a^acnell Hnitietaitg ICtbcatD 3tt|ara. Nmh lorlt BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF HENRY W. SAGE I89t Applied calculus; ,3 1924 031 250 339 Cornell University Library The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031250339 APPLIED CALCULUS NEWTON (1642-1727) "QUI GENUS HUMANUM INGENIO SUPERAVIT" From ifie pnhitin^ by Vatulet'bank { Natiojial Portrait Gallery} APPLIED CALCULUS tAn Introductory Textbook BY F. F. P. BISACRE O.B.E., M.A.(Cantab.), B.Sc.(Lond.), A.M.Inst.C.E. NEW YORK D. VAN NOSTRAND COMPANY EIGHT WARREN STREET 1921 5 KumJ PREFACE This book is intended to provide an introductory course in the Calculus for the use of students of natural and applied science whose knowledge of mathematics is slight. All the mathematics that the student is assumed to know is algebra up to quadratic equations ; elementary trigonometry up to the formulae of sines, cosines, and tangents of compound angles ; the elements of geometry; and the method of graphs. Infinite series are essentially difficult and unconvincing unless treated rigorously — as the old conundrum of Achilles and the tortoise shows — and there is no need to use them in the elementary parts of the subject. They have there- fore been avoided altogether. Definite problems, dealing with actual things, precede the analytical treatment, which I have tried to make simple and convincing; and I hope any reader who pursues the subject further in the standard works will find that he has only to extend and qualify the proofs, not to unlearn them. I have introduced and used limits in the first chapter before defining them, for the same reason that I should show a child a herring and tell him about its habits of life before describing it to him as one of two distinct but closely- allied species of malacopterygian fishes of the genus Clupea. The pictures of celebrated mathematicians and scientists are intended to arouse some human interest in mathemati- cal science and the history of its progress. Some of the founders of the science lived more than ordinarily interesting lives, and if the mathematician ignores the human side of things, he can hardly expect humanity not to ignore him. Perhaps the title of the book needs a word of explana- tion. In applied mechanics it is usual to discuss the theo- retical principles of mechanics as well as their applications. This line has been followed here, the treatment of practical problems being preceded by a fairly full discussion of the necessary theory. vi PREFACE The examples appended to the several chapters have been specially chosen to illustrate particular points, and should all be worked by the student. It has not been thought necessary to supply long sets of examples, such sets being rarely worked through by the student. To help the private student, whose needs I have had specially in mind, solu- tions of all the difficult problems are given, and answers to all exercises and problems. Many of the exercises, especially those in the "Miscellaneous Exercises", are taken by permission from recent University and Army Examination Papers. I gratefully acknowledge my in- debtedness to the authorities of the Universities of Cam- bridge, London, and Glasgow, and to the Comptroller of H.M. Stationery Office, for permissions granted. I wish also to thank Mr. J. Dougall, M.A., D.Sc, F.R.S.E., for the great help he has given me in passing the book through the press. He has read all the proofs, and has made many valuable suggestions. Mr. D. J. Richards, M.A., also kindly read the MS. in the earlier stages, and I am indebted to him for numerous criticisms. The books to which I am most indebted are G. H. Hardy's Pure Mathematics, on critical points, and the treatises of Lamb and Gibson. I have also found Dr. A. N. Whitehead's book. An Introduction to Mathematics, most suggestive. The data used and methods described in the chapter on physical chemistry are based on the classic work of J. H. Van 't Hoff, Lectures on Theoretical and Physical Chemistry, and the treatise of W. C. M'C. Lewis, A System, of Physical Chemistry (2 vols.). I have also con- sulted R. A. Lehfeldt's A Textbook of Physical Chemistry. F. F. P. B. Glasgow, July, 1921. The publishers gratefully acknowledge the kindness of Miss Anna Guldberg and Professor Torup of the University of Christiania in sending them a biography of C. M. Guldberg (p. ^12) and the photo- graph from which the half-tone facing p. jo6 was reproduced. CONTENTS Introduction Pag 1-18 Co-ordinates I Graphs S Ellipse — Geometrical - 8 Equation of Ellipse ir Area of an Ellipse '3 Notation of Integrals 16 An Integral as an Area •7 Notion of a Rate \- 18 Chap. I. General Principles 19-46 Number 19 Notion of Change 22 Variable Quantities 23 Functions Introduced 24 Examples of Functions 25 Approximations - 26 Large and Small Quantities 27 Standard of Approximation 3' Small Quantities 31 Expansion of a Square Plate by Heat 32 Expansion of a Cube by Heat 35 Changes in the Values of Functions 37 Ratio Ay/Ax 38 Notion of " Approach " — Limits 39 Notation of Differential Calculus 40 Derivative of x^ - 42 Derivative of .r**, n a positive integer 44 Exercise i 45 II. Geometrical and Mechanical Meaning of a De- rivative 47-70 Gradient or Slope 47 /iii CONTENTS Chap. Page Equation of a Straig^ht Line 48 Geometrical Meaning of dyjdx 50 Discontinuities 54 Speed - S4 Instantaneous Velocity 59 Extended Idea of Speed 63 Definition of Instantaneous Rate - 65 Graphical Method of finding' Velocity 66 Exercise 2 - - 69 III. Integration - • 71-84 Areas and Integrals 71 Differential and Integral Notations are Inverse 73 Evaluation of Definite Integrals — Graphically 76 Evaluation of Definite Integrals — Arithmetically - 78 Evaluation of Definite Integrals — Analytically - 81 Exercise 3 83 IV. Limits • ... 85-108 Example of a Limit - 85 Expression "in the neighbourhood of" 92 Definition of a Limit 93 Continuous Functions 94 Undetermined Functions 95 Summary of Properties of Limits 97 Limit of a Sum - - 98 Limit of a Product 99 Limit of a Reciprocal 100 Limit of a Quotient loi Circular Measure of Angles 102 To Convert from Radians to Degrees, &c. 103 Trigonometrical Limits 104 TanA: > x > sin* 104 X > sin* > X - — 105 4 I > cos* > I - J*" - 105 X — > 0\ X } Lt Ct.iM\ = , ,06 Progressive and Regressive Limits 107 Exercise 4 107 CONTENTS Chap. V. Practical Applications Page 109-20 Speeds of Trains 109 Area of a Circle - III Motion under the Influence of Gravity- 112 Moment of Inertia "3 An Electrical Problem of Transmission - 116 Exercise 5 118 VI. Differentiation and Integration 121-33 Differentiation of Sines, Cosines 122 Differentiation of Constants .123 Differentiation of Sums and Differences of Functions 1 24 Differentiation of Products of Functions 124 Differentiation of Quotients of Functions 125 Differentiation of Function of a Function 126 Integration, Constants - 128 Integ-ration, Sum of Two Functions - 129 Integration, Product of Two Functions 130 Derivative of «", Rational Index 131 Exercise 6 • 132 VII. Infinitesimals - - - '34-54 Ascending- Powers of A* - 135 The Differential - 137 Infinitesimal Notation 138 The " Equals " Sign 140 Approximate Calculations by Small Differences 142 C. G. of Parabolic Plate - 145 M. I. of Rectangular Plate 148 Expansion of a Gas - 149 Indicator Diagrams 151 The Crosby Indicator 152 Exercise 7 154 VIII. Integration 155-65 Integration by Differentiation 155 Integration by Substitution 156 Integration by Parts - 158 Integration by Reduction 160 Inverse Trigonometrical Functions 161 Exercise 8 163 = /Il\/'+' X CONTENTS Chap. Pae= IX. Mensuration 166-81 Lengths of Curves 166 Use of Parameters in Rectification of Curves 1 72 Determination of Areas i74 Volumes of Solids of Revolution 1 76 Exercise g - 1 79 X. Successive Differentiation 182-7 Notation 183 A Differential Equation 184 A Finite Series 185 Exercise 10 186 XI. Curvature of Plane Curves 188-218 Deviation 188 Curvature of a Circle - 190 DeBnition of Curvature 191 Radius of Curvature - 191 Circle of Curvature 192 Transformation of Formulae 193 Curvature Geometrically Treated 196 Total Length of a Curve: S = | pd 199 Graphical Methods 200 Motion in a Curve 204 The Bending of Beams 209 EI^ = M 213 Exercise 11 216 XII. Maxima and Minima 219-3S Turning Values of Functions '- 219 Condition for a Maximum Point 220 Condition for a Minimum Point 221 Differentiable Functions - 221 Another Form of Test 223 Analytical Test 225 Example on Transformer Losses 226 States of Equilibrium of Physical Systems ' 228 CONTENTS xi Chap. Page Example of Soap Bubble 229 Example in Engineeringf Economics - 230 Exercise 12 233 XIII. Exponential and Logarithmic Functions 236-65 Definitions - 236 Graph of e* 237 Graph of logex 238 Derivative of e^ 240 Derivative of log^jc 242 Why e is chosen as the Base of Natural Logs 245 Rules for Conversion of Log's 246 Notes on Differential Equations 248 dx dx 249 ,254. ax^ + bx + c Partial Fractions 262] Exercise 13 263' XIV. Some Problems in Electricity and Magnetism 266-97 Dynamics of a Fly-wheel 266 The Electric Circuit 268 The Magnetic Field defined 270 A Varying Magnetic Field 272 Magnetic Flux 273 Magnetic Fields due to Coils carrying Currents 275 Right-handed Screw Law - - 276 Faraday's Law of Electro-magnetic Induction 277 Two Fundamental Equations, e = ri, and e = - T -$. dt Self-induction 279 Coefficient of Self-induction. The Henry - 280 Decay of Current with Time 281 Differential Equation L- -f « = o 282 dt Analogy of Rotating Fly-wheel to Electric Circuit 283 Rise of Current in a Circuit 285 E/ -t\ Solution given by z = — I i - o'L I 287 Numerical Example of Rise of Current 288 Damping of Instruments 289 xii CONTENTS Chap. Page Solution of Differential Equation for Damped Motion 290 Law of Diminution of Amplitude 292 Ballistic Galvanometer 293 Exercise 14 296 XV. Some Problems in Chemical Dynamics - 298-318 Numerical Data for Molecules of a Gas 298 Law of Mass Action. Decomposition of Arsine 299 The Gramme-molecule or Mol. Concentration 301 Relation of Concentration to Pressure - 302 Influence of Number of Reacting Molecules 303 Reaction Constant. Equations for Monomolecular and Dimolecular Reactions 305 Guldberg- and Waage's Law 306 Integrals of the Equations of p. 305 - 307 Numerical Exercise on these Integrals 308 Exercise on Decomposition of Phosphine - 309 A Reversible Reaction, 2HI and Hj + I2 - 310 Equilibrium Composition of the Mixture - 311 Composition at any Time, by Integration 312 Numerical Example of same Reaction 314 The same Reaction in Opposite Direction 315 Exercise 15 316 XVI. Some Problems in Thermodynamics 3'9-83 Three Phases of Matter 319 " State ■ of a Body - 320 Functions of Two or more Variables 322 Partial Derivatives of f(x, y) 323 The " Gas Equation " *w = ?« 324 m Value of R. Examples 325 Callendar's Steam Equation. Work done by Ex- panding Gas - 326 Example on Isothermal Expansion 327 Intrinsic (or Internal) Energy 328 Definition of " Cycle ". Joule's Equivalent 329 Conservation of Energy - 330 First Law of Thermodynamics 331 Equation of Energy. Internal Energy a Function of State 332 Integration over a Cycle. The Sign ( / ) 333 CONTENTS xiii Page (f)'iQ = - ipdW 334 Reversible Changes. Exchange of Heat Energy 335 Exchange of Mechanical Energy. Reversible Cycle 336 Adiabatic Expansion of Gas. pi/^ = Constant. Thermodynamics of a Gas 337 Experiment of Joule and Kelvin 338 Internal Energy of a Gas a Function of Temperature only 339 The Entropy 0. -^ = 80. Value of for a Gas - 340 In Reversible Cycle ( / )-^ = o. Practical Engine Cycles 341 The Otto Cycle, and nearest Ideal Cycle - 342 Efficiency of this Ideal Cycle 344 Formula for the Efficiency, ij = i - l-y , Diesel Cycle - 346 Ideal Cycle nearest Diesel Cycle 347 Efficiency of Ideal Diesel Cycle 349 Efficiency of Otto and Diesel Cycles compared 350 Remarks Introductory to Second Law. T and 6 351 Second Law of Thermodynamics. Efficiency of Re- versible Cycle a Maximum 352 Equal Efficiencies of all Reversible Cycles working between Ti and Tj 354 Carnot's Cycle 355 Efficiency of Carnot Cycle 357 T — T Efficiency of a Reversible Cycle is — 1— — ' 358 Ti "Energy-carrier." Radiation - 359 For any Reversible Cycle, S ^ = o. Carnot's Cycle generalized - 360 Second Law expressed in the Result ( / ) -^ = o 362 The Entropy 0. 0b - 0a = f" ^ Definition of Entropy. For Irreversible Cycle < o 365 ij)f._ 364 /:(?),- Irreversible Path. / ( ^r" ) . < "^a ~ '^* " 3^6 CONTENTS Page An Adiabatic Reversible Change is Isentropic - 367 In Irreversible Adiabatic Chang-e, Entropy cannot decrease - 3^8 /: " "zM equal to Change of Entropy only when Path 6 is Reversible - • 3^9 Changes in Entropy of a System 370 Increase of Entropy when Heat is exchanged by Conduction 371 Principleof Maximum Entropy. Steam-engine Cycle 372 Rankine's Cycle 374 Dryness Fraction 376 Efficiency of Rankine Cycle - 377 Practical Calculation of the Rankine Cycle 378 Mollier Diagram. Total Heat - 379 "Heat Drop." Exercise 16 380 Miscellaneous Exercises - 385-401 Biographical Notes 403-'3 Napier, John 403 Kepler, Johann ; Descartes, Ren^ 404 Sir Isaac Newton 405 Leibnitz, Gottfried Wilhelm 407 Volta, Alessandro; Faraday, Michael 408 Carnot, Sadi Nicolas Leonard 409 Henry, Joseph; Graham, Thomas; Joule, James Prescott; Clausius, Rudolf Julius Emmanuel 410 Kelvin, Lord (Sir William Thomson); Clerk Max- well, James 4" Guldberg, Cato Maximilian 412 Gibbs, JosiahWillard; Van't Hoflf, Jacobus-Hendrikus 413 Table I — Indefinite Integrals 414 Table II — Natural Logarithms 415 Table III — Exponential Functions 416 Answers 417-446 PORTRAITS Newton Frontispiece Mathematicians and Philosophers — Kepler, Napier, Descartes, Leibnitz page 132 Physicists — Kelvin, Clerk Maxwell, Faraday, Joule page 284 Physical Chemists — Graham, Van't Hoff, Guldberg, Gibbs - page 306 Physicists — Volta, Carnot, Henry, Clausius page 352 APPLIED CALCULUS Introduction ' ' What a time of it had we, were all men's life and trade still, in all parts of it, a problem, a hypothetic seeking, to be settled by painful Logics and Baconian Inductions! The Clerk in Eastcheap cannot spend the day in verifying his Ready Reckoner; he must take it as verified, true and indisputable ; or his Book-keeping by Double Entry will stand still. ' Where is your Posted Ledger?' asks the Master at night. ' Sir,' answers the other, 'I was verifying my Ready Reckoner and find some errors. The Ledger is 1 ' Fancy such a thing ! "True, all turns in your Ready Reckoner being moderately correct, being not insupportably incorrect. " — Carlyle. In the year 1612, there was an unusually heavy vintage in Germany, and it appeared that there would not be enough barrels to hold all the wine. The amount of wine which would go into a barrel of given shape and dimensions could not be calculated, so the astronomer Kepler at once took the problem in hand. For many years, he had been working at problems which required the finding of areas con- tained by closed curves, and he had invented special methods of his own for solving these problems. These methods he applied to the wine tun, and wrote a book called Nova Stereometria Doliorum (a new method of measuring wine casks), published in 1615. ( D 102 ) 1 2 i APPLIED CALCULUS This book is the earliest work on the subject now called "The Integral Calculus". Before explaining the method Kepler used, it will clear the ground if we deal with a few points of elementary geometry. Position of a Point in a Plane. The position of a point in a plane may be specified in the following way. Suppose the plane in question is the plane of this paper. Rule two straight lines, X'OX, Y'OY, of unlimited length, perpendicular to each other, and intersecting in the point O, called the Origin (fig. i). These lines divide the plane into four quadrants. If we were drawing a map, we should call these quadrants, the N.E. quadrant, N.W. „ S.E. ,, s.w. „ Take a point P (fig. i) anywhere in the N.E. quadrant. Draw PM perpendicular to X'OX, meeting X'OX in M. If we measure the length of OM and of MP we shall know the exact position of P in the N.E. quad- rant with reference to X'OX and Y'OY. • Thus, suppose the length of OM to be i in. and that of MP 2 in., then P is i in. from Y'OY, and 2 in. from X'OX. It is usual to denote OM by x and MP by^. The two lengths denoted by x and y are called the co- ordinates of P. When it is necessary to distinguish INTRODUCTION between them, x is called the abscissa and y the ordinate. The point P is referred to as the point (x, y). In the actual case taken above X = I and j>/ = 2 and P is the point (i, 2). A common unit of length is understood in all measurements — the inch in the above case. The figure is half full size. P ■»---r M' X' Y' R Fig. I , The base lines X'OX and Y'OY are called the axes ; X'OX, the x axis, and Y'OY, the y axis. All measurements are made from these lines, in directions perpendicular to them. Convention of Signs. Every number is associated with a sign, implied or stated. This sign is either plus or minus. In speci- 4 APPLIED CALCULUS fying the point P (i, 2), the positive sign is implied, i.e. P is the point (+ i, + 2). If a measurement is m.ade in one direction when the number expressing it is positive, it must be made in the opposite direction when the number expressing it is negative. This rule means that we regard length as a magni- tude to which a plus or minus sign may be given. The directions usually taken as positive are shown by the arrow-heads at X and Y (fig. i). Thus, a line drawn in the direction from O to X, or from O to Y, is positive, while a line drawn in the direction from O to X', or from O to Y', is negative. This rule at once enables us to specify points in the remaining three quadrants. The point specified by (i, — 2), for instance, is plotted, i.e. drawn, by measuring off OM, equal to I in., and MQ, equal to 2 in., MQ being measured from X'OX in the direction opposite to OY. Likewise, the point (—2,-3) is the point R (see fig. i) where OM' is 2 in. long, M'R is 3 in. long. The point (-2, 3) lies in the X'OY (the N.W.) quadrant. It does not matter which quadrant we take as cor- responding to (+ X, ■^ y) points — it is usual to take the N.E. one — but it is essential to adhere to the sign convention. The rule may be remembered by the mnemonic: Plus, to the right; minus, to the left. Positive, height; negative, depth. INTRODUCTION 5 The signs of x and y in the different quadrants are then {x+,y +), N.E. {x-,y^), N.W. {x-,y-),S.W. {x+,y-),S.^. A Graph. Consider the equation f +^= I (I) 9 4 This equation gives y = 4 (.-?) 9' :>^ = ±2^1 -| (2) This formula enables us to calculate y when x is given, but there is evidently a limit to the admissible values of x. No ordinary number can be equal to the square root of a negative number; for if ^ were such a number equal to the square root of, say, — 4, p = V^M, then /^ = — 4. But the square of any number, be it positive or negative, is a positive number. Hence, if r is to be an ordinary number, (i — — j must be positive. (i — — j will be positive so long as x^ does not 6 APPLIED CALCULUS exceed 9, i.e. so long as x is not greater than 3, numerically. x^ When a: is - 3, the value of— is i, and when x is x^ 9 ' + 3, the value of — is i, and for all values of x 9 _5p2 between — 3 and + 3, the value of — is less than i. The permissible values oi x lie, therefore, between At = - 3 and X = + 3, and on referring to equation (i) it is evident that when X has either of these values, y must be zero. We will now give x a series of values between — 3 and + 3, and calculate^. A few corresponding values are given below. X — 3 1 —2 — I + 1 + 2 + 3 y ±|Vs +fV8 ± 2 + fV8 ±!Vs These numbers are sets of values for x and^ which satisfy the equation between x and y x^ y^ 9 T I. •(3) It will be seen that, corresponding to a given value of X (say, + 2), there are two possible values of y, namely, (+ fx/s) and (- fVs). Now, if the numbers are all supposed to stand for lengths, we can treat these values of x and y (— I, — |x/8, e.g.) as co-ordinates of a set of points in a plane and plot the corresponding points. We shall thus get a picture, so to speak, of pairs of numbers which satisfy the given equation. INTRODUCTION 7 It is convenient to use squared paper for plotting. The points are plotted in fig. 2. A fair curve, suggesting an ellipse, can be drawn through these points. The "graph" of a mathematical equation is a line in a plane so drawn that the co-ordinates of every point in it satisfy the given equation. The line may be either a straight line or a curved one. Thus the curve roughly outlined in fig. 2 is the graph of — +-^ = I. 9 4 We can show that the graph of the equation ^+^ = I 9 4 8 APPLIED CALCULUS is an ellipse, but before we do this we must say what we mean by an "ellipse". Definition and Geometrical Construction of an Ellipse. Suppose a point moves in a plane in such a way that the sum of the distances of the point from two fixed points in the plane is constant. The moving point traces out a path, and this path is called an ellipse. The curve may be drawn in the following way. Suppose P and Q are the given fixed points (fig. 3). F'gr- 3 Fix pins at P and Q, and tie a loop of thread round the pins so that, when stretched out tight, its total length (PQ + QR + RP) is equal to the given sum of the distances of the moving point from P and Q, plus PQ. Since PQ is constant, PR + RQ must be constant, wherever R may be. Place a pencil point at R, as shown in fig. 3, and move it INTRODUCTION 9 round. It will trace out a curve which possesses the property that PR + RQ is constant wherever R may be. The curve is therefore, by definition, an ellipse. We will now use this property to find the equation which must be satisfied by the co-ordinates of the moving point. The Equation of an Ellipse. Join the given fixed points P and Q, and produce the line indefinitely, X'X (fig. 4). X^ Bisect PQ at O and through O draw Y'Y perpendi- cular to PQ of unlimited length. Take O as the origin, and X'X as axis of x and Y'Y as axis of jv. Let R be a point on the ellipse. The co-ordinates of R are :!c = OS >/ = SR. Now PR -I- RQ must be constant (the property of the ellipse). lo APPLIED CALCULUS Let the constant be m. Then PR + RQ = m (4) The length PQ is also given. Let PQ = / .-. OQ = L (5) Also PR = N^PS' + SR' = Vpo + OS^ + SR^ and RQ = '^SQ^ + SR' m by (4). 2 To simplify this equation, we have, on squaring. 2 + X + y^ -\ X +y' a + 2^(i + /+y)(£-.Vy) X^+yi + L-VL 42 INTRODUCTION Simplifying this equation, we get ►,2. —(m^ - /2) _ x%m^ - P) - y^m^ = o (6) 4 When R is at B, PB = BQ and 2 PB = m; hence if OB = b, 2a/— + 6^ = m. 4 f^ = ^^^Hi! (7) 4 Similarly, when R is at A, PA + QA = m = PQ + 2 QA; and, if OA = a, QA = « - 2 m. 2/ 2fl = m. 2 We can therefore rewrite (6) as a = ^ (8) w •(9) This equation is therefore that of an ellipse, a and b are constants which settle the size and shape of the ellipse. O, the origin in fig. 4, is' called the centre; OA, which is equal to a, the semi-major axis; and OB, which is equal to b, the semi-minor axis of the ellipse. APPLIED CALCULUS The graph of ^ +yl = 9 4 is therefore an ellipse whose centre is at O, and whose major axis lies along the x axis and minor axis along thejj/ axis. The semi-major axis is 3, and the semi- Y 9o 7 7 =7 =Sj «cr 1 . X ^, % < ? \ ? < SI ( ^ V X' Ib Y 4 A P, 4 4 4 4 4^„, X \ ^^ D_ - --^ v F!g--S minor axis 2 units long. In fig. 2 the inch is the unit of length. EXAMPLES FOR PRACTICE Draw the graphs of the following mathematical equations : — (i) J)/ = 2:1: + 3. (ii) a:2+y = 4. (ill) y^ = X. The Area of an Ellipse — Kepler's Method. Suppose we have to find the area enclosed by the ellipse shown in fig. 5. INTRODUCTION 13 It is evident from the symmetry of the figure that we need only find the area of the portion lying in the positive quadrant, for if A stands for this area then 4A stands for the required area. Divide OA into a certain number, say 10, of equal parts by the points of division p^, p^, p^, &c. p^ coin- cides with O and ^i„ with A. Through these points of division draw straight lines parallel to Y'OY, meeting the ellipse in points q^,, q^, q^, &c. ^'o coincides with C, and q^^a with/m ^"d A. P, 4 4 4 4 4 4 4 4 4 5< Fig:. 6 Through each of these points q^, q^, q^, &c., draw straight lines parallel to X'OX to meet the neigh- bouring ordinates in v and /, i.e. the line parallel to X'OX through q^ is drawn to intersect the line/^5'4 in l^ and the line p^q^ in Wg. We can thus construct two rectilinear figures, shown in figs. 6 and 7, one of which lies wholly within the given figure OAC, while the given figure lies wholly within the other. There is no difficulty in calculating the areas of these rectilinear figures as each is composed of a set 14 APPLIED CALCULUS of rectangles. The length and the breadth of each of these rectangles are known. The area of the figure {p^, /„, q^, W, q^, 4> • • ■ 4> q^, /g, /„) (fig. 6) is evidently less than the required area, while the area of the figure {p^, q^,, Vi, q^, v^, q^, . . . Vio, pio, pff) (fig. 7) is greater than the required area. Let Aj stand for the area of the rectilinear figure Fig:- 7 shown in fig. 6, and Ag for that shown in fig. 7. Then, if A stands for the true area required, A is greater than Aj and less than Aj, i.e. A lies between Ai and Ag. This fact may be written Ai < A < A2 where < is the sign for " is less than ". The error in taking Aj as the required area does not exceed |( ^ r — -) X 100 i percent. INTRODUCTION 15 Now comes the important point of the method. By making the steps p^pr, Pipi, p^p^, &c., small enough, we can make A^ and Ag differ by as little as we please, i.e. we can make the ewor as small as we please. This statement will be considered more fully in the sequel. Meantime the student may easily verify that in figs. 6 and 7 the difference Ag — A^ is the sum of 10 rectangles, such as hq^^q,^, which have a common breadth a/io.* The sum of the areas of these 10 rect- angles is the product of their common breadth by the sum of their heights. Thus Ag — A^ = ^ab. By taking more divisions, the factor corresponding to yV can be made as small as we please. We will now consider a convenient notation for expressing the areas of the rectilinear figures shown in figs. 6 and 7. Consider any point P (fig. 7) lying on the curve AC. The area of the strip q^p^p^v^ is {p^q^ X {pspe)' lipsq^ and psps are measured in inches, the product is the area in square inches of the rectangle in question. Let the co-ordinates of P be (x, y). When the breadth of the strip (pspg) is small enough for our purpose, we will call it the " difference in X " and write it Sx. Compare Vi which stands for "square root of". Neither 5 nor V stands for a number, and so Sx no more means S X x than -y/x means V X ■*• Sx is simply the symbol for the width of the strips measured in the x direction, it being understood that 5^ is small enough to en- sure that the difference in A2 and Ai is small enough for our purpose. y Stands for the length p^q^, and Sx for the breadth Pipe, .: area of strip = y x Sx = ySx. Now, by construction, the width of each strip is the * a = 0/10 = semi-major axis, b = Oyo = semi-minor axis. i6 APPLIED CALCULUS same, therefore Sx stands for the width of each and every strip. If jVi is the ordinate of g^, jVa M n 92, and so on, the whole area Aj is evidently the sum (yoSx +yiSx +y2^x + . . .jiSx). The co-ordinates of q„ are (o, .^o)- 9i 1, {Sx, jVi). 92 ., i^Sx, jj/2). „ „ 99 „ (9SX, jKg). The sum required is therefore obtained by (i) making P progress in steps, Sx, from x = o to x = a; (2) taking the corresponding value of the ordinate to the curve at the beginning of each step in x, beginning with X = o; and (3) multiplying Sx by each of these ordinates and summing up the products so obtained. This sum is written briefly i:°'ySx where S stands for the words "the sum of terms of type", and o and a indicate the bounds of x between which the sum is to be taken. We have seen that when the steps are made small enough, the error in taking Ag as the required area can be made practically to disappear. In this case, the area A^ is indistinguishable from the area required, A. When Sx is so reduced, a special symbol — an old- fashioned "s" (S) — is used instead of the usual 2 (sigma) and d instead of S, and we write I ydx. Jo INTRODUCTION The mathematical expression ydx 17 /: stands for the area OAC (fig. 7) when y and x are connected by an equation of which the curve AC is the graph. One method of calculating the value of ydx, approximately, is to divide the area up into strips as shown in figs. 6 and 7; sum the areas of these strips; and repeat the process, using smaller and smaller strips, until the inscribed (A^, fig. 6) and Fig. 8 escribed (Ag, fig. 7) rectilinear figures differ by a sufficiently small amount. This method is an easy and useful, though tedious, way of arriving at the required area, within known limits of error. The last words are important. An approximate result may be all that is required, if we know its limits of error. ra The problem of finding the value of I ydx accurately J is the task of the integral calculus. On the other hand, the differential calculus is con- cerned with problems like the following one. Consider the curve OP (fig. 8). Let A stand for the area O/P. ( B 102 ) 3 i8 APPLIED CALCULUS Suppose we divide Op into a large number of equal parts. These parts need not be inches. They can be millionths of inches theoretically, and in actual drawing it is easy to make them ^ in. Let Op increase by one part, i.e. hy pq. The area added is If we write AA for the increase in area, and Aic ,, ,, x, we get.AA = pV x Ax, nearly. AA I.e. -J— = pP = _y, nearly (lo) .'. the ordinate at P is nearly equal to the rate at which the area increases with x in the neighbourhood of P, and by making pq progressively smaller and smaller, v-r-j usually becomes progressively nearer and nearer to y in value. We have, then, to examine in detail the approximate equation AA , , , ^ = ^, nearly (ii) This problem is the role of the Differential Calculus. CHAPTER I General Principles "Observe always that everything is the result of a chang-e, and get used to thinking that there is nothing Nature loves so well as to change existing forms and to make new ones like them." — Marcus AURELIUS. Concrete Number — Quantity. A child sees some apples on a plate, and begins to wonder whether there are enough to go round. The question, ^^ How many apples are there?" arises at once, because the answer is of much interest to him. His interest does not arise from any theoretical im- portance he attaches to the idea of quantity in itself — it is purely practical. He soon learns to count, and finds that if there are 4 apples and 4 people, he may get one, but if there are 2 apples and 4 people, the position is not so satisfactory. He thus comes to associate numbers with groups of things. When he does this, he is beginning to apply mathematics to practical problems. In all practical applications of mathematics, whether by the child or the skilled mathematician, the numbers used refer to definite quantities of a thing of some kind. They are con- crete numbers or numerical quantities. The "thing" need not be a material thing in the sense that it can be weighed, though it may be. The quantity is always measured in terms of a "unit", and the number associated with the unit is the measure of the quantity in terms of the unit employed. 20 APPLIED CALCULUS Suppose we are considering the length of the straight line AB (fig. i). If we use centimetres, the length will be represented by 6 — if we use inches, by 2-36, and so on. The length can therefore be represented by any number, depending on the " unit" length we use. To the purely arithmetical question, "What re- sults when 3 is multiplied by 2?" the answer is 6; but to the question, ' ' What is the area of a rec- tangular field whose sides are 3 and 2?" no answer can be given until we know the "unit", in terms Fig. I of which the sides are measured. If they are measured in furlongs, the answer is 3 fur. x 2 fur. or 6 sq. fur. It is not usually necessary to write after each figure the unit implied in it, but the unit must, nevertheless, be carefully kept in mind. In physical science, the units employed are care- fully chosen so as to keep calculations as simple as possible. For instance, although the area of a rec- tangle is proportional to the product of its length and its breadth, whatever units of length and of area are used, the area of a rectangle 3 in. long and 2 in. wide is represented by 6 only if the unit of area used is the square inch. If the inch is taken as the unit of length, the square inch is the natural unit of area. Abstract Number. In arithmetic, numbers are sometimes considered without any reference to particular objects. 3x2 = 6 GENERAL PRINCIPLES 21 expresses a fact concerning the numbers 3, 2, and 6 considered as mere numbers. It is true no matter what things the numbers refer to. We must distinguish between: 1. The thing or quality we are measuring; 2. The amount of that thing, i.e. the quantity of it; 3. The standard amount of the thing — the unit, i.e. the arbitrary amount we denote by the number i ; and 4. The number which measures the quantity of the thing in terms of the chosen unit. The study of numbers themselves is only of in- direct interest, since the numbers have only artificial meanings so long as they are divorced from a unit. Not till the unit is introduced does the number become r' p' . P R o Fig-. 2 a definite quantity standing for the amount of a thing. It is just as easy to be specific as to what the unit is, and, unless the contrary is stated, we shall sup- pose the unit is a given length, i in., say, of a straight line. Any number, therefore, represents a straight line, i.e. a quantity of length. For instance, if we measure from a starting-point at O, to the right (fig. 2), OP is the quantity represented by 3 if OP is 3 in. long. Similarly, OP' is the quantity represented by —3 if it is 3 in. long and measured in the opposite direction to that of OP. 22 APPLIED CALCULUS Every point, R, R', &c., in the straight line R'OR, when extended indefinitely in each direction, is the end of a straight line, the beginning of which is at O, and this straight line is a quantity of length corre- sponding to some number of units of length. Thus: if OR is X in. long, OR corresponds to the number x. The straight line R'OR is therefore, as it were, a picture of all the numbers, as any point in it corre- sponds to some number. The reader will recall that the whole theory of dimensions in physics is based on the essential differ- ence between "quantity", or the amount of a thing, and the mere number that measures this amount in terms of the chosen unit. The mere number has no dimensions; the quantity has physical dimensions which are the same as those of the unit chosen.* Change. "All the world 's a stage, And all the men and women, merely players;" said Jaques. Some people play dumb parts that they may watch the heroes and the heroines the better — but play some part they must in the ever-changing scenes of the pageant of life and of nature. Change is everywhere — stagnation is anathema. Change seems to be the stimulus that' awakens con- sciousness. The child notices the changing colours on the wall, the moving leaves on the trees, and remembers them. The idea of change or variation is a fundamental one, and is derived from our experience * For a full explanation of this theory, see the chapter on Dimensions in any standard work on physics, e.g. Deschanel, Electricity and Magnetism, pp. 346-50. GENERAL PRINCIPLES 23 of things from earliest infancy. If a thing which is being measured changes, the quantity of it must change too. The idea of variable quantity therefore arises. It arises directly from our fundamental idea of change. On the other hand, there are certain things which we are led to believe do not change. The quantity of anything which does not change must remain constant, and will be represented in any given system of units by a definite unchanging number such as 2 or 3. Variable Quantities — Functions. In elementary mechanics, it is shown that, if a stone falls freely to the earth under the influence of gravity, the distance through which it falls in a given interval is given by s = hS^i' (I) where i' stands for the vertical distance fallen through, in feet; g stands for the acceleration due to gravity, in feet per second per second ; and i stands for the period of flight, in seconds. The acceleration due to gravity does not change, appreciably ; hence it will be measured by a constant number, which is 32 nearly, in the unit stated. So (i) becomes s = y'-t^ = i6t^ (2) A stone may fall for a longer or a shorter period, and may, in consequence, travel a longer or a shorter 24 APPLIED CALCULUS journey. The values of i' and t in (2) are therefore not unalterable numbers. For instance, \i t = i, s = 16, i.e. the stone falls through 16 ft. in i sec. If i is put equal to 2, we find j = 16 x 2^ = 64, i.e. the stone falls through 64 ft. in 2 sec. So we can go on putting any positive number for /, and calculating the corresponding value of s. We can plot a graph of the results. Quantities such as those denoted by J and t in (i) are termed variable quantities, while quantities, such as the acceleration due to gravity, which are essen- tially invariable, are called constant quantities. It is usual to use the later letters of the alphabet, X, y, z, to stand for variable quantities, and the initial ones, a, b, c, &c., for constant quantities. When two quantities are related to each other, so that changes in the value of one are accompanied by changes in the value of the other, each is said to be a function of the other. The value of s, in the example, depends on the value given to t, while if s be given, the correspond- ing value of / depends on the particular value given to J. In fact the graph shows the correspondence which exists between the values of j and t. s is therefore a function of t, while ^ is a function of J. This functional dependence is written ^ = /w or ^ = '^{s), where/, \p- stand for "a function of". In this notation, f{a) stands for the value of the function f{x) when a is put for x; thus '\i f(t) is iSf^, f{a) is i6a^. GENERAL PRINCIPLES 25 To take another example, suppose y = -^y? + 2X + 2. As y has different values, depending on the par- ticular value given to x, we speak of y as being a function of x, and we symbolize the relationship thus: y = <^{x\ and on putting a for x, i, or (i — x) when ^ < i, and write {x ~ i) for the difference taken positively. The same thing is also often written Take the first standard set — an amount of flesh weighing " the twentieth part of one poor scruple". This amount is r-mis lb. If then x stands for the actual weight of flesh cut, in pounds, the difference between x and i, taken positively, must be less than 1 The conditions can be written (^ ~ i) < T^ (3) 30 APPLIED CALCULUS If X satisfies (3), Shylock wins ; if not, Portia wins. But the second test is more severe and less specific — " tlie estimation of a hair ". Hairs vary in weight, but, to give the Jew the benefit, we will take a good long fat hair weighing say j^th part of a grain. This is xVth part of the first standard, so our condition runs (^ ~ i) < y^fijTnr (4) The Jew cannot hope to cut so exactly as this standard requires. He might be sure of his skill to I oz., but to " the estimation of a hair " is out of the question. In other words, the degree of accuracy asked for is higher than the degree practically attain- able. The allowable errors assigned by Portia were both "small", the second, of course, being smaller than the first. We mean by this that a piece of flesh weighing only tttw^ lb. is small enough, in the cir- cumstances, to be ignored, and we regard txt^^jt lb. as a small quantity of flesh. When we describe a quantity as "small " we mean that it is small enough for some purpose we are thinking of ; not that it is necessarily small in itself. Portia's purpose was to set limits to the margin allowed to the Jew, and she decided that -ruh^js lb. was, under the circumstances, small enough to be ignored. Similarly, if we were measuring lines with a micro- scope we might say that a length ^ in. was large — meaning that it is large enough to be measured without a microscope. Again, consider a length j-ott in. The area repre- sented by its square is y^^ sq. in., the volume represented by its cube, lOooo Tny c. in., and so on, GENERAL PRINCIPLES 31 and if we say that ^^ in. is "small" we may mean that it is small enough to permit of the quantities represented by its powers, (T5^r)^ (tw)*> &c., being neglected in our calculations. We shall see an example of this use of "small" later — it is the most common use of the word in this subject. We can regard the quantity x as small when \X\ < K, where k is some specially chosen positive quantity which we decide to regard as our standard of small- ness, or as our standard of approximation. Small Quantities. Consider the finite geometrical series I +r+r2+ . . . +^-"-1, (5) where r is a positive fraction and n a positive integer. We will suppose r is ^V and n is 10, so that the series is ^9" lo 100 10" The series has 10 terms, the last of which is — a- 10" The second term is -fV of the first, and, for some purposes, the second term may measure a quantity, small relative to that measured by the first term, unity. The third term is then small relative to the second, since it is -^ of the second, and is as small relatively to the second as the second is relatively to the first. It is, however, far smaller relatively to the first than to the second, because it involves the square of the "small quantity", yV- 32 APPLIED CALCULUS We may call it a small quantity of the second order. Similarly, the fourth term, T^?y, involves the cube of the small quantity, -^, and is called a small quantity of the third order, and so on for the remaining terms. Our series, then, goes up to a small quantity of the ninth order. In the series (5) if r is " small " relative to unity, r is called a small quantity of the ist order, r' ,, ,, ,, 2ncl ,, ' M »» >» 3™ >> and so on. Whether r w " small " compared to unity B- 0^ 10" ■ 10"- Fig- 3 D^E or not is a question which depends on the particular application we are considering. The importance of this idea of " orders of small- ness " will be seen from an example. Consider the expansion of a square plate of iron by heat. Let the side of the square plate measure 10 in. at 0° C. At 10° C. its sides measure 1000123 in. (fig. 3). The area of the plate at 10° C. can be very simply calculated with accuracy. It is (io-ooi23)^ = 100-0246015129 sq. in. GENERAL PRINCIPLES 33 For most purposes, we do not need as many as 13 significant figures in the calculation. It may be sufficient to be correct to say 7 significant figures. The area of the expanded plate, correct to 7 signifi- cant figures, can be calculated by the following method, which depends on the use of "small quantities", thus: In fig. 3, the area of the expanded plate is AGFE. The original area is ABCD, hence area AGFE = area ABCD + 2 (area CKED) + area HFKC, i.e. area AGFE == [10^ + 2(10 X 0-00123) + (0-00123)^] sq. in. = [100 + 20 X 0-00123 + (0-00123)^] sq. in. = [100 + 0-0246 + O-000001513] sq. in. We have thus expressed the area of the expanded plate as the sum of a series of three terms. The second term depends on 0-00123, and the third term on (0-00123)^ 0.00123 is about ^Trltnrth part of the first term, 100, and may legitimately be re- garded as "small", relatively. The second term is therefore a small quantity of the first order, as it depends on the first power of the small quantity (0-00123); while the third term is a small quantity of the second order. It is evident that the third term can be neglected in calculating the area of the expanded plate to 7 significant figures, for 7 significant figures take us only to the fourth decimal place, and the third term contributes nothing before the sixth place of decimals. It arises from the little square HFKC, and is TTB^innnrTrTTth part of the first term. The area of the expanded plate can therefore be (D102) 4 34 APPLIED CALCULUS obtained as accurately as need be by rejecting the term of the second order of smallness, and retaining only the term of the first order. The result is 1000246 sq. in. Provided the small quantity of the lowest order which we retain is small enough, we can neglect all the small quantities of higher order which occur. To put this calculation into the language of in- equalities : We wished to calculate the area correct to 7 sig- nificant figures. The area must be greater than 100 sq. in. and less than 121 sq. in., hence 7 signifi- cant figures take us to four places of decimals. We can therefore neglect areas less than Tmnnnr sq. in. The true area is [100 + 0-0246 + (o-ooi23)^] sq. in. The approximate area, obtained by rejecting the last term, is 100-0246 sq. in. The numerical difference in these areas is (0-00123)^ sq. in. Now (0-00123)2 < T^rsW- .". (0-00123)^ can be neglected in calculating the area. We cannot neglect the area corresponding to the small quantity of the first order, as this area is 20 X 0-00123 = 0-0246 sq. in., and 0-0246 is not less than x^^nnnr- If we were satisfied with 4 significant figures, we could neglect this term as well as the second, as 4 significant figures require only one decimal place, GENERAL PRINCIPLES 35 and hence we could neglect areas less than ^wu sq. in., and {0-0246 + (0-00123)2} < if^. In this case the expansion would not affect the area of the square plate by amounts within the degree of accuracy required. Expansion of a Cube. Small quantities of the third order arise in the problem of the expansion of a cube. Suppose the Fig. 4 side of the cube measures 10 in. at 0° C. At 10° C. its side measures 10-00123 in. Fig. 4 illustrates this case. The volume of the expanded cube is (10-00123)^ = 1000-369045388860867 c. in.; 36 APPLIED CALCULUS but we do not need 19 significant figures for any practical purpose: 7 figures will be quite enough, as in the case of the square. The expanded cube is got from the original one by making certain additions : 1. A slab like CDGFKJML (fig. 4) is placed on each of the three faces of the cube ABCD, CDGF, and ADGH. The size of each of these slabs in cubic inches is 10^ X 000123 = 0123. The three slabs therefore contribute 3 X 0-123 = 0369 c. in, 2. Three small prisms such as DPONARST will be required to fill up the corners left at the ends of the slabs. The size of each of these in cubic inches is 0-00123^ X 10 = 0-000015129. The three contribute 3 X 0-000015129 = 0-000045387 c. in. 3. A tiny cube will be required to fill up the corner at Q. The size of this cube is (0-00123)' = 0-000000001860867 c. in. The whole growth is the sum of these items, and the volume (V) of the expanded cube is the original volume plus these growths, i.e., in cubic inches, V = [1000 4- (3 X lo^ X 0-00123) + {3 X 10 X (0-00123)2} 4- (0-00123)8] = [1000 4- 0-369 4- 0-000045387 4- 0-000000001860867]. The second term depends on 0-00123, the third on (0-001^3)2, and the fourth on (0-00123)^ GENERAL PRINCIPLES 37 We can consider 0-00123 a small quantity just as in the case of the square. The second term is therefore a small quantity of ihe first order, the third of the second order, and the last of the third order. It is evident that this is the order of importance of the quantities and the order in which they should be taken to get increas- ing accuracy. The three slabs (first order terms) con- tribute most to the growth, the three prisms the next amount, while the little cube contributes the least. We need only retain the first order term to get the 7 significant figures we need, i.e. 1000-369 c. in. is the desired result. Note how rapidly the terms of higher order of small- ness decrease in importance. The Changes in the Values of Functions. Suppose jj/ = f{x). It is important to find how changes in the value of X affect the value ofji^. We may illustrate the general method of dealing with this question by applying a more symbolic treatment to the two practical problems just con- sidered. (i) Find the increment of area when the sides of a square metal plate expand by a given amount. Suppose X stands for the length of'the edge of the plate, and h for the linear expansion (DE or BG, see fig. 3), and J)/ for the area of the plate before expansion, then, if_y' is the area of the expanded plate, y =^ {x + hf and {y' — y) = {x + lif — x^ = 2xh + A^ (6) -•. the increase in area is 2.xh + 1^. 38 APPLIED CALCULUS h. finite increase in x is often written A^. This does not mean A X a:, but is a single symbol standing for "a finite increase in " x. Similarly, a finite increase \ny is written Ly. In this notation we get Ajj/ = zxt^x + (A^)2 (7) This equation gives the information we require, viz. the increment in the area (_>») arising from a given linear expansion {Loc) of the side {pc). The formula has been calculated for an expanding plate, but it is clear that it is true whatever x and y may stand for, provided j>/ = x^ The formula gives the increase in x^ when x increases by Ax, whatever values x and Lx may have. WORKED EXAMPLE Suppose X = <) and t^x = ^, then Ajc = 2X9X2 + 2" on substituting in (7) = 36 + 4 = 40. Check. When ar = 9 x^ = %i X ^ \\ x'^ = 121 Difif. = 40. .'. growth in x^ = 40 when x increases from 9 to 11. The Ratio Ay/ Ax, when y = 3^. \iy stands for the area of the plate in square inches, and X for the length of the side in inches, Ay = 2xAx -^ (Ax)2 by (7) (8) Consider what happens when we cool the plate from say 10° C, towards the original temperature, 0° C. The linear expansion becomes smaller and smaller, and hence Ax becomes smaller and smaller. When the GENERAL PRINCIPLES 39 rise of temperature is 10° C, Ax is 0-00123; when the rise is i" C, Ax is 0-000123; when the rise is xinnr" C., Ax is 0-000000123; and so on. But there must be some rise of temperature, otherwise the problem is not one of the expansion of a warmed plate at all. Hence the value, zero, of Ax does not interest us. We will suppose that Ax can be as small as we like, but not zero. When Ax is not zero, we can divide equation (8) by Ax thus : -^ = 2X + Ax (9) Now, by taking Ax small enough, we can bring 2x + Ax as near to 2x as we please. For instance, {2X + Ax) differs from 2x by less than I billionth, when Ax < i billionth, and so on. We express these ideas by saying that the limit ol (2x + Ax), as Ax approaches zero, is 2X, and we write Lt {2X + Ax) = 2X. Ax — ^ The symbol — >■ reads ^^ approaches". Reverting now to (9), we get Lt (-r=^) = Lt (2.x -\- Ax) = 2X. ...(\o) The fraction Ay\Ax therefore tends to the limit 2X, as Ax approaches zero. The limiting value of Ay/ Ax as Ax -> o is denoted The symbol -5- does not mean d -f- dx, and is not a symbol for a number at all. It is a symbol standing for an operation, which, when applied to y, is the ^^(^) 40 APPLIED CALCULUS operation of taking the limit of the ratio of Ajv to Ax, when A^ — > o, so that (£> -^nm (■■> And if^ = x^ {i)y = ^^ ('"> The symbol -r- may be compared with ^. This sign is a symbol of operation. \/x does not mean y/ X X, but it means the result of the operation of extracting the square root of x. \-f-)y is usually printed -^ for convenience, but when it occurs in this form, it must not be mistaken for a fraction {dy -f dx). Summarizing the notation, we have (^V = i =^t^o(^) ^'^^ or, putting f{x) for y \{ y = f{x), a)/« = ^ -„!i.(^') (-> \dx. It is important that the reader should see clearly the necessity for this treatment by Hmiting values. For instance, it mij^ht be asked, why not put Ar = o in the equation -^ = 2X 4- A;r, Ax and get -^ = 2X and avoid using the idea of the limit? The answer is that Ax cannot be both something and nothing ; it must be something or nothing. If it is nothing, there is no change in x, so AyjAx is meaning- less ; in fact, the fraction cannot arise. GENERAL PRINCIPLES 41 It therefore must be something', but it can approach zero as nearly as we lilie. This possibility suggests our trying to find a number I such that ^-> I as t^x — > o. This number I, if we can find it, is called a limit, and we express the fact that the limit I exists (if it does) by writing Lt {^)=l = f (IS) AX — > as a matter of notation. (2) The symbolic treatment of the second of the two practical problems, viz. to find the volumetric ex- pansion of a metal cube due to a given expansion of the side, will serve to emphasize the foregoing funda- mental principles. Let X stand for the length of the edge of the cube in inches, andjv for the volume in cubic inches. Let A^ stand for the expansion of the side in inches. Then, \iy is the volume of the expanded cube, y = {x-ir ^xf. .'. y' —y = (x -\- i^xf — X?. :. Ly = 3x^Ax + 3x(Axf + (Axf (16) This equation gives us the result required, viz. the volumetric expansion in terms of the linear ex- pansion. EXAMPLES 1. The side of a cube of brass expands from 12 in. at 0° C. to 12(1 4" o-oooiSf) in. at 10° C. Calculate the increase in its volume, and the ratio of increase in volume to increase in length of side. Work to 4. significant figures. 2. Find the change in the value ofy (= x^) when x changes from g to II, and check the result given by the formula (16) by a direct calculation. 42 APPLIED CALCULUS To find dyjdx when y = jr'. We have by (i6) Ay = sx'-Ax + 3x{Axf + (Axf (17) .-. ^ = 3x^ + 3x (Ax) + (Axf. Ax i^ = Lt f^), dx ^ _). oVAjc/ hence we require the limit of —^ as Ax — > o. Ax 3x{Ax) may be made as small as we please by taking Ax small enough. .". 3x(Ax) — >■ o as (Ax) —*■ o. Similarly, (A^)^ — > o as (Ax) — > o. .'• (3^^ + 3xAx + (AxY) — > 2^^ as Aa; — > o; '■'•f =„tl.(^) = 3''- <■«> when ^ = ^. To find 3- (a:"); « a positive integer. ax We have seen, on p. 40, that, iiy = f{x), dy ^ dfjx) ^ ^^ ( Af(x) \ dx dx A* — > o\ Ax / and for two particular functions of x, viz. x^ and x^, we have found the respective limiting values to be 2X -and 3^^. That is -J- = ix when y = x'-, (toe and -f- = 3x^ when y = x*. dx GENERAL PRINCIPLES 43 We will now proceed to find the value of ^ when y = x", dx -^ n being a positive integer. The algebra will be a little long, and it will shorten the expressions to write A, instead of A^, for the increase in x. Multiply a^ + fli + 6^ by fl — b. Result a^ — b^; a^ + a^b + ab'^ + b^ hy a - b. Result a* - d*. In general, {a - b) («"-! + a"-2|& + a»-3^,2 + . . . + 6»-i) = a" - *" (19) This identity can be easily proved by direct multi- plication. It is true for all values of a and*^. Rewrite it "" ~ r = «"-^ + a^'-^b + a^-W + . . . + 6"-i (20) a — There are n terms on the right-hand side. Put a = X + h and b = x. Then a — b — h, and a^ -b^ = {x ^- A)", - 0^. . g" - ^>" _ (^ + /)" - ;c" a — b h (21) Substituting a = x + A and ^ = x in the right- hand side of (20) we get ■ (x + lif-^ + (x + hy^-'^x 4- (x + /i)"-^^^ -I- . . . + x»-i. Now, (x + A)" — x" is the increase which takes 44 APPLIED CALCULUS place in y when the independent variable x changes by the amount h, i.e. by ^x. Therefore Ay _ r {x + hy - x'' ~[ Ai ~ L h J" The limiting value of this expression, as A — >• o, is by definition, the value of ^(■^")- The limiting value we require can be easily found. When A -> o, {x + A)»-i -^ ^"-1 (x + A)"-2x -^ ^"-1 {x + A)"- V _). :x;"-i . . . . — >• . . . Ar"-i -> :)c"-i Adding, (:v + A)"-i + (^ + /O'-^x + . . . + :x:"-i-* « X ^"-1. as there are n terms. " A_>. oL h J = «X"-S '•'• ^("") = «^"-i (22) Derivative of a Function. If jj/ = /(:«), and ^ = V'W* V'W 's called the derwative oi f{x). Equation (22) expresses an important rule which can now be put in words. Rule I. — The derivative of x^ is n times :v""^ when « is a positive integer. The rule is true for all values of x. Thus, the derivative of x^ is 2 x, that of x^ is 3 x^, and so on. GENERAL PRINCIPLES 45 Exercise 1 I. Draw a circle of radius 2 J in. Divide a diameter AB into 10 equal parts, and draw lines through the points of division, cutting the circumference of the circle and perpendicular to the diameter chosen. By the method shown on pages 13, 14, find upper and lower limits for the area of the circle, and express the difference in these limits as a percentage of the lower limit. 2. Repeat (i), dividing AB into 20 equal parts. Notice that by making the strips half as wide as in (i), the area is determined within much closer limits. Find these limits, and compare them with the true area, calculated from -ky^. 3. \l y = jc^, calculate the increase in y corresponding to an increase Ajc in x, when a; = 6, and also the value of -T^. Calculate the values of Ay and — =?- when Ax = 0.1 o-oi, O'Ooi, &c., and fill in the blank spaces in the table below, to show the way in which -^ approaches its limit- ing value 12, as Ax approaches zero. ^x O- I O'OI O'OOl O'OOOI Ay Ay Ax 4. If y = x^ find Ay and -^ corresponding to an in- crease Ax in X when x = 8, and calculate their values when Ax = o-i, o-oi, o-ooi. Fill in the blank spaces in a table similar to that above, to show the way in which -^ approaches its limiting value, 192, as Ax — >■ o. 5. Find from first principles the derivative of x*. 46 APPLIED CALCULUS 6. Differentiate : (i) y = x^. (ii) j> ■= x^. (iii) y = x^'^. 7. A square plate of brass is expanding by heat. When the length of its side is 7 in. it increases by -002 in. in a second. Give an approximate value for the corresponding rate of increase in the area of the plate. CHAPTER II Geometrical and Mechanical Meaning of a Derivative "Nothing' can be more fatal to progress than a too confident reliance on mathematical symbols j for the student is only too apt to take the easier course, and consider the formula and not the fact as the physical reality." — Kelvin and Tait {Natural Philosophy). In Chapter I, we found that \i y = f{x), we are sometimes able to find a limit / to which the fraction -r=^ tends in value as Aa; approaches zero. We ex- press this by writing Lt m = 1 = % lix _> o^Ax/ ax This limit we called the derivative oif{x). In this chapter we shall try to find out the physical significance of this limti. Gradient or Slope ot a Straight Line. Let O'P, fig. I, be any straight line, and let O^, Oy be any pair of rectangular axes. Choose any point P in this line, and let {x, y) be the co-ordinates of P. Then OM = x and MP = y. Let the straight line intersect the Oy axis in the point O' whose co-ordinates are (o, c). Through P draw MP parallel to Oy, and through 48 APPLIED CALCULUS O' draw O'N parallel to Ox, N being the point of intersection of this line with MP. Then the ratio of NP : O'N is called the gradient of the line O'P. If we suppose Ox to be a horizontal ^^^ ^ M * Fig. I line, this definition of "gradient" agrees with the ordinary meaning of the word. Let a be the angle of inclination of O'P to O'N, then NP ~r^ = tana = gradient of the given straight line. Gradient or Slope of a Curve — Definition. The gradient of a curve, at a given point on it, is defined to be the gradient of the tangent to the curve at the point in question. Equation of a Straight Line. Referring to fig. i, we see that, if P is any point on the straight line, MP = MN + NP = OO' + NP = OO' + OM • tana. .', y = X tana + c. MEANING OF A DERIVATIVE 49 Putting m for tana, y = ntx + c, where tn and c are constants, (i) This is the equation of a straight line, and it is an equation of the first degree in x andjv. This equation can be easily transferred into any other form in which a linear equation may be written, thus: y = mx + c. Multiply by B throughout, then Bj/ = 'Qnix + Be. Put ( — A) for the constant Bm, and — C for Be, and we get By = - A.X - C. i.e. Ax +By -[■ C = o. Although Ax + By + C = o appears to have three independent constants, it really only has two, as we can reduce the coefficient of y to unity by dividing the equation by B, Parallel Straight Lines. Parallel straight lines are those having the same gradient, i.e. the same m, hence if y = mx + c represents a given straight line, y — mx + c' ,, a parallel straight line, where c and c' are different numbers. When the straight line is parallel to Ox, its equa- tion becomes y = c, for m = tana = tan 0° = o, and when parallel to Oy, the equation is X = d. In these equations c and d are constants. (D102) 5 50 APPLIED CALCULUS WORKED EXAMPLE Find the equation of a straight line passing through the points (—3. 4) ""'^ (2. — i)- Let the equation be y = mx + c. Then, substituting the given values of the co-ordinales of the two given points for x andjc, we get 4 = — 3'« + c and — I = 2m + c. Subtracting, 5 = — c,m. .'. m = — I, and c = I ; tlierefore the equation of the straight line is y = —x-\- 1, or X -^y = I. Geometrical Meaning of ^. dx Let AB be a portion of a continuous graph (fig. 2). Let_y = i/'- Let the line QPM rotate about P in the clock- wise direction, and the line M'PQ' about P in the MEANING OF A DERIVATIVE 51 counter-clockwise direction, subject to the restriction that Q and Q can approach P as near as desired^ but must not coincide with P. It is evident that, as Q and Q' approach P, the lines QPM and M'PQ' will swing round towards a Bi Fig. 2 limiting position, and that this limiting position will be the tangent line at P, for QPM cannot pass the tangent position, otherwise Q must become coinci- dent with P in the process; and, similarly, M'PQ' cannot pass the tangent position, but the lines QPM and M'PQ' can lie as near together as we like by taking Q and Q' sufficiently near to P. 52 APPLIED CALCULUS It is evident therefore that, as Q (or Q') —*■ P, and 0' -> V' where \}r is the indination of the tangent to the Ox direction ; i.e. tan0 — > tani/r as Q — > P. But tan0 = ^ for tan0 = 58, Ax PR and RQ is the change in the vakie of^ brought about by a change PR in the value oi x. As Q -> P, A;c -> o, .■. -^ —> tani/r as Ax -» o, i.e. tan\/r is the limiting value of -r^ as A;k; — >■ o, at P. Ax i.e. [^1 = tanV. where -r^ is the value of -^^ at P and \I/- is the LdxJx = op ax ^ angle of inclination of the tangent at P, to the Ox axis. On the understanding that both -^ and \/<- are to be measured at P, ^ = tanypr (2) dx The reader is advised to draw out fig^. 2 on a large scale ?nd test, by drawing, the fact that B and 6' —^ ^ as Q and Q' swing towards P. MEANING OF A DERIVATIVE 53 Equation (2) gives us a clear idea of the geometrical meaning of dyjdx. It is the trigonometrical tangent of the angle of inclination of the geometrical tangent at P, to the Ox axis. But this is the gradient of the curve at P, hence dy^ d J., . dx dx-' ^ ' is the measure of the gradient of the curve which is the graph oi y = f(x) at the point at which it is calculated. For example: Let ji/ = x^. Then -^ = 2X. dx At the point x = 10, 2x = 20. \dx/^ = 10 .•. the slope of the graph ofy = x^ at x = 10 is 20. i.e. tan i//- = 20 at ^^c = 10. i.e. i/r = arc tan 20 at :v; = 10.* i.e. i/f = 87°io'about,atA;=io. The tangent to the graph of _y = x^ which touches the curve at the point P[x = 10, y = 100] cuts the Ox axis at an angle of 87° 10'. EXAMPLE Draw the graph of y = x^ to the scale i in. = i horizontally {x axis) and i in. = 10 vertically {y axis). Mark the point P for which x = lo, and therefore y = 100. Draw the tangent at P as accurately as you can, and vieasure the angle at which it cuts Ox, with a protractor. What is this angle? Does it confirm the preceding calculation? * This notation — the Continental one — is now becoming common in Britain instead of the older form tan"' 20. 54 APPLIED CALCULUS The result we obtained in formula (2) is true for all functions of x the graphs of which are smooth curves, Fiff- 3 i.e. are curves free from "breaks" or "kinks". Fig. 3 shows what this statement means. (i) Shows a " break " at jc = a. (ii) Shows a "kink" at x = a, but no "break". Obviously neither of these curves has one definite tangent line at the points A and B, and so there is no definite i/r, and consequently equation (2) gives no definite value for ^, hence any function having a graph like that shown in fig. 3 (i) or (ii) would not have a single definite derivative at x = a. Speed. A train sometimes runs in such a way that it travels equal distances in equal times. We then describe it as running at a uniform or constant speed. Let V be the number of miles travelled in any one hour; then, in t hours, the distance travelled is vi MEANING OF A DERIVATIVE 55 miles, i.e. if j stands for the distance travelled, in miles, •r = vt (3) It is evident from (3) that when a body moves with a uniform or constant velocity, the graph of its motion, Fig-. 4 as a distance-time curve, is a straight line through the origin. The graph is shown in fig. 4. Consider any point P (fig. 4) on the graph. pF = s op = t, hence ^ = - = w by equation (3). ^P But ^ = tani/r, if •\/r is the angle of slope of ^^ the line OP, hence tani/r = v. Consider a neighbouring point Q on the graph. Through Q draw the ordinate Qq, and draw PN parallel' to Ot, and cutting Qq in N. Then, NQ = A.r and PN = M, 56 APPLIED CALCULUS as these lines represent, respectively, the increases of J- and t under the condition of constant speed. Now, g§ = tanQPN = tani/f = V. Ac :, — = V = the constant velocity, no matter over what interval {Lt) the increase (Aj) in J is taken. ^^ "' tani/r = — by equation (2). . ds . . • • -^ = z), a constant, = the velocity of the train. In the case of constant speed, we have then V, the constant speed, — jr}— Yt~ 1 ~ ^^"V'- (4) These different ways of expressing the velocity arise from the fact that it is constant, and the graph of distance-time, a straight line. Variable Speed. It is a fact of everyday life that the speed of a train is not constant, but variable. This is the very kind of obstacle the calculus was invented to overcome. Here, the distances travelled in equal intervals of time are not equal. Motion in a Straight Line with Variable Velocity. Suppose a train moves in a straight line from O to P (fig. 5) in t hours where OP corresponds to i' miles. MEANING OF A DERIVATIVE 57 If the point moved with uniform velocity v m.p.h. we should have s = vt or v = ~ ml./hr. This is s miles _ in t hours Fig. 5 the average velocity of the point during the i hours of its motion from O to P. Definition of Average Velocity. The average velocity of a point during any interval of time is that uniform velocity with which the point would describe the distance travelled in the same time. Instantaneous Velocity. Suppose the graph of distance-time is not a straight line, but is as shown in fig. 6. 58 APPLIED CALCULUS Consider the graph from t = o to ^ = Os. At the end of t hours, the train has moved a distance j miles; thus if P corresponds to (j, t)y Op = if and /P = J. At the end of time (/ + A^) hours, the train has moved a distance {s + Aj) miles from the start, i.e. if Q represents this state of things, O^- = t + ii.t and qQ = s + As. Join P, Q by the straight line PQ. If the actual motion between / and t + Lt were represented by this straight line PQ, it would be a motion at uniform speed. Draw PN parallel to O^ to meet the Q ordinate at N. Then the uniform speed corresponding to the straight line PQ is NQ ^ A£ ; PN At This (uniform) speed, in miles per hour, is the aver- age speed between t and (/ + At). \ But As .^a ' At = *^"^' where 6 is inclination of chord PQ to the O^ direction, hence tan0 is the average speed over the interval At at P. We know that, if As -— — »■ a limit as At -*■ o. At this limit is tani/r, where t/t is the angle of slope of the tangent at P, see equation (2). If we assume that a tangent to the curve can be MEANING OF A DERIVATIVE 59 drawn at any and every point — and it certainly appears as if it could — then As/ At approaches a limit at each point, and this limit is tan i//-, so that -j-, at P, = tani/r, at P. As Now, -T-2 is the average speed over the interval of time from ^ to (/ + At), hence this average speed ap- ds proaches a limit, denoted by -r-, as At — > o. This . . ds . limit, ^, is called the instantaneous velocity at the moment in question, since it is the velocity to which the average velocity over an interval of time (A^) at t tends, as the interval {At) approaches zero. Briefly, it is called *^ the velocity at t", and is the velocity the train possesses at that moment. It will be remem- bered that an average velocity is necessarily a uniform one, by definition, and the limiting velocity is merely the final value, so to speak, of an average velocity, namely, the average velocity over an interval At at t. The instantaneous velocity is therefore measured just as a uniform one is, i.e. if the unit of distance is miles and of time, hours, the proper unit for the instantane- ous velocity is miles per hour. It is numerically the same as the distance the train would travel in the next hour of its journey if its speed were unaltered. Definition of Instantaneous Velocity. The instantaneous velocity at a given moment is dsjdt, calculated at that moment, or the velocity at time t is dsjdt at that instant. 6o APPLIED CALCULUS EXAMPLE Suppose the distance travelled in a given interval is given by s = t^ where s is in feet, and t in seconds. Then .„ dt Hence the velocity at exactly lo sec. from the start {t = o) is 2 X lo = 20 ft. per second, i.e. if the moving body moved for the next I sec. with its speed unaltered, it would move 20 ft. The reader is strongly advised to draw, say, a parabola on a large scale. Take Os say 20 cm. and jS 20 cm., the parabola being s = i^ t"-. Divide Os into 10 equal parts. Draw in the ordinates and the " PQ " straight lines of fig. 6. Determine — for each strip, and plot out a speed-t\me curve on the scale Os ^ 20 cm., and a convenient scale for "speed". This curve will consist of a series of steps. Repeat the experiment to find how fine the divisions must be in order that — {a) The straight lines "PQ" may be practically indistinguishable from the curve. (J) The "steps" in the speed-time curve may practically dis- appear. In the preceding discussion, we have tried to show that one of the properties of a body in uniform motion, namely, velocity, may be measured, in terms of a suitable unit, by a definite and constant number; also that when the motion is not uniform, we can still regard the body as possessing a definite velocity at each and every instant, though this velocity varies from instant to instant. In this discussion, we started from the only information we had, namely, that the body moved along, and so was at different dis- tances from the starting-point at different times. This information gives a distance-time curve, which, MEANING OF A DERIVATIVE 6i if the motion is continuous, must be a smooth curve. If we are prepared to take this instantaneous velocity for granted and also to assume that it is a function of time, whose graph is smooth, the problem of con- necting this velocity with distance and time becomes very easy. In this case we can assume a veloctiy-time V ^ F Q ■^i ^ hv / M <- t ^P g t Figr. 7 curve to start with. Before, we started with a dis- tance-time curve. Let fig. 7 be a graph of such a velocity-time curve where pV, for instance, measures the instantaneous velocity at the instant t, corresponding to Op. Con- sider a later instant, t + A^. The instantaneous velocity at instant ^ + A^ is qQ. Let V = ^P = velocity at P, and V + I^v = qQ = ,, Q, and suppose the velocity is increasing with time at P, as in the figure. Let Ls stand for the increase in s, due to the move- ment which takes place in the time A^. Then A-s- > vM, for the average velocity during M is more than v, and if the velocity during A^ were uniform at v, the distance travelled would be vLt. 62 APPLIED CALCULUS And As < (v + Av)At for similar reasons. . . z» < -— < w + Av. Now when Ai — > o, Af —> o, but v does not change since it is the definite velocity at the time i, and though V varies from time to time, it does not vary at the same instant i, and we are letting A^ change, not i. .'. Lt (v + Av) = V, At — > and hence o\Ai/ ds '■'• di = ^' •(5) ds where both -^ and v are measured at the same (and any) point P. In the figure, v is drawn steadily increasing with t. y Fig. 8 This is not essential. If the curve is as sketched in fig. 8, we can always take a Q near enough to P MEANING OF A DERIVATIVE 63 to ensure that the ordinate steadily increases (or de- creases) from P to Q. Extension of the Idea of Speed. We can extend the idea of speed, and its measure- ment by a derivative, by using the word "rate" instead of speed. One thing changes at a constant rate with respect to another, when equal changes in the one correspond to equal changes in the other. The amount of the change in the second thing chosen is usually a "unit" amount, and therefore is denoted by unity. The amount of the -change of the first thing is then so much of it, per unit change of the second. Let m be the change in the first thing [Y] which corresponds to any unit change in the second thing [X]; then if the change in [X] is X, the corresponding change in [Y] is mX, i.e. if Y stands for the change in [Y] Y = mX, where m is the " rate " in question. \i y and j/q are the final and initial measures of the first thing [Y], and x, x^, of the second [X], then Y = y - y^, and X = X — x^. •*• y -y^ = m{x- Xo). .•. y = mx — (mxQ — y^ But mx^ — ji/q is a constant, — c, say. .'. y = mx + c. This is the equation of a straight line, the gradient of which is m, and which cuts the Oy axis at (o, c). We therefore arrive at a very important result, namely, when one thing changes at a constant 64 APPLIED CALCULUS rate with respect to another, the graph connecting the related quantities of the two things is a straight line. This use of the word " rate " has been current since the end of the sixteenth century. " Six score acres after the rate of 21 foote to every pearche of the sayd acre." — Spencer, 1596. It differs only from "speed" in being more general. "Rate" includes "speed", but "speed", since it refers only to distance and time, does not include " rate ". The equation of any straight line is y = mx + c, where m and c are constants, /. Ay = ml\x. .: -r^ = m. Ax Lt (^) = Lt (m) . dy - dx- ^ (^) since Lt (rri) = m, as m is constant. Aa; — ► But m need not be constant. We deal with a rate which varies from point to point on the same lines as a varying speed. We consider the average rate between x and {x + Aa;), defining the average rate as the uniform rate which produces the change in y, corresponding to the chosen change Ax in x. The average rate, so defined, is measured by AyjAx, just as the average speed is measured by As/Ai. The limit to which AylAx tends, as Ax — > o, is the rate to which the average rate tends, as the MEANING OF A DERIVATIVE 6S interval Ax approaches zero. Hence, if m is this limit, dx m. ■(7) and m is called the rate at x. Definition of Instantaneous Rate. The rate of change of y with respect to x, at x, is the value of dyjdx at x. y •*-! ^ Q> ^^^ 5S: -r 00 y /> XT t (seconds) Fig. 9 The study of gradients, or slopes, can now be com- bined with that of velocity, speed, or rate. The rate of change of ^ with respect to x is -^, at the selected point. But -3^ = tam/r where i/r is the angle of slope of the graph oiy in terms of x. Tan\lr is, however, the gradient of the graph at the selected point, hence the rate of change of y with respect to x, at the selected point, is the same, numerically, as (D102) 6 66 APPLIED CALCULUS the gradient of the graph of y in terms of x, at the same point. Whenjj/ is distance (s) and x, time (i), the speed or velocity at a selected point is the same, numerically, as the gradient of the distance-time graph at that point. For instance, the velocity of a body, whose dis- tance-time curve is shown in fig. 9, is tani/r at P, where PT is the tangent line at P and t/«- the angle of slope. This result gives a graphical method of finding the velocity of a train, for instance. Graphical Method of Finding Velocity. Suppose fig. 10 is the distance-time curve for a suburban train. Fig. 10 To deduce the velocity curve from this curve we have to select a range of points, Pq, Pi, P2, P3, &c. ; draw the tangents at Pq, Pi, Pa, &c., and measure the MEANING OF A DERIVATIVE 67 respective angles of slope t^qi V'h V^z» ^^' Look up the tangents of these angles in a table of tangents, and we get tam/fQ, tan^i, tam/rg, &c. These numbers are the velocities at the points Po, Pi, Pg, &c. A curve of velocity against time can now be plotted by plotting these values for the velocity at i^, 4, 4> 4) &c., which are the times corre- ponding to Pq, Pi, Pj, &c. It is convenient, though not essential, to make the intervals of time equal. Scales. The velocity is tarn/r, when the scales are unit scales, i.e. i unit of distance is represented by i in. s A 'Q 1 .XI N ■0 ^ -^ t (hrs) i ^~~ Fig. II and I unit of time by i in. It is often inconvenient to use unit scales, so we take a scale : m ml. = I in. nhr. = I in. 68 APPLIED CALCULUS Let fig. II represent a portion of a distance-time curve drawn to this scale. Let P be a chosen point on the curve, Q a neigh- bouring point, and PN parallel to O^. Then tani/r, where i/r is the angle of slope as actu- NQ ally drawn, is the limit of p^ as Q — > P. But NO = - in. and PN = — in. n NQ _ £« Aj •• PN ~ m M' ft A.r .•. tani/r is the limit of — v-: as Q — »• P, i.e. as M -> o. , n ds n :. tanu' = r; = — «». ■^ m at m where i/r is the angle of slope of the graph as actually drawn, and — the factor to correct for the scales actually used, v is, of course, measured in the unit corresponding to the units of distance and time used. This method is not a very good one, in practice, as it is difficult to draw the tangent lines accurately. Later on we shall explain a better method, which depends on having the speed-time curve given instead of the distance-time curve (p. 109). We have now obtained two ideas of the nature of the limit denoted by dyjdx. They are: I. It is the measure of the gradient of the graph of jv in terms of x, at the point (x, y). MEANING OF A DERIVATIVE 69 2. It is the measure of the rate of change of jj/ with respect to x, at x. These two ideas are essentially the same, and are of great importance in the applications of the calculus. Exercise 2 1. Write down the equation of the straight line joining the points (2, 3) and ( — 5, 6). What is its gradient? 2. Find the equation of a straight line through the point (3, 7), and making an angle of 30° with OX. 3. Find the equation of the straight line through the point (— 5, 3), which is parallel to the straight line y = 2J[; + 7. 4. What is the gradient of the line 3x + /\.y - "J = o? 5. Determine the value of dyjdx at the point (2, 4) on the parabola x^ = y. Write down the equation of the tangent to the parabola at that point. 6. The velocity of a train is varying from time to time. What is the symbol for the rate at which the velocity changes with time, at any particular instant t? This rate is called the acceleration. T. £1 is invested at compound interest at ?-% per annum. Show that the amount (i.e. the original sum plus accumulated interest) is given by M = R" where M = amount and R = I + -^. 100 Hence show that if the money is undisturbed for 10 years, the amount from a 6 % investment will exceed that from a 5 % investment by approximately 3 shillings. 8. The side of a cube is found by measurement to be 9 in. If there is an error of ^Jir in. in this measurement, ^o APPLIED CALCULUS find an approximate value for the consequent error in the- calculated volume. 9. Two men, A and B, start from a place O to walk in perpendicular directions, A at 3 miles per hour, and B at 4J miles per hour. At what rate are they receding from each other. 10. A cube of copper 10 in. long is heated from 0° C. to 10° C. Find approximately the increase in volume. Take the coefficient of linear expansion for copper as o-ooooiy. CHAPTER III Integration "A snapper-up of unconsidered trifles." — The Winter's Tale. Areas and Integrals. Let RS be a portion of the graph oi y = fix) between x = r and x = s, fig. i (i). Let the area RrjS be required, where rR and jS are parallel to OY and rs is the intercept on the OX axis. This area can be found in many ways. For in- stance, it can be found by a planimeter, which is an instrument for measuring the areas of plane figures. Suppose the area O/PH, where P is any point on the graph and />P the ordinate of P, is denoted by A. Let the ordinate /R', fig. i (ii), represent the area A„ i.e. OrRH, corresponding \.o x = r; and j'S', the area Ag, i.e. OjSH, correspondjng to x = s, and so on. Then ^'P', fig. i (ii) represents the area O/PH, A or Aa;, corresponding to Op = x. A is a function of x, for to each value of x there corresponds a definite value of A. .•, A = area O/PH = ^p-{x), say. Now, if we fix attention on P, the increase in A as p advances a little to the right must be smaller and smaller, as the advance in p is made smaller and smaller; in fact, there must be a gradual growth in A as p advances to the right. We should therefore expect the "area" graph (A, x) to be a smooth curve 72 APPLIED CALCULUS between R' and S', i.e. to represent a continuous function of x. Suppose V increases steadily from rK to jS, fig. i (i), then the area pqQP, which is the increase in area A when x increases from Op to Oq, lies between the S' Fig. I areas of the rectangles pqNF and pqQr', i.e. the areas are in the following order of magnitude ; /^rNP < /^QP < pqQy, i.e. pf.pq < pqQF < qQ.pq, i.e. y^.x < A A < {y + Aj»')A;c, ••e. y < -^ < y + Ay. Since the curve RS is continuous, as Ax A 1 , . AA Ay also o, tween y and y + Ay, o, and since -r— always lies in value be- !->. Ax dA y, i.e. ^ =^- •(I) INTEGRATION 73 The derivative of the "area" function, fig. i (ii), is therefore the ordinate of the (x, y) curve, fig. I (i), at any value of x, between x = r and X = s. The "area" function, '^(x), is therefore a function whose derivaiive, at any point, is f{x), the given function. t/»-(x), therefore, satisfies the equation ^V(^) = /(^). where /(x) is the given function. In the differential calculus we are given '^(x) and asked to findf(x). Here we are given f(x) and asked to find '^(x). Hence we have to work the processes of the differential calculus backwards in order to find areas, for which purpose a special notation is con- venient. If i/f (x) is given, we say that ^^{x) = Ax\ (2) where -j- denotes the operation of finding f(x) from T/r(^) by the rules of the differential calculus. In the present case we write V'W = \Ax)dx, (3) which is exactly the same statement, in another nota- tion, as equation (2). Equation (3) is read i/r(x) = \!a& integral ol f(sc)\ equation (2), f(pc) = the derivative of •^{x). 74 APPLIED CALCULUS The reader will notice that, if then •»/r(x) = l/(x)dx; or putting z for ^{x), if ^ = /(^) = X, say, (4) then = j'Kdx (5) Equations (4) and (5) are not deduciwns, one from the other, but merely different notations for express- ing the same fact, namely, the derivative of z is X, for any value of x at which (4) and (5) hold. If we are dealing with a problem of the differential calculus, notation (4) is convenient; if the problem is one of the integral calculus, notation (5) is convenient. The reason for using the sign I has been mentioned on p. 16. EXAMPLE If Hx) = = ««, where » is a positive integer, then !Cause /(x) = nx^-\ 4-xn = wjc'i-i dx by Rule i, p. 44, and, in the new notation, ar» = jnxn-^dx. Here x^ is the "area" function corresponding to the given function, wjc»— 1. ■\lr{x) is not a fixed quantity, but a function of x. The area under RS is, however, a fixed quantity, ID sq. in., say. INTEGRATION 75 If we know ■ ^^ ^^" P'o*- -^ ~ '^(^)> fig. I (ii), therefore this curve becomes known when \[r{x) is known. At a; = r, \ff(x) = ■\]r(r). At X = s, ■ylr(x) = ^{s). But the area KrsS required is the area A up to X = s^, less the area up to :»£; = r; no matter what value of X (less than s) we start measuring the area from. The area required is (Aj — Ar), i.e. When the integral is to be taken between definite boundary values of x in this way, we write 'f{x)dx = ir{s)-ir{r), (6) /; where r is called the lower limit, and j the upper limit, and ylr{x) = lf{x)dx. Siwh an integral is no longer a function of x. It is a definite quantity, ■^{s) — ^r(r), and it is called a definite integral. When the limits are not specified, the integral is a function of x, and is not a definite quantity. It is called an indefinite integral. The process of finding ■y\r(x)^ given f{po), i.e. of finding jf(x)dx, is called integration. Evaluation of Definite Integrals. The value of the definite integral f{po)dx I J a can be found in at least three ways. I. By equation (6), the integral represents the area of the figure AabB (fig. 3), and this area can be 76 APPLIED CALCULUS TC) FIND ,[ 3x^ dx Scales: Horizontal 2" s unity. Vertical V= unity. .*. I sq. in. of diagram area = unit product, A C B is graph of y=3x2^ .". diagraiTL area of Aa bB =f^^ 3x2 jx Diagram area AabB equals, nearly A. BCD =Ji X Ix 2-60 1-30 A. AFC ='^x IX l-87=0- 93 D. CE = I X 1-87 =1-87 D Ab =2 X 1-50 = 3-00 7- lO ^3x*dac = 7- lO nearly INTEGRATION 77 readily found when the graph of jj/ = f{x) is drawn. This method has already been touched upon in the Introduction. As an example, find graphically (•2 I 2>x^dx. The graph of jv = 3^^ must first be plotted. Take 2 in. horizontally equal to i, and J in. verti- cally equal to i. .". i sq. in. of area in the diagram represents one unit of real area. The figure is drawn in fig. 2. Full details of the calculation are given on the figure. It will be seen that the result is 7-10. 2. The name "integration" suggests adding up, just as differentiation suggests taking differences. We have already seen in the Introduction that there is an addition buried in the preceding work. Divide ab (fig. 3) into any number n of equal parts, and erect ordinates at the points of division. Let Op = Xp, Oq = Xg^, &c. 78 APPLIED CALCULUS Find the area of each elementary rectangle, such as ^PlQ' The area of this rectangle is pV y. pq = f{Xp) X A^. The area of Qqttf = f(Xq) X ii,x, and so on for each area. Hence the sum of all the rectangles between x = Oa and x = Ob is 2^{/(x)Aa;} = area required w^flr/)/ (7) Consider the word " nearly " in (7) a little further. The neglected areas are 2(P^'Q). Translate each of these areas parallel to OX into the final area Vw6B. The sum of them all is manifestly less than the area V»6B. But this area itself is less than f{b) x vb. :. the sum of the neglected areas is less than f{b) xvb. Now, /(b) is finite, hence /{b) X vb —> o as vb —*^ o, i.e. f(b) X Ax — >• o as Ax —*■ o, since all the subdivisions pg, qt, . . . vb are equal by construction. Hence 1.gf{x)Ax —* the area Aa^B as A^ — > o, i.e. Lt 2*/(:'c)Ax = area AadB A* — ^ /: Ax)dx (8) To determine the integral, i.e. the area in question, to any required degree of accuracy, it is necessary to divide ab into parts sufficiently small to ensure that the ratio of [/(*) X Ax\ to 2^{/(^)A^} is less INTEGRATION 79 than the given fraction which defines the degree of accuracy. This example brings out the fact that integration is really addition, more or less disguised, and immedi- ately suggests our finding the value of f{x)dx J J a by direct addition, as this integral is Lt •2lfix)^x (9) Aa; — ► The finding of this limit is simply a matter of arith- metic, using progressively (and sufficiently) dimin- ishing values for Ax, depending on the accuracy required in the result. We can write down a list of all the products /{x)Ax, taking Ax = ^ {b — a), say. At x = a, f{x) = f{a) ; the first product is therefore f{a) (^ - ") = P /^-^) . 100 100 The length of the next strip is the value oif{x) at , . r« + (±=i^)-\ i.e. fL + (^^)l = p.. L 100 J I 100 j ^ The area of the second strip is therefore P (b - ^) . 100 2 ■ There will be, in all, 100 strips. The sum of them all is [P,+ P,+ P3+... +Pl00](^) 100 /^ = z p (^. (10) 1 «\ 100 / ^ ' 8o APPLIED CALCULUS This value is the sum ^J{x)^x, when ^x = (-^)- As we make A^ a smaller and smaller fraction of {b — a), the value of ^\f{x)^x will approach nearer and nearer to the value we want, viz. f{x)dx, I J a which is by definition the limit to which E*/(x)Ax tends as Aair — > o. The arithmetic is evidently lengthy if high accuracy is aimed at. We will take a worked example with fairly large intervals, and compare our result with accurate results which we shall obtain later. WORKED EXAMPLE Find an approximate value of I j^x^x. We have I yi?dx » 2 3Jc'Ax Take A b — a 1/ V I Ax = = — (2 — i) = — . Then/(i) = 3, hence 3 x t\y = 0-300 is the first term of the sum. The calculation can be best pursued in a tabular form. Notice that the factor 3 goes right through the arithmetic as a factor, hence Sj -^x^Ax = 32 x^Ax. INTEGRATION 8i We shall therefore take Sj x^^x, and multiply the answer by 3- ^;!at Interval. Beginning of Interval. P». P«Aa:. I -I>I I-OO I-OO O-IOO I-I-I-2 1-21 1-21 0-121 1-2-1.3 1-44 1-44 0-144 1-3-1.4 I -69 1 -69 0-169 1-4-1.5 1-96 1-96 0- 196 1-5-1-6 2.25 2-25 0-225 1-6-1-7 2-56 2.56 0-256 1-7-1-8 3-89 2-89 0-289 1-8-1-9 3-24 3-24 0-324 1-9-2-0 3.61 3-61 0-361 2P„A;c = 2-185. /, l![x^Ax = 2-185. /. 2^3a;2A^ = 6-555. If, in column 2, we take x^ at the end of each interval, and go through the same calculations, we get 7-455. The mean of these is 7-005, and this, we shall see, is very close to the true value. 3. Yet a third method is available. We have shown that, if V^(^) = jfiP<:)dx, then/(x) = ^ir{x). ••• V'W = ja^^i^)^^- Putting jv = ^(x), we get ^ = /(iK (") (D102) 7 82 APPLIED CALCULUS IfjV = x^, n being a positive integer, ^ = w;c"-i by Rule i. :. X" = jnx'^-^dx (12) We have thus found the value of the integral jnx^'^dx in terms of x. Suppose we wish to find nx'^''^dx. We have fi x^ = Inx^'^dx = our ^{x) of pp. 71 to 75. , .*. the value of the integral between x = 1 and a; = 2 is (2" — I"), by equation (6), I.e. 2" — I" = / nx^-^dx. Generally, nod^-'^dx (13) a It will be seen that this method consists in work- ing backwards from known rules of the differential calculus. It is in this sense that the integral calculus is the inverse of the differential calculus. Rule II. jnx'^-^dx = x^, i.e. the integral of n timte Ar"-^ is jir", when n is a positive integer (excluding n = o). The finding of the integral as a function of x is a great advance on the other two methods. Either of these two methods takes time and is approximate ; INTEGRATION 83 but the third method is accurate and quickly applied, [2, e.g. we can quickly check the result for I 2>x^dx. I :ix^dx = \ x^ \ = 2' — i' = 7, The value obtained by arithmetic was 7-005, and by the graphical method y-i. Exercise 3 I. Find by the arithmetical method, taking- ten steps. /: ■5 1 Find by the graphical method. 3. Find by Rule II, ^x*dx. /:■ Compare the three results. 4. Find by the arithmetical method, taking ten steps, /: 5 e^dx, taking your values of the function e* from a book of tables. Is it approximately equal to (e^ — e^)? What does this result suggest? [Take e = 2-718 in this example.] 5. Find by the graphical method, /: sinxdx. Is it approximately equal to (cos 0° — cos it)? What does this suggest? {x is in radians, see p. 102.) 6. Find by the graphical method, x^{i — xfdx. / 84 APPLIED CALCULUS 7. It was found, p. 56, that ds where v is the velocity at i, and s is the distance travelled by a body moving in a straight path from rest in an interval of time /. Express the distance travelled as an integral of velocity and time. Assuming the velocity is proportional to the square of the time /, and the constant of proportionality to be 3/11 (z> = S/"'^), if distances are measured in feet and times in seconds, show that the distance travelled from rest in T sec. is juT^ ft. Express this result in words. CHAPTER IV Limits "Long calculations or complex diagrams affright the timorous and unexperienced from a second view; but if we have skill sufficient to analyse them into simple principles, it will be discovered that our fear was groundless. " — Samuel Johnson. We have already used the word "limit" to denote a quantity to which another quantity can be made to approach as nearly as we please. We will now examine the nature of a "limit" more closely, as this idea is the very pivot of our subject, and of all higher mathematical analysis. No real progress can be made till the idea is thoroughly grasped. As a simple case, let us seek the limit of %'■ at X = ID (fig. i). At the outset, a warning is necessary. The limit has nothing to do with the value of the function at X = ID. It depends on the values of the function round about x = lo, but not at x = lo. We de- liberately exclude the value ^ = lo and the corre- sponding value of the function in finding the limit required. The reader will soon see why this is so. We have already noticed that, if there is a limit, it is approached as x approaches the given value. Suppose X = 9'9, then x^ = 98-01 X = 9-99, ,, x^ = 99-8001 X = 9-999, ,, x^ = 99-980001 86 86 APPLIED CALCULUS From these figures, it appears that x^ ICX3 as X lo from values oi x less than lo. Now, try some values of x greater than lo. Suppose X = io*i, then ^^ = i02'0i X = lO'OI, ,, X^ = I00'200I X = lO'OOI, x" I0O'020O0I. These values of x^ again approach loo as x -* lo. The figures suggest that loo is the limit we are seeking. Let us test this suggestion. Tabulating these figures. * x^ 1 Jlf^-IOO 1 |^-,o| 9-9 9801 1-99 O-I 9-99 99-8001 0-1999 O-OI 9-999 99-980001 0019999 O-OOI lO-OOI 100-020001 0-020001 O-OOI lOOI 100-2001 0-200I O-OI lO-I 102-01 2-01 O-I we see that: (i) So long as x lies between 9-9 and lo-i (ex- cluding X = 10), the difference between x^ and 100 does not exceed 2-01 numerically, i.e. provided x is not equal to 10, if 9-9 < X < lo-i, then o < \ x^ — 100 | < 2-01. (2) If 9-99 < X < lO-OI, ,, O < I AT^ - 100 I < 0-200I. (3) If 9-999 < ^ < lO-OOI, „ O < \ X^ - ICX> \ < 0-02000I. LIMITS 87 Each of these intervals of x contains the prohibited value X = ID, and as the interval shrinks, the bounds for I x^ — 100 I shrink too — the third and fourth columns in the table show this. It therefore seems likely that, no matter how small a quantity we select for the upper bound of | :!C^ — 100 | , we can find an interval containing the value 10 such that if x lies in this interval, and does not have the value 10, the 1 Y i / \ ">« \ " I \ '^ y ' \ a I V ^ -tXO-^- o value oi \ x^ — 100 | lies between this selected upper bound and zero. We will show that this is so. By looking at the graph of x^ (fig. i) it is obvious that x"^ increases as X increases and x"^ decreases as x decreases, provided X is always positive. Select a number — j^, which would probably denote a small quantity if we were using it with a usual unit, 88 APPLIED CALCULUS and see whether we can choose an interval so that if X lies in this interval, then O < I ^ - IOC I < -^2- Case I. X > lo. Suppose first that x is greater than lo. So long as lO < X < ^IOO+^ \o^' then lOo < ^^ < loo H 75 I.e. o < x^ — 100 < |12' I 12» ID' i.e. ix'' — ico) lies between o and — ^. \ / jq12 If then X lies in the interval between x = \o and ^ = 4-^1 100 + ^ (and does not have either of these "end values"), then the inequality ^.2 o < X'- — 100 < — ID ,12 holds good. Now, suppose, instead of — j-^, we use — jjooj, i.e. 00000 I, with 11,999 noughts between the decimal point and the i at the end. We can still find an interval, viz. from x = 10 to X = +yioo + :^^, such that, if x lies in it, the inequality o < x"^ -\QO < -^ holds good. LIMITS 89 We will now try to put any number, excluding zero, for the fraction. For definiteness, we will first sup- pose the number to be < — ^v ^^ ^ be such a number, we easily see that so long as x lies in the interval between X = 10 and X = + Vioo + e, then o < I ^^ — 100 I < e holds. The difference between x^ and 100 can therefore be made less than any assignable number e (less than — j^ j, excluding zero, by choosing a suitable interval within which x must lie. Case II. x < 10. We have so far supposed x > 10. Now, suppose a; < 10; :x;^ is then less than 100, and the positive difference between 0(? and 100 is 100 — x^. So long as the inequality 10 > ::c > + V'oo \-. holds, then 100 > ^^ > 100 j2, by squaring each term, i.e. > x" — 100 > iH 1 0^2 i.e. > — (100 — X^) > — I Now (100 ■ -x^) is positive, therefore ■ and — I — js are nei B:a( ;ive numbers. (100 — x^) 10" If a given inequality runs in ascending (or descend- ing) order of magnitude, it runs in the reverse order QO APPLIED CALCULUS — descending (or ascending) when the signs of all the terms are changed. For instance 3 > 2 > I, but -3< -2< -I. If then o > — (loo — x^) > — y-j^j then o < ICO — x^ < -r-^- I lO*' Hence so long as x lies in the interval [lo, + yioo- ^aj, (loo — x"^) or \ x^ — ICO I lies in the interval [°' ^ This argument is quite general, and if we put e for any positive number (less than — jjj, we get o < (lOO — x^) < e so long as x lies between + v'loo — e and lo, where e can be any positive number which is less than — jg. We have now proved that 1. \i X > lo, o < \ x^ — ICG I < e, provided x lies between lo and + s/ioo + e, where e can be any finite positive number < — jg. 2. If X < ID, then o < | ^ — lOO | < e, provided x lies between lo and + Vioo — e, where e can be any finite positive number < — j^. LIMITS 91 There is no need to examine specially values of e > — 12, for if e > — j2, we only have to keep x between + y 100 - -Vj2 and + yjioo + ^. excluding x = 10, to ensure that < \ X^ — 100 I < e, for this limitation of x is sufficient to ensure that the inequality o < \ x^ — 100 I < — j2 holds, (i) and if e > — j^ the inequality o < \ x^ — 100 | < e must necessarily hold, for the same interval of x. We can therefore always find an interval including X = ID, Stick that if X lies within this interval, the inequality o < I ^^ — 100 I < e holds, (2) except at the value x = 10, no matter what finite positive value we give to e, except zero. When condition (2) holds, we say, as on p. 39, that I x^ — 100 I tends to zero as x tends to 10, i.e. x^ tends to the limit 100. It is clear that if we choose progressively decreas- ing values of e, the permissible range of x becomes more and more restricted, for the range D is given by D = -s/ioo + e — s/ioo — e, i.e. D^ = 200 — 2\/ioo^ — e^, and if e is given progressively decreasing values, 1 I I J^ ' jq12000 > jqIZOOOOOO 92 APPLIED CALCULUS and so on, (loo^ — e^) takes progressively increasing values, and therefore [200 — 2x/ioo^ — e^] progres- sively decreasing values, hence D progressively decreases. The values of x therefore become restricted to those "in the neighbourhood" of 10, 10 being always in- cluded in the interval. We therefore say that I ^'- 100 I (3) tends to zero in the neighbourhood of 10, i.e. x'- tends to the limit 100 in the neighbourhood of x = 10. As the admissible range of x shrinks more and more, the only value of x we can be sure of in- cluding in the range is 10, just as the only point which certainly remains inside a circle, if we shrink the radius more and more, is the centre of the circle. Now this progressive shrinking of the interval is not only admissible, but is the very essence of the process. The value x = 10 is the nucleus, so to speak, of the shrinking interval, and as it is the only point we are always sure of securing in the interval, it is reasonable to associate the limit we have found with this value of x. We therefore say that the limit of x^ at X = 10 is 100. If we are dealing with quantities we can put the matter thus: — The quantity x^ can be made to differ from 100 by as small a quantity as we please, by bringing x sufficiently near to 10, no matter whether it is more or less than 10. In such circumstances we say that the quantity x'' has the limit 100 at x = 10. The reader will now see that the limit of x^ is a number we associate with a given value of x, here X = 10, and this number is obtained by considering LIMITS 93 the valties of x^ corresponding to values of x round about lo as these values of x progressively approach lo ; it is not dependent in any way on the value of x^ at X = lo, a value which has been deliberately excluded from consideration all through the calcula- tions. Definition of a Limit. f{x) has the limit I at x = a, if, corresponding to any finite positive quantity e, however small we like to make it, we can find an interval of x, includ- ing the value x = a, such that the inequality < I fix) - / I < e holds, so long as the inequality < \ X — a \ < ri holds, where ;; is a finite positive quantity to be found, which depends on the value chosen for e. As e decreases t] decreases. In other words, the limit / exists if 1 f(x) — I I tends to zero as X tends to a. Or we may put it still more briefly, in symbols. If, as X -^ a, f{x) -> /, then / is the limit of f{x) at x = a. This shorter statement is, however, only apparently shorter, as it implies the meaning of — > which is expressed in words in the longer statement. It is, however, a very useful pictorial notation, as the sign — ► suggests vividly the process of finding the limit. It should be noticed that the statement in the definition excludes the value x = a; for the conditional inequality is o < | ^ — a | < ?;, i.e. I a; — a I must not equal zero, i.e. x must not equal a. Further, it implies nothing whatever about the value o(/{x) aX x = a. 94 APPLIED CALCULUS This example has been worked out at length in order to bring out the full details of the idea of a '-'limit". Under ordinary circum- stances, we should not go through the whole of the working, as it is obvious that «" — > lOO The point is that the full working out expresses quite definitely, by an arithmetical method based on inequalities, what appears to be obvious from the graph, and, as the idea of the limit is so important, it is worth while to consider one case at least in detail. Further, limits are not always obvious, and sometimes a graph cannot be drawn. Skill in the use of inequalities is very useful in all branches of mathematics and physics. By using inequalities in a common-sense way we can often set definite limits to approximate calculations, and applied mathematics, physics, and engineering all bristle with these. For example, a glance at fig. 2, p. 76, shows that the area of the figure is greater than f(a) X (4 - a), and less than f(b) X (i - a). Hence /(a) X (4 - a) < f f{x)dx < /(b) X (J - a), J a if f(x) is positive and continually increases as x increases from a to b; and, whether we can integrate /' J a f(x)dx or not, we undoubtedly can work out f(a) X {b — a) and f{b) X (i - a), and thus get bounds to the integral within which it must lie. These bounds are often quite close enough for practical purposes, so that we need not integrate the expression. Values of Functions — Continuous Functions. We have found that the value of the function x^ at ^ = 10 has no bearing on the limit of x^ aX. x = lo. The limit at ;c = lo is loo and the value of x^ at X = 10 is 100. The limit equals the value of the func- tion in this case. As a matter of fact, nearly all the ordinary functions we come across in applied mathe- matics possess the property that the value of the function, for a given value of the variable x, is iden- tical with the limit of the function at that value of x. LIMITS 95 Such functions are called continuous functions. Suppose f{pc) has the limit I sX. x = a, then if f(pc) is continuous at X = a, f{a) = I. It is evident, then, that when f(x) is continuous at ^ "■ ^ /W -> f{a) as jc -> «. This is manifestly the state of things at any point on a smooth graph. \i y = f(x) is continuous at every point between a and b, then \i y and x are the co-ordinates of any point on the graph oif{x) between a and 6, then Ay — > o as Ax — > o. This follows from the fact that f(x + Ax) — > f(pc) as X + Ax — > X, i.e. y ■\- Ly —=*■ y as X -\- Ax — > x, i.e. A^ — > o as Ax — ► o. This property of continuous functions is most im- portant. Undetermined Functions. In the calculus, certain important functions crop up which have no value for certain values of x, and it is just at these values of x that we specially want to know something about the behaviour of the func- tion. Fortunately, it is not necessary for us to know the value of the function at these special values of x; what we want to know is something about the values of the function for values of x near these special values of x. The limit gives this informa- 96 APPLIED CALCULUS tion concisely. As we have already seen, we do not need to know the value of the function at a selected value of X to find the limit of the function at that value of X. WORKED EXAMPLE Consider the function " — ^^ aX x = i. X — I This function lias no value at j; = i, for if we put x = i, we get I — I o /(I) = I — I o This result is quite useless, as we can attach no meaning to it. But we can easily show that has a limit ai x = i, for X — I lies round about the value 2 when x is near unity. When x^ — I I'Oi^ — I I -0201 — I X = I'OI, = = = 2-OI. X — I -oi •01 X'' — 1 0-q80I — I — O'OIQQ X = .99, = — H = 23 = 1.99. x—1 0-99 — I — o-oi These figures point to 2 as the limit, and the reader will have no difficulty in showing, just as we have done for x^, that we can always find a positive value for t; such that \x^-i I O <^ — 2 < e, \ X—1 I so long as O < I AT — I I < 1), no matter what arbitrary finite positive value we give to e. We can, however, see that the limit is 2 at a; = i, without going into the detail of the inequalities. i;2 — — = X -\- I, when ;c — i is not zero, and Lt (^^-^^) = Lt (^ + 1) = We cannot say that = X + 1 for all values of x, LIMITS 97 And that when x = i, x •\- \ =2. .', '- = 2 when X = I, X — \ because the first equation is true only when ji; — i is not zero, i.e. when X is not unity. It is therefore inadmissible to put ;i; = i in the next line of the argfument. When :ir = I , division of*^— i by«— i is division by zero, which is not allowed in the laws of algebra (p. 108, Ex. 9), and, worse, it is division of zero by zero, which is an operation without any meaning;. Further, even if the reasoning were valid, it would give the value of '^— ^^^ — at * = i, and this value has nothing to do with ;c— I the limit at ;c = i which we require. EXAMPLES 1. Show that if y — > b b.s x — > a, the limit at j; = a of {a-\- y) is a + b. 2. Find Lt x?. Important Points about Limits. 1. The limit is not necessarily the value of the function at the value of x under discussion — x = a, say. The function, , e.g. has no value at X = a. 2. The limit is a number from which the values of the function, in the immediate neighbourhood oi x = a, differ by as little as we please. 3. By calculating the value of the function for a value of X sufficiently near to x = a, a value of the function can be found differing as little as we please from the limit. 4. The notation of limits is Lt fix) = /, X = a (D102) 8 98 APPLIED CALCULUS i.e. the limit of f{x) at x = a \s l. This notation can be put f{x) -> / as ^ — > a, which reads f{x) approaches the value / as closely as desired when x approaches the value a sufficiently closely. We will now discuss certain theorems which will be very useful in later work. Theorem i. Limit of a Sum or Difference. The limit of the sum of a finite number of functions of X is equxil to the sum, of their lim,its. /i(^) -> /i as 5C — > a, fi(x) — * 4 as ^ — >• a. When X approaches a, f-^{x) approaches 4 and f^lp^ approaches l^. .'. fi{x) + fi{x) must approach ^ + 4, i.e. fi{x) + /2(^) -> li + l^SiS X —*■ a. We can put this argument thus : Let PQ (fig. 2) be the graph oi fi{x). Y \ Xw^ tr. ...^r*; Q ■ ■ a - M 7 1 > .V O t i / ^ ;: Fig. 2 LIMITS 99 Then, at any point N given by x = ON, NP = NM + MP, i-e. /i(^) = 4 + 0-1, where a-^ = MP. In the figure o-i and l^ are positive; in general, they may be either positive or negative quantities. \if^{x) has the limit l^as x —^ a, fi{x) —> l^ as X —> a, i.e. as fig. 2 shows, cTi — > o as X — > a. Similarly, /2(x) = 4 + 0-2, where 0-2 — > o as x — > a. Adding /i(^) + /aW = A + 4 + 0-1 + <^2- Now, as o-j — >• o and 0-2 —> o when :«; — > a, .'. o"! + 0-2 — »• o when x —>■ a. •'• fi{x) +Mx) -> 4 + 4 as rn; -> a, i.e. Lt [Mx)+Mx)] = 4 + 4. x = a The theorem is thus true for two terms, and can similarly be extended to any number of terms. Theorem 2. Limit of a Product. The limit of the product of a finite number of func- tions of X is equal to the product of their limits, if these limits are all finite. Let f-^x) —> li RS X —^ a, fz{x) -* l^ as X —^ a. loo APPLIED CALCULUS As before, we may put fi{x) = /j + cTi, where (Ti —* o as x —* a, and/2(x) = 4 + o-a, where o-g — > o as x — > a. •'• /ii^)fi{^) "*■ A4 as X — > a, since o-i — >• o, o-^ — > o, and 0-10-2 — > o, under these circumstances, i.e. Lt {A{x)Mx)} = 1,1^ X — > a The theorem can be extended to any number of factors. Theorem 3. Limit of a Reciprocal. The limit of the reciprocal of a function of x is equal to the reciprocal of the limit of the function of x if this limit is not zero, i.e. a; _^ a\f{x)/ I if Lt f{x) = I, I not being zero. As before, put f{x) = I -\- a; where o- — > o as x — ► a. Then f{x) l+cy' and -j— — approaches 7 as o- — > o. t + O- i •• Lt ( ' ) = I X -^ a\f{x)J I if Lt f{x) = I. LIMITS loi Theorem 4. Limit of a Quotient. The hmit of the quotient of two functions of x is equal to the quotient of their limits, if these limits are finite, and if the limit of the denominator is not zero. This theorem follows from Theorems 2 and 3 above, '•^- Lt fM = h X — > a/i(p^) ti if Lt f{x) = 4, X — ► a and Lt f2{x) = l^, 4 not being zero. WORKED EXAMPLES I. Find Lt {«„ + a^x + . . . + anx!>^} X — > n being a finite positive integer, and the coefficients of x being all finite numbers. Lt (flo) = «o a: — > Lt (a^x) = o X — > Lt (a^x^) = o X — > Adding, by Theorem i, Lt {(j|, + ai.r+ . . . + onx"} = «„. a; — > 2. Find the limit y 2X^ + 2x^ + x* 3.r2 + x'^ + xO' Divide numerator and denominator by x^. 2x^ + 3x^ + X* _ 2 + 3.r + x^ yc^ + x* + x^ i + x^ + x*' Lt (2 + s*' + jr^) = 2 by Example i above. X — >. Lt (3 + x^ + x*) = 3 .. « — > /. by Theorem 4, Lt (given function) = f . X — > I02 APPLIED CALCULUS Circular Measure of Angles. In theoretical investigations about angles and func- tions of angles, the angles are measured in circular measure. Let L AOP be any angle (fig. 3). About O as centre, describe a circle APQ of any radius. Fig- 3 By Euclid VI, xxxiii, the angle AOP is propor- tional to the arc AP on which it stands. Hence any other angle AOQ must bear the same ratio to L AOP that the arc AQ bears to the arc AP ; hence, if we make the arc AP = OA = r, and call the angle AOP the unit angle, then Z.AOQ, measured in these units, _ AQ -, if J r r AQ. Putting Q for the angle AOQ, measured in these units, „ e = ;. (4) i.e. J = rQ : (5) This unit angle is called the radian, so is the number which measures the Z.AOQ in radians. LIMITS To Convert from Radians to Degrees. Consider the right angle AOQ (fig. 4). Q 103 Fig. 4 The arc AQ = i x i-wr = ■rr :. d = "^jr by (4) TT 2' i.e. the circular measure of a right angle is — radians. From this formula we can get any other angle by proportion thus: Suppose we want the circular measure of 10°. Let 6 = the required measure, then as 10° : go" : : 6 : -, 2 i.e. 90 X = - X ID, 2 = IP X ^ = ^. go 2 18 In like manner equation (5) can be expressed in degrees. I04 APPLIED CALCULUS The circular measure of i/r° is i8o ^ '"• • Vl V ^ - -^ '•^- ' = iw^"^ ^^) The formula is in its simplest form when the angle is expressed in circular measure, being then simply J = re. The circular measure is the best measure in which to express angles in theoretical work, since mathe- matical formulae involving angles take their simplest form when the angles are so expressed. In practical measurements the "degree" measure is used. Tables for converting angles from one measure to the other are given in books of tables. EXAMPLES 1. Convert 30°, 45°, and 90° into circular measure. 2. Convert 1, J, i -54 radians into degrees and minutes. Trigonometrical Limits. The following relations are important: (a) taxi X > jc > sin jf , (7) where x is the radian measure of a positive angle less than a right angle, i.e. o < X <-. 2 Let PQ be the arc of a circle of radius r (fig. 5), subtending an angle at O of 2X. Let FT, QT be tangents at P and Q, meeting in T. Join PA and QA ; then the quadrilateral OPAQ < sector OPAQ < quadrilateral OPTQ. .•. f* sin X < t^x < 9^ tan x. .', sin x <. X < tan x. LIMITS "05 The reader will notice that this inequality is merely a mathe- matical statement of the axiomatic inequality : — Figr. s The quadrilateral OPAQ < sector OPAQ < quadrilateral OPTQ. (See fig. "s.) (h) x>^va.x> X-—, (8) 4 if X is positive. Suppose X to be less than ir. We have • sin X = 2 sin \x cos \x, i.e. sin ;r = 2 tan \x cos^ \x = 2 tan \x (i — sin^ Ja:) (g) But tan \x > \x, and sin \x < \x. :. sin .r > 2 X \x {i — {,\x)\ i.e. sin x > x — J.r^, and sin x < x \iy inequality (a). .'. a: > sin ;i; > *■ — J.r'. We shall only use this inequality and the following one, when X is less than ir. (c) I > cos X > I — \x^, To prove this, we notice that cos X <. I. .(lo) io6 APPLIED CALCULUS This follows from the definition of cosine x. And cos X = 1 — 2 sin' ^x > I — 2 (i^)^ as sin Jar < ^x x^ > I--. 2 To prove that Lt (?BJ^ = i (n) By inequality {b) above X > sin X > X — \x^. sin X , , /. I > > I - ix\ X ,. Lt C^ = I, since the limit of i is i, and the limit of i — \x^, as X —*■ o is I also. To prove that Lt ( ) = i. tan X sin x X X cos X Lt ft^^) = Lt ('-'^) X Lt (-^). a; — > oV Now Lt COS :v = i by (lo) above, .-. Lt (-^) = I. X _> oVcos x/ r /tan x\ '/ V ,'. Lt ( ) =1X1 = 1 (12) x-^0\ X J How the limit is approached. When Lt /(.r) = /, X — ► a whether x — > a from values of j; < it or from values of x >■ a, if / is a normal limit. LIMITS 107 Sometimes Lt f(x) = h when X approaches a from values of x <. a, and Lt f(x) = h X —^ a when X approaches a from values of x > a. The former is called a progressive limit, the latter a regressive limit. Normally, 4 = 4 "= ^■ The cases are illustrated in Rg. 6. , In all the simple cases dealt with in this book the limit is approached normally. It has not been considered necessary to distinguish between v<^ ^ ->. -.-?!<__ ^.^tf^Z - v** 1 k ■n "^^ -e a >, ' ■ X Fig. 6 the manners in which the limit is approached. Inspection of the proofs employed shows that they are equally valid whatever be the manner of approach. 1. Find 2. Find 3. Find Exercise U Lt xK ■■ — ► a Lt (sin x). Lt .(: x^ + 2X -4 \ ,2X^ + 6x - 8/ ■ 4. What are the circular measures of 43° 12', 52° 5', 107° 21'? 5. What are the measures, in degrees and minutes, of the following angles, given in circular measure : h 3. 5i? io8 APPLIED CALCULUS 6. Show that ^ /arc tan a;\ _ 7. Show that ^^ r sinjr+^- sin^ l ^ ^^^^ [Combine (sin y + x — sin y) into a sine-cosine pro- duct of semi-sum and semi-difference of angles.] 8. Show that ^^ [" tanj^r^-tan^ l ^ ^^^j^^ [tan ji/ + :*r — tan y\ _ X J 9. Point out the flaw in the following argument: Let X = a. .'. x^ = ax. .'. x^ — a^ = ax — a^. (x + a) (x — a) = a(x — a). .'. X + a = a. .'. a + a = a. .'. i = I. CHAPTER V Practical Applications " It's no what we hae, but what we do wi' what we hae, that counts." — (Old Scots Proverb). Beginners often have some difficulty in seeing why the calculus is so useful in the practical problems of physics and engineering. Nature is always chang- ing, and things and their properties vary. Some of these properties are measurable — for instance, the relative positions of the heavenly bodies, the speed of trains, the temperature of bodies, the pressure of the atmosphere, the electric current in a wire, and so on. The quantities which measure these pro- perties change, as the properties themselves change in the course of Nature, and, as we saw in Chapter I, the Calculus is the branch of mathematics which deals especially with changing quantities. The principle underlying its application may be seen in the follow- ing example. In many cases first principles or experience show that some quantity {_y) probably depends on another quantity (x). For instance, the distance a train moves depends on how long it has been moving. If j is the distance in feet travelled in an interval of time i sec, J is a function of i, i.e. ^ =m • (0 Everyone knows that the speed of the train comes into the problem. If the speed is fast, the train will no APPLIED CALCULUS move farther in a given time than if the speed is slow. If the speed is constant, the definition of speed leads at once to j = vt, where v is the speed in feet per second, and so/(^) = vt. If this were all that could happen the problem would be solved, but the speed may vary from point to point in t/ie path. It is just this kind of difficulty that we have already encoun- tered, and that the calculus enables us to overcome. We have, on p. 59, arrived at the result ds dt = ^' where v is the velocity at the instant t. We can express this fact by means of an integral, see p. 75, i.e. = I vdt, J to .(2) where S is the total distance travelled, in feet, in the interval (t^ — t^) sec. Can this integral be evaluated? t (seconds) Fig. I If w is a known function of t, we may be able to integrate I vdt by the rules of the calculus, and get J tfj the result we wish. Or, if we have a graph of v in terms of t (fig. i), then, since | vdt is the measure of PRACTICAL APPLICATIONS iii the shaded area, the answer can be obtained graphi- cally. This is a more powerful method than the first, but is longer. If A is the area of the figure in square inches and the scales of the figure are I in. vertically = V ft. per second, I in. horizontally = T sec, then I sq. in. = VT ft. Hence vdi = [VT] x A = /A, where /is the scale factor [VT]. EXAMPLES 1. If z) = a^, a being a constant S = I atdt J h /■ o, r and Cj do not differ sensibly, as the outer circle shrinks on to the inner one, i.e. c^ — > c as Ax — > o: I.e. -=- = c = 27r:ir. ax Fig. 2 (When we proceed to the limit (Ax — > o), we drop into the d/dx notation.) /. A = 27ri xdx; 27r is constant, and therefore is placed •' " outside the sign of integration. ,', A = 27r -r- J • ''y ^"'^ ^> P- ^^> = 27rR' 2 = :rR2 sq. ft. 3. Motion under the Influence of Gravity. — Required the distance, in feet, that a stone falls from rest in T sec. under the influence of gravity — neglecting air friction. The acceleration of gravity is 32 • 2 ft. per second per second (= g), a constant. Let V be the velocity of the stone at the time t sec, in feet per second, then -j- is the acceleration of the stone at the time at PRACTICAL APPLICATIONS 113 t in feet per second per second, since dvjdt is the time-rate of change of velocity, which is what acceleration is, by definition. , dv _ " ~dt~ ^' :. 11 = jgdt. The velocity at the end of any interval (^sec.) from the start is since the stone starts from rest (t = o) = ^' by Rule 2, p. 82. i.e. V = gi. This formula gives us the velocity as a function of t. ds • Now -J- IS the velocity at t sec .from the start (see p. 59)- • ^ = et " dt ^' V = I gdt, = ytdt, and the distance traversed from rest in T sec. is therefore /•T = J^ gtdt Jo ■T tdt = £[^T2 - 02] by Rule 2, p. 82. = igT". The formula S = ^gT^ = le-iT^, therefore gives the distance traversed, in feet, from rest in T sec. 4. Moments of Inertia. — Moments of inertia arise in problems of the bending of beams, rotational kinetic energy, and so on. Routh defines the Moment of Inertia and Radius of Gyration as follows : — ( D 102 ) g 114 APPLIED CALCULUS " If the mass of every particle of a material system be multiplied by the square of its distance from a straight line, the sum of the products so formed is called the ' mmnent of inertia ' of the system about that line. " If M be the mass of a system and k be such a quantity that MW' is its m.oment of inertia about a given straight line, then k is called the ' radius of gyration ' of the system about that line." In the bending of beams the M. I. of plane areas is required. Let the plane have mass, p lb. per square foot. Let the moment of inertia be required about OY (tig. 3). Fig. 3 Let w be the width (BA), in feet, of the plane at x, BA being parallel to OY. Let I be the moment of inertia (Ib.-ft.^ units) of the area in- cluded between x ^ a and x = x, and let I "= f(x), where _/(jr) is to be found. Al = increase in I, due to an increase A^r in x, = the increase in I, due to the area ABDC. Now wAx < area ABDC < (w + Aw)Ax, if ze/ + Aw is the breadth at {x + Ax), hence pwAx x^ < A\ < p(w + Aw)Ax(x + AxY, i.e. pwx^ < — < p{'w -\- Aw) {x -{■' Axf. PRACTICAL APPLICATIONS 115 •(3) Now when Ax — > o, Aw — > o, and hence p{w + Aw) {x + AxY — > pwx^, i.e. Lt (-7—) = pwx^, rfl 2 I.e. -^ = Bwx'. ax If I = whole moment of inertia, I = p I wx^dx, if /> is a constant. This result is in pound-foot units, if the density is expressed in pounds per square foot, and the dimensions of the surface are expressed in feet. Note. — w cannot be put outside the sigfn of integration as it is not, in general, a constant. No matter what shape of figure we assume, the conclusion is the same, namely, I = p j wx^dx. We have only deduced the formula for fig. 3. j a Special Case. Find the moment of inertia of a uniform plane rectangle of density p lb. per square foot, about a line parallel to its shorter side and passing through the centre of the rectangle. X' "\ X V Fig. 4 Let d be the length in feet, and w the width in feet of the rectangle, d being greater than w (fig. 4). n6 APPLIED CALCULUS Here 6 = -, a = , and w is constant, hence 2 2 /•+? r 3"i"^? \ = pw\ ^x^dx = pa' — ' by Rule 2, p. 82. 2 L J-j = m — + — L24 24J 12 = pwd — 12 = (mass of rectangle) X — , 12 i.e. I = M — (Ib.-ft.") where M is the mass of the '^ rectangle in pounds. This result is a case of Routh's rule, viz. : — M.I. about an axis of symmetry (sum of squares of perp. semi-axes)~| , . 3i 4) 5 -I mass X The denominator is to be 3 for a rectang^le or square, 4 for an ellipse or circle, and 5 for an ellipsoid or sphere. In the case above there is only one semi-axis perpendicular to the axis about which I is required, viz. the one perpendicular to Y'Y, i.e. ^. Routh's rule gives I - MX^^^ = 3 12 which agrees with the result already found. 5. An Electrical Problem. — Required the power lost in re- sistance — " /^/? " loss as it is called — in watts in a twin cable, the current in which is tapped off for lamps at i amperes per mile. The cable may be assumed to be very long compared to the distance between lamps, so that the current may be supposed to be tapped off continuously at i amperes per unit length. A -i 4* _ „ -1 ^ ^H D A R ^ 1 PRACTICAL APPLICATIONS ii? Measure x (fig. s) from the far end. The current passing down AB at P is ix. The current passing down AB at / is i{x + Ax). ^1 n iijr~' i°5 Fig- 5 Let r stand for the resistance in ohms, per mile, of both con- ductors AB and CD; then the resistance of the wires {Pp -{- Rr) is rAx olims. Let w stand for the loss in watts between the far end and the section PR, then the "PR" loss (Aw) in watts in Pp and Rr is given by the inequality xH^rAx < Aze/ < i^(x + AxyrAx, i.e. xy^r < ^ < (x + AxfPr. Ax Now, as Ar — > o, (x + AxY — >• .r^. . dw .1, o ax :. w = i^rTx^dx = P?[—T = i!!2f watts, jo L 3 Jo 3 The whole current fed into the main is h amperes, and the whole resistance is rl ohms. Putting I = li and R = rl, .', the whole loss W = (putting x = I'm -w) = i PR watts. ii8 APPLIED CALCULUS Exercise 5 1. Why does the reference to time occur twice in speci- fying- an acceleration? Compare an acceleration of i mile per hour per second with one of i mile per second per hour. 2. A suburban electric train accelerates during the first 20 sec. of its run at ij miles per hour per second; during the next 40 sec. at J mile per hour per second. It then decelerates at o«o65 mile per hour per second for the next 120 sec. The brakes are then applied, and it de- celerates at 1 1 miles per hour per second until it comes to rest. Find , how long the run takes, in seconds ; find the length of the run, and hence find the average speed of the train during the run. (Draw a velocity-time diagram and integrate it graphi- cally.) 3. In question 2, at what speed is the horse-power taken by the train greatest? If the train weighs 70 tons, what is the maximum horse-power used? 4. An equilateral triangular flat plate is immersed vertically in water so that its upper apex is just sub- merged. Show that the total force on each face, due to hydrostatic pressure, is \wl^, where w = weight of water per cubic foot and / = length of a side of the plate in feet, 5. Show that the moment of inertia of a flat circular plate, about an axis through its centre of gravity and perpendicular to the plate, is Mr^/2 Ib.-ft.^, where M = mass of plate in pounds and r is the radius in feet. (Calculate it from first principles and check your result by Routh's Rule.) 6. Show that the moment of inertia of an equilateral triangular plate, about an axis through an apex, in the plane of the plate and parallel to the opposite side, is f M/^ lb.-ft.2, where M = mass in pounds and / = length of a side in feet. PRACTICAL APPLICATIONS 119 7. Calculate the moment of inertia about XX' of a plate of shape shown in fig. 6, the density of which is i lb. per cubic inch, and the thickness i in. AB = 6 in. AD = I in. KL = 6 in. GH = I in. AK = KB. State clearly the unit in which your answer is expressed. (Use Routh's Rule.) Fig. 6 8. Under the same conditions as in 7, what is the moment of inertia about the axis MN? How do you account for the great difference in the moments of inertia? g. Calculate the moment of inertia about XX' for fig. 7, assuming unit thickness and unit density as in 7. Calculate the moment of inertia about YY'. 10. What general conclusions do you draw from ques- tions 7, 8, and 9 as to the distribution of material required to ensure a high moment of inertia? Does the usual design of a fly-wheel bear out your conclusions? 11. An ordinary fly-wheel weighs i ton and is 6 ft. in diameter. Assuming 85 per cent of the metal is in the I20 APPLIED CALCULUS rim, which is 3 in. wide radially, what is its moment of inertia? Fix bounds within which your answer lies. Figr. 7 12. The field magnet rotor of a large alternator is 5 ft. long-, and 3 ft. in diameter, and may be treated as a solid cylinder weighing j^ tons. What is its moment of inertia, approximately? 13. Taking the earth as a sphere of diameter 8000 miles and of mass 6067 X 10^* tons, what is its moment of inertia about its polar axis? State carefully the unit in which you express your answer. 14. Current is tapped off a pair of street electric mains, of resistance r ohms per yard, at z amperes per yard. If the pressure across the mains at the sending end is E volts, what is the pressure at the far end, if / is the length of the pair of mains in yards (2/ = Mai length of copper conductor)? CHAPTER VI Differentiation and Integration " Data aequatione fluentes quotcumque quantitates involvente fluxiones invenire et vice versa." — ^Newton. In Chapter IV certain properties of limits and some theorems relating to them were deduced. These theorems furnish many important derivatives and integrals, and lead to several general theorems of differentiation and integration which are of the highest importance. Differentiation of Sines and Cosines. I. To find the derivative of sin x. \iy = sin x, dy _ y ^ f sin (x + A^) — sin x \ dx ~ ^^^ ol Ax V Now, sin {x + Ax) — sin x = 2 sin ^Ax cos (x + ^Ax). . sin (x + Ax) — sin x Ax 122 APPLIED CALCULUS Ax y Tsin (x + Ax) — sin x~\ = Lt r^ijil^l X Lt rcos(^ + iA^)l Aa; — > OL 2 ^■''' J A* — > oL J by Theorem 2, p. 99. = I X cos X, by equation (11), Chap. IV. .'. T-(sinx) = cosx, (i) I cosxdx = sin X (2) and 2. To find the derivative of cos x. If y = cos X, fi^ _ y [" cos {x + A-y) — cos jc "! fi?^ AX -> oL '^^ J u [ — 2 sin {x + ^A^) sin ^Aa; "| whence ^(cos x) = — sin x, (3) and I sin xdx = — cos ^ (4) (The reader should supply the details omitted above.) DifTerentiation of Constants. I. Additive Constant. Let^ = f{x) + c where c is a constant number. DIFFERENTIATION AND INTEGRATION 123 If X changes to {x + Ax), y + Ay = f(x + Ax) + c, andji/ = /(x) + c. Subtracting, Ay = f{x + Ax) - f{x). . Ar ^ f{x + Ax)-f{x) "Ax Ax Proceeding to the limit when Ax — >■ o, we find ^ = ^f(x). dx dx Rule III. — An additive constant may be ignored in finding derivatives. 2. A constant multiplier. Let y = c y. f{x) where c = a constant number, y + Ay = ex f{x + Ax). .'. Ay = cf{x + Ax) — cf{x) = c[fix + Ax)-fix)]. . Ar ^ J [fix + Ax)-f(x)] \ Ax { Ax ) ,. Lt (^) = Lt J[/(^ + A^)-/(^)]| ^ _> a\Ax/ Aa — > I Ax ) = ^ X Lt r/(^ + A^)-/(^) ] Aa:_>oL Ax J' -1 = -^- (5) Rule IV. — The derivative of c times a function of :e is equal to c times the derivative of the function of X, or a constant multiplier remains a con- stant multiplier. 124 APPLIED CALCULUS Differentiation of the Sum or Difference of Func- tions. Let y = 0(x) + ■yjr{x), {x) and ^{x) being differentiable functions. Then, as in proof of Rule 3, it easily follows that ^ = Lt f M(^ + MM] dx ^x—^o\_ Ax Ax J Aa!->oL ^^ jAa->0 Ax by Theorem i, p. 98. i.e. £{{x) ; and suppose bothy(^) and . oAx A« — > oV Ax/ A* — > o\ Ax/ + Lt (av^\ i _>, o\ Ax/ Since ?:) is a continuous function, A© — > o when A:>c — > o, whence by Theorems i and 2, p. 99, dx ~ dx dx ^'' Rule VI. — The derivative of the product of two functions is (ist function x derivative of 2nd func- tion) plus (2nd function x derivative of ist function). The result can also be put I dy _ I dv i_ du y dx V du u dx' on dividing (7) hy y = uv. This theorem can evidently be extended to the pro- duct of any finite number of functions. EXAMPLE y = sin X cos x, sin X = 1st function, cos X = 2nd function. , , -f^ = — sm X sin X + cos x cos x ax = cos^ x — sin^AT = cos 2X. Differentiation of Quotients of Functions. 7/ Let y = -, where ^, u, and v are all differentiable functions of x, and v must not be zero. Then u = yv. 126 APPLIED CALCULUS By the last result, equation (7), we have du _ dv dy dx ~ dx dx Multiply by - and we get \ du _ y dv dy V dx ~ V dx dx dx = I du V dx y dv ~ V dx I du u dv ~ vdx v^dx = ^r^ du dv' dx dx (8) Rule VII. — To differentiate a quotient, multiply the denominator into the derivative of the numerator, and the numerator into the derivative of the de- nominator; take the latter product from the former and divide by the square of the denominator. EXAMPLE ^ = ?HL£ = tan x. cos;); Here u = sin x,v=^ cos x. . dy^ _ cos X cos x -|- sin x sin x ' ' dx cos'':!; _ cos'j; + sin'j; cos'jc = sec^:r. Differentiation of a Function of a Function. Suppose y = f{u) and u = (p{x). Let f{u) and o\AxJ = Lt Ax — > ^Ay\ X Lt Ax — > oVaJ' by Theorem 2, p. gg. Now, when Ax — > (p{x) is continuous. 0, Au -* since M = 0(:x;) and dx = Lt Au — > dy du (Ay\ o\Au/ X — dx' X — dx (9) Rule VIII. — If j; is a differentiable function of u, and u a differentiable function of x, the derivative of y with respect to x is the product of the derivative of y with respect to u into the derivative of u with respect to x. EXAMPLE y = cos (ax). Here u = ax. ,', v = cos u, and J-- = — sia u X a = —a sin a^c. Reciprocals. rr.. • Ay Ax The equation ~ x -^ = i is formally true. Now Lt r^ X f^l = Lt r#^) X Lt (Pi, provided both limits are finite. 128 APPLIED CALCULUS Suppose jj/ is a continuous function of x, then Ay — > o when Ax — > o, hence Lt r^ X ^1 = Lt m X Lt m ~ dx dy' Also Lt T^ X X- = i> since x^ x -r- = i. dy dx ''^'dx= V^ ('°-' provided -r- is finite and not zero. Constants in Integration. (i) Suppose the integral j f{x)dx is required, and that /{x) is given. This integral is a function of x, {x) + -^ir{x))dx. and \-T~\lr{x)dx = i/'(a;). (D102) 10 I30 APPLIED CALCULUS Rule IX. — The integral of the sum (or difference) of (any finite number of) functions is the sum (or difference) of the separate integrals. Integral of Product of Two Functions. This theorem is highly important, and is known as "Integration of Products by Parts", or briefly " Integration by Parts "• By equation (7) d , , _ dv du where u and v are differ- dx^ ' ~ dx dx' entiable functions of x. .•, \u-j-dx = uv — jv-T-dx (16) Now V = I J- dx and m = I t- dx. J dx J dx Put w ior -r^, a function of x, dx ' then, by (16), juwdx = ujzvdx— j i-r- x jwdxjdx. (17) Rule X. — The integral of the product of two functions of x is equal to the first function x in- tegral of second function, minus the integral of (the derivative of the first function multiplied by the integral of the second function). EXAMPLE Required Ix cos* dx = j/. XT • • dv T Here u = x, v == sin x, since -=- = cos at, if j; = sm x; dx DIFFERENTIATION AND INTEGRATION 131 hence, regarding x as the first function and cos x as the second function, the rule gives directly y = * sin j; — / 1 X sin xdx = X sin x + cos X. (Check this result by differentiating it. Its derivative should, of course, be x cos x.) Derivative of ,«:", when « is a fractional or negative index. We have proved (p. 43) that ^(x") = wa;""', when n is a positive integer. (18) Suppose n is a. positive fraction = -. Put y = xi, I.e. ^ IS a function of x, :. y = x". ••• Ty^y) X fx = Tx^^')- since/ and q are integers, and so (18) applies. •■•>-xi = P-x^-\ 9 (-•)•- X 1 . dy ^txJ dx q = - X!i . . _^ f^»^ = «^»-i ^hen x is a positive " dx^ ' ' integer or fraction. 132 APPLIED CALCULUS Suppose « is a negative integer or fraction = — w, then x-'^x'^ = i. :.^^{x-x-) = o. .'. ^"^(a;-") + OTX-"" a;"-' = o. >-e. ^(x») = nx''-' (19) when « is a negative integer or a negative fraction. We have now extended Rule I to make it apply to any rational index n, either positive or nega- tive. Exercise 6 DiflFerentiate the following expressions : — 1. 2,x^. 2. X^ -^ X^ + X + 1. 3. (x -\- a){x + b), a and b constants. 4. x{i + x^) (i + ^). 5. x-\ 6. {x + a)l{x + b), a and b constant. 7. {xl{x 4- i)}", m constant. 8. n/i + x^. g. {2«a; + x^y^, a and w constant. 10. jc'/n/i + Jf*. 11. tan a;. [Put tan at = — — ■. Also direct from limit, Exercises, p. 108.] ^"^^ 1 2. sec X. KEPLER (1571-1630) NAPIER (1550-1617) rroin ihc /H^rtnut in Hit' University of Kdnihiir-h DESCARTES (1596-1650) From ihe portrait by Franz Hah i the Louvre LEIBNITZ (1646-1716) From the port7-ait in the Florence Gallery MATHEMATICIANS AND PHILOSOPHERS DIFFERENTIATION AND INTEGRATION 133 13. cosec X. 14. cot X. 15. sin 3^. 16. sin 2^x cos 2X. 17. sin (cos x). 18. (a + sin x)j(b + cos iic), a and 6 constants. 19. (a + sin a;) (6 + cos x), a and b constants. 20. Va + b sin ^, a and b constants. Integrate the following functions of x\ — 21. ax^. 0.] 22. m 23- ax"^. 24. {ax'" + d)™^"-'. [Put ax-^ + 6 25- («^ + 6)"- 26. 3 sin X. 27. sin a: cos X. 28. sec ^ tan x. 29. 10 sec^A;. 30- sin 2a:. CHAPTER VII Infinitesimals " Great fleas have little fleas upon their backs to bite "em, And little fleas have lesser fleas, and so ad infinitum. And the great fleas themselves in turn have g'reater fleas to go on, While these again have greater still, and greater still, and so on." — Prof, de Morgan. Infinitesimals. Up to the present -^ has been defined as one symbol standing for and we have treated -3- as a single symbol just as V is a single symbol. In applied mathematics and in natural science generally, "^" (sometimes §) is used in a different sense, so that dy, dx have definite meanings when standing separately. This new usage arises from the fact that when we are dealing with quantities as distinct from numbers it is usually easy to see when one quantity is rela- tively small and small enough to be ignored. In applied mathematics — indeed in all "concrete" mathematics as distinct from the mathematics of ab- stract numbers — mere numbers are not of much 134 INFINITESIMALS 135 interest. It is not much use a farmer telling us that the stock on his farm is 100, unless he tells us whether he is speaking of bullocks or chickens. Ascending Powers of Ax. We have found that, if jj/ = f{x), Ay can often be expressed as an ascending power series in Ax. For instance, if ;>/ = x*, Ay = 4x^Ax + 6x\Axf + ^{Axf + {Axf. ...(i) For any given value of x, the coefficient of powers of Ax reduce: to common multipliers. Thus, if X = 2, ' Ay = 32 A^ + 24(Ax)2 + 8(Axf + (Ax)'. ...(2) This equation is the exact expression for the in- crease (Ay) in y arising from an increase (Ax) in x, when the relation between x and y is y = x*, and x has the special value 2 (see fig. i). Now, suppose we are considering change^ in x, of the order lo"^^ i.e. Ax = io~i^ say. Ay is then the sum of 32 X 10-" = 32 X lo*" X lo-*" 24 X 10-" = 24 X lo"^ X 10-^^ 8 X iQ-'" = 8 X 10" X 10-"= I X 10-^ = I X IQ-'". The sum of the last three terms is about i billionth of the first term. If the first term (32 x 10'* x lo"^*) measures a " small quantity "* it is an infinitesimal of the first order. * For discussion of this phrase, see p. 27. 136 APPLIED CALCULUS The second term (24 x 10''* x 10*') is an infinitesi- mal of the second order. It is less than one billionth of the first order term. The third term (8 x 10'^ X lo"**) is an infinitesimal of the third order. It is only one three-billionth Fig. I part of the second order term, and is one four- billion-billionth part of the first term, and is there- fore quite negligible compared to the first order term. The fourth term (i x 10"*) is an infinitesimal of the fourth order, and is quite negligible compared to any of the others. The result is that, in actually calculating Ay as a INFINITESIMALS 137 practical problem, we need only retain infinitesimals of the lowest order, provided A:'c is small enough. In the above case, y = X* and ^ = 4^. dx ^y = [£+/(ax)ax]a^. -dx .-. Ajj/ = ^Ax + [/(Ax)A^]A^. ...(3) It is evident from (i), then, that when Lx is very small, the important part of Ajk is the \-j-Lx\ part. The coefficient of the other term, /(Aj£;)Ax, ap- proaches zero as b^x — > o, and For this reason, -^ ^x is given a special name, and is called the differential of y, and when Ax is small enough to make /(Aa;)A:'c relatively negligible, Lx is called the differential of x. &y is written for the differential oiy, and Sx for the differential oi x. Hence ^y = ^^x, (4) where 8x is any small increase oix. We can proceed to the limit with ^x if we wish, and get, dividing both sides of (4) by &x, 8y _ dy 8x ~ dx and Lt m = 1^ = Lt (^) (5) 138 APPLIED CALCULUS The term in equation (3), [/(AAr)AA;;]A^, plays no part in forming the limit of Ay/Ax, for • Lt [AAx)Ax] = o, (6) ^z — > and it contributes nothing of any importance to the change in y which accompanies a sufficiently small change in x. It can therefore be omitted, and when it is omitted the differential or infinitesimal notation is used. Fig. I shows clearly the principle involved in the use of infinitesimals. AjV = RQ = PR tan i/' + SQ - Jj' + SQ (7) and SQ/Ax — > o as A^ — ^o. Hence we may put change in^ ~ ^ X (change in jc), ...(8) when Ax is small enough. In applied mathematics and physics, it is usually not considered necessary to distinguish between (S and d in the notation, dx is often used instead of ?iX when an infinitesimal change in x is referred to. The beginner is advised to use one notation: A for any change — large or small; 5 ,, an infinitesimal change; and T- ,, the limit notation. d should not be used except in the limit notation. INFINITESIMALS 139 The matter can be concisely put by using the formal definition of a limit given on p. 93. Av If -r- has a limit as Ax Ax o, this limit is written dx Assuming the limit to exist, we must be able to find a finite positive number yu such that the in- equality Ay _ dy^ Ax dx o < < K holds so long as o<\Ax— o\<^L holds; where k can be any finite positive number not zero. Now if Ay _ dy Ax dx where e is a positive or negative number, then ■(9) Ay _ dy Ax dx .: O < I e I < k: so long sis o < \ Ax \ < fi. These inequalities are the conditions that e may have the limit zero as Ax — > o, therefore if Lt { -r^ ) exists and equals -r-t + e we may put ^^ _ ^^ Ax dx where Lt (e) = o. Ax — >■ I40 APPLIED CALCULUS Now, in any practical application of numbers to measure quantities we can always select a quan- tity q, such that any quantity less than q is negligible. Hence, if we make | e | < | ^ | , we can write (9) % - t Co) and, by sufficiently decreasing the interval in which I Ax I must lie, we can always make | e | < \q\ ^^'^ I e I -^ o as I Ax I -^ o, i.e. we write ^ « ^ (11) when Ax is small enough. Equation (11) must be read to mean that ^ differs from ~ by less than the ^arbitrary "tolerance" set when Ax is made sufficiently small, or, more briefly but more loosely, -~- is nearly equal to -j-, when Ax is small enough. The "equals" Sign in Practical Calculation. When physical or chemical measurements are in question, a slightly different interpretation is put on the sign " = " from that given to it in pure mathe- matics. In pure mathematics, y = P (12) INFINITESIMALS 141 means that y is exactly equal to p. The number y is neither more nor less than the number p, in fact " = " means xS\aX y and p are identical numbers. Now, suppose we say that / stands for the number which expresses the length of a straight line AB in terms of a unit of length, say the centimetre. If the line is measured and is found to measure say 10 cm., we say that / = 10. But it is almost certain that / does not equal 10 in the pure mathematical sense. Suppose the line to have been measured with a centimetre ruler. It is certain that the length is more than 99 cm. and less than ID- 1 cm., so if we say that / = 10 in the pure mathe- matical sense we cannot be far wrong. If, later on, we found that a possible error of ± o- 1 cm. (which is ± o-i cm. in 10 cm., i.e. ± i per cent) is more than we ought to allow, then we should measure the line more accurately, say with a carefully-constructed travelling microscope fitted with vernier scales. We might then be sure that the length of our line is more than 9-99 cm. and less than looi cm. The possible error in taking I = 10 is then limited to ± ooi cm. in 10, i.e. to i part in 1000; and so the process can be carried on, each approximation coming nearer to the true measurement as the possible error de- creases. The difficulty of making any physical measurement increases seriously as the permissible error decreases. To measure the line correct to ± o- 1 cm. may take one minute and require an appliance costing three- pence. The same measurement correct to + o-oi cm. may take an hour and require apparatus costing many pounds ; while, when a high degree of accuracy 142 APPLIED CALCULUS is required, such, for instance, as is attained in the measurement of base lines in geodesy (i part in 2,000,000 say), the work takes years and becomes an international matter. Poincare writes: "... this history (of geodesy) certainly teaches us what pre- cautions must surround any serious scientific opera- tion, and what time and trouble are involved in the conquest of a single new decimal ". It is unwise therefore to attempt to make a measure- ment to a higher degree of accuracy than is actually required. When we write / = 10 cm., we mean that 10 cm. can be taken as equal to /, the length of AB for the purposes of our problem, and not that 10 cm. is the actual length of the line. It is in this sense that we may write, instead of (1 1), A^ = %Sx = Sy, (13) i.e. when Ax is small enough, it is unnecessary to distinguish between Ajj/ and Sy. Approximate Calculations by Small Differences. Suppose we wish to calculate the weight of a brass tube, per foot, correct to 5 per cent, the internal diameter being 2 in. and the thickness ^V •"• di, internal diameter, = 2 in. £?2i external diameter, = 2^V in. W = 1 2p X (area of cross-section of metal) = I2pA, where W is the weight per foot in pounds, p, the density in pounds per cubic inch, and A, the area in square inches. INFINITESIMALS 143 Let A be the area of cross-section of tube in square inches, Ci, the circumference of inner circle in inches, Cj, „ ,, outer ,, ,, (fig. 2). F'g. 2 Then, if ^ is the thickness in inches, Ci^ < A < Cg?. /. Trdii < A < ird^t, irdj: < A < Ti-ditii +^)» since d^ = di + 2t. Now, 2/ _ 2 X /y _ I di 2 64' .•. ird^t < A < 7rc?i/i+ -p-j 64; X is not known accurately, but it is given by 3.I4I < TT < 3-142. .-. 3-141 < TT < 3-141(1 + ^)- 144 APPLIED CALCULUS p is not linown accurately; it varies with the com- position of the brass, between 0-30 and 0-31. .'. 0-30 < p < 0-31, i.e. 0-30 < p < 0-30(1 + ^). Combining all these inequalities, 12 X 0-30 X 3-141 X 2 X j!- < W 64 <(,.Xo.3oX3-.4.X.xi)(.+jL)(,+_±.)(,+^). ''-(■+^)(' + 3-lV)('+^ = (-+^> If, therefore (12 x 0-30 x 3-141 x 2 x ^-j is taken as the value of W, we can be sure that the result is within the permissible error of 5 per cent. For tubes which are sufficiently thin compared to the diameter, the weight per foot may be calculated by W = I2pirdit, where p is the weight per cubic inch in pounds, diy the internal diameter in inches, and t, the thickness in inches. The exact formula is W = i2p'^{d^-d^) 4 = 3p7r(^2^ — d^), where d^ is the external diameter, and £?, the internal diameter = 3 px(fi?2 + d^ (^2 — '^i) = \2pir{dx + t)t. There are undoubtedly many problems which can be calculated exactly without much difficulty; but INFINITESIMALS 145 there are also very many which cannot. Even if absolute exactness is possibly attainable, it is often only attainable at the cost of immense labour. To quote again from Poincare's writings: "There we meet the physicist or the engineer who says, ' Will you integrate this differential equation* for me; I shall need it within a week for a piece of construction work that has to be completed by a certain date?' 'The equation,' we answer, 'is not included in one of the types that can be integrated, of which you know there are not very many.' 'Yes, I know; but then, what good are you?' More often than not a mutual understanding is sufficient. The engineer does not really require the integral (complete solu- tion); he merely wants a certain figure which would be easily deduced from this integral if we knew it. Ordinarily we do not know it, but we could calculate the figure without it, if we knew just what figure and what degree of exactness the engineer required." The above example (p. 142) has shown how, when the degree of exactness required is specified, the necessary calculation can be simplified. In the ex- ample, we reached a formula which gives sufficieptly accurate results with a slide rule, whereas the exact formula cannot be calculated directly on a slide rule. EXAMPLES FROM MECHANICS I. Find the position of the centre of gravity of a flat parabolic plate of uniform thickness, shaped as shown in fig. j. Let AOB represent the plate. Choose the axes as shown in the figure, OX being the axis of symmetry. By symmetry the * An example of a "differential equation " will be found on p. 184. We " integrate ' a differential equation when we find the relation between the variables from which it arises. (D102) 11 146 APPLIED CALCULUS C. G. must lie somewhere on OC. Let x stand for the distance of G, the C. G., from OY; then by taking moments, about OY, of the forces acting perpendicular to the plane of the plate, we have p X A X ^ = sum of moments of the forces on the pieces Y /tI / ^A* A f R F ;' G C X Q C '- B Fig- 3 of matter in AOB, about OY, where p is the weight per square inch in pounds and A the area in square inches. This follows from the definition of the centre of gravity as the point through which the resultant force of gravity acts. Let X be the distance of any line PQ, at right angles to OC, from OY, and ^x the width of the strip PQQ'P'. Then the area PQQ'P' is greater than QP X Aat, and less than Q'P' X ^. Since OA is parabolic, RP = y = ^^ax, where x = OR. R'P' = jy + iiy. .-. 2{y + Aj^)AAr > PQQ'P' > 2jf^. •'• PQQ'P' = ^(y + kAt)^^. where k is a positive fraction such that o < K < I. Hence the weight of the strip = zpijf + kAj/)Ax lb. INFINITESIMALS 147 Now the effective distance {d) of this strip from OY is given ^^ (x + Ax) > d> X. .'. d = X -\- XA:tr, wliere o < X < i. .'. moment of the weight of this strip = 2p{y + Kiiy)Ax{x + XA^i;). If M = moment of the portion of the plate OPQ, AM = 2p(y + kA.j/) (x + 'KAx)Ax = 2p{xy + KxAy -\- \yAx + KkAyAx]Ax. Malting the changes all infinitesimals, and dropping infini- tesimals of second and higher orders, we get SM = 2pxySx. , 8M *' Sx = 2pyx. ,-. Lt o\Sxl = 2pyx. . dU " dx = 2pyx. Hence, ifM is the whole moment of the plate, M __ 2p 1 yxdx Jo 2p 1 '^^ax X dx Jo 2p V4a j x^dx Jo fpV4a//2 f P CA P. Equating this quantity to phx, we get pM = ipCKlK 148 APPLIED CALCULUS Now the area of the plate = 2 / ydx Jo ft J = 2 V4a I x^dx Jo = 2V^[|^5]^ = 2V^§/5 = i^/^l = iCAl. .; JiCAlx = icKlK .: X = i I, i.e. the position of G is such that OG = f of OC. 2. To find the moment of inertia of a uniform thin rectangular plate about an axis of symmetry. (Compare p. 115.) Fig. 4 illustrates this problem, Y'Y being the axis about which Y X' b \/D Sx -X- X B C Y' Figr. 4 the M. I. is required. Take the centre of the plate as origin and the axes of symmetry as the x and y axes. Let x be tiie distance of AB, in feet, from the axis Y'Y. Con- sider the strip ABCD, which is Sx ft. wide. The area of it is INFINITESIMALS 149 aSx sq. ft. and the mass paSx lb.,* if p is tlie density in pounds per square foot. This mass contributes to the M. I. something be- tween paSx x^ and padx {x + SxY, say paSx (x -\- k SxY, where k is a positive fraction between o and i. This expression is paSx(x^ + 2KxSx + K-Sx^), i.e. paSx x^ + 2paKx{SxY -\- paK^(SxY. Retaining only the infinitesimal of the first order, we get, it I is the moment of inertia of the portion of the plate EFBA, 81 = pax^Sx. . dl •• dx paxK The whole M. I. is given I = by pa 1 x^dx J >> 2 a palfi 12 = Mass X —1 lb. L 12J -it: Expansion of a Gas. Let the pressure be related to the volume of a confined gas by a curve as shown in fig. 5. When the volume is v c. ft. the pressure is p lb. per square ybo/. When the volume is {v + Aw) c. ft. the pressure is {p + Ap) lb. per square ybo^. Ap may be positive or negative; it is negative in the figure, if Av is positive. Suppose the gas expands by pushing a piston out very slowly against a pressure which is just less than the pressure of the gas. ISO APPLIED CALCULUS Let A be the area of the piston in square feet. The average pressure on the piston during the expansion Av is something between p and p + Ap, p / i b p ST <-^'o-'- P B T "i i a 1 » V i i V Fig-.S say p + kAP, where k is a positive fraction between o and I. The force on the piston is {p + kA/)A lb. The displacement is Ax ft. The work done is (/> + kAP)KAx ft.-lb. But AAi« = At). .•. {p + kAP)Av is the work done in the expansion Av. Suppose W is the work done in foot-pounds in ex- pansion to volume v, then AW, the work done in foot- pounds in the expansion Av, is given by AW = ip + kAP)Av = pAv + KApAv. If the changes are infinitesimals, KApAv is of the second order and pAv of the first. INFINITESIMALS 151 Dropping the second order term, ^W = p?,v, and proceeding to tlie limit, and W = \''pdv (14) = work done in expanding from z^o to »i in foot-pounds. .*. the work done in foot-pounds = (the area of AaAB) X /, where / is a scale factor. This scale factor/ can be found as follows: — Suppose the p, v diagram — as a pressure graph, in terms of v, such as fig. 5 is called — is drawn to scales I in. vertically = P lb. per square inch. I in. horizontally = V c. ft. Then i sq. in. of area on the diagram is equivalent to (P X 144 X V) ft.-lb. Hence the "work" represented by AadB is (a X P X 144 X V), where a is the area of AabB in square inches, i.e. Work done in foot-pounds = (a in square inches) x / where/ = 144 PV. Indicator Diagrams. Formula (14) is extremely important in engineering. An instrument called an Indicator is used to draw mechanically a.p,v diagram for the steam on one side of a piston in an engine cylinder. A is a small cylinder (fig. 6), B an accurately fitted piston of I sq. in. (say) in area, C a Spring, D a joint 152 APPLIED CALCULUS by means of which the instrument can be attached to the engine cyUnder, E a link mechanism by which the movement of the piston B is repeated at the pencil G, and F a drum on which a piece of glazed paper is carried. The drum F is given a rotational to and fro motion by means of a driving cord H, the free end of which is attached to the crosshead Figf.e of the engine. The string is kept taut by working against a spiral spring I contained in the drum F. The motion of F therefore repeats the outward and inward motion of the engine piston on a small scale; while the vertical motion of the pencil G, across the paper on F, repeats the motion of the piston B on a magnified scale. The pressure in the engine cylinder is the pressure on the steam side of B. The piston INFINITESIMALS 153 B therefore responds, against the spring, to changes in the cylinder pressure, and the pencil G rises and falls as this pressure rises and falls. The pencil therefore traces out a graph like that shown in fig. 7. The curve is closed since the changes are cyclic and steady. The area enclosed by this curve is by (14) a measure F'&- 7 of the actual work done by the steam in the cylinder on one side of the piston. The appropriate scale factor / is known when the conditions under which the instrument is used are known, and hence the work done in foot-pounds = / X (area enclosed by the curve in square inches). Of course any consistent units can be used by giving different values to /. 154 APPLIED CALCULUS Exercise 7 1. Show that the moment of inertia of a thin rod of length / about an axis- passing- through one end and perpendicular to its length is MP , where M = mass of rod. 3 Hence find the M. L of a rod weighing lo lb. and 7 ft. long. 2. Find the position of the centre of gravity of a semi- circular flat plate. 3. Find the position of the centre of gravity of a solid hemisphere of uniform density. 4. The period of oscillation of a simple pendulum is given by /^ Show that the error in T introduced by a i-per-cent error in ^ is equal and opposite to the error in T caused by a i-per-cent error in /. [Find the small change in T corresponding to a small change in / and in ^. Make SI = o-oi / and Sg- = o-oi^, and find the ratio of the corresponding small changes in T.] CHAPTER VIII Integration "Nova stereometria doliorum; accessit stereotnetriae Archimedeae supptementum." — The Title of the first Book on the Integral Calculus by Kepler (1615). Some important methods for effecting the integra- tion of functions have already been discussed. We propose to collect in this chapter the more important methods, and to give some examples of their applica- tions. I. By Differentiation. When the derivative of a function is found, an integral formula follows at once. E.g., -3-(cos X sin x) = cos^ x — sin^ :r ; .*. j(cos^x — sin^x)dx = cos:!c sin x + c. (i) c is the arbitrary constant discussed on p. 128. Sometimes it is easy to guess a function whose derivative is the given function to be integrated. If ^{x) is the given function and f{x) the guessed function, l{x)dx = f{x) + c (2) /« for^(^) = ^J{x) (3) 156 APPLIED CALCULUS In this way certain standard elementary integrals are found. Many expedients have been devised for transforming integrals into simpler integrals which come within these standard forms. Skill in integra- tion largely depends on ability to find readily suitable transformations which make the given integral de- pend on simpler known integrals. Each success adds another integral to the list of known forms, but in spite of the great number of integrals that are now known, functions still crop up which defy all attempts at integration. .Sometimes, indeed, it can be foreseen that it is impossible to express a given integral exactly in terms of known functions, though we can always find the value of the integral approximately. Some of the simpler and more useful rules are given below. 2. Integration by Substitution. Complicated expressions to be integrated can often be simplified by expressing them in terms of a new variable. Suppose / d){x)dx is required. J a We may suppose jjc to be a function of a new vari- able 0, provided that the new function of z chosen can assume all the values of x required in the integra- tion. For instance, if I xl^dx is required, we could not put x = sin z because sin z cannot be g^reater than i. But we could put x = tan z, because a "sr" can be found to give tan z equal to any number between i and 3. It would not simplify the work, in this instance, to do this — the example is merely intended to show why certain functions are admis- sible while others are not. INTEGRATION 157 Let I = \f{x)dx. Then -t- = fix) = ^(z) if x is a function of z. Let a; be a differentiable function of z, say 9^(0), ^, dl d\ dx . , V = f^^)fz :. I ^ i[,riz)^]dz (4) This is the transformation formula for substitution. EXAMPLES (i) Evaluate I sin {2x)dx. Put 2X = s. , dx I .'. I = jsinz X i k dz = ^ I sin zdz = — ^ cos z = — ^ cos {zx) + c* (ii) Evaluate I sin {mx + w)<«!r, »? and n being constants. Put mx -\-n = z. _ z — n m J dx I and -=- = — dz m * In all cases an arbitrary constant must be added to an indefinite integral. It will be understood in future. 158 APPLIED CALCULUS /. I = f/'sin z X -\(iz = - [sin zdz = — — cos « J \ m) mj m — — ? cos (mx + «). m (iii) Evaluate | ' rfar, provided — a < jc < + o. ] Scfl — at' Under these conditions it is admissible to put .V = asin^, (S) for sin z may assume any value from — i to +1. dx Also -J- = a COBS', by Rule 4, p. 123, dz and a^ — x"^ = cfi — cfi Av? z. .', y/d^ — x^ = a cos z. :, { '^^ = (Hz)fdz = '(^LEBEJdz j \/a^ — x^ J dz J a cos z = jdz = z. But z = angle whose sine is -, by (5) . X = arc sm — . a I dx sJa^ — x"^ Example. — Evaluate \ ^ j^ i > "^ being unlimited in value. (Put X = a\.a.nz.) J " + ^ These are important integrals and should be remembered. 3. Integration by Parts. This theorem has been discussed on p. 130. The rule is: The integral of the product of two functions of X is the first function multiplied by the integral of the second, minus the integral of the derivative INTEGRATION 159 of the first function multiplied by the integral of the second. I.e. /{" ^ ©K ^ "^ - /S^"^^ ^^> EXAMPLES (i) Evaluate I x zosx dx. . \x { Let I x cosa: (i!r = I. The first function is a: = «. The second function is cosj; = ^- (sia*), i.e. z> = sin*. dx X svc\x — / 1 X sinjr dx, since -— = i when u ^= x J Hence I = I ^ j- (sin^r) dx sinxdx, since , dx = X sinAT — I sinx dx = :f sin* + cos*. (ii) Evaluate I sin'xdx. Isin^^rf* = I (sin*) ^ ~ ^dx = sin* (— cos*) — I cos*(— cos*)(fj = — sin* cos* + I (i — sin^*)rf* = — sin* cos* + * — I sin**rf*. .'. 2 I sin'''* X. (iv) (v) Vl — x^' I ^ < I. I + x'^' I (vi) , „ „ . X > a. X \x^ — a' 2. Integrate by parts the following expressions :- (i) x^ cos.r. (ii) X cos2Ar. 3. Show by reduction that (iii) x^ arc sin;r. /. > X arc sinAr (i-:i;2)i" f. ^' i • - • when n is odd. 5 3 n n — 2 n — 4 4. Show that f sin^^ cos^edd = - si"''-'ecos«^'g^/rJ [ sinP-^o cos^Ode; J P + 7 P + r2)}* 7. (a*3 + *)V. 8. [Put^-' + l=*l] /' 17. /tan'»^ according as m and n are unequal or equal positive integers. [This result has important applications in Fourier's Theorem.] 23. Prove example 22 for cosine instead of sine. 24. I* sec* Odd = i. INTEGRATION 165 25. If {a + x), show that I 4>(x)dx = n I {x)dx. [Function has period a, whence result graphically.] ri 26. Show that 0(a) (b — a) < {x) increases steadily from a to b, and is positive. What is the condition that (t>(x) shall increase steadily from a to b? 27. Examine 26, when ip{x) decreases steadily from a to b. 28. Examine 26 for the case when ^{x) is negative between a and b, and when it changes sign between a and b. 29. If ^(x) is continuous, so that the graph of 0(jf) is a continu- ous curve between x = a and x ^ b, show that X Fig. I at each point of division. Let pP, gQ be any two ordinates, distant A^ ft. apart. Then the chord PQ is nearly equal in length to the length of the curve between these ordinates; in fact, in the figure, the chord is scarcely distinguishable from 166 MENSURATION 167 the curve between P and Q. If ab were divided into 100 equal parts instead of g, the chord PQ would be closer still to the curve between P and Q, and the more numerous the divisions the more difficult it becomes to distinguish the curve from the chord between any pair of ordinates. The reader can soon convince himself of this by doing a little geometrical drawing, and finding how fine the division must be, for any given case, for the chord and the curve to become indistinguish- able. Fig. 2 shows the portion of the curve PQ on a larger scale. Let PS and TQ be the tangent lines at P and Q, T being the point of intersection of these tangents. Fig. 2 Let QS be perpendicular to PS and let PQ make an angle Q with PS, then chord PQ < arc PQ < PT + TQ < PT + TS + SQ < PS + SQ. ••• ' < chord% < cos + sin 0. i68 APPLIED CALCULUS Now as PR — > o, Q — > P, and PQ approaches the tangent line PT, .•. — > o. . / arc PQ \ _ •■pRZioVchordPQ/ ~ '• If P is the point on the graph corresponding to x, PR = Ax. . Lt f^I^pQ_^ = , Aa; — > oVchord PQ/ ^°^' arcPQ _ arcPQ chord PQ PR ~ chord PQ ^ PR ' and if f = length of curve from A to P, arc PQ = As, and PR = Ax. . As ^ arc PQ chord PQ A^ chord PQ Ax ... Lt (^) = Lt f-^;^) X Lt (?h2^). A* — > oVAiic/ ^ _>. o^chord PQ/ ^a -> o^ Ax / .. 1^ = I X Lt f t^hord PQ \ £?x ^ _>, 0^ Ax / But chord PQ = y/{AyY + {Axf. '• A^ - V' + Va^;- and Lt (£h2^) = Lt JT^Wf Ai -> o\ Ax / Aa -> ' \Ax/ MENSURATION 169 ds dx i.e. ^ ■ - (i) ■ -ilV'+d)""^ (-) when J is the length of curve between A and B, in feet — or any other unit of length which may be used. Unfortunately, in many practical cases, this integral cannot be found by elementary methods, and even in simple cases devices are desirable to simplify the integral. We will take one case which can be easily inte- grated from formula (i), and which is of some interest physically — the length of the cycloid between two cusps. Note on Cycloid. — The cycloid is the curve generated by a fixed point on the circumference of a circle as the circle rolls without slip- ping- along a straight line uv. It will therefore consist of a series of loops like uOv (fig. 3), with cusps at u, v, and so on. The length of the curve uOv is, then, the length between two cusps, measured along the curve. Let 6 be the angle through which the radius QP has rotated in rolling from R to g'. I70 APPLIED CALCULUS Then Rg' = a0, where a is the radius of the rolling' circle — the g;enerating' circle as it is called, g'p' = a sin 6, p'P' = rt(i + cos fl). Takings O as origin, OR and a line through O, perpendicular to OR, as axes of x and y>, then if x and y are the co-ordinates of the point P' on the cycloid, y = R^' = a(e + sin d), X = 2a - a(i + cos 6) = «(i - cos S). Eliminating 6, we get y = a arc cos ^-^ h "^(zax — x^). (2) WORKED EXAMPLE Find the length of a cycloid defined by the equation y = aarc cos °~^ + '^{2ax — x^) a between x = o and x = 2a. The curve is as sketched in fig. 4. ^ = af (arc cos ^^=-i) + f ( ^(2ax - x^A dx dx \ a ' dx \ / To differentiate the first term, we put a — X z = arc cos . a a — X cos s = a a cos z = a — x. .'. X = a{i — cos z). . dx . . ^- = a sin z. dz . dz _ I _ ' ' dx a sin z .'. a 4- (arc cos ^^1^) = dx\ a / V«a- a^ cos^z I Va2^ («- xf 1 V2ajf -x^ a '^2ax — : MENSURATION 171 For the second term, we get ■^ ( -Jiax - x') = I X ' dx 2 \l2ax — x^ X (2a — 2X') V2 Adding, we get dy a dx Fig. 4 'sizax — x^ + Vzax — x^ ■^2ax - -x^ V2a — X V2a — X Vm — X Vi /-: — X •■V'+^'-y? 172 APPLIED CALCULUS By (i), the integral giving the length we require is U' + (g)" dx. This integral is /.v?- = = Via ia X 2^X = -■o 40. The whole length UOV N OC II II (length OV) 1 (3) The length UOV is therefore four times the diameter of the generating circle. The interest of the cycloid lies in the fact that if a particle moves on a smooth cycloid under gravity, the vertex (O) of the cycloid being at the lowest point, the vibrations of the particle are truly isochronous, whence Lord Kelvin's name for harmonic motion, cycloidal motion. This example has been given to illustrate formula (1). Problems on the cycloid can be more easily worked by keeping the equations for x and y in terms of Q. We shall now discuss this method. Use of a "Parameter", i.e. of a New Variable on which X and y depend. Suppose X and y are each given as differentiable functions of another variable t. ■'■ ^ - V" + (eT' J ds _ ds dx dt dx dt -4 .+(1?)''^ dx/ dt' Ay Now Ay _ Ai Ax ~ Ax' At MENSURATION 173 and Ay and A^ -> o as A^ — >■ o, since x = f{i), and y = qVA^/ <^ _ dytdx dx ~ dtl dt' = /{V-^(f/g)'> EXAMPLE To find the perimeter of a circle. With the origin and co-ordinate axes, as sliown in fig. 5, the co-ordinates of P are y ■= a sin 6, X = a cos 0, where a is the radius of the circle. Our / is here 6, and g = g = acose, 1 dx dx . n ^""^Tt^dd^-"^'''^' ^= " COS g_ cote, afcr — a sin a 174 APPLIED CALCULUS and -775 = — a sin 6. au = I -J- X (- a sin e)de J sin d = - afdd. When X = o, 9 = -; when x = a, 9 = o. 2 the length of a quadrant is -Hr-- the perimeter of the whole circle is zira (5) Y Fig- 5 The student should now rework by this method the problem of finding the length of the cycloid. Areas. Let AB (fig. 6) be a portion of a curve whose equation is ^ ^ ^^^^^ and let the area under AB (AabB) be required. MENSURATION J75 This area is j ydx. .(6) The problem of finding areas is therefore a direct problem of integration. Y A *0 B I I X by. Fig. 6 WORKED EXAMPLE Find the area of the ellipse : + *2 We have dealt with the geometry of the ellipse— so far as we need it — on pp. 8 to 12. The ellipse has O as centre, and is shown in fig. 7. It is symmetrical about OX and OY, so if the area of a quadrant is found, four times this area is the area required. Area of quadrant = | ydx. J By the given equation /7' a .'. the area of a quadrant is - / Va^ _ x^dx. 176 APPLIED CALCULUS This integral has been given on p. 162, Ex. i (iii). It is /: + — arc sm - I. L 2 2 oj + — arc sin 2 xj \ the area of the quadrant = -^ ^ I _ irab ~ 4 ' and the area enclosed by the ellipse is irab (7) Y P Fig. 7 Volumes of Solids of Revolution. Suppose the volume of a figure (fig. 8) like ABCD is required, which is enclosed by two parallel planes and a curved boundary which is everywhere perpendi- cular to the end planes. The base can be divided into unit areas and includes, say, loj of these. If the planes are 9 units of length apart, the volume is clearly io| x 9 units of volume, i.e. volume = area x height. MENSURATION 177 It follows that the volume of a cylinder is the area of the circular base multiplied by the length of the cylinder. Fig. 8 A surface of revolution is the surface generated by a curve as it rotates about a fixed line in its plane. Thus, if AB (fig. 9) rotates about ab so that every point in AB describes a circle whose plane is per- pendicular to OX, and whose centre lies on OX, the surface swept out by AB is a surface of revolution. Consider the volume enclosed by the surface formed by the revolution of AB (fig. 9) about OX and by the end planes x = x^ and x = x^. Y A <-• C 1— 5 t Ax 'Ay B a * X A' ''1 - - J c '■"i "' , B' (D102) Fig. 9 13 178 APPLIED CALCULUS Choose the origin and axes as shown. The dimen- sions shown on the figure then follow. Let V be the volume AA'C'C between the planes X = Xf, and x = x. Then the increase of volume, AV, in going from a; to :« + Ax lies between -ny^Ax and ir^y + AyfAx, i.e., for the figure taken, iry^Ax < AV < iriy + AyfAx. AV vy < ^ < -^{y + Ayf. t-^o\AxJ Try, I.e. dx = iry^. /. V = ttI y^dx .(8) WORKED EXAMPLE J^ind the volume of a sphere. The volume of a sphere is enclosed by the surface of revolution of a circle about a diameter. Fig. lo Let O, the centre of the sphere, be taken as origin, and OX as the axis of revolution (fig. lo). fa = 27r I (a Jo MENSURATION 179 The equation of the revolving circle Is ^•i -|-j/2 = fl2 vjriiere a is the rad. of sphere. .•, _y = Va^ — X-. Let the circle spin about axis OX, then vol. of sphere = 2 X vol. of ^ sphere ra = 2 / TryVji; Jo ■ x^)dx /a ra c^dx — 27r I *-Va; y = 27rl b > o). 6. Show that the area contained by the curve j^ = Ao + \x + A^^ + A^x^ the OX axis, and the ordinates at x^ and x^, is given accurately by Area = - (A + 2B + 4C), 3 where A, B, C, and 5 are obtained as follows: Divide (x^ — aTq) into an even number (n) of equal parts, each equal to s. A = sum of the end ordinates ; B = sum of the other odd ordinates (3rd, 5th, &c.); C = sum of the even ordinates (2nd, 4th, &c.). [Take n = 2, then prove the rule by repeated applica- tion of this case. ] This rule is known as Simpson's Rule. When the rule is applied to find I ydx where V = f(x), «0 the error, if any, is due to the deviation of /{x) from an expression of type, Ap + A^x + A^^ + A3JC8, between the limits of integration. This deviation is usually quite small. MENSURATION i8i 7. Find the area of the curved surface of a cone, of slant length / and radius of base r. 8. Find the area of the surface of a sphere of radius r. 9. Find the volume of a cone of height h and radius of base r. 10. Find the volume of the cap of a sphere of radius r and height h. 11. Find the volume of a paraboloid of revolution of height h and radius of base r. 12. Find the volume of a prolate spheroid of 2a major diameter and 26 minor diameter. 13. Find the volume of an oblate spheroid of 2a major diameter and 2b minor diameter. 14. Find the area of the surface of a sphere lying be- tween two parallel planes distant a apart, the radius of the sphere being r. CHAPTER X Successive Differentiation "... an ancient tale new told And in the last repeating' troublesome." — King John. We saw in Chapter I that, if we start with a function of x, given by y = /H, we can usually differentiate f{pc) and obtain a new function of x, y}r{x), which is called the derivative or derived function. It is usual to denote this function by an accent, thus /'(x) is written for i/'(^)> i-^- ^/(-) =/'(-) (0 This is, of course, a symbolic equation, and does not have any specific meaning until we specify /(x). ^f fix) = ^, fix) = 4^, and so on for other functions. Now, there is no reason why we should not try to differentiate the new function just as we did the original one. For instance, if f{x) = x*, /' (x) = 4X^, and ^(4^) = 12^^ SUCCESSIVE DIFFERENTIATION 183 Equation (i) then gives We can denote the new function ^f\x) byf"{x) in conformity with the notation /'(^), so that i{iM-f"^^^ (3) In the illustrative case above /"(a;) would be 12^^. As a matter of notation, we can write -r-ifix) for hence ^2/W = /"W (4) This equation is merely a matter of notation. We have not said that — ^ ~ ^ ^ ~j~' T^"'^ equation would hold if dldx were a sym- bol for a. number, but it is not. It is a, symbol for an operation, see p. 40. f"{x) is, by convention, the notation we use for the function obtained by differentiating /(;<:) twice. The function f {x) is called the second derivative oif{x). In the same way we can express the third derivative either as ^3., goras/"'(^), and the «th derivative either as 5^; or /W(a;). dx'^ ' ^ ' i84 APPLIED CALCULUS This process is what is known as successive dif- ferentiation. Sometimes it is convenient to use the following notation : — the ist derivative is written ^j, M 2nd ,, ,, jVai >> nth. „ ,, yn. •(5) WORKED EXAMPLES 1. Find yn, if y = sin(ax-\-b). y.^^ = a cos (ax ■\-h) = a sin ( a:«r + 5 + -), j/j = f^ sin \ax + 6 + — ). j/j = a^ sin (ax + S + 3EV and yn = a" sin ( aa: + ^ + — )• 2. ^^ = a:i^ sin x, prove that We are given jv = a^ sin x. .'. y.^ =1 a sin ar + o:r cos Jf, and j/j = 2a cos x — a;r sin :r. .*. .T^2 — 2a;r^ cos x — aafi sin *, — 2xyy = — 2a;i; sin x — 2ax^ cos x, x^y = flA^ sin x, 2y = 2a:c sin x. Adding, x^y2 — 2xy^ + x'^y + zy = o, i.e. x^ -j^ — 2X-^ + (x^ -\- 2)y = o. UrX dX This last equation is called a differential equation.* * For some further information on differential equations, see p. 248. SUCCESSIVE DIFFERENTIATION 185 A solution of it is clearly y = ax sin x, for this value oi y satisfies the equation. 3. Suppose f(x) can be expressed as the sum of a series Aq + A-^x + Aj^x:^, wliere Aj, A,, and Ag are constants. It is required to express Ao, Aj, and Ag in terms of derivatives of any orders which may be necessary. Put f(x) = Ao + AjX + A2X^ {a) then /(o) = A„. Differentiate equation (a), then f\x) = A1 + 2A2A; {b) .-. /'(o) = A,. Differentiate (3), then /"(^) = 2A,. .-. /''(O) = 2A,. •••/U) =/(o)+/'{o)^+4^^^ W We have thus expressed the function as a power series in x, the coefficients of the terms of which de- pend on the values of the successive derivatives of f(x) sX X = o. Thus, if /(.r) = {a-^x)\ then f'{x) = 2(a -\- x), and /"(x) = 2. .: /(o) = a^ /'(o) = 2a, /"(o) = 2, and (a + x)^ = a^ + 2«.r + :if2 on substituting these values of _/(o), f'(o), and /"(o) in equation (c). i86 APPLIED CALCULUS The appearance of the higher derivatives in (a) Differential Equations, and (i) Series, should be carefully noted. (See Exer. lo.) They are constantly ap- pearing- in these subjects, and a clear idea of their nature and of why they appear is essential. Exercise 10 1. Let y = ji^ + jc^ + jf^ + ^+i. Show that — ^ = o. 2. \i y = x^ sin ax, a being constant, find^g. 2,. \i y = sin mx cos nx, m and n being constants, find y^. 4. Prove that the wth derivative of x^, when » is a posi- tive integer, is n\ (i.e. factorial n = n{ii — i) (« — 2) . . . 2.,). 5. \{ y = A sin mx + B cos mx, show that' g + ^^^ = o. where m is a given constant and A and B arbitrary con- stants. This is a most important result, and should be remem- bered. It is the differential equation of harmonic motion, and is perpetually occurring in physics and engineering. 6. Show that tan x satisfies the equation dx^ dx show that ^ = (-i)» 1.2.5. . .«;c-(» + i) dx^ ' ^ = (-i)"w! AT-C + i). SUCCESSIVE DIFFERENTIATION 187 8. Uj> = A + Ba; + C;«;2 + D^ + . . . Njc", n finite, show that A = j/^, \dxJo' C = JL (^) , 1 . 2 \dx^/o' N = -Lf^V where the suffix o denotes that the value of each deriva- tive is to be taken at x = o. [Differentiate y = A + Bx + Cx^ . , . successively, and put ;<; = o in the equations ^ = B + 2C^ + . . ., ax &.C.] 9. We can therefore write (Example 8), 2! 3! «! [/W(o) stands for the value of -^, when x is put equal to zero], provided /{x) can be expanded in a finite ascend- ing series of powers of x. f/n + 1 1; The necessary condition for this to hold is that <, and hence all higher derivatives, must be zero. Examine the case wheny(ji;) = x^. 10. Put x = a + y in the expansion in (9) where a is a constant, and find the expansion for/(«+_y) in powers of ^, by the method of p. 185. 11. Show, by using the expansion found in (10), that {a + x)^ = a* + ^a^x + 6a^x^ + ^ax^ + x^. 12. If _)' = (i 4- jf + x^y, expand y in ascending powers of ^. CHAPTER XI Curvature of Plane Curves " Curvature is the ' moreness ' of slope per inch of ' go forward '." Wp shall suppose that all the curves discussed in this chapter lie in one plane — the plane of the paper. A curve has, as a rule, a definite direction at every point in it, and, at any particular point, may be said to " point " in a direction which is the same as that of the tangent to the curve at that point. The direction usually changes from point to point, and the tangent line rotates as we move along the curve from a selected initial point. If J is the distance of the point in question from the initial point, measured along the curve, and ^ the angle the tangent at this point (s) makes with any fixed line in the plane, then is a function of J, (i) and -^ is the rate at which changes with j at the point (s) of the curve at which it is calculated. This derivative will in general vary from point to point. Deviation. It is convenient to have a word for the total change in direction which takes place as we go from one CURVATURE OF PLANE CURVES 189 point on the curve to another. Thus, in fig. i, if AQ, the tangent at A, is taken as the datum line, then ■yfr is the total change of direction between A and B, where i/r is the angle at which the tangent at B cuts Figr. I AQ. We will call the total change of direction the deviation, then, in fig. I, the deviation from A to B is y^r. Rate of Change of Deviation. Suppose we measure the length of the curve itself from A (fig. i). Select any point P between A and B, and draw the tangent at P (MN). The angle ^ at which this line cuts AQ, the tangent at A, is the deviation between A and P. Let J be the length of the curve AP, then d^ ds measures the rate of change of 96 with i- at the point (s) (see p. 65), i.e. it measures the rate of change of deviation with the distance along the curve at P. We will now seek a geometrical meaning of dlds to x, y co-ordinates. Choose axes as shown in fig. 4. (D102) 14 194 APPLIED CALCULUS Then tan = , > dx ^"'i'f^'^^-"^ = V'l + tan^0 = sec<#>. Also, by differentiating tan <^ = -^, we get dx dx^ , , d^ d ds I , Now -J- = -J- -r = - sec 9, dx ds dx p when p is the radius of curvature at P. sec* = _ aoif^o fl^ .^ ^ . p dx^ Also sec^ <^ = I + tan^ <^ ■■• J - g/[' + ©']' <* = curvature at P (x, y). Sign. The right-hand side of equation (3) may be taken with the positive or negative sign, as it involves the square root of Ti + i^) 1 , which is always positive. ^ CURVATURE OF PLANE CURVES 195 The correct sign to use can be easily seen from the shape of the graph, thus, in fig. 4, for instance, we have i = ^ p ds and in going from P to Q the tangent rotates in the counter-clockwise direction, hence A is positive, and, if we measure j in the direction from P to Q, Aj is positive, hence A/As is positive and must P be positive, so we take the positive sign with g/[- + m'f WORKED EXAMPLES I. Find the radius of curvature of a parabola x^ = ^ay at the vertex (o, o). Fig. s shows the graph of the given parabola. Y Fig:. 5 y = x^ 4a dy _ dx X za d^y I dx^ 196 APPLIED CALCULUS {■ + ©? I 2a (■+5) ■%\\ 2ai (4a2 + :^)f = — when X =■ o. 2a ,'. p = 2a when x = o. Geometrical Treatment of Curvature. Let PQ be a portion of a curve AB (fig, 6). Let PD be the tangent at P, and QC the tangent at Q. Let C be the intersection of the tangents at P and Q. Through P draw PO ± PD, and through Q draw QO JL QC, and let these lines meet in O. Let the tangent at P and the normal at Q (OQ produced) meet in D. Then .=i:i OPD has a right angle at P. Draw PE parallel to CQ, meeting OQ in E. Let the angle POQ be A o as Aj -*• o. CURVATURE OF PLANE CURVES 197 It is evident from the figure that PD > PC + CQ > Aj > chord PQ > PE. /. PD > Aj > PE. _L < i_ < J_ FD As PE' But Fig. 6 PD = OP tan A<^ = r tan A, and PE = OP sin A^ = r sin A^. • r I tan A^ < I A^ < I r sin A4'' • • A<^ < A o, then Lt r- r^l = Lt (i) X Lt (-^). A. — > oLr tan A. o\''/ a« — > ovtan A4>/ and as Aj — > o, A^ — > o. ... Lt (-^) = Lt (-^i~^ = I, by p. io6. A, _>, o\tan A<^/ A* -). oVtan A^/ •' *^ ., Lt P -^1 = Lt (i). AS _> oLr tan AJ a« — > o\^/ Similarly, Lt r- -^1 = Lt (i) As -> oL^ sm A oV'^/ by using the formula (ii) on p. io6, and j^^ A^ ^ ^_ AS — >. 0^ (i^' Hence, since the bounds in the inequality (a) both have the same limit, viz. Lt ( -), AS -^ o\r/ ... f = Lt (i) as As — >. 0^''/ = As^o(i) (4) Le. c?j/rf0 is the value of the limiting length to which OP approaches, as the length of the arc, PQ, between the points at which the normals OP and OQ are drawn, shrinks towards zero. This length is the radius of curvature of the curve at P, for we have defined the radius of curvature at P as the reciprocal of the curvature, d4>lds. CURVATURE OF PLANE CURVES 199 If As is sufficiently small we can write Ss = pS4>, (5) with sufficient accuracy for all practical purposes. Total Length of a Curve. The equation j enables us to express the total length of a curve between any two points A and B, as .(6) where <^o is the deviation at A and <^ atj B. If we measure the angle <^ with reference to the tangent at A, <^o = c. pcl4>. J ■(7) Graphical Methods. A graphical example will help to crystallize the conclusions so far reached. Draw any smooth curve ACDB, about 12 in. long, on a sheet of drawing paper. Step off points Pj, Pa, P3, &c., with dividers set with the points J in. apart. Draw the chords AP^, If these chords are practically indistinguishable from the given curve, we can take the length of the curve between any pair of points as the length of the chord between the points. Draw tangents at A, Pj, Pa, P3, &c., and draw lines 200 APPLIED CALCULUS through A, Pj, P^, Pg, &c., perpendicular to these tangents, each to each. These tangents must, in general, be drawn by eye. With care, they can be drawn with fair accuracy. These straight lines are the normals to the given curve at the respective points. Let the normals at A and P^ intersect at Qj. Measure PiQi; then the length of PiQi is, approximately, the radius of curva- ture at Pj. Strike the arc APjPg with Qj as centre and radius PiQi. This circular arc should be indis- tinguishable from the actual curve between A and Pj, if the length PiQi is a good approximation to the true radius of curvature at Pj. Plot this value of p on a graph of p in terms of s; this is the value of p for i- = chord APj, if we measure the length of the arc from A. Repeat this process for each of the points Pg, Pj, P4, . . . P„, and plot the values of pi, p2> • • • /»« in terms of J. See fig. 8. A curve giving the radius of curvature (p) for given values of s, between Pj and P„, is thus obtained. In plotting- this graph, meantime omit the values of p at the end points A and B, as these points require special treatment, since the curve begins at A and ends at B, and the curve does not have an ordinary tangent at these points. CURVATURE OF PLANE CURVES 201 This curve (fig. 8) shows clearly how the radius of curvature changes as we go along the curve. At certain values of i- (OEi, OE2) the radius gets very +p Fig. 8 great, and changes sign in the neighbourhood of these points. This means that as the corresponding points El, E2 (fig. 7) are approached and passed, on the actual curve ACDB the deviation of the curve 202 APPLIED CALCULUS ceases to increase (decrease) and begins to decrease (increase). We can avoid these very great values of p by plotting, not p, but -, i.e. by plotting curvature and not radius of curvature. The curvature can be obtained from flg. 7 directly. Suppose A^ is the ang'le between the tangents at Pi and P:. Then the deviation in traversing the distance PiP: is A^i, and tlie average curvature between Pi and Pj is ( "^ j radians per unit length. If the step P1P2 is small enough, this average value may be taken as the value of the curvature at Pi. If P1P2 is i in., for instance, then the curvature at P] is approxi- mately I -X? 1 = 2 A^i radians per inch. The angle (A0i) is easily measured in radians by striking the arc of a circle of radius i in. with the centre at the point of intersection of the tangents at Pi and Pj. Measure the arc of this circle which is subtended by A0i; then, if 5i is the length of this arc in inches, A(^i = Si. The chord corresponding to the arc S, is sufficiently nearly equal to the arc Si, if the angle A0i is sufficiently small, as it will be, if the steps PiPj, P2P3, &c., are sufficiently small. The curvature at P is therefore given approximately by - = (2 X chord Si) radians per inch, P in this instance. If the steps P1P2, PjPs, &c., are all equal, the curva- tures at successive points Pi, P2, P3, &c., are proportional to Si, S3, S3, &c. In fig. 9, - is plotted from fig. 8. The work can now be checked by integration. Let T be any point in the graph of - in terms of P s (P in fig. 7 corresponds to this point). CURVATURE OF PLANE CURVES 203 The area shaded, reckoned positive when above the axis and negative when below it, is J OV.P But 'op,P d I ds P OP -ds = (< OP, P .(8) Fig-. 9 If, then, the work is accurate, the area shaded should be algebraically equal to the total deviation between Pj and P. This method is, at best, only roug-hly accurate as it is difficult to draw the tangents at Pi, Pj, Ps, &c. (fig. 7), accurately with the eye as the only guide. See p. 68, where the same diificulty is referred to in connection with the g-raphical problem of determining instantaneous velocities from a distance-time curve. The method shows, however, the general way in which the curvature I - ) varies from point to point, and can be applied to anjy smooth curve, whether we know its mathematical equation or not. The mathematical methods already described cannot be applied till we know the algebraical equation ao4 APPLIED CALCULUS to the curve, and even when we do, the finding- of curvature, analyti- cally, is often a tedious proceeding-. But, as a compensation, the analytical method is theoretically accurate. Physical Applications. We shall conclude this chapter with some examples of physical applications of the preceding theory. I. Motion in a Curve. — (a) Velocity. Suppose a particle P moves on a curved line AQ. Let s denote the distance of the particle, measured along the curve from some selected point A, so that f is the length of the arc AP. Fig. lo Let the co-ordinates of P be x, y. Suppose the particle moves from P to P' in time M. On p. 57 we discussed the average velocity of a rectilinear motion. In rectilinear motion there is no change in the direction of motion except such as can be dealt with by the sign of the speed (or magni- tude) of the motion. The speed is positive when in CURVATURE OF PLANE CURVES 205 one direction, and negative when in the only other possible direction — the opposite direction. In general, the direction of a velocity must be specified, as well as its magnitude, just as the direction of a dis- placement must be specified, as well as its amount. A uniform velocity must therefore be not only con- stant as regards its speed (magnitude), but must also maintain a constant direction. With this extended idea of uniform velocity, we can deal with the velocity of a particle on a curve. , The average velocity between P and P' is evidently a velocity of magnitude /chord PP\ in the direction 6, the angle 9 being the angle PP' makes with the OX axis. At P draw the tangent, cutting OX at the angle 0. Now, let iU — > o. As A^ — ► o, the chord becomes infini- tesimal, and the chord PP' and the arc PP' differ only by infini- tesimals of higher order than the first, so we may put 8j instead of PP', when A^ — >■ o. Further, as A^ -> o, fi — > <)>, the direction of the tangent at P. Hence Lt (PP') = % Ae y 0^ A< / at and Lt {6) = ^, At — > and these limits define the instantaneous velocity of the particle. The instantaneous velocity of the particle at P is ds\dt, and its direction is that of the tangent at P. (b) Acceleration. Choose the tangent and normal at P as axes, as shown in fig. II. At P, let the velocity be v along OX, and at P', (z» -f- Ad) along TP', the tangent at P'. 2o6 APPLIED CALCULUS Tangential Acceleration. The component of {v + hv) along OX is (v + Az>) cos ^. Hence the acceleration at P, along the tangent at P, (/j), is given by f _ T. Viv + At)) cosA^ — v "] •^' At^oL A^^ J- Now, I > cosA0 > I — i(A<^)^ seep. 105. .*. we can put cosA^ = i — ^/c(A0)^, where k is a positive fraction. . (v + Att) cosA0 — V _ (v+^ Av) ( i — ^ k(A^)') — v ■ ■ At ' At _ v + Av — ^ /ro(A^)'— ^ KAv(Atl>Y — z; ~ At + Av Fig. II If we retain only infinitesimals of the first order, this expres- ^'°" &'^«^ (^ + At,)cosA^-z> _ At, Lt I At At' (v + Av) cos A0 - - v' = dv At -1 dt' and since v = ds dt' dv d^s dt ~ dt^- dv d^s Hence the acceleration along the tangent at P is ^r or -j-rr. at at CURVATURE OF PLANE CURVES 207 Normal A cceleration. Now, resolve (f + Azi) along the normal at P. The normal component is [v + kv) sinA^. Hence the normal acceleration at P is /• = Lt r (^ + Ap) sinA0 'l " At -> oL bd J _ T . Til + Ax» sin A0 A0 AjT At — > oL i* A0 At A^J' Taking the limit of each of these four products in turn, we get Lt {■"Jr^\ = I, At — > oV ■V I Lt (v) = V, At — > Lt CilE^) = ., At — > 0^ A0 / Lt ( T^) = -, the curvature at P, At — >. oV Aj/ (0 and Lt (^) = I = z-. At > o^Ar/ ar •'• /n = I X n X I X - X zi = -. P /> Hence the normal acceleration is the square of the speed divided by the radius of curvature. The results are : , ,, f _ dv _ d^s •'' ~ ~dt ~ dfi' p There is therefore always a normal acceleration when a particle moves along a curve. There is also a tangential acceleration if its speed changes from point to point. The normal acceleration arises from the curvature of the path, and therefore from the change of direction 2o8 APPLIED CALCULUS of the motion ; the tangential acceleration from the change in speed. Example. — A particle moves at a uniform speed round an ellipse. Find how the normal acceleration varies from point to point. Let the centre of the ellipse be the origin, the major axis the OX axis, and the minor axis the OY axis ; then (see fig. 4, p. 9) the equation of the ellipse is S+P='' (9) where a is the semi-major axis and b the semi-minor axis. Equation (9) gives jr _ x' Differentiating both sides with regard to x (y being a function of :>;), we get , ' " 2ydy__2x Fdx~ ^" . dy b'^ X " dx a^ y and ^ = - ^'i - ^(-}-'t\ dx^ a^y a'' V y^ dx) = _ il _ ^(- 1 X - - -^ a^y a^ V _j/' a^ y) c^V L a^yU ly \. ay'. _ _ ^ r ay -I- 5V [ a^y L a^y'^ J" a^yi -j. l^x^ = (^jfl by equation (g). • ^ = — Z ^ dx^ c^y c^y"^ ay CURVATURE OF PLANE CURVES 209 Now, i = ^___byequation(3), p. 194. • I _ ay a*4* If the speed of the particle is v, we get ^" ~ (^^vTsM* and this acceleration is directed towards the interior of the ellipse. At the end of the major axis (a, o) /I - -p-' and at the end of the minor axis ^■^ ~ ^• The normal acceleration anywhere else {x, y) is given by equation (10). The tangential acceleration is zero as — is constant, by hypothesis. 2. The Bending of Beams. Let AB be an I girder bent in one plane by forces in that plane, as shown in fig. 12 (a), and supported by knife edges at A and B. Choose rectangular axes as shown in the figure. Let 00"0' be the line passing through the centroid of each cross-section of the girder. This line is called the axis or centre line of the beam. (D102) 15 APPLIED CALCULUS Suppose the cross-section of the girder is as shown in fig. 12 (b). When the girder " hogs", filaments of material originally parallel to the axis (supposed Fig. J2 initially straight) stretch if they are above the axis and shrink if they are below it, provided the axis itself does not stretch nor shrink. The portion of the beam to the left of PQ is held in equilibrium by the applied system of forces acting on that portion, plus the stresses acting in the cross- section PQ, arising from the other portion of the beam. These stresses can be reduced to (i) a force T, tangential to the axis; (2) a shearing force N CURVATURE OF PLANE CURVES 211 — these forces acting at the centroid of the cross- section; and (3) a couple M. These three quantities can be found by considering the statics of the portion of the beam on the left of PQ. The stress system at any cross-section, distant j from O (measured along the axis), is therefore determinate. One of the problems we often wish to solve is to find the equation of the deflected axis. This equation will clearly depend on the dimensions and the elastic constants of the beam. To determine it we have to express the couple M — the bending-moment, as it is called — in terms of the dimensions of the beam and the elastic constants. We shall suppose that T is so small that the axis itself does not change sensibly in length. Let C be the centre of curvature of the axis at P. Consider a small length of the beam Ss (PQSR), and let us calculate what happens to the filaments of material distant jj/ from an axis through the centroid of the section and perpendicular to the plane of bending (fig. 12(6)). Let ^S be the cross-sectional area of the filament at y. The original length of this filament is Ss; the stretched length is (p + j')S'i>, ignoring infinitesimals of higher order than the first. But Ss = pS to the same order of accuracy. Hence the strain of the filament is (,P + y)S - pS ^ y_ pS p If E is Young's Modulus, the stress in this element ^2-. Hence the tension in it is — ^S. P P 212 APPLIED CALCULUS Taking moments about an axis perpendicular to the plane of bending and passing through the centroid of the section PQ, we see that the contribu- tion this filament makes to the total couple is P The whole couple is therefore i^-^-l P J -I P where w is the thickness of the beam at y (fig. 1 2 (b) ). .-. G = -f^'wy^dy, pJ -? where G is the total couple. But I " Tvy^dy is the moment of inertia of the cross- '' . section, assuming it to be of unit surface density. Calling this quantity I, we get G = EI. P This couple must be numerically equal to M. .-. EI = , M, P dy dx^ 'him i.e. EIt ^^r-rr-, = ± M. If the deflection is never very great, dyjdx is small CURVATURE OF PLANE CURVfiS 213 relative to unity, and much more is {dyjdxf. Hence we can neglect (dy/dxy and put EI^ = ± M. Sign. The sign to be given to the right-hand side of the equation ^2^ dx^ is settled by a convention. If the beam bends as shown in the figure, the bending-moment, M, arising from the portion of the beam on the rig'/ii of PQ is clearly clockwise, i.e. negative, with the usual convention that counter- clockwise couples are positive. But g is negative too, since g = ^(g), and the inclination of the axis of the beam \-j-) decreases as we proceed along the beam in the direction of x d?v increasing; thus EI-^ and M have the same sigh, i.e. EI^ = M. dx'' This is the equation that gives the deflected axis of the beam. The above is the Bernoulli-Euler theory of the bending of beams. A state of strain is assumed defined by the properties : 1. The axis is unaltered in length. 2. The cross-sections remain plane, undistorted, and perpen- dicular to the axis. 3. There is no extension or contraction of fibres, perpen- dicular to their length. 2I4 APPLIED CALCULUS The mathematical theory of elasticity shows that the actual state of strain, even under simple distributions of load, does not exactly fulfil the above conditions. In calculating the moment of resistance of the beam arising from these strains, we further suppose that the straining of the longitudinal filaments is not resisted by shearing stresses along the sides of them ; in other words, we suppose the beam to be equivalent to a bundle of free wires placed parallel to the axis. The exact theory shows that when any linear dimension of the cross-section is not more than, say, ^ of the length of the beam, the formula for the deflection obtained on the simple theory given above is sufficiently correct. Example. — A uniformly loaded beam, supported at both ends, carries a load of P lb. per foot run. Find the maximum de- flection. Let / be the length of the beam (fig. 13), OX, OY, the axes chosen in the usual way, where OX coincides with the axis of the unbent beam. Y>. R. i>R/ >P Vx A/\/V\/\/-./\/\/\/\/\/\/\A/\/\/\A O Q f^ ni-A Fig. 13 The forces keeping the beam at rest are the loads acting ver- tically downwards, plus the reactions at O and O'. Symmetry. — From the symmetry of the figure we see that: (a) The reactions at O and O' must be equal, each being \Vl, and they must act vertically upwards. (5) The maximum deflection takes place at the mid-point of the beam, and is negative (downwards). (c) At the mid-point (where the deflection is a maximum) ^ = 0. dx CURVATURE OF PLANE CURVES 215 Boundary Conditions. — It is also clear that j' = o when x =^ o, or when ar = /. Consider any section PQ at x. The moments of the forces to the right of this section are, in the counter-clockwise direction, 2 .-. M = [\vi{i-x)-\ni-xn Integrate this equation, and we get Ei^ = ^pr_ /(/-^)^ + (ir^n + c. uX l» 2 ^ J where C is a constant. We can determine C by noting that -i = o at the mid-point, i.e. when x = -. dx 2 dx = ipr-?4 24. 24' C_P/' • T7T^ - i,X>[- Kl-xf , (l-xf \ , P/3 dx * L 2 ^ 3 J ^ 24" Integrating this equation again, we get Ei^ = iPrfc^-(izi^n + p^,+D, U o 12 J 24 where D is a constant. To determine D we use the fact ihaiy = o when *■ = o. 2i6 APPLIED CALCULUS Let 5 be the maximum deflection, which is the value of y PI* corresponding to j: = _ 2 •••- = K5-5|]-S = -4- p^*- 384 If W is the total load carried, W = ?/, and S = §_ . 384 EI The negative sign shows that the beam is sagging. Exercise 1 1 1. In the cubical parabola, where a^y = X?, show that the radius of curvature is 6a*x 2. In the semi-cubical parabola, aj/^ = x^, show that the radius of curvature is (4fl + gx)i r~ 3. Show that the radius of curvature of the curve y = A(, + Ai^r + k^^ + . . . A„jc", n finite, where the A's are constants, at it = o is [i + A,'']! 2A2 CURVATURE OF PLANE CURVES 217 4. Find the radius of curvature of the hyperbola at the end of the major axis. 5. Show that if a particle of mass th lb. is attached to the end of a string and swings round in a circle of radius a ft., at constant angular velocity w radians per second, the tension in the string is lb., where g is the acceleration due to gravity (32-2 ft. -sec. 2). 6. The equations of a cycloid referred to the vertex as origin, and the tangent and normal at the vertex as axes OX, OY respectively, are X = « (61 + sin^), y = a{i — COS0), where 6 is the angle turned through by the generating circle of radius a. Assuming these equations, show that the Inclination of the tangent (^) to the OX axis at any point {x, y) on the cycloid is given by tan^ = tan-. 2 Hence show that the arc of the cyloid (s), measured from the vertex, is given by .r = 4« sin^ = xj^ay. Show that the radius of curvature at P is given by ^ = ^V' - ^' which is twice the distance of P from the base of the cycloid, when measured along the normal at P. [The base is the line uv in fig. 3, p. i6g.] 2i8 APPLIED CALCULUS 7. A particle of mass in oscillates under g-ravity in a cup. The cup is formed by the revolution of a cycloid about the normal at its vertex. Show that if the particle starts from rest at a distance A (measured on the surface of the cup) from the lowest point, the particle oscillates according to the equation J = A cosfc)/, where w is a constant. Show that 0) is equal to y/{gj^a). Find the reaction on the cup at any point, using the results of the previous example. 8. A straight beam of uniform moment of inertia and negligible weight is bent by a couple applied at each end, the couples being coplanar. Show that the centre line deflects into the arc of a circle. Find an approximate equation for the deflected axis in the usual rectangular co-ordinates. State generally how the shape of the curve is modified, if the weight of the beam is not negligible. 9. A uniform beam is bent in one plane by a single load TO placed at its centre. Find the approximate equation of the deflected axis. What is the central deflection? 10. A straight wire of circular cross-section is clamped horizontally in a vice. A load w is applied at the free end. Assuming this to be the only load to be considered, find the approximate equation of the deflected axis and the deflection at the end. CHAPTER XII Maxima and Minima "And so, from hour to hour, we ripe and ripe, And then, from hour to hour, we rot and rot ; And thereby hang^s a tale." — As You Like It. The graphs of certain functions of x, say sinx, rise and fall as we proceed along them in the direction OX, and exhibit points at which the value of the function is greater than, or less than, its value at neighbour- ing points. Such values of the function are called " turning values ". Turning Values of Functions. Fig. I Let ABCDE (fig. i) be the graph of a continuous function oi x, say/(x); then jv = f{x) (i) 219 220 APPLIED CALCULUS At the point A corresponding to X = a, the ordinate y ceases to increase, and begins to decrease immediately afterwards, as we go along the graph in the direction of "x increasing", i.e. the positive direction. At B, the ordinate ceases to decrease, and begins to increase thereafter. The ordinate steadily increases in going through the point C. At D, we have the same state of things as at A, so far as decrease or increase of ^ is concerned. Points such as A, B, and D are "turning points", and the values of jj/at these points are called " turning values " of the function. Points such as A and D are called maximum points, those such as B, m,inimum points. These terms are used in a relative sense only; for instance, if we say that A is a maximum point, we do not mean that j)/ is an absolute maximum at this point, for it is greater at D. What we do mean is that at A we can choose an interval within which the value x = a \s included, such that for any value of X (other than a) lying within this interval, the value f{x) is less than the value f(a). We must clearly exclude the value x = a, 2is, aX this value f{pc) of course equals y(a). Condition for a Maximum Point. We can put these ideas very simply into symbols. \i y is a maximum aX x — a, we can choose a positive number 5 so that so long as f{x)^ = o at such maximum or minimum values oif{x). The reader should note that -^ may equal zero at points which are neither maximum nor minimum points, e.g. at the point C (fig. i). He should also observe what takes place at points like D and E. At the maximum point D the two branches have the same tangent, the gradient of which (tan 90°) is infinite; at the mini- mum point E the two branches have different tangents. Both at D and E, therefore, the gradient of the tangent is discontinu- ous. The rules now to be given will enable us to find all turning points, such as A and B. If maximum and minimum points like D or E occur, which is extremely unusual, they have to be found by special methods. EXAMPLE / Find the turning value of 3^(x — i) between x = o and x = \, excluding these values. Method a. Ifj, = ^2(;,_ I) =/(;,), '^y = 2x^- 2X. ax Put 3*^ — 2:r = o, whence :r = o or ;i; = f , i.e. between X = o and ar = i, -ji = o when x = i. ax The value x = J may therefore correspond to a turning value of x^x — I ). To test this, /(I 4. 5) = (I + 8)2 (I + a - i) = s'+s'-A, y(«) = - 5V. .-.yd + «)-/(«) = s^ + «^ MAXIMA AND MINIMA 223 and 82 + S' is positive for all values of 8 > — i, i.e. so long as o < |*-f I < I, We have therefore found an interval including x = %, such that /(x) > /(I) in this interval of x, i.e./{x) is a minimum eX x = %. Method b. — We can arrive at this result by another method. 1. y = o, when x = o and x = 1. 2. Let X ^ K, where k is a positive fraction, then /(.) = .\. - I), K^ is positive and (/c — i) is negative. Hence /"(t) is negative, i.e. f(x) is negative for all values of j; between o and i. 3. There is one value only of x between o and i which makes dyjdx equal to zero, i.e. the tangent to the graph is parallel to OX at :i- = I only. 4. On the other hand, there must be one value of x, say f, at least at which x\x — i) is a minimum between at = o and jr = i, since x\x — i) is zero at j; = o and x = 1, and negative between these values of *■. 5. Consequently this value of x must be that at which -2 = o, i.e. f = f . .". x^x — i) is a minimum aX x = %. Another Form of the Maximum-Minimum Test. Consider the tangent at P (fig. i), and trace what happens to it as the point of contact P moves along the curve through A to Q. The tangent line evi- dently rotates in a clockwise direction, becoming parallel to OX at A. Clockwise rotation means a decrease in the angle of inclination 0; hence tan must decrease as P moves through A to Q, i.e. ^ must decrease steadily between P and Q. Q is dx clearly positive at P, zero at A, and negative at Q; hence 224 APPLIED CALCULUS At a maximum point, ^ changes sign from ax positive to negative, as x increases through its critical value, i.e. \i f{x) is a maximum when x = a, then -^ is positive when X < a and negative when x > a. ^ Similarly, at B, we see that: At a minimum point, ^ changes sign from negative to positive, as x increases through its critical value. EXAMPLE 11 Let y ^= (x — \)(x — 2)^. /.g = (.-.) (3.. -4). which is zero when ar = 2 and when x ■= ^. We are therefore led to consider these values of x as possible maximum or minimum points. When ^ < X < 2, (x — 2) {3Jr — 4) is negative. When :ir > 2, (x — 2) (3^; — 4) is positive ; hence at a: = 2, -^ changes from negative to positive as x increases, y is therefore a minimum at j: = 2. Similarly, when x < ^, (x — 2)(3.r — 4) is positive, and when ^ < X < 2, {x — 2) (3^ — 4) is negative, whence dyjdx changes from positive to negative as x increases through the value x = ^. .'. y is a maximum at :i: = J. The reader should draw the graph of y = (-^ — 1) (^ — 2)', and confirm these conclusions. MAXIMA AND MINIMA 225 Analytical Test of Maxima and Minima. There is yet another way in which this test may be put. We have already seen that -^ = o at a turning point. Suppose that at the turning point {x = a), -3^ is positive. Then 3- {-^) is positive, and when the derivative of a function is positive, for x = a say, the function itself increases as x increases, through a. Therefore -f- must increase as x increases from ax a — h to a + A, provided k is sufBciently small. But dx is zero when x = a, hence -y- must be ax negative when x is just less than a, and positive when X is just greater than a, i.e. -^ changes sign from negative to positive as x increases through a, and hence jj/ is a minimum. The rule is, therefore: If dyjdx = at X = a, and d^y/dx^ is positive at ^ = a, then 3; is a minimum at x = a. Similarly: If dy/dx = at X = a, and ePy/dx^ is negative at X == a, then y is a maximum at x = a. If d^yjdx^ is zero when x = a, this test fails, and we then have to revert to the general tests already given. (D102) 16 226 APPLIED CALCULUS EXAMPLE /// Let y = x^— 15* + 3, then -^ = 2x — 15, which equals zero when x = 7^, ax -j^ = 2, which is always positive. .*, the minimum value of x^ — i5:f + 3 occurs at a: = 7 J. EXAMPLE IV Let _j/ = A^, dy die = y<:\ and ^, = 6x. ax- Both derivatives are zero at a: = o, and the analytical test fails. But we can see from the following considerations that there is no maximum nor minimum. It is evident that x^ steadily in- creases as X increases from zero, and steadily decreases as x de- creases from zero, ifi has therefore no maximum nor minimum value for any value of x. EXAMPLE V Find the fraction which exceeds its square by the greatest amount. Let x be the required fraction; then y = x — x^fnust be a maximum. This occurs when x = \. We shall now discuss some problems in maxima and minima which arise in physics and engineering. EXAMPLE VI In an electrical transformer the losses of energy arise in the heating of the copper conductors (the "copper losses") and of the iron of the core (the "core losses"). Assuming that the total copper losses are proportional to the square of the secondary current, and that the core losses are constant at all currents, MAXIMA AND MINIMA 227 show that the transformer works at its maximum efficiency when the core loss is equal to the copper loss, assuming the power-factor of the load to be constant. [The power-factor of an alternating-current circuit is the factor by which the product of pressure and current must be multiplied to get the power taken by the circuit ; for instance, if a coil of wire carries an alternating current, I amperes as read by an ammeter, and if the pressure across the coil is E volts as read by a voltmeter, the power consumed by the circuit is not EI watts, EI cos<^, where cos

-26n I 30 — 1/ J approximately, between the limits of 26 in. and 30 in. of vacuum- MAXIMA AND MINIMA 231 We will suppose, for the sake of getting a first approximation, that the power taken by the auxiliary plant, both in the boiler- and engine-houses is constant at a per cent, and that the evaporative power of the boilers is constant at q lb. of steam per pound of coal. The effect of vacuum on the steam consumption of the turbine is given in a set of curves published by W. H. Wallis. These curves show that between 26 in. and 29 in. of vacuum, the per- centage reduction in steam consumption {p), is given by p = ^(v — 26) nearly. If TV is the steam consumption per kilowatt-hour at vacuum V, and 7f'2i; at 26 in. of vacuum, then w = w^al I — 0'04(z< — 26)}. It must be clearly understood that this formula is an empirical one which holds between 26 in. and 29 in. of vacuum approxi- mately. Between 29 in. and 30 in. the curve of w against v must clearlv bend round and become more or less horizontal because a point will be reached, with increasing v, when the specific volume of the exhaust steam is so great that the exhaust end of the turbine will become choked, and any further increase of vacuum simply goes in overcoming increased exhaust pipe and condenser friction. We will suppose the coal costs £K per ton, and that the plant is to run at full load only and to be so run for H hours per annum. The interest and maintenance charges in the condens- ing plant can be put at r per cent per annum. 1. The capital charges amount to r\( ^ o-56(z,-26)|^ ^^^^^ 100 1. (30 — 11) J 2. The units of electrical energy (Kelvins) available per annum are \ lOO' 3. The steam consumption per kilowatt-hour is ■w = Wii;{i — 0'04{v — 26)} \h. , 232 APPLIED CALCULUS and the cost bill is r P..^{i-o.o4(^>-26)}HK-|^ ^„„„,„ I- 2240^ J 4. If we put, for brevity, _ o-fi6 X 240^. PHfi- -^)x 100 \ 100/ and P = '"J"^^^ , (l - -^)2240^ \ 100/ we get for tlie cost, per unit of electricity available, arising from these items (i) and (3), in pence (^), 0-56 1 (30 — ") ■' 5. At the most economic vacuum, y must be a minimum, whence , i.e. af-i— + ^^1:141-0.04^ = o L30 — V (30 — vY-i o-oip i.e. (30 — 7)) = + . / — °^—-. V must be less than 30. •^ V O.QI/S Substituting again for a and /S we get V o^oi^ ^ ^^"^Va'aeHKP and. = 30-3S-4y^^;f^. It must be borne in mind that this example takes account of MAXIMA AND MINIMA 233 two factors only — the economy arising from a reduction in steam consumption following an improvement in vacuum, and the cost of providing that improvement in condensing plant and its main- tenance. There are many other factors which would be taken into account in engineering practice. Some of these are : 1. The reduction in water rate which follows an improvement in vacuum. This brings about a reduction in the size of the boiler plant and of the main steam piping. 2. A smaller circulating water system and smaller circulating water pumps would follow an improvement in vacuum. 3. More power would be taken by the air-pumps or their equivalent at higher vacua. 4. The low-pressure end of the turbine would be larger and more costly at higher vacua. g. The boiler evaporative power would fall off with higher vacua. The first two points would lead us to use a higher vacuum, the last three, a lower one. The problem can only be adequately thrashed out — like most economic problems — with a complete and accurate knowledge of the data and circumstances of each case, and the investigation is best done arithmetically and graphically. These complications notwithstanding, it has become the usual practice to work at a vacuum of about 29 in. in turbine-driven power stations where there is a copious supply of cooling water for the condensers. Exercise 12 1. Discuss the turning values of (i) jj' = ^ — 6x^ -\- gx — 6. (ii) y = xfi - is,x^ -I- 3. 2. What is the radius of the largest cone that can be put inside a sphere made in the form of two accurately- fitting hemispherical cups? 3. How should a rectangular piece of paper be cut so as to fold into a box, without a lid, of maximum volume? 4. Find for what values of d, sin^ -f- COS26 has turning values, and ascertain which are maxima and which minima. 234 APPLIED CALCULUS 5. The corner of a leaf of paper is turned back so as just to reach the other edge of the page. Find when the length of the crease is a minimum. 6. Two ships are sailing uniformly, with velocities u and V, along straight lines converging at an angle 6. Show that if a and b be their initial distances from the point of intersection of the courses, the least distance of the ships apart is [av — lm)s\nd (u^ + v^ — 2UVCOS&)^' 7. The path of a projectile is given approximately by 4A cos^^' where y is the altitude of the projectile, x the horizontal distance of it from the point at which it is fired, and 6 is the angle of inclination of the axis of the gun to the horizontal. A is a constant, depending on the initial velocity of the projectile. Find for what value of 6 the horizontal range is a maximum. This equation ignores, among other things, air resistance. Would you expect your result to be greater or less than the more correct result, in deriving which air resistance is allowed for? 8. In the example of the soap bubble (p. 229), suppose the bubble carries a charge of electricity e. How would this affect the internal pressure and by how much? [The potential energy of a sphere of radius r carrying a charge e is where a is a constant.! r g. Referring to the condenser problem worked out on p. 230, show that the best vacuum to employ is that at which the rate of increase of the capital charges, per annum, with vacuum equals the rate of decrease of the coal bill, per annum, with vacuum. Work out completely the case when the capita! charges and maintenance of condenser plant are put at 10 per MAXIMA AND MINIMA 235 cent per annum ; the cost of the condensing plant for a io,ooo-kw. plant, ;^25,ooo at 26 in. of vacuum ; the evapora- tive power of the boiler plant, 7-5 lb. of steam per pound of coal ; the steam consumption per kilowatt hour at 26 in. of vacuum, 13 lb.; the hours worked 4380 per annum ; and the cost of coal £1 per ton. 10. The velocity, v, of the cross head of a reciprocatingf single-cylinder steam-engine is connected with the angular velocity of the crank by the approximate equation, V = (Df [s'lnoii + — - sin2(o/], where to is the constant angular velocity, r is the crank radius, I the length of the connecting rod, and i is the time measured from the moment when the piston is at the back of the cylinder and just beginning to move outwards. Find for what value of i the velocity of the cross head is a maximum. Calculate this maximum velocity when the speed of the engine is 75 r.p.m., the length of the connecting rod 12 ft., and the crank throw 2 ft. What is the position of the crank at the moment of maximum velocity of the cross head? CHAPTER XIII Exponential and Logarithmic Functions " In order fully to appreciate the brilliancy of Napier's invention and the merit of the work of Briggs and Vlacq, the reader must bear in mind that even the exponential notation and the idea of an ex- ponential function, not to speak of the inverse exponential function, did not form a part of the stock-in-trade of mathematicians till long after- wards. The fundamental idea of the correspondence of two series of numbers, one in arithmetic, the other in geometric progression, which is so easily represented by means of indices, was explained by Napier through the conception of two points moving on separate straight lines, the one with uniform, the other with accelerated velocity. If the reader, with all his acquired modern knowledge of the results to be arrived at, will attempt to obtain for himself in this way a demon- stration of the fundamental rules of logarithmic calculation, he will rise from the exercise with an adequate conception of the penetrating genius of the inventor of logarithms. " — Chrystal, In many books of mathematical tables, tables are given of (a) the exponential function, C, (b) the logarithmic function, loge^. A short table of each function is given at the end of this book. The exponential function of x \s the number ob- tained by raising a certain irrational number e = 2'7i828 . . ., to the power x, for real values of x. The logarithmic function of x, or the natural loga- rithm of X, is the index of the power to which e must be raised to give the number x. LOGARITHMIC FUNCTIONS 237 In elementary algebra, we have if JK = logeX, then X = e'o (i) The theory of these functions is difficult, and proved a formidable stumbling-block in the history of mathe- matics. The theory is best worked out by methods beyond the scope of this book, and for the present we shall take the tables for granted, and endeavour to ascertain the most important properties of the func- tions from these tables. For any pair of values x and y in the logarithmic table there is a pair of values in the exponential (anti- logarithmic) table, such that \i y = \ogeX, (2) then X = e", (3) in fact the two tables are inverse to each other; i.e. in the one, y is regarded as a function of x, which is written \ogeX, and in the other, x is regarded as a function of ^, written e^, where jv is an index obeying the ordinary laws of indices and e a certain number — the base of the natural logarithm. Graph of e^. A glance at the table of e" will show that: (i) whenjv = o, ev = ^ = i; (2) when y is positive, e" is positive and greater than i; (3) when jv (= — p) is negative, e'P = — = a positive fraction. Plot a few values of e^ for different values of y, taking values of e" from the table. 238 APPLIED CALCULUS The graph will be found to be as shown in fig. i X 9 10 / 8 / 7 1 1 6 5 / X 4 1 c e" 3 / 2 / f 1 y / — ■ -^ -^ 1" - Fig. I 3Y From the tables or the graph it appears that e^ is never negative. Graph of loge:v. The table of loge^ shows that: (i) when X = 1, \ogeX = y = o; (2) when X > I, logeX is positive; (3) when X < I, loggX is negative. These facts are also evident from fig. i. By plotting a few values of loge-ac for different values of x, the graph will be seen to be as" shown LOGARITHMIC FUNCTIONS 239 in fig. 2, from which it appears that we cannot have the logarithm of a negative number. Draw these two graphs on tracing paper, and show that they can be superimposed on each other. They are therefore the Y ^ — - ^ _^ ^ j> ^ ^y ^'09, .X / / / / 1 2 3 4 ^x 6 7 8 9 X / ' -2 Fig. 2 same curve, but the orientation of the graphs with regard to the axes is different. These graphs are drawn as smooth regular curves, and, on looking over the numbers given in the tables, there is nothing to suggest that irregularity will be found in any properties possessed by the curves. One property which a curve has, as a rule, is a definite slope at each point, and it appears that both 240 . APPLIED CALCULUS the graph of e" and of loge^sc have definite tangent lines and therefore definite slopes at every point on them. We are thus led to consider the derivatives of e^ and \ogeX. Derivative of e?'. Let X = e", then Ax = ev + ^v - e» = gve^y — gw = eyie'^y - i). . ^ - M^^" - ') 1 •• Ay ~ 'I Ay J .-. Lt (^) = ..xLt P-^1 (4) We therefore require the value of which the reader will see is simply the value of j- at the point (i, o), by the definition of the derivative. This limit is difficult to find rigorously. We can easily see what its value must be from elementary considerations. I. Plot a few points of the curve ^ = e'' at intervals ofo-02from^ = — o-ito^ = + o-i, and draw the curve through these points. Draw the tangent to the curve aXy = o, i.e. at the point (i, o). It will be found that the tangent line makes an angle of 45° with the axes OX and OY, /dx' I.e. . (f) = tan 45° = I. LOGARITHMIC FUNCTIONS 241 But, as we have already remarked, the value of is simply the value of dxjdy at the point (i, o), which we have just seen is i ; ■(5) 2. Alternatively, we can deduce the value of the limit thus. A glance at a table of values of e" shows that, when y < 1-5, I +y < ev < 1 +y + y\ :. y < &> — I < jv + y. e" - I .■. I < < \ -y y. y From this inequality it is evident that Lt {'^^\ = Lt (?^) = I. y _> oV y / A!/ -> o\ ^y I ,. Lt i^\ = evxi. Ay _> oVAjJ// .*. -^ = e" = X. ay &i therefore satisfies the equation I = -' (6) and the derivative of e" is e". This property is highly important, and may be put in words: the exponential function merely repeats itself by differentiation. No other function has this remarkable property, which has led to the use of e" and logejv in theoretical calculations. (D102) 17 242 APPLIED CALCULUS Derivative of logex. Reverting to equation (4), Ax Ay . Ajj/ Ax e« •^ [e'^y - ij Now Ajv — ► o when Ax — > o, because jc (= e") is a continuous function of jv- Lt m = 1 X Lt X I. dy _ }_ " dx e« dx ~ x' (^^ i-e. ^(logex) = ^ (8) Also logcA; = j he (9) ••• \y''^A = £f ■ .: loge{b) - loge(a) = j^— • Put a = I, then loge(a) = loge(i) = o. .-. logeb = j^^ (10) /^ dx — is a func- 1 ^ tion of b. We have therefore succeeded in expressing the logarithmic function as a definite integral. It is better to keep to x as our variable in logeX, and putting 6 = ^ we get , f dx LOGARITHMIC FUNCTIONS 243 The latter form of the integral is open to the objection that the upper limit has the same symbol as the variable. The upper limit x stands for the extreme value of X occurring in the range of integration. It is more convenient to avoid confusion thus: must be the same as Fig. 3 (*) (Compare figs. 2>^ and 36.) Hence \ogeX = I f .(II) The two important properties of the logarithmic function have now been found. I. j"^ = logeX {X > Oy 2. Jx^logex) = ' X * The curve y =■ - or yt I is a hyperbola, the asymptotes being: the straight lines ^ = o, j/ = o. The logarithm of any particular value of t, say x, is the area shown in fig. 3 (a). For this reason natural logarithms are often called "hyperbolic logarithms". 244 APPLIED CALCULUS To verify graphically that e = 2 '718 . We have y = logeX = j —, X = ev. Put J/ = i; then X = e. dt _ r — ~ Ji T To find the value of e approximately, we have to find the position of E (fig. 4), so that the area PAEjE F!§r-4 measures unit area. OEj then gives the value of e approximately. There is no difficulty in getting e = 272 in this way. This figure is very near to the truer value e = 2-718. LOGARITHMIC FUNCTIONS 245 Why e is chosen as the base of the natural logarithms. The reader may still be at a loss as to why the curious number e — 2-71828 is used as the base of the logarithms used in theoretical calculations, whereas it would seem so much simpler to stick to the usual base 10. The answer is that 10 can be used if preferred, but its use leads to clumsy multipliers. Briefly the reasons are these : 1. We found that (p. 240) in finding' the derivative of the expo- nential function, we required the limit Lt (f^^'V Suppose we had not chosen base e, but base 10. Everything, in our deduction, would have been the same down to Al/ — >■ 0\ Ay ' To find this limit we should have used the table of ordinary anti- logarithms. On looking up the table, we should find that we cannol now say that T +y < lo* < 1 +y+y\ when y is less than a certain value yo- This can be said only of the exponential table constructed to base e. 2. For many purposes we require the value of fdt ("dt JiT' and the value of this is logeX, a fact which we have ascertained from the table of the exponential function constructed to base e. Common Logarithms. There are formulas corresponding to (5), (6), (8), (11), for logio^ and 10", but they are not quite so simple as formulae (5), &c. Conversion from natural logarithms to common logarithms and vice versa is quite simple. 246 APPLIED CALCULUS The rules are: 1. To convert logex into logio^, multiply by 0-43429, i.e. logioA; = logeX x 0-43429 • • 2. To convert logiox into loge^, multiply by 2-30259, i.e. loge^ = logioJif X 2-30259. . The proofs of these rules will be found in any book on elementary algebra. From these fig-iires, it follows that ^(log,,.*) = ^{log«* X 0.43429. •} = ?^3^, (.2) by pp. 123, 242. and j -7 = 2-30259. ..X logiox. (13) Also 10"^ = (glogeio)* = e^'sn^to and rf^/ 2-30259.. I\ ^ ri / 2 30259..j;-l ^ ^2 -30259 . . A:) rf.r\ / rf(2 ■ 30259 .. *)\ / dx = ^2-30259.. X X 2-30259 . . .'. ^('°*) = 2.30259.. X Itf= (14) so that 10^ satisfies the equation ^ = 2.30259 Xjr (15) By talcing e as the base, instead of lo, we get rid of these rather clumsy multipliers in our equations. WORKED EXAMPLES /h f. , T: + '. the given integral = / J ca + [-ico + a log,^ ~ica + d = loge{cb + ^) — \oge{ca + d) ^ \ca + d) 2. Differentiate ~loge(cosx) with respect to x. d { — log«(cos;«:)} = i- . ■ rf cos;i: Put — loge(cos:f) = «, then du _ du dcosx dx dcosx dx .'. -i-{— logeCOS:r) = — X (— sin:f) = tanjf. dx COSAT Note. — From this result, it follows that lta.nxdx = — logfe(cosji;). 3. Show that y = AeF^ inhere A is an arbitrary constant) satisfies the equation -£+--S+--vi+«o- = ° ('^) where (7„, Cn-ii a^-^, . . . Oq are constants, provided m is a root of the equation a„»2" + a„-i7»»— +. . .OiJ« + a„ = o (17) Uy = ke'^, then -^ = Ame/^, dx dx^ ^'"^ • ^ = Knfie^, dx' ^ = Km^^'. dx^ 248 APPLIED CALCULUS Substituting in (16) we get KW + an-i/W-i + . . . + a„)Ae'»' = o, and any value of m which makes rt„W2» + On-im"-^ + . . . flo = o makes Ae"" a solution of the equation, whence the result stated. A = o also solves the equation, but the solution is useless. In the solution retained, A can be a?iy arbitrary constant. This result is highly important, and occurs constantly in science. A^ote. — Equation (i6) is a differential equation with constant co- efficients. Note on Solutions of Differential Equations. — The following facts regarding differential equations may be found useful. They are dis- cussed in full detail, in works devoted to differential equations. A. If J/ is a function of x, any equation which expresses a relation between y, the derivatives of ^ of any order whatever, and x is called a differential equation. E.g. §?' + X;. = o. ax where X is a constant. B. The general solution of a differential equation is the most general relation between x and y, ■without any derivatives occurring in it, which is consistent with the given differential equation. C. A function of x which satisfies the given differential equation may or may not be the general solution. Whether it is the general solution or not depends on the number of arbitrary constants it con- tains and the type of the differential equation, e"^ is a solution of the equation in (A), but it is not the general solution. It is a par- ticulat solution. D. The order of a differential equation is the same as the order of the highest derivative it contains when rationalized. Thus, the equation in (A) is of the first order, while (a), p. 249, is of the second. E. It is proved in works on differential equations that the general solution of a differential equation of order n must contain n arbitrary and independent constants and can have no more. F. In seeking the general solution, therefore, we try to find an equation in x, y, and the requisite number of arbitrary and inde- pendent constants, such that when these constants are eliminated we obtain the given differential equation. Points A to D are definitions of terms. Point E is a theorem which LOGARITHMIC FUNCTIONS 249 requires proof and has been proved, but the proof is difficult. Point F summarizes the nature of the problem of solving a differential equation. 4. As a particular case of (id), solve S-sg+^ = ° w Putj* = Ae*™ as in example (3), p. 247, and substitute. We get nfi — SOT + 6 = o, (b) i.e. {m — 2) (/« — 3) = o. .'. m = 2 or m = ^, i.e. Ae^'" or Be'* satisfies the equation. By adding- these results, we get y = Ae^^-f Be^. Find -^ and -Ji for this value of v and substitute in (a). It ax dx^ will be found that this value of ^ satisfies the given equation, whatever values be given to A and B. Note there are two arbitrary constants in the solution, A and B, which arise from the t-wo roots of the quadratic equation for m, namely nfi — ^m + 6 = 0. This condition is characteristic of linear differential equations of the second order. If -^ were involved, there would be three arbitrary constants ; the equation to find m would be a cubic one and the solution would be the sum of three terms, each with an arbitrary constant. /dr -^~ — j, a being a positive number. I = l-a 3 = lydx where y = -5-? — ,. ix^ — a^ r -^ x^ — a^ The graph of -^ — ^ is shown in fig. 5. It is traced as follows : 1. Plot the graph of j/ = x^ — a'' and take the reciprocals of the ordlnates of this graph. 2. When x"^ = cfi, the denominator of \l(x^ — a^) becomes zero, 2SO APPLIED CALCULUS HI X 1 O"/ «>. u^ i , /,. ^ < to . OQ 5 S / ^^ >• ' ., ' ' 1 V 1 k o \ « S \ n > ' ■ -_S . \ \ 1 u. "x LOGARITHMIC FUNCTIONS 251 and the fraction becomes indefinitely great. The straight lines X ^ ± a, i.e. X = + a and x = — a, which are straight lines parallel to OY, and passing through the points (+ a, o) and (— a, o) respectively, never cut the curve, as there is no point on the curve at a finite distance from OX for which x = + a. These lines are called asymptotes. The line ^ = o is another asymptote, since no matter how large x is taken, numerically, —^ ^ is still more than zero and positive. 3. When x"^ > a^, the denominator is positive ; when jf^ < cfi, the denominator is negative. The numerator (+ i) is always constant and positive. .•. when x passes through + a, in- creasing or decreasing, the value of i/jc^ — a^ changes sign, i,e. the graph disappears on one side of the asymptotes *■=+«, and reappears on the other side with the opposite sign. 4. When x''' > a^, the values of -.--^^ — „ are positive. .'. the x'' — a' graph lies on the positive side of OX, when x > + a and < — a. 5. When j;^ is very large —^ — ^ = -^ nearly, which is a very small positive number. .'. the graph approaches OX (positive side) when X — > large positive or negative values. In fact (2 above) y = o is an asymptote. 6. When x"^ < a', x"^ — a^ is negative and -5 -„ is negative. x^ — a'' ,'. between the asymptotes .r = + a, the graph lies on the nega- tive side of OX, and approaches — 00 when x — > -\- a or —a. 7. When x^ < a^, x^ — a" = a^ — x^ numerically, and a^ — x'^ is greatest numerically when x^ is least, i.e. when x = o. .', ! — - is least numerically when x = o. x^ — a' /. -g-^ — , must approach nearest to OX for the value a: = o, x^ — a^ i.e. :<: = o corresponds to a turning point in the graph. There are therefore three branches, AE, BD, and CF. We must be careful to exclude the values x = +a from the range of integration, as when x = +a, the value of —^-~ — j is indefinitely great as the graph shows, and there is not a definite strip corresponding to a Sx which includes x = +a. 252 APPLIED CALCULUS Confine ourselves first to branch A E. We seek a formula for any area such as P^?Q. x > a every- where on AE. ' = ir_i !_.T x'' — (fi 2cn-X — a .r -f- flj dx r_^^ = j_rr_E -y^ J x^ — a' 2aJ \_x — a x-\-aJ __ I f dx i_ f dx 2aJ x — a 2aJ x -\- a — :^ ['°S«(^ — a) — loge(.T + a)] 2a = ±log.(^^) (i8) 2a \x + af itegral, , where b and c are both positive numbers. ■ + Take now the definite integral, J iX^ — d When b and c axe both greater than a, the definite integral is J i,x^ — a^ 2a L{c + a)(b — a) J Now, suppose b and c are not both greater than a. We then pass to the other two branches of the graph. Branch CF. If — c < — b < — a, m / -rzr~v ^^ '^ negative going from —b to —c, but 2 _ 2 is positive, x^ being > a^, within the range of integration ; therefore the summation indicated by the integral must be negative, i.e. r_^^ = r_Liog.(^)r J .1, x^ — a^ L2a ^x-\-a/-i-i, = JL log/:n£^) - -L log/^l^") 2a \ — c + a/ 2a \—6 + a/ = raM^a)-raM^) 2a^'l(c + a)(b-a)J LOGARITHMIC FUNCTIONS 253 Branch BD. On this branch — a < x < + a, and if we used the formula [ dx _ I , (x — a\ (x — a)l(x + a) would be a negative number, and we have no logarithm of a negative number. We can easily avoid this difficulty thus: I^^ = - !^2' ^here - a < x < + a. a' — x' 2a\-a-\-x a — x-i .-. f^^= Lfr^ + ^idx J a' — x' 2aJ La + X a — xj _ I f dx , I C dx 2aj a -{• X 2a} a — x = ^ [log8(a + x)- loge(« - x)] = i.iog/^±£). 2a \a — xJ With definite limits we get Jta^-x^ 2a ^l(a - c) (a + *) J' where — a<6 4ac the two roots are real. Let a and /S be these roots. Then, when ar = a and x = p, ax' -\- bx-{- c = o. The graph of ax^ -\- bx-\- c is a parabola whose axis is parallel to OY. To show this, transfer the origin from O to O' where O is the point ( — — , o), and let O'X' and CY' be parallel to OX and OY 2a respectively. If y, y be the co-ordinates of any point P on the parabola referred to the new axes, xf =■ X and y' = y. 2a .'. substituting these values in y = ai^ -\- bx -{-c, we get \ 2a/ V 2a' = aix"-l=^+^\+bxf-^+c \ a 4av 2a = ax''-b3^+^ + bxf- — + c 4a 2a = ax''-^ + c LOGARITHMIC FUNCTIONS 255 This equation gives the same value for y' whether x! is posi- tive or negative. The graph is therefore symmetrical about the new axis of v, i.e. about the line j; = = --^ referred 2a 2 to the original axes (figs. 6, 7, 8). The position of the vertex is found from the fact that O'V is the value of^' when x' is zero, i.e. O'V = - (^l^-i^Y 4a {!}'■ — 4ac) is positive when the roots are real. Also the para- bola cuts the OX axis when y = ax^ -\- bx -\- c = o, i.e. at X — a. and X = p, where «■ and j3 are the real roots of ax' -\- bx ■\- c = o. Case I. — Real Roots. ,, . 0^ > 4ac. When the roots of ax' + bx + c = o are real, the graph of ax' -\- bx -{- c is as shown (dotted line) in fig. 6, when a is posi- tive. When a is negative, the graph is inverted with respect to OX. Taking the reciprocal of the ordinate of this graph, for every value of X, we get the heavy line which is the graph of ax' -\- bx -\- c This graph is of the same nature as that shown in fig. 6, for: 1. When ax' -\- bx -\- c = o, i.e. when x = a and x — p, the function 2.\ , > oo- ax' -\- bx -\- c .', the graph has two asymptotes, x = a and x = p. y =z o is also an asymptote. 2. When j; < a or > |9, the denominator of the fraction is positive (see parabola graph). .'. the graph lies above the axis X'OX when x > p and < o. 3. When a < X < p, the denominator is negative (see para- bola graph). .*, the graph lies below the axis X'OX between these limits of x. 4. Numerically, — ^— p- ., is least when ax' -\- bx-\-c is €tX "^ OX "|~ C 2s6 APPLIED CALCULUS Y 1^ 1 III I -b 1 ^2a \ 1 / \ Y _^ \ !o' # ^ X' o -y "3 fac X *-P (\ n Y' Fig-. 6 greatest numerically. This occurs when x = ° "*" , since the parabola is symmetrical about the line x = 5-i-P. 2 5. The line x = 5LX£ is an axis of symmetry of the graph, for it is an axis of symmetry of the parabola. LOGARITHMIC FUNCTIONS 257 The graph has therefore three branches which must each be considered separately ; the similarity between the graph of this example and the preceding one suggests breaking the expression down to one of the form ,^^ j-5. ax^ + bx + c = a\x' + -x+-\ \. a a) Since the roots are real, 5^ > i^ac. is a positive number, which can be written A^. ' — ^ac Put^ + — = X 2a :. ax' + dx + c = a{X2-A2). • f '^^ = ' f t/j; _ 1 V T ■• jax^ + bx + c ~ ai"^?^^^ ~ a ' where I is the integral already discussed (5 above). If X > A or < - A, .'. the given integral becomes VS2 — ^ac " 2ax + 6 + V*2 _ j^ac^ on substituting for X and A. If — A < X < + A, the given integral is I , f V32 _ Aac — 2ax — b\ ; , v ^W^^c ''^' Ub^-\ac + 2ax + b ) ^P- ^53)- Case II. — Imaginary Roots. i" < 4ac. In this case the graph of >> = ax^ -\-bx -\- c does not cut the OX axis, but its axis still remains parallel to OY (see fig. 7). (D102) 18 2s8 APPLIED CALCULUS If a is positive, y is positive for every value of x, and the graph of - is positive. If a is negative, y is negative for every y value of X, and - is also negative for every value of x. y Y , * 2a / \ / '\ ^ V, / c' ^ \D D' X' o O' X Y' Fig-. 7 The general shape is easily seen to be as shown in fig. 7. - is numerically a maximum, when jy is numerically a mini- ^ ■ <. b mum, I.e. at jc = . 2a LOGARITHMIC FUNCTIONS 259 The graph is also symmetrical about the line x = — — , and ^ = o is an asymptote as before. By plotting an actual case as an example, say,)/ = x^ — 4^=' + 13. it will be found that there are points C and D where the curvature changes from positive to negative. This means that there must be points on i for which y the curvature is zero. These points are called points of inflec- tion. If a = f(x), the curvature is given by 'P^ &\V^\dx))- Seep. 194- This expression equals zero, when d^2 I Points of inflection occur where -3-2 = o, and - changes sign.* Differentiating z = (ax^ + bx + c)~^ twice with respect to ;c and equating the result to zero, we get ;iaV -{- 2cibx -\- (b'^ — ac) = o (ig) The roots of this equation are the x co-ordinates of the points of inflection. The roots of this quadratic are real if {^abY — 4 X 3a^ (5^ — ac) is positive, i.e. gd'b^ — i2a^b'' + i2a^c, i.e. I2a^c — zd'b'', i.e. 3a^{4ac — b'}. But 4ac > b^ is the case we are considering and a^ is positive, ,". 332{4ac — i^} is positive, and the roots of (19) are real, i.e. * A point of inflection is a point on a curve where its curvature changes sign. The gradient of the tangent has therefore a turning value at such a point. In the example, this condition is satisfied when = o and changes sign. dx' 26o APPLIED CALCULUS there are two definite points, C and D, at which the curvature is zero. ax'' -\-bx ■\- c can now be put equal to "{('+i)'+'TS^)- 4«' the given integral becomes I — ' r_ ^^ I X2 + A2' where X = * + — and A = — 4^^JZ — , a positive number. The integral /: x^.-i'"«"5 a and (III) X < V,. LOGARITHMIC FUNCTIONS 261 Y III I , b ' 2a . ' , I \ j 1 \ \ / ^ ^\ /'\ ^^ X' X -^a=i 1*4 Y' Fig. 8 Put X = X -\ as before, then 2a /- J a. dx rfX x'^ + bx + c aj X^ I fd aX a.r + 6 \2ax+b)' a formula which may be used over any range of integration which does not include x = a. I = ). V 2a' 262 APPLIED CALCULUS Summarizing : I. When the roots of ax^ -\- bx + c = o are real, b'^ > ^ac J f dx I , f2ax -\- b — y/b'^ — 4acl , , ^"d / — j-p-= — = ;loge-j ■ ^ > (a) J ax^ +bx + c ViS — 4ac Uax + b+ Vi^ - ^c) or = - ' lngJ ^-:_4''^ + ^^^ + ^l (5) V i^ _ ^ac I V J^ — nac — zax — b) according as (x + — ) does not or does lie between \ 2a/ _(_ V&''' - 4ac ^„j _ Va^ - 4ac 2(j 2a n. When the roots of ax^ + bx + c = o are imaginary, J^ < 4ac and idx ^ 2 ^^^ tan( f ^ + ^ )■ in. When the roots of aj;^ + i^ + c = o are equal, i^ = 4ac, f ^£ =_c_^_V J fljr* + &jr + c \2a.r + 6/ provided the value of :r = — — -is not included in any range of integration; i.e. /■« dx ^ _ r__2_^i« J p ax'' -\-bx -\- c L 2ax + bJ^ provided p and q are both less than or both greater than _b_ """ 2(1 Integration by Partial Fractions. Examples 5 and 6 are examples of a method of integration known as Integration by Partial Fractions. PW If R(.r) = j=r)-T, where P and Q stand for polynomials in x, it is often possible to express R(:r) as the sum of a series of simpler fractions. LOGARITHMIC FUNCTIONS 263 An example will make the method clear. (x - 2) (.r - 3) Assume that R(;r) can be expressed thus — 3-- ^ + 2 = A. _|_ B (;i: — 2) (.r — 3) X — 2 X — 2 where A and B are constants to be found. This relation, if it exists, is an identity, i.e. true for a// values oi x. :. 3-r + 2 = A(jf - 3) + B(:i; - 2), i.e. 3.r + 2 = (A + B);<: - (3A + 2B), These are equal for all values of :tr, if A+B = 3 and 3A + 2B = — 2, whence A = — 8 and B = 11, so that { 3X + 2 ^^ ^ /■/^£_ \ ^^ _ /■C_8_\ ^^ J{x — 2){x — 2) JVjr — 3/ J\x—2/ = 1 1 loge(^ — 3) — 8 \0ge(x — 2) + C. Exercise 13 I . Differentiate with respect to x. [a) \oSe{x + >/i + x^}. [b) log-e[{^ + V'^M^'}/«]. [c) {\0gexY- (d) log:e(tan x). / \ 1 / /i — sin;<:\ + siaxj \ + COSJtr/ {g) e" cosjc. (A) gsina; cos^. (t) loge V sinjc +logeVCOS*. 264 APPLIED CALCULUS 2. What are the integral formulae corresponding to (b), (e), and (/) above? 3. Show that provided x^ > a^. 4. Show that j^W + l?dx = W:^l + Z + ^ loge (^ + ^J + "« ^). 5. Show that ^^^^^7^dx = £^^^^ - ^ log,(^ + ^f - '^-), provided ^ > a. 6. Prove that \'^2S^dx = ilog/*)log.(«6). 7. Integrate . ^ j(^+2)(^+3)2 8. Integrate , ^2 9. Integrate , « ^a:. dx. h + 2X^ dx. 10. A pendulum swings under gravity in accordance with the equation dfi^ I provided the arc of swing is small, where B is the inclina- tion of the pendulum to the vertical, g is the acceleration due to gravity, and / is the length of the pendulum. Find the complete solution of this equation. If the pendulum is deflected through an angle 9 when LOGARITHMIC FUNCTIONS 265 t = o, and is just beginning to move, determine the constants A and B (p. 186) in the complete solution of the equation. Show that it executes harmonic oscillations of period 27r./-. ^ g 11. A small hollow cubical box floats in a lake. It is ballasted with a layer of lead at the bottom so that one half is immersed. Show that if it is pressed downwards a little, and then left to itself, it will execute vertical harmonic oscillations, to a first approximation. If M is the mass of the box and a the length of the side, find the period of oscillation. Why would the actual oscillations not necessarily be truly isochronous? 12. In his investigations of the laws of motion of falling bodies, Galileo first assumed that the velocity is propor- tional to the distance moved through from the start. Discuss the consequences of this assumption. 13. A copper condenser tube | in. outside diameter, and jJg- in. thick, is 10 ft. long. Calculate the period of transverse vibration, approximately, assuming the axis of the tube bends into a parabolic form, and that the potential energy of the tube when bent is proportional to the square of the displacement of the axis of the tube from its mean position. If water is flowing through the tube, what eflfect does the water "have on the period of vibration ? CHAPTER XIV Some Problems in Electricity and Magnetism " The little cheesemites held a debate as to who made the cheese. Some thought they had no data to g-o upon, and some that it had come together by a solidification of vapour, or by the centrifugal attraction of atoms. A few surmised that the platter might have something to do with it, but the wisest of them could not deduce the existence of a cow." — Sir A. Conan Doyle. In this chapter, we shall deal with some applica- tions of the Calculus to electricity and magnetism, but shall first consider a simple mechanical problem which calls for the same kind of mathematical treat- ment as the electrical problems require. Consider a fly-wheel of moment of inertia I (Ib.-ft.^) spinning about its axis at angular velocity fl radians per second. Suppose the bearings and windage re- sistances introduce a retarding torque proportional to the actual angular velocity of the wheel. It is convenient to measure time from the moment when the motion begins, i.e. when the angular velo- city (eo) is Q, i.e. when i = o, w = Q. The angular acceleration of the wheel at time t seconds from the start is dw dt' The bearing friction, &c., introduce a retarding ELECTRICITY AND MAGNETISM 267 torque proportional to w, say km poundals-feet where K is a constant. The accelerating torque is theretore hence where I is the moment of inertia about the axis of rotation, by Newton's Laws of Motion applied to a rotating rigid body; i.e. I -J- + Kft) = o, ^^ do) , K ^ "•^ ^ + r = °' where k/1 is a (known) constant, say X. The equation , aw , - , . ^ + ^" = °' (') therefore determines the angular velocity « in terms of t, subject to the fact that ft) = Q, when i = o. Equation (i) is a differential equation with constant coefficients. Its solution has been indicated on p. 248. The solution is ^ ^ ^^_,,^ ^^^ where A is an entirely arbitrary constant. This is the most general function of t that satisfies equa- tion (i). But we are dealing with a real problem, and a result with an arbitrary constant, to which we can give any value we like, is not of much use. 268 APPLIED CALCULUS The conditions of the problem require that to = Q (a known number), when ^ = o. .•. Q = Ae-''^" by substituting in equation (2). i.e. A = Q. To keep our result consistent with the facts, we must therefore give A the particular value fi (A being now no longer arbitrary). .-. o) = Q.e-^K In this result, Q, and X are known from the data, hence to is calculable for any given value of t. We have therefore found a formula which tells us how the velocity falls as time goes on. The Electric Circuit. Two experimental laws govern the flow of currents in closed circuits. (i) Ohm's Laiio. — When an electric current is flow- ing along a wire, it experiences a resistance to its flow, which is analogous to ordinary frictional resist- ance, and electrical pressure is required to drive the current against this resistance. We can measure currents in amperes and pressure in volts. It is found experimentally that the pressure re- quired to drive a current along a given wire is proportional to the streng^th of the current. If e is the pressure and i the current, then e oc «, i.e. e = ri, where >- is a constant for the wire in question, r is called the resistance of the wire. ELECTRICITY AND MAGNETISM 269 It is different for different wires, and depends on the physical state of the wire, e.g. its temperature. For a given wire, the resistance is the same for all values of «", except in so far as the stronger currents may affect the physical state of the wire, e.g. by heating it and raising its temperature. The experimental discovery that e 2 is this flux <^2 oc i, where i is the current in the generating coil, i.e. <^2 = Kt, where /c is a constant which depends on the shapes, sizes, and relative position of the coils and the number of turns on the generating coil. It follows as a corollary that the flux <#>j passing through the generating coil itself is proportional to i, the current in the coil, i.e. <^i = Kii, where k^ is a constant which depends only on the shape, size, and number of turns of wire on the coil. The important point for us to remember is that, for given coils in given positions, the magnetic flux through any given one of them is proportional to the current in the coil which generates the magnetic field. Right-handed Screw Law. Suppose H (fig. 4) is a straight line giving the positive direction of some quantity which has a con- stant direction, e.g. suppose H is a line of force, in a uniform magnetic field. Let P be any plane per- pendicular to H in which any closed circuit AB lies. Suppose AB encircles the line H. Then the positive direction in this circuit is the clockwise direction when we look at the plane of the circuit from the side E in the direction of H. ELECTRICITY AND MAGNETISM 277 If P is regarded as a thin piece of board, and if a screw is screwed into it at O from the side E, the screw travels in the direction OH. The screw rotates Fig- 4 and travels positively. This is the standard conven- tion known as the Right-handed Screw Law. Faraday's Law. We can now deal with Faraday's Law of Electro- magnetic Induction. Suppose a coil lies in a magnetic field (fig. 5). Faraday found by experiment that if the field is changing in strength with time, then the coil A has an electro-motive force induced in it by the action of the changing magnetic field. This induced e.m.f. is proportional to T— the number of turns in the coil — and to the rate of change of the magnetic flux through the coil. Further, it is in the nega- tive direction in the circuit, as it tends to circulate current in the coil in the negative direction of cir- culation with regard to the field. If e is the induced e^m.f., 4> the magnetic flux 278 APPLIED CALCULUS passing through the coil A, i.e. "linked" by the coil A, and T the turns on A, then numerically, i.e. , , numerically, where /c is a constant; but as it is in the negative direction of circulation we must give it the negative sign, i.e. ^, e = — kT-tt- dt If is measured as described above, and is in H Fig- 5 [dynes per unit pole] X [square centimetres], and if e is measured in volts, then k has the value lo"*, so that ^ (volts) = - io-«T^ (3) . where -^ is the rate of increase of the magnetic fiux through the coil. Equation (3) is the mathematical expression of Faraday's Law of Electro-magnetic Induction, in prac- tical units. The magnetic field may be due to a magnet or it may be set up by a current in another coil of wire. ELECTRICITY AND MAGNETISM 279 For instance, if the coil A carries a clockwise current, when viewed from E, it sets up a magnetic field in the direction shown (the corresponding posi- tive direction) in fig. 6. The e.m.f. induced in B is in the counter-clockwise direction (viewed from the same point E). The two equations e = ri (4) and e = — T^ (•()* are of supreme importance in electrical engineering, and underlie the action of all electrical machinery. % Fig:. 6 It should be noticed that they are both the mere sym- bolical expressions of hard experimental facts, which can be very easily verified with the simplest home- made apparatus. Self Induction. But why should the coil B (fig. 6) be affected by changes in the magnetic field produced by A any more than the circuit A itself? * The reader will bear in mind that in engineer's units (volts, am- peres, &c.) this formula must always have lo"^ as a factor on the right-hand side. 28o APPLIED CALCULUS The answer is that the coil A is also affected. The coil A has an e.m.f. induced in it. If (pi is the magnetic flux through A, ^ ^^ di when Ti is the number of turns in coil A, in series. This e.m.f. of self induction, as it is called, acts (negative sign) so as to tend to stop the current in A, which is producing the magnetic flux i. The magnetic flux linked with the circuit A, multi- plied by Ta, per unit of current in A, is known as the coefficient of self induction of A, and is denoted by L. Hence, if i is the current in A, L« is the magnetic flux passing through A and linked with A, multiplied by Ta ; for the magnetic field set up by A is every- where proportional to the current in A (an experi- mental fact). .-. Ta X (^1 = Li. . rj. d^K Compare equation (2). The similarity of the mathematical result is in- teresting. It suggests that (i) L is a kind of an electric inertia corresponding to the mechanical moment of inertia I; (2) r is an electrical resistance analogous to mechanical resistance; and (3) the electric current is analogous to angular velocity. We know by mechanics that the kinetic energy of a fly-wheel of moment of inertia I is JIa,^ where w is the angular velocity. By analogy, the kinetic energy of an electric cur- rent should be it ^ This analogy is found to be tenable. We can work out all the electrical phenomena of circuits by the principle of the conservation of energy if we sup- pose that: 1. The k.e. of the circuit is |L?^ 2. (Resistance x current) is equivalent to a re- tarding torque which deprives the system of its kinetic energy. 3. Current is equivalent to angular velocity. 284 APPLIED CALCULUS The "retarding torque" takes energy out of the system at a rate (ri X i) watts. (Compare koo y. w = torque x angular velocity, in mechanics.) Gain of energy per second = — ri'^, as it is a loss and so negative. at . 1 di , .. L-n + rt = o. at We thus arrive at the same equation by mechanical principles. These analogies have been examined very closely by physicists, and have been found to be more than mere analogies. The whole theory of ordinary mechanics and electricity can be brought under one treatment of "generalized mechanics". These mechanical analogies, being true ones, are most useful in forming clear ideas of the behaviour of electric currents. For instance, it should be difficult to stop suddenly a current in a highly inductive circuit, just as it is dangerous to try to stop suddenly a heavy fly-wheel rotating at high speed. Every electrical engineering student knows that a bad arc may take place if the field coils of a direct-current dynamo are suddenly opened, and that the usual field switch is designed to throw a "kicking" coil, i.e. a resistance, into circuit with the field coils when the source of excitation is switched off. KELVIN (182';-1907) From a pIiott'gTrpIi by Atuui CLERK MAXWELL (1831-1879) From ike J>ortrait by Loivcs Dickiicspa ( Trinity Colb-go, Cambridge) FARADAY (1791-186?) From the portrait hy T. Phillifs, R.A. (National Portrait Gallery) JOULE (1818-1E From tlie portrait by JobtiL Collier (Royal Society) PHYSICISTS ELECTRICITY AND MAGNETISM 285 The reason for the arc is obvious from Faraday's Equation. The induced e. m.f. is proportional to -^, i.e. to ^, and if the current is suddenly broken, ^, i.e. the rate at which the current changes, is very large, and so the induced e.m.f. is very heavy. Several thousand volts can easily be induced by the sudden breaking of a highly inductive field circuit, and these pressures may break down the insulation of the coils and put them out of action until they are rewound — hence the use of the kicking coil to avoid breaking the circuit suddenly, and to dissipate the kinetic energy of the field current gradually — not explosively as an arc does. Rise of Current in a Circuit., Consider the circuit sketched in fig. 8. C is a cell; L, a coil of self inductance L henrys; R, a resistance of r ohms (supposed large compared r^^WVWV B Fig-. 8 with the internal resistance of the battery and the resistance of the rest of the circuit, so that we may suppose the total resistance concentrated in R). Let 286 APPLIED CALCULUS E be the constant pressure given by the cell. Sup- pose the switch D is closed. The current begins to rise. Let i be its value at time ^ (^ = o at instant of closing D). di Then the induced pressure opposing E is L-i^. p, _ J ^« _ ("pressure available for driving dt ~\ current against resistance. .-. E - di _ ^dt ~ ri. , di •'• ^dt + ri = E, dt , r . '•^■di + Z' = E L One solution of this equation 1 is i = E 7' (lo) for -J- = o if / = — , a constant, dt r B, r E and — X ^ = =-. r L, h Suppose E were zero, then, by p. 267, we know that i = Ae~^ solves the equation, r di , hj^+rt = o. Add the two solutions ; t = Ae-i' + ^. r On differentiating this expression for i with respect ELECTRICITY AND MAGNETISM 287 to t, and substituting in (10), we see that tliis equation is a solution of tlie differential equation (10). But i = o when t = o. A + r •• ^ = ^v - ^ "- } and every constant in this equation is known. The graph of ;' in terms of t is as shown in fig. 9. As t increases, e~£ = — clearly decreases, and the higher the resistance the quicker the current attains, practically, its maximum value, while the t (seconds i Fig. 9 higher the self inductance, the slower is the rise in current. If the coil is wound round an iron core the flux produced by a given current is very much greater than it is in air. Highly inductive circuits are there- fore to be expected from coils wound on iron cores. 288 APPLIED CALCULUS EXAMPLE Suppose a ring is made from a round bar of soft iron, 2 in. diameter, so that the internal diameter of tlie ring is 12 in. Sup- pose there are 500 turns of No. 18 B.W.G. copper wire wound closely on it. Let this coil be connected through a key to the terminals of a secondary cell of negligible internal resistance and giving about 1-5 volts. The resistance of such a coil is about I '3 ohms, and the maximum current -^ = !• 15 amperes. This current is not sufficient to cause "saturation" of the iron, so that for all currents between o and i • 15 amperes, the magnetic induction will be nearly proportional to the current producing it for steady currents. If the iron is of low hysteresis, we may suppose that changes in magnetic induction follow closely changes in current and are proportional thereto. We can thus give an equivalent self inductance to the coil. In the case stated, it would be about i henry. Suppose the circuit is completed. — ( I — e'h' y at time t, L = -r = '•3' r 1-3 i = i'i5 (i — g-i") amperes. Plot this curve and show that the current attains 95 per cent of its maximum value in about 2-3 sec. This example has been chosen to illustrate the order of magnitude of the different quan- tities involved in a particular case where the current would rise slowly. Coils with iron cores, such as that described, are used as "choking coils" in alternating-current engineering. Another example of coils in which currents change slowly when "made" or " broken " is the field coils of dynamos. Slow rise or fall of current is always associated with highly inductive circuits. In actual fact, the current would not rise exactly in accordance with the formula obtained above, because : ELECTRICITY AND MAGNETISM 289 {a) Though the iron is unsaturated, the magnetic flux is not quite proportional to the current producing it even under steady conditions. If the iron is saturated when the full current is attained, the flux is not even approximately proportional to the current. (b) The question is further complicated by (i) Hysteresis, which causes the change in flux to lag behind the change in current producing it. (ii) Eddy circuits, which are induced in the body of the iron and have a similar effect. The formula is practically correct for coils in air, but r/L, always has a much higher value than unity for such coils, and so the current rises or falls very quickly indeed — under .^ sec, say. These time efliects go by the general name of transieiit pheno- mena. Damping of Instruments. Consider a small body, such as the small magnet carrying the reflecting mirror of a galvanometer. Suppose, further, that the body is suspended by a thread, so that when it is displaced from its position of equilibrium the thread brings into play a restor- ing torque proportional to the angular displacement. Then the rate of angular deceleration is proportional to the displacement. Accordingly, if Q is the dis- placement, — -^ = <)?Q {(It? a positive constant). • • ^ + " ^ = °-' A solution of this equation we know to be = A sino)^, where A is any constant. This is the most general solution which permits 6 to (D102) 20 ago APPLIED CALCULUS be zero when t is zero. The motion is a simple har- monic motion. Now, suppose there is, in addition to the restoring torque, a drag on the needle, which for small velocities is proportional to the angular velocity of the needle (compare pp. 266, 267). This drag will contribute an additional retarding torque proportional to the angular velocity, say equal to 2K-r.' The equation for the angular deceleration of the needle therefore becomes rf^e dQ ^ 2. '■^■dr^ + ^'dt + "'^ = ° ('') Suppose that k is small, so that w > k. Try as a solution of this equation,* e = Ae-"' sins/ft)* - K^i. (12) It will be found, on differentiating this expression dO d^d to find -jj and -^ and on substituting in (11), that it does satisfy the equation. This is the equation of damped harmonic motion. K is called the " attenuation " constant. The graph is of the type shown in fig. 10. It is evident from (12) that the oscillations are isochronous and of period 2 7r/V«2 _ ^2^ i,e_ oc" = c"c"' = &c., where Oc" = , ^"^ = T'. * A tentative solution, such as this, is sug-g-ested by more advanced considerations. It is not obvious and is not meant to be. ELECTRICITY AND MAGNETISM 2gi Damping makes the time of swings a little longer and gradually suppresses them. These conclusions are obviously reasonable. At the points B, B', B", &c., where the growth of Fig. lo Q reverses its sense, the tangent to the graph is parallel to O/. ••• ^ = o at B, B', B", &c. Another way of looking at this point is that Q in Q = Ae-«' sinVco^ - K^t (13) is the displacement of the needle from the zero line, and dQldt is therefore the measure of its velocity. When the needle is not moving any farther out- wards, its velocity outwards must be zero. .-. ^ = o at B, B', B", &c. From (13) dQldt = o gives Ag-"' {Vftj^ — K^ cosVco^ — K^t — K sin\/(B^ — K^ t] = o. 292 APPLIED CALCULUS Hence, if t' is the value of t which makes this equation hold, ^'^^'^r^cosj^'^^t' = /c sins/a? - K^f. .. tanvto — K t = — K •• ^ = V«^=zt|^^ + ^"-^ t^" — ^ — J' ('4) where m is any positive or negative integer or zero. This formula gives the values of Ob, Ob', Ob", &c. (fig. lo). Let t have the value t^ = Ob, or i^ = Ob'. By (14) /fa = 4 + , ^-—-^ , on putting m = o and I respectively. ^'^ "^ Hence, bB = Ae""'' sinVto' — /c^^i, and ^'B' = Ae-''' sinVw^ - «:% <''+vi^)<:; Ae"H"-V»T SI "p"'-4'+7^}] -K(, = Ae e ^•''-''' sin{x/»'^ - /c^^j + tt} -««, = -Ae e ^"'-"''sinx/w^- /c^ifi. .'. -To" (numerically) = e ^Z"'-"' = e , X positive (15) •■• '°§^(||) = V^^2 = X (positive). X is evidently the same for any other pair of half amplitudes, e.g. b'B'/b"B". ELECTRICITY AND MAGNETISM 293 From equation (13) it is clear that the "damped" period is hence the undamped period would be T = ^^ (put /c = o in (13)). (0 1 to Aft) .-. T = , "^ X T' (16) The object of deliberately "damping" an instru- ment needle is to enable the instrument to be used again after a reading, without having to wait a long time for the needle to come to rest. Since the damped motion is definitely related to the corresponding undamped motion by the equations already derived, we can easily calculate the undamped motion of the needle from observation of its damped motion. Ballistic Galvanometer. In the Ballistic Galvanometer, for instance, the quantity of electricity held by a condenser can be measured by discharging the electricity through the instrument and observing the ensuing swing of the needle on each side of the zero (neutral) position. In this instrument Q = /"Ta, where Q is the quantity of electricity discharged in coulombs, w, a constant for the instrument. 294 APPLIED CALCULUS T, the undamped period of oscillation of the needle, and a, the maximum undamped swing of the needle following the discharge. Let a be the first observed deflection (damped) on one side of the zero, and /3' be the first observed deflection (damped) on the other side of the zero, and let T' = period of oscillation of the needle (damped), a', j8', and T' are all observable quantities. To connect these with T and a, we may proceed thus: -4^ The motion of the needle is given by = Ae-^'sinx/ffi^ - ,^t. Let ^1 be the time when the first maximum swing takes place. Then „' = Ae-'' sin v/o?^^' i?i = Ae""*! sin -j arc tan [ putting w = o in (14). .-. a = Ae-'''^-^^ = -e-"'', to ft) where Q is the initial value of dQjdt (last line of p. 291), which is the same whether there is damping or not. To find a, we put k = o, .. a = — , or a = ae''i. ft) ELECTRICITY AND MAGNETISM 295 •*• Q = yuTa We now have to get rid of e'K a' '' We have -^^ = e , where X = , "^ , by (15). N/ft)2- .2 A 1 ^ I ^ y/io^ - I? Also 4 = / „ arc tan X ^ It = — arc tan^. KTT X . . y = yu 1 a C'f * If — is very small, we have, since ;r is very large, TT X arc tan- = - nearly. X 2 -^ .-. Q = (KTde\ = /* T' a' y-^ nearly (17) since a/^ = e^ and T is very nearly equal to T'. This result gives Q in terms of the measurable quantities r^., ' o' and /u., the constant of the instrument. The only approximation made in this calculation is that X is small compared to ir, i.e. the damping is slight. 296 APPLIED CALCULUS Exercise 1A 1. A fly-wheel of moment of inertia I is spinning about a shaft at angular velocity i2. It is then brought to rest by a retarding couple propor- tional to its angular velocity. Show that its speed falls in accordance with the equation I- + ^a. = o, where w is the angular velocity at time / and /i is a constant. Solve this equation, and find the value of the arbitrary constant in the solution, in terms of fl. If the fly-wheel is 6 ft. in diameter and has a mass of I ton, find the value of /* necessary to bring the fly-wheel to half-speed in 5 min., if fi = 150 r.p.m. 2. Show that, if an electric circuit lies in a magnetic field so that it embraces a magnetic flux <^, the change in the magnetic flux taking place in an interval of time {/<, — /,) is proportional to the amount of electricity dis- placed round the electric circuit, and is given in absolute electromagnetic units by "^2 - <^] = q^Q. where <^, is the magnetic flux at time t^, ^j at /j, Q the quantity of electricity which flows past any section of the circuit in the interval (4 — ^i)> '' the resistance, and T the number of turns of the circuit. 3. Using the result of question 2, show that, if a bal- listic galvanometer is used to measure the amount of elec- tricity displaced in the circuit, then the change of magnetic flux = /itt where o is the first observed deflection on one side of the zero, and /i is a constant for the circuit, in- cluding the galvanometer. 4. A straight solenoid lies with its length in the direc- tion of a uniform magnetic field in air of 100 lines per ELECTRICITY AND MAGNETISM 297 square centimetre. The number of turns on the solenoid is 500 and its diameter is 7-5 cm. The solenoid is connected to a ballistic galvanometer and is turned suddenly through a right angle. The first fling observed on one side is 65 divisions. Find the change of flux per scale division. If the re- sistance of the circuit were halved, what would be the observed fling, assuming no damping eff'ect. 5. A coil of wire consisting of 100 turns of fine wire, the diameter of the coil being 4 cm. and its resistance 50 ohms, is situated in a magnetic field of 100 lines per square centimetre. This field is reduced to zero at a uniform rate, the total time taken to reduce it being a tenth of a second. Calculate the magnitude of the current induced in the small coil in amperes. 6. A coil of 20 turns of wire in the form of a circle 30 cm. in diameter, rotates 2000 times per minute about a vertical diameter. Find the maximum e.m.f. in volts induced in the coil. (Horizontal component of the earth's magnetic field = -iS lines per square centimetre.) 7. In the ballistic galvanometer, when there is no damping, the equation of the motion of the needle (com- pare pp. 289, 290) is where the notation is the same as in equation (n) above, h being a constant and i the current. Integrate this equation, subject to the conditions that i is constant and that Q and -y- are o when t is o. Prove at that /' is proportional to the extreme deflection, and that the needle always keeps to the same side of its undisturbed position. CHAPTER XV Some Problems in Chemical Dynamics " The idea of an atom has been so constantly associated with in- credible assumptions of infinite strength, absolute rigfldity, mystical actions at a distance, and indivisibility, that chemists and many other reasonable naturalists of modern times, losing: all patience with it, have dismissed it to the realms of metaphysics, and made it smaller than ' anything we can conceive '. But if atoms are inconceivably small, why are not all chemical actions infinitely swift?" — Lord Kelvin (1870). The extract at the head of this chapter, which is taken from an article Lord Kelvin wrote on Atoms (Nature, March, 1870), gives an idea of the state of atomic theory in 1870. The position to-day is: Number of molecules in i c. c. of any\_ ,., w ,9 gas at 0° C. atid 760 mm J Number of molecules in i gm. molecule"! _ f; . w ,n23 of a gas / Average distance apart of adjacent"! ^. » ,°, ^ , '^ •' l-=3Xio° cm. molecules at 760 mm. ] Mass of the hydrogen molecule ... = 3.3 X lo"^* Rrn Averagevelocity of hydrogen molecules"! _ ,5 ^ ^j, -g^ sec ato°c / y • H Radius of hydrogen molecule (con-"!_ ,.2x10-8 cm sidered as a sphere) ... ... .../ PROBLEMS IN CHEMICAL DYNAMICS 299 These numbers* are taken, with slight modifica- tions, from Jean's Dynamical Theory of Gases (1916), and show the great strides that have been made in fifty years in atomic theory. It is interesting to compare these figures with Lord Kelvin's estimate in 1870. He found that {a) The radius of a molecule of a gas could not be less than io~^ cm. (b) The number of molecules per normal centi- metre could not be greater than 6 x 10^^. Subsequent research has evidently confirmed both of these predictions. The first measurement of the velocity of a chemical reaction was made about 1850 by Wilhelmy, who measured the rate of inversion of cane sugar by acid. The first step towards a satisfactory explanation of what governs the velocity of chemical reactions was taken by Guldberg and Waage when they established their celebrated law — The Law of Mass Action. Many compound gases are found to decompose into their elements if they are left standing sufficiently long. The decomposition is measurable within a reasonable time if the gases are kept within suitable temperature limits. Among the many instances which have been carefully studied is arsine, AsHs (Van 't Hoff, Studies in Chemical Dynamics). * Such numbers are often curiously printed — for instance, 2'75 X lois appears as 2-75 X 10''. The important figfure is the index, and the memory is helped by thinking' in a. logarithmic notation. Thus — Log. number of molecules per normal c.c. ... ... =19 Log. radius of molecule ... ... ... ... = — 8, and so on. It is natural then to find log, mass of molecule ... = — 24. 300 APPLIED CALCULUS Suppose a vessel A contains arsine (fig. i). The vessel is connected to a pressure-gauge CF, which can Fig:. I be raised or lowered bodily so as to maintain the gas at constant volume. The pressure-gauge reads any change of pressure which may occur. The vessel A is heated to a fairly high constant temperature, say in the vapour of boiling diphenylamine (310° C). Ini- tially A contains only arsine. As time goes on, the pressure is found to rise (at constant volume and temperature), indicating that something must have happened to the molecules in A. If the number of molecules in A had not changed, there could be no rise in pressure at constant volume and temperature. PROBLEMS IN CHEMICAL DYNAMICS 301 Further, free hydrogen is found to make its appear- ance and there is no evidence of atomic hydrogen. A mirror of arsenic is deposited on the walls of the vessel. The simplest molecular equation which would ac- count for these experimental facts is zAsHg = 2As + 3H2. Arsenic cannot exist as a vapour at 310° C. and approximately atmospheric pressure, and therefore it condenses out and is deposited as a mirror on the walls of the vessel. The equation indicates that for every 2 molecules of arsine which disappear we get 3 molecules of hydrogen. Let n be the number of molecules of a gas X per litre, m, the molecular weight of the gas, ■w, the mass of the hydrogen atom, in grammes; then nniw is the mass of the gas X per litre in grammes. /, nw = the mass of the gas X per litre in gramme-mole- cules, i.e. when the molecular weight in grammes is taken as the unit mass instead of the gramme. The gramme -molecule is often called the "mol". Hence the mass of X per litre in "mols" divided by the mass of the hydrogen atofti (a constant) is the number of molecules of X per litre. It is usual to measure all masses in chemical dynamics in mols, i.e. we use a different unit of mass for different substances, the unit mass used being the molecular weight in grammes. The masses so measured are proportional to the number of molecules present per litre, and the constant of proportionality (w) is the same for all sub- stances. The mass of X present in mols per litre is called the con- centration of X. This numberjs proportional to the number of mole- cules in unit volume. At a given temperature and pressure, all gases have the same concentration, by Avogadro's Theory. Let a be the original concentration of AsHj at time t = o, and X, the amount (in mols per litre) of AsHg transformed in time t. 302 APPLIED CALCULUS Then {a — x) is the concentration at time t, and -'^ + ^ - ^ = t, (0 where p is the pressure at time t, and/>o ,, „ t = o (i.e. at start), since, by the kinetic theory of gases, the pressure at constant temperature and volume is proportional to the number of molecules present. X = 2a{^ — ^\ (2) That is to say, the rise of pressure is directly propor- tional to the amount of AsHg decomposed. The change in pressure therefore gives a direct measure of the progress of this reaction. Dissociation might arise in two ways: (a) By collision of pairs of AsHg molecules. (b) By a kind of molecular explosion. In the former case, at least two molecules must take part in each collision; while in the latter, a single molecule is alone concerned in each explosion. In the former case, the dissociation could be perhaps expressed by 2ASH3 -)• 2As + 3H2 (3) while, in the latter, we might have AsHa ^ AS + 3H (4) the free hydrogen atoms uniting in accordance with the equation, 6H -^ 3H, (5) As there is no experimental evidence of the presence of atomic hydrogen, it must be supposed to become PROBLEMS IN CHEMICAL DYNAMICS 303 molecular hydrogen with great rapidity if it is formed at all, and the reaction as a whole must be dominated by AsHa -> As + 3H, the reaction 6H -> 3H2 going on much more quickly. The essential difference between equation (3) and equation (4) is that in the former two molecules of AsHg take part in each discrete transformation, whereas in equation (4) one molecule of AsHj only is involved. The resulting products are the same in both cases, namely, deposited arsenic and hy- drogen, mixed with undecomposed arsine. The two cases can be distinguished by considering the in- fluence the number of reacting molecules has on the speed of the reaction. Number of Reacting Molecules. Suppose there are Ua molecules of a gas A and n\, of a gas B per litre. The number of collisions between pairs of molecules per second must depend on «„ and fii. Suppose Ma is doubled, without altering the mean velocity of the A and B molecules (i.e. suppose the partial pressure of the A molecules is doubled). There will then be twice as many molecules for each of the B molecules to run into, and if the ratio which the number of actual collisions leading to dissociation bears to the number of possible collisions does not change, there will be twice as many collisions with accompanying dissociations per second. Now, suppose rii, is also doubled. There will now be four times as many collisions per second, i.e. the number of collisions with dissociation per second is proportional to Ua x «&. 304 APPLIED CALCULUS There is no reason why A and B should be essen- tially different molecules. The point is that the number of bimolecular collisions with dissociation per second is proportional to «„ x n},. If fia = nb, the number of bimolecular collisions with dissociation per second is proportional to «a^. In the same way, the trimolecular collisions with dissociation per second would be proportional to Ha X «6 X nc, and the wz-molecular collisions to Ha X ni, X ... to m factors. Now, the number of collisions with dissociation per second gives at once the amount of A (or B or C) which appears (or disappears) per second. Let X be the concentration of A at time i. Then nw = x, where n is the number of molecules of A per litre, and w the mass of the hydrogen atom, in grammes. .'. n = —, w being constant. /Tff Now, -J- is the rate at which molecules of A make their appearance, hence -r. new molecules of A ap- pear in the gas per second, and if each one arises from a collision, then ^ is the number of collisions with dissociation per second. Since n = —, w dn I dx , . ^ ^ ^=- = — ,-., w being constant. dl w dt ^ dx But -j7 is the rate at which the concentration of A at , increases per second, i.e. -r-. measures the speed at PROBLEMS IN CHEMICAL DYNAMICS 305 which the reaction goes forward, and it is propor- tional to the number of collisions with dissociation per second. As the number of collisions with dis- sociation per second must be proportional to Wa X M6 X ... to m factors, for a reaction depending on a wj-molecular collision, we have I dx — J- oc «a X «(, X . . . to w factors, mat i.e. -r. = K{na X nj, y. ... torn factors), ..(6) where /c is a constant which is called the reaction constant. In the case of arsine, then, if the dissociation is bi molecular, and if x is the concentration of arsine at time t, we get -ft = ^^' (7) (The negative sign arises because arsine is dis- appearing.) If the reaction goes forward as a series of mono- molecular explosions, the number of explosions per second will depend on the number of molecules per litre, and it is reasonable to assume that by doubling the number of molecules present, we should double the number of explosions, i.e. if x denotes the number of mols of A present per litre at time t, then — —-, the speed of the reaction, is proportional to «„; dx • i.e. — -77 is proportional to x; i.e. ^ = Xx where X is a constant (8) at (D102) , 21 3o6 APPLIED CALCULUS The equatibns (7) and (8) are the mathematical ex- pressions for the two cases, and are governed by the one rule: Velocity of reaction is proportional to Ma X «(, X »c X •••••(9) This relationship is known as the Law of Mass Action. It was enunciated by Guldberg and Waage. In words, it may be put: The velocity of reaction is proportional to the product of the concentrations of the reacting molecules. The law can be deduced directly from the principles of thermodynamics. Equation (9) is the simplest way in which the law can be stated, and is the key to many formulcc of chemical dynamics. Collecting our results, we see that if AsHg decom- poses, by collisions, in accordance with 2ASH3 -> 2As + 3H2, i.e. AsHs + AsHs -> 2As + 3H2, the law of mass action gives the velocity of the reaction as /Va- S = --^ (-) where x is the concentration of arsine at time t, and /c is a constant (for the particular temperature of the experiment). On the other hand, if the reaction is governed by AsHg -^ As + 3H, we should have dx , Si = -^''' ^") GRAHAM (1805-1669) From ,1 liiho^^rapk by II'. Boshy VAN't HOFF (1852-1911) From a /'/loUn:'','/'/! GULDBERG (1836-1902) Fj-o/fi a photograph UIBBS (1839-1903) From a photograph PHYSICAL CHEMISTS PROBLEMS IN CHEMICAL DYNAMICS 307 where \ is a constant (for the particular temperature of the experiment). Equations (10) and (11) are different, hence the Law of Mass Action gives us a means of testing whether the reaction is monomolecular or bimole- cular. If equation (10) holds, we get dt_ _ _ I dx ~ x^' Kt = - j^^dx + c, i.e. Kt = — \- c. X To find c, we know that when t = o, x = a, the initial concentration. . Hence o = — \- c. a I c = , a and kI = ) X a i.e. t = l(i-i) (12) K\x a) ^ ' If equation (11) holds, we get --dt-^""' ^dt I '•^- -^Tx^ X :. —\t = \og^x + c. 3o8 APPLIED CALCULUS To find c, we know that when i = o, x = a. .'. o = \ogea + c. .'. C = — logeO. •-^ = ^M3 ('3) We thus get the two equations, K = - ( ) if (lo) holds, t \x aJ '■'■^ = i-M-') (H) andX = ^ loge(-) if(ii) holds. To test these equations, we require data of x and t. These data have been obtainei experimentally (see textbooks of Physi- cal Chemistry or Van 't Hoff's book already referred to). An experiment gave Time in Hours. Pressure (mm.)** O 784-84 3 878-50 4 904-05 5 928'02 6 949-28 7 969-08 8 987-19 EXAMPLE / From these data calculate : (i) The ratio ajx by equation (2), (3)ilog.©. * Lehfeldt, A Textbook of Physical Chemistry. PROBLEMS IN CHEMICAL DYNAMICS 309 and see whether (2) or (3) is more nearly constant over the range of the experiment. What conclusion do you draw from your result? EXAMPLE // As another example of a gas reaction, we will consider the decomposition of phosphine. The molecular reaction would be 4PH3 -> P4 + 6H2, if phosphorus is present as vapour. Molecules of phosphorus are tetra-atomic at temperatures considerably above the boiling- point of the element. They become diatomic at very high tem- peratures. If, however, the temperature is not very high, the reaction might go forward as PH3^ P + 3H; (is) then 4P -^ and 2H — > h'J '-' the latter two reactions being assumed to be very rapid compared to the first. Test which set of equations is the more likely from the following data: — Time in Hours. -Pressure of Gaseous Mixture.* 7-83 24-17 .... 786.61 „ 6'?'i7 89-67 855-50 „ Assume that when 4 molecules of PHs disappear we get 6 molecules of hydrogen. Then equation (2) gives xja in terms of ^. Use equation (13) on the assumption of a monomolecular reaction, and to get the formula for a quadrimolecular reaction, use -— = — /«f*, where |U is a constant, (17) at and integrate it. * Konig. (Quoted from Lewis, A System of Physical Chemistry.) 3IO APPLIED CALCULUS A Reversible Reaction. As an example of another, and more complicated, type of reaction, take the decomposition of hydriodic acid. When hydriodic acid is left standing it dissociates into hydrogen and iodine, and the dissociation can be measured if the gas is kept at a suitable tempera- ture, so that the dissociation is neither too fast nor too slow. If the gas initially consists solely of hy- driodic acid, the reaction proceeds until a certain percentage of the acid is decomposed. It then ceases. Similarly, if we start with hydrogen and iodine vapour in molecular proportions, these gases combine until a certain definite percentage of hydriodic acid is present. This percentage is the same as that attained by dissociation, and depends only on the temperature of the reaction. The molecular equation is 2HI = H,+ U ('8) If a is the initial concentration of the hydriodic acid and X the amount transformed in time t, (a — oc) is the concentration of hydriodic acid at time t. Hence, if equation (18) represents the reaction, we h^^e 2HI -» Ha + I2 as the equation of dissociation, and f = ^(«--)^ <'9> by the law of mass action. But at time t we have molecules of hydrogen and iodine present, and there appears to be no reason PROBLEMS IN CHEMICAL DYNAMICS 311 why these molecules should not recombine in accord- ance with the law of mass action. The experimental fact that the reaction is reversible supports this idea. When X mols of HI have dissociated we have ~ mols of H2 and — mols of Ig present, and by recombination they unite to decrease x, so that the time rate of increase of x is not «(« — x),^ but Tt = Kia-x)^-— (20) Equilibrium would be reached when dx di = °' i.e. if ^ = Xe for equilibrium, K{a — Xef = , I.e. -Ei ^ — '— = fi where yu = - = a constant, or Xc" ' K a — Xe _ VyU. Xe 2 i.e.— = a constant (21) a At a given temperature, then, a constant fraction of the original HI should be transformed, and this frac- tion is independent of the pressure of the gas used in the experiment (i.e. of a). It is easy to show that the final or equilibrium composition will be the same, whether we start with pure hydriodic acid or with a mixture of hydrogen and iodine vapour in mole- cular proportion. 312 APPLIED CALCULUS Bodenstein found at 448° C. : Original pressure, HI i I ij 2 atmos. x,ja 20-19% 21-43% 22.25% 23-06% These figures show a 15 per cent increase in Xe/a for a 300 per cent increase in pressure. EXAMPLE Taking jTe/fl = 0-2200, calculate/* = (-) at 448° C* We now have to consider what happens before the equilibrium condition is reached. We have, from equation (20), dx K{a — xf 4 Since / = o when x =^ o, this gives t = P '^^ .(22) 4 (Here we have written down the value of t at once as a definite integral, instead of first finding t as an indefinite integral with an undetermined constant added, and then determining the constant from the condition that t =■ o when x = o. The value of t in equation (22), besides giving the proper value of -=-, ob- dx viously vanishes when ;r = o.) Now, write p2 for — , so that = i*/". and we get from 4* ^ K equation (22) ''' = /o (a-4^-A' ^"^^ * Lehfeldt's Physical Chemistry. PROBLEMS IN CHEMICAL DYNAMICS 313 This integral can be evaluated by substitution in one of the general formulae given on p. 262, but it is more instructive to work it out independently, by the method of partial fractions. D ^ I A , B Put ^ =-^ = + -j- (a — xy — pV a — X — px a — x + px To find A, multiply both sides by a — x — px, and then put a — X — px = o, i.e. give x the value a/{p + i). The term in B vanishes, and we get, A = value of 1 , when x Similarly, B 2ap a — x -\- px C+i 2ap ' p— I Hence from equation (23), j 2ap a — {p -\r i)x J 2ap a + (p— i)x' Now, at the start, {a — xY — pV has the value a^, and the reaction goes on till it has the value o. /. (a — x)^ — p'^x^ never becomes negative, so that «—(/) + i)x and a + (p — i)x sire positive. Thus Kt = -I- I _ log{a - J+'ix) + log(a + J^^ix) j ^ = -^ log ^ + i^I'i^ - 2a a — (p + i)x (The student may easily verify by differentiation that this result leads to the value of -=- given above equation (22) ; he will see at ax a glance that the result also gives ^ = o when x = o.) Thus/ = -L= log / + '•;- ')^ (24) oVkX ^'a-(p+i)x where p = i ^-• This equation gives the time required for x mols per litre of HI to be transformed into H2 and I^. 314 APPLIED CALCULUS Numerical Calculation. Some experiments by Bodenstein gave, at 440° C, the time being in seconds, K = 0-00503, X = 0-365. •'• " = ^\/^ = 4-26. Suppose we start with a litre of HI vapour at 440° C, and 760 mm. pressure. I litre of hydrogen at 0° C. and 760 mm. weighs 0-0899 gm. .'. I litre of hydrogen at 440° C. and 760 mm. weighs 0.0899 X Hi g"i- The concentration of hydrogen, and therefore of HI also, at 440° C. and 760 mm., is thus 0-0899 X f^J X J or 0-0172 mols per litre. .". a = 0-0172, and — — = = 1360. The formula (24) now reduces to t = 1360 log, 0^0172 + 3:265 ( ) •^ ^'0-0172 — s-26;r This is the equation we require connecting the time {t sec.) with the amount of HI decomposed {x mols per litre). The equation gives a finite value of t for any value of x up to that value which makes the denominator on the right zero. The final state (only reached, theoretically, after an infinite time) is given by the equation 0-0172 — 5-26^; = o, or jr = 0-0033. From this value of x, we get £ X 100 = °-°°33 X 100 = 19, X X. 100 = 0-0172 so that when equilibrium is reached 19 per cent of the original HI has been decomposed. (Compare with the experimental figures for a slightly different temperature on p. 312.) The time required for a given fraction of the HI, say 10 per cent, to decompose can be found readily from equation (25). PROBLEMS IN CHEMICAL DYNAMICS 315 If X = ^ X 0-OI72, we have i = ,36olog,i±^^ = 1360 X 2-303 logit,!^ 0-474 = 1400. The time is therefore 1400 sec. or 23 min. 20 sec. Calculating similarly, we get the table : Per cent Decomposi- tion of HI. Time. O O 5 10 min. 20 sec. 10 23 min. 20 sec. 15 44 min. 16 sec. 19 (equilibrium). EXAMPLE Trace the reaction in the opposite direction, starting from a mixture of hydrogen and iodine vapour in molecular proportions, and noting that the physics of the reaction in either direction is contained in the equation ^fCni) = XCh,Ci, - k{CbiY. at Prove that if a is the initial concentration of hydrogen, and therefore also of iodine, and if ;i; is the amount of each transformed in time t, Jll = \(a - xf - iLKx"^, at whence- = \ — ^ = (26) 2 70 X(a — ar)^ — 4 k.x'' Taking X = 0-365 and 1 = 0-00503, at 440° C, express ^ as a function of x, and plot a curve of the amount of HI present as time progresses. Find the "equilibrium " percentage of HI and compare it with the value previously obtained. 3i6 APPLIED CALCULUS Exercise 15 1. Rutherford and Soddy found that the activity of Thorium X fell off with time in accordance with the following table : — Time in Days from Start. Activity of Th X. 2 lOO 3 88 4 72 6 53 9 29-5 lo 25-2 13 '5-2 15 "•' The unit of "activity" is an arbitrary one, but the readings are proportional to the concentration of Th X. Show that this reaction is a monomolecular one. 2. Cane sugar decomposes, in aqueous solution in the presence of a catalytic acid, into dextrose and laevulose, in accordance, quantitatively, with the equation C12H22O11 + HgO — > C^HjaOg + CgHjjO,,. dextrose. laeviilose. The substances on the right and left respectively ro- tate a beam of polarized light in opposite directions, and hence, as the reaction proceeds, the observed rotation of the beam of light decreases. The reaction can therefore be studied with the polarimeter. The following are the results of an experiment (Lewis's Physical Chemtstty): — Time (min.). Concentration. O O 30 I'OOI 6o' i'946 90 2*770 130 3-726 180 4.676 The initial concentration of cane sugar was io«o23. PROBLEMS IN CHEMICAL DYNAMICS 317 What conclusion do you draw as to the number of reacting molecules, and how do you reconcile your con- clusion with the chemical equation? 3. An aqueous solution of dibromosuccinic acid trans- forms monomolecularly in accordance with the equation C^H.O^Br^ -> C^HgO^Br + HBr. The reaction constant at 15° C. is 0'00ooog67; at 101° C, 0'03i8, when time is measured in minutes. Show that the time required to transform half the dibromosuccinic acid initially present is about seven weeks at 15° C, and less than half an hour at 101° C. (Van 't Hoflf). 4. Phosphine decomposes quantitatively in accordance with the equation 4PH3 -^ P, + 6U,. An experiment by Konig gave Pressure of Gaseous Time (hr.). Mixture. o 7i5'2i mm. 7-83 730-13 M 24-17 759-45 .. 41-25 786.61 ,, 63-17 819.96 ,, 89-67 855-50 „ Examine these data with a view to settling whether the reaction is quadrimolecular, as the chemical equation sug- gests, or monomolecular. Write equations which you suggest represent the true course of the reaction. 5. It was shown by Fick that the "concentration gradi- ent" is the quantity that determines the drift or diffusion of molecules in a solution. He found that the amount of the solute that would drift in unit time, normally across unit area, under a "concentration gradient" of unity, is a constant for a given temperature. This law is called Pick's Law, and the constant, the diffusion constant, D. 3i8 APPLIED CALCULUS Assuming this law, show that the differential equation governing the change in concentration of a solution which arises from diffusion is S = °g ■<■) where c is the concentration and / the time, supposing the diffusion to take place in one direction x. (a) Assuming D is constant, show that one solution of the equation is Cq + ae-Dii2( smnx, where c^, a, and n are constants. (b) Find the values of n for which the concentration at X = o and :*; = / is independent of /. m (c) Show that Cq + 2 Orfi-^'^'- sinnx is also a solution of the equation, where 2 stands for the sum of all the sine terms got by putting n = i, 2, 3, . , . m. Equation (i) also occurs in the theory of the conduction of heat. The solution given above is one of Fourier's. CHAPTER XVI Some Problems in Thermodynamics " Economy of fuel is only one of the conditions a heat-engine must satisfy; in many cases it is only secondary, and must often give way to considerations of safety, strepgth, and wearing qualities of the machine, of smallness of space occupied, or of expense in erecting. To know how to appreciate justly in each case the considerations of convenience and economy, to be able to distinguish the essential from the accessory, to balance all fairly, and finally to arrive at the best result by the simplest means, such must be the principal talent of the man called on to direct and co-ordinate the work of his fellows for the attainment of a useful object of any kind. " — Sadi Carnot, Reflexions sur la Puissance Motrice du Feu (1824). There is no hard and fast line between the three conditions or phases in which it is convenient to regard matter as existing, namely, the solid, the liquid, and the gaseous. Thus, at ordinary tempera- tures, a lump of steel is solid; it can support a tensile stress, within limits, without any tendency to permanent change of shape. A liquid, such as water, cannot do this. A gas, such as hydrogen, shows no tendency to become liquid at ordinary temperatures, no matter how much pressure is ap- plied to it. On the other hand, if the temperature of the hydrogen is very low, it tends to liquefy if the pressure is sufficiently increased, and in the so- called "critical" state, it appears to be very hard to decide whether the substance is definitely liquid or definitely gaseous. 320 APPLIED CALCULUS Steam is another instance of the intermediate stage. It is difficult to determine whether a volume of "steam" is gaseous steam or atomized water. A substance which is definitely gaseous under all conditions which we are considering, we shall call "a gas". Such gases are hydrogen, oxygen, nitro- gen — so-called "permanent" gases — under ordinary conditions of temperature. Others, such as water, when in the gaseotis form, we shall call vapours, thereby implying that the substance does not obey the simple gas equation pv = Kd, and that lique- faction or even solidification may take place. State. In thermodynamics, we frequently refer to "the state of a body ". This phrase denotes the physical and chemical condition of the body. To have a clear idea of the physical and chemical condition of a body we need information on the following points: — (i) What is "the body" made of? Is it iron, hydrogen, &c.? (2) Is the chemical nature of the body permanent? If it were hydrochloric acid vapour, for instance, it might change into hydrogen and chlorine. (3) How much of it is there (its mass, as measured by its weight)? (4) Where is it (its position in space)? (5) Is it in the solid, liquid, or gaseous form, or partly one, partly another, as in a partially condensed vapour? (6) What are its dimensions? What is its volume, if it is a gas? (7) Is it at rest or in motion? PROBLEMS IN THERMODYNAMICS 321 (8) Is it in a state of stress? What is its pressure if it is a gas? (9) Is it magnetized? (10) Is it electrified? (11) How hot is it (its temperature)? (12) Is it self-luminous, or sending out radiation of any known kind? It is evident that, in general, a great deal of infor- mation is required in order to know the "state of a body ". We can greatly simplify the problem in certain cases: (i) The body we are thinking about may be a definite amount of a specified gas, say nitrogen. (2) We may be examining what happens to it in the gaseous state only. The equation pv = R'0 then holds very closely. (3) It may not be subject to any electrical or magnetic influences, or to influences arising from radiation of any kind, so far as we can tell. (4) Experiments show that ordinary nitrogen remains ordinary nitrogen over a very wide range of change of temperature and pres- sure. (5) The body of gas considered may not alter its position appreciably relative to the earth, so that its potential energy, due to its weight, does not change appreciably: (6) The body may be at rest as a whole. These restrictions eliminate all the questions except (6), (8), and (11), and its "state" is therefore known (D102) 22 322 APPLIED CALCULUS when its dimensions (volume v), its state of stress (pressure />), and its temperature 6 are specified. We may therefore say that the state of this body (a given mass of a permanent gas) depends only on its pressure, volume, and temperature. Functions of Two or More Variables. In the preceding chapters we have dealt exclusively with func- tions of a single variable. In this chapter, we shall have to consider some simple functions of more than one variable. The area of a rectangle is given by the product of its length and breadth. If A stands for the area in square inches, I for the length and b for the breadth in inches, A = lb. The area depends on both the length and the breadth. These quantities have no relation to each other since we can vary either of them without necessarily affecting the other. They are said to be independent variables. Keeping the length constant, if we vary the breadth, we vary the area; keeping the breadth constant, if we vary the length, we also vary the area. If we vary the length and the breadth, we may or may not vary the area (e.g. halve the breadth, double the length, and we get no change in the area ; but double both length and breadth and we get four times the area). If a quantity w depends on the independent quantities x, y, s, &c., in such a way that changes in the values of any one of the quantities x,y, s, &c., are accompanied by changes in the value of w, w is said to be a function of x, y, z, &c. , and in conformity with the notation for functions of one variable, we write w =/(■*> J>'j ^. &c.); w = f(x, y) = x^ -^y^, say, then and /(at, a) = x^ + a^, and so on. (Compare p. 25.) We can find the derivative of a function of several variables. and if PROBLEMS IN THERMODYNAMICS 323 with respect to anyone of them, by the methods already given for a single variable ; for in variations with respect to one of the variables, the others, being independent, are necessarily con- stant. So far, then, as differentiation with respect to any one variable is concerned, the function becomes one of a single variable. For instance, if then^*= Lt r/(-r + A.,^)-/(.r.^)-| ax Ax — >• oL A-t -• EXAMPLE Let K) = :c^ -f" J'^ = y (^1 J')- Let the change in x be A.r. jC, of course, does not change as we are changing x, notjy, and X and J/ are supposed to be independent variables. .'. change iny = {{x -{■ ^x)"^ ■\- y^} — {x^ ■{■ y') = (at + Aa:)2 - AT^ = 2,X^X + (A;i:)2. change in /" , . -5 5 . J. =1 2X + Ax. change in x .: Lt (e|2HI£4^) = Lt {2X + &X) A* — > o\ change m x / ax — >. = 2.r. . . -f^ = 2X. dx We should obtain this result if we applied our ordinary rules for a single variable to ^2 _l «,2 regarding y as constant, i.e. regarding x'^ -\-y^ as a function of X only. Similarly, ^ ^ A function of two variables has thus two first derivatives, one with regard to each variable. -i is an abbreviation for — fix, y). dx dx-" •" 324 APPLIED CALCULUS A function of ti variables has n first derivatives. A special notation is usually used for these several first derivatives. Thus, if ^=y(^,^), we write |/ instead of ^, and ox ax 3/ d/ dy " dy The derivatives are called partial derivatives. Thus -J. is 3:1; called the first partial derivative oi f(x,y) with respeet to x, and i-=^, the first partial derivative of /(jc, y) with respect to y. Characteristic Equation. It is found experimentally that there is a definite relation between the pressure, volume, and tempera- ture of a given mass (say i lb.) of any substance. This relation takes its simplest form for a gas. It is of the type po = RU (I) where p is pressure, v volume, temperature, and R' is a constant for a particular gas, but is different for different gases. If we write R' = — , where m is the molecular ■m weight of the gas in question, R is a constant for all gases. Its value depends on the units we use for p, v, and e. Formula (i) becomes pv = -d, (2) where p is the pressure per unit area, v the volume per unit mass, 6 the absolute temperature, m the PROBLEMS IN THERMODYNAMICS 325 molecular weight of the gas, and R the gas constant. The value of R is given in the following table: — Quantity. British. Scientific. P /Pounds per square I foot. Dynes per square centimetre. V Cubic feet per pound. /Cubic centimetres per \ gramme. e /(461 + t\ t being in \ degrees F. (273 + t), t being in degrees C. in Molecular weight. Molecular weight. R ^15-45 (foot-pounds per pound mol per 1 degree F. abs.). [ (O = 16.) 83.15 X 108 (ergs per gramme mol per degree C. abs.). (O = 16.) EXAMPLES I. Find the " gas equation " for air in Engineer's Units. For oxygen let ^ = Rj, for air <^ = Rj. Suppose standard p and B taken in both these equations. Then, obviously, R2 _ volume of air _ density of oxygen Rj volume of oxygen density of air . D _ p V density of oxygen density of air = ^ X ' = ^545 32 -9047 32 X -9047 = S3-4- •"• P'v = 53-4^. for air, where p is in pounds per square foot, v in cubic feet per pound, and in degrees F. absolute. 326 APPLIED CALCULUS 2. Find the " gas equation " for nitrogen in scientific units. m for nitrogen is 28. 20 .-. pv = 2-97 X 10°^, where p is in dynes per square centimetre, v in cubic centimetres per gramme, and 6 in degrees C. absolute. For steam, the characteristic equation is much more complicated. Callendar's equation is V = 85-65| - 0-4213(^7^7^' + o-o,6, where V is in cubic feet per pound, 6 in degrees F. absolute, and p in pounds per square foot. This equation holds for dry or superheated steam. To find the work done by a gas, expanding slowly and isothermally. By p. 151 the work done is pdv, where p stands for the pressure in pounds per square foot and v for the volume in cubic feet. The work done is then measured in foot-pound units. For isothermal expansion, the characteristic equa- tion gives pv = R'0 = a constant, since 6 is constant. .'. p = — • .: / pdv = R'e ^ = R'eiogJ = R'0 \oger, (3) where r is the ratio of expansion, i.e. — . ^0 PROBLEMS IN THERMODYNAMICS 327 The work done in expansion obviously depends on the amount of gas expanding. It is usual, in engineering, to take a standard amount of gas, i lb. For other amounts, the work done in expansion is proportional to the mass. Equation (3) therefore becomes, _^?' air, Work done, w = 53-4 6 loge{vi/va) per lb — (4) For isothermal expansion, the work done is there- fore proportional to the logarithm of the ratio of expansion r {^= Vi/v^. The; graph of w = 53 4 logr is the logarithmic curve (fig. 2, p. 239). When r = i, ro = o, and when r < I, zo is negative, i.e. the gas is being com- pressed by an outside agency, and work is being done on it, i.e. energy is being added to it. EXAMPLE Assume air at 200° F. expands slowly and isoihermally from an initial volume v^ to a final volume of i c. ft. at atmospheric pres- sure. Let the ratia^f expansion he 6. What is the initial pressure of the air, and how much work is done in the expansion ? ( Take 13 c. ft. of air at atmospheric pressure and 60° F. as weighing I lb.) I c. ft. of air at 200° F.. and atmospheric pressure weighs A^ X SJt = -0606 lb. "I by the gas The initial pressure is given by/oXj = iXi. i equation. ,', pa =•= 6 atmospheres = 88 Ib./square inch. loge6 = 1 .7918. .'. 'VJot)s., per lb. = 53-4 X 661 X i-^giS ft.-Ib. .*. Actual work done = o-o6o6 X 53-4 X 661 X i'79i8 ft.-lb. = 3833 ft.-lb. 328 APPLIED CALCULUS Intrinsic (or Internal) Energy. It is well known that matter, especially of certain kinds, possesses large amounts of stored energy, which we can tap by suitable treatment. One ounce of gunpowder, for instance, may give out 3 or 4 foot-tons of energy if exploded by a spark. This energy cannot come from the spark. The energy required to produce the spark can be easily measured electrically, and is quite trifling in com- parison with the energy given out by the gunpowder. The energy, which shows itself when the stimulus of the spark is applied, must therefore have been stored in the powder. On the other hand, matter can be made to absorb energy, as when water is heated. These facts leads us to suspect that every portion of matter has a definite amount of energy associated with it. This energy may be called the Internal or Intrinsic energy of the body or system of bodies. In some cases we can add to or take from the store by suitable processes, e.g. by sparking the gunpowder, or by heating the water. It is interesting to consider how this energy may be stored in the body. We will consider a certain volume of a gas. By the molecular theory, this volume of gas consists of a multitude of molecules confined within the walls of the containing vessel, and all in a state of violent agitation. Each of the molecules has a perceptible mass and a velocity, and therefore each possesses both kinetic energy and momentum. Each, again, might attract the others as do the stars and planets. Again, each molecule may itself be a minute system PROBLEMS IN THERMODYNAMICS 329 of some kind, possessing an amount of energy proper to itself. It is evident that in a gas the intrinsic energy may consist of: (i) Kinetic energy of molecular motion. (2) Strain energy due to the forces between the molecules. (3) Energy locked up in each molecule. We may regard the intrinsic energy of a body as the energy associated solely with the matter within the boundary of the body. This capacity which the body possesses for doing work must, as we shall see later, arise from, and therefore be dependent upon, its actual physical and chemical state. A Cycle. A series of physical or chemical changes which a body may undergo is called a "cycle" if the final "state" of the body is exactly the same as the initial "state". Joule's Equivalent. It is common knowledge that heat can be produced by expending mechanical energy. A piece of iron gets hot when hammered on an anvil. Joule found that to produce i B.Th.U. of heat, 772 ft. -lb. of mechanical work must be supplied either from the internal energy of the body or from external sources, and this number is quite independent of the initial state of the body, or of the manner in which the mechanical energy is supplied. The number now used, 778, is called Joule's Mechanical Equivalent of Heat. 330 APPLIED CALCULUS Extension of Joule's Discovery. After this discovery was made, it was soon recog- nized tliat all forms of energy are "equivalent", each pair of forms having an appropriate conversion factor. For instance, in electrical engineering, I Kelvin, or K.W.H. = 3412 B.Th.U.'s. The forms of energy which have so far been recog- nized are*: — 1. Kinetic energy, 2. Potential ,, 3. Heat ,, 4. Strain 2 ;3tram ,, (i and 4 are also combined in sound.) 5. Light and radiant energy. 6. Electric energy. 7. Magnetic ,, 8. Chemical ,, Conservation of Energy. Then followed the principle that the total amount of energy of all kinds in an isolated body or system of bodies is constant, and that it is only transformation, not creation nor destruction, that is possible. By " isolated body or system of bodies " is meant a system screened from all physical or chemical action arising from bodies not in the system under con- sideration. The system may be imagined to be enclosed in an "adiabatic" shell, through which forces cannot act nor energy penetrate (a, not; ^ja, through; ^alveiv, to go). *Poynting: & Thomson's ffeat, p. 155. PROBLEMS IN THERMODYNAMICS 331 First Law of Thermodynamics. Consider such an isolated system consisting of a hot body H, a non-conducting cylinder C, a fric- tion less piston P, a compressed spring S, and a gas Adiabatic Coat, Fig. I A inside the cylinder. Suppose the piston is held in position by a trigger. Consider A as one body and the rest of the ap- paratus as another body B. Let the trigger be released and we will suppose that the trifling amount of energy required to release the trigger is included in the total energy of the system at the moment of release. When equilibrium is restored: (i) The spring has compressed the gas and thereby done work on it. Suppose this work is 5W, positive when work is done on the body A. (2) The hot body H has warmed the gas, and therefore pumped heat energy into it. Suppose ^Q 332 APPLIED CALCULUS is the heat energy pumped into A, positive when going into A. A has clearly gained energy. It can only have absorbed this energy by having its internal energy increased. Let E be the excess of the body's internal energy over its internal energy in a chosen standard state. Let E increase by ^E. .-. ^E = 5Q + p|^ (5) where SE and SQ are measured in heat units and SW in units of work. If we measure ^W in heat units, putting 778 ft.-lb. = I B.Th.U., the factor t^f can be omitted. Equation (5) is a direct result of the principle of the conservation of energy. The internal energy of a body depends on its physical state. The truth of this statement can be seen by considering the consequences of its denial. If it were not true, then we could pump heat energy into A, by heating it, without affecting its physical or chemical state. This cannot be done with any natural body. The addition or subtraction of energy always affects in some way the physical or chemical state of the body. Once the numbers specifying the state of the body are set', the internal energy is a unique and definite amount, depending only on these numbers. This experimental fact, namely, that the internal energy is a function of the state of the body, is known as the First Law of Thermodynamics. If then E = ^ (x, y, 0, -m, &c.), where x, y, z, w, &c., are numbers (co-ordinates) which define the state of the body, and we denote the state of the body PROBLEMS IN THERMODYNAMICS 333 before the trigger is released by x^, y^, z^, Wi, &c., and, after, by X2, y^, z^, w^, &c. , tlien E2 - El = 0(^2, jVa, Zi, W2, . ■ .) - ^{Xi,yi, Zi, ^1, ■ • •) = /\e If the initial state be described briefly as " the state A", and the final state as "the state B ", we may conveniently write the latter integral in the form /I(^0 + 778/ Now, suppose the body undergoes a cyclic change. The total change in E must be zero, since by hy- pothesis the initial and final states of the body are identical. (/) ('^Q + 778) = °' ^^^ where (I) means integration carried out for a cyclic change. We shall suppose that mechanical energy is measured in heat units. We can then ignore 778 in the equations and put Eb — Ea = fl(dQ + dw), (7) and E is a function of the numbers that define the state of the body. Since the internal energy is unchanged in any cyclic process, any net work which the substance going through the cycle — the "working" substance 334 APPLIED CALCULUS — may do can only be done at the expense of the net heat received by the body from outside sources. That is to say, the body acts merely as a vehicle of energy, its own energy not being permanently al- tered. In fact, we have ( f ) {dQ + d\M) = o, by equation (6), hence d)dQ = - ([)rfW, i.e. the heat received by the body in the cycle = work done by the body on the outside world during the cycle. (Notice the sign of ( I ) rf W. Work done on the body in question is positive, so work done by the body is negative.) For this reason, it is not surprising to find that the thermodynamic properties of cycles are in some important respects independent of the nature of the working substance, even though some materials are good vehicles, in that they can convey large amounts of energy per pound, while others are bad ones. Sometimes the principle of the conservation of energy is reg'arded as the First Law of Thermodynamics. The facts may be put as follows : — I. In no natural process can energy be created or destroyed. If x units of energy of kind A disappear (appear) then an exactly equiva- lent amount of energy of some other kind appears (disappears). Energy of every kind can therefore be expressed in any unit of energy, say the foot-pound or the B.Th.U. or the Board of Trade unit of electrical energy, and the total amoi;;it of energy in any isolated system is constant. This principle leads to SE = 8Q + 8W, where E, Q, and W are measured in terms of the same u^it of energy. PROBLEMS IN THERMODYNAMICS 335 i!. The internal energy of a body depends on its state. It is im- possible to add energy to a body or subtract energy from it without affecting its state. This may be regarded as the First Law of Thermodynamics. It leads to the equations E = a function of the state of body, (8) and (ArfE 3. It will be seen later that the body possesses another property — its entropy — which depends solely on its state, and which cannot be altered without affecting its state. The Second Law of Thermo- dynamics leads directly to this conclusion. If

0) cannot depend on v, for, if it did, it would change with v and therefore have a derivative r^, which would not be identically zero. .'. E = /(O), a function of 6 only. It is thus an experimental fact that, for a "per- manent" gas, E depends only on the temperature. The function f{d) can be determined as follows : Let a change take place at constant volume. Then the gas does no mechanical work. .-. (5E = 8Q. But,5Q = C„Se, where Co is the specific heat of the gas at constant volume. dE /. oE = C/sou, or -5^ = Ci,. C» is therefore a function of 9 only, and we have E = jCvdO + constant. By experiment it is found that C^ is practically con- stant over a wide range of temperature, so that this equation becomes E = C„0 + constant (13) This is the function of 9 required. 340 APPLIED CALCULUS Reverting now to a reversible change with ex- pansion, in which the working substance does work pSv, we get 5E = 5Q - p8v, i.e. ^Q = C^e+pSv. . ^Q _ c-se ,pSv " T ~ "0 ^^' But I = 1^, by (I). = C,5(loge0) + R'^(l0g,Z.), (p. 243) = ^{C, log,0 + R' log.t;} = S, if we introduce a function ^ defined by the equation ^ = Cv \ogeO + R'loge?? + constant (14) The function thus defined is a function of Q and V — i.e. of the state of the gas — and is just as definitely a property of the gas in a given state as E is. The function <^ is calculable, except for the addi- tive constant, if 6 and v are given. If we measure = o. But ^ = 50 (p. 340); hence (J)^ = o (16) The function is called the entropy of the gas, and is reckoned per pound of the gas from ^ = o at normal temperature and pressure. It should also be carefully noted that all the changes discussed so far are reversible ones, i.e. the tempera- tures of the bodies exchanging heat are equal (T = 0), and the driving and resisting mechanical forces are balanced. (See p. 335.) Practical Engine Cycles. All engine cycles are irreversible both as regards mechanical and thermal exchanges; otherwise the cycle would be carried out indefinitely slowly. In expansion of the working substance, for instance, the friction of the piston must be overcome, and this is an irreversible resistance. Also, the working sub- stance is heated and cooled very rapidly, therefore T > when heating up and T < when cooling, by a considerable amount. The exchanges of heat are therefore irreversible. Further, in the case of gas- or oil-engines, the process which goes on in the cylinder is not even a "cycle" as defined on p. 329; for the working substance burns irreversibly in the 342 Applied calcOlUS process, and therefore its initial and, final "state" cannot possibly be the same. The best we can do, theoretically, is to construct a " cycle "—reversible if possible — which resembles, as closely as may be, the actual process that goes on in the cylinder or engine. The two most important internal combustion engine cycles are the Otto and Diesel cycles, though these cycles are not truly cycles in the strict thermodynamic sense. Otto Cycle. 1. A combustible mixture of gas and air is sucked into a cylinder at approximately atmospheric pres- sure. 2. It is compressed, approximately adiabatically, to a small volume at a high pressure and a fairly high temperature. 3. It is exploded and the pressure rises more or less abruptly according to the speed of the explosion. 4. The high-pressure mixture (no longer the same, chemically, as it was in stages i, 2, and 3) expands approximately adiabatically, doing external work and getting cooler in the process. 5. The burnt mixture at a low pressure and a fairly low temperature is expelled from the cylinder into the atmosphere. In this cycle, we get one explosion per two revolu- tions of the engine, and the engine is known as a four-stroke engine, i.e. four strokes go to each cycle. The programme is repeated in the next two revolu- tions of the engine, and so on indefinitely. The nearest ideal cycle to this one is as follows: — Start with a volume Wj of any permanent gas at PROBLEMS IN THERMODYNAMICS 343 pressure />i, and temperature 6^, in an ideal cylinder. The suffixes refer to the states at the corresponding points on the diagrams. Trace out the following changes on the p, v diagram (fig. 3) : 1. Heat the gas at constant volume by applying hot bodies at progressively increasing temperatures (reversible) from 61 to 62.. Its pressure will be in- creased. 2. Expand the high-pressure gas adiabatically and reversibly against a resisting pressure F = p (re- F'g:- 3 versible). The gas does work, and therefore its tem- perature drops to ^3. Let p^ and v^ be the pressure and volume at the end of the expansion. 3. Cool the gas at constant volume ©3 from 63 to 6^, by applying cold bodies of progressively decreasing temperatures (reversible). The cooling is stopped at the temperature 6^, which is such that the point 4 is on the adiabatic curve through the point i. 4. Compress the gas adiabatically from p^, ©4, d^ to A> ®i> ^1- This is possible since the point 4 was taken on the adiabatic through the point i. The 344 APPLIED CALCULUS temperature rises, of course, during this compression, as external work is done on the gas and heat does not enter or leave it. This cycle yields a definite amount of external work which is absorbed in a mechanical form. The /, V diagram shows a definite area which measures this available work. Now, the efficiency of the cycle is the ratio number of units of work done mechanical equivalent of heat taken in from hot bodies' and it is very important to know what this fraction (ri) is, as it measures the efficiency of the cycle as a 2 means of transforming heat energy into mechanical energy. The numerator we know from the/, v diagram and the formula W = il)pav. (Seep. 153.) PROBLEMS IN THERMODYNAMICS 345 We can calculate the denominator by using the fact that ^ {j)di> = o, where 96 is the entropy per pound of gas. Trace the change in entropy as the gas goes through the cycle beginning at state i. (See fig- 4-) Stage. Heat into Gas. Increase of Entropy. I C„(0, - 6,) „ [''-de ^ , 62 2 3 - c„(03 - e,) -C„log| 4 _ Ci,(02 — 6i — $3 + 84) _ _ 6s — 6^ Ci,(0a — 0i) Qi — Qi f 9 6 a.lso{j)d^=o = Cvloge^ - C^loge^- 61 6i .62 , _ 63 ■■6,-' -6, '• . 62 — 61 _ 63 — 04 01 04 -04 0., 02 — 01 01 02 I _ ^3 ~ 04 _ I _ ^ _ ^3 ~ 03 _ 01 — 04 02 — 01 02 0a 01 346 APPLIED CALCULUS Now, in fig. 3, the curved boundary lines of the cycle are adiabatic lines, i.e. 01,7-1 = constant (p. 337). ••■ I = ©'■' = &"■ where r = — = ratio of expansion, and '=■-©' (■« This is the formula for the efficiency of the ideal gas cycle which most nearly corresponds to the actual Otto cycle. The nature of the gas does not affect the result. It is the same for all gases which obey the gas law pv = R'd. Oil-engines — Diesel Cycle. 1. A charge of air is sucked into a cylinder at approximately atmospheric pressure and temperature. 2. It is highly compressed, more or less adia- batically, so that its temperature is sufficiently high for oil to burn when squirted into it. 3. By means of a pump, oil is squirted into the cylinder and burns. While burning proceeds, ex- pansion is allowed to take place so that the pressure does not rise, i.e. the heat energy of combustion, except such as is used in the expansion referred to, goes into the gas at constant pressure. 4. When combustion is complete, the burnt mix- ture is allowed to expand more or less adiabatically, thereby doing work and getting cooler. 5. When the piston gets to the end of its stroke, the cylinder is connected to the atmosphere, and the con- PROBLEMS IN THERMODYNAMICS 347 tents of the cylinder, at a fairly low pressure and temperature, are expelled from the cylinder into the atmosphere during the ensuing stroke. The ideal gas cycle nearest to this one is as follows : Imagine a volume of gas, initially at pressure /i, 1 2 \ \ p \ \. _: ^4 o V Fig. 5 volume Vi, and temperature 6i, in an ideal cylin- der, and subject it to the following cycle (see fig- 5):- 1. Heat it at constant pressure, by means of a set of hot bodies at progressively increasing tempera- tures, from pi, Vi, di to p2> ^if ^2 (A = Pi)- 2. Expand it adiabatically to ps, v^, 63. 3. Cool the gas at constant volume, by applying cold bodies at progressively decreasing temperatures, from p3, V3, 63 to Pi, Vi, Oi (wg = W4). 4. Compress the gas adiabatically to its initial state, the point 4 being chosen upon the adiabatic line through I. The T, (p diagram is shown in fig, 6. The cycle is reversible, hence ( / )d o. 348 APPLIED CALCULUS Fig. 6 The changes of entropy are given in the following table :— stage. Heat into Gas. Increase of Entropy. I 2 3 4 Cp(0, - e,) o - C„(03 - 6.) o C.log| o - C, log.| o {j)d^ = C,log| 0, o gives C.log^« e: yC„ log^ = C» logg^, where y 9p. PROBLEMS IN THERMODYNAMICS .•.ylog| = log|. • f^y = ^ The efficiency of the cycle is given by _ Cp(02 — 6j) — Ci,(03 — ^4) 349 1 = Cp($2 — 61) I ~ ' ~ y 63 — 6i 62 — Oi I y f(l)'-- is- But^f» 62' . 0, = ^S as A = Pi = p, say, and 04 and 61 are values of 6 on an adiabatic line. " e, as proved in connection with equation (17). ©'■' l\^(Sf"- •■•(■«) y ^p This formula gives the value of the efficiency required, where r is the expansion ratio vjvi and p is the expansion ratio at constant pressure, Vz/v^. Since p > i, py — i > y(p — i). 350 APPLIED CALCULUS This inequality is obviously equivalent to (7—1) > (tp — P*)- Now (7 — i) is the value of the right-hand expression when p = I. If, therefore, we put/(/)) for {yp — p^), then (r- I) = AO- Now, -x/(p) = 7(1 — P^~^), which is negative for all values dp ofp > I. Hence /(/I)) decreases as p increases from 1. Hence, ■ - Q- > . - (i)'"{^)}. This result would suggest that a higher efficiency is attainable with an Otto engine than with a Diesel. But, in practice, a very much greater ratio of com- pression can be used in the Diesel than in the Otto engine, for the reason that in the Diesel engine a non-explosive substance — pure air — is compressed. In consequence of this the Diesel is somewhat more, not less, efficient than the Otto engine. On the other hand, its mechanical efficiency is less, so that, on the whole, there is not a great deal of difference in the net thermal efficiencies of these engines. It must always be borne in mind that the theoretical cycles are merely the best approximations to the actual "cycles", and that these latter are limited by many practical con- siderations which are ignored (because they do not arise) in the theoretical treatment. As already re- marked, the internal combustion engine "cycles" are not true cycles (thermodynamically). The practical considerations are, of course, of first importance, as the cycles have to be actually carried out with engineering materials and appliances. Though one cycle be more efficient than another, theoretically, it PROBLEMS IN THERMODYNAMICS 351 by no means follows that the practical cycle corre- sponding to it will be more efficient than the other. It may be less efficient because of practical limita- tions. Thus far, we have considered gases and have made some progress in discussing the thermodynamics of gases, using only: 1 . The characteristic equations of perfect gases. 2. The conservation of energy and first law of thermodynamics. 3. The idea of entropy as the integral, along a reversible path, of heat taken in to a body divided by the temperature of the body. When we proceed to deal with vapours we meet a new difficulty. Vapours can pass from the gaseous to the liquid, and even to the solid, phase and vice versa. This new difficulty must be overcome by a new principle. This new principle is The Second Law of Thermodynamics. Before considering the second law, it may be well to make the foUowingf remarks : — 1. Entropy is a measurable property of the body itself that we are considering- as the working substance. It depends on the temperature of the body and on its volume (see p. 340, equation 14). 2. On the other hand, the Second Law of Thermodynamics is a theorem about the absorption of heat energy from and delivery of heat energy to other bodies by the working substance. It involves the temperatures of these other bodies. These temperatures we shall de- note by T's, whereas the temperatures of the working substance will be denoted by B's. 3. When heat exchanges are reversible, the temperature of the working substance absorbing heat (say) is identical with that of the body supplying the heat — i.e. if Ti is the temperature of the body supplying the heat and 61 of the working substance, Ti = 9i. 4. If the heat exchanges are irreversible, Ti is not equal to Sj. 352 APPLIED CALCULUS The Second Law of Thermodjmamics. The second law of thermodynamics may be put in this way. It is impossible, by any cyclic operation on a working substance, to transfer heat from a colder to a hotter external body unless energy is absorbed from, external sources. This conclusion is drawn from experience and must be accepted as an experimental fact. A further corollary follows from this fact, namely, With given limits of temperature, no cycle is more efficient than a reversible cycle, no matter what working substance is used. To prove this corollary we may proceed as follows: Let C and C be ideal cylinders (fig. 7), with fric- tion less pistons P and P'. Let a hot body be Removable non-conducting end^~~ -Adiabatic cylindei^- O' J [Irreversible Engine Perfectly conducting end Fig-. 7 available at Tj and a cold body at Tg. Let B and A be volumes of any working substances (a gas, for example, though the working substance need not necessarily be a gas). Assume it is possible, by means of the hot and cold bodies, to cause A and B to go through a VOLTA (1745-18^7) From the portrait by Niath Bi-ttoiii CARNOT (1796-1S32) from au ru^rai'ing HENRY (1797-1878) From a steel engraving CLAUSIUS (1822-1888) From a photograph PHYSICISTS PROBLEMS IN THERMODYNAMICS 353 cycle which is reversible as regards B. How this can be done practically will be seen later. Let the cycle to which A is subjected be reversible or not, as the case may be. Let B, when working directly, absorb Qj heat units from the hot body and reject Qa heat units to the cold body. Then the amount of heat which dis- appears is Qi — Q2, and this quantity, in heat units, must be the equivalent of the mechanical work done by the engine B, i.e. work is (Qi — Qa), and the efficiency of B is ( ^'q ^' )- Let A, when working directly, absorb H^ heat units from the hot body and reject Hg heat units to the cold body. The work done is (Hj — Hj) and the efficiency Now, let the engine A drive engine B backwards, so that the net mechanical work done by A per cycle must equal the net mechanical work done on B per cycle. Since B is reversible, the result of reversing the B cycle is that Q^ units of heat are rejected from the working substance to the hot body, and the working substance absorbs Q2 heat units from the cold body. Now, Qi — Qa = Hi — Ha (= W, say) since the mechanical work done bjy A is equal to that done on B. Also, if A has a higher efficiency than B, Hi -Ha ^ Qi-Q > 'f Hi Q . W W '■'■ Hi > Qi- .-. Hi < Qi, and Ha < Qa (since Qi - Qa = Hi - Ha). (D102) 24 354 APPLIED CALCULUS No mechanical energy from an external source is supplied to this apparatus, yet it transfers (Qi — H^) units of heat to the hot body, and absorbs (Q2 — Hg) units of heat from the cold body, and each of these qtiantities is positive, since Qi > Hi and Q^ > H^. Therefore tfie hot body is getting hotter and the cold body colder, and during tlie process no energy is being supplied from an external source. This is contrary to the Second Law of Thermo- dynamics. Therefore the efficiency of A is not greater than the efficiency of B, and j;a must be either equal to or less than jjb, i.e. The result obtained is quite independent of the work- ing substance employed. If A is reversible, then, by using B to drive A backwards, we can show by the same reasoning as before that if A is less efficient than B, heat will be transferred from the cold body to the hot body without any work being done on the system from outside. It follows that A is not less efficient than B. Therefore all reversible cycles working between the same temperatures T^ and Tg have equal effi- ciencies. This efficiency is the maximum efficiency possible between the given temperature limits, and is independent of the properties of the working sub- stance. To find what this efficiency is, it is only necessary to study in detail the simplest case. PROBLEMS IN THERMODYNAMICS 355 Carnot's Cycle. The simplest reversible cycle, using one upper tem- perature and one lower temperature, is Carnot's. (The theoretical Otto and Diesel cycles require a series of hot and cold bodies at progressive temperatures.) The simplest working substance is a gas. Suppose a volume of gas is initially in the state specified by pa, "Va, and Qa, where da, = T^, and is contained in a cylinder whose sides are imper- vious to heat, and whose end is a perfect conduc- tor of heat. The piston is supposed to be impervious to heat and to be friction less (fig. 8). I. Let the gas expand isother- mally at Qa from A to B by apply- ing the hot body Tj to the end of the cylinder, and very gradually reducing the balancing force F. The gas must expand to prevent the temperature from rising, and must therefore do mechanical work. There is no change in the internal energy E, and the mechanical work done by the gas is measured by the area AB^a. (_ Non-conducting r ffictionless piston Fig. 8 356 APPLIED CALCULUS Therefore, by the conservation of energy, heat must have been supplied. Suppose Qi units of heat were supplied at Tj, the temperature of the hot body. 2. Let the gas expand adiabatically from B to C, by placing I, a heat insulator (fig. 8) against the back, and very gradually reducing the balancing force F. No heat is supplied. The mechanical work of expansion done by the gas is measured by the area BCc^. Hence the internal energy must have dropped, i.e. the temperature must have fallen. Suppose the expansion is such that the temperature falls to Tg, the temperature of the cold body. 3. Let the gas be compressed isothermally from C to D by increasing the external force slowly, D being on the adiabatic through A. The cold body T2 is applied to the end of the cylinder during this compression. The work done on the gas is UCcd. There is no change in internal energy, hence the gas must have given out heat, say Q2 units at Tg. 4. Let the gas be compressed adiabatically, with I (fig. 8) in contact with the end of the cylinder, from D to A by increasing the external force gradually. No heat exchange takes place, but mechanical work is done on the gas equal to ADda. Its internal energy must have increased, i.e. its temperature must have risen. Since the adiabatic through D passes through A, the gas will, after a certain amount of adiabatic compression along DA, reach its initial con- dition A, and we have a cycle of operations. The d, (p diagram is simply a rectangle, the width of which is R' logef^ ) and the height (Oa - 6d). PROBLEMS IN THERMODYNAMICS 357 The net (reversible) mechanical work done by the gas is represented by the area ABCD. The internal energy of the gas has not changed because the gas has gone through a cycle, and hence this work could only have been obtained at the ex- pense of the heat energy of which the gas is the vehicle. This heat energy is (Qi - Qs) = area ABCD (in consistent units), and this amount of heat has been transformed into mechanical work. The efficiency of the cycle is - ^„ ^^ We can calculate this quantity in the following way : All the heat exchanges are reversible, hence Tj = Qa, the temperature of the gas along AB, and T2 = Be ,, ,, „ CD. The mechanical work done, A to B, = R'Ti logeg) (p. 326). Similarly, Q^ = R'T^ \oge(~\ But since AD and BC are portions of adiabatic and Ti^ftT-i = Taws'*" S by equation (11). . -Vb ^ "ih ^ . QlhQ? = RXTi - T,) log,r ■• Qi R'T,log,r ' 358 APPLIED CALCULUS Hence the efficiency (jj) is given by n = Ti This is the fact to which the Second Law leads by considering a Carnot cycle, namely, that no cycle working between temperatures T^ (constant) and Tj (constant) can have a greater efficiency than Since Q^~Q^ = ^\~ \ it follows that 2* = ^ numerically.' If we take heat entering the gas as positive, Qa 's negative. Hence \ - m = - i.e 2^ = o (19) round the cycle. In this result it should be carefully noted that T is the absolute temperature of the external body to which, or from which, heat passes. In a reversible cycle this temperature is the same as the tempera- ture of the working substance when heat is being exchanged. We have shown that the properties of the working substance do not affect the efficiency of the cycle. The efficiency depends simply on temperature — it is the same for the same temperatures, whether one or another substance is used as the working sub- stance. PROBLEMS IN THERMODYNAMICS 359 What part, then, does the working substance play? It must play some part, otherwise we could do with- out it altogether. It plays the part of a carrier of energy. Carnot's theorem comes to this : If mechanical work is done by subjecting an "energy-carrier" to a cycle of changes in which the initial and final states are the same, and in which heat is absorbed at one constant temperature from a hot body, and rejected at a lower constant temperature to a cold body, then the maximum efficiency of energy conversion is ^^~^^ ; and Z^ = o; where Tj is the temperature of the hot body, Tg the temperature of the cold body, and Q is heat entering the energy-carrier. This idea opens the road to the Thermodynamics of Radiation. In 1875 Bartoli pointed out that space containing radiation is an energy-carrier, and therefore that such space can be treated as the "working substance" of a heat engine. The idea has been a most fruitful one in the hands of Rayleigh, Jeans, Boltzmann, Wien, and Planck, and is the starting point in the modern theory of radiation. Value of E^ for any Reversible Cycle. The Second Law requires the efficiency of all re- versible cycles, working between given limits of 36o APPLIED CALCULUS temperature, to be equal. Hence, for any reversible cycle, using any working substance, Qi ~ Q^ = efficiency of Carnot's cycle Qi and 2^ = o, taking heat into the working substance as positive. This result is true for any reversible simple cycle, i.e. for a cycle using one upper temperature and one lower temperature only. Generalization of Carnot's Cycle. Suppose a working substance undergoes any re- versible cyclic change, such as AB (fig. 9). Such a cycle can only be reversible if we suppose we have at our disposal sources and sinks of heat at every temperature between the maximum and mini- mum temperatures of the working substance.* We may then suppose that every exchange of heat is between the working substance and a body at the same temperature, i.e. T = 6 at every point of the cycle. We can draw over the diagram in fig. 9 a system of isothermal and adiabatic lines, a few of which are shown in the figure. * Inasmuch as the conditions of reversibility are external conditions, any series of chang-es of temperature, pressure, and volume in a body can be imagined to be carried out reversibly by suitably adjusting the external surroundings. — Poynting and Thomson's Heat, Fifth Edition, p. 274. PROBLEMS IN THERMODYNAMICS -^Si Any one of the strips, such as abed, is bounded by isothermals and adiabatics, and hence the ele- mentary cycle, abed, is a Carnot cycle, and if (^Q)i is V Fig. 9 the heat absorbed from the external body at Tj and (JQ)2 is the heat passing from the working substance to the external body at Tj, then (<5Q)i _ T, - T, by Carnot's Theorem, i.e. u.«f- 02 ' since T = everywhere because the cycle is re- versible. (This is the point in the argument where the tem- perature of the working substance is introduced instead of the temperature of the external body.) We have therefore 362 APPLIED CALCULUS But, if we take heat passing into the working sub- stance as positive, ((5Q)2 must be written - («SQ)2, and then (iQ)i + (^2 = o. A similar equation holds for the neighbouring Carnot strip. Two neighbouring strips combined are equivalent to a cycle represented by the boundary line of the combined strip, because the common boundary between the strips is traversed in one direction when considered as belonging to one strip and in the other direction for the other strip. In the same way the cycle represented by the zigzag boundary line of all the strips may be considered as the sum of all the elementary Carnot cycles into which the cycle is divided, and for these cycles we have ^Q 2f = o, where S stands for a summation for all the Carnot cycles. By making 5Q — > o, i.e. by making the width of the elementary Carnot cycles — > o, we can make the diagram of the compound Carnot cycle differ as little as we please from that of the given cycle, and hence for the given cycle ^^ _ 0^ o (20) This result is the simplest mathematical statement of the Second Law of Thermodynamics. In any Reversible Cycle. (/)f = (/)f (3.) PROBLEMS IN THERMODYNAMICS 363 Consequences of (1)-^ = for a Reversible Cycle. Let A and B (fig. 10) represent two states of the working substance, which can be reached by different reversible paths I and II. V Fig. 10 Then AIBII is a reversible cycle, and , is called the entropy. We are directly led to it by the Second Law of Thermodynamics applied to reversible cycles. It is here proved to exist ^r any working' substance, and to be a definite function of the state of the work- ing substance. In fact, for steam, COj, SO2, &c., PROBLEMS IN THERMODYNAMICS 365 tables of ^ exist, so that we can look up the value of for any one of these substances in any specified state. Definition of Entropy. Since any series of changes of pressure, volume, and temperature in a body can be imagined to be carried out reversibly by suitably adjusting the ex- ternal surroundings (p. 360), the increase in entropy between state A and state B is I -^ taken along any reversible path from state A to state B. An Irreversible Cycle. If I {-^j be taken along an irreversible path it no longer necessarily measures (0b — ^a). The efficiency of an irreversible cycle cannot be greater than the efficiency of Carnot's cycle, assuming that each cycle works between Tj and Tj. Carnot proved this (p. 352). •• V— q;— A < T, ' where the suffix i refers to an irreversible cycle. T, > t; ^1 .Q2 366 APPLIED CALCULUS or, regarding heat put into the working substance as positive, o > s2i for a simple cycle, and (/)f < ° .(23) for any cycle (cp. p. 362). We use the "equals" sign for a reversible cycle and the combined sign for an irreversible cycle. An Irreversible Path. If the states A and B are connected by a reversible path r and an irreversible path i (fig. 11), we have V Fig. II (/)f 7 o. /:m-/:(f),<°- PROBLEMS IN THERMODYNAMICS 367 ■■■i:m. and ^j. Then, it is an experimental fact (Joule and Thomson's experiments, p. 338) that 6i is very nearly equal to 0. PROBLEMS IN THERMODYNAMICS 369 The expansion is clearly irreversible, and since 6 = ^1 it is iso- thermal, and it is evident from the /, v and 6, 4> diagrams that we can change from state A to state B by a reversible isother- mal expansion in which the entropy increases. In fact, the increase of entropy, per pound, is J e J V V . ■v. i.e. o -I- R' loge-1, which is positive. V 2. When steam expands in a turbine, the expansion is irre- versible and nearly adiabatic. The entropy of the steam, per pound, when it leaves the turbine is greater than when it enters it. This is an experimental fact. 3. When air is cbmpressed in an air compressor, the compres- sion of the air is irreversible and nearly adiabatic. The entropy, per pound, of the air when it leaves the compressor is greater than when it enters the compressor. This, again, is an experi- mental engineering fact. We will now consider changes of state when no mechanical work is done. A Change of State brought about by a Reversible Communication of Heat. The heat added to the working substance is ^Q say. It is added reversibly at 0, the temperature of (50 the working substance, hence -^ is the increase of entropy, i.e. for the complete change where the integral is carried out with reference to the actual heat exchange. (D102) 25 370 APPLIED CALCULUS A Change of State brought about by an Irreversible Communication of Heat. We cannot now integrate along the actual heat exchange to find the change of entropy. The substance changes in the irreversible process from state A {p^, v^, Q^) to state B (/>i, -v^, 6i), and to f^dO find the increase in entropy, we must integrate I -^ along any reversible path which will bring the sub- stance from state A to state B. This integral is not necessarily equal to I -~ for the actual path as the increments of heat might be different. Changes in the Entropy of a System. So far, we have confined ourselves to one body, the working substance. But every other body of the system may undergo changes of state, and has a definite entropy which may or may not change. Suppose two bodies a and b interact in any way, so that the state of a changes from A to A' and of b from B to B'. 1. Suppose it is merely an exchange of heat, then it may be reversible or irreversible. If it is reversible da, the temperature of a, equals 6h, the temperature of b. The gain of entropy of A, due to the passage of 5Q of heat from A to B, is — -^, and the gain of entropy of B is + -^. Va "& As 9a, = 6b, the net gain of entropy is nil. If the exchange of heat is irreversible, the gain in entropy by one body may exceed the loss of entropy by the other, and, consequently, the exchange of heat under irreversible conditions between the bodies of the system may increase the total entropy PROBLEMS IN THERMODYNAMICS 371 of the system. As an instance, we will talie the irreversible exchange of heat by conduction which takes place when i lb. of water is heated from a temperature 6„ to di by means of a hot body of very great mass at Tj = di. The heat exchanged is c{6i — 9(,) heat units, where c is the specific heat of water, viz. unity. The hot body therefore loses (^i — ^0) heat units at constant temperature Ti. Hence it loses entropy I — -~ — '' ) units by the equivalent reversible change. ' On the other hand, the water gains entropy equal to /: -9 = ^°H' by the equivalent reversible change, and 01 — &a _ Oi — Og since T, = Oi- But ^l^li." < loge^ (see below), hence the gain of entropy by the water is greater than the loss of entropy by the hot body, and the algebraic gain of entropy is rio^^i _ ^1 ~ ^0 ] for the system consisting of the two bodies. '-g- is greater than it would be if 2, under the integral sign, were replaced by its least value -^. (See p. 72.) Hence we conclude that by an irreversible exchange of heat we may increase the total entropy of the system consisting of the body being heated and the body heating it. 2. Suppose the exchange is one of mechanical energy without exchange of heat, i.e. it is an adiabatic change. Then one of the bodies expands doing work on the other which contracts. In either case, by p. 368, the entropy of each body must either remain unchanged or increase. 372 APPLIED CALCULUS We have seen that in the two examples considered, viz. an exchange of heat without mechanical work being done and an exchange of mechanical work without heat exchange, the entropy of the system cannot decrease. It may increase if the changes are irreversible; otherwise it is unaltered. These con- clusions have been found to hold good for all cases which have been examined in detail by physicists and chemists, and have led to a principle of physics. This principle, which is generally accepted, is as follows: — In all natural changes of state which may take place in the bodies of an isolated finite system, the total entropy of the system never decreases and generally increases. Suppose that the entropy of a system of bodies is a maximum. This means that any hypothetical slight change of state in the system diminishes the entropy ; but we have just seen that, in natural spon- taneous changes, the entropy can never diminish. Hence the hypothetical slight change cannot occur. The system is therefore in stable equilibrium. The tendency of a natural system of bodies is therefore always towards a state of greater en- tropy. The Principle of Maximum, Entropy, as it is called, is the starting point of much of the work on the applications of thermodynamics to chemistry. The Steam-engine Cycle. We shall suppose the engine to be a "condensing" engine, i.e. one in which the steam, after passing through the engine, is condensed and returned to the boiler. The working substance — sometimes steam. PROBLEMS IN THERMODYNAMICS 373 sometimes water — goes through the following cycle of operations. After leaving the condenser, the water is pumped by the boiler-feed pump into the boiler. It is there heated and turned into steam. It then passes through the engine and enters the condenser. It is there condensed by means of cold water, which removes the latent heat of the low-pressure steam. In the boiler furnace, the heat is generated by com- bustion at a very high temperature, say 2000° F., whereas the highest temperature attained by the steam is comparatively small, say 800° F., even when a "superheater" is used. The transference of heat from the "hot body" — the furnace gases — to the working substance is therefore irreversible. The lowest temperature of the working substance depends upon the vacuum in the condenser. If the condenser pressure is i in. Hg. say, the temperature is about 80° F., the saturation temperature at that pressure. The lowest temperature of the cooling water — the "cold" body — is about 15° F. lower than the lowest temperature of the working substance, hence the passage of heat from the working substance to the cold body is irreversible. The external work is also done by the working substance irreversibly. The actual cycle is therefore essentially an irreversible one. We can, however, imagine a reversible cycle which would correspond to the actual cycle, if all the exchanges of heat and mechanical energy could be carried out reversibly. This ideal reversible cycle was proposed by Rankine and is named after him. It is a compound Carnot cycle. Before describing this cycle we must deal with a point of difference between the steam cycle and any cycle we have hitherto considered. 374 APPLIED CALCULUS all in the Same State Working Substance not Simultaneously. The working substance is not all in the same state at any given moment; for instance, the steam in the boiler is in a very different state from the water in the condenser. We must therefore consider an element ot the working substance of mass Sm, and follow its history as- it passes round the cycle outlined above. Rankine's Cycle. The steam when it leaves the boiler may be wet, dry, or, if the boiler is fitted with a superheater, superheated, i.e. at a temperature higher than the saturation temperature corresponding to the boiler pressure. These three possibilities lead to three dif- ferent sets of equations, but for simplicity we shall take the single case when the steam is initially dry when it leaves the boiler. The principles upon which the calculations are based are exactly the same whether there is initial wetness or initial superheat or neither. Suppose fig. 12 is a. p, v diagram repre- senting the state of a given mass, i lb. say, at dif- ferent points of the ideal cycle. K A B ■iG}^ \ ■«k, ^V^ 1 L '2 ^--^.^ D C E I e V c J Fig. 12 PROBLEMS IN THERMODYNAMICS 375 The point A represents the state of the substance in the boiler before its latent heat is imparted to it. We shall denote this state by/i, o), 61. The point B represents the state after evaporation (A. ■"!< Oi). The point C represents the state after reversible adiabatic expansion in the engine {p2, ^2, O^)- The point D represents the state after condensa- tion to water in the condenser {p2> »> 02)' We can now suppose that the mass, i lb., goes through this cyclic change of state in a cylinder, i.e. in a "one-organ " engine. The actual engine has several organs, namely, boiler, condenser, &c; indeed, the idea of using several organs and keeping one organ for one pur- pose was Watt's great contribution to the practical development of the steam - engine. In the ideal theory, it makes no difference in principle whether we suppose the engine to have one or several organs, but the calculation is easier to follow if we suppose the cyclic change to take place in a cylinder. The steps in the cycle are as follows: 1. The water (at state D) is heated up from 62 to 6^, the pressure on the piston being increased so as to correspond to the saturation pressure of the steam at each temperature. 2. Reversible evaporation under constant pressure pi by a hot body at temperature 61. 3. Reversible adiabatic expansion from pressure /^ to pressure p2) with consequent cooling from 61 to 62, and some condensation. 4. Reversible complete condensation under con- stant pressure p2 by a cold body at temperature 62- 376 APPLIED CALCULUS The figure ABCD (fig. 12) is the indicator diagram for this cycle, and represents the work done per pound per cycle. The work done by the element Sm is then Sm X area ABCD x /, where / is a scale factor. The area ABCD can be obtained by integration, but it is much easier to obtain it from the theorem that {\)d(j> = o. Steps 2, 3, and 4 above are re- versible. A reversible step corresponding to step i is obtained if we suppose the feed-water heating is done by means of a series of progressively hot ex- ternal bodies, each of which is at the temperature of the water it is heating. The increase of entropy along this path is then \-^, where the integral is taken from the state corresponding to D to that corre- sponding to A (fig. 12). But 8Q = 80 for water and T = as the step is reversible, . ("^dQ f^'dd , Oi •• i. T = i,T = '^S%- We get, then, for the increases of entropy in the different steps: I- + log.gi. 2. + Li/01.' 3. None. 4. - XzLzlOi, where Xg is the dryness fraction of the steam at state C, hence ^" = lr + "'4: (»3) PROBLEMS IN THERMODYNAMICS The entropy diagram is shown in fig. 13. 377 Equation (23) gives us x^ as all the other quantities are known, i.e. -^ = rXfe + ^°4] ^^4> The heat taken in, per pound, by the working substance is m - 6^) + Li] B.Th.U. The heat given up to the condenser is x-^L.^ B.Th.U., per pound, i.e. 'It, + ^°^4:] ^•'^^•^•' per pound, on substituting the value of X2 given in equation (24). Hence the efficiency {>]) of the cycle is given by no, - e,) + L, - 0.(ioge| + ^)- ^ L (a _ a\ . T„ _ (6), - e,) + L, •(25) 378 APPLIED CALCULUS The work done, per pound (W), by the working substance is the numerator of this fraction, i.e. W = [(01 - e,) + U- ea(loge| + ^J] B.Th.U. Suppose M is the "water-rate" of the engine, i.e. the number of pounds of steam flowing past the boiler stop valve ^er second. Then a mass M goes through the cycle per second. Now, a mass Sm in going through the cycle does work, WSm B.Th.U., there- fore a mass M does (MW) B.Th.U. of external work, i.e. (MW) B.Th.U. of work are done per second, i.e. the H.P. developed = ^^ >< 778 " 550 = 1-415 MW, where W is the net external work, measured in B.Th.U.'s, done by a pound of steam in going through the cycle. The p, v and B,

) (v + Av) — pv; hence 81 ^ SE + ApSv + AvSp = 0-1- Av8p, by the equation 8Q = SE -\- ApSv, since 8Q = o. ,-. Ib - lo = A\ vdp. Now, I vdp = area KBCL (fig. 12), ■' c hence the area ABCD (fig. 12) is given by area ABCD = ^^^^" - (A - AV .'. the area ABCD in heat units is given by (Ib - Ic) - A(A - /aV. 380 APPLIED CALCULUS In all but the most refined calculations, A{pi — pi)a, is negligible compared to Ib — Ic, hence the required area in heat units is simply the difference in the total heats at states B and C. This quantity is called the ' ' heat drop " between B and C. Tables of heat drops, for different initial and final conditions of temperature and pressure, have been calculated from Callendar's equation for steam (p. 326), by H. Moss {Heat Drop Tables, Arnold). Exercise 16 1. Calculate the ideal efficiency of an Otto gas-engine, the compression ratio of which is 8. 2. Calculate the ideal efficiency of a Diesel cycle for which the compression ratio is 12, and p = 3. 3. Calculate the ideal efficiency of a Rankine cycle for a steam-engine working at 200 lb. per square inch abs. pressure, dry steam, and a condenser pressure of i in. Hg. 4. A pound of ice at 0° C. is put into a bucket of water at 10° C. The thermal capacity of the bucket and the water in it is equal to 20 lb. of water. Calculate the change in the entropy of the system due to the irreversible conversion of the ice into water. Take the pound calorie as the heat unit. The latent heat of liquefaction of ice is 80 C.H.U. 5. A system consists of a non-conducting cylinder, two gases, and a conducting piston between them. The sur- faces of the piston in contact with the gases are non-con- ducting. The piston is a tight fit, so that the gas A, initially at a high pressure, expands very slowly against the gas B at the lower pressure. Show that the entropy of the system when equiUbrium is reached must be greater than in the initial state of the system. 6. Show that the theoretical work, in foot-pounds, which must be done by an air-pump to take in, com- PROBLEMS IN THERMODYNAMICS 381 press adiabatically, and expel i c. ft. of air against a pressure p^ lb. per square inch, the initial pressure of air being ^2 !''• P^'' square inch, is ^^^i^AW^-^ P2' where pv^ = a constant is the equation of the adiabatic change of state. 7. Show that when the difference in pressure ( p^ — p.^ in question 6 is small, the H.P. required, ideally, to com- press V c. ft. per minute of air at pressure /j 's 0-00435 V8/2. 8. Show that, if the initial pressure in question 6 is atmospheric, the H.P. formula becomes "°H(sy"- 4^' where V is the volume of atmospheric air dealt with in cubic feet per minute (n = i'4o8). 9. A gas-engine works on an ideal cycle, with adiabatic compression and expansion, receiving and rejecting heat at constant volume. The piston displacement per stroke is I c. ft., the clearance volume 0-2 c. ft., and at the beginning of compression the temperature of the cylinder contents is 600° F. absolute, the pressure being atmo- spheric. The engine receives o-oS c. ft. of gas per cycle (calorific value 600 B.Th.U. per cubic foot). Atmospheric pressure = 14-7 lb. per square inch. Find — (i) Weight of cylinder contents, (ii) Pressure and temperature at end of compression (7 = 1.38). (iii) Rise of temperature during explosion (neglect jacket loss and take C^ = o-iS). (iv) Pressure at end of explosion. (v) Temperature and pressure at end of expansion. 382 APPLIED CALCULUS (vi) Efficiency of the cycle. (vii) Efficiency of an engine working on a Carnot cycle between the same highest and lowest tempera- tures. (Mech. Sc. Tripos, 1906.) 10. The diagram is the indicator diagram taken on one side of the piston of a double-acting steam-engine. There was iV lb. steam present during expansion. Find, with Fig. 14 the help of steam tables, its dryness fraction at the points A and B, also the change in internal energy and ex- ternal work done between these points. Hence deduce the exchange of heat that must take place between the steam and the cylinder walls during the expansion from A to B. The pressure and volume corresponding to the point A are 65 lb. per square inch absolute and 0-5 c. ft. II. A boiler evaporates 30,000 lb. of water per hour at a coal consumption of 7 lb. of steam per pound of coal. The coal, in burning, uses i7"5 lb. of atmospheric air, per pound of coal. The products of combustion enter an electrically-driven chimney fan at 300'' F. and i in. water- gauge of induced draught. The fan has an efficiency of 60 per cent. Calculate the output from the electric motor. PROBLEMS IN THERMODYNAMICS 383 12. A torpedo air-chamber contains initially 80 lb. of air at a pressure of 1700 lb. per square inch absolute and 15° C, and at the end of the run the pressure is 500 lb. per square inch and the temperature 2° C. How much of the heat of the air which is left in the chamber has been abstracted from the sea? {Mech. Sc. Tripos, igi i.) MISCELLANEOUS EXERCISES I. Find the indefinite integrals of ; ^Kfilogx; (x^ — ^x + 2)-i; x(x — i)^ and prove that /■ J dx IT 5 + 4 COS;<: 3 {Ini. B.Sc. Hons. Land. igo4.) 2. Find the following integrals : C dx f li — Xr f 2x^ — Jtr + lo , ji^qri=-x: JVr+^'^''= j {x^ + ^) (x - 2^- (B.Sc. Pass, Lond. igoj.) 3. Differentiate with regard to x the function Site tan n/(i -x^) Effect the simplification of this function which the form of your answer suggests. Of what function is (i — x^)-* the second derived function? {B.Sc. Eng. Lond. igo6.) 4. Find the following indefinite integrals : /—^, r', ie'sinxdx; [sfie'dx. x\x - i) ] 1 (B.Sc. Eng. Lond. igo^.) (D102) 386 26 386 APPLIED CALCULUS 5. Find the value of -j- at any point x, y of the curve and write down the equation to the tangent to the curve at the point. If the tangent cut the axes Ox, Oy in P and Q respec- tively, show that op + OQ is constant and equal to a. Take a equal to 4 in., and draw a number of straight lines cutting the axes and having the sum of their in- tercepts equal to this. Show that the curve may then be sketched in to touch these lines. [Qual. Exam. Mech. Sc. Trip. igo8.) 6. Show by integration by parts that if I„ denotes the integral r/s j ^ swf'xdx, Jo where B = arc tan-, we have a I _ "("- i) t n' + a' {B.Sc. Pass, Land, igoj.) 7. Prove that if m and n be positive integers, [2 sin™^ cos'^xdx = ^ ~ ' [2 sin'"-*jt; cos^'xdx, Jo 'Jo and evaluate / 2 cos^* sin'xdx and I ^ cos"^ sin^xdx. Jo Jo (B.Sc. Eng. Land. 190^.) 8. What properties of /(x) may you infer from the graph of /'(x)? Amongst all right cones which have the same volumes, show that that one which has the least curved surface may be constructed from a circular sector whose angle contains 3-6 radians approximately. {B.Sc. Eng. Lond. jgo6.) MISCELLANEOUS EXERCISES 387 9. Prove that ("jx -a)(x-b)[x-i{a+ V)^dx = ^" '^^^ • Apply Simpson's Rule to calculate an approximate value of logfiio from the formula logcio = I — . /l X {B.Sc. Eng. Lond. igo6.) 10. Integrate the functions JC^ + 2JC — I I 1 x\x — i)^ ' >^x{x — 2) a^ s\a^x + b^ cos^^' Show that, if c > « > o, I ^^^-^ ^ dx = ir(c — \/c^ — «^)/2C. {Maths. Trip. I. igoS.) 11. Show that, when/(^) is of the form A + Bjc + Cx"^^ f'/{x)dx = ^{/(o) + 4/(i)+/(i)}. Jo Show also that, if <^(:v) be any polynomial of the fifth degree, C(x)dx = xV{5-^W + 8, the inclination of the tangent to the axis of X, is equal to ^6, and that the length of the arc measured from the origin is given by j = 4a sin<^. Prove also that the radius of curvature at any point is twice the intercept on the normal between the curve and the line y = 2a. (B.Sc. Pass, Land, igii.) 14. Evaluate the integral /: where a, /3, and 7 are real constants, and ^ is not equal to 7. Prove that /"<» dx I /: o(2«2 _ Ar2)S 3a4" {Int. B.Sc. Hons. Land, igrj.) 15. Apply the method of integration by parts to evaluate llogexdx, ixlogexdx, txe^dx. Show that /: X loge(i + ix)dx = f (i - 2 log4), and prove in any way that this is less than C^x^dx. Jo (Maths. Trip. I. 191 j.) 16. Show how to find maximum and minimum values of a function of one variable. MISCELLANEOUS EXERCISES 3% Sketch the form of the curve y = — ; ■ — „, and deter- 1 + X -\- x^ mine the maximum and minimum values of jj/. (Qual. Exam. Mech. Sc. Trip. 1914.) 17. State shortly rules for finding the maximum and minimum values of a function of x, and give reasons for the rules. In the theory of transformers, the foUowring expression for cos^ occurs. . _ (i — 0-) sina^ •^"^^ "" {2(1 +0-2) - 2(1 - 0-2) COS26'}i' where o- < i ; prove that when cos^ is a maximum cos 20 = = COSl|t. I + o- (B.Sc. Eng. Glasgow, 1917-) 18. Evaluate the integrals J sm:x; ; sm-'jc prove the reduction formulae, n(Sn+l — Sn) = s'lnznx, Vn+l — Wn = Sn+l, and show that if Vn is taken between the limits o and ^tt, its value is ^mr, when n is an integer. (Maths. Trip. I. 1914.) 19. Sketch the curve 2y^ = x(i — x)^, and find the volume of the solid obtained by rotating the loop about the axis of X. (Qual. Exam. Mech. Sc. Trip. 1919.) 20. Investigate carefully the conditions that f(a) should be a minimum value of the function /(jc). The total work done in a compound air compressor is given by the equation 390 APPLIED CALCULUS where p, the pressure, is the variable ; show that the work done is a minimum when p = >/p\p2- {B.Sc. Eng. Glasgow, 1915.) 21. A spherical raindrop, whose radius is 0-04 in., begins to fall from a height of 6400 ft. , and during the fall its radius grows by precipitation of moisture at the rate of io~* in. per second. Prove that, if its motion is unre- sisted, its radius on reaching the ground will be o* 04205 . . . in., and that it will have taken about 20 '5 sec. to fall. {Int. B.Sc. Hons. Land. 1904.) 22. Trace the portion of the curve;)/ = 8 — ;ic^ for which both X and y are positive. Without further calculation draw the tangent which is equally inclined to the axes. Measure the intercepts made by the tangent on the axes. Find the equation of the tangent at the point whose co- ordinates are a, b. Calculate to two significant figures the values of a, b, for which the tangent is equally inclined to the axes, and the values of the intercepts it makes upon the axes. Compare these values of the intercepts with those found from your drawing. Find by integration the area enclosed by the curve and the axes, and check your result by counting squares. [Army Entrance, Higher Maths. ipiS-) 23. The displacement of the piston of an engine from the mid position is given by y = r cosW ^(i — C0S2W), 4/ where w is the angular velocity of the crank, r the crank radius, and / the length of the connecting rod. If M is the mass of the, piston, find the maximum inertia force arising from it. 24. In the last example, state the percentage of the maximum inertia force which arises from the obliquity of MISCELLANEOUS EXERCISES 391 the connecting rod. Why is the effect of obliquity so much greater in disturbing the inertia force of the piston than it is in affecting its velocity? 25. A fly-wheel, of weight i ton and radius of gyration 3 ft. 6 in., is rotating once every second. What is its kinetic energy, and how long will it take to come to rest under a frictional torque round the axis of 40 lb. -ft.? {Qual. Exam. Mech. Sc. Trip, igig.) 26. The angular displacement of a pendulum is given by e = e^e-^t Slant. Show that the successive maxima of d form a series in geometrical progression. If the time of a complete oscillation is i sec, and if the ratio of the first and the fifth angular displacements on the same side is 4:1, show that the time taken in swinging out from the position of equilibrium to an extreme dis- placement is o- 241 sec. [Maths. Trip. 1. 1917.) 27. Prove that the most economical section of copper to employ in a transmission line is such that the interest on the capital invested in the copper is equal to the cost of the energy consumed in resistance, no other costs being supposed to influence the choice. If the cost of copper is ;^ioo per ton, the interest on the capital invested 6 per cent, the cost of electrical energy 0'75 pence per kilowatt hour, the density of copper 9 gm./cm.^, the specific resistance of copper 1-6 x 10"^ ohm-cm. units, calculate the most economical section of conductors to carry 500 amperes to an outlying station for 12 hr. per day throughout the year. {Mech. Sc. Trip, igis.) 28. A smooth hemispherical bowl of radius a contains a rod of n times its weight, and is suspended from a point on its rim. Show that, if in the position of equilibrium the rod lies entirely within the bowl, it must be horizontal. Prove also that the inclination of the plane of the rim to the horizontal is tan"i2(w -f- i)- 392 APPLIED CALCULUS [The centroid of a hemispherical bowl bisects the radius to its vertex.] {B.Sc. Pass, Land. 1914.) 29. A motor-car of mass fn is driven by a constant force at all speeds, the resistance varying as the square of the velocity. Prove that the time taken to get up a velocity V from rest is 2P '°Hv,-vr where Wq is the full speed and P is the power required to maintain this speed against the resistance. A loaded car of 15 h.p. weighs 2000 lb., and its full speed is 45 m.p.h.; prove, on the hypothesis stated above, that it can attain a speed of 30 m.p.h., starting from rest, in a little under half a minute. [It may be assumed that ^ = 32 in foot-second units and that log«io = 2'303.] {Inter. B.Sc. Hons. Lond. 191 j.) 30. Show how a ballistic galvanometer may be used to measure flux, and how the constant in such a case may be determined. A ballistic galvanometer was connected in series with the secondary of an electromagnetic flux standard and a current of 2 amperes was reversed in the primary of the standard. The first fling observed in the galvanometer scale was 65 divisions, and the tenth swing on the same side 12 divisions. The details of the flux standard were : Turns in primary coil 500, length 40 cm., diameter 8 cm. Turns in secondary coil 500, diameter 7-5 cm. Find the change of flux per scale division. If the resistance of the secondary circuit were halved, what would be the observed fling, assuming no damping effect due to friction? (Mech. Sc. Trip. igi2.) MISCELLANEOUS EXERCISES 393 31. By considering the air enclosed in a cylinder fitted with a piston, prove that the work done by the air in ex- panding from a volume v^ to a volume Wg 's /: pdv, where p is the pressure per unit area. If a quantity of air expands in such a manner that no heat is lost or gained, then p x v" remains constant, where c = I '41. Prove that the work done when the initial pressure and volume are pj^ and v^ and the final pressure and volume are p2 and Wj is Find p2 and the work done if v^ = 10 c. in., Wj = 20 c. in., and /j = 15 lb. per square inch. (Army Entrance Exam. igiS.) 32. Show that in an atmosphere of uniform temperature the pressure, p, at a height above the ground, is given by the equation ^ ^ Ae"''^ where /* = &S and p„ and pf. are the density and pressure A of the air at z = o. A hollow gas-tight sphere containing hydrogen requires a force mg' to prevent it from rising when the lowest point touches the ground ; the total mass of sphere and hydro- gen is M. Show that the sphere can float in equilibrium with its lowest point at a height h above the ground, "^^^'^ ^ I, M+m (Maths. Trip. I. 1915.) 33. If a drop of water is confined in a vessel, show that the increase in vapour pressure due to surface tension, in the state of equilibrium, is given by 8^ = ^^ I, cr — pa 394 APPLIED CALCULUS where Sp is the increase in pressure, cr the density of water and p of its vapour, T the surface tension, and a the radius of the drop. Show that the corresponding' change in vapour density arising from the electrification of the drop is given by — ' P ^^ Rdtr - p Sttka*' where e is the electric charge, k the dielectric constanti R the gas constant, and 6 the temperature. What bearing has this result on the size of raindrops we may expect to accompany thunderstorms? Assume the potential energy of a sphere carrying a . . e^ ~\ charge e is given by 34. Show that the efficiency of the theoretical Diesel cycle is equal to If py-i \ r^-Ayip - i)/' where y is the ratio of the specific heats and r and p the ratios of the maximum volume and cut-off volume respec- tively to the clearance volume. Taking y = i'4, find the value of this efficiency for an engine in which r = 14 and p = i'6. Find also the efficiency of the standard engine of comparison for the same compression ratio. [Meek. Sc. Trip. 1914.) 35. Show how to find the dryness fraction at any point in the adiabatic expansion of a liquid and its vapour, the condition of the mixture being known at one point of the expansion. The specific heat may be assumed constant. If the dryness fraction remains constant during adia- batic expansion, show that the relation between the latent heat and the temperature must be of the form T = ae -T, where « is a constant and o- is the specific heat of the liquid. MISCELLANEOUS EXERCISES 395 [L is the latent heat, T the temperature, and q the dry- ness fraction,] (Mech. Sc. Trip, igij.) 36. In the case of a perfect gas evaluate the definite integral j-=- between any two states defined by v^, i^ and T/j, 4> where dii. is the heat received per pound of the gas at absolute temperature T, irreversible processes being excluded. Also, if the relation between pressure and volume is given by the equation /«" = constant, show that the total heat received is proportional to Tj — Tj, and find the complete expression for it. [Mech. Sc. Trip, igog.) 37. Show by thermodynamical reasoning that if L is the latent heat of vaporization at absolute temperature T of a substance, p the pressure, V the specific volume of the saturated vapour, and la that of the liquid, then (V - h)) ^ L/9T\ J T\dp)v — j is the rate of increase of temperature with pressure, the volume being kept constant. 38. It is known that the discharge of electrons by hot platinum in a vacuum follows the law i = DTie'^, where t is the saturation thermionic current per square centimetre of platinum surface, T the absolute tempera- ture of the platinum, P the work which must be done in enabling as many electrons to escape from the metal as there are molecules in i gm. molecule of platinum, and D is a constant. Show that if the emission of the negative electrons be assumed to be analogous to the evaporation of a liquid. 396 APPLIED CALCULUS the result of example 37 leads directly to the experi- mental law. (H. A. Wilson.) 39. Show thermodynamically that the electromotive force of a reversible electric cell is given by e = x + t(|). where E is the electromotive force, A, the heat evolved or absorbed by the chemical changes accompanying the passage of unit quantity of electricity through the cell, and T is the absolute temperature. 40. Show on thermodynamic grounds that the surface energy of a liquid is not equal to S, the surface tension, but to S — T— =. in r TT T , \ 8T [B.Sc. Hons. Land. 1^14.) The Thermo-electric Circuit. Consider a circuit formed of a copper and an iron wire soldered together into a ring. It is well known that if one junction is heated, a current of electricity will flow in the circuit. This current is called a " thermo-electric " current. The work done in carrying the electric charges round the circuit is provided by the excess of the heat absorbed by the circuit over that liberated by it. The absorption and liberation of heat takes place in two "■^^'ons: (a) At the junctions. (5) In the metals. Peltier (1834) discovered effect (a). The Peltier law is that the absorption (or liberation) of heat is proportional to the quantity of electricity that has passed the junction. The constant of proportionality (the Peltier coefficient) depends on the tempera- ture of the junction and the sign of the heat absorption, on the direction of the flow of electricity at the junction, i.e. Heat liberated per second in passing from B to A at temperature Tj = IIjj. Heat absorbed per second in passing from A to B at temperature Tj = —Vi^. MISCELLANEOUS EXERCISES 397 Kelvin worked out the second effect. He supposed electricity to possess a specific heat a, different for different metals, so that, if o-A is the specific heat of electricity in metal A, the heat absorbed in moving a quantity of electricity (e) from a point at temperature T to one at temperature (T + i) is at^e. Both the Peltier and Thomson Effects are Reversible. By considering the passage of a quantity of electricity e round the thermo-electric circuit, and applying the thermodynamic laws to the reversible interchanges of work and heat, we can arrive at the relationship of n and a to the temperature and electromotive force of the circuit. 41. Show by thermodynamical reasoning that on Kelvin's hypothesis of "the specific heat of electricity" the Peltier ^E effect P at a thermojunction is equal to T—-, and the oT difference of the specific heats of electricity in the two metals of the circuit is — T -~r, where T is the absolute dv'' temperature and E the electromotive force of the circuit. (B.Sc. Hons. Land. igi4.) 42. A volume of an ionized gas loses its ions by diffusion and by recombination. If the effect of diffusion is negligible in a particular experiment, and if there are, initially, as many positive as negative ions, show, by the principle of mass action, that J--J- =. et, where Wj is the initial number of ions, «2 the final number of ions, i the lapse of time, and 9 a constant. This constant is called the coefficient of recombination, and it has been experimentally determined. It is nearly constant for air, oxygen, hydrogen, and carbon dioxide. 43. There is a method of determining the number of 398 APPLIED CALCULUS molecules which take part in a chemical reaction, which depends on measuring the velocity of the reaction at widely different concentrations. The formation of cyamelide from cyanic acid takes place according to the equation «CNOH = C„N„0„H„. An experiment showed the following results : The concentration of the cyanic acid changed from 188-84 to i53'46 in 23 hr., the volume being v. It changed from 79 '01 to 76-04 in 20 hr., the volume being V. Show that , , . ,. . _ ■°^-(l).-Mg)v logeC„ — logeCv and hence deduce that the reaction is a trimolecular one. 44. Deduce the differential equation of mass action for a trimolecular reaction in the form g = k{A-x){B-x)(C-x), where A, B, and C are constants and x is the concentra- tion of one of the reacting substances. Interpret these constants, and show that the integral of the equation is ~ 4C-A)\A-B '°^'A(B^ + C-B '°^'B(C-;«r)/- 45. If A = B = C, the solution in example 44 breaks down. Show that then _ x{2k — x) " " 2A'''(A - xfi Experiments were made on the reaction between ferrous chloride, hydrochloric acid, and potassium chlorate at 20°C. The quantity x of ferrous chloride used up was measured by titration with permanganate solution. MISCELLANEOUS EXERCISES 399 The hydrochloric acid was decinormal, and the ferrous chloride and potassium chlorate of corresponding strength, so that A = B = C = o-i. The quantitative reaction is 6 FeClj + KClOj + 6 HCl = 6 FeClg + KCl + 3 H^O. The reaction proceeded as follows : Time in Minutes. tox. O O'OOO 5 0-048 15 0'122 35 0-238 60 0-329 no 0-452 170 0-525 Show that the reaction is trimolecular and k is nearly unity. How do you reconcile this result with the chemical equation stated above? 46. Consider the formation of ethyl-acetate from ethyl- alcohol and acetic acid at 25° C. C3HP2 + C2H5OH = C.HgOa + H^O. (acetic acid) + (alcohol) = (ethyl-acetate) + (water). Initially the concentrations were, in Knoblauch's experi- ment: /-. _ , '-'acid — 1 ) Cwater ^ Calcohol ^= 12- 75"' During the initial stages of the reaction t-^ethyi-acetate _ 0-00303 mols per litre per minute. Find the reaction constant (kj), assuming it is a di- molecular reaction. 47. The reaction in example 46 is reversible, and can be studied by starting with a mixture of ethyl-acetate and water, and measuring the rate of formation of acid. 400 APPLIED CALCULUS In Knoblauch's test, he started with Cethyl-acetate ^^ Ij Calcohol = Cwater ^ 12 "2 15* The experiment gave AC ^"^ = 0-000996 mols per minute. Calculate the reaction constant (kj). 48. The reaction in example 46 was allowed to go on until equilibrium was established, and no further increase in ethyl-acetate took place. It was then found that Cethyl-acetate = 0-7144 mols. Calculate the final concentrations of the acid, water, and alcohol. Show that Cethyl-acetate X Crater __ « ^ o . Oacid X Caioohol Check this result with the experimental data of (47). 49. Experiments were made on the conversion of dibrom- succinic acid, in aqueous solution, C^H.O.Brg = C.HgO.Br + HBr, a monomolecular reaction. The reaction constant was measured at different tem- peratures with the following results: T° C. K (Time being- in Minutes). 15 0-00000967 40 0-0000863 80 0-0046 loi 0-0318 Show that these results are consistent with a formula 'o&io" = a + bT. What conclusions do you draw as to the general effects of temperature in promoting chemical reactions? MISCELLANEOUS EXERCISES 401 50. Cane-sugar, when treated with acidulated water, changes into two sugars of the same composition, but of different optical properties. Cane-sugar rotates a beam of polarized light in the right-handed direction; the "inverted" sugars are, on the whole, laevo-rotatory. The reaction is expressed by the chemical equation, C12H22O11 -I- HgO = CgHigOg -I- CgHjgO,., (cane-sugar) + (water) = (dextrose) + (laevalose) and its progress can be followed with the polarimeter. The reaction is monomolecular, as the water is greatly in excess, and the concentration of it is nearly constant. In one set of experiments, the reaction constants were «■ at 25° C, 0-765 mol per minute. K at 55*^ C, 35 '5 mols per minute. Show that the time taken for the concentration of the cane-sugar to fall to ^jj of its initial value is 3 '01 min. at 25° C. and 0-0648 min. at 55° C. Note. — Examples 43-50 inclusive are based on data given in Van 't Hoff 's Lectures on Theoretical and Physical Chemistry (Arnold, 1898). The Examination Papers of the University of London are published regularly by, and may be obtained from, the University of London Press, Ltd., 18 Warwick Sqiiare, London, E.C.4. (D102) 27 BIOGRAPHICAL NOTES (See Portraits) Napier, John (1550-1617), was born at Merchiston — now a part of Edinburgh. He lived through the years of the Reformation, and took a prominent part as a zealous protestant in the religious and political discussions of his day. He wrote a com- mentary on the Book of Revelation with propositions and mathematical demonstrations, in which proposition 21, for example, is a proof that the Pope is Anti-Christ, and proposition 36 shows that the apocalyptical "locusts" are the Turks and Mohammedans. There was a craze in his day for enormously long numerical calculations. One contemporary, for instance, practically devoted his life to finding a numerical approxi- mation to the value of it, and finally obtained it correct to 35 places of decimals — in fact, the character Sir Walter Scott gives to David Ramsay in the Fortunes of Nigel is exactly that of a typical sixteenth century mathe- matician. Napier was impressed by the enormous labour of these calculations, and set about trying to find a means for simplifying them. The result was his invention of logarithms, which seems all the more marvellous when one bears in mind that Napier had no index notation to help him. The index notation was invented years afterwards by Descartes. Napier thought of the relative motion of two points connected with one another in the following way: as one point moved in a straight line with uniform velocity. 404 APPLIED CALCULUS the second point moved in another straight line with an accelerated velocity, so that as one point moved in a series of steps in arithmetical progression, the other point would move in a series of steps in geometrical progression. He put these ideas into numbers and developed his system of logarithms. Kepler, Johann (i 571 -1630), a German astronomer, was born near Stuttgart. His main contribution to astronomy is the group of three laws known as Kepler's Laws. He analyzed an enormous mass of astronomical observations, and dis- covered empirically the three laws in accordance with which the planets revolve round the sun. Newton showed that his law of gravitation was the only one consistent with Kepler's Laws. Ill-luck dogged him all his life. He first succeeded Tycho Brahe as astronomer to Rudolph II, Emperor of the Holy Roman Empire, but this empire proved bankrupt and could not pay him his wages. Then his wife lost her reason and died ; his second marriage was also unfortunate; and, finally, he was deprived of an appointment he held for heresy, and narrowly escaped with his life. He cast horoscopes and told fortunes, for which he charged heavily ; but his conscience seems to have pricked him for imposing on the credulity of his age, for he wrote: " Nature, which has conferred upon every animal the means of existence, has designed astrology as an adjunct and ally to astronomy ". Out of it all, however, he gave the world the laws of planetary motion and the idea of the integral calculus. Descartes, Rend (1596-1650), a French mathematician and philosopher, and a contemporary of Galileo, was born near Tours. As a lad he suffered from ill-health. " On account of his delicate health he was permitted to lie in bed BIOGRAPHICAL NOTES 405 till late in the morning's ; this was a custom which he always followed, and when he visited Pascal in 1647 he told him that the only way to do good work in mathematics and to preserve his health, was never to allow anyone to make him g'et up in the morning before he felt inclined to do so; an opinion which I chronicle for the benefit of any schoolboy into whose hands this work may fall."^ He became a soldier by profession, and he is said to have thought of the method of co-ordinate geometry — Cartesian Geometry, as it is now named after him — in three dreams which he dreamt while campaigning on the Danube. He does not appear to have been a particularly amiable man. His appearance was forbidding", and his personality seems to have been even more forbidding than his ap- pearance. He despised learning and art, except for their utility. He is celebrated in mathematics for his invention of co-ordinate geometry, and in philosophy for his dictum, " Cogito, ergo sum"; a dictum which called forth from Carlyle the comment: ^^ Cogito, ergo sum. Alas, poor Cogitator, this takes us but a little way. Sure enough, I am; and lately was not : but Whence? How? Where to? The answer lies around, written in all colours and motions, uttered in all tones of jubilee and wail, in thousand- figured, thousand-voiced, harmonious Nature." Sir Isaac Newton {1642-1727), the English mathe- matician and natural philosopher, was born near Grantham in Lincolnshire on Christmas Day. His father was a yeoman farmer, and it was intended that Isaac should carry on the farm. At the age of fourteen he showed distinct mechanical ability, qnd his widowed mother decided -to give him a chance to develop his mechanical instincts. He was sent to Trinity College, Cambridge, and went through the ordinary college course 1 Ball's History of Mathematics. 4o6 APPLIED CALCULUS in classics and mathematics. He accidentally came across a book on astrology which interested him, but he could not understand it on account of the geometry and trigo- nometry in it. So he bought a copy of Euclid, and was surprised to find how plain it all seemed. On account of the great plague, he left Cambridge in 1665-6 and went to live at his home in Grantham. It was during this visit that his ideas on gravitation took definite shape. He also invented the differential calculus about this time. In i66g he was appointed Lucasian professor in Cam- bridge, and gave much of his time to the study of optics. His writings were attacked by many of his contemporaries with,an arrogant confidence born of ignorance, and Newton had an uphill task from 1672 to 1675, and more than once determined to give up philosophy except for his own satisfaction. " I see I have made myself a slave to philosophy; but if I get rid of Mr. Linus's business, I will resolutely bid adieu to it eternally, excepting what I do for my private satisfaction, or leave to come out after me; for I see a man must either resolve to put out nothing new, or to become a slave to defend it." Of Newton's absent-mindedness many stories have been told. A typical one narrates how, when he was enter- taining some friends to dinner and went to draw some wine, he stayed away so long that his friends set off to seek him. They found him in the wine cellar, busy solving a mathematical problem, his friends forgotten, and the jug still unfilled. He was no "eight hours' day" man, and would often spend eighteen or nineteen hours out of the twenty- four at the most exhausting kind of work. Newton had the strongest sense of duty. When he was appointed Master of the Mint, he stoutly refused to engage in any of his favourite scientific pursuits, as he felt that his whole energies should be given to the service of the State. He broke away from this rule once, for BIOGRAPHICAL NOTES 407 Leibnitz apparently was too much for his patience. The German mathematician had been working at a set of problems and was unable to solve them. He thereupon concluded they were insoluble, and published them as a challenge to the world. Newton received the problems one evening on his return from the mint, and at the end of five hours had solved them all. He spent nearly half his time studying chemistry and theology. Once when Halley made some flippant remark on a question of religion, Newton fell upon him with the rebuke: " I have studied these things: you have not! " His greatest contributions to physical science were his theory of gravitation, by which he accounted for the motions of the bodies of the solar system, and of the moons around the planets ; and the differential calculus. He made many other important, though minor, contri- butions to mathematics and physics. Here is Newton's own estimate of himself: "I do not know what I may appear to the world ; but to myself I seem to have been only like a boy, playing on the sea shore, and diverting myself, in now and then finding a smoother pebble, or a prettier shell than ordinary, while the great ocean of truth lay all undiscovered before Leibnitz, Gottfried Wilhelm (1646-1716), one of the most brilliant men of his time, was born at Leipzig. He was educated first at Leipzig University, which refused him the degree of Doctor of Laws, on account of his youth. He thereupon removed to Nuremberg, where he continued the study of the law and subsequently entered the diplo- matic service. In 1674 Leibnitz became • attached to the Court of Brunswick, and from that time he had greater leisure to devote to his favourite pursuits of mathematics and philosophy. He propounded his views on the differential and integral calculus between 1674 and 1677, and an 4o8 APPLIED CALCULUS acrimonious debate ensued among contemporary mathe- maticians as to who was really the inventor of the differential calculus, Newton or Leibnitz. The dispute seems to have been largely a matter of words. Newton appears to have discovered for himself the view of the differential calculus as "a rate of growth", and he expressed a rate of growth or a fluxion, as he called it, by putting a dot over the letter, just as is done in dynamics at the present day. The dot is satisfactory enough when "time" is the independent variable, and when the rate of growth is a speed, but it is not very convenient when other independent variables are used. Newton used a method akin to the modern method of limits to arrive at these fluxions. Leibnitz, on the other hand, invented the dyjdx notation, which is used to-day. This notation is not altogether free from objection. It suggests first of all that the quantity represented by dyjdx is a fraction equal to dy divided by dx, and secondly, the first thing the beginner wants to do is to cancel the d's. However, Leibnitz's notation appears to be now standard- ized, except in dynamics, where the Newtonian notation with dots is still commonly used — and convenient. Leib- nitz's reputation rests as much on his philosophical as on his mathematical work. He was the propounder of a system of philosophy known as Monadology. It is curious that the practical experimenter, Newton, developed the calculus on more philosophical lines than Leibnitz, whereas the philosopher, Leibnitz, gave the world the practical notation which is used to-day. Volta, Alessandro (1745-1827), an Italian physicist, is chiefly celebrated as a pioneer in electrical science. He was born at Como. The practical unit of electrical pressure — the volt — is named after him. Faraday, Michael (1791-1867), an English physicist and chemist, was born at Newington, Surrey, of York- BIOGRAPHICAL NOTES 409 shire stock, and was the son of a blacksmith. He was apprenticed to a bookbinder, and educated himself. He was " discovered" by Sir Humphry Davy, who made him a laboratory assistant at the Royal Institution, and took him on a tour through France, Italy, and Switzerland. He occupied diflferent posts at the Royal Institution, and finally became professor of chemistry there. He was an experimenter of the highest skill, and pos- sessed great powers of physical reasoning. Sir J. J. Thomson says: "Faraday, who possessed, I believe, almost unrivalled mathematical insight, had had no training in analysis, so that the convenience of the idea of action at a distance for purposes of calculation had no chance of mitigating the repugnance he felt to the idea of forces acting far away from their base and with no physical con- nection with their origin". His reputation rests chiefly on his electrical discoveries. He would have nothing to do with " action at a distance ", and insisted on the " axiom " that matter cannot act except where it is. This led him to picture space as filled with a medium — ether — and to ascribe the electric and magnetic field to states of stress and strain in this medium. He pictured the medium as filled with "tubes of electric and magnetic force", and proved that the phenomena of electromagnetic induction — • which he had discovered — could be accounted for as an effect produced by a change in the number of " Faraday Tubes " passing between the two circuits concerned. The same theory accounted for the induction of currents in closed circuits by magnets in motion relative to them. Carnot, Sadi Nicolas Leonard (1796-1832), was a French army officer and a contemporary of Fourier. He was the son of an eminent geometrician. In 1824 he published a paper " Reflexions sur la puissance motrice du feu ", which laid the foundations of modern thermo- dynamics. Though his paper is based on what is now considered a wrong notion, namely, that heat is a material 4IO APPLIED CALCULUS substance, only a slight modification Is needed to bring his work into line with modern views on this point. His methods are simple, and yet most original and profound, and they form the basis of all later work on the relations between heat and mechanical energy. Henry, Joseph (1797-1878), an American physicist, was born in Albany, N.Y. He made several practical electrical inventions, and carried out important experiments on electrical induction. The practical unit of electromag- netic induction — the Henry — is named after him. Graham, Thomas (1805-69), a Scottish chemist, was born in Glasgow. He was the son of a merchant, and was educated at Glasgow University. He gave special study to the subject of molecular physics, and made im- portant contributions to our knowledge of the diffusion of gases and of the properties of colloids. The subject of colloids is becoming important because of its industrial applications. Joule, James Prescott (1818-89), ^n English physicist, was born at Manchester. His chief scientific work consisted in the proof of the relation between mechanical energy and heat. He demonstrated that if mechanical energy disappears in a system of bodies and no other physical change takes place except a change in the temperatures of the different bodies of the system, then for every 778 (roughly) foot-pounds of mechanical energy that disappear, i B.Th.U. of heat makes its appearance. Thus heat and mechanical work are interchangeable forms of physical energy. Clausius, Rudolf Julius Emmanuel (1822-88), a German physicist, was one of the pioneers of the science of thermodynamics. He was a contemporary of Lord Kelvin, and these two physicists covered much the same BIOGRAPHICAL NOTES 411 ground in thermodynamics by different methods. Clausius also did very important work in the kinetic theory of gases and in the theory of electrolysis. Kelvin, Lord (Sir William Thomson) (1824-1907), was born in Belfast. His father was Professor James Thomson, an Irishman, and his mother, Margaret Gardiner, the daughter of a Glasgow merchant. The family settled in Glasgow in 1832, and William Thomson was educated at Glasgow University and later at Peterhouse, Cambridge. Lord Kelvin's main scientific work consisted in the contributions he made to the theory of the conservation of energy. He lived in an age of great mechanical develop- ments. Watt's steam-engine had been invented before he was born, but the theory underlying its action was only beginning to be understood about 1840. Electrical de- velopments were taking place rapidly, and these industrial applications of electric power depended, in their scientific aspects, mainly upon the principle of conservation of energy. He applied himself to thermodynamics, and did more perhaps than any other man to place the theory on its present basis. He also laid the foundations for the exact measurement of electrical quantities, and so brought the science of electricity and magnetism within the range of exact treatment. In the realm of theoretical physics, Lord Kelvin made important contributions to the theories of hydrodynamics and elasticity. He also constructed a hydrodynamical theory of the constitution of matter, based on vortex atoms in an infinite fluid. In collaboration with Professor Tait of Edinburgh he wrote a Treatise on Natural Philosophy, which entirely changed the standpoint from which the subject was viewed in this country. He took part in the laying of the first submarine cable to America, and acted as adviser in many engineering and industrial enterprises. Clerk Maxwell, James (1831-79), a Scottish mathe- matician and physicist, was born in Edinburgh. He was 412 APPLIED CALCULUS educated at Edinburgh Academy, Edinburgh University, and Cambridge. He was Professor of Natural Philosophy in Aberdeen from 1856 to i860, and subsequently became Cavendish Professor of Physics at Cambridge. Working on Faraday's physical ideas of the electro- magnetic field, he investigated them mathematically with the highest skill, and showed that an electromagnetic theory of light could also be based on the Faraday-Max- well equations. These equations are still the "last word" in electrical theory, and are the starting point of the modern theory of relativity, which centres round Einstein's name. He was a man of great intellectual power. P. G. Tait writes in the EncyclopcBdia Britannica: "In private life Clerk Maxwell was one of the most lovable of men, a sincere and unostentatious Christian. Though perfectly free from any trace of envy or ill-will, he yet showed on fit occasion his contempt for that pseudo-science which seeks for the applause of the ignorant by professing to reduce the whole system of the universe to a fortuitous sequence of uncaused events." Guldberg, Cato Maximilian (1836-1902), a Norwegian mathematician and physicist, was born at Christiania. He was the eldest son of Carl August Guldberg, a clergyman. He was educated in the Royal Norwegian University at Christiania, and subsequently studied in France, Switzer- land, and Germany. In 1869 he was appointed Professor of Applied Mathematics in the Royal University, a position which he held until his death. He devoted himself from his student days to the study of chemical dynamics. At the age of twenty-eight he contributed a paper (in associa- tion with Professor Waage), on the Law of Mass Action, to the Society of Science in Christiania. This was followed by a second paper in 1867 entitled: " fitudes sur les affinit^s chimiques ", and by a third paper in 1879, in which the theory of Mass Action was given in its final BIOGRAPHICAL NOTES 413 form. He contributed many papers to learned societies on Physical Chemistry. He was a keen hunter and fisher- man, and was for a long time President of the Norwegian Hunting and Fishing Society. Gibbs, Josiah Willard (1839- 1903), an American mathematical physicist, was born in New Haven, Connec- ticut, U.S. He was educated at Yale, and later became professor of mathematical physics there. He studied specially the applications of thermodynamics to chemistry, and has been called " the founder of chemical energetics". He wrote important papers on Vector Analysis. His name is especially associated with a theorem in physical chemistry known as the "Phase Rule". Van 't Hoff, Jacobus-Hendrikus (1852-191 1), a Dutch chemist, was born at Rotterdam. He was the founder, with Le Bel, of Stereochemistry, or the chemistry of space, in which the formula of the substance is represented not in one plane, as the usual graphical formulae are, but in the solid; for instance, CH^ is represented by a carbon atom at the centroid of a tetrahedron with a hydrogen atom at each corner. He also did a great deal of work on chemical dynamics and the law of Mass Action. The reader who is interested in the history of mathe- matics and its devotees, should read A Short Account of the History of Mathematics by W. W. Rouse Ball. 414 APPLIED CALCULUS TABLE I LIST OF INDEFINITE INTEGRALS f , ^""*^ 1 ix^dx = — ; — , unless » = — i. / « + i' U = logex, or \ose{-x). 3. jel^dx = eF. I siaxdx = — cosa;. 5. I cos;*;^ = sin^;. 12 13 14, IS 17 18 lsec';c 'f ^^ < 'i^^- 2a a + X j\/{a^ — x^)dx = \x^J{a^ — x?) -\- \a^ arc sin-. fV(^ + k)dx = ^x^[x? + k) + \k \oge{x + y/x^ + k). I e"* cos6;k dx = -= -Ja cosbx + b sinSjc). i (V -\- 0' |e"* sinbxdx = -3 r^(fl smbx — 5 cos6:x:). J a^ + V l— — = logetanijc. i6. j — ~ = logfetanl— H — ). J sinji; <=> ^ J cos^ \4 2/ lsIn^.«r^:K = lj(i — cos2x)dx = Jjc — ^ sin2jt:. Icos'^.r^Ji: = I J(i + cos2a:)^:«; = ^jc + ^^ sin2j[r. These are all standard integrals which are constantly occurring' in the applications of the calculus. The first ten at least should be com- mitted to memory. APPLIED CALCULUS 415 TABLE II NATURAL LOGARITHMS X 1 . I 1.2 1-3 1-4 1-5 1.6 1-7 1.8 1-9 2.0 2. I 2-2 2-3 2.4 2-5 2.6 2.7 2.8 2.9 3-0 3-1 3.2 3-3 10.5 II. "-5 12.0 12.5 13.0 13-5 loge* X 3-4 3-5 3-6 3-7 3-8 3-9 4.0 4-1 4-2 4-3 4-4 4-5 4.6 4.7 4.8 4.9 5-0 5-1 5-2 5-3 5-4 5-5 5-6 14.0 14-5 15.0 15-5 i6.o 16.5 17.0 log«« X log-e^ X 8.0 8.1 8.2 8.3 8.4 8-5 8.6 8.7 8.8 8.9 9.0 9.1 9.2 9-3 9-4 9-5 9.6 9-7 9.8 9-9 10. 22.0 23.0 24.0 25.0 26.0 27.0 28.0 loge^ •0953 .1823 .2624 -3365 -405s .4700 -5306 •5878 •6419 .6931 .7419 .7885 -8329 •8755 .9163 •9555 •9933 1.0296 1.0647 1.0986 1.1314 1 . 1632 I . 1939 2238 2528 2809 3083 3350 3610 3863 4II0 4351 45S6 4816 5041 5261 5476 5686 5892 6094 6292 6487 6677 6864 7047 7228 5-7 5-8 5-9 6.0 6.1 6-2 6-3 6.4 6-5 6.6 6.7 6.8 6.9 7.0 7.1 7-2 7-3 7-4 7-5 7.6 7-7 7.8 7-9 1-7405 I • 7579 1.7750 I. 7918 1.8083 1-8245 1.8405 1-8563 I. 8718 I. 8871 I .9021 I. 9169 I -9315 I -9459 I .9601 I .9741 1.9879 2.0015 2.0149 2.0281 2.0412 2.0541 2 • 0669 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0794 0919 1041 1163 1282 I4OI 1518 1633 1748 1861 1972 2083 2192 2300 2407 2513 2618 2721 2824 2925 3026 2-3513 2-3979 2.4430 2.4849 2.5262 2.5649 2-6027 2.6391 2.6740 2.7081 2.7408 2.7726 2.8034 2.8332 17-5 18.0 18.5 19.0 19-5 20-0 21 .0 2.8621 2 . 8904 2.9173 2.9444 2.9703 2.9957 3-0445 3.091 I 3-1355 3.I781 3.2189 3-2581 3-2958 3-3322 4i6 APPLIED CALCULUS TABLE III EXPONENTIAL FUNCTIONS X e» e-'' -t e" e-'^ •05 1-0513 ■9512 •55 1-7333 •5769 • I I • 1052 .9048 -6 I -8221 -5488 •15 I •1618 •8607 •65 1-9155 -5220 •2 I -2214 .8187 •7 2-0138 -4966 •25 I • 2840 .7788 'P 2- 1 170 •4724 •3 1.3499 -7408 • 8 2-2255 •4493 •35 1-4191 .7047 •85 2.3396 •4274 •4 1. 4918 •6703 •9 2-4596 -4066 •45 1-5683 •6376 •95 2-5857 .3867 •5 1-6487 .6065 I-O 2-7183 .3679 I'l 3-0042 3329 3-1 22-198 • 0450 I -2 3.3201 3012 3-2 24-533 -0408 1^3 3-6693 2725 3-3 27-113 •0369 1.4 4^o552 2466 3-4 29-964 •0334 i-S 4-4817 2231 3-5 33-115 -0302 1.6 4 •9530 2019 3.6 36-598 -0273 1.7 5-4739 1827 3-7 40-447 -0247 1.8 6-0496 1653 3-8 44-701 -0224 1-9 6-6859 1496 3-9 49-402 -0202 2.0 7-3891 1353 4-0 54-598 -0183 2-1 8-1662 1225 4-1 60-340 -0166 2-2 9-0250 1 108 4-2 66-686 -0150 2-3 9-9742 1003 4-3 73.700 -0136 2.4 11-023 0907 4.4 81-451 -0123 ^•5 12-183 0821 4^5 90-017 -OIII 2-6 13-464 0743 4-6 99-484 -OIOI 2.7 14-880 0672 4-7 109-95 -0091 2.8 16-445 0608 4-8 121-51 -0082 2-9 18-174 0550 4-9 134.29 -0074 3.0 20 - 086 0498 5-0 148-41 -0067 ANSWERS TO EXERCISES AND HINTS FOR SOLUTIONS OF THE MORE DIFFICULT PROBLEMS Page 12 (i) A straight line passing througli the points (o, 3), (— f , o). (ii) A circle of radius 2 units, (iii) The curve lies entirely on the right-hand side of Y'OY, and is syrhmetrical about OX. It passes through the origin and is called a parabola. Page 25 I. 8. 2. 219. 3. I + 4a. 4. I. Page 41 I. 0-9696 c. in.; 432-1 c. in. per inch. 2. 602. Exercise 1 I. The lower limit Ai is 16-50 sq. in. The upper limit Aj is 21-50 sq. in. The difference is 30-3 per cent of A,. 2. A^ is 18-15 ^l- '"•> ^2 is 20-65 sq. in. The difference is 13-8 per cent of Aj. The true area is 19-635 sq. in. Ax o-i o-bi O-OOI O-OOOI -» Ay I -21 0-1201 0-012001 0-00120001 — > Ay Ax I2-I 12-01 I2-OOI I2-OOOI — > 12 (D102) 28 4i8 4. APPLIED CALCULUS ^x O'l O-OI O-OOI — >■ Ay 19-441 I -922401 O-I92O2400I — > Aj/ 194-41 192-2401 192-024001 — > 192 5. 4:1:*. 6. (i) sar*, (ii) 8a:', (iii) ii:r>». Page 53 7. 0-028 sq. in. per sec. 63° nearly. The effect of the scale is to make tan^ = 10 tan^, if is the actual angle on the graph. Exercise 2 I. 3* + yj)/ = 27; tanif' = - f i.e. v^ = 156° 48'- 2. j/ = -i-.^ + (7 — ^3)- 3. jf = 2X+ 13. 4. tanif' = — |, .•. i^ = 143° 8' nearly. 9. 5-41 m.p.h. dj/ dx 4,^ = 4x— i). 6.1^. dt 8. 2-43 c. in. 10. 0-51 c. in. Exercise 3 I. 2540, 3780; mean 3160, 2. 3160. 3. 3124. 4. 116, 142; mean, 129; yes; that the integral of e^"^ is e*. 5. 1-98; yes; that the integral of sin^r is — cosj;. 6. 0-033. 7. J' = / vdt + c, where c is a constant ; s = j ^lit^dt = yu | 2^^dt — /nT'. The distance travelled is proportional to the cube of the duration of the journey. Page 97 I. If jc — > b when ar — > «, then an 11 can be found so that o < Ij — b\ < e when o < | :r — a | < 7; whatever e may be, and \y — b\ = \ (y + a) — (a + b) \ , whence result. 2. Let X = a -{■ 5 where 8 may be positive or negative. Then x^ — a^ = S{3a^+3a5 + S^), i.e. | ^ - aM < I 8 I M, where M is the greatest numerical value of 33^ + 3^^ + ^^ Page 104 I. 0-5236; 0-7854; I-S7I. 2. 57° 18'; 28° 39'; 88° 14'. ANSWERS 419 Exercise 4 I. fl'. 2. slna. 3. J. 4. 0-7540; o-gogi; I-8736. 5. 19° 6'; 171° S3'; 315° 8'. 6. If arc tanx = 0, then ^'"'^ ^^"^ = -^. Now apply equation (12), p. 106. 9. The error occurs in the fourth line. We divide by {x — a) to get (.r + a) = fl and (x — a) is zero if ;ir = «. Exercise 5 I. Because acceleration is rate of change of velocity with time, while velocity itself is rate of change of distance or displacement with time, i.e. velocity is distance per second, and acceleration is velocity per second, i.e. (distance per second) per second. The accelerations are the same, for i mile per hour per second is 3j\yg mile per second per second, and i mile per second per hour is jf^xs mile per second per second. 2. 205 sec. ; 1-95 miles; 34-39 m.p.h. 3. At 25 m.p.h. Pv = mfv, hence H P = I po X 2240 X 1-25 X 88 X 25 X 88 -] ^ SSoL 32-2 X 60 X 60 J 4. F = —7=1 ^ x^dx = ^wP. 5. I = I p2irrdrr^ (j> being the surface density) = 2irpl r^dr = where M = trt^p. 6. Divide the plate into sti-ips parallel to the axis, and integrate. 7. 74 Ib.-in.^. 8. 224 Ib.-in.''. In the second case a large part of the total mass is between 5 and 6 in. from the axis, and hence contributes heavily to the moment of inertia, which depends on the square of the distance of the mass from the axis. 9. 447 lb.-in.2; 36-8 Ib.-in.2. 10. The matter should be as far as possible from the axis about which the moment of inertia is to be taken. Yes. II. 85 % of the mass is more than 2' 9" from the axis and less than 3' o" from the axis. Hence the moment of inertia of this part of the mass is less than 0-85 X 3^ = 7-65 ton-ft.^ and is more than 0-85 X 2-75^ = 6-43 ton-ft.^. The remaining mass contributes moment of inertia less than 0-15 X 2-752 = I -16 ton-ft.^ and more than zero. Hence max. I is 8-81 ton-ft.^ and min. I is 6-43 ton-ft.^; mean I is 7-62 ton-ft. The error in this figure cannot exceed 15-6 %. 12. 8-44 ton-ft. 2. 13. 38-82 X lo^^ ton-mile^. 14. E — irP. 420 APPLIED CALCULUS Exercise 6 I. jxi. 2. 2x^ + 2:r + I. 3. 2X + a + b. 4. I + 3^^ + 4^ + 6^. B. - nx—K 6. ^A^. 7. '"^"'"' 8. . ^ - 9. 2m(2ax + x^)'«-^{a + x). 10. — 3£— r— "■ sec'';!;. 12. sear tanjr. (Rule 7.) (i + x^)i 13. — cosecj: cot-r. (Rule 7.) 14. — cosec^. 15. 3 0033^:. 16. cos3.r cos2j;+2 coss:r. 17. — sirtr cos (cos:r). (Put ^ = cos.r.) ,3 j+asiux+bcosx_ (^^,g y ,g_ ^ ^^^^ _ ^ ^j^^ {b + cosjr)^ + cos2:(;-sin2;<;. (Rule6.) 20. j ^ cosj: (Rules2and8.) 21. ?£^. 22. -«. 23. ^-a;."^". 24. ("^" + f >"'". 4 JT m -\- n na(m + i) 25. ("^ + f)"*"^' . 26. - 3 cosjc. 27. i sin^j;. a(?;z +1) 28. secj;. (Ex. 12 above.) 29. 10 tan:v. (Ex. 11 above.) 30. — \ cos2jr. Exercise 7 I. I =p/ x'dx where p is the linear density; 163-3 Ib.-ft.^. 2. 0'4244r from the centre of the circle along the axis of sym- metry, r being the radius. 3. %r from the centre of the sphere, along the axis of symmetry, r being the radius. Example, p. 158 Examples, p. 162 I. (i) — 2 cos '/x (put z = Vx) ; (ii) Vi -)- x^ (put z = i -{• x'); (iii) -| arc sin- (put x = a sin^, whence integral 22a depends on jcos^Sde; integrate this by parts); (iv) arc sin*- (put X = sin9); (v) arc tan;i; (put x = tanff); (vi) i arc sec- (put a a X = a sec 6). 2. (i) x^ sin:r + 2x cosar — 2 sin^r; (ii) - sin2;r + Jcos2*-; (iii) — arc sinj; -|- — Vi — aj^ + fVi — x'^\ 2 3 9 ANSWERS 421 (iv) — Vi — ;tr2 arc sin^r + x. 3. Apply formula (7), p. 160. 4. By integration by parts make I sinP^ cos«^ dd depend on jsin''-''^ cosiOdd; sin^-'^ cos« + '^ vanishes when ^ = o or 2 hence (^ slnod cosid dd — ^ (^ s'lnV-'O cos^O dd ; hence Jo P + lJo [2 sin'ifl cos^e de = H i^cosW de = m\ sin6l cos6l + 6^ ' -'0 L Jo Exercise 8a '• — !=■ 2. — sinj; cos(cos at). 3. cos3;ir cos2;r + 2 0055^:. 4. (i — 2jr — x')l(i + x^f. 5. sin2;r. 6.^^,. 7. a2/(a2 _ :r2)5. 8. ./i^. I + ^"^ V I +^ 9 2 _L 2 • 2 • '0- tanlr. II. , ' cos2;tr + ?j2 sm^'a: ^(i — ;«r2) (l — x) "2- ;7^^- "3- ,, 15. ^V(i + cosecx). Vi — .r^ V(i — 2X — x'-). ^ ' 14. (a + .r)»-i(i + Ar)»-i{;»(5 + :»■) + «{« + x)}. 16. "^'' (i + j;)"*!' 17. arc sina: + _ ^ .. 18. |. 19. 2^ arc tan- + a. VI — x' a X^ Vl—X^ V « 20. sec2:r arc tana: + i^^. 21. -?. . 22. arc tan 'i—x^ 23. -f-,. 24. , ] ,, . 25. - ^"-^""'. 26. — !— 27. sin2:r(3-4sin2j;). 28. ^-^„ ^ -- ■ 29. a(sin.r + cosx). I + a:'' Vi + .r* 30. J' -= ^_f = -^j-^. Express -if as a function of ^, and dx dz dx dzl dz dx then eliminate z by means of .r = a(x — cos^). Exercise 8b j — a;i; » . 2. *■ , , \ . 3. —7= arc tan,' x / 3 \ W2 + » a(ra + i) V6 \ V 2/ 422 APPLIED CALCULUS [Consider integral /^-^,^.r = i/-^- Put. = ^l., hence l[_iilL- = ' f^^. Now apply Ex. i (v), p. 162.] a 5. Remove brackets. Result is ^=^ + 6. Assume 3 4 g + 1 = ^^ whence integral is ^,^J-j-^, - 7. ^"^ J ^^' . 8. - ^^(i!+l). 9. tan^r + t^!^ (put tan;c = t). X 3 10. siar — ar cos:r (integrate by parts). II. i(i + ar^) arc tana; — \x (integrate by parts). 12. (« + *.r + cxfl^. 13. Result is ?^ + *^. 14. J- ' - ' 4 5 ' ' \^-xf (i-*)' 15. 2f — 2 sin;r + ^^^ (put i — cos^r = 2 sin^?, or expand 2 4 2 /^jl^ _1_ Mm+l (i — cosjr)^ and integrate term by term). 16. ^ — ^ ' . «fl(OT + I) (put a:r" + S = a). 17. / tan^S = / tan^fl tan2"-=e dB, whence the result. — (—1)" means that the sign of tan* is negative if n is even and positive if n is odd. 18. +£ — £ r. 9 9 9V-3 arc tanj;^ ^. (Repeated integration by parts, the integral finallydepending on question 3 above, or by division.) 19. x tan-. 20. (■^ + — )— r — 2 — I arc tan^r. 21. J{2arctan arc tana;}. L(x^ + i) (x^ + 4) (^^ +1) x^ + 4 tegrate these fractions. 22. sin»«^ sin«:r = ^ cos(»2 — n)x — J cos(ot + n)x. Hence ii m =i= n, I sinmx s'mnx dx ^ ^ jcos{m — n)x dx — \ jcos{m + n)x dx. If m = n, I sinmx sin«.r dx = Ism^mx dx ^ ^ /(i — cos 2»z:i:) rf:r. The result follows directly from these integrals when taken between the limits given in the question. 23. Express the ANSWERS 423 given product as a sum or difference of cosines of multiple angles. 24. This result follows from Example 9 above. 25. If 4'{x) = ^(a + x) for all values of x, ^{x) must be a periodic function of period u, whence the result is obvious graphically. 26. (!>), whence the result. -j-it>(x) must be positive for all values of x between (and in- dx n eluding) a and J. 27. {a)(b — a) > / 'P{x)dx > ^(i) (5 — «). -=-{x) must be negative when a < x < b. 28. If (x) ax is negative between a and b and steadily increases, then I 0(a)(3 -a)\ > \fl 4>ix) dx \ > \ m (^ - «) I • If 0W is negative between a and b and steadily decreases, then I ^(a)(b - a) I < I I* ,f{x) dx \ < \ ^{b) (b-a) | . When 0(.r) changes sign between a and b, the integral may be posi- tive, zero, or negative. If (p(x) decreases steadily from a to b, ^i"') (?""')> / ^W '^x, where f is the point where (p{x) cuts the Ox axis. If ^(jf) increases steadily from a to b, 0(a) (J — a) < / ^{x) dx. 29. Let ^{x) be a function con- tinuous between a and J. Join the points {a, ^(a)} and {b, ^(b)} by a straight line. The gradient of this straight line is r\ )~r( 1 ^ jj. jg gyijent from a figure that there must be at b — a least one point between a and b where the gradient of f{x) is the same as the mean gradient of the straight line, hence if this point is the point x = ^, f (f) = '''(^) - '/'(«). (b — a) .-. ^(5) - ^(a) = {b- fl)^'(€) (i) Now i'{x) is the integral of i/^x), .'. ^(b) — ^(a) = I fix) dx. :. (''f(x)dx = ^'(€) {* — «}• Putting fix) = 'p(x) and g = a -f- $(b — a) where ^ is a number between o and i, we get / ^{a-ire(b-a)} (2) Equation (i) is also important. If we put J = a -|- A we get ^(a + h) = ^{a) + ^\!i)h = ^(a) + r{''+eh}h (3) 424 APPLIED CALCULUS This is a very useful form of tlie theorem. 30. sin'^ is always positive. /. Vi — J sin^^ lies between i and ^/ -■ :. /' de > /' Vi - i sin^e de > /^ a/^ dd. i.e. 1-571 > I > I •485. If we take the mean, I •528, the error cannot exceed '*'•'„ = 2-8%. 1-528 Exercise 9 I- -^j\r+~c^ — ci] where Vca = f. 2. The integral required is I = jg Vi + A^dx where A = Va X f^. Put I + Arf = z^, when the indefinite integral becomes -^li.^ — z^)dz, and I = ^rii + MI^ - (i+A£4)i _ ^"1 ^,^^^^ a = Va X ?f . A^L 5 3 J 3. The curves intersect above the x axis where jr = o and X = a, hence the area required is / ( ■^2ax — x^ — "Jax')dx. Put j; = a(i — cos S) in the first integral and we get a^MT — ? I for the area required. 4. The graph is like L4 3-1 a figure 8 on its side, 00, the x axis passing through the centres of the two loops, and the y axis through the point where the loops meet. Tlie area required is therefore given by /a /cfi tr2 . „ . . , A / -s— ; — «xdx when a is 10. Put:r^ = z and rationalize the V « + ^ 2 numerator, and the indefinite integral is Ja^ arc sin-; + \ Va* — x^, whence the required area is ioo(ir — 2), when a is 10. 5. The in- Ca \J2ax — x^ dx. Rationalize the numerator and we get as the indefinite integral "•Jiax — x^-^-af ^ — The latter integral is arc cosfi — -)> hence / ^"^^ ~ ^ dx ANSWERS 425 = Vaaj; — x^ -\- a arc cos( i — -), and the required definite integral is 2al '^2ax — x^-{- a arc cos(i 1 I . 6. Integrate the expression for j/ between x^ and Xi to find the area accurately. Compare the result with those obtained by the rule, taking first the simple case when n = 2. 7. irrl. 8. 4irr^. 9. ^irr^k. \0. iirh-^zr-A). II. Jn-^A. i2. ^ttWo. \3. liraH. 14. 2irra. Exercise 10 2. 6a cosax — 6a^x sinax — x^a^ cosax. 3. n^ sin;«;ir sinnx — 3>w»'' cosw2:c cosrear + ^m^n sinmx sinnx — m^ cosmx cosnx, i.e. (n^ + 3»»^») sinmx sinnx — (m^ + ^mn') cosmx cosnx. 5. Dif- ferentiate J/ twice with respect to x, and add m^ times j. The result will be found to be zero, as stated. 9. When f(x) 5= x'' every derivative is zero when x = o except the fifth, which is 5 ! whence the series reduces to x^ as it should. 10. /{a -\-y) = tt>{jy) since a is constant, and 4'(y) = 0(0) + ^'{o)j' + ^^ + . . . + »-", i.e. Aa +y)= Aa) + f(a)y + />)y + . . . + -^'"'(f)-^ . 12. Put :ir + x2 = z and ex- 2! n] pand by Ex. 11 above. Result is i + 4:1: + io;i;^ + i6a;* + 19;!:* + i6:i:S + ioar« + 4^;^ + ^r*. Exercise 11 4. The result follows from the working on pp. 208, 209, put- ting — b^ = b^. At the end of the major axis (a, o), p = — . a 6. (i) Differentiate y and *■ with respect to d and divide the derivatives. The result is the value of dyjdx, i.e. tani/'. The result reduces at once to the value given by expressing it in terms of-. (2) Use the formula f = I °*/l + (j^) ■ -jtx The tangential force along the same direction is — mg sin ^. 426 APPLIED CALCULUS .'. ^+^sin^ = o is the equation which determines the motion, by Newton's Second Law. But sin ^ = — for a cycloid ' ■' 4(1 (Ex. 6 above), .•. —, + £^s = o. If the body starts from rest at''' 4a (i.e. has zero velocity when < = o) at j = A, it mi^kt move in accordance with s = A cosiat, because this equation gives J = A when t = o and ds/dt = o when t = o. By differentiat- ing this equation, it will be found to satisfy the equation of motion, if «2 = j£ irrespective of the value of A. The radius 4a of curvature at the point where the tangent makes angle f with OX is 4a cos^, whence the normal mass-acceleration is where v is the velocity of the particle at f. The re- 4a cosifi ^^2 action is therefore ( 1- mff cos ^ J allowing for the gravity reaction too. 8. In M = -=- ; M, E, I are all constant, H. hence R is constant, and the centre line deflects into a circle. J/ = J=^ {x — l)x, if _y = o at both ends of the beam. If the HI weight is not negligible, we have to add (algebraically) the deflection for a uniformly distributed load (p. 214) to the deflection arising from the couples. Q. y ■= -^=^\ — —) 2EI\ 4 6 / I W/3 W/3 ,- p, W(/ - xf 'WPx , WP Wl' — T ^- i Tend = — • 26 3 Exercise 12 I. (i) A minimum when x = ^, a. maximum when :>; = i by applying the analytical test, (ii) djz/dx = o when *• = o, o, -\- 3, — 3. The zero values of x make d^yjdx^ zero, and these are not turning values. Maximum value when x = — 3 ; mini- mum value when j; = -f- 3. 2. If R is the radius of the sphere and r of the base of the cone, h being the height of the cone, r = h tzxiO = R sin2e. ,-, vol. of cone = ^ !il2^. 3 tand ANSWERS 427 The function of 6 is the only variable part of this formula, and f, ^ = a'^ — 2x^ + x^ where x = sin0, whence we find that 8 tano this is a maximum when x^ = i, i.e. ^'" „ = 31 and the volume ' tan^ 27' of the cone is |?7rR^ and the radius (r = Rsina^) is ? — ? R. 3 3. A square should be cut out of each corner, the side of the square being — i — — , where a and b are the 6 length and breadth of the piece of paper. 4. Differentiating we get i>'{6) = cos^ — 2 sin2^. This expression equals zero when cos^ = 4 sin^ cos6', i.e. when cos^ = o or sin6 = J, i.e. when 6 = 2irn + - (i) or ^ = wtt + (— i)"^i (2), where 6, 2 is the angle lying between o and - whose sine is i. On plotting 2 the graph of sinO + cos25, it is easily seen that the values of 6 given by (i) and (2) interlace. The first maximum occurs at O^, after which the turning values are a minimum and a maxi- mum alternately. 5. The corners of the triangle folded over and the corner of the unfolded page lie on a circle, whence the point of folding at the lower edge should be at the point | along the lower edge. 6. Express the distance between P and Q at time ^ and differentiate it to find the turning value. This method is long and clumsy. The following alternative method is briefer. Superimpose on the two bodies the velocity of one of the bodies A reversed. Then the body A is brought to rest, and the other body B moves with its own velocity, compounded with the velocity added, along the diagonal of the velocity parallelogram. The velocity relative to A (at rest) is then Vk" -|- ''^ — 2UV cos^. Now apply the principle of moments to the resultant velocity and the component velocities of B. 7. If r is the horizontal range, the given equation shows that r = 2h sin25, whence drjdd is zero when cos20 = o, i.e. d = 45° for a turning value. That 6 = 45° gives a maxi- mum value of r is obvious from the physics of the problem. 8. The potential energy due to the charge of electricity is — z= w, and -^ = — ^, hence the equation for equilibrium r dr r^ becomes Sw = 2T Ss — pSv — ^Sr. .; o = 2T X SirrSr — p 428 APPLIED CALCULUS X Airr^Sr — ^Sr, i.e. * = 4_ — — -,, i.e. the equilibrium pres- r^ ^^2 r 4irr* ^^ sure is reduced by ,. 9. 28" vacuum. 10. -;- = wV ■' 47rr'' dt (cosw/ + - cos2t = o. Express CQS2iat in terms of cos'w^ and we get cosu^ = — f— i ±^i + ^ |. The negative sign to the root is inadmissible as it would make | cosu/ | > unity, in general. Hence lot = arc cos — -( — ' + » /i + — ) |, and maximum velocities occur when ^ = i. I 2»7r + arc cos— / / S^^T "*- '^'' (^—1+^1 + -n)\'< 7-88 f.p.s.; angle turned through is 85°. Exercise 13 y/i+x^ ^x^ + a' X ' "■ ' sin2.r' {e) — sec:r; (/)cosec;>r; {g-)e'{cosx — s\nx); {h) e^'^icos^x — sinx); (?) cot2;i;. 2. / ^^a , ^a = log«{^ + V^^ + a^}; /secj: «^;r = '°g' (\ /rT^ ) = '°^' '""(^ ~ 0= \cos^cxdx = — loge . / ' ~" ^°^^ = — log, tan-. 3. Differentiate the right- hand side of the equation. 4 and 5. By integration by parts, and the integrals obtained in questions (3) and (4). 6. In- tegrate by parts. 7. The integrand equals - — 3 — -|- (^-1-3)' {x + i) — , — ? — r, hence the integral is — -. — 2 — . -U 2log.( '^ "j" -3). {x + 2) (* + 3) ^ + 2/ 8. - /^+'2« + l°g«(^^)'- 9- -^. = W-2x x' + bx -^^ S \x -\- 2/ I -\- 2X^ ^, whence the integral is — — £- -|- 1 logeVi -j- 2X^. ' (i + 2*2)' ^ 8 8 ' 8 10. ^ = A cos./^t -J- B sin. /^t Prove by substitution. When / = o, e = e and ^ = o. ^ = - k ft sin, l^t dt dt SJ I '\ 1 ANSWERS 429 + B^^ "^os-y/f '• ••• o = - A^f sin 0° + By/| cos 0°. ,•. B must be zero and 6 = A coso°. .'. A = 6, i.e. 6 = Q COS. / ^t is the solution required. The period of this function is 2ir. If then r is the periodic time,' /^r = ztt. .', r = 27r. /— . II. Show that the force due to buoyancy is a restoring force and proportional to the depth of immersion, approximately. = 27r / S-, where m is the density of water. 12. (i) If V = Ks, V = o when j = o, and how can the body begin to move at all? (2) If z» = KS, then logjj = Kf, i.e. loge(-' j = k(/i — ^2). i.e. the logarithm of the ratio of distances travelled should be proportional to the difference in times of transit. This is not true for motion due to gravity. 13. Taking one of the sup- ported ends as origin, the undeflected axis as axis of x, and the perpendicular through O in the plane of bending as the axis ofj/, the equation of the deflected axis is (-^)" = >--) (■) where A is the deflection at the centre, initially. The total kinetic energy when the tube is crossing the x axis trans- versely is n \pa^\y''dx, (2) where p is the mass per foot, w, the "angular velocity" of the harmonic motion, i.e. 2Trn where n is the frequency, and y the initial co-ordinate of the deflected centre line of the tube. In- tegrating (2) by means of (i) and equating the result to M^, the initial potential energy where jn is a constant, we get ^7- (3) T = 3.25 where t is the required period. We can get p. approximately thus : When the tube is bent by a central concentrated load W, the potential energy is ^WA^ ft.-poundals, where g- is the acceleration due to gravity. But in this case, the central A a ^- A • ■ K A W/3 . 48EIA ^, deflection A is given by A = . .•. W = ^ — — and 48 h.i / 430 APPLIED CALCULUS In the example given / = lo ft., p = 0-52 lb. per foot, EI^ = 24000, whence t = 0-309 sec. It is a good practice to check "dimensions" of a formula of this kind [T] = [O] [L^] [M L-i]» [ML-2L^LT-2]-», i.e. [T] = [T]. The more accu- rate formula is t = o-6'i6l^^ I „^ when the correct deflection •* V EI^ curve for a distributed load is used. (See Morley, Strength of Materials, p. 397, 1908 edition.) The effect of water in the tube is to increase the mass per unit length, without increasing the flexional rigidity. The period is therefore lengthened. Exercise 14 I. The question leads at once to I-^ = — /iu (see p. 267), u = Oe 1,1 = 2240 X 9, w/0 = 0'5, whence /x = 46-5 poundal ft. per radian per second. 2. e = T-^, p. 278. .\ ri = T^. at at :. d-p = ^idt. :. 0, - 01 = ^(Qa - Qi). 3. Q2 - Qi = /a ^Z^. P- 295. .-. 02 — ^1 = A*"" V I where p." = %^. 4. 0,-0, = (100 X '^^y-S') lines. .-. ,." = '°°X'^X7-5' = gg. ^ 4 ^ 4X65 T-. ui ^ ^ 100 X TT X 4^ , d^ Double. 5. = ^^-i- = 400 tt. .*. -=^ 4 at = 40OO7r. .'. e = 100 X 4000^- X 10-' volts [e = T-^ X 10"' dt volts] = 47r X io-» volts. /. j = f = 4!LAi2_ = 25 X lo-^ '•SO ampere. 6. Clearly the flux through the coil varies sinus- oidally, hence, in the usual notation, = cosw^. .". -f dt = — *« sinu^. /. ^max. = *uT X 10-° volts numerically. 4> = [o- 18 X '^^l^'J.T = 20, « = 2T?^. /. emax. = S'S X 'O"' volt. 7. Consider a uniform magnetic field of strength H, and a perfect conductor carrying a current « and of length /, lying in the field perpendicular to it and moving with velocity v perpendicular to its length / and to the direction of the field. ANSWERS 431 Clearly -£ = Hlv. .', ei, the rate of working, is Hlvi e = T-f . at L dtA But this quantity is F/z;, where F is the resultant transverse force on the conductor per centimetre. .'. F/z/ = Hz/d, i.e. F = Hi. Hence, in a uniform field, the force on a conductor is pro- portional to the field strength and the current. A ballistic gal- vanometer is so constructed that the coil moves in a uniform radial field. Hence the force on a vertical element of the coil is H/i, and the total torque is therefore SH/iV, where r is the radius of the coil. By symmetry every element produces the same torque, .". SH/m- = HL«>, where L is the total length of the elements, i.e. the electromagnetic torque is h'i, where K = HLr, a constant ; hence the torque applied to the needle is not now zero but hi, and we get, dividing h' by the moment of inertia of the needle, -^ -|- i^Q = hi, where A = -, I being the moment of inertia of the needle. The complete solution of the hi equation is ^ = A sinw^-|- B coswf -|- -5, if i is constant. Whence Q = o when t = o gives B = — -J, and -— = o when ^ = o a'- at gives A = o. .*. ^ = -| [i — cosw^]. Exercise 15 I. If a is original concentration, and x the amount transferred in time t, then i log.t ) = k if the action be mono- t \a — x' molecular (p. 308). Using this formula, we get the table t intervals a — x k 1 88 (0-128) 2 72 o-i68 4 S3 0-159 7 29-s 0-174 8 25-2 0-172 II 15-2 0-171 13 ii-i 0-169 which shows that k is practically constant. (Lewis, A System of Physical Chemistry, Vol. I, p. 433.) 2. If reaction is mono- molecular, we have - logj — - — J = t. Testing the numbers, t \a — X' we find K is nearly constant at 15 X lO"". (Lewis, A System 432 APPLIED CALCULUS of Physical Chemistry, Vol. I, p. 436.) The chemical equa- tion merely gives the proportions in which the cane sugar changes into dextrose and laevulose. 3. The formula is \ '°^'(§) = *«■ ^^ ^' = **^"' Q = ^- ■'• ' " ^^' '^''^"^ /j = 7.1 wk. and t^ = 21 -8 min. 4. If the formula is mono- molecular, the formula is "1 = ]^°s.^^ W but if it is quadrimolecular, "^ = 1 aicJ^' ~ A ''^ P- 3°9 (^) The numbers work out as follows : i (hr.) Ki K4 X 108 X Io2 o — 7-83 2-36 1-73 24-17 2-37 2'OI 41-25 2-35 2-29 63-17 2.38 2-88 89-67 2-41 3-85 (£[ is more nearly constant than k^ is. The equations may be I PH3 -^ P -H 3 H. ■* * J-extremely rapidly. 2 XT. f ri2-' In this case, the first (monomolecular) reaction would dominate the rate of the complete reaction. (Lewis, A System of Physical Chemistry, Vol. I, p. 443.) 5. (a) Differentiate the given ex- pression and substitute. The two sides of the given equation will be found to be identical, (b) When x = o, sinnx = o and c = Cq ; when ;i; = /, sin«/ = o. ,'. nl = mir (m = o, i , z, 3 . . .). .". c = Ca + ae i^ smt-y-x\. (c) Each term separately satisfies the given differential equation, which is of the first degree. Exercise 16 I. 0-57(7 = 1-4)- 2. 0-52(7 = 1-4). 3. 0-32. The available heat energy is 371 B.Th.U., and the heat absorbed is 1158 B.Th.U. 4. The temperature of the mixture is ANSWERS 433 (T ' fl \ -g + ^°SeJ), i-e. 0.3I3 unit of entropy. The water loses 20 log^-^^^- = 0.308 unit of entropy. 278-7 Net gain of whole system =3 0-005 ""'' °f entropy. 5. Con- sider the gas A. At any moment it is slowly expanding against a force p, due to the piston. This expansion is reversible, for if p is changed to ^ + S^, the expansion would cease and compression would begin. The expansion is also adiabatic ; it is therefore isentropic. Similarly for gas B. The friction of the piston balances the difference in the gas pressures, and work has been done irreversibly against friction and converted into heat. Hence the entropy of the piston must have increased. Hence the total entropy of the system has increased. 6. Work area is [/, ^"1 + / '' pdv — p^ V2] where pv" = k. Whence work done is Pi'Vi] (■^j " — I I ft.-lb., i.e. work done per cubic foot (©2 = i) is 144 pA {—] » — I I if /, is in pounds per n — I L\^2 -■ square inch. 7. Work per minute = 144 — ^ — ^p\ ( )^^ — ' |. If A = A + ^A> I ( , ) " — 1 = r -^ nearly, whence H.P. = 2M-Vp^. 8. Follows at once from Ex. 6. 33000 9. z'a = 1-2 c. ft.; v^ = 0-2 c. ft.; r = 6; 62 = 600" F. abs.; p2 = 14 "7 Ib./sq. in.; gas used is o-o6 c. ft. per cycle. Cal. Val. = 600 B.Th.U. per cubic foot = 36 B.Th.U. per cycle. I c. ft. of air at 32° F. and 14-7 lb. per square inch weighs 0-0807 lb. Hence: (i) Weight of gas in cylinder is -0807 X ^23 600 X 1-2 = 0-0793 lb. (ii) During compression pv^'^ = a con- stant. .•. Pcompressiori = 14-7 X 6''^ = 174-2 Ib./sq. in., and, since -=- = constant, ^compression = 600 X ^^ X — = ii8s°F. abs. 1 14-7 1-2 (iii) The rise of temperature Ad, due to combustion, is given by 0-0793 X 0-18 X A0 = 36. .-. Ad = 2523° F. /. ^expansion = 2523 + 1185 = 3708° F. abs. The corresponding pressure is: (iv) Pexpansion = 174-2 X ^^ = S43 Ib./sq. in. (v) The pres- sure at release {pv^ = «) is given by Prelease = |^ = 46 Ib./sq. (D102) 29 434 APPLIED CALCULUS in. The corresponding temperature (^ = constant) is given by ^release = 37o8 X ^ X i^ = 1878° F. abs. (vi) The ef. 543 *^ ficiency is {i — (if} = 49-4 %• (v") The Carnot efficiency is 3708 — 600 ^ 83-8%. 10. 0"7S. 0-52; internal energy at 3708 A, 89-1 ; at B, 66-4 B.Th.U.; work done, 2040 ft.-lb.; heat lost to walls, 20-I B.Th.U. II. 6-3 h.p. 12. Volume of the reservoir is 9-05 c. ft. 9-05 c. ft. of air at 500 Ib./sq. in. and 275° C. abs. weigh 24-7 lb.; hence 24-7 lb. of air remain in the reservoir. This air occupies 2-8 c. ft. at the initial state. If this air expanded adiabatically, the -final temperature would be ^2 = 288 X 3 -23 -"■58 = 184° C. abs., for 3-23 is the ratio of the change of volume. The actual temperature is 275° C. abs., hence heat from sea is 24-7 X 0'i8 X (275 — 184) = 404 C.H.U., i.e. 404 pound-calories. MISCELLANEOUS I. (i) Put . ' ., = .—^2 + -^ + -■ To find A, x{x — if (x — ly X — 1 X multiply both sides by (x — i)^, and then put x = 1; thus A = i. To find C, multiply by x and put :c = o; thus C = 1. Then B is easily found to be — i. Integral is + logA; — log(ar — i). (ii) Integrate by parts. Integral is ^x^ logj; — tV**- ('") P"^* X — i = z- Integral is log{x — f + y/x^ — -^x -\- 2). (iv) Put tan- = z; indefinite integral be- 2 2. (i) Put ^ = z. Integral is arc tan(e*). u = arc sin;t; + V(i - x^. (iii) \ { -^ , ', ) dx 3 \x — 2 x^ + ^/ = 2 log(:i; — 2) ~\ Site tan^x. 3. (i) Derivative is ' =■ . v(i — x') (ii) The function is arc sinj:. (iii) Integrate (i — x'')'^ once, result is arc sin;c + Cj integrate this again, by parts, function ANSWERS 435 required is x arc sinj; + \/(i — x^) + Cx + D. 4. (i) Express integrand in form ^ + 4 + - + -^— Integral = ii- + - — log*' + log(:c — i). (ii) J«*(sin* — cosj;). (iii) From re- peated integration by parts, integral = e^(x^ — jx^ -{- 6x — 6). 5. -^ = I = — —.. Equation of tangent at (k, k) is ax x* x' :£ + Z = aK OP + OQ = aW + am = a. 6. [e'^ sin":ir dx = ifi"* sin"^ — l-e'^n sin"-'j; cosj; dx, by parts, = -e" sin"* a J a a — — e"* sin"-'* cos* + —^ \e^{— sin"*- + (« — i) sin""'* cos^*}^^^. The integrated terms vanish for * = o and x = p. In last integral write sin""'* cos'* = sin"-='*(i — sin'*) = sin"-'* — sin"*. Thus [^e<^ 310"*^* = -2 re^{{n — i) sin"-'* — n sin"*} dx. Hence (i +^\ {^ e^ sin"* dx = "^" ~ '^ P e°» sin"-'* dx. \ a^J jo a? JO 7. (i) fsin™* cos"* dx = fsin™-!* -^ (— ) ?2!!!^ rfa: = ?— sin™-'* cos"+'* -|- — ; — /cos""'''* (m — i) sin""-'* cos* dx. n -\- 1 J In last integral write cos"'*''* = cos"* — cos"* sin'*, and pro- ceed as in Ex. 6. (ii) By (i), pcos** sin'* — i{a — bf. (ii) Take lo ordinates, viz. at j; = i, i-g, 2-8, &c. The rule then gives 2-312; the correct value is 2 -303. '0- (0 /((^. - ^ i - j^- (") /v{(.-o^-x) = log(ar — I + Vjf2 _ 2j;). (iii) Put tanj: = «. Integral = — arc tanf" *^"^V (iv) Put ^ = a sin^, then tan(9 = t. ab \ / The indefinite integral = /-jf-£2^^ = j {-^, - (.^-%V" + C- }'^^ = -tan^-^^arctan^:^. 1 1, (i) r /(;<:) rf:r = A + iB + iC,/(o) = A, f(\) = A + ^B + JC. J /(i) = A + B + C, and result follows at once, (ii) Put x = \-{- u, and express <)>(x) in powers of u. 12. (i) -; or -; — -. sinjf cos;r sin2;i; (ii) e^'+i ^ „ . (iii) Find ^ and ^, and either test result (jc^ + i)'' ax dx^ by substitution, or else eliminate a, b from the equations giving dy d'y .3 /.» dy _ dy _^dx _ sing _ 2sin^gcos^9 ^' Si' ^2' ^^' dx~ W ' dd 1 + cosg 2 cos^^e ft ^e ft ft = 2aHi + cos 6) = 4a* cos^-. Hence -57= = 2a cos-, j = 4a sin- 2 aC7 2 2 V- = 4a sin0. (iii) p = -=- = 40 cos0, intercept on normal 2a — y all + cosg) _ 2a cos^iA _ ,„ ^. ^ ri = —5 ! i =: = 2a cos 9. COS^ COS0 COS0 14 /" -r + i(^ + 7) + {» - M/3 + 7)} a^ ^ ,(^ 4- BUx + y)\i '+{<^-W + y)} ^og[x + W + y) + {{x + P)(x+ y)n 15. (i) X logx — X. (ii) ^x^ log^t: — ^xK (iii) -xe^ -^. (iv) From the graph of logX it can be seen at once that logX < X — i ; for the line Y = X — i is the tangent to the graph at (i, o), and the log graph lies below this tangent. Put i + ^x for X ; then log(i + ix) < ^x and x log(i + ^x) < ^x\ from which the result stated follows. 16. It is easier here to deal with ANSWERS 437 - = X + I -i- - := z say, and find the minimum and maximum y X values of z. For x = — \, y ■= ^ \, a minimum; and for X = I, J' = J, a maximum. 17. Deal with the reciprocal of cps,^, proving (i - cf secV = ""? + '^T/^ = _L_« + ^ sm^C* cos^C' cos^^ sm''^ = I + tan2^ + . + i)^ - sin(2>» - i)^ ^^ ^ ^;„^^^_ ^j^j^j^ j sin-r is proved at once from the trigonometrical identity sin(2» + i)x — sin(2» — i)x = 2 cos2»a: sin^i;. For second formula use the identity sin^(» + i)^ ~ sin^w*' = sln(2« + i)x sinj;. For the definite integrals, first formula gives Jn+i — ^n = o, so that Jn = •fi. or Sn = — . The second formula then gives »« — 'Z'n-i = -. 2 2 In this result put « = i, 2, 3, &c., and add. 19. wj y^dx J = — 20. Put {p, pi, p^ -^ respectively = q, q^, q^ Theh 24 ( ^ + i — 2 is to be a minimum. 21. If r is the radius and v 1 9i the velocity at time t, then the law that change of momentum = applied force gives T-(r't') = r^g: Also y = r„ + M, where dt k = —, a constant. Hence —-(r^v) = rr-^(r^v), or r^v dt dr- ' kdV ' = iTr'dr = ^Jr* -»-„*). Ii V = '^ = k"^, this gives x' k]r„ ^k^ "', dt dr ^ = ^^T (r — ^\dr = ^^(^2 + ^ - 2»o2). This equation will give fj, the value of r when the drop reaches the ground, where X = h. The quadratic for Tj is {r^ — r^Y — —< so that 433 APPLIED CALCULUS Using foot-second units, we have k = 'f^-io-*, h = 6400, g- = 32, »-j = ^ X 0-04. Thus the final rad. = 0-04205 in., and then from the equation r = ra + kl, we find the time = 20-5 sec. 22. (i) Equation of tangent at (a, b) is y — b = — 2a^(x — a). (ii) a = 0-58, b = 7 '8, intercepts = 8 '4. (iii) Area = 12. 23. My = — Mrio'{cosut + - cos2(i)/). This is numerically greatest when the cosines are both equal to i, i.e. when / = o, — , 3^, &c. The maximum value of the inertia force is .*. fa) w M!f!r(i + ^) lb., if M is in pounds. 24. (i)-^-r (i + ^). or — y— nearly, (ii) For two reasons : (a) the velocity coefficient is -, and the inertia coefficient y, (b) sin2fa)/ is very small when the velocity is a maximum, whereas cos2u/ is equal to i when the acceleration is a maximum. 25. 7-56 ft.-tons or 16,900 ft. -lb. 26. (i) See p. 292. The T.P. occur where ^- = o or — ^ sinn/ -}- « cos»^ = o. Hence tan»^ = ^. If at k we take an acute angle o so that ^ = tana, then nt = a, o + t, a 4- 2jr, o -f- 37r, &c. , give the T. P. The maxima and minima occur alternately, d being -\- at the maxima, and — at the minima. Thus maxima occur when t = ", - -|- — n n n - ■\- 4^, . . . The values of Q, or Q^~^* saxnt, for these values n n of t are in G.P. with common ratio e~'''^''^\ (ii) Given — = 1, and («-ft"2T/»)* = i; if T is time required in seconds, n tanreT = y But e^* = 4, ^ = J log«4 = '3466, tanajrT X . I i8.i3, 2!rT = - — arctan-^^— , T = J — •3466 •^' 2 18-13' 18-13 X27r nearly = 0-241. 27. Energy consumed is proportional to the resistance R, and annual charges for copper are proportional to the cross-section, i.e. inversely proportional to R. Hence annual charges = a + — , value of energy lost = cR, where ANSWERS 439 a, b, c are constants. =- + cR is a minimum when — ^„^ c A = o or — =: cR. Let section = x sq. cm. Interest on cost R r JT X Q X 2-2046 X 6 X 240 . c of copper = — ^ 3 — 1^ O — a. pence, cost of enerev *^^ 1000 X 2240 ^ "•' consumed = i X i-6 X 10-' X 5°o X 500 X 12 X 365 X 3 e. X X 1000 X 4 Equating, we getJc^=io3and;i: = iO'is. /. section=iO'is sq.cm. 28. Take moments about centre of bowl. 29. ot-=- = F — ai/^. At full speed v^, % = °- .•. F = «V. '"^ = «(V - ""% £ ^ = V— 2- -t= - loge^^^^tJ!. But Fz-o = P, .-. P = «V- Hence t = ^ loge'^^i-^ = ^ log^^^ti^. In 2av^ Dj — ■z' 2P V, , . 2000 X 66 X 66 , 4S + 30 33 , „ 33 example, t = loge^3_!_i = i2J logcS = =2=2 2 X IS X SSo X 32 ^ 45 - 30 2 ^ •' 2 X i'6i = 26-6 sec. 30. See pp. 292-5, and p. 296, Exs. 2, 3. When a current of i amperes is reversed, the formula for change of flux is — X 4ir X turns in primary X turns in secondary 10 . . area of i turn of secondary , ., X = i — ^ — -. i = I -6 X IT X 500 X 500 X IT length of prmiary X (15/4)^ -j- 40 = 1,390,000 lines. We have now to find the number of divisions, undamped, corresponding to 65 divisions, damped. Now each maximum swing = preceding maximum swing X e-^. .'. loth maximum swing : ist = e-^^K Hence fi| = filSA. But ist undamped swing : ist damped swing = e2 = (|J)A. /. 1st undamped swing = 68- 1. /. Flux per undamped scale division = ^>j9°'°°° — 20,400 lines. If resist- 68 • I ance were halved, first fling would be doubled, 136-2 divisions. 31. (i) See p. 151. (ii) If pv' = A, work done = (iii) ^2 = a(-)' = 41 = S"^4 "'■ P^"" ^1- ^"- ^"""^ ''°"^' by (ii), = 7-54 ft.-lb. 32. (i) If ^ + S;> = pressure at eighth 440 APPLIED CALCULUS z + Sz, then 5p = — g-pSz nearly, when hs is small, p being in absolute units. Hence -f = — gp = — ff^P, since p is pro- portional to p when temperature is constant. -^ = — ghdz gives log^ + C = — ffhz; and if ^ = ^q when z = o, then log^ = —gAz and :2. = g-cAs^ ^;;) Upward thrust = weight Po Po of air displaced. This weight of air, when lowest point is at fh+d height h, is / pSdz, where S = cross-section of sphere at height z, and d = diameter of sphere; also p = p^'i^', so that [h+d weight of air displaced = Poj Se-i^dz. Put z = h -{- {, and /d rd S'e-i^e-f'fdt = Pffi-i^^ X / S'e-i^idt = e-M* X weight of air displaced when sphere rests on ground = e-/»*{M + m)g: (S' is the cross-section at height f above lowest point of sphere.) Sphere will rest at height h if this last expression = weight of sphere = M^. Hence e-^h = M — ^ ^ loge^L+J^. 33. If a drop of water is introduced into an enclosure containing vapour and water in equilibrium, evaporation will diminish the surface of the drop and therefore its surface energy, but will increase the remaining part of the energy of liquid and vapour. Equilibrium will be reached when the total potential energy is a minimum. Now consider i lb. of a vapour obeying Boyle's Law. If we reckon its potential energy from a standard density p^, the increase in potential energy when it expands to density p is given hyW = - r pdv = - RoT — = - Re log«ii = R^ J Vo J Vo V v„ loge— ; since vp = v^p^ = i. Consider i lb. of water. We suppose it to be incompressible, so that its density (e\ for 1, = M - f and ^ = - I. Now I = V - 5 ^/'?\ _ I cn^ . S * _ , _ P p I -. ii^ = -' a"d ••• - ^ = I - -• Thus, from (i). "■ rff \P/ ass from the initial to the final state by first raising the water to 6 without evaporation, and then evaporat- ing q lb. of it at 6. Thus ^ = o- loge^ -)- ^. Suppose the condition of the mixture is given at 0^ and we wish the value of g at ^j- Since 4> is constant, we have a logeO^ + ^^^ = T logc^i + 2l_!, which gives q^. (ii) If ip and q are constant, "i a logeS — «■ loge^o + ^ = «■" , a constant. /. a loge^ = — ^ T -2i + a', logc^ = _ £^ + logefl, e = ae "». 36. (i) See p. 340. (ii) dQ = CvdO + pdv = CvdO + ^. /. idQ J 1)" J = C»(^2 — ^i) + (A^'i ~ Pt"^ (see Ex. 31 above) = CrKO^ - 6,) + -^—{Rd, - Re^) = (c„ - -B-) (^2- 0,). n — I \ n — 1/ 37. Consider the cycle for 1 lb. of water: (i) Heat it as water from T to T + ST. (2) Expand it to vapour at p -\- Sp. (3) Cool it as vapour from T + ST to T. (4) Condense it to water at T. Then work done per cycle = [{p + (g-~) ST} — p] /?i*\ \oT/v (V - «)) or (V - w)f|2.j ST. By Carnot's Theorem (p. 359) this = — - X heat taken in = -=- X L, L being in work units. Divide by ST, noting that T^j and (j-\ are reciprocals, and the result follows. 38. Apply the result of Ex. 37 to I gramme-molecule (p. 301). Let v^ = volume of vapour and ANSWERS 443 neglect the volume of the liquid. Then, taking 6 for tempera- ture. L = v,ef^, v, = M, so that L = ^ ^. If P is the internal work of evaporization, L = P + pv, ,'. P + R^ logfl = R log/ + C. .-. R log^ - R log|2 = -| + |. /. R logMs = ?('_'"). Now, by the Kinetic Theory of Pi"i ^"1 "2' Gases, p is proportional to NV^, where N is the number of molecules leaving the surface per square centimetre per second. Thus R log ^i^^^i = pfi- - i-V In the electric case, if e = charge on an electron, then i = N« and R = 2. Hence iwrw) ""*f -*l - '««^-*f.' '"" •■■ •■"■ " constant. /. — ^ = /Je^fl or » = D0% 28 {^Electrical Properties t of Flames and Vapours, H. A. Wilson). 39. The vehicle of energy is the unit quantity of electricity. Consider a cycle: (i) Allow unit quantity of electricity to pass through the cell under electric force E+ (^=JST, at constant temperature T + ST. (2) Allow the temperature of the cell to fall to T. (3) Pass unit quantity back through cell under force E. (4) Bring cell back to initial temperature T + 5T. The electrical work done = E + (||)8T - E or I^ST ergs, and this = ^J X (heat taken in) by Carnot's Theorem (compare Ex. 37) or (heat taken in) = T==. Now some of this work is derived internally by 3E chemical action, so that net input of heat = E + 5~5T — X, or E - X approximately. Hence E - X = T?^. 40. Put a film through the following cycle: (i) Stretch it at temperature T + 8T to area A + 5A. (2) Cool it at A + 8A to T. (3) Con- 444 APPLIED CALCULUS tract it at T to A. (4) Warm it at A to T + ST. Surface tension in (i) is S + (|=] 5T. Work done hy surface tension in (3) is 2S5A (considering both sides of film); in (i) it is _ 2(S + ^ST)BA. /. Net work done = - a^STSA. If ST 01 H + SH is the total heat absorbed at T + 8T and H the heat rejected at T, then (p. 360) ^ ^ ^^ ^ T" •' ST ~ T' ^""^ 5H = net work done = - aUsTSA. .'. H = tI^ or H = — 2T~3A. In forming a film of area A we have there- oT fore to supply energy (as work and heat) = 2SA — 2T=A. Hence surface energy per square centimetre (one side) = S — T^^;. 41. Let specific heats in A, B be cTj, a^; Peltier oT effect coefficients P at hot, P^ at cold junction; temperature of hot junction T, of cold Tp. Suppose unit quantity of electricity to pass from the hot to the cold junction through A. Measure all energy in ergs. Then work done = E. Heat absorbed at junctions = P — Pj. Heat absorbed by A = j (tjdT = — r