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MATHEMATICS FOR ENGINEERS PART II The Directly-Useful Technical Series Detailed Prospectus sent on Application. Mathematics for Engineers By W. N. ROSE, B.Sc. Eng. (Lond.) Part I. Demy 8vo. 510 pages. 257 Figures, with over 1200 set and worked examples. Price 9/6 net. Contains chapters on : Aids to Calculations ; Equations ; Mensuration ; Graphs ; Advanced Algebra ; Plane Trigono- metry ; Calculation of Earthwork Volumes ; Plotting of Difficult Curve Equations; Determination of Laws; Con- struction of Practical Charts, etc., etc. "The book teems with practical applications of mathematics to engineering problems ... an excellent book." Mechanical World. " A book which will be of great service to engineers of every class. To the young engineer it will be a God-send." Managing Engineer. The two volumes of " Mathematics for Engineers " form a most comprehensive and practical treatise on the subject, and will prove a valuable reference work embracing all the mathematics needed by engineers in their practice, and by students in all branches of engineering. CHAPMAN & HALL, LTD., LONDON. * The Directly-Useful Technical Series FOUNDED BY THE LATE WILFRID J. LINEHAM, B.Sc., M.Inst.C.E. Mathematics for Engineers PART II BY W. N. ROSE B.Sc. ENG. (LOND.) Late Lecturer in Engineering Mathematics at the University of London Goldsmiths' College Teacher of Mathematics, Borough Polytechnic Institute LONDON CHAPMAN & HALL, LTD. 11 HENRIETTA STREET, W.C. 2 1920 PRINTED IN GREAT BRITAIN BY RICHARD CLAY AND SONS, LIMITED, BRUNSWICK STREET, STAMFORD STREET, S.E. AND BUNGAY, SUFFOLK. EDITORIAL NOTE THE DIRECTLY-USEFUL TECHNICAL SERIES requires a few words by way of introduction. Technical books of the past have arranged themselves largely under two sections : the Theoretical and the Practical. Theoretical books have been written more for the training of- college students than for the supply of information to men in practice, and have been greatly filled with problems of an academic character. Practical books have often sought the other extreme, omitting the scientific basis upon which all good practice is built, whether discernible or not. The present series is intended to occupy a midway position. ,The information, the problems and the exercises are to be of a directly-useful character, but must at the same time be wedded to that proper amount of scientific explanation which alone will satisfy the inquiring mind. We shall thus appeal to all technical people throughout the land, either students or those in actual practice. AUTHOR'S PREFACE CONCERNING the aim and scope of this work nothing need be added here to the statement made in the Preface to the former volume.. It is there asserted that the subject-matter has been so chosen, and, through the examples, so applied to practical problems, that the two volumes " embrace all the mathematical work needed by- engineers in their practice, and by students in all branches of engineering science." As with the first volume, much thought has been given to the elimination of all rules and processes of academic interest only ; but the fact of the importance and necessity of logical reasoning has not been overlooked. With the exception of the chapters on Spherical Trigonometry and Mathematical Probability, this volume is devoted to the study of the Calculus, both Differential and Integral. Whilst it is wise, and even imperative, if this subject is to be presented in an intelligible manner, that much attention should be paid to the graphic interpre- tation of its rules, care must be taken to ensure that the graphic methods do not become other than auxiliaries. Accordingly the treatment throughout is based upon algebraic principles ; but whenever graphic proofs or constructions have been found to amplify or explain the subject, they have been utilised to the fullest extent. Thus from the commencement the connection between the rate of change of a quantity and the slope of a curve is clearly demonstrated : and this correlation of the algebraic and the graphic methods is con- tinued through all the stages of the development of the subject. The conception of " limiting values," mentioned briefly in Part I, is further discussed in Chapter I, a familiar example from Dynamics being chosen as the illustration : in this chapter also two methods of graphic differentiation are given ; the second of which, and the less familiar, being the simpler to apply. viii AUTHOR'S PREFACE The various rules for the differentiation of both algebraic and trigonometric functions are explained in detail in Chapter II ; and Chapter III, containing the rules for the differentiation of a function of a function, a product of functions, etc., together with an introduction to partial differentiation, may be regarded as comple- mentary to Chapter II. From the abstract reasoning required for comprehension of such an idea as that of " limiting values/' the practical mind turns with relief to the applications of differentiation found in Chapter IV ; the determination of maximum and minimum values making a particularly strong appeal. In view of the importance of this branch of the subject, a very varied selection of practical examples is presented, in the choice of which the method of solution has been a determining factor. In this chapter also the use of Taylor's theorem in cases of interpolation from steam tables is demonstrated. Chapters V and VI contain the rules required for the integration of functions occurring in engineering theory and practice. The former chapter serves as an introduction to integration, the signifi- cance of the symbols f and dx being explained by reference to a graph ; whilst in the latter chapter the various types of integrals, many of them of a somewhat complicated character, are discussed. At this stage also the reduction formulae are introduced, and mention is made of the Gamma function and its uses. Instances of the application of the rules of integration are to be found in the processes enumerated in Chapter VII ; and special features of this chapter are the determination of the perimeter of the ellipse, the graphic method for fixing the position of the centroid vertical, the drawing of ist and 2nd moment curves and the evalua- tion of the moment of inertia of a compound vibrator. The utility of polar co-ordinates to the electrical engineer is shown by the inclusion of examples on the candle-powers of lamps, and the employment of the Rousseau diagram to find the mean spherical candle-power ; and Dr. Fleming's graphic method for determina- tion of root mean square values of currents is here inserted, since it involves polar plotting. Differential equations occur so frequently that the methods of solution demand most careful study. Chapter IX presents the most common types, and the selection of examples based upon these, both worked and set, emphasises the need for a proper appreciation of the method of solution. Chapter X, with its applications of the Calculus to problems encountered in the study of Thermodynamics, Strength of Materials, AUTHOR'S PREFACE ix Applied Mechanics, Applied Electricity and Hydraulics, provides further illustration of the need of a sound knowledge of the subject to the engineer desirous of equipping himself at all points. The last two chapters contain much of interest to the surveyor, the examples chosen being such as arise in his practice ; and par- ticular attention is directed to the investigation relating to the corrections following errors of observation. The Author greatly deplores the fact that the inspirer of this work, the late Mr. W. J. LINEHAM, B.Sc., M.I.C.E., does not see its com- pletion : to his enthusiasm for his ideals in education, and for his many personal kindnesses to the Author, tribute is here paid. Sincere thanks are also tendered to Messrs. J. L. BALE and C. B. CLAPHAM, B.Sc., for much valuable assistance. Great care has been taken to produce the book free from errors, but some may remain, notification of which will be esteemed a great favour. W. N. ROSE. Borough Polytechnic Institute, S.E. i, December, 1919. CONTENTS PAGE INTRODUCTORY i Abbreviations. CHAPTER *I INTRODUCTION TO DIFFERENTIATION 3 Historical note Rates of change Average and actual rates of change Slopes of curves Graphic differentiation by two methods. CHAPTER II DIFFERENTIATION OF FUNCTIONS 26 Differentiation of ax" Differentiation of a sum of terms Proof of construction for slope curves Beam problems Lengths of sub-tangents and sub-normals of curves Differentiation of ex- ponential functions Differentiation of log* Differentiation of sinh x and cosh x Differentiation of the trigonometric functions Simple harmonic motion. CHAPTER III ADDITIONAL RULES OF DIFFERENTIATION .... 63 Differentiation of a function of a function Differentiation of a product Differentiation of a quotient Differentiation of inverse trigonometric functions Partial differentiation Total differential Logarithmic differentiation. CHAPTER IV APPLICATIONS OF DIFFERENTIATION ...... 88 Maximum and minimum values Point of inflexion Calculation of small corrections Expansion of functions in series Theorems of Taylor and Maclaurin. xii CONTENTS CHAPTER V PAGE INTEGRATION 115 Meaning of integration Graphic integration Application of in- tegration to " beam " problems Coradi integraph Rules for integration of simpler functions Integration of powers of x Integration of exponential functions Integration of trigonometric functions Indefinite and definite integrals Method of determining the values of definite integrals Proof of Simpson's rule. CHAPTER VI FURTHER METHODS OF INTEGRATION . . . . . . 146 Integration by the aid of partial fractions Integration by the resolution of a product into a sum Integration by substitution Integration by parts Reduction formulae Gamma function List of integrals. CHAPTER VII MEAN VALUES, ETC. . . . . . . . . 180 Determination of mean values Root mean square values Volumes Volumes of solids of revolution Length of arc Peri- meter of ellipse Area of surface of solid of revolution Centre of gravity Centroid " Double sum curve " method of finding the centroid vertical Centroids of sections by calculation Centroids found by algebraic integration Centre of gravity of irregular solids Centre of gravity of a solid of revolution Centre of pressure Moment of inertia Swing radius The parallel axis theorem Theorems of perpendicular axes Moment of inertia of compound vibrators Determination of ist and 2nd moments of sections by means of a graphic construction and the use of a planimeter. CHAPTER VIII POLAR CO-ORDINATES 257 Polar co-ordinates Spirals Connection between rectangular and polar co-ordinates Use of polar co-ordinates for the determination of areas The Rousseau diagram Dr. Fleming's graphic method for the determination of R.M.S. values Theory of the Amsler plani- meter. CONTENTS xiii CHAPTER IX PAGE SIMPLE DIFFERENTIAL EQUATIONS . - 270 Differential equations, definition and classification Types : ~- dy given as a function of x : -~ given as a function of y : General linear equations of the first order : Exact differential equations : Equa- tions homogeneous in x and y : Linear equations of the second order Use of the operator D Useful theorems involving the operator D Equations of the second degree. CHAPTER X APPLICATIONS OF THE CALCULUS 300 Examples in Thermodynamics : Work done in the expansion of a gas Work done in a complete theoretical cycle Entropy of water - Efficiency of engine working on the Rankine cycle Efficiency of engine working on the Rankine cycle, with steam kept saturated by jacket steam Examples relating to loaded beams, simply supported or with fixed ends, the loading and the section varying Shearing stress in beams Examples on Applied Electricity Examples on Strengths of Materials Loaded struts Tension in belt Friction in a footstep bearing Schiele pivot Examples on Hydraulics Centre of Pressure Transition curve in surveying. CHAPTER XI HARMONIC ANALYSIS . .- . 342 Fourier's theorem Analysis by calculation Harrison's graphic method of analysis Analysis by superposition. CHAPTER XII THE SOLUTION OF SPHERICAL TRIANGLES . . . 355 Definition of terms Spherical triangle Solution of spherical triangles Solution of right-angled spherical triangles Napier's rules of circular parts The "ambiguous" case Applications in spherical astronomy Graphic solution of a spherical triangle. CHAPTER XIII MATHEMATICAL PROBABILITY AND THEOREM OF LEAST SQUARES 370 Probability Exclusive events Probability of the happening together of two independent events Probability of error Theorem of least squares Error of the arithmetic mean Weight of an observation. xiv .CONTENTS PAGE ANSWERS TO EXERCISES . 387 TABLES : Trigonometrical ratios 397 Logarithms 398 Antilogarithms 400 Napierian logarithms 402 Natural sines 404 Natural cosines 406 Natural tangents 408 Logarithmic sines 410 Logarithmic cosines , .412 Logarithmic tangents . . . . . . . . . .414 Exponential and hyperbolic functions 416 INDEX ...... 417 MATHEMATICS FOR ENGINEERS PART II INTRODUCTORY THE subject-matter of this volume presents greater difficulty than that of Part I. Many of the processes described herein de- pend upon rules explained and proved in the former volume ; and accordingly it is suggested that, before commencing to read this work, special attention should first be paid to Part I, pp. 452- 460, 463-467, 469-472, and pp. 273-299 ; whilst a knowledge of the forms of the curves plotted in Chapter IX should certainly prove of great assistance. The abbreviations detailed below will be adopted throughout. -> stands for " approaches." " equals " or " is equal to." + ,, ,, " plus." ,, " minus." X " multiplied by." 4- " divided by." " therefore." " plus or minus." ,, " greater than." " less than." ,, " circle." ce ,, ,, " circumference." ,, " varies as." co ,, " infinity." L. ,, ,, " angle." A >, ,, triangle " or " area of triangle." Li_ or 4 ! " factorial four " ; the value being that of the product 1.2.3.4 or 2 4- "P, " the number of permutations of n things taken two at a time." "C 2 " the number of combinations of n things taken two at a time." n. 3 n (n i) (n 2). B 1 2 MATHEMATICS FOR ENGINEERS t\ stands for " efficiency." " angle in degrees." 6 ,, ,, " angle in radians." I.H.P. ,, " indicated horse-power." B.H.P. ,, " brake horse-power." m.p.h. ,, " miles per hour." r.p.m. ,, ,, " revolutions per minute." r.p.s. " revolutions per second." I.V. ,, ,, " independent variable." F. ,, " degrees Fahrenheit." C. ., ,, " degrees Centigrade." E.M.F. ,, ,, " electro-motive force." 1 ,, .., " moment of inertia." E " Young's modulus of elasticity." S n ,, " the sum to n terms." S^ ,, " the sum to infinity (of terms)." 2 ,, ,, " sum of." B.T.U. " Board of Trade unit." B.Th.U. " British thermal unit." T ,, ,, " absolute temperature." M ,, ,, " coefficient of friction." sin" 1 x ,, ,, " the angle whose sine is x." e ,, ,, " the base of Napierian logarithms." g ,, the acceleration due to the force of gravity." cms. ,, ,, " centimetres." grms. ,, ,, " grammes." -Ly ,, " the limit to which y approaches as x approaches the value a." C. of G. ,, ,, " centre of gravity." C. of P. ,, " centre of pressure." k ,, ,, " swing radius," or " radius of gyration." M.V. ,, " mean value." R.M.S. " root mean square." f'(x] ,, ,, " the first derivative of a function of x." f"( x ] ,, ,, " the second derivative of a function of x." dy -j- ,, ,, " the differential coefficient of y with regard to x." I ydx ,, ,, " the integral of y with respect to x as the I.V." 8 ,,..," difference of." ,, " the operation -3-.." " candle-power." M.S.C.P. ,, " mean spherical candle-power." p ,, ,, " density." CHAPTER I INTRODUCTION TO DIFFERENTIATION THE seventeenth century will ever be remarkable for the number of great mathematicians that it produced, and still more so for the magnitude of the research accomplished by them. In the early part of the century Napier and Briggs had introduced their systems of logarithms/ whilst Wallis and others directed their thoughts to the quadrature of curves, which they effected in some instances by expansion into series, although the Binomial Theorem was then unknown to them. In 1665 Newton, in his search for the method of quadrature, evolved what he termed to be a system of " fluxions " or flowing quantities : if x and y, say, were flowing quantities, then he denoted the velocity by which each of these fluents increased by x and y respectively. By the use of these new forms he was enabled to determine expressions for the tangents of curves, and also for their radii of curvature. At about the same time Leibnitz of Leipsic, also concerned with the same problem, arrived at practically the same system, although he obtained his tangents by determining " differences of numbers." To Leibnitz is due the introduction of the term " differential," and also the differential notation, viz., dx and dy for the differentials of x and y : he also in his expression for the summation of a number of quantities first wrote the symbol f, his first idea being to employ the word " omnia " or its abbre- viation " omn~." Thus, if summing a number of quantities like x, he first wrote " omnia x," which he contracted to " omn. x," and later he modified this form to fx. Great controversy raged for some time as to the claims of Newton and Leibnitz to be called the inventor of the system of the " Calculus," which is" the generic term for a classified collection of rules; but it is now generally conceded that the discoveries were independent, and were in fact the natural culmination of the research and discoveries of many minds. 3 MATHEMATICS FOR ENGINEERS The Calculus was further developed by Euler, Bernoulli, Legendre and many others, but until a very recent date it remained merely " a classified collection of rules" : its true meaning and the wide field of its application were for long obscured. Nowadays, however, a knowledge of the Calculus is regarded, particularly by the engineer, as a vital part of his mental equip- ment : its rules have been so modified as to become no serious tax on the memory, and the true significance of the processes has been presented in so clear a light that the study of the Calculus presents few difficulties even to the ultra-practical engineer. This revolution of thought has been brought about entirely through the efforts of men who, realising the vast potentialities of the Calculus, have reorganised the teaching of the subject : they have clothed it and made it a live thing. The Calculus may be divided into two sections, viz., those treating of differentiation and integration respectively. Differentia- tion, as the name suggests, is that part of the subject which is concerned with differences, or more strictly with the comparison of differences of two quantities. Thus the process of differentia- tion resolves itself into a calculation of rates of change; but the manner in which the rate of change is determined depends on the form in which the problem is stated. Thus, if the given quantities are expressed by the co-ordinates of a curve, the rate of change of the ordinate compared with the change in the abscissa for any particular value of the abscissa is measured by the slope of the curve at the point considered. Differentiation is really nothing more nor less than the deter- mination of rates of change or of slopes of curves. The term " rate of change " does not necessarily imply a " time rate of change," i. e., a rate of change with regard to time, such as the rate at which an electric current is changing per second, or the rate at which energy is being stored per minute; but the change in one quantity may be compared with the change in any other quantity. As an illustration of this fact we may discuss the following example The velocity of a moving body was measured at various distances from its starting point and the results were tabulated, thus s (distance in feet) . o 5 12 v (velocity in feet per sec.) 10 M 15 INTRODUCTION TO DIFFERENTIATION 5 To find the values of the " space rate of change of velocity " for the separate space intervals. Considering the displacement from o to 5 ft., the change in the velocity corresponding to this change of position is 1410, *. e., 4 ft. per sec. change of velocity 1410 4 Hence * .^ = - - = - = -8 change of position 5 o 5 or, the change of velocity per one foot change of position = -8 ft. per sec., and rate of change of velocity = -8 ft. per sec. per foot. Again, if s varies from o to 12, the change of v = 1510 = 5 ; or, the rate of change of velocity (for this period) = T 5 ^ ft. per sec. per foot. Similarly, the rate of change of v, whilst s ranges from 5 to 12, 1514 i ,, , -^ - = - ft. per sec. per foot. 12- 5 7 The rates of change have thus been found by comparing differ- ences. The phrase " change of" occurs frequently in this investi- gation, and to avoid continually writing it a symbol is adopted in its place. The letter thus introduced is 8 (delta), the Greek form of d, the initial letter of the word " difference " : it must be regarded on all occasions as an abbreviation, and hence no operation must be performed upon it that could not be performed if the phrase for which S stands was written in full. In other words, the ordinary rules applying to algebraic quantities, such as multiplication, division, addition or subtraction, would be incorrectly used in conjunction with 8. Thus, mv (the formula for momentum) means m multiplied by v, or a mass multiplied by a velocity, whilst Sv represents " the change of v," or if v is the symbol for velocity, 8v change of velocity. Again, 8/ = change of time or change of temperature, as the case may be. Using this notation our previous statements can be written in the shorter forms : thus (1) As s changes from o to 5 Sv = 1410 = 4 8s = 5-0 = 5 H-* (2) As s changes from o to 12 Sv = 15 10 = 5 8s = 12 o = 12 and !_ y = A Ss 12 * * MATHEMATICS FOR ENGINEERS (3) As s changes from 5 to 12 Sv i and ~=s s= !<: 8v = 15 14 = i Ss = 12 5 = 7 It must be noted that we do not cancel S from the numerator and denominator of the fraction ^-. The final result in (i), viz., ^- = -8, as s changes from o to 5, needs further qualification. From the information supplied we cannot say with truth that the change in the velocity for each foot from o to 5 ft. is -8 ft. per sec. : all that we know with certainty is that, as s changes from o to 5 ft., the average rate of change of velocity over this space period is -8 ft. per sec. Supposing the change of velocity to be continuous over the period considered, the value of ^ already obtained would be the actual rate of change of velocity at some point or points in the period considered. It is usual to tabulate the values of the original quantities and their changes, and unless anything is given to the contrary the average values of the rate of change are written in the middle of the respective periods. The table is set out thus Sv s V Ss Sv Ss IO 5 4 i = -8 5 14 7 i r = '143 12 15 ~~~" ~~ _. To distinguish in writing between average and actual rates of change the notation employed is slightly modified, d being used in dv place of 8 ; -j- thus representing an actual rate of change of velocity, and ^7 representing an average rate of change of velocity. Once again it must be emphasised that d must be treated strictly in association with the v or t, as the case may be, and dt does not mean , , j dv v dxt, nor does -j- give . INTRODUCTION TO DIFFERENTIATION Another example can now be considered to demonstrate clearly the distinction between an average and an actual rate of change. For a body falling freely under the influence of gravity the values of the distances covered to the ends of the ist, 2nd and 3rd seconds of the motion are as in the table t (sees.) . o I 2 3 s (feet) . 16.1 64.4 144.9 Find the average velocities during the various intervals of time, and also the actual velocities at the ends of the ist, 2nd and 3rd seconds respectively. The average velocities are found in the manner described before, i. e., by the comparison of differences of space and time, and the results are tabulated, thus ' ' 5s st v- Ss o 16-1 i 16-1 I 16-1 48-3 i 48-3 2 64-4 80-5 i 80-5 3 144-9 ~~~"~ ~ The average velocities, viz., the values in the last column, are written in the lines between the values of the time to signify that they are the averages for the particular intervals. As also it is known that in this case the velocity is increased at a uniform rate, it is perfectly correct to state that the actual velocities at the ends of -5, 1-5 and 2-5 seconds respectively are given by the average velocities over the three periods and are 16-1, 48-3 and 80-5 ft. per sec. We have thus found the actual velocities at the half seconds, but not those at the ends of the ist, 2nd and 3rd seconds. The determination of these velocities introduces a most important process, illustrating well the elements of differentiation, and in consequence the investigation is discussed in great detail. . The student of Dynamics knows that -the law connecting space and time, in the case of a falling body, is s = \gt z = i6-i/ 2 , and 8 MATHEMATICS FOR ENGINEERS a glance at the table of values of s and t confirms this law ; thus, when t = 2, s = 64-4, which = i6-iX2 2 or i6-i^ 2 . To find the actual velocity at the end of the first second we must calculate the average velocities over small intervals of time in the neighbourhood of I sec., and see to what figure these velocities approach as the interval of time is taken smaller and smaller. Thus if t i s = 16-1 x i 2 = 16-1 t = i-i s = 16-1 x i-i 2 = 19-481 8s = 19-481 16-1 = 3-381 U = i-i i = -i \ &s 3-381 and (average) v = ^ = ^- = 33-81 i. e., the average velocity over the interval of time i to i-i sec. is 33-81 ft. per sec. This value must be somewhere near the velocity at the end of the first second, but it cannot be the absolute value, since even in the short interval of time, viz., -i sec., the velocity has been increased by a measurable amount. A better approximation will evidently be found if the time interval is narrowed to -01 sec. Then t i s = 16-1 t = i-oi s = 16-1 X i-oi 2 = 16-42361 Ss = -32361 St = -oi 8s -32361 (average) v = ^ = -^~ = 32-361 A value still nearer to the true will be obtained if the time interval is made -ooi sec. only. t = i s = 16-1 t = i-ooi s = 16-1 X i-ooi 2 = 16-1322161 8s = -0322161 8^ = -ooi 8s -0322161 and (average) v = ^ = ~^^ = 32-2161 By taking still smaller intervals of time, more and more nearly correct approximations would be found for the velocity ; the values of v all tending to 32-2, and thus we are quite justified in saying that when / = i, v = 32-2 ft. per sec. Or, using the language of p. 458 (Mathematics for Engineers, Pt. 1), we state that the limiting value of v as t approaches i is 32-2 ; a result expressed in the shorter form (average) v - 32-2 as 8/ -> o when t = i where the symbol -> means " approaches " but (average) v = , and thus ~ -> 32-2 as St -> o when t = i ot ot INTRODUCTION TO DIFFERENTIATION Again, an actual velocity is an average velocity over an extremely small interval of time ; or, in other words, an actual velocity is the limiting value of an average velocity, so that . (actual) v = (average) v i.e., ds dt st-^-o 8t By similar reasoning it could be proved that the actual velocity at the end of the 2nd second was 64-4 ft. per sec., and at the end of the 3rd second the velocity was 96-6 ft. per sec. This example may usefully be continued a step further, by calculating the values of the acceleration; this being now possible since the velocities are known. Tabulating as before 1 V Sv Si &v a = sl I 32-2 32-2 I 32-2 2 64-4 32-2 I 3 2-2 3 96-6 and we note that the average acceleration is constant and is thus the actual acceleration. Our results may now be grouped together in one table, in which some new symbols are introduced, for the following reason. A velocity is the rate of change of displacement, and is found by " differentiating space with regard to time," and an acceleration is the rate of change of velocity, and hence it is a rate of change of the change of position, and so implies a double differentiation. Thus whilst -T- is called the first derivative or differential coefficient dl) of s with regard to t, -^ is the first derivative of v with regard to t and the second derivative of s with regard to t. TU ds , dv dfds\ ,,.,,, , n Then v = -, , and = TT = -T, -57 , this last form being usually at at at\atj d 2 s written as ^ (spoken as d two s, dt squared) ; and it denotes that the operation of differentiating has been performed twice upon s. 10 MATHEMATICS FOR ENGINEERS The complete table of the values of the velocity and the accelera- tion reads s i Ss st _Ss --GO st Sv S*s ~ St ~ SP O o ' 16-1 i 16-1 16-1 i 32-2 i 32-2 48*3 i 48*3 64-4 2 32-2 i 32-2 80-5 i 80-5 144-9 3 _ ~~~ ' The next example refers to a similar case, but is treated from the graphical aspect. Example i. Experiments made with the rolling of a ball down an inclined plane gave the following results t (sees.) o I 2 3 s (cms.) o 20 80 180 Draw curves giving the space, velocity and acceleration respectively at any time during the period o to 3 sees. By plotting the given values, 5 vertically and t horizontally, the " space-time " curve or " displacement " curve is obtained; the curve being a parabola (Fig. i). Select any two points P and Q on the curve, not too far apart, and draw the chord PQ, the vertical QN and the horizontal PN. Then the slope of the chord PQ = Now PN may be written as St since it represents a small addition to the value of t at P : also QN = 8s, 8s so that slope of chord PQ = ^ but ^ average velocity between the times OM and OR, hence Of the average velocity is measured by the slope of a chord. Now let Q approach P, then the chord PQ tends more and more to lie along the tangent at P, and by taking Q extremely close to P the chord PQ and the tangent at P are practically indistinguishable INTRODUCTION TO DIFFERENTIATION ir the one from the other; whilst in the limit the two lines coincide. go Then since the slope of the chord PQ gives the value of ~ , and ot the limiting value of 7 is -^ , it follows that the slope of the tangent ot (it expresses -^ ; but the slope of a curve at any point is measured .30 1-5 2-O Values of hmc FIG. i. 2-5 by the slope of its tangent at that point, and hence we have evolved the most important principle, viz., that differentiation is the deter- mination of the slopes of curves. [Incidentally it may be remarked that here is a good illustration of the work on limiting values ; for the slope of a curve, or of the tangent to the curve, is the limiting value of the slope of the chord, *'. e., the value found when the extremities of the chord coincide; and this value does not take the indeterminate form C) , as might at o first sight be supposed, but is a definite figure.] 12 MATHEMATICS FOR ENGINEERS Thus the slope of the tangent at any point on the space-time curve measures the actual rate of change of the space with regard to time at that particular instant; or, in other words, the actual velocity at that instant. Hence by drawing tangents to the space- time curve at various points and calculating the slopes, a set of values of the velocity is obtained : these values are then plotted to a base of time and a new curve is drawn, which gives by its ordinate the value of the velocity at any time and is known as the " velocity-time " curve. Since this curve is obtained by the calculation of slopes, or rates of change, it is designated a derived or slope curve; the original curve, viz., the space-time curve, being termed the primitive. In the case under notice the velocity-time curve is a sloping straight line, and in consequence its slope is constant, having the value 40. Hence the derived curve, which is the acceleration-time curve, is a horizontal line, to which the ordinate is 40. There are thus the three curves, viz., the primitive or space-time curve, the first derived curve or the velocity-time curve, and the second derived curve or the acceleration-time curve. Graphic Differentiation. The accurate construction of slope curves is a most tedious business, for the process already described necessitates the drawing of a great number of tangents, the calcu- lations of their respective slopes and the plotting of these values. There are, however, two modes of graphic differentiation, both of which give results very nearly correct provided that reasonable care is taken over their use. Method i (see Fig. 2). Divide the base into small elements, the lengths of the elements not being necessarily alike, but being so chosen that the parts of the curve joining the tops of the con- secutive ordinates drawn through the points of section of the base are, as nearly as possible, straight lines. Thus, when the slope of the primitive is changing rapidly, the ordinates must be close together; and when the curve is straight for a good length, the ordinates may be placed well apart. Choose a pole P, to the left of some vertical OA, the distance OP being made a round number of units, according to the horizontal scale. Erect the mid-ordinates for all the strips. Through P draw PA parallel to ab, the first portion of the curve, and draw the horizontal Ac to meet the mid-ordinate of the first strip in c. Then dc measures, to some scale, the slope of the chord ab, and therefore the slope of the tangent to the primitive INTRODUCTION TO DIFFERENTIATION 13 curve at m, or the average slope of the primitive from a to b, with reasonable accuracy. Continue the process by drawing PM parallel to bl and Ms horizontal to meet the mid-ordinate of the second strip in s : then cs is a portion of the slope or derived curve. Repeat the operations for ah 1 the strips and draw the smooth curve through the points c, s, etc. : then this curve is the curve of slopes. FIG. 2. Graphic Differentiation, Method i. Indicate a scale of slope along a convenient vertical axis and the diagram is complete : the scale of slope being the old vertical scale divided by the polar distance expressed in terms of the horizontal units. E. g., if the original vertical scale is i" = 40 ft. Ibs. and the horizontal scale is i" = 10 ft. : then, if the polar distance p is taken as 2", i. e., as 20 horizontal units, the new vertical scale, or scale of slope, is i" = ^ ' 2 Ibs. 20 ft. Proof of the construction. The slope of the primitive curve at m = slope of curve ab = bf = OA. = cd ~af~ p " p 14 MATHEMATICS FOR ENGINEERS or, the ordinate dc, measured to the old scale, = p x the slope of the curve at m. If, then, the original vertical scale is divided by p the ordinate dc, measured to the new scale, = slope of the curve at m. The great disadvantage of this method is that parallels have to be drawn to very small lengths of line and a slight error in the setting of the set square may quite easily be magnified in the draw- ing of the parallel. Hence, for accuracy, extreme care in draughts- manship is necessary. It should be observed that this method of graphic differentiation is the converse of the method of graphic integration described in FIG. 3. Graphic Differentiation, Method 2. Chapter VII (Part I), and referred to in greater detail in Chapter V of the present volume. Method 2. Let ABC (Fig. 3) be the primitive curve. Shift the curve ABC forward to the right a horizontal distance sufficiently large to give a well-defined difference between the curves DEF and ABC ; but the horizontal distance, denoted by h, must not be great. From the straight line base OX set up ordinates which give the differences between the ordinates of the curves ABC and DEF, the latter curve being treated as the base : thus ab = a'b'. Join the tops of the ordinates so obtained to give the new curve G6H, and shift the curve G6H to the left a horizontal distance = -, this operation giving the curve MPN, which is the true slope or derived curve of the primitive ABC. Complete the diagram by adding a scale of slope, which is the old vertical scale divided by h (expressed in horizontal units). INTRODUCTION TO DIFFERENTIATION This method can be still further simplified by the use of tracing paper, thus : Place the tracing paper over the diagram and trace the curve ABC upon it; move the tracing paper very carefully forward the requisite amount, viz., h, and with the dividers take the various differences between the curves, such as a'b'. Step off these differences from OX as base, but along ordinates - units re- 45 moved to the left of those on which the differences were actually measured : then draw the curve through the points and this is the slope curve. Tamp (C*) JO 4O SO 6O O h" 20 H "Tung (mins.) FIG. 30. Variation of Temperature of Motor Field Coils. Examples on the use of these two methods now follow. Example 2. The temperature of the field coils of a motor was measured at various times during the passage of a strong current, with the following results Time (mins.) . o 5 10 15 20 25 3 35 40 45 50 55 60 65 Temperature (C.) 20 26 32-5 4i 46 49 52-5 54-5 56-5 58 59-5 61 61-7 62 Draw a curve to represent this variation of temperature, and a curve to show the rate at which the temperature is rising at any instant during the period of 65 mins. The values of the temperature when plotted to a base of time give the primitive curve in Fig. 30. i6 MATHEMATICS FOR ENGINEERS To draw the slope curve we first divide the base in such a way that the portions of the curve between consecutive ordinates have the same inclination for the whole of their lengths, i. e., the elements of the curve are approximately straight lines. Thus, in the figure, there is no appreciable change of slope between A and a, or a and d. There is no need to draw the ordinates through the points of section for their full lengths, since the intersections with the primitive curve are all that is required. Next a pole P is chosen, 20 horizontal units to the left of A, and through P the line PB is drawn parallel to the portion of the curve Aa. A horizontal B6 cuts the mid-ordinate of the first strip at b, and 6 is a point on the slope or derived curve. The processes repeated for the second strip, PC being drawn parallel to ad and Cc drawn horizontal to meet the mid-ordinate of the second strip in c, which is thus a second point on the slope curve. A smooth curve through points such as b and c is the slope curve, giving by its ordinates the rate of increase of the temperature ; and it will be observed that the rate of increase is diminished until at the end of 65 mins. the rate of change of temperature is zero, thus indicating that at the end of 65 mins. the losses due to radiation just begin to balance the heating effect of the current. Since the polar distance = 20 units, the scale of slope _ original vertical scale 20 and in the figure the original vertical scale is i" = 20 units ; hence the scale of slope is i" = i unit; and this scale is indicated to the right of the diagram. Example 3. Plot the curve y = x"-, x ranging from o to 3, and use Method 2 to obtain the derived curve. The values for the ordinates of the primitive curve y x"- are as in the table X o I 2 3 y o I 4 9 and the plotting of these gives the curve OAB in Fig. 4. Choosing h as -5 horizontal unit, the curve is first shifted forward this amount, and the curve CG results. The vertical differences be- tween these two curves are measured, CG being regarded as the base curve, and are then set off from the axis of x as the base. Thus when x 3, the ordinate of the curve OAB is 9 units, and that of CG is 6-25, so that the difference is 2-75, and this is the ordinate of the curve MN. By shifting the curve MN to the left by a distance = , i. e., -25 hori- zontal unit, the true slope curve ODE is obtained : this is a straight INTRODUCTION TO DIFFERENTIATION 17 line, as would be expected since the primitive curve is a " square " parabola. As regards the scale of slope, the new vertical scale _ old vertical scale, and since h = -5, the new vertical scale, or scale of slope, which is used when measuring ordinates of the curve ODE, is twice the original vertical scale. The derived curve supplies much information about the primi- tive. Thus, when the ordinate of the derived curve is zero, i. e., when the derived curve touches or cuts the horizontal axis, the O -5 / 1-5 2. 2-5 FIG. 4. Graphic Differentiation. slope of the primitive is zero; but if the slope is zero the curve must be horizontal, since it neither rises nor falls, and this is the case at a turning point, either maximum or minimum. Hence turning points on the primitive curve are at once indicated by zero ordinates of the slope curve. Again, a positive ordinate of the derived curve implies a positive slope of the primitive, and thus indicates that in the neighbourhood considered the ordinate increases with increase of abscissa. Also a large ordinate of the slope curve indicates rapid change of ordinate of the primitive with regard to the abscissa. This last fact suggests another and a more important one. By a careful examination of the primitive curve we see what is actually c iS MATHEMATICS FOR ENGINEERS happening, whilst the slope curve carries us further and tells us what is likely to happen. In fact, the rate at which a quantity is changing is very often of far greater importance than the actual value of the quantity; and as illustrations of this statement the following examples present the case clearly. Example 4. The following table gives the values of the displacement of a 21 knot battleship and the weight of the offensive and defensive factors, viz., armament, armour and protection. From these figures (2000 _ TOGO 2OOOO 84OOO 60OO E8OOO 3OOOO Values of P FIG. 5. Displacement and Armament of Battleship. calculate values of Q (ratio of armament, etc., to displacement) and q (rate of increase of armament, etc., with regard to displacement). Find also the values of .. Displacement \ P tons J 18000 2000O 22OOO 24000 2600O 28000 30000 Armament, etc., \ p tons / 6880 7850 88 3 9820 10810 11820 12845 INTRODUCTION TO DIFFERENTIATION The values of Q are found by direct division and are p 18000 20000 22OOO 24000 26OOO 28OOO 3OOOO Q 383 392 40! 409 416 422 428 dp Values of q, i. e., -,, may be found by (a) construction of a slope curve, or (6) tabulating differences. (a) By construction of a slope curve. Plotting p along the vertical axis and P along the horizontal axis (see Fig. 5), we find that the points lie very nearly on a straight line. Hence the slope curve is a hori- zontal line, whose ordinate everywhere is the slope of the original line. By actual measurement the slope is found to be -498 and thus 498. This di) the average value for ~~, over the range considered, is average value of the rate of change does not, however, give as much information for our immediate purpose as the separate rates of change considered over the various small increases in the displacement. (6) By tabulation of differences, as in previous examples p P sp SP , Sp ? SP 18000 6880 _ _ _ 97 2OOO 485 2OOOO 7850 980 2OOO .490 22OOO 8830 990 2OOO 495 24000 9820 '. 990 2OOO 495 26OOO I08IO 1010 2OOO .505 28OOO II820 1025 2OOO 5125 3OOOO 12845 JV * Now sfs = rate of increase of armament compared with displace- oJr ment ; as the displacement increases it is seen from the table of values that this ratio increases, and the questions then arise : " Does this increase coincide with an increase or a decrease in the values of Q, and if with one of these, what is the relation between the two changes ? " By tabulating the corresponding values of q and Q and calculating the values of i, we obtain the following table (the values of Q at 19000, 21000, etc., being found from a separate plotting not shown here) 20 MATHEMATICS FOR ENGINEERS p 19000 21000 23000 25000 27000 29OOO p 485 490 495 495 505 5125 Q 387 396 405 .412 419 426 q g 1-253 1-236 1-223 1-202 1-203 I-2O4 It will be seen by examination of this table that the fraction Q decreases as ships are made larger : in other words, while the arma- ment increases with the displacement, the increase is not so great as it should be for the size of the ship, since the weight of the necessary engines, etc., is greater in proportion to the weight of armament and protection for the larger than for the smaller ships. Thus, other things being equal, beyond a certain point it is better to rely on a greater number of smaller ships than a few very large ones. Example 5. Friend gives the following figures as the results of tests on iron plates exposed to the action of air and water. The original plates weighed about 2-5 to 3 grms. Plot these figures and obtain the rate curves for the two cases, these curves being a measure of the corrosion : comment on the results. Time in days . ' . 2 7 13 19 26 32 37 In the light : loss of \ weight in grms. . J 0048 031 0645 08 093 126 In the dark : loss of) weight in grms. . / 0032 0208 037 058 0674 0816 0916 The two sets of values are plotted in Fig. 6, the respective curves being LLL for the plates exposed in the light, and DDD for those left in the dark. The effect of the action of light is very apparent from an examination of these curves. Next, the slope curves for the two cases are drawn, Method 2 being employed, but the intermediate steps are not shown. The curve /// is the slope curve for LLL, and ddd that for the curve DDD. It will be observed that in both cases the rate of loss is great at the commencement, but decreases to a minimum value after 20 days exposure in the case of t the curve ///, and after 25 days in the case of the curve ddd. After these turning points have been reached the rate quickens, INTRODUCTION TO DIFFERENTIATION 21 the effect being very marked for the plates exposed to the light ; and for these conditions the slope curve /// suggests that the corrosive action is a very serious matter, since it appears that the rate of loss must steadily increase. A further extremely good illustration of the value of slope curves is found in connection with the cooling curves of metals. In the early days of the research in this branch of science, the cooling curve alone was plotted, viz., temperatures plotted to a base of o s 10 15 O 25 30 FIG. 6. Tests on Corrosion of Iron Plate. time. Later investigations, however, have shown that three other curves are necessary, viz., an inverse rate curve, a difference curve and a derived differential curve ; the co-ordinates for the respective curves being (a) Temperature (0) time (f) curve; t horizontal and vertical. (b) Inverse rate curve : -, horizontal and vertical. To obtain this curve from curve (a), the slopes must be very carefully calcu- lated, and it must be remembered that these slopes are the measures of the inclinations to the vertical axis and not to the horizontal, dt ^ ft i. e., are values of -^ and not ,,. at 22 MATHEMATICS FOR ENGINEERS (c) vertical and X horizontal : 00! being the difference of temperature between the sample and a neutral body cooling under identical conditions. (d) vertical and JA horizontal : this curve thus being the inverse rate curve of curve (c). Exercises 1. On Rates of Change and Derived Curves 1. What do the fractions ^- and -,- actually represent (s being a r V (f( displacement, and t a time) ? Take some figures to illustrate your answer. 2. Further explain the meanings of -~- and -r by reference to a graph. 3. When an armature revolves in a magnetic field the E.M.F. produced depends on the rate at which the lines of force are being cut. Express this statement in a very brief form. 4. For a non-steady electric current the voltage V is equal to the resistance R multiplied by the current C plus the self-inductance L multiplied by the time rate at which the current is changing. Express this in the form of an equation. 5. At a certain instant a body is 45-3 cms. distant from a fixed point. 2-14 seconds afterwards it is 21-7 cms. from this point. Find the average velocity during this movement. At what instant would your result probably measure the actual velocity ? 6. At 3 ft. from one end of a beam the bending moment is 5 tons ft. At 3' 2\" from the same end it is 5-07 tons ft. If the shear is measured by the rate of change of bending moment, what is the average shearing force in this neighbourhood ? 7. Tabulate the values of q, i. e., ^ for the following case, the figures referring to a battleship of 23 knots. p 18000 2OOOO 22OOO 24OOO 26OOO 28OOO 3OOOO 32OOO p 6170 7080 8OOO 8930 9890 10855 II820 I28lO 8. Tabulate the values of ^r for the case of a battleship of 25 knots from the following P 18000 2OOOO 22OOO 24000 26OOO 28OOO 3OOOO 32OOO p 5210 6050 6910 7790 8660 9550 10460 II370 = sp and Q = F INTRODUCTION TO DIFFERENTIATION 9. Tabulate the values of the velocity and the acceleration for the following case Space (feet) i 2-4 4'4 6 7-6 II-2 15-6 2O'4 Time (sees.) 2 4 6 7 8 I 1-2 i-4 10. Plot the space-time curve for the figures given in Question 9 and by graphic differentiation obtain the velocity -time and the accelera- tion-time curves. 11. Plot the curve y -$x 3 from x = 2 to x = +4 and also its derived curve. What is the ordinate of the latter when x 1-94? 12. Given the following figures for the mean temperatures of the year (the average for 50 years), draw a curve for the rate of change of temperature and determine at what seasons of the year it is most rapid in either direction. Time (intervals of J month) o i 2 3 4 5 6 7 8 9 10 ii Temperature 38-6 37'9 38-4 39-8 38-5 39-5 4-3 40-7 4 I- 5 45-5 45-5 48-5 12 49-3 13 14 15 16 I? IS 19 20 21 22 23 24 25 26 52 55 57-2 58-4 60-5 61-4 62-5 62-9 62-2 62-5 61-1 59-8 58-2 55-8 27 54-2 28 51 29 48-8 30 46-8 3* 43-5 32 42-1 33 40-6 34 39-8 35 38-8 36 38-6 13. s is the displacement from a fixed point of a tramcar, in time / sees. Draw the space-time, velocity-time and acceleration-time curves. t .1 2 3 4 5 6 7 8 9 10 s o 4 ii 21 34 50 69 9i 116 144 175 The scales must be clearly indicated. 14. The table gives the temperature of a body at time / sees, after it has been left to cool. Plot the given values and thence by differ- entiation obtain the rate of cooling curve. What conclusions do you draw from your final curve ? Time (mins.) o I 2 3 4 5 6 7 8 9 Temp. (F.) 136 134 132 130 128 126-5 124-8 123-3 122 120-5 10 ii 12 13 14 15 16 17 18 19 119-3 118 II6-8 II5-5 114-5 II3-5 112-5 111-5 110-5 109-5 MATHEMATICS FOR ENGINEERS 15. The following figures give the bending moment at various points along a beam supported at both ends and loaded uniformly. Draw the bending moment curve, and by graphic differentiation obtain the shear and load curves. Indicate clearly the scales and write down the value of the load per foot run. Distance from one \ end (ft.) . . / 2 4 6 8 10 12 14 16 18 20 Bending moment \ (tons ft.) . . J o 3'5 6-3 8-4 9-6 10 9-6 8-4 6-3 3'5 O 16. By taking values of in the neighbourhood of 15 find the actual rate of change of sin 6 with regard to (0 being expressed in radians). Compare your result with the value of cos 15. In what general way could the result be expressed ? 17. If the shear at various points in the length of a beam is as in the table, draw the load curve (i. e., the derived curve) and write down the loading at "3^ ft. from the left-hand end. Distance from left-hand end (ft.) I 2 3 4 5 6 Shearing force (tons) .... i 3 6 i i'5 2-1 18. An E.M.F. wave is given by the equation E = 150 sin 314^ + 50 sin 942^. Derive graphically the wave form of the current which the E.M.F. will send through a condenser of 20 microfarads capacity, assuming the condenser loss to be negligible. dE Given that C = K~j7' where C is current and K is capacity. 19. If momentum is given by the product of mass into velocity, and force is defined as the time rate of change of momentum, show that force is expressed by the product of mass into acceleration. 20. The following are the approximate speeds of a locomotive on a run over a not very level road. Plot these figures and thence obtain a curve showing the acceleration at any time during the run. Time (in mins. and sees.) o i-o 2-15 6-15 9-22 n-45 14-26 16-33 20-52 23-10 Speed (miles per hour) . start 6 10 18-2 22-8 25-5 28 29-2 28-6 26-1 21. Taking the following figures referring to CO 2 for use in a re- frigerating machine, draw the rate curve and find the value of -- when ctt t=i8 F. t F. . ' 5 o 5 10 15 20 25 30 35 40 >lbs.per\ sq. in J 285 310 335 363 392 423 456 491 528 567 INTRODUCTION TO DIFFERENTIATION 22. The weight of a sample of cast iron was measured after various heatings with the following results; the gain in weight being due to the external gases in the muffle. Number of heats . . o 2 6 12 22 23 24 25 26 Weight .... 146-88 146-94 147-04 H7-54 148-02 I48-II I 4 8.2 7 148-36 148-46 - 27 30 35 39 45 148-61 149-18 150-49 152-36 156-44 Plot a curve to represent this table of values, and from it construct the rate curve. 23. The figures in the table are the readings of the temperature of a sample of steel at various times during its cooling. Plot these values to a time base, and thence draw the " inverse rate " curve, i. e., the curve in which values of -. are plotted horizontally and the temperatures along the vertical axis. Time in sees. (I) . . 75 j 9 i5 120 135 150 165 180 195 210 225 Temperature in C. c (0) 1 850 848 844-7 8 4 2 839-5 838-5 838-2 838-1 838 837-9 837-5 240 255 270 285 292-5 300 315 330 345 360 367-5 375 350 45 836 833 829 825 8-'3- 3 822-2 821-7 821-5 821-3 821-1 819 8i5 813 8II-6 CHAPTER II DIFFERENTIATION OF FUNCTIONS Differentiation of ax". It has been shown in Chapter I how to compare the changes in two quantities with one another, and thus to determine the rate at which one is changing with regard to the other at any particular instant, for cases in which sets of values of the two variables have been stated. In a great number of instances, however, the two quantities are connected by an equation, indicating that the one depends upon the other, or, in other words, one is a function of the other. Thus if y = 5# 3 , y has a definite value for each value given to %, and this fact is expressed in the shorter form y =f(x). Again, if z ijx^y^xy 3 +5 log y> where both x and y vary, z depends for its values on those given to both x and y, and z f(x, y). To differentiate a function it is not necessary to calculate values of x and y and then to treat them as was done to the given sets of values in the previous chapter. This would occasion a great waste of time and would not give absolutely accurate results. Rules can be developed entirely from first principles which permit the differentiation of functions without any recourse to tables of values or to a graph. We now proceed to develop the first of the rules for the differen- tiation of functions ; and we shall approach the general case, viz., that of y ax n , by first considering the simple case of y = x 3 . Our problem is thus to find the rate at which y changes with regard to x, the two variables (y the dependent and x the independent variable or I.V.) being connected by the equation y = x s . The rate of change of y with regard to x is given by the value dv of -j-, and this is sometimes written as Dy when it is clearly understood that differentiation is with regard to x : the operator D having many important properties, as will be seen later in the book. If y is expressed as f(x), then -p is often written JfQ fix dx or /'(*). 26 DIFFERENTIATION OF FUNCTIONS 27 ~, Dy, -^-* or f'(x) is called the derivative or differential coefficient of y with respect to x; and the full significance of the latter of these terms is shown in Chapter III. We wish to find a rule giving the actual rate of change of y with regard to x, y being = x 3 , the rule to be true for all values of x. As in the earlier work, the actual rate of change must be determined as the limiting value of the average rate of change. Let x be altered by an amount Sx so that the new value of x = x -f- 8x ; then y, which depends upon x, must change to a new value y + Sy, and since the relation between y and x is y x 3 for all values of x (new value of y) = (new value of x) 3 or y+Sy = (x+Sx) 3 = ,r 3 +3# 2 . 8x+$x . (Sx) 2 +($x) 3 (i) but y = x 3 ........... (2) Hence, by subtraction of (2) from (i) y+Sy-y = 3 * 2 . Sx+3x(Sx) 2 +(Sx) 3 and Sy = $x 2 . Sx+3x(Sx) 2 +(Sx) 3 . Divide through by Sx, and Thus an average value for the rate of change over a small interval Sx has been found; and to deduce the actual rate of change the interval Sx must be reduced indefinitely. Let Sx=-ooi; then ^ = 3* 2 + (3* X-ooi) + -000001 -oooooi whilst if 8x = -ooooi jh_j :- = 3# 2 -f -00003^;+ -oooooooooi . (3) Evidently, by still further reducing &x the 2nd and 3rd terms of (3) can be made practically negligible in comparison with the ist term. Then, in the limit, the right-hand side becomes 3* 2 , and thus- ^ = T f = 3 * 2 dx i ^Sx dx- = * X 28 MATHEMATICS FOR ENGINEERS This relation can be interpreted graphically in the following manner : If the curve y = x s be plotted, and if also its slope curve be drawn by either of the methods of Chapter I, then the equation to the latter curve is found to be y = $x 2 . The two curves are plotted in Fig. 7. 10 01234 X FIG. 7. Primitive and Slope Curves. Example i. Find the slope of the curve y x 3 when x = 4. dy dx 3 o The slope of the curve = -v 1 = -j = 3* CLX CLX and if x = 4 -,- = 3 X 4 2 = 48. Meaning that, in the neighbourhood of x 4, the ordinate of the curve y = x 3 is changing 48 times as fast as the abscissa ; this fact being illustrated by Fig. 7. Working along the same lines, it would be found that -jj = dx 5 -y- himself). and -- = 5# 4 (the reader is advised to test these results for DIFFERENTIATION OF FUNCTIONS 29 Re-stating these relations in a modified form dx 3 - = 4^ = 4* ft* We note that in all these cases the results take the form dx n - = nx n ~ \ dx Thus the three cases considered suggest a general rule, but it would be unwise to accept this as the true rule without the more rigid proof, which can now be given. Proof of the rule dx n -M Vtt ~~ * - /(vV ax Let y = x n , this relation being true for all values of x . (i) If x is increased to x-}-8x, y takes a new value y-f-Sy, and from (i) it is seen that y-|-Sy = (x+8x) n . Expand (x-\-8x) n by the Binomial Theorem (see p. 463, Part I). Then Subtract (i) from (2), and Divide by 8^ 8v , , n(ni) 0/c . , , (w i)(n 2) M ,/ \ 9 , --=^n-i4- v --- 'x n - 2 (8x)-\ s 'x n - 3 (8x) 2 -t- terms 8,r |_2 |_3 containing products of (Sx} 3 and higher powers of (8x). Let 8* be continually decreased, and then, since Sx is a factor of the second and all succeeding terms, the values of these terms can be made as small as we please by sufficiently diminishing Sx. 30 MATHEMATICS FOR ENGINEERS Thus in the limit f- -> n x n - * Sx dv or -- Id Hence the first rule for differentiation of functions is established, viz. i. e., differentiation lowers the power of the I.V. by one, but the new power of x must be multiplied by the original exponent. The reason for the multiplication by the n can be readily seen, for the bigger the value of n the steeper is the primitive curve and therefore the greater the change of y for unit change of x. The n actually determines the slope of the primitive (cf. Part I, p. 340), and it must therefore be an important factor in the result of differentiation, since that operation gives the equation of the slope curve. To make the rule perfectly general, aUowance must be made for the presence of the constant multiplier a in ax n . It will be agreed that if the curve y = x 3 had been plotted, the curve y = ^x 3 would be the same curve modified by simply multiplying the vertical scale by 4. Hence, in the measurement of the slope, the vertical increases would be four times as great for the curve y = 4# 3 as for the curve y = x 3 , provided that the same horizontal increments were considered. Now the slope of the curve y = x 3 is given by the equation dy - ix* dx~ 3X so that the slope of the curve y = ^x 3 is given by ^ = 4x3*2 = I2x 2 . In other words, the constant multiplier 4 remains a multiplier throughout differentiation. This being true for any constant factor j-ax a = nax H-l dx* Accordingly, a constant factor before differentiation remains as such after differentiation. We can approach the differentiation of a multinomial expression DIFFERENTIATION OF FUNCTIONS by discussing the simple case y = $x 2 + 17 (a binomial, or two- term expression). The curves y = $x 2 and y = $x 2 + 17 are seen plotted in Fig. 8, and an examination shows that the latter curve is the former moved vertically an amount equal to 17 vertical units, i. e., the two curves have the same form or shape and consequently their slopes at corresponding points are alike. Thus if a tangent is drawn to each curve at the point for which x = 2-5, the slope of each tangent is measured as , i.e., 25; and con- sequently the diagram informs us that the term 17 makes no difference to the slope. JOO. FIG. 8. Thus dx- Now, by differentiating 5# 2 dx^ 17 term by term, we have since 17 is a constant and does not in any way depend upon x, and therefore its rate of change must be zero. It is seen that in this simple example it is a perfectly logical procedure to differentiate term by term and then add the results; and the method could be equally well applied to all many-term expressions. 32 MATHEMATICS FOR ENGINEERS Hence j x (ax n +bx n ~' i +cx n - 2 + . . . d) = nax n ~ l +b(n i)x n ~ 2 +c(n 2)x n ~ 3 + . . . and ^ x (ax a -{-b) = nax"- 1 . To apply these rules to various numerical examples : Example 2. Differentiate with respect to x the function = (gx i-6;r 6 ) = i4'4#- 8 +#^' 5 or IA-AX-*-\ _ V x Example 3. If y = -Sx A A, find the value of ^. ' * y = -8x A /L = -8x~, = -8^~^ or V ^s ^ so that in comparison with the standard form a = -8 and n = 1-5. Then or _ Example 4. If /w 1 ' 41 = C, the equation representing the adiabatic expansion of air, find --. In this example we have to differentiate p with regard to v, and before this can be done p must be expressed in terms of v. Now pv 1 -* 1 = C, so that p = ^ = Cw- 1 ' 11 . Hence ft = ^-Cz;- 1 - 11 = Cx -i-4iy- 2 - 11 = - i-4iCw- 2 - 41 dv dv and this result can be put into terms of p and v only, if for C we write its value pv l ' tl . Thus = -i-4ix/>w 1 - 4l x- 8 ' ll = i-4ipv- 1 = - -'. DIFFERENTIATION OF FUNCTIONS 33 Example 5. The formula giving the electrical resistance of a length of wire at temperature t C. is where R = Resistance at o C. Find the increase of resistance per i C. rise of temperature per ohm of initial resistance, and hence state a meaning for . The question may be approached from two standpoints ; viz. (a) Working from first principles. "D _ "O "D [ i. e., increase in resistance for t C. = but this is the resistance increase for initial resistance R , hence T? a increase in resistance per i C. per ohm initial resistance = -^- = a. (&) By differentiation. Rate of change of R with regard to t = -~ a = R a and consequently the rate of change of resistance per i C. per i ohm initial resistance = o. The symbol a is thus the " temperature coefficient," its numerical value for pure metals being -0038. Example 6. Find the value of -^-(45* \+6sl 1-8 4 ). CIS \ S ' Write the expression as 45* 3s~ 2 +6s- 5 1-8*. Then 2f ' n-l\ Example 7. If x = a n \i a n /, a formula referring to the flow dx of a gas through an orifice, find an expression for -5-. *( n ^\ As it stands a n \l a n ) is a product of functions of the I.V. (in this case a), and it cannot therefore be differentiated with our D 34 MATHEMATICS FOR ENGINEERS present knowledge. We may simplify, however, by removing the brackets, and then 2 n-l 2 2 M+l x a n aT^ n a n a n + l , , / 2 n dx d I = \a n a da da\ _i -i 2 n n-\-I n = - X a --- a n n 2 n -a -- a n n 2-n 7 n . Example 8. Determine the value of 5 -45m 9 - 86 rfm\ 5W To avoid the quotient of functions of m, divide each term by 5m 4 ' 32 , T>7*w75 .^C*M.^36 then the expression = and - (expression) = (3'4X 3'57^~ 4 ' 57 ) - ~ 4 '" -499m 4 - 54 9' Proof of the construction for the slope curve given on p. 14. Let us deal first with the particular case in which the equation of the primitive curve is y = x 2 . Referring to Fig. 4, the equation of the curve OAB is y = x 2 , and the equation of the curve CG is y l (x h) 2 = x 2 -f h 2 2xh. Hence the difference between the ordinates of the curves OAB and CG, the latter being regarded as the base curve = 2xhh z so that the equation of the curve MN is y-5 = 2xhh 2 . DIFFERENTIATION OF FUNCTIONS 35 Now the curve ODE is the curve MN shifted back a distance of - horizontal units, and hence its equation is y 3 = z(x-\ jh h 2 , 2 \ ^/ since for x we must now write f x-\\ Thus the equation of curve ODE is y 3 = 2xh y or ~ = 2x h A/ i. e., if Y be written for ~, Y = 2x or the equation of the curve ODE is that of the slope curve of the curve y ~ x 2 provided that the ordinates are read to a certain scale; this scale being the original vertical scale divided by h expressed in horizontal units. Hence the curve ODE is the slope curve of the curve OAB. Before discussing the general case, let us take the case of the primitive with equation y = x 3 . If the curve be shifted forward an amount = h, the equation of the new curve is yi = (x-W and the equation of the curve giving the differences of the ordinates is y 2 = y y^-x 3 (x h) 3 = x 3 By shifting this curve - units to the left we change its equation, by writing (x-\ J in 'place of x, into the form Dividing by h h 4 or Y = 3* 2 +- 36 MATHEMATICS FOR ENGINEERS h z Now if h is taken sufficiently small, is negligible in comparison with 3# 2 , and we thus have the equation of the curve Y = 3# 2 , which is the slope curve of the curve y = x 3 ; but the ordinates must be measured to the old vertical scale divided by h. We may now consider the case of the primitive y = x n . Adopt- ing the notation of the previous illustrations n(*L-ll) x n-2} l 2_ . _ \ \ Write (*+ 2 ) in place of x, and then h L 2 8 n(n i) 27 (w i)(w 2) n _3, 2 _ ._, _> _ ^C^ 1 ft _ - _ OC ft ~ 2 4 = n X n - 1 -}- terms containing A as a factor. Hence if h is made very small Y or = w*"- 1 . Exercises 2. On Differentiation of Powers of the I.V. 1. Find from first principles the differential coefficient of x*. o 2. Find the slope of the curve y = 2 when x = -5 ^ (a) By actual measurement and (6) by differentiation. 3. The sensitiveness of a governor is measured by the change of height corresponding to the change of speed expressed as a fraction of the speed. Thus if h and v represent respectively the height and dl) speed, the sensitiveness dh -. --- . If the height is inversely pro- portional to the square of the velocity, find an expression for the sensitiveness. Differentiate with respect to x the functions in Exs. 4 to 15. - --* ' 2I5 4. 3* 9 . 5. -. 6. 8i-5*-*. 7. igx. 8. A/-O / " - ^" ' . ,J DIFFERENTIATION OF FUNCTIONS 37 9. 8*' 10. 12. (* 3 - 7 ) 2 - 8 ~ 13. 14. 15. 16. Find the value of -]- when pv 1 - 3 = 570 and v = 28-1. . 17. Find the value of Jv 18. If E = I5+I4T -oo68T 2 , find the rate of change of E with regard to T when T has the value 240. , dH , dtt i f dp , \ 19. Calculate the value of -j- from -T- = -- \ v ^-+yp( when dv <fo 7 1 1 dv -^) pv 1 - 3 = C and y = 1-4. 20. Find the rate of discharge ( -,- J of air through an orifice from a tank (the pressure being 55 Ibs./n") from the following data I44/>V = wRT R = 53-2, V = 47-7, T = 548. Time (sees ) (/) o 60 I35 21$ 3IS Pressure (Ibs. per sq. in.) (p) 63 45 30 15 10 Hint. Plot p against / and find ? when p = 55. 21. If P = load displacement of a ship, p = weight of offensive and defensive factors. Then P = aP+bP*+p. Find the rate of increase of armament and protection in relation to increase of displacement. II \ O / \ I \ *> yai [ I __ y \ *u\t I 4f \ 901 1 \) V I * 22. if M = w(t x w( I+ y)- w (y *> , 2/ \ // 2 constants. \Vv ('/'M 23. If M = ^(/ 2 4y 2 ), find the value of y that makes , = o. 2/ 2 v dy n* Tt c* w((x-}-ny) 2 x 2 } ,. , , , , dS 24. If S = -i s ST-" k find the value of -,-. 2 I / y J ax 25. Find the value of h which makes -jr- = o when dh dM ""' ^ and [h is the height of a Warren girder; and the value found will be the height for maximum stiffness.] 26. If p = -^ A and q -75+ A, find the value of v f in terms of r 3 * r 3 dr p and q. (This question refers to the stresses in a thick spherical shell, p being the radial pressure, and q the hoop tension.) 38 MATHEMATICS FOR ENGINEERS 27. In a certain vapour the relation between the absolute tem- perature T and the absolute pressure p is given by the equation T = 140^1 + 465, and the latent heat L is given by L = 1431 -ST. Find the volume, in cu. ft., of i Ib. of the vapour when at a pressure of 81 Ibs. per sq. in. absolute, from V .n-? J (T TjRl I 44 T dp 28. For a rolling uniform load of length r on a beam of length /, the bending moment M at a point is given by )"!// v\ ieiv% M = If y is a constant, find an expression for the shear (i. e., the rate of change of bending moment). 29. Given that p = electrical resistance in microhms per cu. cm. and x = percentage of aluminium in the steel, then p = 12 + 12# -3# 2 for steel with low carbon content. Find the 'rate of increase of p with increase of aluminium when x = 4 . 30. The equation giving the form taken by a trolley wire is y = and the radius of curvature = 2000 1760 i dx* Find the value of the radius of curvature. Good examples of the great advantage obtained by utilising the rules of differentiation already proved are furnished by the two following examples, which have reference to loaded beams. Example g. Prove that the shearing force at any point in a beam is given by the rate of change of the bending moment at that point. Consider two sections of the beam 8x apart (see Fig. 9). The shear at a section being denned as the sum of all the force to the right of that section, let the shear at b = S, and let the shear at a = S+8S. Also let the moment of all the forces to the right of b (i. e., the bending moment at b) = M, and let the bending moment at a = M+SM. Taking moments about C M+SM = M+(S+8S)**+S( 8 *) or 8M = DIFFERENTIATION OF FUNCTIONS 39 SM , SS Dividing by Sx, ^- S-\ and when 8x is diminished indefinitely, SS becomes negligible and = S. S+SS .^ooooRftooo s H ""* FIG. 9. FIG. 10. Examples on Loaded Beams. The last example should be considered in conjunction with the following : Example 10. For a beam of length /, fixed at one end and loaded uniformly with w tons per foot run, the deflection y at distance x from the fixed end is given by the formula E being the Young's Modulus of the material of the beam, and I being the moment of inertia of the beam section. , dy d*y d?y , d*y Find the values of -~, -j^, , , and -,.. dx? dx v dx 3 dx* y = Differentiating, = {(W X 2x) - ( 4 l x 3^ 2 ) +4* 3 } Differentiating again, g - 40 MATHEMATICS FOR ENGINEERS Differentiating again, * - w d*y _ d(d?y\ _ dJTw ~ ~ ~ ( * ' Carrying the differentiation one stage further d*y _ d(d?y\ 5** ~ ~dx\dx*J Physical meanings may now be found for these various deriva- tives. Referring to Fig. 10, consider a section of the beam distant x from the fixed end. To the right of this section there is a length of beam lx loaded with w tons per foot, so that the total load or total downward force on this length is w(l x) ', and since this load is evenly distributed, it may be all supposed to be concentrated at distance from the section. 2 Now the bending moment at the section = moment of all the force to the right of the section /lx\ w = force X distance = w(lx)x( ) = ~(lx)\ d 2 v Comparing this result with the value found for ~ z> we notice that the two are alike except for the presence of the constants E and I : thus -~ must be a measure of the bending moment. Actually the rule connecting M, the bending moment, and its- dx* 1S M -tPy = i r M = the proof of this rule being given in a later chapter. Again, we have proved in the previous example that the shear is given by the rate of change of bending moment : thus dx dx dx 2 dx 3 = w(x / DIFFERENTIATION OF FUNCTIONS 41 a result agreeing with our statement that the shear at a section is the sum of all the loads to the right of the section. [The reason for the minus sign, viz., (x I), being written where (lx) might be expected need not be discussed at this stage.] Continuing the investigation d* w or v3 = w dtf but w is the loading on the beam and -~d*y d f^d 3 y\ dS EFr4 = j-( El j-4 j- dx* dx\ dx 3 / dx so that the loading is measured by the rate of change of the shear. If now the deflected form is set out, by constructing successive slope curves we obtain, respectively, the slope curve of the deflected form, the bending moment curve, the shear curve and finally the curve of loads. Example n. The work done in the expansion of gas in gas turbines is given by where r is the ratio of expansion. Compare governing by expansion control with governing by alteration of the initial temperature, from the point of view of efficiency. Deal first with the expansion control, i. e., regard Tj as constant and r as variable. Then the rate at which the work is increased with respect to r is -3. Now 3 dW w 42 MATHEMATICS FOR ENGINEERS Now regard r as constant, but T, as variable. /7W *? P V / w-l\ Then- ^ = ^^rV-^) u J-! n i J-o and, expressing the two results in the form of a ratio /7W /fW PVT (w r^T vv . vv _ x- 1 v 1 {n i)J- -( ""^ ^yn\I r n I Lengths of Sub-tangents and Sub-normals of Curves. The projection of the tangent to a curve on to the axis of x is known as the sub-tangent, i.e., the distance "sub" or "under'? the tangent. The projection of the normal on the x axis is called the sub-normal. The slope of a curve at any point, measured by the slope of its tangent at that point, is given by the value of -f- there, or if a = inclination of the tangent to the x axis dy tan a = dx In Fig. ii FIG. n. Sub-tangent and Sub- normal. PA dy - x -~i = tan a = -f- AT dx AT = PA^ ay But AT = sub-tangent and PA = y dx and hence the length of the sub-tangent = y,- DIFFERENTIATION OF FUNCTIONS Again L APN = a, since L TPN = a right angle AN sub-normal tan i. e. t or tan a = sub-normal sub-normal = y X tan a = y-~ To find the length of the tangent PT (PT) 2 = (PA) 2 +(AT) 2 and In like manner PN = FIG. 12. 43 Example 12. Find the lengths of the sub-tangent and the sub- normal of the parabola y 2 = <\ax (Fig. 12). y z = AfO-x and y = 2 Va . #* then or Then length of sub-tangent = y dx dy Va Va 44 MATHEMATICS FOR ENGINEERS This result illustrates an important property of the parabola and one useful in the drawing of tangents. For AT = 2.x 2 X AO, and hence to draw the tangent at any point P, drop PA perpendicular to the axis, set off OT = OA and join TP. The length of the sub-normal AN = y-2- , v- Va _ 2 Va Vx Va y /\ ___ _ . Vx Vx = 2d. i. e., the length of the sub-normal is independent of the position of P, provided that the sub-normal is measured on the axis of the parabola. Example 13. Find the lengths of the sub-tangent and the sub- normal of the parabola y = i$x 2 2xg when x = 2 and also when x 3. The axis of this parabola is vertical, and consequently the sub- normal, which is measured along the x axis when given by the value of y-^, is not constant. 7 dx Now y = i5x*2Xg dy and -r- = 30* 2. dx dx 2 Hence sub-tangent = y-r- = -r- dy $ox 2 dv and sub-normal = y~~ = (i^x z 2xg) x (30^2). ax Thus when x = 2 sub-tangent = - ~ ? _ __g5 um f- s . _ _ _ 60 2 _ 62 sub-normal = (60+4 9) ( 62) = 3410 units. When x = 3 /I35 6 g\ 120 15 sub-tangent = ( g8 j = - 88 = ^ units. sub-normal = 120 X 88 = 10560 units. Example 14. A shaft 24 ft. long between the bearings weighs 2 cwt. per foot run, and supports a flywheel which weighs 3! tons at a distance of 3 ft. from the right-hand bearing. Find at what point the maximum bending moment occurs and state the maximum bending moment. Regarding the shaft as a simply supported beam AB (see Fig. 13), we may draw the bending moment diagrams for the respective systems DIFFERENTIATION OF FUNCTIONS 45 of loading, viz., ADB for the distributed load, being the weight of the shaft, and ACB for the concentrated load. The total distributed load is wl, i.e., 24X-I = 2-4 tons, giving equal reactions of 1-2 tons at A and B; and the bending moment diagram is a parabola with vertex at D, the maximum ordinate DF being -5-, i. e., Q or 7-2 tons ft. If for convenience in the later working the axes of x and y are as shown in the figure, the equation to this parabola is y 2 = <\ax; or taking the value of y as FB and that of x as DF, 1 22 = 40x7- 2, from which 40 = 20 and y 2 = 20*. Scale of Bending Momen tons- f r pal lei to AC FIG. 13. The load of 3-5 tons produces reactions of X3*5 tons at B, 2 4 and x 3-5 tons at A, i. e., R B = 3-06 and R A = -44 tons : thus the 2 4 bending moment at E is 3-06x3 = 9-18 tons ft. Since the total bending moment is obtained by adding the ordinates of the diagram ADB to the corresponding ordinates of ACB, the maximum bending moment will be determined when the tangent to the parabola is parallel to AC, and the position satisfying this condition can readily be found by differentiation. Thus The equation of the curve ADB is y 2 = 20* or y = 4-47**, and the slope of the curve is given by the value of -~. Now if y = 4-47* , = 4-47X^4 e. f tan = _ 4'47 2*3 4 6 Referring to the figure ACB, tan a = = and thus- 4^42 = ^i 2** 9'l8 or x* = 1:47X9-18 42 i.e., (DR) = 4-47X9-18 42 Again, (PR) 2 = 2oxDR, and thus PR = = 4-37 ft. Thus the maximum bending moment occurs at a distance of 12 4-37, i. e., 7-63 ft. from the right-hand bearing. To find the maximum bending moment DR = **/^v*" = -956 \ 42 / PQ = DF-DR = 7-2--96 = 6-24 tons ft. Also ^jLp = X9-i8 = 7-16 tons ft. Hence the maximum bending moment = 7-16+6-24 13-4 tons ft. Exercises 3. On the Lengths of the Sub-tangent and Sub-normal : also Beam Problems. 1. Find the lengths of the sub-normal and sub-tangent of the curve $y = ^x 3 at the point for which x = 3. 2. If y = -^W, V = 117, and g = 32-2, find the value of x that makes the slope of the curve i in 17-4. 3. A parabolic arched rib has a span of 50 ft. and a rise of 8 ft. Find the equation of the tangent of the slope of the rib. What is the slope of the tangent at the end ? 4. Find the equation of the tangent to the curve p = ^- at the v point for which v 5. In Exercises 5 to 7, y is a deflection and x a distance along the beam. Find, in each case, expressions for the Bending Moment, Shearing Force and Load. The beam is of uniform section throughout, and of span /. 5. The beam is supported at both ends, and loaded with W at the centre. W llx z x a \ y = -^r T ( -=- ) {x is the distance from the centre}. 2EI\ 4 6 / 6. The beam is supported at both ends, and loaded continuously with w per ft. run. y ^^^Y^-g j {x is the distance from the centre}. DIFFERENTIATION OF FUNCTIONS 47 7. A cantilever loaded with W at the free end. W/7# 2 x a \ y = -p_( --- -j-\ \x is the distance from the fixed end}. 8. Find the lengths of the projections on the y axis of the tangent and the normal of the parabola, x 2 iob 2 y + ^c, x having the value ga. 9. Prove that the sub-normal (along the axis of the parabola) of the parabola x 2 6y is constant and find the value of this constant. ^ A rr , /- , r> f. A 4.- 10. If El-/ = ---- ,. --- \-C and C = ------- , find the value rf* 4 6 a 2 12 Differentiation of Exponential Functions. The rule for differentiation already given applies only to functions involving the I.V. (usually the x) raised to some power. A method must now be found for the differentiation of exponential functions, viz., those in which the I.V. appears as exponent ; such as e 5x or 4*. When concerned with the plotting of the curve y = e* (see Part I, p. 352) mention was made of the fact that, if tangents are drawn to the curve at various points, the slopes of these tangents are equal to the ordinates to the primitive curve at the points at which the tangents touch the curve. Thus the slope curve of the curve y = e x lies along the primitive, and **-- 6* dx ~ or the rate of change of the function is equal to the value of the function itself. We may establish the result algebraically thus X % 3C e x = i-\-x-\ --- [- 1.21.2.31.2.3.4 Assuming that a series composed of an unlimited number of terms can be differentiated term by term and the results added to give the true derivative (this being true for all the cases with which we shall deal), then by differentiation de x _ ~ f x 3 = e x . Another respect in which the function e? is unique may be 48 MATHEMATICS FOR ENGINEERS dx i noted : the sub-tangerit = y-,- -- e x x = i, i. e., the sub-tangent is constant and equal to unity. The curve y = e x may be usefully employed as a gauge or template for testing slopes of lines; the curve being drawn on tracing-paper and moved over the line to be tested until the curve and line have the same direction, and the ordinate of the curve being then read, any necessary change of scales being afterwards made. The work may now be carried a stage further, so that the rule for the differentiation of e bx may be found. Referring to Part I, p. 354, we note that if the curve y = e x be plotted, then this curve represents also the equation y = e bx if the numbers marked along the horizontal scale used for the curve y = e x are divided by b. If, then, the slope of the construction curve, i. e., that having the equation y = e x , is measured, we can obtain from it the slope of the curve y e bx by multiplying the slope by b, since vertical distances are unaltered, whilst horizontal distances in the case of y = e bx are T X corresponding horizontal distances for y = e x . Hence the slope of the curve y = e bx is b X slope of curve y = e x r\pbx or -p- = be bx It should be noticed that the power of the function remains the same after differentiation, but the multiplier of the I.V. becomes after differentiation a multiplier of the function. This latter rule must be remembered throughout differentiation, viz., any multiplier or divisor of the I.V. in the function to be differentiated must become a multiplier or divisor of the function after differentiation. From e bx we can proceed to ae bx , the result from the differentia- tion of which is given by dae bx dx = abe bx Example 15. If y = yrk* t find the value of -f~ dx -/- = ^~5e~b x = 5 X T&~^ X dx dx j 6 DIFFERENTIATION OF FUNCTIONS 49 Referring to the last example, note that the power of e is exactly the same after differentiation as before : the factor ^ multiplies the I.V. in the original function, and therefore it occurs as a constant multiplier after differentiation; also the constant factor 5 remains throughout differentiation. Example 16. If C = C e L, where C and C are electrical currents, R is the resistance of a circuit, L is the self-inductance of the circuit and t is a time, find the time-rate of change of C. This example illustrates the importance of the rate of change as compared with the change itself; for it demonstrates the fact that for an inductive circuit the change of current is often extremely rapid and consequently dangerous. T" * i- t /-> dC, d ~ 5* Time rate of change of C = -^ = _,XV~ L dt dt - C x V? ^o Ps T~& L -C -~ R^ i. e., the rate of decrease of the current when the impressed E.M.F. is removed is proportional to the current at the instant the circuit is broken. To better illustrate the example, take the case for which the current at the instant of removal of the E.M.F. is 14-5 amps., the resistance of the circuit is 6-4 ohms, and its self-inductance - oo6 henry. Then the rate of change of the current = -- ^ X 14-5 = 15470 amps. *ooo per second, whereas the actual current is only 14-5 amps. The expressions e x and e bx are particular forms of the more general exponential function a x ; to differentiate which we may proceed by either of two methods : (a) Working from first principles. In Part I, p. 470, the expansion for a x is given, viz. log a+ (*log*)%(*joj^)* 50 MATHEMATICS FOR ENGINEERS Differentiating term by term da x . , . n \2 , /, \sx 2 dx = o+log tf+(log a) x+(log a) ~+ . . . . . (x log a) 2 . (x log a) 3 , = log a{i+x log a+~- ' > +^ ' + L II. = log a X a x da* (b) Assuming ihe result for the differentiation of e bx Let a x = e bx so that a e b , and therefore log e a = b. d , d T a = J-* ax ax = logg axa x or a x log a Then T a x = T e bx = be bx = log e ax e bx ,, da x thus -j = a x . log a. d. Example 17. Find the value of ^p- dX In this case = 4 and log e 4 = 1-3863. dA x Hence - = '1-3863 x <\ x . CIX ^^^*^^m**^mm^^^* Note carefully that this result cannot be simplified by combining 1-3863 with 4 and writing the result as 5-5452*, which is quite incorrect. The 4 alone is raised to the x power, and 1-3863 is not raised to this power. TO Example 18. Find the value of , 2 (3'^) S - % Here a = 3-6 and log 3-6 = 1-2809. Thus j(3' 6 )' = lo 3-6 X (3-6)* = 1-2809 x (3-6)*. Then = I -2809 XT- (3-6)* = 1-2809 X 1-2809 x(3-6) j GfS = 1-64(3-6)*. DIFFERENTIATION OF FUNCTIONS 51 Example 19. Given that s = ^e st -\-ye~ 5t t find the value of -^255. s = = 2O0 5 ' 350 ~ 5t . Again ,. z = ' w*~ 25$ = Differentiation of log x. The rule for the differentiation of logarithmic functions can be derived either from the expansion of loge (i + x) into a series, or by assuming the result for the differentiation of e x . Considering these methods in turn (a) Working from first principles. Let y = log x, i. e., log e x. Then if x be increased to become x-\-8x, y takes a new value y+Sy, and y+Sy = log (x-{-8x). (8x\ ( 8x\ i-\ ) = log x+log ( i-j ), X / \ X / therefore / Sx\ (y-j-Sy) y = log #-|-log ( i-\ ) log x \ x / i. e., 8y = log (l-f* ) Also log ( i-j ) can be expanded into a series of the form \ x / A- 2 /yO A' 1 * log (*+x) = x -+--+ . . (see Part I, p. 470) ^ o 4 so that r ~ I \~ ) ^l~/"r\~/ T\^7/~r 8y = _ _ VA;/ 2\* Dividing all through by SA; 8y = i 8x (8x)* (Sx) 3 8x~ x 2x 2 3% 3 4** ' By sufficiently diminishing the value of Sx we may make the 52 MATHEMATICS FOR ENGINEERS second and succeeding terms as small as we please, and evidently the limiting value of the series is - i* c,. , dx LSy _ i 8x x Hence Sx>o d _ dx ~ x (b) Assuming the result = e x Qt% Let y = loge x, so that x e y dx de y and = - = e y . dy dy ^>. . Now ~ = OA/ T r- and consequently by considering the limiting oA/ *y d values of these fractions /- = -r- dx dx dy dv We wish to find ~ and we have already obtained an expression , dx -=-. dy Hence ~ = T- = . = dx ax e y x dy d log e x _ i \JL 5 dx~ x This result can be amplified to embrace the more general form, s thus for, in accordance with the rule given on p. 48, the A which multiplies the I.V. in the original fun6tion must appear as a multiplier after differentiation. All these rules apply to functions involving natural logs, but DIFFERENTIATION OF FUNCTIONS 53 they can be modified to meet the cases in which common logs occur ; for - Iog 10 x = -4343 log e x d log ln x d loe e x and hence -$&- = . 4343 _J3L_ = d . ,. -4343A and -- Iog 10 (Ax+B) - It should be observed that in all these logarithmic functions the I.V. is raised to the first power only : if "the I.V. is raised to a power higher than the first, other rules, which are given later, must be employed. Example 20. If y = \og e jx, find ~. dy d . -r**-T loge 7 dx dx _ or alternatively log e 7* = log e 7+log e x and thus log e 7* = . i i = o+- = - X X Example 21. Differentiate with regard to t the expression Iog 10 (5^14) and find the numerical value of the derivative when / = 3-2. When t = 3- 2 10T Ut- 1^ - ' 4343 X 5 - dt tog], \y i_ r - log,. (51-14) = = "0858. We may check this result approximately by taking values of J 3-19 and 3-21 and calculating the value of -' *! -- Thus When i = 3-19, Iog 10 (5f 14) = Iog lfl (i5'95 14) = Iog 10 1-95 = -2900 when t = 3-21, Iog 10 (5* 14) = Iog 10 (i6-o5 14) = Iog 10 2-05 = -3118 so that 8 Iog 10 (5^14) = -3118 -2900 = -0218 while 8t = 3-21 3-19 = -02 S . , -0218 and lo < * == ~~ = I>O 9- 54 MATHEMATICS FOR ENGINEERS Differentiation of the Hyperbolic Functions, sinh x and cosh x. Expressing the hyperbolic functions in terms of exponential functions e x er x sinh x = and cosh x = 2 e x -\-e- x 2 Thus to differentiate sinh % we may differentiate d sinh x d (e x e~ x \ i, _ Hr^ I _ / x*g I x> \ jncii^c j 7 i ~ / lu ~ i c? i A; A;\ 2 / 2 = cosh x d d(e x +e- x \ i. , also j- cosh = j-( - - ) = -wfg-'i dx dx\ 2 / 2 V = sinh #. Example 22. Find the inclination to the horizontal of a cable weighing \ Ib. per ft. and stretched to a tension of 30 Ibs. weight, at the end of its span of 50 ft. The equation to the form taken by the cable is / x x \ y = -\e c -\-e c ) = c cosh - horizontal tension 30 ,. where c = --- C r -*_ =60. weight per foot -5 We require the slope of the curve when x = 25, this being given cl/\} by the value of - there. dy d , x i . . x > x ~- = J- c cosh = = c x > sinh >- = sinh 7 - dx dx 60 60 60 oo . < When x 25 -/ = sinh -^ = sinh -4167 = g- 4167 _ g .4167 - dx 60 2 = 1-517 659 2 = -429. This value is that of the tangent of the angle of inclination to the horizontal; which is thus tan- 1 -429 or 23 13'. DIFFERENTIATION OF FUNCTIONS 55 Exercises 4. On the Differentiation of a x , Log x and the Hyperbolic Functions. Differentiate with respect to # the functions in Nos. i to 20. 1. e~ 5 *. 2. i-se 4 - 1 *. 3. -. 4. 4-15*. 5. 8-72^. & . 7. ^x e . 8. 14x2*. 9. 4ie foa-iX -jpl-ZX 10. 3 -i4^-5*- + 3-i te +6. 11- ,VX^T*E- 12. log 7*. 13. 3 log (4 5*). 14. iolog 10 8#. 15. c log (4*0 +56). 16. 9-- te -log -2*+^. 17. log 2*(3#-47). 18. log y)"-- 19- (**) 3 +4 cosh 2*- 17 lo glo 2-3*. 20. log 3* 2 +5#-' 7 i -8 (i -8*) + 12. 21. If y = A^-^+A^-^find the value of S+7 2 ' ' 22. Find -5- log (3 4^) when v = 17. Check your result approxi- CL1) mately by taking as values of v 1-65 and 175. 23. Determine the value of -,- 71 log (18 -04^). (I If- 24. Write down the value of -j- Iog 10 18^. at 25. If T = 5oe' Zd , find the rate of change of T compared with change in 6. / _BA 26. If C = CQ^I e L '), C and C being electrical currents, R the resistance of a circuit and L its self-inductance, find the rate at which the current C is changing, / being the time. dl) 27. Given that v = 2-03 Iog 10 (7 i-Su), find ^-. 28. Evaluate -=- 5 cosh - and also -j- p sinh -. ax 4 ay r q 29. An electromotive force E is given by E A cosh Vlr . #+B sinh Vlr . x. Find the value of -3-3 in terms of E. 30. If W = i44J/) 1 (i f log r) rpb], find the value of r that makes -T = o ; W being the work done in the expansion of steam from pressure p t through a ratio of expansion r. 31. Find the value of -r~ + #y if y e~ ax . ax ' * a a 56 32. If y = Ae Zx -\-'Be 3x +Ce-* x > find the value of d 3 y d z y dy d z V Vy x x/ n -*, -J.x 33. Evaluate -7-2 -- when V=A 1 e x ^ r 2 +A 2 e ^ r 2 W^? ? 2 34. Nernst gives the following rule connecting the pressure p of a refrigerant (such as Carbon Dioxide or Ammonia) and its absolute temperature T p = A+B log T+Cr+5 T where A, B, C and D are constants. Find an expression for -J-. Differentiation of the Trigonometric Functions. Before proceeding to establish the rules for the differentiation of sin x and cos x, it is well to remind ourselves of two trigonometric relations which are necessary for the proofs of these rules, viz. (a) When the angle is small, its sine may be replaced by the angle itself expressed in radians, i. e. T sm6 = T (cf part ^ p 458)> (b) sin A-sin B = 2 cos sin - ( c f. p ar t I, p. 285). \ 2 / 2 To find -v- sin x we proceed as in former cases ; thus Let y = sin # and y+8y = sin (#-f 8#) then 8y y -\-8y-y = sin (x-\-8x)sin x (2x+8x\ = 2 cos ( - sin \ 2 / Dividing through by 8x _ 2 COS \ ~~ / OAII i 8y V 2 / \2 /2^+8^\ . /Bx\ I sin ( J 8* 8x (2x+%x\ f8x^ COS ( I sin I t\ (8X\ - ) sm ( I / \2/ DIFFERENTIATION OF FUNCTIONS 57 The limiting value of ?- is --. and that of the right-hand side is cos x, since cos ( x-\ J approaches more and more nearly to cos x as 8* is made smaller and smaller, and the limiting value of /sin \ / 2 \ .,,.,., sin 9 . I, or, as we might write it, ^ , is I. Hence d sin x dy j or * = dx dx L -^ = cos x dsinx dx = cos x. 75 -5 25 O -y -y- .JG - 7 .3G TT. x JG FIG. 14. Curves of y = sin x and y = cos x. By similar reasoning the derivative of cos x may be obtained ; its value being given by The graphs of the sine and cosine curves assist towards the full appreciation of these results. In Fig. 14 the two curves are plotted, and it is noted that the cosine curve is simply the sine curve shifted backwards along the horizontal axis : thus the slope curve and the primitive have exactly the same shape. This condition also holds for the primitive curve y = e? 1 , and so suggests that there must be some connection between these various natural functions; and further reference to this subject is made later in the book. 58 MATHEMATICS FOR ENGINEERS Much trouble is caused by the presence of the minus sign in d COS OC the relation -5 = sin x, it being rather difficult to remember whether the minus sign occurs when differentiating sin x or cos x. A mental picture of the curves, or the curves themselves, may be used as an aid in this respect. The cosine and sine curves differ in phase by period (see Fig. 14), but are otherwise identical. Treating y = sin x as the primitive : when x is small, sin x and x are very nearly alike, and thus the slope of the curve here is i; as x increases from o to - the slope of the curve continually 7T diminishes until at x = - the slope of the curve is zero. Now the ordinate of the cosine curve when x = o is unity, and it diminishes until at x = ~ it is zero. From x = ~ to x = -n- the 2 2 slope of the sine curve is negative, but increases numerically to I, this being the value when x = TT; and it may be observed that the ordinates of the cosine curve give these changes exactly, both as regards magnitude and sign. Thus the cosine curve is the slope curve of the sine curve. Now regard the cosine curve as the primitive. At x = o the curve is horizontal and the slope = o ; from x = o to x = ~ the 2 slope increases numerically, but is negative, reaching its maximum negative value, viz., i, at x = - ; but the ordinates of the sine 2 curve are all positive from x = o to x = ~, so that although these 2 ordinates give the slope of the curve as regards magnitude, they give the wrong sign. In other words, the sine curve must be folded over the axis of x to be the slope curve of the cosine curve, i. e., the curve y = sin x is the slope curve of the curve y = COS X. To summarise, we can say that the derived curve for the sine curve or for the cosine curve is the curve itself shifted back along the axis a horizontal distance equal to one-quarter of the period. Thus we can say at once that the slope curve of the curve y = sin (x-\-b) is the curve y cos (.r+fr), since the curve y = sin (x+b) is the simple sine curve shifted along the horizontal axis an amount given by the value of b, the amplitude and period being unaltered. DIFFERENTIATION OF FUNCTIONS 59 Thus -,- sin (x+b) = cos (x+b) and, in like manner j x cos (x+b) = - sin (x+b). Again, -7- sin (5#+6) = 5 cos (5#+6), since 5 multiplies the I.V. in the original function. Then in general ?- A sin (Bx+C) = AB cos (Bx+C) -? A cos (Bx+C) = AB sin (Bx+C). To differentiate tan x with regard to x. Let y = tan x and (y+8y) then = tan (x-\-8x) 8y = y+Sy y=tan (x+8x) tan x _ sin (x-\-8x) sin # ~ cos (x+8x) cos A; _ sin (x+Sx) cos A; cos (*+SA;) sin x cos #8# cos x sin cos (x+8x) cos x _ sin 8x ~ cos (A; +8x) cos x Dividing through by 8x Sy _ sin 8x I 8x ~ 8x cos (x+8x) cos x Now as 8x approaches zero, - - approaches i and (x + 8x) approaches x. dx ,8* * ^ cos A; cos x cos 2 A; Hence =1 = IX- ~= *~ = sec 2 x. \ _ , 3- tan x = sec 2 x 60 MATHEMATICS FOR ENGINEERS In like manner it can be proved that d cot x 3 = cosec 2 x d seex sinx dx ~ cos 2 x d cosec x cos x and -j~ = -- r-s dx sin 2 x To generalise ^ A tan (Bx+C) = AB sec 2 (Bx+C) J^ A cot (Bx+C) = AB cosec 2 (Bx+C) rf , ,_ _v AB cos (Bx+C) a - A eosee (BX+C) = - . 2 AB . cos 2 (Bx+C) Example 23. Find the slope of the curve representing the equation s = 5-2 sin (40^2-4) when t -07. The slope of the curve . . . 5 ' 2 Sm (4^ 2 '4) = 208 cos (40^2-4). ds d . . . = ~dt = dt 5 ' 2 Sm (4^ 2 '4) = 5-2x40 cos (40^2-4) Hence when t = -07, the slope = 208 cos (2-82-4) = 208 cos -4 (radian) = 208 cos 22-9 = 192. Example 24. Differentiate, with regard to z, the function 9-4 cot (75?). ,- 9'4 cot (7-5*) = 9'4 X -5 X - cosec 2 (7-5?) U6 = 47 cosec 2 (752). Simple Harmonic Motion. We can now make a more strict examination of simple harmonic motion. Suppose a crank of length r (see Fig. 15), starting from the position OX, rotates at a constant angular velocity <o in a right-handed direction. Let it have reached the position OA after t seconds have elapsed from the start; then the angle passed through in this interval of time = AOM = <&t, since the angular distance covered in I sec. = o> radians and the angular distance in t seconds = a>t radians. DIFFERENTIATION OF FUNCTIONS 61 Considering the displacement along the horizontal axis, the dis- placement in time t = s = OM = AO cos AOM = r cos tat. ds Then the velocity = , , = rXta sin tat = rta sin <at and the acceleration = ,, = at cos tat = ta z xr cos wt = 0) 2 S *'. e., the acceleration is proportional to the displacement, but is directed towards the centre : thus, when the displacement from the centre increases, the acceleration towards the centre increases. When the displacement is greatest, the acceleration is greatest : e. g., if the crank is in the position OX, the acceleration has its maximum value wV and is directed towards the centre, just destroying the outward velocity, which at X is zero. At O the acceler- ation = u> 2 x o = o, or the velocity is here a maximum. An initial lag or lead of the crank does not affect the truth of the foregoing FlG r . connection between acceleration and displacement. The equation of the motion is now s = r cos where c is the angle of lag or lead, and the differentiation to find the velocity and the acceleration is as before. Example 25. If s 5 sin 4* 12 cos \t, show that this is the equation of a S.H.M. and find the angular velocity. s = 5 sin ift 12 cos 4^. Then v = , $ = (5 x 4 cos 4*) (12 x 4 X sin 4*). HP = 20 cos 4^+48 sin 4^ and a = " = (20 x 4 X sin 4*) + (48 x 4 cos 4/) = 80 sin 4^+192 cos 4* = 16(5 sin 4^12 cos 4/) = i6s i. e., the acceleration is proportional to the displacement. Now, in S.H.M., the acceleration = 2 s. &> 2 = 16, i. e., w = angular velocity = 4 radians per sec. 62 MATHEMATICS FOR ENGINEERS This last question might be treated rather differently by first expressing 5 sin 4^12 cos 4^ in the form Msin(4^+c) (see Part I, p. 276) and then differentiating. This method indicates that a S.H.M. may be composed of two simple harmonic motions differing in phase and amplitude. Exercises 5. On the Differentiation of Trigonometric Functions. Differentiate with respect to x the functions in Nos. i to 16. 1. sin (4 5'3x). 2. 3-2 cos 5-1*. 3. -16 tan (3X-\-g). 4. 2-15 sin i 1 ^~ 5 ). 5. 8 cot 5%. \ 4 / 6. 43-15 sec (-05 -117*). 7. be cos (dgx). 8. 4 cos $x 7 sin (2^5). 9. sin 5-2^ cos 3~6x 10. 2-17 cos 4-5* cos 1-7*. 11. 9-04 sin (px+c) sin (qxc). 12. 5 sin 2 x. 13. -065 cos 2 3*. 14. cos 2 (7* i-5)+sin 2 (7* 1-5). 15. 3* 1 - 72 5- 14 log (3#-4-i) + -i4 sin (4-31 -195*) + 24-93*. 16. 7-05 sin -015* -23 cos (6-i -23*) + 1-85 tan (4* -07). 17. x, the displacement of a valve from its central position, is given approximately by x = 1-2 cos a/ 1-8 sin ^ where w = angular velocity of crank shaft (making 300 r.p.m.) and t is time in seconds from dead centre position. Find expressions for the velocity and acceleration of the valve. 18. If 5 = 4-2 sin (2-1 -172) -315 cos (2-1 -J7/), s being a displace- ment and t a time, find an expression for the acceleration in terms of s. What kind of motion does this equation represent ? 19. The current in a circuit is varying according to the law C = 3-16 sin (2irft 3-06). At what rate is the current changing when t -017, the frequency / being 60 ? 20. If the deflected form of a strut is a sine curve, what will be the form of the bending moment curve ? 21. If y = deflection of a rod at a distance x from the end, the end load applied being F Bl y=- 8 COS 1 Find the value of EI~^-}-Fy-\-- & cos -j-; y and x being the only variables. 22. The primary E.M.F. of a certain transformer was given by the expression E = 1500 sin pt-\-ioo sin 3^42 cos ^+28 cos 3pt. Find the rate at which the E.M.F. varied. T 2 23. A displacement s is given by s = sin izt sin 13^. Show that the acceleration = 25 sin 12^1695. CHAPTER III ADDITIONAL RULES OF DD7FERENTIATION Differentiation of a Function of a Function. \Vhilst the expression e sin 4 * is essentially a function of x, it can also be spoken of as a function of sin 4*, which in turn is a function of x; and thus it is observed that e aa ** is a function of a function of x. This fact will be seen more clearly, perhaps, if u is written in place of sin 4* : thus e sin ** = e w , which is a function of u, which, again, is a function of x, since u = sin 4*. To differentiate a function of a function the following rule is employed dy_=dy x du dx du dx \ and this rule is easily proved. Let y be a function of u, and let u be a function of x : then y is a function of a function of x. Now increase # by a small amount 8x; then since u depends on x, it takes a new value u + 8u, and also the new value of y becomes y + Sy. Since these changes are measurable quantities, although small, the ordinary rules of arithmetic can be applied, so that 8y __Sy 8u_ Sx~Su X Sx When Sx approaches zero these fractions approach the limiting values -, and - respectively : and thus in the limit ax du dx dy _ dy du dx du dx In like manner, if y is a function of u, u a function of w, and w a function of x, it can be proved that dy_dy..du_dw dx~du x dw x dx 63 64 MATHEMATICS FOR ENGINEERS It will be observed that on the right-hand side of the equation we have dy as the first numerator and dx as the last denominator (these giving in conjunction the left-hand side of the equation) ; and we may regard the other numerators and denominators as neutralising one another. The simple arithmetic analogy may help to impress the rule upon the memory : thus Example i . If y = e 8to tx , find the value of ~ . dit Let u = sin 4*, so that j- = 4 cos 4* and y = e u . Since y is now a function of n, we can differentiate it with regard to u, whereas it is impossible to differentiate with regard to x directly. y = e u and ~ = ~ = e u = e s^ **. du du T,, dy dy du Then, since -, - = -/-x j- dx du dx dv JL e sui tx x ^ cos 4 # = 4 cos *g sin Example 2. Find the value of j~log (cos 2x} 3 . Let v = (cos 2x) 3 and u = cos 2x; and thus y = loge v and u = w 3 . dy dy dv du u = cos Then -~ = -~ x -j- X -j- d log v du 3 d cos 2.x ~~dv~ X ~du X dx X 2 sin -j- 2 sin V = U 3 dv -j- = 3 U du _ 6 sin 2XX (cos 2^r) 2 _ 6 sin 2 (cos 2#) 3 cos 2x = 6 tan 2 Example 3. The radius of a sphere is being decreased at the rate of -02 in. per min. At what rate is (a) the surface, (b) the weight, varying, when the radius is 15 ins. and the material weighs -3 Ib. per cu. in. ? dr If r radius, then - ,- = rate of change of the radius, and is in dt this case equal to -02. ADDITIONAL RULES OF DIFFERENTIATION 65 (a) The surface = 4irf 2 , and thus the rate of change of surface _ dS ~ dt _ d . 4*r z dt dr* = * v '^dt dr* dr = * v -~dr X df dr 5 '02. Hence when r = 15, -5- = 8v x 15 X -02 = 7-53, i. e., the surface is being diminished at the rate of 7-53 sq. ins, per min. (b) The volume = 4 ^ so that the rate of change of volume = = jft-***) and the rate of change of the weight = -rr = -j-.( - X -Sirr 3 j <AV d dr 3 dr 3 dr ~rr =~JT- '4*r '4* -JT = '4* X -j- X T-. -02. When r = 15, -, = -4^x3x225 x -02 = 16-93 or the weight is decreasing at the rate of 16-93 I DS - per min. Example 4. Find expressions for the velocity and acceleration of the piston of a horizontal steam engine when the crank makes n revolutions per second. In each turn the angle swept out = 2ir radians. Hence in i second 2vn radians are swept out, i. e., the angular velocity = 2.im; and this is the rate of change of angle, so that dQ dT= 2vn - From Fig. 16 CD = / sin a and CD = r sin 6. Thus I sin a = r sin 6 / r or sin = - sin a, and sin a = T sin 0. r I 66 MATHEMATICS FOR ENGINEERS Again, cos = Vi sin 2 a If the connecting rod is long compared with the crank, -j is small r z and 72 still smaller, so that our method of approximation can be p applied to the expansion of the bracket, i. e. I y^ cos o = i 72 sin 2 0, very nearly. FIG. 16. Velocity and Acceleration of Piston. Let AB = displacement of the piston from its in-dead-centre position = x = AE+OE BO = l + r-BV DO = l+rl cos o r cos 6 = l+rl (i p sin 2 0)-y cos y+~, sin 2 Y cos = r-\ dx d i cos 20 2 r 2 cos 20 -r cos fl ~ r COS U. r 2 r 2 cos 20 XT . i. i . ,,-. . dx a f JNow the velocity of the piston = -=- = --=- -j , --- -. We cannot differentiate this expression directly, so we writ n r cos TT Hence dx __df r* dx dx dQ dQ COS 20 ~dt = 50 X dt' ,.) dQ rcosQ\x,. J a/ = -jo-f o (~ x 2 sin 20J (rx sin 0) j- x 2irW fy sin 20 . . 1 = 2trwy -j - j -- f-sm j- or if - = m r dx V = ~dt = f sin 20 \ 2m ADDITIONAL RULES OF DIFFERENTIATION 67 dv dv dQ d (sin 20 , _i d0 Also the acceleration = -rr = -j^X^-^,. zvnrl --- f-sm 6 [ x -r- at ay at at) I 2m at fcos 26 , n ) = 2irnr 1 -- 1- COS 6 h X 2irW f Q . COS 26"| = 4ir 2 n 2 ' { cos 0H * I / Example 5. Water is flowing into a large tank at the rate of 200 gallons per min. The reservoir is in the form of a frustum of a pyramid, the length of the top being 40 ft. and width 28 ft., and the corresponding dimensions of the base being 20 ft. and 14 ft. ; the depth is 12 ft. (see Fig. 17). At what rate is the level of the water rising when the depth of water is 4 ft. ? In 12 ft. the length decreases by 20 ft., and therefore in 8 ft. the length decreases by , i. e., 13 J ft., so that the length when the water is 4 ft. deep is 40 13^ = 26| ft. Similarly, the breadth = 28 (f x 14) = i8f ft. i. e., the area of surface = 26f x i8| = 498 sq. ft. 200 ,, 200 gals, per mm. = ^ cu. ft. per mm. 0*24 = 32-1 cu.. ft. per min. i. e., the rate of change of volume = rr = 32-1. dv Now -v- = -v . Ah, where A = area of surface at at and h = depth of water, *dh f since for the short interval of time considered the\ dt \ area of the surf ace may be considered constant./ Hence the rate of change of level = -JT = -jr X -z- dt dt A 32-1 xi , ,. = Q = -0644 ft. per nun. 49 = 773 in. per min. 68 MATHEMATICS FOR ENGINEERS Example 6. If a curve of velocity be plotted to a base of space, prove that the sub-normal of this curve represents the acceleration. d'v The sub-normal of a curve = y~- (see p. 43). CbX In this case, since v is plotted along the vertical axis and s along the horizontal axis the sub-normal = v-^- ds . dv dt V 'dt X ds = vxax-- v = a [for -=r = rate of change of velocity = a\ dt ds dt and -3T = rate of change of space = v I As a further example of this rule, consider the case of motion due to gravity; in this instance v 2 2gs, i.e., the velocity space curve is a parabola. Hence we know that the sub-normal must be a constant, i. e., the acceleration must be constant. The sub-normal = v- r ds , T dv 2 d ds Now ds=ds-^ S = ^-ds = ^ dv z dv z dv dv but , = -j . -T- = 2v j- ds dv ds ds dv 2V-j- = 2g ds dv ds i. e., the sub-normal or the acceleration = g. V ds = Exercises 6. On the Differentiation of a Function of a Function. T-" j A d sin 2x d . d Find 1. -3- . e . 2. -T- log v 2 . 3. ^2cos 2 t. dx dv dt d _ d n d sin 5x dx Sm dx 3 ' 14 (5^ 2 +7^~ 2 )- 6. ^a 7< ^ 1<88 ' 8< dx logl (3 + 7^-9^ 3 ). 9. ^ cos (log s 5 ). and 10. ~ log tan -. a* 2 ADDITIONAL RULES OF DIFFERENTIATION 69 11. In the consideration of the theory of Hooke's coupling it is required to find an expression for , i. e., a ratio of angular velocities. If o> B = r, <" A = -3T and tan <f> = - , find an expression for B in terms of the ratios of 9, </> and o. "A 12. Find an expression for the slope of the cycloid at any point. The equation of the cycloid is x = a (6 + sin 6) y a(i cos 6) the co-ordinates # and y being measured as indicated in Fig. 18. ^Rolling Circle FIG. i 8. 13. Assuming that the loss of head due to turbulent flow of water in a pipe is expressed by h C(AV 2 +BV?), where V = mean velocity of flow in ft. per sec. ; show that the slope of the curve in which log h and log V are plotted with rectangular co-ordinates is given by d log h dlogV 2A 14. If 3x*+8xy+5y 2 = i show that T = T dx z (. 15. A vessel in the form of a right circular cone whose height is 7 ft. and diameter of its base 6 ft., placed with its axis vertical and vertex downwards, is being filled with water at the rate of 10 cu. ft. per min. ; find the velocity with which the surface is rising (a) when the depth of the water is 4 ft. and (b) when 60 cu. ft. have been poured in. 16. If p = (r) K , prove that-j^ = ~ffi( r^ 1 r - 17. If x 3 6x 2 y 6xy z +y 3 constant, prove that dy _ x z 4xyr--* dx = 2X*~ 7 o MATHEMATICS FOR ENGINEERS 18. A ring weight is being turned in a lathe. It is required to find the weight removed by taking a cut of depth ^thj". The material is cast iron (-26 Ib. per cu. in.), the outside diameter of the ring is 3-26" and the length is 2-5'*. Find the weight removed. Find also a general expression for the weight removed for a cut of depth ^J^" at any diameter. 19. Find the value of -rA log tan 20. If P = -^TT, and -.., = u, find -w. (This question has refer- -.., ence to stresses in redundant frames.) 21. Find the angle which the tangent to the ellipse \- = 2 at 4 9 the point x = 2, y 3, makes with the axis of x. 22. Find the slope of the curve 4# 2 +4y 2 = 25 at the point x = 2, y = f, giving the angle correct to the nearest minute. 23. If force can be defined as the space-rate of change of kinetic , , . , . wv z , , wa energy, and kinetic energy == - , prove that force = . o o dx 24. If x 8 log (i2t 3 74), find the value of ,-,. ctt Differentiation of a Product of Functions of x. It has already been seen that to differentiate the sum of a number of terms we differentiate the terms separately and add the results. We might therefore be led to suppose that the differentiation of a product might be effected by a somewhat similar plan, viz., by multiplication together of the derivatives of the separate factors. This is, however, not the correct procedure ; thus d ,, , , . d log x dx z . i j- (log xxx 2 ) does not equal , --X-T , ^. e., X2x or 2. dx v dx dx x The true rule is expressed in the following manner : If u and v are both functions of x, and y = uv, i. e., their product dy d , . du . dv B -*"*-+ Proof. Let x increase by an amount 8x; then since both u and v are dependent on x, u changes to a new value w+Sw and v becomes v -f- 8v. Now y = uv, and hence the new value of y, which can be written y-\-8y, is given by but y = uv ADDITIONAL RULES OF DIFFERENTIATION 71 whence by subtraction Sy = y+8y y = (u-}-8u)(v-}-8v)uv = uv-}- u8v -\-v8u-\-8u . 8v uv {-8u . 8v. Dividing through by 8x 8y_ 8v <^ , s 8v Sx ~ U ^x^ rV ^c r U ' 8x As 8x is decreased without limit, ~, -=- and - approach the 8x 8x 8x ^^ values -/ , -y- and -3- respectively, and the term 8u . . becomes dx dx dx 8x negligible ; so that in the limit dy _ du dv dx dx dx The rule may be extended to apply to the case of a product of more than two functions of x. Thus if u, v and w are each functions of x dluvw) d(wV) , ,, . , v , ' = -j~, where V is wntten for uv dx dx - w dV ^ .j dw dx^ V dx nn d(uv) dw dx bUV dx ( du . dv\ . = w( v-j \-u-r- )+i \ dx dxj ' dw wU "^ dx and thus d(uv w\ du. ' mil i dv du itit Example 7. Find - when y x z . log x. T i , Let u = x* so that -r-= zx dx and let v = log x so that -j- = -. (Kx X , d .uv du dv .. . , / i \ Then = = v^- +Uj~ = (log xx 2x) + \ x z . - I dx dx dx ^ { '^v xJ = x(l+2 log X). 72 MATHEMATICS FOR ENGINEERS Example 8. Find the value of -7,[5e~ 7 ' . sin (6^ 4)] ctt cLi>t Let u = ^e-~ l so that = 5x je-' 1 35e~ 7t and let v = sin (6/ 4) so that -,7 = 6 cos (6^4). ttt d . uv du , df -3T = "-^ +M -^ = [sin (6/- 4 )x -35*- 7 <] + [5*- 7 ' x6 cos (6/- 4 )] = 5g~ 7 *[6 cos (6^4) 7 sin (6^4)]. Example g. If 2q+~ (px z ) o, show that 2q = 2p xf x dx ax p being a function of x. This example has reference to thick spherical shells. If p is a function of x, px* is of the form uv, where u = p and v = x 2 . d 9 dp , dx 2 9 dp , Hence -,- . px 2 = x 2 ,"-\-p- r - = x 2 -/-4-2Xp. dx ' r Hence 2q-\ --- 3- . px z zq+x ,- 1 x dx ' dx dp i.e., o = 2q+x , r Q//V Example 10. Find the value of -y- gx* sin (3^7) log (i 5#). Let u = x*, v = sin (3^7) and w = log (1 5*) ,, du dv dw 5 5 then -,~ = 4^ 3 , -; = 3 cos (3^7) and -y- = = >V 9** sin (3*- 7) log (1-5*) = 9^ ' ' ^ F rfw , dv . dw~] = g{ wv .,--\-wu, -\-uVj L dx dx dx-1 5x) sin (3^-7)4Ar 3 +{log (i 5*)* 4 X3cos (3^7)} sin (3* 7) = 9* 3 [_4 sin (3Af 7) log (1 5*) + 3* cos (31* 7) log (1 5*) 5#sin (3* 7)1 _ + 5*^T~ ADDITIONAL RULES OF DIFFERENTIATION 73 Exercises 7. On the Differentiation of a Product. Differentiate, with respect to x, the functions in Nos. i to 12. 1. x 2 sin 3*. 2. log 5#X 2# 3 - 4 . 3. e 9 * Iog 10 gx. 4. 4Ar- 5 .tan (3-1 2-07*). 5. cos 3-2* cos 6. cos (5 3*) tan 2#. 7. 8* 1 - 6 cos 8. 9\ogx 3 .5 3 *. 9. e*ig*. 10. ^* 11. 6* te +*(5*+2). 12. 7-2 tan ~ log * 7 . o 13. If y = Ae 3 * cos ( + B), find the value of 14. Find the value of ^/~ 5< cosh ( 5/). 15. y = (A + B*)*- 1 *; find the value of ^+8 W^ W 16. If V = 250 sin (jt -116), A = 7-2 sin 7* and W = VA, find , d\V the value of rr. at *17. Differentiate with respect to t the function i^t 2 sin (4 -8tf). 18. Find the value of -r,(4* 3 ' 7 cos 3/). H Differentiation of a Quotient. If u and y are both functions of x, and y = -, then dy V dx u dx dx i> 2 Proof. (a) From first principles. Let y = - : then a change Sx in x causes changes of Sy in y, Su in u, and Sy in y, so that the new value of y = y+Sy = -^pr-. u u uv-\-v8u y uSv , Then Sy = y+Sy y = and, dividing through by Sx Sy i . . y(y+Sy) _ ' Sx~ U ' Sx 74 MATHEMATICS FOR ENGINEERS When 8x becomes very small, P-, .- and ^- approach the values 8x 8x 8x //-Af CL'IA/ Ul) -J-, -j- and -j- respectively, whilst v-{-8v becomes indistinguishable from v. TT . ,, ,. ., dy i ( du dv\ Hence in the limit -r- = (v . -* u . -r- dx vxv\ ax ax/ du dv V-j tt-j- dx ax V 2 (b) Using the rules for a product and a function of a function. u y = - = v TM dy d , ,, ,du , dir 1 Then ~" = ' (UV ^ = V + U ' ~ x dx U-v (( -v I du\ dy- 1 dv du\ . ( 9 dv _ I If V Tit V . j / T^ I M X\ At/ /\ j .v a*/ \ ^ du dv V ~j W ~j~~ dx dx Example n. Differentiate, with regard to 5, the expression 5 cos (35+4)' T , ,-, du .Let w = 45^+75, tnen 3- = i2s^ + 7 as dv and let v = 5 cos (35+4), then -,- = 15 sin (35 + 4). ivS ZM tiy ,, d (u\ ds d Then -.- . I - ) = - ds \v/ v z = [5 cos (35+4) X (i25 2 +7)]-[(4S 3 + 7*) X -i5sin(35+4)] 25 cos 2 (35+4) 5 cos 2 (3^+4) Example 12. If y 9 4 *X, -- -, find the value of -- Let u = g 4 *, then j- = 4 x 9 4 *log e 9 = 4 X 2-1972 x g* = 8-789 X9 4z and let v = log 7*, then -=- = -*- = -. du dv dy d (u\ dx Hence -/- = T-V - ) = , dx dx\v I v z u i dx (log yx x 8 -79 x 9 4z ) (g 4 * X ^ (log 7#) 2 9 4 ^{(8-79^xlog7^)-i} (log FIG. 19. Spring loaded Governor. Example 13. For a spring loaded governor (see Fig. 19) where Q = force to elongate the spring i unit, T = tension in spring, W = weight of i ball, = angular velocity, r radius of path of balls, / = length of each of the 4 arms. If W = 3, g = 32-2 and -j- 80 when o> = 26, r = -25 and / = i, find T and Q. As there are two unknowns, we must form two equations. By simple substitution _ 32'2{T+2Q(i- V 10625)} ~" V 10625 - "968 whence T+-o64Q = 60-96 ....... (i) We are told that -^ must equal 80. dta -vr NOW Also -5- = -,- X j- = 2w T- dr da> dr dr (2) d dr where and r* dr\v u = g{T+2Q(l Vl*r z )} v = 76 MATHEMATICS FOR ENGINEERS Thus to determine -r- and -y- it is first necessary to find the value rJ A//2 4,2 of : to do this let l*-r* = y dr then Thus and du dv , VT U^ ,, aw dr dr Then - - = so that ' zr dr /.,\ Thus, differentiating both sides of the original equation with respect to r, we have from (2) and (3) 2oi T- = , rfy W Substituting the numerical values 2x26x80 = 52X ; ^9375 X- 968 _ .^SQ+T+^Q . 2 g +T whence J4Q7 = 2 Q+T but from (i) 60-96 = -064(3 +T and therefore Q 695-3! and T = 16-4! Differentiation of Inverse Trigonometric Functions. Since inverse trigonometric functions occur frequently in the study of the Integral Calculus, it is necessary to demonstrate the rules for their differentiation; and in view of their importance in the later stages of the work, the results now to be deduced should be carefully studied. The meaning of an inverse trigonometric function has already been explained (see Part I, p. 297), so that a reminder only is ADDITIONAL RULES OF DIFFERENTIATION 77 needed here. Thus sin~ x x is an inverse trigonometric function, and it is such a function that if y = sin" 1 x, then sin y = x. To differentiate sin~ l x with regard to x. Let y = sin -1 x so that, from definition, sin y = x then but and hence or milarlv dx ~ dx~ d sin y d sin y dy dx dy dx dy i = cosyx^ dy T. i * d# ~~ cos y Vi sin 2 y ~ Vix 2 d in 1 y 1 -? sin jc / dx Vi y 2 d 1 ^ COS 1 Y / (x being supposed to vary between o and -). Example 14. Find the value of -5- tan- 1 -. Let y = tan- 1 -, a *'. e., tan y and Now but sec 2 y = d tan y i+tan* d (x\ i a 2 a 2 d# d tan y dx \a' d tan y a v rf y d* dy a^ Hence i a sec 2 yx S tfy i i a 2 d# a sec 2 y a a 2 +# 2 a 78 MATHEMATICS FOR ENGINEERS Example 15. Find the value of ~ cosh" 1 -. ax a, Let y , - X = cosh" 1 a then cosh y a' So fli'if d cosh y d (x\ i dx d!# \a) a i nit d cosh y d cosh y dy rf# dy X dx hence i = sinh y X ^ a ^ rf* (i) Now cosh 2 y sinh 2 y = i X s whence sinh 2 y = cosh 2 y i = -g i a* and sinh y = - \/# 2 a 8 Then, substituting this value for sinh y in (i) 2 y a x dy , i or -/ = >-=- d , x . v- cosh" 1 = 2 2 Exercises 8. On the Differentiation of a Quotient and the Differentiation of Inverse Functions. Differentiate with respect to x the functions in Nos. i to 12. 1 5^1 2 log ( 2 ~7*) e lx ~ 5 ' cos (2 7*)' 3. . ^X _. O^f . 5 sin . Tf* cos ji. 7 rf 2 _ 5 2 - to _ cosh 1-8^ & - g9^i' 4 l-8a: - I *+3 7 cos- 1 3^ 7 - VT'^W- 8 ' Vf^^ 2 ' q ^o(a x)x ._ ^ a ' 2(6-* cot B)* 1U> a 2 (a 2 +Ar 2 )i' .. / 3 6l z x -j- 1 2/^r 2 7* 3 (an expression occurring in the solution of " 3^4-^ a b 63 - 111 problem). 12. ADDITIONAL RULES OF DIFFERENTIATION 79 e sin (l-ar+1-7) log (8**-7*+3)' 13. Assuming the results for T- cosh # and T- sinh #, find the value of -j- tanh #. d# Nos. 14 and 15 refer to the flow of water through circular pipes; v being the velocity of flow, Q the quantity flowing, and being the angle at the centre subtended by the wetted perimeter. i T* I sm 14. If, = 13-1(1 Q sin 26 ^0 17. If (a velocity) = r* (sin 6 +Z and = "' find 15. Given that Q - 132-4 l. find <. 0s <* 16. Differentiate, with respect to y, the expression '-tan-iy. 2 sin 26 di the acceleration (-57 ); find also the acceleration when is very small. 40 -it sin , dQ - , , , , ., (d<i>\ , 18. If sm <t> -- , and -j-. = a>, find the angular velocity I -if) of m at J \dt J a connecting-rod and also the angular acceleration / j. 19. Given that ^ = TT - . . f find J-R and hence the value of (pq) tanO a0 tan that makes -^ = o. ao o/> rs- j A*. j; <^M ,, WX(l X)(l 2X) ,, . 20. Find the value of -= when M = , , ? -. M is a dx 2(3/ 2x) bending moment, I is the length of a beam and x is a portion of that length. 21. Differentiate, with respect to /, the quotient - ;,- -- -. Partial Differentiation. When dealing with the equation PV = CT in connection with the theory of heat engines, we know that C alone is a constant, P, V and T being variables. If one of these variables has a definite value, the individual values of the others are not thereby determined ; e. g., assuming that C and T are known, then so also is the product PV, but not the individual values of P and V. If, now, the value of one of these is fixed, say of P, then the value of V can be calculated : therefore V depends on both P and T, and any change in V may be due to a change in either or both of the other variables. To find the change in the value of V consequent on changes in values of P and T, 8o MATHEMATICS FOR ENGINEERS the change in V due to the change in P (assuming that T is kept constant) is added to the change in V due to the change in T (P being kept constant). Rates of change found according to this plan are spoken of as partial rates of change, or more usually partial derivatives, and the process of determining them is known as partial differentiation. When only two variables occur, a plane curve may be plotted to depict the connection between them, but for three variables a surface is needed. The three co-ordinate axes will be mutually at FIG. 20. right angles, two in the plane of the paper, and the other at right angles to it. If x, y and z are the variables, we can say that z is a function of x and y, or, in the abbreviated form z = f(x, y). Similarly x = f(y, z) and y = f(x, z). Dealing with the first of these forms, and assuming the axes of x and y to be horizontal (Fig. 20), let us examine, from the aspect of the graph, the significance of this form. Giving any value to x, we know the distance of the point in front of or behind the paper : the value of y determines the distance to the right or left of the axis of z, i. e. t the vertical on which the point lies is ADDITIONAL RULES OF DIFFERENTIATION 81 determined and the actual height up this vertical is fixed by the value of z. If z is kept constant whilst values of x and y are chosen, a number of points are found all lying on a horizontal plane, and if all such points are joined we have what is known as a contour line. Therefore, if one of the quantities is constant our work is confined to one plane; but we have already seen that when dealing with a plane, the rate of change of one quantity with regard to another is measured by the slope of a curve, hence we can ascribe a meaning to a partial derivative. To illustrate by reference to a diagram (Fig. 20). The point P on the surface is fixed by its co-ordinates x, y and z, or SQ, OS and QP. If x is kept constant, the point must lie on the plane LTND. The slope of the curve LPT, as given by the tangent of the angle PMN, must measure the rate of change of z with regard to y when x is constant; and this is what we have termed the partial derivative of z with regard to y. This partial derivative may be expressed by -, or, more conveniently, by ( -j- ) , and if there is no possibility of ambiguity as to the quantity kept constant the suffix x may be dispensed with. fdz\ nn TXT (the slope being negative, since z (-=-)= tan L PMN v \ay] decreases as y increases). Similarly, the slope of the curve KPH _ /fe\ \dx)' If the variables are connected by an equation, the partial derivatives can be obtained by the use of the ordinary rules of differentiation. Example 16. Given that z \ (dz )' (dy dz\ ld z z )> W To find (j ), i. e., to find the rate of change of z with regard to x when y is constant, differentiate in the ordinary way, but treating y as a constant. Thus K- = (5>> x 2x) (zy z x 3**) + 2oye xy = loxy 6 r 2 y 2 + 2oye ry and yj = (loy x i) - (6y 2 A 2x) + (2oy x ye*") = loy i zxy 2 + 2oy z e xy . 82 MATHEMATICS FOR ENGINEERS To find \-f-} and f j ^J x must be kept constant. j z = $x z y'2x 3 y z -}-2oe xy then (3^- J = (5# 2 X i) (2* 3 x 2y) + 20* . 3- and Example 17. If z 6 log #y i8x 5 y 2 , find the values of [-3 =-) \CLX . ^Z^ ' / rf 2 ^ \ and 1 , -=- ), and state the conclusion to be drawn from the results. \dy . dx> To find (~j -j-} we must first find the value of (3-), x being regarded as a constant : then if Y be written for this expression the value of (^ ,- j must next be determined, y being treated as a constant, / d z z \ and this is the value of ( , , ). \dx . dyl XT (dz\ 6xx , 6 - - ^ T Now I T- ) = - -- i8x 5 X2y = --- $6x 5 y = Y, say. \dxl xy y Differentiating this expression with regard to x, y being regarded as a constant or and \ ) = / . dyi \dy . dxi Hence the order of differentiation does not affect the result. Total Differential. If y is a function of x, then y f(x) dy d fl . j.,, \ =- i. e., dy = f'(x)dx. dy and dx are spoken of as differentials, and f'(x) is the coefficient of the differential dx; hence we see the reason for the term differential coefficient. ADDITIONAL RULES OF DIFFERENTIATION 83 If z is a function of x and y, i. e., z f(x, y), the total differential dz is obtained from the partial differentials dx and dy by the use of the following rule fdz dy dy. The reason for this is more clearly seen if we work from the fundamental idea of rates of change, and introduce the actually measurable quantities like Sz, Sx and Sy. FIG. 21. Thus or total change in z = change in z due to the change in change in z due to the change in y. The change in z due to the change in x must be measured by the product of the change in x multiplied by the rate at which z is changing with regard to x ; and this fact can be better illustrated by reference to a diagram (Fig. 21). Let P be a point (x, y, z) on a surface, and let P move to a new position Q near to P. The change of position is made up of (a) A movement 8x to P' on the surface (y being kept constant) . (b) A movement 8y to Q on the surface (x being kept constant) . 84 MATHEMATICS FOR ENGINEERS In (a) z increases by MP' and i t dz \ = Sxx mean value of I -3- ). \dx In (b) the change in z = NQ = Sy X mean value of ( j \dyl If P, P' and Q are taken extremely close to one another, the mean or average slopes become the actual slopes and the total change in z = 8z -MP'+NQ = (*) +*(*). YYIV Example 18. If Kinetic Energy = K = -- , find the change in the energy as m changes from 49 to 49-5 and v from 1600 to 1590. From the above rule, the change in K = 8K s ,dK\ . s /rfK\ = 8m (dm)+ 8v U I Now 8m = 49-549 = '5 and Sy = 1590 1600 = 10. fdK\ ,. .. d /v z \ v z Also I j ) (i. e., v being constant) = -j { X m } = x i \dm) v dm\2g ] 2g (d~K.\ . , . , . d / m ,\ m and \ WJ (m bemg constant ) = dv\2> X V ) = 2 X 2V - vm _20 xi6oo X49 ~ 64 r 4~ 64-4 = 19880 24380 = 4500 units. Example 19. A quantity of water Q is measured by If r l = the probable error of D, a diameter, r. 2 = the probable error of H, a head, and R = the probable error of Q, where ( jM) an( i (^u) are P ar tial derivatives. Find an expression for R. ADDITIONAL RULES OF DIFFERENTIATION 85 Also I TrCD 2 / = IX Vlg Hence R = V R <r i. e., if the probable error of D is 3% and that of H is i% that of Q = V 4 (-03)+ i(-oi)~ = -0602, i. e., is about 6%. Logarithmic Differentiation. Occasionally it is necessary to differentiate an expression which can be resolved into a number of factors; and in such a case, to avoid repeated applications of the rules for the differentiation of products and quotients, we may first take logs throughout, and then differentiate, making use of the rule for the differentiation of a function of a function. By the judicious use of this artifice much labour can often be saved. Example 20. Find the value of - Ax Let- y = (3*~4)(4*-_ (zx-g) then log y = log (3* 4) +log (4*+ 7) -log (2* 9). Differentiating with regard to x dlogy _ 3 , 4_ 2 , I / i ^_\ h,,t d _ l gy dlogy dy i dy LJUt , r ^ - - . ~= dx dy dx y dx i so that (3* -4)^(4* +7) (2* -9) I *?. = 3 4 2 y ** (3*-4) (4^+7) (2^9) ^ _ (3^ -4) (4* + 7) Y rf* ~ (2^-9) /24* 2 66* 189 + 24* 2 + 144 140* 24* 2 10*4-56! l~ (3^ -4) (4* +7) (2^ -9) I 86 MATHEMATICS FOR ENGINEERS [As an exercise, the reader should work this according to the following plan. Write y = -. ^ ^r , and then use the rule for the differentiation of a quotient.] It is with examples in which powers of factors occur that this method is most useful. T- j d y hx+2) 3 (x i) Example 21. Find when y = v/ - --. *-*-rs '. dx (2X 5) 2 Taking logs throughout logy = 3 log (7* + 2)+log (#i) 2 log (2* 5) Then djog y _ 3x7 i_ 2x2 dy ~ ~ (?x+2y(x-i) (2X-5) I47^r + i4^ 2 31^ 10 28^ 2 +2O^r+8 (*-i)(2^r 5) js&ar + 103 ~ ^i)(2X - 5) 28* 2 158^+103 y' dx~ \yx+2)(x i)(2x 5) *(x-i) 28^-158^+103 Exercises 9. On Partial Differentiation and Logarithmic Differentiation. 1. In measuring the sides of a rectangle, the probable errors in the sides were Y and r 2 . If A = area and a and b are the sides, find the probable error R in A. ~(dA\* . . Given that- R = ^ &) +>> the derivatives being partial. 2. If * = a-** 5 '*, find and 3. If 5 = /-#/+log (5^-3) X, find - and . 4. If v = ( 4 -w) 2 ( 3 + 8w)3, find ^. , <iy 2wy show that - 2 - ADDITIONAL RULES OF DIFFERENTIATION 87 6. If y = 8*(i7 + -2*), find ^. 7. Differentiate, with respect to x, ( 8. Find the rate of discharge j- of air from a closed reservoir when m -- , m, p, v and r all being variables. CT 9. If x = r cos Q, y r sin 0, and u is a function of both x and y, prove that tdu\ t\(du\ i . Jdu T-) = cos OU- ) - sin dx' \dr >o r and T- \dx du CHAPTER IV APPLICATIONS OF DIFFERENTIATION HAVING developed the rules for the differentiation of the various functions, algebraic and trigonometric, we are now in a position to apply these rules to the solution of practical problems. By far the most important and interesting direction in which differentiation proves of great service is in the solution of problems concerned with maximum and minimum values ; and with these problems we shall now deal. Maximum and Minimum Values. Numerous cases present themselves, both in engineering theory and practice, in which the value of one quantity is to be found such that another quantity, which depends on the first, has a maximum or minimum value when the first has the determined value. E. g., suppose it is desired to arrange a number of electric cells in such a way. that the greatest possible current is obtained from them. Knowing the voltage and internal resistance of each cell and the external resistance through which the current is to be passed, it is possible by simple differentiation to determine the relation that must exist between the external resistance and the total internal resistance in order that the maximum current flows. Again, it might be necessary to find the least cost of a hydraulic installation to transmit a certain horse-power. Here a number of quantities are concerned, such as diameter of piping, price of power, length of pipe line, etc., any one of which might be treated as the main variable. By expressing all the conditions in terms of this one variable and proceeding according to the plan now to be demonstrated, the problem would become one easy of solution. A graphic method for the solution of such problems has already been treated very fully (see Part I, pp. 183 et seq.). This method, though direct and perfectly general in its application, is somewhat laborious, and unless the graphs are drawn to a large scale in the neighbourhood of the turning points, the results obtained are usually good approximations only. In consequence of these failings APPLICATIONS OF DIFFERENTIATION 89 of the graphic treatment, the algebraic method is introduced, but it should be remembered that its application is not so universal as that of the solution by plotting. The theory of the algebraic method can be simply explained in the following manner : The slope of a curve measures the rate of change of the ordinate with regard to the abscissa; and hence, when the slope of the FIG, 22. Maximum and Minimum Values. curve is zero, the rate of change of the function is zero, and the function must have a turning value, which must be either a maximum or a minimum. But it has already been pointed out that the slope of a curve is otherwise denned as the derivative or the differential coefficient of the function ; therefore the function has a turning value whenever its derivative is zero. Hence, to find maximum or minimum values of a function we must first determine the derivative of the function, and then find MATHEMATICS FOR ENGINEERS the value or values of the I.V. which make the derivative zero; the actual maximum or minimum values of the function being found by the substitution of the particular values of the I.V. in the expression for the function. The rule, stated in a concise form, is : To find the value of the I.V. which makes the function a maximum or minimum, differentiate the function, equate to zero and solve the resulting equation. The full merit of the method will be best appreciated by the discussion of a somewhat academic problem before proceeding to some of a more practical nature. Example i. Find the values of x which give to the function y = 2# 3 + 3# 2 36^ + 15 maximum or minimum values. Find also the value of x at the point of inflexion of the curve. This question may be treated from two points of view, viz. (a) From the graphical aspect. We first plot the primitive curve y 2# 3 +3# 2 36^+15 (see Fig. 22), the table of values for which is : X A 2 X s 2* s +3* 2 -36#+i5 y -4 16 -64 128+48 + 144 + 15 79 -3 9 -27 - 54 + 27+108 + 15 96 -2 4 - 8 - 16 + 12+ 72 + 15 83 I i - i 2+ 3+ 36 + 15 52 O o o o+ o- + 15 15 I i i 2+ 3 36+15 -16 2 4 8 16+12 72 + 15 29 3 9 27 54+27-108+15 12 4 16 64 128+48-144+15 47 5 25 125 250+75-180+15 1 60 This curve has two turns and two turns only, and consequently y has two turning values, one being a maximum and one a minimum. By successive graphic differentiation the first and second derived curves may be drawn, these being shown on the diagram. Now for values of x less than 3 the slope of the primitive curve is positive, as is demonstrated by the fact that the ordinates of the first derived curve are positive. At x 3 the primitive curve is horizontal and the first derived curve crosses the #-axis ; and since dv the ordinates of the first derived curve give the values of , , we see dx that when the primitive curve has a turning value, the value of dy f- = o. For values of x between 3 and +2 the slope of the primitive is negative; when x = +2 the slope is zero, and from that APPLICATIONS OF DIFFERENTIATION 91 point the slope is positive. Thus y has turning values when x = 3 and when x = +2; these values being a maximum at x 3 and a minimum at x = +2 as observed from the curve. This investigation proves of service when we proceed to treat the question from the algebraic aspect; in fact, for complete understanding the two methods must be interwoven. (b) From the algebraic point of view. Let y = 2x 3 + $x 2 36*4-15 then * = 6x*+6x-36 = 6(x z +x-6). Now in order that y may have turning values we have seen that dy it is necessary that -^- o. But ^ = o if 6(# 2 + *-6) = o i. e., if 6(^+3) (x 2) = o i. e., if x = 3 or 2 and hence y has turning values when x = 3 and x +2. We do not yet, however, know the character of these turning values, so that our object must now be to devise a simple method enabling us to discriminate between values of x giving maximum and minimum values to y. An obvious, but slow, method is as follows : Let us take a value of x slightly less than 3, say 3-1; then the calculated value of y is 95 "85. Next, taking a value of x rather bigger than 3, say 2-9, the value of y is found to be 95-85. Therefore, as x increases from 3-1 to 3 and thence to 2-9, y has the values 95'85, 96, and 95-85 respectively. Thus the value of y must , be a maximum at x = 3, since its values on either side are both less than its value when x = 3. In like manner it can be shown that when x -\-2, y has a minimum value. The arithmetical work necessary in this method can, however, be dispensed with by the use of a more mathematical process, now to be described. Referring to the first derived curve, the equation of which is y 6# 2 + 6# 36, we note that as x increases from 4 to 3 the ordinate of the derived curve decreases from 36 to o; from x = 3 to x = .5 the ordinate is negative but increasing numerically, i. e., in the neighbourhood of x = 3 the slope of the second derived curve, which is the slope curve of the first derived curve, is negative (for the ordinate decreases as the abscissa or the I.V. increases). But the slope of the first derived 92 MATHEMATICS FOR ENGINEERS curve, and thus the ordinate of the second derived curve, must d 2 v be expressed by -y-^, so that we conclude that in the neighbourhood dsC of a maximum value of the original function the second derivative of it is a negative quantity. In the same way we see that in the neighbourhood of a minimum value of the function, its second derivative is a positive quantity. Hence a more direct method of discrimination between the turning values presents itself : Having found the values of the I.V. causing turning values of the original function, substitute these values in turn in the expression for the second derivative of the function ; if the result is a negative, then the particular value of the I.V. considered is that giving a maximum value of the function and vice-versa. This rule may be expressed in the following brief fashion : Let y = f(x) and let the values of x that make -j~(x) orf'(x) = o be #! and x z . d 2 y Find the value of -~^ or f"(x), as it may be written, and in this expression substitute in turn the values x^ and x z in place of x: the values thus obtained are those of f"(x-^} and /"(# 2 ) respec- tively. Then if f"(x^), say, is negative, y has a maximum value when x x^; and if f"(xj) is positive, y has a minimum value when x = x v Applying to our present example : y = / When x 3 the value of -~ 2 is I2( 3) +6, i. e., /"( 3) = 30; and since /"( 3) is a negative quantity, y is a maximum when x == 3. Similarly, /"( + 2) = 12(2) +6 = +30 and hence y is a minimum when x +2. Referring to the second derived curve, i. e., the curve y = I2X+6, we note that its ordinate is negative for all values of x less than 5 and positive for all values of x greater than -5, the curve crossing the axis of x when x = -5. This indicates that when x = -5 the first divided curve has a turning value ; but the first APPLICATIONS OF DIFFERENTIATION 93 derived curve is the curve of the gradients of the primitive curve, and hence when x 5 the gradient of the primitive must have a turning value, which may be either a maximum or a minimum. In other words, if we had placed a straight edge to be tangential in all positions to the primitive curve, it would rotate in a right- handed direction until x = -5 was reached, after which the rotation would be in the reverse direction. A point on the curve at which the gradient ceases to rotate in the one direction and commences to rotate in the opposite direction is called a -point of inflexion of the curve. Thus points of inflexion or contra-flexure u d *y occur when -=-4 = o. dx 2 A useful illustration of the necessity for determining points of contra-flexure is furnished by cases of fixed beams. We have of Confraf lexure. [ filll 1 /^ ^\ I -21 ll ! FIG. 23. already seen that the bending moment at any section is propor- d 2 v tional to the value of -j- z there; hence there must be points of CLX* contra-flexure when the bending moment is zero. Example 2. Find the positions of the points of contra-flexure of a beam fixed at its ends and uniformly loaded with w units per foot; the deflected form having the equation i fwlx 3 wl 2 x z _wx*\ = El\ 12 24 24 ')' We may regard this question' from either the graphic aspect or the physical. According to the former we see that it is necessary to determine the points of inflexion, and therefore to find values of x ^ ^d*y . for which , 2 is zero. Reasoning from the physical basis we arrive at the same result, by way of the following argument : the bending moment, which is d z y expressed by EI- , changes sign, as is indicated by the change in OLX the curvature of the beam (see Fig. 23), and therefore at two points the bending moment must be zero, since the variation in it is uniform 94 MATHEMATICS FOR ENGINEERS and continuous; but the bending moment is zero when -~ 2 is zero, v since M = El j-^. dx 2 AT w fix 3 I 2 x 2 x*\ Now y = =pp ( ----------- V El \i2 24 24/ dy w T( I ,\ / / 2 \ 4 ^ hence ^ = -ey ( X3* 2 ) ( X2# ) *- a* El LA 1 2 / \24 / 24 w __ 12 ~~ 6 d 2 y w r / 1 \ i l z \ 3^ 2 -i and -,- -2 = ,-, f ( xzx ) ( XII V ^ 2 El L\4 / \I2 / 6 J (** l * _^ 2 \ . El Va i2~ 2"]' /72 V ./v 72 ;2\ Now the bending moment M = EI.^=a;[- -- - ) *" \ 2 12 2 / /^ / 2 AT 2 and M = o if ------ , i. e., 6lxl 2 6x z = o, 2 12 2 i. e., if 6x z 6lx+l z = o 6l or 12 = -789? or -2 1 iL Hence the points of inflexion occur at points distant -211 of the length from the ends. Example 3. A line, 5 ins. long, is to be divided into two parts such that the square of the length of one part together with four times the cube of the length of the other is a minimum. Find the position of the point of section. Let x ins. = the length of one part, then 5 x = length of the other part. Then (5 #) 2 +4# 3 is to be a minimum. * Let y = ( 5 -*) 2 + 4 * 3 Then Hence ~- o if x ^ or i (the latter root implying external dx o cutting) . APPLICATIONS OF DIFFERENTIATION 95 To test for the nature of the turning value 10 dx d*y and g = 24*4-2. When x = J 6 -5-^ = ( -2 5 J_|_2 = a positive quantity. Therefore y is a minimum when x = and the required point of section is % in. from one end. Example 4. If 5 detrimental surface of an aeroplane S = area of planes K = lifting efficiency KS then/, the " fineness," is obtained from the formula / 2 = ~-. Also the thrust required for sustentation = C (I'acX where C is a constant and i is the angle of incidence of the plane (expressed in radians). Taking S = 255 and K = -4, find the angle of incidence for the case in which the least thrust is required. p = JS . 4 X2 5 = J -o8s -08 5 ' The thrust T = C (*4-js:.) and since i is the only variable in this \ J *%/ expression, we must differentiate with regard to it. Thus - dT ., and = if 1- = i. e., if i 2 = 1 = . / 2 125 Thus i = -0895 or the thrust required is either a maximum or minimum when the angle of incidence is -0895 radian. To test whether this turning value is a maximum or a minimum, let us find the second derivative ^- dT = ^T _ / 2_ & - - u V +/ 2 z 3 /' ^2-p When i -0895, -^ must be positive, and hence T has its minimum value when i = -0895. 96 MATHEMATICS FOR ENGINEERS Example 5. Find the dimensions of the greatest cylinder that can be inscribed in a right circular cone of height 6 ins. and base 10 ins. diameter. FIG. 24. Assume that the radius of the base of the cylinder = x ins. (Fig. 24) and the height of the cylinder y ins. Then the volume V nx^y. We must, then, obtain an expression for y in terms of x before differentiating with regard to x. From the figure, by similar triangles, taking the triangles ADC and EFC 6 _ y or Hence V 5 2 6 . . 67T , g 3 , it* X-(5 ~~ ~c '** , and Thus dV d\' 5 = o if x(io 3^) = o i. e. t or if x = o (giving the cylinder of zero if 10 3*, i. e., x $\ ins. volume) Then y = (5-3*) = 2 ins - and the volume of the greatest cylinder = wX ( ) X2 = 69-8 cu. ins. Example 6. The total running cost in pounds sterling per hour of a certain ship being given by v 3 C =4.5+ ^2100 where v = speed in knots, find for what speed the total cost for a journey is a minimum. The total cost for the journey depends on (a) The cost per hour; and (6) The number of hours taken over the journey. APPLICATIONS OF DIFFERENTIATION 97 Item (b) depends inversely on the speed, so that if the journey j t K were 2000 nautical miles the time taken would be hours; or. in general, the number of hours = Then the total cost for a journey of K nautical miles K / v 3 \ = Q = xU-5+- V V* ' ' 2IOO/ i , U M 5 V -\ ) 2IOO/ Differentiating with regard to the variable v dC t Then 1050 ., * a v (s . . IX "4-5 - ' *J - 1050 v or v 3 = 4-5 X 1050 = 4725 hence *v = 16-78 knots. Example 7. A water main is supplied by water under a head of 60 ft. The loss of head due to pipe friction, for a given length, is proportional to the velocity squared. Find the head lost in friction when the horse-power transmitted by the main is a maximum. If v = velocity of flow, then Head lost = Kv 2 , where K is some constant, i. e., the effective head = 60 Kv 2 = H e . TT -r, , Quantity (in Ibs. per min.) x effective head (in feet) H.P. transmitted = J v * < - v ' 33000 _ area (in sq. ft.) x velocity (ft. per min.) x 62-4 x H g 33000 = CvHe, where C is some constant = Gv{6o-Kv*) = C(6ov-Kv 3 ) Then / (H.P.) = C(6o-3Kz; 2 ) Ctl) = C(6o i8o + 3H e ) or T (H.P.) = o when 3H C = 120 i. e., H e = 40. In general, then, the maximum horse-power is transmitted when the head lost is one-third of the head supplied, i. e., the maximum 2 efficiency is - or 66-7%. 98 MATHEMATICS FOR ENGINEERS Example 8. The stiffness of a beam is proportional to the breadth and the cube of the depth of the section. Find the dimensions of the stiffest beam that can be cut from a cylindrical log 4 ins. in diameter. From hypothesis or S = Kbd 3 . Both breadth and depth will vary, but they depend on each other; and from Fig. 25 we see that b 2 = i6 d 2 . Hence we can substitute for b its value in terms of d and then differentiate with regard to d\ according!}? S = As it stands this would be a rather cumbersome expression to differentiate, and we therefore employ a method which is often of great assistance. Since we are dealing with positive quantities throughout, S 2 will be a maximum when S is a maximum,* and hence we square both sides before differentiating. Thus S 2 = K 2 d 6 (i6-d 2 ) = <fS 2 and -, ~ FIG. 25. '>-8d'>) = 8d 5 (i2-d 2 ) Hence ,-r = o if d 5 o, i. e., d = o (giving zero stiffness) Ct.(t or if i. e., Hence d 2 = 12 d = 3-464 ins. b V 1612 = 2 ins. * If we were dealing with negative quantities it would be incorrect to say that the quantity itself had a maximum value when its square was a maximum, for suppose the values of the quantity y in the neighbourhood of its maximum value were 13, 12, u, 10, ii, 12, etc., corresponding values of y 2 would be +169, +144, + 121, +100, +121, +144, so that if y = 10 (its maximum value) when x = 4, say, then y 2 TOO when x = 4, and therefore a minimum value of y 2 occurs when x 4, and not a maximum. Example g. Find the shape of the rectangular channel of given sectional area A which will permit the greatest flow of water ; being given that Q = Av, v c Vmi, m = hydraulic mean depth = . . f^ rea . wetted perimeter and i is the hydraulic gradient ; Q being the quantity flowing. APPLICATIONS OF DIFFERENTIATION 99 Let the breadth of the section be b and the depth d; then, by hypothesis bd = A. whence b = -j. a w = . . , -. = , j and therefore v = c Vi\/ ; wetted perimeter o+2a = cVAl . Hence Q = Av = Ac VAi . ~ /, , = K . -^Jsas where K = Vb+2d Q will be a maximum when Q 2 is a maximum, hence we shall find the value of b for which Q 2 is a maximum. b+2d~ ' 6+ ?A' Also Q 2 is a maximum when the denominator of this fraction is a minimum. Let this denominator be denoted by D dD d I, , 2A\ 2A then -jjj- = ^1+ r / = I ~~^F rfD ., 2A . / and -n- = o if i = T-, . e., if o = V2A. Now d = = -= = \ O \/2A 2 /. the dimensions would be /A = - depth = V - and breadth = \/2A. Example 10. For a certain steam engine the expression for W, the brake energy per cu. ft. of steam, was found in terms of r, the ratio of expansion, as follows /i+log r\ ( r S-)-27 I2o \V = - _ - _ - _ 00833 , JJ.J-. 000903 Find the value of r that makes W a maximum. Before proceeding to differentiate, we can put the expression in a somewhat simpler form. Thus- W = "O 00833 + and W is a quotient = - where u = i2o(i+log r}2jr ioo MATHEMATICS FOR ENGINEERS du 1 20 so that -y = ---- 27 dr r and v -00833 + -000903? dv so that -f- -000903. du dv J\*T V J -- M J~ TT dW dr dv Hence -y = , dr v z (00833 + 000903*') ( 27^ [iao(i+log r) 27/1-000903 (-00833 +-000903*-) 2 Now - = o if the numerator of the right-hand side = o i . e., if ( -22 L\_ (27 x -00833) + (120 x -000903) (27 x -0009037) (120 x -000903) (120 x -000903 log r) + (27 x -ooogo^r) = o i.e., if. -225 -1084 log r = o. This equation must be solved by plotting, the intersection of the curves y = -1084 log r and y z = --- 225 being found; the value of r here being 2-93. Hence r = 2-93. Example n. The value of a secondary electric current was given by the formula _- _ y = - e L+M e L-M where L = inductance of primary circuit R = resistance of primary circuit M = coefficient of mutual inductance I = steady current. Find for what value of /, y has a maximum value. T / iu m y = -(e~L+M g~L M dv R ?L R - and - = o if ^^r.e L-M __ rf/ L M L+M Transposing the factors L M BI(L-M-L-M) T _ L+M -TM, ~ e L-M. = or eLM. L M APPLICATIONS OF DIFFERENTIATION 101 In order to find an expression for t, this equation must be changed to a log form, thus /L+M\ _ 2 MRf g \L M/ ~ L 8 -M2 L 2 -M 2 , / * = - log If three variables are concerned, say x, y and z, the relation between them being expressed by the equation z=f(x, y), then in order to find the values of x and y for turning values of z, it is necessary to determine where the plane tangential to the surface is horizontal. The algebraic problem is to find the values of x and y that satisfy simultaneously the equations (-T-) = o and f-^-J = o, these \(4'Z / \dZ / derivatives being partial. Example 12. The electric time constant of a cylindrical coil of wire (i. e., the time in which the current through the coil falls from its full value to a value equal to -632 of this) can be expressed approximately by K = . y where z is the axial length of the coil, y is the difference between the external and internal radii and x is the mean radius ; a, b and c representing constants. If the volume of the coil is fixed, find the values of x and y which make the time constant as great as possible. The volume V of the coil = cross section x length V *. e., V = 2Tc#xy Xz and z = - 2-nxy K = m \ r <--- I and is a maximum (ax+by+cz) ax+by+cz a , b , c '. when - or -\ is a minimum. xyz vz xz xv f , a. . b , c Let p = --- z xz x - - --- xyz yz xz xy a. . b , c --- yz xz xy . c + yV xV x~y __ zivcya mxyb c '' ~ + " Now (^ (i. e., with y constant) = 2 y U +(- X - 2TOI C 102 MATHEMATICS FOR ENGINEERS c- -11 (dp\ 27T& C Similarly \f) = -~- 2 . \dy) x V xy z Both \j) an< ^ vft mus ^ k e equated to zero, so that ^jr- V x 2 y i. e., x*y = (i) , 2TI& C V ' xy 2 i. 6. xy z = .... (2) 2TT& To solve for x and y cV From (2) x = j-~ 2 . Substituting in (i) cV cVa whence y 3 = r, 27T0 2 or __2TU6 2 also Exercises 10. On Maximum and Minimum Values. 1. If M = 15* -oix*, find the value of x that makes M a maximum. 2. Find the value of x that makes M a maximum if M = 3-42* -ix 2 . 3. M is a bending moment and x is & length ; find x in terms of / so that M shall be a maximum, and find also the maximum value of M. M = < 4. As for No. 3, but taking 5. The work done by a series motor in time t is given by ,, wx ,, M = (l 2 V R where e = back E.M.F. E = supply pressure R = resistance of armature. a The electrical efficiency is ^. Find the efficiency when the motor so runs that the greatest rate of doing useful work is reached. In Nos. 6 to 8 find values of x which give turning values to y, stating the nature of these turning values. APPLICATIONS OF DIFFERENTIATION 6. y = 4 * 2 +i8*- 4 i. 7. y = 5^- 8. y = x 3 + 6x 2 15^+51 (find also the value of x at the point of inflexion). 9. Sixteen electric cells, each of internal resistance i ohm and giving each I volt, are connected up in mixed circuit through a resistance of 4 ohms. Find the arrangement for the greatest current [say rows with x cells in each row]. x 10. If 40 sq. ft. of sheet metal are to be used in the construction of an open tank with square base, find the dimensions so that the capacity of the tank is a maximum. fj A 11. Given that W = 4C 2 +',7, find a value of C that gives a turning value of W, and state the nature of this turning value. // _ x \ 12. M (a bending moment) = W v . ; (x+y] Wy. For what value of x is M a maximum ? {W, / and y are constants.} 13. The cost C (in pounds sterling per mile) of an electric cable can be expressed by C - ^+636* x where x is the cross section in sq. ins. Find the cross section for which the cost is the minimum, and find also the minimum cost. 14. A window has the form of a rectangle together with a semi- circle on one of its sides as diameter, and the perimeter is 30 ft. Find the dimensions so that the greatest amount of light may be admitted. 15. C, the cost per hour of a ship, in pounds, is given by c3 C = 3 -2 + - 2200 where s = speed in knots. Find the value of s which makes the cost of a journey of 3000 nautical miles a minimum. At speed 10% greater and less than this compare the total cost with its minimum value. 16. An isolated load W rolls over a suspension bridge stiffened with pin-jointed girders. When the load is at A, distant x from the \V# centre, the bending moment at this section M A = (I 2 4# 2 ). For what value of x is M A a maximum ? 17. A riveted steel tank of circular section open at the top has to be constructed to contain 5000 gals, of water. Find the dimensions so that the least possible amount of steel plate is required. 18. A canister having a square base is cut out of 128 sq. ins. of tin, the depth of the lid being i in. Find the dimensions in order that the contents of the canister may be as large as possible. 19. The stiffness of a beam of rectangular section is proportional to the breadth and the cube of the depth. Find the ratio of the sides of the stiffest beam of rectangular section with a given perimeter. 104 MATHEMATICS FOR ENGINEERS 20. A load uniformly distributed over a length r rolls across a beam of length I, and the bending moment M due to this loading at a point is given by -. wry f, r} wx z M=-j* {/-?+*--,- . For what value of x is M a maximum ? 21. Find the value of V (a velocity) that makes R (a resistance) a maximum when = yi 3 (V-i2) 54 V+I2 ' 22. If L = Vr z x z *(r z x*), find the value of x that makes L a y v maximum. 23. A jet of water, moving with velocity v, impinges on a plate moving in the direction of the jet with velocity u. The efficiency TJ = -- 3 - Find values of u for maximum and minimum efficiency, and find also the maximum efficiency. . ^2.tt(l} _ / vC\ 24. If v *-= ', find the value of u for maximum value of i?. v z 25. Given that O = K//T! (cos 6 sin 6), find values of 6 between o and 360 that make Q a maximum, treating K, p. and T x as constants. 26. A cylinder of a petrol engine is of diameter d and length /. . , , d area of exposed surface Find the value of the ratio -, which makes / capacity a minimum. The volume must be treated as a constant. 27. If the exposed surface of a petrol engine cylinder is given by S 27W 2 +2:rr/+-2y 2 , I being the length and r the radius, find the value of the ratio - that makes the ratio ex P osed surface r capacity a minimum. The volume must be treated as a constant. 28. Given that ? = K2 - , find values of K for turning values of y. z#R 2 / 1 \ / /3\ 29. IfM = ^|^ -sin 2 0)--934wR 2 ^cos0- /s /|j, for what values of is M a maximum ? [M is the bending moment at a section of a circular arched rib loaded with a uniform load w per foot of span, and R is the radius of the arch.] 30. An open channel with side slopes at 45 is to have a cross section of 120 sq. ft. Determine the dimensions for the best section (i. e., the section having the smallest perimeter for a given area). _ 31. If M = -^TTZ - ni\ * fi n d the value of x which makes M, a O ( I .- t ) bending moment, a maximum. The final equation should be solved by plotting, a value being assumed for /. APPLICATIONS OF DIFFERENTIATION 105 32. In connection with retaining walls the following equation occurs p _ ph 2 _ i M tan 6 72 i M*+2/* cot 6. Find an expression, giving the value of 6 (in terms of tan 6), that makes P a maximum. {M, p and h are constants.} 33. Assuming that the H.P. of an engine can be expressed by the relation H = C(fnl 3 -Kpn 3 l*) where C is a constant, / = stroke, p = pressure in piston rod due to the pressure on the piston, p average density of the material of the engine, K = constant depending upon the mode of distribution of the mass of the engine parts, n = R.P.M., and / = safe stress in the material, find an expression for I giving the maximum H.P. for engines of different sizes. 34. Find the turning point of the probability curve T * and also the points of inflexion. 35. In a two-stage compressor, neglecting clearances, if P x and V x are the initial pressure and volume of the L.P. cylinder, P 2 the pressure in the intercooler, and P 3 the discharge pressure of the H.P. cylinder, the total work for the two cylinders is given by For what value of P 2 is W a minimum, P^ V^, P 3 and n being regarded as constants ? 36. Find the height A of a Warren girder to give the maximum stiffness, the stiffness being given by the expression tln , Id { W\ 2 EA r 4 d being the length of one bay and I the span, whilst f c , ft and E are constants for the material. 37. The efficiency of a reaction wheel may be expressed by 2( i) For what value of n has >> its maximum value ? 38. The weight W of steam passing through an orifice, from pressure P! to pressure Pg, is given by n+Pj" If n 1-135, nn( i the value of ^ for which W is a maximum. *i 39. Find the height of the greatest cylinder that can be inscribed in the frustum of a paraboloid of revolution cut off by a plane perpendicular to the axis and distant 6 units from the origin. The paraboloid is generated by the revolution about the axis of x of the parabola y z = yc. 106 MATHEMATICS FOR ENGINEERS f 1% _!_ y\ ~\ 40. If M = \V j# - 1*-\ w here y and I are constants, find the value of x that makes M a maximum. 41. If T, / and T/ are the tight, slack and centrifugal tensions respectively in a belt passing round a pulley, and v = speed of the belt in feet per sec., then H.P. transmitted - ^1=3. 550 TJOV^ Being given that T/ = , the maximum permissible tension in the belt T m = T+T/, ^ = coefficient of friction between belt and T pulley, = angle of lap of belt in radians, and . ev e , find the value % of T/ in terms of 1 m so that the maximum H.P. is transmitted. 42. If y 3X*-\-2x 3 j8x 2 24OX-\-54, find the values of x which give turning values to y, stating the nature of these turning values ; and find also the values of x at the points of inflexion. 43. The radial stress in a rotating disc in which expression x is the only variable. Find the value of x which gives to p x its maximum value, and state this value of p x . 44. A pipe of length / and diameter D has at one end a nozzle of diameter d through which water is discharged from a reservoir, the level of the water in which is maintained at a constant head h above the centre of the nozzle. Find the diameter of the nozzle so that the kinetic energy of the jet may be a maximum ; the kinetic energy being expressed by 7T V_2gD'A \t [Hint. If K = kinetic energy, writ _ P 7T = and find the value of d for the maximum value of K*.] 45. Prove that the cuboid of greatest volume which can be inscribed in a sphere of radius a is a cube of side 46. The velocity of the piston of a reciprocating engine can be expressed by /sin 20 .. Q \ 2nnr\ -- hsm ) \ zm I where is the inclination of the crank to the line of stroke. T ,. connecting-rod length _ , . If m = - TxrS ---- 1 - = 8, find the values of between length of crank o and 360 that make the velocity a maximum. APPLICATIONS OF DIFFERENTIATION 107 Calculation of Small Corrections. Differentiation finds another application in the calculation of small corrections. Thus an experiment might' be carried out, certain readings being taken, and results deduced from these readings; then if there is a possibility of some slight error in the readings and it is required to find the consequent error in the calculated result, we may proceed to find that error in the manner now to be explained. Suppose we have two quantities A and B connected with one another by a formula A = KB ; then if the value of B is slightly inaccurate the error in A will depend on this error in B, and also on the rate at which A changes with regard to B. E. g., if A changes three times as fast as B and the error in B is !%, then the consequent error in A must be 3X-I or -3%. We might also look upon this question from a different point of view. Suppose that a reading, instead of being x, as it should have been, was slightly larger, say x-}-8x, i. e., the measured value of x would be represented by OB and not OA (Fig. 26), then the error is Sx or 8x XIOO%. X This error causes an error in the value of y, so that the calculated value of y is BQ and not AP, i. e., the error is Sy. To compare these errors we proceed as follows : -- the slope of the chord PQ, and if Sx is very small (as it should be, for otherwise the experiment would be repeated), then this would also be the slope of the tangent at both P and Q, or, approximately Sy dy 8x dx ^. e., dx or, error in y = rate at which y changes with regard to x X error in x. Example 13. In the measurement of the diameter of a shaft, of which the actual diameter was 4 ins., an error of 2% was made; what was the consequent error in the weight ? io8 MATHEMATICS FOR ENGINEERS Here W = ~d 2 lp, where p is the density 4 - Kd 2 , where K = %. 4 Now the error in the diameter = Sd = --'- X4 = -08 in. too also - - - 2Kd CLLJ\J i i - 7 i - J.Vli- d.d d.d or the percentage error = ^ x 100 = - ^9g^- X 100 Example 14. If some torsion experiments are being made on shafts varying in diameter from i in. to 5 ins. ; then, allowing a maximum error of -5% in the measurement of the diameters, what is the range of the errors in the stress ? Given that T = f-fd 3 . The stress / = x -^ TC a 3 df i6T hence 13- Now the error in the diameter 8.d is -5% i. g., 8d = 5 X rf. 100 Hence the "error in / = /-. X 8d = 4 -7r X d.d -jzd* 100 i. e. t the percentage error in / 8f 48T - 5 d = loo X J f - = loo X r, X ~ X / IT a 4 zoo = -3- Thus the smallest error = '03 x smallest stress \ and the largest error = -03 x largest stress /' If the error in the measurement of the diameter is on the high side, then the stress, as calculated, will be too low. Expansion of Functions in Series. Theorems of Taylor andMaclaurin. Many of the simpler functions, suchaslog (i-f-#), sin x, cos x, etc., can be expressed as the sums of series. These functions can be expressed in terms of these series by the use of a theorem known as Maclaurin's. APPLICATIONS OF DIFFERENTIATION 109 Let f(x) stand for the function of x considered, and let f(x) a+bx+cx^+dx 3 -^ . . ., to be true for all values of x, i.e., /(o) = We assume that the differentiation of the right-hand side term by term gives the derivative of /(#). Differentiate both sides with regard to x. Then ^or/'(*) = b+2cx+idx*+ . . . This must be true for all values of x ; thus, let x = o then f'(x) when x = o dr/'(o) = b /'(o) implying that f'(x) or *-* is first found and then the flwv value o substituted for x throughout. d*f(x) Differentiating again, -4V or f"( x ) = 2c-}-6dx-{- . . . and /"() = zc /"(o) t. e., c= J J . 2 Similarly, /'"(*) = 6^+ terms containing x and higher * powers of x, whence /'"() = 6W or 1.2.3.^. or 1.2.3 |JL Accordingly we may write the expansions This is Maclaurin's Theorem. By a similar investigation we might obtain Taylor's Theorem, which may be regarded as a more general expression of the foregoing. Taylor's Theorem. In this the expansion is of f(x-\-h] and not /(*); thus \A or, as it is sometimes written, to give an expansion for/(*) no MATHEMATICS FOR ENGINEERS If in either of these two expansions we make h = o, then Maclaurin's series results. We may now utilise these theorems to obtain series of great importance. Example 15. To find a series for cos x. Let f(x] = cos x then /(o) cos o = i. A i rn \ d COS X Also / (x), i. e., -, -- = sin x ctx so that /'() = sin o o. Again /"(*) = ^~(~ s ^ n x ) ~ cos x so that /"() = cos o = i and /"'(#) = -3- (cos x) sin x so that /'"() = sin o = o. Now f(x] = /(o Therefore cos x i --- 1 12. [4 Example 16. To find a series for log e Let then Now so that so that and so that Hence /(*) /() /'(o) /"(o) f"'lv\ J (*/ /'"(o) log e (i+#) = log e (i+#) = log 1=0. rflog(l+#) I ~ i i d/ -i \ 2 ^\(l+^) 2 / (l+^) 3 2 = - 2. ^r 2 Ar 3 (Compare with the series found by an entirely different method in Part I, p. 470.) Example 17. Prove that &>* = cos #+_;' sin ^, where /= V I. This equation is of great importance, since it links up the exponential and the trigonometric functions. APPLICATIONS OF DIFFERENTIATION in To find a series for sin x. Let . f(x) = sin x then /(o) = sin = f'(x] = cos x /'(o) = cos o i f"(x) = sin x /"() sin o = o f"'(x) = cos x f "'(o) = cos o = i. Hence sin # = x 1 . . . LI LL and jsinx = i(x |--j . . .) v Li U To find a series for ei x . Let f(x) = ei x /(o) = e = i f"(x) = j z ei x /"(o) = j z e = j z = i -7 2 A 2 ^ ^X^ Hence ei x i+jx+- \- J - . . . 11 11 Now cos x+j sin # (the series for cos x having been found in Example 15). Ivllj| + " { Fory : : -_\^ \ = ei x . ( i* = +i, etc. J Use might be made of Taylor's Theorem to determine a more correct solution to an equation when an approximate solution is known ; for, taking the first two terms of the expansion only /(*+*)=/(*)+*/'(*) or interchanging x and h, as a matter of convenience, then /(*+*)=/(*)+'(*) If A is small compared with x, the assumption that two terms of the series may be taken to represent the expansion is very nearly true. Suppose that a rough approximation for the root has been found (by trial and error) ; denote this by x. Let the true solution be x-\-h; then by substitution in the above equation the value of h can be found, and thence that of x-}-h. H2 MATHEMATICS FOR ENGINEERS As an illustration, consider the following case : A rough test gives 2-4 as a solution of the equation x* i -5^+3 -7% = 21-554. It is required to find a solution more correct. Here x = 2-4 and f(x) = # 4 1-5# 3 +37# 21-554 so that 7(2-4) = 33-17 20-73+8-88 21-554 = -'234- If the correct value of h is found, thenf(x-\-h) must = o. Hence f(x+h) =f(x)+hf'(x) i. ., o = 234+A/'(2-4). [Now /(*) = * 4 i-5* 3 +37* 21-554 so that 7' (2-4) = 55-30-25-92+3-7 = 33-o8.] Hence o = 234+^x33-08) or h= -^o = -0071. 33-o8 Hence a more correct approximation is 2-4+ -0071 *. e. t x = 2-407 is the solution of the equation. This method may thus be usefully employed in lieu of the graphic method when extremely accurate results are desired. The following example illustrates the process of interpolation necessary in many cases where the tables of values supplied are not sufficiently detailed for the purpose in hand; and in view of the importance of the method, every step in the argument should be thoroughly understood. Example 18.' It is desired to use some steam tables giving the pressures for each 10 difference of temperature, to obtain the accurate d-b value of . when t = 132 C. The figures in the line commencing with i = 130 C. (the nearest to 132) are as follows : 1 dp : V d*p ip d* P dp" d'p dt* 130 2025-7I7 60-5995 1-47051 026392 0002738 Calculate, very exactly, the value of -^ when t = 132 C. Taylor's theorem may here be usefully employed, using the form APPLICATIONS OF DIFFERENTIATION 113 Let f(x) = -- when t 130 andf(x + h) = - when t = 132, so that h = 2 Then /'(*) = = and /"(*) = , etc., / having the value 130. Thus the expansion may be re-written as _fdp and substituting the values from the table = 60. 5995 + (2 x.i-4705i) + (2 X -026392) + (4 x -0002738) '133 V 3 = 63-59367- Exercises 11. On the Calculation of Small Corrections and Expansion in Series. 1. If R = R (i+a/+6* 2 ) when R (the resistance of a conductor at o C.) is 1-6, a (the temperature-resistance coefficient of the material) = -00388 and b = -000000587, find the error in R (the resistance at temperature t C.) if t is measured as 101 instead of 100. 2. The quantity Q of water flowing over a notch is given by o - Q = X-64X V^g.H*, where H is the head at the notch. What is the percentage error in Q caused by measuring H as -198 instead Of -2 ? 3. If y = 4* 1 ' 76 , y = 17-3 when x = 2-3. What will be the change in y consequent on a change of x to 2-302 ? 4. A rough approximation gives x = 2-44 as a solution of the 2j equation 10 3 = 16+4^ x z . Find a more correct root. 5. Determine the value of x to satisfy the equation x 1 ' 5 3 sin x = 3, having given that it is in the neighbourhood of 2-67. 6. The height A of a Porter governor is expressed by w ' n* where n is the number of revolutions per minute. If W = 100, w = 2 and / = 10, find the change in the height due to a change in the speed from 200 to 197 r.p.m. 7. In calculating the co-ordinates of a station in a survey it was thought that there was a possibility of an error of 3 minutes (*. e., i either way) in the reading of the bearing. If the bearing of a line was read as 7 12' and the length of the line was 2 chains 74 links, I n 4 MATHEMATICS FOR ENGINEERS find the possible errors in the co-ordinates of the distant end of the line. [Co-ordinates are length X cos (bearing) and length x sine (bearing) .] 8. Find by the methods of this chapter a series for a x . 9. Using the figures given in Example 18, p. 112, calculate very exactly the pressure p at 133 C. 10. The equation d 3 +-6$d -5 = o occurred when finding the sag of a cable. A rough plotting gives the solution to be in the neigh- bourhood of -5 : find a more exact root. CHAPTER V INTEGRATION HAVING discussed the section of the Calculus which treats of differentiation, we can now proceed to the study of the process of integration, this having a far more extensive application, and being, without doubt, far more difficult to comprehend. As with the differentiation, it is impossible fully to appreciate this branch of the subject unless much careful thought is given to the fundamental principles; and accordingly the introduction to the Integral Calculus is here treated at great length, but in a manner which, it is hoped, will commend itself. Meaning of Integration. The terms integer and integral convey the idea of totality; an integer being, as we know, a whole number, and thus the sum of its constituent parts or fractions. The process of integration in the same way implies a summation or a totalling, whereas that of differentiation is the determination of rates of change or the comparison of small differences. Differentiation suggests subtraction or differencing, whilst integration suggests addition ; differentiation deals with rates of change, integration with the results of the total change ; differen- tiation involves the determination of slopes of curves, and integra- tion the determination of areas of figures. Integration is, in fact, the converse to differentiation, and being therefore a converse operation is essentially more difficult to perform. [As instances of this statement contrast the squaring a quantity with the extraction of a square root, or the removal of brackets with factorisation.] A converse operation is rather more vague as concerns the results than a direct; for when performing a direct operation one result only is obtainable, but the results of a converse operation may be many, as we shall find, for example, when dealing with indefinite integrals. To illustrate the connection between differentiation and integra- tion, consider the familiar case of velocity and acceleration. Suppose values of v and t are given, as in the table : "5 n6 MATHEMATICS FOR ENGINEERS Then t I 15 20 25 -30 V 28-4 297 30-5 33'4 36-5 8v i'3 8 2-9 3'i 8t 05 05 05 05 Bv a = w 26 16 58 62 The accelerations are here found by comparing differences of velocity with differences of time. Regard the question from the other point of view : assume that these accelerations are given and we wish to determine the total change in the velocity in the given period of time. The total change must be given by the sum of the changes in the small periods of time; in the first period of -05 sec. the average acceleration was 26, i, e., the velocity was being increased at the rate of 26 units per sec. each sec. ; and therefore the change in the velocity in -05 sec. = 26 x -05 units per sec. = 1-3 units per sec. In the successive periods the changes in velocity are -8, 2-9 and 3-1 respectively. Hence the total change in the velocity over the period -2 sec. = i-3-f--8+2-9+3'i == 8-1 units per sec., or if the initial velocity was 28-4, the final velocity was 28-4+8-1 = 36-5. Note that the acceleration is given by the fraction ^, whilst a small change in ot the velocity is of the nature aU, or the total change of velocity = sum of all small changes = ^aSt. We can thus find integrals by working through the processes of differentiation, but in the reverse order. If a function, expressed in terms of symbols, has to be integrated, it is an advantage to transform the rules for differentiation into forms more readily applicable; the method, however, being entirely algebraic. If numerical values alone are given, the integration resolves itself into a determination of an area. Hence Considered from an algebraic standpoint Differentiation implies the calculation of rates of change ; Integration implies the summation of small quantities. INTEGRATION 117 From the graphic standpoint Differentiation is concerned with the measurement of slopes of curves ; Integration is concerned with the measurement of areas under curves. Just as special symbols are used to denote the processes of differentiation, so also there are special symbols for expressing the processes of integrations. Regarding an integral as an area, it must be of two dimensions, a length and a breadth; and we have seen in an earlier chapter (Part I, Chap. VII) that in order to ascertain an area correctly its base must be divided up into small elements, the smaller the better, these elements not necessarily being of the same length, but all being small. Thus, to find the area ABCD- (Fig. 27) we can suppose it divided up into small strips, as EFGH, then find the area of each of these and add the results. The portion EH of the curve is very nearly straight, so that EFGH is a trapezoid, and hence its area = mean height X width. Now its mean 6JG I F G FIG. 27. height FE and GH are practically the same, so that any one of them can be denoted by y; also the width FG of the strip is a small element of the base, *'. e., is 8x. Hence, the area of the strip EFGH y x 8x, and the total area between the curve, the bounding ordinates and the axis of x must equal the sum of all products like y8x, or, as it might be expressed Area = 2yS# (approximately). However small the width of the strips are made, this sum only gives the area approximately, but as 8x is diminished the result approaches the true more and more closely. Therefore, bearing in mind our previous work on limits, we can say that the limiting value of 2ySx must give the area exactly. To this limiting value of the sum different forms of symbols are n8 MATHEMATICS FOR ENGINEERS attached, the 2 and S being replaced by the English forms / and d respectively, so that the area between the curve and the axis of x = fydx. There is no limit placed to this area in any horizontal direction, so that the area is not denned by the given formula. Hence fydx is spoken of as an indefinite integral. The x is again the I.V., and the size of the area will depend on the values given to it. Suppose that when y = AB, x = a, and when y = CD, x = b ; then the range of x is from a to & if it is the area ABCD that is considered. Accordingly we can state that the area ABCD =fydx, the value of this integral being found between x = a and x = b, or, as it is written for brevity, fx=b fb I ydx, or, more shortly still, I ydx, it being clearly understood J x=a J a that the limits a and b apply to the I.V., i. e., that quantity directly associated with the " d." It is evident that ABCD is a definite area, having one value fb only, and thus I ydx is termed a definite integral. J a The most convenient method for determining areas (provided that a planimeter is not handy) is undoubtedly the " sum curve " method treated in Part I, Chap. VII ; the great virtue of it being that the growth of the area is seen, and that either any portion or the whole of the area of the figure can be readily found by reading a particular ordinate. In view of the great usefulness of the process of integration by graphic means, the method is here explained in detail, following exactly the plan adopted in Part I, Chap. VII. Graphic Integration is a means of summing an area with the aid of tee and set square, by a combination of the principles of the "addition of strips" and "similar figures." An area in Fig. 28 is bounded by a curve a'b'z', a base line az and two vertical ordinates aa' and zz'. The base is first divided in such a way that the widths of the strips are taken to suit the changes of curvature between a' and z' ', and are therefore not necessarily equal ; and mid-ordinates (shown dotted) are erected for every division. Next the tops of the mid-ordinates are projected horizontally on to a vertical line, as BB'. A pole P is now chosen to the left of that vertical; its distance from it, called the polar distance p, being a round number of horizontal units. The pole is next joined to each of the projections in turn and parallels are drawn across the INTEGRATION 119 corresponding strips so that a continuous curve results, known as the Sum Curve. Thus am parallel to PB' is drawn from a across the first strip; mn parallel to PC' is drawn from m across the second strip, and so on. The ordinate to the sum curve through any point in the base gives the area under the original or primitive curve from a up to the point considered. Referring to Fig. 28 Area of strip abb' a' = ab x AB Pole FIG. 28. Graphic Integration, but, by similar figures B'a or BA bin ab whence AB x ab = p X bm , area of strip , , . , , i. e., bm = - or area of strip = px bm P i. e., bm measures the area of the first strip to a particular scale, which depends entirely on the value of p. , area of second strip In the same way nm = - P 120 MATHEMATICS FOR ENGINEERS and by the construction nm' and bm are added, so that area of ist and 2nd strips en P or area of ist and 2nd strips p x en Thus, summing for the whole area Area of aa'z'z = p x zL Thus the scale of area is the old vertical scale multiplied by the polar distance ; and accordingly the polar distance should be selected in terms of a number convenient for multiplication. E. g., if the original scales are i" = 40 units vertically and i" = 25 units horizontally and the polar distance is taken as 2", i. e., 50 horizontal units; then the new vertical scale = old vertical scale x polar distance = 40x50 = 2000 units per inch. If the original scales are given and a -particular scale is desired for the sum curve, then the polar distance must be calculated as follows new vertical scale Polar distance in horizontal units = , , ,-. , , old vertical scale E. g., if the primitive curve is a " velocity-time" curve plotted to the scales, i" 5 ft. per sec. (vertically) and i" = -i sec. (hori- zontally), and the scale of the sum curve, which is a " displacement- time " curve, is required to be i" = 2-5 ft., then 2*S Polar distance (in horizontal units) = = -5 and since i" = -i unit along the horizontal, the polar distance must be made 5". Integration is not limited to the determination of areas only; true, an integral may be regarded as an area, but if the ordinate does not represent a mere length, but, say, an area of cross section, the value of the integral will in such cases measure the volume of the solid. Our standard form throughout will be for the area of the figure as plotted on the paper, viz., Jydx, where y is an ordinate and 8x an element of the base, but y and x may represent many different quantities. Thus, suppose a curve is plotted to represent the expansion of INTEGRATION 121 a gas; if, as is usual, pressures are plotted vertically and volumes horizontally, the ordinate is p and an element of the base is Sv; hence the area under the curve = \ z pdv (if the J n initial and final volumes are v t and v 2 respectively), and since this /6s is of the nature pressure X volume, i. e., 7^3 X (ft) 3 or ft. Ibs., the \7*J area must represent the work done in the expansion. To illustrate such a case : Work 106560 8 10 la 14- 16 18 SO 2 24 26 FIG. 29. Expansion of Steam. Example i. It is required to find the work done in the expansion of i Ib. of dry saturated steam from pressure 100 Ibs. per sq. in. to pressure 15 Ibs. per sq. in. From the steam tables the following corresponding values of p and v are found : s v (cu. ft. per Ib.) 4.44 5-48 7-16 10-50 I3-72 20 26-4 p (Ibs. per sq. in.) IOO 80 60 40 30 2O 15 By plotting these values, p vertically, the expansion curve is obtained (Fig. 29) ; this being the primitive curve. Selecting a polar distance equivalent to 10 horizontal units, we proceed to construct the sum curve, the last ordinate of which measures to a certain scale the work done in the expansion. Now the new vertical scale = old vertical X 10, since the polar distance = 10 ; and also we must multiply by 144, since the pressures are expressed 122 MATHEMATICS FOR ENGINEERS in Ibs. per sq. in. and must be converted to Ibs. per sq. ft., so that the work done may be measured in ft. Ibs. According to this modified scale the last ordinate is read off as 106560 ; thus the work done = 106560 ft. Ibs. or, as it would be written in more mathematical language T26-4 f J4 pdv 106560. Example 2. The diameters of a tapering stone column, 20 ft. long, at 6 equidistant places were measured as 2-52, 2-06, 1-54, i'i5, '80 and -58 ft. respectively. Find its weight at 140 Ibs. per cu. ft. 29\5 6 8 (O 12 14- IG IQ 20 FIG. 30. Problem on Stone Column. The volume will be obtained by plotting the areas against the length and summing. Now the area of any section = -d z , and the total volume will be the sum of the volumes of the small elements into which the solid may be supposed to be divided. /"20 pO^ Thus the volume = J Adi = J Q ~d*.dl. and the weight = 140 I -d 2 .dl. J o 4 Since - is a constant multiplier, it can be omitted until the end, for its effect is simply to alter the final scale ; hence a constant factor before integration remains so after. Hence the weight- = 1097 j d*.dl. INTEGRATION 123 The integral will be of the standard form if for d z we write y and if for / we write x, so that we see that ordinates must represent d 2 and abscissae lengths, and hence the table for plotting reads : I o 4 8 12 16 20 y or d 2 6-34 4-24 2-36 1-32 64 336 Plotting these values and thence constructing the sum curve (see Fig. 30), we find the last ordinate to be 47-15, and this is the value f20 of J Q d*.dl. [20 Weight = 109-7 1 d z .dl = 109-7x47-15 = 5180 Ibs. Application of Integration to " Beam " Problems. At an earlier stage (see p. 38) it has been demonstrated that the shear at any point in the length of a beam loaded in any way whatever is given by the rate of change of the bending moment in the neighbourhood considered, this being the space rate of change. Conversely, then, the bending moment must be found by summating the shearing force ; and hence, if the shear curve is given, its sum curve is the curve of bending moment. In the majority of problems the system of loading is given, from which the curve of loads can be drawn. Then, since the shear at any section is the sum of all the forces to the right or left of that section, the sum curve of the load curve must be the shear curve; continuing the process, the sum curve of the shear curve, i. e., the second sum curve from the load curve as primitive, is the curve of bending moment and the fourth sum curve is the deflected form. Expressing these results or statements in the notation of the calculus ; L, S and M being the respective abbreviations for loading, shear and bending moment S =fLdx M =f$dx =f(fLdx)dx = ffL(dx) z [ff~L(dx) z being termed a double integral] and the deflection y = ff M.(dx) 2 or ffff L(dx)*. If the loading is not uniform, but continuous, the summation must be performed graphically. [The link polygon method largely used obviates half these curves, e. g., the link polygon for the loads gives at once the curve of bending moment.] 124 MATHEMATICS FOR ENGINEERS Example 3. The loading on a beam, 24 ft. long, simply supported at its ends varies continuously, as shown in the table. Draw diagrams of shearing force and bending moment, stating clearly the maximum values of the shearing force and the bending moment. Distance from one end (ft.) o 4 7 10 12 J 4 17 20 24 Load in tons per ft. 44 58 86 1-06 I-I i -06 86 58 '44 The curve of loads is first plotted, as in Fig. 31. By sum-curving this curve, we obtain the curve of shearing force, although no measurements can be made to it until account has been taken of the support reactions. 4 68 1C 12' ' 14 >6 <8 FIG. 31. Problem on Loaded Beam. To find the reactions at the ends : We know that these must be equal, since the loading is symmetrical, each reaction being one-half of the total load. Now the last ordinate AB of the sum curve of the load curve is 19; thus the reactions are each 9-5. Bisecting AB, or, in other words, marking off a length AC to represent the reaction at A, we draw a horizontal, and this is the true base line for the curve of shear; any ordinate to the curve of shear from this base giving the shear at the point in the length of the beam through which the ordinate is drawn. We observe that the shear changes sign and is zero at the centre of the beam; we can conclude from this that the bending moment must have its maximum value at the centre, since shear rate of change of bending moment, and if the shear is zero, the bending moment must have a turning value. By sum-curving the shear curve from CD as base, the resulting curve is that of bending moment. INTEGRATION 125 It is well carefully to consider the scales, for it is with these that difficulties often arise. The scales given here apply to the original drawing, of which Fig. 31 is a reproduction somewhat under half full size. For the length i in. = 2-5 ft. For loads i in. = -4 ton per ft. Polar distance for the first sum curve, i. e., the curve of shear = 4 ins. = 4X2-5, or 10 horizontal units. Hence the scale of shear = -4 x 10, or 4 tons to i in. B FIG. 32. Shearing Force and Bending Moment on Ship's Hull. Polar distance for the second sum curve = 4 ins. = 10 horizontal units. Hence the scale of bending moment = 4 X 10 = 40 or i in. (vertically) = 40 tons. ft. Reading according to these scales The maximum shear = 9-5 tons ^ and the maximum bending moment = 68 tons, ft. J Example 4. In Fig. 32 AAA is the curve of weights or load distribution, and BBB the curve of buoyancy or upward water thrust 126 MATHEMATICS FOR ENGINEERS for a ship whose length is 350 ft., the scale of loads being indicated on the diagram. Draw diagrams of shearing force and bending moment on the hull of the vessel and measure the maximum values of these quantities. It is first necessary to construct the curve of loads to a straight line base, and to do this the differences between the curves AAA and BBB are set off from a horizontal, taken in our case below the original base line. In this way the curve of loads LLL is obtained, the scale being shown to the left of the diagram. By sum-curving this curve, the curve of shear SSS is obtained ; the polar distance (not shown on the diagram) being taken as 50 horizontal units, so that the scale for the shear is 50 times the scale for the loads. Sum-curving the curve SSS, the curve MMM, that of bending moment, is obtained (again the polar distance is 50 horizontal units). Sectio'ns such as K, where the upward thrust of the water balances the downward force due to the weights, are spoken of as water-borne. Reading our maximum values according to the proper scales, we find them to be Maximum shear = 246 tons ~\ Maximum bending moment = 14,300 tons ft. / It should be noted that the last ordinate of both the shear curve and the curve of bending moment is zero ; these results we should expect since the areas under the curves AAA and BBB must be equal, so that the shear at the end must be zero, and also the moments of these areas must be alike. [In practice the maximum bending moment is found by such a formula as- Weight X length Maximum bending moment = 7? Constant the constant for small boats being between 30 and 40, and for larger between 25 and 30.] The Coradi Integraph. A brief description of the Integraph, an instrument devised to draw mechanically the sum curve, can usefully be inserted at this stage. It consists essentially of a carriage running on four milled wheels A (Fig. 33), a slotted arm C carrying the tracer B which is moved along the primitive curve, and the arm D which carries the pencil E which draws the sum curve. As B is moved along the primitive curve, the slotted arm C slides about the pins G and P, thus altering its inclination to the horizontal. A parallel link motion ensures the movement of E parallel to the instantaneous position of C, the sharp-edged wheel F assisting in guiding the tracer bracket. INTEGRATION 127 The principle of the instrument is not difficult to understand, and can be explained in a very few words. The pole is at P, and the tops of the mid-ordinates are pro- jected to the vertical through G by the horizontals like BG; parallels are then drawn to PG by the pencil E, the motion being continuous. The polar distance can be varied as desired, by altering the position of the bracket carrying the pin P along the horizontal arm; and if an extremely small polar distance is found to be advisable, the pin H may be utilised instead of G. Rules for Integration of Simpler Functions. Since integration is the reverse of differentiation, many functions can be integrated by reversing the order of the steps in differentiation. Integration of powers of x. The first rule given in the work on differentiation of functions was d x n = nx n ~ v . dx 128 MATHEMATICS FOR ENGINEERS To change into the integration form, we transpose -5- : the " d" on the one side becomes f on the other side, to indicate the change differencing to summing, and the " dx " occurs on the top line of the other side of the equation. Thus x n = fnx n ~ 1 dx or fx n ~ 1 dx = - x n -\-C % the reason for the presence of the constant term C being explained later. It is a trifle simpler to write n in place of ni, and therefore +i in place of n, so that Whereas, when differentiating a power of the I.V., the power was reduced by i in the process of differentiation ; when integrating, the power is increased by i. fa <y-5 tiv^ - g " dx x whereas fx 5 dx = ^ x 6 -\-C. A special case occurs for which the above rule does not apply : for, let n = i, then fx~ l dx should, according to the rule just given, be - x, but to this fraction no definite meaning can be assigned. We know that -3- log x = - = x~* dx x fx- 1 dx = log x+C or, as it is sometimes written dx , A constant multiplier before integration remains as such after; thus faxndx = T % W+1 +C. Also an expression composed of terms can be integrated term by term, and the results added. INTEGRATION 129 Thus, f (ax n -\-b) dx can be written fax n dx -\-fbdx i. e., fax*dx-\-fbx?dx f or x = i its value being -x n+l -\-bx+C. Note. Differentiation of a constant term gives zero, but the integration gives that constant multiplied by the I.V. The reason for this will be apparent if we consider the state- ments from the graphical standpoint. The curve representing the equation y = b is a horizontal straight line, and therefore the slope is zero (i.e., -v- = o) ; but the area under the curve = the area of a rectangle = base x height = xxb (i. e.,fbdx = bx). Exponential Functions. We have already proved that dx de x - e* ( S ee p. 47) ; then by transposition of d and dx to the other side of the equation we obtain the statement e x = j and since -, corresponds to f we may write this as fe x dx = e x -\-C. Thus if we either differentiate e x or integrate it we arrive at the same result ; and e x is the only function for which the differential coefficient and also the integral are the same as the function itself. Carrying this work a step further, let us consider the integration of e bx , and hence a x : Now f-ae bx = abe bx .*. fae bx dx = ?J*+C. dx b To avoid confusion as to the placing of a and b we must reason in the following manner : The a is a constant multiplier of the whole function, and therefore remains so after integration; the b multiplies the I.V. only; and thus differentiation would cause it to multiply the result, whereas after integration it becomes a divisor. Great attention should be paid to the application of this rule, for unless care is exercised mistakes are very apt to creep in. Example 5. Find the value of -^(15** 7*' 9 +83) and also of 130 MATHEMATICS FOR ENGINEERS Differentiating the first expression d. Integrating the second given expression 4 --i + i = i 5 t*- 7 f+C. Notice that although a function has been differentiated and the derivative integrated, the final expression is not exactly the same as the original, the constant term being represented only by C, where C may have any value. Further reference will be made to this point on p. 137. Example 6. If pv 1 ' 32 = C, find the value oi/pdv. To express p in terms of v fpdv = fCir- 1 -** . dv = Cfv~ l - 3Z dv (K being any constant) = CX- u-l-38+1 -1-32+1 = 32 This result can be written in a slightly different form, if for C we write its value pv 1 ' 32 ; then F., pv l '**xir-** . v ipdv = \-K 32 pv = 3* 32 a Exa'.-nple 7. Find fpdv when pv = C. In this case p = Cv~* fpdv = fav - C/^ = C log v + K = pv log v Example 8. Find the value of Note that 17 is a constant multiplier throughout; 2 multiplies the I.V. and therefore appears as a divisor after integration ; also the power of e remains exactly the same. INTEGRATION Example 9. Find the value of f(^oe &v +v 5 -*)dv. f(4oe'*>+v s -*)dv = f^oe &v dv+fu 5 -*dv (separating the terms) -e 5 131 Example 10. Find an expression for fa x dx, and apply the result to determine the value of /i2 x From our previous work we know that d dx a x = a x . log a. fa x dx = . -.* * log a Hence fiz x 5 tx dx = (i2X- X . ^ X 5 4 *)+C loge 5 = 1-609 = 1-864 X5 4g +C. Afofe. It would be quite incorrect to multiply 1-864 by 5 and express the result as g-^2* x . Alternatively, the result might have been arrived at in the following manner = 12 x, H X(5 4 )*+C log 625 VJ ' = g; X5 4ar +C (log 625 = 6-44) = 1-864 X5 4 *+C. Exercises 12. On Graphic Integration. 1. The acceleration of a slider at various times is given in the table. By graphic integration obtain the velocity and displacement curves to a time base, indicating clearly your scales. j Time . o 008 016 02 028 036 44 048 06 ! Acceleration o 75 87-5 87-5 87-5 87-5 87-5 83 O 068 072 084 10 108 12 78 85 87-5 87-5 83 O 132 MATHEMATICS FOR ENGINEERS 2. An acceleration diagram on a time base has an area of 4-7 sq. ins. The base of the diagram is 2-5 ins. and represents 25 sees. The acceleration scale is i in. =3 ft. per sec. 2 . If the velocity at the beginning is n ft. per sec., find the velocity in ft. per sec. at the end of the 25 sees. 3. A rectangular barge is loaded symmetrically in still water. The curve of loading is a triangle with apex at the centre, and the curve of buoyancy is a rectangle. Draw diagrams of shearing force and bending moment on the barge. 4. The curves of loads for a ship 350 ft. long is as given in the table. Plot this, and by graphic integration obtain the curves of shearing force and bending moment. Distance from one end (ft.) 7 10 35 56 84 102 load (tons per foot) . 3 o -2-6 -3-i -2-3 O 112 133 161 196 210 237 260 280 315 330 350 1-6 2-3 3-15 5'2 57 -4-5 -4-95 o 9 5. The table gives the values of the pressure and volume for the complete theoretical diagram for a triple expansion engine. V o I 2 4 6 8 10 12 p 240 240 I2O 60 40 30 24 20 Find the initial pressure in each cylinder in order that the work done per cycle may be the same for each. (Hint. Divide the last ordinate of the sum curve into three equal parts, draw horizontals through these points of section to meet the sum curve, and from these points of contact erect perpendiculars to cut the expansion line.) 6. A body weighing 3000 Ibs. was lifted vertically by a rope, there being a damped spring balance to indicate the pulling force F Ibs. of the rope. When the body had been lifted x ft. from its position of rest, the pulling force was automatically registered as follows : X 20 40 65 75 95 no 140 F 8000 7950 7850 7500 7400 6800 6400 4000 Find the work done on the body when it has risen 80 ft. How much of this is potential energy and how much is kinetic energy ? Find, also the work done when it has risen 140 ft. INTEGRATION 133 7. The current from a battery was measured at various times, with the following results : Time (hours) . I 3 6 9 10 12 14 15 Current (amperes) 25 28 37 39'5 32 29 24 25-3 2? If its capacity is measured by fCdt, find the capacity in ampere hours. 8. The following are the approximate speeds of a locomotive on a run over a not very level road. Draw a curve showing the distance run up to any time. Time (mins. and sees.) I 2-15 6.15 9.22 11-45 14.26 16.35 20.52 Speed (miles per hr.) 6 10 18-2 22-8 25-5 28 29-2 28-6 9. The load curve at a large central station can be constructed from the following data : Time (hours) . . , I 2 3 4 5 6 14 7 7-5 8 9 10 ii Load (1000 amperes) 3'5 i I 2 I 6-4 i7 17-8 16-4 ii'3 8-7 8-2 12 I 2 3J4 5 i 5'5 6 7 8 9 10 II j 12 | 7 -8 8 7-6 8-7 12-5 19 23-3 21 12-4 ii 10-5 9-6 9 ! 6 Find the total number of ampere hours supplied in the 24 hours. 10. The velocity of a three-phase electric train, with rheostatic control, at various times, was found as in the table : Time (sees.) . . . | o 26-6 66-6 80- 1 99 Velocity (ft. per sec.) . o 40 40 37-3 Draw the space-time curve and find the total distance covered in the 99 seconds. On the Integration of the Powers of x and of the Exponential Functions. 11. What is the significance of the symbols f and dx in the expression fx z dx ? Integrate, with respect to x, the functions in Examples 12 to 27 12. 4* 1 -". 13. 70-15. 14. 15. 16. e'. 17. *- +14. x 18. 19. 134 MATHEMATICS FOR ENGINEERS 20. i2*~. 21. -g^. 22. -i 7 e". 23. 24. 2 . 54 *--i- 8 . 2 *-i++i-i3. 25. 26. ^.*_*--3+ 27. -94*' 18 cos - Find the values of the following 2B.fv 5 dv. 29. /"-". 30. /35^. 31. /g"-<fc 32. //xfo when ^y 1 ' 17 = C. 33. /i4 X 2*^5. -8)d*. 35. /3- 1'. 36. 17 /^g. J r 38. $-z-2(dt) z . 39. 2-ix~ 5 dx 40. Solve the equation -/- = w -. rfy y 41. In connection with the flow of air through a nozzle, if x is the distance outwards from the nozzle and v is the velocity there, v oc - 1 . Also SA (an element of area of flow) = ~K8x Vx. The added momentum for the small element considered = SM = v8A. Show that M the D total increment to the momentum, can be written C 7= where C vx and D are constants. Trigonometric Functions. We have previously seen that the derived curve of either the sine curve or the cosine curve is the primitive curve itself transferred back a horizontal distance of one-quarter of the period. Conversely, then, we may state that the sum curve of either the sine or the cosine curve is the curve itself moved forward for a distance corresponding to one-quarter of the period. In other words, integration does not alter the form of the curve. Taking the case of the sine curve as the primitive, we see, on reference to Fig. 34, that if this curve is shifted forward for one-quarter period the resulting curve is the cosine curve inverted ; or expressing in algebraic language, whilst the equation of the primitive curve is y = sin x, that of the sum or integral curve is y = cos x. Thus, fsmxdx = cos x. In like manner it could be shown that feosxdx = sin*. For emphasis, the dif- ferentiation and the integration of sine x and cosine x are repeated here -T- sin x = cos x /sin* dx = cos #4-C dx -T-COS x sin x /cos* dx = sin x-\- C. dx J INTEGRATION 135 Note. When differentiating the cosine the minus sign appears in the result ; when integrating the sine the minus sign appears ; it is important to get a good grip of these statements, and the con- sideration of them from the graphic aspect is a great help in this respect. To extend the foregoing rules fcosxdx = sin x-\-C /cos (ax+b)dx = sin (ax+b)+C fsinxdx = cos x-\-C /sin (ax+b)dx = cos (ax+b)+C. 1 75 5 25 O 25 5 75 1 ^ \/ J? N \ / s~ A \ . // -si n.. rr / / \ Y / I \ \ / JG % 1 \ / * 24\ -y -a / is.< r\ \ / / \ y / \ V s A V s / FIG. 34. Thus the angle remains the same after integration just as it would after differentiation, but the constant multiplier a of the I.V. becomes a divisor. To integrate sec 2 -AC with regard to x we call to mind the differentiation of tan x, viz., , tan x = sec 2 x. Accordingly /sec 2 xdx = tan x+C. Extending this to apply to the more general case /sec 2 (ax+b)dx - - 1 tan (ax+6)+C In like manner /cosec 2 (ax+b)dx = - cot (ax+6)-{-C. I 3 6 MATHEMATICS FOR ENGINEERS Other two standard integrals are added here, the derivation of which will be considered in the next chapter. /tan* dx log cos x+C fcotx dx = log sin x+C. To verify these we might work from the right-hand side and differentiate. Dealing with the former d , , . _ _ d , if u = cos % dx^ dx whence d log u du du _ _ v Qin v - 1 S\ 7 7 ' Oil! Jv du dx dx * i = x sin x u sin x = - = tan x cos* /tan* dx = log cos x+C. Example n. Find the value of f(5~ sin^t)dt. f$dt fsin^tdt = ^ ~X cos V 4 = 5'+- cos 4*+C. 4 Example 12. Evaluate /sin (5 ^t)dt. /sin (54t)dt = --x -cos (5 4/)+C = ] cos (5 40 + C. 4 4 Example 13. If a force P is given by P = 36-4 sin (1005 -62), find the value oifPds. = -364 cos (1005 62) + C. Example 14. Find the value of 12 cos (4- The expression E = 7-25* 3s~ 1 +i5s -8 + 12 cos (4 35) = (7-2X^)- 3 logs+ i ;|si-8+(i2X-^sin ( 4 - 35 ))+C = I-44S 5 3 Jog 5 + 8-33S 1 - 8 4 sin (4 3 fPds =/36'4 sin (loos -62)^5 = ^X cos (1005 -62) + C INTEGRATION 137 Example 15. If R = n sec 2 (34-71;), find /Rdv. fRdv =/n sec 2 (3 4-?v)dv = - tan (3 4-7t>)+C = 2-343 tan (34- Exercises 13. On Integration of Trigonometric Functions. Integrate, with respect to x, the functions in Nos. i to 10. 1. 3 sin 4*. 2. 5-18 cos (3 3^). 3. 7 sec 2 (^3 x). 4. x 12 -14 cos (-05 -117*). 5. 05-4*4-5 s j n ^b+ax). 6. 9-45 sin 8^. 7. 3-08 sin 2(2-16* 4-5). 8. 9^^+^-1-83 tan x. 9. 4-27 sin (--- J + -2 cos gx 4# 1 - 74 +3 2 *+ 5 . 10. 2 sin 2 #2-91 sin ( 3'7#)+2 cos 2 #14-2 cosec 2 ----- . 11. The acceleration of a moving body is given by the equation a = 49 sin (jt -26). Find expressions for the velocity and the space, the latter being in terms of the acceleration. . T f d z x o o / , cos 20\ _ , ,, , dx 12. If T- 4 = 47c 2 wV^cos 0-i J, find the values of , and x. (x is a displacement of the piston in a steam-engine mechanism.) 13. Find the value of /b^+cos (y77"2p)}dp. 14. If v 117 sin 6^29-4 cos 6t, find the value offvdt. - Indefinite and Definite Integrals. The integrals already given, although correct, are not complete. If "an integral is to denote an area some boundaries must be known ; and nothing was said about the limits to be ascribed to x (or s, as the case might be) in the foregoing, so that we were in reality dealing with indefinite areas or integrals. To indicate that a portion of the area may be dispensed with in certain cases (when the boundaries are stated) a constant C is introduced on the R.H.S. of the equation, x* i. e.,fx 3 dx would be written +C. As soon as the integral, and therefore the area, is made definite it will be observed that C vanishes. 138 MATHEMATICS FOR ENGINEERS If fx 3 dx is to equal -x*+C, -, (-# 4 +Cj should equal x 3 ; and this is the case for d -=- dx ^A 4 c \ ^ dx\4 J dx (C being independent of x). Cf. Example 5, p. 129 ; the constant in that case being 83. It is therefore advisable to add the constant in all examples on integration; in many practical examples the determination of the value of the constant is an important feature, and therefore its omission would invalidate the results obtained. In the list of a few of the simpler standard integrals collected together here for purposes of reference and by way of revision the constant is denoted by C. f(ax n +b)dx a fax n dx Jbdx [dx J x [ dx J ax+b = bx+C = log x+C = ^ log (*+&) +C fae bx dx - a b ^+c fe*dx = e*+C faFdx = rog* x *+ c /sin (ax-{-b)dx - cos (ax+b)-\-C f sin xdx = cos x-\-C /cos (ax-\-b}dx f cos xdx /sec 2 (ax-\-b}dx f sec 2 xdx /cosec 2 (ax-\-b}dx = - sin (ax-}-b)-{-C & sin x-\-C - tan (a^ = tan -- cot (ax+b)+C d INTEGRATION 139 / cosec 2 xdx = cot x-\-C /tan (ax+b)dx = log cos (ax-}-b)-\-C 8 / tan xdx = log (cos #)+C /cot (ax+b)dx = log sin (ax+b)-{-C f cot xdx = log (sin A;) +C. Method of Determining the Values of Definite Integrals. We may regard the area of a closed figure as the difference between two areas, viz., all the area to the left, say, of one boundary, minus all the area to the left of the other parallel boundary. Hence to find the value of a definite integral, the value of the integral must be found when the I.V. has its higher limiting value, and from this must be subtracted its value when the lower limiting value is substituted for the I.V. fx 2 dx = -x 3 -\-C is an indefinite integral, but if to C we give a definite value, it becomes definite and unique. /4 Thus I x 2 dx is a definite integral, because the limits to be J 2 applied to x are indicated. To evaluate it We know that fx z dx = -x 3 -\-C. The value of this integral when x = 4 is -f-C o and the value of this integral when x = 2 is |-C the constant being the same in the two cases. The difference = -^+C - -+C = ^ V 3 / \3 / 3 meaning that if the curve y = x z were plotted, and the area between the curve, the x axis and the ordinates through x = 2 and x = 4 found, its value would be i8| sq. units. It will be noticed that C vanishes, and hence when dealing with definite integrals it is usual to omit it altogether. ) MATHEMATICS FOR ENGINEERS /" 4 /# 3 \ 4 For brevity, I x 2 dx is written I ) , which on expansion .'2 \3/2 r reads __ e 56 3 3/'* '3' Example 16. Find the value of the definite integral I * -i f V**d* = f^ g3*Y 4 = 4 J .,' \3 /-i 3 3 v = | (3-3201-1-3499) = 2-62 <). Example 17. Evaluate the definite integral V f 2 / (50034^+7) rf^ .' 5 I (50054^+7) dx = (- si > o H sn 7?r = ' or ii. 2 Example 18. Find the value of / J o The expression = n-2 4 INTEGRATION 141 Notice that no cancelling takes place, beyond that concerning the constant multiplier 5, until the values (4 and 2) have been substituted in place of x. In other words, it would be quite wrong to say c 4/i Example ig. The total range of an aeroplane in miles can be C Wt obtained from the expression / dq where m = pound-miles per Ib. of petrol, and q = loading at any time initial loading Taking q = -6 and m = 4000, find the total range. = m (log q log i) = m log q. Now if q = -6, log q = 1-4892 = -5108. Hence the range = 4000 x -5108 = 2043 miles. c- A F B O L FIG. 35. Proof of Simpson's Rule. Proof of Simpson's Rule for the Determination of Areas of Irregular Curved Figures. This rule, given on p. 310 of Part I, states that length of one division of the base f first + last ordinate + = \42 even ordmates -f- [22 odd ordinates. 142 MATHEMATICS FOR ENGINEERS It is now possible to give the proof of this rule. Let us deal with a portion of the full area to be measured, such as ABCD in Fig. 35. Let the base AB = 2c. Let the equation of the curve DEC be y = A4-B#+C# 2 , so that DEC is a portion of some parabola. We can assume that the origin is at F, and therefore the abscissae of D, E and C are c, o and +c respectively. Hence AD = y x = A+B(-c)+C(-c) 2 = A-Bc+Cc 2 FE = y 2 = A+B(o)+C(o) 2 = A BC = y 3 = A+B(c)+C(c) 2 = A+Bc+Cc 2 . f+e Now the area ABCD = I ydx J -c = i + / - . . Be 2 , Cc 3 , . Be 2 , Cc 3 . A /_!_ _L A r _ "^ 3~ + "2 3" A , - 2AC+ = -{6A+2Cc 2 } = -{A-Bc+Cc 2 +4A+A+Bc+Cc 2 } Imagine now another strip of total width 2c added to the right of BC; the double width being chosen, since there must be an even number of divisions of the base. Then if GH = y 4 and LK = y 6 Area of BLKC = - O or area of ALKD = -{ O = ~{ If a strip of width = 2c is added to the right Area = Z INTEGRATION 143 Or, in general Area = -{first+last+42 even+22 odd}. o Exercises 14. On the Evaluation of Definite Integrals. Find the values of the definite integrals in Nos. i to 7. i. rw 2 . /"* 3 . ./ 1-02 J 1-7 W 4. P 5 -i sin -26d. 5. p.7 6. I s-2*' 1 ^. 7. -- . I x-"dx 8. The change in entropy of a gas as the absolute temperature f 775 dr changes from 643 to 775 is given by I -85. Find this change. * 643 T IT 9. If H = ^ 1 2 sin 6d6, find the value of H. pJ o 10. The average useful flux density (for a 3-phase motor) i fis*' = B = - B mar sin 6d6. Find B in terms of ~Bmax- itJ _ 12 11. Express sin at cos bt as the sum of two terms and integrate with regard to /. If a is -=? and b is 30, what is the value of the integral between the limits o and T ? 12. If h = find h. g J RI r 3 13. Given that EI^ = ^-^-P*. Also that ^ = o when a^r 2 22 d# x = I, and y = o when .# = o and also when .* = /; find the value of P and an expression for y. wx 2 M d'ty dy 14. If M = - , T = ET~-,, ~-= o and also y o when x = I, 2 I dx 2 dx find an expression for y. 15. Given that M - -(--* 2 )-K, ^ = f? Also ^ = o when 2 \4 / IE dx 2 dx I I x = -, and y = o when x = - ; find an expression for y. (The case of a fixed beam uniformly loaded.) 16. Find the value of J (/# -**)***. 144 MATHEMATICS FOR ENGINEERS 17. Evaluate 3 I jX(l z zlx-\-x z )dx, an integral occurring in a beam J "2* problem. 18. If Q = / qdx and q = ; ; -wx, find Q, the total horizontal J ^ i+sin <t> thrust on a retaining wall of height h, w being the weight of i cu. ft. of earth, and <J> the angle of repose of the earth. 19. Find the area between the positive portion of the curve y = 3# 4# 2 +n and the axis of x, and compare with the area of the surrounding rectangle. /15-8 pdv when pv 1 - 37 = 594. 4*6 d i/ ^ d'V 21. If ~ z = 6* 1 ' 4 \ : -,- 10-5 when x = i, and y = 14 when Q/X X dsG x = 2, find an expression for y in terms of x. 22. Evaluate / - fi Xf J 1-47 13 $X 23. Find the value of n, given by the relation n = I - -. . J n t r ,'r 24. The total centrifugal force on a ring = / 1 - '- ^ ; find J E2 -t^i an expression for the force. 25. The area of a bending moment diagram in a certain case was f'/i 3 \ given by J [-at -- -,)da', find the value of this area. 26. H, the horizontal thrust on a parabolic arch, i Find an expression for H. 27. The work done by an engine working on the Rankine cycle with steam kept saturated = | l dr. j \ T Find the work done if the temperature limits are 620 F. and 800 F. (both absolute), and L = 143771-. 28. Evaluate T2.4 J [>- 5 * sin (25 29. Find the value of n, the frequency of transverse vibrations of a beam simply supported at its ends and uniformly loaded with w tons per foot run, when the equation of the deflected form is y = and Ci , ydx. INTEGRATION 145 30. From Dieterici's experiments we have the following relations If s specific volume of liquid ammonia and c = specific heat of liquid ammonia then for temperatures above 32 F. c i-n8 + -ooii56( 32) and s = / cdt. J o Find s when * = 45 F. 31. If p = ^-j\s 2 -x z }dx, Q being the leakage of fluid past a WJJ o well-fitting plug, find its value. 32. The total ampere conductors per pole due to the three windings /- / A J in a railway motor CV2 / Ai sin -rdx. 2 Jo* I Evaluate this integral. 33. For a viscous fluid flowing through a narrow cylindrical tube of radius v, the quantity Q is given by the formula O prc/fo 2 1* 2 /* where /* is the coefficient of viscosity. Find the value of Q. CHAPTER VI FURTHER METHODS OF INTEGRATION BY the use of the rules enumerated in the previous chapter it is possible to perform any integration by a graphic method and the integration of the simpler functions by algebraic processes. Whilst the graphic integration is of universal application, it at times involves much preliminary arithmetical work, which it is tedious to perform, so that it is very frequently the better plan to resort to a somewhat more difficult, though shorter, algebraic method. For the more complex functions, then, a choice has to be made between the two methods of attack ; the fact being borne in mind that only in cases where definite integrals are concerned does the graphic method of integration compare favourably with the algebraic. It is therefore advisable to introduce new processes and artifices to be employed for the algebraic integration of difficult functions; and whilst it is not absolutely essential that all these forms should be remembered, it is well that the various types should be considered, so that they may be recognised when they occur. It is impossible to deal here with every kind of integral likely to be encountered ; all that can be done is to develop the standard forms which cover a wide range, and to leave them to suggest forms for particular cases. Integration by the Aid of Partial Fractions. Many com- plex fractions can be split up into simpler or partial fractions, to which the simple rules of integration may be applied. Thus if we Q/ OQ are asked to integrate, with respect to x, the fraction 2 _ ~ , we soon discover that we are unable to perform this operation with only the knowledge of integration acquired from the previous chapter. If, however, we break the fraction up, in the manner explained in Part I, Chap. XII, we find that the integration resolves itself into that of two simple fractions. 146 FURTHER METHODS OF INTEGRATION 147 Thus , 8 *~ 3 ? o = - h- 4 - (see Part I, p. 453). 2 x 4 2x 7 Hence- f**^?** = f- 2 - d*+ l J 2x 2 15^+28 J x 4 J 2# 7 = 2 log (* 4) +4 log (2* 7) + log C = log (*- 4 ) 2 +log (2*- 7 )2+log C = log {C(*- 4 ) 2 (2*-7) 2 }. [Note that log C may be written to represent the constant in place of C alone ; and it can then be combined with the other logs.] /dx ~i 2 x z a 2 i A B *2_ a *- (*_ a ) + Equating numerators Let ^ = a, then i = A(2a)+o and A = . 20 Let x = a, then I = o + B( 2a) and B = -2 J i = JL(_J_. _i_l x*a z 2a\xa x-\-a' _, /" dx i f f dx f dx 1 Hence -= = i / J x 2 a z 2.a\-' xa J x+a) = ^{log (^ ) log (#+a)+log C} = ^; lo g This is a standard form. A rather more general result may be deduced from it. Example 2. To find J , Let (x+a) = X - f -T= -. " / g A Ilcll"" r / . \ A v a ^ !*-.- A , i** ~r~ fc* i j rf^r Explanation. x+a = X (X+6) C(^+a &) and thus for dx we may write rfX. 148 MATHEMATICS FOR ENGINEERS Integration by the Resolution of a Product into a Sum. A product cannot be integrated directly; but when the functions are trigonometric the product can be broken up into a sum or difference and the terms of this integrated. Before proceeding with the work of this paragraph the reader would do well to study again pp. 273 to 286, Part I. Example 3. Find the value of /4 sin 5^.3 cos2ldt. 4 sin 5^.3 cos 2t = 12 sin 5^ cos 2t = 6 X 2 sin 5* cos 2t = 6{sin 7/+sin 3*} (cf. p. 286, Part I). Hence /4 sin 5^.3 cos 2tdt = 6[f sin jtdt+fsin $tdi\. = 6J cos jt cos 3/+C J = 6C cos 7^2 cos Example 4. Find /sin 2 xdx. cos 2.x = i 2 sin 2 x, so that sin 2 x = i -cos 2* (cf> p 2g = -5* -25 sin2*+'5C. Example 5. Find f ta,n 2 xdx. We know that sec 2 x = i + tan 2 x. .'. fta,n 2 xdx =f(sec z xi) dx fsec z xdxfi dx. = tan xx-\-C. Integration by Substitution. At times a substitution aids the integration, but the cases in which this happens can only be distinguished after one has become perfectly familiar with the different types. y- is a type to which this method applies. In this fraction it will be observed that the numerator is exactly the differential of the denominator. Hence if u be written for the FURTHER METHODS OF INTEGRATION 149 denominator, the numerator may be replaced by du, so that the integral reduces to the simple form I , i. e., log w+log C. J u For if or Hence u = ax z +bx-\-c du -j- = 2ax-\-b dx du = (2ax-\-b)dx . ((^x+b^x = fdu J ax z -}-bx-\-c J u = log w+k>g C = log Cu = log C(ax*+bx+c). In many cases integration may be effected by substitution of trigonometric for the algebraic functions ; and Examples 6 to 10 illustrate this method of procedure. Example 6. To findyVa 2 -X FIG. 36. Let then x a sinw, as illustrated by Fig. 36 a 2 x 2 = a 2 a 2 sin 2 u = a 2 (i sin 2 w) = a 2 cos 2 u and Va 2 x* = a costt, as will be seen from the figure. A , dx d(a sinw) Also -,- -*s - = a cosu du du i. e., dx = a cosw . du. J'-\/a 2 x 2 dx fa cosu . a COSM du = a 2 f cos 2 u du a? = f(i + cos2ii)du, since cos 2A = 2 cos 2 A i = ( u-\ sin 2M + CJ. Although this result is not expressed in terms of x, it is left in form convenient for many purposes. To express the result in terms of x x x sin u = -, so that u = sin- 1 - a a and also cos u / a 2 _ x % = \f - = . n& 150 MATHEMATICS FOR ENGINEERS Hence - sin aw sinw cosu = - x - Va 2 x 2 z a a a z -x*dx = X sin- 1 - +(T X ^2 Va 2 -* 2 +K V 2 / \ 2 = - sin- 1 -+- Va^^+K. 2 a 2 Example 7. To find J ^^^ Let AT = a sinw i. e., u = sin- 1 - a rf# ^(a sin u) then -j = -- T - - = a cosw du du also Va z x 2 = a cosw, as before. /" dx fa cosu.du __ . J Va*=^~ J~a~^r-J ldu = u+C = sin- 1 -+C. a Example 8. To find /"- X In this case let AT = a sinh u, i. e., u = sinh" 1 - a dx d , . . . then j- = j- (a sinhw) = a coshM. du du ^ Now cosh 2 u sinh 2 u = i (cf. p. 291, Part I) and thus cosh 2 w = i +sinh 2 w or a 2 +Ar 2 = a 2 cosh 2 ^ and Va 2 +# 2 = a coshw. [ dx fg coshu.du _ r -, , r J Va*+^~* ~ ' a coshtT ~ J u = sinh- 1 *-+C. Referring to p. 298, Part I, we see that , . x . (x+ Vx*~d*} cosh" 1 - = logi - a & l a J and also sinh- * = I ~ - sinh- 1 *+C 2 a or FURTHER METHODS OF INTEGRATION 151 r fl x Example 9. Find the value of I - ._ and thence the value of ' vx 2 a 2 f dx J Dealing with the first of these X let x = a coshw, i. e., u = cosh" 1 - a dx . , j = a sinhw du or dx = a sinhw du and x 2 a 2 = a 2 cosh 2 w a 2 = a 2 (cosh 2 w i) = a 2 sinh 2 ** fasinhudu sinhw f dx f dx fasin J Vx 2 a z ~ J a sinhw J a si = fdu u+C cosh" 1 -+C To evaluate the second integral, let x-\-a = X then dx = ^X and J = f Example 10. Find the value of I 0-77 ~a J Ct ~j X The substitution in this case is a tanw for x x i. e., x a tanw or w = tan -1 - dx d . . 2 then j- = j- (a tanw) = a sec^w du dx ^ and A- 2 +a 2 = a 2 tan 2 w + a 2 = a 2 (i+tan%) = a 2 sec 2 ^ sec 2 w rfw i r , = - /aw /" ^ /"a s _ ' ' J a?+x* ~ J a 2 sec 2 w ~ = -M + C a = tan- 1 - + C. 152 MATHEMATICS FOR ENGINEERS f dx By an extension of this result such an integral as I J may be evaluated; for let x-\-a = X dX d(x+a) then -s = v , r ; - = i dx dx i. e., rfX = dx hence ( dx - f ** X - ' tan- 1 X 4-C J x + a z + b* ~ J X^+F 2 ~ b tc F^ i. e., I tan- 1 ^~+C. b b The following examples are illustrative of algebraic substitution or transformation. r Example n. To find the value of I Our plan in this case is so to arrange the integral that the method of a previous example may be applied. z z z Hence ( dx - f - -** __ = [ d * J V2ax-x z J Va*-(x-a) 2 J V 2 -X 2 the change from dx to rfX being legitimate, since , = -, ' = i ctx dx and by Example 7, p. 150, the value of this integral is seen to be f J Example 12. To find J dx . x a = sin- 1 x z a dx In this case the substitution is entirely algebraic. i , , du i Let u then , = x dx x 2 or dx = x 2 du. Then a*+x*S dx f x z du f du !\4 J I !,.! I T \T vx ~3 J ^ [ udu = ~r, FURTHER METHODS OF INTEGRATION 153 To evaluate this integral we must introduce another substitution. Let y = 2 2 +i then -^ = 2a 2 u au or udu = - 9 dy. za z ' Hence the integral - ~ = ~~^ + C + C __ I -4-0 = _ - __ I-C ~ 2 ~ L +c. dz At Example 13. The equation -,. . occurs in the statement dt \/ j2 j of the mathematical theory" of fluid motion, which is of value in connection with aeroplane design. Solve the equation for z. To obtain z from -^ we must integrate with regard to t; and to ctt effect the integration let u t 2 i, so that -^ = , . - = zt du or dt r. 21 , Atdt fAtdu [Adu I hen z = 2U* or AVt 2 i + C. Many difficult integrals of the form / can be evaluated J x(a+bx n ) by the substitution z x~n. For if z = x~n log z = n log x d log z _ d log x j ^ 3 ' dz dz i _ J. log x ^ dx ^ jj ^ ~3~ z dx dz i i dx 154 MATHEMATICS FOR ENGINEERS The integral f x(a + bx ^ thus reduces to -!j ( _, the value of which is log (az-\-b) or log / , , A na ' na \a-\-bx n J /dx For x* write z~ l , so that in comparison with the standard form n = 7. _i ~ 28 I0g Example 15. Find the value of / r . J (i 2x)* It will be observed that the denominator is a surd quantity; and in many such cases it is advisable to choose a substitution that rationalises the denominator. Thus in this case let u* i 2X. ,_, QA& Ct/14/ GLIfi Ctr'VC ~dx = ~du X dx = 2U J and -, (i 2x) = 2 du so that 2M j- = 2 or dx dx iu z (i u z \* Also i 2X = u-, whence - = x and x* (-- j . 2 \ 2 / Expanding by the Binomial Theorem x* ~ (i 4 2 +6M 4 4 6 + w 8 ). /" A;*^ Hence J (l-2^)i - _L /"(i 4 M2 +6M 4 4M 6 +M 8 )X udu ~ i6J u i ( 4U 3 6u 5 ^u 1 . u* = ~^( u - T + T 7 + 4 tf ,^_i- 3 5 7 16 which result could be further simplified if desired. FURTHER METHODS OF INTEGRATION 155 The next example introduces the substitution of an algebraic for a trigonometric function. /dx -. sm x Since sin 2 A = 2 sin A cos A, then sin x = 2 sin - cos - . 2 2 f dx if Hence : = - / sm x 2 / . J J Cl Now let then or dx i dx . x . X XI sm 2 .1 -' sin - cos - X COS 2 - 2 2 X 2 COS - , 2 9 #J sec 2 a* I 2 ~~ 2 X tan - *7 u = tan 2 du i x - = sec 2 dx 2 2 sec 2 - 2 sn x 2 x sec 2 . = J|' = log +C = log tan -+C. Integration by Parts. When differentiating a product, use is made of the rule d i x du . dv dx (u ^ =V dx +U dx {Refer p. 70.! If this equation be integrated throughout, with respect to x uv = fvdu+fudv or, transposing fudv = uvfvdu. Many products may be integrated by the use of this rule. 156 MATHEMATICS FOR ENGINEERS Example 17. To find f^x.e x dx. Let = 4#, i- e-, du = ^dx and let dv = e x dx, i. e., v = e x . Then f^x.e x dx = fudv uvfvdu Example 18. Find f^.e^dx. Let u = 5# 2 , i. e., du = loxdx and dv e* x dx, i. e., v -e 4 *. 4 Then f$x 2 .e* x dx = fudv = uvfvdu = sx* . T e* x r 4 M = ^x 2 .e* x 5 I xe ix dx 4 -2.1 ^-r r A-TJ i n f 1 in rwhere u = x -i Now fxe* x dx = x.-e tx e* x dx 4 J 4 [ and v - VJ = ^.e 4a; -4e 4a: 4 16 + C Example 19. To find Te * sin (bx+c)dx and also fe^ cos (&Ar+c)^Ar. [The two integrals must be worked together.] Dealing with the first, which we shall denote by M Let u = sin (bx-\-c), then du b cos (bx-\-c}dx and dv e^dx, so that v fe^dx = e *. Then M = - e sin (6Ar+c) - e ax b cos (&#+c)d* flS ^ & = - e^sin (6^+c) fe cos (bx+c}dx N ......... (i) (I (t where N stands for the second integral whose value we are finding. FURTHER METHODS OF INTEGRATION 157 By developing this second integral along similar lines we arrive at the value N = -e a:r cos (bx+c) + -M. ...... (2) We have thus a pair of simultaneous equations to solve. Multiplying (i) by b and (2) by a and transposing b b z 6M = - e ax sin (bx+c) -- N a 'a feM = e ax cos (&#-|-c)+aN. Subtracting o = e ax f~- sin (bx+c) -\-cos (bx+c) \ N(-4-a) \__Qt J \ fl ' t, -NT nr rb sm (&#+c)+a cos whence N == e ax \ ' - z , a u jit. . HT /r^R* sin (&#+c) 6 cos (fof+c)~i and, by substitution, M = e ax \ a 8 +6 8 y^ao; s i n (bx-{-c)dx = 2 , 3 [a sin (bx-\-c) b cos ~ and ye^ cos (bx+c)dx = 2 a [6 sin (6^+c)+a cos Example 20. An electric current i whose value at any time t is given by the relation i = I sin pt is passed through the two coils of a wattmeter; the resistances of the two coils being R x and R 2 respectively, and their respective inductances L x and L 2 . Then to find the separate currents in the two branches it is necessary to evaluate the integral fQe*'dt where P = and Q = J-^TT~ sin pt+-. ^ cos Evaluate this integral. j j Q = f rT~(Ri sm pt + pLi COS pt) = = where c = tan- 1 ^~ (see Part I, p. 276) ** or Q = M sin (^>^+c), where M = T-^ Then fQedt = fe^M sin (pt+c)dt = Mfe sin (pt+c)dt and this integral is of the type just discussed; its value being [P sin (pt+c}p cos and in this form it is convenient to leave it, since in any numerical application it would be an easy matter to evaluate P, M and c before substituting into this result. 158 MATHEMATICS FOR ENGINEERS Some miscellaneous examples now follow, involving the use of the methods of this chapter. Example 21. Find the value of Let . e., 5 = Let x = 5, then 5 = o 2B B = -2-5. Let # 3, then 5 = ^_o._B x+5 *+3). A = 2-5, i. e., = 2-5 2-5 p J j x 5 ^# * 2 +8*+i5 x+3 x+5 = ft 2-5 dx f* 2-5 dx = 2-5. flog (^+3) log = 2-5 [log 5 log 7 log 4+log 6] = 2-5 [1-6094 1-9459 1-3863 + 17918] = 2-5 X -069 = '1724. As an alternative method of solution, the graphic process of integration possesses certain advantages in a case such as this. It might even be advisable, in all cases of definite integrals where the algebraic integration involves rather difficult rules, to treat the question both algebraically and graphically, the latter method serving as a very good check on the accuracy of the former. f 2 In this example / where 5 x - V = / J i =51 ydx -' hence it is necessary to plot the curve y = -. . . . and find the ( X ~r3)( x ~r5) area between it, the axis of x and the ordinates through x = i and x = 2. The table for the plotting reads X i 1-2 1-4 1-6 1-8 2 y 04167 03841 0355 03294 03064 02857 and from these values the curve AB is plotted in Fig. 37. FURTHER METHODS OF INTEGRATION 159 The sum curve for AB is the curve CD, the last ordinate of which, measured according to the scale of area, is -0095. This figure is the area between the curve AB, the ordinates through x = i and x = 2, and the base line through y = -025; and hence the full area under the curve AB = -oo95 + area of a rectangle -025 by I, i. e., -0095 + -025 or -0345. dx = ' 345 Thus = 5 X -0345 - -1725. 040 y 035 03O 05 D Sum Curve polar distance '8 010 OO95 008 OO6 OO4 002 o FIG. 37. Graphic Integration of Ex. 21. dx Example 22. Find the value of I 2 . XT Now f dx i . _ (x a) , I -5 , = log C 7 ( (see Example i, p. 147) / # 2 a 2 20 (x-\-a) ' cM) i i , V 2/ rf* i r ^ T r^r a^ A i *-^l *"& *1 / = log V 12 & or (ix+3) 160 MATHEMATICS FOR ENGINEERS Example 23. Find the value of I 4 ----- , dx. J 4 gx 2 #+i2 gx In this case the numerator is of the first degree in x, whilst the denominator is of the second degree. Also we notice that the derivative of the denominator is i8# 5, and the numerator is 4(18^5). Thus the derivative of the denominator and the numerator are alike except as regards the constant factor 4. Hence the substitution will be u for gx 2 If u = 9# 2 5*4-12, -r- = i8# 5 or du = (i8x$)dx (IX so that (j2X2o)dx = 4(i8x5)dx = ^du f 72* -20 J tx 2 :=4 U Example 24. To find the value of I This is evidently of the type I = 4 (loge 418 loge 136) = 4 (loge 4-18 loge 1-36) = 4 (I-4303 3075) = 4-49- dx dx (^4 for so that a = 3 and b = V&. r dx i.i x-\-a _ . , _ = r tan- 1 \-C (cf. Example 10. p. 151) J x z +6x+i$ b b Example 25. For a single straight wire at a potential different from that of the earth, if v radius of wire in cms., / = length of wire in cms., a surface density of charge in electrostatic units per sq. cm., then the potential P at any point on the axis of the wire due to the charge on a length 8x is given by ft dx so that the potential at the middle point = mrffj l , z - Evaluate this integral. FURTHER METHODS OF INTEGRATION 161 /dx , Va 2 +# 2 ;. f^== = log(* + V * 2+r *\ (cf. Example 8, p. 150) P m = 27WO- r (' + J*:. i ir- = 27Wo- log- \/5+*-; or TT^O- log- * where d is the diam. of wire. The following example involves the use of three of the methods of this chapter. Example 26. Find the value of /Vs ^-- L ; i- I ( x \~ 'ZX ~\~ 1 1 \X~\~ ^ ) The fraction under the integral sign should first be resolved into partial fractions. t. e., Let x = 3, then ii = C(9 6+7) = loC i. e.. 10* Values of A and B can be found by equating coefficients of x and also those of x 2 . By equating coefficients of x z o = A+C and hence A = - 162 MATHEMATICS FOR ENGINEERS oo 22 By equating coefficients of x 5 = 3A + B+2C = _- + B . 10 10 10 II *+3 i f n*+39 i Hence the fraction- We can make the numerator of the first of these fractions into some multiple of the derivative of the denominator; thus The derivative of the denominator and the numerator = and if u then du = 2(x+i)dx and = du-\-28dx frr. J (x i - loJ x z +2x+j x+3) iidx i / [(iix+3Q)dx fi IO\J X z + 2X+7 f X i_(fii, f 28dx ~~ 1O\J 2U J 28 ^4-1^ \ rx 2 +2X+? / V6 - g log (x* + 2x + 7 ) + 6 tan- -~ log or log (*!?*7).** .14 \ rx2X? -, =^ 2 +2^+i + 6 } L^^ I 22j V6 Example 27. An integral required in the discussion of probability /-co is / e~ x *dx. Find a value for this. J o /GO e~ x *dx. Replace x by ax and thus dx by adx. Then I = I e-^&adx. J o FURTHER METHODS OF INTEGRATION 163 Multiply all through by e~ at . f Then Ie~ at = I e-**()-+&').adx and integrate throughout with regard to a; thus /<*> rx=x> ra=<x> le-a'da = e-a^+x^adadx. J x=~Q J o=0 i~<*> rco r> r But / Ie-a'da = 1 1 er^da = 1x1 since / e~&dx and / J ^0 J ' have the same value, hence I 2 = /*" ' ** e-+ .' x=Q J o=0 (i) The value of the double integral on the right-hand side will be found by integrating first with regard to a and then with regard to x. Dealing with the " inner " integral /<j=oo a=0 let A = a 2 , then dA. = zada, and let M represent Then P~ e -<K l +'V . ada = f J 0=0 J o 2M\ /o o and oo also i / _ AM \ since the limits for A are 2M / _ _ ~2M."> ~ 2M- Referring to equation (i) and substituting this value therein 1 / \ = - /tan" 1 x ) (cf. Example 10, p. 151) = - (tan- 1 oo t 2 v I/7T \ 7C = -I -- O) = -. 2\2 } 4 = - (tan- 1 oo tan- 1 o) 2 v As an extension of this result it could be proved that- Reduction Formulae. Many of the exceedingly difficult integrals which arise in advanced problems of thermodynamics, 164 MATHEMATICS FOR ENGINEERS theory of stresses, and electricity may be made by suitable sub- stitutions to depend upon standard results obtained by a process of reduction. To grasp thoroughly the underlying principles on which the process is based, it is well to commence with a study of the simpler types. 7T 7T [z /"a We desire to evaluate the integrals I sin"0 dQ, I cos"0 dQ -' o -'o and I sin m cos n dQ, where m and n have any positive integral J n values. Taking the case in which n = m = o, we have the results IT fz reducing to the form / idQ, the value of which we know to J . 7T be - 2 If m = n = i ir w /"2 / \ 2 I sin 0^0 = (cos0) = (cos 90 cos o) = I . . (2) Jo /o IT If I Z cos dQ = ( sin 0Y = (sin 90 sin o) = i ... (3) J o /o from which pair of results we may say 7T 7T 7T / 2 sin dQ = r cos dQ = ] sin ( ?r 0)^9 / / / \ 2 / or more generally j a f(x)dx = j*f(a- X )d X (4) result of great usefulness. 7T W If Also \ sin cos ^0 = - \ sin 20^0 = I cos 20 ) ./o 2./o 2X2\ /o i , , = (COS 7T COS O) 4 =J .... (5) By the process of reduction of powers we may express the integral to be evaluated in such a way that it depends on results (i). (2), (3) or (5). FURTHER METHODS OF INTEGRATION 165 Thus 77 tr ; 77 f 2 sin 2 6 dQ=* I* (i-cos 20)^0 = - I (\dQ- f* cos 20 do] J o 2 J o 2 \_J y I/IP N = - ( O 2 \2 / cos 2 8 rf0 = 2 sin 2 rf0 = -. o Jo 4 and from equation (4) 77 ; f ft Now let n = 3, i. e., we wish to evaluate I sin 3 dQ. J o It 77 77 Then ] sin 3 dQ = \ sin 2 . sin 0^0 = ] (i cos 2 0) sin dQ J o Jo Jo IT = I 2 (I M 2 ) rf 'o (o and ^ being the limits for J u being written for cos and du for sin dQ, since -^ = sin and thus -^ = sin 0. 77 17 . , T f "~ ^ / W 3 \ ~ 2 / COS 3 \2 Now I (i w 2 ) aw = I M- = cos ) Jo = o v V 3/e = o \ 3/0 Thus f 2 sin 3 dQ = I * (i u 2 ) du = +- ^o /-o 3 and if n be written for 3 we note that the result may be expressed 77 77 in the form I sin M ^0 = and also I cos w dQ = - J o n Jo n being an odd integer. 77 fz Let n = 4, then I sin 4 ^0 is required. J o Now sin 4 dQ = sin 3 . sin QdQ = udv, where u = sin 3 0, and dv = sin dQ or v = cos 0. 166 MATHEMATICS FOR ENGINEERS 7T 7T 7T Hence ] sin 4 dQ = (cos 6 sin 3 0Y 1 2 cos . <Z sin 3 J / J 7T 77 / \ 2 /*2 = ( cos sin 3 ) + I cos 8 . 3 sin 2 cos dQ \ /o J o 7T = 0+3 1 sin 2 cos 2 dQ J o 77 / \2 since ( cos sin 3 ) = (o X i) (i X o) = o. 'o 7T 7T Now I sin 2 cos 2 ^0 = ] sin 2 0(i sin 2 0)^0 Jo J o = (sin 2 0-sin 4 0)<f0 o It 77 IT Hence \ sin 4 dQ = 3 ] sin 2 dQ3 ( 2 sin 4 dQ J o J o J o [I [I or 4 sin 4 ^0 = 3 1 sin 2 dQ J o Jo 7T 7T f 2 q /" and I sin 4 dQ = ^ sin 2 dQ. J o 4J o We have thus reduced the power by 2, and knowing the result 7T If for / sin 2 dQ, we can finally state the value for / sin 4 dQ. Jo Jo 7T TVinc cin^fl /7A _ V _ ^ /^r- V S/ _ lllUo oill U Ct-U ^ /\ -^ x- Ul x\ /\ J 4 4 16 422 fi (n i) ( ^ or I sin ^0 = '- X, * - , n being an even integer. J Q W (W 2j 2 In like manner it could be shown that 7T 77 /<j tf T / sin 5 ^0 = - / sin 3 dQ. Jo 5 J o Thus sin 5 0^0 = ^x" or ~ 5 5-2 5-3 FURTHER METHODS OF INTEGRATION IT which is of the form f 2 sin"0^6 = ^^ . ^ J n n2 n being an odd integer. Similarly JT It JO 6 lfZ . 61 6 3 6 5 TT sm 6 6 dQ = - I sin 4 ^ = ^ X z - X - X - 6 ; 6 62 64 2 5.3.1 TT or B-^ x - 6.4.2 2 which is of the form OBfl A V H* V . . y . o n(n 2)(n 4) 2 w being an ;<? integer. Summarising our results sn = cos = ^o n if is an evew integer. P sin w ^6 = P co S *0 dQ = etza^K-nSj^^a 7 Jo n(n2)(n4) . . . I if is an o^ integer. JT /2 sin 9 ^</^. o In this case w = 9 and is odd. Hence sin 9 d9 = / . su J o 8.6.4.2 _ 128 9.7-5-3 315 w fz Example 29. Evaluate / cos 10 dd. J o Here n = 10 and is even. Hence Jo . . 10 .8.6. 4. 2 2 512 Example 30. The expression I ^3 dt gives the theo- 7C-|~4-' * retical thrust on a plane moving through air. Evaluate this. V/=i = *t z i-~ = tVi-sin 2 u = t if sin u is written in place of -. 168 MATHEMATICS FOR ENGINEERS i d sin u dtr- 1 dt~ l dt Then since sm u = - t~ l ^ = -3 = -rr X T- or cos u = -5 j- t 2 du du cos u whence dt = = ^ . Again, when t = x> - = o i. e., sin u = o or u = o t and when t = i 7=1 i- &> sin u = i or u = -. t 2 Hence TT It f 1 Vt 2 !,, f 2 cos u sin'u cos udu f 2 , TT is <# = I = r- = = / cos 2 waw = . J ao t 3 Jo sinttsm 2 w Jo 4 Thus -^ / '-^ -8 = -^F- X -- = -^-. 4 ^ ~r~4 We can now direct our attention to the determination of the r\ value of I sin m cos"0 dft, where m and n are positive integers. ' o It is convenient to discuss a simple case first, viz. TT [z sin 2 cos dQ. ' o T , . on , , du d sin 2 d sin Let u = sm 2 so that -^ = 5^-5 X ^~ ^0 rf sin <?0 = 2 sin cos and du = 2 sin cos ^0 also let dv = cos <0 so that v = sin 0. Then, by integrating by parts 7T IT 7T f 2 sin 2 cos ^0 = (sin 3 0V f 2 2 sin 2 cos ^0 J o \ /o J o /2 / \2 sin 2 cos ^0 = ( sin 3 ) = i o /o n [2 I or sin 2 cos ^0 = - ^o 3 j . , , , ... and might be written as . m+n FURTHER METHODS OF INTEGRATION 169 Example 31. H, the horizontal thrust on a circular arched rib carrying a uniformly distributed load w per foot run of the arch, is obtained from ( X -si V * sin 2 e (cos 0--S66) dQ 2R 3 I 6 (cos 866) dQ J o if the span is equal to the radius of curvature (see Fig. 38). If w = -5 ton per foot, and the span = 60 ft., find the value of H. Here w = -5, R = span = 60. IT [ 8 ( J -sin 2 6) (cos 6- -866) dQ Hence -5X60 (cos 0- -866) <Z0. FIG. 38. Circular Arched Rib. Dealing with the numerator separately, as this alone presents any difficulty (^ sin 2 0) (cos -866) = - cos -2165 sin 2 cos0+-866 sin 2 0. \4 4 Then [(- sin 2 0] (cos - -866) dQ = l~ cos 0d0 -f-2 165 d0-/sin 2 cos dQ+f-866 sin 2 dQ but, as proved on p. 148, /sin 2 dQ = ~fidQ-fcos2Q dQ = sin 20 2 4 and as proved on p. i68ysin 2 cos dQ - sin 3 170 MATHEMATICS FOR ENGINEERS w thus r (i-sin 2 0) (cos 0--S66) dQ J o H ff TT n- TT = (-sin 0)~- (-21650)-* (sin 3 0)V 4 33 (-^ sin 28)* -o)-f (<. S ).-o) + . 43 3 (Hx -866-0+0) = -125 -1134 -04I7+-2267 -1875 == -0091. Dealing now with the denominator IT n (* (cos 8 -866) dQ = (sin 0- -8660Y = -5- -866 X ~ = -5 4534 = -0466. J o ' TT 15 X -0091 Hence H = -* - ^ = 2-93 tons. 0466 yj Carrying the investigation a step further, let us discuss the case of I sin 2 cos 2 dQ. Jo. T It IT IT f 2 sin 2 cos 2 dQ = I * (i cos 2 0) cos 2 dQ = f 2 cos 2 dQ 1 2 cos 4 dQ Jo J o Jo J o and from the previous result this value = ^ - 4 4.22 _ I 7T I I 7T ~82 4*2'' 2 This result might be regarded as obtained by first reducing the power m by 2, and next that of cos 2 by 2. Thus for the first step 7T 7T f 2 sin 2 cos 2 rf0 = ^^ f 2 Jo 4 7 cos 2 ^0 and for the second step IT idQ f 2 cos 2 ^0 = ^^ f 2 Jo 2 J 21 TT ni TT - -,-r. _ /"\T* \JL 2 ' 2 W+ w 2 m, for the second integral being zero. FURTHER METHODS OF INTEGRATION 171 rr ft In a similar fashion we might reduce I sin 4 0cos 3 0^0 as ' o follows : First reduce m by 2, then 1C IT ! * sin 4 6 cos 3 ^0 = 4 -^ (* sin 2 cos 3 dQ. J o 4+3 J o Next reduce the power of sin 2 again by 2. 7T T Thus P sin 4 cos 3 dQ = f* . *^ I * Jo 4+3 2+3 J cos 3 dQ. Now reduce the power of cos 3 by 2, and remember that m is now = o. Then sin'e cos'erfe _ 1=5 . ?=I . 3=1 f* o 4+3 2+3 3+o^o . 7-5-3 /2 In the evaluation of the integral I sui"'0cos n 0^0 we thus * o proceed by reducing by 2 the powers of m and n in turn until they become i or o. The various cases that arise are (a) m and n both even : in which case the final integral is f * rf0 = ? ; 2 (6) w and w both odd : in which case the final integral is IT [2 ! sin cos dQ = -. J o 2 (c) m even and n odd or vice-versa : in which case the final a it fz [2 integral is either / cos 0^0 or I sin0^0, the value of either J o J o being i. The results for the three cases can be thus stated (a) m and n both even Psi J sin m o (n-i)(-3)x . . . i * n(n 2)x ... 2 2 I 7 2 MATHEMATICS FOR ENGINEERS (b) m and n both odd . ( m _ I ) x (w 3)x ... 2 sm w 8 cos n 6 <ft = / , v . . ' * : :' ? ; r (m+) X (m 2+n) X ... (+3) (tt-i) x(n-3) ... 2 i X (+I)X(-I)X . . . 4 X 2 since, after reducing the power of sin m by 2 at a time, we must be left with sin cos n 0, so that the value of m to be used in the reduction of cos M must be taken as i. (c) m even and n odd fl l I sm w cos re ^0 = .' n (m+n) X (m2+ri) X ... (w+2) -ix-3)X . . . 2 X 7T ft Example 32. Evaluate I sin ? cos 10 rf0. / This is case (a), i. e., with both m and w even. 7T Hence 2 sin0 cos"0 dQ = - -' - -- -'^- X f 2 si / - - -- - .10.14.12 10.0.6.4.2 2 1179648' 7T [z Example 33. Find the value of I sin 3 cos 5 0d0. J This is case (6), i. e., with both m and n odd. 7T Hence / 2 sin 3 cos 5 dQ = \ X J X - = - . Jo 8 6.4 2 24 IT Example 34. Find the value of / 2 sin 7 cos*0 ^0. / This is case (c), but with m odd and n even. it Hence T sin 7 cos 4 d0 = .^' j'^ X ^ X i = l6 ' n-9-7 5-3 ii55_ The value of the foregoing formulae is found in their employ- FURTHER METHODS OF INTEGRATION 173 ment in the evaluation of difficult algebraic functions, which may often be transformed by suitable trigonometric substitution. Thus to evaluate / ^-^(ix^^dx, known as the First Eu- J o lerian Integral and usually denoted by the form B(m, n), we may substitute sin 2 for x. Then, since when x = o sin 2 must = o and thus = o, and when x = i sin 2 must = i and thus / x m - 1 (ix) n - 1 dx = sin 2n *- 2 0(i si ~dx d sin 2 6 _ d sin 2 6 d sin 6 ~ d sin 6 ' ^0 _ = 2 sin cos = ] sin 2 *- 2 cos 2n ~ 2 x 2 sin cos 0^0 -'o ir fz J n m and this can readily be evaluated. Example 35. Evaluate / x*(i x z )?dx. J o Let x = sin then i x z = i sin 2 = cos 2 dx d sin ft , and Also when x = o sin = o and thus = o and when x = i sin = i = -. Then v 5 I * 4 (i-x z )*dx = f 2 sin 4 cos 5 dQ cos = f 2 sin 4 cos 6 dQ -' J .' 10. 86. 4. 22 Another important result obtained by the process of reduction r* is the value of I e~ x x n dx. This is termed a Gamftia Function; J o this particular integral being the (+i) Gamma Function denoted Thus f e~*xdx = J o /-co and I e-t&^-dx = Y(n) J o the latter integral being also called the Second Eulerian Integral. 174 MATHEMATICS FOR ENGINEERS To evaluate r(w+i) let u = x n and dv = e~ x dx so that v = e~ x and du = nx n ~ l dx. ,00 XOO / Then I e~ x x n dx = ( e~ x x n ) I e~ x .nx n ~ l dx Jo \ /o J o xoo /. = ( er x x n ) 4-n\ e~ x x n ~ l dx. \ /o J o Now when # = oo e"^" = ^ , which can be proved = o I X O and when x = o e~ x x n = - - = o. e /oo / Hence / e~ x x n dx = n I e~ x x n ~ 1 dx Jo Jo or r(-{-i) = nT(n). In like manner it could be shown that T(n) = (ni)T(ni), and so on. If n is an integer it will be seen that we finally reduce to r(i), f i. e., I e~ x dx, the value of which is (e~ y> e) = (o i) = i. Hence r(-fi) = n(ni)(n2) . . . i = |_. This last relation will not hold, however, if n is not an integer, but the general method of attack holds good; and tables have been compiled giving the values of T(n) for many values of n, whether n be an integer or fractional. Thus if an integral can be reduced to a Gamma function or a combination of Gamma func- tions, its complete evaluation may be effected by reference to the tables. /oo. _a^ e~h'dx by the aid of the Gamma function. Let X - ^ then ^ = -f- = * 2 a# a# h 2 , , Y _ 2^^r ~ "P~ A 2 rfX A 2 dK. hdX. or dx = = ; 7=^ = 2 x 2 h\/x. 2\/X' Then f.TW, = /" e - x x^|- = *r.-*X-ta = *xr(?) ^0 / 2 VX 2-' 2 \2J and the value of T ( \ * s VTT^ FURTHER METHODS OF INTEGRATION r > _a* fr _ Hence I e h*dx = - Vie. JO 2 175 On comparing with Example 27, p. 162, where a rather simpler form of the integral is evaluated, we see the great saving effected by the use of the Gamma function. LIST OF INTEGRALS LIKELY TO BE OF SERVICE f(ax+b) to fae^dx fba^dx fa cos (bx+d) dx fa sin (bx+d) dx . n log a = sin (bx+d)+C. = ~ cos (bx+d)+C. fa tan (bx-\-d) dx = , log cos (bx-\-d)-{-C. fa cot (bx+d) dx = ^ log sin (bx+d)+C. = | tan (6*+<Q+C. = cot (bx+d)+C. fa sec 2 (bx+d) dx /a cosec 2 (6*-H) fa cosh (6^+^) to /a sinh (bx+d) dx = ~ cosh (bx+d)+C = T sinh fa sech 2 (6^+^) to = | tanh /a cosech 2 (bx+d) dx = | coth (bx+d)+C. * sin (6^+^) to = - f"c0s (bx+d) dx *= sn sn cos cos f ax \ +/~2 ' J V U X (-T JVx dx dx 176 -MATHEMATICS FOR ENGINEERS r dx i ( bx+d\ . n I ~~'/i~ ~r~j\ == i 1S ( tan +C. J sin (bx+d) b \ 2 / f dx i ]0x+b =-log(**+fc)+C. /tan (bx-}-d) dx = -? log cos (bx+d)+C. /cot (bx-{-d) dx = r log sin (bx+d)+C. r dx i , , x . r I -o-j = - tan * -+C. J 2 +A; 2 a dx 1,1 ^+ , /- = * tan T +c - r-r^i = X log^+C. J a 2 x z 2a f dx i_ , J (x-\-a) z b z ~ 2b dx = sm~ 1-7===, = log ^~r v -* -* j + c. J v x a r dx }' r (ax-\-b)ax i . , J ^tf^bx+d) ~ 2 g (a Tcosec (ax-\-b) dx = - log (tan ' -)+C. ** d \ 2 / / sec (a^+&) <fo = - log ta /^ i , x . . = - sec" 1 -+C. x\x z u z & a f dx , x . r , x r I . = vers" 1 -+C or i cos x -+C. J V2axx z a l x z dx = -y 2 x z + 2 sm __ x e z jVx z -a z dx = ~Vx z -a z - FURTHER METHODS OF INTEGRATION 177 fV(x+a)*-Pdx = I (x+a)V(x+a)*-b*-~ cosh- 1 *^ *2 -a? dx = ~Vx 2 4-a 2 4- sinbr 1 - 2 2 a /V/(*+a)M-6* <& = ^ -5 /" sin 2 ;*; <fa = sin 2^+C. 2 4 . A; I / cos 2 x dx = -4 sin 2x-\-C. 2 4 /* sin m x cos n A; ^ = /" sin m ~ 2 x . cos" ^ dx J m-\-n J m-\-n [ dx J (aTI^ji I ^-*i^ * o /oo * I e^w dx Jo 2 tr T sin 2 ^0 = | 2 cos 2 0^ = "". J o 4 IT -7 s i n m-l ^ , cOS n+1 # r f-L/. sin6 & = cos6 <*0 = ~ ( 2)( 4) ... 2 2 if w is an even integer. sin6 ^0 = cos0 ^0 = *n7- . . 2 fsi Jo . o ( 2)( 4) . . . I if w is an odd integer. (-l)(-3) . . . I 7T X IT r 1 I SI J A if w and w are both even. sin w cos w dQ= y\ /^ (w+w)(w+w 2) . . . (w+3) (w+i)(w i) ... 4 2 if w and n are both odd. 178 MATHEMATICS FOR ENGINEERS It (* I sin'" ' o e cose rfe = -= x = ' ' ' 2) . . . (n-\-2) ( 2) . . . i if m is even and n is odd. Pdv (where pv n = C) = (v^~ n v^ I pdv (where pv = C) = C log e - 2 . = a log p sin" 1 a#*fo = x sin~ 1 a#+;_ Vi a 2 x z -\-C. tan- 1 ax dx = x tan" 1 ax log 20, b f (a 2 -^ 2 ) 1 ^ = -(a 2 -* 2 ) + J 4 3 - sn Exercises 15. On Further Integration. Evaluate the integrals in Nos. i to 18. dx_ 9 f dx f (x-i)dx **' J *x*+6x+2i' 'Jc>x 2 -i8; 4. - | VJ 3 , which occurred when finding stresses in a crane hook. 5. t 95-5 1 (6h}*dh, referring to the time for emptying a tank. 7T 2 /"2 6. ^3~ W 2 dy. 7 - sin 5 ^ cos ' ft ^ 4 A /"sWR/i i . rtN-, _ , ft 9. ^^ . 10> / -Frrl --- sin 6 R 2 cos6d6. sin 2 5^ -' o El \TT 2 11. /4 tan 5/ <tt. 12. /sin- 1 ^r dx. 13. 7T 14. /5e 3:c sin 2Ar rf^r. 15. J ,^. * 2 . 2 . 16. J 5 ^ sin ^r ^. 17. /cos 6 sin 2 6rf0. 18. h = f [-*-&* ,j~,rr & , relating to the J o 2gn 2 (d+kx) 5 flow of water through a pipe of uniformly varying diameter; the diameter at distance x from the small end being = small end diam. -f-**. FURTHER METHODS OF INTEGRATION 179 19. The time taken (t sees.) to lower the level of the liquid in a certain vessel having two orifices in one side can be found from nodh at = 12 + Vh Find this time if the limits of h are o and 10. 20. Express e~ l * as a series, and thence find the value of 2 (* VrcJ o 21. The maximum intensity of shearing stress over a circular F r r i section of radius r = S = j. / 2(r 2 y z )*ydy. If I = , find a simple expression for S. 22. Evaluate the integral I - _____ [Hint. Rationalise the J t vr 4 denominator.] 23. Write down the value of | 2 cos 8 QdQ. J o 24. If the value of log r(i-85) is given in the tables as 1-9757, ra> find the value of I e~ x x l '**dx. -' o w 25. Write down the value of 1 2 sin 2 6 cos 7 0^0. J o 26. When finding the forces in a circular arched rib it was found necessary to find the value of ir 2j 6 R 3 (cos 2 6 Vs cos 6+^)^0. Evaluate this integral. 27. Evaluate the indefinite integral fVzi y( x z dx. 28. The attraction F of a thin circular disc of radius r on a body on the axis of the disc, and distant z from the centre, is given by rdr - where a is the density of the disc, and k is a constant. Find the value of F for this case. CHAPTER VII MEAN VALUES: ROOT MEAN SQUARE VALUES: VOLUMES : LENGTH OF ARC : AREA OF SURFACE OF SOLID OF REVOLUTION : CENTROID : MOMENT OF INERTIA Determination of Mean Values. It is frequently necessary to calculate the mean value of a varying quantity : thus if a variable force acts against a resistance, the work done will be dependent on the mean value of the force; or to take an illustra- tion from electrical theory, if we can find the values of the current and electro-motive force at various instants during the passage of the current, then the mean rate of working is the mean value of their product. The mean value of a series of values is found by adding the values together and then dividing by the number of values taken. If, however, a curve is drawn to give by its ordinate the magnitude of the quantity at any instant, the mean value of the quantity is determined by the mean height of the diagram, which is the area divided by the length of the base. This really amounts to the taking of an exceedingly large number of ordinates and then calculating their average. The area may be measured by the planimeter, in which case the instrument may be set to record the mean height directly, or by any of the methods enumerated in Part I, Chap. VII. A clear conception of the idea of mean values can be gained by consideration of the examples that follow; the first example being merely of an arithmetical nature. Example i. The corresponding values of an electric current and the E.M.F. producing it are as in the table : c o 1-8 3'5 5 6-1 6-8 6-1 5 3'5 1-8 E 9 17-5 25 30'5 34 30-5 25 17-5 9 1 80 MEAN VALUES 181 Find the mean value of the power over the period during which these values were measured. current x E.M.F., and 61-25 16-2 The power is measured by the product thus the values of the power are o 16-2 61-25 125 186-05 231-2 186-05 I2 5 the sum = 1008-2 the number of values = 10 1008-2 hence the average or mean value = = 100-8. A better result would probably be obtained if the values of the power were plotted and the area of the diagram found. Thus in Fig, 39 the base is taken as 10 units (merely for con- venience), and the area is found to be 1013 sq. units. Then the mean height of the diagram, which is = 101-3, is also the mean value of the power; and a line drawn at a height of 101-3 units divides the figure in such a way that the area B is equal to the areas A+A. eoo Rawer -""Mean hcighf Ol E3456789IO FIG. 39. Let it next be required to find the mean value of a function say, the mean value of y when y = 4x z -\-jx$, the range of x being i to 6. We have seen that it is really necessary first to find the area under the curve y = 4# 2 +7# 5 within the proper boundaries and then to divide by the length of the base; and since the relation between y and x is stated, it is possible to dispense with the graph and work entirely by algebraic integra- tion; thus also ensuring the true result, for in reality the mean is automatically taken of an infinite number of ordinates. In the case taken as an illustration, the base is the axis of x, or, more strictly, the portion of it between x = i and x = 6, so that the length of the base = 61 = 5 units, and the area between the curve, the axis of x and the ordinates through x = i f 6 f 6 and x = 6 is given by the value of I y dx, i. e., I (4x z -\-jx 5) dx, 182 MATHEMATICS FOR ENGINEERS so that the mean value (for which we shall write m.v.) is *2+*_ dx m.v. = _[3 +2 J"~ 5 *]'i 5 It is instructive to compare this result with the results obtained by the use of the mid-ordinate rule : (a) Taking 5 ordinates only, we have the values X 4# 2 +7*- 5 y 'i 2* 3t 4i 5i 9+10-5 5 25 + I7-5-5 49+24-5-5 81+31-5-5 121 + 38-5-5 14-5 37-5 685 107-5 154-5 Their sum = 382-5 and the average = 76-5. (b) Taking 10 ordinates, viz., those at x = ij, if, 2|, etc., the values of y are 10, 19-5, 31, 44-5, 60, 77-5, 97, 118-5, 142 and 167-5 their sum = 767-5 and the average = 76-75. Therefore, by increasing the number of ordinates measured, a better approximation is found. The curve is a parabola with axis vertical, and hence Simpson's rule should give the result accurately if 3 ordinates only are taken, viz., at x = i, 3-5 and 6. Thus, A = 6, M = 68-5, B = 181. -.. 6+(4x68-5)+i8i Hence the mean height = - = = 76-83. MEAN VALUES 183 If, then, the law connecting the two variables is known, the mean value of the one over any range of the other can be found by integrating the former with regard to the latter between the proper limits and then dividing by the range; or to express in symbols, if y = f(x), the mean value of y, as x ranges from a to b, is given by fb , y dx f J " b-a Example 2. Find the mean value of e 5x between x = -2 and x = /? g . a m.v. = 7-'2 '5 t-5 2 ~ 5 '* = ^{33-12-272} = 12-16 i. e., if the curve y = e 5x were plotted between x = -2 and x = 7 its mean ordinate would be 12-16 units. Example 3. If C = 5 sin 3^, find the m.v. of C, when , > , . , 27C . (a) t vanes from o to 3 (b) t varies from o to -. Whenever dealing with the integration of trigonometric functions it is advisable first to determine the period of the function, since much numerical work may often be saved in this way. The sine and cosine curves are curves symmetrical about the axis of the I.V. (i, e., x or t, as the case may be) ; hence there is as much area above this axis as below it, if a full period or a multiple of periods be considered. Therefore, regarding the area above the x axis as positive and the area below this axis as negative, the net area over the full period is zero, so that the mean height of the curve,* and therefore the m.v. of the function, must be zero. For the case of C = 5 sin 3/, the period = ^ ^ = coeff <. of / 3 and hence if the m.v. is required for t ranging fiom o to ' . 4^ o to , etc. 3 the result is zero. 184 MATHEMATICS FOR ENGINEERS Hence, whenever the analysis shows that the full period is involved, there is no need to go through the process of integration. In this case, however, the integration is performed for purposes of verification. 27T 27T , _ fa 5 sin -$tdt 3 x 5 / i ,\- (a) m.v. of C. = / * 3 = - - ( cos 3/ J o 271 27T \ 3 - / 1 3 - -o 3 - (COS 27C COS O) 27T V = ('-I) 27U V ' IT /i.\ t r fissinitdt (b) m.v. of C. = / - = Jo n 3 = - (COS 7T COS O) 7T V 5 / \ I0 = *-* ( r i) = . 7T V 7T Comparing this with the amplitude, which is the maximum ordinate of the curve and has the value 5 sin = 5, we note that Mean ordinate for ^ period 10 _ 2 _ Maximum ordinate for \ period ~~ TT X 5 ~~ n ~ '' The average height of a sine curve is always -637 x the maximum height. Example 4. If an alternating electric current is given by the relation C = -5 sin i207T/+-o6 sin 6007^ find the mean value of C. The graph is of great assistance in this evaluation ; and consequently the curves c x -5 sin 1207^, c 2 = -06 sin GOOTT/, and C = c t + c 2 = -5 sin i2O7t/+-o6 sin 6007^ are plotted in Fig. 40. The period of c, = -5 sin I2O7T/ is or >-, and of c = -06 sin 6007^ is I2O7C 6O - 01 - -, so that the period of the compound curve must be ^-. 60O7T 30O 6O From the pievious reasoning it is seen that the mean height of the curve c t = -5 sin i207r/ must be -637 x amplitude = -637 X -5 = -3185; and the mean height of the curve c 2 = -06 sin 6007^ considered over the same period, viz., o to , must be the mean height of the wave A, 120 since the positive and negative areas are otherwise balanced, but this MEAN VALUES 185 must be spread over five times its usual base ; now the mean height of the wave A is '637 X -06, so that the mean height of the curve c 2 -06 sin 6oo7c/ over the period o to -637 x -06 is or -0076. 120 5 Then the mean value of C = -3185 + -0076 = -3261. The nature of the average should be clearly understood ; for it is possible that the same quantity may have two different averages according to the way those averages are considered. Suppose a piston is pushed by a variable force; then the average value of that force might be found by taking readings of it at every foot of the stroke and dividing by the number of readings taken, in 40 X^ IO i- -Oesin gOOTTf FIG. 40. Problem on Alternating Current. which case the average would be termed a space average ; or the force might be measured at equal intervals of time, whence the time average of the force would result. To take another instance : Suppose a bullet penetrates a target to a depth of s ft. ; the average value of the force calculated from the formula Ps = mv* where P is the force, would be a mv space average, but P calculated from the formula Pt= , i. e., the 6 force causing change of momentum in a definite time, would be a time average. To illustrate further : A body starts from rest and its speed increases at the uniform rate of 4 ft. per sec. 2 . Find the time and space averages of the velocity if the motion is considered to take 186 MATHEMATICS FOR ENGINEERS place for 6 sees. {Use the relations s = - at z , v = at and v 2 = 2as. s = - at z = - x 4 X 6 2 = 72 2 2 and if the limits of t are o and 6 then the limits of s are o and 72. To find the time average of the velocity 8 atdt "6^0 ~ /e re I vdt c f __. * J = 12 f.p.S. To find the space average of the velocity V 2 = 20S = 8s or v= V&Vs and the mean value of v = mean value of V8 / VS a rfS i ,-(2 sV 2 Hence the space average = /T/T _ rt ~^ x ^^\^ sl? j 72 = = Example 5. The electrical resistance R of a rheostat at temperature t C. is given by RJ = 38(1 + -004^). Find its average resistance as t varies from 10 to 40 C. (This will be a temperature average.) m.v. = 30 38x33 30 41-8. MEAN VALUES 187 Example 6. If V = V sin qt and C = C sin (qtc), find the average value oi the power, i. e., the average value of VC. VC = V sin qt . C (sin qtc) = {2 sin qt . sin (qt c)} V C = {COS C COS (20* C)} also the period = . Hence the m.v. of VC r [cos c cos (zqtc)}dt ex/-- sin (zqt-c]\ q 20 '\ cos c- sin ( 4TC -c)-o+ sin (-c) j 20 20 V 'J 275 2 20 20 _ VoCo? 27c |~for sin (4* e) = sin ( e)~| A - COS C , / \ * / \ 47T q Land sin ( c)+sin ( c) = o = ~V C cos c i. e., the mean value of the power = one-half the products of the maximum values or amplitudes with the cosine of the lag. This is a most important result. If c = 90, i. e., if the lag is -, then cos c o and the mean value of the watts = - V C X o = o ; this being spoken of as the case of wattless current. Exercises 16. On Mean Values. 1. If a gas expands so as to follow the law pv 120, find the average pressure between the volumes 2 and 4. 2. Find the mean value of e 2 ' 5 v as y varies from o to -4. 3. The mean height of the curve y = 3* 3 +5# 7 is required between the limits x 2 and x +3; find this height. 4. Find the mean value of 2-i8sin](3/ 1-6) as / ranges from 14 to 1-6. 5. What is the mean value of 4-5 +2 sin 6o/, t ranging from o to ? Discuss this question from its graphic aspect. i88 MATHEMATICS FOR ENGINEERS 6. Find the mean value of p, when pv 1 ' 37 = 550, for the range of v from 4 to 22. . 7. The illumination I (foot candles) of a single arc placed 22 ft. above the ground, at d ft. from the foot of the lamp is given by I = i -4 -oid. Find the mean illumination as d varies from J ft. to 10 ft. 8. An alternating current is given by C = -2 sin IOOTT/+-OI sin 300^. Find the mean value of C for the range of t, o to -02 sec. 9. Taking the figures in Question 8, find the mean value of C when t ranges from o to -01 sec. 10. Find the mean value of 5 sin 6t X 220 sin ^t, t ranging from 7C o to -. 3 11. The table gives the values of the side thrust on the piston of a 160 H.P. Mercedes aero engine for different positions of the crank; the positive values being the thrust on the right-hand wall, and the negative values being the thrust on the left-hand wall of the piston. Angle of crank froml top dead centre | Total side thrust) (Ibs.) / O 40 80 IOO 120 140 160 1 80 20O 240 280 300 320 330 o + 210 O -170 -175 -I8 5 IOO +95 +240 + IOO o -30 o 345 360 10 40 80 1 20 160 1 80 2OO 2 2O 240 260 280 290 320 360 +20 o 600 -* 720 -580 -170 + QO + 170 +160 + 150 +50 210 Plot these values (treating them all as positive) and thence determine the mean side thrust throughout the cycle. Root Mean Square Values. A direct electric current maybe measured by its three effects chemical decomposition, magnetic effect or heating effect. An alternating current (A.C.), however, flowing first in one direction and then in the reverse, cannot be measured by either the first or the second of these effects, because the effect due to the flow in one direction would be neutralised by that due to the opposite flow ; hence an A.C. must be measured by its heating effect. The heating effect of a current expressed by the heat units H may be measured by H = where C is the current, so that it will be seen that H oc C 2 . The measuring instruments are graduated to give the root of the mean value of this heating effect, i. e., the square root of the ROOT MEAN SQUARE VALUES 189 mean value of the squares (called the root mean square value and written R.M.S.) ; in other words, the instrument records what are termed virtual amperes, a virtual ampere being the current that produces the same heat in a resistance as a steady or direct current of i ampere in the same time. In place of the measurement of the current by its heating effect, the ammeter might be of the electro-dynamometer type, and in such the instrument records the mean value of C 2 , and the square root of this value is called the effective current. Similar remarks apply also to the measurement of alternating E.M.F. It is therefore necessary to determine R.M.S. values of functions likely to be encountered, and to compare the virtual with the steady. While the determination of R.M.S. values is of greatest import- ance from the application to electrical problems, it is also occasion- ally of use in problems of mechanics ; thus the calculation of what is known as a swing radius (see p. 240) is in reality a determination of a R.M.S. value. In order to convey the full interpretation of the term R.M.S. value, we shall discuss first a simple arithmetical example, then some algebraic examples, leading up finally to the trigonometric functions. Example j. The values of an alternating electric current at various times are given in the table t 01 02 03 04 05 06 07 08 09 i C o 1-8 3'5 5 6-1 6-8 6-1 5 3'5 1-8 o Find the mean value and also the R.M.S. value of the current, and compare the two values. The mean value, as before explained, is the average of the given values and is 3-96. We must now tabulate the values of the squares; thus o 3-24 12-25 25 37-21 46-24 37-21 25 12-25 3' 2 4 and o. The sum of the squares = 201-64 the mean of the squares or M.S. = for we must only reckon the end values as half-values when adding IQO MATHEMATICS FOR ENGINEERS up the ordinates, since the end values belong equally to the sequences on either side. Thus M.S. = 20-16 and the square root of the result = R.M.S. = V2o-i6 = 4-5. This question might have been worked entirely by graphic methods, according to the following plan Plot the values of C to a base of /, giving the curve ABD in Fig. 41 ; find the area under the curve and divide by the base, thus obtaining the mean height; plot also the values of C 2 against those of t, giving the curve EFG. By graphic summation determine the area under the curve EFG and draw the line MM at the mean height of the diagram. Make MN = i unit on the scale of C 2 and on PN describe a semicircle ; produce MM to cut the semicircle in R. Then MR = 4-5 is the R.M.S. value. Thus the mean value = 3-96 and the R.M.S. value = 4-5, and the ,. R.M.S. value 4-5 ratio - - - = -2-i = 1-1 mean value 3-96 factor. ., / this ratio being termed the form a _ FIG. 41. R.M.S. Value of an Alternating Current. Example 8. Find the R.M.S. value of the function 2X 1 ' 5 3* as x ranges from 2 to 5. The square = (2x l ' 5 ^x) 2 = 4 Explanation. dx and the M.S. = 3 _ i ~ 3 = 13-27- Hence the R.M.S. value 5-2 -959)-(i6+2 4 - 3 8-8)] = Vi3'27 = 3-64- No. Log. 5 3-43 959 2 3-43 38-8 6990 3'5 34950 20970 2-44650 5353 2-9818 3010 3'5 15050 9030 1-05350 5353 V5883 ROOT MEAN SQUARE VALUES 191 Example 9. Suppose that an alternating electric current at any time follows the sine law, i. e. C = C sin qt where C is the instantaneous value of the current at any time t, and C is the maximum value of the current. Find the R.M.S. value of the current. As we have already seen, the determination of the R.M.S. value implies that first the square of the function at various times must be calculated, then the mean value of these squares found, and finally the square root of this average extracted. To assist in the study of this important problem, the curve y sin x, the simple sine curve, is shown in Fig. 42, and also the FIG. 42. R.M.S. Value of an Alternating Current. curve y sin 2 x, which is obtained from the former curve by squaring its ordinates. It will be observed that whilst for the curve y = sin x there are both positive and negative ordinates, in the case of the curve of squares all the ordinates 'are positive. Also the period of the curve of squares is noted to be one-half that of the simple curve; and therefore when calculating the mean height of this curve, it is immaterial whether the full or the half period of the simple sine curve is taken as the range. In the case with which we are here particularly concerned the C 2 2Tc square = C 2 = C 2 sin 2 qt = (i cos 2qt), and the period = , so that the integration may be performed either over the range o to - or o to -. 192 MATHEMATICS FOR ENGINEERS Taking the latter range a I q C 2 sin 2 qt dt Mean of the squares = M.S. = q = & f q (i -cos 2qt) dt* 7C Jo 2q oC 2 TTC I .in = * - --- sin 27c o-{ sin o 27T \_ 2 2q J or R.M.S. value = -707 x maximum value i. e., virtual value of current or E.M.F. = -707 x maximum value of current or E.M.F. Hence if a meter registers 10 amps., the maximum current is - , 707 i. e., 14-14 amps., or there is a variation between + 14-14 and 14-14. * In the evaluation of the integral t q (i cos iqf) dt, it should be it ir noted that it can be written q dt I 9 cos -zqtdt, and the value of Jo Jo the second term is zero, because it is the area under a cosine curve taken over its full period. Hence the integral reduces to / 9 dt, and there is no need to say anything further about the second term. Example 10. Find the R.M.S. value of a-\-b sin 4*. For this function the peiiod = -, 4 2 Then S = (a-\-b sin 4/) 2 = 2 +6 2 sin 2 4^ + 2^6 sin 4^ 52 = 2 + (i cos 8t)-\-2ab sin 4^. 7 2 fa 6 a 6 2 Hence M.S. = - I (a z -\ ---- cos 8t+zab sin 4*) 7T J 22 _ TT n 2 F T2 / fc 2 \ T2 fe2 = ~ I \ aZ +2) dt J - dt cos tt + 2a sn VOLUMES 193 2 2 2 R.M.S.= Exercises 17. On Root Mean Square Values. Find the R.M.S. values of the functions in Nos. i to 7. 1. # 3 +2 (x ranging from i to 3). 2. e~" x (x ranging from -i to +-65). 3. 3-4 sin 5*i/. 4. -165 cos (-07 2/). 5. 1-4 tan 2t (t ranging from o to -43). 6. i -14 +-5 cos -8t. 7. -72 sin (3 4/) ; compare with the mean value. o -c- A .LI. t * /R.M.S. value\ , ,, 8. Find the form factor -, ) of the wave \ mean value / e = E x sin pt+~E a sin 3pt. 9. Compare the " effective " values of two currents, one whose wave form is sinusoidal, having a maximum of 100 amperes, and another of triangular shape with a maximum of 150 amperes. 10. An A.C. has the following values at equal intervals of time : 3. 4. 4'5, 5'5. 8, io- 6, o, 3, 4, 4-5, 5-5, 8, 10, 6, p. Find the R.M.S. value of this current. 11. A number of equal masses are attached to the ends of rods rotating about one axis. If the lengths of these rods are 10, 9, 5, 8, 4, 13 and 15 ins. respectively, find the effective radius (called the swing radius) of the system. (This is the R.M.S. value of the respective radii.) 12. Find the R.M.S. value of the function sin 2 cos 3 6 over the period o to -. (Refer to p. 178.) 13. The value of the primary current through a transformer at equal time intervals was 20 -05 -07 -ii -14 -19 -2i -04 -08 '12 -15 -18 -2i -08 -04 Find the R.M.S. value of this current. Volumes. If a curve be drawn, the ordinates being the values of the cross sections of a solid at the various points along its length, then the area under the curve will represent, to some scale, the volume of the solid. For, considering a small element of the length, 81, if the mean area over that element is A, the o 194 MATHEMATICS FOR ENGINEERS volume SV of the small portion of the solid is A.U, and the total volume of the solid is the sum of all such small volumes, i. e., 2AS. If the length 81 is diminished until infinitely small, 2AS/ becomes fAdl, and hence Volume of solid = / Adi. J o Comparing this with the formula giving the area of a closed figure, viz.,fydx, it is seen to be of the same form, and it is made identical if A is written in place of y (i. e., areas must be plotted vertically) and / in place of x (i. e., lengths must be plotted horizontally) . When values of A and / are given the curve should be drawn and integrated by any of the given methods ; and if it is preferred to find the actual area of the figure in sq. ins. first, the number of units of volume represented will be deduced from a consideration of the scales used in the plotting. Thus if i" (horizontally) repre- sents x ft. of length, and i" (vertically) represents y sq. ft. of area, then i sq. in. of area represents xy cu. ft. of volume. To take a numerical example : If i" = 15" of length, and i" = 5 sq. ft. of area, then i sq. in. of area = x 5 = 6J cu. ft. of volume. 12 If the area is found by the sum curve this conversion is unnecessary, as the scales are settled in the course of the drawing. The Example on p. 122 is an illustration of the determination of volumes by graphic integration; in that case the actual areas of sections are not plotted directly, but values of d 2 , the multiplication by the constant factor being left until the end. Had the solid not been of circular section the actual areas would have been plotted as ordinates and the work carried on as there detailed. We thus see that the determination of the volume of any irregular solid can be effected, if the cross sections at various distances from the ends can be found, by a process of graphic integration. If, however, the law governing the variation in section is known, it may be more direct to perform the integration by algebraic methods. To take a very simple illustration : Example n. The cross section of a certain body is always equal to (5# 3 + 8) sq. ft., where x ft. is the distance of the section from one end. If the length of the body is 5 ft., find its volume. VOLUMES 195 The body might have an elevation like Fig. 43, and its cross section might be of any shape ; the only condition to be satisfied being that the area of a cross section such as that at BB must = 5* 3 +8. Thus the area at AA must = (5 X o) + 8 = 8 and area at CC = (5 X 5 3 ) + 8 = 633. f 5 C 5 Then the volume = J Q Adx = ] Q (5x 3 +8)dx +40 = 821-25 cu. ft. ~"~^^ 4 Volumes of Solids of Revolution. A solid of revolution is generated by the revolution of some closed figure round an axis which does not cut the figure. Thus, dealing with familiar solids, the right circular cone, the cylinder and the sphere are solids of revolution, being generated respectively by the revolution of a right-angled triangle about one of the sides including the right angle, a rectangle about one of its sides, and a semicircle about its diameter ; and of the less well-known solids of revolution the most important to the engineer is the hyperboloid of revolution which is generated by the revolution of a FIG hyperbola about one of its axes and occurs in the design of skew wheels. The axis about which the revolution is made in all these examples lies along a boundary of the re- volving figure; whereas an anchor ring is generated by the revolution of a circle about an axis parallel to a diameter but some distance from it. The revolving figure may have any shape whatever, the only conditions being, for the following rule to hold, that the axis about which the revolution is made does not cut the figure and that the cross section perpendicular to this axis is always circular. Imagine the revolving cross section to be of the character shown in the sketch (ABCD in Fig. 44) ; and let the revolution be about the axis of x. It is required to find the volume of the solid of revolution generated. Working entirely from first principles, i, e., reverting to our idea of dealing with a small element and then summing : if the strip MN of height y and thickness 8x revolves about OX it will generate a cylinder. The radius of this cylinder will be y and its height or length Sx ; and hence its volume = iry 2 Sx. Accordingly the volume swept out by the revolution of ABCD ig6 MATHEMATICS FOR ENGINEERS will be ^ny z Sx (the proper limits being assigned to x) approximately, or fny 2 dx accurately. Again, it will be seen that such a volume can be measured by the area of a figure, for, writing Y in place of Try 2 , the volume = fYdx, which is the standard form for an area. Hence, if values of y are given, corresponding values of Try 2 must be cal- culated and plotted as ordinates, and the area of the resulting figure found. It should be noted that fny 2 dx might be written as itfy*dx t thus saving labour by reserving the multiplication by TC until the M B X X! FIG. 44. Volume of Solid of Revolution. area has been found, i. e., the values of y 2 and not those of Try 2 are plotted as ordinates. The following example will illustrate : Example 12. The curve given by the figures in the table revolves about the axis of x; find the volume of the solid generated, the bounding planes being those through x = 2 and x = j, perpendicular to the axis of revolution. X 2 3 4 5 6 7 y 44 42 44 46 45 38 VOLUMES 197 n Values of y 2 must first be calculated, since the volume = / jty z dx = Hence the table for plotting reads X 2 3 4 5 6 7 y* 1936 1764 1936 2116 2025 1444 and the values of y 2 are plotted vertically, the curve ABC (Fig. 45) resulting. This curve is next integrated from the axis of x as base, the curve DEF resulting; the polar distance being taken as 3, so that the new O 6 / FIG. 45. Volume of Solid of Revolution. vertical scale = 3 X old vertical scale. It must be remembered, how- ever, that the base from which the summation has been made in the figure is not the true base, since the first value of the ordinate is 1400 and not o; thus a rectangle 1400X5 has been omitted. Hence we must start to number our scale at 7000 ; and according to this f 7 numbering the last ordinate reads 9580, hence / z y z dx = 9580, or /? Volume = TC I y z dx = TT x 9580 = 30,100 cu. ins. In cases in which y and x are connected by a law the integra- 198 MATHEMATICS FOR ENGINEERS tion may be performed in accordance with the rules for the integration of functions. Example 13. Find the volume of a paraboloid of revolution and compare it with the volume of the circumscribing cylinder. o Y B FIG. 46. A paraboloid of revolution is generated by the revolution of a parabola about its axis. Suppose the parabola is placed as shown in Fig. 46; the revolution is therefore about OX. The equation to the curve OB is y 2 = $ax (see Part I, p. 106), i. e., if OA = h, (AB) 2 = 4 ah. Hence the volume of the solid swept out by the revolution of OBA [h [h = Tty z dx = TU I qax dx J J _ 4 a7 L><A a _ 7,2 Now the volume of the circumscribing cylinder = TC X 4ah X h and hence the volume of the paraboloid = - X vol. of circumscribing cylinder. Example 14. The curve y 2 = 64 zx 2 revolves about the axis of y. Find the volume of the solid generated, the limits to be applied to x being o and 5. This differs from the cases previously treated in that the revolution is to be about the y axis and not about the x axis. Hence the volume = fnx 2 dy and not Jwy*dx, the y replacing x and vice versa. Also another point must be noted : the limits given are those for x, whereas the limits in the integral fiix 2 dy must apply to the I.V., which is now y. Therefore a preliminary calculation must be made to determine the corresponding limits of y y 2 = 64- x = o, y x = 5, 2 = 64, y = 8 = 14- y = 374- VOLUMES 199 The double signs occurring here may possibly confuse, but actually the equation given is that of an ellipse, symmetrical about the axes of x and y, and the volume required is the volume generated by the revolution of the two shaded portions (Fig. 47), which will be twice that generated by one of these; hence, taking the upper shaded portion, we use the positive limits, viz., 3-74 and 8. f 8 f 8 / v 2 \ Then the volume = I r:x z dy ni ( 32 - }dy J 3-74 ' J 3-ll\ J 2 / ' f y3\8 = m say if ) 6 /3-74 = 7c(256 85-3-H9-5 + 8-7) = 59 '9^. The solid due to the revolution of the lower portion will be also 59-97^, and hence the total volume generated = 119-871; = 376 cu. units. FIG. 47. If the limits for y were 8 and +8, the volume of the whole solid would be required ; then [8 f8/ y2\ r y3-|3 Volume = / -KX z dy 2:rJ \^ ) ^ ]dy = 2TC [_3 2 )' z~J - 27^(256-85-3) = 341-471 = 1070 cu. units. The solid generated by the revolution of an ellipse about its major axis is known as a prolate spheroid; while if the revolution is about the minor axis the solid is an oblate spheroid. The volumes of these may be necessary, so that they are given in a general form. 200 MATHEMATICS FOR ENGINEERS The general equation of an ellipse is 2 -\-jp = i (cf. P- 344)- Let a >>, i. e., the major axis is horizontal. For a prolate spheroid the revolution is about the major axis, /a fa J2 ny z dx = 2rr/ ~- 2 (a z x z )dx o J o a 27C&V FIG. 48. In like manner, the volume of an oblate spheroid = - na?b ; and it should be noted that if b = a, the spheroid becomes a sphere and its volume = ~KO. S . 3 Example 15. Find the volume of a zone of a sphere of radius r, the bounding planes being those through x = a and x = b. The equation of the circle is x z +y z = r 2 (Fig. 48) whence y 2 = r z x z . :. Volume of a zone = f b ny 2 dx = TT f b (r z x 2 )dx J a J a = 7r[- 2 (6-a)--(6 3 -a 3 )] LENGTH OF ARC 20 1 This can be put in the form given on p. 120, Part I, if for (ba) we write h, and for BE and DF their respective values r : and r 2 . Thus a z = r z -r ] z , b z = r z r z z h z = (b-a) z = b z +a z 2ab = r z r 1 z +r z -r 2 z -2ab , h z +2r z r 1 z r 2 z and hence ab = - 2 So that the volume of the zone Length of [Arc. Consider a small portion of a curve, PQ in Fig. 49, P and Q being points near to one another. FIG. 49. The small length of arc PQ is denoted by 8s, so that a complete arc would be denoted by s. Let PM = 8*, and QM = Sy. Then the arc PQ = the chord PQ very nearly, so that we may say (Ss) 2 = (chord PQ) 2 = (S#) 2 +(8y) 2 . /Ss\ 2 y =i * > v~ = Hence when 8x becomes infinitely small, , - becomes ,-, etc. J 8x dx and ds I J- = A/ dx \ dy ds or - = dx 202 MATHEMATICS FOR ENGINEERS Integrating dz dx r The length of arc can thus be found if the value of the integral on the R.H.S. can be evaluated. In only a few cases is the evaluation of the integral simple; and for most curves an approximation is taken, e. g., to find the perimeter of an ellipse by this method one would become involved in a most difficult integral known as an elliptic integral, this being treated later in the chapter; and hence the approximate rules are nearly always used in practice. To deal with a case of a very simple character : Example 16. If y ax-j-b, find the length of arc between x m and x = n. In this case it is really a matter of determining the length of the line AB (Fig. 50) ; the slope of the line being a. Then if y = ax+b, = a and i + = i+ 2 . /m . __ Vi+a 2 dx = n = Vi + a 2 (m ri). On reference to the figure it will be seen that this is a true result, since AB = \/(AC) 2 +(CB) 2 = V(m-w) 2 +a 2 (w-w) 2 = Vi+a 2 (m n). Example 17. Use this method to determine the approximate length of a cable hanging in a parabola, when the droop is D and the span is 2L. For convenience, put the figure in the form of Fig. 51. Then L 2 = 4D L 2 whence a = -~, so that a must be very large. LENGTH OF ARC 203 The equation of the curve is in reality y being written in place of L, and x in place of D. men dx d = -5 A ~r~ x ay dx dy JC so that or -/- 2V dx dy dx = 4<z AO, 2a ~ zy ~ y u ^L^ whence dx - Z. T dy 20,' i Thus FIG. 51. V = 1 + 5-2 approximately, O# since all the subsequent terms contain a 4 and higher powers of a in the denominator, so that all these terms must be very small. Hence s = 2 1 + . dy - . 3 Span Example 18. Find the length of the circumference of a circle of radius r. The equation of the circle is y 2 +# 2 = y 2 . Thus y 2 = r z -x 2 and ay 4 -^- = a# {differentiating with regard to x} ctx or ^ = -- - ^ 204 MATHEMATICS FOR ENGINEERS Hence i + (;, ) = i+~i s= - Length of circumference 4 X length of J circumference f a r = 4 J o "V^F 2 ** To evaluate this integral, let x = r sin u (cf. Example 7, p. 150). Then r 2 x 2 = r 2 r 2 sin 2 u = r 2 (i sin 2 ) and also = r* cos* u dx = r cos u du dx = r cos u . du. y c FIG. 52. To find the limits to be assigned to u sin u = - ; and when x = o, u = o r and when x = i, u = 90 or Thus the circumference ra dx = * r J oVi*=x dx r cos u du o r cos 7T 2 Example 19. Find an expression for the length of the perimeter of the ellipse whose major and minor axes are 2a and 26 respectively. LENGTH OF ARC 205 Let BPA (Fig. 52) be the ellipse, CQA the quadrant of a circle on the major axis as diameter, and BTD the quadrant of a circle on the minor axis as diameter. Selecting P as any point on the ellipse, draw the lines QPN, PT, TS, QR and OTQ as shown. The point P has the co-ordinates x and y, viz., ON and PN, which are respectively equal to QR and TS. Now QR = OQ sin <j> or x a sin <f> and TS = OT cos <p or y = b cos <p. Thus = a cos* and ~- = or 5 = Va 2 cos 2 <f>+& 2 sin 2 <f> d<f>. x* v 2 Now the equation to the ellipse is ~^+ hz = i, and the eccentricity, d which we shall denote by K, is given by T , distance between foci A/a 2 b 2 OF . K - . ---- . - = ~--i-, F being a focus. major axis a OA Hence K 2 = ^- and K 2 a 2 -a 2 = -6 2 or Z> 2 = a 2 (i-K 2 ) so that a*cos a f +Msin 8 f = a^os^+a^in 2 ^ a 2 K 2 sin 2 f = a 2 (i-K 2 sin 2 $>). Thus our integral reduces to the form and for the quarter of the ellipse the perimeter = I a Vl K^in 2 ^ d$, J o since the limits for $ are obviously o and . it p _ Also the full perimeter of the ellipse = 4 I a vi K^in 2 ^ d<t>. Jo This integral, called an " elliptic integral of the second kind," is extremely difficult to evaluate; but in view of the importance of the perimeter of the ellipse it is well that we should carry the work a little further. Knowing the values of a and K for any particular ellipse, recourse may be made to tables of values of elliptic integrals, but if these are not available, a graphic method presents itself which is not at all difficult to use. According to this plan, various values of $ are jT . _ - chosen between o and , and the calculated values of vi K 2 sin 2 <?> are plotted as ordinates to a base of <f>. Then the area under the resulting curve when multiplied by ^a gives the perimeter of the ellipse. 206 MATHEMATICS FOR ENGINEERS o. ees ur 1-6 1-5 1-4. 1-3. 1-2 1-0 FIG. 53.: Perimeter of Ellipse. ^ ^ X s = \ \ \ \ \ P ^ri meter or Ellipse 2 x major axis x orainct \ 1 c t* ' 1-294 \ = \ i \ \ \ = 1 i \ i A II MM MM II III II 1 II / -2 3 -4- -S -6 -7 -8 -9 / e.cce.n/~ric.tfuK FIG. 54. LENGTH OF ARC 207 Example 20. A barrier before a ticket office in a works was constructed out of sheet metal, which was bent to the form of an ellipse of major axis 21 ins. and minor axis 5 ins. Find the area of sheet metal required if the height of the barrier is 5 ft. T j-u- j tr VlO'5 2 C In this case a = 10-5 and K = 10-5 so that K 2 = -944. The table for the plotting reads = '9714 4> sin <f> sin 1 <p l-Ksin* Vl-K*sin^ o O I I 157 1564 0245 i --0232 = -977 99 314 309 0951 0898 = -91 955 471 '454 206 195 = -805 897 628 5878 345 -326 = -674 821 785 7071 5 472 = -528 726 942 809 652 -616 = -384 62 1-099 891 793 749 = -251 501 1-256 9511 9 -85 = -15 388 i-4 J 3 9877 -976 922 = -078 279 1-570 i I 944 = -056 235 and the values in the extreme columns are plotted in Fig. 53. The area under this curve = 1-0663 SC L- um t and thus the perimeter = 40 x i 4 X 10-5 x 1-0663 = 44-78 ins. Hence the area required = 12 ft ^ It is well to compare this value of the perimeter with those obtained by the approximate rules given in Part I, p. 105 (a) Perimeter TU(+&) = 71(10-5+2-5) = 40-7 ins. (b) Perimeter = 4-443 V 2 +6 2 = 4-443x10-8 = 47-8 ins. (c) Perimeter = 7u{i-5(a+&) Vab} = TTX 14-38 = 45-1 ins. and the perimeter, correct to two places of decimals, is given in the tables * as 44-79 ins. * The tables of complete elliptic integrals give the values of Vi K^sin 2 ^ d<p for various values of 0, being the angle whose sine is K, the eccentricity of the ellipse. Thus to use the tables for this particular case we put sin K = -9714, whence = 76 16'; we then read the values of the integral for 75, 76 and 77, and by plotting these values and interpolation we find that for the required 208 MATHEMATICS FOR ENGINEERS Thus the errors in the results found by the different rules are (a) 9-13 % too small (b) 6-69 % too large (c) -67 % too large showing that the rule of Boussinesq gives an extremely good result in this case of a very flat ellipse, whilst the other approximate methods are practically worthless. Area of Surface of a Solid of Revolution. When a solid of revolution is generated, the boundary of the revolving figure sweeps out the surface of that solid. The volume of the solid depends upon the area of the revolving figure, whilst the surface depends upon the perimeter of the revolving figure. To find the surface generated by the revolution of the curve CD about OX (Fig. 55) we must find the sum of the surfaces swept out by small portions of the curve, such as PQ. Let PQ = a small element of arc = 8s. Then the outside surface of the solid generated by the revolution of the strip PQMN about OX will be equal to the circumference of the base X slant height, i.e., 2ny8s. Hence the total surface will be the sum of all similar elements, i. e., \r\x = b > 2Tcy8s, approximately, or if 8x becomes smaller and smaller ^L_ J iC tt fx = b Surface = I 2nyds. J x = a * For ds we may substitute its value, viz. /b 2ny (J 2nyJ !--} .dx. angle, viz., 1-0664. Multiplication by 40, i.e., 42, gives the result 44-79. For the convenience of readers interested in this question, and who desire a result more exact than that given by the approximate r fi _ rules, a curve is here given (Fig. 54) with values of I v i K^in 2 ^ d$ J o plotted against values of K; and for the full perimeter of the ellipse the ordinates of this curve must be multiplied by twice the length of the major axis. E. g., if the major axis = 16 and the minor axis = 10 K = - g-^ = -7807. Erecting an ordinate at K = -7807 to meet the curve, we read the value 1-294; multiplying this by 32, we arrive at the figure 41-41, which is thus the required perimeter. AREA OF SURFACE 209 Example 21. Find the area of the surface of a lune of a sphere of radius a, the thickness or height of the lune being b. The surface will be that generated by the revolution of the arc CD of the circle about its diameter OX (Fig. 56). From the figure y 2 = a 2 x 2 whence 2V.-r 2X dx dy x or - = . dx y _ fds\ 2 ,(dy\ 2 . x 2 y 2 +x 2 a 2 Thus (j- ) =i+l/) = i-f - - = ZZy- = \dxJ \dx' y 2 y 2 a 2 x 2 fd a Hence the surface = I 27uVa 2 x 2 - J c Va 2 - p - 2Tiaj dx = 2r:a(dc) = but 2TOZ& is the area of a portion of the lateral surface of the cylinder circumscribing the sphere. Thus the surface of a lune of a sphere = the lateral surface of the portion of the cylinder circumscribing the sphere (the heights being the same). Exercises 18. On Volumes, Areas of Surfaces and Length of Arc. 1. The cross sections at various points along a cutting are as follows Distance from one end (ft.) o 40 82 103 134 1 66 192 200 Area of cross section (sq. ft.) o 210 296 205 244 154 50 o Find the volume of earth removed in making the cutting. 2IO MATHEMATICS FOR ENGINEERS 2. Find the weight of the stone pillar shown in Fig. 57. The flanges are cylindrical, whilst the radius of the body at any section 2 is determined by the rule, radius = ;=, where x is the distance of vx the section from the fixed point O. (Weight of stone = 140 Ibs. per cu. ft.) 3. The curve y = 2x 2 $x revolves about the axis of x. Find the volume of the solid thus generated, the bounding planes being those for which x 2 and x = +4. 4. Find, by integration, the surface of a hemisphere of radius r. 5. The curve y = ae bx passes through the points x i, y = 3-5, and x = 10, y = 12-6; find a and b. This curve rotates about the axis of x, describing a surface of revolution. Find the volume between the cross sections at x i and x 10. 6. Find the weight of a cylinder of length / and diameter D, the density of the material varying as the distance from the base. (Let the density of a layer distant x from base = K#.) ^r-o 1 FIG. 57. Weight of Stone Pillar. 7. The rectangular hyperbola having the equation x 2 y z = 25 revolves about the axis of x. Find the volume of a segment of height 5 measured from the vertex. 8. The line 4y 5* 12 revolves about the axis of x. Find the surface of the frustum of the cone thus generated, the limits of x being i and 5. 9. The radius of a spindle weight at various points along its length is given in the table Distance from one end (ins.) o 375 5 i-o i'3 1-6 1-85 1-61 1-61 78 42 4 c Find its weight at -283 Ib. per cu. in., the end portions being cylindrical. 10. Determine by the method indicated in Example 19, p. 204, the perimeter of an ellipse whose major axis is 30 ins. and whose minor axis is 18 ins. Compare your result with those obtained by the use of the approximate rules (a), (b) and (c) on p. 207. 11. The curve taken by a freely hanging cable weighing 3 Ibs. per foot and strained by a horizontal pull of 300 Ibs. weight conforms to the equation , x y = c cosh 300 where c = . 3 Find the total length of the cable if the span is 60 ft., i. e.. x ranges from 30 to -{-30. B CENTROIDS 211 Centre of Gravity and Centroid. The Centre of Gravity (C. of G.) of a body is that point at which the resultant of all the forces acting on the body may be supposed to act, '. e., it is the balancing point. The term Centroid has been applied in place of C. of G. when dealing with areas ; and as our work here is more con- cerned with areas it will be convenient to adopt the term centroid. From the definition it will be seen that the whole weight of a body may be supposed to act at its C. of G. ; and in problems in Mechanics this property is most useful. Thus, movements of a complex system of weights may be reduced to the movement of the C. of G. of these. Or to take another instance : in structural work, in connection with fixed beams unsymmetrically loaded, it is necessary to find the position of the centroid of the bending- moment diagram. It is thus extremely important that rules should be found for fixation of the position of the centroid in ' all cases ; and the methods Jt J JL jt & t i , ,..,,. Tn t ff/2 "?3 III A. iffS adopted may be divided into two classes : (a) algebraic (in- " IG " 5 8. -Centre of Gravity or Centroid. eluding purely algebraic, and partly algebraic and partly graphic), (b) graphic. The rules will best be approached by way of a simple example on moments. In place of areas or solids, afterwards to be dealt with, let us consider the case of a uniform bar loaded as shown in Fig. 58. For equilibrium the two conditions to be satisfied are (1) The upward forces balance the downward forces. (2) The right-hand moments about any point balance the left- hand moments about the same point; or, in other words, the algebraic sum of the moments about any point is zero. If C is the balancing point or fulcrum, the upward reaction of the fulcrum = M = fni-^-m 2 -}-m 3 -{-m t -\-m 5 from condition (i). Taking moments about A, let x (x bar) be the distance AC. Then, by condition (2) or The product of a force into its distance from a fixed point or axis is called its first moment about that point or axis ; whilst the 212 MATHEMATICS FOR ENGINEERS product of a force into the square of its distance from a fixed point is called its second moment about that point. Hence our statement concerning the distance AC can be written 1st moments x = masses To extend this rule to meet the case of a number of scattered masses arranged as in Fig. 59, the co-ordinates of the centroid must be found, viz., x and y. Thus FIG. 59. Centroid. 2mx 2 1st moments about OY and 2m 2 masses _ ^ m y _ ^ 1 s * moments about OX ^ "" 2m 2 masses If the masses are not all in one plane, their C. of G. must be found by marking their positions in a plan and elevation drawing and determining the C. of G. of the elevations and also that of their plan. Thus the C. of G. is located by its plan and elevation. It will be observed that some form of summation is necessary for the determination of the positions of centroids or centres of gravity; and this summation may be called by a different name, viz., integration, all the rules of which may be utilised; the integration in some cases being graphic, in some cases algebraic, and in others a combination of the two. CENTROIDS 213 Rules for the Determination of the Centroid of an Area. Let it be required to find the centroid of the area ABCD in Fig. 60. The area may be considered to be composed of an infinite number of small elements or masses, each being the mass of some thin strip like PQMN ; the co-ordinates of the centre of gravity of which may be determined in the manner already explained. ._.b N M B FIG. 60. Centroid of an Area. To find x, i. e., the distance of the centroid from OY Mass of strip PQMN = area x density (considering the strip as of unit thickness) = y8xxp ist moment of strip about OY = mass X distance = pySx x x = pxySx. ist moments about OY Hence x = masses = _/ %. the limits being a and b _5 pyox and if the strips are made extremely narrow /& fb pxydx I xydx a J a X = /pydx I ydx a ' J a p cancelling from both numerator and denominator. 214 MATHEMATICS FOR ENGINEERS Thus a vertical is found on which the centroid of the area must lie; and this line is known as the centroid vertical. To fix the actual position of the centroid some other line must be drawn, say a horizontal line, the intersection of which with the centroid vertical is the centroid. Thus the height of the centroid above OX must be found; this being denoted by y. To find y. The whole mass of the strip PQMN may be supposed to act at R, its mid-point, because the strip is of uniform density ; and hence the moment of the strip PQMN about OX y = mass X distance = py 8xx- 2 2 b ist moments about OX Hence y = - , 2, masses b 2 f b py dx I y dx a. J a As in previous cases, the integration may be algebraic, this being so when y is stated in terms of x, or graphic, when a curve or values of y and x are given. Suppose the latter is the case, and we desire to find x T~ fxydx Then x = J T Z T - Jydx and the values of the numerator and denominator must be found separately. Each of these gives the area of a figure, for if Y is written in place of xy, the numerator becomes fYdx, which is the standard expression for the area under the curve in which Y is plotted against x; and the denominator is already in the required form. Thus a new set of values must be calculated, viz., those of Y, these being obtained by multiplication together of corresponding values of x and y; and these values of Y are plotted to a base of x. Then the area under the curve so obtained is the value of the numerator, and the denominator is the area under the curve with y plotted against x; and, finally, division of the one by the other fixes the value of x. CENTROIDS 215 Example 22. Find the centroid of the area bounded by the curve given by the table, the axis of x and the ordinates through x 10 and x = 60. X IO 25 40 45 5 60 y 4 5'3 6-2 6-4 6-6 6-8 We thus wish to find the centroid of the area ABCD (Fig. To find x : xy 30Q toa C e ntroid Horn ntal. ' JS JG D ZO 3O 4O FIG. 61. Centroid of an Area. The table for the plotting of Y against x reads X IO 25 40 45 50 . 60 Y or xy 40 132-4 248 288 330 408 From this we get the curve AEF. The area of the figure ABCD /GO ydx = 289 10 and the area of the figure AEFD /GO xydx = 10650 10 The method of integration is not shown, to avoid con- fusion of curves. /GO xydx 10 r60 / ydx J 10 10650 289 = 36-9. 2l6 MATHEMATICS FOR ENGINEERS Thus the centroid vertical, or the line PG is fixed. We need now to find the centroid horizontal, i. e., y must be determined. i r 60 = 2 J 10 Y ^ /where Y in this case\ areaofABCD \ stands for y 2 J Now y = f rf J 10 so that the following table must be compiled X 10 25 40 45 5 60 Y or y 2 . 16 28 38-4 4i 43-5 46-1 FIG. 62. C. of G. of Thin Plate. Plotting from this table, the curve RQ results, and the area of the figure ARQD is 1689. _ I area of ARQD &xi68Q y " : areaofABCD : 289 ~ ^22i The intersection of the centroid vertical and the centroid horizontal at G fixes the centroid of ABCD. A modification of this method is necessary when the actual area is given in place of the tabulated list of values, the procedure being outlined in the following example. Example 23. It is required to find the C. of G. of a thin plate having the shape shown in Fig. 62. Show how this may be done. Draw two convenient axes at right angles and divide up the area into thin strips by lines drawn parallel to OY. Draw in, also, the mid-ordinates of these strips. The area of any strip can be assumed CENTROIDS 217 to be "mean height X thickness "; and therefore measure ordinates such as MN and multiply by the thickness or width of the strip. Repeat for each strip, and the sum of all these will be the area of the figure. To find x. OA X = the distance of the centre of ist strip from OY so that the area of strip X OA V = ist moment of strip about OY. Hence, multiply the area of each strip by the distance of its mid- ordinate from OY and add the results ; then the sum will be the ist moment of the area about OY. T,, - Sum of ist moments 2nd total Then x = r = . . . .. Area ist total To find y. Fix R, the mid-point of MN, and do the same for all the strips. The area of the strip has already been found ; multiply this by AjR and repeat for all strips. The sum of all such will be the ist moment about OX; dividing this by the area of the figure, the distance, y, of the centroid from OX is found. [Note that R is the mid-point of MN and not of NA X , because OX is a purely arbitrary axis.] For this example the calculation would be set out thus Strip Length of mid-ordinate (like MN) Width of Strip Area of Strip Distance of centre from OY (like OAj) Distance of centre from OX (like RAi) ist moment about OX ist moment about OY I i-55 5 775 25 2-0 1-55 19 2 2-79 5 1-395 75 2-O 2-79 1-05 3 3'44 5 1-720 1-25 2-18 3-75 2-05 4 3-85 5 1-925 i-75 2-27 4*37 3-36 5 4-01 5 2-005 2-25 2-32 4-64 4-50 6 3-92 5 1-960 2-75 2-3 4-5i 5-40 7 3-60 5 i -800 3-25 2-18 3-92 5-85 8 3-26 5 1-630 3-75 2-04 3-32 6-II 9 2-66 5 i-33o 4-25 2'*O2 2-68 5-65 10 1-47 5 735 4-75 1-95 i-43 3*49 Totals I5-275 32-96 37-65 and 15- = 6 32-96 y = - - 2-16. 15-28 Thus the position of G is fixed by the intersection of a horizontal at a height of 2-16 with a vertical 2-46 units distance from OY. If the centroid of an arc was required, the lengths of small elements of arc would be dealt with in place of the small areas, but otherwise the procedure would be the same. 2l8 MATHEMATICS FOR ENGINEERS "Double Sum Curve " Method of Finding the Centroid Vertical. This method is convenient when only the centroid vertical is required ; for although entirely graphic, it is rather too long to be used for fixing the centroid definitely. Method of Procedure. To find the centroid vertical for the area APQH (Fig. 63). Sum curve the curve PQ in the ordinary way, thus obtaining the curve AegE; for this construction the pole is at O, and the polar distance is p. Produce PA to O l7 making the polar distance p t = HE = last ordinate of the sum curve of the original curve (viz., PQ). Sum curve the curve AegE from AP as base and with O t as FIG. 63. Centroid i Vertical~of~an Area. pole ; then the last ordinate of this curve, viz., CM, is of length x, so that the vertical through C is the centroid vertical. Proof. Consider the strip abed, a portion of the original area. Then Or and eg are parallel (by construction) P ef ab -*- = - / = -f Ar fg fg and thus or i.e., hnxab = 2hn xab = 2p xfg = pSfg = p . HE. Again, the ist moment of the strip about AP = area X distance = hn x ab X Ah m = hnxabxml = pXfgXml CENTROIDS 219 and hence ist moment of area APQH about AP = plfgxtnl but ist moment of area APQH about AP = area x distance of centroid from A and x = MC. fee/. FIG. 64. Problem on Loaded Beam. Example 24. A beam, 16 ft. long, simply supported at its ends is loaded with a continuously varying load, the loading being as expressed in the table. Distance from left-hand) support (feet) / o 2 4 6 8 10 12 14 16 Load in tons per foot run 12 17 21 25 28 29 3* 34 38 Find the centroid vertical of the load curve, and hence determine the reactions of the supports and the point at which the maximum bending moment occurs. We first plot the load curve from the figures given in the table (Fig. 64) ; and next we sum curve this curve, taking a polar distance of 10 horizontal units ; the last ordinate of this sum curve reads 4-27, so that the total load is 4-27 tons. We now set off AD equal in length to BC, and with this as polar distance we sum curve the curve AEC fiom the vertical axis as base. This sum curve finishes at the point G on the horizontal through C, and a vertical through G is the centroid vertical, distant 9-2 ft. from the end A. For purposes of calculation, the whole load may be supposed to 220 MATHEMATICS FOR ENGINEERS act in this line; the total load is 4-27 tons, and taking moments round A 4-27X9-2 = R B x 16 whence R B = 2-46 tons and R A = 4-272-46 = 1-81 tons. We now set up AH, a distance to represent R A , to the new vertical scale, and then a horizontal through H is the true base line of shear. At the point P the shear is zero; but the shear is measured by the rate of change of bending moment, so that zero shear corresponds to maximum bending moment; and hence, grouping our results Reaction at left-hand support = i 81 tons Reaction at right-hand support = 2-46 tons and the maximum bending moment occurs at a distance of 8-4 ft. from the left-hand end. Centroids of Sections by Calculation (for a graphic method especially applicable to these, see p. 251). Special cases arise in N, r L \( r IfcHJ |S| I L N- -NT N 2 * I T T i. *I^N- FlG. 65. the form of sections of beams, joists, rails, etc., for which a modification of the previous methods is sufficient. If the section is composed of a combination of simple figures, such as rectangles or circles, as in the great majority of cases it is, its centroid can be found by loading each of its portions, into which for purposes of calculation it may be divided, with a weight proportional to its area, and treating the question as one for the determination of the C. of G. of a number of isolated weights. Example 25. Find the position of the centroid of the Tee section shown in Fig. 65. We may consider the section to be made up of two rectangles; then f a 5 SO 24O Area of flange = 6 X = Q sq. ins. = -r sq. ins. o o ^4 and the centroid of the flange is at Gj. CENTROIDS 221 Area of web = 3$ x ~ = -~ sq. ins. o 04 and the centroid of the web is at G 2 . From considerations of symmetry we see that the centroid of the section must lie on the line G^G^, at the point G, say. (of length > + , i. e., 2") as a bar loaded with -~- I O ID 04 Treat G joe units at Gj and ~ units at G 2 . Let GjG = x', then the upward force at G = 7^ + ~^ = =^ units. 64 ,,G, B 847 344-4'5l6 Q G> I G I __ i J\ r\f\ /" t 11 ^ -* I '22 FIG. 66. Centroid of Bridge Rail. (In the further calculation we may disregard the denominators, since they are alike.) Taking moments about whence 375 X^ = 135X2 x = ^^ = -72. 375 ' Hence the distance of the centroid from the outside of the flange ' Example 26. Determine the position of the centroid of the bridge rail section shown at (a), Fig. 66. This example presents rather more difficulty than the one imme- diately preceding it. The plan of procedure is, for cases such as this, 222 MATHEMATICS FOR ENGINEERS that adopted in the work on the calculation of weights, viz., we first treat the section as " solid " and then subtract the part cut away. Neglecting the small radii at the corners, and treating the section as " solid," the section has the form shown at (b), Fig. 66. 2 3 *7 The area of AB = -i-X- = 2-52 sq. ins., and its centroid is at G 2 , 10 4 the intersection of its diagonals 3 tj Similarly the area of CD = |x- = 1-313 sq. ins., and its centroid is at Gj. For the part cut away (see (c), Fig. 66) The area of EHM = -Xy!) = -221 sq. in.; and we know from Part I, p. 130, that its centroid G 8 is distant -424 X radius, i.e., 424 X -375 or *I59* from EM. Again, the area of EF = >X- = -516 sq. in., and its centroid is I o 4 at G 4 . Our problem is thus reduced to that of determining the C. of G. of four isolated weights, two of which act in the direction opposed to that of the others, placed as shown at (d), Fig. 66. Let the centroid of the whole section be at G, distant x from O. Now the upward forces = the downward forces and thus R G +'5i6+-22i = 2-52 + 1-313 whence R G = 3-096. Also, by taking moments about O (3-096 X *) + (-516 X -344) + (-221 X -847) = (1-313 X-i88) + (2-52X1-094) whence x = -855 in. or the centroid of the section is '855" distant from the outside of the flange. Centroids found by Algebraic Integration. Suppose that the equation of the bounding curve is given, then the centroid of the area between the curve, the axis and the bounding ordinates may be determined by algebraic integration. We have already seen that Ixydx ~ly z dx x = -. and y = ^, lydx lydx so that if y is stated as a function of x, xy and y 2 may be expressed in terms of x, and the integration performed according to the rules given. The examples here given should be carefully studied, for there CENTROIDS 223 are many possibilities of error arising due to the incorrect substitution of limits. Example 27. Find the centroid of the area between the curve y = 2X 1 ' 5 . the axis of x and the ordinates through x = 2 and x = 5. The curve is plotted in Fig. 67, and it is seen that the position of the centroid o^E the area ABCD is required. Now y= 2x 1 ' 5 , and thus xy = 2x*xx = 2x* and 2 = x 3 . To find x f 5 1 xy dx -f J * \\d X y 20 c / / CenTpoid / Vcrhcal '/'***' I L f'ydx or the centroid ve i -8 1 units from boundary. To find y f 2X*dX \7* ) t 15 (F): ? X 5| 5 *_ 2 A ,0 {5*-,*} 5^68 5 JS / ^Cenrnoiol Ccn^poid Hopizonrai A | | | | B 7 50-25 rtical is distant the left-hand i T 5 - 1 y*dx 2 J 2 ' 23 4 5 J2 FIG. 67. y , 5 / y dx -X4 / ^ 3 ^ 2 ^ 7 2 /5 3 /2 6\5 ^-^ (V)^ i,,5 (5 4 ~2) 5 6oQ -, 8^50-25 Hence the co-ordinates of the centroid are 3-81, 7-57. Example 28. The bending moment curve for a beam fixed at one end and loaded uniformly over its whole length is a parabola, as 224 MATHEMATICS FOR ENGINEERS shown in Fig. 68. The vertex is at A and the ordinate at B, viz., BC is ; the loading being w units per foot and I being the span. We wish to determine the position of the centroid of the figure ABC so that we may find the moment of the area ABC about AD, and finally the deflection at A. From the equation to a parabola, y 2 = ^ax, we see that Wl 2 , 22 l z = 4 . , whence 40, = or y 2 = x ^2 w w i. e., (ND) 2 = - AD. w The distance of the centroid from AD = y r xydy j x&y D f l / FIG. 68. Area of ABCD = - of surrounding rectangle = X/X _ wl 3 All this area may be supposed to be concentrated at its centroid, and hence the moment of ABC about AD = ^ x - I - c 04 o Now the deflection at A = ^ x moment of the bending moment diagram about the vertical through A i wl* ~ El X IP Hence the deflection at A = ^^ oJil W/ 3 = 3^=^, where W = total load. Example 29. Find the position of the centroid of a quadrant of a circle of radius Y. The equation of the circle is x z -\-y 2 = r 2 hence y = 2 * so that xy = xVr z x 2 . CENTRE OF GRAVITY 225 Thus x (and consequently y) *-x 2 dx Try 2 The value of the denominator is , for it is the area of the 4 quadrant. (This integral would be evaluated as shown on p. 149.) To evaluate the numerator, let u = r z x z then du -zxdx du or xdx = . 2 X * fr r du i Then I x Vr 2 -x* dx = I ~ "5 = ~[o-(+' 2 )*] i , i/J f 3 i 3 4 r y = Z = X Qr -424^. 4 Centre of Gravity of Irregular Solids. The methods given for the determination of the centroids of irregular areas apply equally well when solids are concerned. For if A is the area of the cross section of a solid at any point along its length, distant x, say, from one end, and the length is increased by a small amount 8x (and if this is small there will be no appreciable change in the value of A), then the increase in the volume == AS* or the increase in the weight = pA8x, p being the density. The moment of this element about the end = pAS* x x so that x f l ist moments _ J Q 2 weights /i pAdx Axdx 226 MATHEMATICS FOR ENGINEERS As before, two cases arise, viz., (a) when values of A and x are given, and (b) when A is denned in terms of x. To deal with these In case (a) plot one curve in which A is the ordinate and x is n the abscissa and find the area under it ; this is the value of I A.dx. Jo Plot a second curve whose ordinates are the products of corresponding values of A and x and find the area; this is the value of the numerator, and division of the latter area by the former gives the value of r. Thus the centroid vertical is found, and if the solid is symmetrical about the axis of x, this is all that is required; otherwise the centroid horizontal must be found, the procedure being exactly that previously described when dealing with areas in place of volumes. An example on the application of this method is here worked. A 6 5 4 5 2 1 < \ \ c .A * ^* AJC 4 20 16 )2 8 4 O \ X x * -*' ~~^~ s. \ . ^> ^ ^ ^ i ^^ / < i ^-~, y ^^- - ~-~^ *^-^_ F 9 JG C ^ 4 6-8 1O 1 14 1( FIG. 69. Problem on C.I. Column. Example 30. The circumference of a tapering cast iron column, 16 ft. long, at 5 equidistant places is 9-43, 7-92, 6-15, 4-74 and 3-16 ft. respectively. Find its volume and the distance of its C. of G. from the larger end. The areas must first be found from the circumferences. Now the area of a circle = 47U So that the table for plotting reads x = distance from larger end (ft.) o 4 8 12 16 A = area of cross section (sq. ft.) 7-09 4-98 3-o 1-78 79 By plotting these values the curve EF (Fig. 69) is obtained. CENTRE OF GRAVITY 227 The figure here given is a reproduction of the original drawing to rather less than half-size, and since the measurements were made on the original, the results now stated refer to that. In the original drawing the scales were : i* vertically = 2 sq. ft., and i* horizontally = 2 ft., so that i sq. in. of area represented 4 cu. ft. of volume. The area under the curve EF was found, by means of the planimeter, to be 13-66 sq. ins., and accordingly the volume = 13-66x4 = 54-64 cu. ft. The curve BCD results from the plotting of values of Ax as ordinates, the table for which plotting reads X o 4 8 12 16 Ax I9'9 24 21-4 12-6 The area under this curve was found to be 19-06 sq. ins., which FIG. 70. C. of G. of Solid of Revolution. represented I9~o6x 16 units of moment, since for the plotting of BCD i* vertically = 8 units of Ax, and i* horizontally = 2 units of x. area BCDG i6x 19-06 Hence area BEFG 54-64 For case (b), when A is stated in terms of x, the integration is entirely algebraic. Thus if A is a function of x, integrate Ax and also A with regard to x, and divide the former integral by the latter to determine the value of ~x. Example 31. The area of cross section of a rod of uniform density varies as the cube root of the distance of the section from one end ; find the distance of the C. of G. from that end, being given that the area at a distance x from the end = '^ / #. 228 MATHEMATICS FOR ENGINEERS Consider a strip distant x from the stated end and of thickness S.v. Then, from hypothesis, the area of section 4- $3/ x, and thus the volume = area X thickness = 4 - 5^ / ^xS^. Also the mass of the strip = volume X density and the moment of the strip about the end mass x distance 2 ist moments of small elements Hence x = - ... 2 their masses = 4* 7 or the C. of G. is distant ^ o f the length from the given end. C. of G. of a Solid of Revolution. Suppose that the curve BC in Fig. 70 rotates round OX as axis ; and we require to find the position of the C. of G. of the solid so generated. Consider a small strip of area MN; its mean height is y and its width is 8x, so that the volume generated by the revolution of this is Tiy 2 8x, or the mass = p-n:y 2 8x. The ist moment of this strip about OY = mass X distance = p-n:y 2 8x X x = pnxy 2 8x. Thus the total ist moment about OY = /, pnxy 2 8x *~~^a ^\b and the total mass = > pny 2 8x ib fb I piixy 2 dx I xy 2 dx J_ a J_a /* ( b pny 2 dx I y 2 dx. J a J a As before, the two cases arise, viz. (a) When values of x and y are given. For this case make a table of values of x x y 2 and also one of values of y 2 . Plot the values of xy 2 against those of x and find the area under the resulting curve This area = fxy 2 dx . . . . . . (i) CENTRE. OF GRAVITY Plot the values of y z against those of x Area of figure so obtained = fy z dx 229 (2) and (2)'. Also we know that y must be zero, for the axis of x is the axis of rotation; and thus the C. of G. is definitely fixed. (b) When y is expressed as a function of x. In this case find both xy z and also y z in terms of x, integrate these functions algebraically and thence evaluate the quotient. Example 32. The curve given by the tabulated values of y and x revolves about the .ar-axis; find the position of the C. of G. of the solid thus generated. X o i 2 3 4 y 8 10 21 26-4 25 For the first curve, values of xy z are required, and for the second curve, values of y 2 ; these values being X o i 2 3 4 y* 64 IOO 441 696 625 xy 2 o IOO 882 2088 2500 The curve AB (Fig. 71) is obtained by plotting the values of xy z as ordinates; and the area under this curve is 4323; this being thus the value of / xy 2 dx. J o By plotting the values of y- as ordinates the curve CD is obtained ; /i y z dx = 1699. o /4 xy z dx _o T* I y*dx J o i. e., the C. of G. is at G, the point (2-55, o). Example 33. The curve x 5V 2 Vy revolves about the axis of y. Find the position of the centre of gravity of the solid generated, the solid being bounded at its ends by the horizontal planes distant i and 5 units respectively from the axis of x. 4323 1699 = 2-55 units 230 MATHEMATICS FOR ENGINEERS Since the revolution is about the axis of y and not that of x, y must take the place of x in our formulae and x the place of y; therefore the limits employed must be those for y. In Fig. 72 AB is the curve x = $y2Vy, and we see that it is 2000 5oo_ ~S~ FIG. 71. required to find the height of the centroid above the axis of x of the solid generated by the curve AB about the axis of y. Then to find y (5 / yx 2 dy J i /5 x 2 dy i Now x = 5y2Vy, and thus x 2 = 2 and y# 2 2$y 3 2oy*-\-4y 2 . f5 f5 . 4 Then / y^ 2 ay I (2^y 3 2oy-}-^y 2 )dy = y = = 2 454 f 5 j T 5 and / x 2 dy= J i I i j T25V 3 , 4V 2 20X2 |"15 dy = \ ^-~+~ -- ^ y \ L 3 * 5 ->i = 639 /: 2454 639 = 3' CENTRE OF GRAVITY 231 Then since the centroid must lie along the axis of y, its position is definitely fixed at the point G, viz., (o, 3-84). Example 34. Find the mass and also the position of the C. of G. of a bar of uniform cross section a and length I, whose density is proportional to the cube of the distance from one end. Let us consider a small length 8x of the bar, distant x from the end mentioned above; the density of the material here = Kx 3 , where K is some constant; hence o e 4 6 a o IE H- 16 IB 20 FIG. 72. Mass of small element = volume X density = a8x-X K# 3 = Kax 3 8x. Thus the total mass = [ ' Kax 3 dx = Ka f V J o v 4 * " Ka/ 4 Also the ist moment of the element about the end = mass x distance = Kax 3 8xxx. Ka/ 5 Total ist moment -7 1 . Kax*dx = and if x = distance of C. of G. from the lighter end Ka/ 5 5 L Example 35. Find the position of the C. of G. of a triangular lamina whose density varies as the distance from the apex. (Let the thickness of the lamina = /.) 232 MATHEMATICS FOR ENGINEERS Consider a small strip of width 8x, distant x from the apex (Fig- 73)- The area of the strip = y>8x, and thus its volume = yt8x. Now the density r x or density = Kx and also, by similar triangles, Bx So that the mass of the strip H ytSx x Kx BKt H ^ and the ist moment of the strip about OY jr ri S.W.S.L. r / - '\ 1 \6x? 1 A- -y _J_^ / i ; '\ rE- *J 7,- u p J ? f^* ^U D FIG. ; Hence '3- _ / x - 2 /o (f FIG. 74 T _ x 3 dx ti BK/ )o E H-,3 (v3\Ji A T~T^ x \ 4 rz !/ Centre of Pressure. If a body is immersed in a liquid, then the pressure per sq. in. of surface is not uniform over the solid, for the pressure is proportional to the depth. The point at CENTRE OF PRESSURE 233 which the total pressure may be supposed to act is known as the centre of pressure (C. of P.). To find positions of centres of pressure we are, in effect, finding centres of gravity of solids whose density is proportional to the distance from some fixed axis. The C. of G. found in the example last worked is in reality the C. of P. of a triangular lamina immersed vertically in a liquid, with OY as the level of the top of the liquid. Just as, when discussing the stability of solids in air, we have supposed the whole mass to be concentrated at the C. of G., so now, when the solid is immersed in a liquid, the total pressure may be assumed to act at the one point, viz., the C. of P. To find the positions of the C. of P. for various sections and solids we must start from first principles, dealing with the pressure on small elements, and then summing. Example 36. Find the whole pressure on one side of a rectangular sluice gate of depth 5 ft. and breadth 3 ft., if the upper edge is 10 ft. below the level of the water (which we shall speak of as the still water surface level or S.W.S.L). Find also the depth of the centre of pressure. Consider a strip of the gate Sx deep and x ft. below S.W.S.L. (Fig. 74)- Then the area of the strip =3x8* and the pressure per sq. ft = K x depth. Now at a depth of x ft. the pressure per sq. ft. = weight of a column of water x ft. high and i sq. ft. in section, i. e., wt. of x cu. ft. of water or 62-4^ Ibs. Also the pressure is the same in all directions; and thus the pressure on the strip = 38x^x62-4^ and the moment of the pressure on the strip about S.W.S.L. f!5 Hence the total pressure = I i8j-2xdxlbs. J 10 o ( x *\ = 187-2 - ) \2/10 = 187-2 XJC25 lbs 2 11700 lbs. or 5-23 tons. Again, the total ist moment about S.W.S.L. ri5 /x 3 \ = i8rzx 2 dx = i87-2(- ) Jio \3/ io = 62-4x2375. 234 MATHEMATICS FOR ENGINEERS '-hus the depth of the C. of P. below S.W.S.L. = 62-4 X 2375 ft 11700 = 12-65 ft. Hence C. of P. is at the point P, at a depth of 12-65 ft- below the surface of the liquid. The more general investigation for the position of the C. of P. is given in Chap. X. Mi FIG. 75. Centroids. Exercises 19. On the Determination of the Positions of Centroids and Centres of Gravity. 1. The density of the material of which a right circular cone is composed varies as the square of the distance from the vertex. Find the position of the centre of gravity of the cone. CENTROIDS 235 2. The equidistant half-ordinates of the load water plane of a ship are as follows, commencing from forward : -6, 2-85, 9-1, 15-54, J 8, 18-7, 18-45, 17-6, 15-13 and 6-7 ft. respectively. Find the area of the load water plane and the longitudinal position of its centroid. The length of the ship on the load water line is 270 ft. 3. A triangular plate of base 5" and height 8" is immersed in water, its base being along the S.W.S.L. Find the total pressure on the plate and the depth of the centre of pressure if the plate is vertical. 4. A vertical retaining wall is 8 ft. wide and 15 ft. deep. Find the depth of the centre of pressure of the earth on the wall. 5. Draw the quadrant of a circle of 4" radius, and by the double sum curve method determine the position of its centroid. 6. The portion of the parabola y = 2x z gx below the x axis revolves about that axis. Find the volume of the paraboloid so generated, and the distance of its C. of G. from the axis of y. 7. Find the position of the centroid of the area bounded by the curve y = 1-7 2X Z , the axis of x and the ordinates through x = i and x = +4. 8. Reproduce (a), Fig. 75, to scale (full size), and find the position of the centroid of the section represented, employing the method outlined in Example 23, p. 216. 9. Draw a segment of a circle of diameter = 6* on a base of 5-92", and find by the method of Example 23, p. 216, the height of the centroid above the base. (Take the segment that is less than a semicircle.) Find the distance of the centroid from the line AB for the sections in Nos. 10, ii and 12. 10. Channel Section, (&), Fig. 75. 11. Unequal Angle, (c), Fig. 75. 12. Tee Section, (d), Fig. 75. 13. Make a careful drawing of (a) Fig. 76, which represents the half- section of the standard form of a stream, line strut for an aeroplane, taking t as 2", and by the method of Example 23, p. 216, determine the distance of the centroid from the leading edge. 14. Find the position of the centroid of the pillar shown in Fig. 57, p. 210, of which further explanation is given in Question 2 on p. 210. [Deal with the flanges and the body as three separate portions.] 15. One end of a horizontal water main 3 ft. in diameter is closed by a vertical bulkhead, the centre of the main being 35 ft. below the level of the water. Find the total pressure on the bulkhead. 16. A semicircular plate is immersed vertically in sea water, its diameter being along the water surface. Find the total pressure on the plate if its diameter is 12 ft. and the weight of i cu. ft. of sea water is 64 Ibs. ; find also the depth of the centre of pressure. [Note. The reduction formulje given on p. 178 assist in the evaluation of the integrals.] 17. The parabola y 2 = 6x revolves about the axis of x. Find the distance from the vertex of the C. of G. of the paraboloid thus generated, if the diameter of the end of the paraboloid is 18. 236 MATHEMATICS FOR ENGINEERS (a) (b) FIG. 76. MOMENT OF INERTIA 237 18. The diameter of a spindle at various distances along its length was measured with the following results Distance from end (ins.) o i 2 83 3 4 5 6 7 8 2 Diameter (ins.) . i'5 I-I2 85 1-18 i'5 1-78 1-96 Find the distance of the C. of G. from the smaller end. 19. Find, by means of the double sum curve method, the distance from AB of the centroid of the rail section shown at (a), Fig. 75. 20. An aluminium right circular cone is of height 7 ins. and the diameter of its base is 10 ins. Find (a) its mass, the density of aluminium being -093 Ib. per cu. in. ; (6) the height of its centroid above the base. 21. Use the double sum curve method to find the distance from AB of the centroid of the area shown at (6), Fig. 76. 22. A segment of a parabola is of height h and stands on a base b Find the height of the centroid above the base. 23. A triangular plate of height h is immersed in water, its vertex being at the water surface, and its base being horizontal. Find the depth of the centre of pressure of the plate. Moment of Inertia. The product of a mass into the square of its distance from some fixed point or axis is called its second moment about that point or axis; and for a number of masses the sum of their respective second moments becomes the second moment, or moment of inertia of the system. When the number of masses is infinite, i. e., when they merge into one mass, the limiting value of the sum of the second moments is spoken of as the moment of inertia of the body. The moment of inertia of a section or body determines to a large extent the strength of the section or body to resist certain strains ; the symbol I, which always stands for moment of inertia, occurs in numerous engineering formulae ; also when dealing with the formulae of angular movement the mass is replaced by I, and so on, so that it is extremely important that one should be able to calculate values of I for various sections or solids. A few examples will emphasise the frequent recurrence of the letter I. Consider first the case of a loaded beam : Let the figure (Fig. 77) represent the section of a beam loaded in any way. Then it is customary to make the following assumptions (a) There is to be no resultant stress over the section, i. e., the sum of the tensions = the sum of the compressions. 238 MATHEMATICS FOR ENGINEERS (b) That the stress varies as the strain, and that the Young's modulus for the material is the same for tension as for compression. (c) That the original radius of curvature of the beam is exceed- ingly great compared with the dimensions of the cross section of the beam. The surface of the beam which is neither compressed nor stretched is spoken of as the neutral surface, and the line in which this cuts any cross section of the beam is known as the neutral axis. Referring to Fig. 77, let NN be the neutral axis, and let o- be FIG. 77. the stress at unit distance from NN, i. e., a-y = the stress at a distance y from NN. Thus the stress at y on a section of breadth b and depth 8y = a-y, and the force = stress X area = b8y X <ry. Now the forces on one side of NN must balance those on the other (by hypothesis). rr, bdya-y = o. but I 1 a-bdy x y = total ist moment of the forces and the line about which this is zero must pass through the centroid of the section ; hence the line NN passes through the centroid. The tensile and compressive forces form a couple, the moment of which = 2 force x distance = ;x \ b8y<ry x y MOMENT OF INERTIA 239 i. e., in the limit the moment of resistance of the internal forces = o- / bdyxy 2 , i. e., a- /area x (distance) 2 J ~ Y 2 i. e., a- (2nd moment of section about NN) = o-I. If M is the bending moment at the section, i. e., the moment of the external forces, it must be exactly balanced by the moment of the internal forces, so that M = oT. Also if /j = maximum tensile stress and = o-Yj / 2 = maximum compressive stress and = o-Y 2 A ft M then o- = = f- = T *i *2 L M / or, in general, T ^ v~ Hence, in considering the strength of a beam to resist bending, it is necessary to know the moment of inertia of its section; knowing this and the bending moment, we can calculate the maximum skin stress. As a further illustration of the importance of I in engineering formulae let us deal with the following case : If a magnet is allowed to swing in a uniform field, the time T of a complete oscillation is given by where I = moment of inertia of the magnet M = magnetic moment of the magnet H = strength of the uniform field in which the magnet swings. In this case the I of a cuboid is required; and it will be seen that no mention of the mass is made in this fortnula. Actually the I takes account not only of the mass, but also of its disposi- tion, the latter being a most important factor in all questions of angular movement. Thus for a mass of i Ib. swinging at the end of an arm of 10 ft. the energy would be io 2 , i. e., 100 times that of the same mass placed at a radius of i ft. only, although the angular velocities in the two cases were the same. The reason for the presence of I in formulae concerning the energy of rotation will be better understood if the next Example is carefully studied. 240 MATHEMATICS FOR ENGINEERS Example 37. A disc revolves at n revs, per sec. ; find an expression for its energy of rotation, or its kinetic energy. If the total mass = ra, let a small element 8m of mass be considered, distant r from the axis of rotation (Fig. 78). Now the linear velocity at the rim = V = 2?rwR and the angular velocity &> = number of radians per sec. then or Ra, = 2TCWR = V V thus co is constant throughout, whilst V depends on the radius. Kinetic Energy of mass 8m _ massx (veloc.) 2 _ 8mxv 2 FIG. 78. Hence the total K.E. of the disc = = r z 8m. -r 2 dm _ co a f n 2 Jo massx (distance) 2 = w Xl for disc. 2 Thus the K.E. = Io> 2 . Comparing this formula with the cor- responding one for linear motion, viz., K.E. = mv z , we see that ~6 when changing from linear to angular movement, I takes the place of m and <a the place of v. Suppose that the average velocity v t = r^ then my, 2 = Io> 2 2g i. e., mr or Hence I is of the nature of mass x (distance) 2 , so that if the whole mass were concentrated at the distance r t from the axis, the K.E. of the system would be unaltered. Hence the distance r t (which is usually denoted by k) is referred to as the swing or spin radius, or radius of gyration, i. e., it is the effective radius as regards all questions of rotation. MOMENT OF INERTIA 241 [Note that k is not the arithmetic mean of the various radii, but the R.M.S. value for h = J * (radius)* number considered J number considered In general, I can be written as mk 2 (if dealing with a mass) or Ak 2 (if concerned with an area). Method of Determination of the Value of I for any Section. Whilst it is found desirable to commit to memory the values of I for the simpler sections, it is not wise to trust entirely to this plan. It is a far better policy to understand thoroughly the meaning of the term "moment of inertia," and to derive its value for any section or solid by working directly from first principles. Thus, knowing that the moment of inertia is obtained by summing up a series of second moments, we divide the area or mass into a number of very small elements, find the area or mass of each of these and multiply each area or mass by the square of its distance from the axis or point about which moments are required ; the sum of all such products being the value of I. If the length of the swing radius is required, it can be deter- mined from the relation I = Ak 2 (for an area) or I = Mk 2 (for a solid) ; the area or mass being obtained by the summation of the areas or the masses of the separate elements. T , /2 second moments of elements Thus k = \i= , , . \ 2 areas or masses of elements Confusion often arises over the units in which I is measured; and to avoid this it is well to think of I in the form Ak 2 or M& 2 , when it is observed that I is of the nature area x (length) 2 , *'. e., (inches) 2 x (inches) 2 or (inches) 4 for a section, and massx (length) 2 or Ibs. X (inches) 2 for a solid. The moment of inertia must always be expressed with regard to some particular axis; and it is frequently necessary to change from one axis to another. To assist in this change of axis the following rules are necessary : The Parallel Axis Theorem. By means of this theorem, if I is known about an axis through the C. of G., the I about an axis parallel to the first can be deduced. In Fig. 79 NN is the neutral axis of the section; and the moment of inertia is required about AB, i. e. IAB is required. R 242 MATHEMATICS FOR ENGINEERS Dealing with the strip indicated I AB of the strip = pb8y x y 2 . Hence the total I AB = pfbdyxy 2 '= P fbdyx(Y-d) 2 = pfbdy X Y 2 + P /My X d 2 -2 P fbdyYd. Now fpbdy X Y 2 = the total I NN and fpbdy X d z = d 2 Jpbdy = d 2 X total mass = md 2 also 2dfpbdyxY 2d x total ist moment about NN = 2d X o (for the moments on the strips on one side' of NN balance those on the other) = o. Hence I AB = I Nti -{-md 2 i. e., to find the moment of inertia about any axis, find the moment of &/ FIG. 79. inertia about an axis through the G. of G. parallel to the axis given, and to this add the product of the mass into the square of the distance between the axes. e. g., if I NN = 47, mass = 12-4 and d (between AB and NN=2'3) then I AB =I NN +rf=47+(i2-4X2-3 2 ) = 47+ 6 57 = II2 7- Since I AB = Ij then w& AB = n or k" AB and this relation is represented by Fig. 79, which suggests a graphic method of finding &AB when NN is known. MOMENT OF INERTIA 243 Theorem of Perpendicular Axes. We require to find I about an axis perpendicular to the plane of the paper and passing through ; such being spoken of as a polar second moment. To distinguish between the moment about an axis perpendicular to the plane of the paper and that about an axis in the paper, we shall adopt the notation I for the former and I ox or I OY , as the case may be, for the latter. To find I o : Consider a small element of mass 8m at P (Fig. 80). Then I ox of this element = 8m x y 2 , I OY 8mx x 2 , r 2 = x 2 +y 2 and I = 8m xr 2 . 8m. r 2 = 8m.x 2 +8m.y 2 fdm . r 2 = fdm . x 2 +fdm . y 2 total I = total I OY + total I ox I = FIG. 80. Now hence and *'. e., or so that if the moments of inertia about two perpendicular axes in the area are known, the sum o! these is the moment of inertia about an axis perpendicular to the area and through the point of intersection of these axes. In special cases for which I ox = I oy then I = 2l ox To find the Relation between the Moment of Inertia about a Point in a Solid Body and the Moments of Inertia about three mutually Perpendicular Axes meeting in that Point. Thus, referring to Fig. 81, it is desired to connect I o with I ox> I OY and I oz . Consider a small element of the mass 8m placed at the point P. Then if PS = x PT = y PM = z OP = r (ON) 2 +(NM) 2 +(PM) 2 = (OP) 2 , and ON = PS, NM = PT *. e., x 2 -\-y 2 -\-z 2 = r 2 . Now I ox of the element = Sw(PN) 2 = 8m(z 2 +y 2 ) and in like manner I OY = 8m(x 2 +z 2 ) and I oz = 8m(y 2 +x 2 ) also I = Sw(QP) 2 = 8mr 2 . Thus I = Smr 2 = Sw(* 2 +y 2 +* 2 ) = 8m / 244 MATHEMATICS FOR ENGINEERS And for the total mass or total I = -( We may now apply the principles already enunciated to the determination of the moments of inertia of various sections and FIG. 81. solids ; and we take as our first example the case of a rectangular section. Example 38. To find the moments of inertia of a rectangle about various axes. (a) To find I NN (Fig. 82), NN being the neutral axis. Dealing with the small strip, of thickness Sx I N N of strip bSx x x 2 i. e., area X (distance) 2 h h Hence the total I NN = P bx z dx = b(- 3 } = J * \3/ h I2 ~ A N, D M c N | ai I B .lb-~ -_i IN, FIG. 82. = area X 12 but where A is the area of the section AZ_ _ A/fc \2 A^ = 12 or MOMENT OF INERTIA 245 By symmetry it will be seen that (6) To find I AB . _ Ab 2 1 N 1 N 1 ~ : -^ and the distance between AB and hence I AB = b z = A X . = b 2 = A = A6 2 3 ' I AB is larger than IN^. as would be expected, for the effective radius must be greater if the plate swings about AB than if it swings about N^NV I c ! rTT" ; b 1 s*1 i ... NLJ- ._ P-J&. ^ZJfrpJNi n *4 I $ N ; In like manner if FIG. 83. Ah z L AD The rule for the moment of inertia of a rectangle is required very frequently, since many sections can, be broken up into rectangles. Example 39. To find I NN of the Tee section shown in Fig. 83. The neutral axis NN is distant 1-03* from AB (cf. Example 25, p. 220). Dealing with the flange i i /S\ 3 I NiNj = bh 3 = ^ x6x (J] : : * 122 in> also the distance G Z G = 72". - Hence by the parallel axis theorem 246 MATHEMATICS FOR ENGINEERS I NN of the flange = I NiNj + [Ax (G X G) 2 ] (A being the area of the flange) 5,, 2 ) = -I22+I-94 = 2-06 ins. 4 For the web i Nrfl> = ^6A = -L X ^ )* x |- = 2 ins. 4 and also G 2 G = 1-28". Hence I NN of the web = I N2 N 2 + A x x (G 2 G) 2 (A x is area of the web) = 2 + (fxfxi-28 2 ) = 5'45 ins. 4 Hence the total I NN of the section = 2-06+5-45 = 7-51 ins. 4 Example 40. Find the polar 2nd moment of a circular disc of radius R; and also the moment of inertia about a diameter. Consider a ring of width 8r, distant r from the centre (Fig. 84). Then I o of the annul us = mass (or area) x (distance) 2 Hence the total I Now and FIG. 84. To find the respective swing radii AQ = T = ~^~ TUR 4 i. e., ox (Cf. with the R.M.S.l D or k Q , i. e., -TofR. -! value of a sine function V V * ( of amplitude R. J Also MOMENT OF INERTIA 247 ox 4X:rR 2 = - = -sR. R 2 4 To find the swing radius about a tangent (distance) 2 (oxtoTT) = R 2 IYT = Iox+AR 2 hence or Example 41. To find the moment of inertia of a right circular cylinder of length h and radius R, about various axes. (a) About the axis of the cylinder. (b) About an axis through the C. of G. perpendicular to the axis of the cylinder. (c) About an axis parallel to that in (b), but through one end. (a) The 2nd moment about the axis of the thin cylinder of length Sx (Fig. 85) R 2 = mass X from Example 40 R 2 = pnR 2 8x x p being the density of the material . Hence the total 2nd moment about the axis .R 2 = r J dx = R 2 2 R 2 = m - 2 where m = the mass of the cylinder. 248 MATHEMATICS FOR ENGINEERS (b) The 2nd moment of the strip about AA, which is parallel to NN R 2 = mass x - (see Example 40) R 2 Hence I NN of the strip = I AA of the strip + (its massx* 2 ) since AA is an axis through the C. of G. of the strip. Thus I N vr of the strip = Sx + uTiR 2 x 2 8x 4 h h f2 oTtR 4 /"2 and thus the total I NN = j dx-\- I pnR 2 x 2 dx J _h 4 J _ h N (c) To find I FF , FF being parallel to AA and NN. The distance between FF and NN = - 2 Also it has just been proved that I N1 j = m ( Hence T / R2 , h * ! = ml \4 12 /R 2 , h 2 \ = m(- + --). Example 42. Find an expression for the moment of inertia of a large pulley wheel of outside radius R and thickness of rim /. Neglect the arms or spokes of the wheel. Let Y inside radius of wheel, i. e., r = R t. Then, using the result of Example 40, p. 246, we know that the TD4 moment of inertia of the wheel as solid = ; from this must MOMENT OF INERTIA 249 7W 4 be subtracted the moment of inertia of a disc of radius r, viz., Xpb (p being the density of the material and b the breadth of the rim along the face). , , /T ,. Hence I Q - P b-~ P b = -~ (R*-y 4 = (R 2 +r>) (R 2 -* . . . , where M is the mass of the wheel. M M Writing R-/ forn! = ^ (R 2 +R 2 +/ 2 -2R/) = ( 2 R 2 -2R*+* 2 ). From (i) it will be seen that in order to get I o as large as possible, R and r must be very nearly equal, i. e., t must be very small compared with R. Thus for an approximation 2 2 may be neglected in the M expression for I Q , so that I Q = -xaR (R t) = MR(R t). /R 2 +y 2 \ 2 /R 2 +y 2 \ Referring once again to (i), I o = M ( *-)* i- &-, M& o = M ( - ! J or k Q = - and k o = -jojVRt+r 2 . As an approximation for this the rule k Q = - (R-\-r) is often used; k Q being thus taken as the average radius. Moment of Inertia of Compound Vibrators. To find the modulus of rigidity of a sample of wire by the method of torsional oscillations, various forms of vibrators may be used. In the calculations which follow the experiments, the moment of inertia of the vibrator occurs, so that it is necessary to understand how to obtain this. To illustrate by an example of one form of compound vibrator, suppose that the I about an axis through the C. of G. of the one shown in Fig. 86 is required. Let m { = mass of AB, r l be its radius and ^ its length m 2 = mass of C and also of D, r z be its radius and / 2 its length. fr 2 / 2 \ Then I NN of AB = w a ( +- 1 ) (from Example 41, p. 248) c r i r z z , ^2 2 \ rf (f r tne inner radius\ and IofC^m- 1 = ^ 250 MATHEMATICS FOR ENGINEERS by the parallel^ - m z\^ -r I2 y-r w 2 t - m z ^ \ This is also the L \4 of D. 2 4 V axis theorem. ) L NN total I NN = Maxwell's needle is a very convenient form of compound vibrator, and is utilised to determine the modulus of rigidity of the sample of wire by which it is supported. It consists essentially of a tube along which weights may be moved from one position to another, the movement being a definite amount. Referring to Fig. 87 m^ = the mass of each of the movable weights w 2 = the mass of each of the fixed weights. FIG. 87. FIG. 88. Then the time of torsional oscillations is measured when the movable weights are placed as shown, and again when they are moved to the centre; and it can be proved that the modulus of rigidity depends upon the difference between the moments of inertia under the two sets of conditions. Thus, since a mass m t is shifted from the position AB to the position NA, the only difference in the moments of inertia is that due to the changing of the C. of G. of a mass (m^ m 2 ) from a distance fa from the axis of oscillation to Ja; for I NN of m. 2 is unaltered. q r i6 Hence the change of I \ , T considering one mass onlyJ " l 2 ~~ ^ m i~ i "'^\- L ^r ^ (j \ a 2 ) = (w x m 2 )a 2 . Example 43. Find the moment of inertia of a sphere of radius R about its diameter. MOMENT OF INERTIA 251 Consider the thin disc (Fig. 88) of radius y, and thickness 8x. I of the strip about a diameter parallel to OY y*p8x (cf. Example 40, p. 246). Hence I of the strip about OY (distant x from the diameter considered) = T 5 " 1 Now y* = R 2 x*. Thus I OY of disc = TTP J ""^ ~^ v " "^ ^ 18* r I 4 ' / and hence T OY of sphere = I (R 4 ; = 27rp [ R (R* - 4 J o = R^ I 4 L 5 i 27uo i6R 5 8 = - X = 4 15 ! 5 2 ( m being the mass\ - 7/fr /\ ~ JLV \ tii i 5 \ of the sphere. / and Determination of 1st and 2nd Moments of Sections by means of a Graphic Construction and the Use of a Planimeter. The graphic construction now to be described is extremely simple to understand, and has the additional merit of being utilised to give 3rd, 4th and higher moments if desired. It being required to find the ist and 2nd moments about MM of the rail section shown in Fig. 89, and also the position of the neutral axis, the procedure is as follows : Construction. Divide the half -area into a number of strips by means of horizontal lines; the half-area only being treated, since the section is symmetrical. At a convenient distance h from MM draw M X M X parallel to MM. From P, the end of one of the horizontals, draw PR per- pendicular to MM, and from P 1 , the other end of the same horizontal, drop PKR 1 perpendicular to M X M X ; join R X R and note Q, its point of intersection with P 1 ?. Repeat the process for all the other horizontals (of which only three are shown in the diagram) and 252 MATHEMATICS FOR ENGINEERS join up all the points like Q, thus obtaining the curve CQLS, which is termed the ist moment curve. To obtain the 2nd moment curve treat the area CPKXSLQ in the same way as the original area was treated, i. e., drop QR" perpendicular to M. 1 M. l and join RR"; join up all points like Q 1 and the 2nd moment curve is obtained. Calculation. Find by the planimeter the areas of the original M M I^Momenf Curve M, FIG. 89. Moments of Sections by Graphic Construction. half-section, CPKXSLQ and CPKXTWQ 1 ; call these AC, A x and A 2 respectively. Then ist moment of the section about MM = 2Xhh.^ (for Aj is for the half-section only). Distance of the centroid of the section from MM = -r~ 2nd moment of section about MM = i. e., (swing radius MM ) 2 = -^ - and by the theorem of parallel axes, I can be found about NN. MOMENT OF INERTIA 253 In this case the actual results are as follows : h = 3 ins. A = i-n sq. ins. A x = -573 sq. in. A 2 = -39 sq. in. Hence h = 3 X ^ 73 = 1.55 ins. i ist moment of section about MM = 2 X 3 X -573 = 3-44 ins. 3 . 2nd moment of section about MM = 2X3 2 X -39 = 7-02 ins. 4 . Swing radius about MM = \ = 1-78 ins. \ 2-22 ' - N.B. To distinguish which area is to be read off by the planimeter the following rule should be observed : Read the area between the ist or 2nd moment curve, as the case may be, and the side of the original contour from which we dropped perpendiculars on the line about which we required moments. Proof. Consider P 1 ? as the centre line of a thin strip (such as the one indicated). Then the area of the strip = P^xS*, and ist moment about MM = P X P x 8x X RP. From the similar triangles RPQ and RJR 1 RP _ J*l h QP ~ JR 1 ~~ PP 1 whence RP x PP 1 = h X QP and RPxPP x xS* = AxQPxS* i. e., ist moment of the strip about MM = h x the area of which QP is the centre line. Then, by summing Total ist moment of the half-area about MM = h X the area between ist moment curve and right-hand boundary of section = h\ 1 . Again, the 2nd moment of the strip about MM = area x (distance) 2 = PP 1 xS*x(RP) 2 and 1 RJ ~~ JR" ~ PQ RP PQ 1 . or> hPO 1 _ - _ ^_ a T?P - _ S h ' ~ PQ PQ ' Hence the 2nd moment of the strip about MM = P 1 PxRPxRPx8* PO 1 ^ Sx = h*xPQ l x8x = h 2 X area of which PQ 1 is the centre line. 254 MATHEMATICS FOR ENGINEERS And the total 2nd moment of the half-area about MM = A 2 x area between the 2nd moment curve and the right- hand boundary of the section. Exercise 20. On Moment of Inertia. 1. Find the swing radius about the lighter end of a rectangular rod of uniform section and breadth and length I, for which the density is proportional to the square root of the distance from that end. 2. The swing radius of a connecting rod about its centre of suspen- sion was found to be 35-8 ins., and the distance of the C. of G. from the point of suspension was 31-43 ins. Find the swing radius about the neutral axis. If the connecting rod weighed 86-5 Ibs., find its moment of inertia about the neutral axis. FIG. 91. 3. A circular disc, 7" diameter, has a circular hole through it, of diameter 3", the centre of the hole being \" distant from the centre of the disc. Find the swing radius of the disc about an axis through its centre of gravity, perpendicular to the face of the disc. 4. Find the moment of inertia of a rectangle (5" by 3") about a diagonal as axis. 5. Find the swing radius of a triangular plate (of height h) (a) When swinging about its base. (b) When swinging about an axis through the vertex, parallel to the base. 6. By dividing into strips, by lines parallel to AB, find the moment of inertia, about AB, and also the swing radius, of the section shown at (a), Fig. 75, p. 234. 7. Find the radius of gyration about the axis of rotation, of the MOMENT OF INERTIA 255 rim of a flywheel, of outside diameter 5' 2", the radial thickness of the rim being 4". Find the moment of inertia about the neutral axis of the sections in Nos. 8, 9 and 10. 8. Channel Section, (&), Fig. 75. 9. Unequal Angle, (c), Fig. 75. 10. Tee Section, (d), Fig. 75. 11. Find the swing radius, about the axis, of a paraboloid, the diameter of the bounding plane, which is perpendicular to the axis, being d. 12. The flexural rigidity of a beam is measured by the product of the Young's Modulus E for the material into the moment of inertia of the section. Compare the flexural rigidity of a beam of square section with that of one of the same material but of circular section, the span and weight of the two beams being alike. 13. A cylinder 6" long and of i \" diameter is suspended horizontally by means of a long wire attached to a hook, and the wire is then twisted to give an oscillatory movement to the cylinder. Find the moment of inertia of the cylinder about the hook. 14. Determine the moment of inertia and also the swing radius about AB of the rectangular section shown at (a). Fig. 90. 15. Calculate the moment of inertia and also the swing radius of the box section shown at (b), Fig. 90, both about NN and about AB. 16. Find the position of the neutral axis of the section shown at (c), Fig. 91, and then calculate the moment of inertia and also the swing radius about this axis. 17. Determine the swing radius of the -section shown at (d), Fig. 91, about the axis NN. 18. The moment of inertia of the pair of driving wheels of a locomotive connected by a crank axle was found by calculation to be 34133 Ibo. ft. 2 . If the total weight of the two wheels and the axle was 8473 Ibs., and the diameter of the driving wheels was 6 ft. i in., A 2 find the swing radius of the wheel and also the ratio -^, where r is the radius of the wheel. 19. Find the swing radius about the axis of a right circular cone of uniform density, the radius of the base being 5 ins. 20. Employing the method explained on p. 251, determine (a) the ist moment about AB, (b) the 2nd moment about AB, (c) the distance of the centroid from AB, and (d) the swing radius about AB, of the area shown at (b) Fig. 76, p. 236. 21. A steel wire, -15 in. in diameter, hangs vertically; its upper end is clamped, and its lower end is secured to the centre of a horizontal disc of steel, which is 6 in. in diameter and g in. thick. If the length of the wire is 3 ft., and if C, the modulus of transverse elasticity of the steel, has the value 12,540,000 Ibs. per sq. in., find the time of a torsional oscillation of the wire, from the formula * = 402-5 256 MATHEMATICS FOR ENGINEERS where I = moment of inertia of the disc about the axis of suspension in Ibs. ins. 2 , / = length of wire in feet, d = diameter of wire in inches. 22. An anchor ring is generated by the revolution of a circle of radius r about an axis distant R from the centre of the circle. Find the moment of inertia of the ring about this axis. (Hint. Commence with the polar moment, i. e., the moment about the given axis, of an annulus made by a section at right angles to this axis, finding an expression for the inner and outer radii of the annulus in terms of the distance from the central annulus, and then sum up.) 23. Find the swing radius about the major axis of the ellipse whose equation is CHAPTER VIII POLAR CO-ORDINATES Polar Co-ordinates. A point on a plane may be fixed by its distances from two fixed axes, or by its distance along a line which makes a definite angle with some fixed axis. In the former case we are concerned with rectangular co-ordinates and the point is written as the point (x, y) ; whilst in the latter case the co- ordinates are polar and the point is denoted by (r, ff), r being the length along the ray inclined at an angle 6 to the fixed axis. It is really immaterial as to what line is taken as the fixed axis : in many cases the horizontal axis is taken, but in order to agree with the convention adopted for the measurement of angles (see Part I, Chapter VI) we shah 1 here consider the N. and S. line, i. e., a vertical line, as the starting axis and regard all angles as positive when measured in a right-handed direction from that axis. A point is next fixed on that line from which all the rays or radii vectors originate, and this point is spoken of as the pole for the system : thus the reason for the term polar is seen. To illustrate this -method of plot- ting, let us refer to Fig. 92. Taking OY as the starting axis and O as the pole, the point (2, 35) is obtained by drawing a line making 35 with OY and then stepping off a distance OP along it to represent 2 units, i. e., r=2 and 0=35. In like manner Q is the point (17, 289) ; whilst R is the point (2-4, 20). One advantage of this method of plotting is that it is not necessary to classify into quadrants and to remember the arrange- s 257 FIG. 92. 258 MATHEMATICS FOR ENGINEERS ment of the algebraic signs ; all lengths measured outwards from the pole being reckoned as positive. Example i . The following table gives the candle power of an arc lamp for various positions below the lamp : plot the polar diagram. Angle below horizontal . . o 10 20 3 40 5 60 70 80 90 1800 800 600 480 In reality we have to plot a number of polar co-ordinates, the lengths representing the values of the candle power ; but since the horizontal axis is specified, we shall take that as the main axis. Draw rays making 10, 20, 30, etc. (Fig. 93), with the horizontal axis, and along these lines set off distances to represent the respective candle powers, always measuring outwards from the centre. Join the ends of the rays and the polar diagram is completed. The Archimedean spiral and the logarithmic or equiangular spiral, important in connection -IOOO ^ with the forms of cams and gear wheels respectively, may be easily plotted from their polar equations. Thus the equation to the Ar- chimedean spiral is r=aO, and the equation to the equiangular spiral is r=ae be ; indicating that in the former case the rays, for equal angular intervals, are the con- secutive terms in an arithmetic progression, whilst in the latter case the rays are in geometric progression. To illustrate the forms of these curves by taking numerical examples : Example 2. Plot the Archimedean spiral ^=-573^, showing one convolution. FIG. 93. Candle Power of Arc Lamp. In the equation d must be in radians, but to simplify the plotting we can transform the equation so that values of a (in degrees) may replace 6 (radians). Thus r = -5736 = . (degrees) = -oia. O / O POLAR CO-ORDINATES Then the table for the plotting reads : 259 a o 3 60 9 120 150 180 210 240 270 300 33 360 ' 6 2-4 2-7 3' 3*3 3-6 and the plotting is shown in Fig. 94. aro izo" 24O 1.50* eio FIG. 94. Example 3. Plot one convolution of the equiangular spiral 25a In the log form log r = log -5 + -004360 log e = T-6990+ (-00436 X -4343 X a) = 1-6990-!- -ooi894a and thus the table of values reads : a. . . o 3 60 90 120 150 1 80 210 240 270 3 33 36o 00189401 0568 1136 1705 2273 2841 3409 '3977 4546 5114 5682 625 6818 log r . 1-6990 7-7558 1-8126 f-8695 1-9263 1-9831 0399 0967 1536 2104 2672 324 3808 r . . . 5 5699 6495 7407 8439 9618 1-096 1-249 1-424 1-624 1-85 2-109 2-403 26o MATHEMATICS FOR ENGINEERS The curve is drawn in Fig. 95. It will be seen that the ratio of second ray _ -5699 _ first ray -5 third ray = -6495 _ ^ second ray -5699 so that this spiral might alternatively have been defined as one for which the rays at equiangular intervals of 30 form a geometric pro- gression in which the common ratio is 1-14, the first ray being '5". Comparing the given equation r '$e' 259 with that connecting the tensions at the ends of a belt passing round a pulley, viz., T = te* 8 , we observe that the forms are identical, or in other words the equiangular spiral might be used to demonstrate the growth of the ten- sion as the belt continuously em- braces more of the pulley. Selecting any point P on the spiral, and drawing the tangent PT there and also the ray OP which makes an angle < with the tangent, it is found that cot < = -25 = co- efficient of 6 in the original equation. This relation would hold wherever the point P was taken on the spiral, so that the angle between the ray and the curve is constant : and thus the spiral is called " equiangular." If cot < = i, < = 45 and r = ae 6 , or taking a = i, r = e 6 and log e r = 6. Thus a spiral could be constructed in which the angles (in radians) would be the values of the logs of the rays : this spiral, however, is extremely tedious to draw, and its value consists merely in its geometric demonstration of the relationship between the natural logarithms and their numbers. Connection between Rectangular and Polar Co- ordinates. Let P be a given point, with rectangular co-ordinates x and y and with polar co-ordinates r and 0. Then referring to Fig. 96 ON y OP , 180 FIG. 95. so that y = r cos and so that and also POLAR CO-ORDINATES OM x 26l x y x = r sin r sin r cos = tan 0. Use of Polar Co-ordinates for the Determination of Areas. Polar co-ordinates may be usefully employed to find areas of certain figures. It is stated in the previous work on mensuration that Area of sector of circle = where = angle of the sector in radians. N P V X FIG. 96. FIG. 97. Let P and Q (Fig. 97) be the two points (r, 6} and (r+8r, 0+ SO) and close to one another. Then, since r and r+&r differ very slightly Area POQ = and the total area AOB = or re? r z d0 J e l approximately exactly. For the evaluation of this integral the working may be either graphic or algebraic, according to the manner in which the relation between r and is stated. As a simple illustration we may take the case of a circle of radius a. The area of the circle was found at an earlier stage (see p. 225) by evaluating fydx, i. e., by expressing the integral in terms of the rectangular co-ordinates. To evaluate the integral, however, it was found necessary to make the substitution x = a sin 6, the change thus being from rectangular to polar co-ordinates. 262 MATHEMATICS FOR ENGINEERS Evidently the rotating ray is constant in length and equal to a, the radius of the circle, and the limits to 6 are o and 2ir, if the full area is required ; hence fZir [2* Area of circle = I \d*d6 = Ja 2 1 d& = Aa 2 . 2ir Jo Jo = TO?. Example 4. Find the area of the cardioid given by the equation Y = a (i+cos &], 6 ranging from.o to ZTT. In this case r is of variable length, but there is a definite connection between r and 6, so that the integration is algebraic. and r 2 = a 2 (i-f-cos <9) 2 = a 2 (i + J cos 2(9+J+2 cos 0) = a 2 (1-5 + \ cos 20+2 cos 0). j'Zn- Hence area = j \v z d6 COS 20 + 2 COS0) dd 2 / \ TT = ( i'50+J sin 20+2 sin j 2 2 The Rousseau Diagram. The use of the Rousseau diagram simplifies the determination of the mean spherical candle power of a lamp. The candle power of the lamp varies according to the direction in which the illumination is directed (cf . Fig. 93) ; in the case there discussed, however, we considered the illumination in one plane only. If we imagine the polar curve to revolve round the vertical axis we see that a surface is obtained by means of which the illumination in any direction can be measured. The mean of all these candle powers is spoken of as the mean spherical candle power of the lamp. If the arc is placed at the centre of a spheri- cal enclosure, of radius R, then, if IM is the mean spherical candle power (M.S.C.P.) of the lamp, the total illumination is expressed by 47rR 2 lM : this total might be arrived at, however, by summing the products of the candle power in any direction into the area of the zone over which this intensity is spread; and putting this statement into the form of an equation, 47rR 2 I M = 2IA, where I is the intensity on a zone of surface area A. POLAR CO-ORDINATES 263 To find the M.S.C.P. proceed as follows: Suppose that the lamp is at O (Fig. 98). With centre O and any convenient radius R describe a semicircle ; also let the polar diagram be as shown (the curve OPQMC) . The greatest candle power is that given by OC ; draw a horizontal through N, the point in which the line OC meets the circumference of the semicircle, and make ab = OC. Through a and b draw verticals and through A and B draw horizontals, thus obtaining the rectangle DE ; draw a number of rays, OP, OQ, OS, etc., and also horizontals through the points p, q, s, etc., marking along these lines distances equal to OP, OQ, OS, etc., working from DF as base. By joining up the points so obtained the curve FL6D is obtained, known as the Rousseau FIG. curve ; then the mean height of this curve (which can readily be obtained by means of a planimeter) gives the M.S.C.P. of the lamp. Proof of this Construction. Let IM = M.S.C.P. of the lamp 2 area of zonexC.P. then 4irR*. Consider the zone generated by the revolution of TN ( = 8s) about AB ; its area is of the form 27rySs and the intensity of illumination is OC, say. The length y is the projection on the horizontal axis through O of either the line OT, the line ON, or the line midway between these (for these differ in length but slightly if 8s is taken as very small), i. e., y = OT cos 6 or y .= Rcos0. 264 MATHEMATICS FOR ENGINEERS Hence, for this zone, the illumination = candle power x area = OCX27rRcos0Ss = a&X27rRxfl'ifor-r- = cosflf v OS ) Hence the total illumination = 27rR X area under the curve FL&D and thus Total illumination 2?rR X area under the curve FL&D M = 47rR 2 47rR 2 _ area under the curve FL&D ~^zKT = mean height of the curve FL&D since 2R is the base of the curve. Dr. Fleming's Graphic Method for the Determination of R.M.S. Values. The determination of R.M.S. values is of some importance to electrical engineers, and the subject, previously dis- cussed in Chapter VII, is here treated from a different aspect. Instead of squaring the given values of the current and then extracting the square root of the mean of these squares, we may, by a simple graphic construction, obtain the mean of the squares very readily. Let the values of an alternating current at various times be as in the table : time t . . o 001 002 003 004 005 006 007 008 Current C 5 8 12 7 -6 -8-3 -3 5 then, to find the R.M.S. value of C we proceed as follows : Treat the given values as polar co-ordinates, taking t for the angles and C for the rays. Select some convenient scale for t, say 20 -ooi sec., and a scale for C, say i"=4 units, these being the scales chosen for the original drawing of which Fig. 99 is a copy to POLAR CO-ORDINATES 265 about one-half scale; and set out a polar diagram as indicated, making OA = +5, Oa = 6, etc. Join the extremities of the rays, so obtaining, with the first and last rays, the closed figure ABCDaE. Measure the area of this figure by means of a planimeter in this case the area was found to be 4-23 sq. ins. Now the area of the figure = %fr z d& = %fC 2 dt, or so that if we divide twice the area by the range in t, the mean value of the squares is determined. In this case the range of t = 160 = 2-79 radians, and also J"C 2 dt = 2 Xi6x 4-23 = 135-5, f r I "=4 units, and thus i sq. in. = 16 sq. units. Then M.S. = 279 = 48-65, and hence R.M.S. = = 6-98. The rule for the area of a figure, viz., %J"r 2 dO, may be usefully employed to find the height of the centroid of an area above a certain base. Example 5. Find the height above the base OX of the centroid of the irregular area OABX [(a) Fig. 100]. fir 10 a r B 8 12 16 EO (O FIG. 100. Centroid of Irregular Area by Polar Diagram. 266 MATHEMATICS FOR ENGINEERS To do this, first divide the base into a number of equal divisions and erect mid-ordinates in the usual way. Measure these mid-ordinates and set off lengths to represent them as radii vectors from A in (b) Fig. 100, the angles at which the rays are drawn representing the values of x, i. e., the lengths of the divisions of the base. Join the ends of these rays and measure the area of the polar diagram thus obtained ; divide this area by the area of the original figure and the result is the height y required. For the area of the polar diagram = %fr 2 dd = \fy z dx, since rays represent values of y and the angles represent values of x. Also the area of the original figure =fy dx, so that area of polar diagram _ \fy z dx _ _ area of original figure fydx ~ For the particular case illustrated (the scales referring to the original drawing) : For (a) Fig. 100 i" = 5 units vertically i* = 4 units horizontally so that i sq. in. represents 20 units of area. The area was found by the planimeter to be 16-82 sq. ins., so that the actual area is 336-4 sq. units, i. e., fy dx = 336-4. For (b) Fig. 100 i" = 5 units radially and each angular interval = 20, so that the total range = 180 or 3-14 radians. Hence 3-14 radians represent 20 units, the length of the base in (a) Fig. 100, or i radian = 6-36 units. 3-14 Now the area is of the nature r 2 X0, i. e., (length) 2 x angle, hence i sq. in. of area = 5 2 X 6-36 or 159 units. Area of the polar figure (found by the use of the planimeter) = 18-34 sc l- ms - = J 8'34 X 159 units = 2920. Hence y = ^|. = 8-68, 336-4 i. e., the centroid horizontal is found. Theory of the Amsler Planimeter. The principle upon which the planimeter is based may be explained quite simply, in the following way. In Fig. 101 let PP" be a portion of the outline of the figure whose area is to be measured, and let the fixed centre of the instrument be at O. Then in the movement of the tracing point P from P to P" along the curve, the tracing arm changes from the position AP to A'P". This movement may be regarded as made up of two distinct parts : firstly, a sliding or translational move- ment from AP to A'P', and next, a rotation round A' as centre, POLAR CO-ORDINATES 267 from A'P' to A'P". In the former of these movements the record- ing wheel moves from W to W, but part of this movement only, viz., that perpendicular to the axis of the wheel, is actually recorded, so that the wheel records the distance p. The area swept out by the tracing arm AP during the small change from P to P" = APP'P"A' = APP'A'+P'P'A' Hence for the whole area, area swept out = 2APx/>+2(AP) 2 S<9 Now the net angular movement is zero, so that 280 = o . Hence area swept out = AP2/>, or if / = length of the tracing arm, area = / X travel of wheel and hence the reading of the wheel _ area of figure Thus the length of the tracing arm determines the scale to which the area is measured. Hence by suitable adjustment of this length of arm the area of a figure may be read in sq. ins. or sq. cms. as may be neces- sary. If the average height of the figure is required, the length of the tracing arm must be made exactly equal to the length of the figure. This is done by using the points LL (Part I, Fig. 301), and not troubling about the adjustment at A. The difference between the first and last readings gives, when multi- plied or divided by a constant, the actual mean height of the figure. If the ordinary Amsler is used, then the mean height in inches is obtained by dividing the difference between the readings by 400 ; thus if the first reading was 7243 and the last 7967, the mean height would be the difference, viz., 724, divided by 400, i.e., 1-81 ins. The area of the figure = average height X length, but the area of the figure = length of tracing arm X wheel reading, hence if the length of the tracing arm = the length of the diagram, the wheel reading must be the average height of the diagram. FIG. 101. Theory of Amsler Planimeter. 268 MATHEMATICS FOR ENGINEERS [It should be noticed that the area recorded by the instrument is really the difference between the areas swept out by, the ends A and P of the arm AP, but as A moves along an arc of circle, coming back finally to its original position, no area is swept out.] Exercises 21. On Polar Co-ordinates. 1. Plot a- polar curve of crank effort for the following case, the connecting rod being infinitely long. e o 15 3 45 5'i 60 75 90 7'3 105 1 20 i35 . 150 165 1 80 Crank Effort (Ibs.) . . o 2-4 3-9 6-4 7-1 7 6-1 4'9 3'3 *5 2. As for Question i, but taking the connecting rod = 5 cranks. e 15 3 45 60 75 90 i5 1 20 5-8 '35 '5 165 172-5 i So C.E.flbs.) .... 2< 45 4'4 6-1 TO 7'4 7'4 6-6 4'4 3' i'S 6 3. An A.C. is given by C = 7-4 sin ^oirt. Draw the polar curve to represent the variation in C and hence find its R.M.S. value. 4. What is the polar equation of a circle, the extremity of a diameter being taken as the centre from which the various rays are to be measured ? Of what curve (to Cartesian ordinates, i.e., rectangular axes) is the circle the polar curve ? 5. Plot the polar diagram for the arc lamp, from the table. Angle (degrees) . . . o TO 2O 3 40 50 60 70 80 90 (vertical) Candle Power .... 800 T2OO l6<X) 2OOO 2200 2200 2300 2500 2300 1800 6. Plot the Rousseau diagram for the arc lamp in Question 5 and from it calculate the M.S.C.P. of the lamp. 7. An A.C. has the following values at equal intervals of time : 3, 4, 4'5> 5'5> 8, 10, 6, o, 3, 4, 4-5, 5-5, 8, 10, 6, o. Find by Dr. Fleming's method (cf. p. 264) the R.M.S. value of this current. 8. Eiffel's experiments on the position of the centre of pressure for a flat plane moved through air at various inclinations gave the follow- ing results : Inclination to horizontal .... 5 10 15 3 45 60 75 90 T>p Ratio (see Fig. 102) ..... 263 Plot a polar diagram to represent the variation of this ratio. POLAR CO-ORDINATES 269 9. Draw the polar curve to represent the illuminating power of a U.S. standard searchlight from the following figures : Angle (degrees) o (vertical) 10 20 3 40 5 60 70 80 90 Candle Power . 3000 10000 20500 33000 41500 4*5o 43000 43000 30000 24000 10 (above horizontal) 20 30 40 50 60 7 80 90 9000 6000 5000 5000 2OOO 3000 1500 1500 I50O C 15 centre of pressure FIG. 102. CHAPTER IX SIMPLE DIFFERENTIAL EQUATIONS Differential Equations. An equation containing one or more derived functions is called a " differential equation." Thus a very simple form of differential equation is dy = dx ? dv 5 = o x , dy and +2x -= and 4 -;- + 7/ 5 dx 2 dx are more complex forms. Differential equations are classified according to " order " or " degree " ; the order being fixed by that of the highest differential /V coefficient occurring in it. Thus ~ is a differential coefficient of d s y the first order. - is of the third order, and so on. dx 3 dv Hence 4-2 -\-y-jr = 5 '34 is an equation of the first order w% d*y and 8-^j+.y = 7 -I 6 is an equation of the fourth order. wsv The "degree" of an equation is fixed by that of the highest derivative occurring when the equation is free from radicals and fractions. d 2 y Thus T-TJ = c is of the second order and of the first degree whilst 4(3-^) +(-r"o) = 7 is f the second order and of the 2 2 second degree. Much has been written concerning the solution of the many types of differential equations, but it is only possible here to treat 270 SIMPLE DIFFERENTIAL EQUATIONS 271 the forms that are likely to arise in the derivation of the proofs of engineering formulae ; the plan being to discuss the solution accord- ing to the types of equation. Type : ~ given as a function of x. With the solution of such simple forms we have already become familiar, for the equations connecting the bending moment at various sections of a beam with the distances of those sections from some fixed point are of this character. Thus taking the case of a simply supported beam carrying a load W at the centre dfy = \V dx* ~ El dy Wx whence -- = ^F +C dx El which is of the type under consideration. Evidently this equation can be solved by integration through- out, attention being paid to the constants which are necessarily introduced. Expressing in algebraical symbols, !=/<*> ; then by integrating throughout with regard to x or y = \f(x}dx-\-C. *v Example i. If -~ = ^x z + r jx 2 and y = 5 when x=i, find an a% expression for y in terms of x. This equation is of the type with which we are now dealing, since 2 =/(*) Integrating y = r - + * -- 2X+C. The value of C must now be found : thus y = 5 when x = i so that 5 = A_(_i_ 2+C or C = 2-iy. Hence y= i-33# 3 +3-5# 2 2^+2-17. 272 MATHEMATICS FOR ENGINEERS Type : f- given as a function of y. i. e., -^ f(v). ax dx This type of equation differs somewhat from the preceding in that a certain amount of transposition of terms has to be effected before the integration can be performed. dy The equation may be written 7 = dx the transposition being spoken of as " separating the variables," and thence by integration I -,-,--. \dx-\-C = x-\-C. jj(y) J dy Example 2. If ~ = y 3 , find an expression for y. ax Separating the variables ^ = dx. Integrating / -^ = / dx-\-C or \y-* = x +C whence x-\ - + C = o. y _ The two foUowing examples are really particular cases of the type discussed generally on p. 275, but they may also be included here as illustrations of this method of solution. dy Example 3. Solve the equation -j- + ay = b. (tX dy , Here -g = b ay dy J = dx so that ^- = tdx+C J T b ay ^- bay i. e., loge (bay) = x+C or loge (bay) (ax-}-aC) whence e~ ax - aC = (b-ay). Now let A = e~ aC : then e -*-c = e ~ ax x e~ aC = Ae~< and Ae~ aa: & = ay b A _ or "V = ^ a a SIMPLE DIFFERENTIAL EQUATIONS 273 dy Example 4. Solve the equation 4-^- = u-f-7y. Separating the variables i.e., or loge (u+jy) = lx+ whence e* x+ * C = (11+77) 1C and if A be written in place of e 11+77 = Ae* z A \x ii or y = e --- * 1 7 Example 5. The difference in the tensions at the ends of a belt subtending an angle of dd at the centre of a pulley = dT = Tpdd, where /* is 'the coefficient of friction between the belt and pulley. If the greatest and least tensions on the belt are T and t respectively, whilst T the lap is 6, find an expression for the ratio . The equation dT = Tfjdd is of the type dealt with in this section ; to solve it we must separate the variables, thus : Integrate both sides of the equation, applying the limits t and T to T and o and 6 for the angle. /"T dT r e Then = M / 06. J t i Jo / \ T / \ fl Mr <('). T But log e - = loge T - loge / T Hence * O ' x ^ or A word further might be added about Example 3, or a modifi- cation of it. T 2 Let -f- = ay. dx * T 274 MATHEMATICS FOR ENGINEERS If a = i, then -~ =y, i. e., the rate of change of y with regard to ctoc x, for any value of x, is equal to the value of y for that particular value of x. Now we have seen (Part I, p. 353) that this is the case only when y = e x . If a has some value other than i, y must still be some power of e, for the rate of change of y is proportional to y ; actually, if -V = 6*, -- = ae ax = ay, so that y = e * would be one solution of dx the equation -. = ay, but to make more general we should write d% the solution in the form y = e ax -\-C or y = Ae ax , whichever form is the more convenient. Whenever, therefore, one meets with a differential equation expressing the Compound Interest law (*'. e., when the rate of change is proportional to the variable quantity) one can write down the solution according to the method here indicated. Example 6. Find the equation to the curve whose sub-normal is constant and equal to za. dv The sub-normal =)>-/- (See p. 43.) dx dv Thus y~- = ia, J dx or, separating the variables, fydy =fzadx. V 2 Hence = 2ax+C (Integrating) or -y2 This is the equation of a parabola ; if y = o when x = o, then K = o and y 2 = ^ax, i. e., the vertex is at the origin. Example 7. Find an expression giving the relation between the height above the ground and the atmospheric pressure ; assuming that the average temperature decrease is about 3-5 F. per 1000 feet rise, and the ground temperature is 50 F. Let T be the absolute temperature at a height h, then, from hypothesis r = 460+ 50 =- h 1000 = 510 0035/1 (i) Now we know that pv = CT (2) and also that if a small rise 8h be considered, the diminution in the SIMPLE DIFFERENTIAL EQUATIONS 275 pressure, viz., 8p, is due to a layer of air 8h feet high and i sq. ft. in section, and thus From equations (i) and (2) pv = (510 -0035^) and substituting for v its value from equation (3) or, in the limit dp i pdh ~ (510 0035/1) Separating the variables dp _ dh p "(510 0035/1) Integrating, the limits to p being p and p, and those to h being o and h, (-logpY = x -[log (510 0035 A) -log 5 IO | \ /p C -00351- J whence log p log p = ~ [log (510 0035/1) -6-234J or log p = log Po+^ [_ lo g (5 IO -'35 A ) 6-234J which may be further simplified by substituting the values for p and C. General Linear Equations of the First Order, i. e., equations of the type where a and b may be either constants or functions of x. The solution of this equation may be written as y = The proof of this rule depends upon the rule used for differen- d(uv) du , dv tiatmg a product, viz., ~- =v,--\- u~r- ; the reasoning being as dx dx &x follows : Let us first consider the simplest case in which this type of equation occurs, viz., the case of the solution of the equation dy where a is a constant. 276 MATHEMATICS FOR ENGINEERS Multiplying through by * the equation becomes dy , e^~--\- aye** = o, dx dy dv which can be written as v^- + v = o (where v = e* an( j thus dx dx *-- But v-r 4- V-T- = ~r(yv) s that -4-- (vv) = o ; hence w must be dx dx dx^ Wf w a constant, since the result of its differentiation is to be zero. Accordingly yv = C, or y = Ctr 1 = Ce~ ax . Extending to the case in which 6 is not zero, whilst a remains a constant, i. e., the equation is dy +"* we find that after multiplying through by &** the result arrived at is d Integrating both sides with regard to x, ye** = fbe^dx+C or y=e- flJC {fbe ax dx-\-C}. This may be evaluated if the. product of b and e** can be integrated. For the general case, that in which a and b are functions of x, the multiplier or integrating factor is e^ adx , for after multiplication by this the equation reads dv e fadx y J^. ae jadxy = fefadx and this may be written whence by integration we find that y e fadx = or y = SIMPLE DIFFERENTIAL EQUATIONS 277 dv Example 8. Solve the equation j^+izy = e* x . The equation may be written dy 12 i j-H y = -e** ** 7* 7 so that in comparison with the standard form a = and b = -e**. 7 7 Hence y = -/ J T***{ fie**. e s ' -7 47 47 Example g. If -j- y = 2x+i, find an expression ior y. dv so that a= i, 6 = Hence y = e+f<'*{f(2x+i)e-J<**dx+C} * The value of the integral f(2x+i)e~*dx is found by integrating by parts. Thus, let (2x+i)e~ x = udv where dv = e~ x , i. e., v = e~ x and . du U=2X+I, t. e.,-j- = 2, then fudv = uvfvdu = [(2x+i)x(-e-*)]-f-e-*2dx /FT Example 10. Solve for T the equation -3 |-PT = P(tcx) (referring to the transmission of heat through cylindrical tubes) ; P, t and c being constants. 278 MATHEMATICS FOR ENGINEERS The equation -^ [-PT P(tcx) is of the type -/--{-ay = b, where (A/X dX a = P and b = P(tcx). Hence the solution may be written T = e-S Pd *(fP(t-cx)ef pdx dx+K}, the integrating factor being e^ 7dx , i. e., e^ x . Hence T = e - Vx {fP(t-cx)e Px dx+K} ; and to express this in a simple form the integral fP(t cx)e Px dx must first be evaluated. Let fP(tcx}e px dx=fudv where u=P(tcx) and dv = e Px dx so that v ^f? x ', and also du = Pc dx, then fP(tcx)e px dx =fudv = uvfvdu = P(t-cx) x * Pa: +i x e**.Pcdx r rp -, T = e- px \(t-cx)e**+^- +M+K I since e~ fx xe Px e = i and L M+K. Example 1 1 . When finding the currents x and y in the two coils of a wattmeter we arrive at the following differential equation : dy , R,+R 2 where R x and Lj are respectively the resistance and inductance of the one coil and R 2 and L 2 are the resistance and inductance of the other coil ; I being the amplitude of the main current. Solve this equation for y. /1<\) Comparing with the standard form of equation, viz., -^--{-ay b, we ii . -"-^-l I JLVrt . . J - / l A' *- , t i - LV 1 * , i see that a x * and o = ~_, cos ^>^+ T _^ T sin / > ^- Hence /Rl~f~R2j. / r / T J^T TI> T T , T at I // -Lipl , . . -tC,! l/l + jjj -( I I ii r.OV5<-l i \J VLi+L, /2 - / T J,T J / /y_- cospt.e IJ Lf-f-1/ (Ri+R 2 X r Tj r i 2 -, + / * sinpt . e L i+ L 2 dt+C \ J L, i -\-L, 2 =e~ At l fB cospt . e At dt+ /D sin^ . e At dt+C\. SIMPLE DIFFERENTIAL EQUATIONS 279 Now, as proved on p. 157, I B0A cos pt dt = * (p sin pt-\- A cos pt) -f C x and j ~De A* sin pt dt = A> 1 *, (A sin * cos *) + C 2 Hence - At x e At ( _ sin ^i+ AB cos pt+DA sin ptDp cos A2 . . a = * ] R,+R 2 lL.pl R.I Exact Differential Equations. An exact differential equation is one that is formed by equating an exact differential to zero ; thus Pdx-}-Qdy = o is the type, Pdx-{-Qdy being an exact differential. The term exact differential must first be explained. Pdx -\-Qdy is said to be an exact differential if = ^S, the 3y dx derivatives being partial, or, to use the more familiar notation, dy / \dx. To solve such an equation proceed as follows : If the equation is exact, integrate Pdx as though y were constant, integrate the terms in Qdy that do not contain x, and put the sum of the results equal to a constant. [For, let Pdx+Qdy = du. Now, du is the total differential, ( -=- }dx and ( -j- }dy \dxJ \dyJ ' being the partial differentials (see p. 82) ; (du\ (du\ i. e., du = ( -=- MX+ I -j- )dy. \dx/ \dy/ " TU -t ^ ( du \j i ( du \j Then if du = o, ( ]dx4-\ -3- jay = o, \dx' \ay/ ' and this is exactly the same as the original equation ., (du\ idu\ if ( -j- } = P and ( -=- } = Q, \dx/ \dy/ ^y/ \iy/\dx/ \dy.dx \ (d\(dii\ (dQ I -= )\ -. I dx.dyJ \dx/ \dy / \dx 2 8o MATHEMATICS FOR ENGINEERS Our equation thus reduces to du = o, or, by integrating, to u C, but u= I Pd% (y being constant) + I Qdy (x being constant), and hence we have the rule as given.] Example 12. Solve the equation (x 2 4%y 2y 2 ) dx+(y 2 4xy2x 2 ) dy = o. Here P = x z -^xy-2y 2 (^ P ) = -^x-^y \dy 1 Q = y*-4xy-2x* and thus the equation is exact. I P dx (as though y were a constant) /X^ 4X 2 V (x 2 4xy2y z ) dx = 2 -- i 2y*#. I Qdy (as though x were a constant) = {(y*4xy2x z ) dy = ^* ^ 2Ar 2 y ; but of this only ^ 3 must be taken, since the other terms have been obtained by the integration of terms containing x. Hence \x* 2x 2 y zxy z +$y s = C or x 3 6x*y6xy*+y* = K. Example 13. Solve the equation v du u dv = o. If this equation is multiplied through by - 2 we have a form on the left-hand side with which we are familiar, viz., v duu dv _ v 2 ~~' for the left-hand side is d ( - V It Then by integrating, - = C or u = Cv. I This equation might have been regarded as one made exact through multiplication by the integrating factor 2 . SIMPLE DIFFERENTIAL EQUATIONS 281 Equations Homogeneous (/. e.. of the same power throughout) in x and y. Rule. Make the substitution y vx and separate the variables. Example 14. Solve the equation (x z +y z ) dx = 2xy dy. Let y = vx, dy dx , dv . dv then -f- = V-J-+X-T- = v-\-x- r ..... -. . (i) dx dx dx dx Now (x z +y z ) dx = 2xy dy dy _ x z +y z _ x*+x z v z _ i+v z ~~" -- - - SO , ~~" dx 2xy 2x z v Substituting for -- from (i) or dx -2V dv i j r v z 2v 2 i 3- - --- dx -2V 2V Separating the variables, and integrating, fzv dv _ i~dx J iv z ~J x i. e., log (i v z ) log#+log C rthe substitution being u = i v 2 or log#(i v z ) = logC = log K I du=2vdv i.e., x(iv z ) = K or and x z y* = Linear Equations of the Second Order. Let v - **, then - Ae^ and - X ^ 2 so that XV^+aAe^+fce** = o or X 2 +X+6 = o for jy = o would be a special case. There are three possible solutions to this quadratic. 282 MATHEMATICS FOR ENGINEERS The general solution is : X = = ^ i v a z 46 and let X, = We shall now discuss the three cases. Case (i). If # 2 >4&, then Xj and X 2 are real quantities and unequal. d V tt"V Now if v = A, e^ x . -r4 + a-r- + by will equal o, as would be the 1 dx z dx case also if y = A 2 e^, so that to complete the solution the two must be included (for the equation is true if either or both are included). Thus y = Ajtf*!* -\- A%e^ 2X the constants Aj and A 2 being fixed by the conditions. Case (2). If A 2 = 46, then X 1 = X 2 . According to the preceding case we might suppose that the solution was y = Ae^. This, however, is not the complete solution, which is y (A+B*)^. Case (3). If a z <4b. This means that Va 2 46 is the square root of a negative quantity, i. e., it is an imaginary. Now, a 2 4& = 1(46 a 2 ), (46 a 2 } being positive ; hence Va 2 46 = V^i Vtfa 2 a+j\/4ba z and a A Use might be made of the solution to C<zs0 (i), adopting these values of X a and X 2 , but this does not give the most convenient form in which to write the solution. Let c=V4& a z then Xj = x rt ;c and X, = i- SIMPLE DIFFERENTIAL EQUATIONS 283 Then y = A^i^-f- A 2 e*& from Case (i) (-a+jc)x (-a-jc)x Developing one of these only, viz., the first, and neglecting for the time being, -ax+jcx _ax jcx e 2 = e ~ 2"x~2". Now ei* = cos x -\-j sin x (see p. no), and by writing for x & cx, . cx e 2 = cos \-ism 2 2 Hence "" ' cx , . cx\ , -/ cx . . cx\ cos +7 sm + A. 2 e 2 / cos j sin ) < 2 2 / \ 2 2 / where A = V(A 1 +A a ) a +y*(A 1 A 2 ) 2 = 2 VA^A Z . (see Part I, p. 277) and Taking as the standard equation and grouping our results, we have the following (i) If a 2 >4&: the solution is y == A l6 (2) If a 2 = 46 : the solution is _ax y= (A+Bx)e ~*. (3) If a z <4b: the solution is 284 MATHEMATICS FOR ENGINEERS The last of these forms occurs so frequently that very careful consideration should be given to it, and to the equation of which it gives the solution. Example 15. Solve the equation d z y , dy ^- 2V = O. dx f This can be written (after dividing through by 5) d z y , dy -3-4+2-4-^ 4V = o, dx 2 dx so that a 2-4 and b -4 (in comparison with the standard form). -2-4+v / 5 7 76+l : 6, | . -2'4-v / 6-76+l-6. y = A lS It is really easier to work a question of this kind from first principles rather than to try to remember the rule in the form given ; thus the values of X will be the roots of the equation 5\ 2 + I2X 2=0. Then, calling these roots Xj and X 2 respectively If the values of A t and A 2 were required, two values of y with the corresponding values of x would be necessary. Example 16. A body is moving away from a fixed point in such a way that its acceleration is directed towards that point, and is given in magnitude by 64 times the distance of the body from that point. Find the equation of the motion and state of what kind the motion is. The motion is Simple Harmonic. (See p. 60.) If s = displacement at time t from the start = acceleration and = 645 (the reason for the minus sign being obvious). d z s d z y dy Thus ^+645=0, which is of the type -^-}-a~-\-b=o, where a=o and 6=64. If s = e* X 2 + 64 = o so that X = V 64= 8j. SIMPLE DIFFERENTIAL EQUATIONS 285 Hence the solution (according to Case (3)) is _?? y Ae 2 sinf -- \-p\ where a = o and c V^ba* 16 so that s = A sin (8t+p) . The general equation of S.H.M. is s = A sin (o>t-}-p) &> being the angular velocity, so that the angular velocity in this case is 8 and the amplitude is A. Example 17. Solve the equation Let y = e** then X 2 +8X+i6 = o i.e., (X+ 4 )2 = o i. e., the roots are equal. Hence y = (A+~Bx)e>&, (Cf. Case (2), p. 282) where X = 4. hence y = (A-\-~Bx)e~* x . Example 18. Solve the equation d z v dy This differs from the preceding examples in the substitution of a constant in place of o. The equation can be written d z y , dy . 5^+7i+io(y--5)=. Let (y~5) = e* x then ( Q. = \ e te and ^ dx dx z and X 2 -|-7X-j-io = o whence X = 5 or 2 then y- -5 = A^ - fa + A z e or = Ag- //y /-y In other words, the solution is that of -=^j + 7 - + ioy = o plus tt# i**V 5 . the constant dy j ^ , t. e., - ^^ -- T-. This is correct because, if v = -5, -^ and -v4 10 coefficient of y dx dx 2 each equals o, and thus one solution is y -5. The complete solution is the sum of the two solutions. 286 MATHEMATICS FOR ENGINEERS The Operator D. The differential coefficient of y with respect to x may be expressed in a variety of forms : thus either -, -j , f'(x) or Dy might be used to denote the process of differentiation. The last of these forms, which must only be used when there is no ambiguity about the independent variable, proves to be of great advantage when concerned with the solution of certain types of differential equations. It is found that the symbol D has many important algebraic properties, which lend them- selves to the employment of D as an " operator." The first derivative of y with regard to x = Dy, and the second d*y derivative of y with regard to x -j^, which is written as D 2 y ; D 2 indicating that the operation represented by D must be performed twice. This is in accordance with the ordinary rules of indices, so the fact suggests itself that the operator D may be dealt with according to algebraic rules. Thus D 3 must equal D.D.D (this implying not multiplication, but the performance of the operation three times) ; for .. dx* dx\dx 2 / dx dx dx Our rule then holds, at any rate, so long as the index is positive, or the operation is direct ; and for complete establishment we must test for the case when the index is negative. If Dy = d j? D = f:let^ = m, i. *., Dy = m. Then by integration y = I mdx, but if Dy = m and the rules of algebra can be applied to D m i y must = ^r or =- . m. Hence -^m Imdx ) W ~ \i or =- indicates the operation of integration. Again, if the rules of indices are to hold, D.D" 1 ^ must = Dy or y, hence D- 1 must represent the process of integration; since if we SIMPLE DIFFERENTIAL EQUATIONS 287 differentiate a function we must integrate the result to arrive at the original function once again. Hence D' 1 = ^. Having satisfied ourselves that the ordinary rules of indices may be applied to D, we may now prove that the rules of factorisation apply also. Taking the expression D 2 120+32, we can easily show that it can be written in the factor form (D 4)(D 8) : for let y = jx 2 5x, then Dy = 14* 5 and D 2 jy = 14. Also (D 2 120+32)? = D^y i2Vy+32y = 14 i68#-f 6o+224# 2 160* and 5 56*2+40*) 4(14* 5 = 14 II2#+40 56#+20+224# 2 I6OX so that (D2-I2D+32) = (D-4)(D-8). These properties make D of great usefulness in the solution of certain types of differential equations : e. g., Suppose 5^+7^ + ioy = M ..... . (i) then this equation may be re-written as M5D 2 +7D+io)=M M y = 5 D 2 + 7 D+io and the solution of equation (i) may be found by this artifice. Many differential equations occurring in electrical theory may be solved in a very simple manner by the treatment of D as a " quasi- algebraic" quantity: before proceeding to these, however, we must enunciate the following theorems. 288 MATHEMATICS FOR ENGINEERS Useful Theorems, involving the Operator D : (1) (p+qD) operating on the function a sin (bt+c) gives the result aVp 2 +b 2 q 2 sin f^+c+tan" 1 ^ ) \ p ' (2) rp. a sin (bt+c) = _ - sin (bt+c tan- 1 -). p+qV ^*+b 2 * P' Proof of (i). (p+qD)a sin (bt+c) = ap sin (bt+c)+aqb cos (bt+c} = aVp*+q*b z sin (bt+c+ia.^^ (See Part I, p. 277.) \ p/ Proof of (2). sn J j> sin (U+c)bq cos J "1 sin a sin ( 6^+c tan- 1 - V As a test of the correctness of the above rules the combination of the two operations should give the original function. Thus p-\-qu sn = a sn A third theorem might thus be added. D sin (bt+c) = 6 cos D 2 sin (&^+c) = Z> 2 sin or D 2 = -6 2 hence p z q 2 D z = p 2 +q z b z . SIMPLE DIFFERENTIAL EQUATIONS 289 Application of these Rules to the Solution of Differential Equations. d^y dy Example 19. Solve the equation -5-^+7^4-12^ = e Sx . This equation might be written S0that ^ The solution of this equation gives the particular integral, whilst the complementary function, as it is termed, will be obtained by the solution of the equation dx* dx [The solution of this equation we know from the previous work to be Now T>e 5x = 5e Sx , DV 5 * = 250^, i. e. t D = 5 and D 2 = Hence the particular integral is ~ 25+35+12 ^g5* ~7*' Hence the general solution is To test this by differentiation of the result : 2 e-* ;t + e &x = e* x . d z s ds Example 20. Solve the equation ,- +4^7+45 = 5 sin 7/. (This type at dt of equation occurs frequently in electrical problems and in problems on forced vibrations of a system.) U 290 MATHEMATICS FOR ENGINEERS /72 c /7c The solution of --2+4-^7+45 = (It Cli is s = (A+Bt)e~ zt (See p. 283.) To find the particular integral : (D 2 + 4 D + 4 ) 5 = 5 sin jt. _5sin jt -DM-4D+4 D sin jt = j cos jt and D 2 sin jt = 49 sin jt. (Note that D 2 = 49, but D does not = j.) We must thus eliminate D from the denominator: to do this, multiply both numerator and denominator by D 2 +4 40. Then _5(P 2 +4-4P)sin 7 / (D 2 + 4 ) 2 -i6D 2 _ 5( 49 sin 7^+4 sin jt28 cos jt) (-49+4) 2 -(i6x- 4 9) _ 5(45 s i n 7^+28 cos jt) 2809 _ 28 45 2809 Hence the complete solution is 53 A' ' in ~ 1 4-5> 53 ~~V ' " dn 45A The particular integral might have been found in a more direct fashion by the use of Theorem 2, p. 288. For sin 7* i (D+2) 2 -(D+ 2 ) (7\ (4) = 2 jt tan" 1 -), _,, _ 2j from Theorem 2, J ? = sinf jt tan -1 -J p. 288 I c = o U = 7 I I / 7 7 = 7= X 7= sin 1 7i tan" 1 - tan -1 - V 53 V 53 V 2 2 = sin ( jt 2 tan" 1 - Y 53 V 2} SIMPLE DIFFERENTIAL EQUATIONS 291 The two results do not appear at first sight to be the same, but the can be reconciled in the following manner : tan" 1 = tan" 1 -6223 = 31 54' tan- 1 ? =74 3' Thus sin (7/4- tan- 1 ) = sin (7/4-31 54') \ 45' also sin (7* 2tan~ 1 -J = sin (jt 2x74 3') = sin (jt 148 6') = sin (1804-7^ 148 6') = - sin (7/4-31 54') = sin ( 7/4- tan" 1 ). v 45/ IT Tl, -.-' T^T&y , T- , B/ 1TX Example 21. The equation EI- r ^4-Fy4--^- cos-=- = o occurs in dx o I Mechanics, y being the deflection of a rod of length /, and F being the end load. Solve this equation. ,, T d z y , F B/ TT* We may rewrite the equation as -r^+^fV = B^F cos -; I 1 ) dx z EF 8EI / The solution of +-^y = oisy = Asui/\ j^x+p ... (2) Reverting to form (i) B/ nx B/ or 8EI C<S B/ irX 8- cos F B/ Hence the complete solution is y = A sin ( *J ~pX-{-p j4- 8 COS T 292 MATHEMATICS FOR ENGINEERS Example 22. A pin-jointed column, initially bent to a curve of cosines, has a vertical load W applied to it. Find an expression for the deflection at any point. Given that the equation of the initial bent form is irX y being the deflection at distance x from the centre of the column, which is of length /. Also this equation being obtained from a consideration of the bending moment at distance x from the centre. , and thus +--A cos W AW fnx iLi y = -ET COS firX\\ (y J) = o fnx\ (T> Now, as shown on p. 283, the solution of the equation d* W W To find the particular integral, viz., the solution of the equation d* . W AW write the equation as AW AW /irX\ -^-= COS ( ) so that = - \ l > AW l-nx El I* Combining the two results (r). SIMPLE DIFFERENTIAL EQUATIONS 293 The following example combines the methods of solution employed in Examples 17, 18, 19 and 20. Example 23. Solve the equation d 2 s ds = -+ sin 607T/+5. (a) The solution of -5^ 12-^ + 205 = is s = A. 1 e lot +A z e zt . (b) The particular integral for d?s ds d7*- 12 di may be thus found : D 2 I2D+20 D 2 e~ 5t = 2$e- i 25+60+20 (c) To find the particular integral for d*s ds -jr. i2^-+2os = sinooiri. dt z dt D 2 I2D + 20 I ^ sin6o7r< D 10 D 2~ ^ ~/' = \ ==,; sin [6o7rf tan" 1 (3071-)] \q = i J- D-io V4+36007T 2 l& = 6o^J i i /- -.- , -^ ' -/ ^~z a sin [607T/ tan" 1 (30^) tan" 1 ( 6rr)l v4+3ooon- a Vioo+3ooo7r i sin [6orr< tan^ 1 (3077) tan" 1 (677)] 2oV(l+9007T 2 ) (I + 367T 2 ) (d) The particular integral for d z s ds is 20 4 Hence the complete solution is the sum of those in (a), (b),-(c) and (d), viz., s = A e"'+A g ^| g ~ 5 1 2 105 _ 2oV(i+90on 2 ) (I+367T*) 2 9 4 MATHEMATICS FOR ENGINEERS Equations of the Second Degree. The treatment of these equations is very similar, up to a certain point, to that employed in the solution of ordinary quadratic equations ; particularly the solution by factorisation. Example 24. Solve the equation ^Y- 8^-33=0 dx/ dx Let Y= then Y 2 -8Y~33=o dx or (Y-ii)(Y+ 3 ) = o. Thus Y = 1 1 or Y = 3 dy dy i.e., / = ii or ^-= 3, dx dx whence y = i ix + C t or y = $x + C 2 or y i ix C x = o y + 3, C 2 = o, and the complete solution is the product of these two solutions, since the equation is of " degree " higher than the first. Thus the solution is (yiixCJ (y+3x-C z )=o. (dv\ 2 Example 25. Solve the equation 5! - ) 8y 8 =o. \ax/ 3\ A v I * Dividing by 5 Factorising \-~^- 1-26^ )(-/- 1-265^) = - \CiX / \CLX / Hence -^-+1-265^^ = or -^-1-265^ = 0. Separating the variables and integrating /^+i'265/^ = o or -% 1-265 I dx = J yl J J yi J whence the complete solution is, Two further examples are added to illustrate methods of solu- tion other than those already indicated. SIMPLE DIFFERENTIAL EQUATIONS 295 Example 26. Solve the equation -j-^ -^ 14-^+2 AV = o. dx 3 dx* dx The equation may be written (D 3 D 2 i.fD-f 24)^ = o or (D-2)(D-3)(D+ 4 )y = o, whence y = Example 27. The equation j^= m*y occurs in the discussion of the whirling of shafts. Solve this equation. i. e., D 4 = m* or (D 2 m 2 ) (D 2 +m 2 ) = o, whence D= m or jm. Hence y = a 1 e mx -^-a z e~ mx -\-a 3 e jmx -\-a t e- intx . But e' x = cos x+j sin x, e jnue = cos mx+j sin mx and e x cosh AT+ sinh #, i.e., e nix = cosh w# + sinh mx. y = ^ (cosh w#+ sinh m#) +a(cosh mx sinh w#) +a 3 (cos w# +y sin ra#) + 4 (cos mxj sin w#) = ( a a+ a 4) cos wwr-f (a 3 a 4 ); sin w#+ (i+^ 2 ) cosh mx + (! a 2 ) sinh WAT = A cos m# -f- B sin mx-\-C cosh m^+D sinh mx. The constants A, B, C and D are found by consideration of the conditions; four equations must be formed, these being found by successive differentiation and by substituting for f- , -^ and ?? their dx dx z dx 3 values for various values of x. Exercises 22. On the Solution of Differential Equations. 1. If -p = 5# a 2-4 and y = 1-68 when x = 2-29, find y in terms of x. d z s ds 2. Given that = 16-1 ; -=7 = 4-3 when t = 1-7 and s = 9-8 when f = -2, find s in terms of t. 3. If ^ = 8y+5, find an expression for y. 296 MATHEMATICS FOR ENGINEERS dv 4. Given that 8-76-^- +9* 15^ = 76-4 and also that ^ = 2-17 when G/X x = o, find an expression for y in terms of x. 5. A beam simply supported at its ends is loaded with a concentrated load W at the centre. The bending moment M at a section distant x from the centre is given by W/ 1 \ M = 4 - x }. 2 \2 / M d z v If == = -j^, find the equation of the deflected form, y being the deflection. 6. For the case of a fixed beam uniformly loaded, M, the bending (w(P ,\ ) T , M d*y dy I moment, =1-1 x 2 1 K I . If ==- = ^; -f- = o when x = - and also 1 2X4 / El dx z dx 2 when x = o ; and y = o when x = - , find an expression for y. 7. Solve the equation dx i the limits to T being T x and T 2 , and to x being o and / ; the remaining letters representing constants. 8. If -j- = 1- and v = o when x s, find v in terms of x. ax 4//x 9. If T! = the absolute temperature of the gases entering a tube of length / and diam. D, r 2 = the absolute temperature of the gases leaving this tube, 6 = temperature of the water, Q = amount of heat transmitted through the tube per sq. ft. per sec. per degree difference of temperature on the two sides, w = weight of gases along the tube per sec., and s = specific heat of gases, , dr QnT^dx then T H = o. r ws Find an expression for Q, x being the distance from one end of the tube. 10. Find Q if Q^-nDdx-^-wsdr = o (the letters having the same meanings as in No. 9, and the limits being the same). 11. If T-J and r 2 are the inside and outside temperatures respectively of a thick tube of internal radius r t and external radius r z , then rfr H dx ~~ 27rK I I is the length of the tube, H is a quantity of heat, and the limits to are o and r z r v Find an expression for H, K being a constant. SIMPLE DIFFERENTIAL EQUATIONS 297 12. A compound pendulum swings through small axes. If I = moment of inertia about the point of suspension, h = the distance of C. of G. from point of suspension, then I X angular acceleration f i. e., 1^) = mhd. Find an expression for 6. If p. = couple for unit angle mh, prove that t, the time of a com- plete oscillation, = 2/r \/ (in Engineers' units). r O 13. To find expressions for the stresses p and q (hoop) in thick cylinders it is necessary to solve the equation Solve this equation for p. 14. For a thick spherical shell, Up radial pressure, Find an expression for p, a being a constant. 15. If ~K. tt vdp = ~K. p pdv, prove that pv = constant, y being the ,. the specific heat at constant pressure .. K D ratio -rr - r= r - of a gas, i. e., ==? the specific heat at constant volume K B 16. Solve for z in the equation dz . w w 2 3" H --- 7- zx - dx g f 17. Solve the equation d z y dy ^-i7^ and thence the equation d z y dy d^- l7 Tx d z s 18. Solve the equation 875. at d z s 19. Solve the equation ^73+875 = o. (tt 20. Find the time that elapses whilst an electric condenser of capacity K discharges through a constant resistance R, the potential difference at the start being t^ and at the end v z , being given that K x rate of change of potential = = jf* 21. If V = RC+L-r- and V = o, find an expression for C; C being the initial current, i. e., the value of C when t o. J/~* 22. If V = RC-f L -- , and V = V sin qt, find an expression for C. ctt 298 MATHEMATICS FOR ENGINEERS 23. If -y- = -Vy z +2ay, find x in terms of y, a being a constant. dx a 24. An equation occurring when considering the motion of the piston of an indicator is ffix _a^ pa. M ^ + SM M Solve this equation for x ; M, a, S and p being constants. 25. If - Py = Eig. (an equation referring to the bending of struts), find y; given that x = o when y o, and v = Y when x = 26. To find the time t of the recoil of a gun, it was necessary to solve the equation = n Vx z a 2 . dt If a = 47-5, n 3-275 and the limits to x are o and 57-5, find t. 27. Solve the equation -^-\-2/~-\-n z x o. (Ait (Jit dx Take w 2 = 200, /= 7*485 ; also let x = o and = 10 when t = o. Cut 28. The equation ^2"+ 2 /^i +n z x = a sin qt (It Q/v expresses the forced vibration of a system. If n 2 = 49, / = 3, q = 5, find an expression for x. d z V r 29. Solve the equation -,-= V = o. dx 2 r z 30. If H is the amount of heat given to a gas, p is its pressure and v its volume (of i lb.), then -,~ = 7 -Av^+np). Assuming that dv (n i) V dv */ there is no change of heat (i.e., the expansion is adiabatic and -5 = o J, find a simple equation to express the connection between the pressure and volume during this expansion. 31. Newton's law of cooling may be expressed by the equation 5" *<- where k is a constant, and 6 a is the temperature of the air. If Q = Q when t = o, find an expression for 6. 32. The equation 1-04-^ +12-3 +13^634=0 (It Ctt occurred in an investigation to find 6, the angle of incidence of the main planes of an aeroplane. If t = o when 6 = i and = o when t = o, find an expression for 6. at SIMPLE DIFFERENTIAL EQUATIONS 299 33. A circular shaft weighing p Ibs. per ft. rotates at o> radians per second, and is subjected to an endlong compressive force F. The deflection y can be found from the equation d*y IF dfy _ p u d^ + mfa* # Solve this equation for y. 34. An equation relating to the theory of the stability of an aero- plane is dv = g cos a kv Civ where v is a velocity ; g, a and k being constants. Find an expression for the velocity, if it is known that v = o when t o. CHAPTER X APPLICATIONS OF THE CALCULUS THE idea of this chapter is to illustrate the use of the Calculus as applied to many Engineering problems ; and the reader is supposed to be acquainted with the technical principle involved. The various cases will be dealt with as though examples. Examples in Thermodynamics. Example i. To prove that (V w) = =5-, an equation occurring T GtfT in Thermodynamics, where L = latent heat at absolute temperature T, V = vol. of i Ib. of steam at absolute temperature T, w = vol. of i Ib. of water = -016 cu. ft., P = pressure. A quantity q of heat taken in at r-f-8r and discharged at r will, according to the Carnot cycle, give out work = q r r - or approxi- T-j-OT 4. 1 8f mately q . T Hence for i Ib. of steam at the boiling temperature, Sr 87- work = q = L , T T but the work done = volume of steam in the cylinder x change in pressure Hence (V-w)8P and thus V w= r^, T oir or, as Sr becomes infinitely small, X7 . L,dr V w= T dP Now -Tp is the slope of the pressure temperature curve (plotted from 300 APPLICATIONS OF THE CALCULUS 301 the tables) and can be easily found for any temperature r. Hence V can also be found. A numerical example will illustrate further. It is required to find the volume of i Ib. of dry steam at 228F., i. e., at 20 Ibs. per sq. in. pressure. From the /When P= 19, * = 225-3, r= 460+225-3 = 685-3. x 1.1 -^ .. P = 20,* = 228, T = 688. steam tables ' .. - ^ P = 2i, t= 230-6, T = 690-6. Plotting these temperatures to a base of pressures, we find that the portion of the curve dealt with in this range is practically straight. r 69O 689 688 687 686 685 / 3 f / / 2-65 t s i H ! s /* / S 7 /9 eo p e FIG. 103. Problem in Thermodynamics. The slope of this line = 2-65 (Fig. 103), and this is the value of -==, P being given in Ibs. per sq. in. and the latent heat in thermal units. To change the formula to agree with these units, 778L dr v = Also L at 228 F. = 953. V = -oi6-f dP' 778x953x2-65 144x688 = 19-82 cu. ft. Example 2. To prove that the specific heat of saturated steam (expanding dry) is negative. Let Q = the quantity of heat added. H = total heat from 32 F. I = internal energy of the steam. 302 Then i.e., or Now Then and MATHEMATICS FOR ENGINEERS H = internal + external energy I = H-PV 81 = 8H-8(PV)=SH-(P8V+VSP). = SH-PSV-V8P+PSV = 8H-V8P <\T r -*-* <\ 8r L r L (from. Example i, neglecting^ \ w, which is very small J * H = dT^' 305 Now the specific heat = heat to raise the temperature i hence L. = -305 - - and since - is greater than -305, 5 is a negative quantity. E. g., if t = 300 F., i. e., r = 761 F. absol., = 1115 210 = 905. 95 .'. s = '305 ^-^ 3 J 761 = - -882. Work Done in the Expansion of a Gas. Example 3. Find the work done in the expansion of a gas from volume v l to volume v a . There are two distinct cases, which must be treated separately; but for both cases the work done in the expansion is measured by the area ABCD = % areas of strips like MN (Fig. 104) = ^,fp &v or I p dv more exactly. J Vl Case (a), for which the law of the expansion is pv = C. /'2 /"2 / \"2 done = / pdv= I Cv~*dv = Cl log v } Jit* J Vi \ h Jvi Work or C log r r being the ratio of expansion, and = . Thus the work done = pv log r. APPLICATIONS OF THE CALCULUS 303 Case (b), for which the law of the expansion is pv n C, n having any value other than i . 1 ** 1 M FIG. 104. Work Done in the Expansion of a Gas. V FIG. 105. Work done = / 2 pdv = \ *Cv~ndv J t7j Dj i n i in i (p( i n Pz v z Pi in Work Done in a Complete Theoretical Cycle. Example 4. Find the work done in the complete cycle represented by the diagram FGAB in Fig. 105. The work done = area GABF = ABCD+GADH FBCH in 304 MATHEMATICS FOR ENGINEERS Note that, if n = ^| 16 work done == If the expansion is adiabatic, and n is calculated according to Zeuner's rule, n = (q being the initial dryness fraction). If q = i, then n = i-O35+-i = i - i35, so that the work done To Find the Entropy of Water at Absolute Temperature r. Example 5. When a substance takes in or rejects heat (at tempera- ture T) the change in entropy 8$ = (8q = heat taken in) . Let a = specific heat, then o-Sr = 8q. dq Change in entropy from r to T = f J TT dr . T o r_ - /i.e., in the change from\ For steam, the heat taken in at T = L { ). \_ \ the liquid to the gas / Hence the change of entropy = Efficiency of an Engine working on the Rankine Cycle. Example 6. Find the efficiency of an engine working on the Rankine cycle ; using the T(f> diagram for the calculation. Work done = area of ABCD (Fig. 106.) = ABCK+ADMN-DKMN = * x (TJ T 2 )+heat taken in from TI Tjtorj (r t x'DK). Now DK = the change in entropy from water at T Z to water at T X = loge as proved above. APPLICATIONS OF THE CALCULUS 305 Hence the work done = ^-^ (T I r 2 ) + (r 1 T 2 ) r a log The heat put in and thus the efficiency TJ = r r being the dryness fraction at FIG. 106. FIG. 107. Efficiency of an Engine. Efficiency of an Engine working on the Rankine Cycle, with steam kept saturated by jacket steam. Example 7. Find the efficiency of the engine whose cycle is given by abcf in Fig. 107. Work done = area abcf _ f'aLj /the summation being of\ ~~ j TI T \ horizontal strips ) [for L = 1115 -7/1 dr =1437 7r = a+br = alog- a +6(r 2 TJ) Total heat received = L 2 +r 2 rj total heat rejected = Lj Hence the work done = (i) (z). . . (i) (2) 306 MATHEMATICS FOR ENGINEERS from which H,- = a log +6(r 2 TJ} (L 2 L x ) (r 2 T i = a log -+6(T 2 T X ) (a+&T 2 a 6 - T l Now the total heat received = L 8 +r a = L 2 +r a Hence where a = 1437 and b = "j. Example 8. To prove that the equation for adiabatic expansion of air is pv y = C, where _ specific heat at constant pressure _ K p ' specific heat at constant volume K w Dealing throughout with i Ib. of air, let the air expand under constant pressure from conditions p 1 v 1 r x to p^ v r. Then the heat added = K p (r r x ) = K/^- ^~ Now keep the volume constant at v, and subtract as much heat as was previously added : then the pressure falls to p z and the tem- perature to r a . The heat subtracted = K,(i T,) = K v (^- Now, if the changes are regarded as being very small, we may write for v v l and 8p for pip z and thus K v v&p = K p p8v , (dp Kp rdv whence / -~ = ~ / / P KJ i; log p = y log w + lg (constant) i. e., /> = Cy-v or v~i = C. APPLICATIONS OF THE CALCULUS 307 Examples relating to Loaded Beams. Example g. Prove the most important rule M_E_ d*y I R~ dx z applied to a loaded beam ; M, I and E having their usual meanings, and R being the radius of curvature of the bent beam. Assuming the beam to be originally straight, take a section of length / along the neutral lamina, and let l-\-8l be the strained length at distance y (Fig. 108). A - li\ FIG. 108. Problem on Loaded Beam. Then, if R = radius of curvature, 1+81 _ whence or but and thus I R 1 + 81 R I ~R stress / / /R 1-4 - - _ . J "i. _ - . **_ _ strain 3/. y y I R R M / but it has already been proved (see p. 239) that y = Hence M = E i ~R' 308 MATHEMATICS FOR ENGINEERS The total curvature of an arc of a curve is the angle through which the tangent turns as its point of contact moves from one end of the arc to the other ; and the mean curvature is given by the total curva- ture divided by the length of arc. In Fig. 108 8$ = total curvature for the arc 65, and the mean curva- S</> ture = -5-- 8s We know that the slope of the tangent is given by ~ flwv .-. tand> = dy -f- dx Now , , tan $ = sec 2 $ and , ds 2 sk ^^ ^ i^y SCC O ~^ ~^ ~5~ I ~~5 tis as \a^ i d (dy\ sec 2 X s = x I T^ ) R as Va^r/ dx\dx) ds dx* ds Hence ^ = -^~ X - x cos 2 R dx* ds d*y dx* d<b X --= ds dy tan (b = - dx When, as for a beam, -~ is very small, (-*- j may be neglected in comparison with i, and hence This result may be arrived at more briefly, but approximately, in the following manner : 80 = 8 tan very nearly (when the angle is very small) . Hence ~- = -- -- = - tan <6 = rate of change of the tangent (for 8s 8s 8x PM and PQ are sensibly alike). APPLICATIONS OF THE CALCULUS 309 Thus ^^A.^^^y ds dx dx dx* i d 2 y and _ = _. R dx z M = / = E_ F ^ I y R dx* In the use of this rule there should be no difficulty in finding expressions for y in terms of x in cases in which the beam is simply supported; for an expression is found for the bending moment at distance x from one end, or the centre, whichever may be more con- pi venient, and then the relation M d 2 y =- = E -n FIG. 109. Beam Uniformly n v& T j j w;t Loaded, is used ; whence double integration from the equation so formed gives an expression for the deflected form. A few harder cases are here treated, the beam not being simply supported. Example 10. A beam is fixed at one end and supported at the other ; the loading is uniform, w being the intensity. Find the equation of the deflected form. We must first find the force P (part of the couple keeping the end fixed) and then combine this force with the reaction at B calculated on the assumption that the beam is simply supported. Referring to Fig. 109: If the beam is simply supported, the bending moment at distance x t -n Wl WX 2 from B = x 2 2 Hence the actual bending moment ,, wlx wx 2 _ = M = Px 2 2 -~~d 2 y wlx wx 2 .._. i.e., El ^-4= Px dx* 2 2 whence, by integration, _ T dy wlx 2 wx 3 Px 2 . ~ EI-j^- = hQ dx 4 62 3io MATHEMATICS FOR ENGINEERS but -f- = o when x = I ax for the deflected form is horizontal at this end. wl 3 wl 3 PI* , o = f-C, 4 62 r _Pl 2 wl 3 1 ~~ 2 12 -rfrdy _ wlx z wx 3 Px z .Pl z wl 3 dx 4 6 2 2 12 Hence T , , . _ ze># 4 P# 3 , Pl 2 x wl 3 x . Integrating, Ely = ^-C 2 . 12 24 6.2 12 In this equation there are the two unknowns P and C 2 , and to evaluate them we must form two equations from the statements y = o when x = o and y = o when x = 1. If y = o when x = o, then it is readily seen that C a = o. wl* wl* PI 3 , P/ 3 wl* Also if x = l o = ----- ? ---- 12 24 6 2 12 1, T, W l whence P = - 5 - o Point' of Conhraflexure /_ Fixed End FIG. no. Deflected Form of Beam. If the beam were simply supported, the upward reaction would be wl , ,, ,, , . wl wl 3 . , and thus the net reaction = 5- = ^wl. 2 288 Substituting -5- in place of P in the expression for y, we arrive at o the equation of the deflected form the curve for which is shown in Fig. no. We may now proceed to find where the maximum deflection occurs, and also the position of the point of contraflexure. dy w dx 48EI ' dy and T X = the solution of which, applicable to the present case, is x -423^ APPLICATIONS OF THE CALCULUS 311 064 -423) The maximum deflection is thus wl* 005 4 wl* ~EI~ To find C the point of inflexion or contraflexure and 3-^ = o if x = o or if x = /. Example n. A beam is fixed at one end and supported at the other, the loading and the section both varying. Find the equation of the deflected form. Let m = bending moment at a point distant x from B if the beam were simply supported, and let P= the force of the fixing couple. (See Fig. 109.) Then M = mPx i. e., Elf^ = W _ P * , . dx z .~d z y m Px E ^ = I-T the equation being written in this form since I is now a variable. By integrating E^= [ X ^dx-P F^dx+Cj. ...... (i) t*x J I J 1 dy Now -j ss o when x = 1. dx i. e., Cj can be found, for the two integrals may be evaluated. By integrating (i), Ey = f f2(rf*)-pr f \ (d^+c.x+c,. J o .' o L JoJp 1 But j/ = o when x = I and also when x = o, and thus C 2 o and J oJ o i. e., P can be found. The integrations must be performed graphically and with extreme care, or otherwise very serious errors arise. Example 12. A beam is fixed at both ends and the loading and the section both vary. Find the equation of the deflected form. 3 i2 MATHEMATICS FOR ENGINEERS Let Wj and m 2 be the end fixing couples ; then to keep the system in equilibrium it is necessary to introduce equal and opposite forces P (Fig. in), i. e., P/+w 2 = Wj. Let m = the " simply supported " bending moment at section distant x from the right-hand end. Then M = m m z Px d z v and consequently El-r^ = mm z Px d*m m Px Integrating, _ E dy c x m, fxdx ^f x x /- = / -=-dxm t l -V- P/ f dx . o I *J I J I ,4 -I FIG. in. Now -jL*mQ when x = o or dx hence C x = o (taking ^ = o) and also, taking x = I, Integrating again, Ey = / r^w-m. r r j w-p r ff w J QJ * J OJ i J 0-' X Now y = o when # = o and also when x = I. Then taking x = o, C 2 = o, and taking # = /, From equations (i) and (2) the values of w a and P (and hence Wj) may be found, the integration being graphical (except in a few special cases) ; and again it must be emphasised that the integration must be performed most accurately. APPLICATIONS OF THE CALCULUS 313 Example 13. A uniform rectangular beam, fixed at its ends, is 20 ft. long, and has a load of 10 tons at its centre and one of 7 tons at 5 ft. from one end. Find the fixing couples and the true B.M. diagram. This is a special case of Example 12 since the section, and therefore I, is constant. The B.M. diagram for the beam if simply supported would be as ABCD (Fig. 112). The bending moment diagram, due to the fixing couples only, would have the form of a trapezoid, as APQD. Unless the integration, explained in the previous example, is done extremely carefully, there will be serious errors in the results ; and since there are only the two loads to consider, it is rather easier to FIG. 112. Fixing Couples and B.M. Diagram of Loaded Beam. work according to the Goodman scheme. [See Mechanics applied to Engineering, by Goodman.] According to this plan : (i) the opposing areas (i. e., of the free and fixing bending moment diagrams) must be equal ; and (2) the centroids of the opposing areas must be on the same vertical, i. e., their centroid verticals must coincide. To satisfy condition (i), Area of ABCD = (^X5X 5 i-2 5 )+( 5I ' 25 + 67 ' 5 X5)+(^Xiox67-5) = 762-5. Area of APQD = - x 20 x (w 1 +m 2 ) where m^ = AP\ and w = D -> 10 Equating these areas, m 1 -{-m t 76-25 (i) 314 MATHEMATICS FOR ENGINEERS To satisfy condition (2), taking moments about AP, r Moment of ABG = -X5I-25X-X5 = 427 Moment of BGH = -X5i - 25X/5H X5) = 8 54 r / 2 \ Moment of BHC = -x 67-5x1 5H X5j =1405 Moment of DCH = X 67-5 x ( io-| x 10 )= 4500 2 V 3 / i. e., total moment of ABCD about AP = 7186 , . _._ 20 20 200 Moment of APD = x w, X = n. 2 3 3 Moment of DPQ = x m z X = m Hence (2) 200 .400 ^1 + m 2~ The solution of equations (i) and (2) for m^ and w 2 gives the results m i = 44-7 1 and m 2 = 31-54. Thus PQ is the true base of the complete bending moment diagram, AP being made equal to 44-71, and DQ equal to 31-54. Shearing Stress in Beams. Example 14. To find an expression for the maximum intensity of shearing stress over a beam section. The shearing stress at any point in a vertical section of a beam is always accompanied by shearing stress of equal intensity in a hori- zontal plane through that point. FIG. 113. Shearing Stress in Beams. FIG. 114. We require to know the tangential or shearing stress / at E on the plane CEC' (Fig. 113) ; this must be equal to the tangential stress in the direction EF on the plane EF at right angles to the paper. Suppose that the bending moment at CC'=M and that at DD'=M+SM. Then the total pushing forces on DF > total pushing forces on CE, the difference being the tangential forces on EFE/. APPLICATIONS OF THE CALCULUS 315 Let P = the total pushing force on ECE' ,RC (stress = M "j then P = / (stress) x area 1 y IV J RE I area = bdyj RE J- M /" RC M = v / by dy = =- x ist moment of area ECE'. I. 'RE Now the tangential force on EFE' = stress x area =/xEE'x8* and this must equal the difference in the total pushing forces on DF and CE, i. e., 8P. Hence 8P = /x EE' x bx i. e., -=r- x ist moment of area ECE'=/ x EE' ; but = rate of change of B.M.=shear=F (say). Hence the maximum intensity of shearing stress/ ~F T = ist moment of area ECE'x ? X: I "BE* or, as it is usually written, ~ bl where S = an area such as CEE' and y = distance of its centroid from the neutral axis. Example 15. Find the maximum intensity of shearing stress, when the section is circular, of radius r (see Fig. 114). For this section I = V. 4 Applying the rule proved above : F ,-r since zr corresponds to EE' the maximum intensity = -^ J by dy in Example ^ du ('where u = r*y z dy~~ = - X mean intensity. 316 MATHEMATICS FOR ENGINEERS Example 16. A uniformly tapered cantilever of circular cross section is built in at one end and is loaded at the other. The diam. at the loaded end is D ins. and the taper is t ins. per in. of length. Find an expression for the distance of the most highly stressed section from the free end of the beam due to bending moment only. Neglect the weight of the cantilever. Let / be the length in ins. and W the load at the free end. Consider a section distant x ins. from the free end ; then the diam. here is D tx, and the bending moment W#. Also the value of I for the section considered is . _ My /D tx\ 64 Hence the skin stress f = -^- = Vfx( - IX -7-^--. I V 2 / 7r(D tx) K _ 32W D 3 - F ' ~ x (a constant), and / is a maximum when the denominator is a minimum. Let N = denominator then =_ dx dN and -j = o when 2t 3 x 3 ^Dt 2 x 2 -\-D 3 o d 'X i. e., when 2t 3 x 3 zl)t*x 2 D/ a ^ 2 +D 8 = o 2t*x z (tx-~D)-I)(t 2 .v 2 -~D z ) = o (te-D)(2^-Dte-D) = o (txT>)(2tx+T>)(tx'D) = o D D t. e., when x or --- . t it Thus the stress is maximum at a section distant ins. from the t free end. Example 17. To find the deflection of the muzzle of a gun. This is an instructive example on the determination of the deflection of a cantilever whose section varies. The muzzle is divided into a number of elementary discs, the volumes of these found (and hence the weights) so that the curve of loads can be plotted. Integration of this curve gives the curve of shear, and integration of the curve of shear gives the B.M. diagram. APPLICATIONS OF THE CALCULUS 317 The values of I must next be calculated for each disc, and a new curve plotted with ordinates equal to : then double integration of this gives the deflected form. It is necessary to use the ordinates of this curve to find, first the time of the fundamental oscillation and thence the upward velocity due to the deflection. Call the deflection at any section y, and the load or weight of the small disc w. Find the sum of all products like wy? and also find the sum of all products like wy. (Suitable tabulation will facilitate matters.) Then T (time of oscillation) If Y = maximum deflection, assuming the motion to be S.H.M. then the upward velocity v is obtained from vT = 27rY or v = 27rY . Examples on Applied Electricity. Example 18. Arrange n electric cells partly in series and partly in parallel to obtain the maximum current from them through an external resistance R. (Let the internal resistance of each cell = r, and let the E.M.F. of each cell = v.) Suppose the mixed circuit is as shown in Fig. 115, i. e., with x cells per row and there- 7t fore - rows. x Then the total E.M.F. of i row = ;n; and total internal resistance of i row = xr, 1 but as there are rows, the total internal x I-!' I'l H FIG. 115. Maximum Cur- rent from Electric Cells. resistance is - that of i row, i. e., the total x rx* internal resistance = ; but the E.M.F. is n unaltered : in reality the effect being that of one large cell, the area being greater and thus the resistance less. 3i8 MATHEMATICS FOR ENGINEERS total E.M.F. xv Hence the current C = total resistance rx n v ** , R" n x and C is maximum when the denominator is a minimum. Let D = the denominator, then dD r R , dD rx* -= = --- and -r- o when R = . ax n x z dx n i. e., external = internal resistance. Example 19. To find an expression for the time of discharge of an electric condenser of capacity K, discharging through a constant resistance R. Let v = potential difference between the coatings at any time t. v Then, by Ohm's law, the current C = = . But the current is given by the rate of diminution of the quantity q and q = Kv. dq dKv dv Hence C = -^ = -^-= -K- v Tjr dv and thus ^ = K-j- K at If V x = the difference of potential at the start, i. e., at / = o, and V 2 = the difference of potential at the end of T sees, Separating the variables and integrating, whence log = V t i or =J = e KR * 2 V 2 _i . e., ^~ = e KR or the time taken to lower the voltage from Vjto V 2 = Example 20. If R = the electric resistance of a circuit, L = its self -inductance, C = the current flowing and V = the voltage, then V = RC+L- APPLICATIONS OF THE CALCULUS 319 Solve this equation for the cases when V = o, C = C sin qt and For the case of a steady current V = RC since L is zero, and this corresponds to the equation of uniform motion in mechanics, whilst j/-\ the equation V = RC+L-j- may be compared with that for accelerated dt motion, V being the force. Thus the second term of the equation may be regarded as one expressing the " inertia " or " reluctance to change," and since the current may vary according to various laws, the rate of change can have a variety of values. Clt Dealing with the cases suggested : (i) V = o, then o = RC+L^- dt RC =~ L f and this equation is solved by separating the variables and integrating. Th l dC Thus p j whence = log e C+log e A = log e AC JLrf _Rt and AC e L , A being a constant. (2) If C = C sin 9* j/-* then = qC cos qt. dt V = RC sin ^+L^C cos qt = C VR 2 +L V sin U/+tan - 1 - then V sin qt = RC+L dl i. e., V sin qt = (R+LD)C where' D= . at Hence c V^ingf and, using Theorem 2 of p. 288, C = . V = s VR 2 +LV / dC - To this must be added the solution of o = RC+L , viz., C= Ke L ; at hence the complete solution is V 320 MATHEMATICS FOR ENGINEERS Example 21. To find expressions for the potential and the current at any points along a long uniform conductor. At a distance x from the " home " end let the steady potential to the ground = E and the steady current be i. Let the resistance of i unit of length of the conductor r and let the leakage of i unit of length of the insulation = /. Consider a small length of conductor 8x. Its resistance = r8x and the leakage = I8x. Hence the drop in potential ixr8x (i) and the drop in the current = E x I8x (2) i. e., from (i) 8E = ir8x - - ' and from (2) oi = El8x or -^ = -El (4) Writing (3) and (4) in their limiting forms dE _ _ . di _ dx dx .,- d 2 E d . . . di , . Differentiating, TT = j~\ tr i ~ ~ y ^T ~ r ^ '5' ct,\ d% dx and -^r- r= - 7 -( E/) = /-j = ril (6) dx z dx* dx To solve these equations, let D , ax then D 2 E = rlE i. e., D 2 = rl D = Vrl. dx dx Separating the variables, dE .- dE , - = Vrldx or = Vrldx. Integrating, log E = Vrlx+Ci or log E = X/y/tf+Cg or, if the constants are suitably chosen, E = A cosh Vrlx+'B sinh Vrlx. APPLICATIONS OF THE CALCULUS 321 In like manner, i = C cosh Vrl x+~D sinh Vrl x. When x = L E = A cosh Vr/L+B sinh Vr/L. When x = o E = A and hence the constants can be found. Examples on Strengths of Materials. Example 22. To find the shape assumed by a chain loaded with its own weight only ; the weight per foot being w. To find also expres- sions for the length of arc and the tension at any point. Let s = the length of the arc AB (Fig. 116) : then the weight of this portion = ws. \IVS 7 FIG. 116. FIG. 117. Draw the triangle of forces for the three forces T, T and ws (Fig. 117). Let it be assumed that T (the horizontal tension) we, where c is some constant. Then, from Figs. 116 and 117 dy _ dx = Now, as proved on p. 201, .. ws ws s , = tan 6 = =- = = - ax TO we c ds dx* sds 322 MATHEMATICS FOR ENGINEERS To integrate the left-hand side, let u = c 2 +s 2 du then r = 2s ds ( sds rsdu and / -7= = = / 1 = u* = V c 2 -1- s 2 ./ Vc 2 + S 2 J 2SU$ Thus, by integration of equation (i), Now at the point A (Fig. 116) 5 = and y = o hence Vc z = Cj or C x = c. Thus Vc 2 +s 2 = y + c. Squaring c 2 +s 2 = y*-\-c z +2yc or s z = y z +2yc and s Vy z +2yc s dy dy but as proved above -= -f- or s c^- c ax dx dv / = hence c-^-= V y z -\-2yc. Separating the variables dy dx _ __ ^_ e " c 2 ~ c r dy [ Integrating / /. , ., = = / J V+c 2 c 2 J /. , . V(y+c) c c and this integral is of the type discussed on p. 151 ; the result being c J c Now x o when y = o, x being measured from the vertical axis through A, and thus logf- j = C 2 or C 2 = o. Thus f = or in the exponential form X ce c = y-\-c-\- Vy z + zyc. Isolating the surd ce c (y+c) = Vy*+2yc. 2x x Squaring c z e c +y 2 +c 2 +2yc 2(y-\-c) ce* = y*+2yc 2x x_ or c z e c -2.ee* (y+c)+c 2 = o. APPLICATIONS OF THE CALCULUS 323 Dividing through by ce c ce c +ce c 2 (y-\-c), i.e., (y+c)= C -(e+e~). If now the axis of x be shifted downwards a distance c, then the c I X ' ~\ x new ordinate Y = y-\-c and Y - \e c +e c ) c cosh - v: Scale FIG. 1 1 8. Catenary Form of a Cable. Again, since Y = y +c = dx dx dx and also Then d ,x c . , x . .x ^-c cosh - = - sinh - = smh - ax cec c dy .X = sinh - dx c but it has already been proved that dy_ s henee dx c s . , x -=smh- or s = c c 324 MATHEMATICS FOR ENGINEERS To find the tension T at any point T 2 == w z s z +w z c z from Fig. 117 = w z (s z +c z ) = w z (c+y) z = w z Y 2 or T = wY. Thus the form taken by the chain is that for which the equation is fx\ ._^ Y = ccosh(-J, the equation of the catenary: the length of arc is given by 5 = c sinh -, and the tension c at any point is measured by the product of the ordinate at that point and the weight per foot of the chain. Fig. 118 shows the catenary for a cable weighing 3-5 Ibs. per foot and strained to a tension of 40 Ibs. weight, and the method of calculation for the con- struction of this curve is explained on p. 358 of PartT. The tension at 10 ft. from the centre = 3-5 x 15-9 FIG. 119. = 55-6 Ibs. weight, since the ordinate there is 15-9. Example 23. To find the time of oscillation of a compound pendu- lum swinging through small arcs. Let I be the moment of inertia of the pendulum about an axis through the point of suspension O (Fig. 119), and let h = the distance of the C. of G. from the point of suspension. Then the couple acting, to produce the angular acceleration, = moment of inertia x angular acceleration. [Compare the rule for linear motion, Force = mass x acceleration.] Now the angular velocity = -j- dt and hence the angular acceleration = <8 ' Thus the couple acting = 1-^ and this couple is opposed by one whose arm is h sin 6, as is seen in the figure. d z d Thus *dt*~ = mhsvn.6 mhd since 6 is supposed to be small, and consequently sin 6 = 6 or d z d _ mh . _ z ., 2 _ mh dt*~~ ~~T~ " ~T~ and +*-. APPLICATIONS OF THE CALCULUS 325 This equation is of the type dealt with in Case (3), p. 283, and the solution is 6 = A sin (<ef-f B). 2-7T The period of this function is ; also the couple for angular dis- co placement Q = mh6 ; hence the couple for unit angular displacement (denoted by /*) = mh. 27T /T~ /I Hence t = = 2rr \/ r = 2ir \f - co ^ mh v p. or / = 2n A / if engineers' units are used. k i*g This might be written in the easily remembered form, unit moment of inertia per unit twist' If this formula is to be used in the determination of the modulus of rigidity of a sample of wire by means of torsional oscillations, h must be replaced by /, the length of the wire. If /= skin stress, T = torque, and C = modulus of rigidity, d dia. of wire; then whence but and 0=g and T = A 32T/ M ~ 6 ~ yl t=2w \/ / 32/1 hence C = == - and thus C can be determined. As regards the units, if / is in feet, I is in Ibs. ft. 2 , t in sees., and d in feet, feet x Ibs. x ft. 2 x sec. 2 then C = feet x ft. 4 x sec. 2 Ibs = ~ffT' ** e -> C * s m Ibs. per sq. foot. If I is in Ibs. ins. 2 and d is in ins., then C will be in Ibs. per sq. in. Example 24. To find formulae giving the radial and hoop stresses in thick cylinders subjected to internal stress. We may attack this problem by either of two methods : Method i. In (a) Fig. 120 let the outside radius = r t and the inside radius = r ; also let the internal pressure be p, and the crushing stress at right angles to the radii, or the hoop stress, = q. 326 MATHEMATICS FOR ENGINEERS It is rather easier to consider the stress on the outside to be greater than that on the inside : thus for an annulus of radius r and thickness $r, we take the internal stress as p and the external stress as p-\-8p. Considering the element RS of the annule (subtending an angle of 80 at the centre), and dealing with the radial forces, Total radial force = (p-\-8p) x outer arc px inner arc = (p+8p) X (r+8r)86-pr86 = (pr+p8r+r8p+8p.8r-pr)86 . = (p8r+r8p+8p.8r) 86 (for a unit length of the cylinder) . This is balanced by two forces each q 8r . 86, for 2 . 86 86 = sin - = nearly q8r 2 2 [(6) Fig. 120] i. e., x = q8r.86 x being the radial force. FIG. 1 20. Stresses in Thick Cylinders. Thus (p8r+r8p+8p.8r)86 = q8r.8d or, when 8r is very small, p dr-\-r dp = q dr. Assume each longitudinal fibre to lengthen the same amount due to the secondary strains. Then if <r = Poisson's ratio and E = Young's modulus for the material, p i the extension due to p will be ^ X - and the extension due to q will be ^ X - Si, (T then, since the total extension is to be constant, i. e., ~r^ constant, (T-tL p-\-q = 2A, say, for a and E are constants. APPLICATIONS OF THE CALCULUS Hence v dp -\-pdr = qdr = (2 A p}dr i. e., rdp = 2(Ap)dr. Separating the variables and integrating, f dp _ fdr J 2(A-p) J r i.e., | log (A p) + log C = log r C 327 or r = (A-pfl C r -r> or p = A + - 2 but and hence The constants A and B are found from the conditions stated in any example. Method 2. According to this scheme q is taken as a tensile stress. By the thin cylinder theory ; consider the equilibrium of the half elementary ring of unit length [(c) Fig. 120]. Then (pX2r) (p+8p)2(r+8r) = 2q8r whence qdr = pdrrdp. From this point the work is as before except that 2 A is written for p q and not for p-\-q as in Method, i. Example 25. To find expressions for the stresses in Thick Spherical Shells. FIG. 121. Let p = the radial pressure, q = the hoop tension. Take an elementary shell at radius r, the thickness being 8r (Fig. 121). Then 7rr z pn(r+8r) 2 (p+&p) = When 8r is very small this equation reduces to 28r r8 = 2q8r and hence 2q = dp - 328 MATHEMATICS FOR ENGINEERS Assuming the volumetric strain to be the same everywhere, where l v = the circumferential strain, and thus 2/2,= the superficial strain, and l x = the radial strain, then it follows that ^-- P.-C <rE o-E" i. e., 2( Ip Constant = 3A (say). Now 2<7= 2* r~ Separating the variables and integrating, [-$L J T i. e., log r = - log (A+) -f log (^ whence or i _ _ 2B_ A Also zq 3 . 2 B _B Euler's Formulae for Loaded Struts. Example 26. To obtain a formula giving the buckling load for a strut of length L and moment of inertia I. Applying the ordinary rule M =E d*y APPLICATIONS OF THE CALCULUS Bending moment at Q = M = Py [(a] Fig. 122] 329 i. e., Let then dx 2 IE/' IE and the solution of this equation is, according to Case (3), p. 283, y A sin ( fW (c) FIG. 122. The various conditions of end fixing give rise to the following solutions : Case of ends rounded. When x o, y = o then o = A sin (o+B) and A is not zero so that B = o. When i. e., x = , y= Y [(a) Fig. 122] xr . wL Y = A sin 2 Obviously Y is the amplitude, i. e., A = Y 0)L or Thus we may write whence and i sin 2 7T . 0)1 2 2 0>L = 7T 7~P _ V fE XL = 7T 2 IE , L 2 ' (- being the simplest angle \2 \ having its sine = i ) 330 MATHEMATICS FOR ENGINEERS Case of both ends fixed. -The form taken by the column is as at (&) Fig. 122. The half-period of the curve is evidently in this ease. 27T But the period = 27T IE whence 4 7T 2 IE L 2 Case of one end fixed. The form taken by the column is as at (c) Fig. 122. The half-period in this case is f L, but, as before proved, . , . 27T the period is given by O) Hence -L 2co Now /* " V IE Q7T 2 IE 4P 97T 2 IE 4L 2 ' L 2 = P = Tension in Belt passing round a Pulley. Example 27. To compare the tensions T x and T 2 at the ends of a belt passing round a pulley ; the coefficient of friction between the belt and pulley being p., and the angle of lap being 6 radians. T+ST FIG. 123. FIG. 124. Consider a small element of belt subtending an angle of dd at the centre of the pulley (see Fig. 123) : then the tensions at the ends are respectively T-f-oT and T. APPLICATIONS OF THE CALCULUS 331 Resolving the forces horizontally sa s/i /T> I T\ "" T> " T- (T+8T) cos -- Tcos uP 2 2 f./\ i.e., 8T cos = uP 2 or in the limit dT = /J? ......... (i) 80 for cos -- > cos o, z. 0., i. 2 Resolving the forces vertically, Sj/1 P=(T+8T+T)sin~ ' . 80 , ._ . 80 = 2T sin -- f-8T sin 2 2 _, . = 2T -- h8T (for sin = 22 22 when the angle is small) In the limit P = TdO .......... (2) Then, combining equations (i) and (2), dT = pTdS Separating the variables, / -=- = p. / dd J T 2 i Jo T Integrating, log-f^ = \iB *-2 T, n8 or =*= tr *i Friction in a Footstep Bearing. Example 28. To find the moment of the friction force in a footstep bearing; the coefficient of friction being p., R = radius of journal and W= total load. (a) Assume that the pressure is uniform over the bottom surface, i. e., W = 7rR 2 ^>, where p is the intensity of the pressure. Take an annulus at radius r, and of thickness 8r (Fig. 124). Area of the annulus = 2nr8r Pressure on the annulus = 2-nrdrp Friction force on the annulus -z-nrbrpp and hence the moment of the friction force on the annulus = 2irr 8r pp x r 332 MATHEMATICS FOR ENGINEERS /R Up 2nr z dr R T W = - i. e., the moment is the same as it would be if the whole load were supposed concentrated at a distance of two-thirds of the radius from the centre. (&) Assume that the intensity of pressure varies inversely as the velocity, i. e., p= K X - v the velocity at radius r v r = so that p = K x Then the pressure intensity on an annulus distant r from the centre so that p = K x = - (say). r v and the total pressure on the annulus also the friction force = p X this pressure. Hence the moment of the friction force on the annulus and the total moment of the friction force = v Now the total load W = / intensity x area J o /* = I p X 2nrdr J o =r Jo APPLICATIONS OF THE CALCULUS 333 Hence the moment of the friction force = 27rwR X 2 -Wxf i.e., the effective radius is now and not f, as in Case (a). Example 29. To find an expression for the moment of the friction force for a Schiele Pivot. Assume that the pressure is the same all over the rubbing surface, that the wear is uniform and that the normal wear is proportional to the pressure p and to the speed v. Referring to Fig. 125, and thus = normal wear oc pv, i. e., 8n oc pr or &n = Kpr. Let the tangent at the point P make the angle 6 with the axis, then if t = length of tangent, t sin 6 = r. Now 8h vertical drop = - ^ sintf Also 8n = Kpr = Kpt sin 6 whence bh = Kpt. Now 8h is constant, p and K are constant ; hence t must be constant and the curve is that known as a tractrix (i. e., the length of the tangent from the axis to any point on the curve is a constant). To find the moment of the friction force : On a small element of surface, the friction force = lirr 8s X p X p. and the moment of the friction force = 2nr8s fip x r. Now 87 = 8s sin 6. Hence the total moment of the friction force dr FIG. 125. sin 6 tsinti but X W Hence the total moment of the friction force = /*\V/. 334 MATHEMATICS FOR ENGINEERS Examples on Hydraulics. Example 30. To find the time to empty a tank, of area A sq. ft., through an orifice of area a sq. ft., the coefficient of discharge being C d . If the height of the water above the orifice at any time is h, then the velocity of discharge = v = V^gh. Hence the quantity per sec. = C d av and the quantity in time 8t C d av8t. This flow will result in a lowering of the level in the tank by an amount &h, so that the volume taken from the tank in time 8t = A 8h. Hence A8h = C d aV^ghSt. Here we have a simple differential equation to solve, and separating the variables and integrating Adh where h z = initial height A x = final height o A FIG. 126. Triangular Notch. If A 1 = o, then the time to completely empty the tank Example 31. To gauge the flow of water by measurements with a triangular notch. Let the height at the notch be H, and consider a small element of width b, thickness 8h, and height above the apex of the notch (H A). b G From Fig. 126, - = (H h) tan - where 6 = the angle of the notch, whence 6 =2(H h) tan-. APPLICATIONS OF THE CALCULUS 335 Now the area of the element = b 8h and the velocity of water at that height = VzgX height = A Hence the actual quantity flowing = C d X2(H h) tan - and the total quantity flowing for the height H fi = FzVzgCa tan-(H J o tan /) /~n -J ^H 1 / i J 5 2 a ~Li 6 = 90 (a common case), tan- = i, Q and then the discharge = V2gC d H* = 2-66 H ? if C d = -62. Example 32. To estimate the friction on a wheel disc revolving in a fluid. Let the friction per sq. ft. fv x and let the disc (of inside radius R 2 and outside radius R x ) revolve at n revs, per sec. The velocity of an annulus at radius r = 2nnr and thus the friction force per sq. ft. on this annulus = (2nnr) x f. Hence the moment of the friction force on the annulus = /(2 nnr) x X -Zirrbr X r and the total moment of the friction force on one side of the disc = M Rg The total moment (i. e., on the two sides) = aM and H.P. lost in friction = . 550 If x = 2 = 49 .6/n{R 1 -R f }. 336 MATHEMATICS FOR ENGINEERS Example 33. To establish a general rule for determining the depth of the Centre of Pressure of a section below the S.W.S.L. (still water surface level). Suppose the plate (representing a section) is placed as shown in Fig. 127. Consider a small element of area 8a, distant x from OY, the vertical distance being h. Let X = distance of the C. of G. from OY, X = distance of the C. of P. from OY, and let H and H be the corresponding vertical distances. Let P = total pressure and A = total area, k = swing radius about OY, and p weight of i cu. ft. of water. S.W.S.L. IX FIG. 127. Centre of Pressure. The whole pressure on the element = intensity of pressure X area = ph X 8a Whole pressure on surface = 2px sin a8a approx. or fpx sin a da actually, i.e., P = psinaxfxda p sin aX ist moment of area about OY = p sin a X AX, but X sin a = H. To find the position of the C. of P., take moments about OY. Then PxX = 2 moments of the pressures on the elements = Sp sin a x8a X x = psina2# 2 Sa approx. = psinafx z da actually = p sin a X and moment about OY = p sin a X A& 2 . APPLICATIONS OF THE CALCULUS 337 Now P = pHA so that pHAX = psinaAA 2 , i. e., HX = sin a . k z but H = X sin a hence XX = A 2 . Thus if X is known, X can be calculated. If the body is not symmetrical, then Y (the distance of the C. of P. from OX) must be found by taking moments about OX. In a great number of cases a = 90, so that sin a = i, and thus Example 34. A triangular plate is placed with its base along the S.W.S.L., the plate being vertical. Find the depth of the centre of pressure below the surface. For this section I about S.W.S.L. = bh 3 and thus k 2 = also H = -h. ' 3 - k 2 A 2 X3 h Hence H = = -^j^ = - H 6x& 2 Example 35. A circular plate has its upper edge along the S.W.S.L Find the depth of its Centre of Pressure below the S.W.S.L. For a circle, Idiam. = 7- d* 64 and thus k 2 about diam. = 4 and hence, by the parallel axis theorem, A 2 about S.W.S.L. = H ~ ) = also H = - 2. = k z 16 ^d Hence H = = 5-= "-3- H d 338 MATHEMATICS FOR ENGINEERS Example 36. Forced vortex (i. e., water in a tube rotated round a vertical axis). To find the form taken by the surface of the water. Let the rotation be at n R.P.S. Consider an element at P (Fig. 128). tan 6 = th tan 6 = the slope of the curve taken = -^ dn vertical force _ weight *of particle horizontal force centrifugal force on particle _mxgr FIG. 128. Forced Vortex. Separating the variables Integrating dh 4tr 4n 2 n 2 rdr = gdh. ^ =gh . A = ^xr*. Now is a constant, and thus h = constant xr 2 , this being the equation of a parabola. Hence the surface of the liquid will be that of a paraboloid of revolution. An Example from Surveying. Example 37. Prove that a cubic parabola is a suitable " transition " curve. In order that the full curvature of a railway curve may be approached gradually, a curve known as a transition curve is inter- posed between the straight and the curve. It must be so designed that the radius of curvature varies inversely as the distance from the starting point (on the straight, because there the radius is infinite) . i d 2 y As before proved, dx 2 or more exactly ffiy dx 2 APPLICATIONS OF THE CALCULUS 339 For the cubic parabola we may assume an equation y = px 3 , where x is the distance from the straight, along a tangent, and y is the offset there (obviously p must be very small) . If y = px 3 , -r- = ipx 2 , ~ = 6px. Hence = = - as a first approximation, or = 6px nearly (for p 3 is very small) . T T T K" Hence R or K = X - = 6p x x Roc -. x Exercises 23. 1. A cylindrical tank is kept full of water by a supply. Show that the time required to discharge a quantity of water, equal to the capacity of the tank, through an orifice in the bottom equals half the time required to empty the tank when the supply is cut off. A tank 10 ft. high and 6 ft. diam. is filled with water. Find the theoretical time of discharge through an 8" diam. orifice in the bottom. 2. A tank empties through a long pipe discharging into the air. If Kv 2 the head lost in the pipe is written hi = - , show that K can be found from the expression, where A x is the level of the water in the tank at the time t t and h z is the level of the water in the tank at the time t 2 measured from the centre of the discharge end of the pipe. A = area of cross section of the tank. a = area of cross section of the pipe. An experiment with a tank 15-6 sq. ft. in cross section and a 4" diam. pipe gave the following results : Time, t inins. Level in tank, h. I 2 38-35 32-84 3 4 23-19 I9.O2 Find the value of K for the pipe. 340 MATHEMATICS FOR ENGINEERS 3. Use the following table to obtain -= and thence find the volume of i Ib. of steam at 160 Ibs. absolute pressure per sq. in. Absolute press. (Ibs. per sq. in.) 159 1 60 161 Temperature (F) 363-1 363-6 364-1 The latent heat of i Ib. of steam at 160 Ibs. per sq. in. pressure is 858-8 B.Th.U. 4. Find the " fixing moments " for a beam built in at its ends and 40 feet long, when it carries loads of 8 tons and 12 tons, acting 15 feet and 30 feet respectively from one end. 5. A tank of constant cross section has two circular orifices, each 2" diani., in one of its vertical sides, one of which is 20 ft. above the bottom of the tank and the other 8 ft. Find the time required to lower the water from 30 ft. down to 15 ft. above the bottom of the tank. Cross section of the tank =12 sq. ft. Coefficient of discharge = -62. 6. A hemispherical tank 12 ft. in diam. is emptied through a hole 8" diam. at the bottom. Assuming that the coefficient of discharge is 6, find the time required to lower the level of the water surface from 6 ft. to 4 ft. 7. A vertical shaft having a conical bearing is g" in diam. and carries a load of 3^ tons ; the angle of the cone is 120 and the co- efficient of friction is -025. Find the horse power lost in friction when the shaft is making 140 revolutions per minute. Assume that the intensity of pressure is uniform. 8. A circular plate, 5 ft. diam., is immersed in water, its greatest and least depths below the surface being 6 ft. and 3 ft. respectively ; find (a) the total pressure on one face of the plate, (b) the position of the centre of pressure. < 9. An annular plate is submerged in water in such a position that the minimum depth of immersion is 4 ft. and the maximum depth of immersion is 8 ft. If the external diam. of the plate is 8 ft. and the internal diam. 4 ft., determine the total pressure on one face of the plate and the position of the centre of pressure. 10. One pound of steam at 100 Ibs. per sq. in. absol. (vol. = 4-45 cu. it.) is admitted to a cylinder and is then expanded to a ratio of 5, according to the law pv 1 - 06 = C ; it is then exhausted at constant pressure. Find the net work done on the piston. 11. Find the loss of head h in a length I of pipe the diameter of which varies uniformly, being given that 4fLv* 4Q H= J , , and v = ^~. 2gd Ti-d 2 {Let diam. at distance x from entry end = rf e +K# where d e = diam. at entry}. APPLICATIONS OF THE CALCULUS 341 12. Taking the friction of a brass surface in a fluid as -22 Ib. per sq. ft. for a velocity of 10 f.p.s. and as proportional to w 1 ' 9 , find the horse power lo"st in friction on two sides of a brass disc 30" external and 15* internal diam. running at 500 r.p.m. 13. A rectangular plate 2 ft. wide by 5 ft. deep is immersed in water at an inclination of 4O C to the vertical. Find the depth of the centre of pressure, if the top of the plate is 6 ft. below the level of the water. CHAPTER XI HARMONIC ANALYSIS Fourier's Theorem relates to periodic functions, ot which many examples are found in both electrical and mechanical engi- neering theory and practice : it states that any periodic function can be expressed as the sum of a number of sine functions, of different amplitudes, phases and periods. Thus, however irregular the curve representing the function may be, so long as its ordinales repeat themselves after the same interval of time or space, it is possible to resolve it into a number of sine curves, the ordinates of which when added together give the ordinates of the primitive curve. This resolution of a curve into its component sine curves is known as Harmonic Analysis ; and in view of its importance, the simpler and most direct methods employed for the analysis are here treated in great detail. Expressed in mathematical symbols, Fourier's theorem reads or y = +Bj cos tf+B 2 cos 2qt+~B 3 cos 3qt-\- . . . the latter form being equivalent to the first, since Aj sin gtf+Bj cos qt B sin (qt-\-Cj) provided that B and c are suitably chosen. For the purposes of the analysis the expression may appear simpler if we write 6 in place of qt. Thus y = A +A 1 sin^+A 2 sin2^+A 3 sin3<9+ . ..-. +B 1 cos0+B 2 cos20+B 3 cos30-|- . . . Of the various methods given, three are here selected and explained, these being easy to understand and to apply. Dealing with the three processes in turn, viz., (a) by calculation, (b) by a graphical interpretation of method (a), and (c) by super- position, we commence with the study of method (a). 342 HARMONIC ANALYSIS 343 Method (a): Analysis by Calculation. Before actually pro- ceeding to detail the scheme of working, it is well to verify the following statements. ft* cos = 0, this being self-evident, since the area under a .'o cosine curve is zero, provided that the full period is considered. /** I cos mO cos nOdO o ............ (i) .'o for cos mO cos nO = ^{cos (m-\-n)6 cos (m n)0} and hence /2ir /2ir r2n rZv I cos m0cosn9d& = || cos(m + n)OdO 1| cos(mn)6d0 Jo Jo Jo = o o (for both are cosine curves over the full period or a multiple of the full period). rZv f2jr /-2ir I cos m6 sin nO dO = | sin (m+n)B dO% sin (mn)6dO Jo J o Jo = .... ........ (2) fZir ,fZir rZir cos z 6d6 =4 cos20dO + % dd Jo Jo Jo = + (27r-o)=7r ....... (3) /2ir f /-2ir /-2n- ^ sin mO sin M^ ^ = $\ cos (mn)6 dd cos (m+ri)0d6 \ Jo U o y o J = |{o-o}-o ......... (4) and /2ir /-2ir fZir I sin 2 ^^ =J ^ | cos20d0 Jo Jo Jo = \\2ir o} O = ............ (5) To proceed with the analysis : We are told that y = A +A 1 cos0+A 2 cos20+A 3 cos30+ ... +Bj sin 6+B 2 sin 20+B 3 sin 3^+ and we wish to find the values of the coefficients A , Aj . . ., Bi, B 2 , etc. 344 MATHEMATICS FOR ENGINEERS If we integrate throughout (with the limits o and 2ir), every term on the right-hand side, except the first, will vanish, i.e., r2w f yd9 = A, Jo J /2ir or I ydO A X (2-* o) 3 o I P* whence A = - I y dO = the mean value of y (Cf . p. 183) so that A is found by averaging the ordinates ; but in the majority of cases an inspection will show that A is zero. To find A 1 : multiply all through its coefficient, viz., cos0, and integrate, then fZir r'2ir rZir rZ I ycos$dO= I A cosOdO-\- I A 1 cos 2 0^+| Jo ./o Jo ,' o rZn _ rZir Jo lC( Jo f 2 I y cos 6 dO = o+TrAj+o+o . . . Jo or J o " +0+0 . . . [from (3), (2) and (i)] 2 fjr whence A x = / y cos 6 dO = twice the mean value of (y cos 6} i. e., a certain number of values of y must be taken, each being multiplied by the cosine of the angle for which y is the ordinate, the average of these found, and the result multiplied by 2. The values of A 2 , A 3 , etc., may be found in like manner by multiplying through by cos 26, cos 3$, etc., in order, and performing the integration as above. To find B! : multiply throughout by its coefficient, viz., sin 6, and integrate, then /2ir [Z* ,-Zir fZir ysinBdO = I A sm@d@-\- I A,sin0cos0^+ / A o J o J o Jo ,"Zir _ rZir + Jo BlS1 + Jo B2 = o+7rBj [from (2), (4) and (5)] _, 2 r 27r ' Bj = 1 y sin dQ = 2 X mean value of (y X sin 6} so that the values of B,. B 2 , B 3 , etc., may be found HARMONIC ANALYSIS 345 Actually, the values of the coefficients A v A 2 , B r , B 2 , etc., are found by dividing the base into ten or. eight divisions and averaging the mid-ordinates for these divisions. To determine the absolute values, an infinite number of ordinates should be taken, but this would of course be quite out of the question as far as an ordinary calculation is concerned. The work is made clearer by suitable tabulation, as will be seen from the following example. Example i. Resolve the curve ABCD (Fig. 129) into its component curves : it being understood that no higher harmonic than the first occurs. e O -36 72 K>8 144 FIG. 129. Harmonic Analysis. [The term containing 6 is spoken of as the fundamental, and that containing 26 as the first harmonic.] Thus y = A +B sin (6+cJ +C sin or y= A -|-A 1 cos0+A 2 cos20+B 1 sm0+B 2 sin20 will represent the function in this case. A glance at the figure will show that the curve is symmetrical about the axis of 6 ; thus we observe that the average ordinate = o, or A ft = o. 346 MATHEMATICS FOR ENGINEERS Divide the base into 10 equal parts, erect the mid-ordinates and tabulate the values as follows : (a) (/) (c) (d) (e) (/) (g) Ordinate No. 9 1 sin cos e sin 26 cos 26 I 1-56 18 309 951 588 809 2 375 54 809 588 95 1 -309 3 4 90 I- o o I 4 2-91 126 809 588 951 -309 5 1-13 162 309 -951 588 809 6 -1-13 198 309 -95 1 588 809 7 2-91 234 -809 588 95 1 309 8 -4 270 I o T 9 -3'75 306 809 588 951 -309 10 -1-56 342 309 951 588 809 Then A = mean value of y = o. A! = 2 X mean value of (y cos 6} . To obtain the values of y cos 6, corresponding figures in columns (6) and (e) must be multiplied. Then 10 1 +2-913-75) = o -9i\ ) Similarly, B x = X sum of the products of columns (&) and (d) io i-i3 + i- 5 6) + 1(4+4) \ 9i + 2-9i+3-7 5 ) / = -2x20-43=4-086 2 A 2 = X sum of products of columns (6) and (g) -xol = o 3'75 B 2 = X sum of products of columns (6) and (/) Hence or _ lol -2-91+3-75) = -2x2-104 = -421. y = (o x cos 6) + (o x cos 26) + (4-09 sin ff) + (-42 sin 20) y = 4-09 sin 0+ -42 sin 26 (the cosine terms being absent). Analysis by Method (6): The Graphical Interpretation of (a) (due to Professor Harrison). To employ this method we must take at least twice as many ordinates as the largest multiple HARMONIC ANALYSIS 347 of ', thus if we suppose that the second harmonic is the highest occurring we might take 6 ordinates as the minimum, although it would be better if 8 or 10 were taken. The method can best be illustrated by applying it to an example. Example 2. Resolve the curve ABC [(a) Fig. 130] into its com- I8O 24O 3OO 360* FIG. 130. ponent curves (the second being the highest harmonic) ; i. e., find the values of the constants in the equation y = AO+AJ cos 6+ A 2 cos 20+ A 3 cos 30+Bj sin 0+B 2 sin 20+ B 3 sin 3$. To arrange that all the ordinates shall be positive, take a base line DE entirely below the curve. Divide the base into 8 equal divisions and number the ordinates y , y lt y 2 , etc. The angular intervals are thus ^5, since a full period corresponds to 360. 348 MATHEMATICS FOR ENGINEERS Draw a new figure [(&) Fig. 130], the lines OM and ON making 45 with the principal axes ; number these lines : o, i, ..... 8, as shown in the figure. Along line o set off a distance equal to y , along line i set off a distance equal to y t and so on. Drop perpendiculars from the points o, I, etc., on to the principal axes, calling the projec- tions on these axes h (this particular projection being zero), A x ..... h a , and v , v l ...... v a respectively. Then to find Aj and Bj : As already proved A! = f (sy cos 0) = "J{y -y 4 + (y!-y 3 -y & +y 7 ) cos 45 + (y t -y,) cos 90} = MVO-^VI+V^VS-VS+VJ and similarly B t = Jj^+Ag h 5 h 7 +h 2 h 6 } i. e., the lengths v ...... v 7 , and h t ....... / 7 can be read off from the figure and then the values of A 1 and B t are calculated as above. Jn this example "0=13-7 *o= o v v =14 h 1= 14 v 2 = o h 2 = 20-4 v 3 ii h a = ii v t = 7'3 h t = o v 6 = '5 hs= '5 v 6 = o A 6 = 1-5 v,= 5-4 h 7 = 5-4 * andB 1 = X38 = 9-5 By the aid of a strip of paper a great amount of this arithmetical work might be obviated, the procedure being as follows : Mark off along the edge of a strip of paper lengths to represent the various ordinates of the original curve, viz., y , y v etc., and number the points so obtained o, i, 2 .... 8 as shown at (c) Fig. 130. Thus Po == y , P 4 = y t and so on. We have seen that in order to find the value of A a it is necessary to evaluate y cos 6 for the various angles ; i. e., we must find the values of y v cos 45, y z cos 90, y a cos 135 and so on. Now y s cos 135 = y 3 x cos 45 = y a cos 45, so that the one line, viz., that at 45, serves also for 135 provided that the ordinate is stepped off in a negative direction. Thus, for example, yj. cos 6+y 3 cos ^6=y 1 cos 45 y s cos 45 = cos 45(;y 1 y s ) and the value of this expression depends upon the difference between the lengths on the strip Pi and P3, or the distance 3 to i. Evidently, then, the work is shortened by grouping the ordinates in pairs to give differences ; thus y\y% = Pi ?3 i to 3 on the strip y 7 y & Py P5 = 5 to 7 on the strip and so on for other pairs of ordinates. HARMONIC ANALYSIS 349 Having found these differences, we multiply by 00345, by setting these lengths along the line Of in (d) Fig. 130 and then projecting to the horizontal axis OX ; the resultant of these projections being the value of 4Aj. Thus in (d) Fig. 130 : Make Oo = i to 3 (on the strip) and ab 5 to 7. Drop be perpendicular to OX. Then Oc = (y^ y 3 y^+y^ cos 45. Step off cd = o to 4 (i. e., y^y^, then measure Od ; this is the value of 4A X , since Od = Oc+cd = (yiy 9 y t +y 7 ) cos 45+(y -y t ). A!= 3-43 For the value of Bj the strip must be used according to the following plan. A line is drawn at 45 [(d) Fig. 130] and distances marked off along it as follows Oe = i to 7 on the strip, ef= 3 to 5 on the strip. {for 46! = (y n +y t ) sin o+(y 1 +y 3 -y 6 -yj sin 45+(> ; 2-^6) sin 9<> } A perpendicular to OY gives the point g. To Og must be added a distance = 2 to 6 on the strip, but to avoid extending the diagram this distance is set off from O giving the length Oh. Thus Og = 19-5, Oh = 18-2, the sum = 37-7. Then 4 B X = 37-7 and B!=: 9-43. To find the values of A 2 and B 2 : The terms containing 26, i. e., 90, will now occur and so there will be no lines at 45. Ag |{yo cos +yi cos go+y 2 cos 180+ . . . y 1 cos 630} Similarly B 2 = ^{y!- Hence set off Ok = o to 2, i. e., (y 9 y%) and kl = 4 to 6, i. e., (y t y 6 ) along OX and the resultant is O/ = -8. -Hence 4A 2 = -8 and A 2 = -2. Set oft Om =* i to 3, i. e., (y t y a ) and mn = 5 to 7, i. e., (y^y-j) along OX and the resultant is On 2-6 Hence 4B 2 = 2-6 and B 2 = -65. To find the values of A 3 and B 3 : A =-( y COS +yi COS I 35+> / 2 cos 270-f-y 3 cos 4O5+>' 4 cos 540 3 1 +^ 6 cos675+>/ 6 cos8io +^ 7 cos945 -{y9yt+\yyi+yty,} cos 45 }. 4 350 MATHEMATICS FOR ENGINEERS Set off Op = o to 4 on the strip, along OX, and Ob = (y>i y a y 5 +y 7 ) (which has already been done when finding Aj). Project b to c on OX ; then Oc = (y a y-a-^y^y-i) cos 45. Hence cp = 4A3 but cp = i -3 and thus A 3 = 33. To find B s : B _ j>o sin +>'i sin I 35+3 / 2 sin 270 +y 3 sin 405+^ sin 540 3 4*- +n sin6 75 +y 6 sin8io +;y 7 sin945 = -{yr-yt+fa+ytyv-yi) sin 45}- Set off Oh = 6 to 2 along OY. O/= (^i+^a y 6 y 7 ) (which has already been done when finding Bj) . Project / to g on OY, then gh = 463 but gh = 1-2 Hence B 3 = .3 Also == 8 (using the trapezoidal rule given on p. 307, Part I), , * A 6 '75 + I 9-8+20-4+i6+7-3+-9+i-9+7-6+6-75 **., A _ -g 87-4 ^=10-93- y= 10-93+3-43 cos $+9-43 sin 6 -2 cos 26 ~- -65 sin 20 33 cos 3$+ -3 sin 3$. There should be no difficulty in the understanding of this method if method (a) is first carefully studied. All that this method (b) adds is the multiplica- A tion of lengths by the cosines or sines of angles by regarding the products as projections on fixed axes (i. e., if OA = R (Fig. 131) and the angle AOB = 30, then OB = OA cos 30 = R cos 30, and OC = OA sin 30 = R sin 30) . FIG. 131. The beauty of the method consists in the use of the strip of paper for the grouping together of pairs of ordinates which have to be multiplied by the same quantity. HARMONIC ANALYSIS 351 In the example just discussed, the angular intervals were taken as 45, this choice being made as a matter of great convenience, since cos 45 = sin 45 and projections may thus be made on either a horizontal or a vertical axis. For greater accuracy, more ordinates should be taken, and then care must be observed as to the axis on which the projections are made. Thus if the angular intervals were taken as 18, say, the lines corresponding to OM, ON and Of in (b] and (d) Fig. 130 would be drawn making angles of 18 with the horizontal axis ; then for the values of A lt A 2 , etc., the projections along OX would be measured, whilst the values of B 1 , B 2 , etc., would be determined from the projections on OY. Method (c): Analysis by Superposition. This method is much used in alternating current work, for the problems of which it is specially suited. It is not difficult to employ, nor to under- stand, although the proof of the method is long and is in consequence not treated here. In order to present the method in as clear a fashion as possible, the rules of procedure are here set out in place of a detailed explanation. The method is as follows ; the case of a curve containing the second as the highest harmonic being treated, although the process can readily be extended if necessary : (1) Divide the curve into two 'equal parts and superpose the second part upon the first, using dividers and paying attention to the signs. If the resultant curve approximates to a sine curve there is no need to further subdivide. (This gives terms containing 26, 4$, 60, etc., but if this curve is a sine curve, probably only terms containing 2$ occur.) Put in a base line for this new curve (by estimation) ;^ then the height of this from the original base line = 2A . (2) Divide the original curve into three equal parts and super- pose (first, the second on the first, and then to this result add the third). (This gives the terms containing 3$, 60, 'gO, etc.) The height of the base line of the resulting curve from the original base line = 3A . (The two values of A may be compared, and of course they should be alike ; but if not, take the average of these and draw a new base line distant A from the original ; this line we shall speak of as the true base line.) (3) Subtract corresponding ordinates of the 20 curve (divided by^2) and the 36 curve (divided by 3), paying attention to the 352 MATHEMATICS FOR ENGINEERS signs, from the ordinates of the original curve ; the resultant curve is approximately a sine curve symmetrical about the true base line. To calculate the values of the constants, if +A 2 sin (20+c 2 ) A is already found. Select two convenient values of 6 and work from the ordinates of the curve to find Aj and c x ; proceed similarly, using the 26 curve to find A 2 and c 2 . Note that in alternating current work only terms of the order 6, 36, $0, etc., occur, so that the curve would need to be divided into 3, 5, etc., equal divisions and the parts superposed. There is thus no need to divide into 2, 4, etc., equal parts; also it is evident that the value of A must be zero. Example 3. The curve ABCD, Fig. 132, gives by its ordinates the displacement of a valve actuated by a Gooch Link Motion. It is required to find the constants in the equation +A 2 sin etc. The original curve is divided into two equal parts, the second being placed over the first, with the result that Curve 2 is obtained. The estimated base line for this is B 2 ; the height of B 2 above the original base line being -29, i. e., HARMONIC ANALYSIS 353 2Q the height of the true base line is -- or -145 unit. This base line can now be put in, and is indicated as the true base line. By division into 3 and 4 equal parts and superposition the curves 3 and 4 respectively are obtained. B 3 , the base line for 3, is at a height of -43 ; this figure divided by 3 gives -143, which agrees well with our former result. Curve 2 really represents the first harmonic with double amplitude ; therefore we subtract ordinates of Curve 2 (to half scale, i. e., we use proportional compasses) from the corresponding ordinates of the original curve. Similarly we subtract J of the ordinates of Curve 3 from the original curve, and since those for Curve 4 are too small to be taken into account, the net result is Curve i, which represents the funda- mental, and is a sine curve symmetrical about the true base line. To find the constants A t and c : in the equation y l = A l sin(6+c 1 ). When 6 = 0, ;Vi = 2-i75 (measured from the true base line to Curve i). At 6 90, y = o. c, = 90 or 2 At 180 y t = 2-135, t. e., Aj = the mean of 2-175 an d 2>I 35 * e -> 2 ' 1 5> y = 2-15 sin (0+90 3 ) = 2-15 costf. To find A 2 and c 2 . *2Q The amplitude of Curve 2 is , . e., -145. and since the curve has its maximum ordinate when 6 = we have again c g = 90, or the curve is a cosine curve. Hence y^ = -145 cos 20. Beyond this first harmonic we need not proceed as the amplitudes of Curves 3 and 4 are exceedingly small. Hence y = y*+y* = 2-Ij COS ^+-145 COS 20. This method of superposition is to be recommended in cases of A.C. work, as one can so readily tell by its aid which harmonics are present. If the actual constants in the equation are required it may be easier to proceed according to method (a) or method (b) . A A 354 MATHEMATICS FOR ENGINEERS Exercises 24. On Harmonic Analysis. 1. Show how to analyse approximately the displacement x of a point in a mechanism on the assumption that it may be represented by a limited series of sine and cosine terms, and obtain general expres- sions for the values of the coefficients in the series x 2 n (A n cos 0+B n sin n0) + A where n = 3 and 6 is the angular displacement of an actuating crank which revolves uniformly. Apply your results to obtain the values of the coefficients for the values of x and 6 given in the accompanying table, where the linear displacement of a point in a mechanism is given for the corresponding angular displacement of a uniformly revolving crank. Angular displacement of crank in degrees . o 60 90 120 180 240 300 x (in ins ) I'll . 8 1-6 67 , f 2-Q3 _ 2. A part of a machine has an oscillating motion. The displace- ments y at times t are as in the table. t 02 4 06 08 i 12 M 16 18 2 y 64 I-I 3 i-34 95 92 -i-33 1-16 -66 Find the constants in the equation y = A sin (iO7r^+a 1 ) + Bsin (2O7r/+a 2 ). 3. Analyse the curve which results when the following values are plotted. x 45 90 135 1 80 225 270 315 360 y o 21-5 31-25 11-25 9 30 26-5 o 4. The values of the primary E.M.F. of a transformer at different points in the cycle are as follows (6 being written in place of pt for reasons of simplicity) . e o 3 60 90 120 150 1 80 2IO 240 270 300 330 360 E -14 886 1293 1400 130? 814 -70 -886 -1293 - 1400 -1307 -814 70 If 6 and E are connected by the equation E = A sin 0+B sin 3#+C cos 0+D cos find the values of the constants A, B, C and D. CHAPTER XII THE SOLUTION OF SPHERICAL TRIANGLES THE curvature of the earth's surface is not an appreciable factor in the calculations following a small survey, and is therefore not regarded, but when the lengths of the boundaries of the survey are great, as in the case of a " major triangulation," the effect of the curvature must be allowed for, if precision is desired. It is there- fore necessary to use Spherical Trigonometry in place of the more familiar Plane Trigonometry, and accordingly a very brief chapter is inserted here, dealing mainly with the solution of spherical triangles. Definitions of Terms used. The earth may be considered as a sphere of radius 20,890,172 feet, this being the mean radius. A great circle on a sphere is a circle traced by the intersection of the sphere by a plane passing through its centre ; if the plane does not pass through the centre of the sphere, its intersection with the sphere is called a small circle. Thus all meridians are great circles, whilst parallels of latitude, except for the equator parallel, are all small circles. A straight line on the earth's surface is in reality a portion of a great circle ; hence a parallel of latitude is not a straight line, or, in other words, a movement due East or West is not a movement along a straight line. A triangle set out on the earth's surface with straight sides is what is termed a "spherical triangle," its sides being arcs of great circles. The lengths of these sides might be measured according to the usual rules, viz., in miles, furlongs, etc., but it is more usual to measure them by the sizes of the angles subtended by them at the centre of the sphere. In this connection it is convenient to remember that an arc of one nautical mile (6076 feet) subtends an angle of i' at the centre of the earth ; hence a length of 80 nautical miles would be spoken of as a side of 80', i. e., i 20'. In Fig. 133 is shown the difference between great and small circles ; and AB, BC and CA being portions of great circles form a 355 356 MATHEMATICS FOR ENGINEERS spherical triangle (shown cross hatched). The length BA would be expressed by the magnitude of the angle BOA. A spherical triangle ABC is shown in Fig. 134, O being the centre of the sphere. The arc AB is proportional to the angle AOB, and therefore, instead of speaking of AB as a length, it is quite legitimate to represent it by L AOB. c would thus stand for /.AOB, b for ^.COA, and a for <iCOB. As regards the angles of the triangle, the angle between CA and AB is that between the planes AOC and AOB and is, therefore, the angle between the tangents AD and AE. Spherical triangles should be regarded as the most general form of plane triangles ; for if the radius of the sphere becomes infinite the spherical triangle becomes a plane triangle. FIG. 133. Spherical Triangles. Many rules with which we are familiar in connection with plane triangles hold also for spherical triangles, as, for example, "Any two sides of a triangle are greater than the third," or, again, " If two triangles have two sides and the included angle of the one respectively equal to two sides and the included angle of the other, the triangles are equal in ah 1 respects " ; " The greater side of every triangle is subtended by the greater angle." There is one important difference between the rule for a plane triangle and a corresponding rule for a spherical triangle: viz., whilst the three angles of a plane triangle add up to 180 the sum of those in a spherical triangle always exceeds 180, the sum in fact lying between 180 and 540 ; and the difference between the sum of the three angles and 180 is known as the " spherical excess." THE SOLUTION OF SPHERICAL TRIANGLES 357 The magnitude of this can be found from the rule 360 X area of triangle spherical excess = ^~ 2irT z (27rr 2 being the area of the surface of the hemisphere). This spherical excess is a small quantity for the cases likely to be considered in connection with surveys. E. g., consider the case of an equilateral triangle of side 68 miles. r = 20,900,000 ft. approx. = 3960 miles. The area of the triangle is about 2000 sq. miles. Then the spherical excess 360x60x2000 = minutes = -437 minute. 27rX 3960x3960 -"^ A good approximation for the spherical excess of a triangle on the earth's surface is : area of spherical triangle in sq. miles spherical excess (seconds) = 78 Solution of Spherical Triangles. The most widely used rule in connection with the solution of plane triangles is the " sine " rule which states that the sides are proportional to the sines of the angles opposite. In the case of spherical triangles this becomes modified and reads " The sines of the angles are proportional to the sines of the sides opposite. Therefore, adopting the notation of Fig. 134, sin a sin b sin c sin A ~ sin B sin C it being remembered that sin a is really sin L BOC, etc. Other rules are . A /sin (s b) sin (s c) sin = A/ - v . ' i- .... (2) 2 > . 2 > Sln f) sln c A /sin s sin (s a) cos = \/ : = .... (3) 2 v sin & sin c A /sin (s 6) sin (s c) tan =v ^ : , v . . . . . (4) 2 v sin s sin (s a) T> /- and corresponding forms for and - , obtained by writing 2 2 the letters one on in the proper sequence, a b c a. s in these formulas = - and is, therefore, an angle (in plane trigonometry, s = - , but is a length). 358 MATHEMATICS FOR ENGINEERS It is of interest to compare these with the corresponding rules in connection with plane triangles, which are A /(sb)(sc) sin - = \/ '- '- 2 v be A A '*=V- 's(s COS be 8(9 -a) It will be seen that, as in the previous case, sides occurring in the formulae of plane trigonometry are replaced by their sines in the corresponding formulae of spherical trigonometry. Other rules are : cos A+cos B cos C cos a = . . (5) sin B sin C cos a cos b cos c cos A : =. (6) sin b sin c cot A sin B = cot a sin c cos B cos c . . . (7) (a-V) cos- tan ~lT~ ~7^R>\ COt 2 (8) COS - J \ 2 / (a-b) . sin A B 2 C tan = - -r- cot .....-.. (q) 2 . fa-\-o\ 2 sin - V 2 / Solution of Right Angled Spherical Triangles. In the case of a right angled spherical triangle these rules can be put into somewhat simpler forms. Assume that the triangle is right angled at C. = 90, .'. cosC = o, and sinC = i. cose cos a cos b From (6) cos C = sin a sin b but cos C = o, cos c cos a cos b = o, i.e., cos c = cos a cos b (10) THE SOLUTION OF SPHERICAL TRIANGLES 359 cos a cos b cos c Also cos A = - . . . from (6) sin sin c cos c = f COS COS C COS0 , from (10) sin sin c cos c i cos cos c \s _^_, w sin c sin cos sin sin c if i cos 01 ~ tan c\sin b cos b sin J : cos 2 &1 cot ex sin 2 = cote] T Isi sin cos 0J sin cos b sino = cot c X , COS0 = cot c X tan or tan ox tan (90 c) i.e., cos A = tan b tan (90 c)l , > also cos B = tan a tan (90 c) J COStf COS COS C Again cos A = - ; ^. sin & sin c COS 2 C cos a cos a , . . from (10) sin sin c cos 2 cos 2 c sin 2 c sin 2 a = -- or ~ir- (12) cos a sin sin c cos a sin sin c . . sin a And from (i) sin A = -, (13) sine In plane trigonometry sin A = - . Napier's Rules of Circular Parts. The equations (10), (n) and (13) and their modifications may be easily remembered by Napier's two rules of circular parts, which may almost be regarded as a mnemonic. For the application of these two rules the five parts of the spherical triangle, other than the right angle at C, are 'regarded as a, b, (go A), (90 c), and (90 B) respectively, the complements of A, c and B being taken instead of the values A, c and B in order that the two rules may embrace all the cases. 360 MATHEMATICS FOR ENGINEERS These five parts are written in the five sectors of a circle in the order in which they occur in a triangle : thus in Fig. 135, com- mencing from the side a and making the circuit of the triangle in the direction indicated, the parts in turn are a b A (for which we write 90 A), c (for which is written 90 c) and B (for which is written 90 B). These parts are set out as shown in Fig. 136. Then Napier's rules state : Sine of the middle part = product of tangents of adjacent parts. Sine of the middle part = product of cosines of opposite parts. The terms middle, adjacent and opposite have reference to the mutual position of the parts in Fig. 136. Thus if b is selected as the middle part, the adjacent parts are those in immediate contact FIG. 135. FIG. 136. with b, viz., a and (90 A), whilst (90 c) and (906) are the opposite parts. Hence sin b = tan a x tan (90 A) = tan a cot A sin b = cos (906) X cos (90 c) = sin B sin c . sin b ., ,. , . or smB = - . (Cf. equation (13), p. 359.) olll C/ Again if (906) is selected as the middle part, the adjacent parts are (goc) and, and the opposite parts are (90 A) and b. Hence sin (906) = tan (90 c) tana or cos B = tan (900) tan a (cf . equation (n), p. 359), and sin (90 B) = cos (90 A) cos b or cos B = sin A cos b. These rules, being composed of products and quotients only, lend themselves well to logarithmic computation. The Ambiguous Case in the Solution of Spherical Triangles. In the solution of a plane triangle, if two sides and the angle opposite the shorter of these is given, there is the possibility THE SOLUTION OF SPHERICAL TRIANGLES 361 of two solutions of the problem ; the best test for which, as pointed out in Chap. VI, Part I, being the drawing to scale. A similar difficulty occurs in the solution of spherical triangles, when two sides and the angle opposite one of them is given. E. g., let a b and B be the given parts. Then from equation (i), p. 357. sin A sin B sin a sin b sin a sin B or sin A = : -, . sino Now sin A = sin(i8o A) and thus the right-hand side of this last equation may be the value of either a particular angle or its supplement. Without going into the proof it may be stated that there will be one solution only if the side opposite the given angle has a value between the other given side and its supplement. Thus in the case in which a b and B are given, there will be one solution only if b lies between a and (180 a}. If b is not between a and (180 a], then the test must be applied that the greater angle must be opposite the greater side : thus for the case of a b and B given, if a > b then A must be > B. The possible cases may be best illustrated by numerical examples, a b and B being regarded as the given parts throughout. (a) Given a = 144 40', b = 87 37', B = ng' to find A. Using equation (i) of p. 357 A _ sin a sin B _ sin 144 40' X sin 11 9' Sin A : f ; = -. sm b sin 87 37 and log sin A = log sin 144 4o'-flog sin 11 9' log sin 87 37' = i 7622 + 1 28641" 9996 = I 0490 so that it is possible that A = either 6 26' or 173 34'. Now a > b and therefore A must be > B, and this condition is only satisfied if A = 173 34', since 6 26' is not > n9'. It will be noted that the case chosen is that in which b, viz., 87 37', lies between a, i. e., 144 40', and (180 a), i. e., 35 20', and therefore only one solution is expected. . (b) b = 44 35', a = 55 10' and B = 38 46'. Here b does not lie between a and (180 a), so that two solutions are possible. 362 MATHEMATICS FOR ENGINEERS As before log sin A = log sin a + log sin B log sin b = log sin 55 io'+ log sin 38 46' log sin 44 35' = I 9142+1 79661 8463 = I 8645 log sin 47 3' so that possible values of A are 47 3' and 132 57' and we must test each of these values. Now a > b, and hence A must be > B ; but 47 3' and 132 57' are both > 38 46', so that we have two triangles satisfying the con- ditions, and for complete solution the two values of A, C and c must be determined. Example i. In a spherical triangle ABC, having given = 30, 6 = 40, C = 70, find A and B. Given also that Lsin 5 =8-9402960 L tan 12 14' 38" = 9-3364779 L sin 35 = 9'75 8 59i3 L tan 60 4' 3" = 10-2397529 L cos 5 9-9983442 L cos 35 = 9-9133645 In this case two sides and the included angle are given ; we there- fore use equations (8) and (9) . fa b\ A+B DS V 2 / ,C , ,. tan = cot - . . from (8) 2 (a+b\ 2 cos f j = -5_: cot35 . {*<^f } cos 35 \ cos 5 cos 35 _ cos cos 35 sin 35 ~ sin 35- Taking logs of both sides A+B Ltan = L cos 5 Lsin 35+io A+B (or, alternatively, log tan = log cos 5 log sin 35) Ltan A+B I9-9983442 = = L tan 60 4' 3* A+B -^- = 6o 4 ' 3 " A+B = 120 8' 6" ....... (a) THE SOLUTION OF SPHERICAL TRIANGLES 363 From equation (9) fa-b\ . -_. sin I 1 A B \ 2 / , C tan -j- cot 2 . fa+b\ 2 sin - V 2 / A B_ sin 5 00535 tan sin 35" sin 35 B A _ sin 5x cos 35 sin 2 35 taking logs throughout. B A L tan = Lsin 5-fLcos35 2Lsin35+io 18-9402960 9-9I33645 = 28-8536605 19-5171826 =- Ltan 12 14' 38* B-A=2 4 29'i6* ...... . (6) By adding (a) and (b) 2B = 144 37' 22* B = 72 i8'4i* and A = 120 8' 6" 72 18' 41" = 47 49'25*. [Note that A+B+C = 47 49' 25"+72 i8'4i*+7o = 190 8' 6' so that the spherical excess = 10 8' 6".] Example 2. Solve the spherical triangle ABC, having given c = gii8', a 72 27', and = 90. In this case the triangle is right angled, and therefore rules (10) to (13) may be used. To find A : . . sin a From equation (13), p. 359, sin A -: -- sin o L sin A = L sin a L sin c-\- 10 = Lsin 72 27' Lsin 91 i8'+io 19-97930 9-97941 = L sin 72 29' 45*. A = 72 29' 45'. 364 MATHEMATICS FOR ENGINEERS To find b : From equation (10), p. 358, cos c cos a cos b cos c whence cos 6 = - cos a cosoii8 / cos 88 42' * cos b _ _ _ -- - _ cos 72 27' cos 72 27' cos 88 42' or cos (180 b) = cos 6 = - 5-^ > cos 72 27' Hence we shall work to find the supplement of 6. Taking logs * log cos (180 6) log cos 88 42' log cos 72 27' = 1147934 2-87644 = log cos 85 41' 7*. 1806 = 85 41' 7" * It is rather easier to work in terms of the logs in preference to the logarithmic ratios. One must remember, however, that the L sine A = log sin A+io, so that if a L sin A reading is 9-97941, then the reading for log sin A would be 1-97941. If the logarithmic ratios are used the addition of the 10 must not be overlooked. To find B : From equation (n), p. 359, cos B = tan a tan (90 c) i. e, t cos (180 B) =tanatan (c 90) = tan 72 27' x tan i 18'. Iogcos(i8o B) log tan 72 2 7'+ log tan i 18' 4990 = 2-3559Q 2-85586 -log cos 85 53' 6". 1806 = 85 53' 6" B = 9 4 6'54*. Hence, grouping our results, a = 72 27' A = 72 29' 45' 6-94i8'53* B = 946'54* C = 90 Example 3. At a point A, in latitude 50 N., a straight line is ranged out which runs due E. at A. This straight line is prolonged for THE SOLUTION OF SPHERICAL TRIANGLES 365 60 nautical miles to B. Find the latitude of B, and if it be desired to travel due N. from B so as to meet the 50 parallel again at C, find the angle ABC at which we must set out and also the distance BC. In Fig. 137 let A be the point on latitude 50 N. and ABD be a great circle passing through A : thus AB is a straight line running due E. from A. Let NB be the meridian through B, and NA that through A. The sides NA, AB and BN are straight lines, because they are parts of great circles and therefore they together form a spheri- cal triangle. In this triangle we know -the side AB (its value being 60', for i nautical mile subtends an angle of i' at the centre) ; the angle at A (90) ; and the side NA (90 latitude, i.e., 40). Thus two sides and the included angle are given and we require to solve the triangle ; hence we use rule (10), p. 358, from which cos NB cos NA cos AB = cos 40 cos 60' or logcosNB = logcoS4O-f log cos i = 1-88425 +1-99993 = 1-88418 i.e., NB = 4oo'38' 1 ' or the latitude of B is Q0 40 o' 38* = 49 59' 22*. Now C is at the same latitude as A, so that BC is 38", corresponding to f- nautical miles ; i. e., BC -633 nautical mile. 60 To find the angle ABC, we use rule (13).. p. 359. . __ sin NA sin 40 SI* L. ABC = ,-== = . ^-fQ, sin NB sin 40 o 38* log sin L. ABC log sin 40 log sin 40 o' 38* i -90807 I -90817 = I -99990 whence L ABC = 88 45'. For the surveyor, spherical trigonometry has an important application in questions relating to spherical astronomy. Thus in the determination of the latitude of a place by observation to a star, the calculations necessary involve the solution of a spherical triangle. This triangle is indicated in Fig. 139, the sides AB, BC 366 MATHEMATICS FOR ENGINEERS and CA representing the co-latitude of the place, i. e., (90 latitude), (90 declination) and (90 altitude) respectively ; whilst the angles A, B and C measure respectively the azimuth, the hour angle and the parallactic angle. The terms just mentioned are denned as follows: Fig. 138 represents a section of the celestial sphere at the meridian through the point of observation O. RDT is the celestial equator, CEX is the horizon, Z is the zenith of the point of observation, i. e., the point on the celestial sphere directly above O, and S marks the position of the heavenly body to which observa- FIG. 138. Determination of Latitude. tions are made. Also PSD, ZSC, RDT and CEX are portions of great circles. The altitude of a heavenly body is the arc of a great circle passing through the zenith of the point of observation and the heavenly body ; the arc being that intercepted between the body and the horizon. We may thus compare the altitude in astro- nomy with the angle of elevation in surveying. Referring to Fig. 138, ZSC is the great circle passing through Z and S, and SC is the altitude. ZS, which is the complement of SC, is called the zenith distance. THE SOLUTION OF SPHERICAL TRIANGLES 367 The azimuth of a heavenly body is the angle between the meridian plane through the point of observation and the vertical plane passing through the body. It can be compared with the 'bearing" of plane trigonometry. In Fig. 138, the angle PZS is the azimuth of S. The hour angle of a heavenly body is the angle at the pole, between the meridian plane through the point of observation and the great circle through the pole and the body. Thus, in the figure, P is the pole, and PSD is the great circle passing through P and S ; this being known as the " declination circle." Then ^_ZPS = the hour angle of S, and it is usually expressed in terms of time rather than in degrees, etc. The declination of a heavenly body is the arc of the declination FIG. 139. circle intercepted between the celestial equator and the heavenly body : thus DS is the declination of S. The method of calculation can be best explained by working through a numerical example ; and in order to ensure a clear conception of the problem, it is treated both graphically and analytically. Example 4. At a certain time at a place in latitude 52 13' N. the altitude of the Sun was found to be 48 19' and its declination was 15 44' N. Determine the azimuth. As explained before, a spherical triangle can be constructed with sides as follows: a 90 declination = 74 16', b = 90 altitude 41 41', and c = co-latitude 37 47'. Then the angle A is the a?imuth (Fig. 139). 368 MATHEMATICS FOR ENGINEERS Graphic construction. With any convenient radius OD describe an arc of a circle DABF. Draw OA, OB and OF, making the angle DOA = 6 = 4i4i', LAOB = c = 37 47', and L BOF = a == 74 16'. Draw DCE at right angles to OA, and FGC at right angles to OB, intersecting at the point C. Note that C lies outside the triangle AOB. With centre E and radius ED construct the arc of a circle DH : draw CH perpendicular to DE to meet this arc at H and join EH. Then the angle REH is the value of the required angle A, and is found to be in the neighbourhood of 141. [If C had fallen the other side of OA, the angle CEH would have been measured.] The actual spherical triangle ABC is formed by the circular arc BA and the elliptical arcs AC and BC. Proof of the construction The side b is such that it subtends an angle of 41 41' at the centre of the sphere. Thus DA measures the actual length of b, but does not represent it in its true position. In like manner BF gives the length of a, but again does not give its position on the sphere. Let the circular sector OAD be rotated about OA as axis, and the sector OBF about OB as axis, and let the rotation of both be continued until they have a common radius OC, i. e., OC is the intersection of the two revolving planes. Then evidently C is the third angular point of the spherical triangle ABC, since the given conditions concerning the lengths of the sides are satisfied by its position. We observe that in this case the rotation of OBF has to be continued beyond OA, from which fact we gather that the angle at A must be obtuse. The line OA is in the plane of the paper, and taking a section along DE and turning this down to the plane of the paper, we observe that the actual height of C above the paper is CH. Thus EH is a line on the plane OAC, also ER is a line in the plane AOB, both lines being perpendicular to the line of intersection, and the angle REH therefore measures the inclination of the plane AOC to the plane AOB, this angle being by definition the angle A of the spherical triangle ABC. By calculation. Here we have the three sides given, and we wish to find an angle which may be done by use of equation (4), p. 357, viz., A /sin (sb) sin (s c) tan = A / - : - : - - - r -- 2 N sin s sin (sa) Now , = <*+*>+<> _ 74 i6'+4i 4i'+37 47' _ 153 44' ^ 6 o , 222 so that s a 76 52' 74 16' = 2 36' s -b = 7652'-4i 4i' = 35 ii' Hence tan A = /^ 2 v sin 76 52' X sm 2 36 THE SOLUTION OF SPHERICAL TRIANGLES 369 A i r(logsin35Ti / +logsin395') ~1 and log tan - = - [_ _ (log sin ?6 o 52 / + i og sin 2 3 g/)J i'7 6 57\ 1-79965 J _ f 2-65670 I- 56022 / = x -91503 = H5752 thus = 70 46' 30* and A = 141 33'. Exercises 25. On the Solution of Spherical Triangles. (4-figufe log tables only have been used in the solution of these problems.) In Nos. i to 6, solve the spherical triangle ABC, when 1. =^50 = 90 & = 32i7'. 2. = 90 a = 4543' A = 6ii5'. 3. a = 72 14' & 43 47' c = 29 33'. Find also the value of B by the graphic method explained on p. 368. 4. Z> = 525' a = 58 25' C = 6 4 . 5. b = 27 13' c = 5ii8 / 6 = 85 9' and the spherical excess is 2 14'. 6. c = 7949' b = 28 5' B=i5i8' 7. If the sun's altitude is 17 58', its declination is 28i6'N., and its azimuth is N. 65 43' W., find the latitude of the place of observation. 8. The spherical excess of a triangle on the earth's surface is i 15': taking the earth as a sphere of radius 3,960 miles, find the area of the triangle in square miles. 9. Given that the azimuth of the sun is 10, and its zenith distance is 24 50' when its declination is 22 15', find the latitude of the place and also the hour angle. B B MATHEMATICAL PROBABILITY AND THEOREM OF LEAST SQUARES WHEN extremely accurate results are desired, these results being derived from a series of observations, the possibility of error in each or all of the observations must be considered. The correct result, or what is termed " the most probable result," is usually found by combining the mean of the observations with " the pro- bable error of the mean." The work that is to follow is concerned primarily with the establishment of a rule enabling us to find this probable error ; and as a preliminary investigation, a few simple rules of probability will be discussed. Supposing that an event is likely to happen 5 times and to fail 7 times, then the probability that it will happen on any specified occasion is r \, whilst its probability of failing is r 7 ^, because, considered over a great range, it only happens 5 times out of 12. It is important to note the significance of the phrase *' considered over a great range " ; we could not say with truth that the event was bound to happen 5 times out of the first 12, 10 times out of the first 24, and so on ; it might be doubtful whether it would happen 50 times out of 120. If, however, say, 12,000 opportunities offered, it would be fairly correct to say that the happenings would be 5,000 and the failures 7,000, for when a large number of occasions were considered, all " freaks " would be eliminated. To take another illustration : the probability that a man will score 90 per cent, of the full score or over on a target is /y indicates that he is rather more likely to score 90 per cent, than not (in the proportion 6 to 5) if he fires a great number of shots. In general terms, if an event may happen in a ways and fail in b ways, and all these are equally likely to occur, then the pro- bability of its happening is -j^-, and of its failing =- ; and if 37 MATHEMATICAL PROBABILITY 371 a = b, then it is as likely to happen as not, i. e., its probability of either happening or failing is \. Probability of happening _ a a+b _ a Probability of failing ~ a+b b ~ b' i. e., the odds are a to b for the event, or b to a against it, the first form being used if a > b and vice versa. E.g., if the odds are 10 to i against an event, the probability of its happening = = ; or it will probably happen once only out of eleven attempts. Exclusive Events. Let us now consider the case of two exclusive events, viz., the case in which the happenings do not concur. / Suppose the probability of the happening of the first event =- / and the probability of the happening of the second event = -T-. Then for purposes of comparison each of these fractions may be expressed with the same denominator : if this common denominator is c, write the fractions as and respectively. c c Now out of c equally likely ways the first event may happen in a ways and the second in 2 ways, and since the two events are exclusive, i. e., the happenings of the one do not coincide with the happenings of the other, the two events together may happen in !-}-, ways. Hence the probability that one or the other will happen is -. which may be written in the form -f- , i. e., as the sum c c c of the separate probabilities. E. g., suppose that one event happens once out of 8 times, and a second event happens three times out of 17, and that there is no possibility of the two events happening together ; then, the common denominator of 8 and 17 being 136, the first event happens 17 times out of 136 and the second event happens 24 times out of 136, and hence, either the one or the other happens 41 times out cf each 136. Probability of the Happening together of Two Inde- pendent Events. Suppose that one event is likely to happen once out of every 6 times, whilst another is likely to happen twice out of every 17 times ; then the probability that the two will happen together must be smaller than the probability of the 372 MATHEMATICS FOR ENGINEERS happening of either in fact, it must be the product of the separate probabilities ; i. e., the probability of the two events happening 122 I together = ;rX = or : or out of every 10.200 times the 6 17 102 51 first will probably happen 1,700 times, the second will probably happen 1,200 times, whilst the two would happen together 200 times only. Probability of Error. Bearing in mind these fundamental theorems, we can proceed to a study of the question of probability of error ; with particular reference to its application in precision surveying. It will be admitted that, for any well made series of observa- tions, the following assumptions may be regarded as reasonable : (1) That small errors are more likely to occur frequently than large errors, and hence extremely large errors never occur. (2) That positive and negative errors are equally likely, i. e., we are as likely to give a result that is -ooi too high as one that is ooi too low. Hence the probability of the occurrence of an error of a given magnitude, which is denoted by the number of errors of that magnitude total number of errors depends in some way upon the magnitude of that error. Our first idea, therefore, might be that the probability of the occurrence of an error of magnitude x could be expressed as f(x), i. e., as some linear function of x. It will be seen, however, that this is not in accordance with assumption (2) ; for assumption (2) demands that if a curve be plotted, the ordinates showing probabilities and the abscissae indicating errors, it must be symmetrical about the y axis. The function must therefore be of an even power of x, and taking the simplest power we say that the probability of occurrence of an error of magnitude x =y = f(x 2 ). Now, from assumption (i) we note that the coefficient of x z must be negative, because y must decrease as x increases. The probability of an error of magnitude x being included in the range x to x +S# must thus depend on x z , and also on the range Bx ; hence it would be reasonable to say that it =f(x z }8x, because the greater the range the more is the chance of happening increased. Therefore, the probability that an error of magnitude x falls MATHEMATICAL PROBABILITY 373 between any assigned limits, a and -\-a, must be the sum of the probability f(x z ) 8x extended over the range a to -\-a, i.e., P = ^ f(x*)dx this being the probability that the error does not exceed a. Hence the probability that the error may have. any value whatever (i. e., the probability is i) must be expressed by for the range is unlimited, so that /+ I f(x 2 )dx must = i. .' -oo It has been proved by Lord Kelvin (see his " Natural Philo- sophy ") that f(x 2 ) must be such that and since e?* x 0* = and e 1 ** X e*v* = this condition will be satisfied if _2 or Ae~* 2 the minus sign being inserted in accordance with assumption i on p. 372 ; and the coefficient k being written as ^ f r the reason that ft is explained later. A value can now be found for the constant A. r+oo r+oo It is known that I f(x 2 )dx = i, J -00 / + _Z2 hence A / e &dx i. Now it has already been proved (see p. 163) that r -z* I O e ' * = ''~z~ /_?: " h^/ir and I e h *dx = Jo 2 >r2 r2 /OO _ Z^ ,-flO _ *" also I e h ' 2 dx = 2 / e A2 rfA; = y -oo J o 374 MATHEMATICS FOR ENGINEERS hence AxWir = I, or A = Thus y =/(*) = the law being known as the Normal Error Law. The curve representing this equation is called the probability curve and also Gauss's Error curve. Two such curves are plotted in Fig. 140, to show the effect of the variation of the parameter h. In the one case h = -2, and for the second curve h = -5 ; and it will be noticed on comparing the curves that for the smaller value of h the probability of the occurrence of small errors is greater, i. e., the set of observations for which h = -2 would be more nearly correct than that for which h = -5. It will be seen that the curve is in agreement with the axioms stated on p. 372 ; for the probability of error is greatest when the error is least, the probability of a large error is very small, and there is as much likelihood of an error of + -2, say, as of -2. The probability that the error does not exceed -i is given by the area ABCD in the one case, and ABEF in the other. Theorem of Least Squares. If a number of observations are made upon a quantity, and the errors in each of these noted, i. e., as nearly as can be estimated ; then from a knowledge of these errors it is possible to find the most probable or likely value of the quantity. Let n observations be taken and let the errors be x^ x z Xn '. also suppose that all the measurements are equally good, i. e., the "fineness" of reading is the same throughout; h in the formulae above being a measure of the fineness. The probability of the error x being within a certain range 8x MATHEMATICAL PROBABILITY 375 will be the probability of an error of magnitude x l multiplied by the range *, i.e., i . and for error x 2 P 2 = 8xX -=.e~ * a and so on. Now #! x 2 etc. are quite independent, so that the probability of all the errors falling within the range &x will be the product of the separate probabilities, i.e., P = P 1 XP 8 X . . .P B &x -^ Sx -** x =e 2 X . . hVi We have thus obtained an expression which gives us the prob- ability of all the errors falling within a certain range. We might say that this range was -i, for instance, or -05. Evidently if all the errors were kept within the range -05 to +-05 the calculated result would be a nearer approximation to the truth than if the range were double the amount stated. Our object then is to find when the probability of a small error (8x may be reduced as we please) is greatest, *'. e., to find when P is a maximum. _ .. -!(*-) K Now P = K e h& -; and the smaller the denominator is made, the larger will P become. But the only variable in the denominator is 2# 2 , and hence, in order that P may have its maximum value, 2# 2 must be the least possible. Hence the most probable value of the quantity to be determined is that which makes the sum of the squares of the errors the least. The fact can now be established that the arithmetic mean of the observed values is the most probable value of the quantity. 376 MATHEMATICS FOR ENGINEERS Thus, if n observations are made, let ! 2 # 3 . . . On be the respective observations a the A.M. of these values a the value most probably correct then (! a) (a 2 a) etc. are known as residual errors. Now the probability of making this system of errors T> A ~ ro or P = Ae h ~ - A - To differentiate P with regard to a, put u = du then i_ = o+2tfw fl# Thus P = A*~^ rfP dP du _ __ _ \/ _ da du da A --- = F5 h 2 dP and -3-=- =o if 2an zSa, = o w or if = -za-, n but -2#i = a = A.M. of the observations n and hence <z = a or the most probable value is the A.M. of the observations. Again, if x is the error of the A.M. and x l x z x 3 etc. are the respective errors of the observations, By squaring X = - ft = l 2 (2V)+i< M 2V I / l n z\ MATHEMATICAL PROBABILITY 377 then, since it is assumed that all the observations are equally good, and that positive and negative errors are equally likely to occur, v 2 -y 2 A* 2 ,,- QTtin ^? ^ v {\ vj ^'2 3 * " r^ ciuA-i ^H-vi-vo "~~ vj for all the errors are small and their products, two at a time, are still smaller. i u 2 Also x 2 = (u 2 ) = * or # = ~= Vn . . probable error of a single observation or the probable error of the A.M. = , Vn and thus, other things being equal, the possibility of a large error in the final result is greatly reduced by taking a great number of observations. Also in a set of well made observations, if a sufficient number are made, the arithmetic mean cannot differ from any of the observations to any very great extent, and accordingly the residual errors and .the actual errors are very nearly alike. We are now in a position to summarise the results of the investigation so far as we have pursued it ; thus (a) The arithmetic mean of the series of observations, which are supposed to have been made with equal care, is the most probable value of the quantity. (b) The sum of the squares of the residual errors must be the least. (c) The probable error of the A.M. is equal to the probable error of a single observation divided by the square root of the number of observations. Example i. Seven observations of a certain quantity, all made with equal care, were 12, n, 14, 12, 11-2, 11-7, and 12-1. Find the most probable value of the quantity. The most probable value = A.M. = = 12, 7 and it can readily be shown, by actual calculation, that this value makes the sum of the squares of the residual errors the least. The residual errors are (12 12), (11 12), (1412), (12 12), (11-2 12), (11-712) and (12-1 12) or o, i, 2, o, -8, -3, -i and 2 squares of residual errors 0+1+4+0+ -64+ -09+ -01 = 574- 378 MATHEMATICS FOR ENGINEERS To test whether this is the least, let us suppose that the most probable value is 11-5; then the residual errors are: -5, -5, 2-5, -5, 3, -2 and -6 respectively. 2 squares = -25 + -25+6-254- -25+ -09+ -04+ -36 = 7-49. Similarly, if we assume, say, 12-2 as the most probable value, Z (residual error) 2 = (. 2 ) 2 +(i-2)?+(i-8) 2 +(-2) 2 +(i) 2 +(-5) 2 +(-i) 2 = -04+1-44+3-24+ -04+1+ -25+ -oi = 6-02 both of which totals exceed 5-74. To find the Probable Error of the Arithmetic Mean. Let r = the probable error of any one of the observations ; then if this is an "average" error, i. e., if errors greater are as likely to occur as errors smaller, the probability that the error is less than r is \. Now, the probability that an error lies within the range r to -\-r T f+r -I 2 2 [' - r - is =1 e h *dr = -=.1 e h2 dr T f+r -I 2 [' =1 e h *dr = -=.1 e hV-n-J -r flVirJ - /r /for dr=hd( r ) and the limits are now those for T and not those for r\ \ \hj h I There must be some connection between the amount of error and the fineness of measurement, i. e., between r and h, and this we must now find. If X = h A/V' o x x V ir-i o and we see from the above statement that the value of this integral is to be . 9 o XT , X * , X 3 . N o w e i i- jf -i u i 2 6 and thus e~ X2 i X 2 -| ^-+ . 2 o and if X 2 is small we may perform the integration by way of expansion in series : if X 2 is not small the value of the integral would be read from probability tables which give the values of the 2 /-x _ X 2 integral ^ I e ^X : these tables being given in the Transactions VWo of the Royal Society of Edinburgh, Vol. xxxix. For the present MATHEMATICAL PROBABILITY 379 application of the integral, however, X is a small quantity, and a sufficiently correct result is obtained by expanding in a series and calculating from a few terms in this series. Thus L _T o fh -X 2 -2 / rh --' ~ -41 dX = - 2 ( (i VTT\J FIG. 141. Hence and this equation may be written v5 = /r_j:L . ' or if for j- we again write X X 3 X 5 X 7 44SI = X 1 h terms which are very small. 3 10 42 By selecting values of X and plotting, the solution of this equation is found, the final plotting being represented in Fig. 141, where it is seen that the solution is X = -4769. 3 8o MATHEMATICS FOR ENGINEERS Y Thus T = '4769 or r = -4769^. Y [If solved to a greater degree of accuracy, the value of r is found to be -47696^, and this figure will be used in the work that follows.] Again, if n equally good measurements have been made, each will have what is termed a weight of unity, i. e., none is better or worse than any other, and when working towards the result to be deduced from the measurements, equal consideration must be paid to each measurement ; also the A.M. is said to have a weight of n since on the average n observations of equal weight must be made to give a result as true as the A.M. r Knowing that r m = ~/^ Vn where r m = probable error of the A.M. and r = probable error of any observation weight of A.M. n and also r-^-r ?- *r- r- = - weight of one observation i , . , .. v which we can write as = - w i we can link up w m and w with r m and r, for ^!-^i r* n w m or the weight varies inversely as the square of the probable error. Thus the determination of the probable error, whilst a useful guide to the accuracy of the one set of observations, is more use- ful in fixing the relative weights that must be given to different sets of observations. Thus, if three sets of observations have been made on a certain length with the results that the probable errors of the A.M. are 45, -29, and -51 respectively ; then the weights to be given to these sets are - ^ -. -^ - r^ respectively (.45)2 (. 29 ) 2 (-5i) 2 or -494 1-19 -384. Then in assessing for the final result, by far the most reliance would be placed on the second set of observations, less on the first, and least on the third set ; this fact being well illustrated by MATHEMATICAL PROBABILITY 381 Fig. 142, the resultant weight being nearer to the weight 1-19 than to either of the other weights. 1R.2-068 494 1-19 394 FIG. 142. To return to the object of this paragraph : If #! # 2 *s are the actual errors of observation, then the probability that each falls within the small range 8* is i l^ 2 i -(^Y P! = ^j V*/ 8* P 2 = -7=e ^ h) (&) etc., hVir hVir and the probability that they all fall within this range at the same time will be less than either of the separate probabilities ; it will actually be the product of these. Thus P = P 1 xP 2 X ...... i ( ~ hVir hV-jr -(*,+*,+ h We wish to find for what value of h P has its greatest value hence differentiate P with respect to h. P xe -- where K = = UXV = and h n dh v = g-A2< 2a: i 2 ) or if w = TtCZx-, 2 ) v = h? and thus ~ = (2^ 2 ) X -2h - 3 ah dv dv dw Also = -- x -3- an aw dh _de-" dw_ ^ : dw X dh~ 382 MATHEMATICS FOR ENGINEERS Then dP du . dv ~rr= v-r; +U-T; dh dh dh =o if 22V \/^ or h== 1-414 Now it has already been proved that r = -47696/1 so that r = -47696 Vs /(W) - -6745V ^ Also we have previously stated that the sum of the squares of the actual errors differs very little from the sum of the squares of residual errors ; this being true if a great number of observations are taken. The difference in the two sums may be expressed rather more accurately by the relation Yl ^Xj 2 = - - 2 (residual error) 2 . w r Hence if for 2 (residual error) 2 we write Applying these results to Example I on p. 377, W = 574 n = 7 then ^ * - '6745^ = -2475 MATHEMATICAL PROBABILITY 383 / 2n^r e z /2X574 c also h = v 7 v == v =i '3 v (n i)Xn 6 *. e., h has a very high value ; and this would be expected, for the "fineness " of reading, as judged by the results, is not at all good (one error being as much as 2 in 12). Example 2. In a chain survey four measurements of a base line gave 867-35, 867-51, 867-28 and 867-62 links respectively. Find the best length and the probable error in this length. The best result is the A.M. of these, i. e., 867-35+867-5I+867-28+867-62 4 867-44 links and whilst this is the best result it contains a probable error. Probable error in A.M. = ^=-6745\/^Yy = -6745 /( Q9) 2 +(-Q7) 2 +( 4X3 - -6745 V ^ = -0517 i. e., the base line measurement (867-44) is subject to an error of 0517 link, and as this result could not be bettered it would be unnecessary to repeat this portion of the survey. The probable error in any one observation would be '='6745 \-- = -io3, so that there is a decided gain in accuracy obtained by increasing the number of observations. (Cf. "repetition," when working with the theodolite.) It is of interest to find h for this example. 2X-07I _ -J = ' 2176 and as this is a small quantity we are confirmed in our conclusion that the observations were well made. Example 3. The mean values of the three angles of a spherical triangle were calculated from the actual observations to be 75 40' 21 -6", 39 1 i' 47-3", and 65 7' 56-2"; and these values were subject to probable errors 2-9", 3-6*, and 4-3* respectively. From a knowledge MATHEMATICS FOR ENGINEERS of the area of the triangle, the spherical excess of the triangle was found to be 3-3*. Make the necessary adjustments to the angles to satisfy this condition. The actual spherical excess = (75 40'2i-6"+39 ii / 47-3"+657 / 56-2")-i8o There is thus (5-1 3-3) to be divided among the angles, according to the respective weights ; and these weights are in the proportion or -119 -077 -054, the sum of the weights being -250. IIO *O77 Hence the corrections to be applied are Xi-8, xi-8. and 250 -250 ^ xi '8 to the respective angles; all these corrections being sub- 250 tracted, since the observed angles give a spherical excess greater than should actually be the case. These corrections are -857, -555, and -389. Hence the true angles are (75 40' 21-6* -86*), (39 ii'47'3* -56") and (65 / 56-2*- -39*), or 75 40' 20-74", 39 II/ 46'74' / and 65 7' 55-81*. Example 4. Measurements of an angle in a traverse survey were made by two different observers, with the following results : Readings by A. Readings by B. 76 50' 20* 76 50' 55" 76 50' 50* 76 50' 35* 76 50' 30" 76 51' 15" 76 51' 10* 76 51' 20* 76 5 0' 3 0* 76 51' o" 76 51' o" 76 50' 45* 76 50' 40* 76 50' 25* 76 50' 30* 76 50' 40" Compare the two results from the point of accuracy, and find the most probable value of the angle. We must first find the arithmetic mean of each set of observations, and then, by subtracting this from each reading, we determine the residual errors. The A.M. of set A = 76 50' 4 1-25* and A.M. of set B = 76 50' 5 1 -88*. MATHEMATICAL PROBABILITY 385 Since the differences are of seconds only, we need not concern ourselves for the present with the degrees and minutes ; and thus the table of residual errors and their squares becomes Residual Error. (Residual Error) 2 . Residual Error. (Residual Error)*. 21-25 451.4 + 3-12 9-7 + 8-75 76-6 16-88 284-9 11-25 126-6 + 23-I2 534-6 +28-75 826-8 + 28-12 790-7 -11-25 .126-6 + 8-12 66-0 + 18-75 35 1' 6 - 6-88 47-3 - 1-25 1-6 -26-88 722-4 11-25 126-6 -H-88 141-2 sum o 2087-8 o 2596-8 In case A f / 2087-8 '"745 <v 8x7 In case B I 745 V 8x7 " weight of observations by A (4-594) 2 1-244 (4-119)' Thus A's readings can be relied on before those of B ; the former being roughly ij times as good as the latter. The most probable value of the angle, taking into account the two sets of readings, will be obtained by the calculation of the " weighted mean," i. e., the mean of the two arithmetic means already found, determined with due regard to the respective weights to be given to A's readings and B's readings. Dealing only with the seconds, the most probable value (41-25 x i -244) + (5 1 -88 x i) 1 + 1-244 51-31+51-88 _ 103-19 2-244 2-244 = - - = 46 seconds. Hence the most probable value of the angle = 76 50' 46* Exercises 26. On the Calculations of Errors of Measurements. 1. One surveying party measured a certain base line as 6 chains 42-7 links, 6 chains 53-5 links, 6 chains 46-4 links, and 6 chains 41-9 links ; and a second party measured the same line as 6 chains 38-4 links, 6 chains 39-7 links, 6 chains 46-9 links, and 6 chains 43 links. State which of the two parties is the more dependable, and find the most probable length of the line. C C 3 86 MATHEMATICS FOR ENGINEERS i -*_? 2. Plot the probability curve y = /- e AZ the value of h being n v TT 1414, and find the probability that an error lies within the limits -6 and + -6. 3. The following are the values of the determination of the azimuth of Allen from Sears, Texas, the results of a U.S. Coast and Geodetic Survey; the values of the seconds only being stated after the first reading: 98 6'4i-5*, 42-8, 43-4, 43-1, 39-7, 42-7, 41-6, 43-3, 40-0, 45-0, 43-3 and 40-7. Find the A.M., the probable error of a single obser- vation and the probable error of the mean. 4. Find the weighted mean of the following observations : 95-8, 96-9, 97-2, 95-4, 95'7, 97 g ii 96-5, 96-7 and 97 ; the probable errors in the measurements being -2, -4, -i, -9, -7, 1-2, -8, -3 and 1-5 respectively. ANSWERS TO EXERCISES Exercises 1 3*7 AC* 3. E = constant x -~ 4. V == RC + L ^ at at 5. 11-03 cms. per sec. : 1-07 seel;, from start 6. -336 ton 11. 5.65 12. Middle of May : middle of October 15. Loading is -2 ton per foot run 16. -966 : =^ = cos 6 uv 17. -42 ton per foot 21. 6-3 Exercises 2 1. 4# 3 2. -128 3. 2* 4. 27*" 5. g 18-75 0086 i'ii -pl.23 #v 1 1 982 19 -fi?* 8 ' 3 13 /<car 2 /i/i Si' 1 - 8 4- X 2 ' 3 X* 14. * /pi -6 347 17. JIV 3 ' Si - , -84 . 1-29 52'5 12-48 ._ .- 20. 073 21. i - a - &P~* 22. w(^ + X jr~j- x 23. 289* w "* d / I 24. : (Arv + ny z xl) 25. = - \/ i + -; V/ 2 w 26. -2(/> + /*"f \ 9) 27.7-85 28. w(-j--x\ 29.9-6 30. -7333 Exercises 3 1. Sub-normal 466 ; sub-tangent = i 2. 25-7 3. y = -0256^ ; (# is distance from centre) : -64 + 142-5 5. M-= ( -- xy. S = - -: L, o 6. M = -(-- 2\4 7. M = W (l-x) : S = - W : L = o 8. Sub-tangent = : sub-normal = 9. 3 10. -~ (wl -wx- aP) 387 388 MATHEMATICS FOR ENGINEERS Exercises 4 1. -5*~ 5 * 2. 6-I5**' 1 * 3. -^=| 4. 1-423 (4-15)* & 5. 4-33 (8-72) 2x 6. ge 9 * 35e~ 7x 7. 5-44* 1 ' 718 8. 9-7 (2)* 9. io-25<r 25 *- 10. 11. i2-6e*' 2 * 12. - 13. ^ 14. x 5* 4 4 ac^ _!.&,-. 2* 1 O. r A u. ioe i^ X 1 4 17. I 4 --- ? - or *2L-VT- [Use the rule log AB = logA+logB] * 3* 47 *(3* 47) 18. -> -- \ -- - -- \ -- ^ 19. 30 3jc + 8 sinh zx '^~ 5* + 4^3* -2 7-4^ * 20. - X - 1-057 (1-8)* 21. o 22. 1-052 23. 24.^43 25> . 2T 26.^,- 27.^7- t -L* i *ow 7 28. ^ sinh- :^ cosh y - 29. frE 30. & 31. 44?? ^fc 32. o 33. o 34. - + C - -, - Exercises 5 1. 5'3 cos (4 5-3*) 2. 16-3 sin 5-1* 3. -48 sec 2 (3* -f 9) 5-05sin(-05 -117*) 4. -QI4COS (-425^ 1-25) 5. 4ocosec 2 # 6. , : cos 2 (-05 -117*) 7. gbc sin (rf gx) 8. 20 sin 5^ 14 cos (zx 5) 9. 4-40038-8^+ -S cos 1-6^ {Usetherule : 2 sinAcosB = sin( -(- sin (A B)} 10. 6-74 sin 6-2X 3-04 sin 2-8 11. 4'5 2 V ~^ sin (P x 1 X + 2C ) + ., sin (px \O 'U "U T" U 12. 5 sin 2x. {Use the rule: cos 2A = i 2 sin 2 A} 13. -195 sin 6x 14. o 15. S-I6A" 72 -5-2 -0273 cos (4-31 -195^) + 24-93 3# 4' 1 16. -1056 cos -015^ -0529 sin (6-1 -23*) + 7-4 sec 2 (4* -07) 17. Velocity = 37-7 sin 31-4* 56-56 cos 31-4* : acceleration =1184 cos 31-4/4- J 777 si n 3 I- 4' 18. Acceleration -02895 : S.H.M. 19. 1162 20. Sine curve (i.e., second derived curve). f B/ 21. o I Treat as a constant the portion 22. 1500^ cos pt + $oop cos $pt + 42^ sin pt 8^p sin ANSWERS TO EXERCISES 389 Exercises 6 1. 2 cos 2.x . <? 8in2 * 2. - 3 2 sin 2* 4. 24*2. cos x 3 v 5. 3-14 (iox + 7) sec 2 (5# 2 + 7* 2) 6. 3 log a . cos 3* . a" in3 * 3 ~T" i ~j % ~~~~ Q# 5 sec 2 ._ sin<9 (dy dy dd\ 10. COSCCA- 11. 12. - . { = 3^ - X 3 sec z cp cos a i + cosy la^r aw a^J 13. \ ,, ,j. slope of curve = j^ X -=- X ^ =- , IrflogV rfA dV rflogV) 15. i -08 ft. per min. : -377 ft. per min. 18. -033215.: 01020' 19. 7 cosec 7* + 45*2 20.=^ 21. 56 19' 22. 53 7' 24. Exercises 7 1. x (2 sin 3* -f $x cos 3*) 2. 2^ 2 ' 4 (i + 3-4 log 3. ( 2-07 5 sin (3-1- 2-07*) \ Icos -1 2-0^ x J 4 Ar 5 cos (3-1 2-07.*) cos (3-1 2-07^) 5. - {2-575 sin (5-15* + 4) + -625 sin (1-25* - 4)} or - {3-2 sin 3-2* cos (i-95* + 4) + i'95 cos 3-2^ sin (1-95* + 4)} 6. sec 2 2x {2 cos (5 3#) + i -5 sin (5 3*) sin 4*} or 3 tan 2.x sin (5 3*) + 2 sec 2 2# cos (5 3*) 7. I2-8*' 6 (cos (3 + 8x) + 2 5* sin (3 + 8*)} 8. 27 (5) 3z J4-83 log x + - 9. (i + log AT) e xlo x 10. 11. 3 oe (5* + 4)(5* + 2) 12. /tan -125^ log^ \ 3 ^\ x 8cos 2 -i25Ar/ 13. o 14. 5e- lot 15. o 16. 12600 sin (14* -116) 17. 6t {5 sin (4 -8t) 2t cos (4 -8*)) 18. 4* 2 ' 7 (3-7 cos 3^ 3* sin 3*) Exercises 8 1. 5 * 2( 7 3 ~ 7;y) 2. - -. (-^ - log (2 - 7*) tan (2 - 7*)} e 7 *- 5 cos (2 7#) \7# 2 3 20 4 & 5 5-46 (5) 2 ^ '" Vd* 6. TCsin^n-^-^S^coshi-S^ or 21 i _ * 2 + 6^ + 15 (i - Q wb (ab -zbx + x 2 cot B) ~2~(&-*cotB) 2 3 9 o MATHEMATICS FOR ENGINEERS e s\n(i-2x + l-7) 12< {loi^^T^Ts)? x j I -2 COS (l-2X -{- I -7) log (8x z JX + 3) + -~- . 6-55 (sin cos 0) 13. sech 2 * 14. J3 . v 5 (0 sin 0)^ . 66-2 (0 sin 0)* (20 30 cos + sin 0) 4 y z 15. lo. 17. rw 2 cos -1 L (m 2 - sin 2 0)* d(f> a> cos . d z (f> _ a> z sin (i m 2 ) '' ~dt ~ \/m 2 - sin 2 ' ^ 2 (m a -sin a 0)^ 19. ^^ a : J* 20. (# ?)sm 2 ^ > w Exercises 9 2. *pt + 4e . log ( 5 p - 3) 4. io( 4 -w)(9-4w)(3 + 8w) 2 6. 8(1-7 ?_(*__ ' CrVdt r'dt 7 _ 31 o ?_(*P__t dr\ ' ' Exercises 10 1. 750 2. 17-1 3. ^ : ^- 4. -577/ : -I28W/ 5. -5 6. 2-25 : minimum 7. -278 : maximum 8. maximum at x = 5 : point of inflexion at x = 2 minimum at x = i 9. 2 rows of 8 10. base = 3-652 ft. : height = 1-826 ft. 11. 2-1 : minimum 12. - 13. -496: 631 14. width = height = 8-4' 15. 15-2 knots: 956, 948, 957 16. x = -289; 17. h = 6-34 ft. : d = 12-68 ft. 18. base = 4* : height = 5* 19. depth = 3 x breadth. 20. ^ 21. 6 22. -866 r w 23. v : | : 24. u = -.51; 25. 135 or 315 ANSWERS TO EXERCISES 391 26. d = l 27. I = 2'o6sr 28. 20-15: -45 29. o or 20 56' 30. height = 8-1 ft. : base = 6-72 ft. 31. x = -4* 32. tan- (-*^?$\ 38. / _ i fi z 34. maximum at x = o : points of inflexion at x j= . V2 35. VP 36. /(4/+d 37. K i + K Let r = ^ and find ^] 39. 3 units 40. * = ^^ P! dr J 2 41. T/ = ^T m 42. maximum at * = 2 minimum at * = 4 and at * = 2-5 points of inflexion when * = 2-26 and 1-92 / 38. -58 ( \ 43. x = VRjRg : / (maximum) = - fam + i) (R x R 2 ) 2 44. d = \/^j 46. 83 i' or 276 59' ' ojt Exercises 11 1. -006 2. 25%toolow 3. -0264 4. 2-45 5. 2-66 6. decrease of -00135 7. -03 link; -237 link , (#loga) 2 , (#loga) 3 , 8. i + x log a + * - ! : i- + . . . 1.2 1.2.3 9. 2214-2525 10. -536 Exercises 12 2. 152 5. 240 ; 205 ; 64 6. 621,000 ft. Ibs. : potential energy = 240,000 : kinetic energy = 381,000 : 987,000 ft. Ibs. 7. 480 9. 238,000 10. 3006 12. i-526# 2 - 68 +C 13. 70-15* + C 14. - +C 15. 17. -i* 10 10 log x + 14* + C 18. 3-32* 1 ' 04 2-5* 2 + C 19. i-o74* 3 ' 718 + ie* + C 20. i-33 9 *- f + C 21. 6-54-*'- 1 - + C 22. -689* + C 23. -^ - + C C 24. 3-025*' 84 8-2 log x 2-7ie- 2> * + 1-13* + C 25. -0234*- 10 ' 2 * + C 26. i-g6e' 61x 1-297*-" + -674* 8 ' 04 + C 27. -797 cos 0* 1 -" 2-2g-8'-i*+C 28. Ju 6 + C 29. -- ^+C 30. 35< + C 31. ie^-+C 32. - 5-88^ + C 33. 20-2 (2) + C 34. - * + 2-5^2 + 4-25** - 8x + C ^T 35. -88 5 ( 3 -i)t + C 36. _ + C 392 MATHEMATICS FOR ENGINEERS 37. _ _ z-jiSt + C 38. i6-i* 2 + C 39. - + C 45 x 40. Write the equation in the form -f w and then integrate : p v pv n C Exercises 13 1. -Jcos 4 * + C 2. i -73 sin (3 - 3*) + C 3. 49 tan (3 }x) + C 4. i-oix' 9 ** + 1-195 sin (-05 -117^) + C 5. -I8540 5 ' 4 * -cos (b + ax) + C 6. 9-45* . sin 8t + C & 7. -713 cos 2(2-i6x 4-5) + C 8. i2-85e' 7 * ^ r + i -83 log cos # + C A# 2-8\ 9. 9-95 cos ( ) + '022 sin 9^r i'46^ 2 ' 74 + -455(3) 2x + 6 + C \ 7 / 10. 2* -787 cos ( - 3-7* ] + 7-55 cot. ^ + C V4 / 5 (i 11. v = 7 cos (7^ -26) + C ; s = 49 12. -5^ = 47r 2 nV( sin ^ + Sin 2 \ + C; * = 47r 2 nV( cos ^ + C S 2 ) + C D \ 2m / \ 4m / 13. 3iH5 8p ) - -139 sin (3-7 - 7-2^) + C 14. 19-5 cos 6t 4-9 sin 6t + C Exercises 14. 1. -182 2. -345 3. 1-7 4. -561 5. -0626 2! 6. 1-218 x io 7 7. 2-62 8. -1589 9. - P 10. -6i6B mai 11. - - | C (a J"^ + C ^ S (fl ~ - f W 2 I a + 6 a 6 ^o -n 2 13 - p = 14. { ^4 - 4 /3^ + 3 ,4 } 1 - 24EI ( 16. ^ 17. .20 4 6/* 18. I ^ SiU ^ Wk * 3 i + sin $ 2 19. 26-24 {Limits must first be found} ; f 20. 334 21. y = -736*?' + 5 log # + 3^ 3-25 22. i -087. . . ._ 2 6.H = g 27 . 240 3 ; 8 ., 9I 29. 8 30. 49-82 31. ^f 32. -8596CA.W 33. ANSWERS TO EXERCISES 393 Exercises 15 1. -158 log - 2. 5 (* +1-583) 3 V6 V6 3. log (9*2 - 18* + 17)" -f C 4. -1919 5. 605 {Let u 6 h] 6. yV-j$ y 2 + -75 sin" 1 1 -154^ + C 7. -106 8. $sini2* + 4/ + C 9. tcot5#+C .. WR 3 /i i\ -0683 WR 3 10. -==- --- ) or -- - El \ir 47 El 11. -8 log cos 5* + C {Let u = cos 5*} 12. # sin" 1 # + Vi ^ + C _ -g&Z 13. V (a 2 AT 2 ) 3 + C 14. 3 {3 sin 2* 2 cos -zx] 15. tan- 1 * H -- ~ + C (Let u - tan- 1 AT) 16. -081 i + x z 17 tsin<9 18 18> k 19. 183 sees. : {Rationalise denominator of right hand side by multiplying top and bottom by Vh + 12 Vh ; then integrate, making the substitution u = A + 12.} 21. 10 42 22. ^ + t - 1 cosh- 1 1 + C 23. 355 24. i -749 12 8 2 2 256 25. 26. -oiR 3 315 27. (-5* + 1-25) 2 T^5* -x* + 13-63 sin- 1 + C 28. F= V a 2 Exercises 16 1.. 41-59 2. 1-718 3. 10-85 4 - <688 5 - 4'5 6. 23-05 7. 1-348 ft. candles. 8. o 9. -1294 10. 273 11. 205 Ibs. Exercises 17 1. 14-14 2. -6215 3. 2-4 4. -1165 5. -833 6. 1-194 394 MATHEMATICS FOR ENGINEERS Exercises 18 1. 1450 cu. yds. 2. 5977 Ibs. 3. 1070 4. 4nr 2 c KTrD 2 / 2 x _ . Kl 5. a 3-036, b = -1423 ; 1617 6. ^ or Vol. x 7. 524 (limits are 5 and 10} 8. 271-6 9. 1-2 Ibs. 10. true == 76-62 : (a) 75-41 : 1-58 % low (b) 77-73 : 1-45 % high (c) 76-60 : correct. 11. 60-9 ft. Exercises 19 1. | height from vertex 2. area = 7200 sq. ft. : centroid is 158' from forward end 3. 3-84 Ibs. ; 4" 4. 10 ft. from top 5. x = y = i -7* (taking the centre as origin) 6. 771 ; 2-25 7. (o, -95) 8. 2-35* from AB 9. 1-055* 10. -935" 11. -877* 12. 1-02* 13. 2-68* i. e., 1-34* 14. 5-12 ft. from O 15. 30880 Ibs. 16. 18432 Ibs. : 3-534 ft. 17. 9 units. 18. 5-1 ins. 19. 2-35 ins. 20. (a) 17 Ibs. (b) 1 1* 21. 1-48 ins. * 22. \h 23. |A Exercises 20 1. -655/ 2. 17-11 ins.: 785 Ibs. ins. 2 3. C. of G. is -1125* distant from centre of large circle: 2-68" 4. 16-6 ins. 4 5. (a) -408^ ; (6) -707^ 6. IAB = 80-7 ins. 4 : & A B = 3'O4* 7. 29-1* 8. 377 9. 9-86 10. 7-35 11. -2887^. (Divide into strips by planes perpendicular to the axis and sum the polar moments of these) . IE of circular 3 12. T _ , - = - = -956 13. 33.3 inch units IE of square TT 14. 681-6: 17-1 15. I NN = 169-4: NN = 2-44: I AB = 570 : k^ = 4-47 16. NN is 3-99* from bottom of lower flange : INN = 461 : NN = 3'77 17. 5-04* 18. 2-023': -444 19. 2-74* 20. (a) 13-9 (6) 31-1 (c) 1-48" (d) 1-82* 21. -28 22. massR* + 23. -56 Exercises 21 3. 5-23 4. = 2rsin0: the sine curve 6. 892 7. 5-01 Exercises 22 1. y = i-6jx* 2-4* 12-82 2. s = 8-05^ 23-1* + 14-1 3. y = Ae** 4. y = 8-35 -1490 -i-o** W fdv I 5. y = - 2 3 3 = = = = - ANSWERS TO EXERCISES 395 wl z w 7. log = - 8. v = (5 - 9- lg = 10. ws 11. H = , 7- 2 ) . / ftnh J 12. = Asm(/ - , , B 13. ^ = A+- 2 17. y = A^ 10 - 6 + Ajg 7 ' : y = A^ 10 ' + A^ 7 * + i 18. 5 = A^ 9 ' 33 ' + A^- 9 ' 33 ' 19. 5 = A sin (9-33^ + B) _ 20. KRlog- 1 21. C = C Ae 22. C= 23. ^ = alo or a= (Separate the variables and use the / -5 - ^ form J Ysin /y/ ==, 25. =- - 26. Mf = log^ + -):-i945 27. AT= 28. x = Ae~ 3t sin (632/ + c) + -026 a sin (5^ tan -1 1-25) J\ - /5. 29. V - A x e V r * + A 2 e V r * 30. /w - C 31. 6 = 6 a + e- kt (6 -6 a } 32. 5= - 53-636-^^+ 5 -8 3 e where = V V 2EI e 2EI or y = B^* + B 2 - -f B 3 sin <f>x + B 4 cos (f>x -^ 34. = ' 396 MATHEMATICS FOR ENGINEERS Exercises 23 1. 103 sees, if coefficient of discharge is taken as -62 2. 7-37 3. 2-83 cu. ft. 4. 69-5 : 95-5 (Draw in the " simply supported " bending moment diagram and work on the Goodman plan, see page 313) 5. Find the time to lower to level of upper orifice (183 sees.) with both orifices open; then the time for the further lowering of 5 ft., through the one orifice (180 sees.). Total time = 363 sees. Note that * 6. 57-7 sees. 7. 1-4 12 8. 5500 Ibs. : 4-71 ft. below S.W.S.L. 9. 14100 Ibs. : 6-65' below S.W.S.L. 10.761 11. '^f G;.- ,^),) (Hint. let u = d e +Kx) 12.1-23 13. Vertical depth = 8-07 ft. Exercises 24 1. x = 2-31 1-231 cos 5 1-55 sin 6 -16 cos 20 -022 sin 20 -004 cos 3# -04 sin 3$ 2. A 1-29, oj = O, B = -I4, o 2 = rr 3. ^ = 16-97+6-49 cos #+ -002 sin x 12-66 cos 2* i -46 sin 2.x 1-75 cos 3# -7 sin $x 4. E = 1500 sin 0+too sin 3042 cos #+28 cos 3$ Exercises 25 1. B = 393i / ; A = 6 5 5i'; c = 57 5' 2. c - 54 44i' ; * = 34 14' ' B = 43 32^' 3. A= 161 8'; B= 13 35'; C = 9 38' 4. A = 76 36'; B = 648'; = 52 5. B = 35 43' 40"; A = 6i 21' 20"; a = 43 25' 23" 6. C= 33 29' or 146 3 1' a = 103 28' or 55 28' A= 146 58' or 27 30' 7. 55 10' 8. 342,200 9. Latitude = 43 54' : hour angle = 4 31-3' Exercises 26 1. 2nd set better than ist set in the proportion 1-943 to i : 6 elms. 43-4 links. 2. Just under i. 3. 98 6' 42-26" : y m -307", v 1-062" 4. 96-93 MATHEMATICAL TABLES TABLE I. TRIGONOMETRICAL RATIOS Angle. Chord. Sine. Tangent. Co- tangent. Cosine. De- grees. Radians. o o o GO i 1-414 1-5708 90 i 2 3 4 0175 0349 0524 0698 017 035 052 070 0175 0349 0523 0698 0175 0349 0524 0699 ' 7'290O 28-6363 I9-08II I4-3007 9998 9994 9986 9976 1-402 1-389 1-377 1-364 1-5533 1-5359 1-5184 1-5010 89 88 87 86 5 0873 087 0872 0875 II-43OI 9962 l'35l I-4835 85 6 8 9 1047 1222 1396 1571 105 122 I 4 157 1045 1219 1392 1564 1051 1228 1405 1584 9-5I44 8-H43 7-H54 6-3I38 9945 9925 9903 9877 1-338 1-325 1-312 1-299 1-4661 1-4486 1-4312 1-4137 84 83 82 81 10 1745 174 1736 1763 5-67I3 9848 1-286 1-3963 80 ii 12 13 14 15 1920 2094 2269 2443 192 209 226 244 1908 2079 2250 2419 1944 2126 2309 2493 5-I446 4-7046 4-33I5 4-OIO8 9816 9781 9744 9703 1-272 1-259 1-245 1-231 1-3788 1-3614 1-3439 1-3265 79 78 77 76 2618 261 2588 2679 3-7321 9659 1-218 1-3090 75 16 17 18 19 2793 2967 3*42 33i6 278 296 313 330 2756 2924 3090 3256 2867 357 3249 3443 3-4874 3-2709 3-0777 2-9042 9613 9563 95" 9455 1-204 1-190 1-176 1-161 1-2915 1-2741 1-2566 1-2392 74 73 72 7i 20 3491 347 3420 3640 2-7475 9397 1-147 1-2217 70 21 22 23 24 3665 3840 4014 4189 364 382 399 416 3584 3746 3907 4067 3839 4040 4245 4452 2-6051 2-4751 2-3559 2-2460 9336 9272 9205 9135 1-133 1-118 1-104 1-089 1-2043 1-1868 1-1694 1-1519 69 68 67 66 25 4363 433 4226 4663 2-1445 9063 1-075 I-I345 65 26 27 28 29 4538 4712 4887 5061 45 467 484 501 4384 454 4695 4848 4877 5095 5317 5543 2-0503 1-9626 1-8807 1-8040 8988 8910 8829 8746 i -060 1-045 1-030 1-015 1-1170 1-0996 1-0821 1-0647 64 63 62 61 30 5236 518 5000 5774* I-732I 8660 I-OOO 1-0472 60 31 32 33 34 54ii 5585 '5760 5934 534 551 568 585 5150 5299 5446 5592 6009 6249 6494 6745 1-6643 1-6003 1-5399 1-4826 8572 8480 8387 8290 985 970 954 939 1-0297 1-0123 9948 9774 59 58 57 56 35 6109 601 5736 7002 1-4281 8192 923 9599 55 36 37 38 39 6283 6458 6632 6807 618 635 651 668 5878 6018 6i57 6293 7265 7536 7813 8098 1-3764 1-3270 1-2799 1-2349 8090 7986 7880 7771 908 892 877 861 9425 9250 9076 8901 54 53 52 51 40 6981 684 6428 8391 1-1918 7660 845 8727 5 41 42 43 44 7156 7330 7505 7679 700 717 733 749 6561 6691 6820 6947 8693 9004 9325 9657 1-1504 1-1106 1-0724 1-0355 7547 7431 7314 7193 829 813 797 781 8552 8378 8203 8029 4 i 48 47 46 45 7854 765 7071 I-OOOO I-OOOO 7071 765 7854 45 Cosine Co-tangent Tangent Sine Chord Radians Degrees Angle 397 398 MATHEMATICAL TABLES TABLE II. LOGARITHMS J 1 2 3 4 5 6 7 8 9 123 456 789 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 9 13 4 8 12 17 21 26 16 20 24 30 34 38 28 32 37 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 12 4 7 11 15 19 23 15 19 22 27 31 35 26 30 33 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 11 3 7 10 14 18 21 14 17 20 25 28 32 24 27 31 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 7 10 3 7 10 13 16 20 12 16 19 23 26 30 22 25 29 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 369 369 12 15 18 12 15 17 21 24 28 20 23 26 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 369 368 11 14 17 11 14 16 20 23 26 19 22 25 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 358 358 11 14 16 10 13 15 19 22 24 18 21 23 17 2304 2330 2355 2380 2406 2430 2465 2480 2504 2529 368 267 10 13 15 10 12 15 18 20 23 17 19 22 18 2553 2677 2601 2625 2648 2672 2695 2718 2742 2766 267 257 9 12 14 9 11 14 16 19 21 16 18 21 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 247 246 9 11 13 8 11 13 16 18 20 15 17 19 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 246 8 11 13 15 17 19 21 22 23 21 3222 3424 3617 3802 3243 3444 3636 3820 3263 3464 3655 3838 3284 3483 3674 3856 3304 3502 3692 3874 3324 3522 3711 3892 S345 3541 3729 3909 3365 3560 3747 8927 3385 3579 3766 3945 3404 3598 3784 3962 246 246 246 245 8 10 12 8 10 12 7 9 11 7 9 11 14 16 18 14 15 17 13 15 17 12 14 16 25 979 3997 4014 4031 4048 4065 4083 4099 4116 4133 235 7 9 10 12 14 16 26 27 28 29 4150 4314 4472 4624 4166 4330 4487 4639 4183 4346 4502 4654 4200 4362 4518 4669 4216 4378 4533 4683 4232 4393 4548 4698 4249 4409 4664 4713 4265 4425 4579 4728 4281 4440 4594 4742 4298 4456 4609 4757 235 235 235 134 7 8 10 689 689 679 11 13 15 11 13 14 11 12 14 10 12 13 30 4771 4786 4800 4814 4829 4843 4867 4871 4886 4900 134 679 10 11 13 31 32 S3 34 4914 5051 5185 5315 4928 5065 6198 5328 4942 5079 6211 5340 4955 5092 5224 5353 4969 5105 6237 5366 4983 5119 5250 5378 4997 6132 5263 6391 6011 5145 6276 6403 6024 5159 5289 6416 6038 5172 6302 6428 134 134 134 134 678 678 568 568 10 11 12 y 11 12 9 10 12 9 10 11 36 5441 5453 6465 5478 5490 6502 6514 6527 5639 6551 124 567 9 10 11 86 37 38 39 5563 5682 5798 6911 5576 5694 5809 5922 5587 5705 5821 5933 5599 6717 5832 5944 6611 6729 6843 5955 6623 6740 5855 5966 5635 6752 5866 6977 6647 6763 6877 6988 5658 6775 5888 5999 5670 5786 5899 6010 124 123 123 123 667 667 667 457 8 10 11 8 9 10 8 9 10 8 9 10 40 6021 6031 6042 6053 6064 6076 6086 6096 6107 6117 123 466 8 9 10 41 42 43 44 6128 6232 6335 6435 6138 6243 6345 6444 6149 6253 6355 6454 6160 6263 6365 6464 6170 6274 6375 6474 6180 6284 6385 6484 6191 6294 6395 6493 6201 6304 6405 6603 6212 6314 6416 6513 6222 6325 6425 6522 123 123 123 123 456 456 456 456 789 789 789 789 45 6532 6542 6551 6561 6671 6580 6590 6599 6609 6618 123 456 789 46 47 48 49 6628 6721 6812 6902 6637 6730 6821 6911 6646 6739 6830 6920 6656 6749 6839 6928 6665 6758 6848 6937 6675 6767 6857 6946 6684 6776 6866 6955 6693 6785 6875 6964 6702 6794 6884 6972 6712 6803 6893 6981 123 123 123 123 4 5 < 465 445 445 778 678 678 678 50 6990 6898 7007 7016 7024 7033 7042 7060 7059 7067 1 2 S 345 678 MATHEMATICAL TABLES TABLE II. (contd.) 399 1 2 3 4 5 6 7 8 9 123 456 789 51 52 53 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 7152 7235 7316 7396 123 122 122 122 345 345 345 345 678 677 667 667 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 122 345 567 58 67 58 59 7482 7559 7634 7709 7490 7566 7642 7716 7497 7574 7649 7723 7505 7582 7657 7731 7513 7589 7664 7738 7520 7597 7672 7745 7528 7604 7679 7752 7536 7612 7686 7760 7543 7619 7694 7767 7551 7627 7701 7774 122 122 112 112 345 345 344 344 567 567 567 567 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 112 344 566 61 62 63 64 7853 7924 7993 8062 7860 7931 8000 8069 7868 7938 8007 8075 7875 7945 8014 8082 7882 7952 8021 8089 7889 7959 8028 8096 7896 7966 8035 8102 7903 7973 8041 8109 7910 7980 8048 8116 7917 7987 8055 8122 112 112 112 112 344 334 334 334 566 566 & 5 6 556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 112 334 556 66 67 68 69 8195 8261 8325 8383 8202 8267 8331 8395 8209 8274 8338 8401 8215 8280 8344 8407 8222 8287 8351 8414 8228 8293 8357 8420 8235 8299 8363 8426 8241 8306 8370 8432 8248 8312 8376 8439 8254 8319 8382 8445 112 112 112 112 $ 3 4 334 334 234 556 556 456 456 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1 1 2 234 456 71 72 73 74 8513 8573 8633 8692 8519 8579 8639 8698 8525 8585 S645 8704 8531 8591 8651 8710 8537 8597 8657 8716 8543 8603 8663 8722 8549 8609 8669 8727 8555 8615 8675 8733 8561 8621 8681 8739 8567 8627 868G 8745 112 112 112 112 234 234 234 234 455 455 455 455 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 112 2 3 3 455 76 77 78 79 880S 8865 8921 8976 8814 8871 8927 8982 8820 8876 8932 8987 8825 8882 8938 8993 8831 8887 8943 899S 8837 8893 8949 9004 8842 8899 8954 9009 8848 8904 8960 9015 8854 8910 8965 9020 8859 8915 8971 9025 112 112 112 112 2 S 3 233 233 233 455 445 445 445 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 112 233 446 81 82 83 84 9085 9138 9191 9243 9090 9143 9196 9248 9096 9149 9201 9253 9101 9154 9206 9258 9106 9159 9212 9263 9113 9165 9217 9269 9117 9170 9222 9274 9122 9175 9227 9279 9128 9180 9232 9284 9133 9186 9238 9289 112 112 112 112 233 233 233 233 446 446 446 445 65 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 112 233 445 86 87 88 89 9345 9395 9445 9494 9350 9400 9450 9499 9355 9405 9455 9504 9360 9410 9460 9509 9365 9415 9465 9513 9370 9420 9469 9518 9375 9425 9474 9523 9330 9430 9479 9528 9385 9435 9484 9533 9390 9440 9489 9538 112 Oil Oil Oil 233 223 223 223 445 344 344 344 90 9542 9547 9552 9557 9562 9566 9671 9576 9581 9586 Oil 223 344 91 92 93 94 9590 9638 9585 9731 9595 9643 9689 9736 9600 9647 9694 9741 9605 9(352 9699 9745 9609 9657 9703 9750 9614 9661 9708 9754 9619 9666 9713 9759 9624 9671 9717 9763 9628 9675 9722 9768 9633 9680 9727 9773 Oil Oil Oil Oil 223 223 223 223 344 344 344 344 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 Oil 223 344 96 97 95 99 9823 9868 9912 995 9827 987-; 9917 9961 9832 9877 9921 99C5 9836 9881 9926 9969 9841 9886 9930 9'J74 9845 9890 9934 9978 9850 9894 9939 9983 9854 9899 9943 9987 9859 9903 9948 9991 9863 9908 9952 9996 Oil Oil Oil Oil 223 223 223 223 344 344 344 334 400 MATHEMATICAL TABLES TABLE III. ANTILOGARITHMS 1 2 3 4 5 6 7 8 9 123 456 789 00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 001 111 212 01 02 03 04 1023 1047 1072 10-J6 1026 1050 1074 1099 1028 1052 1076 1102 1030 1054 1079 1104 1033 1057 1081 1107 1035 1059 1084 1109 1038 1062 1086 1112 1040 1064 1089 1114 1042 1067 1091 1117 1045 1069 1094 1119 001 001 001 Oil 111 111 ill 112 222 222 222 223 05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 Oil 112 222 06 07 03 09 1143 1175 1202 1230 1151 1178 1205 1233 1153 1180 1208 1236 1156 1183 1211 1239 1159 1186 1213 1242 1161 1189 1216 1245 1164 1191 1219 1247 1167 1194 1222 1250 1169 1197 1225 1253 1172 1199 1227 1256 Oil Oil Oil Oil 112 112 112 112 222 222 223 223 10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 Oil 112 223 11 12 18 14 1288 1318 1349 1380 1291 1321 1352 1384 1294 1324 1355 1387 1297 1327 1358 1390 1300 1330 1361 1393 1303 1334 1365 1396 1306 1337 1368 1400 1309 1340 1371 1403 1312 1343 1374 1406 1315 1346 1377 1409 Oil Oil Oil Oil 122 122 122 122 223 223 233 233 15 1413 1416 1419 1422 1426 1429 1432 1435 1439 1442 Oil 122 233 16 17 18 19 1445 1479 1514 1549 1449 1483 1517 1552 1452 1486 1521 1556 1455 1489 1524 1560 1459 1493 1528 1563 1462 1496 1531 1567 1466 1500 1535 1670 1469 1503 1638 1674 1472 1607 1542 1578 1476 1510 1545 1581 1 1 Oil Oil Oil 122 122 122 122 233 233 233 333 20 1685 1589 1592 1596 1600 1603 1607 1611 1614 1618 1 1 1 2 2 333 21 22 23 24 1623 1660 1698 1738 1626 1663 1702 1742 1629 1667 1706 1746 1633 1671 1710 1750 1637 1675 1714 1754 1641 1679 1718 1758 1644 1683 1722 1762 1648 1687 1726 1766 1652 1690 1730 1770 1656 1694 1734 1774 1 1 1 1 Oil 1 1 222 222 222 222 333 333 334 334 25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 Oil 222 334 26 27 28 29 1820 1862 1905 1950 1824 1866 1910 1954 1828 1871 1914 1959 1832 1875 1919 1963 1837 1879 1923 1968 1841 1884 1928 1972 1845 1888 1932 1977 1849 1892 1936 1982 1854 1897 1941 1986 1858 1901 1945 1991 Oil Oil Oil Oil 223 223 223 223 334 334 344 3 4 i 30 1995 2000 2004 2009 2014 2018 2023 9028 2032 2031 Oil 223 344 31 82 33 34 2042 2089 2138 2188 2046 2094 2143 2193 2051 2099 2148 2198 2058 2104 2163 2203 2061 2109 2158 2208 2065 2113 2163 2213 2070 2118 2168 2218 2076 2123 2173 2223 2080 2128 2178 2228 2084 2133 2183 2234 Oil Oil Oil 112 223 223 223 233 344 344 344 446 35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 112 233 446 36 37 33 39 2291 2344 2399 2455 2296 2360 2404 2460 2301 2355 2410 2466 2307 2360 2415 2472 2312 2366 2421 2477 2317 2371 2427 2483 2323 2377 2432 2489 2328 2382 2438 2495 2333 2388 2443 2500 2339 2393 2449 2506 112 112 112 112 233 233 233 233 445 446 446 455 40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 112 234 455 41 42 43 44 2570 2630 2692 2754 2576 2636 2698 2761 2582 2642 2704 2767 2588 2649 2710 2773 2594 2655 2716 2780 2600 2661 2723 2786 2606 2667 2729 2793 2612 2673 2735 2799 2618 2679 2742 2805 2624 2685 2748 2812 112 112 112 112 234 234 334 334 455 456 456 466 45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 112 334 656 46 47 48 49 2884 2951 3020 3090 2891 2958 8027 8097 2897 2966 3034 3105 2904 2972 3041 3112 2911 2979 3048 3119 2917 2985 3055 S126 2924 2992 3062 3133 2931 2999 3069 3141 2938 3006 3076 3148 2944 3013 3083 3155 112 112 112 112 334 334 344 344 566 666 666 666 MATHEMATICAL TABLES TABLE III. (contd). 401 1 2 3 4 5 6 7 8 9 123 456 789 50 3162 3170 3177 318 3192 3199 3206 3214 3221 322 112 344 5 6 t 5 5 53 54 3236 3311 3388 3467 3243 3319 3396 3475 3251 3327 3404 3483 3258 3334 3412 3491 3266 3342 3420 3499 3273 3350 3428 3508 3281 3357 3436 3516 3289 3365 3443 3524 3296 3373 3451 3532 3304 338 345 3540 122 133 122 122 345 346 345 345 567 667 667 667 55 3548 3556 3565 35J3 3581 3589 3697 3606 3614 3622 122 346 677 56 57 58 59 3G31 3715 3802 3890 3639 3724 3811 3899 3648 3733 3819 3908 3656 3741 3828 3917 3664 8750 3837 3926 3673 3758 3846 3936 3681 3767 3855 3945 3690 3776 3864 3954 3698 3784 3873 3963 3707 3793 3882 3972 123 123 123 123 346 346 445 455 678 678 678 678 60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1 2 3 456 *,, 61 62 63 64 4074 4169 4266 4365 4083 4178 4276 4375 4093 4188 4285 4385 4102 4198 4295 4395 4111 4207 4305 4406 4121 4217 4315 4416 4130 4227 4325 4426 4140 4236 4335 4436 4150 4246 4345 4446 4159 4256 4355 4457 123 123 123 123 466 466 466 456 789 789 789 789 65 4467 4477 4487 4498 4508 4519 4629 4539 4550 4560 123 456 789 66 67 68 69 4571 4677 4786 4898 4581 4688 4797 4909 4592 4699 4808 4920 4603 4710 4819 4932 4613 4721 4831 4943 4624 4732 4842 4955 4634 4742 4853 4966 4645 4753 4864 4977 4656 4764 4875 4989 4667 4775 4887 5000 1 2 3 123 123 123 456 457 467 567 7 9 10 8 9 10 8 9 10 8 9 10 70 6012 6023 5035 5047 6058 6070 6082 6093 6105 6117 124 567 8 9 11 71 72 73 74 5129 6248 6370 6495 5140 6260 6383 6608 5152 5272 6395 5521 6164 528 i 5408 5534 6176 6297 6420 6546 5188 5309 5433 6559 5200 5321 5445 5572 5212 6333 6458 6685 5224 5346 6470 6598 5236 5358 6483 6610 124 124 134 134 567 567 568 568 8 10 11 9 10 11 9 10 11 9 10 12 75 5623 5636 5649 6662 6675 5689 5702 6715 5728 5741 134 578 9 10 12 76 77 78 79 6754 5888 6026 6166 5768 6902 6039 6180 6781 5916 6053 6194 6794 5929 6067 6209 6808 5943 6081 6223 5821 5957 6095 6237 6834 6970 6109 6252 6848 6984 6124 6266 6861 5998 6138 6281 5875 6012 6152 6295 134 134 134 134 578 678 678 679 9 11 12 10 11 12 10 11 13 10 11 13 80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 134 679 10 12 13 81 82 83 64 6457 6607 6761 6918 6471 6622 6776 6934 6486 6637 6792 6950 6501 6653 6808 6966 1 6516 6668 6823 6982 6631 6683 6839 6998 6546 6699 6855 7015 6561 6714 6871 7031 6577 6730 6887 7047 6592 6745 6902 7063 235 235 236 235 689 689 689 6 8 10 11 12 14 11 12 14 11 13 14 11 13 15 85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7228 336 7 8 10 12 13 15 86 67 83 69 7244 7413 7586 7762 7261 7430 7603 7780 7278 7447 7621 7798 7295 7464 7638 7816 7311 7482 7656 7834 7328 7499 7674 7852 7345 7516 7691 7870 7362 7534 7709 7889 7379 7551 7727 7907 7396 7568 7745 7925 235 236 345 245 7 8 10 7 9 10 7 9 11 7 9 11 12 13 15 12 14 16 12 14 16 13 14 16 80 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 7 9 11 13 15 17 91 92 93 S4 8128 8318 b5U 8710 8147 8337 8531 8730 : 8166 8356 ' 8551 8750 8185 8375 8570 8770 204 395 590 790 8222 8414 ! 8610 8810 8241 8433 8630 8831 8260 8453 8650 8851 8279 8472 8670 8872 8299 8492 8690 8892 246 246 246 246 8 9 11 8 10 12 8 10 12 8 10 12 13 15 17 14 15 17 14 16 18 14 16 18 95 8913 8933 8954 8974 995 9016 9036 9057 9078 9099 246 8 10 12 15 17 1 86 97 98 99 9120 9333 9550 '772 9141 9354 9572 9795 9162 9376 9594 9817 9183 9397 9616 9840 204 419 9638 9863 9226 9441 ! 9661 9S86 9247 9462 9683 9906 9268 9484 9705 9931 9290 9506 9727 9954 9311 9528 9750 9977 246 347 247 267 8 11 13 9 11 13 9 11 13 9 11 14 16 17 19 15 17 20 16 18 20 16 18 20 D D 402 MATHEMATICAL TABLES TABLE IV. NAPIERIAN, NATURAL, OR HYPERBOLIC LOGARITHMS Number. 1 1 2 3 4 5 6 7 8 9 01 3-6974 7927 8797 9598 0339 1029 1674 2280 2852 3393 0-2 2-3906 4393 4859 5303 5729 6i37 6529 6907 7270 7621 03 7960 8288 8606 8913 9212 9502 97830057 0324 6584 0-4 1-0837 1084 1325 1560 1790 2015 2235 2450 2660 2866 05 3068 3267 3461 3651 3838 4022 4202 4379 4553 4724 06 4892 5057 5220 538o 5537 5692 5845 5995 6i43 6289 Mean Differences. 0-7 6433 6575 67 J 5 6853 6989 7 I2 3 7256 7386 7515 7643 0-8 7769 7893 8015 8137 8256 8375 8492 8607 8722 8835 09 8946 9057 9166 9274 938i 9487 9592 9695 9798 9899 123 456 789 1-0 o-oooo OIOO 0198 0296 0392 0488 0583 0677 0770 0862 1-1 0953 1044 "33 1222 1310 1398 1484 157 1655 1740 9 17 26 35 44 52 61 70 78 1-2 1823 1906 1989 2070 2151 2231 2311 2390 2469 2546 8 1624 32 40 48 56 64 72 1-3 2624 2700 2776 2852 2927 3001 375 3148 3221 3293 7 15 22 30 37 45 52 59 67 1-4 3365 3436 357 3577 3646 37 l6 3784 3853 3920 3988 71421 28 35 41 48 55 62 1-5 455 4121 4187 4253 43i8 4383 4447 45" 4574 4637 613 19 26 32 39 45 52 58 1-6 4700 4762 4824 4886 4947 5008 5068 5128 5188 5247 6 12 18 24 30 36 42 48 55 1-7 5306 5365 5423 548i 5539 5596 5653 57io 5766 5822 6 ii 17 24 29 34 40 46 52 1-8 5878 5933 5988 6043 6098 6152 6206 6259 6313 6366 5 " 16 22 27 32 38 43 49 19 6419 6471 6523 6575 6627 6678 6729 6780 6831 6881 5 1015 20 20 31 36 41 46 2-0 693i 6981 73i 7080 7129 7178 7227 7275 7324 7372 5 ioi5 2O 24 29 34 39 44 2-1 7419 7467 75M 756i 7608 7655 7701 7747 7793 7839 5 9 14 19 23 28 33 37 42 22 7885 7930 7975 8020 8065 8109 8i54 8198 8242 8286 4 913 l8 22 27 3i 36 40 23 8329 8372 8416 8459 8502 8544 8587 8629 8671 8713 4 913 17 21 20 3 34 38 2-4 8755 8796 8838 8879 8920 8961 9002 9042 9083 9123 4 8 12 16 20 24 29 33 37 2-5 9163 9203 9243 9282 9322 9361 9400 9439 9478 9517 4 8 12 16 20 24 2731 35 26 9555 9594 9632 9670 9708 9746 9783 9821 9858 9895 4 8 ii 15 1923 26 30 34 2-7 9933 9969 0006 0043 0080 0116 0152 6188 0225 0260 4711 15 l8 22 26 29 33 2-8 1-0296 0332 0367 0403 0438 0473 0508 0543 0578 0613 4 711 14 18 21 25 28 32 2-9 0647 0682 0716 0750 0784 0818 0852 0886 0919 0953 3 7 I0 14 17 2O 242731 30 0986 1019 1053 1086 1119 1151 1184 1217 1249 1282 3 7 10 13 16 20 23 26 30 31 1314 1346 1378 1410 1442 1474 1506 1537 1569 1600 3 6 10 13 16 19 22 25 29 3-2 1632 1663 1694 1725 1756 1787 1817 1848 1878 1909 369 12 15 18 21 25 28 33 1939 1969 2OOO 2030 2060 2090 2119 2149 2179 2208 369 12 15 18 21 24 27 3-4 2238 2267 2296 2326 2355 2384 2413 2442 2470 2499 369 12 14 17 2O 23 20 35 2528 2556 2585 2613 2641 2669 2698 2726 2754 2782 368 II 14 17 2O 22 25 36 2809 2837 2865 2892 2920 2947 2975 3002 3029 3056 3 5 8 ii 14 16 19 22 25 3-7 3083 3110 3137 3164 3191 3218 3244 3271 3297 3324 3 5 8 ii 13 16 19 21 24 38 335 337 6 343 3429 3455 348i 3507 3533 3558 3584 3 5 8 10 13 16 18 21 23 3;9 3610 3635 3661 3686 37 12 3737 3762 3788 3813 3838 3 5 8 1013 15 1 8 20 23 40 3863 3888 3913 3938 3962 3987 4012 4036 4061 4085 257 IO 12 15 17 2O 2 2 4-1 4110 4134 4159 4183 4207 4231 4255 4279 4303 4327 257 10 12 14 17 1922 4-2 435i 4375 4398 4422 4446 4469 4493 45i6 454 4563 257 9 12 14 16 19 21 4-3 4586 4609 4633 4656 4679 4702 47 2 5 4748 4770 4793 257 9 II 14 16 18 21 44 4816 4839 4861 4884 4907 4929 495i 4974 4996 5019 247 9 II 13 16 18 20 45 5041 5063 5085 5107 5129 5151 5173 5195 5217 5239 247 9 II 13 15 18 20 46 5261 5282 53<>4 5326 5347 5369 5390 5412 5433 5454 246 9 II 13 15 17 19 4-7 5476 5497 55i8 5539 556o 558i 5602 5623 5644 5665 2 4, 6 8 II 13 15 17 19 48 49 5686 5892 5707 5913 5728 5933 5748 5953 57 6 9 5974 5790 5994 5810 6014 5831 6034 5851 6054 5872 6074 246 246 8 10 12 8 IO 12 14 16 19 14 16 18 50 6094 6114 6i34 6i54 6174 6194 6214 6233 6253 6273 246 8 IO 12 14 16 18 MATHEMATICAL TABLES TABLE IV (contd.) 403 6 8 9 Mean Differences. 123456 789 51 52 53 54 55 56 57 5-8 59 60 6-1 62 63 64 65 66 6-7 68 69 7-0 7-1 72 73 7-4 75 7-6 7-7 7-8 79 80 8-1 8-2 8-3 84 85 8-6 8-7 8-8 8-9 90 91 92 93 94 95 96 97 98 99 10 1-6292 6312 6332 6351 6371 6390 64 6487 6506 6525 6544 6563 6582 66oi 6677:6696.671516734167526771 6840 6883 690116919 6938 6956 6975|6993|7Oi i 7029 7048! 7066! 7084 7102 7120 7228,7246 7263 7281 775017767 7596 7613 7630J647 7 66 4 7 682 7440 091642964486467 2 662066396658 2 68o8|6827,6846| 2 6975'6993'7OII 7O29 2 71387156717471927210 2 7299 7317 7334 7352,7370 7387 7457 7475 7492 7509 7527 7544 7561 7783 7800,7817 7834 7851 769977167733 78687884,7901 7968 7984 8001 8017 8034 8o5O ! 8o67 8o83 ! 8o99 811618132 8148 8165 8181 8197,8213 8229 2 262'8278,8294 ! 83io 8326 8342 8358 8374 8390 8563^579 8594 ! 86io 862518641 8656'8672|8687'8703 2 7i8,87338749 8764^779 8795 88io8825j8840 ( 8856 2 887l'8886 8goi 89l6|893I 8946 8g6l 8976 8991 9006 2 2 041204250438045104640. 477 0490,050310516 0528 ' 1580 0592 0605 0618 0631 o66g'o68 1 ;o6g4j07O7'o7igJQ732 0744,075 7:0769 078 0794 0807 0819 0832 0844 0857 0869,0882 p8g4 0906 0919 O93i|og43og56 i og68 0980 ogg2 1005 1017 1029 1114 1 126)1138 1151 12471259,1270 1041 1163 1282 1401 1518 1529 154 1633 164511656 1668 167911691 I748I759I77 2192 io54jio66 107811090 1102 1175 1187 H99'i2ii 1294 1300)1318 1330 1342 1436144814591471 1861 1972 2083 209412105 2ii6J2i27 2138 1552 1668 1782 1223 1235 187211883 i894 l i905 1917 1928 1939 1950 1961 I9&3JI994 2bo6|2Oi7 2028 2039 2050 2061 2072 49215921702181 46 2257 226&2279J2289 tf A s*lf\Z T O *7 C 2203 2214 2225!2235 ~J~~ "O" -J--|-OJ-"pJ i tO *OJt -^O^ 2407 2418 2428 2439 2450 2460 2471 2481 2492 2502 2513 2523 2534 2544 2555 2565 2576 2586 2597 2607 2618 2628 2638 2649 2659 2670 2680 2690 2701 2711 2721 2732 2742 2752J2762 2773 2783 2793 2803 2814 2824 2834 2844 2854 2865 2875 29251293512946 2956 2966 2976 2986 2-3020 1702 1793^1804 1816 1827 1838 1849 1483 1494 1506 171317251736 2885 2895 2905 2915 2996 3006 3016 4 6 4 6 4 6 9169 9184 gigg 9213 9228 9243^257^9272 9286 9301 2 9315 933 9344l9359!9373 9387J94 02 94i6|943i!9445 I 9459 9473 9488,9502,95i6|g530 9545 9559 9573 9587 i g6oi|g6i5 9629 9643 ^657^671 g685 g6gg[g7i3 9727 i 974IJ9755 9769 9782J9796 g8io g824 ^838 g85i 9865 i 9879 g8g2g9o6;9920 | g9339g47gg6ijg974|gg88 oooi i 2-0015:002800420055,006900820096010901220136 0149^162 0x76 Ol8g O2O2 O2l6 0229^242^255 O268 0281 0295 0308,0321 0334 347 0360 0373 0386 03gg 3 4 8 IO 12 8 IO 12 g ii 7 9 ii 4 5| 7 9 ii 2 4 5:7 9 4 57 9 10 3 57 9 10 3 5 7 8 10 3 5 7 8 10 6 8 9 5 7 5 6 7 467 14 16 18 13 15 17 13 15 17 13 15 17 13 14 16 12 14 16 12 14 16 12 I 4 15 12 13 15 12 13 15 II 13 15 II 13 14 II 13 14 II 12 14 II 12 14 II 12 14 IO 12 13 IO 12 13 IO 12 13 10 ii 13 10 ii 13 IO II 12 IO II 12 9 II 12 9 II 12 9 II 12 9 IO 12 9 10 II 9 10 ii 9 10 ii 9 10 ii 9 10 ii 8 10 ii 8 9 ii 8911 9 10 9 10 9 10 9 10 9 10 9 10 9 10 9 10 9 10 8 10 404 MATHEMATICAL TABLES TABLE V. NATURAL SINES. E 9 Q 0' 00 6' 0-1 12' 0-2 18' 0-3 24' 0-4 30' 0-5 36' 06 42' 0-7 48' 0-8 54' 09 Mean Differences. 1' 2' 3^4' 5' oooo 0017 0035 0052 0070 0087 0105 OI22 0140 0157 3 6 9 12 15 1 0175 0192 0209 0227 0244 0262 0279 0297 0314 0332 3 6 9 12 15 2 0349 0366 0384 0401 0419 0436 454 47 I 0488 0506 3 6 9 12 15 3 0523 054 1 0558 0576 0593 0610 0628 0645 0663 0680 3 9 12 15 4 0698 0715 0732 0750 0767 0785 0802 0819 0837 0854 3 6 9 12 14 5 0872 0889 0906 0924 0941 0958 0976 0993 IOII 1028 3 6 9 12 14 6 1045 1063 1080 1097 1115 1132 1149 1167 1184 1201 3 6 9 12 14 7 1219 1236 1253 1271 1288 1305 1323 1340 1357 !374 3 6 9 12 14 8 1392 1409 1426 1444 1461 1478 H95 1513 1530 1547 3 6 9 12 14 9 1564 1582 1599 1616 1633 1650 1668 I68 5 1702 1719 3 6 9 12 14 10 1736 1754 1771 1788 1805 1822 1840 1857 1874 1891 3 6 9 ii 14 11 1908 1925 1942 1959 1977 1994 2OII 2O28 2045 2062 3 6 9 ii 14 12 2079 2096 2113 2130 2147 2164 2181 2198 2215 2233 3 6 9 ii 14 13 2250 2267 2284 2300 2317 2334 2351 2368 2385 2402 3 6 8 ii 14 14 2419 2436 2453 2470 2487 2504 2521 2538 2554 2571 3 6 8 ii 14 15 2588 2605 2622 2639 2656 2672 2689 2706 2723 2740 3 6 8 ii 14 16 2756 2773 2790 2807 2823 2840 2857 2874 2890 2907 3 6 8 ii 14 17 2924 2940 2957 2974 2990 3007 302 4 3040 3057 3074 3 6 8 ii 14 18 3090 3107 3123 3MO 3156 3i73 3190 3206 3223 3239 3 6 8 ii 14 19 3256 3272 3289 3305 3322 3338 3355 337 1 3387 3404 3 5 8 ii 14 20 3420 3437 3453 34 6 9 3486 3502 35i8 3535 355i 3567 3 5 8 ii 14 21 3584 3600 3616 3633 3649 3665 3681 3697 37M 3730 3 5 8 ii 14 22 3746 3762 3778 3795 3811 3827 3843 3859 3875 3891 3 5 8 ii 14 23 3907 3923 3939 3955 397i 3987 4003 4019 4035 45i 3 5 8 ii 14 24 4067 4083 4099 4"5 4131 4*47 4163 4179 4i95 4210 3 5 8 ii 13 25 4226 4242 4258 4274 4289 4305 4321 4337 4352 4368 3 5 8 ii 13 26 4384 4399 4415 4431 4446 4462 4478 4493 4509 4524 3 5 8 10 13 27 454 4555 4571 4586 4602 4617 4633 4648 4664 4679 3 5 8 10 13 28 4 6 95 4710 4726 4741 4756 4772 4787 4802 4818 4833 3 5 8 10 13 29 4848 4863 4879 4894 4909 4924 4939 4955 4970 4985 3 5 8 10 13 30 5000 5015 5030 5045 5060 5075 5090 5105 5120 5135 3 5 8 10 ij 31 5150 5165 5180 5195 5210 5225 5240 5255 5270 5284 2 5 7 10 12 32 5299 53H 5329 5344 5358 5373 5388 5402 54i7 5432 2 5 7 10 12 33 5446 546i 5476 5490 5505 5519 5534 5548 5563 5577 2 5 7 10 12 34 5592 5606 5621 5635 5650 5664 5678 5693 5707 5721 2 5 7 10 12 35 5736 5750 5764 5779 5793 5807 5821 5835 5850 5864 2 5 7 9 12 36 5878 5892 5906 5920 5934 5948 5962 5976 5990 6004 2 5 7 9 12 37 6018 6032 6046 6060 6074 6088 6101 6115 6129 6i43 2 5 7 9 12 38 6157 6170 6184 6198 6211 6225 6239 6252 6266 6280 2 5 7 9 ii 39 6293 6307 6320 6334 6347 6361 6 374 6388 6401 6414 247 9 ii 40 6428 6441 6455 6468 6481 6494 6508 6521 6534 6547 247 9 ii 41 6561 6574 6587 6600 6613 6626 6639 6652 6665 6678 2 4 7 9 ii 42 6691 6704 6717 6730 6743 6756 6769 6782 6794 6807 2 4 6 9 ii 43 6820 6833 6845 6858 6871 6884 6896 6909 6921 6934 246 8 ii 44 6947 6959 6972 6984 6997 7009 7022 734 7046 759 246 8 10 45 -7071 7083 7096 7108 7120 7133 7M5 7157 7169 7181 246 8 10 MATHEMATICAL TABLES 405 TABLE V. (contd.) g 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. 1 00 0-1 02 03 04 0-5 06 0-7 0-8 09 1' 2 3' 4' 5' 45 7071 7083 7096 7108 7120 7133 7M5 7i57 7169 7181 246 8 10 46 7193 7206 7218 7230 7242 7254 7266 7278 7290 7302 246 8 10 47 73M 7325 7337 7349 736i 7373 7385 7396 7408 7420 246 8 10 48 743i 7443 7455 7466 7478 7490 75oi 7513 7524 7536 2 4 6 8 10 49 7547 7559 7570 758i 7593 7604 7615 7627 7638 7649 24689 50 7660 7672 7683 7694 7705 7716 7727 7738 7749 7760 24679 51 7771 7782 7793 7804 7815 7826 7837 7848 7859 7869 2 4579 52 7880 7891 7902 7912 7923 7934 7944 7955 7965 7976 24579 53 7986 7997 8007 8018 8028 8039 8049 8059 8070 8080 23579 54 8090 8100 8111 8121 8131 8141 8151 8161 8171 8181 23578 55 8192 8202 8211 8221 8231 8241 8251 8261 8271 8281 23578 56 8290 8300 8310 8320 8329 8339 8348 8358 8368 8377 23568 57 8387 8396 8406 8415 8425 8434 8443 8453 8462 8471 23568 58 8480 8490 8499 8508 8517 8526 8536 8545 8554 8563 23568 59 8572 8581 8590 8599 8607 8616 8625 8634 8643 8652 13467 60 8660 8669 8678 8686 8695 8704 8712 8721 8729 8738 13467 61 8746 8755 8763 8771 8780 8788 8796 8805 8813 8821 13467 62 8829 8838 8846 8854 8862 8870 8878 8886 8804 8902 13457 63 8910 8918 8926 8934 8942 8949 8957 8965 8973 8980 13456 64 8988 8996 9003 9011 9018 9026 9033 9041 9048 9056 13456 65 9063 9070 9078 9085 9092 9100 9107 9114 9121 9128 12456 66 9135 9M3 9150 9157 9164 9171 9178 9184 9191 9198 12356 67 9205 9212 9219 9225 9232 9239 9245 9252 9259 9265 12346 68 9272 9278 9285 9291 9298 9304 93ii 9317 9323 9330 12345 69 9336 9342 934 8 9354 936i 9367 9373 9379 9385 9391 12345 70 9397 943 9409 9415 9421 9426 9432 9438 9444 9449 12345 71 9455 9461 9466 9472 9478 9483 9489 9494 9500 9505 12345 72 95ii 95i6 9521 9527 9532 9537 9542 9548 9553 9558 12334 73 9563 9568 9573 9578 9583 9588 9593 9598 9603 9608 12234 74 9613 9617 9622 9627 9632 9636 9641 9646 9650 9655 12234 75 9659 9664 9668 9673 9677 9681 9686 9690 9694 9699 11234 76 973 9707 9711 97*5 9720 9724 9728 9732 9736 9740 11233 77 9744 9748 9751 9755 9759 97 6 3 9767 9770 9774 9778 11233 78 9781 9785 9789 9792 9796 9799 9803 9806 9810 9813 i 223 79 9816 9820 9823 9826 9829 9833 9836 9839 9842 9845 i 223 80 9848 9851 9854 9857 9860 9863 9866 9869 9871 9874 122 81 9877 9880 9882 9885 9888 9890 9893 9895 9898 9900 122 82 9903 9905 9907 9910 9912 9914 9917 9919 9921 9923 122 83 992.5 9928 9930 9932 9934 9936 9938 9940 9942 9943 112 84 9945 9947 9949 995i 9952 9954 9956 9957 9959 9960 I I I 2 85 9962 9963 9965 9966 9968 9969 997 1 9972 9973 9974 I I I 86 9976 9977 9978 9979 9980 9981 9982 9983 9984 9985 I I I 87 9986 9987 9988 9989 9990 9990 9991 9992 9993 9993 O O O I I 88 9994 9995 9995 9996 9996 9997 9997 9997 9998 9998 O O O O 89 9998 9999 9999 9999 9999 I'OOO I'OOO i-ooo i-ooo i-ooo O O O O O 90 i-ooo 406 MATHEMATICAL TABLES TABLE VI. NATURAL COSINES 1 B, 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mezn Differenci's. 3 00 01 02 03 04 0-5 06 0-7 0-8 09 1' 2' 3' 4' 5' I'OOO ooo ooo ooo I'OOO I'OOO 9999 9999 9999 9999 o o o o o 1 9998 9998 9998 9997 9997 9997 9996 9996 9995 9995 O O O O 2 9994 9993 9993 9992 9991 9990 9990 9989 9988 9987 I I 3 9986 9985 9984 9983 9982 9981 9980 9979 9978 9977 O O II 4 9976 9974 9973 9972 997 1 9969 9968 9966 9965 9963 O O II 5 9962 9960 9959 9957 9956 9954 9952 9951 9949 9947 01 12 6 9945 9943 9942 994 9938 9936 9934 9932 9930 9928 01 12 7 9925 9923 9921 9919 9917 9914 9912 9910 9907 9905 O I 22 8 9903 9900 9898 9895 9893 9890 9888 9885 9882 9880 01 22 9 9877 9874 9871 9869 9866 9863 9860 9857 9854 9851 O 22 10 9848 9845 9842 9839 9836 9833 9829 9826 9823 9820 I 223 11 9816 9813 9810 9806 9803 9799 9796 9792 9789 9785 I 223 12 9781 9778 9774 977 9767 9763 9759 9755 9751 974 s I 233 13 9744 974 9736 9732 9728 9724 9720 97*5 9711 9707 I 233 14 9703 9699 9694 9690 9686 9681 9677 9673 9668 9664 1 234 15 9659 9655 9650 9646 9641 9636 9632 9627 9622 9617 12234 16 9613 9608 9603 9598 9593 9588 9583 9578 9573 9568 12234 17 9563 9558 9553 9548 9542 9537 9532 9527 9521 95i6 12334 18 95" 955 95oo 9494 9489 94 8 3 9478 9472 9466 9461 12345 19 9455 9449 9444 9438 9432 9426 9421 9415 9409 943 12345 20 9397 9391 9385 9379 9373 9367 9361 9354 9348 9342 12345 21 9336 9330 93^3 9317 93" 9304 9298 9291 9285 9278 12345 22 9272 9265 9259 9252 9245 9239 9232 9225 9219 9212 12346 23 9205 9198 9191 9184 9178 9171 9164 9157 9150 9M3 12356 24 9135 9128 9121 9114 9107 9100 9092 9085 9078 9070 2456 25 9063 9056 9048 9041 9033 9026 9018 9011 9003 8996 3456 26 8988 8980 8973 8965 8957 8949 8942 8934 8926 8918 3456 27 8910 8902 8894 8886 8878 8870 8862 8854 9846 8838 3457 28 8829 8821 8813 8805 8796 8788 8780 8771 8763 8755 3467 29 8746 8738 8729 8721 8712 8704 8695 8686 8678 8669 1 3 4 6 7 30 8660 8652 8643 8634 8625 8616 8607 8599 8590 8581 13467 31 8572 8563 8554 8545 8536 8526 8517 8508 8499 8490 23568 32 8480 8471 8462 8453 8443 8434 8425 8415 8406 8396 23568 33 8387 8377 8368 8358 8348 8339 8329 8320 8310 8300 23568 34 8290 8281 8271 8261 8251 8241 8231 8221 8211 8202 23578 35 8192 8181 8171 8161 8151 8141 8131 8121 8111 8100 23578 36 8090 8080 8070 8059 8049 8039 8028 8018 8007 7997 23579 37 7986 7976 7965 7955 7944 7934 7923 7912 7902 7891 24579 38 7880 7869 7859 7848 7837 7826 7815 7804 7793 7782 24579 39 7771 7760 7749 7738 7727 7716 7705 7694 7683 7672 24679 40 7660 7649 7638 7627 7615 7604 7593 758i 7570 7559 24689 41 7547 7536 7524 7513 75i 7490 7478 7466 7455 7443 2 4 6 8 10 42 7431 7420 7408 7396 7385 7373 736i 7349 7337 7325 246 8 10 43 73H 7302 7290 7278 7266 7254 7242 7230 7218 7206 246 8 10 44 .7193 7181 7169 7157 7145 7133 7120 7108 7096 7083 246 8 10 45 .7071 7059 7046 7034 7022 7009 6997 6984 6972 6959 2 4 6 8 10 1 MATHEMATICAL TABLES TABLE VI (contd.) 407 1 Q 0' 00 6' 01 12' 02 18' 03 24' 0-4 30' 0-5 36' 06 42' 0-7 48' 08 54' 0-9 Mean Differences. 1' 2' 3' 4' 5' 45 7071 759 7046 7034 7022 7009 6997 6984 6972 6959 246 8 10 46 6947 6934 6921 6909 6896 6884 6871 6858 6845 6833 246 8 ii 47 6820 6807 6794 6782 6769 6756 6743 6730 6717 6704 246 9 ii 48 6691 6678 6665 6652 6639 6626 6613 6600 6587 6574 2 4 7 9 ii 49 6561 6547 6534 6521 6508 6494 6481 6468 6455 6441 247 9 II 50 6428 6414 6401 6388 6374 6361 6347 6334 6320 6307 247 9 II 51 6293 6280 6266 6252 6239 6225 6211 6198 6184 6170 257 9 II 52 6157 6i43 6129 6115 6101 6088 6074 6060 6046 6032 257 9 12 53 6018 6004 5990 5976 5962 5948 5934 5920 5906 5892 2 5 7 9 12 54 5878 5864 5850 5835 5821 5807 5793 5779 5764 5750 2 5 7 9 12 55 5736 5721 577 5693 5678 5664 5650 5635 5621 5606 2 5 7 10 12 56 5592 5577 5563 5548 5534 55i9 5505 5490 5476 5461 2 5 7 10 12 57 5446 5432 5417 542 5388 5373 5358 5344 5329 53i4 2 5 7 10 12 58 5299 5284 527 5255 524 5225 5210 5195 5180 5165 2 5 7 10 12 59 515 5135 5120 5105 5090 575 5060 5045 5030 5015 3 5 8 10 13 60 5000 4985 4970 4955 4939 4924 4909 4894 4879 4863 3 5 8 10 13 61 4848 4833 4818 4802 4787 4772 4756 4741 4726 4710 3 5 8 10 13 62 4695 4679 4664 4648 4633 4617 4602 4586 4571 4555 3 5 8 10 13 63 454 4524 4509 4493 447 s 4462 4446 4431 4415 4399 3 5 8 10 13 64 4384 4368 4352 4337 4321 4305 4289 4274 4258 4242 3 5 8 ii 13 65 4226 4210 4195 4179 4163 4M7 4131 4115 4099 4083 3 5 8 ii 13 66 4067 4051 435 4019 4003 3987 3971 3955 ' 3939 3923 3 5 8 ii 14 67 3907 3891 3875 3 8 59 3843 3827 3811 3795 3778 3762 3 5 3 ii 14 68 3746 3730 37M 3697 3681 3665 3649 3633 , 3616 3600 3 5 8 ii 14 69 3584 3567 3551 3535 35i8 3502 3486 3469 3453 3437 3 5 3 ii 14 70 3420 344 3387 3371 3355 3338 3322 3305 3289 3272 3 5 8 ii 14 71 3256 3239 3223 3206 3190 3173 3156 3140 3123 3107 3 6 8 li 14 72 390 3074 3057 3040 3024 3007 2990 2974 2957 2940 3 6 8 ii 14 73 2924 2907 2890 2874 2857 2840 2823 2807 2790 2773 3 6 8 ii 14 74 2756 2740 2723 2706 2689 2672 2656 2639 2622 2605 3 6 8 ii 14 75 2588 257 1 2554 2538 2521 2504 2487 2470 2453 2436 3 6 8 ii 14 76 2419 2402 2385 2368 2351 2334 2317 2300 2284 2267 3 6 8 ii 14 77 2250 2233 2215 2198 2181 2164 2147 2130 2113 2096 3 6 9 ii 14 78 2079 2062 2045 2028 201 1 1994 1977 1959 1942 1925 3 6 9 ii 14 79 1908 1891 1874 1857 1840 1822 1805 1788 1771 1754 3 6 9 ii 14 80 1736 1719 1702 1685 1668 1650 1633 1616 1599 1582 3 6 9 12 14 81 1564 1547 1530 1513 1495 1478 1461 1444 1426 1409 3 6 9 12 14 82 1392 1374 1357 134 1323 1305 1288 1271 1253 1236 3 6 9 12 14 83 1219 1201 1184 1167 1149 1132 i"5 1097 1080 1063 3 6 9 12 14 84 1045 1028 IOII 0993 0976 0958 0941 0924 0906 0889 3 6 9 12 14 85 0872 0854 0837 0819 O8O2 0785 0767 0750 0732 0715 3 6 9 12 14 86 0698 0680 0663 0645 0628 0610 0593 0576 0558 0541 3 6 9 12 15 87 0523 0506 0488 0471 0454 0436 0419 0401 0384 0366 3 6 9 12 15 88 0349 0332 0314 0297 0279 0262 0244 0227 0209 0192 3 6 9 12 15 89 0175 0157 0140 OI22 OIO5 0087 0070 0052 0035 0017 3 6 9 12 15 90 oooo 4S MATHEMATICAL TABLES TABLE VII. NATURAL TANGENTS. & ID c 0' 00 6' 01 12' 02 18' 03 24' 0-4 30' 05 36' 0-6 r 42' 0-7 48' 08 54' 09 Mean Differences. 1' 2' 3' 4' 5' oooo 0017 0035 0052 0070 0087 0105 0122 0140 oi57 3 6 9 12 15 1 0175 0192 0209 0227 0244 0262 0279 0297 0314 0332 3 6 9 12 15 2 0349 0367 0384 0402 0419 437 454 0472 0489 0507 3 6 9 12 15 3 0524 0542 0559 0577 594 0612 0629 0647 0664 0682 3 6 9 12 15 4 0699 0717 0734 0752 0769 0787 0805 0822 0840 0857 3 6 9 12 15 5 0875 0892 0910 0928 0945 0963 0981 0998 1016 1033 3 6 9 12 15 6 1051 1069 1086 1104 1122 H39 "57 "75 1192 I2IO 3 6 9 12 15 7 1228 1246 1263 1281 1299 1317 J 334 *352 137 1388 3 6 9 12 15 8 1405 1423 1441 1459 H77 1495 1512 1530 1548 1566 3 6 9 12 15 9 1584 1602 1620 16^8 1655 1673 1691 1709 1727 1745 3 6 9 12 15 10 1763 1781 1799 1817 1835 1853 1871 1890 1908 1926 3 6 9 12 15 11 1944 1962 1980 1998 2016 2035 2053 2071 2089 2IO7 3 6 9 12 15 12 2126 2144 2162 2180 2199 2217 2235 2254 2272 229O 3 6 9 12 15 13 2309 2327 2345 2364 2382 2401 2419 2438 2456 2475 3 9 12 15 14 2493 2512 2530 2549 2568 2586 2605 2623 2642 2661 3 6 9 12 16 15 2679 2698 2717 2736 2754 2 773 2792 2811 2830 2849 3 6 9 13 16 16 2867 2886 2905 2924 2943 2962 2981 3000 3019 3038 3 6 9 13 16 17 3057 3076 3096 3H5 3134 3153 3172 3I9T 3211 3230 3 6 10 13 16 18 3249 3269 3288 3307 3327 3346 3365 3385 344 3424 3 6 10 13 16 19 3443 3463 3482 3502 3522 3541 356i 358i 3600 3620 3 7 10 13 16 20 3640 3659 3679 3699 3719 3739 3759 3779 3799 3819 3 7 I0 J 3 17 21 3839 3859 3879 3899 3919 3939 3959 3979 4000 4O2O 3 7 I0 T 3 17 22 4040 4061 4081 4101 4122 4142 4i63 4183 4204 4224 3 7 10 14 17 23 4245 4265 4286 4307 4327 4348 4369 4390 4411 4431 3 7 10 14 17 24 4452 4473 4494 45i5 4536 4557 4578 4599 4621 4642 4 7 ii 14 18 25 4663 4684 4706 4727 4748 4770 4791 48i3 4834 4856 4 7 ii 14 18 26 4877 4899 4921 4942 4964 4986 5008 5029 5051 5073 4 7 ii 15 18 27 5095 5"7 5139 5161 5184 5206 5228 5250 5272 '5295 4 7 ii 15 18 28 5317 5340 5362 5384 5407 5430 5452 5473 5498 5520 4 8 ii 15 19 29 5543 5566 5589 5612 5635 5658 5681 5704 5727 5750 4 8 12 15 19 30 5774 5797 5820 5844 5867 5890 5914 5938 596i 5985 4 8 12 16 20 31 6009 6032 6056 6080 6104 6128 6152 6176 6200 6224 4 8 12 16 20 32 6249 6273 6297 6322 6346 6371 6395 6420 6 445 6469 4 8 12 16 20 33 6494 6519 6 544 6569 6594 6619 6644 6669 6694 6720 4 8 13 17 21 34 6745 6771 6796 6822 6847 6873 6899 6924 6950 6976 4 9 13 17 21 35 7002 7028 754 7080 7107 7 J 33 7*59 7186 7212 7239 4 9 13 18 22 36 7265 7292 7319 7346 7373 7400 7427 7454 748i 7508 5 9 14 18 23 37 7536 7563 7590 7618 7646 7673 7701 7729 7757 7785 5 9 14 18 23 38 7813 7841 7869 7898 7926 7954 7983 8012 8040 8069 5 9 14 19 24 39 8098 8127 8156 8185 8214 8243 8273 8302 8332 8361 5 10 15 20 24 40 8391 8421 8451 8481 8511 8541 8571 8601 8632 8662 5 10 15 20 25 41 8693 8724 8754 8785 8816 8847 8878 8910 8941 8972 5 10 16 21 26 42 9004 9036 9067 9099 9131 9163 9195 9228 9260 9293 5 ii 16 21 27 43 9325 9358 9391 9424 9457 9490 9523 9556 9590 9623 6 ii 17 22 28 44 9657 9691 9725 9759 9793 9827 9861 9896 9930 9965 6 ii 17 23 29 45 i-oooo 0035 0070 0105 0141 0176 O2 1 2 0247 0283 0319 6 12 18 24 30 MATHEMATICAL TABLES 409 TABLE VII. (contd.) 1 I 0' 00 6' 01 12' 0-2 18' 03 24' 04 30' 0-5 36' 0-6 42' 07 48' 0-8 54' 09 Mean Differences. 1' 2' 3' 4' 5' 45 I-OOOO 0035 0070 0105 0141 0176 O2I2 0247 0283 0319 6 12 18 24 30 46 1-0355 0392 0428 0464 0501 0538 0575 0612 0649 0686 6 12 18 25 31 47 1-0724 0761 0799 0837 0875 0913 0951 0990 1028 1067 6 13 19 25 32 48 1-1106 H45 1184 1224 1263 1303 1343 1383 1423 1463 7 13 20 27 33 49 1-1504 1544 1585 1626 1667 1708 1750 1792 1833 1875 7 14 21 28 34 50 1-1918 1960 2OO2 2045 2088 2131 2174 2218 2261 2305 7 14 22 29 36 51 1-2349 2393 2437 2482 2527 2572 2617 2662 2708 2753 8 15 23 30 38 52 1-2799 2846 2892 2938 2985 3032 3079 3127 3i75 3222 8 16 24 31 39 53 1-3270 33i9 3367 34i6 3465 35H 3564 3613 3663 3713 8 16 25 33 '41 54 1-3764 3814 386 5 39i6 3968 4019 4071 4124 4176 4229 9 17 26 34 43 55 1-4281 4335 43 88 4442 4496 4550 4605 4659 47i5 4770 9 18 27 36 45 56 1-4826 4882 4938 4994 5051 5108 5166 5224 5282 5340 10 19 29 38 48 57 1-5399 5458 5517 5577 5637 5697 5757 5818 5880 594i 10 20 30 40 50 58 1-6003 6066 6128 6191 6255 6319 6383 6447 6512 6577 ii 21 32 43 53 59 1-6643 6709 6775 6842 6909 6977 745 7H3 7182 7251 II 23 34 45 56 60 1-7321 739i 7461 7532 7603 7675 7747 7820 7893 7966 12 24 36 48 60 61 1-8040 8115 8190 8265 8341 8418 8495 8572 8650 8728 13 26 38 51 64 62 1-8807 8887 8967 9047 9128 9210 9292 9375 9458 9542 14 27 41 55 68 63 1-9626 9711 9797 9883 997 0057 oi45 0233 0323 0413 15 29 44 58 73 64 2-0503 0594 0686 0778 0872 0965 1060 "55 1251 1348 16 31 47 63 78 65 2-1445 1543 1642 1742 1842 1943 2045 2148 2251 2355 17 34 51 68 85 66 2-2460 2566 2673 2781 2889 2998 3109 3220 3332 3445 18 37 55 73 92 67 2-3559 3673 3789 3906 4023 4142 4262 4383 4504 4627 20 40 60 79 99 68 2-4751 4876 5002 5129 5257 5386 5517 5649 5782 5916 22 43 65 87 108 69 2-6051 6187 6325 6464 6605 6746 6889 734 7179 7326 24 47 71 95 119 70 2-7475 7625 7776 7929 8083 8239 8397 8556 8716 8878 26 52 78 104 131 71 2-9042 9208 9375 9544 97 J 4 9887 0061 0237 0415 0595 29 58 87 116 145 72 3-0777 0961 1146 1334 1524 1716 1910 2106 2305 2506 32 64 96 129 161 73 3-2709 2914 3122 3332 3544 ! 3759 3977 4197 4420 4646 36 72 108 144 1 80 74 3-4 8 74 5105 5339 5576 5816 6059 6305 6554 6806 7062 41 81 122 163 204 75 3-7321 7583 7848 8118 8391 8667 8947 9232 9520 9812 46 93 139 1 86 232 76 4-0108 0408 0713 IO22 1335 1653 1976 2303 2635 2972 77 4-33I5 3662 4 OI 5 4374 4737 5107 5483 5864 6252 6646 78 4-7046 7453 7867 8288 8716 9152 9594 0045 6504 0970 79 5-I446 1929 2422 2924 3435 3955 4486 5026 5578 6140 80 5-67I3 7297 7894 8502 9124 9758 0405 1066 1742 2432 81 6-3138 3859 4596 535 6122 6912 7720 8548 9395 6264 Mean differences are 82 7' IJ 54 2066 3002 3962 4947 5958 6996 8062 9158 0285 no longer suffici- 83 8-1443 2636 3863 5126 6427 7769 9152 0579 2052 3572 since the differ- 84 9-5I4 9-677 9-845 IO-O2 IO-2O 10-39 10-58 10-78 10-99 II-2O ences vary con- 85 "43 n-66 11-91 12-16 12-43 12-71 13-00 13-3 13-62 13-95 siderably along each line. 86 14-30 14-67 15-06 I5-46 15-89 16-35 16-83 17-34 17-89 18-46 87 19-08 19-74 20-45 21-20 22-02 22-90 23-86 24-90 26-03 27-27 88 28-64 30-14 31-82 33-69 35-8o 38-19 40-92 44-07 47-74 52-08 89 57-29 63-66 71-62 81-85 95-49 114-6 143-2 191-0 286-5 573-0 90 00 MATHEMATICAL TABLES TABLE VIII. LOGARITHMIC SINES. Q 0' 00 6' 0-1 12' 02 18' 24' 03 0-4 30' 05 36' 06 42' 0-7 48' 0-8 54' 0-9 Mean Differences. 1' 2' 3' 4' 5' 00 3-2419 54 2 9 7190 8439 9408 O2OO 6870 1450 1961 1 2-2419 2832 3210 3558 3880 4 J 79 4459 4723 497 1 5206 2 5428 5640 5842 6035 6220 6397 6567 6731 6889 7041 3 7188 7330 7468 7602 773i 7857 7979 8098 8213 8326 4 8436 8543 86 47 8749 8849 8946 9042 9i35 9226 93i5 16 32 48 64 80 5 9403 9489 9573 9655 9736 9816 9894 997 0046 OI2O 13 26 39 52 65 6 1-0192 0264 0334 0403 0472 0539 0605 0670 0734 0797 ii 22 33 44 55 7 0859 0920 0981 1040 1099 H57 1214 1271 1326 I38l 10 19 29 38 48 8 1436 1489 1542 1594 1646 1697 1747 1797 1847 1895 8 17 25 34 4 2 9 1943 1991 2038 2085 2131 2176 2221 2266 2310 2353 8 15 23 30 38 10 2397 2439 2482 2524 2565 2606 26 47 2687 2727 2767 7 M 20 27 34 11 2806 2845 2883 2921 2959 2997 3034 3070 3107 3143 6 12 19 25 31 12 3179 3214 3251 3284 3319 3353 3387 3421 3455 3488 6 ii 17 23 28 13 3521 3554 3586 3618 3650 3682 3713 3745 3775 3806 5 ii 16 21 26 14 3837 3867 3897 3927 3957 3986 4015 4044 473 4IO2 5 10 15 20 24 15 4130 4158 4186 4214 4242 4269 4296 4323 4350 4377 5 9 14 18 23 16 4403 4430 4456 4482 4508 4533 4559 4584 4609 4 6 34 4 9 13 17 21 17 4659 4684 4709 4733 4757 4781 4805 4829 4853 4876 4 8 12 16 20 18 4900 4923 4946 4969 4992 5015 5037 5060 5082 5104 4 8 ii 15 19 19 5126 5H8 5I7 5192 5213 5235 5256 5278 5299 5320 4 7 ii 14 18 20 5341 536i 5382 5402 5423 5443 5463 5484 5504 5523 3 7 10 14 17 21 5543 5563 5583 5602 5621 5641 5660 5679 5698 5717 3 6 10 13 16 22 5736 5754 5773 5792 5810 5828 5847 5865 5883 59oi 3 6 9 12 15 23 -5919 5937 5954 5972 5990 6007 6024 6042 6059 6076 3 6 9 12 15 24 -6093 6110 6127 6144 6161 6177 6194 6210 6227 6243 3 6 8 ii 14 25 6259 6276 6292 6308 6324 6340 6356 6371 6387 6403 3 5 8 ii 13 26 6418 6434 6449 6465 6480 6495 6510 6526 6541 6556 3 5 8 10 13 27 6570 6585 6600 6615 6629 6644 6659 6673 6687 6702 2 5 7 10 12 28 6716 6730 ! 6744 6759 6 773 6787 6801 6814 6828 6842 2 5 7 9 12 29 6856 6869 6883 6896 6910 6923 6937 6950 6963 6977 2 4 7 9 ii 30 6990 7003 7016 7029 7042 755 7068 7080 793 7106 2 4 6 9 ii 31 7118' 7131 7M4 7156 7168 7181 7193 7205 7218 7230 2 4 6 8 10 32 7242 7254 7266 7278 7290 7302 7314 7326 7338 7349 2 4 6 8 10 33 736i ' 7373 7384 7396 7407 7419 7430 7442 7453 7464 2 4 6 8 10 34; -7476 | 7487 7498 7509 7520 753i 7542 7553 7564 7575 24679 35 7586 7597 7607 7618 7629 7640 7650 7661 7671 7682 24579 36 7692 773 7713 7723 7734 7744 7754 7764 7774 7785 23579 37 '7795 7805 7815 7825 7835 7844 7854 7864 7874 7884 23578 38 7893 7903 7913 7922 7932 794i 7951 7960 797 7979 23568 39 7989 7998 8007 8017 8026 8035 8044 8053 8063 8072 23568 40 8081 8090 8099 8108 8117 8125 8i34 8i43 8152 8161 13467 41 8169 8178 8187 8i95 8204 8213 8221 8230 8238 8247 13467 42 8255 8264 8272 8280 8289 8297 8305 8313 8322 8330 13467 43 -8338 8346 8354 8362 8370 8378 8386 8394 8402 8410 13457 44 -8418 8426 8433 8441 8449 8457 8464 8472 8480 8487 13456 45 -8495 8502 8510 8517 8525 8532 8540 8547 8555 8562 12456 1 MATHEMATICAL TABLES 411 TABLE VIII. (contd.) E 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. t Q 00 0-1 02 03 04 05 06 07 08 09 1' 2' 3' 4' 5' 45 ^8495 8502 8510 8517 8525 8532 8540 8547 8555 8562 12456 46 8569 8577 8584 8591 8598 8606 9613 8620 8627 8634 12456 47 8641 8648 8655 8662 8669 8676 8683 8690 8697 8704 1 2 3 5 6 48 8711 8718 8724 8731 8738 8745 8751 8758 8765 8771 2 346 49 8778 8784 8791 8797 8804 8810 8817 8823 8830 8836 2 345 50 8843 8849 8855 8862 8868 8874 8880 8887 8893 8899 2345 51 8905 8911 8917 8923 8929 8935 8941 8947 8953 8959 2345 52 8965 8971 8977 8983 8989 8995 9000 9006 9012 9018 2345 53 9023 9029 9035 9041 9046 9052 9057 9063 9069 9074 2345 54 9080 9085 9091 9096 9101 9107 9112 9118 9123 9128 2345 55 9134 9139 9144 9149 9155 9160 9165 9170 9175 9181 2334 56 9186 9191 9196 9201 9206 9211 9216 9221 9226 9231 2334 57 9236 9241 9246 9251 9255 9260 9265 9270 9275 9279 2234 58 9284 9289 9294 9-298 9303 9308 9312 9317 9322 9326 2234 5 S 933i 9335 9340 9344 9349 9353 9358 9362 9367 937 1 1234 60 '9375 9380 9384 9388 9393 9397 9401 9406 9410 9414 1234 61 9418 9422 9427 9431 9435 9439 9443 9447 945i 9455 1233 62 '9459 9463 9467 9471 9475 9479 9483 9487 9491 9495 233 63 '9499 9503 9507 95io 95H 95i8 9522 9525 9529 9533 233 64 '9537 9540 9544 9548 955i 9555 9558 9562 9566 9569 223 65 "9573 9576 958o 9583 9587 9590 9594 9597 9601 9604 223 66 9607 9611 9614 9617 9621 9624 9627 9631 9634 9637 i 223 67 9640 9643 9647 9650 9653 9656 9659 9662 9666 9669 i 223 68 9672 9675 9678 9681 9684 9687 9690 9693 9696 9699 I I 2 2 69 9702 9704 9707 9710 97*3 9716 9719 9722 9724 9727 O I I 2 2 70 '973 9733 9735 9738 9741 9743 9746 9749 975i 9754 O I I 2 2 71 9757 9759 9762 9764 9767 977 9772 9775 9777 9780 I I 2 2 72 9782 9785 9787 9789 9792 9794 9797 9799 9801 9804 I I 2 2 73 9806 9808 9811 9813 9815 9817 9820 9822 9824 9826 O I I 2 2 74 9828 9831 9833 9835 9837 9839 9841 9843 9845 9847 I I I 2 75 '9849 9851 9853 9855 9857 9859 9861 9863 9865 9867 I I I 2 76 9869 9871 9873 9875 9876 9878 9880 9882 9884 9885 O I I I 2 77 9887 9889 9891 9892 9894 9896 9897 9899 9901 9902 I I I I 78 9904 9906 9907 9909 9910 9912 9913 9915 9916 9918 O I I I I 79 9919 9921 9922 9924 9925 9927 9928 9929 993i 9932 O I I I 80 '9934 9935 9936 9937 9939 994 9941 9943 9944 9945 I I I 81 9946 9947 9949 9950 9951 9952 9953 9954 9955 9956 O O I I I 82 9958 9959 9960 9961 9962 9963 9964 9965 9966 9967 O I I I 83 9968 9968 9969 997 9971 9972 9973 9974 9975 9975 O O O I I 84 9976 9977 997 s 997 s 9979 9980 9981 998i 9982 9983 O O O I 85 9983 9984 9985 9985 9986 9987 9987 9988 9988 9989 O O O O 86 9989 9990 9990 9991 9991 9992 9992 9993 9993 9994 O O O O O 87 '9994 9994 9995 9995 9996 9996 9996 9996 9997 9997 O O O O O 88 '9997 9998 9998 9998 9998 9999 9999 9999 9999 9999 O O O O O 89 9999 9999 oooo oooo oooo oooo oooo oooo oooo oooo O O O O O 90 O'OOOO 4T2 MATHEMATICAL TABLES TABLE IX. LOGARITHMIC COSINES 8 & 0' 6' 12' 18' 24' 30' j 36' 42' 48' 54' Mean Differences. of 00 0-1 0-2 03 04 05 0-6 0-7 08 09 1' 2' 3' 4' 5' o-oooo oooo oooo oooo oooo oooo oooo oooo oooo 9999 O O O O 1 1-9999 9999 9999 9999 9999 9999 9998 9998 9998 9998 o o o o o 2 9997 9997 9997 9996 9996 9996 9996 9995 9995 9994 O O O O O 3 9994 9994 9993 9993 9992 9992 9991 9991 9990 9990 O O O 4 9989 9989 9988 9988 9987 9987 9986 9985 9985 9984 o o o o o 5 9983 9983 9982 9981 9981 9980 9979 9978 997 s 9977 O O O I 6 9976 9975 9975 9974 9973 9972 9971 9970 9969 9968 O O I I 7 9968 9967 9966 9965 9964 9963 9962 9961 9960 9959 O I I I 8 995 8 9956 9955 9954 9953 9952 9951 9950 9949 9947 I I I 9 9946 9945 9944 9943 9941 9940 9939 9937 9936 9935 O O I I I 10 9934 9932 9931 9929 9928 9927 9925 9924 9922 9921 I I I 11 9919 9918 9916 9915 9913 9912 9910 9909 9907 9906 O I I I I 12 9904 9902 9901 9899 9897 9896 9894 9892 9891 g88q I I I I 13 9887 9885 9884 9882 9880 9878 9876 9875 9873 9871 I I I 2 14 9869 9867 9865 9863 9861 9859 9857 9855 9853 9851 I I I 2 15 9849 9847 9845 9843 9841 9839 9837 9835 9833 9831 I I I 2 16 9828 9826 9824 9822 9820 9817 9815 9813 9811 9808 I I 2 2 17 9806 9804 9801 9799 9797 9794 9792 9789 9787 9785 O I I 2 2 18 9782 9780 9777 9775 9772 9770 9767 9764 9762 9759 I I 2 2 19 9757 9754 975i 9749 9746 9745 974 1 9738 9735 9733 I I 2 2 20 9730 9727 9724 9722 9719 9716 9713 9710 9707 9704 I I 2 2 21 9702 9699 9696 9693 9690 9687 9684 9681 9678 9675 I I 2 2 22 9672 9669 9666 9662 9659 9656 9653 9650 9647 9643 II223 23 9640 9637 9634 9631 9627 9624 9621 9617 9614 9611 II223 24 9607 9604 9601 9597 9594 9590 9587 9583 958o 9576 II223 25 9573 9569 9566 9562 9558 9555 955i 9548 9544 9540 II223 26 9537 9533 9529 9525 9522 95i8 9514 95io 9507 9503 1 i 2 3 3 27 9499 9495 9491 9487 9483 9479 9475 9471 9467 94 6 3 II233 28 9459 9455 945i 9447 9443 9439 9435 9431 9427 9422 1 i 2 3 3 29 9-1 1 8 9414 9410 9406 9401 9397 9393 9388 9384 9380 11.234 30 9375 9371 9367 9362 9358 9353 9349 9344 934 9335 II234 31 9331 9326 9322 9317 9312 9308 9303 9298 9294 9289 12234 32 9284 0279 9275 9270 9265 9260 9255 9251 9246 9241 12234 33 9236 9231 9226 9221 9216 9211 9206 9201 9196 9191 2 334 34 9186 9181 9175 9170 9165 9160 9155 9149 9144 9139 2 334 35 9134 9128 9123 9118 9112 9107 9101 9096 9091 9085 2345 36 9080 9074 9069 9063 9057 9052 9046 9041 9035 9029 2 345 37 9023 9018 9012 9006 9000 8995 8989 8983 8977 8971 2 345 38 8965 8959 8953 8947 8941 8935 8929 8923 8917 8911 2345 39 8905 8899 8893 8887 8880 8874 8868 8862 8855 8849 1 2 3 4 5 40 8843 8836 8830 8823 8817 8810 8804 8797| 8791 8784 1 2 3 4 5 41 8778 8771 8765 8758 8751 8745 8738 8731 8724 8718 12356 42 8711 8704 8697 8690 8683 8676 8669 8662 8655 8648 1 2 3 5 6 43 8641 8634 8627 8620 8613 8606 8598 8591 .8584 8577 12456 44 8569 8562 8555 8547 8540 8532 8525 8517 8510 8502 12456 45 8495 8487 8480 8472 8464 8457 8449 8441 8433 8426 13456 MATHEMATICAL TABLES 413 TABLE IX (contd.) V 1 Q 0' 00 6' 01 12' 18' 0-2 03 24' 0-4 30' 05 36' 42' 06 0-7 48' 08 54' 0-9 Mean Differences. 1' 2' 3' 4' 5' 45 1-8495 8487 8480 8472 8464 8457 8449 8441 8433 8426 13456 46 8418 8410 8402 8394 9386 8378 8370 8362 8354 8346 13457 47 -8338 8330 8322 8313 8305 8297 8289 8280 8272 8264 1 3 4 6 7 48 -8255 8247 8238 8230 8221 8213 8204 8195 8187 8178 1 3 4 6 7 49 -8169 8161 8152 8143 8134 8125 8117 8108 8099 8090 13467 50 8081 8072 8063 8053 8044 8035 8026 8017 8007 7998 23568 51 7989 7979 7970 7960 795i 794i 7932 7922 79i3 7903 23568 52 7893 7884 7874 7864 7854 7844 7835 7825 7815 7805 23578 53 7795 7785 7774 7764 7754 7744 7734 7723 7713 : 7703 23579 54 7692 7682 7671 7661 7650 7640 7629 7618 7607 7597 24579 55 7586 7575 7564 7553 7542 7531 7520 7509 7498 7487 24679 56 7476 7464 7453 7442 7430 7419 7407 7396 7384 7373 246 8 10 57 7361 7349 7338 7326 7314 7302 7290 7278 7266 7254 246 8 10 58 7242 7230 7218 7205 7*93 7181 7168 7156 7H4 7i3i 246 8 10 59 7118 7106 793 7080 7068 7055 7042 7029 7016 7003 2 4 6 9 II 60 6990 6977 6963 6950 6937 6923 6910 6896 6883 ' 6869 247 9 II 61 6856 6842 i 6828 6814 6801 6787 6773 6759 6744 6730 2 5 7 9 12 62 6716 6702 6687 6673 6659 6644 6629 6615 6600 6585 2 5 7 10 21 63 6570 6556 6541 6526 6510 6495 6480 6465 6449 6434 3 5 8 10 13 64 6418 6403 6387 637 1 6356 6340 6324 6308 6292 6276 3 5 8 it 13 65 6259 6243 6227 6210 6194 6177 6161 6144 6127 6110 3 6 8 ii 14 66 6093 6076 6059 6042 6024 6007 5990 597 2 5954 5937 3 6 9 12 15 67 5919 5901 5883 5865 5847 5828 5810 5792 5773 5754 3 6 9 12 15 68 5736 5717 5698 5679 5660 5641 5621 5602 5583 5563 3 6 10 13 16 69 5543 5523 5504 5484 5463 5443 54 2 3 5402 5382 536i 3 7 10 14 17 70 534 1 5320 5299 5278 5256 5235 5213 5192 5170 5M8 4 7 ii 14 18 71 5162 5 I0 4 5082 5060 5037 5015 4992 4969 4946 4923 4 8 ii 15 19 72 4900 4876 4853 4829 4805 4781 4757 4733 4709 4684 4 8 12 16 20 73 4659 4 6 34 4609 4584 4559 4533 4508 4482 4456 4430 4 9 13 17 21 74 443 4377 4350 4323 4296 4269 4242 4214 4186 4158 5 9 14 18 23 75 4130 4102 473 4044 4i5 3986 3957 3927 3897 3867 5 10 15 20 24 76 3837 3806 3775 3745 3713 3682 3650 3618 3586 3554 5 ii 16 21 26 77 3521 3488 3455 3421 3387 3353 3319 3284 3250 3214 6 it 17 23 28 78 3179 3143 3io/ 3070 3034 2997 2959 2921 2883 2845 6 12 19 25 31 79 2806 2767 2727 2687 2647 2606 2565 2524 2482 2439 7 14 20 27 34 80 2397 2353 2310 2266 2221 2176 2131 2085 2038 1991 8 15 23 30 38 81 1943 1895 1847 1797 J 747 1697 1646 1594 1542 1489 8 i? 25 34 42 82 1436 1381 1326 1271 1214 "57 1099 1040 0981 0920 10 19 29 38 48 83 0859 0797 734 0670 0605 539 0472 0403 0334 0264 ii 22 33 44 55 84 0192 OI2O 0046 9970 9894 9816 9736 9655 9573 9489 13 26 39 52. 65 85 2-9403 9315 9226 9135 9042 8946 8849 8749 8647 8543 16 32 48 64 80 86 8436 8326 8213 8098 7979 7857 7731 7602 7468 733 87 -7188 7041 6889 6731 6567 6397 6220 6035 5842 5640 88 -5428 52O6 4971 4723 4459 4*79 3880 3558 3210 2832 89 2419 I96l M50 0870 O200 9408 8439 7190 5429 2419 90 CO 414 MATHEMATICAL TABLES TABLE X. LOGARITHMIC TANGENTS. o 0' 00 6' 01 12' 02 18' 0-3 24' 04 30' 05 38' 06 42' 0-7 48' 08 54' 0-9 Mean Differences. 1' 2' 3' 4' 5' 00 3-2419 5429 7190 8439 9409 02 oo 0870 M5o 1962 1 2-2419 2833 3211 3559 3881 4181 4461 4725 4973 5208 2 5431 5643 5845 6038 6223 6401 657 1 6736 6894 7046 3 7194 7337 7475 7609 7739 7865 7988 8107 8223 8336 4 8446 8554 8659 8762 8862 8960 9056 915 9241 933i 16 32 48 64 81 5 9420 9506 959i 9674 9756 9836 9915 9992 0068 0143 13 26 40 53 66 6 1-0216 0289 0360 0430 0499 0567 0633 0699 0764 0828 ii 22 34 45 56 7 0891 0954 1015 1076 H35 1194 1252 1310 1367 1423 10 20 29 39 49 8 1478 1533 1587 1640 1693 1745 1797 1848 1898 1948 9 17 26 35 43 9 1997 2046 2094 2142 2189 2236 2282 2328 2 374 2419 8 16 23 31 39 10 2463 2507 2551 2594 2637 2680 2722 2764 2805 2846 7 14 21 28 35 11 2887 2927 2967 3006 3046 3085 3123 3162 3200 3237 6 13 19 26 32 12 3275 3312 3349 3385 3422 3458 3493 3529 3564 3599 6 12 18 24 30 13 3634 3668 3702 3736 3770 3804 3837 3870 3903 3935 6 ii 17 22 28 14 3968 4000 4032 4064 495 4127 4158 4189 4220 4250 5 10 16 21 26 15 4281 43ii 4341 4371 4400 4430 4459 4488 4517 4546 5 10 15 20 25 16 4575 4603 4632 4660 4688 4716 4744 477i 4799 4826 5 9 14 19 23 17 4853 4880 4907 4934 4961 4987 5014 5040 5066 5092 4 9 13 18 22 18 5118 5143 5169 5195 5220 5245 5270 5295 5320 5345 4 8 13 17 21 19 537 5394 5419 5443 5467 549i 55i6 5539 5563 5587 4 8 12 16 20 20 5611 564 5658 5681 5704 5727 5750 5773 5796 5819 4 8 12 15 19 21 5842 5864 5887 5909 5932 5954 5976 5998 6020 6042 4 7 ii 15 19 22 6064 6086 6108 6129 6151 6172 6194 6215 6236 6257 4 7 ii 14 18 23 6279 6300 6321 6341 6362 6383 6404 6424 6 445 6465 3 7 10 14 17 24 6486 6506 6527 6547 6567 6587 6607 6627 6647 6667 3 7 10 13 17 25 6687 6706 6726 6746 6765 6785 6804 6824 6843 6863 3 7 10 13 16 26 6882 6901 6920 6939 6958 6977 6996 7i5 734 753 3 6 9 13 16 27 7072 7090 7109 7128 7146 7^5 7183 7202 7220 7238 3 6 9 12 15 28 7257 7275 7293 73H 7330 7348 7366 7384 7402 7420 3 6 9 12 15 29 7438 7455 7473 749i 7509 7526 7544 7562 7579 7597 3 6 9 12 15 30 7614 7632 7649 7667 7684 7701 7719 7736 7753 7771 3 6 9 12 14 31 7788 7805 7822 7839 7856 7873 7890 7907 7924 794i 3 6 9 ii 14 32 7958 7975 7992 8008 8025 8042 8059 8075 8092 8109 3 6 8 ii 14 33 8125 8142 8158 8i?5 8191 8208 8224 8241 8257 8274 3 5 8 n 14 84 8290 8306 8323 8339 8355 8371 8388 8404 8420 8436 3 5 8 n 14 35 8452 8468 8484 8501 8517 8533 8549 8565 8581 8597 3 5 8 ii 13 36 8613 8629 8644 8660 8676 8692 8708 8724 8740 8755 3 5 8 ii 13 37 8771 8787 8803 8818 8834 8850 8865 8881 8897 8912 3 5 8 10 13 38 8928 8944 8959 8975 8990 9006 9022 9037 9053 9068 3 5 8 10 13 39 9084 9099 9H5 9130 9146 9161 9176 9192 9207 9223 3 5 8 10 13 40 9238 9254 9269 9284 9300 9315 9330 9346 936i 937 3 5 8 10 13 41 9392 9407 9422 9438 9453 9468 9483 9499 9514 9529 3 5 8 10 13 42 9544 9560 9575 9590 9605 9621 9636 9651 9666 9681 3 5 8 10 13 43 9697 9712 9727 9742 9757 9773 9788 9803 9818 9833 3 5 8 10 13 44 9848 9864 9879 9894 9909 9924 9939 9955 997 9985 3 5 8 10 13 45 o-oooo 0015 0030 0045 0061 0076 0091 0106 0121 0136 3 5 8 10 13 MATHEMATICAL TABLES TABLE X. (contd.) g> Q 0' 00 6' 01 12' 18' 02 03 24' 30' 04 05 36' 06 42' 0-7 48' 08 54' 09 Mean D;ffer;uces. 1' 2' 3' 4' 5' 45 oooo 0015 0030 0045 0061 0076 0091 OIO6 OI2I OI36 35 8 10 13 46 0152 0167 Ol82 0197 O2I2 0228 0243 0258 0273 0288 35 8 10 13 47 0303 0319 0334 0349 0364 379 395 0410 0425 0440 35 8 10 13 48 0456 0471 0486 0501 0517 0532 0547 0562 0578 0593 35 8 10 13 49 0608 0624 0639 0654 0670 0685 0700 0716 0731 0746 35 8 10 13 50 0762 0777 0793 0808 ; 0824 0839 0854 0870 0885 0901 35 8 10 13 51 0916 0932 0947 0963 0978 0994 IOIO IO25 1041 1056 35 8 10 13 52 1072 1088 1103 nig 1135 1150 1166 1182 1197 1213 35 8 10 13 53 1229 1245 1260 1276 1292 1308 1324 1340 1356 I37 1 35 8 ii 13 54 1387 1403 1419 H35 1451 1467 1483 1490 1516 1532 35 8 ii 13 55 1548 1564 1580 1596 1612 1629 1645 1661 1677 1694 35 8 ii 14 56 1710 1726 1743 ! 1759 1776 1792 1809 1825 1842 1858 35 8 ii 14 57 1875 1891 1908 1925 1941 1958 1975 1992 2008 2025 36 8 ii 14 58 2042 2059 2076 2093 2110 2127 2144 2161 2178 2195 3 6 9 ii 14 59 2212 2229 2247 2264 228l 2299 2316 2 333 2351 2368 3 6 9 12 14 60 2386 2403 2421 2438 2456 2474 2491 2509 2527 2545 3 6 9 12 15 61 2562 2580 2598 2616 2634 2652 2670 2689 2707 2725 3 6 9 12 15 62 2743 2762 2780 2798 2817 2835 2854 2872 2891 2910 3 6 9 12 15 63 2928 2947 2966 2985 3004 3023 3042 3061 3080 3099 36 9 13 16 64 3"8 3i37 3i57 3176 3196 3215 3235 3254 3274 3294 3 6 10 13 16 65 3313 3333 3353 3373 3393 3413 3433 3453 3473 3494 3 7 I0 13 17 66 35H 3535 3555 357 6 3596 3617 3638 3659 3679 3700 3 7 10 14 17 67 3721 3743 3764 3785 3806 3828 3849 3871 3892 3914 4 7 ii 14 18 68 3936 3958 398o 4002 4024 4046 4068 4091 4"3 4136 4 7 ii 15 19 69 4158 4181 4204 4227 4250 4273 4296 4319 4342 4366 4 8 12 15 19 70 4389 44U ' 4437 4461 4484 4509 4533 4557 458i 4606 4 8 12 16 20 71 4630 4 6 55 4680 475 4730 4755 4780 4805 4831 4857 4 8 13 17 21 72 4882 4908 4934 4960 4986 5013 5039 5066 5093 5120 4 9 13 18 22 73 5M7 5174 5201 5229 5256 5284 5312 5340 5368 5397 5 9 14 19 23 74 54 2 5 5454 5483 5512 5541 5570 5600 5629 5659 5689 5 10 15 20 25 75 5719 5750 578o 5811 5842 5873 5905 5936 5968 6000 5 10 16 21 26 76 6032 6065 6097 6130 6l6 3 6196 6230 6264 6298 6332 6 II 17 22 28 77 6366 6401 6436 6471 6507 6542 6578 6615 6651 6688 6 12 18 24 30 78 6725 6763 6800 6838 6877 6915 6954 6994 7033 773 6 13 19 26 32 79 7"3 7154 7*95 7236 7278 7320 7363 7406 7449 7493 7 14 21 28 35 80 7537 758i 7626 7672 77 l8 7764 7811 7858 7906 7954 8 16 23 31 39 81 8003 8052 8102 8152 8203 8255 8307 8360 8413 8467 9 17 26 35 43 82 8522 8577 8633 8690 8748 8806 8865 8924 8985 9046 10 20 29 39 49 83 9109 9172 9236 9301 9367 9433 9501 9570 9640 97" ii 22 34 45 56 84 9784 9857 9932 0008 0085 0164 0244 0326 0409 0494 13 26 40 53 66 85 1-0580 0669 0759 0850 0944 1040 1138 1238 1341 1446 16 32 48 64 81 86 I-I554 1664 1777 1893 2OI2 2135 2261 2391 2525 2663 87 1-2806 2954 3106 3264 3429 3599 3777 3962 4155 4357 88 1-4569 4792 5027 5275 5539 5819 6119 6441 6789 7167 89 1-7581 8038 8550 9130 9800 0591 1561 2810 457 1 758i 90 -00 4i 6 MATHEMATICAL TABLES TABLE XI. EXPONENTIAL AND HYPERBOLIC FUNCTIONS X e x 6-x cosh x e x +e-* sink x e x e~ x tanh x _e*-e~* 2 2 e x +e~* 1 1-1052 9048 1-0050 i 002 0997 2 1-2214 8187 I-020I 2013 1974 3 1-3499 7408 1-0453 3045 2913 4 1-4918 6703 1-0811 4108 3799 5 1-6487 6065 1-1276 5211 4621 6 1-8221 5488 1-1855 6367 537 7 2-0138 4966 1-2552 7586 6044 8 2-2255 4493 1-3374 8881 6640 9 2-4596 4066 I-433I 1-0265 7163 10 2-7183 3679 I-543I 1-1752 7616 1-1 3-0042 3329 1-6685 1-3357 8005 12 3-3201 3012 1-8107 1-5095 8337 13 3-6693 2725 1-9709 1-6984 8617 1-4 4-0552 2466 2-1509 1-9043 8854 15 4-4817 2231 2-3524 2-1293 9051 1-6 4-953 2019 2-5775 2-3756 9217 17 5-4739 1827 2-8283 2-6456 9354 1-8 6-0496 i653 3-1075 2-9422 9468 19 6-6859 1496 3-4I77 3-2682 9563 20 7-3891 1353 3-7622 3-6269 9640 2-1 8-1662 1225 4 -I 443 4-0219 9704 22 9-0251 1108 4-5679 4-457 1 9758 2-3 9-9742 1003 5-037 2 4-937 9801 2-4 11-0232 0907 5-557 5-4662 9837 25 12-1825 0821 6-1323 6-0502 9866 26 13-4638 0743 6-7690 6-6947 9890 2-7 14-8797 0672 7-4735 7-4063 9910 28 16-4446 0608 8-2527 8-1919 9926 29 18-1741 0550 9-1146 9-0596 9940 30 20-0855 0498 10-068 10-018 9951 3-1 22-1980 0450 11-122 11-076 9959 32 24-5325 0408 12-287 12-246 9967 33 27-1126 0369 13-575 I3-538 9973 34 29-9641 0334 I4-999 14-965 9978 35 33-II55 0302 16-573 16-543 9982 36 36-5982 0273 I8-3I3 18-285 9985 3-7 4 -4473 0247 20-236 2O-2II 9988 3-8 44-7012 0224 22-362 22-339 9990 39 49-4024 O2O2 24-711 24-691 9992 40 54-5982 0183 27-308 27-290 9993 4-1 60-3403 0166 30-178 30-162 9995 4-2 66-6863 0150 33-351 33-336 9996 4-3 73-6998 0136 36-857 36-843 9996 4-4 81-4509 OI23 40-732 40-719 9997 4-5 90-0171 01 1 1 45-014 45-003 9997 4-6 99-4843 OIOI 49-747 49-737 9998 4-7 109-9472 0091 54-978 54-969 9998 4-8 121-5104 0082 60-759 60-751 9999 4-9 134-2898 0074 67-149 67-141 9999 5-0 148-4132 0067 74-210 74-203 9999 INDEX Abbreviations, i Altitude, 366 " Ambiguous " case in the solution of spherical triangles, 360 Amsler planimeter, theory of, 266 Analysis, harmonic, 342 Anchor ring, moment of inertia of, 256 Applications of the Calculus, 300 et seq. Applications of Differentiation, 88 ' et seq. Applied electricity, examples in, 317 Arc, length of, 201 Archimedean spiral, 258 Area of cardioid, 262 Areas by polar co-ordinates, 261 Areas by sum-curve method, 118 Arithmetic mean, probable error of, 378 Arrangement of electric cells, 317 Azimuth, 367 Contraflexure, point of, 93 Cooling curves, 21 Coradi integraph, 126 Curvature of an arc, 308 Cycloid, equation of, 69 D, the operator, 26, 286 d?s d p , meaning of, 9 Declination, 367 Definite integral, 118, 137 Deflection of muzzle of a gun, 316 Derivative, 9 Derived curve, 12 Differential coefficient, 9 Differential equations, exact, 279 -- , homogeneous, 281 -- , solution of, 270 et seq. - of type - = /(*), 271 B Beam problems, 38, 93, 123, 307 Belt round pulley, tension in, 330 Bending moment on ship, 126 Buoyancy, curve of, 125 Cable, approximate length of, 202 Calculation of small corrections, 107 Cardioid, area of, 262 Catenary, 321 Centre of Gravity, 211 et seq. of irregular solids, 225 of solids of revolution, 228 Centre of Pressure, 232, 336 Centroid, 211 et seq. of sections by calculation, 220 by drawing, 251 Centroid vertical, determination of, by double sum curve method, 218 Circular parts, Napier's rules of, 359 Complementary function, 289 Compound pendulum, time of oscilla- tion of, 324 dv =b > 2 75 of the second degree, 294 Differential, total, 82 Differentiation, applications of, 88 et seq. , graphic, 12 , logarithmic, 85 Differentiation of ax n , 26 exponential functions, 47 - function of a function, 63 - hyperbolic functions, 54 - inverse trigonometric functions, 76 - log x, 51 - product, 70 quotient, 73 - trigonometric functions, 56 Differentiation, partial, 79 Double integral, 123 Double sum curve method for fixing the centroid vertical, 218 MATHS. FOR ENG. E E 417 4i8 INDEX E Efficiency of engine working on the Rankine cycle, 304 Electric condenser, time of discharge of, 318 Ellipse, perimeter of, 205 Elliptic integral, 205 Entropy of water, 304 Equations, differential, 270 Euler's formula for struts, 328 Exact differential equations, 279 Expansion in series, 108 Exponential functions, differentia- tion of, 47 , integration of, 129 Integration by partial fractions, 146 by parts, 155 by substitution, 148 Integration, graphic, 118 Integration, meaning of, 115 Integration of exponential functions 129 powers of x, 127 trigonometric functions, 134 Interpolation, using Taylor's theorem, 112 K k, symbol for swing radius, 240 Kinetic energy, 240 First moment, 211 Fixed beams, deflection of, 309 Fleming's graphic method of finding R.M.S. values, 264 'Footstep bearing, friction in, 331 Forced vortex, 338 Fourier's theorem, 342 Friction on wheel disc in fluid, 335 Least squares, theorem of, 374 Leibnitz, 3 Length of arc, 201 Length of cable, approximate, 202 List of integrals, 175 Logarithmic differentiation, 85 Logarithmic functions, differentia- tion of, 51 Logarithmic spiral, 258 Gamma function, 173 Gauss's error curve, 374 Goodman scheme for fixed beams, 313 Governor, problem on, 75 Graphic differentiation, 12 et seq. Graphic integration, 1 1 8 et seq. Graphic solution of spherical triangles, 368 H Harmonic analysis, 342 et seq. Harrison's method of harmonic analysis, 346 Homogeneous differential equations, 281 Hour angle, 367 Hydraulics, examples on, 334 et seq. Inertia, moment of, 237 et seq. Inflexion, point of, 93 Integral, definite, 118, 137 , double, 123 , indefinite, 118, 137 Integrals, list of, 175 Integraph, the Coradi, 126 M Maclaurin's theorem, 108 Maximum and minimum values, 88 Maximum intensity of shear stress, 3H Maxwell's needle, 250 Mean spherical candle-power, 262 Mean values, 180 et seq. Modulus of rigidity, determination of, 325 Moment of inertia, 237 et seq. Moment of inertia of anchor ring, 256 circle, 246 compound vibrator, 249 cylinder, 247 pulley wheel, 248 rectangle, 244 sphere, 250 Tee section, 245 Moments, ist and 2nd, by construc- tion, 251 Muzzle of gun, deflection of, 316 N Napier's rules of circular parts, 359 Neutral axis, 238 Newton, 3 Notch, triangular, 334 INDEX 419 Oblate spheroid, volume of, 200 Parallel axis theorem, 241 Partial differentiation, 79 Pendulum, time of swing of, 324 Perimeter of ellipse, 205 Perpendicular axes theorem, 243 Planimeter, theory of, 266 Point of inflexion, 93 Polar co-ordinates, 257 et seq. Pressure, centre of, 232, 336 Probability, 370 Probability of error, 372 Probable error of arithmetic mean, 378 Prolate spheroid, volume of, 200 K Radius of gyration, 240 Rankine cycle, efficiency of, 304 Reduction formulae, 163 Right-angled spherical triangles, solu- tion of, 358 Root mean square values, 188 et seq. , Fleming's gra- phic method for, 264 Rousseau diagram, 262 Strength of materials, examples on, 321 et seq. Stresses in thick cylinders, 325 spherical shells, 327 Struts, formulae for, 328 Sub-normal, length of, 42 Sub-tangent, length of, 42 Sum curve, 119 Surface of solid of revolution, 208 Swing radius, 240 Taylor's theorem, 108 Tee section, centroid of, 220 , moment of inertia of, 245 Thermodynamics, examples in, 300 et seq. Thick cylinders, stresses in, 325 Time to empty a tank, 334 Total differential, 82 Tractrix, 333 Transition curve, 338 Triangle, spherical, 355 , solution of, 357 Triangular notch, measurement of flow by, 324 Trigonometric functions, differentia- tion of, 56 , integration of, 134 Trigonometry, spherical, 355 Schiele pivot, 333 Second moment, 211 Shear stress in beams, 314 Simple harmonic motion, 60 Simpson's rule, proof of, 141 " Sine " rule for spherical triangles, . 357 Solid of revolution, centre of gravity of, 228 Solid of revolution, volume of, 195 Solution of right-angled spherical triangles, 358 Solution of spherical triangles, 357 Sphere, moment of inertia of, 250 , volume of zone of, 200 Spherical excess, 356 Spherical triangle, 355 Spherical trigonometry, 355 et seq. Spheroid, volume of, 200 Spiral, Archimedean, 258 , logarithmic, 258 Values, mean, 180 , root mean square, 188 Velocity of piston, 65 Volume of oblate spheroid, 200 prolate spheroid, 200 solid of revolution, 195 Vortex, forced, 338 W Wattless current, 187 Wheel disc in fluid, friction on, 335 Work done in complete theoretical cycle, 303 expansion of a gas, 302 Zenith, 366 Zone of sphere, volume of, 200 s 0> 52? & <Di 0> pi -R University of Toronto Library DO NOT REMOVE THE CARD FROM THIS POCKET Acme Library Card Pocket Under Pat "Ref. Index File" Made by LIBRARY BUREAU Ill,, >Xw iPll|P