^. 86. Uv = Aiof'^y^^ + A^y^^ + &;c., where aj + 6i = Oa + 62 = &c. = n, V is called a homogeneous function of x and y of n dimensions. Show that fl7(-7^)+v(-T-)= ^^- Illustrate this when 1; = — ^ and V = V ^* + 3/"- 87. In general if u =/ (1/ 4- a^) + i" (y — ax), where / and -P are any functions whatsoever, prove that dhi,_^(Pu dx^^'' dy'' the differentiation of course being partial. 88. If v = (x^ + f + ^)-i, prove that g + + g = 0. 89. If 5 = ae--* sin y8^ satisfies ^ + 2/ j, + w^s = 0, find / and n^ in terms of a and /3, or find a and ^ in terms of / and n^, 90. If y = €** is a solution of find a. As an example take ^__2^^-^^ + 2^ = da^ da? dx^ dx ' and find its solution. Answer : y — ae^-\' be~^ + ce^ + e, where a, 6, c, e are any constants whatsoever. 219. To integrate any fraction of the form Ax"^ + Bx"^-"- + Cx"^-^ + &c. ax"" + 6a?^-i + ca;"-2 + &c. where ?^ and n are positive integers. (1). PARTIAL FRACTIONS. 291 If m is greater than or equal to n, divide, and we have a quotient together with a remainder. The quotient is at once integrable and we have left a fraction of the form (1) in which m is less than n. Now the factors of the denominator can always be found and the fraction split up into partial fractions. For every factor of the denominator of the shape x — a assume that we have a partial fraction ^ ; for every factor of the shape x^ -\- oix-\- ^ assume that we have a partial of the shape — 77; if there are n equal factors each of them being x— a assume that we have the corresponding partial fractions Thus for example, suppose we have to deal with a fraction which we shall calK^^^ and that F{pii) splits up into factors a? — a, X — y8, a?- + ax + h, {x — 7)** ; we write /(^__4_ ^ Gx^B E F(x) " x-a'^ x-^'^ af + ax-{-b'^ (^-7)" + 7— ^^-T^+&c (2). Now multiply by F{x) all across and we can either follow certain rules or we can exercise a certain amount of mother wit in finding A, B, C, D, E, F, G, &c. Notice that as we have an identity, that is, an equation which is true for any value of x, it is true if we put a; = a or X— ^ or a? = 7 or x'^ + ax-hb—0. Do all these things and we find that we have obtained A, B, E, C and D. To find G we may have first to differentiate our identity and then put ^ = 7 and so on. You will have found it more difficult to understand this description than to actually carry out the process. Having split our given fraction into partials the integra- tion is easy. 19—2 292 CALCULUS FOR ENGINEERS. ^^ a? A B Gx-\-D Q1 ^^ _______ _| J Hence x'' = ZX^^ -f I) + 5 (^ - 1) (^=^ + 1) + {Cx -{-D)(x- If. Let dr^+l = 0, and we have with not much difficulty (7= - J. i) = 0. Put a? = 1, and we have ^ = J. To find B, make x = 0, and we find B=^. Hence we have to inte- grate 11 11 1 X 2(07-1)2 2^-1 2l+af' and the answer is -| -^ + ilog(«-l)-Jlog(^= + l). When there are r equal quadratic factors, we assume the partials (x' + cuc + ^y {x' + axi-^y-^'^ °* It is not difficult to see how all the constants are deter- mined. We seldom, however, have complicated cases in our practical work. 92. Integrate ^-^^-^ or -^^^^-^—^^; assume it to be equal to M F P X "^a;+3"^a;-2' so that x^-\-x-l^M{x-\-^){x-'2,)-^Nx (a;- 2)+ P^ {x^- 3). As this is true for all values of x, put x = and find M, put ^ = — 3 and find N, put x—l and find P. Thus we find that the given fraction splits up into Qx 3a7+3 2a;-2' so that the integral is Jloga?-i-ilog(a^ + 3) + ilog(^-2). EXERCISES. 298 5^2 = — + 15a; - 6 log (a; - 1) + 41 log {x - 2). 94. 1^-7^3-6'^ /■/^u.^^1 1 32 1 ^243 1 \, 95 [ ^-^^ = J tan-1^4- i log(H-a^')- i log(l 4-^). ^3-5a;24-3^ + 9 a; + l"^(a;-3)2 ^--3' and we find ^ = - 8, A = - 5, ^3== 17 ; so that the integral is -81og(^4-l)+-^ + l71og(a;~3). ^^^- /^ + i- 3 = ^ l"g (^' + 3) + i log i^ - !)■ 294 CALCULUS FOR ENGINEERS. 107 r (2a; - 5) dx __ 7 ^ + 1 j(a; + 3)(a; + i> 2(a; + l)'^^^''^^+3* 108. - = -7=tan^ — 7=—. a;+4 220. Maxima and Minima. If we draw any curve with maxima and minima points, and also draw the curve showing the value of -~ in the first curve, we notice that ; — where y is a maximum, -f- = and -~ is negative; whereas, 1 • • • ^V /x J d^y • :• T<. • ^vnere y is a mmimum, t =^ and -^^ is positive, it in any practical example we can find no easier way of discrimi- nating, Ave use this way. Notice, however, that what is here called a maximum, value, means that y has gradually increased to that value and begins to diminish, y may have many maximum and MAXIMA AND MINIMA. 295 minimum values, the curve beinsr wavy. Notice that -^ (Py . . ^^ may be and ~- = so that there is neither a maximum a 3c nor a minimum value, y ceasing to increase and then begin- ning" to increase again. See M, fig. 6. 1. Find the maximum and minimum values of . Answer : ^ and — J. 2. Find the createst value of .— : t-tt-^ r . Answer : {a + hy 3. Prove that a sec 6-\-h cosec ^ is a minimum when 3 tan 6 = V5 4. When is , a maximum ? Answer : x = ^. 5. When is x'^ {a — xy^ a maximum or minimum ? . ma Answer : x = , a maximum. m + n 6. Given the angle (7 of a triangle, prove that sin^ A + sin^ 5 is a maximum and cos'^ A + cos^ J5 is a mini- mum when A=B. 7. y = asmx-\-h cos x. What are the maximum and tninimum values of y ? Answer : maximum is y = \/aFTb^, minimum is - Va^ + ¥. 8. Find the least value of a ton + bcot 6. Answer : 2 ^ah. 9. Find the maximum and minimum values of ^^^ + 2^ + 11 i»2 4- 4^ + 10 ' Answer : 2 a maximum and f a minimum. Students ought to plot the function as a curve on squared paper. 296 CALCULUS FOR ENGINEERS. 10. Find the maximum and minimum values of a^-x + 1 x^ + os-l' Answer : maximum, — 1. 11. Find the values of x which make y = ztr — ^ a;- 10 a maximum and a minimum. Answer : x=A gives a maximum, a;= 16 a minimum. 12. What value of c will make v a maximum if ?; = - logo? Answer : c = e. 13. If^ = ^^ -^ yt^Nah gives a mmimum value t 1 A sin" 6 ^ IT . . , 14. X — , ^ , u —- £(ives a maximum value to x. 1 - cos ^ 3 ° 15. What value of c will make v a minimum if V = q , ? Answer: c = i. 16. When is 4^— 15a^4- 12a; — 1 a maximum or minimum ? Answer : a; = J a maximum ; a; = 2 a minimum. 17. tan'^ a? tan** (a — a;) is a maximum when tan (a — 2a;) = tan a. n-\-m 1 ft _ ^^ ^ = 3 a maximum, 9 + ^v ^ = — 3 a minimum. 19. Given the vertical angle of a triangle and its area, find when its base is a minimum. 20. The characteristic of a series Dynamo is ^=T^a W. MAXIMA AND MINIMA. 297 where a is a number proportional to the angular velocity of the armature, and a and s depend upon the size of the iron, number of turns &c., E is the E.M.F. of the armature in volts and (7 the current in amperes. If r is the internal resistance of the machine in ohms and i^ is an outside resistance, the current ^'=4^ (2), and the power given out by the machine is P:=OR (3). What value of R will make P a maximum ? Here (2) and (1) give --^ -^ = (7. So that 1+5(7=-^, C=if-^-lV r-\-R' s \r -\-R J '-7(;:fji-')'.-<"'S-». we have {-^ - iT + 2i2 {-^- \\ {- ,— ^o,.! = 0- Rejecting ^— 1 = because it gives (7=0, we have — r-15 — 1 = 7 7^ > and from this R may be found if r and r-\-R (r + Ry ^ a are given. Take a = l'2, 5 = 0'03, r = *05 and illustrate with curves. 21. A man is at sea 4 miles distant from the nearest point of a straight shore, and he wishes to get to a place 10 miles distant from this nearest point, the road lying along the shore. He can row and walk. Find at what point he ought to land, to get to this place in the minimum time, if he rows at 3 miles per hour and walks at 4 miles per hour. Assume that he can equally well leave his boat at one place as at another. 298 CALCULUS FOR ENGINEERS. Fig. 100, ^C=4, CT=10 . Let him land at D where CD = X. Then AD = Vl6+a-^ and D^ = 10 - a;. Hence the total time in hours = r 1 -. — . 3 4 This is a minimum when ^^(16 +^)~*=J, or J^'=16+a;*, or a; = 4*535 miles. 22. The candle power c of a certain kind of incandescent lamp X its probable life I in hours, was found experimentally to approximate on the average to Iq _ ]^Qll-897— 00764et> where v is the potential difference in volts. The watts w expended per candle power were found to be tt; = 3-7 + 108 O07-07667W The price of a lamp being 2^., the lamps being lighted for 560 hours per year, and one electrical horse-power (or 746 watts) costing £2 for this year of 560 hours, find the most economical v for these lamps, so that the total cost in lamps and power may be a minimum. —J- lamps are needed per year, each costing £01. Cost . 56 . per year is then -y- in pounds, and this is for c candles, so that . 56 cost per year in pounds per candle is y- . Now £1 per year 746 means —^ watts, so that the cost per year per candle is 56 746 ,, _x— watts. This added to w gives total cost in watts. INDETERMINATE FORMS. 299 We have Ic and lu as functions of v. Hence R« y 74fi <j\j /^ -x\j -|^Q_ii.697+o 07545W i g-^ _I_ ]^Q8 007— 07667V It is to be made a minimum. Answer : i; = 101*15 volts. 221. Sometimes when a particular value is given to ^ a function takes an indeterminate form. Thus for example in Art. 43, the area of the curve y=mx~^^ between the ordinates /* TYl mx~^^ . dx was z (6^~"^ — a}~^). Now when «,= 1 the area becomes ^ (1 — 1) or ^, and this may obviously have any value whatsoever. fix) In any such case, say 4r/—'J > '^^ fipd — ^ and F {a) = 0, we £ {X) proceed as follows. We take a value of x very near to a and find the limiting value of our expression as x is made nearer and nearer to a in value. Thus \q\} x — a-\- hx. Now as hx is made smaller and smaller it is evident dfix)^ that f{x + hx) is more and more nearly f{x) + hx . — — . If in this we put x=a, f{x) or f(ci) disappears, and conse- quently our fraction*— ^ becomes more and more nearly The rule then adopted is this : — Differentiate the numerator oidy and call it a new numerator; differentiate the denomi- nator only and call it a new denominator; now insert the critical value of x, and we obtain the critical value of our fraction. The process may need repetition. loP" X Example 1. Find the value of ° when x = l, X ~~' L 300 CALCULUS FOR EiNGINEERS. First try, and we see that we have 0/0. Now follow the 1 oc above rule, and we have - , and inserting in this a; = 1 we get 1 as our answer. 2. rmd ^—„ — ^, r— when a? = c. First try a? = c, and we get 0/0. Now try ^, _^ , , and again we get 0/0. Now repeating our process we get r - sc—l 1 3. Find ^- when x=l. Answer: -. a?" - 1 n 4. Find when x = 0. Answer : log j- . 5. Try the example referred to above. The area of a curve is ^j (6^~" — a^"") = A. If m, b, a are constants, what is the value of A when n = 1? Writing it as m — = , 1—71 differentiate both numerator and denominator with regard to n, and we have, since A(5i-»)=6^-«.log6x(-l), 6^~" log b - a^~" log a 7)1 - , and if we insert n = 1 in this, we get m (log b — log a) or m log - , CL which is indeed the answer we should have obtained if instead of taking our integral p~^ . dx as following the rule i a;~" . dx — + c, — 7l-\-l GLOSSARY. 301 we had remembered that in this special case ]a;~^ . dx—logiv. I' 222 J Glossary and Exercises. Asymptote. A straight line which gets closer and closer to a curve sls x ov y gets greater and greater without limit. Thus y = - Jx^ — a^ is a Hyperbola. Now as x gets greater ^ a . and greater, so that - is less and less important, the equation X T approaches more and more y=- x, which is the asymptote. a The test for an asymptote is that -^ has a limiting value for points further and further from the origin, and the inter- dx cept of a tangent on the axis of x, x — y -^ , has a limiting value, or the intercept on the axis oiy.y — x -j-, has a limit- ing value. ^^ Point of Inflection. A point where -^^ changes sign. Point of Osculation. A point where there are two or more equal values of — . Cusp. Where two branches of a curve meet at a common tangent. Coi^Jugate Point. An isolated point, the coordinates of which satisfy the equation to the curve. Point d' AiTet. A point at which a single* branch of a curve suddenly stops. Example, the origin in y = a? log x. The Companion to the Cycloid. x-a{\ -cos (/>), The Epitrochoid. x={a + h) cos <^ - m6 cos f ^ + 1 J </>, y = (a -f- 6) sin — m6 sin ( T- -f 1 J (f>, 302 CALCULUS FOR ENGINEERS. where h is the radius of the rolling circle, a is the radius of the fixed circle, and mb = distance of tracing point along radius from centre of rolling circle. Make m=l, and this is the Epicycloid. The H3rpotrochoid. ^ = (a— fe)cos<^+m6cos f ^-1 J<^, y= (rt— 6)sin<^— ??i6sin(7- — 1 ]</). Make ??i = 1, and we have the Hypocycloid. Take a = 46, and obtain a Hypocycloid in the form a;* + y* = ai Take a = 26, and obtain the Hypocycloid which is a straight line. In obtaining the Cycloid, Art. 11, let the tracing point be anywhere on a radius of the rolling circle or the radius pro- duced and obtain a? = a (1 — m cos <^), y = a (<^ + m sin <^). If wi > 1, or < 1, we have a prolate or a curtate Cycloid. The Lemniscata (a^ + y^y= a^ (jx^ ^y^) becomes in polar coordinates r* = a* cos 26, and taking successively ^ = 0, ^ = '1, &c., we calculate r and graph the curve easily. The Spiral of Archimedes, r = ad. The Logarithmic or Equiangular Spiral, r = ae*^ The Logarithmic Curve. y=a\ogba)-{- c. The Conchoid a^ = (a + xy (¥ - x') becomes r = a + 6 sec 6. The CiBBoid y^ = a^/{2a — x) becomes r = 2a tan 6 . sin d. The Cardioide. r = a (1 — cos 0). The Hyperbolic Spiral. r6=a. The I.ituusis?^2^ = a2. The Trisectrix. r = a (2 cos ^ + 1). 1. In the curve y=—^ ^, show that there are points of a ~i~ X inflexion where a? is and a VS; the axis of x is an asymptote on both sides ; there are points of maxima where x = a and — a ; the curve cuts the axis of ac at 45°. EXAMPLES. 303 2. In aj^y = 36^?^ — a^ show that there is a point of inflexion where x = b, y= —r - 3. If y^x = 4a'^ (2a — a?), show that there are two points of , „ . , 3a ^ 2a inflexion when oo= -zr- , y=^±-j=-. ^ V3 4. If 2/^ {x^ — a^) = ^, show that the equations to the asymptotes are y=^-\-x and y=- w. 5. The curve a^ — y^ — a^ cuts the axis of x at right angles at a? = a where there is a point of inflexion. 6. Show that y = a^xjiah + ^) has three points of in- flexion. 7. Prove again the statements of Exercise 2, Art. 99, and work the exercises there. 8. Find the subtangent and subnormal to the curve y = e*** Answer : subtangent - , subnormal ae^^^. 9. Find the subnormal and subtangent to the catenary. =!{-■■• c i subtangent = c Answer : subnormal = -J6<'— e "l c - c 10. Find the subtangent of the curve x^ — Saycc + y* = 0, Hence ^ ^ay-j^ dx y^ — ax Subtangent at point x, y is y-^ = y ^ — . (ty cty — ocr 304 CALCULUS FOR ENGINEERS. 11. In the curve 2/^ = find the equation to the asymptote. Here i/^ = a?^ I — \=ia^ f 1 -\ 1- --- + &c. j by division. As x is greater and greater, — &c. get smaller and smaller and in the limit (see Art. 3) 3/=±^(l+^). y=±{x^-a\ So we have a pair of asymptotes y = x-\-a, y = — x — a. Again, the straight line x = a,a> line parallel to the axis of y, is also an asymptote, y becoming greater and greater as X gets nearer and nearer to a in value. 12. Find the tangent to y* + a^ = a*. Hence at the point x^, y^ the tangent is ~ — ^ = — a/^\ 13. In the curve 3/ — 2 = (a? — 1) Va; — 2, where is -^ = x ? At what angle does the curve cut the axis ? This is infinity where x — 2 and then 3/ = 2 ; that is, the tangent at (2, 2) is at right angles to the axis of x. Where y=0/\t will be found that a? = 3 and -t- = 2. 14. In the curve y^ — ax^ ■\- a? , find the intercept by the tangent on the axis of 3/, that is, find y — cc-^. Sy^ -^ = 2ax + Sx\ JSXAMPLES AND EXERCISES. 305 bo tnat we want y — x — ^r— — or -^ r- and 3i/2 %f this will be found to be 3 \a +xj ' 15. The length of the subnormal at x, y, is 20^^, what is the curve ? Here 2/-,— = 2a V. Hence ^y^ = ^a^x* or y = ax^ is the dx equation to the curve, a parabola. The subtangent is ^ y , ^^^•2f^^3>or2^,oria;. 16. Show that the length of the normal to the catenary^ is - y\ 17. Show that y^ — x^-\- 2hx^y = has the two asymptotes b , b y^x-^andy=-x--^. 18. Show that the subtangent and subnormal to the cii'cle y^==2ax — x^, are — — — and a — x respectively, and to the IT 2 ^\^ 2X4.1. 2ax-x' ,b\ . ellipse y^ = —Azax — x^) they are and -- (a — x\ a? 19. Find the tangent to the cissoid y* = 2a — X Answer : y = j ^^^^^y l {(3a -x)x^- ax\, 20. What curve has a constant subtangent ? dx ^ dy , ^? 2/ -J- = a or a^ = a -^ , or x= a log y + c or y = Ce^y the logarithmic curve. 21. Show that a? — y^-\- ax^ = has the asymptote P. 20 306 CALCULUS FOR ENGINEERS. 22. Show that a curve is convex or concave to the axis of X as y and -— have the same or opposite signs. See Art. 60. 223. The circle which passes through a point in a curve, which has the same slope there as the curve, and which has also the same rate of change of slope, is said to be the circle of curvature there. If the centre of a circle has a and h for its co-ordinates, and if the radius is r, it is easy to see that its equation is (a;-a)» + (i/-6)» = ?-« (1). Differentiating (1) (and dividing by 2) and again dif- ferentiating we have a.-a + (y-6)J=0 (2), and i+(,_,)g+(|J = (3). writing p for -~ and q for j^ we have from (3) y-^ = — ^ (4); using this in (2) we have X'-a=^—^p (5). Now p and q and x and y at any point of the curve being known, we know that these are the same for the circle of curvature there, and so a and h can be found and also r. If the subject of evolutes were of any interest to engineers, this would be the place to speak of finding an equation connecting a and 6, for this would be the equation of the evolute of the curve. The curve itself would then be called the involute to the evolute. Any practical man can work out this matter for himself. It is of more interest to find r the radius of curvature. Inserting (4) and (5) in (1) we find the curvature 1 = ^_ (6) .(8> CURVATURE. 307 A better way of putting the matter is this: — A curve turns through the angle Bd in the length Bs, and curvature is defined as the limiting value of hd 1 dO B^'^^'r^dS (^)- Now tan = -~ = p, say, so that 6 = tan~^ j^. Hence dd _ 1 f^ ds ~ 1 -i-p- ' ds ' Now ^ = ViT^^=-.'^ = -.^. dw ^ dp dx dp dx" H-~ \-f.-%i[Htr\' Exercises. 1. The equation to a curve is ic^ - 1500a;2 + 30000^ - 3000000y = 0. Show that the denominator of - in (8) is practically 1 from ;r = to iz; = 100. Find the curvature where x — 0. 2. In the curve y = ar* — 4a'3 — ISar^ find the curvature at the origin. Answer : 36. 3. Show that the radius of curvature at x=^a,y = Ooi the ellipse — + ^r = 1, is — . 4. Find the radius of curvature where a; = 0, of the parabola, y^ = 4aa;. Answer : 7' — 2a. 5. Find the radius of curvature of y = 6e"^. (1 + a^h-6^^)^ Answer : q = a^be^"^, p = ci&e«^, r = -'^ ,, „^ , so that where a; = 0, r = ^ -j — ^ . a-'o 6. Find the radius of curvature of y = a sin bx. . (l + a262cos2 6a?)5 .,,, . Answer : /• = ^^ j-—. — , — — . Where x = 0, r=QC, or — ab^ sin ox curvature is : where 6a; = - , ?■ = r, ♦ 2 — ao^ 20—2 308 CALCULUS FOR ENGINEERS. 7. Find the radius of curvature of the catenary Answer -. r=.^ . At the vertex where a; = 0, 1/ = c, r = c. 8. Show that the radius of curvature of 7/^ {x — 4m) = 7nx {x — Sm), at one of the points where y = 0, is -^ , and at the other, ^r- . 9. Find the equation of the circle of curvature of the curve y* = 4m'^ — a^, where x=0, 3/ = 0. 10. The radius of curvature of Sa^y = x^, is ?' = ^ . . 11. In the ellipse show that the radius of curvature is (a' — e-x"^)^ -^ ah, where e^=\ ^ , e being the eccentricity. 12. Find the radius of curvature oi xy = a. Answer: (a;^ 4- 2/")* -i- 2a. 13. Find the radius of curvature of the hyperbola Answer: (eV — a^)^ -r- ab, where e- = l-\ — . a^ 14. In the catenary the radius of curvature is equal and opposite to the length of the normal. 15. Find the radius of curvature of the tractrix, the dx equation to which is y -7- = Ja^ — y^. 224. Let f(x,y,a) = (1) be the equation to a family of curves, a being a constant for each curve, but called a variable parameter for the family, ENVELOPES. 309 as it is by taking different values for a that one obtains different members of the famil}^ Thus f(a;,y,a + Ba) = (2) is the next member of the family as Ba is made smaller and smaller. Now (2) may be written (see (1) Art. 21) /(^,2/,a)+Sa.^/(^,2/,a) = (3), and the point of intersection of (2)f and (1) is obtainable by solving them as simultaneous equations in oo and y; or again, if we eliminate a from (1) and ^J(^.y,a) = (4). we obtain a relation which must hold for the values of x and y, of the points of ultimate intersection of the curves formed by varying a continuously ; this is said to be the equation of the envelope of the family of curves (1) and it can be proved that it is touched by every curve of the family. Example. If by taking various values of a in m , ^ a we have a family of straight lines, find the envelope. Here f(x, y, a) — is represented by y ow? = (1)*, and differentiating with regard to a we have + ^■-^^ = (4)*, 1 a" , m or - == — or a^ = _ , X m X Using this in(l)*we have y — ^mx — X A / — = 0, or y — 2 \/nix = 0, or y^ = 4>mx, a parabola. 310 CALCULUS FOR ENGINEERS. Example. In Ex. 3, Art. 24, if projectiles are all sent out with the same velocity F, at different angular elevations a, their paths form the family of curves, Fsin a 2/ = \9 Fcosa ^^F^cos^a' or . y — xa-^ma^{a^'\-\) = 0, where a stands for tan a and is a variable parameter, and 1 9 Differentiating with regard to a, — x-\- 2ma^a = or a = + - — : 2mx 1 ,/ 1 ,4fm^x^ is the equation to the envelope, or 1 + 1 ).. -y == — rtiOi^ -f 4>m This is the equation to a parabola whose vertex is j — or F- — above the poii^t of projection. 225."^ Polar Co-ordinates. If instead of giving the position of a point P in .r and y co-ordinates, we give it in terms of the distance OP called r, the radius vector, and the angle QOP (fig. 101) called (9, so that what we used to call x is r cos and what we used to call y is r sin 6, the equations of some curves, such as spirals, become simpler. If the co-ordinates of P' are r + Sr and 6 + hO, then ^ig- 101- in the limit PSP' may be looked upon as a little right-angled triangle in which PS = r . BO, SP' is Sr, PP' or 8.9 = Jr'(8dy + (8ry so that dd y ' ^ \de) POLAR CO-ORDINATES. 311 Also the elementary area POP' is in the limit ^r"^ .hd, and the area enclosed between a radius vector at ^j and another at ^2 is i I r"^ ' dO, so that if ?' can be stated in terms of 6 it is easy to find the area of the sector. Also the angle </> between the tangent at P and ?' is evidently such that dO tan cj) = PS/P'S or, t -r- — tan <f). This method of dealing with curves is interesting to students who are studying astronomy. If r = a*** (the equiangular spiral) — = ha^^ log a, and so, ?' -7-; , or ?'-f- -^ = 1/6 log a, so that tan <^ is a constant ; that is the curve everywhere makes the same angle with the radius vector. Let ^ = rcos 6 so that x is always the projection of the radius vector on a line, x — a^^ cos 6. Now imagine the radius vector to rotate with uniform angular velocity of — ^ radians per second starting with ^ = when ^ = 0, so that 6 — — qt, then X = a~^9^ cos qt. Thus we see that if simple harmonic motion is the pro- jection of uniform angular motion in a circle; damped simple harmonic motion is the projection of uniform angular motion in an equiangular spiral. See Note, Art. 112. Ex. 1. Find the area of the curve r = a (1 + cos 0). Draw the curve and note that the whole area is I r^ . dO, or ^iral Ex.2. Find the area of ?^= a (cos 2^+ sin 2^). Answer: Tral Ex. 3. Find the area between the conchoid and two radii vectores. Answer : b^ (tan e^ - tan d^) + 2ab log (tan (7r/4 - i^a) -f- tan (7r/4 - ^6,)}, 226. Exercises. 1. Find the area of the surface gener- ated by the revolution of the catenary (Art. 38) round the axis of y. 2. Prove that the equation to the cycloid, the vertex being the origin, is x=:a(0 + sinO) y = a{l- cos 6), if (fig. 102) PB = x, PA= y, OCQ = 6, 312 CALCULUS FOR ENGINEERS. Show that when the cycloid revolves about OF it generates (0_.2 Q\ •~9 — q) > and when it revolves about OX it Y F -)V y.^^ o A X Fig. 102. generates the volume ttW If it revolves about EF it generates the volume bir^a^. 3. Find the length of the curve ^ay^ — 4a^. Answer, . = /J^ 1 + f . <te = fa j(l + ^^ - 1 4. Find the length of the curve y^ = 2ax — x^. Answer : s = a vers~^ x/a. 5. Find the length of the cycloid. See Art. 47. Answer : 5 = 8a (1 — cos ^(f>) = 8a — 4 V4a^ — 2ay. parabola y — ^x + Va + X 6. Find the length of the parabola y — ^4iax, from the vertex. Answer : 5 = Nax -^ x^ -{■ a log Va 7. Show that the whole area of the companion to the cycloid is twice that of the generating circle. 8. Find the area of r = he^l'^ between the radii n and ra, •^2 r«2 using A — \ ^r^ . d6. J 0, Answer: -r{r.^ — r^). EXERCISES. SltS y 9. Show that in the logarithmic curve x — ae'^, 5 = clog J + Vc^ + a?-^ + C^. c + Vc^ + a-^ 10. Show that in the curve r = a (1 + cos ^), using s = 4a sm ^ . 11. Show that in the curve r = ^e*/*, 5 = rVl + c2+(7. 12. Show that in the cycloid, ds \l 2a' and consequently s = 4 Va^ — ^ay — 2a. 13. Show that in the curve x^ + 2/* = a*, s = ^ciSx^. x" f 14. The ellipse, — + r^ = 1 revolves about the axis of x. ^ a- 0^ Prove that the area of the surface generated is where e^=l -. x^ 1/^ 15. Show that the whole area of the curve, -^ + fi = 1 IS fTrao. 16. Find the area of the loop of the curve, y=xtJ —33- . Answer : 2a=^ ( 1 — -7 j . 314 CALCULUS FOR ENGINEERS. 17. Find the whole area of y = a; + Va^ — {jc\ Answer: Tral 18. Find the area of a loop of the curve r^- = a^ coa 20. Answer : Jal 19. Find the area of the ellipse — „ + r;: = 1 ; that is, find four times the value of the integral •«6 /, « 20. Find the area of the cycloid in terms of the angle <p (Art. 11). Answer : a^ (|</) — 2 sin </> + J sin 2<f)) ; and if the limits are (f> = and </) = 27r we have the whole area equal to 3 times that of the rolling circle. 227. A body of weight W acted upon by gi-avity, moves in a medium in which the resistance = a?^", where V is the velocity and a and n are constants. 9 dt What is the velocity when acceleration ceases? Let Vi be this terminal velocity, av^^ = W, or our a = Wvr\ dt I 1 ^di^ so that Thus let n= 2, t=^ log ^-^^ , 2g ^ Vi — v ^ , at dx or v=^Vi tann -- = ^rr • V. dt EXAMPLES. 315 If a; is the depth fallen through, Vi^. 1 Qt X — — losf . cosh ^- . 228. Our old Example of Art. 24. A point moves so that it has no acceleration horizontally and its acceleration downward is ^ a constant. Let y be measured upwards and x horizontally, then d'x dt^ "' dt' ff- dx dt~ --C, dt dy dx '~di ' It'' ' dx ) d'y dt'~ d^y dx ^'^dx'' ~dt = c2 d^ dhj d^~ .-9 dy dx Hence y-^-lS'^^ + o.c+h... (I), which is a parabola. Compare Art. 24. If Ave take y— when .r=0, 6=0. Also we see that a is the tangent of the angle which the path makes with the hori- zontal at d? = and c is the constant horizontal velocity. If a projectile has the initial velocity V with the upward inclina- tion a, then c = Fcos a, and tan a = a, so that (1) becomes 1 gx^ y= — ^ TF^ — ;; — f- oc tan a. 229. Exercises on Fourier. 1. A periodic function of x has the value /(^) = mx, from iv — to a? = c where c is the period, suddenly becoming 316 CALCULUS FOR ENGINEERS. and increasing to mc in the same way in the next period. 27r Here, see Art. 133, 7 is — , c mju = cto -\- ttj sin qx + ^1 cos qx + &e. + ttg sin sqx + hg cos sqx + &c. tto is ^mc, Gg = - I mx . sin sqx . dx, hg=- I mx . cos sqx . dx. c J c J Answer : 7}hC mx = ^mc (sin qx +i sin 2qx + J sin Sqx -f i sin iiqx -f&c.). 2. Expand a? in a series of sines and also in a series of cosines. Answer : x = 2 (sin ii" — J^ sin 2a; + J sin 3a; — &c.) from — tt to tt; also a* = — (sin a? — ^ sin Sx -f ^ sin 5x — &c.) from to ^ , TT 2 TT 4 and ''^" = 9 (<^os A' 4-^ cos 3.r -i-^ cos 5a; + &c.). TT 3. Prove t = sin x -\- 1 sin 3./; + i sin 5a^ + &c. 4 4. Show that 5. Integrate each of the above expansions. 230. 1. The radius of gyration of a sphere about a diameter being k and the radius a, prove that Here, since a;' + 3/^ = a", and the moment of inertia of a circular slice of radius y and thickness hx about its centre, is |y.Sa;. Taylor's theorem. 317 The moment of inertia is I iriny'^ .da)Xi/'=l irmy^ . dx = m^Tra*, Jo Jo and the mass is m^TraK 2. In a paraboloid of height h and radius of base a, about the axis, P = Ja^. About the diameter of the base k^ — ^ (a^ 4- h^). 3. In a triangle of height h, about a line through the vertex parallel to the base, k^ = ^h\ About a line through centre of triangle parallel to base k^=^^h\ 231. Taylor^ s Theorem. If a function of « + h, be differentiated with regard to ic, h being supposed constant, we get the same answer as if we differentiate with regard to h, x being supposed constant. This is evident. Call the function f{u) where u = x-\-h. Then -y- f(u) = y- fiu) x ^- = -j- fCu) as -y- is 1, and this is dx-^ ^ ^^ dir ^ ' dx du-^ ^ ^ dx the same as ifrfi}^) because Assume that f(x-\-h) may be expanded in a series of ascending powers of h. f(x + h) = Xo + XJi + X,h'^+XJi'-^&c (1), where Xq, X^, X^ Sac. do not contain k ^^!i^tA) = o + X,+ 2Z^ + 3X3/1^+ &c (2), an, §f<^)=i^+^\h + ^.h' + &c (3). dx dx dx ax As (2) and (3) are identical ^ _dXo ldX,_ 1 d'Xo '" dx ' ^"^2 rf;^; "1.2 dx' ' Z.= 1 dX^ 1 d'Xo ' S dx 1.2,2 dx" ' 318 CALCULUS FOR ENGINEERS. A.lso if h = in (1) we find that Xo =f(x). If we indicate -r-^f(^) ^y f'{^)y <ihen Taylors Theorem is f (X + h) = f (X) + hf (x) + ^ i" (x) ^' r'(x) + &c (4). 1.2.3 After having differentiated f{x) twice, if we substitute for x, let us call the result /" (0) ; if we imagine substituted for X in (4) we have /W=/(O) + A/(0) + j^/"(0) 1.2.3 /"(0) + &c (5). Observe that we have no longer anything to do with the quantity which we call x. We may if we please use any other letter than h in (5) ; let us use the new letter x, and (5) becomes /(^•)=/(0) + ^/(0) + j^/'(0) r'(0) + &c (6); 1.2.-3 which is called Maclaurin's Theorem. The proof here given of Taylor's theorem is incomplete, as we have used an infinite series without proving it con- vergent. More exact proofs will be found in the regular treatises. Note that if x is time and s =f(t) means distance of a body from some invariable plane in space ; then if at the present time, which we shall call to, we know s and the velocity and the acceleration and -5-, &c.; that is, if we know all the circumstances of the motion absolutely correctly at the present time, then we can predict where the body will be at any future time, and we can say where the body was at any past time. It is a very far-reaching theorem and gives food for much speculation. TAYLOR AXD MACLAURIN. 319 232. Exercises on Taylor. 1. Expand (^4-/^)'^ iu powers of h. Here f{x) = x^, f {x) = nx''-\ f {x) = n {n - V^x'^'S &c, and hence {x 4- A)" = X'' + n/t^"-i + ^^ Y ~ -^ fe"-2 + &c. This is the Binomial Theorem, which is an example of Taylor. 2. ExjDand log (.r + h) in powers of A. Here /(..) = log (..), /(.;) = 1 X /'(4 = -*-=,/"(.■) = + 2*-', 1 A/^ 1 /i^ 1 and hence log (a; + A) = log ^ + A -^ -3 + -^ -5 — &c; ^ X A X o Qj If we put a; = 1 ,we have the useful formula log(l+A) = + A-| + ^-&c. 8. Show that sin (x + h) = smx-\- h cos ^ — y— « sin os — ■= — ^— s cos a} + &c. 4. Show that cos (a? + A) = cosa; — Asini»— = — jrcos^+ z — ^— ^ sina? + &c. 6. What do 3 and 4 become when ^ = ? 233. Exercises on Maclaurin. 1. Expand sin x in powers of x, fix) ^ sin X, -/(0) = 0, /(^) = cos^, /(0)=1. /'(^) = -sina;, /'(0)=0, /» = -cos^, r{0)=-l, /X^) = sin^j, /v(0)=0, &c. /'(0)=1. />io /^>5 /j,-»7 Hence sin a; = i^* — -r^ + — — -c + &c. |3 • 1 5 \7_ 320 CALCULUS FOR ENGINEERS. ».2 or 2. Similarly cos ^ = 1 — 'r^ -I- 77 — 77, + &;c. \A ]i E. Calculate from the above series the values of the sine and cosine of any angle, say 0*2 radians, and compare with what is given in books of mathematical tables. 3. Expand tan~^ w. Another method is adopted. The differential coefficient oHair^xisr—, — , , and by actual division this is 1 — a;^ + a;* — a.* -{- x* — &c. Integrating this, term by term, we find tsin-^a; = x- J^ + ^a* - -f a;' + J^ - &c. We do not add a constant because when a? = 0, tan~* ic =« 0. 4. Expand tan (1 — x) directly by Maclaurin. 5. Show that a*= 1 +0; loga+ 1^ (logay + ^ (log ay+ &c 6. Show that tan a; = a? + 7r4-T^ + &c. 6 15 234. Expand e'^, compare with the expansions of sin Q and cos Qy and show that e-** = cos — i sin 6, cos (9 = J (e^'^ + e-'^), sin^ = -i(e^'^-e-^). Evidently (cos Q ±i sin ^)" = cos nd + % sin nB^ which is Demoivre's Theorem. In solving cubic equations when there are three real roots, we find it necessary to extract the roots of unreal quantities by Demoivre. To find the gth root of a + hi where a and h are given numerically. First write a + ln = r (cos -\-i sin 0)» DIFFERENTIAL EQUATIONS. 321 Then r cos = a, r sin. 6 = b, r = Va* + b^, tan 6 = ^- Cal- 6 culate r and 6 therefore. - / 1 1 \ Now the qth roots are, r^ f cos - ^ + 1 sin - ^ J , r9 jcos - (27r + (9) + i sin - (27r + 0)] , r^ ]cos - (47r + ^) + t sin - (4>7r + ^) [ (Sec. We easily see that there are only q, qth. roots. Exercise. Find the three cube roots of 8. Write it 8 (cos 0-\-i sin 0), 8 (cos 27r + i sin 27r), 8 (cos 47r + { sin 47r) and proceed as directed. 235. The expansion of e^^ is 1+he + ^h'0' + j-|-^A3<9» + &c. Now let 6 stand for the operation -7- , and we see that h — f(w +h)—e^^f(ai) ; or e ^f{x), symbolically represents Taylor's Theorem. 236. An equation which connects x, y and the differ- ential coefficients is called a Differential Equation. We have already solved some of these equations. The order is that of the highest differential coefficient. The degree is the power of the highest differential coeffi- cient. A differential equation is said to be linear, when it would be of the first degree, if y (the dependent variable), and all the differential coefficients, were regarded as unknown quantities. It will be found that if several solutions of a linear equation are obtained, their sum is also a solution. Given any equation connecting x and y, containing constants; by differentiating one or more times we obtain sufficient equations to enable" us to eliminate the constants. P. 21 322 CALCULUS FOR ENGINEERS. Thus we produce a differential equation. Its primitive evidently contains n arbitrary constants if the equation is of the nth. order. Exercise. Eliminate a and b from y^ax^+bx (1), Hence (1) becomes ^'i^-^4x^^y-' (2>- If we solve (2) we find y = Aa^ + Bx, where A and B are any arbitrary constants. 237. In the solution of DifiPerential Equations we begin with equations of the First Order and the First Degree. These are all of the type M^-N-^ —0, where M and N are functions of x and y. We usually write this in the shape M. dx + N.dy=0. Examples. 1. (a + x) (b + y) dx -i-dy=0 or (a + x) dx -\- r dy — 0. Integrating we have the general solution a^ + i;z;2 4-log(6 + 2/) = (7, where C is an arbitrary constant. It is to be noticed here, as in any case when we can separate the variables^ the solution is easy. Thus if /(^) F{y).dx-^^ {x) .'f{y).dy = Oy^Q have f{x).dx ^{y).dy <^(.') ^ F{y) -"' and this can be at once integrated. FIRST ORDER AND FIRST DEGREE. 323 {l+x)i/.dx-\-(l -y)x.dy={), \ogx + x + \ogy-y=G, log xy — G+y — x. Integrating, we have sin"^ x + sin~^ y — c. This may be put in other shapes. Thus taking the sines of the two sides of the equation we have X Vl-y2 ^ y Vl_^ = (7. '^^ (^ " ^ • iTv ^^^^"^^^ (2/ + 2/') dy = (x + ^) (^^. Answer : ^y^ ^ j^^s _ |^ ^_ ^^ _j_ constant. 5. ^±5") = ^. Answer: (1+^) (1+2/^) = c^. l-¥y^ ay ^ f \ a f 6. sin ^ . cos y . c^o; — cos a? . sin y . (ii/ = 0. Answer: cos y = c cos ij?. 7. {f ■\- xy'')dx -^ {cc" - yo(P')dy =^. Answer : loff - = c + ^ . y ^y ^' -^. + a/t-^=^- Answer: VlT^^ + VIT? == C. 238. Sometimes we guess and find a substitution which answers our purpose. Thus to solve dy _y^ — x dx Ixy ' . dx we try y — Nxv, and we find \-dv=^0, leading to log a? + — = c. Solve (y - x) Vr+^ ^ = ?i ( 1 + 7/2)1 21—2 324 CALCULUS FOR ENGINEERS. 239. If M and N are homog^eneous functions of ic and y of the same degree : assume y = vx and the equation reduces to the form of Art. 237. Example 1. ydx + (2 ^fxy — a;) dy — 0. Assume y — vx, dy==v.dx-{-x.dv, vx.dx-^ {2x s/v — x)(v.dx + x. dv) = 0, (2^^) dx + (2.X-2 VS - x") dv = 0, 2dx 2 Vti - 1 , ^ X \v V^J 2 log .r 4- 2 log^; + 2v-* = 0, log XV + v~* = (?, Answer: y = C€ V », where c is an arbitrary constant. Let y — vx and we find the answer x^ — y^ + log((;^-yi)(.^-y)i)-=0. Answer: a^ ^'^.Ay ^-A^. ' Remember that two answers may really be the same although they may seem to be altogether different. 4. Solve (x^ + Sxy"") dx + (y^ -f 3^y) dy = 0. Answer : ^-^ 4- 6.r'y^ + y*=C. V HOMOGENEOUS EQUATIONS. 325 5. Solve 3^24.(^^4.^)^ = 0. Answer : Zxy + x" + 2^^ ^ q^ (2y + xf. 6. Solve [x — y cos ^ J c?^' 4- a; cos - .dy =0. Answer : x = ce * _x 7. (y/ — x)dy-\-y. dx = 0. Answer : 1y = ce K 8. j;c?2/ — y-dx — ^x^ 4- y^ . cZ.:^ = 0. Answer : x^ = c^ + 2cy. 9. x-\-y~ ^%j. Answer : (./; - y ) e^ " ^ = (7. QjX 240. Of the form {ax + by-{-c) dx + (a'i^- + b'y + c') (/y = 0. Assume x = w + a, y = v + y^, and choose a and /3 so that the constant terms disappear. Thus if (3^' - 2y + 4>) dx -{■ (2x - y -^l)dy ==0 ', •a.sdx^dw and dy = dv, we have (Sw-{-S^ji-2v-2^-\-4>)dw + (2w-\-2oL-v~^ + l)dv = 0. Now choose a and /9 so that 3a- 2^ + 4 = and 2a-/34-l = 0, or -a4-2 = 0, or a = 2, /3=5. Therefore the substitution ought to be x=tu-^2, y = v-\-b, and the equation becomes a homogeneous one. Exercise. (3y - 1 x -{■ 1) dx -^ {1y - Sx + 3) dy=^0. ^ 'n.'- Answer : (y — x + 1)^ (y + x — ly — c. r, • dy 2x — y-\-l _ Exercise. -f + ^ ^ — ^ = 0. ax zy — x—1 Answer : x^ — xy-{-y'^-\-x —y=c. 241. Exact Differential Equations are those which have been derived by the differentiation of a function of x and y, not being afterwards multiplied or divided by any function of x and y. Consult Art. 83. .326 CALCULUS FOR ENGINEERS. Mdx 4- Ndy = is an exact differential equation if .dM\ (dN\, ,, df{x,y) ,^ d., . the primitive being /(^, y)=c. It will be found that (x" '-Safy)dx-\-(y^- a-') dy = 0, is an exact differential equation. Then x'-Sx^y=^ ^-^'^f''^\ OjCO Integrating therefore, as if y were constant, and adding Y an unknown function of y, instead of a constant, f{x,y) = la^-x^y^Y. Differentiating as if /c were constant, and equating to Ny we have dV -J- = y^ and hence Y=^y^ + c. Hence ^x" - afy + ^y^ + c = 0, where c is any arbitrary constant. 242. Any equation M . dx-\- N .dy = may be made exact by multiplying by some function of x called an Inte- grating Factor. See Art. 83. For the finding of such factors, students are referred to the standard works on differential equations. 243. Linear equations of the first order. These are of the type s+^y=« W' where P and Q are functions of x. The general solution is this. Let IP.c^a* be called X, then y=e^{/e^.Q.dx+cJ (2), where C is an arbitrary constant. LINEAR EQUATIONS. 327 No proof of this need be given, other than that if the vakie .of y is tried, it will be found to satisfy the equation. Here is the trial : — (2) is the same as ye^ = I e^Qdx-\- G (3). and dy J XT Diiferentiating, and recollecting that -y— = P, (3) becomes dx e^ + ye^P=e^Q (4), ov -~- + Py — Q, the original equation. To obtain the answer (2) from (1), multiply (1) by e^ and we get (4) ; integi-ate (4) and we have (3) ; divide by e^ and we have (2). We have, before, put (1) in the form (e + P)y=Q or y^ie + PrQ. and now we see the general meaning of the inverse operation (d + P)-\ In fact if I P.dwhe called X, {d + P)-i Q means, e^^ || e=^ . Q . dx + c| . . .(5). Thus if Q is 0, (0 + P)-i = Ce-^. Again, if P is a con- stant a, and if Q is 0, then (6 + a)~^ = Ce~"^, where C is an arbitrary constant. We had this in Art. 168. Again, if Q is also a constant, say n, {6 4- a)-' n = e""^ I ( He'*^ . dx -[- C {!■ = a€-«* + - (6), ft where G is an arbitrary constant. See Art. 169. " Again, if Q = e^^, ((9 + a)-' e^ = e-^'-^ | j €'«+^) "^ . dx ■{■ g\ 1 .(7). 328 CALCULUS FOR ENGINEERS. It is easy to show that when a = — h,y = (G-i-a;) €~"* li Q — h sin (ex + e), (6 + a)-' h sin {ex -\-e) = e-«* j h | €*»* sin {ex + e) . dx + C = Ce-«^ + , sin fco; + e - tan-^ -^ (8). 244. Example. In an electric circuit let the voltage at the time t be F, and let (7 be the current, the resistance being R and the self-induction L. We have the well-known equation F=ie(7 + X^, or ^^^G=\rV. Now dt L L --t (1 f-t ) and hence C=6 ^ ]j j^^ F . cZ^ 4- constant -4 [ (1). Of this we may have many cases. 1st. Let F at time 0, suddenly change from having been a constant Fj, to another constant V^. Put F=F2 therefore in the above answer, and we have 0=r^'UF,e^ + 4| V -it y To determine A we know that = ^ when^=0; there- in! V V V — V fore -^^ = ^^ 4- ^ so that A = — -„ — ^ and hence ±1 li K c=Yi-Y^^I-^r^' (2) ELECTRICAL EXAMPLE. 329 K/ --t\ Thus if Fi was 0, ^=^(1--^^) (3)> showing how a current rises when a circuit is closed. V -^t Again if F2 is 0, ^=:^^^ ^^)' showing how a current falls when an electromotive force is destroyed. Students ought to plot these values of C with time. Take as an example, F^=100 volts in (3), i2 = 1 ohm, L = -01 Henry. Again take Vi= 100 volts in (4). Compare Art. 169. 2nd. Let V at time suddenly become Vq sin qt, a=e ^ ^(F V f -t ~ le^ .smqt.dt-\-A itfR . R \ ^r ^^ f -f sin qt — q cos qt ) -^t V This becomes C = Ae ^ + , " sin (qt -e).. .(5), where tan e = ~ . M The constant JL, of the evanescent term Ae ^ , depends upon the initial conditions ; thus if (7 = when ^ = 0, = ^.-^=^ sing, or A = VoLql(R' + Df). Students ought to plot curves of several examples, taking other initial conditions. 330 CALCULUS FOR ENGINEERS. 245. Example. A body of mass M, moving with velocity V, in a fluid which exerts a resistance to its motion, of the amount fv, is acted upon by a force whose amount is F at the time t. The equation is M%+fv = F. Notice that this is exactly the electrical case, if M stands for Z,/for R, F for Volts V, v for C; and we have exactly the same solutions if we take it that F is constant, or that F alters from one constant value Fi to another F.., or that F follows a law like Fo sin qt This analogy might be made much use of by lecturers on electricity. A mechanical model to illustrate how electric currents are created or destroyed could easily be made. The solutions of Linear Differential Equations with constant coefficients have such practical uses in engineering calculations that we took up the subject and gave many examples in Chap. II. Possibly the student may do well now to read Art. 151 over again at this place. 246. Example, x ^ = ay-^x + \, dy a ^ 1 OC X X I- — . dx=X= — a loff a-. X ^ Observe that e-'*^^'^'" = x'"', e^^^K'^ = x^. Hence y = x^\ L"" ( 1 + -\ dx -\- c\ y y = x^ ||(^-« + x-^-^) dx-^c\, " \ — a a the answer, where C is an arbitrary constant. LUBRICATION OF JOURNAL. 331 247. There being continuous lubricating liquid between the surfaces AB and EF as of a brass and a journal. OC=/iQ the nearest distance between them. At the distance ^, measured along the arc OAj let the thickness be h. Anywhere in. the normal line there, representing the thickness, let there be a point in the liquid at the distance y from the journal, and let the velocity of the liquid there be 2t. Then if p be the pressure it can be shown '^- 541 (^)' if /u, is the coefficient of viscosity of the lubricant, and Wq the linear velocity of the journal, and ic is the velocity of the liquid at any place; we have no space for the reasoning from (1) G^Uq dh h^ duo Fig. 103. leading to d'^p 3 dh dp dx^ h ' dx dx .(2). Let dp _ dx 0, then d4 3 dx h dh 6/iWg ^^^ _ /^ dx ' ^ A3 dx'' This is of the shape (1) Art. 243, h in terms of x being given. Let X= JP . dx= f f . ^ . dx=3\ogh, e^=h\ Hence (t> = h \l -A3 QjiUq dh h^ dx dx + & }.-<^=-|= A-'(6;.«„A + = ^?+f3. The solution depends upon the law of variation of A. The real case is most simply approximated to by h = hQ + ax^: using this we find p = C' GixUq ~2h V C'-C If students were to spend a few weeks on this example they might be induced to consult the original paper by Prof. 0. Reynolds in the Phil. Trans, vol. 177, in which he first explained to engineers the theory of lubrication, t Numerical Exercise. Let OB = 2-59, OA = 11 -09 centimetres. fi = 2-l6, Aoor 0C=0-001135, a- -0000082, W(, = 80cm. per second. 332 CALCULUS FOR ENGINEERS. Calculate C and C\ assuming jo = at B and at A. Now calculate the pressure for various values of x and graph it on squared paper. The friction per square cm. being /ij- at y=0, the total friction F will be found to be 11 between the limits A and B. The total load on the bearing is \p.dx bearing is / ^ . ( between the limits, if AB covers only a small part of the journal, and may be calculated easily in any case. The bearing is supposed to be infinitely long at right angles to the pai)er in fig. 103, but forces are reckoned i)er cm. of length. 248. Example. Solve Writing this in the form (^ - 4^ + 3) 7/ = 2e^, Now (e^ - 4^ + 3)- = i (^-^^ - ^ :^ i) • Indeed we need not have been so careful about the J as it is obvious that the general solution is the sum of the two (^— 3)~^ and (^ — 1)~^ each multiplied by an arbitrary constant. Anyhow, y = ^__^_^, and this by (7) Art. 243 is or y=(C^+a')^-hG.je\ 249. Equations like ^ 4- Py = Qy", where P and Q, as before, are functions of oo only. Divide all across by y* and substitute z = y^~'\ and the equation becomes linear. Example. (1 — w"^) -^ — cci/ = ax\f. DIFFERENTIAL EQUATIONS. S38 Substituting z = y~^ we find dz xz ax dx \—x^ \-x'^' Answer : 2/-1 = (1 _ ^)*{_ a(l -j;2)-^ + 0} = - a + OVr^T^^. Exercise. ^~j -^V =y^ log iz*. Answer : - = 1 + Oa; + los: a?. 2/ ^ 250 . 1. Given C^Y-ay = 0. This is an equation of the first order and second degree. Solve for -^ and we find two results, dx dt/ -^ — ai/ = 0, so that log y—ax—A^ = 0, -^ 4- a^ = 0, so that log y + ax— A2 = 0. Hence the solution is (log y - ax - A,) (log y + ax-A^)= 0. It will be found that each value of y only involves one arbitrary constant, although two are shown in the equation. 2. Given i + (^^Y = ^, This is an equation of the first order and third degree. Hence 2 = (.^-l)*, and y = |(.r-l)HG 334 CALCULUS FOR ENGINEERS. 3. Given (^y_7^ + 12 = 0. \dxj ax This is an equation of the first order and second degree. (2-*)g-»)=»^ (3/ - 4^ + Ci) (y - S.'Zr + Co) = 0. 251. Clairaut's equation is of the first order and of any degree y^xp-Vf{p) (1), where p is -^ and f{p) is any function of ~ . Differentiate with regard to x, and we find ^■^iMt=' <^)- So, either ^ = (3) ^ + ^fiP) = 0.... (4) will satisfy the equation. 1=0., =0. Substituting this in (1) we have y = CX+f{c) (5), which is the complete solution. Eliminating^ now between (1) and (4) we obtain another solution which contains no arbitrary constant. Much may be said about this Singular Solution as it is called. It is the result of eliminating c from the family of curves (5), and is, therefore, their Envelope. See Art. 224. Example of Clairaut's equation. m We have the general solution (5) ... y — cxA — , a family c HIGHER ORDER THAN FIRST. 385 of straight lines the members of which differ in the values of their c. Hence y = 2 Jmx or y^ = 4ma;, a parabola which we found to be the envelope of the family in Art. 224. This curve satisfies the original equation, because in any infinitesimal length, the values of x, y and -^ are the same for it as for a member of the family of straight lines. 252. If a differential equation is of the form it can be at once solved by successive integration. We have had many examples of this in our work already. 253. Equations of the form -^^=f{y); multiply by 2 ^ cix dx and integrate and we have Extracting the square root, the equation may be solved, as the variables are separated. Thus let g = ay Proceeding as above, y -=^dx. Integrating we find x=^^\og{ay + JaY-^C}-\-C' (1). 336 CALCULUS FOR ENGINEERS. If this equation (1) is put in the shape ce'^^ = ay-\- Ja^ + C, it becomes c^e^'' - ^ayce*"^ = C, ^ la or ?/=^6«^ + ^e-«^ (2), which looks different from (1) but is really the same. (2) is what we obtain at once if we solve according to the rule for linear equations, Art. 159. 254. Solve -7^= a -s~, an equation of the third order and first degree. Let, g = ?, theng = a5, 9 = 6e«* so that t^ = - «"^ + G, ^ a? or y == Ae"^ •\- Coo+ C\ where A, 0, C are arbitrary constants. This also might have been solved by the rule for linear equations, Art. 159. 255. Solv( a. dp ^^=:dx SO that -=log {p+ V«2+ 1) +C", or (7e<»-jp = Vp2 + l. 1 -^ d. 20 ~dx - 1 -- dti Squaring, we find p = ^Ce** — ^r^ e " EXERCISES. 337 Integrating this we have X X where G and c are arbitrary constants. 256. General Exercises on Differential Equations. (1) {a? + y-) dx + Ja^^^ .dy = 0. x X y Answer : sm"^ - + — tan"^ ^ = c. a a- a /^\ dy X 1 ^^ dx^l-\-x^'^ 2x(l+x')' . , _ 1 + JU^'] Answer 1 f ,, l-fVl+^^'l (3) _ + ycOS*--^. Answer : i/ = sin i» — 1 + Ce" ^^"'^. Answer : (2y -x'^-c) (log (a; + y - 1) + ^ - c} = 0. (6) J+2^y = 2a^y. Answer : y = (Ce^^' + ^a {2x^ + l)}"*. (6) 1 = f ^V + ?^ ^ . Answer : y' = 2c^ + c\ ^ ^ \dxj y dx (7) x-^ + y = y^logx. Answer: y == {ex + log x + 1)-\ (8) y = 2x^- + y^ C^Y . Answer : y^ = 2c^ + c\ (9) Solve (l + 5)c^^-2|c^y=.0. 1st, after the manner of the Exercise of Art. 241. 2nd, as a homogeneous equation. Answer: af — y"==cx, P 22 338 CALCULUS FOR ENGINEERS. (10) Solve — '- f- 1 -^ J c?y = after the manner of Art. 241. Answer : x^ — y^ = cif. Answer : y = {x-[-\y [^ {x + 1)^ + C). (12) (a; + 2/)2 ^ = a\ Answer : y-a tan-^ ^^ = c. (13) xy (1 4- a??/2) ^ = 1. Answer : - = 2 - 2/^ + ce-^J/'. Answer : (y —a log a; — c) (?/ + a log a; — c) = 0. Answer : y = (ao + aio; + aoa?'' + a^) e~^. Answer : a? = (ai + dj^) e-^' + /-^r— Tvj • Answer : 3/ = (a? + 1)" (e* + c). (19) (1 —x^)-f- — xy = axy\ Answer : - = c Jl — a^ - a. (1^) yi = sin (^ - 6). Answer : cot |j - ^-^ I = <^ + c. Answer: 3/ = C^ie'^^ + e^ Jf Og-^j cos a; + fOg + ^jsinajl . (21) Change the independent variable from a; to i in (1 — a?) -~ — a? -^ = 0, if a; = sin f, and solve the equation. ELLIPTIC INTEGRALS. 339 (22) Also in {a- +x'')-j^^-\-2x -^ = 0, ii x=a tan t, and solve the equation. % «■ '■ '-■• S=-£/©'. , „ d's [dt dH „fdHV\ fdty *""* ^- -dt= = -\ds-d^-^W)\^\ds)- 3. Prove as i^ = -^ / tt, so we have also ax dt I dt d^y _ fdx (Py _ d'x dy\ ^ /dxV dx^ ~ [dt df dt^ ' dt) ' \dt) ' and find the equivalent expression for -^^ . 4. If X = e* show that b,sx^ — x-^-~-t: this equals -^ . dx dt dt ^ dt Also a^^^Jl-^^^-fl-l]^ ^^^"^ "^ da?~ dP dt~\dt V dt' and a^^-(l-2\(^-l]^ ^"""^ "^da--\dt V\dt ^Jdt' 5. Change the independent variable from x to t if x = €*, -^--\-x^ dx^ dx in x^ ;r 2 + ^' ;/ + ^^^y = ^> ^^^ ^^^^^ *^® equation. 258. If we try to find by the method of Art. 47 the length of the arc of an elhpse, we encounter the second class Elliptic Integral which is called E (k, x). It may be evaluated in an infinite series. Its value has been calculated for values of k and x and tabulated in Mathemati- cal tables. When the angle through which a pendulum swings is not small, and we try to find the periodic time, we encounter the first class Elliptic Integral which is called F (k, x). It can be shown that the integral of any algebraic expression involving the square root of a polynomial of the third or fourth degree may be made to depend on one or more of the three integrals 22—2 340 CALCULUS FOR ENGINEERS. rx /i _ z.2«.2 re . f 1 \ i"" ^ 7r(w, h. 6)=\ ;o(l+7ism2^) do \Jl-k^Hm^6' k, which is always positiA e and less than 1, is called the tnodulus. n, which is any real number, is called the parameter. The change from the x form to the 6 form is effected by the substi- tution, :r = sin 6. When the limits of F and E are 1 and in the x case, or - and in the B case, the integrals are called complete, and the letters K and E m erely are used for them. B is called the amplitude and \/\ — ^2sin2^ is called by the name A^. If u = F{k, x) = F{ky B\ then in dealing with fimctions which have the same k if we use the names ^ = amw, x=Bnu (in words, x is the sine of the amplitude of w), *^l — x^=cmi (or Vl — -^ is the cosine of the amplitude of u), Vl - k^x^=dnu (or Vl - k'^^^ is the delta of the amplitude of u), it is found that sn^w-f cn2?< = l, dn'^u+k^.Bn^u=l, -=-(amw)=dnw, &c. Also am { — u)=— am . ti, &c. ., / . \ sn w. en V. dn v + cni/. snv.dn . -M Also sn {u±v) = Yx-== r, , and similar relations for en {u±v) and dn {u±v). Expressions for sn (tc + v)+sn(u- v), &c. follow. Also for sn2u, cn2w, dn2u. So that there is as complete a set of formulae connecting these elliptic functions, as connect the Trigonometrical functions, and there are series by means of which tables of them may be calculated. Legendre published tables of the first and second class integrals, and as they have known relations with those of the third class, special values of these and of the various elliptic functions may be worked out. If complete tables of them existed, it is possible that these functions might be familiar to practical men. 259. To return to our dififerentiation of IHinctions of two or more variables. HEAT CONDUCTION EQUATION. 341 1. If u = z'^ + 7/^ -\- zy and ^ = sin a; and y = e^, du and hence ^ = (Si/^ + ^) e* + (2^ + y) cos a;. If this is expressed all in terms of x we have the same answer that we should have had, if we had substituted for y and z in terms of x in u originally, and differentiated directly. 2. If u= a/ — — „ , where v and w are functions of «?, V v^ + w- nnd -Y . ax 3. If sin (xy) = ma?, find -^ . 4. If %i — sin~^ - , where z and y are functions of x, find da?* 5. If \i = tan~^ - , show that du, = ^-^^ — -^ . y f + 2^ 260. Exercise. Try if the equation dx^~Kdt ^ ■ has a solution like v = e*^ sin (qt + yx), and if so, find a and y, and make it fit the case in which v=0 when x = oo , and v = asm qt where x = 0. We leave out the brackets of 'dv"" C) &c. dv -7- = «€"* sin (qt + yx) + e*^ 7 cos (qt + ^cij), — — = a^e*^ sin (g^ + yx) + a7€*^ cos (qt + 7.<c) + a7e** cos (qt + yx) — 6**7^ sin (qt + 70;). dv Also -^ = ge*^ cos (qt + 7^), 342 CALCULUS FOR ENGINEERS. SO that to satisfy (1) for all values of t and oc a"-rf=:0 or a= ±y, and ay + ay _2 As - is not zero, a = + 7 only, 2a=» = ^, a= + K Hence we have v= A e"'* sin (qt -f cue) -f i^e"** sin (qt — ax) where A and B are any constants, and ol= /u -x— or a/ if ^ = 27r/2. Now if i; = when x= oc , obviousl}^ A=0. If V = a sin qt where x=0, obviously B = a. Hence the answer is Trn V — ae V " sin (27r?^^ - .'?? W K ) .(2). 261. Let a point P be moving in a curved path ^PQ; let^P = a^,j5P = y, Cm *ir (ill -^ and -~ being the velocities in the directions OX and OY^ -j~ and -7^ being the accelerations in the directions OX and OY. Let OP=r, BOP —6,x = r cos ^, ?/ = r sin ^. The acceler- ation or velocity of P in any direction is to be obtained just as we resolve forces. Thus the velocity in the direction r is Jcos^4-Jsin^...(l), and in the direction PT which is at right angles to r, the velocity is Pig^ 104 -t«in^ + ^!cos^ (2). Y A t\ \ ; p o\ I 3 dt dt POLAR CO-ORDINATES. 343 Now diiferentiating a; and y, as functions of the variables r and 0, since ( ^— j = cos and f -^ J = — ?* sin 6, dx dr ^ . ^ dO ,.-,. 5«=*'=°^^-'-^'°^rfi ^^^' dy dr . ^ ^dO ,^. Solving (3) and (4) for j- and r -j- , we find dr dx ^ dy . ^ = ;i- cos ^ +-^ sm d (o), (^^ dt dt dO dx . ^ a dt dt dt dO dx . ^ dy ^ ,_. From (1) and (2) we see therefore that -j- is the velocity de dt in the direction OP and that r -^ is the velocity in the direction PT. Some readers may think this obvious. Now if we resolve the x and y accelerations in the direction of OP and PT, as we did the velocities, and if we again differentiate (3) and (4) with regard to t, we find d^x d^y Acceleration in direction OP = -i- cos ^ + -i^ sin ^ . . .(7), d^x d^y Acceleration in direction PT— — ^ sin ^ + -j^ cos ^...(8). And S={£-'©"}-*-('Sf-S) '"•■■<'>. S=(S-(S)l"»»-(4JS-S)~«-» And hence, the acceleration in the direction r is (and this is not very obvious without our proof), dt' '\dt) (iiX 344 CALCULUS FOR ENGINEERS. and the acceleration in the direction PT is ^dtdt^"^ dt^'°' rdtK dt) ^^^^- r^ -J- is usually called A. It is evidently twice the area per second swept over by the radius vector, and (12) is - -^ . 262. If the force causing motion is a central Force, an attraction in the direction PO, which ia a function of r per unit mass of P, say /(r) ; or mf{r) on the mass m at P; then (12) is 0, or r^ -^- = constant, or h constant. Hence under the influence of a central force, the radius vector sweeps out equal areas in equal times. Equating mf{r) to the mass multiplied by the accelera- tion in the direction PO we have /<')"©-!?■ "»> But r"^ -^ = h a. constant. As r is a function of 6 dt dr _dr dd _dr h di~dd'di~der^' ^ ^\^h _ h (dA^^)h df'~\d6'7'' 7^\dd)]r'' and we can use these values in (13) to eliminate t. If we use - for r, (13) simplifies into f(r)=h2u2(^+u) (14). If f(r) = a?'~" or a^*", an attraction varying inversely as the ?ith power of the distance, -^ + ^^ = I^ '^^~^ — ^w"~^ say. Multiplying by 2 -^ and integrating, we have (sr— «-?!— <->• CENTRAL FORCE. 345 Thus let the law be that of the inverse square; f{r) — ar~^ or ait^ ; (14) becomes (Pu a , Let lu — iL — h, then The solution of this is, -m; = ^ cos (^ + B), and it may be written i,= l = ^{H-ecos(^-a)l (IG). This is known as the polar equation to a conic section, the focus being the pole. The nature of the conic section depends upon the initial conditions. (15) enables us, when given the shape of path, to find the law of central force which produces it. Thus if a particle describes an ellipse under an attraction always directed towards the centre, it will be found that the force of attraction is proportional to distance. It is easier when given this law to find the path. For if the force is proportional to PO, the X component of it is proportional to oo, and the y component to y. If the accelerations in these directions are written down, we find that simple harmonic motions of the same period are executed in these two directions and the composition of such motions is well known to give an elliptic path. If the law of attraction is the inverse cube or f{r) = (7?'~^ = au^, (14) becomes -j^- + ?^ = r^ ^^• If ^-l = a-, w = ^6*« + 56-*^ h If l-~ = ^2 u = Asm0 + Bcos0e, fi- giving curiously different answers according to the initial conditions of the motion. 346 CALCULUS FOR ENGINEERS. 263. If a? = r cos 0,y = r sin 6, so that if y is a function of x, r must be a function of ^ ; if i* is any function of x and y, it is also a function of r and 6. Express f -7- ) and f -^ j in terms of the polar co-ordinates r and 6. \drJ~\dx)dJi^^\d^)d^ ^^^' 6 being supposed constant, and fdu\ _ fdu\ dx /dic\ dy \de)~\Tx)de^\d^jlde ^^^' r being supposed constant. dx Now -77; if r is constant = — ?- sin 0, au -M. if r is constant = r cos ^, -7- if ^ is constant = cos ^, — ^ if ^ is constant = sin ^. ar Treating (-7- J and (-7-) in (1) and (2) as unkno\vn, and finding them, we have (|) = .„..g).Jeos..(|) (4). Notice that in [-,— ], the bracket means that y is supposed to be constant in the differentiation. In ( -7- j , it is ^ that is supposed to be constant. r, ^, <^ CO-ORDINATES. 347 In (o) or (4) treat ( i-l or ( -, ) as it is treated, and find ^i— and Ti^ . However carefully one works, mistakes are da^ dy^ *^ likely to occur, and this practice is excellent as one must think very carefully at every step. Prove that d2u d2u d^ di^ d^'^dy2"dr^ "^^r dr'*"?d9^' 1 du r dr 1 d2u .(5). 264. Sometimes instead of x, y and z, we use r, 6, (f) co-ordinates for a point in space. Imagine that from the centre of the earth (Fig. 105), we have OZ the axis of the earth, OX a line at right angles to OZ, the plane ZOX being through Greenwich ; OF a line at right angles to the other two. The position of a point P is defined by x its distance from the plane ZOY, y its distance from the plane ZOX, z its distance from the equatorial plane YOX. Let r be OP the distance of the point from 0. Let <^ be the west longitude or the angle between the planes POZ and XOZ\ or if Q be the foot of the perpendicular from P upoii XOY^ the angle QOX is <^. Let Q be the co-latitude or the angle POZ. Then it is easy for anyone who has done practical geometry to see that, drawing the lines in the figure, QfiO is a 348 CALCULUS FOR ENGINEERS. right angle and OR = x, QR — y, also PQO is a right angle, a; = rsm6. cos </>, y = r sin 6 .sincf), z = r cos 6. If u is a given function of x, y and z, it can be expressed in terms of r, 6 and ^, by making substitutions. It is an excellent exercise to prove du . ^ , du cos 6 . cos <f) cZm sin ci c??^ -y- = sin ^ cos </> ;t- H ^ -ja ^n -ji , dx ^ dr r do r sm 6 d<f> du . ^ . , du cos 6 ,sin<f> du cos <f> du dy ^ dr r dd r sm 6 d<f> du _ ^ cZ?^ sin ^ du dz dr r dO' It will be noticed that we easily slip into the habit of leaving out the brackets indicating partial differentiation. The average student will not have the patience, possibly he may not be able to work sufficiently accurately, to prove that d2u d^ d2u_d2u l^d^ 33? + dy2 ■*■ dF " dp" "*" r2 d(92 1 d2u 2du cotg du ■^r2sin2^'d^ + 7d?"^"J2-d9 ^ ^• This relation is of very great practical importance. 265. The foundation of much practical work consists in understanding the equation d^u dhc d^u__ldu dx^'^df'^'d?~'K~dt ^ ^' where t is time. For example, we must solve (1) in Heat du Conduction Problems if u is temperature, or in case ;t- = and u is electric or magnetic potential, or velocity potential, in Hydrodynamics. (1) is usually written ^^-4S (^>- ZONAL HARMONICS. 349 We see then in (A) the form that V"u takes, in terms of r, and </> co-ordinates. We know that if u is symmetrical about the axis of z, that is, if u is independent of </>, the above expression becomes _, dhi 1 d-ii 2 da cot 6 du ,^. ^'''=d;^-^?w^-^rd^+-^d0 <^>- 266. Students are asked to work out every step of the following long example with great care. The more time taken, the better. This example contains all the essential part of the theory of Zonal Spherical Harmonics^ so very useful in Practical Problems in Heat, Magnetism, Electricity, Hydrodynamics and Gravitation. When to is independent of </> we sometimes write (2) in the form du K id f „du\ 1 d / . ^du 'hMei^-'U ^^>' dr u being a function of time t, r and 6, The student had better see if it is correct according to (3). dxi If -^ = 0, show that the equation becomes „d^u - du ^ ndu d^u _ ,^. ,._ + 2r^^ + cot^^^ + ^, = (2). Try if there is a solution of the form u = RP where R is a function of r only, and P is a function of 6 only, and show that we have T^d^R 2rdR ,AdP 1 d'P ,^, R^^^Rd^=-''''^^p~de-pd¥ (^^- Now the left-hand side contains only r and no 0, the right-hand side contains only 6 and no r. Consequently each of them must be a constant. Let this constant be called C and we have drR ldR_RG_ dr' +r dr r^ ~" ^^^' %^.oie%^PG = (5). 360 CALCULUS FOR ENGINEERS. There is no restriction as to the value of 0, and it must be the same in (4) and (5), and then the product of the two answers is a value of u which will satisfy (2). The solutions of many linear Partial Differential Equations are obtained in the form of a product in this way. There are numberless other solutions but we can make good practical use of these. We have then reduced our solution of the Partial Differ- ential Equation (1), to the solution of a pair of ordinary differential equations (4) and (5). Now a solution of (4) may be found by trial to be r*", and when this is the case we have a method (see Art. 268) of proving the general solution to be ii = ^r»« + 5r-(»«+i' (6), where G is m (m + 1); anyhow (6) will be found by trial to answer. Using this way of writing G in (5) and letting cos = fjb, we find that we have an equation called Legendre's Equation, an ordinary linear equation of the 2nd order We now find it convenient to restrict m. Let m be a positive integer, and try if there is a solution of (7) in the form P = 1 + Aj/M + A^^ + A^^ + &c. Calling it P^ (/it) or P,^ (0), the answers are foimd to be Pq (0) = 1, if m is put 0, Pi (0) = ^, if m is put 1, Pg (0) = 3^2 _ ^, if ^ is put 2, Ps (6) = f /!« - f /x, if m is put 3, P, (0) = ^fi* - ^fi^ + 3, if m is put 4. A student will find it a good exercise to work out these to Pg. My pupils have worked out tables of values of Po, Pi, P2, &c., to F7 for every degree from ^ = to ^=180°. See the Proceedings of the Physical Society, London, Nov. 14, 1890, where clear instructions are given as to the use of Zonal Harmonics in solving practical problems. We see then that (Ar- + j^,)p„(<)) (8) BESSEL FUNCTION. 351 is a solution of (1). A practical problem usually consists in this : — Find u to satisfy (1) and also to satisfy certain limit- ing conditions. In a great number of cases terms like (8) have only to be added together to give the complete solution wanted. In the present book I think that it would be unwise to do more in this subject than to set the above very beautiful exercise as an example of easy differentiation. (8) is usually called The Solid Zonal Harmonic of the mth degree^ P^ (6) is called the Surface Zonal Harmonic of the mth degree. 267. In many axial problems, u is a function only of time and of r the distance of a point from an axis, and we require solutions of (1) which in this case becomes d^u \ du ^\ du dr^ r dr k di * Let us, as before, look for a solution in the form u = BT (2), where i? is a function of r only and J' is a function of t only. (1) becomes rpd^_^lrpdE^lj^dT dr^ r dr k dt' Dividing by RT Rdr'^r Rdr~ kT dt~ ^"^ ^^^' where /^^ is a constant. Jim Then -m = - /cfi^dt or log T= — k/jlH + c, or T=Ge-'^f^'i (3), where G is an arbitrary constant. We must now solve d'R IdR ,^ ^ ,,, ^+-r;^+^^^=^ (^)- Let r = - and (4) becomes d2R 1 dR . „ ^ ,^^ dF + xdJ + ^ = ^ (^>- 352 CALCULUS FOR ENGINEERS. Assume now that there is a solution of (5) of the shape we find that A = G = E= G=0 and in fact that x^ X^ x^ x^ This is an important series first used by Fourier, although it has Bessels name. It is called the Zeroth Bessel and the symbol Jo (*") is used for it. Tables are published which enable us for any value of x to find Jo(x). Thus then R = Jo {^r) is a solution of (4), and hence u=Ge-'^|'''J,{^lr) (7) is a solution of (1). Any solution of (1) needed in a practical problem is usually built up of the sum of terms like (7), where different values of iju and different values of G are selected to suit the given conditions. 268. In the linear differential equation when P and Q are functions of x, if we know a particular solution^ say y=^v, we can find the general solution. Substitute y = vii, and we get "d^+^id-. + ^Vd^-"' (2). Calling -p= u\ (2) becomes U V or log u' + log v^+ I P .dx = constant. LINEAR EQUATIONS. S53 Let I F .dx = X then u' or -^ = A ~ e-^, J ax v^ u^B + Al^^e-'^.dx (3). Thus we lind the general solution y = Bv-\- Av I — €~^ .dx .(4), where A and B are arbitrary constants. Even if the right-hand side is not zero, the above substitution will enable the solution to be found, if v is a solution when the right-hand side is 0. Easy Example. One solution of -r^ + a^x = is ?/ = cos ax. dx^ "^ Find the general solution. Here P = so that jP.dx^X^ 0. f dx Hence y = B cos ax + A cos ax I — -i, — , •^ J cos^ax and as I — - — = - tan ax, J cos^ ax a we have as the general solution y = B cos ax+G sin ax. Exercise. We find by trial that y = x^"' is a solution of ic* ^ + 2a; ^ - m (m + 1) y = 0, see Art. 266. Show that the general solution is y=Ax^-^ — ;^^. X Exercise. We find by trial that y — e^^ is a solution of d^v -T^ = a^y, show that the general solution is y = Ae^^ + i?e~^*. P. 23 S54 CALCULUS t'OR I:NGINEERS. E.rercise. We saw that u — P^n{t^) is a solution of Legcndre's equation Art. 2GG, prove that xl =APin(fi)-\-BQmi^) is the general solution, where Q,„W=P,„(^)/^j-^^- Wl" Qm(H') or Qui(0) is called the Surface Zonal Harmonic of the second kind. Exercise. We saw that Jo(x) was a solution of the Bessel equation (5), show that the general solution is AJq {x) -f BKq {x)y where Kq (x) is called the Zeroth Bessel of the second kind. 269. Conduction of Heat. If material supposed to be homogeneous has a plane face AB, If at the point P which is at the dis- tance X from AB, the temperature is v, and we imagine the temperature the same at all points in the same plane as P parallel to -4^ (that is, we are only con- sidering flow of heat at right angles dv to the plane AB), and if t- is the rate of rise of temperature per centimetre at P, then — A; -7- is the amount of heat flowing per second through a square centimetre of area like PQ, in the direction of increas- Fig. 106. -^g ^ rpj^-g -g ^^^ definition of k, the conductivity. We shall imagine k constant. k is the heat that flows per second per square centimetre, when the temperature gradient is 1. Let us imagine PQ exactly a square centimetre in area. Now what is the flow dv across TS, or what is the value of — A; -j- at the new place, dx dv which is x + hx from the plane ABl Observe that —k-j- is a function of x\ call it f{x) for a moment, then the HEAT CONDUCTIVITY. 355 space PQTS receives heat f(x) per second, and gives out heat f(x + Bx) per second. Now f(x-\- Bw) -f{x) = Bx i^^ . These expressions are of course absolutely true only when Bx is supposed to be smaller and smaller without limit. We have then come to the conclusion that —Bx-j- f(x) is being added to the space PQTS every second : this is — Bx^ri— f^' 1-] or +k.Bx-j—. ax \ ax J oar But the volume is 1 x Bx^ and if w is the weight per cubic centimetre, and if s is the specific heat or the heat required to raise unit weight one degree in temperature, then if t is time in seconds, 5j dv W .ox .S-Ti dt also measures the rate per second at which the space receives heat. Hence 1 ^ d'^v ^ dv tc, ox. -j-- — W.OX.S.-Tly da? at dH ws dv /,v d^^T-dt ^^)- This is the fundamental equation in conduction of heat problems. Weeks of study would not be thrown away upon it. It is in exactly this same way that we arrive at the fundamental equations in Electricity and Hydrodynamics. If flow were not confined to one direction w^e should have k the equation (1) of Art. 265. — is often called the diffusivity ^ ws for heat of a material, and is indicated by the Greek letter K', wsis, the capacity for heat of unit volume of the material. Let us write (1) as dh^ __ 1 ^ /.^K dx^'Kdi ^"^' 23—2 356 CALCULUS FOR ENGINEERS. 270. It will be found that there are innumerable solutions of this equation, but there is only one that suits a particular problem. Let us imagine the average temperature every- where to be (it is of no consequence what zero of tempera- ture is taken, as only differences enter into our calculations), and that F= a sin 27r;i^, or a sin ^'^ (3), is the law according to which the temperature changes at the skin where a; is ; w or ^ means the number of com- plete periodic changes per second. Now we have carefully examined the cycle of temperature of steam in the clearance space of a steam-cylinder, and it follows sufficiently closely a simple harmonic law for us to take this as a basis of calculation. Take any periodic law one pleases, it consists of terms like this, and any complicated case is easily studied. Considering the great complexity of the phenomena occurring in a steam-cylinder, we think that this idea of simple harmonic variation at the surface of the metal, is a good enough hypothesis for our guidance. Now we take it that although the range of temperature of the actual skin of the metal is much less than that of the steam, it is probably roughly proportional to it, so we take a to be proportional to the range of temperature of the steam. We are not now considering the water in the cylinder, on the skin and in pockets, as requiring itself to be heated and cooled ; this heating and cooling occurs with enormous rapidity, and the less there is of such water the better, so it ought to be drained away rapidly. But besides this function of the water, the layer on the skin acts as creating in the actual metal, a range of temperature which approaches that in the steam itself, keeping a large. Our n means the number of revolutions of t lie engine per second. To suit this problem we find the value of v everywhere and at all times to be what is given in (2) of Art. 260, V = ae "V^sin (2',mt-x^^'j (4). This is the answer for an infinite mass of material with one plane face. It is approximately true in the wall of a thick PENETRATION OF HEAT 357 cylinder, if the outside is at temperature 0. If the outside is at temperature v and the thickness of the metal is h 1)' we have only to add a term y ^ to the expression (4). This shows the effect of a steam-jacket as far as mere con- ductivity is concerned. The steam-jacket diminishes the value of a also. Taking (4) as it stands, the result ought to be very carefully studied. At any point at the depth x there is a simple harmonic rise and fall of temperature every revolution of the engine ; but the range gets less and less as the depth is greater and greater. Note also that the changes lag more as we go deeper. This is exactly the sort of thing observed in the buried thermometers at Craigleith Quarry, Edinburgh. The changes in temperature were 1st of 24 hours period, 2nd of 1 year period ; we give the yearly periodic changes, the average results of eighteen years* observations. Depth in feet below surface Yearly range of temperature Fahrenheit Time of highest temperature 3 feet 6 feet 12 feet 24 feet 16138 12-296 8-432 3-672 August 14 „ 26 Sept. 17 Nov. 7 Observations at 24 feet below the surface at Calton Hill, Edinburgh, showed highest temperature on January 6th. Now let us from (4) find the rate at which heat is flowing through a square centimetre; that is, calculate —kj for any instant ; calling a/ — = a. dv dx — aae~*^ sin {2irnt — ax) — aae~'^^ cos {27rnt — ax), where x = 0, that is, at the skin, it becomes when multiplied 358 CALCULUS FOR ENGINEERS. hy —k; l—k -J-] = + kaa {sin 27mt + cos 27mt] = kaa V2 sin (2'Trnt + ^] , by Art. 116. This is + for half a revolution when heat is flowing into the metal, and it is — for the other half revolution when heat is flowing out of the metal. Let us find how much flows in ; it will be equal to the amount flowing out. It is really the same as kaa V2 | sin 27rnt . dt = kaa^/2 . - , where t = - , Jo IT n = «\/ 2kws nir That is, it is inversely proportional to the square root of the speed and is proportional to the range of temperature. We have here a certain simple exact mathematical result ; students must see in what way it can be applied in an engineering problem when the phenomena are very compli- cated. We may take it as furnishing us with a roughly correct notion of what happens. That is, we may take it that the latent heat lost by steam in one operation is less with steam jacketing, and with drying of the skin ; is pro- portional to the range of temperature of the steam, to the surface exposed at cut off, and inversely proportional to the square root of the speed. Probably what would diminish it more than anything else, would be the admixture with the steam of some air, or an injection of flaming gas, or some vapour less readily condensed than steam. When we use many terms of a Fourier development instead of merely one, we are led to the result that the heat lost in a steam cylinder in one stroke is {0,-0,){h + ^^AI>Jn, where 6^ is the initial temperature and 6^ the back pressure temperature, r the ratio of cut ofi", n the number of revolu- tions per minute, A the area of the piston h and c constants LIST OF INTEGRALS. 359 whose values depend upon the type of engine and the arrangements as to drainage and jacketing. 271. Students will find it convenient to keep by them a good list of integrals. It is most important that they collect such a list for their own use, but we have always found that it gets mislaid unless bound up in some book of reference. We therefore print such a list here. Repetition was unavoidable. Fundamental cases : 1. Ix"',dx = V x'"+\ J m + 1 2. j - .dx= log X. 4. la^.dcc = -, a*. J log a 5. /cos mo) .dx= — sin mx. J m sin mx .dx = cos mx. m 6. /si 7. I cot x.dx = log (sin x). 8. I tan X .dx= - log (cos x), 9. I tan X . sec x.dx = sec x. 10. I sec^ x.dx — tan x. cosec^ X .dx = — cot x. I „. /, , dx 1 . 12. I = - tan ax. cos^ ax a 360 CALCULUS FOR ENGINEERS. 13. / -^—z — = — cot ax. J sm^ax a 1 ^ f dx • , ^ 14. I — . = sin~* - . ^ V a^ - ar" « f dx 1 , X J a'^ + x^ a a li 10. / — _ =-sec' I dx 1 X —- zzi SGC~~ ~ X ^a?-a? a a 17*. I cosh ax .dx — - sinh ax. J a * From 17 to 23 we have used the symbols (called Hyperbolic sines, cosines &c. ) Binhx=i (c* - e-*), and cosech x— . . , cosha' = i(c*+6~*),8echa:= — ^^ , ^^ ' coah« ^ , sinh a; e^-1 ^, 1 tanh x= — i— = -iiT, — - , coth a; = cosha: e'^ + l' tanhx' Also if 2/ = sinh x, a: = sinh"^ »/. It is easy to prove that sinh (a + h) = sinh a . cosh h + cosh a . sinh h, cosh (a + &) = cosh a cosh h + sinh a . sinh t, sinh {a-b) = sinh a cosh & - cosh a sinh Z>, cosh (a - ?^) = cosh a cosh t - sinh a sinh t, sinh 2a = 2 sinh a . cosh a, cosh 2a = cosh^ a + sinh^ a = 2 cosh'^ a - 1 , = 2sinh2a + l. If we assume that aJ -1, or i as we call it, submits to algebraic rules and t2= - 1, t^= -t, 1*^1, 1^ = 1 &c. we can write a + bi as r (cos 6 + i sin ^), where r^=a'^+h^, and tan ^=- . It is easy to extract the 7tth root of a + bi: being r^ " I cos - + 1 sin - j , and by adding on 27r to 6 as many times as we please, we get n, 7ith roots. ■ We also find that c** = cos a + i sin a, — ia . . e = cos a - 1 sm a. If z = a + bi-r (cos + i sin ^) = r . e**, log 2 = log r + id = ii log {a^ + b"^) + z tan"' - . HYPERBOLIC FUNCTIONS. 361 18. I sinh ax.dx = - cosh ax, J a 19. I sech^ ax .dx = - tanh ax. J a 20. I cosech^ ax.dx = — coth ax. J a 21. [ --M^ = sinh-^ - = log [x + V^M^^K 22. 1 -^^— = cosh-^ - = log {x + ^/^^~^'l • 23. f,^,==ltanh-f=llog J a^ — x- a a za ° a-\-x a —X This is indeterminate because tan~i - may have any number of times 2ir in it, and indeed the indeterminateness might have been expected as e '"=1. Evidently cosh a;= cos ix, sinh x= -i sin ix. sinh is usually pronounced shin, tanh is usually pronounced tank. Prove that if if = sinh~ia; or a; = |(e" - e~«), only positive values of u being taken, then e" = x + ^/l + x^, and therefore u = sinh"^ x = log {x + >^1 + x"^). Similarly cosh-ix = log {x+ Jx- - 1), tanh-ia; = ^log^— |, sech-ia: = logQ+yi-l), cosech-i a; = log (^ + ^^^ + 1) • Now compare / -7==5= 8in"^a;, / , = sinh-i rr = log (a; + Vl+^). y aJi-x^ j ^1+x^ f dx I I - .- = cosh-^a; = log {x + ^Jx"^ - 1). j-j^,= tan i.r. j j--,= tanh-»a:=ilog ^~^. 362 CALCULUS FOR ENGINEERS, dx 1 , a; — a 24 ( ^ = -i- log 25. f^^ = ver«-.^, f- , _... dx ,x[ dx i . ,x — a — - SIQ" 27. I \/^Ta* . rfa? = ia; VaM^ + ia' log {a; -f Vi»Ta«}. 28. I Var' - a^ . da? = Ja; v/^t^ _ ^2 _ |^2 i^g [_^ _^ \/.x"2 - a^}. c?a; 1 . , 0. 1 , a ■ - = — sm~^ - or = - Na?- — a? ^ ^ ^ 29. I — ^'^^ = - - sin"^ - or = - cos~* - . ±0^ a + Vo^T^ 30. f ,j^-=^iog ; J x\/a^±a? a a + V «' ± a^ 31. |-\/aM^.d'a; = \/anr^-a 32. I - si a? -a? . rfa; = Va;^ - a^ - a cos"^ - . J a? X r x.dx , r x.dx , 35. f a? V^To^ . rfa; = J ^{¥Tc^y. 36. l*a;\/a2-a;2.(Za; = -iV(a2-a^)». 37. I V2aa; - a?^ . rfa? = ^^ sJ2ax-a^ + ^sin-*^^ ] 2 2 a 33 r ^ ^ fx-\ i(a; + l)V^-l Va; + X* LIST OK INTEGRALS. 40. I A / , . dx = sin"^ x — s/\— a?. J V 1 — « ^1- \sJ"'-^^'dx=^J{x^a){x^h) 363 4- (a — 6) log {sjx 4- a + -s/a; + 6). 42. I a;"^-i (a + 6a;")9 . c^a?. 1st. If ^/g' be a positive integer, expand, multiply and integrate each term. 2nd. Assume a + hx^ — y^, and if this fails, 3rd. Assume ax~'^ ^-h — y^. This also may fail to give an easy answer. 43. I sin~^ X .dx — x sin~^ x-{- Jl— x"^. 44. I X log x.dx= — (log a? — 1^). 45. I xe*^"^ dx = - 6«* ( ic j . 46. I a?"6«^ . c^a; = - aj'^e"^ - - / a;«-i6«* . c^a--. J a a] Observe this first example of a formula of reduction to reduce n by successive steps. ^^' j^'^^~"^^r:n.^— l■^7;^^ja;--^'^''• ^ 1 1 r e"^ 48. I 6**^ log a? . (^a? = - ^'^ log a; I — dx. 49. rJ^=jiogl±£!^=loglcot(^-f)l. j cos a; ^ ^ 1 - sm a; ^ ( V4 2/J 1. 364 CALCULUS FOR ENGINEERS. 51- — 7T = / r *^^ 7 ; it a > 6, ^ Jb—a tan ^ + Jb+a log if a < 6. Jb^ - «=» /j; ^ ^ /,— - Jb — a tan ^ — V^ + fi ^^ r . , e*''^ (c sin aa? — a cos cw;) o2. e*'* sm ax .ax= — ^ r- — ^ . J a^ ^ c- f _ , ^^ (c Go^ ax ■\- a sin aa;) 53. I e<^ cos aa; . rfa; = — ^^ :— — r ^ . f . „ , cos X . sin"~^ a; w — 1 f . „__ , o4. sin** a? . c?a? = + sin"-^ x . dx. J n n J [ dx _ cos a? 71 — 2 f dx J sin** x~ (w — 1) sin**"* a; ri. — 1 j sin""*-^ x ' r 6^* . . 56. I €"* sin" x.dx= ~- sin**~^ x (a sin x — n cos a?) J a^ + n^ ^ n(n-l) r .^^,^^^^ a2 + ?i=' j -K,* r . . , sin (7?t — w) •''■ sin(m + 7i)a? 57*. I sin 7«a; . sm ?ja; . aa; = — ^r-^ r ^ ^ . J 2(m-n) 2(m + n) -o f 7 sin (?/i — ?i) a; sin(m + w)a? 58. I cos 7?ia7 . cos 7ix . aa? = -^) f- H — ~ ^- . J 2(m-n) 2(m + ?i) -^ r . 7 cos(m+?i)a; cos(7ti—n)x 59. I sin mx . cos ??a; . ax = -^ ^ — -^^ f- . ; z (??i + n) 2 (m — n) 60. I sin- na? . dx =:la; — -— sin 2na;. j 4fn 61. I cos^ ?ia; .dx — — sin 2wa; + Jo;. * In integrating any of these products 57 to 61 we must recollect the following formulae : 2 sin vix . sin nx = cos (m -n) x- cos (w + 7j) a;. 2 cos wa; cos 7/a; = cos (m-w)a: + cos (m + n) x. 2 sin wx cos jja: = sin (m + n) a: + sin (m - n) a;, cos 2«x = 2 cos^ ;ia; - 1 = 1 - 2 siu^ nx. FORMULAE OF REDUCTION. 365 In the following examples 62 to 67, m and n are supposed to be unequal integers. rrr or 2n- 62. I sin 1710) . sin nx .dx= 0. Tir or 27r 63. I COS ??!« . COS iix . dx = 0. Jo 64. j sin^ nx,dx = - , I cos^ nx. dx = ^ , if /j is an Jo -^ Jo ^ I si] Jo integer •2jr 65. I sin mx . cos nx. dx= 66. / sin ?7ZA' . cos nx. dx=0 if 7?i — n is even. Jo 67. I sin 771^ . cos nx .dx— „ if vi — n is odd. 17V — ?r Jo '-/ 68. I sin"* ic . cos x ,dx = sin"*+i a; m+1 ^r. r «. • 7 cos'"-^"a; 69. I cos"* a; . sin a; . ttic = =— J m + 1 Hence any odd power of cos x or sin x may be integrated, because we may write it in the form (1 — sin'^a^)"'cos x or (1 — cos^ xj^ sin a?, and if we develope we have terms of the above shapes. Similarly sin^ x . cos^ x may be integrated when either p or q is an odd integer. 70. I a?"* sin x.dx = — x^ cos x-^-mi x^"'~^ cos x . c?a7. 71. I ic*" co^ x.dx = x"^ sin x —m j x'^~^ sin a; . c?a;. H-^ fsina; , 1 sin a; 1 fcosx , * 2, I — — - . dx = = — -— H :, I — — -; dx. »_-, fcosx y 1 cos a; 1 Tsina;, J a;"* m-la;"*-^ m-lja;^"~^ 74. [tan** x.dx = (^^^y~' ^ |'(tan x^-' . dx 366 CALCULUS FOR ENGINEERS. 75. A'" Sin-l X.dx = = :r I ,- . J 71+1 n+lj l-^-a? 77. — -r = , tan-i ,.==^ , if 4ac > 6^ ja-\-bx+ cx^ V4ac - 6" v 4ac - 6^ 1 , 2cx + h- ^J¥ - Aiac ... ,, log . , it 4ac < o-'. Vft^ - 4ac 2ca; + 6 + Vt" - 4ac 2cx + 6 If X = a + bx + cx^ and 5' = 4<ac — 6^, then ^^ fc^a; 2cx-{-h 2c fdx „ tdx _ 2cx + b ( 1 3c\ ic^ ("da; '^- JX^- q [iX'-'^qX)'^ q'jX- Q^ r^' . dx bx+2a b [dx 01 fdx 1 . a^ b [dx ^^' J^ = 2-J''Sx-Tajx' ^^ fdx b , X 1 ^(¥ c\ fdx 83. f__^__ = 2sin-A/^. N{x-a){b-x) V b-a 84. f_=iL==_=-Asin-.y^pp Js/{a + bx){oL-px) V6^ V c^^ + 6a 85. I , = — sec^ I — . Jx\lx''-a'' «^ \^/ ob. r rig? _ j_ , vv_+^^ a 87. LIST OF INTEGRALS. 367 f dx ^ \/a-\-hx-\- caf = -p log f c^ + 2 + Vc (a^bx + cx'^yj . „„ f dx 1 • -1 ^cx-b J Na-\-hx — ca? Vc v 4ac + 6^ 89. i£±g^ may be altered to •^ (^ + 2ca?) 2pc-gh J^ 2c y/a + bx + ex"" 2c V^ 4- hx + c«2 ' and so integrated. 90. Any integral of the form 1 ^ — ^ — T^r^^> where P, Q, ii and S are rational integral functions of x, can be rationalized by the substitution oi ax-\-b = v^. 91. Any integral of the form 1 ,_ dx, where U is a+bx-\- cx^, can be rationalized, (1) when b^ — 4>ac is positive and c negative, by the substitution , ^ = - — L. (2) When ¥ — 4iac is positive and c positive by Vc"VF _ 2y \/62-4ac~l~2/'* (3) When 6- — 4ac is negative and a positive by VcVF _ 1 +y ' A/4ac-62 ~ 1-2/' ' 4c If ?7 = a 4- 6a; 4- ca;^, g = 4ac — ¥, S = — , 92 f cZa;_^ 2(2ca;_j-6) [_dx__ 2(2cx-{b)\/ U 2S(n-l) f dx JU^s/U (2n-l)qU'' "^ 2n-l JU^-i^' 368 CALCULUS FOR ENGINEERS. ^. r, (2ca;+6)\/|7 1 [dx nr r /- (2c^ + 6)VF/.. 3\ 3 [dx ra;rfa; _ J^ _ ^ [ dx J W^ s/a "^ I X 2ja) 1 . _ / bx + 2a \ .. ^„ V - a \x V 6^ - 4ac/ -2\fU ., . = — = if a = 0. f dx ^ 'JU b C dx J x-^U~~ d^ 2aj^VF" 272. The solutions of many Physical Problems are given in terms of certain well-known definite integrals some of which have been tabulated. The study of these is beyond the scope of this book. I say a few words about The Gamma Function which is defined as [ e-^.x^'-Kdx^Tin) (1). By parts, I e~^ . a" ,dx=- e'^x^ +n\e-^. a;"-^ . dx. Putting these between limits it is easy to prove that — e~^x'^ vanishes when x = and a; = x . And therefore / e-^x'^ ,dx = nj e~^ . x^~^ .dx .. .(2). THE GAMMA FUNCTION. 369 Hence T (?i4-l) = n r(w) (3), so that if n is an integer r(n + l) = 1.2.3.4, &c. n=\n (4). Notice that j/i has a meaning only when n is an integer, whereas F (ii) is a function of any value of n. Tables of the values of F (n) have been calculated, we need not here describe how. The proof that ra) = V^ (5), as given in most books, is very pretty. The result enables us through (3) to write out r(f) or F(— f), &;c. A very great number of useful definite integrals can be evaluated in terms of the Gamma Function. V IT Thus 1. [' sin« e,dd=j cos*^ d.dO o r «.-./i ^«-lJ r '»'"+' (fo T(m)V(n) 4. j a;«e-«* dx = a-<'*+^) F (?i + 1 ). Jo 2a Vy 2a ^^' 1. i £ i/»'-> (1 - y)""-' dy = r (I) r (m) - 2r (I + m) . 2r (£+? + !) 24 fsi sinP e cos9 e,d0 = 370 APPENDIX. The following notes are intended to be rccod in connection with the text on the page whose number appears before the note. The exact position on the page is indicated by a t. Page 3. The ordinary propositions in Geometry ought to be illustrated by actual drawing. The best sign of the health of our race is shown in this, that for two generations the average British boy has been taught Euclid the mind destroying, and he has not deteriorated. Euclid's proofs are seemingly logical; advanced students know that this is only in appearance. Even if they were logical the Euclidian Philosophy ought only to be taught to men who have been Senior Wranglei-s. 95 per cent, of the schoolboys, whose lives it makes miserable, arc as little capable of taking an interest in abstract reason- ing as the other five per cent, are of original thought. Page 43. There is a much more accurate method for finding ^ described in my book on the Steam Engine. Page 48. This rule is not to be used for values of .r greater than 16. Page 63. In Art. 39 I have given the flue investigations as usually given, but prefer the following method, for I have for some time liad reason to believe that in a tube of section A, X)erimeter P, if irib. of gases flow per second, the loss of heat per second per unit area of tube is proportional to vejt, if t is the absolute temperature of the gases and v the velocity of the gases. Now v oc Wt-^A^ so that the loss of heat per second is pro- portional to weiA, Proceeding as before -w,de=cwe.dsiA, or -A^=G.dS, a or -A\oge-\-c=CS, where c and G are constants. As before, this leads to c=^log^i, >^=^log|. A 6 . Whole >S'=7j log ^-, so that the efficiency becomes E^\-e-^^l^. APPENDIX. 371 Now if it ia a tube of perimeter F and length ^, JS/A becomes PI I A or l/m where m is known as the hydraulic mean depth of the flue, so that This makes the efl&ciency to be independent of the quantity of stuff flowing, and within reasonable limits I believe that this is true on the assumption of extremely good circulation on the water side. This notion, and experiments illustrating it, were pubhshed in 1874 by Prof. Osborne Eeynolds before the Manchester Philosophical Society. Mr Stanton has recently {Phil. Trans. 1897) published experiments which show that we have in this a principle which ought to lead to remarkable reductions in the weights of boilers and surface condensers; using extremely rapid circulation and fine tubes. Page 95. It ought not to be necessary to say here that the compressive stress at any point in the section of a beam such as AC AC, fig. 47, is -jz, if z is the distance (say JB) of a point on the compression side of the neutral line A A from the neutral line. The neutral line passes through the centre of the section. If z is negative the stress is a tensile stress. The greatest stresses occur where z is greatest. Beams of uniform strength are those in which the same greatest stress occurs in every section. Page 111. Mr George Wilson (Proc. Royal Soc., 1897) describes a method of solving the most general problems in continuous beams which is simpler than any other. Let there be supports at ^, ^, C, Dj E. (1) Imagine no supports except A and E, and find the deflections at B, C and D. Now assume only an upward load of any amount at B, and find the upward deflections at B, (7, and D. Do the same for C and B. These answers enable us to calculate the required upward loads at J5, C and B which will just bring these points to their proper levels. Page 139. For beginners this is the end of Chap. I. Page 146. In all cases then, dll=l'.dt + t(^\dv (23*). Exercise 1. By means of (23) express K, I, Z, P and V in terms of k and write out the most general form of equation (3) in terms of k. Show that among many other interesting statements we have what Maxwell calls the four Thermodynamical relations, U/ \dpJt' WJp \dp)i' \dt) \dv)t' \d(t>), [dvU Exercise 2. Prove that in fig. 55, Area ABCB=AE. AF, =A U. AJ==AO.AM=AQ. AB, and show that these are the above four relations. 24—2 372 APPENDIX. Exercise 3. Show by using (23) with (7), (8), (11) and (14) that for any substance and that (20) becomes (dk\ d^ so that k =1-0 -{-t j( ^ j dv, where I'q ls a function of temperature only. Bividing dR=K.dt+L.(' differential, show that we are Dividing dH=K .dt+L.dp by t and stating that it is a complete led to K=K,-tj where Kq is a function of temperature only. Page 152. Another way is merely to recognize (8) as being the same as bH=^k.ht + l.bv, for l — t(~\, and when a pound of stuff of volume s^ receives bH=L at constant temperature (or bt = 0) in increasing to the volume «2 (so that dv = «2-«i) we have, since dpjdt is independent of v, 8Zr=Z = + i.^.(52-«i). Page 188. Sine functions of the time related to one another by linear operators such as a-{-h6-\-c€^-\-etc. + e6~^+fB~^-\-Qic.^ where 6 means -y- , are represented by and dealt with as vectors in the manner here described. Eepresenting an electromotive force and a current in this way, the scalar product means Power. Dr Sumpner has shown {Proc. Roy. Soc, May 1897) that in many important practical problems more complicated kinds of periodic functions may be dealt with by the Vector Method. Page 190. The symbol tan~i means " an angle whose tangent is." Page 195. To understand how we develope a given function in a Fourier Series, it is necessary to notice some of the results of Art. 109, very important for other reasons ; indeed, I may say, all-important to electrical engineers. Page 195. Article 126 should be considered as displaced so as to come immediately before Art. 141, p. 210. Page 202. The problem of Art. 125 is here continued. APPENDIX. 373 Page 208. The beginner is informed that 2 means "the sum of all such terms as may be written out, writing 1 for s, 2 for 5, 3 for 5, and so on." Page 209. See Ex. 23, p. 184. Page 213. The student ought to alter from v to (7 or to ^ in (9) as an exercise. Page 213. See Art. 152. Page Ml. Remember that the effective value of as.\n{nt-\-e) is a^sj2. Page 254. Here again a student needs a numerical Example. Page 256. After copper read " and their insulations." Page 259. Then e-^-\-e.^=E [sin {nt + a) + sin (nt — a)] = 2i7 cos a sin w^. {See Art. 109.) Page 266. All other statements about this subject that I have seen are of infantine simplicity, but utterly wrong. Page 269. Insert " and in consequence." Page 278. In the same manner show that if y=«^ ^=«*ioga. Page 281. See the eighth fundamental case, Art. 215. Page 299. See (1) Note to Art. 21. Page 301. Article 225 should be considered as displaced so as to precede Art. 222. Page 305. See Ex. 8, Art. 99. Page 309. Or (3). Page 310. Art. 225 should be read before Art. 222. Page 331. I have taken an approximate law for h and so greatly shortened the work. Page 359. Viscosity. All the fluid in one plane layer moves with the velocity v\ the fluid in a parallel plane layer at the distance bx moves with the velocity v-\-bv in the same direction; the tangential force per unit area necessary to maintain the motion is /* ^ or /* ^- , where /x is the viscosity. Example 1. A circular tube is filled with fluid, the velocity v at any point whose distance from the axis is r being parallel to the axis. Consider the equilibrium of the stuff contained between the cylindi-ic spaces of radii r and r+5r of unit length parallel to the axis. The 374 APPENDIX. tangential force on the inner surface is 2wrfi. ,~ and on the outer surface it is what this becomes when r is changed to r-\-hr or ^TTfi -J- (i'-f) 5^1 the difference of j^ressure between the ends gives us a force - 2wr ~ dr if x is measured parallel to the axis. The mass of the stuflF is 27rr . 6r . m if dr is very small and if m is the mass per unit volume ; its acceleration is -^ if ^ is time ; and hence, equating force to mass X acceleration and dividing by 27r/x8r, d / dv\ r dp _r.m dv dr \ dr) fi dx~ fi dt ' Example 2. Let -^ be constant; say that we have a change of pressure P in the length I so that t^ = y • Let a state of steady flow have been reached so that -f=Oy then dr\ dry fi I If r -7- be called u and if i'/^u be called 2a. then -^ + 2ar = 0, so that dr '^ dr ' du 4- 2ar . dr = 0, or u + ar^ = constant c. rj^ + ar^=^c, or ^ + ar=- (1), dv-\- {ar-i^dr=0, c vi-j^ar' + ^-, = C (2). Evidently as there is no tangential force where r=0, -f- = thei-e, c must be 0. Hence v + ^ar^ = C (2). Now v=0 where r^r^, the outer radius of the fluid, and hence (2) becomes v=^a {r^^-r^). The volume of fluid per second passing any section is 27r / rv .dr=^ - aro* = nrQ*P/8lfx. This enables us to calculate the viscosity of a fluid passing through a cylindrical tube if we know the rate of flow for a given difference of pressure. INDEX. The References are to pages. Academic exercises, Chap. iii. Acceleration, 24, 30, 188, 220 Adiabatic law, 92, 148, 167 Advance, 186 Air in furnaces, 65 Air turbine, 128 Alternating current formulae, 183 — 5, 189, 209, 239 Alternator, 178 Alternators in parallel and series, 259 Amplitude, 172 Analogies in beam problems, 108 — in mechanical and electri- cal systems, 213 Angle between two straight lines, 16 Angular displacement, 33 Angular vibrations, 212 Apparent power, 209 Approximate calculations, 2 Arc, length of, 170 Archimedes, spiral of, 302 Area, centre of, 85 — of catenary, 171 — of curves, 69 — of parabola, 71 — of ring, 80 — of sine curve, 197 — of surfaces of revolution, 75, 78 Asymptote, 301 Atmospheric pressure, 166 Attraction, 87, 344 Average value of product of sine functions, 185 Ayrton, 53, 199 Ballistic effects, 181 Basin, water in, 130 Beams, 48 — fixed at the ends, 100—108 — of uniform strength, 102 — shear stress in, 115 — standard cases, 97 — 99 Beats in music, 194 Belt, slipping of, 165 Bending, 94—121 — in struts, 262 Bessels, 205, 352, 354 Bifilar suspension, 179 Binomial Theorem, 34 Boiler, heating surface, 63 Bramwell's valve gear, 193 Bridge, suspension, 61 Calton HUl, 357 Cardioide, 302 Capacity of condensers, 162, 236, 240 Carnot cycle, 145, 152 Carnot's function, 145 Catenary, 62, 170 Central force, 344 Centre of gravity, 73, 85 Centrifugal force, 123 Centrifugal pump, 131 Chain, hanging, 61 Change of state, 150 — of variable, 339 Characteristic of dynamo, 296 Circle, 10 — moment of inertia of, 82 Circuitation in electricity, 134 Cissoid, 302 Cisterns, maximum volume of, 49 376 INDEX. The References are to pages. Clairaut's equation, 334 Clearance in gas engines, 150 — in pumps, 131 Commutative law, 231 Companion to cycloid, 301 Complete differential, 143—145, 153 Compound interest law, 161, 164 Concavity, 306 Conchoid, 302 Condensation in steam cylinders, 54, 153, 358 Condenser, electric, 162, 212, 236 — annulling self-induction, 247 — with induction coil, 257 Conductivity of heat, 341, 354 Conductors, network of, 237 Cone, 73, 78 Conjugate point, 301 Connecting rod, 191 Constraint, 179 Continuous beams, 111 Convexity, 306 Cooling, Newton's law of, 163 Co-ordinate geometry, 6 Co-ordinates, r, 0, <p, 347 — polar, 310, 342 Cosines, development in, 207 Cos-la;, 277 Cot-la;, 277 Cot X, 275 Coupling rod, 265 Craigleith quarry, 367 Crank, 173 Crank and connecting rod, 12, 191 Curl, 134 Current, effective, 200—202 — electric, 33, 168, 189, 239 Curvature, 96, 120, 169, 306 — of beams, 96—121 — of struts, 262 Curves, 43 — ' area of, 69 — lengths of, 77, 312 Cusp, 301 Cycloid, 12, 276, 302, 312 — companion to, 301 Cylinder, heat conduction in sides of steam-engine, 356 — moment of inertia of, 82, 85 Cylinders, strength of thick, 88 Cylindric body rotating, 90 Damped vibrations, 10, 211, 225— 228 Definite integral, 68 Deflection of beams, 96, 118 Demoivre, 320 Development, Fourier, 202 — in cosines, 207 Differential coefficient, 21, 28, 268 — complete, 143, 153 — equation, 220-225 — equations, general exer- cises on, 337 — equations, partial, 341, 346—351 — partial, 139 Differentiation of function of more than one varia- ble, 340 — of product, 269 — of quotient, 270 Diffusivity for heat, 355 Discharge of condenser, 156 Displacement, angular, 33 Distributive law, 231 Drop in transformers, 255, 257 Dynamics of a particle, 344 Dynamo, series, 296 df>' dx dy ^( = ^.^,155.271 dx dz dx e<^, 10 e<" sin bx, 10, 285 e'^cosbx, 286 Earth, attraction of, 87 Earthquake recorder, 215 Economy in electric conductors, 65, 57—59 — — lamps, 294 — hydraulic mains, 58 Eddy currents, 209 Effective current, 200—202 Efficiency of gas-engine, 150 — of heat-engine, 41 — of heating surface, 64 Elasticity, 93, 141 Electric alternator, 178 — circuit, 33, 168, 178, 189, 208, 212, 236, 239, 247 INDEX. 377 The References are to pages. Electric condenser, 162, 236, 240—245 — — and Ruhmkorf coil, 257 — — as shunt, 243, 246 — conductor, 168 — — economy in, 55, 57—59 — current, efifective, 200—202 — illustration, alternating cur- rents, 199 — lamps, economy in, 294 — make and break curve, 201, 205 — power meter, 209 — time constant of coil, 60, 160 — traction, 59 — transformer, 33, 249, 253 — vibrations, 156, 212, 213, 225 — voltage, effective, 202, 247 — voltaic cells, 51, 52 Electricity, partial differential equa- tions in, 349 — problems in, 33, 51, 52, 55, 57—59, 60, 134— 136, 156, 157, 162, 168, 178,182—186,189,195, 202,205,208—210,212, 225, 236, 239—261 — gelf-induction in, 135 Electrodynamometer, 200 Electromagnetic theory, fundamental laws of, 134 Electromotive force, 134 — — in moving coil, 178 Ellipse, 8, 10, 11, 83, 158, 276 Ellipsoid of revolution, 76 Elliptic functions, 339 — integrals, 339 Empirical formulae, 17 Energy, intrinsic, 143 — kinetic, 31, 180 — of moving body, 156 — potential, 180 — stored in compressed spring, 33 Engineer, 1 Entropy, 143, 152 Envelopes, 309 Epicycloid, 302 Epitrochoid, 301 Equality of forces, 217 Equations, differential, 220—225 — partial differential, 341, 346—351 — solving, 51 Equiangular spiral, 186 Euler's law for struts, 265 Evanescent term, 169, 190, 208, 240 Exact differential equation, 339 Exercises, 38 — general, on differential equations, 33/ — — on integration and differen- tiation, 279 — maxima and minima, 295 — on curves, 169, 311 — on integration of sin x and cos^, 182 — on Maclaurin, 319 Expansion of functions, 317 — 320 — of gas, 17 Experiments, 7 Explosions, 48 Exponential and trigonometrical for- mulae, 177, 185, 190, 222, 320 — theorem, 161 e«, 161 Factor, integrating, 328 Factorial fractions, 235 Factorials and gamma function, 369 Falling body, 21, 30 Feedwater missing in steam cylinder, 358 Ferranti effect, 247 Field, magnetic, about straight wire, 134 — — rotating, 195, 251 Flow, maximum, of gas, 128 — of gas, 54 — of liquid, hypothetical, 133 Fluid friction, 167 — motion, 125 — pressure, 121 — whirhng, 123 — work done by, 66 Flywheel, 84 — stopped by friction, 167 Force, central, 344 — due to jet, 27 — — pressure of fluids, 121 — lines of, 124 378 INDEX. T/te References are to pages. Force of blow, 26 — of gravity, 87 — unit of, 26 Forces on moving bodies, 180 Forms, indeterminate, 299 Formulae of reduction, 284, 286 Fourier development, 208 — exercises on, 315 — proof, 183, 184, 197, 201 — rule, 202, 204 — theorem, 195 Fractions, partial, 291 Frequency, 186 Friction at pivot, 94 — fluid, 167 — solid, 168 Frustrum of cone, 78 Fuel on voyage, 49 Function, 8 — average value of sine, 185 — Bessel, 352 — elliptic, 339 — gamma, 369 — hyperbolic, 172, 360 — of more than one variable, 137, 341 Fundamental integrals, 278 — rules on electricity, 134 Furnaces, air in, 65 «7, 26 Gamma functions, 235, 369 Gas, 38 — elasticity of, 94 — engine, 91 — — diagram, heat in, 272 — — formulae, 17, 91, 147, 150, 272 — flow of, 55, 127 — perfect, 136, 150 — work done by, 66, 72 Gauge notches, 133 Gaussage, 134 Gear, valve, 193 General case of two coils, 249 — exercises in differential equa- tions, 337 — — in differentiation and integration, 279 — rule, 271 — — for differential equa- tions, 224 General rule for operators, 237 Girders, continuous, 111 — shear stress in, 117 Glossary, 301, 302 Gordon's rule for struts, 264 Gradient, temperature, 354 Graph exercises, 8 — 10 Graphical Fourier development, 204 — work in beams, 108 Gravity, 87 — motion of centre of, 230 Groves' problem, 257 Guldinus' theorems, 80 Guns, 90 Hammer, 26 Hanging chain, 61 Harmonic functions, 172, 186 Harmonics, zonal, 205, 354 — — spherical, 349 Heat conductivity, 354 — — equations, 341, 347—352 — equations, 138 — experiments at Edinburgh, 357 — latent, 41, 43 — lost in steam cylinders, 356 — reception in gas-engine, 272 — specific, 93, 141 Heating surface, 63 Hedgehog transformer, 244 Henrici, Professor, 205 Henry, 136 Hertz, 215 Horse-power and steam, 41 Hydraulic jet, 26 — press, strength of, 89 — transmission of power, 58 Hydraulics, 133 Hyperbola, 10, 11 Hyperbolic functions, 360 — spiral, 302 Hypocycloid, 277, 302 Hypotrochoid, 302 Hysteresis, 209, 255 Idle current in transformers, 243 Imaginaries, 185 Incandescent lamp, economy in, 298 Independent variables, change of, 338 — — more than one, 56, 136, 154 INDEX. 379 Ttie References are to payes. Indeterminate forms, 299, 300 — multipliers, 158 Index law, 231 Indicator diagram, 53, 67 — — gas-engine, 91, 27*2 — vibration, 215 Induction, 136 — coil and condenser, 257 — in transformers, 256 — mutual, 249 — self-, 33, 60 — self-, and capacity, 240 Inertia, moment of, 81 — — of cylinder, 81, 85 — — of ellipse, 83 — — of flywheel, 84 — — of rectangle, 86 — — of rod, 84 Illustrations of meaning of differen- tiation, 37, 40, 42, 162, 176, 177 Inflexion, point of, 20, 301 Instruments, measuring, 179 Integral, definite, 68 — double, 68 — line, 69 — surface, 69 Integrals, elliptic, 339 — list of, 359—369 Integrating factor, 144, 327 Integration, 23, 35 — by parts, 285 — exercises on, 183 — of fractions, 282 Interest law, compound, 161 Intersection of two straight lines, 16 Intrinsic energy, 143 Isothermal expansion, 92 Joule's equivalent, 43 Journal, lubrication of, 231 Kelvin, 161 Kinetic energy, 31 Labour-saving rule, 237 Lag, 209 Latent heat, 15, 42 Lateral loads, struts with, 264 Lamp, incandescent, 298 Law, adiabatic, 148 — commutative, 231 — of flow of heat, Peclet's, 63 Law of cooling, Newton's, 163 — compound interest, 161 — distributive, 231 — of Entropy, 148, 152 — Euler's, for struts, 265 — of expansion of steam, 17 — of falling bodies, 21 — Index, 231 — of loss of heat in steam cylin- der, 356 — of ^ and t, 18 — of Thermodynamics, first, 142 — — — second, 146 — of vibratory systems, 225 — 230 Lead in branch electric circuit, 247 Leakage of condenser, 162 Legendre's equation, 350 Lemniscata, 302 Length of arc, 170 — of curve, 77, 312 Level surface, 124 Limit, meaning of, 22 Line integral, 69, 134 Linear equations, 220, 326, 353 Link motions, 14, 193 Liquid, flow of, 130 List of Integrals, 359—369 Lituus, 302 Loci, 11 Logarithmic curve, 10, 302 — decrement, 11 — function, 40 — spiral, 302 Logarithms, 2, 161 Log a;, 274 Lubrication of journal Maclaurin, 319 Magnet suspended, 179 Magnetic field about straight wire, 134, 195 — — rotating, 195, 251 — force, 134 — leakage, drop due to, 257 Make and break curve, 201, 205 Mass, 26 — energy of moving, 157 — of body, variable, 76 — vibrating at end of spring, 156 Maxima and minima, 20 — — exercises on, 46, 47, 60, 294 380 INDEX. The References are to pages. Maximum current from voltaic cells, 62 — flow of air, 128 — parcel by post, 53 — volume of cistern, 49 — power from dynamo, 297 Measuring instruments, 179 Merchant and squared paper, 6 Meter, electric power, 209 Minimum, 20 Modulus, 340 Moment of inertia, 80 Momentum, 26 Motion, 30, 157 — angular, 33, 212 — of fluids, 125 — of translation, 344 — in resisting medium, 314 Multipliers, indeterminate, 158 Mutual induction, 250 Natural vibrations, 225, 220 Negative and positive slope, I'J Network of conductors, 237 Newton's law of cooling, 163 Normals, 15, 43, 169 Notches, gauge, 133 Numerical calculations, 2 Octave, 192 Ohm's law, 33 — — modified, 136, 168, 189, 208, 236 Operation, symbols of, 231 Orifice, flow of gas through, 54 Oscillation, 123, 156, 190, 210, 211, 225—30 Otto cycle, 149 Parabola, 8, 11, 27, 31, 61, 71 Paraboloid of revolution, 74 Parallel motion, 13 — alternators in, 261 Parameter, 340 ~ variable, 308 Partial differential equations, 341, 347—352 — differentiation, 39, 137, 155, 269, 341 — fractions, 224, 234, 291, 294 Particle, dynamics of, 344 Parts, integration by, 284 Peclet's law of flow of heat, 63 Pendulum, 179 Percussion, point of, 123 Perfect gas, 38 — — thermodynamics of, 147 — steam-engine, 41 Periodic functions, 194, 203 — motions in two directions, 196 — time, 186 Perpendicular lines, equations to, 16 Pivot, friction at, 94 Point, conjugate, 301 — d'arr^t, 301 — of inflection, 301 — of osculation, 301 — moving in curved path, 342 Polar expressions, 342 Pound, unit of force, 26 Poundal, 26 ^n»(^).350 Positive and negative slope, 19 Potential energy, 32 Power, apparent, 209 — electric, 208 — — transmission of, 58 — meter, electric, 209 — true, 209 Press, hydraulic, 90 Pressure, 136, 273 — atmospheric, 166 — fluid, 121 Primary, transformed resistance of, 254 Product, differentiation of, 155, 269 Projectile, 310, 315 Pulley, slipping of belt on, 165 Pump rod, 164 Quotient, differentiation of, 270 Radial valve gears, 14, 193 Badian, 9 Radius of curvature, 169, 306 — of gyration, 82 Radius vector, 342 Rate of reception of heat, 92 Ratio of spcific heats, 93, 139, 141 Ratios, trigonometrical, 182 Reduction, formulae of, 284, 286 Recorder, earthquake, 215 Rectangle, moment of inertia of, 86 Resistance, electric, 33 INDEX. 381 The References are to pages. Resistance, leakage, 163 — operational, 236 Besistances in parallel, 245 Resisting medium, motion in, 314 Resonance, 215 Resultant of any periodic functions, 197 Revolution, surfaces of, 78 — volume of solids of, 75 Rigid body, 61 — — motion of, 217 Ring, volume and area of, 80 Rod, moment of inertia of, 85 Roots of equations, 224, 321 Rotating field, 195, 251 Rotation in fluids, 132 Ruhmkorff coil, 257 Rule, 4 r and d co-ordinates, 342 r, df <t> co-ordinates, 347 Secant x, 276 Secohm, 136 Self-induction, annulled by con- denser, 247 — — of parallel wires, 135 Series, alternators in, 257 — development in, 207 — dynamo, 296 Shape of beams, 109 Shear stress in beams, 115 Shearing force in beams, 108 Simple harmonic motion, 173 — — — damped, 311 Sina;, 9, 161, 274 Sine curve, area of, 173 — functions, 172 Sines, curve of, 9, 173 — development in, 207 Singular solution, 334 SHpping of belt, 165 Slope of curve, 15, 19, 70 Solution of forced vibration equa- tions, 214 — of linear differential equa- tions, 229 Sound, 94 Specific heat, 139 — heats, ratio of, 93, 141 — volume of steam, 42 Speed, 21 Spherical Harmonics, 205, 349, 354 Spin, 132 Spiral flow of water, 130 — hyperboUc, 302 — of Archimedes (or equi- angular), 302 — line, 173 — logarithmic, 302, 311 Spring with mass, vibrating, 156 Springs, 21, 32, 53 — which bend, 119 Square root of mean square, 200, 202 Squared paper, 6, 7 State, change of, 150 Steady point, 219 Steam, 41 — work per pound of, 53 — -engines, 41 — -engine indicator, vibrations of, 215 — — piston, motion of, 191 Stiffness of beams, 48 Straight line, 14 Strains in rotating cylinder, 90 Stream lines, 126 Strength of beams, 48 — of thick cylinders, 88 — of thin cyhnders, 91 Struts, 261 — with lateral loads, 264 Subnormal, 44, 169 Substitution, 279 Subtangent, 44, 169, 305 Successive integration, 335 Sum, differentiation of a, 268 Surface heating, 63 — integral, 69 — of revolution, 75, 78 — level, 124 Suspension, 179 — bridge, 61 Swinging bodies, 179, 182 Symbols of operation, 233 — — — simplification of, 237 Synchronism, 215 Table of Fundamental Forms, 278 Tangents, 43, 158, 169 Tana;, 275 Tan-la:, 277 Taylor's theorem, 317 Temperature, 136 882 INDEX. The Refererwes are to pages. Temperature, absolate, 145 — gradient, 354 — in rocks, <feo. , 357 Terminal velocity, 314 Theorem, Binomial, 34 Theorem of three moments, 111 — Guldinus', 80 — Demoivre's, 320 — Maclaurin's, 318 — Taylor's, 317 Thermodynamics, 42, 138, 153 Thomson, Professor James, 133 Three moments, 111 Tides, 194 Time constant of coil, 60, 160 t, diagram, 153 Torque, 33 Torsion, 179 Traction, electric, 59 Transformer, 33 Transformers, 252, 257 — condenser shunt, 243 — idle current of, 243 Triangle of forces, 61 Trigonometry, 2 Trigonometric and Exponential Functions, 177, 185, 222, 234 Trigonometric Formulae, 182 Trisectrix, 302 Tuning-forks, 230 Turbine, air, 128 Two circuits, 249 Valve gears, 14, 193 Variables, independent, 57, 136, 341 Variable mass of body, 76 — parameter, 308 Velocity, 21, 30, 188 Vibration, 156, 190, 210, 212, 216, 225, 230, 238 — (electrical), 156, 212, 213 — indicator, 217, 219 — of indicator, 215 Volcanoes, 172 Voltage in moving coil, 178 Voltaic cells, 51, 52 Volume of cone, 74 — of ellipsoid of revolution, 76 — of paraboloid of revolution, 74 — of ring, 80 — of solid of revolution, 76 Voyage, fuel consumed on, 49 Water in steam cylinder, 358 Watt's parallel motion, 13 Wedmore, 205 Weight, 26 Whirhng fluid, 123 Willans, 54 Work, 31—32 — in angular displacement, 35 — per pound of steam, 31, 53 — done by expanding fluid, 66, 72 — — gases, 149 Uniform strength in beams, 102, 103 Uniformly accelerated motion, 29 Unreal quantities, 3, 177, 185, 222, 320 X . e«*, 285 x-K- 6—160 Zonal Harmonics, 205, 349, 354 CAMBRIDGE : PRINTED BY J. & C. 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