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."

0 ,, „ " 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 24- "P, „ „ " the number of permutations of n things taken

two at a time." "C2 „ „ " 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."

0 „ „ " 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."

Sn „ ,, " 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 14—10, *. e., 4 ft. per sec.

change of velocity 14—10 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 = 15—10 = 5 ; or, the rate of change of velocity (for this period) = T5^ ft. per sec. per foot.

Similarly, the rate of change of v, whilst s ranges from 5 to 12,

15—14 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 = 14—10 = 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

0

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) .

0

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

0

—

—

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 = \gtz = 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-iX22 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 i2 = 16-1

t = i-i s = 16-1 x i-i2 = 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-oi2 = 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-ooi2 = 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

32-2

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

d2s 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 ah1 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

8830

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 0 vertical.

(b) Inverse rate curve : -, horizontal and 0 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) 0 vertical and 0— 0X horizontal : 0—0! being the difference of temperature between the sample and a neutral body cooling under identical conditions.

(d) 0 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

rV (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 — -$x3 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

4I-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

0

.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.) . . /

0

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 0 in the neighbourhood of 15° find the actual rate of change of sin 6 with regard to 0 (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.)

0

I

2

3

4

5

6

Shearing force (tons) ....

0

•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 CO2 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

I48.27

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°

i°5

120

135

150

165

180

195

210

225

Temperature in C.c (0)

1 850 848

844-7

842

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

4°5

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—^xy3 +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 — axn, by first considering the simple case of y = x3. 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 = xs.

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 = x3, 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 — x3 for all values of x —

(new value of y) = (new value of x)3

or y+Sy = (x+Sx)3 = ,r3+3#2 . 8x+$x . (Sx)2+($x)3 (i) but y = x3 ........... (2)

Hence, by subtraction of (2) from (i) —

y+Sy-y = 3*2 . Sx+3x(Sx)2+(Sx)3 and Sy = $x2 . 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 = xs 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 = $x2.

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 — x3 when x = 4.

dy dx3 o

The slope of the curve = -v1 = -j— = 3*

CLX CLX

and if x = 4 -,- = 3 X 42 = 48.

Meaning that, in the neighbourhood of x — 4, the ordinate of the curve y = x3 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— =

dx5 -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 — dx3

- = 4^ = 4*

ft*

We note that in all these cases the results take the form —

dxn

— - = nxn ~ \

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 —

dxn

— -M Vtt ~~ *

- /(vV •

ax Let y = xn, 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(n—i) „ 0/c. , , «(w — i)(n — 2) M ,/» \9 ,

-•-=«^n-i4- v --- 'xn-2(8x)-\ — s — — 'xn-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 xn - * 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 axn.

It will be agreed that if the curve y = x3 had been plotted, the curve y = ^x3 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 = x3, provided that the same horizontal increments were considered.

Now the slope of the curve y = x3 is given by the equation —

dy - ix* dx~3X

so that the slope of the curve y = ^x3 is given by — ^ = 4x3*2 = I2x2.

In other words, the constant multiplier 4 remains a multiplier throughout differentiation. This being true for any constant factor —

j-axa = 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 = $x2 + 17 (a binomial, or two- term expression). The curves y = $x2 and y = $x2 + 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— jx(axn+bxn~'i+cxn-2+ . . . d)

= naxn~l+b(n — i)xn~2+c(n — 2)xn~3 + . . . and ^x(axa-{-b) = nax"-1.

To apply these rules to various numerical examples : —

Example 2. — Differentiate with respect to x the function —

= (gx i-6;r6)

= i4'4#-8+#^'5 or IA-AX-*-\ — — _

V x

Example 3.— If y = -Sx AA, 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 /w1'41 = 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 pv1-*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 pvl'tl.

Thus = -i-4ix/>w1-4lx»-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 R0 = 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 R0, 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 = -~

0a = R0a

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 — 1-84).

CIS \ S '

Write the expression as 45*— 3s~2+6s-5— 1-8*. Then—

2f ' n-l\

Example 7. — If x = an\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 an\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 — an — aT^ n — an — a n

+ l

, , / 2 n

dx d I •

— = —\an — 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— -45m9-86

rfm\ 5W

To avoid the quotient of functions of m, divide each term by 5m4'32,

T>7*w»75 .^C*M.^»36

then the expression =

and - (expression) = (3'4X — 3'57^~4'57) — -

~4'" — -499m4-54— 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 = x2.

Referring to Fig. 4, the equation of the curve OAB is y = x2, and the equation of the curve CG is yl — (x — h)2 = x2 -f h2 — 2xh.

Hence the difference between the ordinates of the curves OAB and CG, the latter being regarded as the base curve —

= 2xh—hz

so that the equation of the curve MN is y-5 = 2xh—h2.

DIFFERENTIATION OF FUNCTIONS 35

Now the curve ODE is the curve MN shifted back a distance of - horizontal units, and hence its equation is y3 = z(x-\ — jh — h2,

2 \ ^/

since for x we must now write f x-\—\

Thus the equation of curve ODE is —

y3 = 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 ~ x2 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 = x3.

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 — •

y2 = y— y^-x3— (x— h)3 = x3—

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

hz 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 = x3 ; but the ordinates must be measured to the old vertical scale divided by h.

We may now consider the case of the primitive y = xn. Adopt- ing the notation of the previous illustrations —

n(*L-ll)xn-2}l2_ . _ \

\

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

= nXn-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 pv1-3 = 570 and v = 28-1.

. 17. Find the value of

Jv

18. If E = — I5+I4T — -oo68T2, 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 -^)

pv1-3 = 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 xw(I+y)-w(y *> ,

2/ \ // 2

constants.

\Vv ('/'M

23. If M = — ^(/2 — 4y2), find the value of y that makes ,— = o.

2/2 v dy

n* Tt c* w((x-}-ny)2 x2} ,. , , , , 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 r3 * r3 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

I44T 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? dxv dx3 dx*

y =

Differentiating, = {(W X 2x) - (4l 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 l—x 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

/l—x\ w = force X distance = w(l—x)x( ) = ~(l—x)\

d2v 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 dx2 dx3

= 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 (l—x) 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^d3y\ dS

EFr4 = j-( El j-4 — j-

dx* dx\ dx3/ 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-1v0±1 {n i)J-0

-( ""^

^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 y2 = <\ax (Fig. 12).

yz = 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$x2—2x—g

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. 7dx

Now — y = i5x*—2X—g

dy

and -r- = 30*— 2.

dx

dx 2——

Hence sub-tangent — = y-r- =

-r- —

dy $ox — 2

dv

and sub-normal = y~~ = (i^xz—2x—g) x (30^—2).

ax

Thus when x = — 2

sub-tangent = - ~ — ? _ __g5 umf-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 y2 = <\ax; or taking the value of y as FB and that of x as DF, 1 22 = 40x7- 2, from which 40 = 20 and y2 = 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,

24

and — x 3-5 tons at A, i. e., RB = 3-06 and RA = -44 tons : thus the

24 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 y2 = 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

46

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 = ^x3 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 llxz xa\

y = -^rT( -=- ) {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 xa\ 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, x2 — iob2y + ^c, x having the value ga.

9. Prove that the sub-normal (along the axis of the parabola) of the parabola x2 — 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 e5x 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 = ex 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

ex = 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 —

dex _

~ f

x3

= ex. Another respect in which the function e? is unique may be

48 MATHEMATICS FOR ENGINEERS

dx i

noted : the sub-tangerit = y-,- -•- exx— = i, i. e., the sub-tangent

is constant and equal to unity.

The curve y = ex 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 ebx may be found.

Referring to Part I, p. 354, we note that if the curve y = ex be plotted, then this curve represents also the equation y = ebx if the numbers marked along the horizontal scale used for the curve y = ex are divided by b. If, then, the slope of the construction curve, i. e., that having the equation y = ex, is measured, we can obtain from it the slope of the curve y — ebx by multiplying the slope by b, since vertical distances are unaltered, whilst horizontal

distances in the case of y = ebx are T X corresponding horizontal

distances for y = ex.

Hence the slope of the curve y = ebx is b X slope of curve y = ex

r\pbx

or— -p- = bebx

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 ebx we can proceed to aebx, the result from the differentia- tion of which is given by —

daebx

dx

= abebx

Example 15. — If y = yrk*t find the value of -f~

dx

-/- = ^~5e~bx = 5 X — T&~^X dx dxj 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 = C0e L, where C and C0 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 ex and ebx are particular forms of the more general exponential function ax; to differentiate which we may proceed by either of two methods : —

(a) Working from first principles. In Part I, p. 470, the expansion for ax is given, viz. —

log a+(*log*)%(*joj^)*

50 MATHEMATICS FOR ENGINEERS

Differentiating term by term —

dax . , . n \2 , /, \sx2

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 ax da*

(b) Assuming ihe result for the differentiation of ebx —

Let— ax = ebx

so that a — eb, and therefore loge a = b.

d , d

Ta = J-*

ax ax

= logg axax or ax log a

Then— Tax = Tebx = bebx = loge ax ebx

,, dax

thus — -j— = ax . log a.

d. Example 17. — Find the value of ^p-

dX

In this case « = 4 and loge 4 = 1-3863.

dAx 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 = ^est-\-ye~5tt find the value of -^—255.

s =

= 2O05' — 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 ex. Considering these methods in turn —

(a) Working from first principles. Let y = log x, i. e., loge 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 2x2 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 — —— = ex

Qt%

Let — y = loge x, so that x — ey

dx dey

and — = - = ey.

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 ey x

dy

d loge 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 - Iog10 x = -4343 loge x

d logln x d loee x

and hence -$&- = .4343 _J3£L_ =

d . ,. -4343A

and -- Iog10 (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 = \oge jx, find ~. dy d .

-r**-T loge 7

dx dx _

or alternatively — loge 7* = loge 7+loge x and thus— loge 7* =

. i i

= o+- = -

X X

Example 21. — Differentiate with regard to t the expression Iog10 (5^—14) and find the numerical value of the derivative when / = 3-2.

When t = 3- 2

10£T 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, Iog10(5f— 14) = Ioglfl(i5'95— 14) = Iog10 1-95 = -2900 when t = 3-21, Iog10(5*— 14) = Iog10(i6-o5— 14) = Iog10 2-05 = -3118

so that —

8 Iog10 (5^—14) = -3118 — -2900 = -0218

while 8t = 3-21 — 3-19 = -02

S . , -0218

and lo <— * ==~~ = I>O9-

54 MATHEMATICS FOR ENGINEERS

Differentiation of the Hyperbolic Functions, sinh x and cosh x. — Expressing the hyperbolic functions in terms of exponential functions —

ex— erx

sinh x = and cosh x =

2

ex-\-e-x

2

Thus to differentiate sinh % we may differentiate d sinh x d (ex—e~x\ i, _

H£»r»^£» — I — _ / x*g I x> — £\

jncii^c j — 7 i ~ / — lu ~ i c? i

»A; «A;\ 2 / 2

= cosh x

d d(ex+e-x\ i. ,

also j- cosh « = j-( - - ) = -wf—g-'i

dx dx\ 2 / 2V

= 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 = --- — Cr -••*_— =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 ax, Log x and the Hyperbolic

Functions.

Differentiate with respect to # the functions in Nos. i to 20.

1. e~5*. 2. i-se4-1*. 3. -£. 4. 4-15*. 5. 8-72^.

&

. 7. ^xe. 8. 14x2*. 9. 4ie

foa-iX -jpl-ZX

10. 3-i4«»^-5*»-« + 3-ite+6. 11- °,VX^T*E- 12. log 7*. 13. 3 log (4— 5*). 14. iolog108#. 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 loglo 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- Iog10 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 C0 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 Iog10 (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 pt through a ratio of expansion r.

31. Find the value of -r~ + #y if y — e~ax.

ax ' * a a

56

32. If y = AeZx-\-'Be3x+Ce-*x> find the value of— d3y dzy dy

dzV Vyx x/n-*, -J.x

33. Evaluate -7-2 -- when V=A1ex^ r2 +A2e ^ r2

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 - (cf. part 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 —

jx 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 cos2 A;

Hence — =1 = IX- ~= *~ = sec2 x.

\ _ ,

3- tan x = sec2 x

60 MATHEMATICS FOR ENGINEERS

In like manner it can be proved that —

d cot x

— 3 — = — cosec2 x

d seex sinx dx ~ cos2 x

d cosec x cos x

and — -j~ = -- r-s —

dx sin2 x

To generalise —

^ A tan (Bx+C) = AB sec2 (Bx+C)

J^ A cot (Bx+C) = — AB cosec2 (Bx+C)

rf , ,_ _v AB cos (Bx+C)

a- A eosee (BX+C) = - .2

AB

. cos2 (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 = dt5'2 Sm (4°^— 2'4) = 5-2x40 cos (40^—2-4)

Hence when —

t = -07, the slope = 208 cos (2-8—2-4) = 208 cos -4 (radian)

= 208 cos 22-9° = 192.

Example 24. — Differentiate, with regard to z, the function 9-4 cot (7—5?).

,- 9'4 cot (7-5*) = 9'4 X -5 X - cosec2 (7-5?)

U»6

= 47 cosec2 (7—52).

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 = — tazxr cos wt

= — 0)2S

*'. 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>2x 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 = — «2s.

&>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 i1 ^~5). 5. 8 cot 5%.

\ 4 /

6. 43-15 sec (-05 — -117*). 7. be cos (d—gx).

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 (qx—c).

12. 5 sin2 x. 13. -065 cos2 3*.

14. cos2 (7*— i-5)+sin2 (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=-

8COS1

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 ** = ew, 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_=dyxdu

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~SuXSx

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~duxdwxdx 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 = eu.

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 = eu and ~ = ~ = eu = 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 = w3.

dy dy dv du u = cos

Then -~ = -~ x -j- X -j-

d log v du3 d cos 2.x ~~dv~X~duX dx

X — 2 sin

-j- — — 2 sin

V = U3

dv

-j- = 3U 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 = 4irf2, and thus the rate of change of surface

_ dS ~ dt

_ d . 4*rz dt

dr* = *v'^dt

dr* dr = *v-~drXdf

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 -Sirr3 j

<AV d dr3 dr3 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 IDS- 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 0 = - sin a, and sin a = T sin 0.

r I

66

MATHEMATICS FOR ENGINEERS

Again, cos « = Vi — sin2 a

If the connecting rod is long compared with the crank, -j is small

rz 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 sin2 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+r—l cos o— r cos 6

= l+r—l (i — £p sin2 0)-y cos 0

— y+~, sin2 0— Y cos 0

= r-\ — — dx d

i — cos 20

2

r2 cos 20

-r cos 0

fl

~ — r COS U.

r2 r2 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 0

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 . . 01 = 2trwy -j - j -- f-sm 0 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"|

= 4ir2n2»' { 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'dtXds

= 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 v2 — 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 dv2 d ds

Now ds=ds-^S = ^-ds = ^

dvz dvz 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.

Vds =

Exercises 6. — On the Differentiation of a Function of a Function.

T-" j A d sin 2x d . „ 0 d

Find 1. -3- . e . 2. -T- log v2. 3. ^2cos2 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 s5).

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(AV2+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+5y2 = i

show that T = T

dxz (.

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 x3— 6x2y— 6xyz+y3 — constant, prove that —

dy _ xz—4xy—r--* dx= 2X*~

7o 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+4y2 = 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

, , . , . wvz , , wa

energy, and kinetic energy == - , prove that force = — .

o o

dx

24. If x — 8 log (i2t3— 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 dxz . i

j- (log xxx2) 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^cr 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

- wdV^

.j dw

dx^

V dx

nnd(uv)

dw

dx

bUVdx

( 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 — xz . 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) + \ xz . - 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+~ (pxz) — 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 = x2.

d 9dp , dx2 9dp ,

Hence — -,- . px2 = x2,"-\-p-r- = x2-/-4-2Xp.

dx ' r

Hence — 2q-\ --- 3- . pxz — 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- = — = — —

>V9** sin (3*- 7) log (1-5*) = 9^ ' '

^

F rfw , dv . dw~] = g{ wv .,--\-wu, -\-uVj L dx dx dx-1

— 5x) sin (3^-7)4Ar3+{log (i — 5*)*4X3cos (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. x2 sin 3*. 2. log 5#X 2#3-4. 3. e9* Iog10 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\ogx3.53*. 9. e*i°g*. 10. ^*

11. 6*te+*(5*+2)«. 12. 7-2 tan ~ log *7.

o

13. If y = Ae3* 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^t2 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 Vdx udx

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

V2

(b) Using the rules for a product and a function of a function.

u

y = - = v

TM dy d , ,, ,du , dir1

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/ vz

= [5 cos (35+4) X (i252+7)]-[(4S3 + 7*) X -i5sin(35+4)] 25 cos2 (35+4)

5 cos2 (3^+4)

Example 12. — If y — 94*X, -- -, find the value of -£- Let

u = g4*, then j- = 4 x 94*loge 9 = 4 X 2-1972 x g*

= 8-789 X94z

and let

v = log 7*, then -=- = -*- = -.

du dv

dy d (u\ dx Hence -/- = T-V - ) = — — ,

dx dx\v I vz

— ui— dx

(log yx x 8 -79 x 94z) — (g4* X ^

(log 7#)2 94^{(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 1—0625)} ~"

V 1—0625 - "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*—rz)} 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 . 2g+T

whence J4Q7 = 2Q+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— sin2 y ~ Vi—x2

d »in 1 y 1

-?— sin jc — / — dx Vi y2

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

sec2 y = d tan y

i+tan* d (x\

i

a2

a2

d# d tan y

dx \a' d tan y

a

vrfy

d*

dy

a^

Hence

i a

sec2yx

S

tfy

i

i

a2

d# a sec2 y a a2 +#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 Xdx

hence

i

= sinh y X ^

a

•^ rf*

(i) Now cosh2 y— sinh2 y = i

Xs

whence sinh2 y = cosh2 y— i = -g— i

a*

and sinh y = ±- \/#2— a8

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*)

elx~5' cos (2— 7*)'

3. . ^X » _. O^f

. 5 sin — . Tf* cos — ji.

7 rf2

_ 52-to _ cosh 1-8^

&- g9^i' 4l-8a: -•

I *+3 7 cos-1 3^

7- VT'^W- 8' Vf^^2'

q ^o(a x)x ._ ^

a' 2(6-* cot B)* 1U> a2(a2+Ar2)i'

.. /3 — 6lzx -j- 1 2/^r2 — 7*3 (an expression occurring in the solution of "• 3^—4-^ a b63-111 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 0 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 0 is very small.

40 -it • sin 0 , 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

(p—q) tanO a0

tan 0 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\ ldzz )> 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) — (zyz x 3**) + 2oyexy

= loxy — 6 r2y2 + 2oyery

and yj = (loy x i) - (6y2 A 2x) + (2oy x ye*")

= loy — i zxy 2 + 2oyzexy.

82 MATHEMATICS FOR ENGINEERS

To find \-f-} and f j ^J x must be kept constant.

j

z = $xzy—'2x3yz-}-2oexy then (3^- J = (5#2 X i) — (2*3 x 2y) + 20* .

3-

and

Example 17. — If z — 6 log #y — i8x5y2, find the values of [-3 — •=-)

\CLX . ^Z^ '

/ rf2^ \

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,

/ dzz \ and this is the value of ( , — •, ).

\dx . dyl

XT (dz\ 6xx „, 6 - - ^T

Now I T- ) = - -- i8x5X2y = --- $6x5y = 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 = 49-5—49 = '5

and Sy = 1590 — 1600 = — 10.

fdK\ ,. .. d /vz \ vz

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 ~ 64r4~ 64-4

= 19880 — 24380 = —4500 units.

Example 19. — A quantity of water Q is measured by

If rl = 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 Partial derivatives. Find an expression for R.

ADDITIONAL RULES OF DIFFERENTIATION 85

Also —

I TrCD2 /—

= — 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 = V4 (-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 dy 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 r2. 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

A2

Xs

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- 0 + 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 = 2x3 + $x2— 36*4-15

then *£ = 6x*+6x-36

= 6(xz+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

d2v 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 xz.

d2y Find the value of -~^ or f"(x), as it may be written, and in

this expression substitute in turn the values x^ and xz 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 = xv

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. dx2

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-

d2v 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 fwlx3 wl2xz_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

dzy 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-^. dx2

AT w fix3 I2x2 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

d2y w r / 1 \ i lz \ 3^2-i and -,- -2 = ,-,f ( xzx )— ( XII — V

^2 El L\4 / \I2 / 6 J

— (**— l* _^2\ . El Va i2~ 2"]'

/72V ./v 72 «;2\

Now the bending moment M = EI.^=a;[-— -- - )

«*" \ 2 12 2 /

/^ /2 AT2

and M = o if ------ , i. e., 6lx—l2—6xz = o,

2 12 2

i. e., if 6xz— 6lx+lz = 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 = J£S .4X25 =

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 = 0 if 1- =

i. e., if i2 = 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 +/2z3/'

^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 —

v3 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 / v3 \

= Q = xU-5+-

V V* ' ' 2IOO/

i , UM

5V -\ )•

2IOO/

Differentiating with regard to the variable v dCt

Then

1050

.,

* a

v • (s . . IX

"4-5

- ' *J -

1050 v

or v3 = 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 = Kv2, where K is some constant, i. e., the effective head = 60 — Kv2 = He.

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 Hg

33000

= CvHe, where C is some constant = Gv{6o-Kv*) = C(6ov-Kv3)

Then / (H.P.) = C(6o-3Kz;2)

Ctl)

= C(6o — i8o + 3He)

or T— (H.P.) = o when 3HC = 120

i. e., He = 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 = Kbd3.

Both breadth and depth will vary, but they depend on each other; and from Fig. 25 we see that b2 = i6 — d2. 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, S2 will be a maximum when S is a maximum,* and hence we square both sides before differentiating.

Thus— S2 = K2d6(i6-d2) =

<fS2 and -, ~

FIG. 25.

'>-8d'>) = 8d5(i2-d2)

Hence — ,-r = o if d5 — o, i. e., d = o (giving zero stiffness)

Ct.(t

or if — i. e.,

Hence

d2 = 12 d = 3-464 ins.

b — V 16—12 = 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 y2 would be +169, +144, + 121, +100, +121, +144, so that if y = — 10 (its maximum value) when x = 4, say, then y2 — TOO when x = 4, and therefore a minimum value of y2 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 Q2 is a maximum, hence we shall find the value of b for which Q2 is a maximum.

b+2d~ '6+?A'

Also Q2 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 VJ -- MJ~ TT dW dr dv Hence -y = — , —

dr vz

(•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 yz = --- 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\ _ 2MRf °g \L— M/ ~ L8-M2

L2-M2 , /

* = - 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) = 2yU+(- X -

2TOI C

102 MATHEMATICS FOR ENGINEERS

c- -11 (dp\ 27T& C

Similarly \f) = -~- 2.

\dy)x V xyz

Both \j) an<^ vft mus^ ke equated to zero,

so that —^jr- — —

V x2y

i. e., x*y = — (i)

, 2TI& C

V ' xy2 i. 6. xyz = .... (2)

2TT&

To solve for x and y —

cV From (2) — x = — j-~2.

Substituting in (i) —

cV

cVa whence y3 = — r,

27T02

or

__2TU62

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* — -ix2.

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*-4i. 7. y = 5^-

8. y = x3 + 6x2— 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 = 4C2+',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 — MA = —» (I2— 4#2). For

what value of x is MA 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} wxz

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 = Vrz— xz— *(rz— x*), find the value of x that makes L a

yv

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?.

vz

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 Tx 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 — 27W2+2:rr/+-2y2, I being the length and r the radius,

find the value of the ratio - that makes the ratio exPosed 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#R2/ 1 \ / /3\

29. IfM = ^|^ -sin20)--934wR2^cos0- /s/|j, for what values

of 0 is M a maximum ? [M is the bending moment at