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Full text of "Mathematics for engineers"

MATHEMATICS FOR ENGINEERS 

PART II 



The Directly-Useful Technical Series 

Detailed Prospectus sent on Application. 

Mathematics for Engineers 

By W. N. ROSE, B.Sc. Eng. (Lond.) 

Part I. 

Demy 8vo. 510 pages. 257 Figures, with over 1200 set 
and worked examples. Price 9/6 net. 

Contains chapters on : Aids to Calculations ; Equations ; 
Mensuration ; Graphs ; Advanced Algebra ; Plane Trigono- 
metry ; Calculation of Earthwork Volumes ; Plotting of 
Difficult Curve Equations; Determination of Laws; Con- 
struction of Practical Charts, etc., etc. 

"The book teems with practical applications of mathematics to 
engineering problems ... an excellent book." Mechanical World. 

" A book which will be of great service to engineers of every class. 
To the young engineer it will be a God-send." Managing Engineer. 



The two volumes of " Mathematics for Engineers " form 
a most comprehensive and practical treatise on the subject, 
and will prove a valuable reference work embracing all the 
mathematics needed by engineers in their practice, and by 
students in all branches of engineering. 

CHAPMAN & HALL, LTD., LONDON. 

* 



The Directly-Useful 




Technical Series 



FOUNDED BY THE LATE WILFRID J. LINEHAM, B.Sc., M.Inst.C.E. 



Mathematics for Engineers 



PART II 



BY 



W. N. ROSE 

B.Sc. ENG. (LOND.) 

Late Lecturer in Engineering Mathematics at the 

University of London Goldsmiths' College 

Teacher of Mathematics, Borough 

Polytechnic Institute 




LONDON 
CHAPMAN & HALL, LTD. 

11 HENRIETTA STREET, W.C. 2 
1920 









PRINTED IN GREAT BRITAIN BY 
RICHARD CLAY AND SONS, LIMITED, 
BRUNSWICK STREET, STAMFORD STREET, S.E. 
AND BUNGAY, SUFFOLK. 



EDITORIAL NOTE 

THE DIRECTLY-USEFUL TECHNICAL SERIES requires a few words 
by way of introduction. Technical books of the past have arranged 
themselves largely under two sections : the Theoretical and the 
Practical. Theoretical books have been written more for the 
training of- college students than for the supply of information to 
men in practice, and have been greatly filled with problems of an 
academic character. Practical books have often sought the other 
extreme, omitting the scientific basis upon which all good practice 
is built, whether discernible or not. The present series is intended 
to occupy a midway position. ,The information, the problems and 
the exercises are to be of a directly-useful character, but must at 
the same time be wedded to that proper amount of scientific 
explanation which alone will satisfy the inquiring mind. We 
shall thus appeal to all technical people throughout the land, either 
students or those in actual practice. 



AUTHOR'S PREFACE 

CONCERNING the aim and scope of this work nothing need be added 
here to the statement made in the Preface to the former volume.. 
It is there asserted that the subject-matter has been so chosen, and, 
through the examples, so applied to practical problems, that the 
two volumes " embrace all the mathematical work needed by- 
engineers in their practice, and by students in all branches of 
engineering science." 

As with the first volume, much thought has been given to the 
elimination of all rules and processes of academic interest only ; 
but the fact of the importance and necessity of logical reasoning 
has not been overlooked. 

With the exception of the chapters on Spherical Trigonometry 
and Mathematical Probability, this volume is devoted to the study 
of the Calculus, both Differential and Integral. Whilst it is wise, 
and even imperative, if this subject is to be presented in an intelligible 
manner, that much attention should be paid to the graphic interpre- 
tation of its rules, care must be taken to ensure that the graphic 
methods do not become other than auxiliaries. Accordingly the 
treatment throughout is based upon algebraic principles ; but 
whenever graphic proofs or constructions have been found to amplify 
or explain the subject, they have been utilised to the fullest extent. 
Thus from the commencement the connection between the rate of 
change of a quantity and the slope of a curve is clearly demonstrated : 
and this correlation of the algebraic and the graphic methods is con- 
tinued through all the stages of the development of the subject. 

The conception of " limiting values," mentioned briefly in Part I, 
is further discussed in Chapter I, a familiar example from Dynamics 
being chosen as the illustration : in this chapter also two methods 
of graphic differentiation are given ; the second of which, and the 
less familiar, being the simpler to apply. 



viii AUTHOR'S PREFACE 

The various rules for the differentiation of both algebraic and 
trigonometric functions are explained in detail in Chapter II ; 
and Chapter III, containing the rules for the differentiation of a 
function of a function, a product of functions, etc., together with an 
introduction to partial differentiation, may be regarded as comple- 
mentary to Chapter II. 

From the abstract reasoning required for comprehension of such 
an idea as that of " limiting values/' the practical mind turns with 
relief to the applications of differentiation found in Chapter IV ; 
the determination of maximum and minimum values making a 
particularly strong appeal. In view of the importance of this 
branch of the subject, a very varied selection of practical examples 
is presented, in the choice of which the method of solution has 
been a determining factor. In this chapter also the use of Taylor's 
theorem in cases of interpolation from steam tables is demonstrated. 

Chapters V and VI contain the rules required for the integration 
of functions occurring in engineering theory and practice. The 
former chapter serves as an introduction to integration, the signifi- 
cance of the symbols f and dx being explained by reference to a 
graph ; whilst in the latter chapter the various types of integrals, 
many of them of a somewhat complicated character, are discussed. 
At this stage also the reduction formulae are introduced, and mention 
is made of the Gamma function and its uses. 

Instances of the application of the rules of integration are to be 
found in the processes enumerated in Chapter VII ; and special 
features of this chapter are the determination of the perimeter of 
the ellipse, the graphic method for fixing the position of the centroid 
vertical, the drawing of ist and 2nd moment curves and the evalua- 
tion of the moment of inertia of a compound vibrator. 

The utility of polar co-ordinates to the electrical engineer is shown 
by the inclusion of examples on the candle-powers of lamps, and the 
employment of the Rousseau diagram to find the mean spherical 
candle-power ; and Dr. Fleming's graphic method for determina- 
tion of root mean square values of currents is here inserted, since 
it involves polar plotting. 

Differential equations occur so frequently that the methods of 
solution demand most careful study. Chapter IX presents the most 
common types, and the selection of examples based upon these, 
both worked and set, emphasises the need for a proper appreciation 
of the method of solution. 

Chapter X, with its applications of the Calculus to problems 
encountered in the study of Thermodynamics, Strength of Materials, 



AUTHOR'S PREFACE ix 

Applied Mechanics, Applied Electricity and Hydraulics, provides 
further illustration of the need of a sound knowledge of the subject 
to the engineer desirous of equipping himself at all points. 

The last two chapters contain much of interest to the surveyor, 
the examples chosen being such as arise in his practice ; and par- 
ticular attention is directed to the investigation relating to the 
corrections following errors of observation. 

The Author greatly deplores the fact that the inspirer of this work, 
the late Mr. W. J. LINEHAM, B.Sc., M.I.C.E., does not see its com- 
pletion : to his enthusiasm for his ideals in education, and for his 
many personal kindnesses to the Author, tribute is here paid. 
Sincere thanks are also tendered to Messrs. J. L. BALE and C. B. 
CLAPHAM, B.Sc., for much valuable assistance. 

Great care has been taken to produce the book free from errors, 
but some may remain, notification of which will be esteemed a great 
favour. 

W. N. ROSE. 

Borough Polytechnic Institute, 

S.E. i, 
December, 1919. 



CONTENTS 

PAGE 

INTRODUCTORY i 

Abbreviations. 

CHAPTER *I 
INTRODUCTION TO DIFFERENTIATION 3 

Historical note Rates of change Average and actual rates of 
change Slopes of curves Graphic differentiation by two methods. 

CHAPTER II 
DIFFERENTIATION OF FUNCTIONS 26 

Differentiation of ax" Differentiation of a sum of terms Proof 
of construction for slope curves Beam problems Lengths of 
sub-tangents and sub-normals of curves Differentiation of ex- 
ponential functions Differentiation of log* Differentiation of 
sinh x and cosh x Differentiation of the trigonometric functions 
Simple harmonic motion. 

CHAPTER III 
ADDITIONAL RULES OF DIFFERENTIATION .... 63 

Differentiation of a function of a function Differentiation of a 
product Differentiation of a quotient Differentiation of inverse 
trigonometric functions Partial differentiation Total differential 
Logarithmic differentiation. 

CHAPTER IV 

APPLICATIONS OF DIFFERENTIATION ...... 88 

Maximum and minimum values Point of inflexion Calculation 
of small corrections Expansion of functions in series Theorems 
of Taylor and Maclaurin. 



xii CONTENTS 

CHAPTER V 

PAGE 

INTEGRATION 115 

Meaning of integration Graphic integration Application of in- 
tegration to " beam " problems Coradi integraph Rules for 
integration of simpler functions Integration of powers of x 
Integration of exponential functions Integration of trigonometric 
functions Indefinite and definite integrals Method of determining 
the values of definite integrals Proof of Simpson's rule. 

CHAPTER VI 
FURTHER METHODS OF INTEGRATION . . . . . . 146 

Integration by the aid of partial fractions Integration by the 
resolution of a product into a sum Integration by substitution 
Integration by parts Reduction formulae Gamma function 
List of integrals. 

CHAPTER VII 
MEAN VALUES, ETC. . . . . . . . . 180 

Determination of mean values Root mean square values 
Volumes Volumes of solids of revolution Length of arc Peri- 
meter of ellipse Area of surface of solid of revolution Centre of 
gravity Centroid " Double sum curve " method of finding the 
centroid vertical Centroids of sections by calculation Centroids 
found by algebraic integration Centre of gravity of irregular 
solids Centre of gravity of a solid of revolution Centre of pressure 
Moment of inertia Swing radius The parallel axis theorem 
Theorems of perpendicular axes Moment of inertia of compound 
vibrators Determination of ist and 2nd moments of sections by 
means of a graphic construction and the use of a planimeter. 

CHAPTER VIII 
POLAR CO-ORDINATES 257 

Polar co-ordinates Spirals Connection between rectangular and 
polar co-ordinates Use of polar co-ordinates for the determination 
of areas The Rousseau diagram Dr. Fleming's graphic method for 
the determination of R.M.S. values Theory of the Amsler plani- 
meter. 



CONTENTS xiii 

CHAPTER IX 



PAGE 



SIMPLE DIFFERENTIAL EQUATIONS . - 270 

Differential equations, definition and classification Types : ~- 

dy 
given as a function of x : -~ given as a function of y : General linear 

equations of the first order : Exact differential equations : Equa- 
tions homogeneous in x and y : Linear equations of the second order 
Use of the operator D Useful theorems involving the operator D 
Equations of the second degree. 

CHAPTER X 
APPLICATIONS OF THE CALCULUS 300 

Examples in Thermodynamics : Work done in the expansion of a 
gas Work done in a complete theoretical cycle Entropy of water 
- Efficiency of engine working on the Rankine cycle Efficiency of 
engine working on the Rankine cycle, with steam kept saturated by 
jacket steam Examples relating to loaded beams, simply supported 
or with fixed ends, the loading and the section varying Shearing 
stress in beams Examples on Applied Electricity Examples on 
Strengths of Materials Loaded struts Tension in belt Friction 
in a footstep bearing Schiele pivot Examples on Hydraulics 
Centre of Pressure Transition curve in surveying. 

CHAPTER XI 
HARMONIC ANALYSIS . .- . 342 

Fourier's theorem Analysis by calculation Harrison's graphic 
method of analysis Analysis by superposition. 

CHAPTER XII 
THE SOLUTION OF SPHERICAL TRIANGLES . . . 355 

Definition of terms Spherical triangle Solution of spherical 
triangles Solution of right-angled spherical triangles Napier's 
rules of circular parts The "ambiguous" case Applications in 
spherical astronomy Graphic solution of a spherical triangle. 

CHAPTER XIII 

MATHEMATICAL PROBABILITY AND THEOREM OF LEAST 
SQUARES 370 

Probability Exclusive events Probability of the happening 
together of two independent events Probability of error 
Theorem of least squares Error of the arithmetic mean Weight 
of an observation. 



xiv .CONTENTS 

PAGE 

ANSWERS TO EXERCISES . 387 

TABLES : 

Trigonometrical ratios 397 

Logarithms 398 

Antilogarithms 400 

Napierian logarithms 402 

Natural sines 404 

Natural cosines 406 

Natural tangents 408 

Logarithmic sines 410 

Logarithmic cosines , .412 

Logarithmic tangents . . . . . . . . . .414 

Exponential and hyperbolic functions 416 

INDEX ...... 417 



MATHEMATICS FOR ENGINEERS 



PART II 

INTRODUCTORY 

THE subject-matter of this volume presents greater difficulty 
than that of Part I. Many of the processes described herein de- 
pend upon rules explained and proved in the former volume ; 
and accordingly it is suggested that, before commencing to read 
this work, special attention should first be paid to Part I, pp. 452- 
460, 463-467, 469-472, and pp. 273-299 ; whilst a knowledge of the 
forms of the curves plotted in Chapter IX should certainly prove 
of great assistance. 

The abbreviations detailed below will be adopted throughout. 

-> stands for " approaches." 

" equals " or " is equal to." 
+ ,, ,, " plus." 

,, " minus." 
X " multiplied by." 

4- " divided by." 

" therefore." 
" plus or minus." 

,, " greater than." 

" less than." 

,, " circle." 

ce ,, ,, " circumference." 

,, " varies as." 

co ,, " infinity." 

L. ,, ,, " angle." 

A >, ,, triangle " or " area of triangle." 

Li_ or 4 ! " factorial four " ; the value being that of the 

product 1.2.3.4 or 2 4- 
"P, " the number of permutations of n things taken 

two at a time." 
"C 2 " the number of combinations of n things taken 

two at a time." 
n. 3 n (n i) (n 2). 

B 1 



2 MATHEMATICS FOR ENGINEERS 

t\ stands for " efficiency." 

" angle in degrees." 
6 ,, ,, " angle in radians." 
I.H.P. ,, " indicated horse-power." 
B.H.P. ,, " brake horse-power." 
m.p.h. ,, " miles per hour." 
r.p.m. ,, ,, " revolutions per minute." 
r.p.s. " revolutions per second." 
I.V. ,, ,, " independent variable." 
F. ,, " degrees Fahrenheit." 
C. ., ,, " degrees Centigrade." 
E.M.F. ,, ,, " electro-motive force." 

1 ,, .., " moment of inertia." 

E " Young's modulus of elasticity." 

S n ,, " the sum to n terms." 

S^ ,, " the sum to infinity (of terms)." 

2 ,, ,, " sum of." 

B.T.U. " Board of Trade unit." 

B.Th.U. " British thermal unit." 

T ,, ,, " absolute temperature." 

M ,, ,, " coefficient of friction." 

sin" 1 x ,, ,, " the angle whose sine is x." 

e ,, ,, " the base of Napierian logarithms." 

g ,, the acceleration due to the force of gravity." 

cms. ,, ,, " centimetres." 

grms. ,, ,, " grammes." 

-Ly ,, " the limit to which y approaches as x approaches 

the value a." 

C. of G. ,, ,, " centre of gravity." 

C. of P. ,, " centre of pressure." 

k ,, ,, " swing radius," or " radius of gyration." 

M.V. ,, " mean value." 

R.M.S. " root mean square." 

f'(x] ,, ,, " the first derivative of a function of x." 

f"( x ] ,, ,, " the second derivative of a function of x." 
dy 

-j- ,, ,, " the differential coefficient of y with regard to x." 

I ydx ,, ,, " the integral of y with respect to x as the I.V." 

8 ,,..," difference of." 

,, " the operation -3-.." 

" candle-power." 

M.S.C.P. ,, " mean spherical candle-power." 

p ,, ,, " density." 



CHAPTER I 
INTRODUCTION TO DIFFERENTIATION 

THE seventeenth century will ever be remarkable for the number 
of great mathematicians that it produced, and still more so for the 
magnitude of the research accomplished by them. In the early 
part of the century Napier and Briggs had introduced their systems 
of logarithms/ whilst Wallis and others directed their thoughts to 
the quadrature of curves, which they effected in some instances 
by expansion into series, although the Binomial Theorem was then 
unknown to them. In 1665 Newton, in his search for the method of 
quadrature, evolved what he termed to be a system of " fluxions " or 
flowing quantities : if x and y, say, were flowing quantities, then 
he denoted the velocity by which each of these fluents increased by 
x and y respectively. By the use of these new forms he was enabled 
to determine expressions for the tangents of curves, and also for 
their radii of curvature. At about the same time Leibnitz of Leipsic, 
also concerned with the same problem, arrived at practically the 
same system, although he obtained his tangents by determining 
" differences of numbers." To Leibnitz is due the introduction of 
the term " differential," and also the differential notation, viz., 
dx and dy for the differentials of x and y : he also in his expression 
for the summation of a number of quantities first wrote the symbol 

f, his first idea being to employ the word " omnia " or its abbre- 
viation " omn~." Thus, if summing a number of quantities like x, 
he first wrote " omnia x," which he contracted to " omn. x," and 

later he modified this form to fx. 

Great controversy raged for some time as to the claims of 
Newton and Leibnitz to be called the inventor of the system of the 
" Calculus," which is" the generic term for a classified collection of 
rules; but it is now generally conceded that the discoveries were 
independent, and were in fact the natural culmination of the 
research and discoveries of many minds. 

3 



MATHEMATICS FOR ENGINEERS 



The Calculus was further developed by Euler, Bernoulli, Legendre 
and many others, but until a very recent date it remained merely 
" a classified collection of rules" : its true meaning and the wide 
field of its application were for long obscured. 

Nowadays, however, a knowledge of the Calculus is regarded, 
particularly by the engineer, as a vital part of his mental equip- 
ment : its rules have been so modified as to become no serious tax 
on the memory, and the true significance of the processes has 
been presented in so clear a light that the study of the Calculus 
presents few difficulties even to the ultra-practical engineer. 

This revolution of thought has been brought about entirely 
through the efforts of men who, realising the vast potentialities 
of the Calculus, have reorganised the teaching of the subject : 
they have clothed it and made it a live thing. 

The Calculus may be divided into two sections, viz., those 
treating of differentiation and integration respectively. Differentia- 
tion, as the name suggests, is that part of the subject which is 
concerned with differences, or more strictly with the comparison 
of differences of two quantities. Thus the process of differentia- 
tion resolves itself into a calculation of rates of change; but the 
manner in which the rate of change is determined depends on the 
form in which the problem is stated. Thus, if the given quantities 
are expressed by the co-ordinates of a curve, the rate of change of 
the ordinate compared with the change in the abscissa for any 
particular value of the abscissa is measured by the slope of the 
curve at the point considered. 

Differentiation is really nothing more nor less than the deter- 
mination of rates of change or of slopes of curves. 

The term " rate of change " does not necessarily imply a " time 
rate of change," i. e., a rate of change with regard to time, such as 
the rate at which an electric current is changing per second, or the 
rate at which energy is being stored per minute; but the change 
in one quantity may be compared with the change in any other 
quantity. As an illustration of this fact we may discuss the 
following example 

The velocity of a moving body was measured at various distances 
from its starting point and the results were tabulated, thus 



s (distance in feet) . 


o 


5 


12 


v (velocity in feet per sec.) 


10 


M 


15 



INTRODUCTION TO DIFFERENTIATION 5 

To find the values of the " space rate of change of velocity " for 
the separate space intervals. 

Considering the displacement from o to 5 ft., the change in the 
velocity corresponding to this change of position is 1410, *. e., 
4 ft. per sec. 

change of velocity 1410 4 

Hence * .^ = - - = - = -8 

change of position 5 o 5 

or, the change of velocity per one foot change of position = -8 ft. 
per sec., and rate of change of velocity = -8 ft. per sec. per foot. 

Again, if s varies from o to 12, the change of v = 1510 = 5 ; 
or, the rate of change of velocity (for this period) = T 5 ^ ft. per 
sec. per foot. 

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

1514 i ,, , 

-^ - = - ft. per sec. per foot. 
12- 5 7 

The rates of change have thus been found by comparing differ- 
ences. The phrase " change of" occurs frequently in this investi- 
gation, and to avoid continually writing it a symbol is adopted in 
its place. The letter thus introduced is 8 (delta), the Greek form 
of d, the initial letter of the word " difference " : it must be regarded 
on all occasions as an abbreviation, and hence no operation must 
be performed upon it that could not be performed if the phrase for 
which S stands was written in full. In other words, the ordinary 
rules applying to algebraic quantities, such as multiplication, 
division, addition or subtraction, would be incorrectly used in 
conjunction with 8. 

Thus, mv (the formula for momentum) means m multiplied by v, 
or a mass multiplied by a velocity, whilst Sv represents " the change 
of v," or if v is the symbol for velocity, 8v change of velocity. 

Again, 8/ = change of time or change of temperature, as the 
case may be. Using this notation our previous statements can be 
written in the shorter forms : thus 

(1) As s changes from o to 5 Sv = 1410 = 4 

8s = 5-0 = 5 

H-* 

(2) As s changes from o to 12 Sv = 15 10 = 5 

8s = 12 o = 12 
and !_ y = A 

Ss 12 * * 



MATHEMATICS FOR ENGINEERS 

(3) As s changes from 5 to 12 

Sv i 
and ~=s s= !<: 



8v = 15 14 = i 
Ss = 12 5 = 7 



It must be noted that we do not cancel S from the numerator and 
denominator of the fraction ^-. 

The final result in (i), viz., ^- = -8, as s changes from o to 5, needs 

further qualification. From the information supplied we cannot 
say with truth that the change in the velocity for each foot from 
o to 5 ft. is -8 ft. per sec. : all that we know with certainty is that, 
as s changes from o to 5 ft., the average rate of change of velocity 
over this space period is -8 ft. per sec. Supposing the change of 

velocity to be continuous over the period considered, the value of ^ 

already obtained would be the actual rate of change of velocity at 
some point or points in the period considered. 

It is usual to tabulate the values of the original quantities and 
their changes, and unless anything is given to the contrary the 
average values of the rate of change are written in the middle of the 
respective periods. 

The table is set out thus 











Sv 


s 


V 


Ss 


Sv 


Ss 





IO 

















5 


4 


i = -8 


5 


14 















7 


i 


r = '143 


12 


15 


~~~" 


~~ 


_. 



To distinguish in writing between average and actual rates of 
change the notation employed is slightly modified, d being used in 

dv 
place of 8 ; -j- thus representing an actual rate of change of velocity, 

and ^7 representing an average rate of change of velocity. Once 

again it must be emphasised that d must be treated strictly in 
association with the v or t, as the case may be, and dt does not mean 

, , j dv v 
dxt, nor does -j- give . 






INTRODUCTION TO DIFFERENTIATION 



Another example can now be considered to demonstrate clearly 
the distinction between an average and an actual rate of change. 

For a body falling freely under the influence of gravity the 
values of the distances covered to the ends of the ist, 2nd and 3rd 
seconds of the motion are as in the table 



t (sees.) . 


o 


I 


2 


3 


s (feet) . 





16.1 


64.4 


144.9 



Find the average velocities during the various intervals of time, 
and also the actual velocities at the ends of the ist, 2nd and 3rd 
seconds respectively. 

The average velocities are found in the manner described before, 
i. e., by the comparison of differences of space and time, and the 
results are tabulated, thus 



' 


' 


5s 


st 


v- Ss 


o 




















16-1 


i 


16-1 


I 


16-1 

















48-3 


i 


48-3 


2 


64-4 
















80-5 


i 


80-5 


3 


144-9 


~~~"~ 


~ 





The average velocities, viz., the values in the last column, are 
written in the lines between the values of the time to signify that 
they are the averages for the particular intervals. As also it is 
known that in this case the velocity is increased at a uniform rate, 
it is perfectly correct to state that the actual velocities at the ends 
of -5, 1-5 and 2-5 seconds respectively are given by the average 
velocities over the three periods and are 16-1, 48-3 and 80-5 ft. 
per sec. 

We have thus found the actual velocities at the half seconds, 
but not those at the ends of the ist, 2nd and 3rd seconds. The 
determination of these velocities introduces a most important 
process, illustrating well the elements of differentiation, and in 
consequence the investigation is discussed in great detail. . 

The student of Dynamics knows that -the law connecting space 
and time, in the case of a falling body, is s = \gt z = i6-i/ 2 , and 



8 MATHEMATICS FOR ENGINEERS 

a glance at the table of values of s and t confirms this law ; thus, 
when t = 2, s = 64-4, which = i6-iX2 2 or i6-i^ 2 . 

To find the actual velocity at the end of the first second we 
must calculate the average velocities over small intervals of time 
in the neighbourhood of I sec., and see to what figure these velocities 
approach as the interval of time is taken smaller and smaller. 

Thus if t i s = 16-1 x i 2 = 16-1 

t = i-i s = 16-1 x i-i 2 = 19-481 

8s = 19-481 16-1 = 3-381 U = i-i i = -i 

\ &s 3-381 
and (average) v = ^ = ^- = 33-81 

i. e., the average velocity over the interval of time i to i-i sec. 
is 33-81 ft. per sec. This value must be somewhere near the velocity 
at the end of the first second, but it cannot be the absolute value, 
since even in the short interval of time, viz., -i sec., the velocity has 
been increased by a measurable amount. A better approximation 
will evidently be found if the time interval is narrowed to -01 sec. 

Then t i s = 16-1 

t = i-oi s = 16-1 X i-oi 2 = 16-42361 

Ss = -32361 St = -oi 

8s -32361 
(average) v = ^ = -^~ = 32-361 

A value still nearer to the true will be obtained if the time 
interval is made -ooi sec. only. 

t = i s = 16-1 

t = i-ooi s = 16-1 X i-ooi 2 = 16-1322161 

8s = -0322161 8^ = -ooi 

8s -0322161 
and (average) v = ^ = ~^^ = 32-2161 

By taking still smaller intervals of time, more and more nearly 
correct approximations would be found for the velocity ; the values 
of v all tending to 32-2, and thus we are quite justified in saying that 
when / = i, v = 32-2 ft. per sec. 

Or, using the language of p. 458 (Mathematics for Engineers, Pt. 1), 
we state that the limiting value of v as t approaches i is 32-2 ; a 
result expressed in the shorter form 

(average) v - 32-2 as 8/ -> o when t = i 
where the symbol -> means " approaches " 

but (average) v = , and thus ~ -> 32-2 as St -> o when t = i 

ot ot 



INTRODUCTION TO DIFFERENTIATION 



Again, an actual velocity is an average velocity over an extremely 
small interval of time ; or, in other words, an actual velocity is the 
limiting value of an average velocity, so that . 



(actual) v = (average) v 



i.e., 



ds 
dt 



st-^-o 



8t 



By similar reasoning it could be proved that the actual velocity 
at the end of the 2nd second was 64-4 ft. per sec., and at the end 
of the 3rd second the velocity was 96-6 ft. per sec. 

This example may usefully be continued a step further, by 
calculating the values of the acceleration; this being now possible 
since the velocities are known. 

Tabulating as before 



1 


V 


Sv 


Si 


&v 

a = sl 


I 


32-2 

















32-2 


I 


32-2 


2 


64-4 

















32-2 


I 


3 2-2 


3 


96-6 












and we note that the average acceleration is constant and is thus 
the actual acceleration. 

Our results may now be grouped together in one table, in which 
some new symbols are introduced, for the following reason. A 
velocity is the rate of change of displacement, and is found by 
" differentiating space with regard to time," and an acceleration 
is the rate of change of velocity, and hence it is a rate of change 
of the change of position, and so implies a double differentiation. 

Thus whilst -T- is called the first derivative or differential coefficient 

dl) 

of s with regard to t, -^ is the first derivative of v with regard to t 

and the second derivative of s with regard to t. 

TU ds , dv dfds\ ,,.,,, , n 

Then v = -, , and = TT = -T, -57 , this last form being usually 
at at at\atj 

d 2 s 
written as ^ (spoken as d two s, dt squared) ; and it denotes that 

the operation of differentiating has been performed twice upon s. 



10 



MATHEMATICS FOR ENGINEERS 



The complete table of the values of the velocity and the accelera- 
tion reads 



s 


i 


Ss 


st 


_Ss 


--GO 


st 


Sv S*s 
~ St ~ SP 


O 


o 








' 

















16-1 


i 


16-1 











16-1 


i 











32-2 


i 


32-2 








48*3 


i 


48*3 











64-4 


2 











32-2 


i 


32-2 







80-5 


i 


80-5 











144-9 


3 








_ 







~~~ ' 



The next example refers to a similar case, but is treated from 
the graphical aspect. 

Example i. Experiments made with the rolling of a ball down an 
inclined plane gave the following results 



t (sees.) 


o 


I 


2 


3 


s (cms.) 


o 


20 


80 


180 



Draw curves giving the space, velocity and acceleration respectively 
at any time during the period o to 3 sees. 

By plotting the given values, 5 vertically and t horizontally, the 
" space-time " curve or " displacement " curve is obtained; the curve 
being a parabola (Fig. i). 

Select any two points P and Q on the curve, not too far apart, 
and draw the chord PQ, the vertical QN and the horizontal PN. 

Then the slope of the chord PQ = 

Now PN may be written as St since it represents a small addition 
to the value of t at P : also QN = 8s, 

8s 
so that slope of chord PQ = ^ 

but ^ average velocity between the times OM and OR, hence 

Of 

the average velocity is measured by the slope of a chord. Now 
let Q approach P, then the chord PQ tends more and more to lie 
along the tangent at P, and by taking Q extremely close to P the 
chord PQ and the tangent at P are practically indistinguishable 



INTRODUCTION TO DIFFERENTIATION 



ir 



the one from the other; whilst in the limit the two lines coincide. 

go 

Then since the slope of the chord PQ gives the value of ~ , and 

ot 

the limiting value of 7 is -^ , it follows that the slope of the tangent 

ot (it 

expresses -^ ; but the slope of a curve at any point is measured 




.30 



1-5 2-O 

Values of hmc 
FIG. i. 



2-5 



by the slope of its tangent at that point, and hence we have evolved 
the most important principle, viz., that differentiation is the deter- 
mination of the slopes of curves. 

[Incidentally it may be remarked that here is a good illustration 
of the work on limiting values ; for the slope of a curve, or of the 
tangent to the curve, is the limiting value of the slope of the chord, 
*'. e., the value found when the extremities of the chord coincide; 

and this value does not take the indeterminate form C) , as might at 

o 

first sight be supposed, but is a definite figure.] 



12 MATHEMATICS FOR ENGINEERS 

Thus the slope of the tangent at any point on the space-time 
curve measures the actual rate of change of the space with regard 
to time at that particular instant; or, in other words, the actual 
velocity at that instant. Hence by drawing tangents to the space- 
time curve at various points and calculating the slopes, a set of 
values of the velocity is obtained : these values are then plotted 
to a base of time and a new curve is drawn, which gives by its 
ordinate the value of the velocity at any time and is known as the 
" velocity-time " curve. 

Since this curve is obtained by the calculation of slopes, or 
rates of change, it is designated a derived or slope curve; the 
original curve, viz., the space-time curve, being termed the 
primitive. 

In the case under notice the velocity-time curve is a sloping 
straight line, and in consequence its slope is constant, having the 
value 40. Hence the derived curve, which is the acceleration-time 
curve, is a horizontal line, to which the ordinate is 40. There are 
thus the three curves, viz., the primitive or space-time curve, the 
first derived curve or the velocity-time curve, and the second derived 
curve or the acceleration-time curve. 

Graphic Differentiation. The accurate construction of slope 
curves is a most tedious business, for the process already described 
necessitates the drawing of a great number of tangents, the calcu- 
lations of their respective slopes and the plotting of these values. 
There are, however, two modes of graphic differentiation, both of 
which give results very nearly correct provided that reasonable 
care is taken over their use. 

Method i (see Fig. 2). Divide the base into small elements, 
the lengths of the elements not being necessarily alike, but being 
so chosen that the parts of the curve joining the tops of the con- 
secutive ordinates drawn through the points of section of the base 
are, as nearly as possible, straight lines. Thus, when the slope 
of the primitive is changing rapidly, the ordinates must be close 
together; and when the curve is straight for a good length, the 
ordinates may be placed well apart. Choose a pole P, to the left 
of some vertical OA, the distance OP being made a round number 
of units, according to the horizontal scale. Erect the mid-ordinates 
for all the strips. 

Through P draw PA parallel to ab, the first portion of the 
curve, and draw the horizontal Ac to meet the mid-ordinate of the 
first strip in c. Then dc measures, to some scale, the slope of the 
chord ab, and therefore the slope of the tangent to the primitive 



INTRODUCTION TO DIFFERENTIATION 13 

curve at m, or the average slope of the primitive from a to b, with 
reasonable accuracy. 

Continue the process by drawing PM parallel to bl and Ms 
horizontal to meet the mid-ordinate of the second strip in s : then 
cs is a portion of the slope or derived curve. 

Repeat the operations for ah 1 the strips and draw the smooth 
curve through the points c, s, etc. : then this curve is the curve 
of slopes. 




FIG. 2. Graphic Differentiation, Method i. 

Indicate a scale of slope along a convenient vertical axis and 
the diagram is complete : the scale of slope being the old vertical scale 
divided by the polar distance expressed in terms of the horizontal units. 

E. g., if the original vertical scale is i" = 40 ft. Ibs. and the 
horizontal scale is i" = 10 ft. : then, if the polar distance p is taken 
as 2", i. e., as 20 horizontal units, 

the new vertical scale, or scale of slope, is i" = ^ ' 2 Ibs. 

20 ft. 

Proof of the construction. 

The slope of the primitive curve at m = slope of curve ab 

= bf = OA. = cd 
~af~ p " p 



14 MATHEMATICS FOR ENGINEERS 

or, the ordinate dc, measured to the old scale, = p x the slope of 
the curve at m. 

If, then, the original vertical scale is divided by p the ordinate 
dc, measured to the new scale, = slope of the curve at m. 

The great disadvantage of this method is that parallels have 
to be drawn to very small lengths of line and a slight error in the 
setting of the set square may quite easily be magnified in the draw- 
ing of the parallel. Hence, for accuracy, extreme care in draughts- 
manship is necessary. 

It should be observed that this method of graphic differentiation 
is the converse of the method of graphic integration described in 




FIG. 3. Graphic Differentiation, Method 2. 



Chapter VII (Part I), and referred to in greater detail in Chapter V 
of the present volume. 

Method 2. Let ABC (Fig. 3) be the primitive curve. 

Shift the curve ABC forward to the right a horizontal distance 
sufficiently large to give a well-defined difference between the 
curves DEF and ABC ; but the horizontal distance, denoted by h, 
must not be great. From the straight line base OX set up ordinates 
which give the differences between the ordinates of the curves 
ABC and DEF, the latter curve being treated as the base : thus 
ab = a'b'. Join the tops of the ordinates so obtained to give the 
new curve G6H, and shift the curve G6H to the left a horizontal 

distance = -, this operation giving the curve MPN, which is the 

true slope or derived curve of the primitive ABC. Complete the 
diagram by adding a scale of slope, which is the old vertical scale 
divided by h (expressed in horizontal units). 



INTRODUCTION TO DIFFERENTIATION 



This method can be still further simplified by the use of tracing 
paper, thus : Place the tracing paper over the diagram and trace 
the curve ABC upon it; move the tracing paper very carefully 
forward the requisite amount, viz., h, and with the dividers take 
the various differences between the curves, such as a'b'. Step off 

these differences from OX as base, but along ordinates - units re- 

45 

moved to the left of those on which the differences were actually 
measured : then draw the curve through the points and this is 
the slope curve. 



Tamp (C*) 




JO 4O SO 6O O 

h" 20 H "Tung (mins.) 

FIG. 30. Variation of Temperature of Motor Field Coils. 

Examples on the use of these two methods now follow. 

Example 2. The temperature of the field coils of a motor was 
measured at various times during the passage of a strong current, with 
the following results 



Time (mins.) . 


o 


5 


10 


15 


20 


25 


3 


35 


40 


45 


50 


55 


60 


65 


Temperature (C.) 


20 


26 


32-5 


4i 


46 


49 


52-5 


54-5 


56-5 


58 


59-5 


61 


61-7 


62 



Draw a curve to represent this variation of temperature, and a 
curve to show the rate at which the temperature is rising at any instant 
during the period of 65 mins. 

The values of the temperature when plotted to a base of time give 
the primitive curve in Fig. 30. 



i6 



MATHEMATICS FOR ENGINEERS 



To draw the slope curve we first divide the base in such a way that 
the portions of the curve between consecutive ordinates have the same 
inclination for the whole of their lengths, i. e., the elements of the curve 
are approximately straight lines. Thus, in the figure, there is no 
appreciable change of slope between A and a, or a and d. There is 
no need to draw the ordinates through the points of section for their 
full lengths, since the intersections with the primitive curve are all that 
is required. Next a pole P is chosen, 20 horizontal units to the left 
of A, and through P the line PB is drawn parallel to the portion of the 
curve Aa. A horizontal B6 cuts the mid-ordinate of the first strip 
at b, and 6 is a point on the slope or derived curve. The processes 
repeated for the second strip, PC being drawn parallel to ad and Cc 
drawn horizontal to meet the mid-ordinate of the second strip in c, 
which is thus a second point on the slope curve. A smooth curve 
through points such as b and c is the slope curve, giving by its ordinates 
the rate of increase of the temperature ; and it will be observed that 
the rate of increase is diminished until at the end of 65 mins. the rate 
of change of temperature is zero, thus indicating that at the end of 
65 mins. the losses due to radiation just begin to balance the heating 
effect of the current. 

Since the polar distance = 20 units, the scale of slope 
_ original vertical scale 

20 

and in the figure the original vertical scale is i" = 20 units ; hence the 
scale of slope is i" = i unit; and this scale is indicated to the right of 
the diagram. 

Example 3. Plot the curve y = x"-, x ranging from o to 3, and use 
Method 2 to obtain the derived curve. 

The values for the ordinates of the primitive curve y x"- are as 
in the table 



X 


o 


I 


2 


3 


y 


o 


I 


4 


9 



and the plotting of these gives the curve OAB in Fig. 4. 

Choosing h as -5 horizontal unit, the curve is first shifted forward 
this amount, and the curve CG results. The vertical differences be- 
tween these two curves are measured, CG being regarded as the base 
curve, and are then set off from the axis of x as the base. Thus when 
x 3, the ordinate of the curve OAB is 9 units, and that of CG is 6-25, 
so that the difference is 2-75, and this is the ordinate of the curve MN. 

By shifting the curve MN to the left by a distance = , i. e., -25 hori- 
zontal unit, the true slope curve ODE is obtained : this is a straight 



INTRODUCTION TO DIFFERENTIATION 



17 



line, as would be expected since the primitive curve is a " square " 
parabola. 

As regards the scale of slope, the new vertical scale 

_ old vertical scale, 



and since h = -5, the new vertical scale, or scale of slope, which is used 
when measuring ordinates of the curve ODE, is twice the original 
vertical scale. 

The derived curve supplies much information about the primi- 
tive. Thus, when the ordinate of the derived curve is zero, i. e., 
when the derived curve touches or cuts the horizontal axis, the 




O -5 / 1-5 2. 2-5 

FIG. 4. Graphic Differentiation. 

slope of the primitive is zero; but if the slope is zero the curve 
must be horizontal, since it neither rises nor falls, and this is the 
case at a turning point, either maximum or minimum. Hence turning 
points on the primitive curve are at once indicated by zero ordinates 
of the slope curve. 

Again, a positive ordinate of the derived curve implies a positive 
slope of the primitive, and thus indicates that in the neighbourhood 
considered the ordinate increases with increase of abscissa. Also 
a large ordinate of the slope curve indicates rapid change of ordinate 
of the primitive with regard to the abscissa. 

This last fact suggests another and a more important one. By 
a careful examination of the primitive curve we see what is actually 
c 



iS 



MATHEMATICS FOR ENGINEERS 



happening, whilst the slope curve carries us further and tells us 
what is likely to happen. In fact, the rate at which a quantity is 
changing is very often of far greater importance than the actual 
value of the quantity; and as illustrations of this statement the 
following examples present the case clearly. 

Example 4. The following table gives the values of the displacement 
of a 21 knot battleship and the weight of the offensive and defensive 
factors, viz., armament, armour and protection. From these figures 



(2000 _ 




TOGO 



2OOOO 



84OOO 60OO E8OOO 3OOOO 
Values of P 



FIG. 5. Displacement and Armament of Battleship. 

calculate values of Q (ratio of armament, etc., to displacement) and q 
(rate of increase of armament, etc., with regard to displacement). 

Find also the values of .. 



Displacement \ 
P tons J 


18000 


2000O 


22OOO 


24000 


2600O 


28000 


30000 


Armament, etc., \ 
p tons / 


6880 


7850 


88 3 


9820 


10810 


11820 


12845 



INTRODUCTION TO DIFFERENTIATION 
The values of Q are found by direct division and are 



p 


18000 


20000 


22OOO 


24000 


26OOO 


28OOO 


3OOOO 


Q 


383 


392 


40! 


409 


416 


422 


428 



dp 
Values of q, i. e., -,, may be found by (a) construction of a slope 

curve, or (6) tabulating differences. 

(a) By construction of a slope curve. Plotting p along the vertical 
axis and P along the horizontal axis (see Fig. 5), we find that the points 
lie very nearly on a straight line. Hence the slope curve is a hori- 
zontal line, whose ordinate everywhere is the slope of the original 
line. By actual measurement the slope is found to be -498 and thus 

498. This 



di) 
the average value for ~~, over the range considered, is 



average value of the rate of change does not, however, give as much 
information for our immediate purpose as the separate rates of change 
considered over the various small increases in the displacement. 
(6) By tabulation of differences, as in previous examples 



p 


P 


sp 


SP 


, Sp 
? SP 


18000 


6880 


_ 


_ 


_ 








97 


2OOO 


485 


2OOOO 


7850 

















980 


2OOO 


.490 


22OOO 


8830 

















990 


2OOO 


495 


24000 


9820 




'. 








990 


2OOO 


495 


26OOO 


I08IO 
















1010 


2OOO .505 


28OOO 


II820 
















1025 


2OOO 


5125 


3OOOO 


12845 












JV * 

Now sfs = rate of increase of armament compared with displace- 

oJr 

ment ; as the displacement increases it is seen from the table of values 
that this ratio increases, and the questions then arise : " Does this 
increase coincide with an increase or a decrease in the values of Q, 
and if with one of these, what is the relation between the two changes ? " 
By tabulating the corresponding values of q and Q and calculating 

the values of i, we obtain the following table (the values of Q at 19000, 
21000, etc., being found from a separate plotting not shown here) 



20 



MATHEMATICS FOR ENGINEERS 



p 


19000 


21000 


23000 


25000 


27000 


29OOO 


p 


485 


490 


495 


495 


505 


5125 


Q 


387 


396 


405 


.412 


419 


426 


q 

g 


1-253 


1-236 


1-223 


1-202 


1-203 


I-2O4 



It will be seen by examination of this table that the fraction Q 

decreases as ships are made larger : in other words, while the arma- 
ment increases with the displacement, the increase is not so great as 
it should be for the size of the ship, since the weight of the necessary 
engines, etc., is greater in proportion to the weight of armament and 
protection for the larger than for the smaller ships. 

Thus, other things being equal, beyond a certain point it is better 
to rely on a greater number of smaller ships than a few very large 
ones. 

Example 5. Friend gives the following figures as the results of 
tests on iron plates exposed to the action of air and water. The 
original plates weighed about 2-5 to 3 grms. 

Plot these figures and obtain the rate curves for the two cases, 
these curves being a measure of the corrosion : comment on the results. 



Time in days . ' . 


2 


7 


13 


19 


26 


32 


37 


In the light : loss of \ 
weight in grms. . J 


0048 


031 


0645 


08 


093 


126 





In the dark : loss of) 
weight in grms. . / 


0032 


0208 


037 


058 


0674 


0816 


0916 



The two sets of values are plotted in Fig. 6, the respective curves 
being LLL for the plates exposed in the light, and DDD for those 
left in the dark. The effect of the action of light is very apparent 
from an examination of these curves. Next, the slope curves for the 
two cases are drawn, Method 2 being employed, but the intermediate 
steps are not shown. The curve /// is the slope curve for LLL, and 
ddd that for the curve DDD. 

It will be observed that in both cases the rate of loss is great at 
the commencement, but decreases to a minimum value after 20 days 
exposure in the case of t the curve ///, and after 25 days in the case of 
the curve ddd. 

After these turning points have been reached the rate quickens, 



INTRODUCTION TO DIFFERENTIATION 



21 



the effect being very marked for the plates exposed to the light ; and 
for these conditions the slope curve /// suggests that the corrosive 
action is a very serious matter, since it appears that the rate of loss 
must steadily increase. 

A further extremely good illustration of the value of slope curves 
is found in connection with the cooling curves of metals. In the 
early days of the research in this branch of science, the cooling 
curve alone was plotted, viz., temperatures plotted to a base of 




o s 10 15 O 25 30 

FIG. 6. Tests on Corrosion of Iron Plate. 

time. Later investigations, however, have shown that three other 
curves are necessary, viz., an inverse rate curve, a difference curve 
and a derived differential curve ; the co-ordinates for the respective 
curves being 

(a) Temperature (0) time (f) curve; t horizontal and vertical. 

(b) Inverse rate curve : -, horizontal and vertical. To obtain 

this curve from curve (a), the slopes must be very carefully calcu- 
lated, and it must be remembered that these slopes are the measures 
of the inclinations to the vertical axis and not to the horizontal, 

dt ^ ft 

i. e., are values of -^ and not 



,,. 
at 






22 



MATHEMATICS FOR ENGINEERS 



(c) vertical and X horizontal : 00! being the difference 
of temperature between the sample and a neutral body cooling under 
identical conditions. 

(d) vertical and JA horizontal : this curve thus being 
the inverse rate curve of curve (c). 

Exercises 1. On Rates of Change and Derived Curves 

1. What do the fractions ^- and -,- actually represent (s being a 

r V (f( 

displacement, and t a time) ? Take some figures to illustrate your 
answer. 

2. Further explain the meanings of -~- and -r by reference to a 

graph. 

3. When an armature revolves in a magnetic field the E.M.F. 
produced depends on the rate at which the lines of force are being 
cut. Express this statement in a very brief form. 

4. For a non-steady electric current the voltage V is equal to the 
resistance R multiplied by the current C plus the self-inductance L 
multiplied by the time rate at which the current is changing. Express 
this in the form of an equation. 

5. At a certain instant a body is 45-3 cms. distant from a fixed 
point. 2-14 seconds afterwards it is 21-7 cms. from this point. Find 
the average velocity during this movement. At what instant would 
your result probably measure the actual velocity ? 

6. At 3 ft. from one end of a beam the bending moment is 5 tons ft. 
At 3' 2\" from the same end it is 5-07 tons ft. If the shear is measured 
by the rate of change of bending moment, what is the average shearing 
force in this neighbourhood ? 

7. Tabulate the values of q, i. e., ^ for the following case, the figures 
referring to a battleship of 23 knots. 



p 


18000 


2OOOO 


22OOO 


24OOO 


26OOO 


28OOO 


3OOOO 


32OOO 


p 


6170 


7080 


8OOO 


8930 


9890 


10855 


II820 


I28lO 



8. Tabulate the values of ^r for the case of a battleship of 25 knots 
from the following 



P 


18000 


2OOOO 


22OOO 


24000 


26OOO 


28OOO 


3OOOO 


32OOO 


p 


5210 


6050 


6910 


7790 


8660 


9550 


10460 


II370 



= sp and Q = F 



INTRODUCTION TO DIFFERENTIATION 



9. Tabulate the values of the velocity and the acceleration for 
the following case 



Space (feet) 


i 


2-4 


4'4 


6 


7-6 


II-2 


15-6 


2O'4 


Time (sees.) 


2 


4 


6 


7 


8 


I 


1-2 


i-4 



10. Plot the space-time curve for the figures given in Question 9 
and by graphic differentiation obtain the velocity -time and the accelera- 
tion-time curves. 

11. Plot the curve y -$x 3 from x = 2 to x = +4 and also its 
derived curve. What is the ordinate of the latter when x 1-94? 

12. Given the following figures for the mean temperatures of the 
year (the average for 50 years), draw a curve for the rate of change of 
temperature and determine at what seasons of the year it is most 
rapid in either direction. 



Time (intervals of J month) 


o 


i 


2 


3 


4 


5 


6 


7 


8 9 


10 


ii 


Temperature 




38-6 


37'9 


38-4 


39-8 


38-5 


39-5 


4-3 


40-7 


4 I- 5 


45-5 


45-5 


48-5 






12 

49-3 


13 


14 


15 


16 


I? 


IS 


19 


20 


21 


22 


23 


24 


25 


26 


52 


55 


57-2 


58-4 


60-5 


61-4 


62-5 


62-9 


62-2 


62-5 


61-1 


59-8 


58-2 


55-8 




27 

54-2 


28 
51 


29 
48-8 


30 
46-8 


3* 

43-5 


32 
42-1 


33 
40-6 


34 
39-8 


35 
38-8 


36 
38-6 



13. s is the displacement from a fixed point of a tramcar, in time 
/ sees. Draw the space-time, velocity-time and acceleration-time curves. 



t 





.1 2 


3 


4 


5 


6 


7 


8 


9 


10 


s 


o 


4 ii 


21 


34 


50 


69 


9i 


116 


144 


175 



The scales must be clearly indicated. 

14. The table gives the temperature of a body at time / sees, after 
it has been left to cool. Plot the given values and thence by differ- 
entiation obtain the rate of cooling curve. What conclusions do you 
draw from your final curve ? 



Time (mins.) 


o 


I 


2 


3 


4 


5 


6 


7 


8 


9 


Temp. (F.) 


136 


134 


132 


130 


128 


126-5 


124-8 


123-3 


122 


120-5 





10 


ii 


12 


13 


14 


15 


16 


17 


18 


19 


119-3 


118 


II6-8 


II5-5 


114-5 


II3-5 


112-5 


111-5 


110-5 


109-5 



MATHEMATICS FOR ENGINEERS 



15. The following figures give the bending moment at various 
points along a beam supported at both ends and loaded uniformly. 
Draw the bending moment curve, and by graphic differentiation obtain 
the shear and load curves. Indicate clearly the scales and write down 
the value of the load per foot run. 



Distance from one \ 
end (ft.) . . / 





2 


4 


6 


8 


10 


12 


14 


16 


18 


20 


Bending moment \ 
(tons ft.) . . J 


o 


3'5 


6-3 


8-4 


9-6 


10 


9-6 


8-4 


6-3 


3'5 


O 



16. By taking values of in the neighbourhood of 15 find the 
actual rate of change of sin 6 with regard to (0 being expressed in 
radians). Compare your result with the value of cos 15. In what 
general way could the result be expressed ? 

17. If the shear at various points in the length of a beam is as in 
the table, draw the load curve (i. e., the derived curve) and write down 
the loading at "3^ ft. from the left-hand end. 



Distance from left-hand end (ft.) 





I 


2 


3 


4 


5 


6 


Shearing force (tons) .... 





i 


3 


6 


i 


i'5 


2-1 



18. An E.M.F. wave is given by the equation 

E = 150 sin 314^ + 50 sin 942^. 

Derive graphically the wave form of the current which the E.M.F. will 
send through a condenser of 20 microfarads capacity, assuming the 
condenser loss to be negligible. 

dE 
Given that C = K~j7' where C is current and K is capacity. 

19. If momentum is given by the product of mass into velocity, 
and force is defined as the time rate of change of momentum, show 
that force is expressed by the product of mass into acceleration. 

20. The following are the approximate speeds of a locomotive on 
a run over a not very level road. Plot these figures and thence obtain 
a curve showing the acceleration at any time during the run. 



Time (in mins. and sees.) 


o 


i-o 


2-15 


6-15 


9-22 


n-45 


14-26 


16-33 


20-52 


23-10 


Speed (miles per hour) . 


start 


6 


10 


18-2 


22-8 


25-5 


28 


29-2 


28-6 


26-1 



21. Taking the following figures referring to CO 2 for use in a re- 
frigerating machine, draw the rate curve and find the value of -- when 

ctt 
t=i8 F. 



t F. . ' 


5 


o 


5 


10 


15 


20 


25 


30 


35 


40 


>lbs.per\ 
sq. in J 


285 


310 


335 


363 


392 


423 


456 


491 


528 


567 



INTRODUCTION TO DIFFERENTIATION 



22. The weight of a sample of cast iron was measured after various 
heatings with the following results; the gain in weight being due to 
the external gases in the muffle. 



Number of heats . . 


o 


2 


6 


12 


22 


23 


24 


25 


26 


Weight .... 


146-88 


146-94 


147-04 


H7-54 


148-02 


I48-II 


I 4 8.2 7 


148-36 


148-46 


- 




27 


30 


35 


39 


45 


148-61 


149-18 


150-49 


152-36 


156-44 



Plot a curve to represent this table of values, and from it construct 
the rate curve. 

23. The figures in the table are the readings of the temperature of 
a sample of steel at various times during its cooling. Plot these values 
to a time base, and thence draw the " inverse rate " curve, i. e., the 

curve in which values of -. are plotted horizontally and the temperatures 
along the vertical axis. 



Time in sees. (I) . . 


75 j 9 


i5 


120 


135 


150 


165 


180 


195 


210 


225 


Temperature in C. c (0) 


1 
850 848 


844-7 


8 4 2 


839-5 


838-5 


838-2 


838-1 


838 


837-9 


837-5 






240 


255 


270 


285 292-5 


300 


315 


330 


345 


360 


367-5 


375 


350 


45 


836 


833 


829 


825 8-'3- 3 


822-2 


821-7 


821-5 


821-3 


821-1 


819 


8i5 


813 


8II-6 



CHAPTER II 
DIFFERENTIATION OF FUNCTIONS 

Differentiation of ax". It has been shown in Chapter I 
how to compare the changes in two quantities with one another, 
and thus to determine the rate at which one is changing with 
regard to the other at any particular instant, for cases in which 
sets of values of the two variables have been stated. In a great 
number of instances, however, the two quantities are connected 
by an equation, indicating that the one depends upon the other, 
or, in other words, one is a function of the other. Thus if y = 5# 3 , 
y has a definite value for each value given to %, and this fact is 
expressed in the shorter form y =f(x). Again, if z ijx^y^xy 3 
+5 log y> where both x and y vary, z depends for its values on 
those given to both x and y, and z f(x, y). 

To differentiate a function it is not necessary to calculate 
values of x and y and then to treat them as was done to the given 
sets of values in the previous chapter. This would occasion a 
great waste of time and would not give absolutely accurate results. 
Rules can be developed entirely from first principles which permit 
the differentiation of functions without any recourse to tables of 
values or to a graph. 

We now proceed to develop the first of the rules for the differen- 
tiation of functions ; and we shall approach the general case, viz., 
that of y ax n , by first considering the simple case of y = x 3 . 
Our problem is thus to find the rate at which y changes with regard 
to x, the two variables (y the dependent and x the independent 
variable or I.V.) being connected by the equation y = x s . 

The rate of change of y with regard to x is given by the value 

dv 
of -j-, and this is sometimes written as Dy when it is clearly 

understood that differentiation is with regard to x : the operator D 
having many important properties, as will be seen later in the 

book. If y is expressed as f(x), then -p is often written JfQ 

fix dx 

or /'(*). 

26 



DIFFERENTIATION OF FUNCTIONS 27 

~, Dy, -^-* or f'(x) is called the derivative or differential 

coefficient of y with respect to x; and the full significance of the 
latter of these terms is shown in Chapter III. 

We wish to find a rule giving the actual rate of change of y 
with regard to x, y being = x 3 , the rule to be true for all values 
of x. As in the earlier work, the actual rate of change must be 
determined as the limiting value of the average rate of change. 

Let x be altered by an amount Sx so that the new value of 
x = x -f- 8x ; then y, which depends upon x, must change to a 
new value y + Sy, and since the relation between y and x is y x 3 
for all values of x 

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

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

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

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

Divide through by Sx, and 



Thus an average value for the rate of change over a small 
interval Sx has been found; and to deduce the actual rate of 
change the interval Sx must be reduced indefinitely. 

Let Sx=-ooi; then ^ = 3* 2 + (3* X-ooi) + -000001 



-oooooi 
whilst if 8x = -ooooi 

jh_j 

:- = 3# 2 -f -00003^;+ -oooooooooi . (3) 

Evidently, by still further reducing &x the 2nd and 3rd terms 
of (3) can be made practically negligible in comparison with the 
ist term. 

Then, in the limit, the right-hand side becomes 3* 2 , 



and thus- ^ = T f = 3 * 2 
dx i ^Sx 



dx- = * X 



28 



MATHEMATICS FOR ENGINEERS 



This relation can be interpreted graphically in the following 
manner : If the curve y = x s be plotted, and if also its slope curve 
be drawn by either of the methods of Chapter I, then the equation 
to the latter curve is found to be y = $x 2 . 

The two curves are plotted in Fig. 7. 




10 



01234 

X 

FIG. 7. Primitive and Slope Curves. 

Example i. Find the slope of the curve y x 3 when x = 4. 

dy dx 3 o 

The slope of the curve = -v 1 = -j = 3* 

CLX CLX 

and if x = 4 -,- = 3 X 4 2 = 48. 

Meaning that, in the neighbourhood of x 4, the ordinate of the 
curve y = x 3 is changing 48 times as fast as the abscissa ; this fact 
being illustrated by Fig. 7. 



Working along the same lines, it would be found that -jj = 

dx 5 
-y- 

himself). 



and -- = 5# 4 (the reader is advised to test these results for 



DIFFERENTIATION OF FUNCTIONS 29 

Re-stating these relations in a modified form 
dx 3 



- = 4^ = 4* 

ft* 



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

dx n 

- = nx n ~ \ 

dx 

Thus the three cases considered suggest a general rule, but 
it would be unwise to accept this as the true rule without the 
more rigid proof, which can now be given. 

Proof of the rule 

dx n 

-M Vtt ~~ * 

- /(vV 

ax 
Let y = x n , this relation being true for all values of x . (i) 

If x is increased to x-}-8x, y takes a new value y-f-Sy, and 
from (i) it is seen that 

y-|-Sy = (x+8x) n . 

Expand (x-\-8x) n by the Binomial Theorem (see p. 463, Part I). 
Then 



Subtract (i) from (2), and 



Divide by 8^ 

8v , , n(ni) 0/c . , , (w i)(n 2) M ,/ \ 9 , 

--=^n-i4- v --- 'x n - 2 (8x)-\ s 'x n - 3 (8x) 2 -t- terms 

8,r |_2 |_3 

containing products of (Sx} 3 and higher powers of (8x). 

Let 8* be continually decreased, and then, since Sx is a factor 
of the second and all succeeding terms, the values of these terms 
can be made as small as we please by sufficiently diminishing Sx. 



30 MATHEMATICS FOR ENGINEERS 

Thus in the limit f- -> n x n - * 
Sx 



dv 
or -- 



Id 



Hence the first rule for differentiation of functions is established, 
viz. 



i. e., differentiation lowers the power of the I.V. by one, but the 
new power of x must be multiplied by the original exponent. 

The reason for the multiplication by the n can be readily seen, 
for the bigger the value of n the steeper is the primitive curve 
and therefore the greater the change of y for unit change of x. 
The n actually determines the slope of the primitive (cf. Part I, 
p. 340), and it must therefore be an important factor in the 
result of differentiation, since that operation gives the equation of 
the slope curve. 

To make the rule perfectly general, aUowance must be made 
for the presence of the constant multiplier a in ax n . 

It will be agreed that if the curve y = x 3 had been plotted, 
the curve y = ^x 3 would be the same curve modified by simply 
multiplying the vertical scale by 4. Hence, in the measurement 
of the slope, the vertical increases would be four times as great 
for the curve y = 4# 3 as for the curve y = x 3 , provided that the 
same horizontal increments were considered. 

Now the slope of the curve y = x 3 is given by the equation 

dy - ix* 
dx~ 3X 

so that the slope of the curve y = ^x 3 is given by 
^ = 4x3*2 = I2x 2 . 

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



j-ax a = nax 



H-l 



dx* 

Accordingly, a constant factor before differentiation remains as 
such after differentiation. 

We can approach the differentiation of a multinomial expression 



DIFFERENTIATION OF FUNCTIONS 



by discussing the simple case y = $x 2 + 17 (a binomial, or two- 
term expression). The curves y = $x 2 and y = $x 2 + 17 are seen 
plotted in Fig. 8, and an examination shows that the latter curve 
is the former moved vertically an amount equal to 17 vertical 
units, i. e., the two curves have the same form or shape and 
consequently their slopes at corresponding points are alike. Thus 
if a tangent is drawn to each curve at the point for which x = 2-5, 

the slope of each tangent is measured as , i.e., 25; and con- 
sequently the diagram informs us that the term 17 makes no 
difference to the slope. 

JOO. 




FIG. 8. 



Thus 



dx- 



Now, by differentiating 5# 2 
dx^ 



17 term by term, we have 



since 17 is a constant and does not in any way depend upon x, 
and therefore its rate of change must be zero. 

It is seen that in this simple example it is a perfectly logical 
procedure to differentiate term by term and then add the results; 
and the method could be equally well applied to all many-term 
expressions. 



32 MATHEMATICS FOR ENGINEERS 

Hence j x (ax n +bx n ~' i +cx n - 2 + . . . d) 

= nax n ~ l +b(n i)x n ~ 2 +c(n 2)x n ~ 3 + . . . 
and ^ x (ax a -{-b) = nax"- 1 . 

To apply these rules to various numerical examples : 

Example 2. Differentiate with respect to x the function 



= (gx i-6;r 6 ) 

= i4'4#- 8 +#^' 5 or IA-AX-*-\ _ 

V x 



Example 3. If y = -Sx A A, find the value of ^. 

' * 



y = -8x A /L = -8x~, = -8^~^ or 
V ^s ^ 

so that in comparison with the standard form 
a = -8 and n = 1-5. 



Then 





or _ 



Example 4. If /w 1 ' 41 = C, the equation representing the adiabatic 
expansion of air, find --. 

In this example we have to differentiate p with regard to v, and 
before this can be done p must be expressed in terms of v. 

Now pv 1 -* 1 = C, so that p = ^ = Cw- 1 ' 11 . 

Hence ft = ^-Cz;- 1 - 11 = Cx -i-4iy- 2 - 11 = - i-4iCw- 2 - 41 

dv dv 

and this result can be put into terms of p and v only, if for C we write 
its value pv l ' tl . 

Thus = -i-4ix/>w 1 - 4l x- 8 ' ll = i-4ipv- 1 = - -'. 



DIFFERENTIATION OF FUNCTIONS 33 

Example 5. The formula giving the electrical resistance of a length 
of wire at temperature t C. is 



where R = Resistance at o C. Find the increase of resistance per 
i C. rise of temperature per ohm of initial resistance, and hence state 
a meaning for . 

The question may be approached from two standpoints ; viz. 
(a) Working from first principles. 

"D _ "O "D [ 

i. e., increase in resistance for t C. = 
but this is the resistance increase for initial resistance R , hence 

T? a 

increase in resistance per i C. per ohm initial resistance = -^- = a. 
(&) By differentiation. 
Rate of change of R with regard to t = -~ 



a = R a 

and consequently the rate of change of resistance per i C. per i ohm 
initial resistance = o. 

The symbol a is thus the " temperature coefficient," its numerical 
value for pure metals being -0038. 

Example 6. Find the value of -^-(45* \+6sl 1-8 4 ). 

CIS \ S ' 

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



2f ' n-l\ 

Example 7. If x = a n \i a n /, a formula referring to the flow 

dx 
of a gas through an orifice, find an expression for -5-. 

*( n ^\ 
As it stands a n \l a n ) is a product of functions of the I.V. 

(in this case a), and it cannot therefore be differentiated with our 

D 



34 MATHEMATICS FOR ENGINEERS 

present knowledge. We may simplify, however, by removing the 
brackets, and then 



2 n-l 2 2 M+l 

x a n aT^ n a n a n 

+ l 



, , / 2 n 

dx d I 

= \a n a 

da da\ 



_i -i 

2 n n-\-I n 

= - X a --- a 

n n 



2 n 

-a -- a 

n n 



2-n 
7 n . 



Example 8. Determine the value of 
5 -45m 9 - 86 



rfm\ 5W 

To avoid the quotient of functions of m, divide each term by 5m 4 ' 32 , 

T>7*w75 .^C*M.^36 

then the expression = 



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



~ 4 '" -499m 4 - 54 9' 



Proof of the construction for the slope curve given on p. 14. 

Let us deal first with the particular case in which the equation 
of the primitive curve is y = x 2 . 

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

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



= 2xhh z 

so that the equation of the curve MN is 
y-5 = 2xhh 2 . 



DIFFERENTIATION OF FUNCTIONS 35 

Now the curve ODE is the curve MN shifted back a distance of 
- horizontal units, and hence its equation is y 3 = z(x-\ jh h 2 , 

2 \ ^/ 

since for x we must now write f x-\\ 

Thus the equation of curve ODE is 

y 3 = 2xh 

y 

or ~ = 2x 

h 

A/ 

i. e., if Y be written for ~, Y = 2x 

or the equation of the curve ODE is that of the slope curve of 
the curve y ~ x 2 provided that the ordinates are read to a certain 
scale; this scale being the original vertical scale divided by h 
expressed in horizontal units. 

Hence the curve ODE is the slope curve of the curve OAB. 

Before discussing the general case, let us take the case of the 
primitive with equation y = x 3 . 

If the curve be shifted forward an amount = h, the equation 
of the new curve is 

yi = (x-W 

and the equation of the curve giving the differences of the 
ordinates is 

y 2 = y y^-x 3 (x h) 3 = x 3 



By shifting this curve - units to the left we change its equation, 



by writing (x-\ J in 'place of x, into the form 



Dividing by h 



h 4 

or Y = 3* 2 +- 



36 MATHEMATICS FOR ENGINEERS 

h z 
Now if h is taken sufficiently small, is negligible in comparison 

with 3# 2 , and we thus have the equation of the curve Y = 3# 2 , 
which is the slope curve of the curve y = x 3 ; but the ordinates 
must be measured to the old vertical scale divided by h. 

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



n(*L-ll) x n-2} l 2_ . _ \ 



\ 



Write (*+ 2 ) in place of x, and then 



h L 2 8 

n(n i) 27 (w i)(w 2) n _3, 2 

_ ._, _> _ ^C^ 1 ft _ - _ OC ft ~ 

2 4 

= n X n - 1 -}- terms containing A as a factor. 

Hence if h is made very small 

Y or = w*"- 1 . 



Exercises 2. On Differentiation of Powers of the I.V. 

1. Find from first principles the differential coefficient of x*. 

o 

2. Find the slope of the curve y = 2 when x = -5 

^ 

(a) By actual measurement and (6) by differentiation. 

3. The sensitiveness of a governor is measured by the change of 
height corresponding to the change of speed expressed as a fraction 
of the speed. Thus if h and v represent respectively the height and 

dl) 

speed, the sensitiveness dh -. --- . If the height is inversely pro- 

portional to the square of the velocity, find an expression for the 
sensitiveness. 

Differentiate with respect to x the functions in Exs. 4 to 15. 
- --* ' 2I5 



4. 3* 9 . 5. -. 6. 8i-5*-*. 7. igx. 8. 



A/-O / " - ^" ' . ,J 



DIFFERENTIATION OF FUNCTIONS 



37 



9. 



8*' 



10. 







12. 



(* 3 - 7 ) 2 - 8 
~ 



13. 



14. 



15. 



16. Find the value of -]- when pv 1 - 3 = 570 and v = 28-1. 



. 
17. Find the value of 



Jv 



18. If E = I5+I4T -oo68T 2 , find the rate of change of E with 
regard to T when T has the value 240. 

, dH , dtt i f dp , \ 

19. Calculate the value of -j- from -T- = -- \ v ^-+yp( when 

dv <fo 7 1 1 dv -^) 

pv 1 - 3 = C and y = 1-4. 

20. Find the rate of discharge ( -,- J of air through an orifice from a 

tank (the pressure being 55 Ibs./n") from the following data 

I44/>V = wRT 
R = 53-2, V = 47-7, T = 548. 



Time (sees ) (/) 


o 


60 


I35 


21$ 


3IS 














Pressure (Ibs. per sq. in.) (p) 


63 


45 


30 


15 


10 



Hint. Plot p against / and find ? when p = 55. 

21. If P = load displacement of a ship, 

p = weight of offensive and defensive factors. 
Then P = aP+bP*+p. 

Find the rate of increase of armament and protection in relation 
to increase of displacement. 

II \ O / \ I \ *> 

yai [ I __ y \ *u\t I 4f \ 901 1 \) V I * 

22. if M = w(t x w( I+ y)- w (y *> , 

2/ \ // 2 

constants. 

\Vv ('/'M 

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

2/ 2 v dy 

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

24. If S = -i s ST-" k find the value of -,-. 

2 I / y J ax 

25. Find the value of h which makes -jr- = o when 

dh 



dM 

""' ^ and 



[h is the height of a Warren girder; and the value found will be 
the height for maximum stiffness.] 

26. If p = -^ A and q -75+ A, find the value of v f in terms of 
r 3 * r 3 dr 

p and q. (This question refers to the stresses in a thick spherical 
shell, p being the radial pressure, and q the hoop tension.) 



38 MATHEMATICS FOR ENGINEERS 

27. In a certain vapour the relation between the absolute tem- 
perature T and the absolute pressure p is given by the equation 
T = 140^1 + 465, and the latent heat L is given by L = 1431 -ST. 
Find the volume, in cu. ft., of i Ib. of the vapour when at a pressure 
of 81 Ibs. per sq. in. absolute, from 

V .n-? J (T TjRl 

I 44 T dp 

28. For a rolling uniform load of length r on a beam of length /, 
the bending moment M at a point is given by 

)"!// v\ ieiv% 

M = 



If y is a constant, find an expression for the shear (i. e., the rate 
of change of bending moment). 

29. Given that p = electrical resistance in microhms per cu. cm. 

and x = percentage of aluminium in the steel, 
then p = 12 + 12# -3# 2 for steel with low carbon content. 
Find the 'rate of increase of p with increase of aluminium when 
x = 4 . 

30. The equation giving the form taken by a trolley wire is 



y = 

and the radius of curvature = 



2000 1760 

i 



dx* 
Find the value of the radius of curvature. 

Good examples of the great advantage obtained by utilising 
the rules of differentiation already proved are furnished by the 
two following examples, which have reference to loaded beams. 

Example g. Prove that the shearing force at any point in a beam 
is given by the rate of change of the bending moment at that point. 

Consider two sections of the beam 8x apart (see Fig. 9). The 
shear at a section being denned as the sum of all the force to the right 
of that section, let the shear at b = S, and let the shear at a = S+8S. 
Also let the moment of all the forces to the right of b (i. e., the bending 
moment at b) = M, and let the bending moment at a = M+SM. 

Taking moments about C 

M+SM = M+(S+8S)**+S( 8 *) 



or 8M = 



DIFFERENTIATION OF FUNCTIONS 39 

SM , SS 
Dividing by Sx, ^- S-\ 

and when 8x is diminished indefinitely, SS becomes negligible 

and = S. 



S+SS 




.^ooooRftooo 



s H ""* 

FIG. 9. FIG. 10. 

Examples on Loaded Beams. 

The last example should be considered in conjunction with the 
following : 

Example 10. For a beam of length /, fixed at one end and loaded 
uniformly with w tons per foot run, the deflection y at distance x 
from the fixed end is given by the formula 



E being the Young's Modulus of the material of the beam, and I being 
the moment of inertia of the beam section. 

, dy d*y d?y , d*y 

Find the values of -~, -j^, , , and -,.. 

dx? dx v dx 3 dx* 



y = 



Differentiating, = {(W X 2x) - ( 4 l x 3^ 2 ) +4* 3 } 






Differentiating again, g - 



40 MATHEMATICS FOR ENGINEERS 

Differentiating again, * - 



w 



d*y _ d(d?y\ _ dJTw 

~ ~ ~ ( * ' 



Carrying the differentiation one stage further 

d*y _ d(d?y\ 
5** ~ ~dx\dx*J 



Physical meanings may now be found for these various deriva- 
tives. Referring to Fig. 10, consider a section of the beam distant 
x from the fixed end. To the right of this section there is a length 
of beam lx loaded with w tons per foot, so that the total load 
or total downward force on this length is w(l x) ', and since this 
load is evenly distributed, it may be all supposed to be concentrated 

at distance from the section. 

2 

Now the bending moment at the section 

= moment of all the force to the right of the section 

/lx\ w 
= force X distance = w(lx)x( ) = ~(lx)\ 

d 2 v 
Comparing this result with the value found for ~ z> we notice 

that the two are alike except for the presence of the constants 
E and I : thus -~ must be a measure of the bending moment. 
Actually the rule connecting M, the bending moment, and 



its- 

dx* 1S 



M -tPy 

= i r M = 



the proof of this rule being given in a later chapter. 

Again, we have proved in the previous example that the shear 
is given by the rate of change of bending moment : thus 



dx dx dx 2 dx 3 



= w(x / 



DIFFERENTIATION OF FUNCTIONS 41 

a result agreeing with our statement that the shear at a section is 
the sum of all the loads to the right of the section. [The reason 
for the minus sign, viz., (x I), being written where (lx) might 
be expected need not be discussed at this stage.] 
Continuing the investigation 

d* w 



or v3 = w 

dtf 

but w is the loading on the beam and 

-~d*y d f^d 3 y\ dS 

EFr4 = j-( El j-4 j- 

dx* dx\ dx 3 / dx 

so that the loading is measured by the rate of change of the shear. 
If now the deflected form is set out, by constructing successive 
slope curves we obtain, respectively, the slope curve of the deflected 
form, the bending moment curve, the shear curve and finally the 
curve of loads. 

Example n. The work done in the expansion of gas in gas turbines 
is given by 



where r is the ratio of expansion. 

Compare governing by expansion control with governing by 
alteration of the initial temperature, from the point of view of 
efficiency. 

Deal first with the expansion control, i. e., regard Tj as constant 
and r as variable. Then the rate at which the work is increased with 



respect to r is -3. 

Now 



3 
dW 

w 




42 MATHEMATICS FOR ENGINEERS 

Now regard r as constant, but T, as variable. 

/7W *? P V / w-l\ 

Then- ^ = ^^rV-^) 

u J-! n i J-o 

and, expressing the two results in the form of a ratio 

/7W /fW PVT (w r^T 

vv . vv _ x- 1 v 1 {n i)J- 



-( ""^ 

^yn\I r n I 

Lengths of Sub-tangents and Sub-normals of Curves. 

The projection of the tangent to a curve on to the axis of x is 
known as the sub-tangent, i.e., the distance "sub" or "under'? 
the tangent. The projection of the normal on the x axis is called 
the sub-normal. 

The slope of a curve at any point, measured by the slope of 

its tangent at that point, is given by the value of -f- there, or if 
a = inclination of the tangent to the x axis 

dy 



tan a = 



dx 




In Fig. ii 



FIG. n. Sub-tangent and Sub- normal. 

PA dy 

- x -~i = tan a = -f- 
AT dx 

AT = PA^ 

ay 



But AT = sub-tangent and PA = y 

dx 

and hence the length of the sub-tangent = y,- 



DIFFERENTIATION OF FUNCTIONS 

Again L APN = a, since L TPN = a right angle 

AN sub-normal 



tan 



i. e. t 



or 



tan a = 



sub-normal 



sub-normal = y X tan a = y-~ 



To find the length of the tangent PT 
(PT) 2 = (PA) 2 +(AT) 2 



and 
In like manner PN = 




FIG. 12. 



43 



Example 12. Find the lengths of the sub-tangent and the sub- 
normal of the parabola y 2 = <\ax (Fig. 12). 



y z = AfO-x and y = 2 Va . #* 



then 



or 



Then length of sub-tangent = y 



dx 
dy 



Va 



Va 



44 MATHEMATICS FOR ENGINEERS 

This result illustrates an important property of the parabola and 
one useful in the drawing of tangents. For AT = 2.x 2 X AO, and 
hence to draw the tangent at any point P, drop PA perpendicular to 
the axis, set off OT = OA and join TP. 

The length of the sub-normal AN = y-2- 

, v- Va _ 2 Va Vx Va 

y /\ ___ _ . 

Vx Vx 

= 2d. 

i. e., the length of the sub-normal is independent of the position of P, 
provided that the sub-normal is measured on the axis of the parabola. 

Example 13. Find the lengths of the sub-tangent and the sub- 
normal of the parabola y = i$x 2 2xg 

when x = 2 and also when x 3. 

The axis of this parabola is vertical, and consequently the sub- 
normal, which is measured along the x axis when given by the value 

of y-^, is not constant. 
7 dx 

Now y = i5x*2Xg 

dy 

and -r- = 30* 2. 

dx 

dx 2 

Hence sub-tangent = y-r- = 



-r- 

dy $ox 2 

dv 

and sub-normal = y~~ = (i^x z 2xg) x (30^2). 

ax 

Thus when x = 2 

sub-tangent = - ~ ? _ __g5 um f- s . 
_ _ _ 60 2 _ 62 

sub-normal = (60+4 9) ( 62) = 3410 units. 

When x = 3 

/I35 6 g\ 120 15 
sub-tangent = ( g8 j = - 88 = ^ units. 

sub-normal = 120 X 88 = 10560 units. 

Example 14. A shaft 24 ft. long between the bearings weighs 
2 cwt. per foot run, and supports a flywheel which weighs 3! tons 
at a distance of 3 ft. from the right-hand bearing. Find at what 
point the maximum bending moment occurs and state the maximum 
bending moment. 

Regarding the shaft as a simply supported beam AB (see Fig. 13), 
we may draw the bending moment diagrams for the respective systems 



DIFFERENTIATION OF FUNCTIONS 



45 



of loading, viz., ADB for the distributed load, being the weight of the 
shaft, and ACB for the concentrated load. 

The total distributed load is wl, i.e., 24X-I = 2-4 tons, giving 
equal reactions of 1-2 tons at A and B; and the bending moment 
diagram is a parabola with vertex at D, the maximum ordinate DF 

being -5-, i. e., Q or 7-2 tons ft. If for convenience in the 

later working the axes of x and y are as shown in the figure, the 
equation to this parabola is y 2 = <\ax; or taking the value of y as FB 
and that of x as DF, 1 22 = 40x7- 2, from which 40 = 20 and 
y 2 = 20*. 



Scale of 
Bending Momen 
tons- f r 



pal lei to AC 




FIG. 13. 



The load of 3-5 tons produces reactions of X3*5 tons at B, 

2 4 

and x 3-5 tons at A, i. e., R B = 3-06 and R A = -44 tons : thus the 

2 4 
bending moment at E is 3-06x3 = 9-18 tons ft. 

Since the total bending moment is obtained by adding the ordinates 
of the diagram ADB to the corresponding ordinates of ACB, the 
maximum bending moment will be determined when the tangent to 
the parabola is parallel to AC, and the position satisfying this condition 
can readily be found by differentiation. Thus 

The equation of the curve ADB is y 2 = 20* or y = 4-47**, and 
the slope of the curve is given by the value of -~. 



Now if y = 4-47* , 



= 4-47X^4 



e. f 



tan = 



_ 4'47 



2*3 



4 6 






Referring to the figure ACB, tan a = = 
and thus- 4^42 = ^i 

2** 9'l8 

or x* = 1:47X9-18 

42 

i.e., (DR) = 4-47X9-18 

42 



Again, (PR) 2 = 2oxDR, and thus PR = 

= 4-37 ft. 

Thus the maximum bending moment occurs at a distance of 
12 4-37, i. e., 7-63 ft. from the right-hand bearing. 
To find the maximum bending moment 



DR = **/^v*" = -956 

\ 42 / 

PQ = DF-DR = 7-2--96 = 6-24 tons ft. 
Also ^jLp = X9-i8 = 7-16 tons ft. 

Hence the maximum bending moment = 7-16+6-24 13-4 tons ft. 

Exercises 3. On the Lengths of the Sub-tangent and Sub-normal : also 

Beam Problems. 

1. Find the lengths of the sub-normal and sub-tangent of the 
curve $y = ^x 3 at the point for which x = 3. 

2. If y = -^W, V = 117, and g = 32-2, find the value of x that 
makes the slope of the curve i in 17-4. 

3. A parabolic arched rib has a span of 50 ft. and a rise of 8 ft. 
Find the equation of the tangent of the slope of the rib. What is the 
slope of the tangent at the end ? 

4. Find the equation of the tangent to the curve p = ^- at the 

v 

point for which v 5. 

In Exercises 5 to 7, y is a deflection and x a distance along the 
beam. Find, in each case, expressions for the Bending Moment, 
Shearing Force and Load. The beam is of uniform section throughout, 
and of span /. 

5. The beam is supported at both ends, and loaded with W at the 
centre. 

W llx z x a \ 

y = -^r T ( -=- ) {x is the distance from the centre}. 

2EI\ 4 6 / 

6. The beam is supported at both ends, and loaded continuously 
with w per ft. run. 

y ^^^Y^-g j {x is the distance from the centre}. 



DIFFERENTIATION OF FUNCTIONS 47 

7. A cantilever loaded with W at the free end. 

W/7# 2 x a \ 
y = -p_( --- -j-\ \x is the distance from the fixed end}. 

8. Find the lengths of the projections on the y axis of the tangent 
and the normal of the parabola, x 2 iob 2 y + ^c, x having the value ga. 

9. Prove that the sub-normal (along the axis of the parabola) of 
the parabola x 2 6y is constant and find the value of this constant. 



^ A rr , /- , r> f. A 4.- 

10. If El-/ = ---- ,. --- \-C and C = ------- , find the value 

rf* 4 6 a 2 12 



Differentiation of Exponential Functions. The rule for 
differentiation already given applies only to functions involving 
the I.V. (usually the x) raised to some power. A method must 
now be found for the differentiation of exponential functions, viz., 
those in which the I.V. appears as exponent ; such as e 5x or 4*. 

When concerned with the plotting of the curve y = e* (see 
Part I, p. 352) mention was made of the fact that, if tangents 
are drawn to the curve at various points, the slopes of these 
tangents are equal to the ordinates to the primitive curve at the 
points at which the tangents touch the curve. Thus the slope 
curve of the curve y = e x lies along the primitive, and 

**-- 6* 

dx ~ 

or the rate of change of the function is equal to the value of the 
function itself. 

We may establish the result algebraically thus 

X % 3C 

e x = i-\-x-\ --- [- 




1.21.2.31.2.3.4 

Assuming that a series composed of an unlimited number of 
terms can be differentiated term by term and the results added 
to give the true derivative (this being true for all the cases with 
which we shall deal), then by differentiation 



de x _ 

~ f 



x 3 



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



48 MATHEMATICS FOR ENGINEERS 

dx i 

noted : the sub-tangerit = y-,- -- e x x = i, i. e., the sub-tangent 

is constant and equal to unity. 

The curve y = e x may be usefully employed as a gauge or 
template for testing slopes of lines; the curve being drawn on 
tracing-paper and moved over the line to be tested until the curve 
and line have the same direction, and the ordinate of the curve 
being then read, any necessary change of scales being afterwards 
made. 

The work may now be carried a stage further, so that the rule 
for the differentiation of e bx may be found. 

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

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

distances for y = e x . 

Hence the slope of the curve y = e bx is b X slope of curve y = e x 

r\pbx 

or -p- = be bx 

It should be noticed that the power of the function remains 
the same after differentiation, but the multiplier of the I.V. becomes 
after differentiation a multiplier of the function. This latter rule 
must be remembered throughout differentiation, viz., any multiplier 
or divisor of the I.V. in the function to be differentiated must become a 
multiplier or divisor of the function after differentiation. 

From e bx we can proceed to ae bx , the result from the differentia- 
tion of which is given by 

dae bx 



dx 



= abe bx 



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

dx 

-/- = ^~5e~b x = 5 X T&~^ X 
dx dx j 6 



DIFFERENTIATION OF FUNCTIONS 49 

Referring to the last example, note that the power of e is 
exactly the same after differentiation as before : the factor ^ 

multiplies the I.V. in the original function, and therefore it occurs 
as a constant multiplier after differentiation; also the constant 
factor 5 remains throughout differentiation. 



Example 16. If C = C e L, where C and C are electrical currents, 
R is the resistance of a circuit, L is the self-inductance of the circuit 
and t is a time, find the time-rate of change of C. 

This example illustrates the importance of the rate of change as 
compared with the change itself; for it demonstrates the fact that 
for an inductive circuit the change of current is often extremely rapid 
and consequently dangerous. 

T" * i- t /-> dC, d ~ 5* 

Time rate of change of C = -^ = _,XV~ L 

dt dt 

- C x V? 

^o Ps T~& L 

-C -~ 
R^ 



i. e., the rate of decrease of the current when the impressed E.M.F. 
is removed is proportional to the current at the instant the circuit is 
broken. 

To better illustrate the example, take the case for which the 
current at the instant of removal of the E.M.F. is 14-5 amps., the 
resistance of the circuit is 6-4 ohms, and its self-inductance - oo6 henry. 

Then the rate of change of the current = -- ^ X 14-5 = 15470 amps. 

*ooo 

per second, whereas the actual current is only 14-5 amps. 

The expressions e x and e bx are particular forms of the more 
general exponential function a x ; to differentiate which we may 
proceed by either of two methods : 

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

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



50 MATHEMATICS FOR ENGINEERS 

Differentiating term by term 

da x . , . n \2 , /, \sx 2 

dx = o+log tf+(log a) x+(log a) ~+ . . . 

. . (x log a) 2 . (x log a) 3 , 

= log a{i+x log a+~- ' > +^ ' + 

L II. 

= log a X a x 
da* 



(b) Assuming ihe result for the differentiation of e bx 

Let a x = e bx 

so that a e b , and therefore log e a = b. 

d , d 

T a = J-* 

ax ax 

= logg axa x 
or a x log a 



Then T a x = T e bx = be bx = log e ax e bx 



,, da x 

thus -j = a x . log a. 



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

dX 

In this case = 4 and log e 4 = 1-3863. 

dA x 
Hence - = '1-3863 x <\ x . 

CIX ^^^*^^m**^mm^^^* 

Note carefully that this result cannot be simplified by combining 
1-3863 with 4 and writing the result as 5-5452*, which is quite incorrect. 
The 4 alone is raised to the x power, and 1-3863 is not raised to this 
power. 

TO 

Example 18. Find the value of , 2 (3'^) S - 

% 
Here a = 3-6 and log 3-6 = 1-2809. 

Thus j(3' 6 )' = lo 3-6 X (3-6)* = 1-2809 x (3-6)*. 

Then 



= I -2809 XT- (3-6)* = 1-2809 X 1-2809 x(3-6) j 

GfS 

= 1-64(3-6)*. 



DIFFERENTIATION OF FUNCTIONS 51 

Example 19. Given that s = ^e st -\-ye~ 5t t find the value of -^255. 



s = 



= 2O0 5 ' 350 ~ 5t . 

Again ,. z = 

' w*~ 25$ = 

Differentiation of log x. The rule for the differentiation 
of logarithmic functions can be derived either from the expansion 
of loge (i + x) into a series, or by assuming the result for the 
differentiation of e x . Considering these methods in turn 

(a) Working from first principles. Let y = log x, i. e., log e x. 
Then if x be increased to become x-\-8x, y takes a new value 
y+Sy, and y+Sy = log (x-{-8x). 

(8x\ ( 8x\ 
i-\ ) = log x+log ( i-j ), 
X / \ X / 

therefore 

/ Sx\ 
(y-j-Sy) y = log #-|-log ( i-\ ) log x 

\ x / 

i. e., 8y = log (l-f* ) 

Also log ( i-j ) can be expanded into a series of the form 
\ x / 

A- 2 /yO A' 1 * 

log (*+x) = x -+--+ . . (see Part I, p. 470) 
^ o 4 

so that 

r ~ I \~ ) ^l~/"r\~/ T\^7/~r 



8y = _ _ 

VA;/ 2\* 
Dividing all through by SA; 

8y = i 8x (8x)* (Sx) 3 
8x~ x 2x 2 3% 3 4** ' 

By sufficiently diminishing the value of Sx we may make the 



52 MATHEMATICS FOR ENGINEERS 

second and succeeding terms as small as we please, and evidently 

the limiting value of the series is - 



i* c,. , 

dx 



LSy _ i 
8x x 



Hence 



Sx>o 

d 



_ 

dx ~ x 



(b) Assuming the result = e x 

Qt% 

Let y = loge x, so that x e y 

dx de y 

and = - = e y . 

dy dy 



^>. . 

Now ~ = 

OA/ 



T 

r- and consequently by considering the limiting 

oA/ 

*y d 

values of these fractions /- = -r- 

dx dx 

dy 

dv 
We wish to find ~ and we have already obtained an expression 

, dx 



-=-. 

dy 

Hence ~ = T- = . = 

dx ax e y x 

dy 

d log e x _ i 

\JL 5 

dx~ x 



This result can be amplified to embrace the more general form, 
s 



thus 



for, in accordance with the rule given on p. 48, the A which 
multiplies the I.V. in the original fun6tion must appear as a 
multiplier after differentiation. 

All these rules apply to functions involving natural logs, but 



DIFFERENTIATION OF FUNCTIONS 53 

they can be modified to meet the cases in which common logs 
occur ; for - Iog 10 x = -4343 log e x 

d log ln x d loe e x 

and hence -$&- = . 4343 _J3L_ = 

d . ,. -4343A 

and -- Iog 10 (Ax+B) - 



It should be observed that in all these logarithmic functions 
the I.V. is raised to the first power only : if "the I.V. is raised to 
a power higher than the first, other rules, which are given later, 
must be employed. 

Example 20. If y = \og e jx, find ~. 
dy d . 

-r**-T loge 7 

dx dx _ 

or alternatively log e 7* = log e 7+log e x 
and thus log e 7* = 



. i i 

= o+- = - 

X X 



Example 21. Differentiate with regard to t the expression 
Iog 10 (5^14) and find the numerical value of the derivative when 
/ = 3-2. 



When t = 3- 2 



10T Ut- 1^ - ' 4343 X 5 - 

dt tog], \y i_ r - 



log,. (51-14) = = "0858. 



We may check this result approximately by taking values of J 
3-19 and 3-21 and calculating the value of -' *! -- 
Thus 

When i = 3-19, Iog 10 (5f 14) = Iog lfl (i5'95 14) = Iog 10 1-95 = -2900 
when t = 3-21, Iog 10 (5* 14) = Iog 10 (i6-o5 14) = Iog 10 2-05 = -3118 

so that 

8 Iog 10 (5^14) = -3118 -2900 = -0218 

while 8t = 3-21 3-19 = -02 

S . , -0218 

and lo < * == ~~ = I>O 9- 



54 MATHEMATICS FOR ENGINEERS 

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

e x er x 



sinh x = 
and cosh x = 



2 

e x -\-e- x 



2 

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

Hr^ I _ / x*g I x> \ 

jncii^c j 7 i ~ / lu ~ i c? i 

A; A;\ 2 / 2 

= cosh x 

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

also j- cosh = j-( - - ) = -wfg-'i 

dx dx\ 2 / 2 V 

= sinh #. 



Example 22. Find the inclination to the horizontal of a cable 
weighing \ Ib. per ft. and stretched to a tension of 30 Ibs. weight, at 
the end of its span of 50 ft. 

The equation to the form taken by the cable is 

/ x x \ 
y = -\e c -\-e c ) = c cosh - 

horizontal tension 30 ,. 
where c = --- C r -*_ =60. 

weight per foot -5 

We require the slope of the curve when x = 25, this being given 

cl/\} 
by the value of - there. 

dy d , x i . . x > x 

~- = J- c cosh = = c x > sinh >- = sinh 7 - 
dx dx 60 60 60 oo 



. < 

When x 25 -/ = sinh -^ = sinh -4167 = 



g- 



4167 _ g .4167 



- 

dx 60 2 

= 1-517 659 

2 

= -429. 



This value is that of the tangent of the angle of inclination to 
the horizontal; which is thus tan- 1 -429 or 23 13'. 



DIFFERENTIATION OF FUNCTIONS 55 



Exercises 4. On the Differentiation of a x , Log x and the Hyperbolic 

Functions. 

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

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

& 

. 7. ^x e . 8. 14x2*. 9. 4ie 



foa-iX -jpl-ZX 

10. 3 -i4^-5*- + 3-i te +6. 11- ,VX^T*E- 12. log 7*. 
13. 3 log (4 5*). 14. iolog 10 8#. 15. c log (4*0 +56). 

16. 9-- te -log -2*+^. 17. log 2*(3#-47). 



18. log y)"-- 19- (**) 3 +4 cosh 2*- 17 lo glo 2-3*. 

20. log 3* 2 +5#-' 7 i -8 (i -8*) + 12. 

21. If y = A^-^+A^-^find the value of S+7 

2 ' ' 



22. Find -5- log (3 4^) when v = 17. Check your result approxi- 

CL1) 

mately by taking as values of v 1-65 and 175. 

23. Determine the value of -,- 71 log (18 -04^). 

(I If- 

24. Write down the value of -j- Iog 10 18^. 

at 

25. If T = 5oe' Zd , find the rate of change of T compared with 
change in 6. 

/ _BA 

26. If C = CQ^I e L '), C and C being electrical currents, R the 
resistance of a circuit and L its self-inductance, find the rate at which 
the current C is changing, / being the time. 

dl) 

27. Given that v = 2-03 Iog 10 (7 i-Su), find ^-. 

28. Evaluate -=- 5 cosh - and also -j- p sinh -. 

ax 4 ay r q 

29. An electromotive force E is given by 

E A cosh Vlr . #+B sinh Vlr . x. 

Find the value of -3-3 in terms of E. 

30. If W = i44J/) 1 (i f log r) rpb], find the value of r that makes 
-T = o ; W being the work done in the expansion of steam from 
pressure p t through a ratio of expansion r. 

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

ax ' * a a 



56 



32. If y = Ae Zx -\-'Be 3x +Ce-* x > find the value of 
d 3 y d z y dy 



d z V Vy x x/ n -*, -J.x 

33. Evaluate -7-2 -- when V=A 1 e x ^ r 2 +A 2 e ^ r 2 

W^? ? 2 

34. Nernst gives the following rule connecting the pressure p of a 
refrigerant (such as Carbon Dioxide or Ammonia) and its absolute 
temperature T 

p = A+B log T+Cr+5 

T 

where A, B, C and D are constants. Find an expression for -J-. 

Differentiation of the Trigonometric Functions. Before 
proceeding to establish the rules for the differentiation of sin x 
and cos x, it is well to remind ourselves of two trigonometric 
relations which are necessary for the proofs of these rules, viz. 

(a) When the angle is small, its sine may be replaced by the 
angle itself expressed in radians, i. e. 



T sm6 = T (cf part ^ p 458)> 



(b) sin A-sin B = 2 cos sin - ( c f. p ar t I, p. 285). 

\ 2 / 2 

To find -v- sin x we proceed as in former cases ; thus 



Let y = sin # and y+8y = sin (#-f 8#) 

then 8y y -\-8y-y = sin (x-\-8x)sin x 

(2x+8x\ 



= 2 cos ( - sin 

\ 2 / 



Dividing through by 8x 

_ 2 COS \ ~~ / OAII i 

8y V 2 / \2 



/2^+8^\ . /Bx\ 
I sin ( J 



8* 8x 

(2x+%x\ f8x^ 

COS ( I sin I 



t\ (8X\ 

- ) sm ( I 

/ \2/ 




DIFFERENTIATION OF FUNCTIONS 57 

The limiting value of ?- is --. and that of the right-hand side 

is cos x, since cos ( x-\ J approaches more and more nearly to 
cos x as 8* is made smaller and smaller, and the limiting value of 

/sin \ 

/ 2 \ .,,.,., sin 9 . 

I, or, as we might write it, ^ , is I. 







Hence 



d sin x dy 

j or * = 
dx dx 



L 



-^ = cos x 



dsinx 
dx 



= cos x. 



75 
-5 
25 
O 



-y 



-y- 



.JG 



- 

7 



.3G 



TT. 



x 



JG 






FIG. 14. Curves of y = sin x and y = cos x. 



By similar reasoning the derivative of cos x may be obtained ; 
its value being given by 



The graphs of the sine and cosine curves assist towards the 
full appreciation of these results. In Fig. 14 the two curves are 
plotted, and it is noted that the cosine curve is simply the sine 
curve shifted backwards along the horizontal axis : thus the slope 
curve and the primitive have exactly the same shape. This 
condition also holds for the primitive curve y = e? 1 , and so suggests 
that there must be some connection between these various natural 
functions; and further reference to this subject is made later in 
the book. 



58 MATHEMATICS FOR ENGINEERS 

Much trouble is caused by the presence of the minus sign in 

d COS OC 

the relation -5 = sin x, it being rather difficult to remember 

whether the minus sign occurs when differentiating sin x or cos x. 
A mental picture of the curves, or the curves themselves, may be 
used as an aid in this respect. The cosine and sine curves differ 
in phase by period (see Fig. 14), but are otherwise identical. 
Treating y = sin x as the primitive : when x is small, sin x and x 
are very nearly alike, and thus the slope of the curve here is i; 

as x increases from o to - the slope of the curve continually 

7T 

diminishes until at x = - the slope of the curve is zero. Now 
the ordinate of the cosine curve when x = o is unity, and it 

diminishes until at x = ~ it is zero. From x = ~ to x = -n- the 
2 2 

slope of the sine curve is negative, but increases numerically to I, 
this being the value when x = TT; and it may be observed that 
the ordinates of the cosine curve give these changes exactly, both 
as regards magnitude and sign. Thus the cosine curve is the 
slope curve of the sine curve. 

Now regard the cosine curve as the primitive. At x = o the 

curve is horizontal and the slope = o ; from x = o to x = ~ the 

2 

slope increases numerically, but is negative, reaching its maximum 
negative value, viz., i, at x = - ; but the ordinates of the sine 

2 

curve are all positive from x = o to x = ~, so that although these 

2 

ordinates give the slope of the curve as regards magnitude, they 
give the wrong sign. In other words, the sine curve must be 
folded over the axis of x to be the slope curve of the cosine 
curve, i. e., the curve y = sin x is the slope curve of the curve 

y = COS X. 

To summarise, we can say that the derived curve for the sine 
curve or for the cosine curve is the curve itself shifted back along 
the axis a horizontal distance equal to one-quarter of the period. 

Thus we can say at once that the slope curve of the curve 
y = sin (x-\-b) is the curve y cos (.r+fr), since the curve 
y = sin (x+b) is the simple sine curve shifted along the horizontal 
axis an amount given by the value of b, the amplitude and period 
being unaltered. 



DIFFERENTIATION OF FUNCTIONS 59 

Thus -,- sin (x+b) = cos (x+b) 

and, in like manner 

j x cos (x+b) = - sin (x+b). 

Again, -7- sin (5#+6) = 5 cos (5#+6), since 5 multiplies the 

I.V. in the original function. 
Then in general 

?- A sin (Bx+C) = AB cos (Bx+C) 
-? A cos (Bx+C) = AB sin (Bx+C). 

To differentiate tan x with regard to x. 
Let y = tan x and (y+8y) then = tan (x-\-8x) 

8y = y+Sy y=tan (x+8x) tan x 

_ sin (x-\-8x) sin # 
~ cos (x+8x) cos A; 

_ sin (x+Sx) cos A; cos (*+SA;) sin x 



cos #8# cos x 
sin 



cos (x+8x) cos x 

_ sin 8x 

~ cos (A; +8x) cos x 

Dividing through by 8x 

Sy _ sin 8x I 

8x ~ 8x cos (x+8x) cos x 



Now as 8x approaches zero, - - approaches i and (x + 8x) 



approaches x. 



dx ,8* * ^ cos A; cos x cos 2 A; 



Hence =1 = IX- ~= *~ = sec 2 x. 

\ _ , 



3- tan x = sec 2 x 



60 MATHEMATICS FOR ENGINEERS 



In like manner it can be proved that 

d cot x 

3 = cosec 2 x 

d seex sinx 
dx ~ cos 2 x 

d cosec x cos x 

and -j~ = -- r-s 

dx sin 2 x 

To generalise 

^ A tan (Bx+C) = AB sec 2 (Bx+C) 

J^ A cot (Bx+C) = AB cosec 2 (Bx+C) 

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

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

AB 



. 
cos 2 (Bx+C) 

Example 23. Find the slope of the curve representing the equation 
s = 5-2 sin (40^2-4) when t -07. 



The slope of the curve 

. . . 

5 ' 2 Sm (4^ 2 '4) 

= 208 cos (40^2-4). 



ds d . . . 

= ~dt = dt 5 ' 2 Sm (4^ 2 '4) = 5-2x40 cos (40^2-4) 



Hence when 

t = -07, the slope = 208 cos (2-82-4) = 208 cos -4 (radian) 

= 208 cos 22-9 
= 192. 

Example 24. Differentiate, with regard to z, the function 
9-4 cot (75?). 

,- 9'4 cot (7-5*) = 9'4 X -5 X - cosec 2 (7-5?) 

U6 

= 47 cosec 2 (752). 

Simple Harmonic Motion. We can now make a more strict 
examination of simple harmonic motion. Suppose a crank of 
length r (see Fig. 15), starting from the position OX, rotates at 
a constant angular velocity <o in a right-handed direction. Let it 
have reached the position OA after t seconds have elapsed from 
the start; then the angle passed through in this interval of time 
= AOM = <&t, since the angular distance covered in I sec. = o> 
radians and the angular distance in t seconds = a>t radians. 



DIFFERENTIATION OF FUNCTIONS 



61 



Considering the displacement along the horizontal axis, the dis- 
placement in time t = s = OM 

= AO cos AOM = r cos tat. 

ds 
Then the velocity = , , = rXta sin tat = rta sin <at 



and the acceleration = ,, = 
at 



cos tat = ta z xr cos wt 

= 0) 2 S 




*'. e., the acceleration is proportional to the displacement, but is 
directed towards the centre : thus, when the displacement from 
the centre increases, the acceleration towards the centre increases. 
When the displacement is greatest, the 
acceleration is greatest : e. g., if the 
crank is in the position OX, the 
acceleration has its maximum value 
wV and is directed towards the centre, 
just destroying the outward velocity, 
which at X is zero. At O the acceler- 
ation = u> 2 x o = o, or the velocity is 
here a maximum. 

An initial lag or lead of the crank 

does not affect the truth of the foregoing FlG r . 

connection between acceleration and 
displacement. The equation of the motion is now s = r cos 
where c is the angle of lag or lead, and the differentiation to find the 
velocity and the acceleration is as before. 

Example 25. If s 5 sin 4* 12 cos \t, show that this is the 
equation of a S.H.M. and find the angular velocity. 

s = 5 sin ift 12 cos 4^. 
Then v = , $ = (5 x 4 cos 4*) (12 x 4 X sin 4*). 

HP 

= 20 cos 4^+48 sin 4^ 
and a = " = (20 x 4 X sin 4*) + (48 x 4 cos 4/) 

= 80 sin 4^+192 cos 4* 

= 16(5 sin 4^12 cos 4/) = i6s 

i. e., the acceleration is proportional to the displacement. 
Now, in S.H.M., the acceleration = 2 s. 

&> 2 = 16, i. e., w = angular velocity 
= 4 radians per sec. 



62 MATHEMATICS FOR ENGINEERS 

This last question might be treated rather differently by first 
expressing 5 sin 4^12 cos 4^ in the form Msin(4^+c) (see 
Part I, p. 276) and then differentiating. This method indicates 
that a S.H.M. may be composed of two simple harmonic motions 
differing in phase and amplitude. 

Exercises 5. On the Differentiation of Trigonometric Functions. 

Differentiate with respect to x the functions in Nos. i to 16. 
1. sin (4 5'3x). 2. 3-2 cos 5-1*. 3. -16 tan (3X-\-g). 

4. 2-15 sin i 1 ^~ 5 ). 5. 8 cot 5%. 

\ 4 / 

6. 43-15 sec (-05 -117*). 7. be cos (dgx). 

8. 4 cos $x 7 sin (2^5). 9. sin 5-2^ cos 3~6x 

10. 2-17 cos 4-5* cos 1-7*. 11. 9-04 sin (px+c) sin (qxc). 

12. 5 sin 2 x. 13. -065 cos 2 3*. 

14. cos 2 (7* i-5)+sin 2 (7* 1-5). 

15. 3* 1 - 72 5- 14 log (3#-4-i) + -i4 sin (4-31 -195*) + 24-93*. 

16. 7-05 sin -015* -23 cos (6-i -23*) + 1-85 tan (4* -07). 

17. x, the displacement of a valve from its central position, is given 
approximately by x = 1-2 cos a/ 1-8 sin ^ where w = angular 
velocity of crank shaft (making 300 r.p.m.) and t is time in seconds 
from dead centre position. 

Find expressions for the velocity and acceleration of the valve. 

18. If 5 = 4-2 sin (2-1 -172) -315 cos (2-1 -J7/), s being a displace- 
ment and t a time, find an expression for the acceleration in terms of 
s. What kind of motion does this equation represent ? 

19. The current in a circuit is varying according to the law 
C = 3-16 sin (2irft 3-06). At what rate is the current changing when 
t -017, the frequency / being 60 ? 

20. If the deflected form of a strut is a sine curve, what will be 
the form of the bending moment curve ? 

21. If y = deflection of a rod at a distance x from the end, the 
end load applied being F 

Bl 



y=- 



8 COS 1 



Find the value of EI~^-}-Fy-\-- & cos -j-; y and x being the only 

variables. 

22. The primary E.M.F. of a certain transformer was given by the 
expression 

E = 1500 sin pt-\-ioo sin 3^42 cos ^+28 cos 3pt. 
Find the rate at which the E.M.F. varied. 

T 2 

23. A displacement s is given by s = sin izt sin 13^. Show 

that the acceleration = 25 sin 12^1695. 



CHAPTER III 
ADDITIONAL RULES OF DD7FERENTIATION 

Differentiation of a Function of a Function. \Vhilst the 
expression e sin 4 * is essentially a function of x, it can also be spoken 
of as a function of sin 4*, which in turn is a function of x; and 
thus it is observed that e aa ** is a function of a function of x. 
This fact will be seen more clearly, perhaps, if u is written in place 
of sin 4* : thus e sin ** = e w , which is a function of u, which, 
again, is a function of x, since u = sin 4*. 

To differentiate a function of a function the following rule is 
employed 

dy_=dy x du 

dx du dx 

\ 
and this rule is easily proved. 

Let y be a function of u, and let u be a function of x : then y 
is a function of a function of x. Now increase # by a small 
amount 8x; then since u depends on x, it takes a new value 
u + 8u, and also the new value of y becomes y + Sy. Since these 
changes are measurable quantities, although small, the ordinary 
rules of arithmetic can be applied, so that 

8y __Sy 8u_ 
Sx~Su X Sx 

When Sx approaches zero these fractions approach the limiting 

values -, and - respectively : and thus in the limit 

ax du dx 

dy _ dy du 
dx du dx 

In like manner, if y is a function of u, u a function of w, and 
w a function of x, it can be proved that 

dy_dy..du_dw 
dx~du x dw x dx 
63 



64 MATHEMATICS FOR ENGINEERS 

It will be observed that on the right-hand side of the equation 
we have dy as the first numerator and dx as the last denominator 
(these giving in conjunction the left-hand side of the equation) ; 
and we may regard the other numerators and denominators as 
neutralising one another. The simple arithmetic analogy may help 
to impress the rule upon the memory : thus 



Example i . If y = e 8to tx , find the value of ~ . 

dit 
Let u = sin 4*, so that j- = 4 cos 4* and y = e u . 

Since y is now a function of n, we can differentiate it with regard 
to u, whereas it is impossible to differentiate with regard to x directly. 

y = e u and ~ = ~ = e u = e s^ **. 

du du 

T,, dy dy du 

Then, since -, - = -/-x j- 

dx du dx 

dv 

JL e sui tx x ^ cos 4 # 

= 4 cos *g sin 



Example 2. Find the value of j~log (cos 2x} 3 . 

Let v = (cos 2x) 3 and u = cos 2x; and thus y = loge v and u = w 3 . 

dy dy dv du u = cos 

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



d log v du 3 d cos 2.x 
~~dv~ X ~du X dx 






X 2 sin 



-j- 2 sin 

V = U 3 

dv 

-j- = 3 U 
du 



_ 6 sin 2XX (cos 2^r) 2 _ 6 sin 2 

(cos 2#) 3 cos 2x 

= 6 tan 2 



Example 3. The radius of a sphere is being decreased at the rate 
of -02 in. per min. At what rate is (a) the surface, (b) the weight, 
varying, when the radius is 15 ins. and the material weighs -3 Ib. per 
cu. in. ? 

dr 
If r radius, then - ,- = rate of change of the radius, and is in 

dt 

this case equal to -02. 






ADDITIONAL RULES OF DIFFERENTIATION 65 

(a) The surface = 4irf 2 , and thus the rate of change of surface 

_ dS 
~ dt 

_ d . 4*r z 
dt 

dr* 
= * v '^dt 

dr* dr 
= * v -~dr X df 

dr 
5 

'02. 



Hence when r = 15, -5- = 8v x 15 X -02 = 7-53, i. e., the surface 
is being diminished at the rate of 7-53 sq. ins, per min. 

(b) The volume = 4 ^ 
so that the rate of change of volume = = jft-***) 

and the rate of change of the weight = -rr = -j-.( - X -Sirr 3 j 

<AV d dr 3 dr 3 dr 

~rr =~JT- '4*r '4* -JT = '4* X -j- X T-. 



-02. 

When r = 15, -, = -4^x3x225 x -02 = 16-93 

or the weight is decreasing at the rate of 16-93 I DS - per min. 

Example 4. Find expressions for the velocity and acceleration of 
the piston of a horizontal steam engine when the crank makes n 
revolutions per second. 

In each turn the angle swept out = 2ir radians. 

Hence in i second 2vn radians are swept out, i. e., the angular 
velocity = 2.im; and this is the rate of change of angle, so that 
dQ 

dT= 2vn - 

From Fig. 16 CD = / sin a 

and CD = r sin 6. 

Thus I sin a = r sin 6 

/ r 

or sin = - sin a, and sin a = T sin 0. 

r I 



66 



MATHEMATICS FOR ENGINEERS 



Again, cos = Vi sin 2 a 



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

r z 
and 72 still smaller, so that our method of approximation can be 

p 

applied to the expansion of the bracket, i. e. 

I y^ 

cos o = i 72 sin 2 0, very nearly. 




FIG. 16. Velocity and Acceleration of Piston. 

Let AB = displacement of the piston from its in-dead-centre position 
= x = AE+OE BO = l + r-BV DO 

= l+rl cos o r cos 6 



= l+rl (i p sin 2 0)-y cos 



y+~, sin 2 Y cos 



= r-\ 
dx d 



i cos 20 

2 

r 2 cos 20 



-r cos 



fl 

~ r COS U. 



r 2 r 2 cos 20 



XT . i. i . ,,-. . dx a f 

JNow the velocity of the piston = -=- = --=- -j , --- -. 

We cannot differentiate this expression directly, so we writ 



n 
r cos 



TT 
Hence 



dx __df r* 



dx dx dQ 

dQ 

COS 20 



~dt = 50 X dt' 



,.) dQ 
rcosQ\x,. 
J a/ 



= -jo-f o (~ x 2 sin 20J (rx sin 0) j- x 2irW 

fy sin 20 . . 1 
= 2trwy -j - j -- f-sm j- 



or if 



- = m 
r 

dx 
V = ~dt = 



f sin 20 

\ 2m 



ADDITIONAL RULES OF DIFFERENTIATION 67 

dv dv dQ d (sin 20 , _i d0 

Also the acceleration = -rr = -j^X^-^,. zvnrl --- f-sm 6 [ x -r- 
at ay at at) I 2m at 



fcos 26 , n ) 

= 2irnr 1 -- 1- COS 6 h X 2irW 
f Q . COS 26"| 

= 4ir 2 n 2 ' { cos 0H 
* I / 



Example 5. Water is flowing into a large tank at the rate of 
200 gallons per min. The reservoir is in the form of a frustum of a 
pyramid, the length of the top being 40 ft. and width 28 ft., and the 
corresponding dimensions of the base being 20 ft. and 14 ft. ; the 
depth is 12 ft. (see Fig. 17). At what rate is the level of the water 
rising when the depth of water is 4 ft. ? 




In 12 ft. the length decreases by 20 ft., and therefore in 8 ft. the 
length decreases by , i. e., 13 J ft., so that the length when the 

water is 4 ft. deep is 40 13^ = 26| ft. 

Similarly, the breadth = 28 (f x 14) = i8f ft. 

i. e., the area of surface = 26f x i8| = 498 sq. ft. 

200 ,, 

200 gals, per mm. = ^ cu. ft. per mm. 
0*24 

= 32-1 cu.. ft. per min. 

i. e., the rate of change of volume = rr = 32-1. 

dv 

Now -v- = -v . Ah, where A = area of surface 

at at 

and h = depth of water, 

*dh f since for the short interval of time considered the\ 
dt \ area of the surf ace may be considered constant./ 

Hence the rate of change of level = -JT = -jr X -z- 

dt dt A 

32-1 xi , ,. 

= Q = -0644 ft. per nun. 
49 

= 773 in. per min. 



68 MATHEMATICS FOR ENGINEERS 

Example 6. If a curve of velocity be plotted to a base of space, 
prove that the sub-normal of this curve represents the acceleration. 

d'v 
The sub-normal of a curve = y~- (see p. 43). 

CbX 

In this case, since v is plotted along the vertical axis and s along 
the horizontal axis 

the sub-normal = v-^- 

ds . 

dv dt 
V 'dt X ds 

= vxax-- 
v 

= a 
[for -=r = rate of change of velocity = a\ 



dt 
ds 
dt 



and -3T = rate of change of space = v I 



As a further example of this rule, consider the case of motion due 
to gravity; in this instance v 2 2gs, i.e., the velocity space curve 
is a parabola. Hence we know that the sub-normal must be a 
constant, i. e., the acceleration must be constant. 

The sub-normal = v- r 
ds 

, T dv 2 d ds 

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

dv z dv z dv dv 

but , = -j . -T- = 2v j- 

ds dv ds ds 

dv 

2V-j- = 2g 

ds 



dv 
ds 

i. e., the sub-normal or the acceleration = g. 



V ds = 



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

T-" j A d sin 2x d . d 

Find 1. -3- . e . 2. -T- log v 2 . 3. ^2cos 2 t. 

dx dv dt 

d _ d n d sin 5x 

dx Sm dx 3 ' 14 (5^ 2 +7^~ 2 )- 6. ^a 

7< ^ 1<88 ' 8< dx logl (3 + 7^-9^ 3 ). 9. ^ cos (log s 5 ). 

and 10. ~ log tan -. 
a* 2 






ADDITIONAL RULES OF DIFFERENTIATION 69 

11. In the consideration of the theory of Hooke's coupling it is 
required to find an expression for , i. e., a ratio of angular velocities. 

If o> B = r, <" A = -3T and tan <f> = - , find an expression for 

B in terms of the ratios of 9, </> and o. 

"A 

12. Find an expression for the slope of the cycloid at any point. 
The equation of the cycloid is x = a (6 + sin 6) 

y a(i cos 6) 
the co-ordinates # and y being measured as indicated in Fig. 18. 




^Rolling Circle 
FIG. i 8. 

13. Assuming that the loss of head due to turbulent flow of water 
in a pipe is expressed by h C(AV 2 +BV?), where V = mean velocity 
of flow in ft. per sec. ; show that the slope of the curve in which log h 
and log V are plotted with rectangular co-ordinates is given by 



d log h 
dlogV 



2A 






14. If 3x*+8xy+5y 2 = i 

show that T = T 

dx z (. 

15. A vessel in the form of a right circular cone whose height is 
7 ft. and diameter of its base 6 ft., placed with its axis vertical and 
vertex downwards, is being filled with water at the rate of 10 cu. ft. 
per min. ; find the velocity with which the surface is rising (a) when 
the depth of the water is 4 ft. and (b) when 60 cu. ft. have been 
poured in. 

16. If p = (r) K , prove that-j^ = ~ffi( r^ 1 r - 

17. If x 3 6x 2 y 6xy z +y 3 constant, prove that 

dy _ x z 4xyr--* 
dx = 2X*~ 



7 o MATHEMATICS FOR ENGINEERS 

18. A ring weight is being turned in a lathe. It is required to 
find the weight removed by taking a cut of depth ^thj". The material 
is cast iron (-26 Ib. per cu. in.), the outside diameter of the ring is 
3-26" and the length is 2-5'*. Find the weight removed. 

Find also a general expression for the weight removed for a cut 
of depth ^J^" at any diameter. 

19. Find the value of -rA log tan 



20. If P = -^TT, and -.., = u, find -w. (This question has refer- 



-.., 
ence to stresses in redundant frames.) 

21. Find the angle which the tangent to the ellipse \- = 2 at 

4 9 
the point x = 2, y 3, makes with the axis of x. 

22. Find the slope of the curve 4# 2 +4y 2 = 25 at the point x = 2, 



y = f, giving the angle correct to the nearest minute. 

23. If force can be defined as the space-rate of change of kinetic 

, , . , . wv z , , wa 

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

o o 

dx 

24. If x 8 log (i2t 3 74), find the value of ,-,. 

ctt 

Differentiation of a Product of Functions of x. It has 

already been seen that to differentiate the sum of a number of 
terms we differentiate the terms separately and add the results. 
We might therefore be led to suppose that the differentiation of 
a product might be effected by a somewhat similar plan, viz., by 
multiplication together of the derivatives of the separate factors. 
This is, however, not the correct procedure ; thus 

d ,, , , . d log x dx z . i 

j- (log xxx 2 ) does not equal , --X-T , ^. e., X2x or 2. 

dx v dx dx x 

The true rule is expressed in the following manner : If u and v 
are both functions of x, and y = uv, i. e., their product 

dy d , . du . dv 

B -*"*-+ 

Proof. Let x increase by an amount 8x; then since both u 
and v are dependent on x, u changes to a new value w+Sw and 
v becomes v -f- 8v. 

Now y = uv, and hence the new value of y, which can be 
written y-\-8y, is given by 



but 



y = uv 






ADDITIONAL RULES OF DIFFERENTIATION 71 

whence by subtraction 

Sy = y+8y y = (u-}-8u)(v-}-8v)uv 

= uv-}- u8v -\-v8u-\-8u . 8v uv 
{-8u . 8v. 



Dividing through by 8x 

8y_ 8v <^ , s 8v 
Sx ~ U ^x^ rV ^c r U ' 8x 

As 8x is decreased without limit, ~, -=- and - approach the 

8x 8x 8x ^^ 

values -/ , -y- and -3- respectively, and the term 8u . . becomes 
dx dx dx 8x 

negligible ; so that in the limit 

dy _ du dv 
dx dx dx 

The rule may be extended to apply to the case of a product 
of more than two functions of x. Thus if u, v and w are each 
functions of x 

dluvw) d(wV) , ,, . , 

v , ' = -j~, where V is wntten for uv 
dx dx 





- w dV ^ 


.j dw 






dx^ 


V dx 






nn d(uv) 


dw 




dx 


bUV dx 


( du . dv\ . 
= w( v-j \-u-r- )+i 
\ dx dxj ' 


dw 

wU "^ 

dx 


and thus 
d(uv 


w\ du. 
' mil i 


dv 


du 
itit 






Example 7. Find - when y x z . log x. 



T i , 

Let u = x* so that -r-= zx 

dx 

and let v = log x so that -j- = -. 

(Kx X 

, d .uv du dv .. . , / i \ 

Then = = v^- +Uj~ = (log xx 2x) + \ x z . - I 

dx dx dx ^ { '^v xJ 

= x(l+2 log X). 



72 MATHEMATICS FOR ENGINEERS 

Example 8. Find the value of -7,[5e~ 7 ' . sin (6^ 4)] 

ctt 

cLi>t 
Let u = ^e-~ l so that = 5x je-' 1 35e~ 7t 

and let v = sin (6/ 4) so that -,7 = 6 cos (6^4). 

ttt 

d . uv du , df 

-3T = "-^ +M -^ 

= [sin (6/- 4 )x -35*- 7 <] + [5*- 7 ' x6 cos (6/- 4 )] 
= 5g~ 7 *[6 cos (6^4) 7 sin (6^4)]. 

Example g. If 2q+~ (px z ) o, show that 2q = 2p xf 
x dx ax 

p being a function of x. This example has reference to thick spherical 
shells. 

If p is a function of x, px* is of the form uv, where u = p and 
v = x 2 . 

d 9 dp , dx 2 9 dp , 

Hence -,- . px 2 = x 2 ,"-\-p- r - = x 2 -/-4-2Xp. 

dx ' r 



Hence 2q-\ --- 3- . px z zq+x ,- 

1 x dx ' dx 

dp 

i.e., o = 2q+x , r 

Q//V 



Example 10. Find the value of -y- gx* sin (3^7) log (i 5#). 

Let u = x*, v = sin (3^7) and w = log (1 5*) 

,, du dv dw 5 5 

then -,~ = 4^ 3 , -; = 3 cos (3^7) and -y- = = 

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



^ 

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

5x) sin (3^-7)4Ar 3 +{log (i 5*)* 4 X3cos (3^7)} 
sin (3* 7) 



= 9* 3 [_4 sin (3Af 7) log (1 5*) + 3* cos (31* 7) log (1 5*) 

5#sin (3* 7)1 
_ + 5*^T~ 



ADDITIONAL RULES OF DIFFERENTIATION 73 

Exercises 7. On the Differentiation of a Product. 

Differentiate, with respect to x, the functions in Nos. i to 12. 
1. x 2 sin 3*. 2. log 5#X 2# 3 - 4 . 3. e 9 * Iog 10 gx. 

4. 4Ar- 5 .tan (3-1 2-07*). 5. cos 3-2* cos 

6. cos (5 3*) tan 2#. 7. 8* 1 - 6 cos 

8. 9\ogx 3 .5 3 *. 9. e*ig*. 10. ^* 

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

o 

13. If y = Ae 3 * cos ( + B), find the value of 



14. Find the value of ^/~ 5< cosh ( 5/). 

15. y = (A + B*)*- 1 *; find the value of ^+8 

W^ W 

16. If V = 250 sin (jt -116), A = 7-2 sin 7* and W = VA, find 

, d\V 

the value of rr. 
at 



*17. Differentiate with respect to t the function i^t 2 sin (4 -8tf). 
18. Find the value of -r,(4* 3 ' 7 cos 3/). 

H 

Differentiation of a Quotient. If u and y are both functions 



of x, and y = -, then 



dy V dx u dx 



dx i> 2 

Proof. 

(a) From first principles. Let y = - : then a change Sx in x 
causes changes of Sy in y, Su in u, and Sy in y, so that the new 
value of y = y+Sy = -^pr-. 

u u uv-\-v8u y uSv 



, 

Then Sy = y+Sy y = 



and, dividing through by Sx 
Sy i 



. . 
y(y+Sy) 



_ 
' Sx~ U ' Sx 



74 MATHEMATICS FOR ENGINEERS 

When 8x becomes very small, P-, .- and ^- approach the values 

8x 8x 8x 

//-Af CL'IA/ Ul) 

-J-, -j- and -j- respectively, whilst v-{-8v becomes indistinguishable 

from v. 

TT . ,, ,. ., dy i ( du dv\ 

Hence in the limit -r- = (v . -* u . -r- 

dx vxv\ ax ax/ 

du dv 

V-j tt-j- 

dx ax 



V 2 



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



u 

y = - = 
v 



TM dy d , ,, ,du , dir 1 

Then ~" = ' (UV ^ = V + U ' ~ 



x dx 

U-v (( -v 

I du\ dy- 1 dv 



du\ . ( 9 dv 

_ I If V Tit V 

. j / T^ I M X\ At/ /\ j 

.v a*/ \ ^ 

du dv 

V ~j W ~j~~ 

dx dx 



Example n. Differentiate, with regard to 5, the expression 



5 cos (35+4)' 

T , ,-, du 

.Let w = 45^+75, tnen 3- = i2s^ + 7 

as 

dv 
and let v = 5 cos (35+4), then -,- = 15 sin (35 + 4). 

ivS 

ZM tiy 



,, d (u\ ds d 
Then -.- . I - ) = - 

ds \v/ v z 



= [5 cos (35+4) X (i25 2 +7)]-[(4S 3 + 7*) X -i5sin(35+4)] 
25 cos 2 (35+4) 



5 cos 2 (3^+4) 




Example 12. If y 9 4 *X, -- -, find the value of -- 
Let 



u = g 4 *, then j- = 4 x 9 4 *log e 9 = 4 X 2-1972 x g* 

= 8-789 X9 4z 



and let 



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

du dv 

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

dx dx\v I v z 



u i 
dx 



(log yx x 8 -79 x 9 4z ) (g 4 * X ^ 

(log 7#) 2 
9 4 ^{(8-79^xlog7^)-i} 



(log 




FIG. 19. Spring loaded Governor. 
Example 13. For a spring loaded governor (see Fig. 19) 



where Q = force to elongate the spring i unit, T = tension in 
spring, W = weight of i ball, = angular velocity, r radius of 
path of balls, / = length of each of the 4 arms. 

If W = 3, g = 32-2 and -j- 80 when o> = 26, r = -25 and / = i, 

find T and Q. 

As there are two unknowns, we must form two equations. By 
simple substitution 



_ 32'2{T+2Q(i- V 10625)} 
~" 



V 10625 
- "968 

whence T+-o64Q = 60-96 ....... (i) 

We are told that -^ must equal 80. 

dta 



-vr 
NOW 

Also 



-5- = -,- X j- = 2w T- 

dr da> dr dr 



(2) 



d 
dr 



where 
and 



r* dr\v 

u = g{T+2Q(l Vl*r z )} 
v = 



76 MATHEMATICS FOR ENGINEERS 

Thus to determine -r- and -y- it is first necessary to find the value 

rJ A//2 4,2 

of : to do this let l*-r* = y 



dr 
then 

Thus 
and 



du dv 

, VT U^ 

,, aw dr dr 
Then - - = 



so that ' zr 
dr 




/.,\ 



Thus, differentiating both sides of the original equation with respect 
to r, we have from (2) and (3) 



2oi T- = , 

rfy W 

Substituting the numerical values 

2x26x80 = 



52X ; ^9375 X- 968 _ .^SQ+T+^Q . 2 g +T 

whence J4Q7 = 2 Q+T 

but from (i) 60-96 = -064(3 +T 

and therefore Q 695-3! 

and T = 16-4! 

Differentiation of Inverse Trigonometric Functions. 

Since inverse trigonometric functions occur frequently in the study 
of the Integral Calculus, it is necessary to demonstrate the rules 
for their differentiation; and in view of their importance in the 
later stages of the work, the results now to be deduced should be 
carefully studied. 

The meaning of an inverse trigonometric function has already 
been explained (see Part I, p. 297), so that a reminder only is 



ADDITIONAL RULES OF DIFFERENTIATION 77 

needed here. Thus sin~ x x is an inverse trigonometric function, 
and it is such a function that if y = sin" 1 x, then sin y = x. 

To differentiate sin~ l x with regard to x. 

Let y = sin -1 x so that, from definition, sin y = x 



then 
but 
and hence 
or 

milarlv 


dx ~ dx~ 
d sin y d sin y dy 


dx dy dx 

dy 
i = cosyx^ 

dy T. i * 


d# ~~ cos y Vi sin 2 y ~ Vix 2 

d in 1 y 1 


-? sin jc / 
dx Vi y 2 

d 1 

^ COS 1 Y / 



(x being supposed to vary between o and -). 



Example 14. Find the value of -5- tan- 1 -. 



Let 


y = 


tan- 1 -, 
a 


*'. e., 


tan y 


and 

Now 
but 


sec 2 y = 
d tan y 


i+tan* 
d (x\ 


i 


a 2 


a 2 


d# 
d tan y 


dx \a' 
d tan y 


a 

v rf y 


d* 


dy 


a^ 


Hence 


i 
a 


sec 2 yx 


S 






tfy 


i 


i 


a 2 



d# a sec 2 y a a 2 +# 2 
a 



78 MATHEMATICS FOR ENGINEERS 

Example 15. Find the value of ~ cosh" 1 -. 

ax a, 



Let 


y 


, - X 

= cosh" 1 
a 


then 


cosh y 


a' 


So fli'if 


d cosh y 


d (x\ i 




dx 


d!# \a) a 


i nit 


d cosh y 


d cosh y dy 




rf# 


dy X dx 


hence 


i 


= sinh y X ^ 




a 


^ rf* 



(i) 
Now cosh 2 y sinh 2 y = i 

X s 

whence sinh 2 y = cosh 2 y i = -g i 



a* 

and sinh y = - \/# 2 a 8 

Then, substituting this value for sinh y in (i) 



2 y 

a x 



dy , i 
or -/ = >-=- 



d , x . 
v- cosh" 1 = 




2 2 



Exercises 8. On the Differentiation of a Quotient and the 
Differentiation of Inverse Functions. 

Differentiate with respect to x the functions in Nos. i to 12. 

1 5^1 2 log ( 2 ~7*) 

e lx ~ 5 ' cos (2 7*)' 

3. . ^X _. O^f 

. 5 sin . Tf* cos ji. 

7 rf 2 

_ 5 2 - to _ cosh 1-8^ 

& - g9^i' 4 l-8a: - 

I *+3 7 cos- 1 3^ 

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

q ^o(a x)x ._ ^ 

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



.. / 3 6l z x -j- 1 2/^r 2 7* 3 (an expression occurring in the solution of 
" 3^4-^ a b 63 - 111 problem). 



12. 



ADDITIONAL RULES OF DIFFERENTIATION 79 

e sin (l-ar+1-7) 



log (8**-7*+3)' 
13. Assuming the results for T- cosh # and T- sinh #, find the value 

of -j- tanh #. 
d# 

Nos. 14 and 15 refer to the flow of water through circular pipes; 
v being the velocity of flow, Q the quantity flowing, and being the 
angle at the centre subtended by the wetted perimeter. 



i T* I sm 

14. If, = 13-1(1 Q 



sin 26 ^0 

17. If (a velocity) = r* (sin 6 +Z and = "' find 



15. Given that Q - 132-4 l. find <. 

0s <* 

16. Differentiate, with respect to y, the expression 

'-tan-iy. 
2 

sin 26 

di 

the acceleration (-57 ); find also the acceleration when is very small. 

40 -it sin , dQ - , , , , ., (d<i>\ , 

18. If sm <t> -- , and -j-. = a>, find the angular velocity I -if) of 

m at J \dt J 

a connecting-rod and also the angular acceleration / j. 

19. Given that ^ = TT - . . f find J-R and hence the value of 

(pq) tanO a0 

tan that makes -^ = o. 
ao 

o/> rs- j A*. j; <^M ,, WX(l X)(l 2X) ,, . 

20. Find the value of -= when M = , , ? -. M is a 

dx 2(3/ 2x) 

bending moment, I is the length of a beam and x is a portion of that 
length. 

21. Differentiate, with respect to /, the quotient - ;,- -- -. 

Partial Differentiation. When dealing with the equation 
PV = CT in connection with the theory of heat engines, we know 
that C alone is a constant, P, V and T being variables. If one of 
these variables has a definite value, the individual values of the 
others are not thereby determined ; e. g., assuming that C and T 
are known, then so also is the product PV, but not the individual 
values of P and V. If, now, the value of one of these is fixed, 
say of P, then the value of V can be calculated : therefore V 
depends on both P and T, and any change in V may be due to a 
change in either or both of the other variables. To find the change 
in the value of V consequent on changes in values of P and T, 



8o 



MATHEMATICS FOR ENGINEERS 



the change in V due to the change in P (assuming that T is kept 
constant) is added to the change in V due to the change in T 
(P being kept constant). Rates of change found according to this 
plan are spoken of as partial rates of change, or more usually 
partial derivatives, and the process of determining them is known 
as partial differentiation. 

When only two variables occur, a plane curve may be plotted 
to depict the connection between them, but for three variables a 
surface is needed. The three co-ordinate axes will be mutually at 




FIG. 20. 

right angles, two in the plane of the paper, and the other at right 
angles to it. If x, y and z are the variables, we can say that z is 
a function of x and y, or, in the abbreviated form z = f(x, y). 

Similarly x = f(y, z) 

and y = f(x, z). 

Dealing with the first of these forms, and assuming the axes 
of x and y to be horizontal (Fig. 20), let us examine, from the 
aspect of the graph, the significance of this form. Giving any 
value to x, we know the distance of the point in front of or behind 
the paper : the value of y determines the distance to the right or 
left of the axis of z, i. e. t the vertical on which the point lies is 



ADDITIONAL RULES OF DIFFERENTIATION 81 

determined and the actual height up this vertical is fixed by the 
value of z. If z is kept constant whilst values of x and y are 
chosen, a number of points are found all lying on a horizontal 
plane, and if all such points are joined we have what is known as 
a contour line. Therefore, if one of the quantities is constant our 
work is confined to one plane; but we have already seen that 
when dealing with a plane, the rate of change of one quantity 
with regard to another is measured by the slope of a curve, hence 
we can ascribe a meaning to a partial derivative. 
To illustrate by reference to a diagram (Fig. 20). 
The point P on the surface is fixed by its co-ordinates x, y 
and z, or SQ, OS and QP. 

If x is kept constant, the point must lie on the plane LTND. 
The slope of the curve LPT, as given by the tangent of the angle 
PMN, must measure the rate of change of z with regard to y when 
x is constant; and this is what we have termed the partial 
derivative of z with regard to y. This partial derivative may be 

expressed by -, or, more conveniently, by ( -j- ) , and if there is 

no possibility of ambiguity as to the quantity kept constant the 
suffix x may be dispensed with. 

fdz\ nn TXT (the slope being negative, since z 

(-=-)= tan L PMN v 

\ay] decreases as y increases). 

Similarly, the slope of the curve KPH 

_ /fe\ 
\dx)' 

If the variables are connected by an equation, the partial 
derivatives can be obtained by the use of the ordinary rules of 
differentiation. 



Example 16. Given that z 

\ (dz 
)' (dy 



dz\ ld z z 
)> W 



To find (j ), i. e., to find the rate of change of z with regard to x 

when y is constant, differentiate in the ordinary way, but treating y 
as a constant. 



Thus K- = (5>> x 2x) (zy z x 3**) + 2oye xy 

= loxy 6 r 2 y 2 + 2oye ry 



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

= loy i zxy 2 + 2oy z e xy . 



82 MATHEMATICS FOR ENGINEERS 

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



j 

z = $x z y'2x 3 y z -}-2oe xy 
then (3^- J = (5# 2 X i) (2* 3 x 2y) + 20* . 



3- 



and 



Example 17. If z 6 log #y i8x 5 y 2 , find the values of [-3 =-) 

\CLX . ^Z^ ' 

/ rf 2 ^ \ 

and 1 , -=- ), and state the conclusion to be drawn from the results. 
\dy . dx> 

To find (~j -j-} we must first find the value of (3-), x being 
regarded as a constant : then if Y be written for this expression the 
value of (^ ,- j must next be determined, y being treated as a constant, 

/ d z z \ 
and this is the value of ( , , ). 

\dx . dyl 

XT (dz\ 6xx , 6 - - ^ T 

Now I T- ) = - -- i8x 5 X2y = --- $6x 5 y = Y, say. 
\dxl xy y 

Differentiating this expression with regard to x, y being regarded 
as a constant 



or 



and 



\ 

) = 
/ 



. dyi \dy . dxi 

Hence the order of differentiation does not affect the result. 
Total Differential. If y is a function of x, then y f(x) 

dy d fl . j.,, \ 
=- 



i. e., dy = f'(x)dx. 

dy and dx are spoken of as differentials, and f'(x) is the coefficient 
of the differential dx; hence we see the reason for the term 
differential coefficient. 



ADDITIONAL RULES OF DIFFERENTIATION 83 

If z is a function of x and y, i. e., z f(x, y), the total differential 
dz is obtained from the partial differentials dx and dy by the use 
of the following rule 

fdz 



dy 



dy. 



The reason for this is more clearly seen if we work from the 
fundamental idea of rates of change, and introduce the actually 
measurable quantities like Sz, Sx and Sy. 




FIG. 21. 



Thus 



or total change in z = change in z due to the change in 

change in z due to the change in y. 

The change in z due to the change in x must be measured by 
the product of the change in x multiplied by the rate at which z 
is changing with regard to x ; and this fact can be better illustrated 
by reference to a diagram (Fig. 21). 

Let P be a point (x, y, z) on a surface, and let P move to a new 
position Q near to P. The change of position is made up of 

(a) A movement 8x to P' on the surface (y being kept constant) . 

(b) A movement 8y to Q on the surface (x being kept constant) . 



84 MATHEMATICS FOR ENGINEERS 

In (a) z increases by MP' 

and 



i t dz \ 

= Sxx mean value of I -3- ). 

\dx 

In (b) the change in z = NQ 

= Sy X mean value of ( j 

\dyl 

If P, P' and Q are taken extremely close to one another, the 
mean or average slopes become the actual slopes and 
the total change in z = 8z 



-MP'+NQ = (*) +*(*). 



YYIV 

Example 18. If Kinetic Energy = K = -- , find the change in the 
energy as m changes from 49 to 49-5 and v from 1600 to 1590. 

From the above rule, the change in K = 8K 

s ,dK\ . s /rfK\ 
= 8m (dm)+ 8v U I 
Now 8m = 49-549 = '5 

and Sy = 1590 1600 = 10. 

fdK\ ,. .. d /v z \ v z 

Also I j ) (i. e., v being constant) = -j { X m } = x i 
\dm) v dm\2g ] 2g 

(d~K.\ . , . , . d / m ,\ m 

and \ WJ (m bemg constant ) = dv\2> X V ) = 2 X 2V - 



vm 



_20 xi6oo X49 
~ 64 r 4~ 64-4 

= 19880 24380 = 4500 units. 

Example 19. A quantity of water Q is measured by 



If r l = the probable error of D, a diameter, r. 2 = the probable 
error of H, a head, and R = the probable error of Q, 



where ( jM) an( i (^u) are P ar tial derivatives. 
Find an expression for R. 



ADDITIONAL RULES OF DIFFERENTIATION 85 

Also 



I TrCD 2 / 

= IX Vlg 



Hence R = V 



R 

<r 

i. e., if the probable error of D is 3% and that of H is i% 

that of Q = V 4 (-03)+ i(-oi)~ 

= -0602, i. e., is about 6%. 

Logarithmic Differentiation. Occasionally it is necessary 
to differentiate an expression which can be resolved into a number 
of factors; and in such a case, to avoid repeated applications of 
the rules for the differentiation of products and quotients, we may 
first take logs throughout, and then differentiate, making use of 
the rule for the differentiation of a function of a function. By the 
judicious use of this artifice much labour can often be saved. 

Example 20. Find the value of - 

Ax 



Let- y = (3*~4)(4*-_ 

(zx-g) 

then log y = log (3* 4) +log (4*+ 7) -log (2* 9). 

Differentiating with regard to x 

dlogy _ 3 , 4_ 2 

, I / i ^_\ 



h,,t d _ l gy dlogy dy i dy 

LJUt , r ^ - - . ~= 

dx dy dx y dx 

i 

so that 



(3* -4)^(4* +7) (2* -9) 

I *?. = 3 4 2 

y ** (3*-4) (4^+7) (2^9) 

^ _ (3^ -4) (4* + 7) Y 
rf* ~ (2^-9) 

/24* 2 66* 189 + 24* 2 + 144 140* 24* 2 10*4-56! 
l~ (3^ -4) (4* +7) (2^ -9) I 



86 MATHEMATICS FOR ENGINEERS 

[As an exercise, the reader should work this according to the 

following plan. Write y = -. ^ ^r , and then use the rule for 

the differentiation of a quotient.] 

It is with examples in which powers of factors occur that this 
method is most useful. 

T- j d y hx+2) 3 (x i) 

Example 21. Find when y = v/ - --. *-*-rs '. 
dx (2X 5) 2 

Taking logs throughout 

logy = 3 log (7* + 2)+log (#i) 2 log (2* 5) 
Then 

djog y _ 3x7 i_ 2x2 

dy ~ ~ (?x+2y(x-i) (2X-5) 

I47^r + i4^ 2 31^ 10 28^ 2 +2O^r+8 



(*-i)(2^r 5) 

js&ar + 103 

~ 



^i)(2X - 5) 
28* 2 158^+103 



y' dx~ \yx+2)(x i)(2x 5) 

*(x-i) 28^-158^+103 



Exercises 9. On Partial Differentiation and Logarithmic 
Differentiation. 

1. In measuring the sides of a rectangle, the probable errors in 
the sides were Y and r 2 . If A = area and a and b are the sides, find 
the probable error R in A. 

~(dA\* 



. . 

Given that- R = ^ &) +>> 

the derivatives being partial. 

2. If * = a-** 5 '*, find and 



3. If 5 = /-#/+log (5^-3) X, find - and . 

4. If v = ( 4 -w) 2 ( 3 + 8w)3, find ^. 

, <iy 2wy 

show that - 2 - 



ADDITIONAL RULES OF DIFFERENTIATION 87 

6. If y = 8*(i7 + -2*), find ^. 

7. Differentiate, with respect to x, 

( 

8. Find the rate of discharge j- of air from a closed reservoir 
when m -- , m, p, v and r all being variables. 

CT 

9. If x = r cos Q, y r sin 0, and u is a function of both x and y, 
prove that 

tdu\ t\(du\ i . Jdu 

T-) = cos OU- ) - sin 
dx' \dr >o r 



and 



T- 
\dx 

du 



CHAPTER IV 
APPLICATIONS OF DIFFERENTIATION 

HAVING developed the rules for the differentiation of the various 
functions, algebraic and trigonometric, we are now in a position 
to apply these rules to the solution of practical problems. By far 
the most important and interesting direction in which differentiation 
proves of great service is in the solution of problems concerned with 
maximum and minimum values ; and with these problems we shall 
now deal. 

Maximum and Minimum Values. Numerous cases present 
themselves, both in engineering theory and practice, in which the 
value of one quantity is to be found such that another quantity, 
which depends on the first, has a maximum or minimum value 
when the first has the determined value. 

E. g., suppose it is desired to arrange a number of electric 
cells in such a way. that the greatest possible current is obtained 
from them. Knowing the voltage and internal resistance of each 
cell and the external resistance through which the current is to be 
passed, it is possible by simple differentiation to determine the 
relation that must exist between the external resistance and the 
total internal resistance in order that the maximum current flows. 

Again, it might be necessary to find the least cost of a hydraulic 
installation to transmit a certain horse-power. Here a number of 
quantities are concerned, such as diameter of piping, price of 
power, length of pipe line, etc., any one of which might be treated 
as the main variable. By expressing all the conditions in terms 
of this one variable and proceeding according to the plan now to 
be demonstrated, the problem would become one easy of solution. 

A graphic method for the solution of such problems has already 
been treated very fully (see Part I, pp. 183 et seq.). This method, 
though direct and perfectly general in its application, is somewhat 
laborious, and unless the graphs are drawn to a large scale in 
the neighbourhood of the turning points, the results obtained are 
usually good approximations only. In consequence of these failings 



APPLICATIONS OF DIFFERENTIATION 



89 



of the graphic treatment, the algebraic method is introduced, but 
it should be remembered that its application is not so universal as 
that of the solution by plotting. 

The theory of the algebraic method can be simply explained 
in the following manner : 

The slope of a curve measures the rate of change of the ordinate 
with regard to the abscissa; and hence, when the slope of the 




FIG, 22. Maximum and Minimum Values. 

curve is zero, the rate of change of the function is zero, and the 
function must have a turning value, which must be either a 
maximum or a minimum. But it has already been pointed out 
that the slope of a curve is otherwise denned as the derivative or 
the differential coefficient of the function ; therefore the function 
has a turning value whenever its derivative is zero. 

Hence, to find maximum or minimum values of a function we 
must first determine the derivative of the function, and then find 



MATHEMATICS FOR ENGINEERS 



the value or values of the I.V. which make the derivative zero; 
the actual maximum or minimum values of the function being 
found by the substitution of the particular values of the I.V. in 
the expression for the function. 

The rule, stated in a concise form, is : To find the value of 
the I.V. which makes the function a maximum or minimum, differentiate 
the function, equate to zero and solve the resulting equation. 

The full merit of the method will be best appreciated by the 
discussion of a somewhat academic problem before proceeding to 
some of a more practical nature. 

Example i. Find the values of x which give to the function 
y = 2# 3 + 3# 2 36^ + 15 maximum or minimum values. Find also the 
value of x at the point of inflexion of the curve. 

This question may be treated from two points of view, viz. 
(a) From the graphical aspect. 

We first plot the primitive curve y 2# 3 +3# 2 36^+15 (see 
Fig. 22), the table of values for which is : 



X 


A 2 


X s 


2* s +3* 2 -36#+i5 


y 


-4 


16 


-64 


128+48 + 144 + 15 


79 


-3 


9 


-27 


- 54 + 27+108 + 15 


96 


-2 


4 


- 8 


- 16 + 12+ 72 + 15 


83 


I 


i 


- i 


2+ 3+ 36 + 15 


52 


O 


o 


o 


o+ o- + 15 


15 


I 


i 


i 


2+ 3 36+15 


-16 


2 


4 


8 


16+12 72 + 15 


29 


3 


9 


27 


54+27-108+15 


12 


4 


16 


64 


128+48-144+15 


47 


5 


25 


125 


250+75-180+15 


1 60 



This curve has two turns and two turns only, and consequently 
y has two turning values, one being a maximum and one a minimum. 
By successive graphic differentiation the first and second derived 
curves may be drawn, these being shown on the diagram. 

Now for values of x less than 3 the slope of the primitive curve 
is positive, as is demonstrated by the fact that the ordinates of the 
first derived curve are positive. At x 3 the primitive curve is 
horizontal and the first derived curve crosses the #-axis ; and since 

dv 

the ordinates of the first derived curve give the values of , , we see 

dx 

that when the primitive curve has a turning value, the value of 

dy 

f- = o. For values of x between 3 and +2 the slope of the 

primitive is negative; when x = +2 the slope is zero, and from that 



APPLICATIONS OF DIFFERENTIATION 91 

point the slope is positive. Thus y has turning values when x = 3 
and when x = +2; these values being a maximum at x 3 and 
a minimum at x = +2 as observed from the curve. 

This investigation proves of service when we proceed to treat the 
question from the algebraic aspect; in fact, for complete understanding 
the two methods must be interwoven. 

(b) From the algebraic point of view. 

Let y = 2x 3 + $x 2 36*4-15 

then * = 6x*+6x-36 

= 6(x z +x-6). 

Now in order that y may have turning values we have seen that 

dy 
it is necessary that -^- o. 

But ^ = o if 6(# 2 + *-6) = o 

i. e., if 6(^+3) (x 2) = o 

i. e., if x = 3 or 2 

and hence y has turning values when x = 3 and x +2. We do 
not yet, however, know the character of these turning values, so that 
our object must now be to devise a simple method enabling us to 
discriminate between values of x giving maximum and minimum 
values to y. 

An obvious, but slow, method is as follows : Let us take a value 
of x slightly less than 3, say 3-1; then the calculated value of y 
is 95 "85. Next, taking a value of x rather bigger than 3, say 2-9, 
the value of y is found to be 95-85. Therefore, as x increases from 
3-1 to 3 and thence to 2-9, y has the values 95'85, 96, and 
95-85 respectively. Thus the value of y must , be a maximum at 
x = 3, since its values on either side are both less than its value 
when x = 3. In like manner it can be shown that when x -\-2, 
y has a minimum value. 

The arithmetical work necessary in this method can, however, 
be dispensed with by the use of a more mathematical process, 
now to be described. 

Referring to the first derived curve, the equation of which is 
y 6# 2 + 6# 36, we note that as x increases from 4 to 3 
the ordinate of the derived curve decreases from 36 to o; from 
x = 3 to x = .5 the ordinate is negative but increasing 
numerically, i. e., in the neighbourhood of x = 3 the slope of 
the second derived curve, which is the slope curve of the first 
derived curve, is negative (for the ordinate decreases as the 
abscissa or the I.V. increases). But the slope of the first derived 



92 MATHEMATICS FOR ENGINEERS 

curve, and thus the ordinate of the second derived curve, must 

d 2 v 
be expressed by -y-^, so that we conclude that in the neighbourhood 

dsC 

of a maximum value of the original function the second derivative 
of it is a negative quantity. 

In the same way we see that in the neighbourhood of a 
minimum value of the function, its second derivative is a positive 
quantity. Hence a more direct method of discrimination between 
the turning values presents itself : Having found the values of the 
I.V. causing turning values of the original function, substitute these 
values in turn in the expression for the second derivative of the function ; 
if the result is a negative, then the particular value of the I.V. considered 
is that giving a maximum value of the function and vice-versa. 

This rule may be expressed in the following brief fashion : 

Let y = f(x) and let the values of x that make -j~(x) orf'(x) = o 

be #! and x z . 

d 2 y 
Find the value of -~^ or f"(x), as it may be written, and in 

this expression substitute in turn the values x^ and x z in place of 
x: the values thus obtained are those of f"(x-^} and /"(# 2 ) respec- 
tively. Then if f"(x^), say, is negative, y has a maximum value 
when x x^; and if f"(xj) is positive, y has a minimum value 
when x = x v 

Applying to our present example : 

y = 

/ 



When x 3 the value of -~ 2 is I2( 3) +6, i. e., /"( 3) = 30; 

and since /"( 3) is a negative quantity, y is a maximum when 
x == 3. 

Similarly, /"( + 2) = 12(2) +6 = +30 

and hence y is a minimum when x +2. 

Referring to the second derived curve, i. e., the curve y = I2X+6, 
we note that its ordinate is negative for all values of x less than 
5 and positive for all values of x greater than -5, the curve 
crossing the axis of x when x = -5. This indicates that when 
x = -5 the first divided curve has a turning value ; but the first 



APPLICATIONS OF DIFFERENTIATION 93 

derived curve is the curve of the gradients of the primitive curve, 
and hence when x 5 the gradient of the primitive must have 
a turning value, which may be either a maximum or a minimum. 
In other words, if we had placed a straight edge to be tangential 
in all positions to the primitive curve, it would rotate in a right- 
handed direction until x = -5 was reached, after which the 
rotation would be in the reverse direction. A point on the curve 
at which the gradient ceases to rotate in the one direction and 
commences to rotate in the opposite direction is called a -point of 
inflexion of the curve. Thus points of inflexion or contra-flexure 

u d *y 

occur when -=-4 = o. 
dx 2 

A useful illustration of the necessity for determining points of 
contra-flexure is furnished by cases of fixed beams. We have 

of Confraf lexure. 
[ filll 1 /^ ^\ I -21 ll ! 




FIG. 23. 

already seen that the bending moment at any section is propor- 

d 2 v 
tional to the value of -j- z there; hence there must be points of 

CLX* 

contra-flexure when the bending moment is zero. 

Example 2. Find the positions of the points of contra-flexure of 
a beam fixed at its ends and uniformly loaded with w units per foot; 
the deflected form having the equation 

i fwlx 3 wl 2 x z _wx*\ 
= El\ 12 24 24 ')' 

We may regard this question' from either the graphic aspect or the 
physical. According to the former we see that it is necessary to 
determine the points of inflexion, and therefore to find values of x 

^ ^d*y . 
for which , 2 is zero. 

Reasoning from the physical basis we arrive at the same result, 
by way of the following argument : the bending moment, which is 

d z y 
expressed by EI- , changes sign, as is indicated by the change in 

OLX 

the curvature of the beam (see Fig. 23), and therefore at two points 
the bending moment must be zero, since the variation in it is uniform 



94 MATHEMATICS FOR ENGINEERS 

and continuous; but the bending moment is zero when -~ 2 is zero, 



v 

since M = El j-^. 
dx 2 



AT w fix 3 I 2 x 2 x*\ 

Now y = =pp ( ----------- V 

El \i2 24 24/ 

dy w T( I ,\ / / 2 \ 4 ^ 

hence ^ = -ey ( X3* 2 ) ( X2# ) *- 

a* El LA 1 2 / \24 / 24 



w __ 

12 ~~ 6 



d 2 y w r / 1 \ i l z \ 3^ 2 -i 
and -,- -2 = ,-, f ( xzx ) ( XII V 

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

(** l * _^ 2 \ 
. El Va i2~ 2"]' 

/72 V ./v 72 ;2\ 

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

*" \ 2 12 2 / 

/^ / 2 AT 2 

and M = o if ------ , i. e., 6lxl 2 6x z = o, 

2 12 2 



i. e., if 6x z 6lx+l z = o 

6l 



or 



12 



= -789? or -2 1 iL 

Hence the points of inflexion occur at points distant -211 of the 
length from the ends. 



Example 3. A line, 5 ins. long, is to be divided into two parts 
such that the square of the length of one part together with four 
times the cube of the length of the other is a minimum. Find the 
position of the point of section. 

Let x ins. = the length of one part, then 5 x = length of the 
other part. 

Then (5 #) 2 +4# 3 is to be a minimum. * 

Let y = ( 5 -*) 2 + 4 * 3 

Then 



Hence ~- o if x ^ or i (the latter root implying external 
dx o 

cutting) . 



APPLICATIONS OF DIFFERENTIATION 95 

To test for the nature of the turning value 



10 

dx 

d*y 
and g = 24*4-2. 

When x = J 

6 

-5-^ = ( -2 5 J_|_2 = a positive quantity. 

Therefore y is a minimum when x = and the required point of 
section is % in. from one end. 

Example 4. If 5 detrimental surface of an aeroplane 
S = area of planes 
K = lifting efficiency 

KS 
then/, the " fineness," is obtained from the formula / 2 = ~-. 

Also the thrust required for sustentation = C (I'acX where C is 



a constant and i is the angle of incidence of the plane (expressed in 
radians). 

Taking S = 255 and K = -4, find the angle of incidence for the 
case in which the least thrust is required. 

p = JS . 4 X2 5 = 

J -o8s -08 5 ' 

The thrust T = C (*4-js:.) and since i is the only variable in this 

\ J *%/ 

expression, we must differentiate with regard to it. 
Thus - 



dT ., 

and = if 1- = 



i. e., if i 2 = 1 = . 

/ 2 125 

Thus i = -0895 

or the thrust required is either a maximum or minimum when the 
angle of incidence is -0895 radian. 

To test whether this turning value is a maximum or a minimum, 
let us find the second derivative 

^- 

dT = 

^T _ / 2_ 
& - - u V +/ 2 z 3 /' 

^2-p 

When i -0895, -^ must be positive, and hence T has its minimum 
value when i = -0895. 



96 MATHEMATICS FOR ENGINEERS 

Example 5. Find the dimensions of the greatest cylinder that 
can be inscribed in a right circular cone of height 6 ins. and base 
10 ins. diameter. 




FIG. 24. 

Assume that the radius of the base of the cylinder = x ins. (Fig. 24) 
and the height of the cylinder y ins. 

Then the volume V nx^y. 

We must, then, obtain an expression for y in terms of x before 
differentiating with regard to x. 

From the figure, by similar triangles, taking the triangles ADC 
and EFC 

6 _ y 



or 
Hence 


V 


5 

2 6 . . 67T , g 3 , 

it* X-(5 ~~ ~c '** , 




and 
Thus 


dV 
d\' 


5 

= o if x(io 3^) = o 




i. e. t 
or 




if x = o (giving the cylinder of zero 
if 10 3*, i. e., x $\ ins. 


volume) 



Then 



y = (5-3*) = 2 ins - 



and the volume of the greatest cylinder = wX ( ) X2 = 69-8 cu. ins. 

Example 6. The total running cost in pounds sterling per hour 
of a certain ship being given by 

v 3 
C =4.5+ 

^2100 

where v = speed in knots, find for what speed the total cost for a 
journey is a minimum. 

The total cost for the journey depends on 
(a) The cost per hour; and 
(6) The number of hours taken over the journey. 



APPLICATIONS OF DIFFERENTIATION 97 



Item (b) depends inversely on the speed, so that if the journey 

j t 
K 



were 2000 nautical miles the time taken would be hours; or. 



in general, the number of hours = 

Then the total cost for a journey of K nautical miles 



K / v 3 \ 

= Q = xU-5+- 

V V* ' ' 2IOO/ 



i , U M 

5 V -\ ) 

2IOO/ 

Differentiating with regard to the variable v 
dC t 



Then 



1050 



., 

* a 

v (s . . IX 



"4-5 

- ' *J 
- 



1050 v 

or v 3 = 4-5 X 1050 = 4725 

hence *v = 16-78 knots. 

Example 7. A water main is supplied by water under a head of 
60 ft. The loss of head due to pipe friction, for a given length, is 
proportional to the velocity squared. Find the head lost in friction 
when the horse-power transmitted by the main is a maximum. 

If v = velocity of flow, then 

Head lost = Kv 2 , where K is some constant, 
i. e., the effective head = 60 Kv 2 = H e . 

TT -r, , Quantity (in Ibs. per min.) x effective head (in feet) 

H.P. transmitted = J v * < - v ' 

33000 

_ area (in sq. ft.) x velocity (ft. per min.) x 62-4 x H g 

33000 

= CvHe, where C is some constant 
= Gv{6o-Kv*) 
= C(6ov-Kv 3 ) 

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

Ctl) 

= C(6o i8o + 3H e ) 

or T (H.P.) = o when 3H C = 120 

i. e., H e = 40. 

In general, then, the maximum horse-power is transmitted when 
the head lost is one-third of the head supplied, i. e., the maximum 

2 

efficiency is - or 66-7%. 



98 MATHEMATICS FOR ENGINEERS 

Example 8. The stiffness of a beam is proportional to the breadth 
and the cube of the depth of the section. Find the dimensions of 
the stiffest beam that can be cut from a cylindrical log 4 ins. in 
diameter. 



From hypothesis 
or 



S = Kbd 3 . 



Both breadth and depth will vary, but they depend on each 
other; and from Fig. 25 we see that b 2 = i6 d 2 . Hence we can 
substitute for b its value in terms of d and then differentiate with 
regard to d\ according!}? 

S = 

As it stands this would be a rather cumbersome 
expression to differentiate, and we therefore employ 
a method which is often of great assistance. Since 
we are dealing with positive quantities throughout, 
S 2 will be a maximum when S is a maximum,* and 
hence we square both sides before differentiating. 

Thus S 2 = K 2 d 6 (i6-d 2 ) = 

<fS 2 
and -, ~ 




FIG. 25. 



'>-8d'>) = 8d 5 (i2-d 2 ) 



Hence ,-r = o if d 5 o, i. e., d = o (giving zero stiffness) 

Ct.(t 

or if 
i. e., 

Hence 



d 2 = 12 
d = 3-464 ins. 

b V 1612 = 2 ins. 



* If we were dealing with negative quantities it would be incorrect 
to say that the quantity itself had a maximum value when its square 
was a maximum, for suppose the values of the quantity y in the 
neighbourhood of its maximum value were 13, 12, u, 10, 
ii, 12, etc., corresponding values of y 2 would be +169, +144, 
+ 121, +100, +121, +144, so that if y = 10 (its maximum value) 
when x = 4, say, then y 2 TOO when x = 4, and therefore a minimum 
value of y 2 occurs when x 4, and not a maximum. 



Example g. Find the shape of the rectangular channel of given 
sectional area A which will permit the greatest flow of water ; being given 

that Q = Av, v c Vmi, m = hydraulic mean depth = . . f^ rea . 

wetted perimeter 

and i is the hydraulic gradient ; Q being the quantity flowing. 



APPLICATIONS OF DIFFERENTIATION 99 

Let the breadth of the section be b and the depth d; then, by 
hypothesis 

bd = A. whence b = -j. 
a 



w = . . , -. = , j and therefore v = c Vi\/ ; 
wetted perimeter o+2a 

= cVAl . 



Hence Q = Av = Ac VAi . ~ /, , 

= K . -^Jsas where K = 
Vb+2d 

Q will be a maximum when Q 2 is a maximum, hence we shall find 
the value of b for which Q 2 is a maximum. 



b+2d~ ' 6+ ?A' 

Also Q 2 is a maximum when the denominator of this fraction is 
a minimum. 

Let this denominator be denoted by D 

dD d I, , 2A\ 2A 

then -jjj- = ^1+ r / = I ~~^F 

rfD ., 2A . / 

and -n- = o if i = T-, . e., if o = V2A. 



Now d = = -= = \ 

O \/2A 2 

/. the dimensions would be 

/A 

= - 



depth = V - and breadth = \/2A. 



Example 10. For a certain steam engine the expression for W, 
the brake energy per cu. ft. of steam, was found in terms of r, the 
ratio of expansion, as follows 

/i+log r\ 

( r S-)-27 



I2o 
\V = 



- _ - _ - _ 
00833 , 

JJ.J-. 000903 



Find the value of r that makes W a maximum. 

Before proceeding to differentiate, we can put the expression in a 
somewhat simpler form. 

Thus- W = "O 

00833 + 

and W is a quotient = - where u = i2o(i+log r}2jr 



ioo MATHEMATICS FOR ENGINEERS 

du 1 20 

so that -y = ---- 27 

dr r 

and v -00833 + -000903? 

dv 
so that -f- -000903. 

du dv 

J\*T V J -- M J~ 
TT dW dr dv 
Hence -y = , 

dr v z 

(00833 + 000903*') ( 27^ [iao(i+log r) 27/1-000903 

(-00833 +-000903*-) 2 
Now - = o if the numerator of the right-hand side = o 

i . e., if ( -22 L\_ (27 x -00833) + (120 x -000903) (27 x -0009037) 

(120 x -000903) (120 x -000903 log r) + (27 x -ooogo^r) = o 
i.e., if. -225 -1084 log r = o. 

This equation must be solved by plotting, the intersection of the 
curves y = -1084 log r and y z = --- 225 being found; the value of r 

here being 2-93. 

Hence r = 2-93. 

Example n. The value of a secondary electric current was given 
by the formula 



_- _ 

y = - e L+M e L-M 



where L = inductance of primary circuit 
R = resistance of primary circuit 
M = coefficient of mutual inductance 
I = steady current. 
Find for what value of /, y has a maximum value. 

T / iu m 

y = -(e~L+M g~L M 



dv R ?L R - 

and - = o if ^^r.e L-M __ 
rf/ L M L+M 

Transposing the factors 

L M 



BI(L-M-L-M) T _ 



L+M 

-TM, ~ 

e L-M. = 



or eLM. 

L M 



APPLICATIONS OF DIFFERENTIATION 101 

In order to find an expression for t, this equation must be changed 
to a log form, thus 

/L+M\ _ 2 MRf 
g \L M/ ~ L 8 -M2 

L 2 -M 2 , / 

* = - log 



If three variables are concerned, say x, y and z, the relation 
between them being expressed by the equation z=f(x, y), then 
in order to find the values of x and y for turning values of z, it 
is necessary to determine where the plane tangential to the surface 
is horizontal. 

The algebraic problem is to find the values of x and y that 

satisfy simultaneously the equations (-T-) = o and f-^-J = o, these 

\(4'Z / \dZ / 

derivatives being partial. 

Example 12. The electric time constant of a cylindrical coil of 
wire (i. e., the time in which the current through the coil falls from 
its full value to a value equal to -632 of this) can be expressed 

approximately by K = . y where z is the axial length of the 

coil, y is the difference between the external and internal radii and x 
is the mean radius ; a, b and c representing constants. If the volume 
of the coil is fixed, find the values of x and y which make the time 
constant as great as possible. 

The volume V of the coil = cross section x length 

V 

*. e., V = 2Tc#xy Xz and z = - 

2-nxy 

K = m \ r <--- I and is a maximum 
(ax+by+cz) 

ax+by+cz a , b , c '. 

when - or -\ is a minimum. 

xyz vz xz xv 



f , a. . b , c 

Let p = --- 

z xz x 



- - --- 
xyz yz xz xy 

a. . b , c 
--- 
yz xz xy 



. c 

+ 



yV xV x~y 

__ zivcya mxyb c 
'' ~ + " 



Now (^ (i. e., with y constant) = 2 y U +(- X - 

2TOI C 



102 MATHEMATICS FOR ENGINEERS 

c- -11 (dp\ 27T& C 

Similarly \f) = -~- 2 . 

\dy) x V xy z 

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

so that ^jr- 

V x 2 y 

i. e., x*y = (i) 

, 2TI& C 

V ' xy 2 
i. 6. xy z = .... (2) 

2TT& 

To solve for x and y 

cV 
From (2) x = j-~ 2 . 



Substituting in (i) 

cV 



cVa 
whence y 3 = r, 

27T0 2 



or 



__2TU6 2 

also 

Exercises 10. On Maximum and Minimum Values. 

1. If M = 15* -oix*, find the value of x that makes M a maximum. 

2. Find the value of x that makes M a maximum if M = 3-42* -ix 2 . 

3. M is a bending moment and x is & length ; find x in terms of / 
so that M shall be a maximum, and find also the maximum value of M. 

M = < 

4. As for No. 3, but taking 

5. The work done by a series motor in time t is given by 



,, wx ,, 

M = (l 

2 V 



R 

where e = back E.M.F. 

E = supply pressure 

R = resistance of armature. 

a 

The electrical efficiency is ^. Find the efficiency when the motor 
so runs that the greatest rate of doing useful work is reached. 

In Nos. 6 to 8 find values of x which give turning values to y, 
stating the nature of these turning values. 



APPLICATIONS OF DIFFERENTIATION 



6. y = 4 * 2 +i8*- 4 i. 7. y = 5^- 

8. y = x 3 + 6x 2 15^+51 (find also the value of x at the point of 
inflexion). 

9. Sixteen electric cells, each of internal resistance i ohm and 
giving each I volt, are connected up in mixed circuit through a 
resistance of 4 ohms. Find the arrangement for the greatest current 

[say rows with x cells in each row]. 
x 

10. If 40 sq. ft. of sheet metal are to be used in the construction 
of an open tank with square base, find the dimensions so that the 
capacity of the tank is a maximum. 

fj A 

11. Given that W = 4C 2 +',7, find a value of C that gives a turning 

value of W, and state the nature of this turning value. 

// _ x \ 

12. M (a bending moment) = W v . ; (x+y] Wy. For what value 

of x is M a maximum ? {W, / and y are constants.} 

13. The cost C (in pounds sterling per mile) of an electric cable 
can be expressed by 

C - ^+636* 
x 

where x is the cross section in sq. ins. 

Find the cross section for which the cost is the minimum, and 
find also the minimum cost. 

14. A window has the form of a rectangle together with a semi- 
circle on one of its sides as diameter, and the perimeter is 30 ft. Find 
the dimensions so that the greatest amount of light may be admitted. 

15. C, the cost per hour of a ship, in pounds, is given by 

c3 

C = 3 -2 + - 

2200 

where s = speed in knots. 

Find the value of s which makes the cost of a journey of 3000 
nautical miles a minimum. 

At speed 10% greater and less than this compare the total cost 
with its minimum value. 

16. An isolated load W rolls over a suspension bridge stiffened 
with pin-jointed girders. When the load is at A, distant x from the 

\V# 
centre, the bending moment at this section M A = (I 2 4# 2 ). For 

what value of x is M A a maximum ? 

17. A riveted steel tank of circular section open at the top has to 
be constructed to contain 5000 gals, of water. Find the dimensions 
so that the least possible amount of steel plate is required. 

18. A canister having a square base is cut out of 128 sq. ins. of 
tin, the depth of the lid being i in. Find the dimensions in order 
that the contents of the canister may be as large as possible. 

19. The stiffness of a beam of rectangular section is proportional 
to the breadth and the cube of the depth. Find the ratio of the 
sides of the stiffest beam of rectangular section with a given perimeter. 



104 MATHEMATICS FOR ENGINEERS 

20. A load uniformly distributed over a length r rolls across a 
beam of length I, and the bending moment M due to this loading at 
a point is given by 

-. wry f, r} wx z 

M=-j* {/-?+*--,- . 

For what value of x is M a maximum ? 

21. Find the value of V (a velocity) that makes R (a resistance) 
a maximum when 

= yi 3 (V-i2) 
54 V+I2 ' 

22. If L = Vr z x z *(r z x*), find the value of x that makes L a 

y v 

maximum. 

23. A jet of water, moving with velocity v, impinges on a plate 
moving in the direction of the jet with velocity u. The efficiency 

TJ = -- 3 - Find values of u for maximum and minimum efficiency, 
and find also the maximum efficiency. . 

^2.tt(l} _ / vC\ 

24. If v *-= ', find the value of u for maximum value of i?. 

v z 

25. Given that O = K//T! (cos 6 sin 6), find values of 6 between 
o and 360 that make Q a maximum, treating K, p. and T x as 
constants. 

26. A cylinder of a petrol engine is of diameter d and length /. 

. , , d area of exposed surface 

Find the value of the ratio -, which makes 

/ capacity 

a minimum. The volume must be treated as a constant. 

27. If the exposed surface of a petrol engine cylinder is given by 
S 27W 2 +2:rr/+-2y 2 , I being the length and r the radius, 

find the value of the ratio - that makes the ratio ex P osed surface 

r capacity 

a minimum. The volume must be treated as a constant. 





28. Given that ? = K2 - , find values of K for turning 
values of y. 

z#R 2 / 1 \ / /3\ 

29. IfM = ^|^ -sin 2 0)--934wR 2 ^cos0- /s /|j, for what values 

of is M a maximum ? [M is the bending moment at a section of a 
circular arched rib loaded with a uniform load w per foot of span, and 
R is the radius of the arch.] 

30. An open channel with side slopes at 45 is to have a cross 
section of 120 sq. ft. Determine the dimensions for the best section 
(i. e., the section having the smallest perimeter for a given area). 



_ 

31. If M = -^TTZ - ni\ * fi n d the value of x which makes M, a 

O ( I .- t ) 

bending moment, a maximum. The final equation should be solved 
by plotting, a value being assumed for /. 



APPLICATIONS OF DIFFERENTIATION 105 

32. In connection with retaining walls the following equation 
occurs 

p _ ph 2 _ i M tan 6 

72 i M*+2/* cot 6. 

Find an expression, giving the value of 6 (in terms of tan 6), that 
makes P a maximum. {M, p and h are constants.} 

33. Assuming that the H.P. of an engine can be expressed by the 
relation 

H = C(fnl 3 -Kpn 3 l*) 

where C is a constant, / = stroke, p = pressure in piston rod due to 
the pressure on the piston, p average density of the material of the 
engine, K = constant depending upon the mode of distribution of 
the mass of the engine parts, n = R.P.M., and / = safe stress in the 
material, find an expression for I giving the maximum H.P. for 
engines of different sizes. 

34. Find the turning point of the probability curve 

T * 



and also the points of inflexion. 

35. In a two-stage compressor, neglecting clearances, if P x and V x 
are the initial pressure and volume of the L.P. cylinder, P 2 the pressure 
in the intercooler, and P 3 the discharge pressure of the H.P. cylinder, 
the total work for the two cylinders is given by 



For what value of P 2 is W a minimum, P^ V^, P 3 and n being 
regarded as constants ? 

36. Find the height A of a Warren girder to give the maximum 
stiffness, the stiffness being given by the expression 



tln , Id { W\ 



2 EA r 4 

d being the length of one bay and I the span, whilst f c , ft and E are 
constants for the material. 

37. The efficiency of a reaction wheel may be expressed by 

2( i) 

For what value of n has >> its maximum value ? 

38. The weight W of steam passing through an orifice, from 
pressure P! to pressure Pg, is given by 

n+Pj" 



If n 1-135, nn( i the value of ^ for which W is a maximum. 

*i 

39. Find the height of the greatest cylinder that can be inscribed 
in the frustum of a paraboloid of revolution cut off by a plane 
perpendicular to the axis and distant 6 units from the origin. The 
paraboloid is generated by the revolution about the axis of x of the 
parabola y z = yc. 



106 MATHEMATICS FOR ENGINEERS 

f 1% _!_ y\ ~\ 

40. If M = \V j# - 1*-\ w here y and I are constants, find the 
value of x that makes M a maximum. 

41. If T, / and T/ are the tight, slack and centrifugal tensions 
respectively in a belt passing round a pulley, and v = speed of the 
belt in feet per sec., then 

H.P. transmitted - ^1=3. 
550 

TJOV^ 

Being given that T/ = , the maximum permissible tension in 

the belt T m = T+T/, ^ = coefficient of friction between belt and 

T 
pulley, = angle of lap of belt in radians, and . ev e , find the value 

% 

of T/ in terms of 1 m so that the maximum H.P. is transmitted. 



42. If y 3X*-\-2x 3 j8x 2 24OX-\-54, find the values of x which 
give turning values to y, stating the nature of these turning values ; 
and find also the values of x at the points of inflexion. 

43. The radial stress in a rotating disc 



in which expression x is the only variable. 

Find the value of x which gives to p x its maximum value, and 
state this value of p x . 

44. A pipe of length / and diameter D has at one end a nozzle of 
diameter d through which water is discharged from a reservoir, the 
level of the water in which is maintained at a constant head h above 
the centre of the nozzle. Find the diameter of the nozzle so that 
the kinetic energy of the jet may be a maximum ; the kinetic energy 
being expressed by 

7T V_2gD'A \t 



[Hint. If K = kinetic energy, writ 

_ P 7T 

= 



and find the value of d for the maximum value of K*.] 

45. Prove that the cuboid of greatest volume which can be inscribed 
in a sphere of radius a is a cube of side 



46. The velocity of the piston of a reciprocating engine can be 
expressed by 

/sin 20 .. Q \ 

2nnr\ -- hsm ) 

\ zm I 

where is the inclination of the crank to the line of stroke. 

T ,. connecting-rod length _ , . 

If m = - TxrS ---- 1 - = 8, find the values of between 
length of crank 

o and 360 that make the velocity a maximum. 



APPLICATIONS OF DIFFERENTIATION 



107 



Calculation of Small Corrections. Differentiation finds 
another application in the calculation of small corrections. 

Thus an experiment might' be carried out, certain readings 
being taken, and results deduced from these readings; then if 
there is a possibility of some slight error in the readings and it 
is required to find the consequent error in the calculated result, 
we may proceed to find that error in the manner now to be 
explained. 

Suppose we have two quantities A and B connected with one 
another by a formula A = KB ; then if the value of B is slightly 
inaccurate the error in A will depend on this error in B, and also 
on the rate at which A changes with regard to B. E. g., if A 
changes three times as fast as B and the error in B is !%, then 
the consequent error in A must be 3X-I or -3%. 

We might also look upon this question 
from a different point of view. Suppose 
that a reading, instead of being x, as it 
should have been, was slightly larger, 
say x-}-8x, i. e., the measured value of x 
would be represented by OB and not 
OA (Fig. 26), then the error is Sx or 
8x 



XIOO%. 
X 



This error causes an error 




in the value of y, so that the calculated 
value of y is BQ and not AP, i. e., the 
error is Sy. 

To compare these errors we proceed as follows : -- the slope 

of the chord PQ, and if Sx is very small (as it should be, for 
otherwise the experiment would be repeated), then this would also 
be the slope of the tangent at both P and Q, or, approximately 

Sy dy 

8x dx 



^. e., 



dx 



or, error in y = rate at which y changes with regard to x 
X error in x. 



Example 13. In the measurement of the diameter of a shaft, of 
which the actual diameter was 4 ins., an error of 2% was made; 
what was the consequent error in the weight ? 



io8 MATHEMATICS FOR ENGINEERS 

Here W = ~d 2 lp, where p is the density 

4 

- Kd 2 , where K = %. 
4 

Now the error in the diameter = Sd = --'- X4 = -08 in. 

too 



also - - - 2Kd 

CLLJ\J i i - 7 i - J.Vli- 

d.d d.d 



or the percentage error = ^ x 100 = - ^9g^- X 100 




Example 14. If some torsion experiments are being made on 
shafts varying in diameter from i in. to 5 ins. ; then, allowing a maximum 
error of -5% in the measurement of the diameters, what is the range 

of the errors in the stress ? Given that T = f-fd 3 . 

The stress / = x -^ 
TC a 3 

df i6T 

hence 13- 

Now the error in the diameter 8.d is -5% 

i. g., 8d = 5 X rf. 

100 

Hence the "error in / = /-. X 8d = 4 -7r X 
d.d -jzd* 100 

i. e. t the percentage error in / 

8f 48T - 5 d 

= loo X J f - = loo X r, X ~ X 
/ IT a 4 zoo 

= -3- 

Thus the smallest error = '03 x smallest stress \ 
and the largest error = -03 x largest stress /' 

If the error in the measurement of the diameter is on the high 
side, then the stress, as calculated, will be too low. 

Expansion of Functions in Series. Theorems of Taylor 

andMaclaurin. Many of the simpler functions, suchaslog (i-f-#), 
sin x, cos x, etc., can be expressed as the sums of series. These 
functions can be expressed in terms of these series by the use of 
a theorem known as Maclaurin's. 



APPLICATIONS OF DIFFERENTIATION 109 

Let f(x) stand for the function of x considered, and let 
f(x) a+bx+cx^+dx 3 -^ . . ., to be true for all values of x, 

i.e., /(o) = 



We assume that the differentiation of the right-hand side term 
by term gives the derivative of /(#). 

Differentiate both sides with regard to x. 

Then ^or/'(*) = b+2cx+idx*+ . . . 

This must be true for all values of x ; thus, let x = o 
then f'(x) when x = o dr/'(o) = b 

/'(o) implying that f'(x) or *-* is first found and then the 

flwv 

value o substituted for x throughout. 

d*f(x) 
Differentiating again, -4V or f"( x ) = 2c-}-6dx-{- . . . 

and /"() = zc 

/"(o) 
t. e., c= J J . 

2 

Similarly, /'"(*) = 6^+ terms containing x and higher 

* powers of x, 

whence /'"() = 6W or 1.2.3.^. 



or 



1.2.3 |JL 

Accordingly we may write the expansions 



This is Maclaurin's Theorem. By a similar investigation we 
might obtain Taylor's Theorem, which may be regarded as a more 
general expression of the foregoing. 

Taylor's Theorem. In this the expansion is of f(x-\-h] and not 
/(*); thus 



\A 
or, as it is sometimes written, to give an expansion for/(*) 



no MATHEMATICS FOR ENGINEERS 

If in either of these two expansions we make h = o, then 
Maclaurin's series results. 

We may now utilise these theorems to obtain series of great 
importance. 

Example 15. To find a series for cos x. 

Let f(x] = cos x 

then /(o) cos o = i. 

A i rn \ d COS X 

Also / (x), i. e., -, -- = sin x 
ctx 

so that /'() = sin o o. 

Again /"(*) = ^~(~ s ^ n x ) ~ cos x 

so that /"() = cos o = i 

and /"'(#) = -3- (cos x) sin x 

so that /'"() = sin o = o. 
Now f(x] = /(o 



Therefore cos x i --- 1 

12. [4 



Example 16. To find a series for log e 



Let 
then 

Now 
so that 

so that 
and 
so that 
Hence 


/(*) 

/() 

/'(o) 
/"(o) 

f"'lv\ 

J (*/ 

/'"(o) 
log e (i+#) 


= log e (i+#) 
= log 1=0. 
rflog(l+#) I 


~ i 


i 
d/ -i \ 2 


^\(l+^) 2 / (l+^) 3 

2 
= - 2. 

^r 2 Ar 3 



(Compare with the series found by an entirely different method in 
Part I, p. 470.) 



Example 17. Prove that &>* = cos #+_;' sin ^, where /= V I. 
This equation is of great importance, since it links up the exponential 
and the trigonometric functions. 



APPLICATIONS OF DIFFERENTIATION in 

To find a series for sin x. 

Let . f(x) = sin x then /(o) = sin = 

f'(x] = cos x /'(o) = cos o i 

f"(x) = sin x /"() sin o = o 

f"'(x) = cos x f "'(o) = cos o = i. 

Hence sin # = x 1 . . . 

LI LL 

and jsinx = i(x |--j . . .) 

v Li U 

To find a series for ei x . 

Let f(x) = ei x /(o) = e = i 

f"(x) = j z ei x /"(o) = j z e = j z = i 

-7 2 A 2 ^ ^X^ 

Hence ei x i+jx+- \- J - . . . 

11 11 

Now cos x+j sin # (the series for cos x having been found in 
Example 15). 



Ivllj| + " { Fory : : -_\^ \ 

= ei x . ( i* = +i, etc. J 

Use might be made of Taylor's Theorem to determine a more 
correct solution to an equation when an approximate solution is 
known ; for, taking the first two terms of the expansion only 

/(*+*)=/(*)+*/'(*) 

or interchanging x and h, as a matter of convenience, then 

/(*+*)=/(*)+'(*) 

If A is small compared with x, the assumption that two terms 
of the series may be taken to represent the expansion is very 
nearly true. 

Suppose that a rough approximation for the root has been 
found (by trial and error) ; denote this by x. Let the true solution 
be x-\-h; then by substitution in the above equation the value of 
h can be found, and thence that of x-}-h. 



H2 MATHEMATICS FOR ENGINEERS 

As an illustration, consider the following case : A rough test 
gives 2-4 as a solution of the equation x* i -5^+3 -7% = 21-554. 
It is required to find a solution more correct. 

Here x = 2-4 and f(x) = # 4 1-5# 3 +37# 21-554 

so that 7(2-4) = 33-17 20-73+8-88 21-554 = -'234- 

If the correct value of h is found, thenf(x-\-h) must = o. 

Hence f(x+h) =f(x)+hf'(x) 

i. ., o = 234+A/'(2-4). 

[Now /(*) = * 4 i-5* 3 +37* 21-554 

so that 7' (2-4) = 55-30-25-92+3-7 = 33-o8.] 
Hence o = 234+^x33-08) 

or h= -^o = -0071. 

33-o8 

Hence a more correct approximation is 2-4+ -0071 
*. e. t x = 2-407 is the solution of the equation. 

This method may thus be usefully employed in lieu of the 
graphic method when extremely accurate results are desired. 

The following example illustrates the process of interpolation 
necessary in many cases where the tables of values supplied are 
not sufficiently detailed for the purpose in hand; and in view of 
the importance of the method, every step in the argument should 
be thoroughly understood. 

Example 18.' It is desired to use some steam tables giving the 
pressures for each 10 difference of temperature, to obtain the accurate 

d-b 
value of . when t = 132 C. The figures in the line commencing 

with i = 130 C. (the nearest to 132) are as follows : 



1 


dp 

: V 


d*p 

ip 


d* P 

dp" 


d'p 
dt* 


130 


2025-7I7 60-5995 


1-47051 


026392 


0002738 



Calculate, very exactly, the value of -^ when t = 132 C. 
Taylor's theorem may here be usefully employed, using the form 



APPLICATIONS OF DIFFERENTIATION 113 

Let f(x) = -- when t 130 

andf(x + h) = - when t = 132, so that h = 2 



Then /'(*) = = and /"(*) = , etc., / having the 

value 130. 

Thus the expansion may be re-written as 
_fdp 



and substituting the values from the table 

= 60. 5995 + (2 x.i-4705i) + (2 X -026392) + (4 x -0002738) 

'133 V 3 

= 63-59367- 



Exercises 11. On the Calculation of Small Corrections and Expansion in 

Series. 

1. If R = R (i+a/+6* 2 ) when R (the resistance of a conductor at 
o C.) is 1-6, a (the temperature-resistance coefficient of the material) 
= -00388 and b = -000000587, find the error in R (the resistance at 
temperature t C.) if t is measured as 101 instead of 100. 

2. The quantity Q of water flowing over a notch is given by 

o - 

Q = X-64X V^g.H*, where H is the head at the notch. What 
is the percentage error in Q caused by measuring H as -198 instead 

Of -2 ? 

3. If y = 4* 1 ' 76 , y = 17-3 when x = 2-3. What will be the change 
in y consequent on a change of x to 2-302 ? 

4. A rough approximation gives x = 2-44 as a solution of the 

2j 

equation 10 3 = 16+4^ x z . Find a more correct root. 

5. Determine the value of x to satisfy the equation x 1 ' 5 3 sin x = 3, 
having given that it is in the neighbourhood of 2-67. 

6. The height A of a Porter governor is expressed by 



w ' n* 

where n is the number of revolutions per minute. If W = 100, w = 2 
and / = 10, find the change in the height due to a change in the 
speed from 200 to 197 r.p.m. 

7. In calculating the co-ordinates of a station in a survey it was 

thought that there was a possibility of an error of 3 minutes (*. e., i 

either way) in the reading of the bearing. If the bearing of a line 

was read as 7 12' and the length of the line was 2 chains 74 links, 

I 



n 4 MATHEMATICS FOR ENGINEERS 

find the possible errors in the co-ordinates of the distant end of the 
line. [Co-ordinates are length X cos (bearing) and length x sine (bearing) .] 

8. Find by the methods of this chapter a series for a x . 

9. Using the figures given in Example 18, p. 112, calculate very 
exactly the pressure p at 133 C. 

10. The equation d 3 +-6$d -5 = o occurred when finding the sag 
of a cable. A rough plotting gives the solution to be in the neigh- 
bourhood of -5 : find a more exact root. 



CHAPTER V 
INTEGRATION 

HAVING discussed the section of the Calculus which treats of 
differentiation, we can now proceed to the study of the process 
of integration, this having a far more extensive application, and 
being, without doubt, far more difficult to comprehend. 

As with the differentiation, it is impossible fully to appreciate 
this branch of the subject unless much careful thought is given 
to the fundamental principles; and accordingly the introduction 
to the Integral Calculus is here treated at great length, but in a 
manner which, it is hoped, will commend itself. 

Meaning of Integration. The terms integer and integral 
convey the idea of totality; an integer being, as we know, a 
whole number, and thus the sum of its constituent parts or 
fractions. The process of integration in the same way implies a 
summation or a totalling, whereas that of differentiation is the 
determination of rates of change or the comparison of small 
differences. Differentiation suggests subtraction or differencing, 
whilst integration suggests addition ; differentiation deals with rates 
of change, integration with the results of the total change ; differen- 
tiation involves the determination of slopes of curves, and integra- 
tion the determination of areas of figures. Integration is, in fact, 
the converse to differentiation, and being therefore a converse 
operation is essentially more difficult to perform. [As instances 
of this statement contrast the squaring a quantity with the 
extraction of a square root, or the removal of brackets with 
factorisation.] 

A converse operation is rather more vague as concerns the 
results than a direct; for when performing a direct operation one 
result only is obtainable, but the results of a converse operation 
may be many, as we shall find, for example, when dealing with 
indefinite integrals. 

To illustrate the connection between differentiation and integra- 
tion, consider the familiar case of velocity and acceleration. Suppose 
values of v and t are given, as in the table : 

"5 



n6 



MATHEMATICS FOR ENGINEERS 



Then 



t 


I 


15 


20 


25 -30 


V 


28-4 


297 


30-5 


33'4 36-5 


8v 


i'3 


8 


2-9 


3'i 


8t 


05 


05 


05 


05 


Bv 

a = w 


26 


16 


58 


62 



The accelerations are here found by comparing differences of 
velocity with differences of time. 

Regard the question from the other point of view : assume 
that these accelerations are given and we wish to determine the 
total change in the velocity in the given period of time. The 
total change must be given by the sum of the changes in the small 
periods of time; in the first period of -05 sec. the average 
acceleration was 26, i, e., the velocity was being increased at the 
rate of 26 units per sec. each sec. ; and therefore the change in the 
velocity in -05 sec. = 26 x -05 units per sec. 
= 1-3 units per sec. 

In the successive periods the changes in velocity are -8, 2-9 
and 3-1 respectively. 

Hence the total change in the velocity over the period -2 sec. 
= i-3-f--8+2-9+3'i == 8-1 units per sec., or if the initial velocity 
was 28-4, the final velocity was 28-4+8-1 = 36-5. Note that the 

acceleration is given by the fraction ^, whilst a small change in 

ot 

the velocity is of the nature aU, or the total change of velocity 
= sum of all small changes = ^aSt. 

We can thus find integrals by working through the processes 
of differentiation, but in the reverse order. If a function, expressed 
in terms of symbols, has to be integrated, it is an advantage to 
transform the rules for differentiation into forms more readily 
applicable; the method, however, being entirely algebraic. 

If numerical values alone are given, the integration resolves 
itself into a determination of an area. 
Hence 

Considered from an algebraic standpoint 
Differentiation implies the calculation of rates of change ; 
Integration implies the summation of small quantities. 



INTEGRATION 



117 



From the graphic standpoint 

Differentiation is concerned with the measurement of slopes of 
curves ; 

Integration is concerned with the measurement of areas under 
curves. 

Just as special symbols are used to denote the processes of 
differentiation, so also there are special symbols for expressing the 
processes of integrations. 

Regarding an integral as an area, it must be of two dimensions, 
a length and a breadth; and we have seen in an earlier chapter 
(Part I, Chap. VII) that in order to ascertain an area correctly 
its base must be divided up into small elements, the smaller the 
better, these elements 
not necessarily being of 
the same length, but all 
being small. Thus, to 
find the area ABCD- 
(Fig. 27) we can suppose 
it divided up into small 
strips, as EFGH, then 



find the area of each of 
these and add the results. 
The portion EH of the 
curve is very nearly 
straight, so that EFGH 
is a trapezoid, and hence 
its area = mean height 
X width. Now its mean 




6JG 

I 



F G 

FIG. 27. 



height FE and GH are practically the same, so that any one of 
them can be denoted by y; also the width FG of the strip is a 
small element of the base, *'. e., is 8x. 

Hence, the area of the strip EFGH y x 8x, and the total 
area between the curve, the bounding ordinates and the axis of x 
must equal the sum of all products like y8x, or, as it might be 
expressed 

Area = 2yS# (approximately). 

However small the width of the strips are made, this sum only 
gives the area approximately, but as 8x is diminished the result 
approaches the true more and more closely. 

Therefore, bearing in mind our previous work on limits, we 
can say that the limiting value of 2ySx must give the area exactly. 
To this limiting value of the sum different forms of symbols are 



n8 MATHEMATICS FOR ENGINEERS 

attached, the 2 and S being replaced by the English forms / 
and d respectively, so that the area between the curve and the 
axis of x = fydx. There is no limit placed to this area in any 
horizontal direction, so that the area is not denned by the given 
formula. 

Hence fydx is spoken of as an indefinite integral. 

The x is again the I.V., and the size of the area will depend 
on the values given to it. Suppose that when y = AB, x = a, 
and when y = CD, x = b ; then the range of x is from a to & if 
it is the area ABCD that is considered. Accordingly we can state 
that the area ABCD =fydx, the value of this integral being 
found between x = a and x = b, or, as it is written for brevity, 

fx=b fb 

I ydx, or, more shortly still, I ydx, it being clearly understood 

J x=a J a 

that the limits a and b apply to the I.V., i. e., that quantity 
directly associated with the " d." 

It is evident that ABCD is a definite area, having one value 

fb 

only, and thus I ydx is termed a definite integral. 

J a 

The most convenient method for determining areas (provided 
that a planimeter is not handy) is undoubtedly the " sum curve " 
method treated in Part I, Chap. VII ; the great virtue of it being 
that the growth of the area is seen, and that either any portion 
or the whole of the area of the figure can be readily found by 
reading a particular ordinate. 

In view of the great usefulness of the process of integration 
by graphic means, the method is here explained in detail, following 
exactly the plan adopted in Part I, Chap. VII. 

Graphic Integration is a means of summing an area with the 
aid of tee and set square, by a combination of the principles of 
the "addition of strips" and "similar figures." An area in 
Fig. 28 is bounded by a curve a'b'z', a base line az and two vertical 
ordinates aa' and zz'. The base is first divided in such a way 
that the widths of the strips are taken to suit the changes of 
curvature between a' and z' ', and are therefore not necessarily equal ; 
and mid-ordinates (shown dotted) are erected for every division. 
Next the tops of the mid-ordinates are projected horizontally on 
to a vertical line, as BB'. A pole P is now chosen to the left of that 
vertical; its distance from it, called the polar distance p, being a 
round number of horizontal units. The pole is next joined to 
each of the projections in turn and parallels are drawn across the 



INTEGRATION 



119 



corresponding strips so that a continuous curve results, known as 
the Sum Curve. Thus am parallel to PB' is drawn from a across 
the first strip; mn parallel to PC' is drawn from m across the 
second strip, and so on. 

The ordinate to the sum curve through any point in the base 
gives the area under the original or primitive curve from a up to 
the point considered. 

Referring to Fig. 28 

Area of strip abb' a' = ab x AB 




Pole 



FIG. 28. Graphic Integration, 
but, by similar figures 



B'a or BA 



bin 
ab 



whence 



AB x ab = p X bm 



, area of strip , , . , , 

i. e., bm = - or area of strip = px bm 

P 

i. e., bm measures the area of the first strip to a particular scale, 
which depends entirely on the value of p. 

, area of second strip 
In the same way nm = - 

P 



120 MATHEMATICS FOR ENGINEERS 

and by the construction nm' and bm are added, so that 

area of ist and 2nd strips 
en 

P 

or area of ist and 2nd strips p x en 

Thus, summing for the whole area 

Area of aa'z'z = p x zL 

Thus the scale of area is the old vertical scale multiplied by the polar 
distance ; and accordingly the polar distance should be selected 
in terms of a number convenient for multiplication. 

E. g., if the original scales are 

i" = 40 units vertically 
and i" = 25 units horizontally 

and the polar distance is taken as 2", i. e., 50 horizontal units; 
then the new vertical scale 

= old vertical scale x polar distance 
= 40x50 = 2000 units per inch. 

If the original scales are given and a -particular scale is desired 
for the sum curve, then the polar distance must be calculated as 
follows 

new vertical scale 

Polar distance in horizontal units = , , ,-. , , 

old vertical scale 

E. g., if the primitive curve is a " velocity-time" curve plotted 
to the scales, i" 5 ft. per sec. (vertically) and i" = -i sec. (hori- 
zontally), and the scale of the sum curve, which is a " displacement- 
time " curve, is required to be i" = 2-5 ft., then 

2*S 

Polar distance (in horizontal units) = = -5 

and since i" = -i unit along the horizontal, the polar distance 
must be made 5". 

Integration is not limited to the determination of areas only; 
true, an integral may be regarded as an area, but if the ordinate 
does not represent a mere length, but, say, an area of cross section, 
the value of the integral will in such cases measure the volume of 
the solid. 

Our standard form throughout will be for the area of the figure 
as plotted on the paper, viz., Jydx, where y is an ordinate and 
8x an element of the base, but y and x may represent many 
different quantities. 

Thus, suppose a curve is plotted to represent the expansion of 



INTEGRATION 



121 



a gas; if, as is usual, pressures are plotted vertically and 
volumes horizontally, the ordinate is p and an element of the 

base is Sv; hence the area under the curve = \ z pdv (if the 

J n 
initial and final volumes are v t and v 2 respectively), and since this 



/6s 



is of the nature pressure X volume, i. e., 7^3 X (ft) 3 or ft. Ibs., the 

\7*J 

area must represent the work done in the expansion. 
To illustrate such a case : 



Work 

106560 





8 10 la 14- 16 18 SO 2 24 26 
FIG. 29. Expansion of Steam. 

Example i. It is required to find the work done in the expansion 
of i Ib. of dry saturated steam from pressure 100 Ibs. per sq. in. to 
pressure 15 Ibs. per sq. in. 

From the steam tables the following corresponding values of p 
and v are found : s 



v (cu. ft. per Ib.) 


4.44 


5-48 


7-16 


10-50 


I3-72 


20 


26-4 


p (Ibs. per sq. in.) 


IOO 


80 


60 


40 


30 


2O 


15 



By plotting these values, p vertically, the expansion curve is 
obtained (Fig. 29) ; this being the primitive curve. 

Selecting a polar distance equivalent to 10 horizontal units, we 
proceed to construct the sum curve, the last ordinate of which 
measures to a certain scale the work done in the expansion. Now 
the new vertical scale = old vertical X 10, since the polar distance = 10 ; 
and also we must multiply by 144, since the pressures are expressed 



122 



MATHEMATICS FOR ENGINEERS 



in Ibs. per sq. in. and must be converted to Ibs. per sq. ft., so that 
the work done may be measured in ft. Ibs. 

According to this modified scale the last ordinate is read off as 
106560 ; thus the work done = 106560 ft. Ibs. 

or, as it would be written in more mathematical language 

T26-4 



f 

J4 



pdv 106560. 



Example 2. The diameters of a tapering stone column, 20 ft. long, 
at 6 equidistant places were measured as 2-52, 2-06, 1-54, i'i5, '80 
and -58 ft. respectively. 

Find its weight at 140 Ibs. per cu. ft. 




29\5 

6 8 (O 12 14- IG IQ 20 

FIG. 30. Problem on Stone Column. 

The volume will be obtained by plotting the areas against the 
length and summing. Now the area of any section = -d z , and the 
total volume will be the sum of the volumes of the small elements 
into which the solid may be supposed to be divided. 

/"20 pO^ 

Thus the volume = J Adi = J Q ~d*.dl. 

and the weight = 140 I -d 2 .dl. 
J o 4 

Since - is a constant multiplier, it can be omitted until the end, 
for its effect is simply to alter the final scale ; hence a constant factor 
before integration remains so after. 



Hence the weight- 



= 1097 j 



d*.dl. 



INTEGRATION 



123 



The integral will be of the standard form if for d z we write y and 
if for / we write x, so that we see that ordinates must represent d 2 and 
abscissae lengths, and hence the table for plotting reads : 



I 


o 


4 


8 


12 


16 


20 


y or d 2 


6-34 


4-24 


2-36 


1-32 


64 


336 



Plotting these values and thence constructing the sum curve (see 
Fig. 30), we find the last ordinate to be 47-15, and this is the value 

f20 

of J Q d*.dl. 

[20 

Weight = 109-7 1 d z .dl = 109-7x47-15 = 5180 Ibs. 

Application of Integration to " Beam " Problems. At an 

earlier stage (see p. 38) it has been demonstrated that the shear 
at any point in the length of a beam loaded in any way whatever 
is given by the rate of change of the bending moment in the 
neighbourhood considered, this being the space rate of change. 
Conversely, then, the bending moment must be found by summating 
the shearing force ; and hence, if the shear curve is given, its sum 
curve is the curve of bending moment. 

In the majority of problems the system of loading is given, 
from which the curve of loads can be drawn. Then, since the 
shear at any section is the sum of all the forces to the right or 
left of that section, the sum curve of the load curve must be the 
shear curve; continuing the process, the sum curve of the shear 
curve, i. e., the second sum curve from the load curve as primitive, 
is the curve of bending moment and the fourth sum curve is the 
deflected form. 

Expressing these results or statements in the notation of the 
calculus ; L, S and M being the respective abbreviations for loading, 
shear and bending moment 

S =fLdx 

M =f$dx =f(fLdx)dx = ffL(dx) z 
[ff~L(dx) z being termed a double integral] 

and the deflection y = ff M.(dx) 2 or ffff L(dx)*. 

If the loading is not uniform, but continuous, the summation 
must be performed graphically. [The link polygon method largely 
used obviates half these curves, e. g., the link polygon for the loads 
gives at once the curve of bending moment.] 



124 



MATHEMATICS FOR ENGINEERS 



Example 3. The loading on a beam, 24 ft. long, simply supported 
at its ends varies continuously, as shown in the table. Draw diagrams 
of shearing force and bending moment, stating clearly the maximum 
values of the shearing force and the bending moment. 



Distance from one end (ft.) 


o 


4 


7 


10 


12 


J 4 


17 


20 


24 


Load in tons per ft. 


44 


58 


86 


1-06 


I-I 


i -06 


86 


58 


'44 



The curve of loads is first plotted, as in Fig. 31. 

By sum-curving this curve, we obtain the curve of shearing force, 
although no measurements can be made to it until account has been 
taken of the support reactions. 




4 68 1C 12' ' 14 >6 <8 

FIG. 31. Problem on Loaded Beam. 

To find the reactions at the ends : We know that these must be 
equal, since the loading is symmetrical, each reaction being one-half 
of the total load. Now the last ordinate AB of the sum curve of the 
load curve is 19; thus the reactions are each 9-5. Bisecting AB, or, 
in other words, marking off a length AC to represent the reaction 
at A, we draw a horizontal, and this is the true base line for the curve 
of shear; any ordinate to the curve of shear from this base giving 
the shear at the point in the length of the beam through which the 
ordinate is drawn. 

We observe that the shear changes sign and is zero at the centre 
of the beam; we can conclude from this that the bending moment 
must have its maximum value at the centre, since shear rate of 
change of bending moment, and if the shear is zero, the bending 
moment must have a turning value. 

By sum-curving the shear curve from CD as base, the resulting 
curve is that of bending moment. 



INTEGRATION 



125 



It is well carefully to consider the scales, for it is with these that 
difficulties often arise. 

The scales given here apply to the original drawing, of which 
Fig. 31 is a reproduction somewhat under half full size. 

For the length i in. = 2-5 ft. 

For loads i in. = -4 ton per ft. 

Polar distance for the first sum curve, i. e., the curve of shear 

= 4 ins. = 4X2-5, or 10 horizontal units. 
Hence the scale of shear = -4 x 10, or 4 tons to i in. 



B 




FIG. 32. Shearing Force and Bending Moment on Ship's Hull. 

Polar distance for the second sum curve = 4 ins. = 10 horizontal 
units. 

Hence the scale of bending moment = 4 X 10 = 40 
or i in. (vertically) = 40 tons. ft. 

Reading according to these scales 

The maximum shear = 9-5 tons ^ 
and the maximum bending moment = 68 tons, ft. J 



Example 4. In Fig. 32 AAA is the curve of weights or load 
distribution, and BBB the curve of buoyancy or upward water thrust 



126 MATHEMATICS FOR ENGINEERS 

for a ship whose length is 350 ft., the scale of loads being indicated 
on the diagram. 

Draw diagrams of shearing force and bending moment on the hull 
of the vessel and measure the maximum values of these quantities. 

It is first necessary to construct the curve of loads to a straight 
line base, and to do this the differences between the curves AAA and 
BBB are set off from a horizontal, taken in our case below the original 
base line. 

In this way the curve of loads LLL is obtained, the scale being 
shown to the left of the diagram. 

By sum-curving this curve, the curve of shear SSS is obtained ; 
the polar distance (not shown on the diagram) being taken as 50 horizontal 
units, so that the scale for the shear is 50 times the scale for the loads. 

Sum-curving the curve SSS, the curve MMM, that of bending 
moment, is obtained (again the polar distance is 50 horizontal units). 

Sectio'ns such as K, where the upward thrust of the water balances 
the downward force due to the weights, are spoken of as water-borne. 

Reading our maximum values according to the proper scales, we 
find them to be 

Maximum shear = 246 tons ~\ 

Maximum bending moment = 14,300 tons ft. / 

It should be noted that the last ordinate of both the shear curve 
and the curve of bending moment is zero ; these results we should 
expect since the areas under the curves AAA and BBB must be 
equal, so that the shear at the end must be zero, and also the moments 
of these areas must be alike. 

[In practice the maximum bending moment is found by such a 
formula as- 
Weight X length 

Maximum bending moment = 7? 

Constant 

the constant for small boats being between 30 and 40, and for larger 
between 25 and 30.] 

The Coradi Integraph. A brief description of the Integraph, 
an instrument devised to draw mechanically the sum curve, can 
usefully be inserted at this stage. 

It consists essentially of a carriage running on four milled 
wheels A (Fig. 33), a slotted arm C carrying the tracer B which 
is moved along the primitive curve, and the arm D which carries 
the pencil E which draws the sum curve. 

As B is moved along the primitive curve, the slotted arm C 
slides about the pins G and P, thus altering its inclination to the 
horizontal. A parallel link motion ensures the movement of E 
parallel to the instantaneous position of C, the sharp-edged wheel F 
assisting in guiding the tracer bracket. 



INTEGRATION 



127 



The principle of the instrument is not difficult to understand, 
and can be explained in a very few words. 

The pole is at P, and the tops of the mid-ordinates are pro- 
jected to the vertical through G by the horizontals like BG; 
parallels are then drawn to PG by the pencil E, the motion being 
continuous. 

The polar distance can be varied as desired, by altering the 
position of the bracket carrying the pin P along the horizontal 




arm; and if an extremely small polar distance is found to be 
advisable, the pin H may be utilised instead of G. 

Rules for Integration of Simpler Functions. Since 
integration is the reverse of differentiation, many functions can be 
integrated by reversing the order of the steps in differentiation. 

Integration of powers of x. The first rule given in the 
work on differentiation of functions was 

d 

x n = nx n ~ v . 

dx 



128 MATHEMATICS FOR ENGINEERS 

To change into the integration form, we transpose -5- : the " d" 

on the one side becomes f on the other side, to indicate the change 
differencing to summing, and the " dx " occurs on the top line of the 
other side of the equation. 

Thus x n = fnx n ~ 1 dx 

or fx n ~ 1 dx = - x n -\-C 

% 

the reason for the presence of the constant term C being explained 
later. 

It is a trifle simpler to write n in place of ni, and therefore 
+i in place of n, so that 



Whereas, when differentiating a power of the I.V., the power 
was reduced by i in the process of differentiation ; when integrating, 
the power is increased by i. 

fa <y-5 tiv^ 

- g " dx x 

whereas fx 5 dx = ^ x 6 -\-C. 

A special case occurs for which the above rule does not apply : 
for, let n = i, then fx~ l dx should, according to the rule just 

given, be - x, but to this fraction no definite meaning can be 
assigned. 

We know that -3- log x = - = x~* 

dx x 

fx- 1 dx = log x+C 
or, as it is sometimes written 
dx , 



A constant multiplier before integration remains as such after; 
thus faxndx = T % W+1 +C. 

Also an expression composed of terms can be integrated term 
by term, and the results added. 



INTEGRATION 129 

Thus, f (ax n -\-b) dx can be written 

fax n dx -\-fbdx 
i. e., fax*dx-\-fbx?dx f or x = i 

its value being -x n+l -\-bx+C. 

Note. Differentiation of a constant term gives zero, but the 
integration gives that constant multiplied by the I.V. 

The reason for this will be apparent if we consider the state- 
ments from the graphical standpoint. The curve representing the 
equation y = b is a horizontal straight line, and therefore the 

slope is zero (i.e., -v- = o) ; but the area under the curve = the 
area of a rectangle = base x height = xxb (i. e.,fbdx = bx). 



Exponential Functions. We have already proved that 
dx 



de x 

- e* ( S ee p. 47) ; then by transposition of d and dx to the 



other side of the equation we obtain the statement e x = j 

and since -, corresponds to f we may write this as fe x dx = e x -\-C. 

Thus if we either differentiate e x or integrate it we arrive at the 
same result ; and e x is the only function for which the differential 
coefficient and also the integral are the same as the function itself. 
Carrying this work a step further, let us consider the integration 
of e bx , and hence a x : 

Now f-ae bx = abe bx .*. fae bx dx = ?J*+C. 

dx b 

To avoid confusion as to the placing of a and b we must reason 
in the following manner : The a is a constant multiplier of the 
whole function, and therefore remains so after integration; the b 
multiplies the I.V. only; and thus differentiation would cause it 
to multiply the result, whereas after integration it becomes a 
divisor. Great attention should be paid to the application of this 
rule, for unless care is exercised mistakes are very apt to creep in. 

Example 5. Find the value of -^(15** 7*' 9 +83) and also of 



130 MATHEMATICS FOR ENGINEERS 

Differentiating the first expression 
d. 



Integrating the second given expression 






4 --i + i 
= i 5 t*- 7 f+C. 

Notice that although a function has been differentiated and the 
derivative integrated, the final expression is not exactly the same as 
the original, the constant term being represented only by C, where C 
may have any value. Further reference will be made to this point 
on p. 137. 

Example 6. If pv 1 ' 32 = C, find the value oi/pdv. 
To express p in terms of v 



fpdv = fCir- 1 -** . dv = Cfv~ l - 3Z dv 
(K being any constant) 



= CX- u-l-38+1 

-1-32+1 



= 
32 

This result can be written in a slightly different form, if for C we 
write its value pv 1 ' 32 ; then 

F., pv l '**xir-** . v 

ipdv = \-K 

32 

pv 

= 3* 

32 a 



Exa'.-nple 7. Find fpdv when pv = C. 
In this case p = Cv~* 

fpdv = fav - C/^ = C log v + K 
= pv log v 

Example 8. Find the value of 



Note that 17 is a constant multiplier throughout; 2 multiplies the 
I.V. and therefore appears as a divisor after integration ; also the 
power of e remains exactly the same. 



INTEGRATION 
Example 9. Find the value of f(^oe &v +v 5 -*)dv. 

f(4oe'*>+v s -*)dv = f^oe &v dv+fu 5 -*dv (separating the terms) 

-e 
5 



131 



Example 10. Find an expression for fa x dx, and apply the result 
to determine the value of /i2 x 



From our previous work we know that 

d 



dx 



a x = a x . log a. 



fa x dx = . -.* 

* log a 



Hence fiz x 5 tx dx = (i2X- X . ^ X 5 4 *)+C 



loge 5 = 1-609 



= 1-864 X5 4g +C. 

Afofe. It would be quite incorrect to multiply 1-864 by 5 and 
express the result as g-^2* x . 

Alternatively, the result might have been arrived at in the following 
manner 



= 12 x, H X(5 4 )*+C 

log 625 VJ ' 

= g; X5 4ar +C (log 625 = 6-44) 

= 1-864 X5 4 *+C. 

Exercises 12. On Graphic Integration. 

1. The acceleration of a slider at various times is given in the 
table. By graphic integration obtain the velocity and displacement 
curves to a time base, indicating clearly your scales. 



j Time . 


o 


008 


016 


02 


028 


036 


44 


048 


06 


! Acceleration 


o 


75 


87-5 


87-5 


87-5 


87-5 


87-5 


83 


O 



068 


072 


084 


10 


108 


12 


78 


85 


87-5 


87-5 


83 


O 



132 



MATHEMATICS FOR ENGINEERS 



2. An acceleration diagram on a time base has an area of 4-7 
sq. ins. The base of the diagram is 2-5 ins. and represents 25 sees. 
The acceleration scale is i in. =3 ft. per sec. 2 . If the velocity at the 
beginning is n ft. per sec., find the velocity in ft. per sec. at the end 
of the 25 sees. 

3. A rectangular barge is loaded symmetrically in still water. 
The curve of loading is a triangle with apex at the centre, and the 
curve of buoyancy is a rectangle. Draw diagrams of shearing force 
and bending moment on the barge. 

4. The curves of loads for a ship 350 ft. long is as given in the 
table. Plot this, and by graphic integration obtain the curves of 
shearing force and bending moment. 



Distance from one end (ft.) 





7 


10 


35 


56 


84 


102 


load (tons per foot) . 





3 


o 


-2-6 


-3-i 


-2-3 


O 



112 


133 


161 


196 


210 


237 


260 


280 


315 


330 


350 


1-6 


2-3 


3-15 


5'2 


57 





-4-5 


-4-95 


o 


9 






5. The table gives the values of the pressure and volume for the 
complete theoretical diagram for a triple expansion engine. 



V 


o 


I 


2 


4 


6 


8 


10 


12 


p 


240 


240 


I2O 


60 


40 


30 


24 


20 



Find the initial pressure in each cylinder in order that the work 
done per cycle may be the same for each. 

(Hint. Divide the last ordinate of the sum curve into three equal 
parts, draw horizontals through these points of section to meet the 
sum curve, and from these points of contact erect perpendiculars to 
cut the expansion line.) 

6. A body weighing 3000 Ibs. was lifted vertically by a rope, there 
being a damped spring balance to indicate the pulling force F Ibs. of 
the rope. When the body had been lifted x ft. from its position of 
rest, the pulling force was automatically registered as follows : 



X 





20 


40 


65 


75 


95 


no 


140 


F 


8000 


7950 


7850 


7500 


7400 


6800 


6400 


4000 



Find the work done on the body when it has risen 80 ft. How 
much of this is potential energy and how much is kinetic energy ? 
Find, also the work done when it has risen 140 ft. 



INTEGRATION 



133 



7. The current from a battery was measured at various times, 
with the following results : 



Time (hours) . 


I 


3 


6 


9 


10 


12 


14 


15 


Current (amperes) 


25 28 


37 


39'5 


32 


29 


24 


25-3 


2? 



If its capacity is measured by fCdt, find the capacity in ampere 
hours. 

8. The following are the approximate speeds of a locomotive on a 
run over a not very level road. Draw a curve showing the distance 
run up to any time. 



Time (mins. and sees.) 





I 


2-15 


6.15 


9.22 


11-45 14.26 


16.35 


20.52 


Speed (miles per hr.) 





6 


10 


18-2 


22-8 


25-5 28 


29-2 


28-6 



9. The load curve at a large central station can be constructed 
from the following data : 



Time (hours) . . , 





I 


2 


3 


4 


5 


6 

14 


7 


7-5 8 


9 


10 


ii 


Load (1000 amperes) 


3'5 


i 


I 


2 


I 


6-4 


i7 


17-8 16-4 


ii'3 


8-7 


8-2 



12 


I 


2 


3J4 


5 i 5'5 


6 


7 


8 


9 


10 


II j 12 


| 7 -8 


8 


7-6 


8-7 12-5 


19 23-3 


21 


12-4 


ii 


10-5 


9-6 


9 ! 6 



Find the total number of ampere hours supplied in the 24 hours. 

10. The velocity of a three-phase electric train, with rheostatic 
control, at various times, was found as in the table : 



Time (sees.) . . . | o 


26-6 


66-6 


80- 1 


99 


Velocity (ft. per sec.) . o 


40 


40 


37-3 






Draw the space-time curve and find the total distance covered in 
the 99 seconds. 



On the Integration of the Powers of x and of the Exponential Functions. 

11. What is the significance of the symbols f and dx in the 
expression fx z dx ? 

Integrate, with respect to x, the functions in Examples 12 to 27 



12. 4* 1 -". 13. 70-15. 



14. 



15. 



16. e'. 



17. *- +14. 
x 



18. 



19. 



134 MATHEMATICS FOR ENGINEERS 

20. i2*~. 21. -g^. 22. -i 7 e". 23. 

24. 2 . 54 *--i- 8 . 2 *-i++i-i3. 25. 



26. ^.*_*--3+ 27. -94*' 18 cos - 

Find the values of the following 

2B.fv 5 dv. 29. /"-". 30. /35^. 31. /g"-<fc 

32. //xfo when ^y 1 ' 17 = C. 33. /i4 X 2*^5. 

-8)d*. 35. /3- 1'. 36. 17 /^g. 

J r 

38. $-z-2(dt) z . 39. 2-ix~ 5 dx 



40. Solve the equation -/- = w -. 

rfy y 

41. In connection with the flow of air through a nozzle, if x is the 
distance outwards from the nozzle and v is the velocity there, v oc - 1 . 

Also SA (an element of area of flow) = ~K8x Vx. The added momentum 
for the small element considered = SM = v8A. Show that M the 

D 
total increment to the momentum, can be written C 7= where C 

vx 

and D are constants. 

Trigonometric Functions. We have previously seen that 
the derived curve of either the sine curve or the cosine curve is the 
primitive curve itself transferred back a horizontal distance of 
one-quarter of the period. Conversely, then, we may state that 
the sum curve of either the sine or the cosine curve is the curve 
itself moved forward for a distance corresponding to one-quarter 
of the period. In other words, integration does not alter the form 
of the curve. Taking the case of the sine curve as the primitive, 
we see, on reference to Fig. 34, that if this curve is shifted forward 
for one-quarter period the resulting curve is the cosine curve 
inverted ; or expressing in algebraic language, whilst the equation 
of the primitive curve is y = sin x, that of the sum or integral 
curve is y = cos x. Thus, fsmxdx = cos x. In like manner 
it could be shown that feosxdx = sin*. For emphasis, the dif- 
ferentiation and the integration of sine x and cosine x are repeated 
here 

-T- sin x = cos x /sin* dx = cos #4-C 

dx 

-T-COS x sin x /cos* dx = sin x-\- C. 

dx J 



INTEGRATION 



135 



Note. When differentiating the cosine the minus sign appears in 
the result ; when integrating the sine the minus sign appears ; it is 
important to get a good grip of these statements, and the con- 
sideration of them from the graphic aspect is a great help in this 
respect. 

To extend the foregoing rules 

fcosxdx = sin x-\-C 
/cos (ax+b)dx = sin (ax+b)+C 

fsinxdx = cos x-\-C 
/sin (ax+b)dx = cos (ax+b)+C. 



1 
75 

5 
25 
O 

25 
5 

75 
1 


^ 


\/ 


J? 


N 


\ 












/ 


s~ 




A 






\ 


. // 


-si 


n.. 


rr 




/ 




/ 




\ 






Y 




/ 






I 




\ 






\ 








/ 


JG 








% 




1 


\ 




/ 


* 




24\ 


-y 


-a 


/ 
is.< 


r\ 






\ 




/ 






/ 




\ 






y 






/ 












\ 


V 


s 


A 


V 


s 


/ 





FIG. 34. 

Thus the angle remains the same after integration just as it 
would after differentiation, but the constant multiplier a of the 
I.V. becomes a divisor. 

To integrate sec 2 -AC with regard to x we call to mind the 

differentiation of tan x, viz., , tan x = sec 2 x. Accordingly 

/sec 2 xdx = tan x+C. 

Extending this to apply to the more general case 

/sec 2 (ax+b)dx - - 1 tan (ax+6)+C 
In like manner 

/cosec 2 (ax+b)dx = - cot (ax+6)-{-C. 



I 3 6 MATHEMATICS FOR ENGINEERS 

Other two standard integrals are added here, the derivation of 
which will be considered in the next chapter. 

/tan* dx log cos x+C 
fcotx dx = log sin x+C. 

To verify these we might work from the right-hand side and 
differentiate. Dealing with the former 

d , , . _ _ d , if u = cos % 

dx^ dx whence 

d log u du du 

_ _ v Qin v 

- 1 S\ 7 7 ' Oil! Jv 

du dx dx 

* 

i 

= x sin x 
u 

sin x 

= - = tan x 
cos* 

/tan* dx = log cos x+C. 

Example n. Find the value of f(5~ sin^t)dt. 

f$dt fsin^tdt 



= ^ ~X cos 

V 4 

= 5'+- cos 4*+C. 
4 

Example 12. Evaluate /sin (5 ^t)dt. 

/sin (54t)dt = --x -cos (5 4/)+C = ] cos (5 40 + C. 
4 4 

Example 13. If a force P is given by P = 36-4 sin (1005 -62), 
find the value oifPds. 

= -364 cos (1005 62) + C. 
Example 14. Find the value of 

12 cos (4- 
The expression E = 7-25* 3s~ 1 +i5s -8 + 12 cos (4 35) 

= (7-2X^)- 3 logs+ i ;|si-8+(i2X-^sin ( 4 - 35 ))+C 
= I-44S 5 3 Jog 5 + 8-33S 1 - 8 4 sin (4 3 



fPds =/36'4 sin (loos -62)^5 = ^X cos (1005 -62) + C 



INTEGRATION 137 

Example 15. If R = n sec 2 (34-71;), find /Rdv. 



fRdv =/n sec 2 (3 4-?v)dv = - tan (3 4-7t>)+C 

= 2-343 tan (34- 



Exercises 13. On Integration of Trigonometric Functions. 

Integrate, with respect to x, the functions in Nos. i to 10. 
1. 3 sin 4*. 2. 5-18 cos (3 3^). 3. 7 sec 2 (^3 x). 

4. x 12 -14 cos (-05 -117*). 5. 05-4*4-5 s j n ^b+ax). 

6. 9-45 sin 8^. 7. 3-08 sin 2(2-16* 4-5). 

8. 9^^+^-1-83 tan x. 

9. 4-27 sin (--- J + -2 cos gx 4# 1 - 74 +3 2 *+ 5 . 

10. 2 sin 2 #2-91 sin ( 3'7#)+2 cos 2 #14-2 cosec 2 ----- . 

11. The acceleration of a moving body is given by the equation 

a = 49 sin (jt -26). 

Find expressions for the velocity and the space, the latter being 
in terms of the acceleration. 

. T f d z x o o / , cos 20\ _ , ,, , dx 

12. If T- 4 = 47c 2 wV^cos 0-i J, find the values of , and x. 

(x is a displacement of the piston in a steam-engine mechanism.) 

13. Find the value of /b^+cos (y77"2p)}dp. 

14. If v 117 sin 6^29-4 cos 6t, find the value offvdt. 

- 

Indefinite and Definite Integrals. The integrals already 
given, although correct, are not complete. If "an integral is to 
denote an area some boundaries must be known ; and nothing was 
said about the limits to be ascribed to x (or s, as the case might 
be) in the foregoing, so that we were in reality dealing with 
indefinite areas or integrals. To indicate that a portion of the 
area may be dispensed with in certain cases (when the boundaries 
are stated) a constant C is introduced on the R.H.S. of the equation, 

x* 

i. e.,fx 3 dx would be written +C. 

As soon as the integral, and therefore the area, is made definite 
it will be observed that C vanishes. 



138 MATHEMATICS FOR ENGINEERS 

If fx 3 dx is to equal -x*+C, -, (-# 4 +Cj should equal x 3 ; and 



this is the case for 



d 



-=- 
dx 



^A 4 c \ ^ 

dx\4 J dx 

(C being independent of x). 

Cf. Example 5, p. 129 ; the constant in that case being 83. 

It is therefore advisable to add the constant in all examples 
on integration; in many practical examples the determination of 
the value of the constant is an important feature, and therefore 
its omission would invalidate the results obtained. 

In the list of a few of the simpler standard integrals collected 
together here for purposes of reference and by way of revision 
the constant is denoted by C. 

f(ax n +b)dx a 



fax n dx 



Jbdx 
[dx 
J x 
[ dx 
J ax+b 


= bx+C 
= log x+C 

= ^ log (*+&) +C 


fae bx dx 


- a b ^+c 


fe*dx 


= e*+C 


faFdx 


= rog* x *+ c 


/sin (ax-{-b)dx 


- cos (ax+b)-\-C 


f sin xdx 


= cos x-\-C 



/cos (ax-\-b}dx 
f cos xdx 
/sec 2 (ax-\-b}dx 
f sec 2 xdx 
/cosec 2 (ax-\-b}dx 



= - sin (ax-}-b)-{-C 



& 

sin x-\-C 
- tan (a^ 



= tan 



-- cot (ax+b)+C 

d 



INTEGRATION 139 

/ cosec 2 xdx = cot x-\-C 

/tan (ax+b)dx = log cos (ax-}-b)-\-C 

8 

/ tan xdx = log (cos #)+C 

/cot (ax+b)dx = log sin (ax+b)-{-C 
f cot xdx = log (sin A;) +C. 

Method of Determining the Values of Definite Integrals. 

We may regard the area of a closed figure as the difference 
between two areas, viz., all the area to the left, say, of one boundary, 
minus all the area to the left of the other parallel boundary. 

Hence to find the value of a definite integral, the value of the 
integral must be found when the I.V. has its higher limiting value, 
and from this must be subtracted its value when the lower limiting 
value is substituted for the I.V. 

fx 2 dx = -x 3 -\-C is an indefinite integral, but if to C we give 

a definite value, it becomes definite and unique. 

/4 
Thus I x 2 dx is a definite integral, because the limits to be 

J 2 

applied to x are indicated. 
To evaluate it 

We know that fx z dx = -x 3 -\-C. 

The value of this integral when x = 4 is -f-C 

o 

and the value of this integral when x = 2 is |-C 
the constant being the same in the two cases. 



The difference = -^+C - -+C = ^ 
V 3 / \3 / 3 



meaning that if the curve y = x z were plotted, and the area between 
the curve, the x axis and the ordinates through x = 2 and x = 4 
found, its value would be i8| sq. units. 

It will be noticed that C vanishes, and hence when dealing 
with definite integrals it is usual to omit it altogether. 



) MATHEMATICS FOR ENGINEERS 

/" 4 /# 3 \ 4 

For brevity, I x 2 dx is written I ) , which on expansion 

.'2 \3/2 r 



reads 



__ e 56 
3 3/'* '3' 



Example 16. Find the value of the definite integral I 

* -i 



f V**d* = f^ g3*Y 4 = 4 
J .,' \3 /-i 3 



3 v 
= | (3-3201-1-3499) 

= 2-62 <). 



Example 17. Evaluate the definite integral 

V 
f 2 

/ (50034^+7) rf^ 
.' 



5 

I (50054^+7) dx = (- si 
> o H 



sn 



7?r 
= ' or ii. 

2 



Example 18. Find the value of 
/ 

J o 



The expression 



= n-2 

4 



INTEGRATION 



141 



Notice that no cancelling takes place, beyond that concerning the 
constant multiplier 5, until the values (4 and 2) have been substituted 
in place of x. In other words, it would be quite wrong to say 



c 



4/i 



Example ig. The total range of an aeroplane in miles can be 

C Wt 

obtained from the expression / dq where m = pound-miles 



per Ib. of petrol, and q = 



loading at any time 



initial loading 
Taking q = -6 and m = 4000, find the total range. 



= m (log q log i) 
= m log q. 
Now if q = -6, log q = 1-4892 = -5108. 

Hence the range = 4000 x -5108 = 2043 miles. 




c- 



A F B O L 

FIG. 35. Proof of Simpson's Rule. 

Proof of Simpson's Rule for the Determination of Areas 
of Irregular Curved Figures. This rule, given on p. 310 of 
Part I, states that 

length of one division of the base f first + last ordinate + 
= \42 even ordmates -f- 

[22 odd ordinates. 



142 MATHEMATICS FOR ENGINEERS 

It is now possible to give the proof of this rule. 

Let us deal with a portion of the full area to be measured, 
such as ABCD in Fig. 35. Let the base AB = 2c. 

Let the equation of the curve DEC be y = A4-B#+C# 2 , so 
that DEC is a portion of some parabola. 

We can assume that the origin is at F, and therefore the abscissae 
of D, E and C are c, o and +c respectively. 

Hence AD = y x = A+B(-c)+C(-c) 2 = A-Bc+Cc 2 
FE = y 2 = A+B(o)+C(o) 2 = A 
BC = y 3 = A+B(c)+C(c) 2 = A+Bc+Cc 2 . 

f+e 

Now the area ABCD = I ydx 

J -c 

= i + 

/ - 



. . Be 2 , Cc 3 , . Be 2 , Cc 3 

. A /_!_ _L A r _ 

"^ 3~ + "2 3" 

A , 

- 2AC+ 



= -{6A+2Cc 2 } 

= -{A-Bc+Cc 2 +4A+A+Bc+Cc 2 } 



Imagine now another strip of total width 2c added to the right 
of BC; the double width being chosen, since there must be an 
even number of divisions of the base. 

Then if GH = y 4 and LK = y 6 

Area of BLKC = - 

O 

or area of ALKD = -{ 

O 

= ~{ 

If a strip of width = 2c is added to the right 

Area = Z 



INTEGRATION 143 

Or, in general 

Area = -{first+last+42 even+22 odd}. 

o 

Exercises 14. On the Evaluation of Definite Integrals. 

Find the values of the definite integrals in Nos. i to 7. 

i. rw 2 . /"* 3 . 

./ 1-02 J 1-7 W 



4. P 5 -i sin -26d. 5. 

p.7 
6. I s-2*' 1 ^. 7. 



-- . 
I x-"dx 

8. The change in entropy of a gas as the absolute temperature 

f 775 dr 
changes from 643 to 775 is given by I -85. Find this change. 

* 643 T 

IT 

9. If H = ^ 1 2 sin 6d6, find the value of H. 

pJ o 

10. The average useful flux density (for a 3-phase motor) 

i fis*' 
= B = - B mar sin 6d6. Find B in terms of ~Bmax- 

itJ _ 

12 

11. Express sin at cos bt as the sum of two terms and integrate 
with regard to /. If a is -=? and b is 30, what is the value of the 
integral between the limits o and T ? 



12. If h = find h. 

g J RI r 3 

13. Given that EI^ = ^-^-P*. Also that ^ = o when 

a^r 2 22 d# 

x = I, and y = o when .# = o and also when .* = /; find the value 
of P and an expression for y. 

wx 2 M d'ty dy 

14. If M = - , T = ET~-,, ~-= o and also y o when x = I, 

2 I dx 2 dx 

find an expression for y. 

15. Given that M - -(--* 2 )-K, ^ = f? Also ^ = o when 

2 \4 / IE dx 2 dx 

I I 

x = -, and y = o when x = - ; find an expression for y. (The case 

of a fixed beam uniformly loaded.) 

16. Find the value of J (/# -**)***. 



144 MATHEMATICS FOR ENGINEERS 

17. Evaluate 3 I jX(l z zlx-\-x z )dx, an integral occurring in a beam 

J "2* 

problem. 

18. If Q = / qdx and q = ; ; -wx, find Q, the total horizontal 

J ^ i+sin <t> 

thrust on a retaining wall of height h, w being the weight of i cu. ft. 
of earth, and <J> the angle of repose of the earth. 

19. Find the area between the positive portion of the curve 
y = 3# 4# 2 +n and the axis of x, and compare with the area of the 
surrounding rectangle. 

/15-8 
pdv when pv 1 - 37 = 594. 
4*6 

d i/ ^ d'V 

21. If ~ z = 6* 1 ' 4 \ : -,- 10-5 when x = i, and y = 14 when 

Q/X X dsG 

x = 2, find an expression for y in terms of x. 

22. Evaluate / - fi Xf 

J 1-47 13 $X 

23. Find the value of n, given by the relation n = I - -. . 

J n t 



r ,'r 

24. The total centrifugal force on a ring = / 1 - '- ^ ; find 

J E2 -t^i 

an expression for the force. 

25. The area of a bending moment diagram in a certain case was 

f'/i 3 \ 
given by J [-at -- -,)da', find the value of this area. 



26. H, the horizontal thrust on a parabolic arch, 

i 



Find an expression for H. 

27. The work done by an engine working on the Rankine cycle 

with steam kept saturated = | l dr. 

j \ T 

Find the work done if the temperature limits are 620 F. and 
800 F. (both absolute), and L = 143771-. 



28. Evaluate 



T2.4 

J [>- 5 * sin (25 



29. Find the value of n, the frequency of transverse vibrations of 
a beam simply supported at its ends and uniformly loaded with w 
tons per foot run, when the equation of the deflected form is 

y = 

and 



Ci , 
ydx. 



INTEGRATION 145 

30. From Dieterici's experiments we have the following relations 

If s specific volume of liquid ammonia 
and c = specific heat of liquid ammonia 
then for temperatures above 32 F. 

c i-n8 + -ooii56( 32) 

and s = / cdt. 

J o 

Find s when * = 45 F. 

31. If p = ^-j\s 2 -x z }dx, Q being the leakage of fluid past a 

WJJ o 

well-fitting plug, find its value. 

32. The total ampere conductors per pole due to the three windings 



/- / A J 

in a railway motor CV2 / Ai sin -rdx. 

2 Jo* I 

Evaluate this integral. 

33. For a viscous fluid flowing through a narrow cylindrical tube 
of radius v, the quantity Q is given by the formula 

O prc/fo 2 

1* 2 /* 

where /* is the coefficient of viscosity. 
Find the value of Q. 



CHAPTER VI 
FURTHER METHODS OF INTEGRATION 

BY the use of the rules enumerated in the previous chapter it 
is possible to perform any integration by a graphic method and 
the integration of the simpler functions by algebraic processes. 
Whilst the graphic integration is of universal application, it at 
times involves much preliminary arithmetical work, which it is 
tedious to perform, so that it is very frequently the better plan to 
resort to a somewhat more difficult, though shorter, algebraic 
method. For the more complex functions, then, a choice has to 
be made between the two methods of attack ; the fact being borne 
in mind that only in cases where definite integrals are concerned 
does the graphic method of integration compare favourably with 
the algebraic. 

It is therefore advisable to introduce new processes and artifices 
to be employed for the algebraic integration of difficult functions; 
and whilst it is not absolutely essential that all these forms should 
be remembered, it is well that the various types should be 
considered, so that they may be recognised when they occur. 

It is impossible to deal here with every kind of integral likely 
to be encountered ; all that can be done is to develop the standard 
forms which cover a wide range, and to leave them to suggest 
forms for particular cases. 

Integration by the Aid of Partial Fractions. Many com- 
plex fractions can be split up into simpler or partial fractions, to 
which the simple rules of integration may be applied. Thus if we 

Q/ OQ 

are asked to integrate, with respect to x, the fraction 2 _ ~ , 

we soon discover that we are unable to perform this operation 
with only the knowledge of integration acquired from the previous 
chapter. 

If, however, we break the fraction up, in the manner explained 
in Part I, Chap. XII, we find that the integration resolves itself 
into that of two simple fractions. 

146 



FURTHER METHODS OF INTEGRATION 147 

Thus , 8 *~ 3 ? o = - h- 4 - (see Part I, p. 453). 
2 x 4 2x 7 



Hence- f**^?** = f- 2 - d*+ l 
J 2x 2 15^+28 J x 4 J 



2# 7 
= 2 log (* 4) +4 log (2* 7) + log C 

= log (*- 4 ) 2 +log (2*- 7 )2+log C 
= log {C(*- 4 ) 2 (2*-7) 2 }. 

[Note that log C may be written to represent the constant in 
place of C alone ; and it can then be combined with the other logs.] 



/dx 
~i 2 
x z a 2 

i A B 

*2_ a *- (*_ a ) + 



Equating numerators 



Let ^ = a, then i = A(2a)+o 



and A = . 

20 



Let x = a, then I = o + B( 2a) 

and B = -2 J 

i = JL(_J_. _i_l 

x*a z 2a\xa x-\-a' 

_, /" dx i f f dx f dx 1 

Hence -= = i / 

J x 2 a z 2.a\-' xa J x+a) 

= ^{log (^ ) log (#+a)+log C} 

= ^; lo g 



This is a standard form. 

A rather more general result may be deduced from it. 

Example 2. To find J , 



Let (x+a) = X 



- f -T= -. " / g 



A Ilcll"" r / . \ A v a ^ !*-.- A , i** ~r~ fc* i j 

rf^r 



Explanation. 
x+a = X 



(X+6) 

C(^+a &) and thus for dx we may 
write rfX. 



148 MATHEMATICS FOR ENGINEERS 

Integration by the Resolution of a Product into a 
Sum. A product cannot be integrated directly; but when the 
functions are trigonometric the product can be broken up into a 
sum or difference and the terms of this integrated. 

Before proceeding with the work of this paragraph the reader 
would do well to study again pp. 273 to 286, Part I. 

Example 3. Find the value of /4 sin 5^.3 cos2ldt. 

4 sin 5^.3 cos 2t = 12 sin 5^ cos 2t 

= 6 X 2 sin 5* cos 2t 

= 6{sin 7/+sin 3*} (cf. p. 286, Part I). 
Hence 

/4 sin 5^.3 cos 2tdt = 6[f sin jtdt+fsin $tdi\. 

= 6J cos jt cos 3/+C J 



= 6C cos 7^2 cos 



Example 4. Find /sin 2 xdx. 



cos 2.x = i 2 sin 2 x, so that 
sin 2 x = i -cos 2* (cf> p 2g 



= -5* -25 sin2*+'5C. 

Example 5. Find f ta,n 2 xdx. 

We know that sec 2 x = i + tan 2 x. 

.'. fta,n 2 xdx =f(sec z xi) dx fsec z xdxfi dx. 

= tan xx-\-C. 

Integration by Substitution. At times a substitution aids 
the integration, but the cases in which this happens can only be 
distinguished after one has become perfectly familiar with the 
different types. 

y- is a type to which this method applies. 



In this fraction it will be observed that the numerator is exactly 
the differential of the denominator. Hence if u be written for the 



FURTHER METHODS OF INTEGRATION 



149 



denominator, the numerator may be replaced by du, so that the 

integral reduces to the simple form I , i. e., log w+log C. 

J u 



For if 

or 
Hence 



u = ax z +bx-\-c 

du 

-j- = 2ax-\-b 

dx 

du = (2ax-\-b)dx . 

((^x+b^x = fdu 
J ax z -}-bx-\-c J u 

= log w+k>g C 

= log Cu 

= log C(ax*+bx+c). 



In many cases integration may be effected 
by substitution of trigonometric for the algebraic 
functions ; and Examples 6 to 10 illustrate this 
method of procedure. 



Example 6. To findyVa 2 




-X 



FIG. 36. 



Let 



then 



x a sinw, as illustrated by Fig. 36 
a 2 x 2 = a 2 a 2 sin 2 u = a 2 (i sin 2 w) = a 2 cos 2 u 



and Va 2 x* = a costt, as will be seen from the figure. 

A , dx d(a sinw) 

Also -,- -*s - = a cosu 

du du 

i. e., dx = a cosw . du. 

J'-\/a 2 x 2 dx fa cosu . a COSM du 
= a 2 f cos 2 u du 

a? 

= f(i + cos2ii)du, since cos 2A = 2 cos 2 A i 

= ( u-\ sin 2M + CJ. 

Although this result is not expressed in terms of x, it is left in 
form convenient for many purposes. 
To express the result in terms of x 

x x 

sin u = -, so that u = sin- 1 - 

a a 



and also 



cos u 



/ a 2 _ x % 

= \f - = . 

n& 



150 MATHEMATICS FOR ENGINEERS 

Hence - sin aw sinw cosu = - x - Va 2 x 2 

z a a 



a z -x*dx = X sin- 1 - +(T X ^2 Va 2 -* 2 +K 

V 2 / \ 2 

= - sin- 1 -+- Va^^+K. 
2 a 2 



Example 7. To find J ^^^ 



Let AT = a sinw i. e., u = sin- 1 - 

a 

rf# ^(a sin u) 

then -j = -- T - - = a cosw 
du du 



also Va z x 2 = a cosw, as before. 

/" dx fa cosu.du __ . 

J Va*=^~ J~a~^r-J ldu 

= u+C 



= sin- 1 -+C. 
a 



Example 8. To find /"- 



X 

In this case let AT = a sinh u, i. e., u = sinh" 1 - 

a 

dx d , . . . 

then j- = j- (a sinhw) = a coshM. 

du du ^ 

Now cosh 2 u sinh 2 u = i (cf. p. 291, Part I) 

and thus cosh 2 w = i +sinh 2 w 



or a 2 +Ar 2 = a 2 cosh 2 ^ 

and Va 2 +# 2 = a coshw. 

[ dx fg coshu.du _ r -, , r 

J Va*+^~* ~ ' a coshtT ~ J u 

= sinh- 1 *-+C. 
Referring to p. 298, Part I, we see that 

, . x . (x+ Vx*~d*} 
cosh" 1 - = logi - 

a & l a J 

and also sinh- * = I 



~ - sinh- 1 *+C 
2 a 



or 



FURTHER METHODS OF INTEGRATION 151 

r fl x 

Example 9. Find the value of I - ._ and thence the value of 

' vx 2 a 2 

f dx 
J 



Dealing with the first of these 

X 

let x = a coshw, i. e., u = cosh" 1 - 

a 

dx . , 

j = a sinhw 
du 

or dx = a sinhw du 

and x 2 a 2 = a 2 cosh 2 w a 2 = a 2 (cosh 2 w i) = a 2 sinh 2 ** 

fasinhudu 
sinhw 



f dx f dx fasin 

J Vx 2 a z ~ J a sinhw J a si 
= fdu 

u+C 



cosh" 1 -+C 



To evaluate the second integral, let x-\-a = X 
then dx = ^X and J = f 



Example 10. Find the value of I 0-77 ~a 

J Ct ~j X 

The substitution in this case is 

a tanw for x 

x 
i. e., x a tanw or w = tan -1 - 

dx d . . 2 

then j- = j- (a tanw) = a sec^w 

du dx ^ 

and A- 2 +a 2 = a 2 tan 2 w + a 2 = a 2 (i+tan%) = a 2 sec 2 ^ 

sec 2 w rfw i r , 
= - /aw 



/" ^ /"a s _ 

' ' J a?+x* ~ J a 2 sec 2 w ~ 

= -M + C 

a 
= tan- 1 - + C. 



152 MATHEMATICS FOR ENGINEERS 

f dx 

By an extension of this result such an integral as I 

J 

may be evaluated; for let x-\-a = X 

dX d(x+a) 

then -s = v , r ; - = i 

dx dx 

i. e., rfX = dx 

hence ( dx - f ** X - ' tan- 1 X 4-C 

J x + a z + b* ~ J X^+F 2 ~ b tc F^ 



i. e., I tan- 1 ^~+C. 
b b 

The following examples are illustrative of algebraic substitution or 
transformation. 



r 
Example n. To find the value of I 



Our plan in this case is so to arrange the integral that the method 
of a previous example may be applied. 

z z z 



Hence ( dx - f - -** __ = [ d * 

J V2ax-x z J Va*-(x-a) 2 J V 2 -X 2 

the change from dx to rfX being legitimate, since , = -, ' = i 

ctx dx 

and by Example 7, p. 150, the value of this integral is seen to be 



f 

J 



Example 12. To find J 



dx . x a 

= sin- 1 
x z a 



dx 



In this case the substitution is entirely algebraic. 

i , , du i 

Let u then , = 

x dx x 2 

or dx = x 2 du. 

Then a*+x*S 



dx f x z du f du 

!\4 J I !,.! I T \T vx ~3 J ^ 

[ udu 

= ~r, 



FURTHER METHODS OF INTEGRATION 153 

To evaluate this integral we must introduce another substitution. 

Let y = 2 2 +i 

then -^ = 2a 2 u 

au 

or udu = - 9 dy. 

za z ' 

Hence the integral - ~ = ~~^ + C 



+ C 



__ I -4-0 = _ - __ I-C 

~ 2 ~ 



L +c. 



dz At 

Example 13. The equation -,. . occurs in the statement 

dt \/ j2 j 

of the mathematical theory" of fluid motion, which is of value in 
connection with aeroplane design. Solve the equation for z. 

To obtain z from -^ we must integrate with regard to t; and to 

ctt 

effect the integration let u t 2 i, so that -^ = , . - = zt 

du 
or dt r. 

21 



, Atdt fAtdu [Adu 

I hen z = 

2U* 



or AVt 2 i + C. 

Many difficult integrals of the form / can be evaluated 

J x(a+bx n ) 
by the substitution z x~n. 

For if z = x~n log z = n log x 

d log z _ d log x 

j ^ 3 ' 

dz dz 

i _ J. log x ^ dx 

^ jj ^ ~3~ 

z dx dz 

i i dx 



154 MATHEMATICS FOR ENGINEERS 



The integral f x(a + bx ^ thus reduces to -!j ( _, the value 

of which is log (az-\-b) or log / , , A 
na ' na \a-\-bx n J 

/dx 



For x* write z~ l , so that in comparison with the standard form 

n = 7. 



_i 

~ 28 I0g 



Example 15. Find the value of / r . 

J (i 2x)* 



It will be observed that the denominator is a surd quantity; and 
in many such cases it is advisable to choose a substitution that 
rationalises the denominator. Thus in this case let u* i 2X. 

,_, QA& Ct/14/ GLIfi Ctr'VC 

~dx = ~du X dx = 2U J 
and -, (i 2x) = 2 

du 

so that 2M j- = 2 or dx 

dx 



iu z (i u z \* 

Also i 2X = u-, whence - = x and x* (-- j . 

2 \ 2 / 

Expanding by the Binomial Theorem 

x* ~ (i 4 2 +6M 4 4 6 + w 8 ). 

/" A;*^ 
Hence 

J (l-2^)i 

- _L /"(i 4 M2 +6M 4 4M 6 +M 8 )X udu 
~ i6J u 

i ( 4U 3 6u 5 ^u 1 . u* 

= ~^( u - T + T 7 + 
4 tf ,^_i- 

3 5 7 



16 



which result could be further simplified if desired. 



FURTHER METHODS OF INTEGRATION 



155 



The next example introduces the substitution of an algebraic 
for a trigonometric function. 



/dx 
-. 
sm x 



Since sin 2 A = 2 sin A cos A, then sin x = 2 sin - cos - . 

2 2 



f dx if 

Hence : = - 

/ sm x 2 / . 

J J Cl 



Now let 
then 
or 



dx i 


dx 


. x 


. X XI 


sm 

2 .1 -' 


sin - cos - 


X COS 2 - 


2 2 


X 2 




COS - 


, 


2 




9 #J 




sec 2 a* 


I 


2 


~~ 2 


X 

tan - 

*7 



u = tan 

2 

du i x 
- = sec 2 

dx 2 2 



sec 2 - 

2 



sn x 2 x 

sec 2 . 



= J|' = log +C 



= log tan -+C. 



Integration by Parts. When differentiating a product, use 
is made of the rule 

d i x du . dv 
dx (u ^ =V dx +U dx {Refer p. 70.! 

If this equation be integrated throughout, with respect to x 

uv = fvdu+fudv 
or, transposing 

fudv = uvfvdu. 
Many products may be integrated by the use of this rule. 



156 MATHEMATICS FOR ENGINEERS 

Example 17. To find f^x.e x dx. 

Let = 4#, i- e-, du = ^dx 

and let dv = e x dx, i. e., v = e x . 

Then f^x.e x dx = fudv 

uvfvdu 



Example 18. Find f^.e^dx. 

Let u = 5# 2 , i. e., du = loxdx 

and dv e* x dx, i. e., v -e 4 *. 

4 

Then f$x 2 .e* x dx = fudv 

= uvfvdu 

= sx* . T e* x r 
4 M 

= ^x 2 .e* x 5 I xe ix dx 
4 -2.1 

^-r r A-TJ i n f 1 in rwhere u = x -i 

Now fxe* x dx = x.-e tx e* x dx 

4 J 4 [ and v - VJ 

= ^.e 4a; -4e 4a: 
4 16 



+ C 



Example 19. To find Te * sin (bx+c)dx and also 

fe^ cos (&Ar+c)^Ar. 
[The two integrals must be worked together.] 

Dealing with the first, which we shall denote by M 
Let u = sin (bx-\-c), then du b cos (bx-\-c}dx 

and dv e^dx, so that v fe^dx = e *. 

Then M = - e sin (6Ar+c) - e ax b cos (&#+c)d* 

flS ^ & 

= - e^sin (6^+c) fe cos (bx+c}dx 



N ......... (i) 

(I (t 

where N stands for the second integral whose value we are finding. 



FURTHER METHODS OF INTEGRATION 157 

By developing this second integral along similar lines we arrive at 
the value 

N = -e a:r cos (bx+c) + -M. ...... (2) 

We have thus a pair of simultaneous equations to solve. 
Multiplying (i) by b and (2) by a and transposing 

b b z 

6M = - e ax sin (bx+c) -- N 
a 'a 

feM = e ax cos (&#-|-c)+aN. 
Subtracting o = e ax f~- sin (bx+c) -\-cos (bx+c) \ N(-4-a) 

\__Qt J \ fl ' 



t, -NT nr rb sm (&#+c)+a cos 

whence N == e ax \ ' 



- z , a 

u jit. . HT /r^R* sin (&#+c) 6 cos (fof+c)~i 
and, by substitution, M = e ax \ a 8 +6 8 

y^ao; s i n (bx-{-c)dx = 2 , 3 [a sin (bx-\-c) b cos 



~ 

and ye^ cos (bx+c)dx = 2 a [6 sin (6^+c)+a cos 



Example 20. An electric current i whose value at any time t is 
given by the relation i = I sin pt is passed through the two coils of 
a wattmeter; the resistances of the two coils being R x and R 2 
respectively, and their respective inductances L x and L 2 . Then to 
find the separate currents in the two branches it is necessary to 
evaluate the integral 

fQe*'dt where P = 



and Q = J-^TT~ sin pt+-. ^ cos 

Evaluate this integral. 

j j 

Q = f rT~(Ri sm pt + pLi COS pt) = = 



where c = tan- 1 ^~ (see Part I, p. 276) 

** 

or Q = M sin (^>^+c), where M = T-^ 



Then fQedt = fe^M sin (pt+c)dt = Mfe sin (pt+c)dt 
and this integral is of the type just discussed; its value being 

[P sin (pt+c}p cos 



and in this form it is convenient to leave it, since in any numerical 
application it would be an easy matter to evaluate P, M and c before 
substituting into this result. 



158 



MATHEMATICS FOR ENGINEERS 



Some miscellaneous examples now follow, involving the use of 
the methods of this chapter. 



Example 21. Find the value of 



Let 



. e., 5 = 

Let x = 5, then 5 = o 2B 

B = -2-5. 

Let # 3, then 5 = 



^_o._B 

x+5 



*+3). 



A = 2-5, i. e., 



= 2-5 2-5 



p 

J j x 



5 ^# 



* 2 +8*+i5 x+3 x+5 
= ft 2-5 dx f* 2-5 dx 

= 2-5. flog (^+3) log 



= 2-5 [log 5 log 7 log 4+log 6] 

= 2-5 [1-6094 1-9459 1-3863 + 17918] 

= 2-5 X -069 = '1724. 

As an alternative method of solution, the graphic process of 
integration possesses certain advantages in a case such as this. 

It might even be advisable, in all cases of definite integrals 
where the algebraic integration involves rather difficult rules, to 
treat the question both algebraically and graphically, the latter 
method serving as a very good check on the accuracy of the 
former. 



f 2 
In this example / 

where 



5 x - 



V = 



/ 

J i 



=51 ydx 
-' 



hence it is necessary to plot the curve y = -. . . . and find the 

( X ~r3)( x ~r5) 
area between it, the axis of x and the ordinates through x = i and 



x = 2. 

The table for the plotting reads 



X 


i 


1-2 


1-4 


1-6 


1-8 


2 


y 


04167 


03841 


0355 


03294 


03064 


02857 



and from these values the curve AB is plotted in Fig. 37. 



FURTHER METHODS OF INTEGRATION 



159 



The sum curve for AB is the curve CD, the last ordinate of which, 
measured according to the scale of area, is -0095. This figure is the 
area between the curve AB, the ordinates through x = i and x = 2, 
and the base line through y = -025; and hence the full area under 
the curve AB = -oo95 + area of a rectangle -025 by I, i. e., -0095 + -025 
or -0345. 

dx 

= ' 345 



Thus 



= 5 X -0345 - -1725. 



040 

y 

035 



03O 



05 



D 



Sum Curve 
polar distance '8 




010 

OO95 

008 



OO6 
OO4 
002 

o 




FIG. 37. Graphic Integration of Ex. 21. 
dx 



Example 22. Find the value of I 2 . 



XT 
Now 



f dx i . _ (x a) , 

I -5 , = log C 7 ( (see Example i, p. 147) 

/ # 2 a 2 20 (x-\-a) ' 

cM) 

i i , V 2/ 



rf* i r ^ 



T r^r a^ 

A i *-^l *"& *1 / 

= log V 
12 & 



or 



(ix+3) 



160 MATHEMATICS FOR ENGINEERS 

Example 23. Find the value of I 4 ----- , dx. 

J 4 gx 2 #+i2 



gx 

In this case the numerator is of the first degree in x, whilst the 
denominator is of the second degree. Also we notice that the 
derivative of the denominator is i8# 5, and the numerator is 
4(18^5). Thus the derivative of the denominator and the numerator 
are alike except as regards the constant factor 4. Hence the substitution 
will be u for gx 2 



If u = 9# 2 5*4-12, -r- = i8# 5 or du = (i8x$)dx 

(IX 

so that (j2X2o)dx = 4(i8x5)dx = ^du 
f 72* -20 

J tx 2 



:=4 U 



Example 24. To find the value of I 
This is evidently of the type I 



= 4 (loge 418 loge 136) 
= 4 (loge 4-18 loge 1-36) 
= 4 (I-4303 3075) 
= 4-49- 

dx 



dx 

(^4 

for 



so that a = 3 and b = V&. 

r dx i.i x-\-a _ . , _ 

= r tan- 1 \-C (cf. Example 10. p. 151) 
J x z +6x+i$ b b 



Example 25. For a single straight wire at a potential different 
from that of the earth, if v radius of wire in cms., / = length of 
wire in cms., a surface density of charge in electrostatic units per 
sq. cm., then the potential P at any point on the axis of the wire due 
to the charge on a length 8x is given by 



ft dx 
so that the potential at the middle point = mrffj l , z - 

Evaluate this integral. 



FURTHER METHODS OF INTEGRATION 161 

/dx 
, 
Va 2 +# 2 

;. f^== = log(* + V * 2+r *\ (cf. Example 8, p. 150) 



P m = 27WO- 



r (' + J*:. 
i ir- 



= 27Wo- log- 



\/5+*-; 



or TT^O- log- 



* where d is the diam. of wire. 



The following example involves the use of three of the methods 
of this chapter. 

Example 26. Find the value of /Vs ^-- L ; i- 

I ( x \~ 'ZX ~\~ 1 1 \X~\~ ^ ) 



The fraction under the integral sign should first be resolved into 
partial fractions. 



t. e., 
Let x = 3, then 

ii = C(9 6+7) = loC 

i. e.. 





10* 



Values of A and B can be found by equating coefficients of x and 
also those of x 2 . 

By equating coefficients of x z o = A+C and hence A = - 



162 MATHEMATICS FOR ENGINEERS 

oo 22 

By equating coefficients of x 5 = 3A + B+2C = _- + B . 



10 10 



10 

II 



*+3 
i f n*+39 i 



Hence the fraction- 



We can make the numerator of the first of these fractions into 
some multiple of the derivative of the denominator; thus 

The derivative of the denominator 



and the numerator = 

and if u 

then du = 2(x+i)dx 

and 



= du-\-28dx 



frr. 

J (x i - 



loJ x z +2x+j x+3) 

iidx 



i / [(iix+3Q)dx fi 

IO\J X z + 2X+7 f X 



i_(fii, f 28dx 

~~ 1O\J 2U J 



28 ^4-1^ \ rx 2 +2X+? 



/ 
V6 

- g log (x* + 2x + 7 ) + 6 tan- -~ log 

or log (*!?*7).** .14 



\ rx2X? -, 

=^ 2 +2^+i + 6 
} L^^ I 22j 



V6 



Example 27. An integral required in the discussion of probability 

/-co 

is / e~ x *dx. Find a value for this. 
J o 

/GO 
e~ x *dx. 

Replace x by ax and thus dx by adx. 

Then I = I e-^&adx. 
J o 



FURTHER METHODS OF INTEGRATION 163 

Multiply all through by e~ at . 

f 

Then Ie~ at = I e-**()-+&').adx and integrate throughout with 

regard to a; thus 

/<*> rx=x> ra=<x> 

le-a'da = e-a^+x^adadx. 

J x=~Q J o=0 

i~<*> rco r> r 

But / Ie-a'da = 1 1 er^da = 1x1 since / e~&dx and / 

J ^0 J ' 

have the same value, 



hence I 2 = /*" ' ** e-+ 

.' x=Q J o=0 



(i) 



The value of the double integral on the right-hand side will be 
found by integrating first with regard to a and then with regard to x. 
Dealing with the " inner " integral 

/<j=oo 
a=0 

let A = a 2 , then dA. = zada, and let M represent 



Then P~ e -<K l +'V . ada = f 
J 0=0 J 



o 
2M\ /o o and oo also 



i / _ AM \ since the limits for A are 
2M / 



_ _ 
~2M."> ~ 2M- 

Referring to equation (i) and substituting this value therein 



1 / \ 

= - /tan" 1 x ) (cf. Example 10, p. 151) 

= - (tan- 1 oo t 

2 v 

I/7T \ 7C 

= -I -- O) = -. 

2\2 } 4 



= - (tan- 1 oo tan- 1 o) 
2 v 






As an extension of this result it could be proved that- 



Reduction Formulae. Many of the exceedingly difficult 
integrals which arise in advanced problems of thermodynamics, 



164 MATHEMATICS FOR ENGINEERS 

theory of stresses, and electricity may be made by suitable sub- 
stitutions to depend upon standard results obtained by a process 
of reduction. To grasp thoroughly the underlying principles on 
which the process is based, it is well to commence with a study 
of the simpler types. 

7T 7T 

[z /"a 

We desire to evaluate the integrals I sin"0 dQ, I cos"0 dQ 

-' o -'o 

and I sin m cos n dQ, where m and n have any positive integral 

J n 





values. 



Taking the case in which n = m = o, we have the results 

IT 

fz 
reducing to the form / idQ, the value of which we know to 

J 



. 7T 

be - 
2 



If m = n = i 

ir w 

/"2 / \ 2 

I sin 0^0 = (cos0) = (cos 90 cos o) = I . . (2) 
Jo /o 

IT If 

I Z cos dQ = ( sin 0Y = (sin 90 sin o) = i ... (3) 
J o /o 

from which pair of results we may say 

7T 7T 7T 

/ 2 sin dQ = r cos dQ = ] sin ( ?r 0)^9 

/ / / \ 2 / 

or more generally 

j a f(x)dx = j*f(a- X )d X (4) 

result of great usefulness. 

7T W If 

Also \ sin cos ^0 = - \ sin 20^0 = I cos 20 ) 

./o 2./o 2X2\ /o 

i , , 

= (COS 7T COS O) 

4 

=J .... (5) 

By the process of reduction of powers we may express the 
integral to be evaluated in such a way that it depends on results 
(i). (2), (3) or (5). 



FURTHER METHODS OF INTEGRATION 165 

Thus 

77 tr ; 77 

f 2 sin 2 6 dQ=* I* (i-cos 20)^0 = - I (\dQ- f* cos 20 do] 

J o 2 J o 2 \_J y 

I/IP N 
= - ( O 

2 \2 / 



cos 2 8 rf0 = 2 sin 2 rf0 = -. 
o Jo 4 



and from equation (4) 

77 

; 

f 

ft 
Now let n = 3, i. e., we wish to evaluate I sin 3 dQ. 

J o 

It 77 77 

Then ] sin 3 dQ = \ sin 2 . sin 0^0 = ] (i cos 2 0) sin dQ 
J o Jo Jo 

IT 

= I 2 (I M 2 ) rf 

'o 
(o and ^ being the limits for J 

u being written for cos and du for sin dQ, since -^ = sin 
and thus -^ = sin 0. 

77 17 . 

, T f "~ ^ / W 3 \ ~ 2 / COS 3 \2 

Now I (i w 2 ) aw = I M- = cos ) 

Jo = o v V 3/e = o \ 3/0 



Thus f 2 sin 3 dQ = I * (i u 2 ) du = +- 

^o /-o 3 

and if n be written for 3 we note that the result may be expressed 

77 77 

in the form I sin M ^0 = and also I cos w dQ = - 

J o n Jo 

n being an odd integer. 

77 

fz 

Let n = 4, then I sin 4 ^0 is required. 
J o 

Now sin 4 dQ = sin 3 . sin QdQ = udv, where u = sin 3 0, 
and dv = sin dQ or v = cos 0. 



166 MATHEMATICS FOR ENGINEERS 

7T 7T 7T 

Hence ] sin 4 dQ = (cos 6 sin 3 0Y 1 2 cos . <Z sin 3 

J / J 

7T 77 

/ \ 2 /*2 

= ( cos sin 3 ) + I cos 8 . 3 sin 2 cos dQ 
\ /o J o 

7T 

= 0+3 1 sin 2 cos 2 dQ 
J o 

77 
/ \2 

since ( cos sin 3 ) = (o X i) (i X o) = o. 
'o 

7T 7T 

Now I sin 2 cos 2 ^0 = ] sin 2 0(i sin 2 0)^0 

Jo J o 



= (sin 2 0-sin 4 0)<f0 
o 



It 77 IT 

Hence \ sin 4 dQ = 3 ] sin 2 dQ3 ( 2 sin 4 dQ 

J o J o J o 

[I [I 

or 4 sin 4 ^0 = 3 1 sin 2 dQ 

J o Jo 

7T 7T 

f 2 q /" 

and I sin 4 dQ = ^ sin 2 dQ. 

J o 4J o 

We have thus reduced the power by 2, and knowing the result 

7T If 

for / sin 2 dQ, we can finally state the value for / sin 4 dQ. 
Jo Jo 

7T 

TVinc cin^fl /7A _ V _ ^ /^r- V S/ _ 

lllUo oill U Ct-U ^ /\ -^ x- Ul x\ /\ 

J 4 4 16 422 

fi (n i) ( ^ 
or I sin ^0 = '- X, * - , n being an even integer. 

J Q W (W 2j 2 

In like manner it could be shown that 

7T 77 

/<j tf T / 

sin 5 ^0 = - / sin 3 dQ. 
Jo 5 J o 



Thus sin 5 0^0 = ^x" or ~ 

5 5-2 5-3 



FURTHER METHODS OF INTEGRATION 

IT 

which is of the form f 2 sin"0^6 = ^^ . ^ 
J n n2 

n being an odd integer. 
Similarly 

JT It 

JO 6 lfZ . 61 6 3 6 5 TT 

sm 6 6 dQ = - I sin 4 ^ = ^ X z - X - X - 
6 ; 6 62 64 2 

5.3.1 TT 
or B-^ x - 

6.4.2 2 

which is of the form 



OBfl A V H* V . . y . 

o n(n 2)(n 4) 2 

w being an ;<? integer. 

Summarising our results 



sn = cos = 

^o n 

if is an evew integer. 

P sin w ^6 = P co S *0 dQ = etza^K-nSj^^a 
7 Jo n(n2)(n4) . . . I 

if is an o^ integer. 

JT 

/2 
sin 9 ^</^. 
o 
In this case w = 9 and is odd. 



Hence sin 9 d9 = 



/ . 
su 

J o 



8.6.4.2 _ 128 



9.7-5-3 315 



w 

fz 

Example 29. Evaluate / cos 10 dd. 
J o 

Here n = 10 and is even. 



Hence 



Jo 



. . 

10 .8.6. 4. 2 2 512 



Example 30. The expression I ^3 dt gives the theo- 

7C-|~4-' * 

retical thrust on a plane moving through air. Evaluate this. 



V/=i = *t z i-~ = tVi-sin 2 u = t 
if sin u is written in place of -. 



168 MATHEMATICS FOR ENGINEERS 

i d sin u dtr- 1 dt~ l dt 
Then since sm u = - t~ l ^ = -3 = -rr X T- 

or cos u = -5 j- 

t 2 du 

du cos u 
whence dt = = ^ . 

Again, when t = x> - = o i. e., sin u = o or u = o 

t 

and when t = i 7=1 i- &> sin u = i or u = -. 

t 2 

Hence 

TT It 

f 1 Vt 2 !,, f 2 cos u sin'u cos udu f 2 , TT 

is <# = I = r- = = / cos 2 waw = . 

J ao t 3 Jo sinttsm 2 w Jo 4 

Thus -^ / '-^ -8 = -^F- X -- = -^-. 

4 ^ ~r~4 



We can now direct our attention to the determination of the 

r\ 

value of I sin m cos"0 dft, where m and n are positive integers. 
' o 

It is convenient to discuss a simple case first, viz. 

TT 

[z 

sin 2 cos dQ. 
' o 

T , . on , , du d sin 2 d sin 

Let u = sm 2 so that -^ = 5^-5 X ^~ 

^0 rf sin <?0 

= 2 sin cos 

and du = 2 sin cos ^0 

also let dv = cos <0 so that v = sin 0. 

Then, by integrating by parts 

7T IT 7T 

f 2 sin 2 cos ^0 = (sin 3 0V f 2 2 sin 2 cos ^0 
J o \ /o J o 

/2 / \2 

sin 2 cos ^0 = ( sin 3 ) = i 
o /o 

n 
[2 I 

or sin 2 cos ^0 = - 

^o 3 



j . , , , ... 

and might be written as 



. 

m+n 



FURTHER METHODS OF INTEGRATION 



169 



Example 31. H, the horizontal thrust on a circular arched rib 
carrying a uniformly distributed load w per foot run of the arch, is 
obtained from 



( X -si 
V * 



sin 2 e (cos 0--S66) dQ 



2R 3 I 6 (cos 866) dQ 
J o 

if the span is equal to the radius of curvature (see Fig. 38). 

If w = -5 ton per foot, and the span = 60 ft., find the value of H. 

Here w = -5, R = span = 60. 

IT 

[ 8 ( J -sin 2 6) (cos 6- -866) dQ 



Hence 



-5X60 



(cos 0- -866) <Z0. 




FIG. 38. Circular Arched Rib. 

Dealing with the numerator separately, as this alone presents any 
difficulty 

(^ sin 2 0) (cos -866) = - cos -2165 sin 2 cos0+-866 sin 2 0. 
\4 4 

Then [(- sin 2 0] (cos - -866) dQ 

= l~ cos 0d0 -f-2 165 d0-/sin 2 cos dQ+f-866 sin 2 dQ 
but, as proved on p. 148, /sin 2 dQ = ~fidQ-fcos2Q dQ 

= sin 20 

2 4 

and as proved on p. i68ysin 2 cos dQ - sin 3 



170 MATHEMATICS FOR ENGINEERS 

w 

thus r (i-sin 2 0) (cos 0--S66) dQ 
J o H 

ff TT n- TT 

= (-sin 0)~- (-21650)-* (sin 3 0)V 4 33 (-^ sin 28)* 

-o)-f (<. S ).-o) + . 43 3 (Hx -866-0+0) 



= -125 -1134 -04I7+-2267 -1875 == -0091. 
Dealing now with the denominator 

IT n 

(* (cos 8 -866) dQ = (sin 0- -8660Y = -5- -866 X ~ = -5 4534 = -0466. 
J o ' 

TT 15 X -0091 

Hence H = -* - ^ = 2-93 tons. 

0466 yj 

Carrying the investigation a step further, let us discuss the 



case of I sin 2 cos 2 dQ. 
Jo. 

T It IT IT 

f 2 sin 2 cos 2 dQ = I * (i cos 2 0) cos 2 dQ = f 2 cos 2 dQ 1 2 cos 4 dQ 
Jo J o Jo J o 

and from the previous result this value = ^ - 

4 4.22 

_ I 7T I I 7T 

~82 4*2'' 2 

This result might be regarded as obtained by first reducing the 
power m by 2, and next that of cos 2 by 2. 

Thus for the first step 



7T 7T 

f 2 sin 2 cos 2 rf0 = ^^ f 2 
Jo 4 7 



cos 2 ^0 



and for the second step 

IT 

idQ 



f 2 cos 2 ^0 = ^^ f 2 
Jo 2 J 



21 TT ni TT 

- -,-r. _ /"\T* 

\JL 



2 ' 2 W+ w 2 

m, for the second integral being zero. 



FURTHER METHODS OF INTEGRATION 171 

rr 
ft 

In a similar fashion we might reduce I sin 4 0cos 3 0^0 as 

' o 
follows : 

First reduce m by 2, then 

1C IT 

! * sin 4 6 cos 3 ^0 = 4 -^ (* sin 2 cos 3 dQ. 
J o 



4+3 J o 
Next reduce the power of sin 2 again by 2. 



7T T 

Thus P sin 4 cos 3 dQ = f* . *^ I * 
Jo 4+3 2+3 J 



cos 3 dQ. 



Now reduce the power of cos 3 by 2, and remember that m 
is now = o. Then 



sin'e cos'erfe _ 1=5 . ?=I . 3=1 f* 

o 4+3 2+3 3+o^o 



. 

7-5-3 

/2 

In the evaluation of the integral I sui"'0cos n 0^0 we thus 

* o 

proceed by reducing by 2 the powers of m and n in turn until they 
become i or o. The various cases that arise are 

(a) m and n both even : in which case the final integral is 

f * rf0 = ? 

; 2 

(6) w and w both odd : in which case the final integral is 

IT 

[2 ! 

sin cos dQ = -. 
J o 2 

(c) m even and n odd or vice-versa : in which case the final 

a it 

fz [2 

integral is either / cos 0^0 or I sin0^0, the value of either 

J o J o 

being i. 

The results for the three cases can be thus stated 
(a) m and n both even 

Psi 

J 



sin m 
o 



(n-i)(-3)x . . . i * 
n(n 2)x ... 2 2 



I 7 2 MATHEMATICS FOR ENGINEERS 

(b) m and n both odd 

. ( m _ I ) x (w 3)x ... 2 

sm w 8 cos n 6 <ft = / , v . . ' * : :' ? ; r 

(m+) X (m 2+n) X ... (+3) 

(tt-i) x(n-3) ... 2 i 
X (+I)X(-I)X . . . 4 X 2 

since, after reducing the power of sin m by 2 at a time, we must 
be left with sin cos n 0, so that the value of m to be used in the 
reduction of cos M must be taken as i. 



(c) m even and n odd 

fl l 

I sm w cos re ^0 = 

.' n 



(m+n) X (m2+ri) X ... (w+2) 



-ix-3)X . . . 2 
X 



7T 

ft 

Example 32. Evaluate I sin ? cos 10 rf0. 

/ 

This is case (a), i. e., with both m and w even. 

7T 

Hence 2 sin0 cos"0 dQ = - -' - -- -'^- X 



f 2 si 

/ 



- - -- - 

.10.14.12 10.0.6.4.2 2 



1179648' 



7T 

[z 
Example 33. Find the value of I sin 3 cos 5 0d0. 

J 

This is case (6), i. e., with both m and n odd. 

7T 

Hence / 2 sin 3 cos 5 dQ = \ X J X - = - . 
Jo 8 6.4 2 24 

IT 

Example 34. Find the value of / 2 sin 7 cos*0 ^0. 

/ 

This is case (c), but with m odd and n even. 

it 

Hence T sin 7 cos 4 d0 = .^' j'^ X ^ X i = l6 

' 



n-9-7 5-3 ii55_ 

The value of the foregoing formulae is found in their employ- 



FURTHER METHODS OF INTEGRATION 173 

ment in the evaluation of difficult algebraic functions, which may 
often be transformed by suitable trigonometric substitution. 

Thus to evaluate / ^-^(ix^^dx, known as the First Eu- 
J o 

lerian Integral and usually denoted by the form B(m, n), we may 
substitute sin 2 for x. Then, since when x = o sin 2 must = o 
and thus = o, and when x = i sin 2 must = i and thus 



/ 



x m - 1 (ix) n - 1 dx = sin 2n *- 2 0(i si 
~dx d sin 2 6 



_ d sin 2 6 d sin 6 
~ d sin 6 ' ^0 
_ = 2 sin cos 



= ] sin 2 *- 2 cos 2n ~ 2 x 2 sin cos 0^0 
-'o 

ir 

fz 

J n m 



and this can readily be evaluated. 

Example 35. Evaluate / x*(i x z )?dx. 
J o 

Let x = sin then i x z = i sin 2 = cos 2 
dx d sin ft 



, 

and 



Also when x = o sin = o and thus = o 

and when x = i sin = i = -. 

Then 

v 5 

I * 4 (i-x z )*dx = f 2 sin 4 cos 5 dQ cos = f 2 sin 4 cos 6 dQ 

-' J .' 



10. 86. 4. 22 
Another important result obtained by the process of reduction 

r* 

is the value of I e~ x x n dx. This is termed a Gamftia Function; 
J o 

this particular integral being the (+i) Gamma Function denoted 



Thus f e~*xdx = 

J o 

/-co 

and I e-t&^-dx = Y(n) 

J o 

the latter integral being also called the Second Eulerian Integral. 



174 MATHEMATICS FOR ENGINEERS 

To evaluate r(w+i) let u = x n and dv = e~ x dx 
so that v = e~ x and du = nx n ~ l dx. 

,00 XOO / 

Then I e~ x x n dx = ( e~ x x n ) I e~ x .nx n ~ l dx 

Jo \ /o J o 

xoo /. 

= ( er x x n ) 4-n\ e~ x x n ~ l dx. 
\ /o J o 

Now when # = oo e"^" = ^ , which can be proved = o 



I X O 

and when x = o e~ x x n = - - = o. 

e 

/oo / 

Hence / e~ x x n dx = n I e~ x x n ~ 1 dx 

Jo Jo 

or r(-{-i) = nT(n). 

In like manner it could be shown that 

T(n) = (ni)T(ni), and so on. 
If n is an integer it will be seen that we finally reduce to r(i), 

f 
i. e., I e~ x dx, the value of which is (e~ y> e) = (o i) = i. 

Hence r(-fi) = n(ni)(n2) . . . i = |_. 

This last relation will not hold, however, if n is not an integer, 
but the general method of attack holds good; and tables have 
been compiled giving the values of T(n) for many values of n, 
whether n be an integer or fractional. Thus if an integral can be 
reduced to a Gamma function or a combination of Gamma func- 
tions, its complete evaluation may be effected by reference to the 
tables. 

/oo. _a^ 
e~h'dx by the 

aid of the Gamma function. 



Let X - ^ then ^ = -f- = * 
2 a# a# h 2 

, , Y _ 2^^r 

~ "P~ 

A 2 rfX A 2 dK. hdX. 
or dx = = ; 7=^ = 



2 x 2 h\/x. 2\/X' 
Then f.TW, = /" e - x x^|- = *r.-*X-ta = *xr(?) 

^0 / 2 VX 2-' 2 \2J 

and the value of T ( \ * s VTT^ 



FURTHER METHODS OF INTEGRATION 

r > _a* fr _ 

Hence I e h*dx = - Vie. 

JO 2 



175 



On comparing with Example 27, p. 162, where a rather simpler 
form of the integral is evaluated, we see the great saving effected 
by the use of the Gamma function. 



LIST OF INTEGRALS LIKELY TO BE OF SERVICE 

f(ax+b) to 



fae^dx 

fba^dx 

fa cos (bx+d) dx 

fa sin (bx+d) dx 






. 
n log a 



= sin (bx+d)+C. 

= ~ cos (bx+d)+C. 
fa tan (bx-\-d) dx = , log cos (bx-\-d)-{-C. 

fa cot (bx+d) dx = ^ log sin (bx+d)+C. 

= | tan (6*+<Q+C. 
= cot (bx+d)+C. 



fa sec 2 (bx+d) dx 

/a cosec 2 (6*-H) 

fa cosh (6^+^) to 

/a sinh (bx+d) dx = ~ cosh (bx+d)+C 



= T sinh 



fa sech 2 (6^+^) to = | tanh 

/a cosech 2 (bx+d) dx = | coth (bx+d)+C. 

* sin (6^+^) to = - 

f"c0s (bx+d) dx *= 



sn 



sn 



cos 



cos 



f ax 

\ +/~2 ' 
J V U X 

(-T 

JVx 



dx 



dx 



176 -MATHEMATICS FOR ENGINEERS 

r dx i ( bx+d\ . n 

I ~~'/i~ ~r~j\ == i 1S ( tan +C. 

J sin (bx+d) b \ 2 / 

f dx i 

]0x+b =-log(**+fc)+C. 

/tan (bx-}-d) dx = -? log cos (bx+d)+C. 



/cot (bx-{-d) dx = r log sin (bx+d)+C. 

r dx i , , x . r 

I -o-j = - tan * -+C. 

J 2 +A; 2 a 

dx 1,1 ^+ , /- 

= * tan T +c - 



r-r^i = X log^+C. 

J a 2 x z 2a 

f dx i_ , 

J (x-\-a) z b z ~ 2b 



dx 

= sm~ 



1-7===, = log ^~r v -* -* j + c. 

J v x a 

r dx 

}' 



r (ax-\-b)ax i . , 

J ^tf^bx+d) ~ 2 g (a 

Tcosec (ax-\-b) dx = - log (tan ' -)+C. 

** d \ 2 / 



/ sec (a^+&) <fo = - log ta 

/^ i , x . 

. = - sec" 1 -+C. 
x\x z u z & a 

f dx , x . r , x r 

I . = vers" 1 -+C or i cos x -+C. 

J V2axx z a 



l x z dx = -y 2 x z + 2 sm 

__ x e z 

jVx z -a z dx = ~Vx z -a z - 



FURTHER METHODS OF INTEGRATION 177 

fV(x+a)*-Pdx = I (x+a)V(x+a)*-b*-~ cosh- 1 *^ 



*2 



-a? dx = ~Vx 2 4-a 2 4- sinbr 1 - 

2 2 a 



/V/(*+a)M-6* <& = 

^ -5 

/" sin 2 ;*; <fa = sin 2^+C. 

2 4 

. A; I 

/ cos 2 x dx = -4 sin 2x-\-C. 

2 4 

/* sin m x cos n A; ^ = /" sin m ~ 2 x . cos" ^ dx 

J m-\-n J 



m-\-n 
[ dx 
J (aTI^ji 

I ^-*i^ 
* o 

/oo * 

I e^w dx 

Jo 2 

tr T 

sin 2 ^0 = | 2 cos 2 0^ = "". 

J o 4 

IT 

-7 



s i n m-l ^ , cOS n+1 # r 

f-L/. 



sin6 & = cos6 <*0 = ~ 



( 2)( 4) ... 2 2 

if w is an even integer. 



sin6 ^0 = cos0 ^0 = *n7- . . 2 



fsi 
Jo 



. 
o ( 2)( 4) . . . I 

if w is an odd integer. 



(-l)(-3) . . . I 7T 

X 



IT 

r 1 

I SI 

J A 



if w and w are both even. 



sin w cos w dQ= y\ /^ 

(w+w)(w+w 2) . . . (w+3) (w+i)(w i) ... 4 2 

if w and n are both odd. 



178 MATHEMATICS FOR ENGINEERS 



It 

(* 

I sin'" 
' o 



e cose rfe = -= x = ' ' ' 

2) . . . (n-\-2) ( 2) . . . 



i 



if m is even and n is odd. 
Pdv (where pv n = C) = (v^~ n v^ 

I pdv (where pv = C) = C log e - 2 . 
= a log p 



sin" 1 a#*fo = x sin~ 1 a#+;_ Vi a 2 x z -\-C. 
tan- 1 ax dx = x tan" 1 ax log 

20, b 



f (a 2 -^ 2 ) 1 ^ = -(a 2 -* 2 ) + 
J 4 



3 - sn 



Exercises 15. On Further Integration. 

Evaluate the integrals in Nos. i to 18. 

dx_ 9 f dx f (x-i)dx 

**' J *x*+6x+2i' 'Jc>x 2 -i8; 



4. - | VJ 3 , which occurred when finding stresses in a crane 
hook. 

5. t 95-5 1 (6h}*dh, referring to the time for emptying a tank. 

7T 

2 /"2 



6. ^3~ W 2 dy. 7 - sin 5 ^ cos ' 



ft ^ 4 A /"sWR/i i . rtN-, _ , ft 

9. ^^ . 10> / -Frrl --- sin 6 R 2 cos6d6. 

sin 2 5^ -' o El \TT 2 



11. /4 tan 5/ <tt. 12. /sin- 1 ^r dx. 13. 

7T 

14. /5e 3:c sin 2Ar rf^r. 15. J ,^. * 2 . 2 . 16. J 5 ^ sin ^r ^. 

17. /cos 6 sin 2 6rf0. 18. h = f [-*-&* ,j~,rr & , relating to the 

J o 2gn 2 (d+kx) 5 

flow of water through a pipe of uniformly varying diameter; the 
diameter at distance x from the small end being = small end diam. 
-f-**. 



FURTHER METHODS OF INTEGRATION 179 

19. The time taken (t sees.) to lower the level of the liquid in a 
certain vessel having two orifices in one side can be found from 

nodh 

at = 



12 + Vh 
Find this time if the limits of h are o and 10. 

20. Express e~ l * as a series, and thence find the value of 

2 (* 
VrcJ o 

21. The maximum intensity of shearing stress over a circular 

F r r i 

section of radius r = S = j. / 2(r 2 y z )*ydy. 



If I = , find a simple expression for S. 

22. Evaluate the integral I - _____ [Hint. Rationalise the 

J t vr 4 
denominator.] 



23. Write down the value of | 2 cos 8 QdQ. 

J o 

24. If the value of log r(i-85) is given in the tables as 1-9757, 

ra> 

find the value of I e~ x x l '**dx. 
-' o 

w 

25. Write down the value of 1 2 sin 2 6 cos 7 0^0. 

J o 

26. When finding the forces in a circular arched rib it was found 
necessary to find the value of 

ir 

2j 6 R 3 (cos 2 6 Vs cos 6+^)^0. Evaluate this integral. 

27. Evaluate the indefinite integral fVzi y( x z dx. 

28. The attraction F of a thin circular disc of radius r on a body 
on the axis of the disc, and distant z from the centre, is given by 

rdr 

- 



where a is the density of the disc, and k is a constant. 
Find the value of F for this case. 



CHAPTER VII 



MEAN VALUES: ROOT MEAN SQUARE VALUES: 
VOLUMES : LENGTH OF ARC : AREA OF SURFACE 
OF SOLID OF REVOLUTION : CENTROID : MOMENT 
OF INERTIA 

Determination of Mean Values. It is frequently necessary 
to calculate the mean value of a varying quantity : thus if a 
variable force acts against a resistance, the work done will be 
dependent on the mean value of the force; or to take an illustra- 
tion from electrical theory, if we can find the values of the current 
and electro-motive force at various instants during the passage of 
the current, then the mean rate of working is the mean value of 
their product. 

The mean value of a series of values is found by adding the 
values together and then dividing by the number of values taken. 
If, however, a curve is drawn to give by its ordinate the magnitude 
of the quantity at any instant, the mean value of the quantity is 
determined by the mean height of the diagram, which is the area 
divided by the length of the base. This really amounts to the 
taking of an exceedingly large number of ordinates and then 
calculating their average. 

The area may be measured by the planimeter, in which case 
the instrument may be set to record the mean height directly, or 
by any of the methods enumerated in Part I, Chap. VII. 

A clear conception of the idea of mean values can be gained 
by consideration of the examples that follow; the first example 
being merely of an arithmetical nature. 

Example i. The corresponding values of an electric current and 
the E.M.F. producing it are as in the table : 



c 


o 


1-8 


3'5 


5 


6-1 


6-8 


6-1 


5 


3'5 


1-8 


E 





9 


17-5 


25 


30'5 


34 


30-5 


25 


17-5 


9 



1 80 



MEAN VALUES 



181 



Find the mean value of the power over the period during which 
these values were measured. 



current x E.M.F., and 
61-25 



16-2 



The power is measured by the product 
thus the values of the power are 

o 16-2 61-25 125 186-05 231-2 186-05 I2 5 
the sum = 1008-2 the number of values = 10 

1008-2 
hence the average or mean value = = 100-8. 

A better result would probably be obtained if the values of the 
power were plotted and the area of the diagram found. 

Thus in Fig, 39 the base is taken as 10 units (merely for con- 
venience), and the area is found to be 1013 sq. units. Then the mean 

height of the diagram, which is = 101-3, is also the mean value 

of the power; and a line drawn at a height of 101-3 units divides the 
figure in such a way that the area B is equal to the areas A+A. 



eoo 

Rawer 




-""Mean hcighf 




Ol E3456789IO 

FIG. 39. 

Let it next be required to find the mean value of a function 
say, the mean value of y when y = 4x z -\-jx$, the range of x 
being i to 6. We have seen that it is really necessary first to 
find the area under the curve y = 4# 2 +7# 5 within the proper 
boundaries and then to divide by the length of the base; and 
since the relation between y and x is stated, it is possible to 
dispense with the graph and work entirely by algebraic integra- 
tion; thus also ensuring the true result, for in reality the mean 
is automatically taken of an infinite number of ordinates. 

In the case taken as an illustration, the base is the axis of x, 
or, more strictly, the portion of it between x = i and x = 6, so 
that the length of the base = 61 = 5 units, and the area 
between the curve, the axis of x and the ordinates through x = i 

f 6 f 6 

and x = 6 is given by the value of I y dx, i. e., I (4x z -\-jx 5) dx, 



182 MATHEMATICS FOR ENGINEERS 

so that the mean value (for which we shall write m.v.) is 

*2+*_ dx 



m.v. = 



_[3 +2 J"~ 5 *]'i 
5 



It is instructive to compare this result with the results obtained 
by the use of the mid-ordinate rule : 

(a) Taking 5 ordinates only, we have the values 



X 


4# 2 +7*- 5 


y 


'i 

2* 

3t 
4i 
5i 


9+10-5 5 

25 + I7-5-5 
49+24-5-5 
81+31-5-5 
121 + 38-5-5 


14-5 
37-5 
685 
107-5 
154-5 



Their sum = 382-5 
and the average = 76-5. 

(b) Taking 10 ordinates, viz., those at x = ij, if, 2|, etc., 
the values of y are 10, 19-5, 31, 44-5, 60, 77-5, 97, 118-5, 142 
and 167-5 

their sum = 767-5 

and the average = 76-75. 

Therefore, by increasing the number of ordinates measured, a 
better approximation is found. 

The curve is a parabola with axis vertical, and hence Simpson's 
rule should give the result accurately if 3 ordinates only are taken, 
viz., at x = i, 3-5 and 6. 

Thus, A = 6, M = 68-5, B = 181. 

-.. 6+(4x68-5)+i8i 

Hence the mean height = - 



= = 76-83. 



MEAN VALUES 183 

If, then, the law connecting the two variables is known, the 
mean value of the one over any range of the other can be found 
by integrating the former with regard to the latter between the 
proper limits and then dividing by the range; or to express in 
symbols, if y = f(x), the mean value of y, as x ranges from a to b, 
is given by 

fb , 

y dx 



f 

J " 



b-a 

Example 2. Find the mean value of e 5x between x = -2 and x = 

/? g . 
a 



m.v. = 

7-'2 '5 t-5 

2 

~ 5 '* 
= ^{33-12-272} 

= 12-16 

i. e., if the curve y = e 5x were plotted between x = -2 and x = 7 its 
mean ordinate would be 12-16 units. 

Example 3. If C = 5 sin 3^, find the m.v. of C, when 

, > , . , 27C . 

(a) t vanes from o to 

3 

(b) t varies from o to -. 

Whenever dealing with the integration of trigonometric functions 
it is advisable first to determine the period of the function, since 
much numerical work may often be saved in this way. The sine and 
cosine curves are curves symmetrical about the axis of the I.V. 
(i, e., x or t, as the case may be) ; hence there is as much area above 
this axis as below it, if a full period or a multiple of periods be 
considered. Therefore, regarding the area above the x axis as positive 
and the area below this axis as negative, the net area over the full 
period is zero, so that the mean height of the curve,* and therefore 
the m.v. of the function, must be zero. 

For the case of C = 5 sin 3/, the period = ^ ^ = 

coeff <. of / 3 

and hence if the m.v. is required for t ranging fiom o to ' 

. 4^ 
o to , etc. 

3 
the result is zero. 



184 MATHEMATICS FOR ENGINEERS 

Hence, whenever the analysis shows that the full period is involved, 
there is no need to go through the process of integration. In this 
case, however, the integration is performed for purposes of verification. 

27T 27T 

, _ fa 5 sin -$tdt 3 x 5 / i ,\- 

(a) m.v. of C. = / * 3 = - - ( cos 3/ 

J o 271 27T \ 3 - / 1 

3 



- -o 3 



- (COS 27C COS O) 

27T V 

= ('-I) 

27U V ' 



IT 

/i.\ t r fissinitdt 

(b) m.v. of C. = / - = 

Jo n 





3 

= - (COS 7T COS O) 

7T V 

5 / \ I0 

= *-* ( r i) = . 

7T V 7T 

Comparing this with the amplitude, which is the maximum ordinate 
of the curve and has the value 5 sin = 5, we note that 

Mean ordinate for ^ period 10 _ 2 _ 

Maximum ordinate for \ period ~~ TT X 5 ~~ n ~ '' 

The average height of a sine curve is always -637 x the maximum 
height. 

Example 4. If an alternating electric current is given by the 
relation 

C = -5 sin i207T/+-o6 sin 6007^ 

find the mean value of C. 

The graph is of great assistance in this evaluation ; and 
consequently the curves c x -5 sin 1207^, c 2 = -06 sin GOOTT/, and 
C = c t + c 2 = -5 sin i2O7t/+-o6 sin 6007^ are plotted in Fig. 40. The 

period of c, = -5 sin I2O7T/ is or >-, and of c = -06 sin 6007^ is 

I2O7C 6O 

- 01 - -, so that the period of the compound curve must be ^-. 

60O7T 30O 6O 

From the pievious reasoning it is seen that the mean height of the 
curve c t = -5 sin i207r/ must be -637 x amplitude = -637 X -5 = -3185; 
and the mean height of the curve c 2 = -06 sin 6007^ considered over 

the same period, viz., o to , must be the mean height of the wave A, 

120 

since the positive and negative areas are otherwise balanced, but this 



MEAN VALUES 



185 



must be spread over five times its usual base ; now the mean height 
of the wave A is '637 X -06, so that the mean height of the curve 

c 2 -06 sin 6oo7c/ over the period o to 



-637 x -06 

is or -0076. 

120 5 



Then the mean value of C = -3185 + -0076 = -3261. 

The nature of the average should be clearly understood ; for it 
is possible that the same quantity may have two different averages 
according to the way those averages are considered. Suppose a 
piston is pushed by a variable force; then the average value of 
that force might be found by taking readings of it at every foot 
of the stroke and dividing by the number of readings taken, in 




40 X^ IO 

i- -Oesin gOOTTf 

FIG. 40. Problem on Alternating Current. 

which case the average would be termed a space average ; or the 
force might be measured at equal intervals of time, whence the 
time average of the force would result. 

To take another instance : Suppose a bullet penetrates a 
target to a depth of s ft. ; the average value of the force calculated 



from the formula Ps = 



mv* 



where P is the force, would be a 



mv 



space average, but P calculated from the formula Pt= , i. e., the 

6 

force causing change of momentum in a definite time, would be a 
time average. 

To illustrate further : A body starts from rest and its speed 
increases at the uniform rate of 4 ft. per sec. 2 . Find the time and 
space averages of the velocity if the motion is considered to take 



186 MATHEMATICS FOR ENGINEERS 

place for 6 sees. {Use the relations s = - at z , v = at and v 2 = 2as. 

s = - at z = - x 4 X 6 2 = 72 

2 2 

and if the limits of t are o and 6 
then the limits of s are o and 72. 

To find the time average of the velocity 

8 

atdt 
"6^0 ~ 



/e re 

I vdt c 

f __. * J 



= 12 f.p.S. 

To find the space average of the velocity 

V 2 = 20S = 8s 

or v= V&Vs 

and the mean value of v = mean value of V8 



/ VS a rfS i ,-(2 sV 2 

Hence the space average = /T/T _ rt ~^ x ^^\^ sl? j 



72 



= = 

Example 5. The electrical resistance R of a rheostat at temperature 
t C. is given by 

RJ = 38(1 + -004^). 

Find its average resistance as t varies from 10 to 40 C. 
(This will be a temperature average.) 



m.v. = 



30 



38x33 
30 

41-8. 



MEAN VALUES 187 

Example 6. If V = V sin qt and C = C sin (qtc), find the 
average value oi the power, i. e., the average value of VC. 

VC = V sin qt . C (sin qtc) 
= {2 sin qt . sin (qt c)} 
V C 

= {COS C COS (20* C)} 

also the period = . 
Hence the m.v. of VC 

r 

[cos c cos (zqtc)}dt 



ex/-- sin (zqt-c]\ q 
20 '\ 



cos 



c- sin ( 4TC -c)-o+ sin (-c) j 

20 20 V 'J 



275 2 20 20 

_ VoCo? 27c |~for sin (4* e) = sin ( e)~| 

A - COS C , / \ * / \ 

47T q Land sin ( c)+sin ( c) = o 

= ~V C cos c 

i. e., the mean value of the power = one-half the products of the 
maximum values or amplitudes with the cosine of the lag. 

This is a most important result. 

If c = 90, i. e., if the lag is -, then cos c o and the mean value 

of the watts = - V C X o = o ; this being spoken of as the case of 
wattless current. 

Exercises 16. On Mean Values. 

1. If a gas expands so as to follow the law pv 120, find the 
average pressure between the volumes 2 and 4. 

2. Find the mean value of e 2 ' 5 v as y varies from o to -4. 

3. The mean height of the curve y = 3* 3 +5# 7 is required 
between the limits x 2 and x +3; find this height. 

4. Find the mean value of 2-i8sin](3/ 1-6) as / ranges from 
14 to 1-6. 

5. What is the mean value of 4-5 +2 sin 6o/, t ranging from 
o to ? Discuss this question from its graphic aspect. 



i88 



MATHEMATICS FOR ENGINEERS 



6. Find the mean value of p, when pv 1 ' 37 = 550, for the range of 
v from 4 to 22. . 

7. The illumination I (foot candles) of a single arc placed 22 ft. 
above the ground, at d ft. from the foot of the lamp is given by 
I = i -4 -oid. Find the mean illumination as d varies from J ft. 
to 10 ft. 

8. An alternating current is given by C = -2 sin IOOTT/+-OI sin 300^. 
Find the mean value of C for the range of t, o to -02 sec. 

9. Taking the figures in Question 8, find the mean value of C when 
t ranges from o to -01 sec. 

10. Find the mean value of 5 sin 6t X 220 sin ^t, t ranging from 

7C 

o to -. 
3 

11. The table gives the values of the side thrust on the piston of 
a 160 H.P. Mercedes aero engine for different positions of the crank; 
the positive values being the thrust on the right-hand wall, and the 
negative values being the thrust on the left-hand wall of the piston. 



Angle of crank froml 
top dead centre | 

Total side thrust) 
(Ibs.) / 


O 


40 


80 


IOO 


120 


140 


160 


1 80 


20O 


240 


280 


300 


320 


330 


o 


+ 210 


O 


-170 


-175 


-I8 5 


IOO 





+95 


+240 


+ IOO 


o 


-30 


o 



345 


360 


10 


40 


80 


1 20 


160 


1 80 


2OO 


2 2O 


240 


260 


280 


290 


320 


360 


+20 


o 


600 


-* 


720 


-580 


-170 





+ QO 


+ 170 


+160 


+ 150 


+50 





210 






Plot these values (treating them all as positive) and thence 
determine the mean side thrust throughout the cycle. 

Root Mean Square Values. A direct electric current maybe 
measured by its three effects chemical decomposition, magnetic 
effect or heating effect. An alternating current (A.C.), however, 
flowing first in one direction and then in the reverse, cannot be 
measured by either the first or the second of these effects, because 
the effect due to the flow in one direction would be neutralised by 
that due to the opposite flow ; hence an A.C. must be measured 
by its heating effect. 

The heating effect of a current expressed by the heat units H 
may be measured by 



H = 



where C is the current, 



so that it will be seen that H oc C 2 . 

The measuring instruments are graduated to give the root of 
the mean value of this heating effect, i. e., the square root of the 



ROOT MEAN SQUARE VALUES 



189 



mean value of the squares (called the root mean square value and 
written R.M.S.) ; in other words, the instrument records what are 
termed virtual amperes, a virtual ampere being the current that 
produces the same heat in a resistance as a steady or direct current 
of i ampere in the same time. 

In place of the measurement of the current by its heating 
effect, the ammeter might be of the electro-dynamometer type, 
and in such the instrument records the mean value of C 2 , and the 
square root of this value is called the effective current. 

Similar remarks apply also to the measurement of alternating 
E.M.F. 

It is therefore necessary to determine R.M.S. values of functions 
likely to be encountered, and to compare the virtual with the 
steady. 

While the determination of R.M.S. values is of greatest import- 
ance from the application to electrical problems, it is also occasion- 
ally of use in problems of mechanics ; thus the calculation of what 
is known as a swing radius (see p. 240) is in reality a determination 
of a R.M.S. value. 

In order to convey the full interpretation of the term R.M.S. 
value, we shall discuss first a simple arithmetical example, then 
some algebraic examples, leading up finally to the trigonometric 
functions. 

Example j. The values of an alternating electric current at various 
times are given in the table 



t 





01 


02 


03 


04 


05 


06 


07 


08 


09 


i 


C 


o 


1-8 


3'5 


5 


6-1 


6-8 


6-1 


5 


3'5 


1-8 


o 



Find the mean value and also the R.M.S. value of the current, 
and compare the two values. 

The mean value, as before explained, is the average of the given 
values and is 3-96. 

We must now tabulate the values of the squares; thus 
o 3-24 12-25 25 37-21 46-24 37-21 25 12-25 3' 2 4 and o. 
The sum of the squares = 201-64 
the mean of the squares or M.S. = 



for we must only reckon the end values as half-values when adding 



IQO 



MATHEMATICS FOR ENGINEERS 



up the ordinates, since the end values belong equally to the sequences 
on either side. 

Thus M.S. = 20-16 

and the square root of the result = R.M.S. = V2o-i6 = 4-5. 

This question might have been worked entirely by graphic methods, 
according to the following plan 

Plot the values of C to a base of /, giving the curve ABD in Fig. 41 ; 
find the area under the curve and divide by the base, thus obtaining 
the mean height; plot also the values of C 2 against those of t, giving 
the curve EFG. By graphic summation determine the area under 
the curve EFG and draw the line MM at the mean height of the 
diagram. Make MN = i unit on the scale of C 2 and on PN describe 
a semicircle ; produce MM to cut the semicircle in R. Then MR = 4-5 
is the R.M.S. value. 

Thus the mean value = 3-96 and the R.M.S. value = 4-5, and the 



,. R.M.S. value 4-5 
ratio - - - = -2-i = 1-1 

mean value 3-96 

factor. 



., / 
this ratio being termed the form 




a _ 



FIG. 41. R.M.S. Value of an Alternating Current. 

Example 8. Find the R.M.S. value of the function 2X 1 ' 5 3* as x 
ranges from 2 to 5. 

The square = (2x l ' 5 ^x) 2 = 4 



Explanation. 



dx 



and the M.S. = 



3 

_ i 

~ 3 
= 13-27- 
Hence the R.M.S. value 



5-2 



-959)-(i6+2 4 - 3 8-8)] 



= Vi3'27 = 3-64- 



No. 


Log. 


5 

3-43 
959 

2 

3-43 

38-8 


6990 
3'5 


34950 
20970 


2-44650 
5353 


2-9818 

3010 
3'5 


15050 
9030 


1-05350 
5353 


V5883 



ROOT MEAN SQUARE VALUES 



191 



Example 9. Suppose that an alternating electric current at any 
time follows the sine law, i. e. 

C = C sin qt 

where C is the instantaneous value of the current at any time t, and 
C is the maximum value of the current. 
Find the R.M.S. value of the current. 

As we have already seen, the determination of the R.M.S. value 
implies that first the square of the function at various times must 
be calculated, then the mean value of these squares found, and 
finally the square root of this average extracted. 

To assist in the study of this important problem, the curve 
y sin x, the simple sine curve, is shown in Fig. 42, and also the 




FIG. 42. R.M.S. Value of an Alternating Current. 

curve y sin 2 x, which is obtained from the former curve by squaring 
its ordinates. It will be observed that whilst for the curve y = sin x 
there are both positive and negative ordinates, in the case of the 
curve of squares all the ordinates 'are positive. Also the period of 
the curve of squares is noted to be one-half that of the simple curve; 
and therefore when calculating the mean height of this curve, it is 
immaterial whether the full or the half period of the simple sine curve 
is taken as the range. 

In the case with which we are here particularly concerned the 

C 2 2Tc 

square = C 2 = C 2 sin 2 qt = (i cos 2qt), and the period = , so 

that the integration may be performed either over the range o to - 
or o to -. 



192 MATHEMATICS FOR ENGINEERS 

Taking the latter range 

a 

I q C 2 sin 2 qt dt 
Mean of the squares = M.S. = 

q 



= & f q (i -cos 2qt) dt* 

7C Jo 



2q 



oC 2 TTC I .in 

= * - --- sin 27c o-{ sin o 
27T \_ 2 2q J 




or R.M.S. value = -707 x maximum value 

i. e., virtual value of current or E.M.F. = -707 x maximum value of 
current or E.M.F. 

Hence if a meter registers 10 amps., the maximum current is - , 

707 

i. e., 14-14 amps., or there is a variation between + 14-14 and 14-14. 



* In the evaluation of the integral t q (i cos iqf) dt, it should be 

it ir 

noted that it can be written q dt I 9 cos -zqtdt, and the value of 

Jo Jo 

the second term is zero, because it is the area under a cosine curve 

taken over its full period. Hence the integral reduces to / 9 dt, and 
there is no need to say anything further about the second term. 

Example 10. Find the R.M.S. value of a-\-b sin 4*. 

For this function the peiiod = -, 

4 2 

Then S = (a-\-b sin 4/) 2 = 2 +6 2 sin 2 4^ + 2^6 sin 4^ 

52 
= 2 + (i cos 8t)-\-2ab sin 4^. 



7 

2 fa 6 a 6 2 

Hence M.S. = - I (a z -\ ---- cos 8t+zab sin 4*) 

7T J 22 

_ TT n 

2 F T2 / fc 2 \ T2 fe2 

= ~ I \ aZ +2) dt J - 



dt 



cos tt + 2a sn 



VOLUMES 193 



2 2 2 

R.M.S.= 



Exercises 17. On Root Mean Square Values. 
Find the R.M.S. values of the functions in Nos. i to 7. 

1. # 3 +2 (x ranging from i to 3). 

2. e~" x (x ranging from -i to +-65). 

3. 3-4 sin 5*i/. 4. -165 cos (-07 2/). 

5. 1-4 tan 2t (t ranging from o to -43). 

6. i -14 +-5 cos -8t. 

7. -72 sin (3 4/) ; compare with the mean value. 

o -c- A .LI. t * /R.M.S. value\ , ,, 

8. Find the form factor -, ) of the wave 

\ mean value / 

e = E x sin pt+~E a sin 3pt. 

9. Compare the " effective " values of two currents, one whose 
wave form is sinusoidal, having a maximum of 100 amperes, and another 
of triangular shape with a maximum of 150 amperes. 

10. An A.C. has the following values at equal intervals of time : 
3. 4. 4'5, 5'5. 8, io- 6, o, 3, 4, 4-5, 5-5, 8, 10, 6, p. 
Find the R.M.S. value of this current. 

11. A number of equal masses are attached to the ends of rods 
rotating about one axis. If the lengths of these rods are 10, 9, 5, 
8, 4, 13 and 15 ins. respectively, find the effective radius (called the 
swing radius) of the system. (This is the R.M.S. value of the respective 
radii.) 

12. Find the R.M.S. value of the function sin 2 cos 3 6 over the 
period o to -. (Refer to p. 178.) 

13. The value of the primary current through a transformer at 
equal time intervals was 

20 -05 -07 -ii -14 -19 -2i -04 -08 '12 -15 -18 -2i -08 -04 
Find the R.M.S. value of this current. 

Volumes. If a curve be drawn, the ordinates being the values 
of the cross sections of a solid at the various points along its 
length, then the area under the curve will represent, to some 
scale, the volume of the solid. For, considering a small element 
of the length, 81, if the mean area over that element is A, the 
o 



194 MATHEMATICS FOR ENGINEERS 

volume SV of the small portion of the solid is A.U, and the 
total volume of the solid is the sum of all such small volumes, 
i. e., 2AS. If the length 81 is diminished until infinitely small, 
2AS/ becomes fAdl, and hence 

Volume of solid = / Adi. 
J o 

Comparing this with the formula giving the area of a closed 
figure, viz.,fydx, it is seen to be of the same form, and it is made 
identical if A is written in place of y (i. e., areas must be plotted 
vertically) and / in place of x (i. e., lengths must be plotted 
horizontally) . 

When values of A and / are given the curve should be drawn 
and integrated by any of the given methods ; and if it is preferred 
to find the actual area of the figure in sq. ins. first, the number 
of units of volume represented will be deduced from a consideration 
of the scales used in the plotting. Thus if i" (horizontally) repre- 
sents x ft. of length, and i" (vertically) represents y sq. ft. of area, 
then i sq. in. of area represents xy cu. ft. of volume. To take 
a numerical example : If i" = 15" of length, and i" = 5 sq. ft. of 

area, then i sq. in. of area = x 5 = 6J cu. ft. of volume. 

12 

If the area is found by the sum curve this conversion is 
unnecessary, as the scales are settled in the course of the drawing. 
The Example on p. 122 is an illustration of the determination of 
volumes by graphic integration; in that case the actual areas of 
sections are not plotted directly, but values of d 2 , the multiplication 
by the constant factor being left until the end. Had the solid not 
been of circular section the actual areas would have been plotted 
as ordinates and the work carried on as there detailed. 

We thus see that the determination of the volume of any 
irregular solid can be effected, if the cross sections at various 
distances from the ends can be found, by a process of graphic 
integration. 

If, however, the law governing the variation in section is known, 
it may be more direct to perform the integration by algebraic 
methods. 

To take a very simple illustration : 

Example n. The cross section of a certain body is always equal 
to (5# 3 + 8) sq. ft., where x ft. is the distance of the section from one 
end. If the length of the body is 5 ft., find its volume. 



VOLUMES 195 

The body might have an elevation like Fig. 43, and its cross section 
might be of any shape ; the only condition to be satisfied being that 
the area of a cross section such as that at BB must = 5* 3 +8. Thus 
the area at AA must = (5 X o) + 8 = 8 and area at CC = (5 X 5 3 ) + 8 = 633. 

f 5 C 5 

Then the volume = J Q Adx = ] Q (5x 3 +8)dx 

+40 = 821-25 cu. ft. 

~"~^^ 




4 

Volumes of Solids of Revolution. A solid of revolution is 
generated by the revolution of some closed figure round an axis 
which does not cut the figure. Thus, dealing with familiar solids, 
the right circular cone, the cylinder and the sphere are solids of 
revolution, being generated respectively by the revolution of a 
right-angled triangle about one of the sides 
including the right angle, a rectangle about 
one of its sides, and a semicircle about its 
diameter ; and of the less well-known solids 
of revolution the most important to the 
engineer is the hyperboloid of revolution 
which is generated by the revolution of a FIG 

hyperbola about one of its axes and occurs 
in the design of skew wheels. The axis about which the revolution 
is made in all these examples lies along a boundary of the re- 
volving figure; whereas an anchor ring is generated by the 
revolution of a circle about an axis parallel to a diameter but 
some distance from it. 

The revolving figure may have any shape whatever, the only 
conditions being, for the following rule to hold, that the axis about 
which the revolution is made does not cut the figure and that the 
cross section perpendicular to this axis is always circular. 

Imagine the revolving cross section to be of the character 
shown in the sketch (ABCD in Fig. 44) ; and let the revolution 
be about the axis of x. It is required to find the volume of the 
solid of revolution generated. 

Working entirely from first principles, i, e., reverting to our 
idea of dealing with a small element and then summing : if the 
strip MN of height y and thickness 8x revolves about OX it will 
generate a cylinder. 

The radius of this cylinder will be y and its height or length Sx ; 
and hence its volume = iry 2 Sx. 

Accordingly the volume swept out by the revolution of ABCD 



ig6 



MATHEMATICS FOR ENGINEERS 



will be ^ny z Sx (the proper limits being assigned to x) approximately, 
or fny 2 dx accurately. 

Again, it will be seen that such a volume can be measured by 
the area of a figure, for, writing Y in place of Try 2 , the volume 
= fYdx, which is the standard form for an area. Hence, if 
values of y are given, corresponding values of Try 2 must be cal- 
culated and plotted as ordinates, and the area of the resulting 
figure found. 

It should be noted that fny 2 dx might be written as itfy*dx t 
thus saving labour by reserving the multiplication by TC until the 



M 



B X 




X! 



FIG. 44. Volume of Solid of Revolution. 

area has been found, i. e., the values of y 2 and not those of Try 2 
are plotted as ordinates. 

The following example will illustrate : 

Example 12. The curve given by the figures in the table revolves 
about the axis of x; find the volume of the solid generated, the 
bounding planes being those through x = 2 and x = j, perpendicular 
to the axis of revolution. 



X 


2 


3 


4 


5 


6 


7 


y 


44 


42 


44 


46 


45 


38 



VOLUMES 197 

n 
Values of y 2 must first be calculated, since the volume = / jty z dx 

= 

Hence the table for plotting reads 



X 


2 


3 


4 


5 


6 


7 


y* 


1936 


1764 


1936 


2116 


2025 


1444 



and the values of y 2 are plotted vertically, the curve ABC (Fig. 45) 
resulting. 

This curve is next integrated from the axis of x as base, the curve 
DEF resulting; the polar distance being taken as 3, so that the new 




O 6 / 

FIG. 45. Volume of Solid of Revolution. 

vertical scale = 3 X old vertical scale. It must be remembered, how- 
ever, that the base from which the summation has been made in the 
figure is not the true base, since the first value of the ordinate is 1400 
and not o; thus a rectangle 1400X5 has been omitted. Hence we 
must start to number our scale at 7000 ; and according to this 

f 7 
numbering the last ordinate reads 9580, hence / z y z dx = 9580, or 

/? 
Volume = TC I y z dx 

= TT x 9580 

= 30,100 cu. ins. 

In cases in which y and x are connected by a law the integra- 



198 



MATHEMATICS FOR ENGINEERS 



tion may be performed in accordance with the rules for the 

integration of functions. 

Example 13. Find the volume of a 
paraboloid of revolution and compare it 
with the volume of the circumscribing 
cylinder. 



o 



Y 




B 



FIG. 46. 



A paraboloid of revolution is generated 
by the revolution of a parabola about 
its axis. Suppose the parabola is placed 
as shown in Fig. 46; the revolution is 
therefore about OX. 

The equation to the curve OB is 
y 2 = $ax (see Part I, p. 106), i. e., if 
OA = h, (AB) 2 = 4 ah. 

Hence the volume of the solid swept 
out by the revolution of OBA 



[h [h 

= Tty z dx = TU I qax dx 

J J 



_ 4 a7 L><A a _ 7,2 
Now the volume of the circumscribing cylinder 



= TC X 4ah X h 



and hence the volume of the paraboloid 



= - X vol. of circumscribing cylinder. 



Example 14. The curve y 2 = 64 zx 2 revolves about the axis of y. 
Find the volume of the solid generated, the limits to be applied to x 
being o and 5. 

This differs from the cases previously treated in that the revolution 
is to be about the y axis and not about the x axis. 

Hence the volume = fnx 2 dy and not Jwy*dx, the y replacing x 
and vice versa. Also another point must be noted : the limits given 
are those for x, whereas the limits in the integral fiix 2 dy must apply 
to the I.V., which is now y. Therefore a preliminary calculation 
must be made to determine the corresponding limits of y 



y 2 = 64- 
x = o, y 
x = 5, 



2 = 64, y = 8 
= 14- y = 374- 



VOLUMES 



199 



The double signs occurring here may possibly confuse, but actually 
the equation given is that of an ellipse, symmetrical about the axes 
of x and y, and the volume required is the volume generated by the 
revolution of the two shaded portions (Fig. 47), which will be twice 
that generated by one of these; hence, taking the upper shaded 
portion, we use the positive limits, viz., 3-74 and 8. 

f 8 f 8 / v 2 \ 

Then the volume = I r:x z dy ni ( 32 - }dy 

J 3-74 ' J 3-ll\ J 2 / ' 

f y3\8 

= m say if ) 

6 /3-74 

= 7c(256 85-3-H9-5 + 8-7) 
= 59 '9^. 

The solid due to the revolution of the lower portion will be also 59-97^, 
and hence the total volume generated = 119-871; = 376 cu. units. 




FIG. 47. 

If the limits for y were 8 and +8, the volume of the whole solid 
would be required ; then 

[8 f8/ y2\ r y3-|3 

Volume = / -KX z dy 2:rJ \^ ) ^ ]dy = 2TC [_3 2 )' z~J 

- 27^(256-85-3) 

= 341-471 = 1070 cu. units. 

The solid generated by the revolution of an ellipse about its 
major axis is known as a prolate spheroid; while if the revolution 
is about the minor axis the solid is an oblate spheroid. 

The volumes of these may be necessary, so that they are given 
in a general form. 



200 



MATHEMATICS FOR ENGINEERS 



The general equation of an ellipse is 2 -\-jp = i (cf. 

P- 344)- 

Let a >>, i. e., the major axis is horizontal. 

For a prolate spheroid the revolution is about the major axis, 

/a fa J2 

ny z dx = 2rr/ ~- 2 (a z x z )dx 
o J o a 

27C&V 




FIG. 48. 

In like manner, the volume of an oblate spheroid = - na?b ; 
and it should be noted that if b = a, the spheroid becomes a 

sphere and its volume = ~KO. S . 

3 

Example 15. Find the volume of a zone of a sphere of radius r, 
the bounding planes being those through x = a and x = b. 

The equation of the circle is x z +y z = r 2 (Fig. 48) 
whence y 2 = r z x z . 

:. Volume of a zone = f b ny 2 dx = TT f b (r z x 2 )dx 

J a J a 



= 7r[- 2 (6-a)--(6 3 -a 3 )] 



LENGTH OF ARC 



20 1 



This can be put in the form given on p. 120, Part I, if for (ba) 
we write h, and for BE and DF their respective values r : and r 2 . 
Thus a z = r z -r ] z , b z = r z r z z 

h z = (b-a) z = b z +a z 2ab = r z r 1 z +r z -r 2 z -2ab 

, h z +2r z r 1 z r 2 z 

and hence ab = - 

2 

So that the volume of the zone 






Length of [Arc. Consider a small portion of a curve, PQ in 
Fig. 49, P and Q being points near to one another. 




FIG. 49. 

The small length of arc PQ is denoted by 8s, so that a complete 
arc would be denoted by s. 
Let PM = 8*, and QM = Sy. 
Then the arc PQ = the chord PQ very nearly, so that we may 

say 

(Ss) 2 = (chord PQ) 2 = (S#) 2 +(8y) 2 . 
/Ss\ 2 

y =i 



* > v~ = 

Hence when 8x becomes infinitely small, , - becomes ,-, etc. 

J 8x dx 



and 



ds I 

J- = A/ 
dx \ 



dy 



ds 
or - = 



dx 



202 MATHEMATICS FOR ENGINEERS 

Integrating 

dz 

dx r 



The length of arc can thus be found if the value of the integral 
on the R.H.S. can be evaluated. 

In only a few cases is the evaluation of the integral simple; 
and for most curves an approximation is taken, e. g., to find the 
perimeter of an ellipse by this method one would become involved 
in a most difficult integral known as an elliptic integral, this being 
treated later in the chapter; and hence the approximate rules are 
nearly always used in practice. 

To deal with a case of a very simple character : 

Example 16. If y ax-j-b, find the length of arc between x m 
and x = n. 




In this case it is really a matter of determining the length of the 
line AB (Fig. 50) ; the slope of the line being a. 

Then 



if y = ax+b, = a and i + = i+ 2 . 



/m . __ 
Vi+a 2 dx = 
n 



= Vi + a 2 (m ri). 

On reference to the figure it will be seen that this is a true result, 
since AB = \/(AC) 2 +(CB) 2 = V(m-w) 2 +a 2 (w-w) 2 

= Vi+a 2 (m n). 

Example 17. Use this method to determine the approximate 
length of a cable hanging in a parabola, when the droop is D and 
the span is 2L. 

For convenience, put the figure in the form of Fig. 51. 
Then L 2 = 4D 

L 2 
whence a = -~, so that a must be very large. 



LENGTH OF ARC 203 

The equation of the curve is in reality 

y being written in place of L, and x in place of D. 



men 


dx 


d 


= -5 A ~r~ 

x ay dx 




dy 




JC 


so that 
or 


-/- 2V 

dx 

dy 
dx 


= 4<z 

AO, 2a 
~ zy ~ y u 


^L^ 


whence 


dx 


- Z. 


T 




dy 


20,' 


i 


Thus 






FIG. 51. 



V 

= 1 + 5-2 approximately, 

O# 

since all the subsequent terms contain a 4 and higher powers of a in 
the denominator, so that all these terms must be very small. 



Hence s = 2 1 + . dy 



- 



. 
3 Span 



Example 18. Find the length of the circumference of a circle of 
radius r. 

The equation of the circle is y 2 +# 2 = y 2 . 
Thus y 2 = r z -x 2 

and ay 4 -^- = a# {differentiating with regard to x} 

ctx 

or ^ = -- - ^ 



204 MATHEMATICS FOR ENGINEERS 
Hence i + (;, ) = i+~i s= - 



Length of circumference 



4 X length of J circumference 



f a r 

= 4 J o "V^F 2 ** 



To evaluate this integral, let x = r sin u (cf. Example 7, p. 150). 
Then r 2 x 2 = r 2 r 2 sin 2 u = r 2 (i sin 2 ) 



and also 



= r* cos* u 

dx 

= r cos u 
du 

dx = r cos u . du. 



y c 




FIG. 52. 
To find the limits to be assigned to u 

sin u = - ; and when x = o, u = o 
r 



and when x = i, u = 90 or 



Thus the circumference 



ra dx 
= * r J oVi*=x 



dx 
r cos u du 



o r cos 

7T 
2 





Example 19. Find an expression for the length of the perimeter 
of the ellipse whose major and minor axes are 2a and 26 respectively. 



LENGTH OF ARC 205 

Let BPA (Fig. 52) be the ellipse, CQA the quadrant of a circle 
on the major axis as diameter, and BTD the quadrant of a circle on 
the minor axis as diameter. Selecting P as any point on the ellipse, 
draw the lines QPN, PT, TS, QR and OTQ as shown. 

The point P has the co-ordinates x and y, viz., ON and PN, which 
are respectively equal to QR and TS. 

Now QR = OQ sin <j> or x a sin <f> 

and TS = OT cos <p or y = b cos <p. 

Thus = a cos* and ~- = 



or 5 = Va 2 cos 2 <f>+& 2 sin 2 <f> d<f>. 

x* v 2 
Now the equation to the ellipse is ~^+ hz = i, and the eccentricity, 

d 

which we shall denote by K, is given by 

T , distance between foci A/a 2 b 2 OF . 

K - . ---- . - = ~--i-, F being a focus. 
major axis a OA 

Hence K 2 = ^- and K 2 a 2 -a 2 = -6 2 

or Z> 2 = a 2 (i-K 2 ) 
so that a*cos a f +Msin 8 f = a^os^+a^in 2 ^ a 2 K 2 sin 2 f 

= a 2 (i-K 2 sin 2 $>). 
Thus our integral reduces to the form 



and for the quarter of the ellipse the perimeter = I a Vl K^in 2 ^ d$, 

J o 

since the limits for $ are obviously o and . 

it 

p _ 

Also the full perimeter of the ellipse = 4 I a vi K^in 2 ^ d<t>. 

Jo 

This integral, called an " elliptic integral of the second kind," is 
extremely difficult to evaluate; but in view of the importance of the 
perimeter of the ellipse it is well that we should carry the work a 
little further. 

Knowing the values of a and K for any particular ellipse, recourse 
may be made to tables of values of elliptic integrals, but if these are 
not available, a graphic method presents itself which is not at all 
difficult to use. According to this plan, various values of $ are 

jT . _ - 

chosen between o and , and the calculated values of vi K 2 sin 2 <?> 

are plotted as ordinates to a base of <f>. Then the area under the 
resulting curve when multiplied by ^a gives the perimeter of the 
ellipse. 



206 



MATHEMATICS FOR ENGINEERS 



o. 



ees 



ur 



1-6 



1-5 



1-4. 



1-3. 



1-2 



1-0 







FIG. 53.: Perimeter of Ellipse. 








^ 






























^ 


X s 














= 










\ 


\ 























\ 


\ 























\ 













P 


^ri meter or Ellipse 
2 x major axis x orainct 


\ 






1 




c 


t* ' 


1-294 
\ 




= 




\ 








i 


















\ 


\ 























\ 


= 
















1 
i 




\ 



















i 




A 




II 


MM 


MM 


II 


III 


II 




1 


II 




/ -2 3 -4- -S -6 -7 -8 -9 / 

e.cce.n/~ric.tfuK 



FIG. 54. 



LENGTH OF ARC 



207 



Example 20. A barrier before a ticket office in a works was 
constructed out of sheet metal, which was bent to the form of an 
ellipse of major axis 21 ins. and minor axis 5 ins. Find the area of 
sheet metal required if the height of the barrier is 5 ft. 



T j-u- j tr VlO'5 2 C 

In this case a = 10-5 and K = 

10-5 

so that K 2 = -944. 

The table for the plotting reads 



= '9714 













4> 


sin <f> 


sin 1 <p 


l-Ksin* 


Vl-K*sin^ 


o 


O 





I 


I 


157 


1564 


0245 


i --0232 = -977 


99 


314 


309 


0951 


0898 = -91 


955 


471 


'454 


206 


195 = -805 


897 


628 


5878 


345 


-326 = -674 


821 


785 


7071 


5 


472 = -528 


726 


942 


809 


652 


-616 = -384 


62 


1-099 


891 


793 


749 = -251 


501 


1-256 


9511 


9 


-85 = -15 


388 


i-4 J 3 


9877 


-976 


922 = -078 


279 


1-570 


i 


I 


944 = -056 


235 



and the values in the extreme columns are plotted in Fig. 53. 
The area under this curve = 1-0663 SC L- um t 

and thus the perimeter = 40 x i 4 X 10-5 x 1-0663 = 44-78 ins. 



Hence the area required = 



12 



ft 

^ 



It is well to compare this value of the perimeter with those 
obtained by the approximate rules given in Part I, p. 105 

(a) Perimeter TU(+&) = 71(10-5+2-5) = 40-7 ins. 

(b) Perimeter = 4-443 V 2 +6 2 = 4-443x10-8 = 47-8 ins. 

(c) Perimeter = 7u{i-5(a+&) Vab} = TTX 14-38 = 45-1 ins. 
and the perimeter, correct to two places of decimals, is given in 
the tables * as 44-79 ins. 

* The tables of complete elliptic integrals give the values of 



Vi K^sin 2 ^ d<p for various values of 0, being the angle whose 

sine is K, the eccentricity of the ellipse. Thus to use the tables for 
this particular case we put sin K = -9714, whence = 76 16'; 
we then read the values of the integral for 75, 76 and 77, and by 
plotting these values and interpolation we find that for the required 



208 MATHEMATICS FOR ENGINEERS 

Thus the errors in the results found by the different rules are 

(a) 9-13 % too small (b) 6-69 % too large (c) -67 % too large 
showing that the rule of Boussinesq gives an extremely good result 
in this case of a very flat ellipse, whilst the other approximate 
methods are practically worthless. 

Area of Surface of a Solid of Revolution. When a solid 
of revolution is generated, the boundary of the revolving figure 
sweeps out the surface of that solid. The volume of the solid 
depends upon the area of the revolving figure, whilst the surface 
depends upon the perimeter of the revolving figure. 

To find the surface generated by the revolution of the curve CD 
about OX (Fig. 55) we must find the sum of the surfaces swept out 
by small portions of the curve, such as PQ. Let PQ = a small 
element of arc = 8s. Then the outside surface of the solid generated 
by the revolution of the strip PQMN about OX will be equal to 
the circumference of the base X slant height, i.e., 2ny8s. Hence 
the total surface will be the sum of all similar elements, i. e., 

\r\x = b 

> 2Tcy8s, approximately, or if 8x becomes smaller and smaller 

^L_ J iC tt 

fx = b 

Surface = I 2nyds. 

J x = a * 

For ds we may substitute its value, viz. 



/b 
2ny 
(J 



2nyJ !--} .dx. 



angle, viz., 1-0664. Multiplication by 40, i.e., 42, gives the result 
44-79. For the convenience of readers interested in this question, 
and who desire a result more exact than that given by the approximate 

r 
fi _ 

rules, a curve is here given (Fig. 54) with values of I v i K^in 2 ^ d$ 

J o 

plotted against values of K; and for the full perimeter of the ellipse 
the ordinates of this curve must be multiplied by twice the length 
of the major axis. 

E. g., if the major axis = 16 and the minor axis = 10 



K = - g-^ = -7807. 

Erecting an ordinate at K = -7807 to meet the curve, we read 
the value 1-294; multiplying this by 32, we arrive at the figure 41-41, 
which is thus the required perimeter. 



AREA OF SURFACE 



209 



Example 21. Find the area of the surface of a lune of a sphere 
of radius a, the thickness or height of the lune being b. 

The surface will be that generated by the revolution of the arc CD 
of the circle about its diameter OX (Fig. 56). 
From the figure y 2 = a 2 x 2 

whence 2V.-r 2X 

dx 

dy x 

or - = . 

dx y 

_ fds\ 2 ,(dy\ 2 . x 2 y 2 +x 2 a 2 

Thus (j- ) =i+l/) = i-f - - = ZZy- = 

\dxJ \dx' y 2 y 2 a 2 x 2 




fd a 

Hence the surface = I 27uVa 2 x 2 - 

J c Va 2 - 




p 

- 2Tiaj 



dx 



= 2r:a(dc) = 

but 2TOZ& is the area of a portion of the lateral surface of the cylinder 
circumscribing the sphere. 

Thus the surface of a lune of a sphere = the lateral surface of the 
portion of the cylinder circumscribing the sphere (the heights being 
the same). 

Exercises 18. On Volumes, Areas of Surfaces and Length of Arc. 
1. The cross sections at various points along a cutting are as follows 



Distance from one end (ft.) 


o 


40 


82 


103 


134 


1 66 


192 


200 


Area of cross section (sq. ft.) 


o 


210 


296 


205 


244 


154 


50 


o 



Find the volume of earth removed in making the cutting. 



2IO 



MATHEMATICS FOR ENGINEERS 



2. Find the weight of the stone pillar shown in Fig. 57. The 
flanges are cylindrical, whilst the radius of the body at any section 

2 

is determined by the rule, radius = ;=, where x is the distance of 

vx 

the section from the fixed point O. (Weight of stone = 140 Ibs. per 
cu. ft.) 

3. The curve y = 2x 2 $x revolves about the axis of x. Find 
the volume of the solid thus generated, the bounding planes being 
those for which x 2 and x = +4. 

4. Find, by integration, the surface of a hemisphere of radius r. 

5. The curve y = ae bx passes through the points x i, y = 3-5, 
and x = 10, y = 12-6; find a and b. This curve rotates about the 
axis of x, describing a surface of revolution. Find the volume between 
the cross sections at x i and x 10. 

6. Find the weight of a cylinder of length / and diameter D, the 
density of the material varying as the distance from the base. (Let 
the density of a layer distant x from base = K#.) 



^r-o 1 




FIG. 57. Weight of Stone Pillar. 

7. The rectangular hyperbola having the equation x 2 y z = 25 
revolves about the axis of x. Find the volume of a segment of 
height 5 measured from the vertex. 

8. The line 4y 5* 12 revolves about the axis of x. Find the 
surface of the frustum of the cone thus generated, the limits of x 
being i and 5. 

9. The radius of a spindle weight at various points along its length 
is given in the table 



Distance from one end (ins.) 


o 


375 


5 


i-o 


i'3 


1-6 


1-85 




1-61 


1-61 


78 


42 


4 





c 



















Find its weight at -283 Ib. per cu. in., the end portions being 
cylindrical. 

10. Determine by the method indicated in Example 19, p. 204, 
the perimeter of an ellipse whose major axis is 30 ins. and whose 
minor axis is 18 ins. Compare your result with those obtained by 
the use of the approximate rules (a), (b) and (c) on p. 207. 

11. The curve taken by a freely hanging cable weighing 3 Ibs. 
per foot and strained by a horizontal pull of 300 Ibs. weight conforms 
to the equation 

, x 
y = c cosh 

300 
where c = . 

3 

Find the total length of the cable if the span is 60 ft., i. e.. x 
ranges from 30 to -{-30. 




B 



CENTROIDS 211 

Centre of Gravity and Centroid. The Centre of Gravity 
(C. of G.) of a body is that point at which the resultant of all the 
forces acting on the body may be supposed to act, '. e., it is the 
balancing point. The term Centroid has been applied in place of 
C. of G. when dealing with areas ; and as our work here is more con- 
cerned with areas it will be convenient to adopt the term centroid. 

From the definition it will be seen that the whole weight of 
a body may be supposed to act at its C. of G. ; and in problems in 
Mechanics this property is most useful. Thus, movements of a 
complex system of weights may be reduced to the movement of 
the C. of G. of these. Or to take another instance : in structural 
work, in connection with fixed beams unsymmetrically loaded, it 
is necessary to find the position of the centroid of the bending- 
moment diagram. It is thus 
extremely important that rules 
should be found for fixation of 
the position of the centroid in ' 
all cases ; and the methods Jt J JL jt & 

t i , ,..,,. Tn t ff/2 "?3 III A. iffS 

adopted may be divided into 

two classes : (a) algebraic (in- " IG " 5 8. -Centre of Gravity or Centroid. 
eluding purely algebraic, and partly algebraic and partly graphic), 
(b) graphic. 

The rules will best be approached by way of a simple example 
on moments. In place of areas or solids, afterwards to be dealt with, 
let us consider the case of a uniform bar loaded as shown in Fig. 58. 

For equilibrium the two conditions to be satisfied are 

(1) The upward forces balance the downward forces. 

(2) The right-hand moments about any point balance the left- 
hand moments about the same point; or, in other words, the 
algebraic sum of the moments about any point is zero. 

If C is the balancing point or fulcrum, the upward reaction of 
the fulcrum = M = fni-^-m 2 -}-m 3 -{-m t -\-m 5 from condition (i). 
Taking moments about A, let x (x bar) be the distance AC. 
Then, by condition (2) 



or 



The product of a force into its distance from a fixed point or 
axis is called its first moment about that point or axis ; whilst the 



212 



MATHEMATICS FOR ENGINEERS 



product of a force into the square of its distance from a fixed 
point is called its second moment about that point. 

Hence our statement concerning the distance AC can be 
written 

1st moments 



x = 



masses 



To extend this rule to meet the case of a number of scattered 
masses arranged as in Fig. 59, the co-ordinates of the centroid 
must be found, viz., x and y. 




Thus 



FIG. 59. Centroid. 
2mx 2 1st moments about OY 






and 



2m 2 masses 

_ ^ m y _ ^ 1 s * moments about OX 
^ "" 2m 2 masses 



If the masses are not all in one plane, their C. of G. must be 
found by marking their positions in a plan and elevation drawing 
and determining the C. of G. of the elevations and also that 
of their plan. Thus the C. of G. is located by its plan and 
elevation. 

It will be observed that some form of summation is necessary 
for the determination of the positions of centroids or centres of 
gravity; and this summation may be called by a different name, 
viz., integration, all the rules of which may be utilised; the 
integration in some cases being graphic, in some cases algebraic, 
and in others a combination of the two. 



CENTROIDS 



213 



Rules for the Determination of the Centroid of an 
Area. Let it be required to find the centroid of the area ABCD 
in Fig. 60. 

The area may be considered to be composed of an infinite 
number of small elements or masses, each being the mass of some 
thin strip like PQMN ; the co-ordinates of the centre of gravity of 
which may be determined in the manner already explained. 




._.b 



N 



M 



B 



FIG. 60. Centroid of an Area. 



To find x, i. e., the distance of the centroid from OY 
Mass of strip PQMN = area x density (considering the strip as of 

unit thickness) 
= y8xxp 
ist moment of strip about OY = mass X distance = pySx x x 

= pxySx. 
ist moments about OY 



Hence 



x = 



masses 



= _/ %. the limits being a and b 
_5 pyox 

and if the strips are made extremely narrow 

/& fb 

pxydx I xydx 
a J a 



X = 



/pydx I ydx 
a ' J a 

p cancelling from both numerator and denominator. 



214 MATHEMATICS FOR ENGINEERS 

Thus a vertical is found on which the centroid of the area 
must lie; and this line is known as the centroid vertical. 

To fix the actual position of the centroid some other line must 
be drawn, say a horizontal line, the intersection of which with 
the centroid vertical is the centroid. 

Thus the height of the centroid above OX must be found; 
this being denoted by y. 

To find y. The whole mass of the strip PQMN may be supposed 
to act at R, its mid-point, because the strip is of uniform density ; 
and hence the moment of the strip PQMN about OX 

y 

= mass X distance = py 8xx- 

2 



2 b ist moments about OX 
Hence y = - , 

2, masses 



b 2 f b 

py dx I y dx 

a. J a 



As in previous cases, the integration may be algebraic, this 
being so when y is stated in terms of x, or graphic, when a curve 
or values of y and x are given. 

Suppose the latter is the case, and we desire to find x 

T~ fxydx 

Then x = J T Z T - 

Jydx 

and the values of the numerator and denominator must be found 
separately. Each of these gives the area of a figure, for if Y is 
written in place of xy, the numerator becomes fYdx, which is 
the standard expression for the area under the curve in which Y 
is plotted against x; and the denominator is already in the 
required form. 

Thus a new set of values must be calculated, viz., those of Y, 
these being obtained by multiplication together of corresponding 
values of x and y; and these values of Y are plotted to a base 
of x. Then the area under the curve so obtained is the value of 
the numerator, and the denominator is the area under the curve 
with y plotted against x; and, finally, division of the one by the 
other fixes the value of x. 



CENTROIDS 



215 



Example 22. Find the centroid of the area bounded by the curve 
given by the table, the axis of x and the ordinates through x 10 
and x = 60. 



X 


IO 


25 


40 


45 


5 


60 


y 


4 


5'3 


6-2 


6-4 


6-6 


6-8 



We thus wish to find the centroid of the area ABCD (Fig. 
To find x : 



xy 



30Q 



toa 




C e ntroid Horn 



ntal. 



' 



JS 



JG D 



ZO 3O 4O 

FIG. 61. Centroid of an Area. 



The table for the plotting of Y against x reads 



X 


IO 


25 


40 


45 


50 


. 
60 


Y or xy 


40 


132-4 


248 


288 


330 


408 



From this we get the curve AEF. 
The area of the figure ABCD 

/GO 
ydx = 289 
10 

and the area of the figure AEFD 

/GO 
xydx = 10650 
10 



The method 
of integration 
is not shown, 
to avoid con- 
fusion of 
curves. 



/GO 
xydx 
10 
r60 

/ ydx 
J 10 



10650 

289 



= 36-9. 



2l6 



MATHEMATICS FOR ENGINEERS 



Thus the centroid vertical, or the line PG is fixed. 

We need now to find the centroid horizontal, i. e., y must be 

determined. 

i r 60 

= 2 J 10 Y ^ /where Y in this case\ 
areaofABCD \ stands for y 2 J 



Now 



y = 



f rf 
J 10 

so that the following table must be compiled 




X 


10 


25 


40 


45 


5 


60 


Y or y 2 . 


16 


28 


38-4 


4i 


43-5 


46-1 




FIG. 62. C. of G. of Thin Plate. 

Plotting from this table, the curve RQ results, and the area of 
the figure ARQD is 1689. 

_ I area of ARQD &xi68Q 
y " : areaofABCD : 289 ~ ^22i 

The intersection of the centroid vertical and the centroid horizontal 
at G fixes the centroid of ABCD. 

A modification of this method is necessary when the actual 
area is given in place of the tabulated list of values, the procedure 
being outlined in the following example. 

Example 23. It is required to find the C. of G. of a thin plate 
having the shape shown in Fig. 62. Show how this may be done. 

Draw two convenient axes at right angles and divide up the area 
into thin strips by lines drawn parallel to OY. Draw in, also, the 
mid-ordinates of these strips. The area of any strip can be assumed 



CENTROIDS 



217 



to be "mean height X thickness "; and therefore measure ordinates 
such as MN and multiply by the thickness or width of the strip. 
Repeat for each strip, and the sum of all these will be the area of the 
figure. 

To find x. OA X = the distance of the centre of ist strip from OY 
so that the area of strip X OA V = ist moment of strip about OY. 

Hence, multiply the area of each strip by the distance of its mid- 
ordinate from OY and add the results ; then the sum will be the 
ist moment of the area about OY. 

T,, - Sum of ist moments 2nd total 

Then x = r = . . . .. 

Area ist total 

To find y. Fix R, the mid-point of MN, and do the same for 
all the strips. The area of the strip has already been found ; multiply 
this by AjR and repeat for all strips. The sum of all such will be 
the ist moment about OX; dividing this by the area of the figure, 
the distance, y, of the centroid from OX is found. 

[Note that R is the mid-point of MN and not of NA X , because 
OX is a purely arbitrary axis.] 

For this example the calculation would be set out thus 



Strip 


Length of 
mid-ordinate 
(like MN) 


Width 
of 
Strip 


Area 
of 
Strip 


Distance of 
centre from 
OY 
(like OAj) 


Distance of 
centre from 
OX 
(like RAi) 


ist 
moment 
about 
OX 


ist 
moment 
about 
OY 


I 


i-55 


5 


775 


25 


2-0 


1-55 


19 


2 


2-79 


5 


1-395 


75 


2-O 


2-79 


1-05 


3 


3'44 


5 


1-720 


1-25 


2-18 


3-75 


2-05 


4 


3-85 


5 


1-925 


i-75 


2-27 


4*37 


3-36 


5 


4-01 


5 


2-005 


2-25 


2-32 


4-64 


4-50 


6 


3-92 


5 


1-960 


2-75 


2-3 


4-5i 


5-40 


7 


3-60 


5 


i -800 


3-25 


2-18 


3-92 


5-85 


8 


3-26 


5 


1-630 


3-75 


2-04 


3-32 


6-II 


9 


2-66 


5 


i-33o 


4-25 


2'*O2 


2-68 


5-65 


10 


1-47 


5 


735 


4-75 


1-95 


i-43 


3*49 


Totals 


I5-275 




32-96 


37-65 



and 



15- 



= 6 



32-96 

y = - - 2-16. 
15-28 



Thus the position of G is fixed by the intersection of a horizontal 
at a height of 2-16 with a vertical 2-46 units distance from OY. 

If the centroid of an arc was required, the lengths of small 
elements of arc would be dealt with in place of the small areas, 
but otherwise the procedure would be the same. 



2l8 



MATHEMATICS FOR ENGINEERS 



"Double Sum Curve " Method of Finding the Centroid 
Vertical. This method is convenient when only the centroid 
vertical is required ; for although entirely graphic, it is rather too 
long to be used for fixing the centroid definitely. 

Method of Procedure. To find the centroid vertical for the area 
APQH (Fig. 63). 

Sum curve the curve PQ in the ordinary way, thus obtaining 
the curve AegE; for this construction the pole is at O, and the 
polar distance is p. 

Produce PA to O l7 making the polar distance p t = HE = last 
ordinate of the sum curve of the original curve (viz., PQ). 

Sum curve the curve AegE from AP as base and with O t as 




FIG. 63. Centroid i Vertical~of~an Area. 

pole ; then the last ordinate of this curve, viz., CM, is of length x, 
so that the vertical through C is the centroid vertical. 

Proof. Consider the strip abed, a portion of the original area. 

Then Or and eg are parallel (by construction) 

P ef ab 
-*- = - / = -f 
Ar fg fg 



and thus 



or 

i.e., hnxab = 

2hn xab = 2p xfg = pSfg = p . HE. 
Again, the ist moment of the strip about AP = area X distance 

= hn x ab X Ah 
m = hnxabxml 
= pXfgXml 



CENTROIDS 



219 



and hence ist moment of area APQH about AP 

= plfgxtnl 



but ist moment of area APQH about AP 

= area x distance of centroid from A 



and 



x = MC. 




fee/. 
FIG. 64. Problem on Loaded Beam. 

Example 24. A beam, 16 ft. long, simply supported at its ends 
is loaded with a continuously varying load, the loading being as 
expressed in the table. 



Distance from left-hand) 
support (feet) / 


o 


2 


4 


6 


8 


10 


12 


14 


16 


Load in tons per foot run 


12 


17 


21 


25 


28 


29 


3* 


34 


38 



Find the centroid vertical of the load curve, and hence determine 
the reactions of the supports and the point at which the maximum 
bending moment occurs. 

We first plot the load curve from the figures given in the table 
(Fig. 64) ; and next we sum curve this curve, taking a polar distance 
of 10 horizontal units ; the last ordinate of this sum curve reads 4-27, 
so that the total load is 4-27 tons. We now set off AD equal in 
length to BC, and with this as polar distance we sum curve the curve 
AEC fiom the vertical axis as base. This sum curve finishes at the 
point G on the horizontal through C, and a vertical through G is the 
centroid vertical, distant 9-2 ft. from the end A. 

For purposes of calculation, the whole load may be supposed to 



220 MATHEMATICS FOR ENGINEERS 

act in this line; the total load is 4-27 tons, and taking moments 

round A 

4-27X9-2 = R B x 16 
whence R B = 2-46 tons 

and R A = 4-272-46 = 1-81 tons. 

We now set up AH, a distance to represent R A , to the new vertical 
scale, and then a horizontal through H is the true base line of shear. 

At the point P the shear is zero; but the shear is measured by 
the rate of change of bending moment, so that zero shear corresponds 
to maximum bending moment; and hence, grouping our results 
Reaction at left-hand support = i 81 tons 
Reaction at right-hand support = 2-46 tons 

and the maximum bending moment occurs at a distance of 8-4 ft. 
from the left-hand end. 

Centroids of Sections by Calculation (for a graphic method 
especially applicable to these, see p. 251). Special cases arise in 



N, r 




L \( 


r IfcHJ |S| I 


L 






N- 






-NT 


N 2 


* 


I 






T 


T 






i. 







*I^N- 
FlG. 65. 

the form of sections of beams, joists, rails, etc., for which a 
modification of the previous methods is sufficient. 

If the section is composed of a combination of simple figures, 
such as rectangles or circles, as in the great majority of cases it is, 
its centroid can be found by loading each of its portions, into 
which for purposes of calculation it may be divided, with a weight 
proportional to its area, and treating the question as one for the 
determination of the C. of G. of a number of isolated weights. 

Example 25. Find the position of the centroid of the Tee section 
shown in Fig. 65. 

We may consider the section to be made up of two rectangles; 
then 

f a 5 SO 24O 

Area of flange = 6 X = Q sq. ins. = -r sq. ins. 

o o ^4 

and the centroid of the flange is at Gj. 



CENTROIDS 



221 



Area of web = 3$ x ~ = -~ sq. ins. 

o 04 

and the centroid of the web is at G 2 . 

From considerations of symmetry we see that the centroid of the 
section must lie on the line G^G^, at the point G, say. 



(of length > + , i. e., 2") as a bar loaded with -~- 

I O ID 04 



Treat G 

joe 

units at Gj and ~ units at G 2 . 



Let GjG = x', then the upward force at G = 7^ + ~^ 

= =^ units. 
64 




,,G, 



B 




847 

344-4'5l6 
Q G> I G 



I __ i J\ r\f\ /" 

t 11 ^ -* I 



'22 



FIG. 66. Centroid of Bridge Rail. 
(In the further calculation we may disregard the denominators, since 



they are alike.) 

Taking moments about 



whence 



375 X^ = 135X2 

x = ^^ = -72. 
375 ' 



Hence the distance of the centroid from the outside of the flange 

' 



Example 26. Determine the position of the centroid of the bridge 
rail section shown at (a), Fig. 66. 

This example presents rather more difficulty than the one imme- 
diately preceding it. The plan of procedure is, for cases such as this, 



222 MATHEMATICS FOR ENGINEERS 

that adopted in the work on the calculation of weights, viz., we first 
treat the section as " solid " and then subtract the part cut away. 

Neglecting the small radii at the corners, and treating the section 
as " solid," the section has the form shown at (b), Fig. 66. 

2 3 *7 

The area of AB = -i-X- = 2-52 sq. ins., and its centroid is at G 2 , 
10 4 

the intersection of its diagonals 

3 tj 

Similarly the area of CD = |x- = 1-313 sq. ins., and its centroid 

is at Gj. 

For the part cut away (see (c), Fig. 66) 

The area of EHM = -Xy!) = -221 sq. in.; and we know from 

Part I, p. 130, that its centroid G 8 is distant -424 X radius, i.e., 
424 X -375 or *I59* from EM. 

Again, the area of EF = >X- = -516 sq. in., and its centroid is 
I o 4 

at G 4 . 

Our problem is thus reduced to that of determining the C. of G. 
of four isolated weights, two of which act in the direction opposed to 
that of the others, placed as shown at (d), Fig. 66. 

Let the centroid of the whole section be at G, distant x from O. 
Now the upward forces = the downward forces 
and thus R G +'5i6+-22i = 2-52 + 1-313 

whence R G = 3-096. 

Also, by taking moments about O 
(3-096 X *) + (-516 X -344) + (-221 X -847) = (1-313 X-i88) + (2-52X1-094) 

whence x = -855 in. 

or the centroid of the section is '855" distant from the outside of the 
flange. 

Centroids found by Algebraic Integration. Suppose that 
the equation of the bounding curve is given, then the centroid of 
the area between the curve, the axis and the bounding ordinates 
may be determined by algebraic integration. 

We have already seen that 

Ixydx ~ly z dx 

x = -. and y = ^, 

lydx lydx 

so that if y is stated as a function of x, xy and y 2 may be expressed 
in terms of x, and the integration performed according to the 
rules given. 

The examples here given should be carefully studied, for there 



CENTROIDS 



223 



are many possibilities of error arising due to the incorrect 
substitution of limits. 

Example 27. Find the centroid of the area between the curve 
y = 2X 1 ' 5 . the axis of x and the ordinates through x = 2 and x = 5. 

The curve is plotted in Fig. 67, and it is seen that the position of 
the centroid o^E the area ABCD is required. 

Now y= 2x 1 ' 5 , and thus xy = 2x*xx = 2x* 

and 2 = x 3 . 



To find x 

f 5 
1 xy dx 
-f J * 


\\d X y 

20 


c / 

/ CenTpoid 
/ Vcrhcal 

'/'***' 


I 
L 


f'ydx 

or the centroid ve 
i -8 1 units from 
boundary. 
To find y 


f 2X*dX 

\7* ) t 15 


(F): 

? X 5| 5 *_ 2 A ,0 


{5*-,*} 

5^68 5 


JS / ^Cenrnoiol 
Ccn^poid 
Hopizonrai 

A | | | | B 


7 50-25 

rtical is distant 


the left-hand 

i T 5 
- 1 y*dx 

2 J 2 ' 


23 4 5 J2 
FIG. 67. 


y , 5 

/ y dx 

-X4 / ^ 3 ^ 
2 ^ 7 2 


/5 3 /2 6\5 
^-^ (V)^ 

i,,5 (5 4 ~2) 




5 6oQ -, 



8^50-25 

Hence the co-ordinates of the centroid are 3-81, 7-57. 

Example 28. The bending moment curve for a beam fixed at one 
end and loaded uniformly over its whole length is a parabola, as 



224 MATHEMATICS FOR ENGINEERS 

shown in Fig. 68. The vertex is at A and the ordinate at B, viz., 
BC is ; the loading being w units per foot and I being the span. 

We wish to determine the position of the centroid of the figure 
ABC so that we may find the moment of the area ABC about AD, 
and finally the deflection at A. 

From the equation to a parabola, y 2 = ^ax, we see that 

Wl 2 , 22 

l z = 4 . , whence 40, = or y 2 = x 
^2 w w 



i. e., (ND) 2 = - AD. 
w 

The distance of the centroid from AD = y 




r xydy 



j x&y 



D 



f l 
/ 



FIG. 68. 



Area of ABCD = - of surrounding rectangle = X/X 

_ wl 3 

All this area may be supposed to be concentrated at its centroid, 

and hence the moment of ABC about AD = ^ x - I - c 

04 o 

Now the deflection at A = ^ x moment of the bending moment 
diagram about the vertical through A 



i wl* 
~ El X IP 



Hence the deflection at A = 



^^ 

oJil 



W/ 3 
= 3^=^, where W = total load. 



Example 29. Find the position of the centroid of a quadrant of 
a circle of radius Y. 



The equation of the circle is x z -\-y 2 = r 2 
hence y 



= 2 * 



so that 



xy = xVr z x 2 . 



CENTRE OF GRAVITY 225 

Thus x (and consequently y) 




*-x 2 dx 



Try 2 
The value of the denominator is , for it is the area of the 

4 
quadrant. (This integral would be evaluated as shown on p. 149.) 

To evaluate the numerator, let u = r z x z 
then du -zxdx 

du 

or xdx = . 

2 



X * 

fr r du i 

Then I x Vr 2 -x* dx = I ~ "5 



= ~[o-(+' 2 )*] 
i , 

i/J 

f 

3 
i 

3 4 r 

y = Z = X Qr -424^. 



4 

Centre of Gravity of Irregular Solids. The methods 
given for the determination of the centroids of irregular areas 
apply equally well when solids are concerned. For if A is the 
area of the cross section of a solid at any point along its length, 
distant x, say, from one end, and the length is increased by a 
small amount 8x (and if this is small there will be no appreciable 
change in the value of A), then the increase in the volume == AS* 
or the increase in the weight = pA8x, p being the density. 

The moment of this element about the end = pAS* x x 



so that x 



f l 
ist moments _ J Q 



2 weights 



/i 
pAdx 



Axdx 



226 



MATHEMATICS FOR ENGINEERS 



As before, two cases arise, viz., (a) when values of A and x are 

given, and (b) when A is denned in terms of x. To deal with these 

In case (a) plot one curve in which A is the ordinate and x is 

n 

the abscissa and find the area under it ; this is the value of I A.dx. 

Jo 

Plot a second curve whose ordinates are the products of 
corresponding values of A and x and find the area; this is the 
value of the numerator, and division of the latter area by the 
former gives the value of r. Thus the centroid vertical is found, 
and if the solid is symmetrical about the axis of x, this is all that 
is required; otherwise the centroid horizontal must be found, the 
procedure being exactly that previously described when dealing 
with areas in place of volumes. 

An example on the application of this method is here worked. 



A 
6 

5 

4 
5 
2 

1 



< 


\ 


\ 










c 




.A * 
^* 














AJC 

4 
20 
16 
)2 
8 
4 
O 






\ 


X 


x * 


-*' 










~~^~ 


s. 


















\ 


. 
















^> 


^ 
















^ 


^ 




i 












^^ 




/ 














< 


i 




^-~, 












y 






















^^- 


- 


~-~^ 


*^-^_ 


F 


9 




























JG 


C 


^ 4 6-8 1O 1 14 1( 



FIG. 69. Problem on C.I. Column. 

Example 30. The circumference of a tapering cast iron column, 
16 ft. long, at 5 equidistant places is 9-43, 7-92, 6-15, 4-74 and 3-16 ft. 
respectively. Find its volume and the distance of its C. of G. from 
the larger end. 

The areas must first be found from the circumferences. 
Now the area of a circle = 



47U 



So that the table for plotting reads 



x = distance from larger end (ft.) 


o 


4 


8 


12 


16 


A = area of cross section (sq. ft.) 


7-09 


4-98 


3-o 


1-78 


79 



By plotting these values the curve EF (Fig. 69) is obtained. 



CENTRE OF GRAVITY 



227 



The figure here given is a reproduction of the original drawing to 
rather less than half-size, and since the measurements were made on 
the original, the results now stated refer to that. 

In the original drawing the scales were : i* vertically = 2 sq. ft., 
and i* horizontally = 2 ft., so that i sq. in. of area represented 
4 cu. ft. of volume. The area under the curve EF was found, by 
means of the planimeter, to be 13-66 sq. ins., and accordingly the 
volume = 13-66x4 = 54-64 cu. ft. 

The curve BCD results from the plotting of values of Ax as 
ordinates, the table for which plotting reads 



X 


o 


4 


8 


12 


16 


Ax 





I9'9 


24 


21-4 


12-6 



The area under this curve was found to be 19-06 sq. ins., which 




FIG. 70. C. of G. of Solid of Revolution. 

represented I9~o6x 16 units of moment, since for the plotting of BCD 
i* vertically = 8 units of Ax, and i* horizontally = 2 units of x. 

area BCDG i6x 19-06 
Hence 



area BEFG 54-64 

For case (b), when A is stated in terms of x, the integration is 
entirely algebraic. Thus if A is a function of x, integrate Ax and 
also A with regard to x, and divide the former integral by the 
latter to determine the value of ~x. 

Example 31. The area of cross section of a rod of uniform density 
varies as the cube root of the distance of the section from one end ; 
find the distance of the C. of G. from that end, being given that the 
area at a distance x from the end = '^ / #. 



228 MATHEMATICS FOR ENGINEERS 

Consider a strip distant x from the stated end and of thickness S.v. 
Then, from hypothesis, the area of section 4- $3/ x, and thus the 
volume = area X thickness = 4 - 5^ / ^xS^. 

Also the mass of the strip = volume X density 



and the moment of the strip about the end mass x distance 



2 ist moments of small elements 
Hence x = - ... 

2 their masses 



= 4* 
7 

or the C. of G. is distant ^ o f the length from the given end. 

C. of G. of a Solid of Revolution. Suppose that the curve 
BC in Fig. 70 rotates round OX as axis ; and we require to find 
the position of the C. of G. of the solid so generated. 

Consider a small strip of area MN; its mean height is y and 
its width is 8x, so that the volume generated by the revolution 
of this is Tiy 2 8x, or the mass = p-n:y 2 8x. The ist moment of this 
strip about OY = mass X distance = p-n:y 2 8x X x = pnxy 2 8x. 



Thus the total ist moment about OY = /, pnxy 2 8x 

*~~^a 

^\b 

and the total mass = > pny 2 8x 



ib fb 

I piixy 2 dx I xy 2 dx 

J_ a J_a 

/* ( b 

pny 2 dx I y 2 dx. 
J a J a 



As before, the two cases arise, viz. 

(a) When values of x and y are given. For this case make a 
table of values of x x y 2 and also one of values of y 2 . 

Plot the values of xy 2 against those of x and find the area under 
the resulting curve 

This area = fxy 2 dx . . . . . . (i) 



CENTRE. OF GRAVITY 

Plot the values of y z against those of x 

Area of figure so obtained = fy z dx 



229 



(2) 



and 



(2)'. 



Also we know that y must be zero, for the axis of x is the axis 
of rotation; and thus the C. of G. is definitely fixed. 

(b) When y is expressed as a function of x. In this case find 
both xy z and also y z in terms of x, integrate these functions 
algebraically and thence evaluate the quotient. 

Example 32. The curve given by the tabulated values of y and x 
revolves about the .ar-axis; find the position of the C. of G. of the 
solid thus generated. 



X 


o 


i 


2 


3 


4 


y 


8 


10 


21 


26-4 


25 



For the first curve, values of xy z are required, and for the second 
curve, values of y 2 ; these values being 



X 


o 


i 


2 


3 


4 


y* 


64 


IOO 


441 


696 


625 


xy 2 


o 


IOO 


882 


2088 


2500 



The curve AB (Fig. 71) is obtained by plotting the values of xy z 
as ordinates; and the area under this curve is 4323; this being thus 

the value of / xy 2 dx. 
J o 

By plotting the values of y- as ordinates the curve CD is obtained ; 

/i 
y z dx = 1699. 
o 

/4 
xy z dx 
_o 

T* 

I y*dx 

J o 
i. e., the C. of G. is at G, the point (2-55, o). 

Example 33. The curve x 5V 2 Vy revolves about the axis of y. 
Find the position of the centre of gravity of the solid generated, the 
solid being bounded at its ends by the horizontal planes distant i and 
5 units respectively from the axis of x. 



4323 
1699 



= 2-55 units 



230 



MATHEMATICS FOR ENGINEERS 



Since the revolution is about the axis of y and not that of x, y must 
take the place of x in our formulae and x the place of y; therefore 
the limits employed must be those for y. 

In Fig. 72 AB is the curve x = $y2Vy, and we see that it is 



2000 




5oo_ 



~S~ 

FIG. 71. 

required to find the height of the centroid above the axis of x of the 
solid generated by the curve AB about the axis of y. 

Then to find y 

(5 

/ yx 2 dy 

J i 

/5 
x 2 dy 
i 

Now x = 5y2Vy, and thus x 2 = 2 

and y# 2 2$y 3 2oy*-\-4y 2 . 

f5 f5 . 4 

Then / y^ 2 ay I (2^y 3 2oy-}-^y 2 )dy = 



y = 



= 2 454 



f 5 j T 5 

and / x 2 dy= 

J i I i 



j T25V 3 , 4V 2 20X2 |"15 

dy = \ ^-~+~ -- ^ y \ 
L 3 * 5 ->i 



= 639 

/: 



2454 
639 



= 3' 



CENTRE OF GRAVITY 



231 



Then since the centroid must lie along the axis of y, its position 
is definitely fixed at the point G, viz., (o, 3-84). 

Example 34. Find the mass and also the position of the C. of G. 
of a bar of uniform cross section a and length I, whose density is 
proportional to the cube of the distance from one end. 

Let us consider a small length 8x of the bar, distant x from the 
end mentioned above; the density of the material here = Kx 3 , where 
K is some constant; hence 




o e 4 6 a o IE H- 16 IB 20 
FIG. 72. 

Mass of small element = volume X density = a8x-X K# 3 = Kax 3 8x. 

Thus the total mass = [ ' Kax 3 dx = Ka f V 
J o v 4 * " 

Ka/ 4 



Also the ist moment of the element about the end 

= mass x distance 
= Kax 3 8xxx. 

Ka/ 5 



Total ist moment 



-7 1 . 



Kax*dx = 



and if x = distance of C. of G. from the lighter end 

Ka/ 5 

5 L 



Example 35. Find the position of the C. of G. of a triangular 
lamina whose density varies as the distance from the apex. (Let the 
thickness of the lamina = /.) 



232 



MATHEMATICS FOR ENGINEERS 



Consider a small strip of width 8x, distant x from the apex 
(Fig- 73)- 

The area of the strip = y>8x, and thus its volume = yt8x. 
Now the density r x or density = Kx 



and also, by similar triangles, 



Bx 



So that the mass of the strip 



H 

ytSx x Kx 
BKt 
H ^ 
and the ist moment of the strip about OY 



jr 
ri 



S.W.S.L. 




r 



/ - 


'\ 
1 \6x? 


1 


A- -y 


_J_^ 


/ i 


; '\ 


rE- 


*J 7,- 


u p 


J 


? f^* 
^U 


D 

FIG. ; 
Hence 


'3- 

_ / 

x - 2 
/o 

(f 


FIG. 74 

T _ x 3 dx 
ti 


BK/ 
)o E H-,3 


(v3\Ji A T~T^ 
x \ 4 rz 
!/ 



Centre of Pressure. If a body is immersed in a liquid, then 
the pressure per sq. in. of surface is not uniform over the solid, 
for the pressure is proportional to the depth. The point at 



CENTRE OF PRESSURE 233 

which the total pressure may be supposed to act is known as the 
centre of pressure (C. of P.). 

To find positions of centres of pressure we are, in effect, finding 
centres of gravity of solids whose density is proportional to the 
distance from some fixed axis. 

The C. of G. found in the example last worked is in reality 
the C. of P. of a triangular lamina immersed vertically in a liquid, 
with OY as the level of the top of the liquid. 

Just as, when discussing the stability of solids in air, we have 
supposed the whole mass to be concentrated at the C. of G., so 
now, when the solid is immersed in a liquid, the total pressure 
may be assumed to act at the one point, viz., the C. of P. 

To find the positions of the C. of P. for various sections and 
solids we must start from first principles, dealing with the pressure 
on small elements, and then summing. 

Example 36. Find the whole pressure on one side of a rectangular 
sluice gate of depth 5 ft. and breadth 3 ft., if the upper edge is 10 ft. 
below the level of the water (which we shall speak of as the still 
water surface level or S.W.S.L). Find also the depth of the centre 
of pressure. 

Consider a strip of the gate Sx deep and x ft. below S.W.S.L. 

(Fig. 74)- 

Then the area of the strip =3x8* 

and the pressure per sq. ft = K x depth. 

Now at a depth of x ft. the pressure per sq. ft. = weight of a 
column of water x ft. high and i sq. ft. in section, i. e., wt. of x cu. ft. 
of water or 62-4^ Ibs. 

Also the pressure is the same in all directions; 

and thus the pressure on the strip = 38x^x62-4^ 
and the moment of the pressure on the strip about S.W.S.L. 



f!5 

Hence the total pressure = I i8j-2xdxlbs. 

J 10 

o ( x *\ 

= 187-2 - ) 

\2/10 

= 187-2 XJC25 lbs 

2 

11700 lbs. or 5-23 tons. 
Again, the total ist moment about S.W.S.L. 



ri5 /x 3 \ 

= i8rzx 2 dx = i87-2(- ) 

Jio \3/ 



io 
= 62-4x2375. 



234 



MATHEMATICS FOR ENGINEERS 



'-hus the depth of the C. of P. below S.W.S.L. 
= 62-4 X 2375 ft 
11700 

= 12-65 ft. 

Hence C. of P. is at the point P, at a depth of 12-65 ft- below the 
surface of the liquid. 

The more general investigation for the position of the C. of P. 
is given in Chap. X. 




Mi 



FIG. 75. Centroids. 

Exercises 19. On the Determination of the Positions of Centroids and 
Centres of Gravity. 

1. The density of the material of which a right circular cone is 
composed varies as the square of the distance from the vertex. Find 
the position of the centre of gravity of the cone. 



CENTROIDS 235 

2. The equidistant half-ordinates of the load water plane of a ship 
are as follows, commencing from forward : -6, 2-85, 9-1, 15-54, J 8, 18-7, 
18-45, 17-6, 15-13 and 6-7 ft. respectively. Find the area of the load 
water plane and the longitudinal position of its centroid. The length 
of the ship on the load water line is 270 ft. 

3. A triangular plate of base 5" and height 8" is immersed in water, 
its base being along the S.W.S.L. Find the total pressure on the 
plate and the depth of the centre of pressure if the plate is vertical. 

4. A vertical retaining wall is 8 ft. wide and 15 ft. deep. Find 
the depth of the centre of pressure of the earth on the wall. 

5. Draw the quadrant of a circle of 4" radius, and by the double 
sum curve method determine the position of its centroid. 

6. The portion of the parabola y = 2x z gx below the x axis 
revolves about that axis. Find the volume of the paraboloid so 
generated, and the distance of its C. of G. from the axis of y. 

7. Find the position of the centroid of the area bounded by the 
curve y = 1-7 2X Z , the axis of x and the ordinates through x = i 
and x = +4. 

8. Reproduce (a), Fig. 75, to scale (full size), and find the position 
of the centroid of the section represented, employing the method 
outlined in Example 23, p. 216. 

9. Draw a segment of a circle of diameter = 6* on a base of 5-92", 
and find by the method of Example 23, p. 216, the height of the 
centroid above the base. (Take the segment that is less than a 
semicircle.) 

Find the distance of the centroid from the line AB for the sections 
in Nos. 10, ii and 12. 

10. Channel Section, (&), Fig. 75. 

11. Unequal Angle, (c), Fig. 75. 

12. Tee Section, (d), Fig. 75. 

13. Make a careful drawing of (a) Fig. 76, which represents the half- 
section of the standard form of a stream, line strut for an aeroplane, 
taking t as 2", and by the method of Example 23, p. 216, determine 
the distance of the centroid from the leading edge. 

14. Find the position of the centroid of the pillar shown in Fig. 57, 
p. 210, of which further explanation is given in Question 2 on p. 210. 
[Deal with the flanges and the body as three separate portions.] 

15. One end of a horizontal water main 3 ft. in diameter is closed 
by a vertical bulkhead, the centre of the main being 35 ft. below the 
level of the water. Find the total pressure on the bulkhead. 

16. A semicircular plate is immersed vertically in sea water, its 
diameter being along the water surface. Find the total pressure on 
the plate if its diameter is 12 ft. and the weight of i cu. ft. of sea 
water is 64 Ibs. ; find also the depth of the centre of pressure. [Note. 
The reduction formulje given on p. 178 assist in the evaluation of the 
integrals.] 

17. The parabola y 2 = 6x revolves about the axis of x. Find the 
distance from the vertex of the C. of G. of the paraboloid thus 
generated, if the diameter of the end of the paraboloid is 18. 



236 



MATHEMATICS FOR ENGINEERS 

(a) 




(b) 



FIG. 76. 



MOMENT OF INERTIA 



237 



18. The diameter of a spindle at various distances along its length 
was measured with the following results 



Distance from end (ins.) 


o 


i 


2 
83 


3 


4 


5 


6 


7 


8 

2 


Diameter (ins.) . 


i'5 


I-I2 


85 


1-18 


i'5 


1-78 


1-96 



Find the distance of the C. of G. from the smaller end. 

19. Find, by means of the double sum curve method, the distance 
from AB of the centroid of the rail section shown at (a), Fig. 75. 

20. An aluminium right circular cone is of height 7 ins. and the 
diameter of its base is 10 ins. Find (a) its mass, the density of 
aluminium being -093 Ib. per cu. in. ; (6) the height of its centroid 
above the base. 

21. Use the double sum curve method to find the distance from 
AB of the centroid of the area shown at (6), Fig. 76. 

22. A segment of a parabola is of height h and stands on a base b 
Find the height of the centroid above the base. 

23. A triangular plate of height h is immersed in water, its vertex 
being at the water surface, and its base being horizontal. Find the 
depth of the centre of pressure of the plate. 

Moment of Inertia. The product of a mass into the square 
of its distance from some fixed point or axis is called its second 
moment about that point or axis; and for a number of masses 
the sum of their respective second moments becomes the second 
moment, or moment of inertia of the system. When the number 
of masses is infinite, i. e., when they merge into one mass, the 
limiting value of the sum of the second moments is spoken of as 
the moment of inertia of the body. 

The moment of inertia of a section or body determines to a 
large extent the strength of the section or body to resist certain 
strains ; the symbol I, which always stands for moment of inertia, 
occurs in numerous engineering formulae ; also when dealing with 
the formulae of angular movement the mass is replaced by I, and 
so on, so that it is extremely important that one should be able 
to calculate values of I for various sections or solids. 

A few examples will emphasise the frequent recurrence of the 
letter I. Consider first the case of a loaded beam : 

Let the figure (Fig. 77) represent the section of a beam loaded 
in any way. Then it is customary to make the following 
assumptions 

(a) There is to be no resultant stress over the section, i. e., the 
sum of the tensions = the sum of the compressions. 



238 MATHEMATICS FOR ENGINEERS 

(b) That the stress varies as the strain, and that the Young's 
modulus for the material is the same for tension as for compression. 

(c) That the original radius of curvature of the beam is exceed- 
ingly great compared with the dimensions of the cross section of 
the beam. 

The surface of the beam which is neither compressed nor 
stretched is spoken of as the neutral surface, and the line in which 
this cuts any cross section of the beam is known as the neutral 
axis. 

Referring to Fig. 77, let NN be the neutral axis, and let o- be 




FIG. 77. 

the stress at unit distance from NN, i. e., a-y = the stress at a 
distance y from NN. 

Thus the stress at y on a section of breadth b and depth 8y = a-y, 
and the force = stress X area = b8y X <ry. 

Now the forces on one side of NN must balance those on the 
other (by hypothesis). 

rr, 

bdya-y = o. 



but I 1 a-bdy x y = total ist moment of the forces 

and the line about which this is zero must pass through the centroid 
of the section ; hence the line NN passes through the centroid. 

The tensile and compressive forces form a couple, the moment 
of which 

= 2 force x distance = ;x \ b8y<ry x y 



MOMENT OF INERTIA 239 

i. e., in the limit the moment of resistance of the internal forces 
= o- / bdyxy 2 , i. e., a- /area x (distance) 2 

J ~ Y 2 

i. e., a- (2nd moment of section about NN) 
= o-I. 

If M is the bending moment at the section, i. e., the moment 
of the external forces, it must be exactly balanced by the moment 
of the internal forces, so that M = oT. 

Also if /j = maximum tensile stress and = o-Yj 

/ 2 = maximum compressive stress and = o-Y 2 

A ft M 
then o- = = f- = T 

*i *2 L 
M / 

or, in general, T ^ v~ 

Hence, in considering the strength of a beam to resist bending, 
it is necessary to know the moment of inertia of its section; 
knowing this and the bending moment, we can calculate the 
maximum skin stress. 

As a further illustration of the importance of I in engineering 
formulae let us deal with the following case : If a magnet is allowed 
to swing in a uniform field, the time T of a complete oscillation is 
given by 



where I = moment of inertia of the magnet 
M = magnetic moment of the magnet 
H = strength of the uniform field in which the 
magnet swings. 

In this case the I of a cuboid is required; and it will be seen 
that no mention of the mass is made in this fortnula. Actually 
the I takes account not only of the mass, but also of its disposi- 
tion, the latter being a most important factor in all questions of 
angular movement. Thus for a mass of i Ib. swinging at the end 
of an arm of 10 ft. the energy would be io 2 , i. e., 100 times that 
of the same mass placed at a radius of i ft. only, although the 
angular velocities in the two cases were the same. 

The reason for the presence of I in formulae concerning the 
energy of rotation will be better understood if the next Example 
is carefully studied. 



240 



MATHEMATICS FOR ENGINEERS 



Example 37. A disc revolves at n revs, per sec. ; find an expression 
for its energy of rotation, or its kinetic energy. 

If the total mass = ra, let a small element 8m of mass be considered, 
distant r from the axis of rotation (Fig. 78). 

Now the linear velocity at the rim = V = 2?rwR 

and the angular velocity &> = number of radians per sec. 



then 




or 



Ra, = 2TCWR = V 

V 



thus co is constant throughout, whilst V depends 
on the radius. 

Kinetic Energy of mass 8m 

_ massx (veloc.) 2 _ 8mxv 2 



FIG. 78. 
Hence the total K.E. of the disc = 



= r z 8m. 



-r 2 dm 



_ co a f n 

2 Jo 



massx (distance) 2 



= w Xl for disc. 

2 

Thus the K.E. = Io> 2 . Comparing this formula with the cor- 
responding one for linear motion, viz., K.E. = mv z , we see that 

~6 

when changing from linear to angular movement, I takes the place of 
m and <a the place of v. 

Suppose that the average velocity v t = r^ 
then my, 2 = Io> 2 

2g 

i. e., mr 

or 

Hence I is of the nature of mass x (distance) 2 , so that if the whole 
mass were concentrated at the distance r t from the axis, the K.E. of 
the system would be unaltered. 

Hence the distance r t (which is usually denoted by k) is referred 
to as the swing or spin radius, or radius of gyration, i. e., it is the 
effective radius as regards all questions of rotation. 



MOMENT OF INERTIA 241 

[Note that k is not the arithmetic mean of the various radii, but 
the R.M.S. value for 



h = J * (radius)* 

number considered J 



number considered 

In general, I can be written as mk 2 (if dealing with a mass) or 
Ak 2 (if concerned with an area). 

Method of Determination of the Value of I for any 
Section. Whilst it is found desirable to commit to memory the 
values of I for the simpler sections, it is not wise to trust entirely 
to this plan. It is a far better policy to understand thoroughly 
the meaning of the term "moment of inertia," and to derive its 
value for any section or solid by working directly from first 
principles. 

Thus, knowing that the moment of inertia is obtained by 
summing up a series of second moments, we divide the area or 
mass into a number of very small elements, find the area or mass 
of each of these and multiply each area or mass by the square 
of its distance from the axis or point about which moments are 
required ; the sum of all such products being the value of I. 

If the length of the swing radius is required, it can be deter- 
mined from the relation I = Ak 2 (for an area) or I = Mk 2 (for 
a solid) ; the area or mass being obtained by the summation of 
the areas or the masses of the separate elements. 



T , /2 second moments of elements 

Thus k = \i= , , . 

\ 2 areas or masses of elements 

Confusion often arises over the units in which I is measured; 
and to avoid this it is well to think of I in the form Ak 2 or M& 2 , 
when it is observed that I is of the nature area x (length) 2 , *'. e., 
(inches) 2 x (inches) 2 or (inches) 4 for a section, and massx (length) 2 
or Ibs. X (inches) 2 for a solid. 

The moment of inertia must always be expressed with regard 
to some particular axis; and it is frequently necessary to change 
from one axis to another. To assist in this change of axis the 
following rules are necessary : 

The Parallel Axis Theorem. By means of this theorem, if 
I is known about an axis through the C. of G., the I about an 
axis parallel to the first can be deduced. 

In Fig. 79 NN is the neutral axis of the section; and the 
moment of inertia is required about AB, i. e. IAB is required. 

R 



242 MATHEMATICS FOR ENGINEERS 

Dealing with the strip indicated 

I AB of the strip = pb8y x y 2 . 
Hence the total I AB = pfbdyxy 2 

'= P fbdyx(Y-d) 2 



= pfbdy X Y 2 + P /My X d 2 -2 P fbdyYd. 
Now fpbdy X Y 2 = the total I NN 
and fpbdy X d z = d 2 Jpbdy = d 2 X total mass = md 2 

also 2dfpbdyxY 2d x total ist moment about NN 

= 2d X o (for the moments on the strips on 
one side' of NN balance those on the 
other) 
= o. 

Hence I AB = I Nti -{-md 2 

i. e., to find the moment of inertia about any axis, find the moment of 

&/ 




FIG. 79. 

inertia about an axis through the G. of G. parallel to the axis given, and 
to this add the product of the mass into the square of the distance 
between the axes. 

e. g., if I NN = 47, mass = 12-4 and d (between AB and NN=2'3) 
then I AB =I NN +rf=47+(i2-4X2-3 2 ) 

= 47+ 6 57 = II2 7- 
Since I AB = Ij 

then w& AB = n 



or 



k" AB 



and this relation is represented by Fig. 79, which suggests a graphic 
method of finding &AB when NN is known. 



MOMENT OF INERTIA 



243 



Theorem of Perpendicular Axes. We require to find I 
about an axis perpendicular to the plane of the paper and passing 
through ; such being spoken of as a polar second moment. 

To distinguish between the moment about an axis perpendicular 
to the plane of the paper and that about an axis in the paper, 
we shall adopt the notation I for the former and I ox or I OY , as the 
case may be, for the latter. 

To find I o : 

Consider a small element of mass 8m at P (Fig. 80). 

Then I ox of this element = 8m x y 2 , I OY 8mx x 2 , 



r 2 = x 2 +y 2 



and I = 8m xr 2 . 



8m. r 2 = 8m.x 2 +8m.y 2 
fdm . r 2 = fdm . x 2 +fdm . y 2 
total I = total I OY + total I ox 



I = 



FIG. 80. 



Now 
hence 
and 
*'. e., 
or 

so that if the moments of inertia about two 

perpendicular axes in the area are known, 

the sum o! these is the moment of inertia 

about an axis perpendicular to the area and 

through the point of intersection of these axes. 

In special cases for which I ox = I oy 

then I = 2l ox 

To find the Relation between the Moment of Inertia 
about a Point in a Solid Body and the Moments of Inertia 
about three mutually Perpendicular Axes meeting in that 
Point. 

Thus, referring to Fig. 81, it is desired to connect I o with I ox> 
I OY and I oz . 

Consider a small element of the mass 8m placed at the point P. 
Then if PS = x PT = y PM = z OP = r 
(ON) 2 +(NM) 2 +(PM) 2 = (OP) 2 , and ON = PS, NM = PT 

*. e., x 2 -\-y 2 -\-z 2 = r 2 . 

Now I ox of the element = Sw(PN) 2 = 8m(z 2 +y 2 ) 
and in like manner I OY = 8m(x 2 +z 2 ) and I oz = 8m(y 2 +x 2 ) 
also I = Sw(QP) 2 = 8mr 2 . 

Thus I = Smr 2 = Sw(* 2 +y 2 +* 2 ) 



= 8m 



/ 



244 MATHEMATICS FOR ENGINEERS 

And for the total mass 



or 



total I = -( 



We may now apply the principles already enunciated to the 
determination of the moments of inertia of various sections and 




FIG. 81. 

solids ; and we take as our first example the case of a rectangular 
section. 

Example 38. To find the moments of inertia of a rectangle about 
various axes. 

(a) To find I NN (Fig. 82), NN being the neutral axis. 
Dealing with the small strip, of thickness Sx 

I N N of strip bSx x x 2 i. e., area X (distance) 2 
h h 

Hence the total I NN = P bx z dx = b(- 3 } = 
J * \3/ h I2 

~ 



A 




N, 


D 
M 

c 


N 


| 





ai 




I 








B 


.lb-~ 





-_i 


IN, 

FIG. 82. 



= area X 
12 

but 
where A is the area of the section 

AZ_ _ A/fc \2 



A^ = 

12 



or 



MOMENT OF INERTIA 



245 



By symmetry it will be seen that 
(6) To find I AB . 



_ Ab 2 

1 N 1 N 1 ~ : -^ 



and the distance between AB and 
hence I AB = 



b z 

= A X . 



= b 

2 

= A 

= A6 2 
3 ' 

I AB is larger than IN^. as would be expected, for the effective 
radius must be greater if the plate swings about AB than if it swings 
about N^NV 

I c ! 

rTT" ; b 1 s*1 i ... 

NLJ- ._ P-J&. ^ZJfrpJNi 



n 

*4 


I 

$ 


N ; 



In like manner 



if 

FIG. 83. 

Ah z 



L AD 



The rule for the moment of inertia of a rectangle is required very 
frequently, since many sections can, be broken up into rectangles. 

Example 39. To find I NN of the Tee section shown in Fig. 83. 

The neutral axis NN is distant 1-03* from AB (cf. Example 25, p. 220). 
Dealing with the flange 

i i /S\ 3 

I NiNj = bh 3 = ^ x6x (J] : : * 122 in> 

also the distance G Z G = 72". 
- Hence by the parallel axis theorem 



246 MATHEMATICS FOR ENGINEERS 

I NN of the flange = I NiNj + [Ax (G X G) 2 ] (A being the area of the 

flange) 
5,, 



2 ) = -I22+I-94 

= 2-06 ins. 4 
For the web i Nrfl> = ^6A = -L X ^ )* x |- = 2 ins. 4 

and also G 2 G = 1-28". 
Hence I NN of the web = I N2 N 2 + A x x (G 2 G) 2 (A x is area of the web) 

= 2 + (fxfxi-28 2 ) 

= 5'45 ins. 4 
Hence the total I NN of the section = 2-06+5-45 = 7-51 ins. 4 

Example 40. Find the polar 2nd moment of a circular disc of 
radius R; and also the moment of inertia about a diameter. 

Consider a ring of width 8r, distant r from the centre (Fig. 84). 
Then I o of the annul us = mass (or area) x (distance) 2 




Hence 

the total I 



Now 
and 

FIG. 84. 
To find the respective swing radii 

AQ = T = ~^~ 
TUR 4 



i. e., 




ox 



(Cf. with the R.M.S.l 



D 

or k Q , i. e., -TofR. -! value of a sine function V 

V * ( of amplitude R. J 



Also 



MOMENT OF INERTIA 



247 



ox 



4X:rR 2 
= - = -sR. 



R 2 
4 



To find the swing radius about a tangent 
(distance) 2 (oxtoTT) = R 2 

IYT = Iox+AR 2 



hence 



or 



Example 41. To find the moment of inertia of a right circular 
cylinder of length h and radius R, about various axes. 





(a) About the axis of the cylinder. 

(b) About an axis through the C. of G. perpendicular to the axis 
of the cylinder. 

(c) About an axis parallel to that in (b), but through one end. 

(a) The 2nd moment about the axis of the thin cylinder of 
length Sx (Fig. 85) 

R 2 

= mass X from Example 40 

R 2 
= pnR 2 8x x p being the density of the material . 

Hence the total 2nd moment about the axis 
.R 2 



= r 

J 



dx = 



R 2 

2 



R 2 

= m - 

2 



where m = the mass of the cylinder. 



248 



MATHEMATICS FOR ENGINEERS 



(b) The 2nd moment of the strip about AA, which is parallel 

to NN 

R 2 
= mass x - (see Example 40) 

R 2 



Hence I NN of the strip = I AA of the strip + (its massx* 2 ) 
since AA is an axis through the C. of G. of the strip. 

Thus I N vr of the strip = Sx + uTiR 2 x 2 8x 

4 

h h 

f2 oTtR 4 /"2 

and thus the total I NN = j dx-\- I pnR 2 x 2 dx 

J _h 4 J _ h 




N 



(c) To find I FF , FF being parallel to AA and NN. 
The distance between FF and NN = - 

2 

Also it has just been proved that I N1 j = m ( 



Hence 



T / R2 , h * 
! = ml 

\4 12 
/R 2 , h 2 \ 
= m(- + --). 



Example 42. Find an expression for the moment of inertia of a 
large pulley wheel of outside radius R and thickness of rim /. Neglect 
the arms or spokes of the wheel. 

Let Y inside radius of wheel, i. e., r = R t. 

Then, using the result of Example 40, p. 246, we know that the 

TD4 



moment of inertia of the wheel as solid = 



; from this must 



MOMENT OF INERTIA 249 

7W 4 
be subtracted the moment of inertia of a disc of radius r, viz., Xpb 

(p being the density of the material and b the breadth of the rim 
along the face). 



, , /T ,. 

Hence I Q - P b-~ P b = -~ (R*-y 4 



= (R 2 +r>) (R 2 -* 



. . . , 

where M is the mass of the wheel. 

M M 

Writing R-/ forn! = ^ (R 2 +R 2 +/ 2 -2R/) = ( 2 R 2 -2R*+* 2 ). 

From (i) it will be seen that in order to get I o as large as possible, 
R and r must be very nearly equal, i. e., t must be very small compared 
with R. Thus for an approximation 2 2 may be neglected in the 

M 
expression for I Q , so that I Q = -xaR (R t) = MR(R t). 

/R 2 +y 2 \ 2 /R 2 +y 2 \ 

Referring once again to (i), I o = M ( *-)* i- &-, M& o = M ( - ! J 

or k Q = - and k o = -jojVRt+r 2 . As an approximation for 

this the rule k Q = - (R-\-r) is often used; k Q being thus taken as the 
average radius. 

Moment of Inertia of Compound Vibrators. To find the 
modulus of rigidity of a sample of wire by the method of torsional 
oscillations, various forms of vibrators may be used. In the 
calculations which follow the experiments, the moment of inertia 
of the vibrator occurs, so that it is necessary to understand how 
to obtain this. To illustrate by an example of one form of 
compound vibrator, suppose that the I about an axis through 
the C. of G. of the one shown in Fig. 86 is required. 

Let m { = mass of AB, r l be its radius and ^ its length 

m 2 = mass of C and also of D, r z be its radius and / 2 its length. 

fr 2 / 2 \ 
Then I NN of AB = w a ( +- 1 ) (from Example 41, p. 248) 

c r i r z z , ^2 2 \ rf (f r tne inner radius\ 

and IofC^m- 1 = ^ 



250 MATHEMATICS FOR ENGINEERS 

by the parallel^ 
- m z\^ -r I2 y-r w 2 t - m z ^ \ 

This is also the L 



\4 

of D. 

2 



4 V axis theorem. ) 



L NN 



total I NN = 



Maxwell's needle is a very convenient form of compound vibrator, 
and is utilised to determine the modulus of rigidity of the sample 
of wire by which it is supported. It consists essentially of a tube 
along which weights may be moved from one position to another, 
the movement being a definite amount. 
Referring to Fig. 87 

m^ = the mass of each of the movable weights 
w 2 = the mass of each of the fixed weights. 




FIG. 87. 



FIG. 88. 



Then the time of torsional oscillations is measured when the 
movable weights are placed as shown, and again when they are 
moved to the centre; and it can be proved that the modulus of 
rigidity depends upon the difference between the moments of 
inertia under the two sets of conditions. 

Thus, since a mass m t is shifted from the position AB to the 
position NA, the only difference in the moments of inertia is that 
due to the changing of the C. of G. of a mass (m^ m 2 ) from a 
distance fa from the axis of oscillation to Ja; for I NN of m. 2 is 
unaltered. 



q 

r 

i6 



Hence the change of I \ , T 

considering one mass onlyJ " l 2 ~~ ^ m i~ i "'^\- L ^r ^ 

(j \ 
a 2 ) 

= (w x m 2 )a 2 . 



Example 43. Find the moment of inertia of a sphere of radius R 
about its diameter. 



MOMENT OF INERTIA 251 

Consider the thin disc (Fig. 88) of radius y, and thickness 8x. 
I of the strip about a diameter parallel to OY 

y*p8x (cf. Example 40, p. 246). 

Hence I of the strip about OY (distant x from the diameter 
considered) 

= T 5 " 1 

Now y* = R 2 x*. 

Thus I OY of disc = TTP J ""^ ~^ v " "^ ^ 18* 

r I 4 ' / 

and hence T OY of sphere = I (R 4 ; 

= 27rp [ R (R* - 
4 J o 



= R^ I 

4 L 5 i 

27uo i6R 5 8 

= - X = 
4 15 ! 



5 
2 ( m being the mass\ 

- 7/fr /\ ~ JLV \ tii i 

5 \ of the sphere. / 



and 



Determination of 1st and 2nd Moments of Sections 
by means of a Graphic Construction and the Use of a 
Planimeter. 

The graphic construction now to be described is extremely 
simple to understand, and has the additional merit of being 
utilised to give 3rd, 4th and higher moments if desired. 

It being required to find the ist and 2nd moments about MM 
of the rail section shown in Fig. 89, and also the position of the 
neutral axis, the procedure is as follows : 

Construction. Divide the half -area into a number of strips by 
means of horizontal lines; the half-area only being treated, since 
the section is symmetrical. 

At a convenient distance h from MM draw M X M X parallel to 
MM. From P, the end of one of the horizontals, draw PR per- 
pendicular to MM, and from P 1 , the other end of the same horizontal, 
drop PKR 1 perpendicular to M X M X ; join R X R and note Q, its point 
of intersection with P 1 ?. Repeat the process for all the other 
horizontals (of which only three are shown in the diagram) and 



252 



MATHEMATICS FOR ENGINEERS 



join up all the points like Q, thus obtaining the curve CQLS, which 
is termed the ist moment curve. 

To obtain the 2nd moment curve treat the area CPKXSLQ 
in the same way as the original area was treated, i. e., drop QR" 
perpendicular to M. 1 M. l and join RR"; join up all points like Q 1 
and the 2nd moment curve is obtained. 

Calculation. Find by the planimeter the areas of the original 



M 



M 



I^Momenf 
Curve 




M, 



FIG. 89. Moments of Sections by Graphic Construction. 

half-section, CPKXSLQ and CPKXTWQ 1 ; call these AC, A x and 
A 2 respectively. 

Then ist moment of the section about MM = 2Xhh.^ 
(for Aj is for the half-section only). 

Distance of the centroid of the section from MM = -r~ 
2nd moment of section about MM = 



i. e., (swing radius MM ) 2 = -^ - 

and by the theorem of parallel axes, I can be found about NN. 



MOMENT OF INERTIA 253 

In this case the actual results are as follows : 

h = 3 ins. A = i-n sq. ins. A x = -573 sq. in. A 2 = -39 sq. in. 

Hence h = 3 X ^ 73 = 1.55 ins. 

i 
ist moment of section about MM = 2 X 3 X -573 = 3-44 ins. 3 . 

2nd moment of section about MM = 2X3 2 X -39 = 7-02 ins. 4 . 

Swing radius about MM = \ = 1-78 ins. 

\ 2-22 ' - 

N.B. To distinguish which area is to be read off by the 
planimeter the following rule should be observed : Read the area 
between the ist or 2nd moment curve, as the case may be, and the 
side of the original contour from which we dropped perpendiculars 
on the line about which we required moments. 

Proof. Consider P 1 ? as the centre line of a thin strip (such 
as the one indicated). Then the area of the strip = P^xS*, and 
ist moment about MM = P X P x 8x X RP. 

From the similar triangles RPQ and RJR 1 

RP _ J*l h 

QP ~ JR 1 ~~ PP 1 

whence RP x PP 1 = h X QP 

and RPxPP x xS* = AxQPxS* 

i. e., ist moment of the strip about MM 

= h x the area of which QP is the centre line. 
Then, by summing 
Total ist moment of the half-area about MM 

= h X the area between ist moment curve and right-hand 

boundary of section 
= h\ 1 . 

Again, the 2nd moment of the strip about MM = area x (distance) 2 

= PP 1 xS*x(RP) 2 

and 1 



RJ ~~ JR" ~ PQ 

RP PQ 1 . or> hPO 1 

_ - _ ^_ a T?P - _ S 

h ' ~ PQ PQ ' 

Hence the 2nd moment of the strip about MM 
= P 1 PxRPxRPx8* 

PO 1 

^ Sx = h*xPQ l x8x 

= h 2 X area of which PQ 1 is the centre line. 



254 MATHEMATICS FOR ENGINEERS 

And the total 2nd moment of the half-area about MM 

= A 2 x area between the 2nd moment curve and the right- 
hand boundary of the section. 



Exercise 20. On Moment of Inertia. 

1. Find the swing radius about the lighter end of a rectangular 
rod of uniform section and breadth and length I, for which the density 
is proportional to the square root of the distance from that end. 

2. The swing radius of a connecting rod about its centre of suspen- 
sion was found to be 35-8 ins., and the distance of the C. of G. from 
the point of suspension was 31-43 ins. Find the swing radius about 
the neutral axis. 

If the connecting rod weighed 86-5 Ibs., find its moment of inertia 
about the neutral axis. 




FIG. 91. 

3. A circular disc, 7" diameter, has a circular hole through it, of 
diameter 3", the centre of the hole being \" distant from the centre 
of the disc. Find the swing radius of the disc about an axis through 
its centre of gravity, perpendicular to the face of the disc. 

4. Find the moment of inertia of a rectangle (5" by 3") about a 
diagonal as axis. 

5. Find the swing radius of a triangular plate (of height h) 

(a) When swinging about its base. 

(b) When swinging about an axis through the vertex, parallel to 
the base. 

6. By dividing into strips, by lines parallel to AB, find the moment 
of inertia, about AB, and also the swing radius, of the section shown 
at (a), Fig. 75, p. 234. 

7. Find the radius of gyration about the axis of rotation, of the 



MOMENT OF INERTIA 255 

rim of a flywheel, of outside diameter 5' 2", the radial thickness of 
the rim being 4". 

Find the moment of inertia about the neutral axis of the sections 
in Nos. 8, 9 and 10. 

8. Channel Section, (&), Fig. 75. 

9. Unequal Angle, (c), Fig. 75. 

10. Tee Section, (d), Fig. 75. 

11. Find the swing radius, about the axis, of a paraboloid, the 
diameter of the bounding plane, which is perpendicular to the axis, 
being d. 

12. The flexural rigidity of a beam is measured by the product of 
the Young's Modulus E for the material into the moment of inertia 
of the section. Compare the flexural rigidity of a beam of square 
section with that of one of the same material but of circular section, 
the span and weight of the two beams being alike. 

13. A cylinder 6" long and of i \" diameter is suspended horizontally 
by means of a long wire attached to a hook, and the wire is then 
twisted to give an oscillatory movement to the cylinder. Find the 
moment of inertia of the cylinder about the hook. 

14. Determine the moment of inertia and also the swing radius 
about AB of the rectangular section shown at (a). Fig. 90. 

15. Calculate the moment of inertia and also the swing radius of 
the box section shown at (b), Fig. 90, both about NN and about AB. 

16. Find the position of the neutral axis of the section shown at 
(c), Fig. 91, and then calculate the moment of inertia and also the 
swing radius about this axis. 

17. Determine the swing radius of the -section shown at (d), Fig. 91, 
about the axis NN. 

18. The moment of inertia of the pair of driving wheels of a 
locomotive connected by a crank axle was found by calculation to 
be 34133 Ibo. ft. 2 . If the total weight of the two wheels and the axle 
was 8473 Ibs., and the diameter of the driving wheels was 6 ft. i in., 

A 2 
find the swing radius of the wheel and also the ratio -^, where r is 

the radius of the wheel. 

19. Find the swing radius about the axis of a right circular cone 
of uniform density, the radius of the base being 5 ins. 

20. Employing the method explained on p. 251, determine (a) the 
ist moment about AB, (b) the 2nd moment about AB, (c) the 
distance of the centroid from AB, and (d) the swing radius about AB, 
of the area shown at (b) Fig. 76, p. 236. 

21. A steel wire, -15 in. in diameter, hangs vertically; its upper 
end is clamped, and its lower end is secured to the centre of a 
horizontal disc of steel, which is 6 in. in diameter and g in. thick. 
If the length of the wire is 3 ft., and if C, the modulus of transverse 
elasticity of the steel, has the value 12,540,000 Ibs. per sq. in., find 
the time of a torsional oscillation of the wire, from the formula 

* = 402-5 



256 MATHEMATICS FOR ENGINEERS 

where I = moment of inertia of the disc about the axis of suspension 
in Ibs. ins. 2 , / = length of wire in feet, d = diameter of wire in inches. 

22. An anchor ring is generated by the revolution of a circle of 
radius r about an axis distant R from the centre of the circle. Find 
the moment of inertia of the ring about this axis. (Hint. Commence 
with the polar moment, i. e., the moment about the given axis, of an 
annulus made by a section at right angles to this axis, finding an 
expression for the inner and outer radii of the annulus in terms of 
the distance from the central annulus, and then sum up.) 

23. Find the swing radius about the major axis of the ellipse 
whose equation is 



CHAPTER VIII 
POLAR CO-ORDINATES 

Polar Co-ordinates. A point on a plane may be fixed by 
its distances from two fixed axes, or by its distance along a line 
which makes a definite angle with some fixed axis. In the former 
case we are concerned with rectangular co-ordinates and the point 
is written as the point (x, y) ; whilst in the latter case the co- 
ordinates are polar and the point is denoted by (r, ff), r being the 
length along the ray inclined at an angle 6 to the fixed axis. 

It is really immaterial as to what line is taken as the fixed axis : 
in many cases the horizontal axis is taken, but in order to agree 
with the convention adopted for the measurement of angles (see 
Part I, Chapter VI) we shah 1 here 
consider the N. and S. line, i. e., a 
vertical line, as the starting axis and 
regard all angles as positive when 
measured in a right-handed direction 
from that axis. A point is next fixed 
on that line from which all the rays 
or radii vectors originate, and this 
point is spoken of as the pole for the 
system : thus the reason for the term 
polar is seen. 

To illustrate this -method of plot- 
ting, let us refer to Fig. 92. Taking 
OY as the starting axis and O as the 
pole, the point (2, 35) is obtained by 
drawing a line making 35 with OY 

and then stepping off a distance OP along it to represent 2 units, 
i. e., r=2 and 0=35. In like manner Q is the point (17, 289) ; 
whilst R is the point (2-4, 20). 

One advantage of this method of plotting is that it is not 
necessary to classify into quadrants and to remember the arrange- 
s 257 




FIG. 92. 



258 



MATHEMATICS FOR ENGINEERS 



ment of the algebraic signs ; all lengths measured outwards from 
the pole being reckoned as positive. 



Example i . The following table gives the candle power of an arc 
lamp for various positions below the lamp : plot the polar diagram. 



Angle below horizontal . . 


o 


10 


20 


3 


40 


5 


60 


70 


80 


90 








1800 








800 




600 


480 

























In reality we have to plot a number of polar co-ordinates, the 
lengths representing the values of the candle power ; but since the 
horizontal axis is specified, we shall take that as the main axis. Draw 
rays making 10, 20, 30, etc. (Fig. 93), with the horizontal axis, and 
along these lines set off distances to represent the respective candle 
powers, always measuring outwards from the centre. Join the ends 
of the rays and the polar diagram is completed. 

The Archimedean spiral and the logarithmic or equiangular 

spiral, important in connection 

-IOOO ^ with the forms of cams and gear 

wheels respectively, may be easily 
plotted from their polar equations. 
Thus the equation to the Ar- 
chimedean spiral is r=aO, and the 
equation to the equiangular spiral 
is r=ae be ; indicating that in the 
former case the rays, for equal 
angular intervals, are the con- 
secutive terms in an arithmetic progression, whilst in the latter case 
the rays are in geometric progression. 

To illustrate the forms of these curves by taking numerical 
examples : 

Example 2. Plot the Archimedean spiral ^=-573^, showing one 
convolution. 




FIG. 93. Candle Power of Arc 
Lamp. 



In the equation d must be in radians, but to simplify the plotting we 
can transform the equation so that values of a (in degrees) may 
replace 6 (radians). 

Thus r = -5736 = . (degrees) = -oia. 

O / O 



POLAR CO-ORDINATES 
Then the table for the plotting reads : 



259 



a 


o 


3 


60 


9 


120 


150 


180 


210 


240 


270 


300 


33 


360 


' 






6 








































2-4 


2-7 


3' 


3*3 


3-6 



and the plotting is shown in Fig. 94. 




aro 



izo" 



24O 



1.50* 



eio 



FIG. 94. 
Example 3. Plot one convolution of the equiangular spiral 



25a 



In the log form log r = log -5 + -004360 log e 

= T-6990+ (-00436 X -4343 X a) 
= 1-6990-!- -ooi894a 
and thus the table of values reads : 



a. . . 


o 


3 


60 


90 


120 


150 


1 80 


210 


240 


270 


3 


33 


36o 


00189401 





0568 


1136 


1705 


2273 


2841 


3409 


'3977 


4546 


5114 


5682 


625 


6818 


log r . 


1-6990 


7-7558 


1-8126 


f-8695 


1-9263 


1-9831 


0399 


0967 


1536 


2104 


2672 


324 


3808 


r . . . 


5 


5699 


6495 


7407 


8439 


9618 


1-096 


1-249 


1-424 


1-624 


1-85 


2-109 


2-403 



26o 



MATHEMATICS FOR ENGINEERS 



The curve is drawn in Fig. 95. 
It will be seen that the ratio of 

second ray _ -5699 _ 
first ray -5 

third ray = -6495 _ ^ 
second ray -5699 

so that this spiral might alternatively have been defined as one for 
which the rays at equiangular intervals of 30 form a geometric pro- 
gression in which the common ratio is 1-14, the first ray being '5". 

Comparing the given equation r '$e' 259 with that connecting 
the tensions at the ends of a belt passing round a pulley, viz., 

T = te* 8 , we observe that the forms 
are identical, or in other words the 
equiangular spiral might be used to 
demonstrate the growth of the ten- 
sion as the belt continuously em- 
braces more of the pulley. 

Selecting any point P on the 
spiral, and drawing the tangent PT 
there and also the ray OP which 
makes an angle < with the tangent, 
it is found that cot < = -25 = co- 
efficient of 6 in the original equation. 
This relation would hold wherever 
the point P was taken on the spiral, 
so that the angle between the ray and the curve is constant : and 
thus the spiral is called " equiangular." 

If cot < = i, < = 45 and r = ae 6 , or taking a = i, r = e 6 and 
log e r = 6. Thus a spiral could be constructed in which the angles 
(in radians) would be the values of the logs of the rays : this 
spiral, however, is extremely tedious to draw, and its value consists 
merely in its geometric demonstration of the relationship between 
the natural logarithms and their numbers. 

Connection between Rectangular and Polar Co- 
ordinates. Let P be a given point, with rectangular co-ordinates 
x and y and with polar co-ordinates r and 0. 

Then referring to Fig. 96 

ON y 
OP , 




180 



FIG. 95. 



so that 



y = r cos 



and 
so that 
and also 



POLAR CO-ORDINATES 
OM x 



26l 



x 
y 



x = r sin 
r sin 



r cos 



= tan 0. 



Use of Polar Co-ordinates for the Determination of 
Areas. Polar co-ordinates may be usefully employed to find areas 
of certain figures. 

It is stated in the previous work on mensuration that 

Area of sector of circle = 
where = angle of the sector in radians. 



N 




P 

V 



X 




FIG. 96. 



FIG. 97. 



Let P and Q (Fig. 97) be the two points (r, 6} and (r+8r, 0+ SO) 
and close to one another. 

Then, since r and r+&r differ very slightly 
Area POQ = 



and the total area AOB = 



or 



re? 

r z d0 
J e l 



approximately 
exactly. 



For the evaluation of this integral the working may be either 
graphic or algebraic, according to the manner in which the relation 
between r and is stated. 

As a simple illustration we may take the case of a circle of 
radius a. The area of the circle was found at an earlier stage 
(see p. 225) by evaluating fydx, i. e., by expressing the integral in 
terms of the rectangular co-ordinates. To evaluate the integral, 
however, it was found necessary to make the substitution x = a sin 6, 
the change thus being from rectangular to polar co-ordinates. 



262 MATHEMATICS FOR ENGINEERS 

Evidently the rotating ray is constant in length and equal to a, 
the radius of the circle, and the limits to 6 are o and 2ir, if the full 
area is required ; hence 

fZir [2* 

Area of circle = I \d*d6 = Ja 2 1 d& = Aa 2 . 2ir 
Jo Jo 

= TO?. 

Example 4. Find the area of the cardioid given by the equation 
Y = a (i+cos &], 6 ranging from.o to ZTT. 

In this case r is of variable length, but there is a definite connection 
between r and 6, so that the integration is algebraic. 

and r 2 = a 2 (i-f-cos <9) 2 



= a 2 (i + J cos 2(9+J+2 cos 0) 
= a 2 (1-5 + \ cos 20+2 cos 0). 

j'Zn- 

Hence area = j \v z d6 



COS 20 + 2 COS0) dd 

2 / \ TT 

= ( i'50+J sin 20+2 sin j 

2 2 

The Rousseau Diagram. The use of the Rousseau diagram 
simplifies the determination of the mean spherical candle power of 
a lamp. 

The candle power of the lamp varies according to the direction 
in which the illumination is directed (cf . Fig. 93) ; in the case there 
discussed, however, we considered the illumination in one plane 
only. If we imagine the polar curve to revolve round the vertical 
axis we see that a surface is obtained by means of which the 
illumination in any direction can be measured. The mean of all 
these candle powers is spoken of as the mean spherical candle 
power of the lamp. If the arc is placed at the centre of a spheri- 
cal enclosure, of radius R, then, if IM is the mean spherical candle 
power (M.S.C.P.) of the lamp, the total illumination is expressed 
by 47rR 2 lM : this total might be arrived at, however, by summing 
the products of the candle power in any direction into the area of 
the zone over which this intensity is spread; and putting this 
statement into the form of an equation, 

47rR 2 I M = 2IA, 
where I is the intensity on a zone of surface area A. 



POLAR CO-ORDINATES 



263 



To find the M.S.C.P. proceed as follows: Suppose that the 
lamp is at O (Fig. 98). With centre O and any convenient radius 
R describe a semicircle ; also let the polar diagram be as shown 
(the curve OPQMC) . The greatest candle power is that given by 
OC ; draw a horizontal through N, the point in which the line OC 
meets the circumference of the semicircle, and make ab = OC. 
Through a and b draw verticals and through A and B draw 
horizontals, thus obtaining the rectangle DE ; draw a number of 
rays, OP, OQ, OS, etc., and also horizontals through the points 
p, q, s, etc., marking along these lines distances equal to OP, OQ, 
OS, etc., working from DF as base. By joining up the points so 
obtained the curve FL6D is obtained, known as the Rousseau 




FIG. 



curve ; then the mean height of this curve (which can readily be 

obtained by means of a planimeter) gives the M.S.C.P. of the lamp. 

Proof of this Construction. Let IM = M.S.C.P. of the lamp 

2 area of zonexC.P. 



then 



4irR*. 



Consider the zone generated by the revolution of TN ( = 8s) 
about AB ; its area is of the form 27rySs and the intensity of 
illumination is OC, say. The length y is the projection on the 
horizontal axis through O of either the line OT, the line ON, or the 
line midway between these (for these differ in length but slightly 
if 8s is taken as very small), i. e., y = OT cos 6 
or y .= Rcos0. 



264 MATHEMATICS FOR ENGINEERS 

Hence, for this zone, the illumination 
= candle power x area 
= OCX27rRcos0Ss 

= a&X27rRxfl'ifor-r- = cosflf 

v OS ) 

Hence the total illumination 



= 27rR X area under the curve FL&D 
and thus 
Total illumination 2?rR X area under the curve FL&D 

M = 47rR 2 47rR 2 

_ area under the curve FL&D 
~^zKT 

= mean height of the curve FL&D 
since 2R is the base of the curve. 

Dr. Fleming's Graphic Method for the Determination of 
R.M.S. Values. The determination of R.M.S. values is of some 

importance to electrical engineers, 
and the subject, previously dis- 
cussed in Chapter VII, is here 
treated from a different aspect. 
Instead of squaring the given 
values of the current and then 
extracting the square root of the 
mean of these squares, we may, 
by a simple graphic construction, 
obtain the mean of the squares very 
readily. 

Let the values of an alternating 
current at various times be as in the table : 




time t . . 


o 


001 


002 


003 


004 


005 


006 


007 


008 


Current C 


5 


8 


12 


7 





-6 


-8-3 


-3 


5 



then, to find the R.M.S. value of C we proceed as follows : 

Treat the given values as polar co-ordinates, taking t for the 
angles and C for the rays. Select some convenient scale for t, say 
20 -ooi sec., and a scale for C, say i"=4 units, these being the 
scales chosen for the original drawing of which Fig. 99 is a copy to 



POLAR CO-ORDINATES 



265 



about one-half scale; and set out a polar diagram as indicated, 
making OA = +5, Oa = 6, etc. Join the extremities of the rays, 
so obtaining, with the first and last rays, the closed figure ABCDaE. 
Measure the area of this figure by means of a planimeter in this 
case the area was found to be 4-23 sq. ins. 

Now the area of the figure = %fr z d& = %fC 2 dt, or 



so that if we divide twice the area by the range in t, the mean 
value of the squares is determined. 

In this case the range of t = 160 = 2-79 radians, and also 
J"C 2 dt = 2 Xi6x 4-23 = 135-5, f r I "=4 units, and thus i sq. in. 
= 16 sq. units. 



Then 



M.S. = 



279 



= 48-65, 



and hence R.M.S. = 



= 6-98. 



The rule for the area of a figure, viz., %J"r 2 dO, may be usefully 
employed to find the height of the centroid of an area above a 
certain base. 



Example 5. Find the height above the base OX of the centroid of 
the irregular area OABX [(a) Fig. 100]. 



fir 



10 



a 




r 



B 



8 12 16 EO 



(O 




FIG. 100. Centroid of Irregular Area by Polar Diagram. 



266 MATHEMATICS FOR ENGINEERS 

To do this, first divide the base into a number of equal divisions 
and erect mid-ordinates in the usual way. Measure these mid-ordinates 
and set off lengths to represent them as radii vectors from A in (b) 
Fig. 100, the angles at which the rays are drawn representing the 
values of x, i. e., the lengths of the divisions of the base. Join the ends 
of these rays and measure the area of the polar diagram thus obtained ; 
divide this area by the area of the original figure and the result is the 
height y required. 

For the area of the polar diagram = %fr 2 dd = \fy z dx, since rays 
represent values of y and the angles represent values of x. Also the 
area of the original figure =fy dx, so that 

area of polar diagram _ \fy z dx _ _ 
area of original figure fydx ~ 

For the particular case illustrated (the scales referring to the original 
drawing) : 

For (a) Fig. 100 i" = 5 units vertically 

i* = 4 units horizontally 

so that i sq. in. represents 20 units of area. The area was found by 
the planimeter to be 16-82 sq. ins., so that the actual area is 336-4 
sq. units, i. e., fy dx = 336-4. 

For (b) Fig. 100 i" = 5 units radially 

and each angular interval = 20, so that the total range = 180 or 3-14 
radians. Hence 3-14 radians represent 20 units, the length of the base 
in (a) Fig. 100, 

or i radian = 6-36 units. 

3-14 

Now the area is of the nature r 2 X0, i. e., (length) 2 x angle, hence 
i sq. in. of area = 5 2 X 6-36 or 159 units. Area of the polar figure 
(found by the use of the planimeter) 

= 18-34 sc l- ms - = J 8'34 X 159 units = 2920. 

Hence y = ^|. = 8-68, 

336-4 

i. e., the centroid horizontal is found. 

Theory of the Amsler Planimeter. The principle upon 
which the planimeter is based may be explained quite simply, in 
the following way. 

In Fig. 101 let PP" be a portion of the outline of the figure 
whose area is to be measured, and let the fixed centre of the 
instrument be at O. Then in the movement of the tracing point 
P from P to P" along the curve, the tracing arm changes from the 
position AP to A'P". This movement may be regarded as made 
up of two distinct parts : firstly, a sliding or translational move- 
ment from AP to A'P', and next, a rotation round A' as centre, 



POLAR CO-ORDINATES 



267 



from A'P' to A'P". In the former of these movements the record- 
ing wheel moves from W to W, but part of this movement only, 
viz., that perpendicular to the axis of the wheel, is actually 
recorded, so that the wheel records the distance p. 

The area swept out by the tracing arm AP during the small 
change from P to P" = APP'P"A' = APP'A'+P'P'A' 



Hence for the whole area, 

area swept out = 2APx/>+2(AP) 2 S<9 



Now the net angular movement is zero, so that 280 = o . 

Hence area swept out = AP2/>, 

or if / = length of the tracing arm, 

area = / X travel of wheel 
and hence the reading of the wheel 
_ area of figure 



Thus the length of the tracing arm determines the scale to 
which the area is measured. Hence by suitable adjustment of this 
length of arm the area of a figure may be 
read in sq. ins. or sq. cms. as may be neces- 
sary. If the average height of the figure is 
required, the length of the tracing arm must 
be made exactly equal to the length of the 
figure. This is done by using the points LL 
(Part I, Fig. 301), and not troubling about 
the adjustment at A. The difference between 
the first and last readings gives, when multi- 
plied or divided by a constant, the actual 
mean height of the figure. If the ordinary 
Amsler is used, then the mean height in 
inches is obtained by dividing the difference 
between the readings by 400 ; thus if the first 
reading was 7243 and the last 7967, the mean 
height would be the difference, viz., 724, 
divided by 400, i.e., 1-81 ins. 

The area of the figure = average height X 
length, but the area of the figure = length of 
tracing arm X wheel reading, hence if the length of the tracing 
arm = the length of the diagram, the wheel reading must be the 
average height of the diagram. 




FIG. 101. Theory of 
Amsler Planimeter. 



268 



MATHEMATICS FOR ENGINEERS 



[It should be noticed that the area recorded by the instrument 
is really the difference between the areas swept out by, the ends 
A and P of the arm AP, but as A moves along an arc of circle, 
coming back finally to its original position, no area is swept out.] 

Exercises 21. On Polar Co-ordinates. 

1. Plot a- polar curve of crank effort for the following case, the 
connecting rod being infinitely long. 



e 


o 


15 


3 


45 
5'i 


60 


75 


90 
7'3 


105 


1 20 


i35 


. 150 


165 


1 80 


Crank Effort (Ibs.) . . 


o 


2-4 


3-9 


6-4 


7-1 


7 


6-1 


4'9 


3'3 


*5 






2. As for Question i, but taking the connecting rod = 5 cranks. 



e 





15 


3 


45 


60 


75 


90 


i5 


1 20 

5-8 


'35 


'5 


165 


172-5 


i So 


C.E.flbs.) .... 





2< 45 


4'4 


6-1 


TO 


7'4 


7'4 


6-6 


4'4 


3' 


i'S 


6 






3. An A.C. is given by C = 7-4 sin ^oirt. Draw the polar curve to 
represent the variation in C and hence find its R.M.S. value. 

4. What is the polar equation of a circle, the extremity of a 
diameter being taken as the centre from which the various rays are to 
be measured ? 

Of what curve (to Cartesian ordinates, i.e., rectangular axes) is the 
circle the polar curve ? 

5. Plot the polar diagram for the arc lamp, from the table. 



Angle (degrees) . . . 


o 


TO 


2O 


3 


40 


50 


60 


70 


80 


90 

(vertical) 


Candle Power .... 


800 


T2OO 


l6<X) 


2OOO 


2200 


2200 


2300 


2500 


2300 


1800 



6. Plot the Rousseau diagram for the arc lamp in Question 5 and 
from it calculate the M.S.C.P. of the lamp. 

7. An A.C. has the following values at equal intervals of time : 3, 4, 
4'5> 5'5> 8, 10, 6, o, 3, 4, 4-5, 5-5, 8, 10, 6, o. Find by 
Dr. Fleming's method (cf. p. 264) the R.M.S. value of this current. 

8. Eiffel's experiments on the position of the centre of pressure for 
a flat plane moved through air at various inclinations gave the follow- 
ing results : 



Inclination to horizontal .... 





5 


10 


15 


3 


45 


60 


75 


90 


T>p 

Ratio (see Fig. 102) ..... 


263 







































Plot a polar diagram to represent the variation of this ratio. 



POLAR CO-ORDINATES 



269 



9. Draw the polar curve to represent the illuminating power of a 
U.S. standard searchlight from the following figures : 



Angle (degrees) 


o 
(vertical) 


10 


20 


3 


40 


5 


60 


70 


80 


90 


Candle Power . 


3000 


10000 


20500 


33000 


41500 


4*5o 


43000 


43000 


30000 


24000 



10 


















(above 
horizontal) 


20 


30 


40 


50 


60 


7 


80 


90 


9000 


6000 


5000 


5000 


2OOO 


3000 


1500 


1500 I50O 




C 15 centre of pressure 
FIG. 102. 



CHAPTER IX 
SIMPLE DIFFERENTIAL EQUATIONS 

Differential Equations. An equation containing one or more 
derived functions is called a " differential equation." 
Thus a very simple form of differential equation is 

dy = 
dx ? 



dv 

5 = o 
x 

, dy 

and +2x -= 



and 4 -;- + 7/ 5 

dx 2 dx 



are more complex forms. 

Differential equations are classified according to " order " or 
" degree " ; the order being fixed by that of the highest differential 

/V 

coefficient occurring in it. Thus ~ is a differential coefficient of 

d s y 

the first order. - is of the third order, and so on. 
dx 3 

dv 
Hence 4-2 -\-y-jr = 5 '34 is an equation of the first order 

w% 

d*y 
and 8-^j+.y = 7 -I 6 is an equation of the fourth order. 

wsv 

The "degree" of an equation is fixed by that of the highest 
derivative occurring when the equation is free from radicals and 
fractions. 

d 2 y 
Thus T-TJ = c is of the second order and of the first degree 



whilst 4(3-^) +(-r"o) = 7 is f the second order and of the 

2 2 



second degree. 

Much has been written concerning the solution of the many 
types of differential equations, but it is only possible here to treat 

270 



SIMPLE DIFFERENTIAL EQUATIONS 271 

the forms that are likely to arise in the derivation of the proofs of 
engineering formulae ; the plan being to discuss the solution accord- 
ing to the types of equation. 

Type : ~ given as a function of x. With the solution 

of such simple forms we have already become familiar, for the 
equations connecting the bending moment at various sections of a 
beam with the distances of those sections from some fixed point are 
of this character. 

Thus taking the case of a simply supported beam carrying a 
load W at the centre 

dfy = \V 

dx* ~ El 

dy Wx 

whence -- = ^F +C 
dx El 

which is of the type under consideration. 

Evidently this equation can be solved by integration through- 
out, attention being paid to the constants which are necessarily 
introduced. Expressing in algebraical symbols, 

!=/<*> ; 

then by integrating throughout with regard to x 



or y = \f(x}dx-\-C. 



*v 

Example i. If -~ = ^x z + r jx 2 and y = 5 when x=i, find an 
a% 

expression for y in terms of x. 

This equation is of the type with which we are now dealing, since 
2 =/(*) 



Integrating y = r - + * -- 2X+C. 

The value of C must now be found : thus y = 5 when x = i 
so that 5 = A_(_i_ 2+C 

or C = 2-iy. 

Hence y= i-33# 3 +3-5# 2 2^+2-17. 



272 MATHEMATICS FOR ENGINEERS 

Type : f- given as a function of y. i. e., -^ f(v). 

ax dx 

This type of equation differs somewhat from the preceding in 
that a certain amount of transposition of terms has to be effected 
before the integration can be performed. 

dy 
The equation may be written 7 = dx 

the transposition being spoken of as " separating the variables," 
and thence by integration I -,-,--. \dx-\-C = x-\-C. 

jj(y) J 



dy 
Example 2. If ~ = y 3 , find an expression for y. 

ax 

Separating the variables ^ = dx. 

Integrating / -^ = / dx-\-C 

or \y-* = x +C 

whence x-\ - + C = o. 

y _ 



The two foUowing examples are really particular cases of the 
type discussed generally on p. 275, but they may also be included 
here as illustrations of this method of solution. 

dy 
Example 3. Solve the equation -j- + ay = b. 

(tX 

dy , 
Here -g = b ay 

dy 

J = dx 



so that ^- = tdx+C 

J 



T 

b ay 

^- 

bay 



i. e., loge (bay) = x+C 

or loge (bay) (ax-}-aC) 

whence e~ ax - aC = (b-ay). 

Now let A = e~ aC : then e -*-c = e ~ ax x e~ aC = Ae~< 
and Ae~ aa: & = ay 

b A _ 
or "V = 

^ a a 



SIMPLE DIFFERENTIAL EQUATIONS 273 

dy 
Example 4. Solve the equation 4-^- = u-f-7y. 

Separating the variables 



i.e., 

or loge (u+jy) = lx+ 

whence e* x+ * C = (11+77) 

1C 

and if A be written in place of e 

11+77 = Ae* z 

A \x ii 
or y = e --- 

* 1 7 



Example 5. The difference in the tensions at the ends of a belt 
subtending an angle of dd at the centre of a pulley = dT = Tpdd, where 
/* is 'the coefficient of friction between the belt and pulley. If the 
greatest and least tensions on the belt are T and t respectively, whilst 

T 
the lap is 6, find an expression for the ratio . 

The equation dT = Tfjdd is of the type dealt with in this section ; to 
solve it we must separate the variables, thus : 



Integrate both sides of the equation, applying the limits t and T to 
T and o and 6 for the angle. 

/"T dT r e 

Then = M / 06. 

J t i Jo 

/ \ T / \ fl 

Mr <('). 



T 

But log e - = loge T - loge / 

T 
Hence * O ' x ^ 



or 



A word further might be added about Example 3, or a modifi- 
cation of it. 



T 2 

Let -f- = ay. 

dx * 



T 



274 MATHEMATICS FOR ENGINEERS 

If a = i, then -~ =y, i. e., the rate of change of y with regard to 
ctoc 

x, for any value of x, is equal to the value of y for that particular 
value of x. Now we have seen (Part I, p. 353) that this is the 
case only when y = e x . 

If a has some value other than i, y must still be some power of 
e, for the rate of change of y is proportional to y ; actually, if 

-V = 6*, -- = ae ax = ay, so that y = e * would be one solution of 
dx 

the equation -. = ay, but to make more general we should write 

d% 

the solution in the form y = e ax -\-C or y = Ae ax , whichever form is 
the more convenient. Whenever, therefore, one meets with a 
differential equation expressing the Compound Interest law (*'. e., 
when the rate of change is proportional to the variable quantity) 
one can write down the solution according to the method here 
indicated. 

Example 6. Find the equation to the curve whose sub-normal is 
constant and equal to za. 

dv 

The sub-normal =)>-/- (See p. 43.) 

dx 

dv 

Thus y~- = ia, 

J dx 

or, separating the variables, fydy =fzadx. 

V 2 
Hence = 2ax+C (Integrating) 

or -y2 



This is the equation of a parabola ; if y = o when x = o, then K = o 
and y 2 = ^ax, i. e., the vertex is at the origin. 

Example 7. Find an expression giving the relation between the 
height above the ground and the atmospheric pressure ; assuming that 
the average temperature decrease is about 3-5 F. per 1000 feet rise, and 
the ground temperature is 50 F. 

Let T be the absolute temperature at a height h, 
then, from hypothesis 

r = 460+ 50 =- h 
1000 

= 510 0035/1 (i) 

Now we know that pv = CT (2) 

and also that if a small rise 8h be considered, the diminution in the 



SIMPLE DIFFERENTIAL EQUATIONS 275 

pressure, viz., 8p, is due to a layer of air 8h feet high and i sq. ft. in 
section, and thus 



From equations (i) and (2) 

pv = (510 -0035^) 
and substituting for v its value from equation (3) 



or, in the limit 

dp i 

pdh ~ (510 0035/1) 

Separating the variables 

dp _ dh 

p "(510 0035/1) 

Integrating, the limits to p being p and p, and those to h being 
o and h, 

(-logpY = x -[log (510 0035 A) -log 5 IO | 
\ /p C -00351- J 

whence log p log p = ~ [log (510 0035/1) -6-234J 

or log p = log Po+^ [_ lo g (5 IO -'35 A ) 6-234J 

which may be further simplified by substituting the values for p and C. 

General Linear Equations of the First Order, i. e., 

equations of the type 



where a and b may be either constants or functions of x. 
The solution of this equation may be written as 



y = 

The proof of this rule depends upon the rule used for differen- 

d(uv) du , dv 

tiatmg a product, viz., ~- =v,--\- u~r- ; the reasoning being as 
dx dx &x 

follows : 

Let us first consider the simplest case in which this type of 
equation occurs, viz., the case of the solution of the equation 

dy 
where a is a constant. 



276 MATHEMATICS FOR ENGINEERS 

Multiplying through by * the equation becomes 

dy , 

e^~--\- aye** = o, 
dx 

dy dv 

which can be written as v^- + v = o (where v = e* an( j thus 

dx dx 

*-- 

But v-r 4- V-T- = ~r(yv) s that -4-- (vv) = o ; hence w must be 
dx dx dx^ Wf w 

a constant, since the result of its differentiation is to be zero. 
Accordingly yv = C, 

or y = Ctr 1 = Ce~ ax . 

Extending to the case in which 6 is not zero, whilst a remains a 
constant, i. e., the equation is 

dy 

+"* 

we find that after multiplying through by &** the result arrived at is 

d 



Integrating both sides with regard to x, 

ye** = fbe^dx+C 
or y=e- flJC {fbe ax dx-\-C}. 

This may be evaluated if the. product of b and e** can be 
integrated. 

For the general case, that in which a and b are functions of x, 
the multiplier or integrating factor is e^ adx , for after multiplication 
by this the equation reads 

dv 

e fadx y J^. ae jadxy = fefadx 

and this may be written 



whence by integration we find that 
y e fadx = 

or y = 



SIMPLE DIFFERENTIAL EQUATIONS 277 

dv 
Example 8. Solve the equation j^+izy = e* x . 

The equation may be written 

dy 12 i 
j-H y = -e** 
** 7* 7 

so that in comparison with the standard form 

a = and b = -e**. 
7 7 



Hence y = -/ J T***{ fie**. e s ' 



-7 47 



47 



Example g. If -j- y = 2x+i, find an expression ior y. 



dv 

so that a= i, 6 = 



Hence y = e+f<'*{f(2x+i)e-J<**dx+C} 



* The value of the integral f(2x+i)e~*dx is found by integrating 
by parts. 

Thus, let (2x+i)e~ x = udv where dv = e~ x , i. e., v = e~ x and 
. du 

U=2X+I, t. e.,-j- = 2, 

then fudv = uvfvdu 

= [(2x+i)x(-e-*)]-f-e-*2dx 



/FT 
Example 10. Solve for T the equation -3 |-PT = P(tcx) (referring 

to the transmission of heat through cylindrical tubes) ; P, t and c being 
constants. 



278 MATHEMATICS FOR ENGINEERS 

The equation -^ [-PT P(tcx) is of the type -/--{-ay = b, where 

(A/X dX 

a = P and b = P(tcx). 

Hence the solution may be written 

T = e-S Pd *(fP(t-cx)ef pdx dx+K}, 
the integrating factor being e^ 7dx , i. e., e^ x . 

Hence T = e - Vx {fP(t-cx)e Px dx+K} ; 

and to express this in a simple form the integral fP(t cx)e Px dx must 
first be evaluated. 

Let fP(tcx}e px dx=fudv where u=P(tcx) and dv = e Px dx so 

that v ^f? x ', and also du = Pc dx, 

then fP(tcx)e px dx =fudv = uvfvdu 

= P(t-cx) x * Pa: +i x e**.Pcdx 



r rp -, 

T = e- px \(t-cx)e**+^- +M+K I 



since e~ fx xe Px e = i and L M+K. 

Example 1 1 . When finding the currents x and y in the two coils of 
a wattmeter we arrive at the following differential equation : 
dy , R,+R 2 



where R x and Lj are respectively the resistance and inductance of the 
one coil and R 2 and L 2 are the resistance and inductance of the other 
coil ; I being the amplitude of the main current. 
Solve this equation for y. 

/1<\) 

Comparing with the standard form of equation, viz., -^--{-ay b, we 



ii . -"-^-l I JLVrt . . J - / l A' *- , t i - LV 1 * , i 

see that a x * and o = ~_, cos ^>^+ T _^ T sin / > ^- 
Hence 

/Rl~f~R2j. / r / T J^T TI> T 
T , T at I // -Lipl , . . -tC,! 
l/l + jjj -( I I ii r.OV5<-l i 

\J VLi+L, 



/2 



- / T J,T 

J / /y_- cospt.e 

IJ Lf-f-1/ 



(Ri+R 2 X 



r Tj r i 2 -, 

+ / * sinpt . e L i+ L 2 dt+C \ 

J L, i -\-L, 2 
=e~ At l fB cospt . e At dt+ /D sin^ . e At dt+C\. 



SIMPLE DIFFERENTIAL EQUATIONS 279 

Now, as proved on p. 157, 

I B0A cos pt dt = * (p sin pt-\- A cos pt) -f C x 

and j ~De A* sin pt dt = A> 1 *, (A sin * cos *) + C 2 
Hence 

- At x e At ( _ 

sin ^i+ AB cos pt+DA sin ptDp cos 



A2 . . a 

= * ] 

R,+R 2 lL.pl R.I 



Exact Differential Equations. An exact differential equation 
is one that is formed by equating an exact differential to zero ; thus 
Pdx-}-Qdy = o is the type, Pdx-{-Qdy being an exact differential. 

The term exact differential must first be explained. 

Pdx -\-Qdy is said to be an exact differential if = ^S, the 

3y dx 

derivatives being partial, or, to use the more familiar notation, 



dy / \dx. 

To solve such an equation proceed as follows : If the equation is 
exact, integrate Pdx as though y were constant, integrate the terms 
in Qdy that do not contain x, and put the sum of the results equal 
to a constant. 

[For, let Pdx+Qdy = du. 

Now, du is the total differential, ( -=- }dx and ( -j- }dy 

\dxJ \dyJ ' 

being the partial differentials (see p. 82) ; 

(du\ (du\ 

i. e., du = ( -=- MX+ I -j- )dy. 

\dx/ \dy/ " 

TU -t ^ ( du \j i ( du \j 

Then if du = o, ( ]dx4-\ -3- jay = o, 

\dx' \ay/ ' 
and this is exactly the same as the original equation 

., (du\ idu\ 

if ( -j- } = P and ( -=- } = Q, 

\dx/ \dy/ 



^y/ \iy/\dx/ \dy.dx 



\ (d\(dii\ (dQ 
I -= )\ -. I 



dx.dyJ \dx/ \dy / \dx 



2 8o MATHEMATICS FOR ENGINEERS 

Our equation thus reduces to du = o, or, by integrating, to u C, 
but u= I Pd% (y being constant) + I Qdy (x being constant), and 
hence we have the rule as given.] 

Example 12. Solve the equation 

(x 2 4%y 2y 2 ) dx+(y 2 4xy2x 2 ) dy = o. 

Here P = x z -^xy-2y 2 (^ P ) = -^x-^y 

\dy 1 



Q = y*-4xy-2x* 
and thus the equation is exact. 

I P dx (as though y were a constant) 

/X^ 4X 2 V 
(x 2 4xy2y z ) dx = 2 -- i 2y*#. 

I Qdy (as though x were a constant) 

= {(y*4xy2x z ) dy = ^* ^ 2Ar 2 y ; 

but of this only ^ 3 must be taken, since the other terms have been 
obtained by the integration of terms containing x. 

Hence \x* 2x 2 y zxy z +$y s = C 

or x 3 6x*y6xy*+y* = K. 

Example 13. Solve the equation v du u dv = o. 






If this equation is multiplied through by - 2 we have a form on the 
left-hand side with which we are familiar, viz., 

v duu dv _ 
v 2 ~~' 

for the left-hand side is d ( - V 

It 

Then by integrating, - = C 

or u = Cv. 

I This equation might have been regarded as one made exact 

through multiplication by the integrating factor 2 . 



SIMPLE DIFFERENTIAL EQUATIONS 281 

Equations Homogeneous (/. e.. of the same power 
throughout) in x and y. 

Rule. Make the substitution y vx and separate the variables. 

Example 14. Solve the equation (x z +y z ) dx = 2xy dy. 

Let y = vx, 

dy dx , dv . dv 

then -f- = V-J-+X-T- = v-\-x- r ..... -. . (i) 

dx dx dx dx 

Now (x z +y z ) dx = 2xy dy 

dy _ x z +y z _ x*+x z v z _ i+v z 

~~" -- - - 



SO , ~~" 

dx 2xy 2x z v 

Substituting for -- from (i) 



or 



dx -2V 

dv i j r v z 2v 2 i 



3- - --- 

dx -2V 2V 

Separating the variables, and integrating, 

fzv dv _ i~dx 
J iv z ~J x 

i. e., log (i v z ) log#+log C rthe substitution being 

u = i v 2 
or log#(i v z ) = logC = log K I du=2vdv 

i.e., x(iv z ) = K 



or 

and x z y* = 



Linear Equations of the Second Order. 



Let v - **, then - Ae^ and - 

X ^ 2 

so that XV^+aAe^+fce** = o 

or X 2 +X+6 = o for jy = o would be a special case. 
There are three possible solutions to this quadratic. 



282 MATHEMATICS FOR ENGINEERS 
The general solution is : X = = 



^ i v a z 46 
and let X, = 



We shall now discuss the three cases. 

Case (i). If # 2 >4&, then Xj and X 2 are real quantities and 
unequal. 

d V tt"V 

Now if v = A, e^ x . -r4 + a-r- + by will equal o, as would be the 
1 dx z dx 

case also if y = A 2 e^, so that to complete the solution the two 
must be included (for the equation is true if either or both are 
included). 

Thus y = Ajtf*!* -\- A%e^ 2X 

the constants Aj and A 2 being fixed by the conditions. 

Case (2). If A 2 = 46, then X 1 = X 2 . 

According to the preceding case we might suppose that the 
solution was y = Ae^. 

This, however, is not the complete solution, 
which is y (A+B*)^. 

Case (3). If a z <4b. This means that Va 2 46 is the square 
root of a negative quantity, i. e., it is an imaginary. 

Now, a 2 4& = 1(46 a 2 ), (46 a 2 } being positive ; 
hence Va 2 46 = V^i Vtfa 2 

a+j\/4ba z 



and 



a A 



Use might be made of the solution to C<zs0 (i), adopting these 
values of X a and X 2 , but this does not give the most convenient 
form in which to write the solution. 



Let c=V4& a z then Xj = 

x 

rt ;c 
and X, = i- 



SIMPLE DIFFERENTIAL EQUATIONS 283 



Then y = A^i^-f- A 2 e*& from Case (i) 

(-a+jc)x (-a-jc)x 



Developing one of these only, viz., the first, and neglecting 
for the time being, 

-ax+jcx _ax jcx 

e 2 = e ~ 2"x~2". 
Now ei* = cos x -\-j sin x (see p. no), 

and by writing for x 

& cx, . cx 

e 2 = cos \-ism 

2 2 

Hence 

"" ' cx , . cx\ , -/ cx . . cx\ 
cos +7 sm + A. 2 e 2 / cos j sin ) 

< 2 2 / \ 2 2 / 



where 

A = V(A 1 +A a ) a +y*(A 1 A 2 ) 2 = 2 VA^A Z . (see Part I, p. 277) 

and 



Taking as the standard equation 



and grouping our results, we have the following 
(i) If a 2 >4&: the solution is 



y == A l6 

(2) If a 2 = 46 : the solution is 

_ax 

y= (A+Bx)e ~*. 

(3) If a z <4b: the solution is 



284 MATHEMATICS FOR ENGINEERS 

The last of these forms occurs so frequently that very careful 
consideration should be given to it, and to the equation of which 
it gives the solution. 

Example 15. Solve the equation 
d z y , dy 

^- 2V = O. 

dx f 



This can be written (after dividing through by 5) 

d z y , dy 
-3-4+2-4-^ 4V = o, 
dx 2 dx 

so that a 2-4 and b -4 (in comparison with the standard form). 

-2-4+v / 5 7 76+l : 6, | . -2'4-v / 6-76+l-6. 

y = A lS 



It is really easier to work a question of this kind from first principles 
rather than to try to remember the rule in the form given ; thus the 
values of X will be the roots of the equation 

5\ 2 + I2X 2=0. 
Then, calling these roots Xj and X 2 respectively 



If the values of A t and A 2 were required, two values of y with the 
corresponding values of x would be necessary. 

Example 16. A body is moving away from a fixed point in such a 
way that its acceleration is directed towards that point, and is given in 
magnitude by 64 times the distance of the body from that point. Find 
the equation of the motion and state of what kind the motion is. 

The motion is Simple Harmonic. (See p. 60.) 
If s = displacement at time t from the start 

= acceleration and = 645 

(the reason for the minus sign being obvious). 

d z s d z y dy 

Thus ^+645=0, which is of the type -^-}-a~-\-b=o, where a=o 

and 6=64. 
If s = e* 

X 2 + 64 = o 

so that X = V 64= 8j. 



SIMPLE DIFFERENTIAL EQUATIONS 285 

Hence the solution (according to Case (3)) is 

_?? 

y Ae 2 sinf -- \-p\ 

where a = o and c V^ba* 16 

so that s = A sin (8t+p) . 

The general equation of S.H.M. is 

s = A sin (o>t-}-p) 

&> being the angular velocity, so that the angular velocity in this case 
is 8 and the amplitude is A. 

Example 17. Solve the equation 



Let y = e** 

then X 2 +8X+i6 = o 

i.e., (X+ 4 )2 = o 
i. e., the roots are equal. 

Hence y = (A+~Bx)e>&, (Cf. Case (2), p. 282) 

where X = 4. 

hence y = (A-\-~Bx)e~* x . 

Example 18. Solve the equation 
d z v dy 



This differs from the preceding examples in the substitution of a 
constant in place of o. 

The equation can be written 

d z y , dy . 
5^+7i+io(y--5)=. 

Let (y~5) = e* x 

then ( Q. = \ e te and ^ 

dx dx z 

and X 2 -|-7X-j-io = o 

whence X = 5 or 2 

then y- -5 = A^ - fa + A z e 
or = Ag- 



//y /-y 

In other words, the solution is that of -=^j + 7 - + ioy = o plus 

tt# i**V 

5 . the constant dy j ^ 

, t. e., - ^^ -- T-. This is correct because, if v = -5, -^ and -v4 
10 coefficient of y dx dx 2 

each equals o, and thus one solution is y -5. The complete solution 
is the sum of the two solutions. 



286 MATHEMATICS FOR ENGINEERS 

The Operator D. The differential coefficient of y with 
respect to x may be expressed in a variety of forms : thus either 

-, -j , f'(x) or Dy might be used to denote the process of 

differentiation. The last of these forms, which must only be used 
when there is no ambiguity about the independent variable, proves 
to be of great advantage when concerned with the solution of 
certain types of differential equations. It is found that the symbol 
D has many important algebraic properties, which lend them- 
selves to the employment of D as an " operator." 

The first derivative of y with regard to x = Dy, and the second 

d*y 
derivative of y with regard to x -j^, which is written as D 2 y ; D 2 

indicating that the operation represented by D must be performed 
twice. This is in accordance with the ordinary rules of indices, so 
the fact suggests itself that the operator D may be dealt with 
according to algebraic rules. Thus D 3 must equal D.D.D (this 
implying not multiplication, but the performance of the operation 
three times) ; for 



.. 

dx* dx\dx 2 / dx dx dx 

Our rule then holds, at any rate, so long as the index is positive, 
or the operation is direct ; and for complete establishment we must 
test for the case when the index is negative. 

If Dy = d j? D = f:let^ = m, i. *., Dy = m. 



Then by integration y = I mdx, 



but if Dy = m and the rules of algebra can be applied to D 

m i 
y must = ^r or =- . m. 

Hence -^m Imdx 



) W ~ \i 



or =- indicates the operation of integration. 

Again, if the rules of indices are to hold, 

D.D" 1 ^ must = Dy or y, 
hence D- 1 must represent the process of integration; since if we 



SIMPLE DIFFERENTIAL EQUATIONS 287 

differentiate a function we must integrate the result to arrive at 
the original function once again. 

Hence D' 1 = ^. 

Having satisfied ourselves that the ordinary rules of indices may 
be applied to D, we may now prove that the rules of factorisation 
apply also. 

Taking the expression D 2 120+32, we can easily show that it 
can be written in the factor form (D 4)(D 8) : 

for let y = jx 2 5x, then Dy = 14* 5 and D 2 jy = 14. 
Also (D 2 120+32)? = D^y i2Vy+32y 

= 14 i68#-f 6o+224# 2 160* 



and 



5 56*2+40*) 4(14* 5 

= 14 II2#+40 56#+20+224# 2 I6OX 



so that 

(D2-I2D+32) = (D-4)(D-8). 

These properties make D of great usefulness in the solution of 
certain types of differential equations : e. g., 

Suppose 5^+7^ + ioy = M ..... . (i) 

then this equation may be re-written as 




M5D 2 +7D+io)=M 
M 



y = 



5 D 2 + 7 D+io 



and the solution of equation (i) may be found by this artifice. 
Many differential equations occurring in electrical theory may be 
solved in a very simple manner by the treatment of D as a " quasi- 
algebraic" quantity: before proceeding to these, however, we 
must enunciate the following theorems. 



288 MATHEMATICS FOR ENGINEERS 

Useful Theorems, involving the Operator D : 

(1) (p+qD) operating on the function a sin (bt+c) 

gives the result aVp 2 +b 2 q 2 sin f^+c+tan" 1 ^ ) 

\ p ' 

(2) rp. a sin (bt+c) = _ - sin (bt+c tan- 1 -). 
p+qV ^*+b 2 * P' 



Proof of (i). 
(p+qD)a sin (bt+c) 
= ap sin (bt+c)+aqb cos (bt+c} 

= aVp*+q*b z sin (bt+c+ia.^^ (See Part I, p. 277.) 

\ p/ 

Proof of (2). 



sn 

J j> sin (U+c)bq cos 



J 
"1 



sin 



a sin ( 6^+c tan- 1 - 
V 



As a test of the correctness of the above rules the combination 
of the two operations should give the original function. 



Thus 

p-\-qu 



sn 



= a sn 
A third theorem might thus be added. 



D sin (bt+c) = 6 cos 
D 2 sin (&^+c) = Z> 2 sin 

or D 2 = -6 2 
hence p z q 2 D z = p 2 +q z b z . 



SIMPLE DIFFERENTIAL EQUATIONS 289 

Application of these Rules to the Solution of Differential 

Equations. 

d^y dy 
Example 19. Solve the equation -5-^+7^4-12^ = e Sx . 

This equation might be written 



S0that ^ 

The solution of this equation gives the particular integral, whilst 
the complementary function, as it is termed, will be obtained by the 
solution of the equation 



dx* dx 
[The solution of this equation we know from the previous work to be 



Now T>e 5x = 5e Sx , DV 5 * = 250^, i. e. t D = 5 and D 2 = 
Hence the particular integral is 



~ 25+35+12 

^g5* 

~7*' 
Hence the general solution is 



To test this by differentiation of the result : 



2 e-* ;t + e &x 
= e* x . 

d z s ds 

Example 20. Solve the equation ,- +4^7+45 = 5 sin 7/. (This type 

at dt 

of equation occurs frequently in electrical problems and in problems on 
forced vibrations of a system.) 

U 



290 MATHEMATICS FOR ENGINEERS 

/72 c /7c 

The solution of --2+4-^7+45 = 

(It Cli 

is s = (A+Bt)e~ zt (See p. 283.) 

To find the particular integral : 

(D 2 + 4 D + 4 ) 5 = 5 sin jt. 

_5sin jt 
-DM-4D+4 

D sin jt = j cos jt and D 2 sin jt = 49 sin jt. 

(Note that D 2 = 49, but D does not = j.) 

We must thus eliminate D from the denominator: to do this, 
multiply both numerator and denominator by D 2 +4 40. 

Then _5(P 2 +4-4P)sin 7 / 

(D 2 + 4 ) 2 -i6D 2 

_ 5( 49 sin 7^+4 sin jt28 cos jt) 
(-49+4) 2 -(i6x- 4 9) 

_ 5(45 s i n 7^+28 cos jt) 
2809 

_ 28 
45 



2809 
Hence the complete solution is 



53 A' ' in ~ 1 4-5> 



53 ~~V ' " dn 45A 



The particular integral might have been found in a more direct 
fashion by the use of Theorem 2, p. 288. 



For 



sin 7* i 

(D+2) 2 -(D+ 2 ) 



(7\ (4) = 2 

jt tan" 1 -), _,, 
_ 2j from Theorem 2, J ? = 



sinf jt tan -1 -J 



p. 288 I c = o 

U = 7 



I I / 7 7 

= 7= X 7= sin 1 7i tan" 1 - tan -1 - 
V 53 V 53 V 2 2 

= sin ( jt 2 tan" 1 - Y 
53 V 2} 



SIMPLE DIFFERENTIAL EQUATIONS 291 

The two results do not appear at first sight to be the same, but the 
can be reconciled in the following manner : 

tan" 1 = tan" 1 -6223 = 31 54' 
tan- 1 ? =74 3' 

Thus sin (7/4- tan- 1 ) = sin (7/4-31 54') 
\ 45' 

also sin (7* 2tan~ 1 -J = sin (jt 2x74 3') = sin (jt 148 6') 
= sin (1804-7^ 148 6') 
= - sin (7/4-31 54') 

= sin ( 7/4- tan" 1 ). 
v 45/ 

IT Tl, -.-' T^T&y , T- , B/ 1TX 

Example 21. The equation EI- r ^4-Fy4--^- cos-=- = o occurs in 

dx o I 

Mechanics, y being the deflection of a rod of length /, and F being the 
end load. 

Solve this equation. 

,, T d z y , F B/ TT* 

We may rewrite the equation as -r^+^fV = B^F cos -; I 1 ) 

dx z EF 8EI / 



The solution of +-^y = oisy = Asui/\ j^x+p ... (2) 
Reverting to form (i) 



B/ nx 



B/ 

or 8EI C<S 



B/ irX 

8- cos 



F 

B/ 



Hence the complete solution is y = A sin ( *J ~pX-{-p j4- 



8 COS T 



292 MATHEMATICS FOR ENGINEERS 

Example 22. A pin-jointed column, initially bent to a curve of 
cosines, has a vertical load W applied to it. Find an expression for 
the deflection at any point. 

Given that the equation of the initial bent form is 

irX 

y being the deflection at distance x from the centre of the column, 
which is of length /. 

Also 



this equation being obtained from a consideration of the bending 
moment at distance x from the centre. 



, 

and thus +--A cos 



W AW fnx 

iLi y = -ET COS 



firX\\ 

(y J) = o 
fnx\ 

(T> 



Now, as shown on p. 283, the solution of the equation 
d* W 

W 

To find the particular integral, viz., the solution of the equation 
d* . W AW 



write the equation as 

AW 



AW /irX\ 
-^-= COS ( ) 

so that = - \ l > 



AW l-nx 





El I* 
Combining the two results 



(r). 



SIMPLE DIFFERENTIAL EQUATIONS 293 

The following example combines the methods of solution 
employed in Examples 17, 18, 19 and 20. 



Example 23. Solve the equation 
d 2 s ds 



= -+ sin 607T/+5. 



(a) The solution of 



-5^ 12-^ + 205 = is s = A. 1 e lot +A z e zt . 



(b) The particular integral for 

d?s ds 
d7*- 12 di 
may be thus found : 



D 2 I2D+20 D 2 e~ 5t = 2$e- 

i 



25+60+20 



(c) To find the particular integral for 



d*s ds 

-jr. i2^-+2os = sinooiri. 

dt z dt 



D 2 I2D + 20 

I ^ sin6o7r< 
D 10 D 2~ 

^ ~/' = \ ==,; sin [6o7rf tan" 1 (3071-)] \q = i J- 

D-io V4+36007T 2 l& = 6o^J 

i i 

/- -.- , -^ ' -/ ^~z a sin [607T/ tan" 1 (30^) tan" 1 ( 6rr)l 
v4+3ooon- a Vioo+3ooo7r i 

sin [6orr< tan^ 1 (3077) tan" 1 (677)] 

2oV(l+9007T 2 ) (I + 367T 2 ) 

(d) The particular integral for 
d z s ds 



is 

20 4 



Hence the complete solution is the sum of those in (a), (b),-(c) 
and (d), viz., 

s = A e"'+A g ^| g ~ 5 

1 2 



105 _ 2oV(i+90on 2 ) (I+367T*) 



2 9 4 MATHEMATICS FOR ENGINEERS 

Equations of the Second Degree. The treatment of these 
equations is very similar, up to a certain point, to that employed 
in the solution of ordinary quadratic equations ; particularly the 
solution by factorisation. 

Example 24. Solve the equation 

^Y- 8^-33=0 
dx/ dx 



Let Y= then Y 2 -8Y~33=o 

dx 

or (Y-ii)(Y+ 3 ) = o. 
Thus Y = 1 1 or Y = 3 

dy dy 

i.e., / = ii or ^-= 3, 

dx dx 

whence y = i ix + C t or y = $x + C 2 
or y i ix C x = o y + 3, C 2 = o, 

and the complete solution is the product of these two solutions, since 
the equation is of " degree " higher than the first. 
Thus the solution is 

(yiixCJ (y+3x-C z )=o. 

(dv\ 2 
Example 25. Solve the equation 5! - ) 8y 8 =o. 

\ax/ 



3\ A v I * 

Dividing by 5 

Factorising \-~^- 1-26^ )(-/- 1-265^) = - 

\CiX / \CLX / 

Hence -^-+1-265^^ = or -^-1-265^ = 0. 

Separating the variables and integrating 

/^+i'265/^ = o or -% 1-265 I dx = 

J yl J J yi J 

whence the complete solution is, 



Two further examples are added to illustrate methods of solu- 
tion other than those already indicated. 



SIMPLE DIFFERENTIAL EQUATIONS 295 

Example 26. Solve the equation -j-^ -^ 14-^+2 AV = o. 

dx 3 dx* dx 

The equation may be written (D 3 D 2 i.fD-f 24)^ = o 
or (D-2)(D-3)(D+ 4 )y = o, 

whence y = 



Example 27. The equation j^= m*y occurs in the discussion of 
the whirling of shafts. Solve this equation. 



i. e., D 4 = m* 

or (D 2 m 2 ) (D 2 +m 2 ) = o, 

whence D= m or jm. 

Hence y = a 1 e mx -^-a z e~ mx -\-a 3 e jmx -\-a t e- intx . 

But e' x = cos x+j sin x, e jnue = cos mx+j sin mx 

and e x cosh AT+ sinh #, 

i.e., e nix = cosh w# + sinh mx. 

y = ^ (cosh w#+ sinh m#) +a(cosh mx sinh w#) 

+a 3 (cos w# +y sin ra#) + 4 (cos mxj sin w#) 
= ( a a+ a 4) cos wwr-f (a 3 a 4 ); sin w#+ (i+^ 2 ) cosh mx 

+ (! a 2 ) sinh WAT 
= A cos m# -f- B sin mx-\-C cosh m^+D sinh mx. 

The constants A, B, C and D are found by consideration of the 
conditions; four equations must be formed, these being found by 

successive differentiation and by substituting for f- , -^ and ?? their 

dx dx z dx 3 

values for various values of x. 

Exercises 22. On the Solution of Differential Equations. 

1. If -p = 5# a 2-4 and y = 1-68 when x = 2-29, find y in terms of x. 

d z s ds 

2. Given that = 16-1 ; -=7 = 4-3 when t = 1-7 and s = 9-8 when 



f = -2, find s in terms of t. 

3. If ^ = 8y+5, find an expression for y. 



296 MATHEMATICS FOR ENGINEERS 

dv 

4. Given that 8-76-^- +9* 15^ = 76-4 and also that ^ = 2-17 when 

G/X 

x = o, find an expression for y in terms of x. 

5. A beam simply supported at its ends is loaded with a concentrated 
load W at the centre. The bending moment M at a section distant x 
from the centre is given by 

W/ 1 \ 
M = 4 - x }. 

2 \2 / 

M d z v 
If == = -j^, find the equation of the deflected form, y being the 

deflection. 

6. For the case of a fixed beam uniformly loaded, M, the bending 

(w(P ,\ ) T , M d*y dy I 

moment, =1-1 x 2 1 K I . If ==- = ^; -f- = o when x = - and also 

1 2X4 / El dx z dx 2 

when x = o ; and y = o when x = - , find an expression for y. 

7. Solve the equation 

dx i 



the limits to T being T x and T 2 , and to x being o and / ; the remaining 
letters representing constants. 

8. If -j- = 1- and v = o when x s, find v in terms of x. 

ax 4//x 

9. If T! = the absolute temperature of the gases entering a tube of 
length / and diam. D, 

r 2 = the absolute temperature of the gases leaving this tube, 

6 = temperature of the water, 

Q = amount of heat transmitted through the tube per sq. ft. per sec. 
per degree difference of temperature on the two sides, 

w = weight of gases along the tube per sec., and s = specific heat 
of gases, 

, dr QnT^dx 

then T H = o. 

r ws 

Find an expression for Q, x being the distance from one end of the 
tube. 

10. Find Q if Q^-nDdx-^-wsdr = o (the letters having the same 
meanings as in No. 9, and the limits being the same). 

11. If T-J and r 2 are the inside and outside temperatures respectively 
of a thick tube of internal radius r t and external radius r z , then 

rfr H 

dx ~~ 27rK I 



I is the length of the tube, H is a quantity of heat, and the limits to 
are o and r z r v Find an expression for H, K being a constant. 



SIMPLE DIFFERENTIAL EQUATIONS 297 

12. A compound pendulum swings through small axes. If I = 
moment of inertia about the point of suspension, h = the distance of 
C. of G. from point of suspension, then 

I X angular acceleration f i. e., 1^) = mhd. 

Find an expression for 6. 

If p. = couple for unit angle mh, prove that t, the time of a com- 

plete oscillation, = 2/r \/ (in Engineers' units). 

r O 

13. To find expressions for the stresses p and q (hoop) in thick 
cylinders it is necessary to solve the equation 



Solve this equation for p. 

14. For a thick spherical shell, Up radial pressure, 



Find an expression for p, a being a constant. 

15. If ~K. tt vdp = ~K. p pdv, prove that pv = constant, y being the 

,. the specific heat at constant pressure .. K D 

ratio -rr - r= r - of a gas, i. e., ==? 

the specific heat at constant volume K B 

16. Solve for z in the equation 

dz . w w 2 

3" H --- 7- zx - 
dx g f 

17. Solve the equation 

d z y dy 
^-i7^ 

and thence the equation 

d z y dy 

d^- l7 Tx 

d z s 

18. Solve the equation 875. 

at 

d z s 

19. Solve the equation ^73+875 = o. 

(tt 

20. Find the time that elapses whilst an electric condenser of 
capacity K discharges through a constant resistance R, the potential 
difference at the start being t^ and at the end v z , being given that 

K x rate of change of potential = = 

jf* 

21. If V = RC+L-r- and V = o, find an expression for C; C being 
the initial current, i. e., the value of C when t o. 

J/~* 

22. If V = RC-f L -- , and V = V sin qt, find an expression for C. 

ctt 



298 MATHEMATICS FOR ENGINEERS 

23. If -y- = -Vy z +2ay, find x in terms of y, a being a constant. 

dx a 

24. An equation occurring when considering the motion of the 
piston of an indicator is 

ffix _a^ pa. 
M ^ + SM M 
Solve this equation for x ; M, a, S and p being constants. 

25. If - Py = Eig. 

(an equation referring to the bending of struts), find y; given that 
x = o when y o, and v = Y when x = 

26. To find the time t of the recoil of a gun, it was necessary to 

solve the equation = n Vx z a 2 . 

dt 

If a = 47-5, n 3-275 and the limits to x are o and 57-5, find t. 

27. Solve the equation -^-\-2/~-\-n z x o. 

(Ait (Jit 

dx 
Take w 2 = 200, /= 7*485 ; also let x = o and = 10 when t = o. 

Cut 

28. The equation ^2"+ 2 /^i +n z x = a sin qt 

(It Q/v 

expresses the forced vibration of a system. If n 2 = 49, / = 3, q = 5, find 
an expression for x. 

d z V r 

29. Solve the equation -,-= V = o. 

dx 2 r z 

30. If H is the amount of heat given to a gas, p is its pressure and 

v its volume (of i lb.), then -,~ = 7 -Av^+np). Assuming that 

dv (n i) V dv */ 

there is no change of heat (i.e., the expansion is adiabatic and -5 = o J, 

find a simple equation to express the connection between the pressure 
and volume during this expansion. 

31. Newton's law of cooling may be expressed by the equation 

5" *<- 

where k is a constant, and 6 a is the temperature of the air. 
If Q = Q when t = o, find an expression for 6. 

32. The equation 1-04-^ +12-3 +13^634=0 

(It Ctt 

occurred in an investigation to find 6, the angle of incidence of the main 
planes of an aeroplane. 

If t = o when 6 = i and = o when t = o, find an expression for 6. 
at 



SIMPLE DIFFERENTIAL EQUATIONS 299 

33. A circular shaft weighing p Ibs. per ft. rotates at o> radians per 
second, and is subjected to an endlong compressive force F. The 
deflection y can be found from the equation 

d*y IF dfy _ p u 
d^ + mfa* # 
Solve this equation for y. 

34. An equation relating to the theory of the stability of an aero- 
plane is 

dv 

= g cos a kv 

Civ 

where v is a velocity ; g, a and k being constants. Find an expression 
for the velocity, if it is known that v = o when t o. 



CHAPTER X 
APPLICATIONS OF THE CALCULUS 

THE idea of this chapter is to illustrate the use of the Calculus 
as applied to many Engineering problems ; and the reader is 
supposed to be acquainted with the technical principle involved. 

The various cases will be dealt with as though examples. 

Examples in Thermodynamics. 
Example i. To prove that (V w) = =5-, an equation occurring 

T GtfT 

in Thermodynamics, 

where L = latent heat at absolute temperature T, 

V = vol. of i Ib. of steam at absolute temperature T, 
w = vol. of i Ib. of water = -016 cu. ft., 
P = pressure. 

A quantity q of heat taken in at r-f-8r and discharged at r will, 
according to the Carnot cycle, give out work = q r r - or approxi- 

T-j-OT 

4. 1 8f 

mately q . 

T 

Hence for i Ib. of steam at the boiling temperature, 

Sr 87- 
work = q = L , 

T T 

but the work done = volume of steam in the cylinder x change in pressure 



Hence (V-w)8P 

and thus V w= r^, 

T oir 

or, as Sr becomes infinitely small, 

X7 . L,dr 

V w= 

T dP 

Now -Tp is the slope of the pressure temperature curve (plotted from 

300 



APPLICATIONS OF THE CALCULUS 



301 



the tables) and can be easily found for any temperature r. Hence V can 
also be found. 

A numerical example will illustrate further. 

It is required to find the volume of i Ib. of dry steam at 228F., 
i. e., at 20 Ibs. per sq. in. pressure. 

From the /When P= 19, * = 225-3, r= 460+225-3 = 685-3. 

x 1.1 -^ .. P = 20,* = 228, T = 688. 
steam tables ' .. - 

^ P = 2i, t= 230-6, T = 690-6. 

Plotting these temperatures to a base of pressures, we find that the 
portion of the curve dealt with in this range is practically straight. 



r 

69O 
689 
688 
687 
686 
685 


















/ 
















3 


f 












/ 


/ 


2-65 








t 




s 

i 





H 


! 






s 


/* 










/ 


S 














7 



















/9 eo p e 

FIG. 103. Problem in Thermodynamics. 



The slope of this line = 2-65 (Fig. 103), and this is the value of 
-==, P being given in Ibs. per sq. in. and the latent heat in thermal 
units. To change the formula to agree with these units, 

778L dr 





v = 

Also L at 228 F. = 953. 
V = -oi6-f 



dP' 
778x953x2-65 



144x688 
= 19-82 cu. ft. 



Example 2. To prove that the specific heat of saturated steam 
(expanding dry) is negative. 



Let 



Q = the quantity of heat added. 
H = total heat from 32 F. 
I = internal energy of the steam. 



302 



Then 

i.e., 
or 

Now 



Then 
and 



MATHEMATICS FOR ENGINEERS 
H = internal + external energy 



I = H-PV 
81 = 8H-8(PV)=SH-(P8V+VSP). 



= SH-PSV-V8P+PSV 
= 8H-V8P 

<\T r -*-* <\ 



8r 



L 

r 
L 



(from. Example i, neglecting^ 
\ w, which is very small J * 



H = 



dT^' 305 

Now the specific heat = heat to raise the temperature i 



hence 
L. 



= -305 - - 



and since - is greater than -305, 5 is a negative quantity. 
E. g., if t = 300 F., i. e., r = 761 F. absol., 



= 1115 210 = 905. 

95 

.'. s = '305 ^-^ 
3 J 761 

= - -882. 

Work Done in the Expansion of a Gas. 

Example 3. Find the work done in the expansion of a gas from 
volume v l to volume v a . 

There are two distinct cases, which must be treated separately; but 
for both cases the work done in the expansion is measured by the area 
ABCD = % areas of strips like MN (Fig. 104) 

= ^,fp &v or I p dv more exactly. 

J Vl 

Case (a), for which the law of the expansion is pv = C. 

/'2 /"2 / \"2 

done = / pdv= I Cv~*dv = Cl log v } 

Jit* J Vi \ h Jvi 



Work 



or C log r 



r being the ratio of expansion, and = . 
Thus the work done = pv log r. 



APPLICATIONS OF THE CALCULUS 



303 



Case (b), for which the law of the expansion is pv n C, n having any 
value other than i . 




1 ** 1 



M 

FIG. 104. 
Work Done in the Expansion of a Gas. 



V 

FIG. 105. 



Work done = / 2 pdv = \ *Cv~ndv 

J t7j Dj 



i n 

i 
in 

i 

(p( 
i n 

Pz v z Pi 
in 



Work Done in a Complete Theoretical Cycle. 

Example 4. Find the work done in the complete cycle represented 
by the diagram FGAB in Fig. 105. 

The work done = area GABF = ABCD+GADH FBCH 



in 



304 MATHEMATICS FOR ENGINEERS 

Note that, if n = ^| 

16 

work done == 



If the expansion is adiabatic, and n is calculated according to 
Zeuner's rule, n = 



(q being the initial dryness fraction). 

If q = i, then n = i-O35+-i = i - i35, so that the work done 



To Find the Entropy of Water at Absolute Temperature r. 

Example 5. When a substance takes in or rejects heat (at tempera- 

ture T) the change in entropy 8$ = (8q = heat taken in) . 

Let a = specific heat, 

then o-Sr = 8q. 

dq 



Change in entropy from r to T = f 

J TT 



dr 



. 
T o 

r_ - /i.e., in the change from\ 

For steam, the heat taken in at T = L { ). 

\_ \ the liquid to the gas / 

Hence the change of entropy = 

Efficiency of an Engine working on the Rankine Cycle. 

Example 6. Find the efficiency of an engine working on the 
Rankine cycle ; using the T(f> diagram for the calculation. 

Work done = area of ABCD (Fig. 106.) 
= ABCK+ADMN-DKMN 

= * x (TJ T 2 )+heat taken in from 

TI Tjtorj (r t x'DK). 

Now DK = the change in entropy from water at T Z to 
water at T X 

= loge as proved above. 



APPLICATIONS OF THE CALCULUS 305 

Hence the work done = ^-^ (T I r 2 ) + (r 1 T 2 ) r a log 



The heat put in 



and thus the efficiency TJ = 



r 




r 



being the dryness 

fraction at 



FIG. 106. FIG. 107. 

Efficiency of an Engine. 

Efficiency of an Engine working on the Rankine Cycle, with steam 
kept saturated by jacket steam. 

Example 7. Find the efficiency of the engine whose cycle is given 
by abcf in Fig. 107. 

Work done = area abcf 



_ f'aLj /the summation being of\ 
~~ j TI T \ horizontal strips ) 



[for L = 1115 -7/1 
dr =1437 7r 

= a+br 



= alog- a +6(r 2 TJ) 



Total heat received = L 2 +r 2 rj 
total heat rejected = Lj 
Hence the work done = (i) (z). 



. . (i) 
(2) 



306 MATHEMATICS FOR ENGINEERS 

from which H,- = a log +6(r 2 TJ} (L 2 L x ) (r 2 
T i 

= a log -+6(T 2 T X ) (a+&T 2 a 6 

- 



T l 

Now the total heat received = L 8 +r a 

= L 2 +r a 



Hence 



where a = 1437 and b = "j. 

Example 8. To prove that the equation for adiabatic expansion of 
air is pv y = C, where 

_ specific heat at constant pressure _ K p 
' specific heat at constant volume K w 

Dealing throughout with i Ib. of air, let the air expand under 
constant pressure from conditions p 1 v 1 r x to p^ v r. 

Then the heat added = K p (r r x ) = K/^- ^~ 



Now keep the volume constant at v, and subtract as much heat as 
was previously added : then the pressure falls to p z and the tem- 
perature to r a . 

The heat subtracted = K,(i T,) = K v (^- 



Now, if the changes are regarded as being very small, we may write 
for v v l and 8p for pip z 

and thus K v v&p = K p p8v 

, (dp Kp rdv 

whence / -~ = ~ / 

/ P KJ i; 

log p = y log w + lg (constant) 
i. e., /> = Cy-v 

or v~i = C. 



APPLICATIONS OF THE CALCULUS 307 

Examples relating to Loaded Beams. 

Example g. Prove the most important rule 
M_E_ d*y 
I R~ dx z 

applied to a loaded beam ; M, I and E having their usual meanings, 
and R being the radius of curvature of the bent beam. 

Assuming the beam to be originally straight, take a section of length 
/ along the neutral lamina, and let l-\-8l be the strained length at 
distance y (Fig. 108). 

A - 



li\ 




FIG. 108. Problem on Loaded Beam. 



Then, if R = radius of curvature, 
1+81 _ 



whence 



or 



but 



and thus 



I 



R 



1 + 81 



R 



I ~R 

stress / / /R 

1-4 - - _ . J "i. _ - . **_ _ 

strain 3/. y y 



I R 



R 



M / 
but it has already been proved (see p. 239) that y = 



Hence 



M = E 

i ~R' 



308 



MATHEMATICS FOR ENGINEERS 



The total curvature of an arc of a curve is the angle through which 
the tangent turns as its point of contact moves from one end of the 
arc to the other ; and the mean curvature is given by the total curva- 
ture divided by the length of arc. 

In Fig. 108 8$ = total curvature for the arc 65, and the mean curva- 

S</> 

ture = -5-- 
8s 

We know that the slope of the tangent is given by ~ 

flwv 



.-. tand> = 



dy 
-f- 
dx 



Now 



, , 

tan $ = sec 2 $ and 



, 
ds 



2 sk ^^ ^ i^y 

SCC O ~^ ~^ ~5~ I ~~5 

tis as \a^ 

i d (dy\ 

sec 2 X s = x I T^ ) 

R as Va^r/ 



dx\dx) ds 
dx* ds 



Hence ^ = -^~ X - x cos 2 
R dx* ds 



d*y 
dx* 



d<b 

X --= 
ds 



dy 

tan (b = - 
dx 



When, as for a beam, -~ is very small, (-*- j may be neglected in 
comparison with i, and hence 



This result may be arrived at more briefly, but approximately, in 
the following manner : 

80 = 8 tan very nearly (when the angle is very small) . 

Hence ~- = -- -- = - tan <6 = rate of change of the tangent (for 
8s 8s 8x 

PM and PQ are sensibly alike). 



APPLICATIONS OF THE CALCULUS 309 

Thus ^^A.^^^y 

ds dx dx dx* 

i d 2 y 
and _ = _. 

R dx z 

M = / = E_ F ^ 
I y R dx* 

In the use of this rule there should be no difficulty in finding 
expressions for y in terms of x in 
cases in which the beam is simply 
supported; for an expression is 
found for the bending moment at 
distance x from one end, or the 
centre, whichever may be more con- pi 
venient, and then the relation 




M d 2 y 

=- = E -n FIG. 109. Beam Uniformly 

n v& T j j 

w;t Loaded, 

is used ; whence double integration 

from the equation so formed gives an expression for the deflected 
form. 

A few harder cases are here treated, the beam not being simply 
supported. 

Example 10. A beam is fixed at one end and supported at the 
other ; the loading is uniform, w being the intensity. Find the equation 
of the deflected form. 



We must first find the force P (part of the couple keeping the end 
fixed) and then combine this force with the reaction at B calculated 
on the assumption that the beam is simply supported. Referring to 
Fig. 109: 

If the beam is simply supported, the bending moment at distance x 

t -n Wl WX 2 

from B = x 

2 2 

Hence the actual bending moment 

,, wlx wx 2 _ 
= M = Px 

2 2 

-~~d 2 y wlx wx 2 .._. 

i.e., El ^-4= Px 

dx* 2 2 

whence, by integration, 

_ T dy wlx 2 wx 3 Px 2 . ~ 

EI-j^- = hQ 

dx 4 62 



3io 



MATHEMATICS FOR ENGINEERS 



but 



-f- = o when x = I 
ax 



for the deflected form is horizontal at this end. 

wl 3 wl 3 PI* , 

o = f-C, 

4 62 

r _Pl 2 wl 3 

1 ~~ 2 12 

-rfrdy _ wlx z wx 3 Px z .Pl z wl 3 
dx 4 6 2 2 12 



Hence 



T , , . _ ze># 4 P# 3 , Pl 2 x wl 3 x . 

Integrating, Ely = ^-C 2 . 

12 24 6.2 12 

In this equation there are the two unknowns P and C 2 , and to 
evaluate them we must form two equations from the statements 

y = o when x = o and y = o when x = 1. 
If y = o when x = o, then it is readily seen that C a = o. 

wl* wl* PI 3 , P/ 3 wl* 
Also if x = l o = ----- ? ---- 

12 24 6 2 12 

1, T, W l 

whence P = - 5 - 

o 



Point' of Conhraflexure 
/_ 



Fixed End 




FIG. no. Deflected Form of Beam. 

If the beam were simply supported, the upward reaction would be 

wl , ,, ,, , . wl wl 3 . 

, and thus the net reaction = 5- = ^wl. 

2 288 

Substituting -5- in place of P in the expression for y, we arrive at 

o 

the equation of the deflected form 



the curve for which is shown in Fig. no. 

We may now proceed to find where the maximum deflection occurs, 
and also the position of the point of contraflexure. 

dy w 

dx 48EI ' 

dy 



and 



T X = 



the solution of which, applicable to the present case, is x -423^ 



APPLICATIONS OF THE CALCULUS 311 

064 -423) 



The maximum deflection is thus 
wl* 



005 4 wl* 
~EI~ 

To find C the point of inflexion or contraflexure 



and 3-^ = o if x = o or if x = /. 



Example n. A beam is fixed at one end and supported at the 
other, the loading and the section both varying. Find the equation of 
the deflected form. 



Let m = bending moment at a point distant x from B if the beam were 
simply supported, and let P= the force of the fixing couple. (See Fig. 109.) 
Then M = mPx 

i. e., Elf^ = W _ P * 

, . dx z 

.~d z y m Px 

E ^ = I-T 

the equation being written in this form since I is now a variable. 

By integrating E^= [ X ^dx-P F^dx+Cj. ...... (i) 

t*x J I J 1 

dy 

Now -j ss o when x = 1. 

dx 



i. e., Cj can be found, for the two integrals may be evaluated. 
By integrating (i), 

Ey = f f2(rf*)-pr f \ (d^+c.x+c,. 

J o .' o L JoJp 1 

But j/ = o when x = I and also when x = o, and thus C 2 o 



and 

J oJ o 
i. e., P can be found. 

The integrations must be performed graphically and with extreme 
care, or otherwise very serious errors arise. 

Example 12. A beam is fixed at both ends and the loading and the 
section both vary. Find the equation of the deflected form. 



3 i2 MATHEMATICS FOR ENGINEERS 

Let Wj and m 2 be the end fixing couples ; then to keep the system 
in equilibrium it is necessary to introduce equal and opposite forces P 
(Fig. in), i. e., P/+w 2 = Wj. 

Let m = the " simply supported " bending moment at section 
distant x from the right-hand end. 

Then M = m m z Px 

d z v 
and consequently El-r^ = mm z Px 

d*m m Px 



Integrating, 

_ 
E 



dy c x m, fxdx ^f x x 

/- = / -=-dxm t l -V- P/ f 
dx . o I *J I J I 



,4 




-I 




FIG. in. 

Now -jL*mQ when x = o or 

dx 

hence C x = o (taking ^ = o) 

and also, taking x = I, 



Integrating again, 

Ey = / r^w-m. r r j w-p r ff w 

J QJ * J OJ i J 0-' X 

Now y = o when # = o and also when x = I. 

Then taking x = o, C 2 = o, 

and taking # = /, 



From equations (i) and (2) the values of w a and P (and hence Wj) 
may be found, the integration being graphical (except in a few special 
cases) ; and again it must be emphasised that the integration must be 
performed most accurately. 



APPLICATIONS OF THE CALCULUS 



313 



Example 13. A uniform rectangular beam, fixed at its ends, is 
20 ft. long, and has a load of 10 tons at its centre and one of 7 tons 
at 5 ft. from one end. Find the fixing couples and the true B.M. 
diagram. 

This is a special case of Example 12 since the section, and therefore 
I, is constant. 

The B.M. diagram for the beam if simply supported would be as 
ABCD (Fig. 112). 

The bending moment diagram, due to the fixing couples only, would 
have the form of a trapezoid, as APQD. 

Unless the integration, explained in the previous example, is done 
extremely carefully, there will be serious errors in the results ; and 
since there are only the two loads to consider, it is rather easier to 




FIG. 112. Fixing Couples and B.M. Diagram of Loaded Beam. 



work according to the Goodman scheme. [See Mechanics applied to 
Engineering, by Goodman.] 

According to this plan : (i) the opposing areas (i. e., of the free and 
fixing bending moment diagrams) must be equal ; and (2) the centroids 
of the opposing areas must be on the same vertical, i. e., their centroid 
verticals must coincide. 

To satisfy condition (i), 

Area of ABCD = (^X5X 5 i-2 5 )+( 5I ' 25 + 67 ' 5 X5)+(^Xiox67-5) 
= 762-5. 



Area of APQD = - x 20 x (w 1 +m 2 ) where m^ = AP\ 

and w = D 



-> 10 



Equating these areas, m 1 -{-m t 76-25 



(i) 



314 MATHEMATICS FOR ENGINEERS 

To satisfy condition (2), taking moments about AP, 



r 

Moment of ABG = -X5I-25X-X5 



= 427 



Moment of BGH = -X5i - 25X/5H X5) = 8 54 

r / 2 \ 

Moment of BHC = -x 67-5x1 5H X5j =1405 

Moment of DCH = X 67-5 x ( io-| x 10 )= 4500 
2 V 3 / 

i. e., total moment of ABCD about AP = 7186 

, . _._ 20 20 200 

Moment of APD = x w, X = n. 

2 3 3 

Moment of DPQ = x m z X = m 



Hence 



(2) 



200 .400 

^1 + m 2~ 

The solution of equations (i) and (2) for m^ and w 2 gives the results 

m i = 44-7 1 and m 2 = 31-54. 

Thus PQ is the true base of the complete bending moment diagram, 
AP being made equal to 44-71, and DQ equal to 31-54. 

Shearing Stress in Beams. 

Example 14. To find an expression for the maximum intensity of 
shearing stress over a beam section. 

The shearing stress at any point in a vertical section of a beam is 
always accompanied by shearing stress of equal intensity in a hori- 
zontal plane through that point. 





FIG. 113. Shearing Stress in Beams. 



FIG. 114. 



We require to know the tangential or shearing stress / at E on the 
plane CEC' (Fig. 113) ; this must be equal to the tangential stress in 
the direction EF on the plane EF at right angles to the paper. 
Suppose that the bending moment at CC'=M and that at DD'=M+SM. 
Then the total pushing forces on DF > total pushing forces on CE, the 
difference being the tangential forces on EFE/. 



APPLICATIONS OF THE CALCULUS 315 

Let P = the total pushing force on ECE' 

,RC (stress = M "j 

then P = / (stress) x area 1 y IV 

J RE I area = bdyj 



RE J- 

M /" RC M 

= v / by dy = =- x ist moment of area ECE'. 
I. 'RE 

Now the tangential force on EFE' = stress x area 

=/xEE'x8* 

and this must equal the difference in the total pushing forces on DF 
and CE, i. e., 8P. 

Hence 8P = /x EE' x bx 

i. e., -=r- x ist moment of area ECE'=/ x EE' ; 



but = rate of change of B.M.=shear=F (say). 

Hence the maximum intensity of shearing stress/ 

~F T 
= ist moment of area ECE'x ? X: 



I "BE* 

or, as it is usually written, 



~ bl 

where S = an area such as CEE' and y = distance of its centroid from 
the neutral axis. 

Example 15. Find the maximum intensity of shearing stress, when 
the section is circular, of radius r (see Fig. 114). 

For this section 

I = V. 

4 
Applying the rule proved above : 

F ,-r since zr corresponds to EE' 

the maximum intensity = -^ J by dy in Example ^ 



du ('where u = r*y z 

dy~~ 



= - X mean intensity. 



316 MATHEMATICS FOR ENGINEERS 

Example 16. A uniformly tapered cantilever of circular cross 
section is built in at one end and is loaded at the other. The diam. 
at the loaded end is D ins. and the taper is t ins. per in. of length. 
Find an expression for the distance of the most highly stressed section 
from the free end of the beam due to bending moment only. Neglect 
the weight of the cantilever. 

Let / be the length in ins. and W the load at the free end. Consider 
a section distant x ins. from the free end ; then the diam. here is 
D tx, and the bending moment W#. 

Also the value of I for the section considered is 



. _ 

My /D tx\ 64 

Hence the skin stress f = -^- = Vfx( - IX -7-^--. 

I V 2 / 7r(D tx) 



K _ 32W 

D 3 - F ' ~ 



x (a constant), 

and / is a maximum when the denominator is a minimum. 
Let N = denominator 



then =_ 

dx 

dN 
and -j = o when 2t 3 x 3 ^Dt 2 x 2 -\-D 3 o 

d 'X 

i. e., when 2t 3 x 3 zl)t*x 2 D/ a ^ 2 +D 8 = o 

2t*x z (tx-~D)-I)(t 2 .v 2 -~D z ) = o 

(te-D)(2^-Dte-D) = o 

(txT>)(2tx+T>)(tx'D) = o 

D D 

t. e., when x or --- . 
t it 

Thus the stress is maximum at a section distant ins. from the 

t 

free end. 

Example 17. To find the deflection of the muzzle of a gun. 

This is an instructive example on the determination of the deflection 
of a cantilever whose section varies. 

The muzzle is divided into a number of elementary discs, the 
volumes of these found (and hence the weights) so that the curve of 
loads can be plotted. Integration of this curve gives the curve of 
shear, and integration of the curve of shear gives the B.M. diagram. 



APPLICATIONS OF THE CALCULUS 



317 



The values of I must next be calculated for each disc, and a new 
curve plotted with ordinates equal to : then double integration of 

this gives the deflected form. 

It is necessary to use the ordinates of this curve to find, first the 
time of the fundamental oscillation and thence the upward velocity 
due to the deflection. 

Call the deflection at any section y, and the load or weight of the 
small disc w. 

Find the sum of all products like wy? and also find the sum of all 
products like wy. 

(Suitable tabulation will facilitate matters.) 

Then T (time of oscillation) 



If Y = maximum deflection, assuming the motion to be S.H.M. 
then the upward velocity v is obtained from 

vT = 27rY 



or 



v = 



27rY 



. Examples on Applied Electricity. 

Example 18. Arrange n electric cells partly in series and partly in 
parallel to obtain the maximum current 
from them through an external resistance R. 
(Let the internal resistance of each cell = r, 
and let the E.M.F. of each cell = v.) 



Suppose the mixed circuit is as shown in 
Fig. 115, i. e., with x cells per row and there- 

7t 

fore - rows. 
x 

Then the total E.M.F. of i row = ;n; 
and total internal resistance of i row = xr, 

1 

but as there are rows, the total internal 
x 



I-!' I'l 



H 



FIG. 115. Maximum Cur- 
rent from Electric Cells. 



resistance is - that of i row, i. e., the total 
x 

rx* 

internal resistance = ; but the E.M.F. is 
n 

unaltered : in reality the effect being that of one large cell, the area 
being greater and thus the resistance less. 



3i8 MATHEMATICS FOR ENGINEERS 

total E.M.F. xv 



Hence the current C = 



total resistance rx 
n 
v 



** , R" 

n x 

and C is maximum when the denominator is a minimum. 
Let D = the denominator, then 

dD r R , dD rx* 

-= = --- and -r- o when R = . 
ax n x z dx n 

i. e., external = internal resistance. 

Example 19. To find an expression for the time of discharge of an 
electric condenser of capacity K, discharging through a constant 
resistance R. 

Let v = potential difference between the coatings at any time t. 

v 
Then, by Ohm's law, the current C = = . 

But the current is given by the rate of diminution of the quantity 
q and q = Kv. 

dq dKv dv 

Hence C = -^ = -^-= -K- 

v Tjr dv 
and thus ^ = K-j- 

K at 

If V x = the difference of potential at the start, i. e., at / = o, 
and V 2 = the difference of potential at the end of T sees, 
Separating the variables and integrating, 



whence log = 

V t i 
or =J = e KR 

* 2 

V 2 _i 

. e., ^~ = e KR 

or the time taken to lower the voltage from 
Vjto V 2 = 

Example 20. If R = the electric resistance of a circuit, L = its 
self -inductance, C = the current flowing and V = the voltage, then 

V = RC+L- 



APPLICATIONS OF THE CALCULUS 319 

Solve this equation for the cases when V = o, C = C sin qt and 



For the case of a steady current V = RC since L is zero, and this 

corresponds to the equation of uniform motion in mechanics, whilst 
j/-\ 

the equation V = RC+L-j- may be compared with that for accelerated 
dt 

motion, V being the force. Thus the second term of the equation may 
be regarded as one expressing the " inertia " or " reluctance to change," 
and since the current may vary according to various laws, the rate of 

change can have a variety of values. 
Clt 

Dealing with the cases suggested : 

(i) V = o, then o = RC+L^- 

dt 

RC =~ L f 

and this equation is solved by separating the variables and integrating. 

Th l dC 

Thus 



p j 

whence = log e C+log e A = log e AC 

JLrf 

_Rt 

and AC e L , A being a constant. 

(2) If C = C sin 9* 

j/-* 

then = qC cos qt. 

dt 

V = RC sin ^+L^C cos qt 
= C VR 2 +L V sin U/+tan - 1 - 

then V sin qt = RC+L 

dl 

i. e., V sin qt = (R+LD)C 

where' D= . 

at 

Hence c V^ingf 



and, using Theorem 2 of p. 288, 

C = . V = s 

VR 2 +LV / 

dC - 

To this must be added the solution of o = RC+L , viz., C= Ke L ; 

at 

hence the complete solution is 

V 



320 MATHEMATICS FOR ENGINEERS 

Example 21. To find expressions for the potential and the current 
at any points along a long uniform conductor. 



At a distance x from the " home " end let the steady potential to 
the ground = E and the steady current be i. 

Let the resistance of i unit of length of the conductor r and 
let the leakage of i unit of length of the insulation = /. 

Consider a small length of conductor 8x. 

Its resistance = r8x and the leakage = I8x. 

Hence the drop in potential ixr8x (i) 

and the drop in the current = E x I8x (2) 

i. e., from (i) 8E = ir8x 

- - ' 

and from (2) oi = El8x 
or -^ = -El (4) 

Writing (3) and (4) in their limiting forms 

dE _ _ . di _ 

dx dx 

.,- d 2 E d . . . di , . 

Differentiating, TT = j~\ tr i ~ ~ y ^T ~ r ^ '5' 

ct,\ d% dx 

and -^r- r= - 7 -( E/) = /-j = ril (6) 

dx z dx* dx 

To solve these equations, let D , 

ax 

then D 2 E = rlE 

i. e., D 2 = rl 

D = Vrl. 

dx dx 

Separating the variables, 

dE .- dE , - 

= Vrldx or = Vrldx. 

Integrating, 

log E = Vrlx+Ci or log E = X/y/tf+Cg 

or, if the constants are suitably chosen, 

E = A cosh Vrlx+'B sinh Vrlx. 



APPLICATIONS OF THE CALCULUS 



321 



In like manner, i = C cosh Vrl x+~D sinh Vrl x. 
When x = L 

E = A cosh Vr/L+B sinh Vr/L. 
When x = o 

E = A 
and hence the constants can be found. 



Examples on Strengths of Materials. 

Example 22. To find the shape assumed by a chain loaded with its 
own weight only ; the weight per foot being w. To find also expres- 
sions for the length of arc and the tension at any point. 

Let s = the length of the arc AB (Fig. 116) : then the weight of 
this portion = ws. 





\IVS 



7 

FIG. 116. 



FIG. 117. 



Draw the triangle of forces for the three forces T, T and ws 
(Fig. 117). 

Let it be assumed that T (the horizontal tension) we, where c is 
some constant. 

Then, from Figs. 116 and 117 



dy _ 
dx = 

Now, as proved on p. 201, 



.. ws ws s 
, = tan 6 = =- = = - 
ax TO we c 



ds 



dx* 



sds 



322 MATHEMATICS FOR ENGINEERS 

To integrate the left-hand side, let u = c 2 +s 2 

du 

then r = 2s 

ds 

( sds rsdu 
and / -7= = = / 1 = u* = V c 2 -1- s 2 

./ Vc 2 + S 2 J 2SU$ 

Thus, by integration of equation (i), 



Now at the point A (Fig. 116) 5 = and y = o 
hence Vc z = Cj or C x = c. 

Thus Vc 2 +s 2 = y + c. 

Squaring c 2 +s 2 = y*-\-c z +2yc 

or s z = y z +2yc and s Vy z +2yc 

s dy dy 

but as proved above -= -f- or s c^- 
c ax dx 

dv / = 

hence c-^-= V y z -\-2yc. 

Separating the variables 

dy dx 



_ __ ^_ 
e " c 2 ~ c 



r dy [ 

Integrating / /. , ., = = / 

J V+c 2 c 2 J 



/. , . 
V(y+c) c c 

and this integral is of the type discussed on p. 151 ; the result being 



c J c 

Now x o when y = o, x being measured from the vertical axis 

through A, and thus logf- j = C 2 or C 2 = o. 

Thus f = 

or in the exponential form 

X 

ce c = y-\-c-\- Vy z + zyc. 
Isolating the surd ce c (y+c) = Vy*+2yc. 

2x x 

Squaring c z e c +y 2 +c 2 +2yc 2(y-\-c) ce* = y*+2yc 

2x x_ 

or c z e c -2.ee* (y+c)+c 2 = o. 



APPLICATIONS OF THE CALCULUS 



323 



Dividing through by ce c 

ce c +ce c 2 (y-\-c), 
i.e., (y+c)= C -(e+e~). 

If now the axis of x be shifted downwards a distance c, then the 

c I X ' ~\ x 

new ordinate Y = y-\-c and Y - \e c +e c ) c cosh - 




v: Scale 

FIG. 1 1 8. Catenary Form of a Cable. 



Again, since Y = y +c = 

dx 



dx 



dx 



and also 
Then 



d ,x c . , x . .x 
^-c cosh - = - sinh - = smh - 
ax cec c 

dy .X 

= sinh - 
dx c 



but it has already been proved that 

dy_ s 



henee 



dx c 

s . , x 
-=smh- or s = 
c c 




324 MATHEMATICS FOR ENGINEERS 

To find the tension T at any point 

T 2 == w z s z +w z c z from Fig. 117 
= w z (s z +c z ) 
= w z (c+y) z = w z Y 2 
or T = wY. 

Thus the form taken by the chain is that for which the equation is 

fx\ 
._^ Y = ccosh(-J, the equation of the catenary: the 

length of arc is given by 5 = c sinh -, and the tension 

c 

at any point is measured by the product of the 
ordinate at that point and the weight per foot of the 
chain. 

Fig. 118 shows the catenary for a cable weighing 
3-5 Ibs. per foot and strained to a tension of 40 Ibs. 
weight, and the method of calculation for the con- 
struction of this curve is explained on p. 358 of PartT. 

The tension at 10 ft. from the centre = 3-5 x 15-9 
FIG. 119. = 55-6 Ibs. weight, since the ordinate there is 15-9. 

Example 23. To find the time of oscillation of a compound pendu- 
lum swinging through small arcs. 

Let I be the moment of inertia of the pendulum about an axis 
through the point of suspension O (Fig. 119), and let h = the distance 
of the C. of G. from the point of suspension. 

Then the couple acting, to produce the angular acceleration, 
= moment of inertia x angular acceleration. 

[Compare the rule for linear motion, Force = mass x acceleration.] 

Now the angular velocity = -j- 

dt 

and hence the angular acceleration 

= <8 ' 
Thus the couple acting = 1-^ 

and this couple is opposed by one whose arm is h sin 6, as is seen in 
the figure. 

d z d 
Thus *dt*~ = mhsvn.6 mhd 

since 6 is supposed to be small, and consequently sin 6 = 6 
or d z d _ mh . _ z ., 2 _ mh 

dt*~~ ~~T~ " ~T~ 

and +*-. 



APPLICATIONS OF THE CALCULUS 325 

This equation is of the type dealt with in Case (3), p. 283, and the 
solution is 6 = A sin (<ef-f B). 

2-7T 

The period of this function is ; also the couple for angular dis- 
co 

placement Q = mh6 ; hence the couple for unit angular displacement 
(denoted by /*) = mh. 

27T /T~ /I 

Hence t = = 2rr \/ r = 2ir \f - 

co ^ mh v p. 

or / = 2n A / if engineers' units are used. 

k i*g 

This might be written in the easily remembered form, 



unit moment of inertia per unit twist' 
If this formula is to be used in the determination of the modulus of 

rigidity of a sample of wire by means of torsional oscillations, h must 

be replaced by /, the length of the wire. 

If /= skin stress, T = torque, and C = modulus of rigidity, d dia. 

of wire; then 



whence 
but 
and 


0=g and T = 
A 32T/ 




M ~ 6 ~ yl 

t=2w \/ 


/ 32/1 



hence C = == - and thus C can be determined. 



As regards the units, if / is in feet, I is in Ibs. ft. 2 , t in sees., and 
d in feet, 

feet x Ibs. x ft. 2 x sec. 2 



then C = 



feet x ft. 4 x sec. 2 



Ibs 
= ~ffT' ** e -> C * s m Ibs. per sq. foot. 

If I is in Ibs. ins. 2 and d is in ins., then C will be in Ibs. per sq. in. 

Example 24. To find formulae giving the radial and hoop stresses 
in thick cylinders subjected to internal stress. 

We may attack this problem by either of two methods : 

Method i. In (a) Fig. 120 let the outside radius = r t and the inside 

radius = r ; also let the internal pressure be p, and the crushing stress 

at right angles to the radii, or the hoop stress, = q. 



326 



MATHEMATICS FOR ENGINEERS 



It is rather easier to consider the stress on the outside to be greater 
than that on the inside : thus for an annulus of radius r and thickness 
$r, we take the internal stress as p and the external stress as p-\-8p. 

Considering the element RS of the annule (subtending an angle of 
80 at the centre), and dealing with the radial forces, 

Total radial force = (p-\-8p) x outer arc px inner arc 
= (p+8p) X (r+8r)86-pr86 
= (pr+p8r+r8p+8p.8r-pr)86 . 
= (p8r+r8p+8p.8r) 86 
(for a unit length of the cylinder) . 

This is balanced by two forces each q 8r . 86, for 



2 . 86 86 
= sin - = nearly 
q8r 2 2 



[(6) Fig. 120] 



i. e., x = q8r.86 

x being the radial force. 




FIG. 1 20. Stresses in Thick Cylinders. 



Thus (p8r+r8p+8p.8r)86 = q8r.8d 

or, when 8r is very small, 

p dr-\-r dp = q dr. 

Assume each longitudinal fibre to lengthen the same amount due to 
the secondary strains. 

Then if <r = Poisson's ratio and E = Young's modulus for the 
material, 

p i 
the extension due to p will be ^ X - 

and the extension due to q will be ^ X - 

Si, (T 

then, since the total extension is to be constant, 



i. e., 



~r^ constant, 
(T-tL 

p-\-q = 2A, say, for a and E are constants. 



APPLICATIONS OF THE CALCULUS 

Hence v dp -\-pdr = qdr 

= (2 A p}dr 

i. e., rdp = 2(Ap)dr. 

Separating the variables and integrating, 

f dp _ fdr 
J 2(A-p) J r 

i.e., | log (A p) + log C = log r 

C 



327 



or r = 



(A-pfl 

C 

r 



-r> 

or p = A + - 2 



but 

and hence 



The constants A and B are found from the conditions stated in any 
example. 

Method 2. According to this scheme q is taken as a tensile stress. 
By the thin cylinder theory ; consider the equilibrium of the half 
elementary ring of unit length [(c) Fig. 120]. 



Then (pX2r) (p+8p)2(r+8r) = 2q8r 
whence qdr = pdrrdp. 

From this point the work is as before except 
that 2 A is written for p q and not for p-\-q 
as in Method, i. 

Example 25. To find expressions for the 
stresses in Thick Spherical Shells. 




FIG. 121. 



Let p = the radial pressure, q = the hoop tension. 

Take an elementary shell at radius r, the thickness being 8r (Fig. 121). 

Then 7rr z pn(r+8r) 2 (p+&p) = 



When 8r is very small this equation reduces to 
28r r8 = 2q8r 



and hence 



2q = 



dp 
- 



328 MATHEMATICS FOR ENGINEERS 

Assuming the volumetric strain to be the same everywhere, 

where l v = the circumferential strain, and thus 2/2,= the superficial 
strain, and l x = the radial strain, 

then it follows that 

^-- P.-C 
<rE o-E" 

i. e., 2( Ip Constant = 3A (say). 

Now 2<7= 2* r~ 



Separating the variables and integrating, 
[-$L 

J T 



i. e., log r = - log (A+) -f log (^ 

whence 



or 



i _ _ 2B_ A 



Also zq 

3 . 2 B 



_B 



Euler's Formulae for Loaded Struts. 

Example 26. To obtain a formula giving the buckling load for a 
strut of length L and moment of inertia I. 



Applying the ordinary rule 



M =E d*y 



APPLICATIONS OF THE CALCULUS 

Bending moment at Q = M = Py [(a] Fig. 122] 



329 



i. e., 
Let 
then 



dx 2 



IE/' 



IE 



and the solution of this equation is, according to Case (3), p. 283, 
y A sin ( 




fW (c) 

FIG. 122. 

The various conditions of end fixing give rise to the following 
solutions : 

Case of ends rounded. When x o, y = o 

then o = A sin (o+B) and A is not zero 

so that B = o. 



When 



i. e., 



x = , y= Y [(a) Fig. 122] 

xr . wL 

Y = A sin 

2 



Obviously Y is the amplitude, i. e., A = Y 

0)L 



or 

Thus we may write 
whence 
and 



i sin 

2 

7T . 0)1 

2 2 

0>L = 7T 

7~P _ 
V fE XL = 

7T 2 IE 

, L 2 ' 



(- being the simplest angle 
\2 \ 

having its sine = i ) 



330 



MATHEMATICS FOR ENGINEERS 



Case of both ends fixed. -The form taken by the column is as at 
(&) Fig. 122. The half-period of the curve is evidently in this ease. 



27T 



But the period = 



27T 



IE 



whence 



4 7T 2 IE 

L 2 



Case of one end fixed. The form taken by the column is as at 
(c) Fig. 122. The half-period in this case is f L, but, as before proved, 

. , . 27T 

the period is given by 

O) 

Hence -L 



2co 



Now 



/* 

" V IE 

Q7T 2 IE 

4P 

97T 2 IE 

4L 2 ' 



L 2 = 



P = 



Tension in Belt passing round a Pulley. 

Example 27. To compare the tensions T x and T 2 at the ends of a 
belt passing round a pulley ; the coefficient of friction between the belt 
and pulley being p., and the angle of lap being 6 radians. 



T+ST 





FIG. 123. 



FIG. 124. 



Consider a small element of belt subtending an angle of dd at the 
centre of the pulley (see Fig. 123) : then the tensions at the ends are 
respectively T-f-oT and T. 



APPLICATIONS OF THE CALCULUS 331 

Resolving the forces horizontally 

sa s/i 

/T> I T\ "" T> " T- 

(T+8T) cos -- Tcos uP 

2 2 

f./\ 

i.e., 8T cos = uP 

2 

or in the limit dT = /J? ......... (i) 

80 
for cos -- > cos o, z. 0., i. 

2 

Resolving the forces vertically, 

Sj/1 

P=(T+8T+T)sin~ 

' . 80 , ._ . 80 
= 2T sin -- f-8T sin 

2 2 



_, . 

= 2T -- h8T (for sin = 

22 22 

when the angle is small) 
In the limit P = TdO .......... (2) 

Then, combining equations (i) and (2), 

dT = pTdS 

Separating the variables, / -=- = p. / dd 
J T 2 i Jo 

T 

Integrating, log-f^ = \iB 

*-2 

T, n8 
or =*= tr 

*i 



Friction in a Footstep Bearing. 

Example 28. To find the moment of the friction force in a footstep 
bearing; the coefficient of friction being p., R = radius of journal and 
W= total load. 

(a) Assume that the pressure is uniform over the bottom surface, 
i. e., W = 7rR 2 ^>, where p is the intensity of the pressure. 

Take an annulus at radius r, and of thickness 8r (Fig. 124). 
Area of the annulus = 2nr8r 

Pressure on the annulus = 2-nrdrp 

Friction force on the annulus -z-nrbrpp 
and hence the moment of the friction force on the annulus 

= 2irr 8r pp x r 



332 MATHEMATICS FOR ENGINEERS 

/R 
Up 2nr z dr 

R 

T 



W 



= - 

i. e., the moment is the same as it would be if the whole load were 
supposed concentrated at a distance of two-thirds of the radius from 
the centre. 

(&) Assume that the intensity of pressure varies inversely as the 
velocity, 

i. e., p= K X - 

v 



the velocity at radius r v r = 
so that p = K x 

Then the pressure intensity on an annulus distant r from the centre 



so that p = K x = - (say). 

r v 



and the total pressure on the annulus 



also the friction force = p X this pressure. 
Hence the moment of the friction force on the annulus 



and the total moment of the friction force 

= v 



Now the total load W = / intensity x area 
J o 



/* 

= I p X 2nrdr 
J o 



=r 

Jo 



APPLICATIONS OF THE CALCULUS 333 

Hence the moment of the friction force 

= 27rwR X 

2 
-Wxf 

i.e., the effective radius is now and not f, as in Case (a). 

Example 29. To find an expression for the moment of the friction 
force for a Schiele Pivot. 



Assume that the pressure is the same all over the rubbing surface, 
that the wear is uniform and that the normal wear is proportional to 
the pressure p and to the speed v. 



Referring to Fig. 125, 
and thus 



= normal wear oc pv, i. e., 

8n oc pr or &n = Kpr. 




Let the tangent at the point P make the angle 6 with the axis, then 
if t = length of tangent, t sin 6 = r. 

Now 8h vertical drop = - ^ 

sintf 

Also 8n = Kpr = Kpt sin 6 

whence bh = Kpt. 

Now 8h is constant, p and K are constant ; 
hence t must be constant and the curve is that 
known as a tractrix (i. e., the length of the 
tangent from the axis to any point on the 
curve is a constant). 

To find the moment of the friction force : 

On a small element of surface, the friction force 

= lirr 8s X p X p. 
and the moment of the friction force 

= 2nr8s fip x r. 

Now 87 = 8s sin 6. 

Hence the total moment of the friction force 

dr 



FIG. 125. 



sin 6 



tsinti 



but 



X 

W 



Hence the total moment of the friction force = /*\V/. 



334 



MATHEMATICS FOR ENGINEERS 



Examples on Hydraulics. 

Example 30. To find the time to empty a tank, of area A sq. ft., 
through an orifice of area a sq. ft., the coefficient of discharge being C d . 

If the height of the water above the orifice at any time is h, then 
the velocity of discharge = v = V^gh. 

Hence the quantity per sec. = C d av 
and the quantity in time 8t C d av8t. 

This flow will result in a lowering of the level in the tank by an 
amount &h, so that the volume taken from the tank in time 8t = A 8h. 

Hence A8h = C d aV^ghSt. 

Here we have a simple differential equation to solve, and separating 
the variables and integrating 

Adh where h z = initial height 
A x = final height 



o A 




FIG. 126. Triangular Notch. 
If A 1 = o, then the time to completely empty the tank 



Example 31. To gauge the flow of water by measurements with a 
triangular notch. 

Let the height at the notch be H, and consider a small element of 
width b, thickness 8h, and height above the apex of the notch (H A). 

b G 

From Fig. 126, - = (H h) tan - where 6 = the angle of the notch, 



whence 



6 =2(H h) tan-. 



APPLICATIONS OF THE CALCULUS 335 

Now the area of the element = b 8h 

and the velocity of water at that height = VzgX height = 

A 

Hence the actual quantity flowing = C d X2(H h) tan - 
and the total quantity flowing for the height H 

fi 

= FzVzgCa tan-(H 
J o 



tan 



/) /~n 

-J ^H 



1 / i 

J 5 2 

a 

~Li 6 = 90 (a common case), tan- = i, 

Q 

and then the discharge = V2gC d H* = 2-66 H ? if C d = -62. 

Example 32. To estimate the friction on a wheel disc revolving in 
a fluid. 

Let the friction per sq. ft. fv x and let the disc (of inside radius R 2 
and outside radius R x ) revolve at n revs, per sec. 
The velocity of an annulus at radius r = 2nnr 

and thus the friction force per sq. ft. on this annulus 

= (2nnr) x f. 
Hence the moment of the friction force on the annulus 

= /(2 nnr) x X -Zirrbr X r 



and the total moment of the friction force on one side of the disc = M 



Rg 



The total moment (i. e., on the two sides) = aM 

and H.P. lost in friction = . 

550 
If x = 2 



= 49 .6/n{R 1 -R f }. 



336 MATHEMATICS FOR ENGINEERS 

Example 33. To establish a general rule for determining the depth 
of the Centre of Pressure of a section below the S.W.S.L. (still water 
surface level). 

Suppose the plate (representing a section) is placed as shown in 
Fig. 127. Consider a small element of area 8a, distant x from OY, the 
vertical distance being h. 

Let X = distance of the C. of G. from OY, 

X = distance of the C. of P. from OY, 

and let H and H be the corresponding vertical distances. 
Let P = total pressure and A = total area, 

k = swing radius about OY, 
and p weight of i cu. ft. of water. 



S.W.S.L. 




IX 

FIG. 127. Centre of Pressure. 

The whole pressure on the element 

= intensity of pressure X area = ph X 8a 



Whole pressure on surface = 2px sin a8a approx. 
or fpx sin a da actually, 

i.e., P = psinaxfxda 

p sin aX ist moment of area about OY 

= p sin a X AX, 
but X sin a = H. 



To find the position of the C. of P., take moments about OY. 
Then PxX = 2 moments of the pressures on the elements 

= Sp sin a x8a X x 

= psina2# 2 Sa approx. 

= psinafx z da actually 

= p sin a X and moment about OY 

= p sin a X A& 2 . 



APPLICATIONS OF THE CALCULUS 337 

Now P = pHA 

so that pHAX = psinaAA 2 , 

i. e., HX = sin a . k z 

but H = X sin a 

hence XX = A 2 . 

Thus if X is known, X can be calculated. 

If the body is not symmetrical, then Y (the distance of the C. of P. 
from OX) must be found by taking moments about OX. 

In a great number of cases a = 90, so that sin a = i, and thus 



Example 34. A triangular plate is placed with its base along the 
S.W.S.L., the plate being vertical. Find the depth of the centre of 
pressure below the surface. 

For this section I about S.W.S.L. = bh 3 



and thus k 2 = 



also H = -h. ' 

3 

- k 2 A 2 X3 h 

Hence H = = -^j^ = - 

H 6x& 2 

Example 35. A circular plate has its upper edge along the S.W.S.L 
Find the depth of its Centre of Pressure below the S.W.S.L. 

For a circle, Idiam. = 7- d* 

64 



and thus k 2 about diam. = 



4 

and hence, by the parallel axis theorem, 

A 2 about S.W.S.L. = H ~ ) = 
also H = - 

2. 

= k z 16 ^d 

Hence H = = 5-= "-3- 

H d 



338 



MATHEMATICS FOR ENGINEERS 



Example 36. Forced vortex (i. e., water in a tube rotated round a 
vertical axis). To find the form taken by the surface of the water. 



Let the rotation be at n R.P.S. 
Consider an element at P (Fig. 128). 



tan 6 = 




th 



tan 6 = the slope of the curve taken = -^ 

dn 



vertical force _ weight *of particle 

horizontal force centrifugal force on particle 

_mxgr 



FIG. 128. Forced Vortex. 
Separating the variables 
Integrating 



dh 4tr 
4n 2 n 2 rdr = gdh. 

^ =gh . 

A = ^xr*. 



Now 



is a constant, and thus 



h = constant xr 2 , this being the equation of a parabola. 
Hence the surface of the liquid will be that of a paraboloid of 
revolution. 

An Example from Surveying. 

Example 37. Prove that a cubic parabola is a suitable " transition " 
curve. 

In order that the full curvature of a railway curve may be 
approached gradually, a curve known as a transition curve is inter- 
posed between the straight and the curve. It must be so designed 
that the radius of curvature varies inversely as the distance from the 
starting point (on the straight, because there the radius is infinite) . 

i d 2 y 
As before proved, 



dx 2 



or more exactly 



ffiy 

dx 2 



APPLICATIONS OF THE CALCULUS 



339 



For the cubic parabola we may assume an equation 

y = px 3 , 

where x is the distance from the straight, along a tangent, and y is the 
offset there (obviously p must be very small) . 

If y = px 3 , -r- = ipx 2 , ~ = 6px. 

Hence = = - 



as a first approximation, or = 6px nearly (for p 3 is very small) . 

T T T K" 

Hence 



R 



or K = X - = 
6p x x 

Roc -. 
x 



Exercises 23. 

1. A cylindrical tank is kept full of water by a supply. Show that 
the time required to discharge a quantity of water, equal to the capacity 
of the tank, through an orifice in the bottom equals half the time 
required to empty the tank when the supply is cut off. 

A tank 10 ft. high and 6 ft. diam. is filled with water. Find the 
theoretical time of discharge through an 8" diam. orifice in the bottom. 

2. A tank empties through a long pipe discharging into the air. If 

Kv 2 
the head lost in the pipe is written hi = - , show that K can be found 

from the expression, 



where A x is the level of the water in the tank at the time t t 
and h z is the level of the water in the tank at the time t 2 

measured from the centre of the discharge end of the pipe. 

A = area of cross section of the tank. 

a = area of cross section of the pipe. 

An experiment with a tank 15-6 sq. ft. in cross section and a 4" diam. 
pipe gave the following results : 



Time, t inins. 


Level in tank, h. 




I 
2 


38-35 
32-84 


3 
4 


23-19 
I9.O2 



Find the value of K for the pipe. 



340 MATHEMATICS FOR ENGINEERS 

3. Use the following table to obtain -= and thence find the volume 
of i Ib. of steam at 160 Ibs. absolute pressure per sq. in. 



Absolute press. (Ibs. per sq. in.) 


159 


1 60 


161 


Temperature (F) 


363-1 


363-6 


364-1 











The latent heat of i Ib. of steam at 160 Ibs. per sq. in. pressure is 
858-8 B.Th.U. 

4. Find the " fixing moments " for a beam built in at its ends and 
40 feet long, when it carries loads of 8 tons and 12 tons, acting 15 feet 
and 30 feet respectively from one end. 

5. A tank of constant cross section has two circular orifices, each 
2" diani., in one of its vertical sides, one of which is 20 ft. above the 
bottom of the tank and the other 8 ft. 

Find the time required to lower the water from 30 ft. down to 15 ft. 
above the bottom of the tank. 

Cross section of the tank =12 sq. ft. 
Coefficient of discharge = -62. 

6. A hemispherical tank 12 ft. in diam. is emptied through a hole 
8" diam. at the bottom. Assuming that the coefficient of discharge is 
6, find the time required to lower the level of the water surface from 
6 ft. to 4 ft. 

7. A vertical shaft having a conical bearing is g" in diam. and 
carries a load of 3^ tons ; the angle of the cone is 120 and the co- 
efficient of friction is -025. Find the horse power lost in friction when 
the shaft is making 140 revolutions per minute. 

Assume that the intensity of pressure is uniform. 

8. A circular plate, 5 ft. diam., is immersed in water, its greatest 
and least depths below the surface being 6 ft. and 3 ft. respectively ; 
find 

(a) the total pressure on one face of the plate, 

(b) the position of the centre of pressure. < 

9. An annular plate is submerged in water in such a position that 
the minimum depth of immersion is 4 ft. and the maximum depth of 
immersion is 8 ft. If the external diam. of the plate is 8 ft. and the 
internal diam. 4 ft., determine the total pressure on one face of the 
plate and the position of the centre of pressure. 

10. One pound of steam at 100 Ibs. per sq. in. absol. (vol. = 4-45 
cu. it.) is admitted to a cylinder and is then expanded to a ratio of 5, 
according to the law pv 1 - 06 = C ; it is then exhausted at constant 
pressure. 

Find the net work done on the piston. 

11. Find the loss of head h in a length I of pipe the diameter of 
which varies uniformly, being given that 

4fLv* 4Q 

H= J , , and v = ^~. 
2gd Ti-d 2 

{Let diam. at distance x from entry end = rf e +K# 

where d e = diam. at entry}. 



APPLICATIONS OF THE CALCULUS 341 

12. Taking the friction of a brass surface in a fluid as -22 Ib. per 
sq. ft. for a velocity of 10 f.p.s. and as proportional to w 1 ' 9 , find the 
horse power lo"st in friction on two sides of a brass disc 30" external 
and 15* internal diam. running at 500 r.p.m. 

13. A rectangular plate 2 ft. wide by 5 ft. deep is immersed in 
water at an inclination of 4O C to the vertical. Find the depth of the 
centre of pressure, if the top of the plate is 6 ft. below the level of 
the water. 



CHAPTER XI 
HARMONIC ANALYSIS 

Fourier's Theorem relates to periodic functions, ot which 
many examples are found in both electrical and mechanical engi- 
neering theory and practice : it states that any periodic function 
can be expressed as the sum of a number of sine functions, of 
different amplitudes, phases and periods. Thus, however irregular 
the curve representing the function may be, so long as its ordinales 
repeat themselves after the same interval of time or space, it is 
possible to resolve it into a number of sine curves, the ordinates of 
which when added together give the ordinates of the primitive 
curve. This resolution of a curve into its component sine curves is 
known as Harmonic Analysis ; and in view of its importance, the 
simpler and most direct methods employed for the analysis are 
here treated in great detail. 

Expressed in mathematical symbols, Fourier's theorem reads 



or y = 

+Bj cos tf+B 2 cos 2qt+~B 3 cos 3qt-\- . . . 
the latter form being equivalent to the first, since 
Aj sin gtf+Bj cos qt B sin (qt-\-Cj) 

provided that B and c are suitably chosen. 

For the purposes of the analysis the expression may appear 
simpler if we write 6 in place of qt. 

Thus 

y = A +A 1 sin^+A 2 sin2^+A 3 sin3<9+ . ..-. 

+B 1 cos0+B 2 cos20+B 3 cos30-|- . . . 

Of the various methods given, three are here selected and 
explained, these being easy to understand and to apply. 

Dealing with the three processes in turn, viz., (a) by calculation, 
(b) by a graphical interpretation of method (a), and (c) by super- 
position, we commence with the study of method (a). 

342 



HARMONIC ANALYSIS 343 

Method (a): Analysis by Calculation. Before actually pro- 
ceeding to detail the scheme of working, it is well to verify the 
following statements. 

ft* 

cos = 0, this being self-evident, since the area under a 
.'o 

cosine curve is zero, provided that the full period is considered. 

/** 

I cos mO cos nOdO o ............ (i) 

.'o 
for 

cos mO cos nO = ^{cos (m-\-n)6 cos (m n)0} 

and hence 

/2ir 



/2ir r2n rZv 

I cos m0cosn9d& = || cos(m + n)OdO 1| cos(mn)6d0 
Jo Jo Jo 



= o o 



(for both are cosine curves over the full period or a multiple of the 
full period). 

rZv f2jr /-2ir 

I cos m6 sin nO dO = | sin (m+n)B dO% sin (mn)6dO 
Jo J o Jo 

= .... ........ (2) 

fZir ,fZir rZir 

cos z 6d6 =4 cos20dO + % dd 

Jo Jo Jo 

= + (27r-o)=7r ....... (3) 

/2ir f /-2ir /-2n- ^ 

sin mO sin M^ ^ = $\ cos (mn)6 dd cos (m+ri)0d6 \ 
Jo U o y o J 

= |{o-o}-o ......... (4) 

and 

/2ir /-2ir fZir 

I sin 2 ^^ =J ^ | cos20d0 

Jo Jo Jo 

= \\2ir o} O 

= ............ (5) 

To proceed with the analysis : 
We are told that 

y = A +A 1 cos0+A 2 cos20+A 3 cos30+ ... 
+Bj sin 6+B 2 sin 20+B 3 sin 3^+ 

and we wish to find the values of the coefficients A , Aj . . ., 
Bi, B 2 , etc. 



344 MATHEMATICS FOR ENGINEERS 

If we integrate throughout (with the limits o and 2ir), every 
term on the right-hand side, except the first, will vanish, 



i.e., 



r2w f 

yd9 = A, 
Jo J 

/2ir 

or I ydO A X (2-* o) 

3 o 

I P* 
whence A = - I y dO 



= the mean value of y (Cf . p. 183) 

so that A is found by averaging the ordinates ; but in the majority 
of cases an inspection will show that A is zero. 

To find A 1 : multiply all through its coefficient, viz., cos0, and 
integrate, then 



fZir r'2ir rZir rZ 

I ycos$dO= I A cosOdO-\- I A 1 cos 2 0^+| 
Jo ./o Jo ,' o 

rZn _ rZir 

Jo lC( Jo 

f 2 

I y cos 6 dO = o+TrAj+o+o . . . 
Jo 



or 

J o " 

+0+0 . . . [from (3), (2) and (i)] 

2 fjr 

whence A x = / y cos 6 dO 

= twice the mean value of (y cos 6} 

i. e., a certain number of values of y must be taken, each being 
multiplied by the cosine of the angle for which y is the ordinate, 
the average of these found, and the result multiplied by 2. 

The values of A 2 , A 3 , etc., may be found in like manner by 
multiplying through by cos 26, cos 3$, etc., in order, and performing 
the integration as above. 

To find B! : multiply throughout by its coefficient, viz., sin 6, 
and integrate, then 

/2ir [Z* ,-Zir fZir 

ysinBdO = I A sm@d@-\- I A,sin0cos0^+ / A 
o J o J o Jo 

,"Zir _ rZir 

+ Jo BlS1 + Jo B2 

= o+7rBj [from (2), (4) and (5)] 

_, 2 r 27r 
' Bj = 1 y sin dQ = 2 X mean value of (y X sin 6} 

so that the values of B,. B 2 , B 3 , etc., may be found 



HARMONIC ANALYSIS 



345 



Actually, the values of the coefficients A v A 2 , B r , B 2 , etc., are 
found by dividing the base into ten or. eight divisions and averaging 
the mid-ordinates for these divisions. To determine the absolute 
values, an infinite number of ordinates should be taken, but this 
would of course be quite out of the question as far as an ordinary 
calculation is concerned. 

The work is made clearer by suitable tabulation, as will be seen 
from the following example. 

Example i. Resolve the curve ABCD (Fig. 129) into its component 
curves : it being understood that no higher harmonic than the first 
occurs. 





e 



O -36 72 K>8 144 




FIG. 129. Harmonic Analysis. 



[The term containing 6 is spoken of as the fundamental, and that 
containing 26 as the first harmonic.] 



Thus y = A +B sin (6+cJ +C sin 

or y= A -|-A 1 cos0+A 2 cos20+B 1 sm0+B 2 sin20 
will represent the function in this case. 



A glance at the figure will show that the curve is symmetrical about 
the axis of 6 ; thus we observe that the average ordinate = o, or 
A ft = o. 



346 



MATHEMATICS FOR ENGINEERS 



Divide the base into 10 equal parts, erect the mid-ordinates and 
tabulate the values as follows : 

(a) (/) (c) (d) (e) (/) (g) 



Ordinate No. 


9 


1 


sin 


cos e 


sin 26 


cos 26 


I 


1-56 


18 


309 


951 


588 


809 


2 


375 


54 


809 


588 


95 1 


-309 


3 


4 


90 


I- 


o 


o 


I 


4 


2-91 


126 


809 


588 


951 


-309 


5 


1-13 


162 


309 


-951 


588 


809 


6 


-1-13 


198 


309 


-95 1 


588 


809 


7 


2-91 


234 


-809 


588 


95 1 


309 


8 


-4 


270 


I 


o 





T 


9 


-3'75 


306 


809 


588 


951 


-309 


10 


-1-56 


342 


309 


951 


588 


809 



Then A = mean value of y = o. 

A! = 2 X mean value of (y cos 6} . 

To obtain the values of y cos 6, corresponding figures in columns 
(6) and (e) must be multiplied. 



Then 



10 1 +2-913-75) 

= o 



-9i\ 
) 



Similarly, B x = X sum of the products of columns (&) and (d) 






io 



i-i3 + i- 5 6) + 1(4+4) \ 
9i + 2-9i+3-7 5 ) / 



= -2x20-43=4-086 

2 

A 2 = X sum of products of columns (6) and (g) 



-xol 

= o 



3'75 



B 2 = X sum of products of columns (6) and (/) 



Hence 
or 



_ 

lol -2-91+3-75) 
= -2x2-104 = -421. 

y = (o x cos 6) + (o x cos 26) + (4-09 sin ff) + (-42 sin 20) 
y = 4-09 sin 0+ -42 sin 26 (the cosine terms being absent). 



Analysis by Method (6): The Graphical Interpretation of 

(a) (due to Professor Harrison). To employ this method we must 
take at least twice as many ordinates as the largest multiple 



HARMONIC ANALYSIS 



347 



of ', thus if we suppose that the second harmonic is the highest 
occurring we might take 6 ordinates as the minimum, although it 
would be better if 8 or 10 were taken. 

The method can best be illustrated by applying it to an 
example. 

Example 2. Resolve the curve ABC [(a) Fig. 130] into its com- 



I8O 24O 3OO 360* 




FIG. 130. 

ponent curves (the second being the highest harmonic) ; i. e., find the 
values of the constants in the equation 
y = AO+AJ cos 6+ A 2 cos 20+ A 3 cos 30+Bj sin 0+B 2 sin 20+ B 3 sin 3$. 

To arrange that all the ordinates shall be positive, take a base line 
DE entirely below the curve. Divide the base into 8 equal divisions 
and number the ordinates y , y lt y 2 , etc. The angular intervals are thus 
^5, since a full period corresponds to 360. 



348 MATHEMATICS FOR ENGINEERS 

Draw a new figure [(&) Fig. 130], the lines OM and ON making 45 
with the principal axes ; number these lines : o, i, ..... 8, as 
shown in the figure. Along line o set off a distance equal to y , along 
line i set off a distance equal to y t and so on. Drop perpendiculars 
from the points o, I, etc., on to the principal axes, calling the projec- 
tions on these axes h (this particular projection being zero), A x ..... 
h a , and v , v l ...... v a respectively. 

Then to find Aj and Bj : 

As already proved 

A! = f (sy cos 0) = "J{y -y 4 + (y!-y 3 -y & +y 7 ) cos 45 + (y t -y,) cos 90} 

= MVO-^VI+V^VS-VS+VJ 
and similarly B t = Jj^+Ag h 5 h 7 +h 2 h 6 } 
i. e., the lengths v ...... v 7 , and h t ....... / 7 can be read off 

from the figure and then the values of A 1 and B t are calculated as 
above. 

Jn this example 

"0=13-7 *o= o 

v v =14 h 1= 14 

v 2 = o h 2 = 20-4 

v 3 ii h a = ii 

v t = 7'3 h t = o 

v 6 = '5 hs= '5 

v 6 = o A 6 = 1-5 

v,= 5-4 h 7 = 5-4 * 



andB 1 = X38 = 9-5 

By the aid of a strip of paper a great amount of this arithmetical 
work might be obviated, the procedure being as follows : 

Mark off along the edge of a strip of paper lengths to represent 
the various ordinates of the original curve, viz., y , y v etc., and 
number the points so obtained o, i, 2 .... 8 as shown at (c) Fig. 130. 

Thus Po == y , P 4 = y t and so on. 

We have seen that in order to find the value of A a it is necessary 
to evaluate y cos 6 for the various angles ; i. e., we must find the values 
of y v cos 45, y z cos 90, y a cos 135 and so on. 

Now y s cos 135 = y 3 x cos 45 = y a cos 45, so that the one line, 
viz., that at 45, serves also for 135 provided that the ordinate is 
stepped off in a negative direction. Thus, for example, 

yj. cos 6+y 3 cos ^6=y 1 cos 45 y s cos 45 = cos 45(;y 1 y s ) 
and the value of this expression depends upon the difference between 
the lengths on the strip Pi and P3, or the distance 3 to i. 

Evidently, then, the work is shortened by grouping the ordinates 
in pairs to give differences ; thus 

y\y% = Pi ?3 i to 3 on the strip 
y 7 y & Py P5 = 5 to 7 on the strip 
and so on for other pairs of ordinates. 



HARMONIC ANALYSIS 349 

Having found these differences, we multiply by 00345, by setting 
these lengths along the line Of in (d) Fig. 130 and then projecting to 
the horizontal axis OX ; the resultant of these projections being the 
value of 4Aj. 

Thus in (d) Fig. 130 : Make Oo = i to 3 (on the strip) and ab 5 to 7. 

Drop be perpendicular to OX. Then Oc = (y^ y 3 y^+y^ cos 45. 

Step off cd = o to 4 (i. e., y^y^, then measure Od ; this is the value of 
4A X , since Od = Oc+cd = (yiy 9 y t +y 7 ) cos 45+(y -y t ). 



A!= 3-43 

For the value of Bj the strip must be used according to the following 
plan. A line is drawn at 45 [(d) Fig. 130] and distances marked off 
along it as follows 

Oe = i to 7 on the strip, ef= 3 to 5 on the strip. 
{for 46! = (y n +y t ) sin o+(y 1 +y 3 -y 6 -yj sin 45+(> ; 2-^6) sin 9<> } 
A perpendicular to OY gives the point g. To Og must be added a 
distance = 2 to 6 on the strip, but to avoid extending the diagram this 
distance is set off from O giving the length Oh. 

Thus Og = 19-5, Oh = 18-2, the sum = 37-7. 

Then 4 B X = 37-7 

and B!=: 9-43. 

To find the values of A 2 and B 2 : The terms containing 26, i. e., 90, 
will now occur and so there will be no lines at 45. 

Ag |{yo cos +yi cos go+y 2 cos 180+ . . . y 1 cos 630} 



Similarly B 2 = ^{y!- 

Hence set off 

Ok = o to 2, i. e., (y 9 y%) and kl = 4 to 6, i. e., (y t y 6 ) along OX 

and the resultant is O/ = -8. 

-Hence 4A 2 = -8 

and A 2 = -2. 
Set oft 

Om =* i to 3, i. e., (y t y a ) and mn = 5 to 7, i. e., (y^y-j) along OX 

and the resultant is On 2-6 

Hence 4B 2 = 2-6 

and B 2 = -65. 

To find the values of A 3 and B 3 : 



A =-( y COS +yi COS I 35+> / 2 cos 270-f-y 3 cos 4O5+>' 4 cos 540 
3 1 +^ 6 cos675+>/ 6 cos8io +^ 7 cos945 



-{y9yt+\yyi+yty,} cos 45 }. 

4 



350 MATHEMATICS FOR ENGINEERS 

Set off Op = o to 4 on the strip, along OX, 

and Ob = (y>i y a y 5 +y 7 ) (which has already been done when 
finding Aj). 

Project b to c on OX ; then Oc = (y a y-a-^y^y-i) cos 45. 
Hence cp = 4A3 

but cp = i -3 

and thus A 3 = 33. 
To find B s : 

B _ j>o sin +>'i sin I 35+3 / 2 sin 270 +y 3 sin 405+^ sin 540 
3 4*- +n sin6 75 +y 6 sin8io +;y 7 sin945 

= -{yr-yt+fa+ytyv-yi) sin 45}- 

Set off Oh = 6 to 2 along OY. 

O/= (^i+^a y 6 y 7 ) (which has already been done when 
finding Bj) . 

Project / to g on OY, 
then gh = 463 
but gh = 1-2 
Hence B 3 = .3 

Also 



== 

8 
(using the trapezoidal rule given on p. 307, Part I), 

, * A 6 '75 + I 9-8+20-4+i6+7-3+-9+i-9+7-6+6-75 
**., A _ -g 

87-4 
^=10-93- 

y= 10-93+3-43 cos $+9-43 sin 6 -2 cos 26 

~- -65 sin 20 33 cos 3$+ -3 sin 3$. 

There should be no difficulty in the understanding of this 
method if method (a) is first carefully studied. All that this 

method (b) adds is the multiplica- 
A tion of lengths by the cosines or 
sines of angles by regarding the 
products as projections on fixed 
axes (i. e., if OA = R (Fig. 131) 
and the angle AOB = 30, then 
OB = OA cos 30 = R cos 30, and 
OC = OA sin 30 = R sin 30) . 

FIG. 131. The beauty of the method 

consists in the use of the strip of 

paper for the grouping together of pairs of ordinates which have to 
be multiplied by the same quantity. 




HARMONIC ANALYSIS 351 

In the example just discussed, the angular intervals were taken 
as 45, this choice being made as a matter of great convenience, 
since cos 45 = sin 45 and projections may thus be made on either 
a horizontal or a vertical axis. 

For greater accuracy, more ordinates should be taken, and then 
care must be observed as to the axis on which the projections are 
made. Thus if the angular intervals were taken as 18, say, the 
lines corresponding to OM, ON and Of in (b] and (d) Fig. 130 
would be drawn making angles of 18 with the horizontal axis ; 
then for the values of A lt A 2 , etc., the projections along OX would 
be measured, whilst the values of B 1 , B 2 , etc., would be determined 
from the projections on OY. 

Method (c): Analysis by Superposition. This method is 
much used in alternating current work, for the problems of which 
it is specially suited. It is not difficult to employ, nor to under- 
stand, although the proof of the method is long and is in consequence 
not treated here. 

In order to present the method in as clear a fashion as possible, 
the rules of procedure are here set out in place of a detailed 
explanation. 

The method is as follows ; the case of a curve containing the 
second as the highest harmonic being treated, although the process 
can readily be extended if necessary : 

(1) Divide the curve into two 'equal parts and superpose the 
second part upon the first, using dividers and paying attention to 
the signs. If the resultant curve approximates to a sine curve 
there is no need to further subdivide. (This gives terms containing 
26, 4$, 60, etc., but if this curve is a sine curve, probably only terms 
containing 2$ occur.) 

Put in a base line for this new curve (by estimation) ;^ then the 
height of this from the original base line = 2A . 

(2) Divide the original curve into three equal parts and super- 
pose (first, the second on the first, and then to this result add the 
third). 

(This gives the terms containing 3$, 60, 'gO, etc.) 
The height of the base line of the resulting curve from the 
original base line = 3A . (The two values of A may be compared, 
and of course they should be alike ; but if not, take the average of 
these and draw a new base line distant A from the original ; this 
line we shall speak of as the true base line.) 

(3) Subtract corresponding ordinates of the 20 curve (divided 
by^2) and the 36 curve (divided by 3), paying attention to the 



352 



MATHEMATICS FOR ENGINEERS 



signs, from the ordinates of the original curve ; the resultant curve 

is approximately a sine curve 
symmetrical about the true 
base line. 

To calculate the values of 
the constants, if 




+A 2 sin (20+c 2 ) 
A is already found. 

Select two convenient 
values of 6 and work from 
the ordinates of the curve 
to find Aj and c x ; proceed 
similarly, using the 26 curve 
to find A 2 and c 2 . 

Note that in alternating 
current work only terms of 
the order 6, 36, $0, etc., occur, 
so that the curve would need 
to be divided into 3, 5, etc., 
equal divisions and the parts 
superposed. There is thus 
no need to divide into 2, 4, 
etc., equal parts; also it is 
evident that the value of A 
must be zero. 

Example 3. The curve 
ABCD, Fig. 132, gives by its 
ordinates the displacement of 
a valve actuated by a Gooch 
Link Motion. 

It is required to find the 
constants in the equation 



+A 2 sin 



etc. 



The original curve is divided 
into two equal parts, the second 
being placed over the first, with 
the result that Curve 2 is 
obtained. 

The estimated base line for 
this is B 2 ; the height of B 2 above the original base line being -29, i. e., 



HARMONIC ANALYSIS 353 

2Q 

the height of the true base line is -- or -145 unit. This base line 

can now be put in, and is indicated as the true base line. 

By division into 3 and 4 equal parts and superposition the curves 
3 and 4 respectively are obtained. 

B 3 , the base line for 3, is at a height of -43 ; this figure divided by 
3 gives -143, which agrees well with our former result. 

Curve 2 really represents the first harmonic with double amplitude ; 
therefore we subtract ordinates of Curve 2 (to half scale, i. e., we use 
proportional compasses) from the corresponding ordinates of the 
original curve. 

Similarly we subtract J of the ordinates of Curve 3 from the 
original curve, and since those for Curve 4 are too small to be taken 
into account, the net result is Curve i, which represents the funda- 
mental, and is a sine curve symmetrical about the true base line. 

To find the constants A t and c : in the equation 
y l = A l sin(6+c 1 ). 

When 6 = 0, ;Vi = 2-i75 (measured from the true base line to 
Curve i). 

At 6 90, y = o. 

c, = 90 or 
2 

At 180 y t = 2-135, 

t. e., Aj = the mean of 2-175 an d 2>I 35 * e -> 2 ' 1 5> 

y = 2-15 sin (0+90 3 ) 

= 2-15 costf. 
To find A 2 and c 2 . 

*2Q 

The amplitude of Curve 2 is , . e., -145. 



and since the curve has its maximum ordinate when 6 = we have 
again c g = 90, or the curve is a cosine curve. 

Hence y^ = -145 cos 20. 

Beyond this first harmonic we need not proceed as the amplitudes 
of Curves 3 and 4 are exceedingly small. 

Hence y = y*+y* 

= 2-Ij COS ^+-145 COS 20. 

This method of superposition is to be recommended in cases of 
A.C. work, as one can so readily tell by its aid which harmonics are 
present. If the actual constants in the equation are required it 
may be easier to proceed according to method (a) or method (b) . 

A A 



354 



MATHEMATICS FOR ENGINEERS 



Exercises 24. On Harmonic Analysis. 

1. Show how to analyse approximately the displacement x of a 
point in a mechanism on the assumption that it may be represented 
by a limited series of sine and cosine terms, and obtain general expres- 
sions for the values of the coefficients in the series 

x 2 n (A n cos 0+B n sin n0) + A 

where n = 3 and 6 is the angular displacement of an actuating crank 
which revolves uniformly. Apply your results to obtain the values of 
the coefficients for the values of x and 6 given in the accompanying 
table, where the linear displacement of a point in a mechanism is given 
for the corresponding angular displacement of a uniformly revolving 
crank. 



Angular displacement 
of crank in degrees . 


o 


60 


90 


120 


180 


240 


300 


x (in ins ) 


I'll 


. 


8 


1-6 


67 


, f 


2-Q3 










_ 









2. A part of a machine has an oscillating motion. The displace- 
ments y at times t are as in the table. 



t 


02 


4 


06 


08 


i 


12 


M 


16 


18 


2 


y 


64 


I-I 3 


i-34 


95 





92 


-i-33 


1-16 


-66 






Find the constants in the equation 

y = A sin (iO7r^+a 1 ) + Bsin (2O7r/+a 2 ). 



3. Analyse the curve which results when the following values are 
plotted. 



x 





45 


90 


135 


1 80 


225 


270 


315 


360 


y 


o 


21-5 


31-25 


11-25 





9 


30 


26-5 


o 



4. The values of the primary E.M.F. of a transformer at different 
points in the cycle are as follows (6 being written in place of pt for 
reasons of simplicity) . 



e 


o 


3 


60 


90 


120 


150 


1 80 


2IO 


240 


270 


300 


330 


360 


E 


-14 


886 


1293 


1400 


130? 


814 


-70 


-886 


-1293 


- 1400 


-1307 


-814 


70 



If 6 and E are connected by the equation 

E = A sin 0+B sin 3#+C cos 0+D cos 
find the values of the constants A, B, C and D. 



CHAPTER XII 
THE SOLUTION OF SPHERICAL TRIANGLES 

THE curvature of the earth's surface is not an appreciable factor 
in the calculations following a small survey, and is therefore not 
regarded, but when the lengths of the boundaries of the survey are 
great, as in the case of a " major triangulation," the effect of the 
curvature must be allowed for, if precision is desired. It is there- 
fore necessary to use Spherical Trigonometry in place of the more 
familiar Plane Trigonometry, and accordingly a very brief chapter 
is inserted here, dealing mainly with the solution of spherical 
triangles. 

Definitions of Terms used. The earth may be considered 
as a sphere of radius 20,890,172 feet, this being the mean radius. 

A great circle on a sphere is a circle traced by the intersection 
of the sphere by a plane passing through its centre ; if the plane 
does not pass through the centre of the sphere, its intersection with 
the sphere is called a small circle. Thus all meridians are great 
circles, whilst parallels of latitude, except for the equator parallel, 
are all small circles. 

A straight line on the earth's surface is in reality a portion of a 
great circle ; hence a parallel of latitude is not a straight line, or, in 
other words, a movement due East or West is not a movement 
along a straight line. 

A triangle set out on the earth's surface with straight sides is what 
is termed a "spherical triangle," its sides being arcs of great circles. 
The lengths of these sides might be measured according to the 
usual rules, viz., in miles, furlongs, etc., but it is more usual to 
measure them by the sizes of the angles subtended by them at the 
centre of the sphere. In this connection it is convenient to 
remember that an arc of one nautical mile (6076 feet) subtends an 
angle of i' at the centre of the earth ; hence a length of 80 
nautical miles would be spoken of as a side of 80', i. e., i 20'. 

In Fig. 133 is shown the difference between great and small 
circles ; and AB, BC and CA being portions of great circles form a 

355 



356 



MATHEMATICS FOR ENGINEERS 



spherical triangle (shown cross hatched). The length BA would be 
expressed by the magnitude of the angle BOA. 

A spherical triangle ABC is shown in Fig. 134, O being the 
centre of the sphere. The arc AB is proportional to the angle 
AOB, and therefore, instead of speaking of AB as a length, it is 
quite legitimate to represent it by L AOB. 

c would thus stand for /.AOB, b for ^.COA, and a for <iCOB. 
As regards the angles of the triangle, the angle between CA and 
AB is that between the planes AOC and AOB and is, therefore, the 
angle between the tangents AD and AE. Spherical triangles 
should be regarded as the most general form of plane triangles ; 
for if the radius of the sphere becomes infinite the spherical triangle 
becomes a plane triangle. 




FIG. 133. 



Spherical Triangles. 



Many rules with which we are familiar in connection with plane 
triangles hold also for spherical triangles, as, for example, "Any 
two sides of a triangle are greater than the third," or, again, " If 
two triangles have two sides and the included angle of the one 
respectively equal to two sides and the included angle of the other, 
the triangles are equal in ah 1 respects " ; " The greater side of every 
triangle is subtended by the greater angle." 

There is one important difference between the rule for a plane 
triangle and a corresponding rule for a spherical triangle: viz., 
whilst the three angles of a plane triangle add up to 180 the sum 
of those in a spherical triangle always exceeds 180, the sum in fact 
lying between 180 and 540 ; and the difference between the sum 
of the three angles and 180 is known as the " spherical excess." 



THE SOLUTION OF SPHERICAL TRIANGLES 357 

The magnitude of this can be found from the rule 

360 X area of triangle 
spherical excess = ^~ 

2irT z 

(27rr 2 being the area of the surface of the hemisphere). 
This spherical excess is a small quantity for the cases likely to 
be considered in connection with surveys. 

E. g., consider the case of an equilateral triangle of side 68 miles. 

r = 20,900,000 ft. approx. = 3960 miles. 
The area of the triangle is about 2000 sq. miles. 
Then the spherical excess 

360x60x2000 

= minutes = -437 minute. 
27rX 3960x3960 -"^ 

A good approximation for the spherical excess of a triangle on 
the earth's surface is : 

area of spherical triangle in sq. miles 



spherical excess (seconds) = 



78 



Solution of Spherical Triangles. The most widely used 
rule in connection with the solution of plane triangles is the " sine " 
rule which states that the sides are proportional to the sines of the 
angles opposite. In the case of spherical triangles this becomes 
modified and reads " The sines of the angles are proportional to the 
sines of the sides opposite. 

Therefore, adopting the notation of Fig. 134, 

sin a sin b sin c 

sin A ~ sin B sin C 
it being remembered that sin a is really sin L BOC, etc. 



Other rules are 



. A /sin (s b) sin (s c) 

sin = A/ - v . ' i- .... (2) 

2 > 



. 

2 > Sln f) sln c 



A /sin s sin (s a) 

cos = \/ : = .... (3) 

2 v sin & sin c 

A /sin (s 6) sin (s c) 

tan =v ^ : , v . . . . . (4) 

2 v sin s sin (s a) 

T> /- 

and corresponding forms for and - , obtained by writing 

2 2 

the letters one on in the proper sequence, a b c a. 

s in these formulas = - and is, therefore, an angle 

(in plane trigonometry, s = - , but is a length). 



358 MATHEMATICS FOR ENGINEERS 

It is of interest to compare these with the corresponding rules 
in connection with plane triangles, which are 



A /(sb)(sc) 

sin - = \/ '- '- 

2 v be 



A A 

'*=V- 



's(s 

COS 



be 



8(9 -a) 

It will be seen that, as in the previous case, sides occurring in 
the formulae of plane trigonometry are replaced by their sines in 
the corresponding formulae of spherical trigonometry. 

Other rules are : 

cos A+cos B cos C 

cos a = . . (5) 

sin B sin C 

cos a cos b cos c 

cos A : =. (6) 

sin b sin c 

cot A sin B = cot a sin c cos B cos c . . . (7) 

(a-V) 
cos- 

tan ~lT~ ~7^R>\ COt 2 (8) 

COS - J 

\ 2 / 

(a-b) 

. sin 

A B 2 C 

tan = - -r- cot .....-.. (q) 

2 . fa-\-o\ 2 

sin - 
V 2 / 

Solution of Right Angled Spherical Triangles. In the 

case of a right angled spherical triangle these rules can be put into 
somewhat simpler forms. 

Assume that the triangle is right angled at C. 

= 90, .'. cosC = o, and sinC = i. 

cose cos a cos b 

From (6) cos C = 

sin a sin b 

but cos C = o, 
cos c cos a cos b = o, 
i.e., cos c = cos a cos b (10) 



THE SOLUTION OF SPHERICAL TRIANGLES 359 

cos a cos b cos c 

Also cos A = - . . . from (6) 

sin sin c 

cos c 

= f COS COS C 

COS0 , 

from (10) 
sin sin c 

cos c i cos cos c 

\s _^_, w 

sin c sin cos sin sin c 

if i cos 01 

~ tan c\sin b cos b sin J 

: cos 2 &1 cot ex sin 2 



= cote] T 
Isi 



sin cos 0J sin cos b 

sino 

= cot c X , 

COS0 

= cot c X tan 
or tan ox tan (90 c) 

i.e., cos A = tan b tan (90 c)l , > 

also cos B = tan a tan (90 c) J 

COStf COS COS C 

Again cos A = - ; ^. 
sin & sin c 

COS 2 C 



cos a 



cos a , . . 
from (10) 



sin sin c 

cos 2 cos 2 c sin 2 c sin 2 a 

= -- or ~ir- (12) 

cos a sin sin c cos a sin sin c 

. . sin a 

And from (i) sin A = -, (13) 

sine 

In plane trigonometry sin A = - . 

Napier's Rules of Circular Parts. The equations (10), 
(n) and (13) and their modifications may be easily remembered by 
Napier's two rules of circular parts, which may almost be regarded 
as a mnemonic. 

For the application of these two rules the five parts of the 
spherical triangle, other than the right angle at C, are 'regarded as 
a, b, (go A), (90 c), and (90 B) respectively, the complements of 
A, c and B being taken instead of the values A, c and B in order 
that the two rules may embrace all the cases. 



360 MATHEMATICS FOR ENGINEERS 

These five parts are written in the five sectors of a circle in the 
order in which they occur in a triangle : thus in Fig. 135, com- 
mencing from the side a and making the circuit of the triangle in 
the direction indicated, the parts in turn are a b A (for which we 
write 90 A), c (for which is written 90 c) and B (for which is 
written 90 B). These parts are set out as shown in Fig. 136. 

Then Napier's rules state : 

Sine of the middle part = product of tangents of adjacent parts. 

Sine of the middle part = product of cosines of opposite parts. 

The terms middle, adjacent and opposite have reference to the 
mutual position of the parts in Fig. 136. Thus if b is selected as 
the middle part, the adjacent parts are those in immediate contact 





FIG. 135. FIG. 136. 

with b, viz., a and (90 A), whilst (90 c) and (906) are the 
opposite parts. 

Hence sin b = tan a x tan (90 A) = tan a cot A 

sin b = cos (906) X cos (90 c) = sin B sin c 

. sin b ., ,. , . 

or smB = - . (Cf. equation (13), p. 359.) 

olll C/ 

Again if (906) is selected as the middle part, the adjacent 
parts are (goc) and, and the opposite parts are (90 A) and b. 
Hence sin (906) = tan (90 c) tana 

or cos B = tan (900) tan a (cf . equation (n), p. 359), 

and sin (90 B) = cos (90 A) cos b 
or cos B = sin A cos b. 

These rules, being composed of products and quotients only, 
lend themselves well to logarithmic computation. 

The Ambiguous Case in the Solution of Spherical 
Triangles. In the solution of a plane triangle, if two sides and the 
angle opposite the shorter of these is given, there is the possibility 



THE SOLUTION OF SPHERICAL TRIANGLES 361 

of two solutions of the problem ; the best test for which, as pointed 
out in Chap. VI, Part I, being the drawing to scale. 

A similar difficulty occurs in the solution of spherical triangles, 
when two sides and the angle opposite one of them is given. 

E. g., let a b and B be the given parts. 
Then from equation (i), p. 357. 

sin A sin B 

sin a sin b 

sin a sin B 

or sin A = : -, . 

sino 

Now sin A = sin(i8o A) and thus the right-hand side of this 
last equation may be the value of either a particular angle or its 
supplement. 

Without going into the proof it may be stated that there will 
be one solution only if the side opposite the given angle has a 
value between the other given side and its supplement. Thus in 
the case in which a b and B are given, there will be one solution 
only if b lies between a and (180 a}. 

If b is not between a and (180 a], then the test must be 
applied that the greater angle must be opposite the greater side : 
thus for the case of a b and B given, if a > b then A must be > B. 
The possible cases may be best illustrated by numerical examples, 
a b and B being regarded as the given parts throughout. 

(a) Given a = 144 40', b = 87 37', B = ng' to find A. 

Using equation (i) of p. 357 

A _ sin a sin B _ sin 144 40' X sin 11 9' 

Sin A : f ; = -. 

sm b sin 87 37 

and log sin A = log sin 144 4o'-flog sin 11 9' log sin 87 37' 

= i 7622 + 1 28641" 9996 

= I 0490 
so that it is possible that A = either 6 26' or 173 34'. 

Now a > b and therefore A must be > B, and this condition is 
only satisfied if A = 173 34', since 6 26' is not > n9'. 

It will be noted that the case chosen is that in which b, viz., 
87 37', lies between a, i. e., 144 40', and (180 a), i. e., 35 20', and 
therefore only one solution is expected. 

. (b) b = 44 35', a = 55 10' and B = 38 46'. 

Here b does not lie between a and (180 a), so that two solutions 
are possible. 



362 MATHEMATICS FOR ENGINEERS 

As before 

log sin A = log sin a + log sin B log sin b 

= log sin 55 io'+ log sin 38 46' log sin 44 35' 
= I 9142+1 79661 8463 = I 8645 
log sin 47 3' 

so that possible values of A are 47 3' and 132 57' and we must 
test each of these values. 

Now a > b, and hence A must be > B ; but 47 3' and 132 57' are 
both > 38 46', so that we have two triangles satisfying the con- 
ditions, and for complete solution the two values of A, C and c 
must be determined. 

Example i. In a spherical triangle ABC, having given = 30, 
6 = 40, C = 70, find A and B. 
Given also that 

Lsin 5 =8-9402960 L tan 12 14' 38" = 9-3364779 

L sin 35 = 9'75 8 59i3 L tan 60 4' 3" = 10-2397529 
L cos 5 9-9983442 
L cos 35 = 9-9133645 



In this case two sides and the included angle are given ; we there- 
fore use equations (8) and (9) . 

fa b\ 

A+B DS V 2 / ,C , ,. 

tan = cot - . . from (8) 

2 (a+b\ 2 

cos f j 

= -5_: cot35 . {*<^f } 



cos 35 \ 

cos 5 cos 35 _ cos 
cos 35 sin 35 ~ sin 35- 



Taking logs of both sides 

A+B 
Ltan = L cos 5 Lsin 35+io 

A+B 

(or, alternatively, log tan = log cos 5 log sin 35) 



Ltan 



A+B I9-9983442 
= 



= L tan 60 4' 3* 

A+B 

-^- = 6o 4 ' 3 " 

A+B = 120 8' 6" ....... (a) 



THE SOLUTION OF SPHERICAL TRIANGLES 363 

From equation (9) 

fa-b\ 

. -_. sin I 1 

A B \ 2 / , C 

tan -j- cot 

2 . fa+b\ 2 

sin - 
V 2 / 

A B_ sin 5 00535 



tan 



sin 35" sin 35 

B A _ sin 5x cos 35 

sin 2 35 
taking logs throughout. 

B A 
L tan = Lsin 5-fLcos35 2Lsin35+io 

18-9402960 

9-9I33645 

= 28-8536605 

19-5171826 



=- Ltan 12 14' 38* 
B-A=2 4 29'i6* ...... . (6) 

By adding (a) and (b) 2B = 144 37' 22* 

B = 72 i8'4i* 

and A = 120 8' 6" 72 18' 41" = 47 49'25*. 

[Note that A+B+C = 47 49' 25"+72 i8'4i*+7o 

= 190 8' 6' 
so that the spherical excess = 10 8' 6".] 

Example 2. Solve the spherical triangle ABC, having given 
c = gii8', a 72 27', and = 90. 

In this case the triangle is right angled, and therefore rules (10) to 
(13) may be used. 

To find A : 

. . sin a 

From equation (13), p. 359, sin A -: -- 

sin o 

L sin A = L sin a L sin c-\- 10 

= Lsin 72 27' Lsin 91 i8'+io 
19-97930 



9-97941 

= L sin 72 29' 45*. 
A = 72 29' 45'. 



364 MATHEMATICS FOR ENGINEERS 

To find b : 

From equation (10), p. 358, 

cos c cos a cos b 

cos c 

whence cos 6 = - 

cos a 

cosoii8 / cos 88 42' 

* cos b _ _ _ -- - _ 

cos 72 27' cos 72 27' 

cos 88 42' 

or cos (180 b) = cos 6 = - 5-^ > 

cos 72 27' 

Hence we shall work to find the supplement of 6. 
Taking logs * 

log cos (180 6) log cos 88 42' log cos 72 27' 



= 1147934 
2-87644 

= log cos 85 41' 7*. 
1806 = 85 41' 7" 



* It is rather easier to work in terms of the logs in preference 
to the logarithmic ratios. One must remember, however, that the 
L sine A = log sin A+io, so that if a L sin A reading is 9-97941, then the 
reading for log sin A would be 1-97941. If the logarithmic ratios are 
used the addition of the 10 must not be overlooked. 

To find B : 

From equation (n), p. 359, 

cos B = tan a tan (90 c) 

i. e, t cos (180 B) =tanatan (c 90) = tan 72 27' x tan i 18'. 
Iogcos(i8o B) log tan 72 2 7'+ log tan i 18' 

4990 

= 2-3559Q 
2-85586 

-log cos 85 53' 6". 
1806 = 85 53' 6" 
B = 9 4 6'54*. 
Hence, grouping our results, 

a = 72 27' A = 72 29' 45' 

6-94i8'53* B = 946'54* 

C = 90 



Example 3. At a point A, in latitude 50 N., a straight line is 
ranged out which runs due E. at A. This straight line is prolonged for 




THE SOLUTION OF SPHERICAL TRIANGLES 365 

60 nautical miles to B. Find the latitude of B, and if it be desired to 
travel due N. from B so as to meet the 50 parallel again at C, find the 
angle ABC at which we must set out and also the distance BC. 

In Fig. 137 let A be the point on latitude 50 N. and ABD be a 
great circle passing through A : thus AB 
is a straight line running due E. from A. 
Let NB be the meridian through B, and NA 
that through A. 

The sides NA, AB and BN are straight 
lines, because they are parts of great circles 
and therefore they together form a spheri- 
cal triangle. 

In this triangle we know -the side AB 
(its value being 60', for i nautical mile 
subtends an angle of i' at the centre) ; 
the angle at A (90) ; and the side NA 
(90 latitude, i.e., 40). 

Thus two sides and the included angle are given and we require to 
solve the triangle ; hence we use rule (10), p. 358, 

from which cos NB cos NA cos AB 

= cos 40 cos 60' 
or logcosNB = logcoS4O-f log cos i 

= 1-88425 +1-99993 

= 1-88418 

i.e., NB = 4oo'38' 1 ' 

or the latitude of B is 

Q0 40 o' 38* = 49 59' 22*. 

Now C is at the same latitude as A, so that BC is 38", corresponding 

to f- nautical miles ; i. e., BC -633 nautical mile. 
60 

To find the angle ABC, we use rule (13).. p. 359. 
. __ sin NA sin 40 

SI* L. ABC = ,-== = . ^-fQ, 

sin NB sin 40 o 38* 

log sin L. ABC log sin 40 log sin 40 o' 38* 
i -90807 I -90817 
= I -99990 
whence L ABC = 88 45'. 

For the surveyor, spherical trigonometry has an important 
application in questions relating to spherical astronomy. Thus in 
the determination of the latitude of a place by observation to a 
star, the calculations necessary involve the solution of a spherical 
triangle. This triangle is indicated in Fig. 139, the sides AB, BC 



366 



MATHEMATICS FOR ENGINEERS 



and CA representing the co-latitude of the place, i. e., (90 latitude), 
(90 declination) and (90 altitude) respectively ; whilst the 
angles A, B and C measure respectively the azimuth, the hour 
angle and the parallactic angle. 

The terms just mentioned are denned as follows: 
Fig. 138 represents a section of the celestial sphere at the 
meridian through the point of observation O. RDT is the celestial 
equator, CEX is the horizon, Z is the zenith of the point of 
observation, i. e., the point on the celestial sphere directly above O, 
and S marks the position of the heavenly body to which observa- 




FIG. 138. Determination of Latitude. 

tions are made. Also PSD, ZSC, RDT and CEX are portions of 
great circles. 

The altitude of a heavenly body is the arc of a great circle 
passing through the zenith of the point of observation and the 
heavenly body ; the arc being that intercepted between the body 
and the horizon. We may thus compare the altitude in astro- 
nomy with the angle of elevation in surveying. Referring to 
Fig. 138, ZSC is the great circle passing through Z and S, and SC 
is the altitude. ZS, which is the complement of SC, is called the 
zenith distance. 



THE SOLUTION OF SPHERICAL TRIANGLES 367 

The azimuth of a heavenly body is the angle between the 
meridian plane through the point of observation and the vertical 
plane passing through the body. It can be compared with the 
'bearing" of plane trigonometry. In Fig. 138, the angle PZS is 
the azimuth of S. 

The hour angle of a heavenly body is the angle at the pole, 
between the meridian plane through the point of observation and 
the great circle through the pole and the body. 

Thus, in the figure, P is the pole, and PSD is the great circle 
passing through P and S ; this being known as the " declination 
circle." Then ^_ZPS = the hour angle of S, and it is usually 
expressed in terms of time rather than in degrees, etc. 

The declination of a heavenly body is the arc of the declination 




FIG. 139. 

circle intercepted between the celestial equator and the heavenly 
body : thus DS is the declination of S. 

The method of calculation can be best explained by working 
through a numerical example ; and in order to ensure a clear 
conception of the problem, it is treated both graphically and 
analytically. 

Example 4. At a certain time at a place in latitude 52 13' N. the 
altitude of the Sun was found to be 48 19' and its declination was 
15 44' N. Determine the azimuth. 

As explained before, a spherical triangle can be constructed with 
sides as follows: a 90 declination = 74 16', b = 90 altitude 
41 41', and c = co-latitude 37 47'. Then the angle A is the 
a?imuth (Fig. 139). 



368 MATHEMATICS FOR ENGINEERS 

Graphic construction. With any convenient radius OD describe an 
arc of a circle DABF. Draw OA, OB and OF, making the angle 
DOA = 6 = 4i4i', LAOB = c = 37 47', and L BOF = a == 74 16'. 
Draw DCE at right angles to OA, and FGC at right angles to OB, 
intersecting at the point C. Note that C lies outside the triangle AOB. 
With centre E and radius ED construct the arc of a circle DH : draw 
CH perpendicular to DE to meet this arc at H and join EH. Then the 
angle REH is the value of the required angle A, and is found to be in 
the neighbourhood of 141. [If C had fallen the other side of OA, the 
angle CEH would have been measured.] 

The actual spherical triangle ABC is formed by the circular arc 
BA and the elliptical arcs AC and BC. 

Proof of the construction The side b is such that it subtends an 
angle of 41 41' at the centre of the sphere. Thus DA measures the 
actual length of b, but does not represent it in its true position. In like 
manner BF gives the length of a, but again does not give its position 
on the sphere. 

Let the circular sector OAD be rotated about OA as axis, and the 
sector OBF about OB as axis, and let the rotation of both be continued 
until they have a common radius OC, i. e., OC is the intersection of the 
two revolving planes. Then evidently C is the third angular point of 
the spherical triangle ABC, since the given conditions concerning the 
lengths of the sides are satisfied by its position. 

We observe that in this case the rotation of OBF has to be continued 
beyond OA, from which fact we gather that the angle at A must be 
obtuse. The line OA is in the plane of the paper, and taking a section 
along DE and turning this down to the plane of the paper, we observe 
that the actual height of C above the paper is CH. Thus EH is a line 
on the plane OAC, also ER is a line in the plane AOB, both lines being 
perpendicular to the line of intersection, and the angle REH therefore 
measures the inclination of the plane AOC to the plane AOB, this angle 
being by definition the angle A of the spherical triangle ABC. 

By calculation. Here we have the three sides given, and we wish to 
find an angle which may be done by use of equation (4), p. 357, viz., 

A /sin (sb) sin (s c) 

tan = A / - : - : - - - r -- 

2 N sin s sin (sa) 
Now , = <*+*>+<> _ 74 i6'+4i 4i'+37 47' _ 153 44' ^ 6 o , 

222 

so that s a 76 52' 74 16' = 2 36' 

s -b = 7652'-4i 4i' = 35 ii' 



Hence tan A = /^ 

2 v sin 76 52' X sm 2 36 



THE SOLUTION OF SPHERICAL TRIANGLES 369 

A i r(logsin35Ti / +logsin395') ~1 

and log tan - = - [_ _ (log sin ?6 o 52 / + i og sin 2 3 g/)J 



i'7 6 57\ 

1-79965 J _ f 2-65670 

I- 56022 / 



= x -91503 = H5752 



thus = 70 46' 30* 

and A = 141 33'. 



Exercises 25. On the Solution of Spherical Triangles. 

(4-figufe log tables only have been used in the solution of these 
problems.) 

In Nos. i to 6, solve the spherical triangle ABC, when 

1. =^50 = 90 & = 32i7'. 

2. = 90 a = 4543' A = 6ii5'. 

3. a = 72 14' & 43 47' c = 29 33'. Find also the value of B 
by the graphic method explained on p. 368. 

4. Z> = 525' a = 58 25' C = 6 4 . 

5. b = 27 13' c = 5ii8 / 6 = 85 9' and the spherical excess is 

2 14'. 

6. c = 7949' b = 28 5' B=i5i8' 

7. If the sun's altitude is 17 58', its declination is 28i6'N., and 
its azimuth is N. 65 43' W., find the latitude of the place of observation. 

8. The spherical excess of a triangle on the earth's surface is i 15': 
taking the earth as a sphere of radius 3,960 miles, find the area of the 
triangle in square miles. 

9. Given that the azimuth of the sun is 10, and its zenith distance 
is 24 50' when its declination is 22 15', find the latitude of the place 
and also the hour angle. 



B B 



MATHEMATICAL PROBABILITY AND THEOREM OF 
LEAST SQUARES 

WHEN extremely accurate results are desired, these results 
being derived from a series of observations, the possibility of error 
in each or all of the observations must be considered. The correct 
result, or what is termed " the most probable result," is usually 
found by combining the mean of the observations with " the pro- 
bable error of the mean." The work that is to follow is concerned 
primarily with the establishment of a rule enabling us to find this 
probable error ; and as a preliminary investigation, a few simple 
rules of probability will be discussed. 

Supposing that an event is likely to happen 5 times and to 
fail 7 times, then the probability that it will happen on any 
specified occasion is r \, whilst its probability of failing is r 7 ^, 
because, considered over a great range, it only happens 5 times 
out of 12. It is important to note the significance of the phrase 
*' considered over a great range " ; we could not say with truth that 
the event was bound to happen 5 times out of the first 12, 
10 times out of the first 24, and so on ; it might be doubtful 
whether it would happen 50 times out of 120. If, however, say, 
12,000 opportunities offered, it would be fairly correct to say that 
the happenings would be 5,000 and the failures 7,000, for when a 
large number of occasions were considered, all " freaks " would be 
eliminated. 

To take another illustration : the probability that a man will 
score 90 per cent, of the full score or over on a target is /y 
indicates that he is rather more likely to score 90 per cent, than 
not (in the proportion 6 to 5) if he fires a great number of shots. 

In general terms, if an event may happen in a ways and fail in 
b ways, and all these are equally likely to occur, then the pro- 
bability of its happening is -j^-, and of its failing =- ; and if 

37 



MATHEMATICAL PROBABILITY 371 

a = b, then it is as likely to happen as not, i. e., its probability of 
either happening or failing is \. 

Probability of happening _ a a+b _ a 

Probability of failing ~ a+b b ~ b' 

i. e., the odds are a to b for the event, or b to a against it, the 
first form being used if a > b and vice versa. 

E.g., if the odds are 10 to i against an event, the probability of 

its happening = = ; or it will probably happen once only 

out of eleven attempts. 

Exclusive Events. Let us now consider the case of two 
exclusive events, viz., the case in which the happenings do not 
concur. 

/ 

Suppose the probability of the happening of the first event =- 

/ 

and the probability of the happening of the second event = -T-. 



Then for purposes of comparison each of these fractions may be 
expressed with the same denominator : if this common denominator 

is c, write the fractions as and respectively. 

c c 

Now out of c equally likely ways the first event may happen in 
a ways and the second in 2 ways, and since the two events are 
exclusive, i. e., the happenings of the one do not coincide with the 
happenings of the other, the two events together may happen in 
!-}-, ways. 

Hence the probability that one or the other will happen is 

-. which may be written in the form -f- , i. e., as the sum 

c c c 

of the separate probabilities. 

E. g., suppose that one event happens once out of 8 times, and 
a second event happens three times out of 17, and that there is no 
possibility of the two events happening together ; then, the 
common denominator of 8 and 17 being 136, the first event happens 
17 times out of 136 and the second event happens 24 times out of 
136, and hence, either the one or the other happens 41 times out cf 
each 136. 

Probability of the Happening together of Two Inde- 
pendent Events. Suppose that one event is likely to happen 
once out of every 6 times, whilst another is likely to happen twice 
out of every 17 times ; then the probability that the two will 
happen together must be smaller than the probability of the 



372 MATHEMATICS FOR ENGINEERS 

happening of either in fact, it must be the product of the separate 
probabilities ; i. e., the probability of the two events happening 

122 I 

together = ;rX = or : or out of every 10.200 times the 

6 17 102 51 

first will probably happen 1,700 times, the second will probably 
happen 1,200 times, whilst the two would happen together 200 
times only. 

Probability of Error. Bearing in mind these fundamental 
theorems, we can proceed to a study of the question of probability 
of error ; with particular reference to its application in precision 
surveying. 

It will be admitted that, for any well made series of observa- 
tions, the following assumptions may be regarded as reasonable : 

(1) That small errors are more likely to occur frequently than 
large errors, and hence extremely large errors never occur. 

(2) That positive and negative errors are equally likely, i. e., we 
are as likely to give a result that is -ooi too high as one that is 
ooi too low. 

Hence the probability of the occurrence of an error of a given 
magnitude, which is denoted by 

the number of errors of that magnitude 
total number of errors 

depends in some way upon the magnitude of that error. Our first 
idea, therefore, might be that the probability of the occurrence 
of an error of magnitude x could be expressed as f(x), i. e., as some 
linear function of x. It will be seen, however, that this is not in 
accordance with assumption (2) ; for assumption (2) demands that 
if a curve be plotted, the ordinates showing probabilities and the 
abscissae indicating errors, it must be symmetrical about the y axis. 
The function must therefore be of an even power of x, and taking 
the simplest power we say that the probability of occurrence of an 
error of magnitude x =y = f(x 2 ). 

Now, from assumption (i) we note that the coefficient of x z 
must be negative, because y must decrease as x increases. 

The probability of an error of magnitude x being included in the 
range x to x +S# must thus depend on x z , and also on the range Bx ; 
hence it would be reasonable to say that it =f(x z }8x, because the 
greater the range the more is the chance of happening increased. 
Therefore, the probability that an error of magnitude x falls 



MATHEMATICAL PROBABILITY 373 

between any assigned limits, a and -\-a, must be the sum of the 
probability f(x z ) 8x extended over the range a to -\-a, 

i.e., P = ^ f(x*)dx 



this being the probability that the error does not exceed a. 

Hence the probability that the error may have. any value 
whatever (i. e., the probability is i) must be expressed by 



for the range is unlimited, so that 

/+ 

I f(x 2 )dx must = i. 

.' -oo 

It has been proved by Lord Kelvin (see his " Natural Philo- 
sophy ") that f(x 2 ) must be such that 



and since e?* x 0* = 

and e 1 ** X e*v* = 

this condition will be satisfied if 

_2 

or Ae~* 2 
the minus sign being inserted in accordance with assumption i on 

p. 372 ; and the coefficient k being written as ^ f r the reason that 

ft 

is explained later. 

A value can now be found for the constant A. 

r+oo 



r+oo 

It is known that I f(x 2 )dx = i, 

J -00 
/ + _Z2 

hence A / e &dx i. 



Now it has already been proved (see p. 163) that 



r -z* 

I O e ' * = ''~z~ 

/_?: " h^/ir 

and I e h *dx = 

Jo 2 

>r2 r2 

/OO _ Z^ ,-flO _ *" 

also I e h ' 2 dx = 2 / e A2 rfA; = 

y -oo J o 



374 MATHEMATICS FOR ENGINEERS 

hence AxWir = I, 



or 



A = 



Thus 



y =/(*) = 



the law being known as the Normal Error Law. 
The curve representing this equation is called the probability 
curve and also Gauss's Error curve. Two such curves are plotted 
in Fig. 140, to show the effect of the variation of the parameter h. 
In the one case h = -2, and for the second curve h = -5 ; and it 
will be noticed on comparing the curves that for the smaller value 
of h the probability of the occurrence of small errors is greater, 
i. e., the set of observations for which h = -2 would be more nearly 
correct than that for which h = -5. 




It will be seen that the curve is in agreement with the axioms 
stated on p. 372 ; for the probability of error is greatest when the 
error is least, the probability of a large error is very small, and 
there is as much likelihood of an error of + -2, say, as of -2. 

The probability that the error does not exceed -i is given by 
the area ABCD in the one case, and ABEF in the other. 

Theorem of Least Squares. If a number of observations are 
made upon a quantity, and the errors in each of these noted, i. e., as 
nearly as can be estimated ; then from a knowledge of these errors 
it is possible to find the most probable or likely value of the 
quantity. 

Let n observations be taken and let the errors be x^ x z Xn '. 
also suppose that all the measurements are equally good, i. e., the 
"fineness" of reading is the same throughout; h in the formulae 
above being a measure of the fineness. 

The probability of the error x being within a certain range 8x 



MATHEMATICAL PROBABILITY 375 

will be the probability of an error of magnitude x l multiplied by 
the range *, i.e., 



i . 
and for error x 2 P 2 = 8xX -=.e~ * a and so on. 



Now #! x 2 etc. are quite independent, so that the probability of 
all the errors falling within the range &x will be the product of the 
separate probabilities, i.e., 

P = P 1 XP 8 X . . .P B 

&x -^ Sx -** 

x =e 2 X . . 



hVi 



We have thus obtained an expression which gives us the prob- 
ability of all the errors falling within a certain range. We might 
say that this range was -i, for instance, or -05. Evidently if all 
the errors were kept within the range -05 to +-05 the calculated 
result would be a nearer approximation to the truth than if the 
range were double the amount stated. 

Our object then is to find when the probability of a small error 
(8x may be reduced as we please) is greatest, *'. e., to find when P is 
a maximum. 

_ .. -!(*-) K 
Now P = K e h& -; 



and the smaller the denominator is made, the larger will P become. 
But the only variable in the denominator is 2# 2 , and hence, in order 
that P may have its maximum value, 2# 2 must be the least 
possible. Hence the most probable value of the quantity to be 
determined is that which makes the sum of the squares of the 
errors the least. 

The fact can now be established that the arithmetic mean of 
the observed values is the most probable value of the quantity. 



376 MATHEMATICS FOR ENGINEERS 

Thus, if n observations are made, 

let ! 2 # 3 . . . On be the respective observations 
a the A.M. of these values 
a the value most probably correct 
then (! a) (a 2 a) etc. are known as residual errors. 
Now the probability of making this system of errors 



T> A ~ ro 

or P = Ae h ~ 

- A - 
To differentiate P with regard to a, put u = 

du 
then i_ = o+2tfw 

fl# 

Thus P = A*~^ 

rfP dP du 
_ __ _ \/ _ 

da du da 

A --- 

= F5 
h 2 

dP 
and -3-=- =o if 2an zSa, = o 



w 

or if = -za-, 
n 

but -2#i = a = A.M. of the observations 

n 

and hence <z = a 
or the most probable value is the A.M. of the observations. 

Again, if x is the error of the A.M. and x l x z x 3 etc. are the 
respective errors of the observations, 



By squaring 



X = - 

ft 



= l 2 (2V)+i< 

M 2V I / l n z\ 



MATHEMATICAL PROBABILITY 377 

then, since it is assumed that all the observations are equally good, 
and that positive and negative errors are equally likely to occur, 

v 2 -y 2 A* 2 ,,- QTtin ^? ^ v {\ 

vj ^'2 3 * " r^ ciuA-i ^H-vi-vo "~~ vj 

for all the errors are small and their products, two at a time, are 
still smaller. 

i u 2 

Also x 2 = (u 2 ) = 

* 

or # = ~= 

Vn 

. . probable error of a single observation 
or the probable error of the A.M. = , 

Vn 

and thus, other things being equal, the possibility of a large error 
in the final result is greatly reduced by taking a great number of 
observations. Also in a set of well made observations, if a sufficient 
number are made, the arithmetic mean cannot differ from any of 
the observations to any very great extent, and accordingly the 
residual errors and .the actual errors are very nearly alike. 

We are now in a position to summarise the results of the 
investigation so far as we have pursued it ; thus 

(a) The arithmetic mean of the series of observations, which are 
supposed to have been made with equal care, is the most probable 
value of the quantity. 

(b) The sum of the squares of the residual errors must be the 
least. 

(c) The probable error of the A.M. is equal to the probable 
error of a single observation divided by the square root of the 
number of observations. 

Example i. Seven observations of a certain quantity, all made with 
equal care, were 12, n, 14, 12, 11-2, 11-7, and 12-1. 
Find the most probable value of the quantity. 

The most probable value = A.M. = = 12, 

7 

and it can readily be shown, by actual calculation, that this value 
makes the sum of the squares of the residual errors the least. 

The residual errors are 

(12 12), (11 12), (1412), (12 12), (11-2 12), (11-712) 
and (12-1 12) or o, i, 2, o, -8, -3, -i 
and 2 squares of residual errors 0+1+4+0+ -64+ -09+ -01 

= 574- 



378 MATHEMATICS FOR ENGINEERS 

To test whether this is the least, let us suppose that the most 
probable value is 11-5; then the residual errors are: -5, -5, 2-5, -5, 
3, -2 and -6 respectively. 

2 squares = -25 + -25+6-254- -25+ -09+ -04+ -36 = 7-49. 

Similarly, if we assume, say, 12-2 as the most probable value, 

Z (residual error) 2 = (. 2 ) 2 +(i-2)?+(i-8) 2 +(-2) 2 +(i) 2 +(-5) 2 +(-i) 2 
= -04+1-44+3-24+ -04+1+ -25+ -oi 
= 6-02 
both of which totals exceed 5-74. 

To find the Probable Error of the Arithmetic Mean. 
Let r = the probable error of any one of the observations ; then if 
this is an "average" error, i. e., if errors greater are as likely to 
occur as errors smaller, the probability that the error is less than 
r is \. 

Now, the probability that an error lies within the range r to -\-r 

T f+r -I 2 2 [' - r - 

is =1 e h *dr = -=.1 e h2 dr 



T f+r -I 2 [' 

=1 e h *dr = -=.1 e 

hV-n-J -r flVirJ 



- /r 



/for dr=hd( r ) and the limits are now those for T and not those for r\ 
\ \hj h I 

There must be some connection between the amount of error 
and the fineness of measurement, i. e., between r and h, and this we 
must now find. 

If X = 
h 

A/V' o x x V ir-i o 

and we see from the above statement that the value of this 
integral is to be . 

9 o 

XT , X * , X 3 . 

N o w e i i- jf -i u i 

2 6 

and thus e~ X2 i X 2 -| ^-+ . 

2 o 

and if X 2 is small we may perform the integration by way of 
expansion in series : if X 2 is not small the value of the integral 
would be read from probability tables which give the values of the 

2 /-x _ X 2 

integral ^ I e ^X : these tables being given in the Transactions 
VWo 

of the Royal Society of Edinburgh, Vol. xxxix. For the present 



MATHEMATICAL PROBABILITY 



379 



application of the integral, however, X is a small quantity, and a 
sufficiently correct result is obtained by expanding in a series and 
calculating from a few terms in this series. 
Thus 

L _T 

o fh -X 2 -2 / rh --' ~ 

-41 dX = - 2 ( (i 

VTT\J 




FIG. 141. 



Hence 




and this equation may be written 

v5 = /r_j:L . ' 



or if for j- we again write X 

X 3 X 5 X 7 

44SI = X 1 h terms which are very small. 

3 10 42 

By selecting values of X and plotting, the solution of this 
equation is found, the final plotting being represented in Fig. 141, 
where it is seen that the solution is X = -4769. 



3 8o MATHEMATICS FOR ENGINEERS 

Y 

Thus T = '4769 or r = -4769^. 

Y 

[If solved to a greater degree of accuracy, the value of r is 

found to be -47696^, and this figure will be used in the work that 
follows.] 

Again, if n equally good measurements have been made, each 
will have what is termed a weight of unity, i. e., none is better or 
worse than any other, and when working towards the result to 
be deduced from the measurements, equal consideration must be 
paid to each measurement ; also the A.M. is said to have a weight 
of n since on the average n observations of equal weight must be 
made to give a result as true as the A.M. 

r 

Knowing that r m = ~/^ 

Vn 

where r m = probable error of the A.M. 

and r = probable error of any observation 

weight of A.M. n 

and also r-^-r ?- *r- r- = - 

weight of one observation i 

, . , .. v 

which we can write as = - 

w i 

we can link up w m and w with r m and r, 

for ^!-^i 

r* n w m 

or the weight varies inversely as the square of the probable error. 

Thus the determination of the probable error, whilst a useful 
guide to the accuracy of the one set of observations, is more use- 
ful in fixing the relative weights that must be given to different 
sets of observations. 

Thus, if three sets of observations have been made on a certain 
length with the results that the probable errors of the A.M. are 
45, -29, and -51 respectively ; then the weights to be given to 

these sets are - ^ -. -^ - r^ respectively 

(.45)2 (. 29 ) 2 (-5i) 2 

or -494 1-19 -384. 

Then in assessing for the final result, by far the most reliance 
would be placed on the second set of observations, less on the first, 
and least on the third set ; this fact being well illustrated by 



MATHEMATICAL PROBABILITY 381 

Fig. 142, the resultant weight being nearer to the weight 1-19 than 
to either of the other weights. 

1R.2-068 



494 1-19 394 

FIG. 142. 

To return to the object of this paragraph : 
If #! # 2 *s are the actual errors of observation, then the 
probability that each falls within the small range 8* is 

i l^ 2 i -(^Y 

P! = ^j V*/ 8* P 2 = -7=e ^ h) (&) etc., 

hVir hVir 

and the probability that they all fall within this range at the same 
time will be less than either of the separate probabilities ; it will 
actually be the product of these. 

Thus P = P 1 xP 2 X ...... 

i ( 



~ 

hVir hV-jr 



-(*,+*,+ 
h 



We wish to find for what value of h P has its greatest value 
hence differentiate P with respect to h. 



P xe -- where K = 



= UXV 



= and 

h n dh 



v = g-A2< 2a: i 2 ) or if w = TtCZx-, 2 ) v = 

h? 

and thus ~ = (2^ 2 ) X -2h - 3 

ah 

dv dv dw 
Also = -- x -3- 
an aw dh 

_de-" dw_ ^ 
: dw X dh~ 



382 MATHEMATICS FOR ENGINEERS 

Then 

dP du . dv 
~rr= v-r; +U-T; 
dh dh dh 



=o if 

22V 



\/^ 



or h== 1-414 

Now it has already been proved that r = -47696/1 
so that r = -47696 Vs 



/(W) 
- -6745V ^ 



Also we have previously stated that the sum of the squares 
of the actual errors differs very little from the sum of the squares 
of residual errors ; this being true if a great number of observations 
are taken. The difference in the two sums may be expressed rather 
more accurately by the relation 

Yl 

^Xj 2 = - - 2 (residual error) 2 . 
w r 

Hence if for 2 (residual error) 2 we write 



Applying these results to Example I on p. 377, 

W = 574 

n = 7 

then ^ 



* - '6745^ = -2475 






MATHEMATICAL PROBABILITY 383 

/ 2n^r e z /2X574 c 

also h = v 7 v == v =i '3 

v (n i)Xn 6 

*. e., h has a very high value ; and this would be expected, for the 
"fineness " of reading, as judged by the results, is not at all good 
(one error being as much as 2 in 12). 

Example 2. In a chain survey four measurements of a base line 
gave 867-35, 867-51, 867-28 and 867-62 links respectively. Find the 
best length and the probable error in this length. 

The best result is the A.M. of these, i. e., 

867-35+867-5I+867-28+867-62 

4 

867-44 links 

and whilst this is the best result it contains a probable error. 
Probable error in A.M. 

= ^=-6745\/^Yy 
= -6745 



/( Q9) 2 +(-Q7) 2 +( 



4X3 

- -6745 V ^ 

= -0517 

i. e., the base line measurement (867-44) is subject to an error of 
0517 link, and as this result could not be bettered it would be 
unnecessary to repeat this portion of the survey. 

The probable error in any one observation would be 



'='6745 \-- = -io3, 

so that there is a decided gain in accuracy obtained by increasing the 
number of observations. (Cf. "repetition," when working with the 
theodolite.) 

It is of interest to find h for this example. 



2X-07I _ 

-J = ' 2176 

and as this is a small quantity we are confirmed in our conclusion that 
the observations were well made. 

Example 3. The mean values of the three angles of a spherical 
triangle were calculated from the actual observations to be 75 40' 21 -6", 
39 1 i' 47-3", and 65 7' 56-2"; and these values were subject to 
probable errors 2-9", 3-6*, and 4-3* respectively. From a knowledge 



MATHEMATICS FOR ENGINEERS 

of the area of the triangle, the spherical excess of the triangle was 
found to be 3-3*. Make the necessary adjustments to the angles to 
satisfy this condition. 

The actual spherical excess 

= (75 40'2i-6"+39 ii / 47-3"+657 / 56-2")-i8o 

There is thus (5-1 3-3) to be divided among the angles, according to 
the respective weights ; and these weights are in the proportion 



or -119 -077 -054, 

the sum of the weights being -250. 

IIO *O77 

Hence the corrections to be applied are Xi-8, xi-8. and 

250 -250 

^ xi '8 to the respective angles; all these corrections being sub- 
250 

tracted, since the observed angles give a spherical excess greater than 
should actually be the case. 

These corrections are -857, -555, and -389. 

Hence the true angles are (75 40' 21-6* -86*), (39 ii'47'3* -56") 
and (65 / 56-2*- -39*), 

or 75 40' 20-74", 39 II/ 46'74' / and 65 7' 55-81*. 

Example 4. Measurements of an angle in a traverse survey were 
made by two different observers, with the following results : 



Readings by A. 


Readings by B. 


76 50' 20* 


76 50' 55" 


76 50' 50* 


76 50' 35* 


76 50' 30" 


76 51' 15" 


76 51' 10* 


76 51' 20* 


76 5 0' 3 0* 


76 51' o" 


76 51' o" 


76 50' 45* 


76 50' 40* 


76 50' 25* 


76 50' 30* 


76 50' 40" 



Compare the two results from the point of accuracy, and find the 
most probable value of the angle. 

We must first find the arithmetic mean of each set of observations, 
and then, by subtracting this from each reading, we determine the 
residual errors. 

The A.M. of set A = 76 50' 4 1-25* 

and A.M. of set B = 76 50' 5 1 -88*. 



MATHEMATICAL PROBABILITY 



385 



Since the differences are of seconds only, we need not concern 
ourselves for the present with the degrees and minutes ; and thus the 
table of residual errors and their squares becomes 





Residual Error. 


(Residual Error) 2 . 


Residual Error. 


(Residual Error)*. 




21-25 451.4 


+ 3-12 


9-7 




+ 8-75 


76-6 


16-88 


284-9 




11-25 


126-6 


+ 23-I2 


534-6 




+28-75 


826-8 


+ 28-12 


790-7 




-11-25 


.126-6 


+ 8-12 


66-0 




+ 18-75 


35 1' 6 


- 6-88 


47-3 




- 1-25 


1-6 


-26-88 


722-4 




11-25 


126-6 


-H-88 


141-2 




sum o 


2087-8 


o 


2596-8 


In case A f 


/ 






2087-8 


'"745 <v 


8x7 


In case B I 




745 V 


8x7 " 




weight of observations by A (4-594) 


2 1-244 



(4-119)' 

Thus A's readings can be relied on before those of B ; the former 
being roughly ij times as good as the latter. 

The most probable value of the angle, taking into account the two 
sets of readings, will be obtained by the calculation of the " weighted 
mean," i. e., the mean of the two arithmetic means already found, 
determined with due regard to the respective weights to be given to 
A's readings and B's readings. 

Dealing only with the seconds, the most probable value 

(41-25 x i -244) + (5 1 -88 x i) 



1 + 1-244 

51-31+51-88 _ 103-19 
2-244 2-244 



= - - = 46 seconds. 



Hence the most probable value of the angle = 76 50' 46* 



Exercises 26. On the Calculations of Errors of Measurements. 

1. One surveying party measured a certain base line as 6 chains 
42-7 links, 6 chains 53-5 links, 6 chains 46-4 links, and 6 chains 
41-9 links ; and a second party measured the same line as 6 chains 38-4 
links, 6 chains 39-7 links, 6 chains 46-9 links, and 6 chains 43 links. 
State which of the two parties is the more dependable, and find the 
most probable length of the line. 

C C 



3 86 MATHEMATICS FOR ENGINEERS 

i -*_? 

2. Plot the probability curve y = /- e AZ the value of h being 

n v TT 

1414, and find the probability that an error lies within the limits 
-6 and + -6. 

3. The following are the values of the determination of the azimuth 
of Allen from Sears, Texas, the results of a U.S. Coast and Geodetic 
Survey; the values of the seconds only being stated after the first 
reading: 98 6'4i-5*, 42-8, 43-4, 43-1, 39-7, 42-7, 41-6, 43-3, 40-0, 45-0, 
43-3 and 40-7. Find the A.M., the probable error of a single obser- 
vation and the probable error of the mean. 

4. Find the weighted mean of the following observations : 95-8, 
96-9, 97-2, 95-4, 95'7, 97 g ii 96-5, 96-7 and 97 ; the probable errors in 
the measurements being -2, -4, -i, -9, -7, 1-2, -8, -3 and 1-5 respectively. 



ANSWERS TO EXERCISES 

Exercises 1 

3*7 AC* 

3. E = constant x -~ 4. V == RC + L ^ 

at at 

5. 11-03 cms. per sec. : 1-07 seel;, from start 6. -336 ton 
11. 5.65 12. Middle of May : middle of October 

15. Loading is -2 ton per foot run 16. -966 : =^ = cos 6 

uv 

17. -42 ton per foot 21. 6-3 

Exercises 2 



1. 


4# 3 


2. -128 3. 2* 4. 27*" 5. g 




18-75 


0086 i'ii 




-pl.23 


#v 


1 1 


982 


19 -fi?* 8 ' 3 13 /<car 2 /i/i Si' 1 - 8 4- 




X 2 ' 3 


X* 


14. 







* /pi -6 347 


17. 


JIV 3 ' Si - 


, -84 . 1-29 52'5 12-48 ._ .- 




20. 


073 


21. i - a - &P~* 22. w(^ + X jr~j- x 


23. 


289* 


w "* d / I 
24. : (Arv + ny z xl) 25. = - \/ i + -; 

V/ 2 w 


26. 


-2(/> + 


/*"f \ 

9) 27.7-85 28. w(-j--x\ 29.9-6 


30. 


-7333 





Exercises 3 

1. Sub-normal 466 ; sub-tangent = i 2. 25-7 

3. y = -0256^ ; (# is distance from centre) : -64 



+ 142-5 5. M-= ( -- xy. S = - -: L, o 



6. M = -(-- 

2\4 

7. M = W (l-x) : S = - W : L = o 

8. Sub-tangent = : sub-normal = 

9. 3 10. -~ (wl -wx- aP) 

387 



388 MATHEMATICS FOR ENGINEERS 

Exercises 4 
1. -5*~ 5 * 2. 6-I5**' 1 * 3. -^=| 4. 1-423 (4-15)* 

& 

5. 4-33 (8-72) 2x 6. ge 9 * 35e~ 7x 7. 5-44* 1 ' 718 8. 9-7 (2)* 
9. io-25<r 25 *- 10. 



11. i2-6e*' 2 * 12. - 13. ^ 14. 

x 5* 4 

4 ac^ _!.&,-. 2* 

1 O. r A u. ioe 



i^ 
X 1 4 



17. I 4 --- ? - or *2L-VT- [Use the rule log AB = logA+logB] 
* 3* 47 *(3* 47) 

18. -> -- \ -- - -- \ -- ^ 19. 30 3jc + 8 sinh zx '^~ 
5* + 4^3* -2 7-4^ * 

20. - X - 1-057 (1-8)* 21. o 22. 1-052 23. 



24.^43 25> . 2T 26.^,- 27.^7- 

t -L* i *ow 7 

28. ^ sinh- :^ cosh y - 29. frE 30. & 31. 

44?? ^fc 

32. o 33. o 34. - + C - -, 

- 



Exercises 5 

1. 5'3 cos (4 5-3*) 2. 16-3 sin 5-1* 3. -48 sec 2 (3* -f 9) 

5-05sin(-05 -117*) 

4. -QI4COS (-425^ 1-25) 5. 4ocosec 2 # 6. , : 

cos 2 (-05 -117*) 

7. gbc sin (rf gx) 8. 20 sin 5^ 14 cos (zx 5) 

9. 4-40038-8^+ -S cos 1-6^ {Usetherule : 2 sinAcosB = sin( 
-(- sin (A B)} 10. 6-74 sin 6-2X 3-04 sin 2-8 



11. 4'5 2 V ~^ sin (P x 1 X + 2C ) + ., sin (px 

\O 'U "U T" U 

12. 5 sin 2x. {Use the rule: cos 2A = i 2 sin 2 A} 13. -195 sin 6x 

14. o 15. S-I6A" 72 -5-2 -0273 cos (4-31 -195^) + 24-93 

3# 4' 1 

16. -1056 cos -015^ -0529 sin (6-1 -23*) + 7-4 sec 2 (4* -07) 

17. Velocity = 37-7 sin 31-4* 56-56 cos 31-4* : 

acceleration =1184 cos 31-4/4- J 777 si n 3 I- 4' 

18. Acceleration -02895 : S.H.M. 19. 1162 

20. Sine curve (i.e., second derived curve). 

f B/ 

21. o I Treat as a constant the portion 



22. 1500^ cos pt + $oop cos $pt + 42^ sin pt 8^p sin 



ANSWERS TO EXERCISES 389 

Exercises 6 

1. 2 cos 2.x . <? 8in2 * 2. - 3 2 sin 2* 4. 24*2. cos x 3 

v 

5. 3-14 (iox + 7) sec 2 (5# 2 + 7* 2) 6. 3 log a . cos 3* . a" in3 * 

3 ~T" i ~j % ~~~~ Q# 5 

sec 2 ._ sin<9 (dy dy dd\ 

10. COSCCA- 11. 12. - . { = 3^ - X 3 

sec z cp cos a i + cosy la^r aw a^J 

13. \ ,, ,j. slope of curve = j^ X -=- X ^ =- , 
IrflogV rfA dV rflogV) 

15. i -08 ft. per min. : -377 ft. per min. 18. -033215.: 01020' 
19. 7 cosec 7* + 45*2 20.=^ 21. 56 19' 22. 53 7' 24. 



Exercises 7 

1. x (2 sin 3* -f $x cos 3*) 2. 2^ 2 ' 4 (i + 3-4 log 

3. 



( 2-07 5 sin (3-1- 2-07*) \ 

Icos -1 2-0^ x J 



4 

Ar 5 cos (3-1 2-07.*) cos (3-1 2-07^) 

5. - {2-575 sin (5-15* + 4) + -625 sin (1-25* - 4)} or 

- {3-2 sin 3-2* cos (i-95* + 4) + i'95 cos 3-2^ sin (1-95* + 4)} 

6. sec 2 2x {2 cos (5 3#) + i -5 sin (5 3*) sin 4*} or 

3 tan 2.x sin (5 3*) + 2 sec 2 2# cos (5 3*) 

7. I2-8*' 6 (cos (3 + 8x) + 2 5* sin (3 + 8*)} 

8. 27 (5) 3z J4-83 log x + - 9. (i + log AT) e xlo x 10. 



11. 3 oe (5* + 4)(5* + 2) 12. /tan -125^ log^ \ 

3 ^\ x 8cos 2 -i25Ar/ 



13. o 14. 5e- lot 15. o 16. 12600 sin (14* -116) 

17. 6t {5 sin (4 -8t) 2t cos (4 -8*)) 18. 4* 2 ' 7 (3-7 cos 3^ 3* sin 3*) 

Exercises 8 

1. 5 * 2( 7 3 ~ 7;y) 2. - -. (-^ - log (2 - 7*) tan (2 - 7*)} 

e 7 *- 5 cos (2 7#) \7# 2 

3 20 4 & 5 5-46 (5) 2 ^ 

'" 



Vd* 
6. TCsin^n-^-^S^coshi-S^ or 






21 i 



_ 

* 2 + 6^ + 15 (i - 



Q wb (ab -zbx + x 2 cot B) 
~2~(&-*cotB) 2 



3 9 o MATHEMATICS FOR ENGINEERS 



e s\n(i-2x + l-7) 

12< {loi^^T^Ts)? x 

j I -2 COS (l-2X -{- I -7) log (8x z JX + 3) + -~- 

. 6-55 (sin cos 0) 
13. sech 2 * 14. J3 . v 

5 (0 sin 0)^ 

. 66-2 (0 sin 0)* (20 30 cos + sin 0) 4 y z 
15. lo. 



17. rw 2 cos -1 

L (m 2 - sin 2 0)* 

d(f> a> cos . d z (f> _ a> z sin (i m 2 ) 

'' ~dt ~ \/m 2 - sin 2 ' ^ 2 (m a -sin a 0)^ 



19. ^^ a : J* 20. 
(# ?)sm 2 ^ > w 



Exercises 9 

2. 

*pt + 4e . log ( 5 p - 3) 



4. io( 4 -w)(9-4w)(3 + 8w) 2 6. 8(1-7 

?_(*__ 
' CrVdt r'dt 



7 _ 31 o ?_(*P__t dr\ 

' ' 



Exercises 10 

1. 750 2. 17-1 3. ^ : ^- 4. -577/ : -I28W/ 

5. -5 6. 2-25 : minimum 7. -278 : maximum 

8. maximum at x = 5 : point of inflexion at x = 2 

minimum at x = i 9. 2 rows of 8 

10. base = 3-652 ft. : height = 1-826 ft. 11. 2-1 : minimum 

12. - 13. -496: 631 14. width = height = 8-4' 

15. 15-2 knots: 956, 948, 957 16. x = -289; 

17. h = 6-34 ft. : d = 12-68 ft. 18. base = 4* : height = 5* 

19. depth = 3 x breadth. 20. ^ 21. 6 22. -866 r 

w 

23. v : | : 24. u = -.51; 25. 135 or 315 



ANSWERS TO EXERCISES 391 



26. d = l 27. I = 2'o6sr 28. 20-15: -45 29. o or 20 56' 
30. height = 8-1 ft. : base = 6-72 ft. 31. x = -4* 

32. tan- (-*^?$\ 38. / _ 

i fi z 

34. maximum at x = o : points of inflexion at x j= 

. V2 



35. VP 36. /(4/+d 37. 



K 
i + K 

Let r = ^ and find ^] 39. 3 units 40. * = ^^ 
P! dr J 2 

41. T/ = ^T m 42. maximum at * = 2 

minimum at * = 4 and at * = 2-5 

points of inflexion when * = 2-26 and 1-92 



/ 
38. -58 ( 

\ 



43. x = VRjRg : / (maximum) = - fam + i) (R x R 2 ) 2 

44. d = \/^j 46. 83 i' or 276 59' 

' ojt 

Exercises 11 

1. -006 2. 25%toolow 3. -0264 4. 2-45 

5. 2-66 6. decrease of -00135 7. -03 link; -237 link 

, (#loga) 2 , (#loga) 3 , 

8. i + x log a + * - ! : i- + . . . 

1.2 1.2.3 

9. 2214-2525 10. -536 



Exercises 12 

2. 152 5. 240 ; 205 ; 64 

6. 621,000 ft. Ibs. : potential energy = 240,000 : 

kinetic energy = 381,000 : 987,000 ft. Ibs. 

7. 480 9. 238,000 10. 3006 12. i-526# 2 - 68 +C 



13. 70-15* + C 14. - +C 15. 

17. -i* 10 10 log x + 14* + C 18. 3-32* 1 ' 04 2-5* 2 + C 
19. i-o74* 3 ' 718 + ie* + C 20. i-33 9 *- f + C 





21. 6-54-*'- 1 - + C 22. -689* + C 23. -^ - + C 

C 

24. 3-025*' 84 8-2 log x 2-7ie- 2> * + 1-13* + C 

25. -0234*- 10 ' 2 * + C 26. i-g6e' 61x 1-297*-" + -674* 8 ' 04 + C 

27. -797 cos 0* 1 -" 2-2g-8'-i*+C 28. Ju 6 + C 29. -- ^+C 
30. 35< + C 31. ie^-+C 32. - 5-88^ + C 

33. 20-2 (2) + C 34. - * + 2-5^2 + 4-25** - 8x + C 

^T 

35. -88 5 ( 3 -i)t + C 36. _ + C 



392 MATHEMATICS FOR ENGINEERS 



37. _ _ z-jiSt + C 38. i6-i* 2 + C 39. - + C 

45 x 

40. Write the equation in the form -f w and then integrate : 

p v 

pv n C 

Exercises 13 

1. -Jcos 4 * + C 2. i -73 sin (3 - 3*) + C 

3. 49 tan (3 }x) + C 4. i-oix' 9 ** + 1-195 sin (-05 -117^) + C 



5. -I8540 5 ' 4 * -cos (b + ax) + C 6. 9-45* . sin 8t + C 

& 

7. -713 cos 2(2-i6x 4-5) + C 

8. i2-85e' 7 * ^ r + i -83 log cos # + C 

A# 2-8\ 

9. 9-95 cos ( ) + '022 sin 9^r i'46^ 2 ' 74 + -455(3) 2x + 6 + C 

\ 7 / 

10. 2* -787 cos ( - 3-7* ] + 7-55 cot. ^ + C 

V4 / 5 

(i 

11. v = 7 cos (7^ -26) + C ; s = 

49 

12. -5^ = 47r 2 nV( sin ^ + Sin 2 \ + C; * = 47r 2 nV( cos ^ + C S 2 ) + C 
D \ 2m / \ 4m / 

13. 3iH5 8p ) - -139 sin (3-7 - 7-2^) + C 

14. 19-5 cos 6t 4-9 sin 6t + C 

Exercises 14. 

1. -182 2. -345 3. 1-7 4. -561 5. -0626 

2! 

6. 1-218 x io 7 7. 2-62 8. -1589 9. - 

P 

10. -6i6B mai 11. - - | C (a J"^ + C ^ S (fl ~ - f W 

2 I a + 6 a 6 



^o -n 

2 13 - p = 

14. { ^4 - 4 /3^ + 3 ,4 } 1 - 

24EI ( 

16. ^ 17. .20 4 6/* 18. I ^ SiU ^ Wk * 

3 i + sin $ 2 

19. 26-24 {Limits must first be found} ; f 20. 334 

21. y = -736*?' + 5 log # + 3^ 3-25 22. i -087. 



. . ._ 

2 6.H = g 27 . 240 3 ; 8 ., 9I 29. 8 

30. 49-82 31. ^f 32. -8596CA.W 33. 



ANSWERS TO EXERCISES 393 

Exercises 15 



1. -158 log - 2. 

5 (* +1-583) 3 V6 V6 

3. log (9*2 - 18* + 17)" -f C 4. -1919 

5. 605 {Let u 6 h] 6. yV-j$ y 2 + -75 sin" 1 1 -154^ + C 

7. -106 8. $sini2* + 4/ + C 9. tcot5#+C 

.. WR 3 /i i\ -0683 WR 3 

10. -==- --- ) or -- - 

El \ir 47 El 

11. -8 log cos 5* + C {Let u = cos 5*} 12. # sin" 1 # + Vi ^ + C 

_ -g&Z 

13. V (a 2 AT 2 ) 3 + C 14. 3 {3 sin 2* 2 cos -zx] 

15. tan- 1 * H -- ~ + C (Let u - tan- 1 AT) 16. -081 

i + x z 

17 tsin<9 18 

18> k 

19. 183 sees. : {Rationalise denominator of right hand side by 
multiplying top and bottom by Vh + 12 Vh ; then integrate, making 
the substitution u = A + 12.} 

21. 

10 42 



22. ^ + t - 1 cosh- 1 1 + C 23. 355 24. i -749 

12 8 2 2 256 

25. 26. -oiR 3 
315 



27. (-5* + 1-25) 2 T^5* -x* + 13-63 sin- 1 + C 

28. F= 

V a 2 



Exercises 16 

1.. 41-59 2. 1-718 3. 10-85 4 - <688 5 - 4'5 

6. 23-05 7. 1-348 ft. candles. 8. o 9. -1294 

10. 273 11. 205 Ibs. 



Exercises 17 
1. 14-14 2. -6215 3. 2-4 4. -1165 5. -833 6. 1-194 



394 MATHEMATICS FOR ENGINEERS 

Exercises 18 

1. 1450 cu. yds. 2. 5977 Ibs. 3. 1070 4. 4nr 2 

c KTrD 2 / 2 x _ . Kl 
5. a 3-036, b = -1423 ; 1617 6. ^ or Vol. x 

7. 524 (limits are 5 and 10} 8. 271-6 9. 1-2 Ibs. 

10. true == 76-62 : (a) 75-41 : 1-58 % low (b) 77-73 : 1-45 % high 
(c) 76-60 : correct. 11. 60-9 ft. 

Exercises 19 

1. | height from vertex 

2. area = 7200 sq. ft. : centroid is 158' from forward end 

3. 3-84 Ibs. ; 4" 4. 10 ft. from top 

5. x = y = i -7* (taking the centre as origin) 6. 771 ; 2-25 
7. (o, -95) 8. 2-35* from AB 9. 1-055* 10. -935" 

11. -877* 12. 1-02* 13. 2-68* i. e., 1-34* 

14. 5-12 ft. from O 15. 30880 Ibs. 16. 18432 Ibs. : 3-534 ft. 

17. 9 units. 18. 5-1 ins. 19. 2-35 ins. 

20. (a) 17 Ibs. (b) 1 1* 21. 1-48 ins. * 22. \h 23. |A 

Exercises 20 

1. -655/ 2. 17-11 ins.: 785 Ibs. ins. 2 

3. C. of G. is -1125* distant from centre of large circle: 2-68" 

4. 16-6 ins. 4 5. (a) -408^ ; (6) -707^ 

6. IAB = 80-7 ins. 4 : & A B = 3'O4* 7. 29-1* 8. 377 
9. 9-86 10. 7-35 

11. -2887^. (Divide into strips by planes perpendicular to the axis 
and sum the polar moments of these) 

. IE of circular 3 

12. T _ , - = - = -956 13. 33.3 inch units 
IE of square TT 

14. 681-6: 17-1 

15. I NN = 169-4: NN = 2-44: I AB = 570 : k^ = 4-47 

16. NN is 3-99* from bottom of lower flange : INN = 461 : NN = 3'77 

17. 5-04* 18. 2-023': -444 19. 2-74* 

20. (a) 13-9 (6) 31-1 (c) 1-48" (d) 1-82* 21. -28 



22. massR* + 23. -56 

Exercises 21 

3. 5-23 4. = 2rsin0: the sine curve 6. 892 7. 5-01 



Exercises 22 

1. y = i-6jx* 2-4* 12-82 2. s = 8-05^ 23-1* + 14-1 

3. y = Ae** 4. y = 8-35 -1490 -i-o** 

W fdv I 

5. y = - 2 3 3 = = = = - 



ANSWERS TO EXERCISES 395 

wl z w 



7. log = - 8. v = (5 - 



9- lg = 10. 

ws 



11. H = 



, 7- 2 ) . / ftnh 

J 12. = Asm(/ - 



, , B 

13. ^ = A+- 2 

17. y = A^ 10 - 6 + Ajg 7 ' : y = A^ 10 ' + A^ 7 * + i 

18. 5 = A^ 9 ' 33 ' + A^- 9 ' 33 ' 19. 5 = A sin (9-33^ + B) 



_ 
20. KRlog- 1 21. C = C Ae 22. C= 



23. ^ = alo or a= 



(Separate the variables and use the / -5 - ^ form J 



Ysin /y/ ==, 
25. =- - 



26. Mf = log^ + -):-i945 27. AT= 

28. x = Ae~ 3t sin (632/ + c) + -026 a sin (5^ tan -1 1-25) 

J\ - /5. 

29. V - A x e V r * + A 2 e V r * 30. /w - C 
31. 6 = 6 a + e- kt (6 -6 a } 32. 5= - 53-636-^^+ 5 -8 3 e 



where 



= 



V 

V 



2EI 



e 



2EI 

or y = B^* + B 2 - -f B 3 sin <f>x + B 4 cos (f>x -^ 
34. = ' 



396 MATHEMATICS FOR ENGINEERS 

Exercises 23 

1. 103 sees, if coefficient of discharge is taken as -62 2. 7-37 

3. 2-83 cu. ft. 4. 69-5 : 95-5 (Draw in the " simply supported " 

bending moment diagram and work on the Goodman plan, see page 313) 

5. Find the time to lower to level of upper orifice (183 sees.) with 

both orifices open; then the time for the further lowering of 5 ft., 

through the one orifice (180 sees.). Total time = 363 sees. Note that 

* 

6. 57-7 sees. 7. 1-4 

12 

8. 5500 Ibs. : 4-71 ft. below S.W.S.L. 9. 14100 Ibs. : 6-65' 
below S.W.S.L. 10.761 11. '^f G;.- ,^),) (Hint. 

let u = d e +Kx) 12.1-23 13. Vertical depth = 8-07 ft. 

Exercises 24 

1. x = 2-31 1-231 cos 5 1-55 sin 6 -16 cos 20 

-022 sin 20 -004 cos 3# -04 sin 3$ 

2. A 1-29, oj = O, B = -I4, o 2 = rr 

3. ^ = 16-97+6-49 cos #+ -002 sin x 12-66 cos 2* 

i -46 sin 2.x 1-75 cos 3# -7 sin $x 

4. E = 1500 sin 0+too sin 3042 cos #+28 cos 3$ 

Exercises 25 

1. B = 393i / ; A = 6 5 5i'; c = 57 5' 

2. c - 54 44i' ; * = 34 14' ' B = 43 32^' 

3. A= 161 8'; B= 13 35'; C = 9 38' 

4. A = 76 36'; B = 648'; = 52 

5. B = 35 43' 40"; A = 6i 21' 20"; a = 43 25' 23" 

6. C= 33 29' or 146 3 1' 
a = 103 28' or 55 28' 

A= 146 58' or 27 30' 7. 55 10' 8. 342,200 

9. Latitude = 43 54' : hour angle = 4 31-3' 

Exercises 26 

1. 2nd set better than ist set in the proportion 1-943 to i : 6 elms. 
43-4 links. 2. Just under i. 3. 98 6' 42-26" : y m -307", v 1-062" 
4. 96-93 



MATHEMATICAL TABLES 



TABLE I. TRIGONOMETRICAL RATIOS 



Angle. 


Chord. 


Sine. 


Tangent. 


Co- tangent. 


Cosine. 








De- 
grees. 


Radians. 





o 


o 





o 


GO 


i 


1-414 


1-5708 


90 


i 

2 

3 

4 


0175 
0349 
0524 
0698 


017 
035 
052 
070 


0175 
0349 
0523 
0698 


0175 

0349 
0524 
0699 


' 7'290O 
28-6363 
I9-08II 
I4-3007 


9998 
9994 
9986 
9976 


1-402 
1-389 
1-377 
1-364 


1-5533 
1-5359 
1-5184 
1-5010 


89 
88 
87 
86 


5 


0873 


087 


0872 


0875 


II-43OI 


9962 


l'35l 


I-4835 


85 


6 

8 
9 


1047 

1222 
1396 
1571 


105 

122 
I 4 
157 


1045 
1219 
1392 
1564 


1051 
1228 
1405 
1584 


9-5I44 
8-H43 
7-H54 
6-3I38 


9945 
9925 
9903 
9877 


1-338 
1-325 
1-312 
1-299 


1-4661 
1-4486 
1-4312 
1-4137 


84 
83 

82 
81 


10 


1745 


174 


1736 


1763 


5-67I3 


9848 


1-286 


1-3963 


80 


ii 

12 
13 
14 

15 


1920 
2094 
2269 
2443 


192 
209 
226 
244 


1908 
2079 
2250 
2419 


1944 
2126 
2309 
2493 


5-I446 
4-7046 
4-33I5 
4-OIO8 


9816 
9781 
9744 
9703 


1-272 
1-259 
1-245 
1-231 


1-3788 
1-3614 
1-3439 
1-3265 


79 
78 
77 
76 


2618 


261 


2588 


2679 


3-7321 


9659 


1-218 


1-3090 


75 


16 

17 
18 
19 


2793 
2967 
3*42 
33i6 


278 
296 
313 
330 


2756 
2924 
3090 
3256 


2867 
357 
3249 
3443 


3-4874 
3-2709 

3-0777 
2-9042 


9613 
9563 
95" 
9455 


1-204 
1-190 
1-176 
1-161 


1-2915 
1-2741 
1-2566 
1-2392 


74 
73 

72 

7i 


20 


3491 


347 


3420 


3640 


2-7475 


9397 


1-147 


1-2217 


70 


21 
22 
23 
24 


3665 
3840 
4014 
4189 


364 
382 
399 
416 


3584 
3746 
3907 
4067 


3839 
4040 
4245 
4452 


2-6051 
2-4751 
2-3559 
2-2460 


9336 
9272 
9205 
9135 


1-133 
1-118 
1-104 
1-089 


1-2043 
1-1868 
1-1694 
1-1519 


69 
68 
67 
66 


25 


4363 


433 


4226 


4663 


2-1445 


9063 


1-075 


I-I345 


65 


26 
27 
28 
29 


4538 
4712 
4887 
5061 


45 
467 
484 
501 


4384 
454 
4695 
4848 


4877 
5095 
5317 
5543 


2-0503 
1-9626 
1-8807 
1-8040 


8988 
8910 
8829 
8746 


i -060 
1-045 
1-030 
1-015 


1-1170 
1-0996 
1-0821 
1-0647 


64 
63 
62 
61 


30 


5236 


518 


5000 


5774* 


I-732I 


8660 


I-OOO 


1-0472 


60 


31 
32 

33 

34 


54ii 
5585 
'5760 
5934 


534 
551 
568 
585 


5150 
5299 
5446 
5592 


6009 
6249 
6494 
6745 


1-6643 
1-6003 
1-5399 
1-4826 


8572 
8480 
8387 
8290 


985 
970 
954 
939 


1-0297 
1-0123 
9948 
9774 


59 
58 

57 
56 


35 


6109 


601 


5736 


7002 


1-4281 


8192 


923 


9599 


55 


36 
37 
38 
39 


6283 
6458 
6632 
6807 


618 
635 
651 
668 


5878 
6018 
6i57 
6293 


7265 
7536 
7813 
8098 


1-3764 
1-3270 
1-2799 
1-2349 


8090 
7986 
7880 
7771 


908 
892 
877 
861 


9425 
9250 
9076 
8901 


54 

53 
52 
51 


40 


6981 


684 


6428 


8391 


1-1918 


7660 


845 


8727 


5 


41 
42 
43 

44 


7156 
7330 
7505 
7679 


700 
717 
733 

749 


6561 
6691 
6820 
6947 


8693 
9004 
9325 
9657 


1-1504 
1-1106 
1-0724 
1-0355 


7547 
7431 
7314 
7193 


829 
813 
797 
781 


8552 
8378 
8203 
8029 


4 i 
48 

47 
46 


45 


7854 


765 


7071 


I-OOOO 


I-OOOO 


7071 


765 


7854 


45 








Cosine 


Co-tangent 


Tangent 


Sine 


Chord 


Radians 


Degrees 


Angle 



397 



398 



MATHEMATICAL TABLES 
TABLE II. LOGARITHMS 



J 





1 


2 


3 


4 


5 


6 


7 


8 


9 


123 


456 


789 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 9 13 
4 8 12 


17 21 26 
16 20 24 


30 34 38 
28 32 37 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


4 8 12 
4 7 11 


15 19 23 
15 19 22 


27 31 35 
26 30 33 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


3 7 11 
3 7 10 


14 18 21 
14 17 20 


25 28 32 
24 27 31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


3 7 10 
3 7 10 


13 16 20 
12 16 19 


23 26 30 
22 25 29 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


369 
369 


12 15 18 
12 15 17 


21 24 28 
20 23 26 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


369 
368 


11 14 17 
11 14 16 


20 23 26 
19 22 25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


358 
358 


11 14 16 
10 13 15 


19 22 24 
18 21 23 


17 


2304 


2330 


2355 


2380 


2406 


2430 


2465 


2480 


2504 


2529 


368 
267 


10 13 15 
10 12 15 


18 20 23 
17 19 22 


18 


2553 


2677 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2766 


267 
257 


9 12 14 
9 11 14 


16 19 21 
16 18 21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


247 
246 


9 11 13 
8 11 13 


16 18 20 
15 17 19 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


246 


8 11 13 


15 17 19 


21 
22 
23 
21 


3222 
3424 
3617 
3802 


3243 
3444 
3636 
3820 


3263 
3464 
3655 
3838 


3284 
3483 
3674 
3856 


3304 
3502 
3692 
3874 


3324 
3522 
3711 
3892 


S345 
3541 
3729 
3909 


3365 
3560 
3747 
8927 


3385 
3579 
3766 
3945 


3404 
3598 
3784 
3962 


246 
246 
246 
245 


8 10 12 
8 10 12 
7 9 11 
7 9 11 


14 16 18 
14 15 17 
13 15 17 
12 14 16 


25 


979 


3997 


4014 


4031 


4048 


4065 


4083 


4099 


4116 


4133 


235 


7 9 10 


12 14 16 


26 
27 
28 
29 


4150 
4314 
4472 
4624 


4166 
4330 
4487 
4639 


4183 
4346 
4502 
4654 


4200 
4362 
4518 
4669 


4216 
4378 
4533 
4683 


4232 
4393 
4548 
4698 


4249 
4409 
4664 
4713 


4265 
4425 
4579 
4728 


4281 
4440 
4594 
4742 


4298 
4456 
4609 
4757 


235 
235 
235 
134 


7 8 10 
689 
689 
679 


11 13 15 
11 13 14 
11 12 14 
10 12 13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4867 


4871 


4886 


4900 


134 


679 


10 11 13 


31 
32 
S3 
34 


4914 
5051 
5185 
5315 


4928 
5065 
6198 
5328 


4942 
5079 
6211 
5340 


4955 
5092 
5224 
5353 


4969 
5105 
6237 
5366 


4983 
5119 
5250 
5378 


4997 
6132 
5263 
6391 


6011 
5145 
6276 
6403 


6024 
5159 
5289 
6416 


6038 
5172 
6302 
6428 


134 
134 
134 
134 


678 
678 
568 

568 


10 11 12 

y 11 12 

9 10 12 
9 10 11 


36 


5441 


5453 


6465 


5478 


5490 


6502 


6514 


6527 


5639 


6551 


124 


567 


9 10 11 


86 
37 
38 
39 


5563 
5682 
5798 
6911 


5576 
5694 
5809 
5922 


5587 
5705 
5821 
5933 


5599 
6717 
5832 
5944 


6611 
6729 
6843 
5955 


6623 
6740 
5855 
5966 


5635 
6752 
5866 
6977 


6647 
6763 
6877 
6988 


5658 
6775 
5888 
5999 


5670 
5786 
5899 
6010 


124 
123 
123 
123 


667 
667 
667 
457 


8 10 11 
8 9 10 
8 9 10 
8 9 10 


40 


6021 


6031 


6042 


6053 


6064 


6076 


6086 


6096 


6107 


6117 


123 


466 


8 9 10 


41 
42 
43 
44 


6128 
6232 
6335 
6435 


6138 
6243 
6345 
6444 


6149 
6253 
6355 
6454 


6160 
6263 
6365 
6464 


6170 
6274 
6375 
6474 


6180 
6284 
6385 
6484 


6191 
6294 
6395 
6493 


6201 
6304 
6405 
6603 


6212 
6314 
6416 
6513 


6222 
6325 
6425 
6522 


123 
123 
123 
123 


456 
456 
456 
456 


789 
789 
789 
789 


45 


6532 


6542 


6551 


6561 


6671 


6580 


6590 


6599 


6609 


6618 


123 


456 


789 


46 
47 
48 
49 


6628 
6721 
6812 
6902 


6637 
6730 
6821 
6911 


6646 
6739 
6830 
6920 


6656 
6749 
6839 
6928 


6665 
6758 
6848 
6937 


6675 
6767 
6857 
6946 


6684 
6776 
6866 
6955 


6693 
6785 
6875 
6964 


6702 
6794 
6884 
6972 


6712 
6803 
6893 
6981 


123 
123 
123 
123 


4 5 < 

465 
445 
445 


778 
678 
678 
678 


50 


6990 


6898 


7007 


7016 


7024 


7033 


7042 


7060 


7059 


7067 


1 2 S 


345 


678 



MATHEMATICAL TABLES 
TABLE II. (contd.) 



399 








1 


2 


3 


4 


5 


6 


7 


8 


9 


123 


456 


789 


51 

52 
53 
54 


7076 
7160 
7243 
7324 


7084 
7168 
7251 
7332 


7093 
7177 
7259 
7340 


7101 
7185 
7267 
7348 


7110 
7193 
7275 
7356 


7118 
7202 
7284 
7364 


7126 
7210 
7292 
7372 


7135 
7218 
7300 
7380 


7143 

7226 
7308 
7388 


7152 
7235 
7316 
7396 


123 
122 
122 

122 


345 
345 
345 
345 


678 
677 
667 
667 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


122 


345 


567 


58 
67 
58 
59 


7482 
7559 
7634 
7709 


7490 
7566 
7642 
7716 


7497 
7574 
7649 
7723 


7505 
7582 
7657 
7731 


7513 
7589 
7664 
7738 


7520 
7597 
7672 
7745 


7528 
7604 
7679 
7752 


7536 
7612 
7686 
7760 


7543 
7619 
7694 
7767 


7551 
7627 
7701 
7774 


122 
122 
112 
112 


345 
345 
344 
344 


567 
567 
567 
567 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


112 


344 


566 


61 
62 
63 

64 


7853 
7924 
7993 
8062 


7860 
7931 
8000 
8069 


7868 
7938 
8007 
8075 


7875 
7945 
8014 
8082 


7882 
7952 
8021 
8089 


7889 
7959 
8028 
8096 


7896 
7966 
8035 
8102 


7903 
7973 
8041 
8109 


7910 
7980 
8048 
8116 


7917 
7987 
8055 
8122 


112 
112 
112 
112 


344 
334 
334 
334 


566 
566 
& 5 6 
556 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


112 


334 


556 


66 
67 
68 
69 


8195 
8261 
8325 
8383 


8202 
8267 
8331 
8395 


8209 
8274 
8338 
8401 


8215 
8280 
8344 
8407 


8222 
8287 
8351 
8414 


8228 
8293 
8357 
8420 


8235 
8299 
8363 
8426 


8241 
8306 
8370 
8432 


8248 
8312 
8376 
8439 


8254 
8319 
8382 
8445 


112 
112 
112 
112 


$ 3 4 

334 
334 
234 


556 
556 
456 
456 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


1 1 2 


234 


456 


71 
72 
73 
74 


8513 
8573 
8633 
8692 


8519 
8579 
8639 
8698 


8525 
8585 
S645 
8704 


8531 
8591 
8651 
8710 


8537 
8597 
8657 
8716 


8543 
8603 
8663 
8722 


8549 
8609 
8669 
8727 


8555 
8615 
8675 
8733 


8561 
8621 
8681 
8739 


8567 
8627 

868G 
8745 


112 
112 
112 
112 


234 
234 
234 
234 


455 
455 
455 

455 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


112 


2 3 3 


455 


76 

77 
78 
79 


880S 
8865 
8921 
8976 


8814 
8871 
8927 
8982 


8820 
8876 
8932 
8987 


8825 
8882 
8938 
8993 


8831 
8887 
8943 
899S 


8837 
8893 
8949 
9004 


8842 
8899 
8954 
9009 


8848 
8904 
8960 
9015 


8854 
8910 
8965 
9020 


8859 
8915 
8971 
9025 


112 
112 
112 

112 


2 S 3 
233 
233 
233 


455 
445 
445 
445 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


112 


233 


446 


81 
82 
83 
84 


9085 
9138 
9191 
9243 


9090 
9143 
9196 
9248 


9096 
9149 
9201 
9253 


9101 
9154 
9206 
9258 


9106 
9159 
9212 
9263 


9113 
9165 
9217 
9269 


9117 
9170 
9222 
9274 


9122 
9175 
9227 
9279 


9128 
9180 
9232 
9284 


9133 
9186 
9238 
9289 


112 
112 
112 
112 


233 
233 
233 
233 


446 
446 
446 
445 


65 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


112 


233 


445 


86 
87 
88 
89 


9345 
9395 
9445 
9494 


9350 
9400 
9450 
9499 


9355 
9405 
9455 
9504 


9360 
9410 
9460 
9509 


9365 
9415 
9465 
9513 


9370 
9420 
9469 
9518 


9375 
9425 
9474 
9523 


9330 
9430 
9479 
9528 


9385 
9435 
9484 
9533 


9390 
9440 
9489 
9538 


112 
Oil 
Oil 

Oil 


233 
223 
223 
223 


445 
344 
344 
344 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9671 


9576 


9581 


9586 


Oil 


223 


344 


91 
92 
93 
94 


9590 
9638 
9585 
9731 


9595 
9643 
9689 
9736 


9600 
9647 
9694 
9741 


9605 
9(352 
9699 
9745 


9609 
9657 
9703 
9750 


9614 
9661 
9708 
9754 


9619 
9666 
9713 
9759 


9624 
9671 
9717 
9763 


9628 
9675 
9722 
9768 


9633 
9680 
9727 
9773 


Oil 
Oil 
Oil 
Oil 


223 
223 
223 
223 


344 
344 
344 
344 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


Oil 


223 


344 


96 
97 

95 
99 


9823 
9868 
9912 
995 


9827 
987-; 
9917 
9961 


9832 
9877 
9921 
99C5 


9836 
9881 
9926 
9969 


9841 
9886 
9930 
9'J74 


9845 
9890 
9934 
9978 


9850 
9894 
9939 
9983 


9854 
9899 
9943 
9987 


9859 
9903 
9948 
9991 


9863 
9908 
9952 
9996 


Oil 
Oil 
Oil 
Oil 


223 
223 
223 
223 


344 
344 
344 
334 



400 



MATHEMATICAL TABLES 
TABLE III. ANTILOGARITHMS 








1 


2 


3 


4 


5 


6 


7 


8 


9 


123 


456 


789 


00 


1000 


1002 


1005 


1007 


1009 


1012 


1014 


1016 


1019 


1021 


001 


111 


212 


01 
02 
03 
04 


1023 
1047 
1072 
10-J6 


1026 
1050 
1074 
1099 


1028 
1052 
1076 
1102 


1030 
1054 
1079 
1104 


1033 
1057 
1081 
1107 


1035 
1059 
1084 
1109 


1038 
1062 
1086 
1112 


1040 
1064 
1089 
1114 


1042 
1067 
1091 
1117 


1045 
1069 
1094 
1119 


001 
001 
001 
Oil 


111 
111 
ill 

112 


222 
222 
222 
223 


05 


1122 


1125 


1127 


1130 


1132 


1135 


1138 


1140 


1143 


1146 


Oil 


112 


222 


06 
07 
03 
09 


1143 
1175 
1202 
1230 


1151 
1178 
1205 
1233 


1153 
1180 
1208 
1236 


1156 
1183 
1211 
1239 


1159 
1186 
1213 
1242 


1161 
1189 
1216 
1245 


1164 
1191 
1219 
1247 


1167 
1194 
1222 
1250 


1169 
1197 
1225 
1253 


1172 
1199 
1227 
1256 


Oil 
Oil 
Oil 
Oil 


112 
112 
112 
112 


222 
222 
223 
223 


10 


1259 


1262 


1265 


1268 


1271 


1274 


1276 


1279 


1282 


1285 


Oil 


112 


223 


11 
12 
18 
14 


1288 
1318 
1349 
1380 


1291 
1321 
1352 
1384 


1294 
1324 
1355 
1387 


1297 
1327 
1358 
1390 


1300 
1330 
1361 
1393 


1303 
1334 
1365 
1396 


1306 
1337 
1368 
1400 


1309 
1340 
1371 
1403 


1312 
1343 
1374 
1406 


1315 
1346 
1377 
1409 


Oil 
Oil 
Oil 
Oil 


122 
122 
122 
122 


223 
223 
233 
233 


15 


1413 


1416 


1419 


1422 


1426 


1429 


1432 


1435 


1439 


1442 


Oil 


122 


233 


16 
17 
18 
19 


1445 
1479 
1514 
1549 


1449 
1483 
1517 
1552 


1452 
1486 
1521 
1556 


1455 
1489 
1524 
1560 


1459 
1493 
1528 
1563 


1462 
1496 
1531 
1567 


1466 
1500 
1535 
1670 


1469 
1503 
1638 
1674 


1472 
1607 
1542 
1578 


1476 
1510 
1545 
1581 


1 1 
Oil 
Oil 
Oil 


122 
122 
122 
122 


233 
233 
233 
333 


20 


1685 


1589 


1592 


1596 


1600 


1603 


1607 


1611 


1614 


1618 


1 1 


1 2 2 


333 


21 
22 
23 

24 


1623 
1660 
1698 
1738 


1626 
1663 
1702 
1742 


1629 
1667 
1706 
1746 


1633 
1671 
1710 
1750 


1637 
1675 
1714 
1754 


1641 
1679 
1718 
1758 


1644 
1683 
1722 
1762 


1648 
1687 
1726 
1766 


1652 
1690 
1730 
1770 


1656 
1694 
1734 
1774 


1 1 
1 1 
Oil 
1 1 


222 
222 
222 
222 


333 
333 
334 

334 


25 


1778 


1782 


1786 


1791 


1795 


1799 


1803 


1807 


1811 


1816 


Oil 


222 


334 


26 

27 
28 
29 


1820 
1862 
1905 
1950 


1824 
1866 
1910 
1954 


1828 
1871 
1914 
1959 


1832 
1875 
1919 
1963 


1837 
1879 
1923 
1968 


1841 
1884 
1928 
1972 


1845 
1888 
1932 
1977 


1849 
1892 
1936 
1982 


1854 
1897 
1941 
1986 


1858 
1901 
1945 
1991 


Oil 
Oil 
Oil 
Oil 


223 
223 
223 
223 


334 
334 
344 
3 4 i 


30 


1995 


2000 


2004 


2009 


2014 


2018 


2023 


9028 


2032 


2031 


Oil 


223 


344 


31 
82 
33 

34 


2042 
2089 
2138 
2188 


2046 
2094 
2143 
2193 


2051 
2099 
2148 
2198 


2058 
2104 
2163 
2203 


2061 
2109 
2158 
2208 


2065 
2113 
2163 
2213 


2070 
2118 
2168 
2218 


2076 
2123 
2173 
2223 


2080 
2128 
2178 
2228 


2084 
2133 
2183 
2234 


Oil 
Oil 
Oil 
112 


223 
223 
223 
233 


344 
344 
344 
446 


35 


2239 


2244 


2249 


2254 


2259 


2265 


2270 


2275 


2280 


2286 


112 


233 


446 


36 
37 
33 
39 


2291 
2344 
2399 
2455 


2296 
2360 
2404 
2460 


2301 
2355 
2410 
2466 


2307 
2360 
2415 
2472 


2312 
2366 
2421 
2477 


2317 
2371 

2427 
2483 


2323 
2377 
2432 
2489 


2328 
2382 
2438 
2495 


2333 
2388 
2443 
2500 


2339 
2393 
2449 
2506 


112 
112 
112 
112 


233 
233 

233 
233 


445 
446 
446 
455 


40 


2512 


2518 


2523 


2529 


2535 


2541 


2547 


2553 


2559 


2564 


112 


234 


455 


41 
42 

43 
44 


2570 
2630 
2692 
2754 


2576 
2636 
2698 
2761 


2582 
2642 
2704 
2767 


2588 
2649 
2710 
2773 


2594 
2655 
2716 
2780 


2600 
2661 
2723 
2786 


2606 
2667 
2729 
2793 


2612 
2673 
2735 
2799 


2618 
2679 
2742 
2805 


2624 
2685 
2748 
2812 


112 
112 
112 
112 


234 
234 
334 
334 


455 
456 
456 
466 


45 


2818 


2825 


2831 


2838 


2844 


2851 


2858 


2864 


2871 


2877 


112 


334 


656 


46 
47 
48 
49 


2884 
2951 
3020 
3090 


2891 
2958 
8027 
8097 


2897 
2966 
3034 
3105 


2904 
2972 
3041 
3112 


2911 
2979 
3048 

3119 


2917 
2985 
3055 
S126 


2924 
2992 
3062 
3133 


2931 
2999 
3069 
3141 


2938 
3006 
3076 
3148 


2944 
3013 
3083 
3155 


112 
112 
112 

112 


334 
334 

344 

344 


566 
666 
666 
666 



MATHEMATICAL TABLES 
TABLE III. (contd). 



401 








1 


2 


3 


4 


5 


6 


7 


8 


9 


123 


456 


789 


50 


3162 


3170 


3177 


318 


3192 


3199 


3206 


3214 


3221 


322 


112 


344 


5 6 t 


5 
5 
53 
54 


3236 
3311 
3388 
3467 


3243 
3319 
3396 
3475 


3251 
3327 
3404 
3483 


3258 
3334 
3412 
3491 


3266 
3342 
3420 
3499 


3273 
3350 
3428 

3508 


3281 
3357 
3436 
3516 


3289 
3365 
3443 
3524 


3296 
3373 
3451 
3532 


3304 
338 
345 
3540 


122 
133 

122 
122 


345 
346 
345 
345 


567 
667 

667 
667 


55 


3548 


3556 


3565 


35J3 


3581 


3589 


3697 


3606 


3614 


3622 


122 


346 


677 


56 
57 
58 
59 


3G31 
3715 
3802 
3890 


3639 
3724 
3811 
3899 


3648 
3733 
3819 
3908 


3656 
3741 
3828 
3917 


3664 
8750 
3837 
3926 


3673 
3758 
3846 
3936 


3681 
3767 
3855 
3945 


3690 
3776 
3864 
3954 


3698 
3784 
3873 
3963 


3707 
3793 
3882 
3972 


123 
123 
123 
123 


346 
346 
445 

455 


678 
678 
678 

678 


60 


3981 


3990 


3999 


4009 


4018 


4027 


4036 


4046 


4055 


4064 


1 2 3 


456 


*,, 


61 
62 
63 
64 


4074 
4169 
4266 
4365 


4083 
4178 
4276 
4375 


4093 
4188 
4285 
4385 


4102 
4198 
4295 
4395 


4111 
4207 
4305 
4406 


4121 
4217 
4315 
4416 


4130 
4227 
4325 
4426 


4140 
4236 
4335 
4436 


4150 
4246 
4345 
4446 


4159 
4256 
4355 
4457 


123 
123 
123 
123 


466 
466 
466 
456 


789 
789 
789 
789 


65 


4467 


4477 


4487 


4498 


4508 


4519 


4629 


4539 


4550 


4560 


123 


456 


789 


66 
67 
68 
69 


4571 
4677 
4786 
4898 


4581 
4688 
4797 
4909 


4592 
4699 
4808 
4920 


4603 
4710 
4819 
4932 


4613 
4721 
4831 
4943 


4624 
4732 
4842 
4955 


4634 
4742 
4853 
4966 


4645 
4753 
4864 
4977 


4656 
4764 
4875 
4989 


4667 
4775 
4887 
5000 


1 2 3 
123 
123 
123 


456 
457 
467 
567 


7 9 10 
8 9 10 
8 9 10 
8 9 10 


70 


6012 


6023 


5035 


5047 


6058 


6070 


6082 


6093 


6105 


6117 


124 


567 


8 9 11 


71 

72 
73 
74 


5129 
6248 
6370 
6495 


5140 
6260 
6383 
6608 


5152 
5272 
6395 
5521 


6164 
528 i 
5408 
5534 


6176 
6297 
6420 
6546 


5188 
5309 
5433 
6559 


5200 
5321 
5445 
5572 


5212 
6333 
6458 
6685 


5224 
5346 
6470 
6598 


5236 
5358 
6483 
6610 


124 
124 
134 
134 


567 
567 
568 
568 


8 10 11 
9 10 11 
9 10 11 
9 10 12 


75 


5623 


5636 


5649 


6662 


6675 


5689 


5702 


6715 


5728 


5741 


134 


578 


9 10 12 


76 
77 
78 
79 


6754 
5888 
6026 
6166 


5768 
6902 
6039 
6180 


6781 

5916 
6053 
6194 


6794 
5929 
6067 
6209 


6808 
5943 
6081 
6223 


5821 
5957 
6095 
6237 


6834 
6970 
6109 
6252 


6848 
6984 
6124 
6266 


6861 
5998 
6138 
6281 


5875 
6012 
6152 
6295 


134 
134 
134 
134 


578 
678 
678 
679 


9 11 12 
10 11 12 
10 11 13 
10 11 13 


80 


6310 


6324 


6339 


6353 


6368 


6383 


6397 


6412 


6427 


6442 


134 


679 


10 12 13 


81 

82 
83 
64 


6457 
6607 
6761 
6918 


6471 

6622 
6776 
6934 


6486 
6637 
6792 
6950 


6501 
6653 
6808 
6966 


1 
6516 
6668 
6823 
6982 


6631 

6683 
6839 
6998 


6546 
6699 
6855 
7015 


6561 
6714 
6871 
7031 


6577 
6730 
6887 
7047 


6592 
6745 
6902 
7063 


235 
235 
236 
235 


689 
689 
689 
6 8 10 


11 12 14 
11 12 14 
11 13 14 
11 13 15 


85 


7079 


7096 


7112 


7129 


7145 


7161 


7178 


7194 


7211 


7228 


336 


7 8 10 


12 13 15 


86 

67 
83 
69 


7244 
7413 
7586 
7762 


7261 
7430 
7603 
7780 


7278 
7447 
7621 
7798 


7295 
7464 
7638 
7816 


7311 
7482 
7656 
7834 


7328 
7499 
7674 
7852 


7345 
7516 
7691 
7870 


7362 
7534 

7709 
7889 


7379 
7551 
7727 
7907 


7396 
7568 
7745 
7925 


235 
236 
345 
245 


7 8 10 
7 9 10 
7 9 11 
7 9 11 


12 13 15 
12 14 16 
12 14 16 
13 14 16 


80 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


8091 


8110 




7 9 11 


13 15 17 


91 
92 
93 

S4 


8128 
8318 
b5U 
8710 


8147 
8337 
8531 
8730 : 


8166 
8356 ' 
8551 
8750 


8185 
8375 
8570 
8770 


204 
395 
590 
790 


8222 
8414 ! 
8610 
8810 


8241 
8433 
8630 
8831 


8260 
8453 
8650 
8851 


8279 
8472 
8670 
8872 


8299 
8492 
8690 
8892 


246 
246 
246 
246 


8 9 11 

8 10 12 
8 10 12 
8 10 12 


13 15 17 
14 15 17 
14 16 18 
14 16 18 


95 


8913 


8933 


8954 


8974 


995 


9016 


9036 


9057 


9078 


9099 


246 


8 10 12 


15 17 1 


86 
97 
98 
99 


9120 
9333 
9550 
'772 


9141 
9354 
9572 
9795 


9162 
9376 
9594 
9817 


9183 
9397 
9616 
9840 


204 
419 
9638 
9863 


9226 
9441 ! 
9661 
9S86 


9247 
9462 
9683 
9906 


9268 
9484 
9705 
9931 


9290 
9506 
9727 
9954 


9311 

9528 
9750 
9977 


246 
347 
247 

267 


8 11 13 
9 11 13 
9 11 13 
9 11 14 


16 17 19 
15 17 20 
16 18 20 
16 18 20 



D D 



402 MATHEMATICAL TABLES 

TABLE IV. NAPIERIAN, NATURAL, OR HYPERBOLIC LOGARITHMS 



Number. 1 





1 


2 


3 


4 


5 


6 


7 


8 


9 




01 


3-6974 


7927 


8797 


9598 


0339 


1029 


1674 


2280 


2852 


3393 




0-2 


2-3906 


4393 


4859 


5303 


5729 


6i37 


6529 


6907 


7270 


7621 




03 


7960 


8288 


8606 


8913 


9212 


9502 


97830057 


0324 


6584 




0-4 


1-0837 


1084 


1325 


1560 


1790 


2015 


2235 2450 


2660 


2866 




05 


3068 


3267 


3461 


3651 


3838 


4022 


4202 


4379 


4553 


4724 




06 


4892 


5057 


5220 


538o 


5537 


5692 


5845 


5995 


6i43 


6289 


Mean Differences. 


0-7 


6433 


6575 


67 J 5 


6853 


6989 


7 I2 3 


7256 


7386 


7515 


7643 




0-8 


7769 


7893 


8015 


8137 


8256 


8375 


8492 


8607 


8722 


8835 








09 


8946 


9057 


9166 


9274 


938i 


9487 


9592 9695 


9798 


9899 


123 


456 


789 


1-0 


o-oooo 


OIOO 


0198 


0296 


0392 


0488 


0583 


0677 


0770 


0862 








1-1 


0953 


1044 


"33 


1222 


1310 


1398 


1484 


157 


1655 


1740 


9 17 26 


35 44 52 


61 70 78 


1-2 


1823 


1906 


1989 


2070 


2151 


2231 


2311 


2390 


2469 


2546 


8 1624 


32 40 48 


56 64 72 


1-3 


2624 


2700 


2776 


2852 


2927 


3001 


375 


3148 


3221 


3293 


7 15 22 


30 37 45 


52 59 67 


1-4 


3365 


3436 


357 


3577 


3646 


37 l6 


3784 


3853 


3920 


3988 


71421 


28 35 41 


48 55 62 


1-5 


455 


4121 


4187 


4253 


43i8 


4383 


4447 


45" 


4574 


4637 


613 19 


26 32 39 


45 52 58 


1-6 


4700 


4762 


4824 


4886 


4947 


5008 


5068 


5128 


5188 


5247 


6 12 18 


24 30 36 


42 48 55 


1-7 


5306 


5365 


5423 


548i 


5539 


5596 


5653 


57io 


5766 


5822 


6 ii 17 


24 29 34 


40 46 52 


1-8 


5878 


5933 


5988 


6043 


6098 


6152 


6206 


6259 


6313 


6366 


5 " 16 


22 27 32 


38 43 49 


19 


6419 


6471 


6523 


6575 


6627 


6678 


6729 


6780 


6831 


6881 


5 1015 


20 20 31 


36 41 46 


2-0 


693i 


6981 


73i 


7080 


7129 


7178 


7227 


7275 


7324 


7372 


5 ioi5 


2O 24 29 


34 39 44 


2-1 


7419 


7467 


75M 


756i 


7608 


7655 


7701 


7747 


7793 


7839 


5 9 14 


19 23 28 


33 37 42 


22 


7885 


7930 


7975 


8020 


8065 


8109 


8i54 


8198 


8242 


8286 


4 913 


l8 22 27 


3i 36 40 


23 


8329 


8372 


8416 


8459 


8502 


8544 


8587 


8629 


8671 


8713 


4 913 


17 21 20 


3 34 38 


2-4 


8755 


8796 


8838 


8879 


8920 


8961 


9002 


9042 


9083 


9123 


4 8 12 


16 20 24 


29 33 37 


2-5 


9163 


9203 


9243 


9282 


9322 


9361 


9400 


9439 


9478 


9517 


4 8 12 


16 20 24 


2731 35 


26 


9555 


9594 


9632 


9670 


9708 


9746 


9783 


9821 


9858 


9895 


4 8 ii 


15 1923 


26 30 34 


2-7 


9933 


9969 


0006 


0043 


0080 


0116 


0152 


6188 


0225 


0260 


4711 


15 l8 22 


26 29 33 


2-8 


1-0296 


0332 


0367 


0403 


0438 


0473 


0508 


0543 


0578 


0613 


4 711 


14 18 21 


25 28 32 


2-9 


0647 


0682 


0716 


0750 


0784 


0818 


0852 


0886 


0919 


0953 


3 7 I0 


14 17 2O 


242731 


30 


0986 


1019 


1053 


1086 


1119 


1151 


1184 


1217 


1249 


1282 


3 7 10 


13 16 20 


23 26 30 


31 


1314 


1346 


1378 


1410 


1442 


1474 


1506 


1537 


1569 


1600 


3 6 10 


13 16 19 


22 25 29 


3-2 


1632 


1663 


1694 


1725 


1756 


1787 


1817 


1848 


1878 


1909 


369 


12 15 18 


21 25 28 


33 


1939 


1969 


2OOO 


2030 


2060 


2090 


2119 


2149 


2179 


2208 


369 


12 15 18 


21 24 27 


3-4 


2238 


2267 


2296 


2326 


2355 


2384 


2413 


2442 


2470 


2499 


369 


12 14 17 


2O 23 20 


35 


2528 


2556 


2585 


2613 


2641 


2669 


2698 


2726 


2754 


2782 


368 


II 14 17 


2O 22 25 


36 


2809 


2837 


2865 


2892 


2920 


2947 


2975 


3002 


3029 


3056 


3 5 8 


ii 14 16 


19 22 25 


3-7 


3083 


3110 


3137 


3164 


3191 


3218 


3244 


3271 


3297 


3324 


3 5 8 


ii 13 16 


19 21 24 


38 


335 


337 6 


343 


3429 


3455 


348i 


3507 


3533 


3558 


3584 


3 5 8 


10 13 16 


18 21 23 


3;9 


3610 


3635 


3661 


3686 


37 12 


3737 


3762 


3788 


3813 


3838 


3 5 8 


1013 15 


1 8 20 23 


40 


3863 


3888 


3913 


3938 


3962 


3987 


4012 


4036 


4061 


4085 


257 


IO 12 15 


17 2O 2 2 


4-1 


4110 


4134 


4159 


4183 


4207 


4231 


4255 


4279 


4303 


4327 


257 


10 12 14 


17 1922 


4-2 


435i 


4375 


4398 


4422 


4446 


4469 


4493 


45i6 


454 


4563 


257 


9 12 14 


16 19 21 


4-3 


4586 


4609 


4633 


4656 


4679 


4702 


47 2 5 


4748 


4770 


4793 


257 


9 II 14 


16 18 21 


44 


4816 


4839 


4861 


4884 


4907 


4929 


495i 


4974 


4996 


5019 


247 


9 II 13 


16 18 20 


45 


5041 


5063 


5085 


5107 


5129 


5151 


5173 


5195 


5217 


5239 


247 


9 II 13 


15 18 20 


46 


5261 


5282 


53<>4 


5326 


5347 


5369 


5390 


5412 


5433 


5454 


246 


9 II 13 


15 17 19 


4-7 


5476 


5497 


55i8 


5539 


556o 


558i 


5602 


5623 


5644 


5665 


2 4, 6 


8 II 13 


15 17 19 


48 
49 


5686 
5892 


5707 
5913 


5728 
5933 


5748 
5953 


57 6 9 
5974 


5790 
5994 


5810 
6014 


5831 
6034 


5851 
6054 


5872 
6074 


246 
246 


8 10 12 
8 IO 12 


14 16 19 
14 16 18 


50 


6094 


6114 


6i34 


6i54 


6174 


6194 


6214 


6233 


6253 


6273 


246 


8 IO 12 


14 16 18 



MATHEMATICAL TABLES 
TABLE IV (contd.) 



403 



6 



8 



9 



Mean Differences. 



123456 789 



51 
52 
53 
54 

55 

56 
57 
5-8 
59 
60 

6-1 
62 
63 
64 
65 

66 
6-7 
68 
69 
7-0 

7-1 
72 
73 
7-4 
75 

7-6 

7-7 
7-8 
79 
80 

8-1 
8-2 
8-3 
84 
85 

8-6 
8-7 
8-8 
8-9 
90 

91 
92 
93 
94 
95 

96 
97 
98 
99 
10 



1-6292 6312 6332 6351 6371 6390 64 
6487 6506 6525 6544 6563 6582 66oi 
6677:6696.671516734167526771 
6840 6883 690116919 6938 6956 6975|6993|7Oi i 7029 
7048! 7066! 7084 7102 7120 

7228,7246 7263 7281 



775017767 



7596 7613 7630J647 7 66 4 7 682 



7440 



091642964486467 2 
662066396658 2 
68o8|6827,6846| 2 

6975'6993'7OII 7O29 2 
71387156717471927210 2 



7299 7317 7334 7352,7370 7387 
7457 7475 7492 7509 7527 7544 7561 



7783 7800,7817 7834 7851 



769977167733 
78687884,7901 

7968 7984 8001 8017 8034 8o5O ! 8o67 

8o83 ! 8o99 811618132 8148 8165 8181 8197,8213 8229 2 
262'8278,8294 ! 83io 8326 8342 8358 8374 8390 



8563^579 8594 ! 86io 862518641 



8656'8672|8687'8703 2 



7i8,87338749 8764^779 8795 88io8825j8840 ( 8856 2 

887l'8886 8goi 89l6|893I 8946 8g6l 8976 8991 9006 2 

2 



041204250438045104640. 



477 0490,050310516 0528 
' 



1580 0592 0605 0618 0631 

o66g'o68 1 ;o6g4j07O7'o7igJQ732 0744,075 7:0769 078 
0794 0807 0819 0832 0844 0857 0869,0882 p8g4 0906 

0919 O93i|og43og56 i og68 0980 ogg2 1005 1017 1029 

1114 1 126)1138 1151 
12471259,1270 



1041 



1163 
1282 
1401 

1518 1529 154 

1633 164511656 1668 167911691 



I748I759I77 



2192 



io54jio66 107811090 1102 



1175 1187 H99'i2ii 



1294 1300)1318 1330 1342 

1436144814591471 



1861 
1972 

2083 209412105 2ii6J2i27 2138 



1552 
1668 
1782 



1223 1235 



187211883 i894 l i905 1917 1928 1939 1950 1961 
I9&3JI994 2bo6|2Oi7 2028 2039 2050 2061 2072 

49215921702181 

46 2257 226&2279J2289 

tf A s*lf\Z T O *7 C 



2203 2214 2225!2235 



~J~~ "O" -J--|-OJ-"pJ i tO *OJt -^O^ 

2407 2418 2428 2439 2450 2460 2471 2481 2492 2502 

2513 2523 2534 2544 2555 2565 2576 2586 2597 2607 

2618 2628 2638 2649 2659 2670 2680 2690 2701 2711 
2721 2732 2742 2752J2762 2773 2783 2793 2803 2814 



2824 2834 2844 2854 2865 2875 
29251293512946 2956 2966 2976 2986 
2-3020 



1702 



1793^1804 1816 1827 1838 1849 



1483 1494 1506 



171317251736 



2885 2895 2905 2915 



2996 3006 3016 



4 6 



4 6 
4 6 



9169 9184 gigg 9213 9228 9243^257^9272 9286 9301 2 
9315 933 9344l9359!9373 9387J94 02 94i6|943i!9445 I 
9459 9473 9488,9502,95i6|g530 9545 9559 9573 9587 i 

g6oi|g6i5 9629 9643 ^657^671 g685 g6gg[g7i3 9727 i 

974IJ9755 9769 9782J9796 g8io g824 ^838 g85i 9865 i 

9879 g8g2g9o6;9920 | g9339g47gg6ijg974|gg88 oooi i 

2-0015:002800420055,006900820096010901220136 

0149^162 0x76 Ol8g O2O2 O2l6 0229^242^255 O268 

0281 0295 0308,0321 0334 347 0360 0373 0386 03gg 



3 4 



8 IO 12 
8 IO 12 



g ii 
7 9 ii 
4 5| 7 9 ii 



2 4 5:7 9 



4 57 9 10 

3 57 9 10 

3 5 7 8 10 

3 5 7 8 10 



6 8 9 



5 7 



5 6 7 



467 



14 16 18 
13 15 17 
13 15 17 
13 15 17 
13 14 16 

12 14 16 
12 14 16 

12 I 4 15 
12 13 15 
12 13 15 

II 13 15 

II 13 14 

II 13 14 

II 12 14 

II 12 14 

II 12 14 

IO 12 13 

IO 12 13 

IO 12 13 

10 ii 13 
10 ii 13 

IO II 12 

IO II 12 

9 II 12 

9 II 12 

9 II 12 
9 IO 12 
9 10 II 

9 10 ii 
9 10 ii 

9 10 ii 
9 10 ii 
8 10 ii 
8 9 ii 
8911 



9 10 
9 10 
9 10 
9 10 
9 10 

9 10 
9 10 
9 10 
9 10 
8 10 



404 



MATHEMATICAL TABLES 
TABLE V. NATURAL SINES. 



E 

9 

Q 


0' 
00 


6' 
0-1 


12' 
0-2 


18' 
0-3 


24' 
0-4 


30' 
0-5 


36' 
06 


42' 
0-7 


48' 
0-8 


54' 
09 


Mean Differences. 


1' 2' 3^4' 5' 





oooo 


0017 


0035 


0052 


0070 


0087 


0105 


OI22 


0140 


0157 


3 6 9 12 15 


1 


0175 


0192 


0209 


0227 


0244 


0262 


0279 


0297 


0314 


0332 


3 6 9 12 15 


2 


0349 


0366 


0384 


0401 


0419 


0436 


454 


47 I 


0488 


0506 


3 6 9 12 15 


3 


0523 


054 1 


0558 


0576 


0593 


0610 


0628 


0645 


0663 


0680 


3 9 12 15 


4 


0698 


0715 


0732 


0750 


0767 


0785 


0802 


0819 


0837 


0854 


3 6 9 12 14 


5 


0872 


0889 


0906 


0924 


0941 


0958 


0976 


0993 


IOII 


1028 


3 6 9 12 14 


6 


1045 


1063 


1080 


1097 


1115 


1132 


1149 


1167 


1184 


1201 


3 6 9 12 14 


7 


1219 


1236 


1253 


1271 


1288 


1305 


1323 


1340 


1357 


!374 


3 6 9 12 14 


8 


1392 


1409 


1426 


1444 


1461 


1478 


H95 


1513 


1530 


1547 


3 6 9 12 14 


9 


1564 


1582 


1599 


1616 


1633 


1650 


1668 


I68 5 


1702 


1719 


3 6 9 12 14 


10 


1736 


1754 


1771 


1788 


1805 


1822 


1840 


1857 


1874 


1891 


3 6 9 ii 14 


11 


1908 


1925 


1942 


1959 


1977 


1994 


2OII 


2O28 


2045 


2062 


3 6 9 ii 14 


12 


2079 


2096 


2113 


2130 


2147 


2164 


2181 


2198 


2215 


2233 


3 6 9 ii 14 


13 


2250 


2267 


2284 


2300 


2317 


2334 


2351 


2368 


2385 


2402 


3 6 8 ii 14 


14 


2419 


2436 


2453 


2470 


2487 


2504 


2521 


2538 


2554 


2571 


3 6 8 ii 14 


15 


2588 


2605 


2622 


2639 


2656 


2672 


2689 


2706 


2723 


2740 


3 6 8 ii 14 


16 


2756 


2773 


2790 


2807 


2823 


2840 


2857 


2874 


2890 


2907 


3 6 8 ii 14 


17 


2924 


2940 


2957 


2974 


2990 


3007 


302 4 


3040 


3057 


3074 


3 6 8 ii 14 


18 


3090 


3107 


3123 


3MO 


3156 


3i73 


3190 


3206 


3223 


3239 


3 6 8 ii 14 


19 


3256 


3272 


3289 


3305 


3322 


3338 


3355 


337 1 


3387 


3404 


3 5 8 ii 14 


20 


3420 


3437 


3453 


34 6 9 


3486 


3502 


35i8 


3535 


355i 


3567 


3 5 8 ii 14 


21 


3584 


3600 


3616 


3633 


3649 


3665 


3681 


3697 


37M 


3730 


3 5 8 ii 14 


22 


3746 


3762 


3778 


3795 


3811 


3827 


3843 


3859 


3875 


3891 


3 5 8 ii 14 


23 


3907 


3923 


3939 


3955 


397i 


3987 


4003 


4019 


4035 


45i 


3 5 8 ii 14 


24 


4067 


4083 


4099 


4"5 


4131 


4*47 


4163 


4179 


4i95 


4210 


3 5 8 ii 13 


25 


4226 


4242 


4258 


4274 


4289 


4305 


4321 


4337 


4352 


4368 


3 5 8 ii 13 


26 


4384 


4399 


4415 


4431 


4446 


4462 


4478 


4493 


4509 


4524 


3 5 8 10 13 


27 


454 


4555 


4571 


4586 


4602 


4617 


4633 


4648 


4664 


4679 


3 5 8 10 13 


28 


4 6 95 


4710 


4726 


4741 


4756 


4772 


4787 


4802 


4818 


4833 


3 5 8 10 13 


29 


4848 


4863 


4879 


4894 


4909 


4924 


4939 


4955 


4970 


4985 


3 5 8 10 13 


30 


5000 


5015 


5030 


5045 


5060 


5075 


5090 


5105 


5120 


5135 


3 5 8 10 ij 


31 


5150 


5165 


5180 


5195 


5210 


5225 


5240 


5255 


5270 


5284 


2 5 7 10 12 


32 


5299 


53H 


5329 


5344 


5358 


5373 


5388 


5402 


54i7 


5432 


2 5 7 10 12 


33 


5446 


546i 


5476 


5490 


5505 


5519 


5534 


5548 


5563 


5577 


2 5 7 10 12 


34 


5592 


5606 


5621 


5635 


5650 


5664 


5678 


5693 


5707 


5721 


2 5 7 10 12 


35 


5736 


5750 


5764 


5779 


5793 


5807 


5821 


5835 


5850 


5864 


2 5 7 9 12 


36 


5878 


5892 


5906 


5920 


5934 


5948 


5962 


5976 


5990 


6004 


2 5 7 9 12 


37 


6018 


6032 


6046 


6060 


6074 


6088 


6101 


6115 


6129 


6i43 


2 5 7 9 12 


38 


6157 


6170 


6184 


6198 


6211 


6225 


6239 


6252 


6266 


6280 


2 5 7 9 ii 


39 


6293 


6307 


6320 


6334 


6347 


6361 


6 374 


6388 


6401 


6414 


247 9 ii 


40 


6428 


6441 


6455 


6468 


6481 


6494 


6508 


6521 


6534 


6547 


247 9 ii 


41 


6561 


6574 


6587 


6600 


6613 


6626 


6639 


6652 


6665 


6678 


2 4 7 9 ii 


42 


6691 


6704 


6717 


6730 


6743 


6756 


6769 


6782 


6794 


6807 


2 4 6 9 ii 


43 


6820 


6833 


6845 


6858 


6871 


6884 


6896 


6909 


6921 


6934 


246 8 ii 


44 


6947 


6959 


6972 


6984 


6997 


7009 


7022 


734 


7046 


759 


246 8 10 


45 -7071 


7083 


7096 


7108 


7120 


7133 


7M5 


7157 


7169 


7181 


246 8 10 



MATHEMATICAL TABLES 



405 



TABLE V. (contd.) 



g 


0' 


6' 


12' 


18' 


24' 


30' 


36' 


42' 


48' 


54' 


Mean Differences. 


1 


00 


0-1 


02 


03 


04 


0-5 


06 


0-7 


0-8 


09 


1' 2 3' 4' 5' 


45 


7071 


7083 


7096 


7108 


7120 


7133 


7M5 


7i57 


7169 


7181 


246 8 10 


46 


7193 


7206 


7218 


7230 


7242 


7254 


7266 


7278 


7290 


7302 


246 8 10 


47 


73M 


7325 


7337 


7349 


736i 


7373 


7385 


7396 


7408 


7420 


246 8 10 


48 


743i 


7443 


7455 


7466 


7478 


7490 


75oi 


7513 


7524 


7536 


2 4 6 8 10 


49 


7547 


7559 


7570 


758i 


7593 


7604 


7615 


7627 


7638 


7649 


24689 


50 


7660 


7672 


7683 


7694 


7705 


7716 


7727 


7738 


7749 


7760 


24679 


51 


7771 


7782 


7793 


7804 


7815 


7826 


7837 


7848 


7859 


7869 


2 4579 


52 


7880 


7891 


7902 


7912 


7923 


7934 


7944 


7955 


7965 


7976 


24579 


53 


7986 


7997 


8007 


8018 


8028 


8039 


8049 


8059 


8070 


8080 


23579 


54 


8090 


8100 


8111 


8121 


8131 


8141 


8151 


8161 


8171 


8181 


23578 


55 


8192 


8202 


8211 


8221 


8231 


8241 


8251 


8261 


8271 


8281 


23578 


56 


8290 


8300 


8310 


8320 


8329 


8339 


8348 


8358 


8368 


8377 


23568 


57 


8387 


8396 


8406 


8415 


8425 


8434 


8443 


8453 


8462 


8471 


23568 


58 


8480 


8490 


8499 


8508 


8517 


8526 


8536 


8545 


8554 


8563 


23568 


59 


8572 


8581 


8590 


8599 


8607 


8616 


8625 


8634 


8643 


8652 


13467 


60 


8660 


8669 


8678 


8686 


8695 


8704 


8712 


8721 


8729 


8738 


13467 


61 


8746 


8755 


8763 


8771 


8780 


8788 


8796 


8805 


8813 


8821 


13467 


62 


8829 


8838 


8846 


8854 


8862 


8870 


8878 


8886 


8804 


8902 


13457 


63 


8910 


8918 


8926 


8934 


8942 


8949 


8957 


8965 


8973 


8980 


13456 


64 


8988 


8996 


9003 


9011 


9018 


9026 


9033 


9041 


9048 


9056 


13456 


65 


9063 


9070 


9078 


9085 


9092 


9100 


9107 


9114 


9121 


9128 


12456 


66 


9135 


9M3 


9150 


9157 


9164 


9171 


9178 


9184 


9191 


9198 


12356 


67 


9205 


9212 


9219 


9225 


9232 


9239 


9245 


9252 


9259 


9265 


12346 


68 


9272 


9278 


9285 


9291 


9298 


9304 


93ii 


9317 


9323 


9330 


12345 


69 


9336 


9342 


934 8 


9354 


936i 


9367 


9373 


9379 


9385 


9391 


12345 


70 


9397 


943 


9409 


9415 


9421 


9426 


9432 


9438 


9444 


9449 


12345 


71 


9455 


9461 


9466 


9472 


9478 


9483 


9489 


9494 


9500 


9505 


12345 


72 


95ii 


95i6 


9521 


9527 


9532 


9537 


9542 


9548 


9553 


9558 


12334 


73 


9563 


9568 


9573 


9578 


9583 


9588 


9593 


9598 


9603 


9608 


12234 


74 


9613 


9617 


9622 


9627 


9632 


9636 


9641 


9646 


9650 


9655 


12234 


75 


9659 


9664 


9668 


9673 


9677 


9681 


9686 


9690 


9694 


9699 


11234 


76 


973 


9707 


9711 


97*5 


9720 


9724 


9728 


9732 


9736 


9740 


11233 


77 


9744 


9748 


9751 


9755 


9759 


97 6 3 


9767 


9770 


9774 


9778 


11233 


78 


9781 


9785 


9789 


9792 


9796 


9799 


9803 


9806 


9810 


9813 


i 223 


79 


9816 


9820 


9823 


9826 


9829 


9833 


9836 


9839 


9842 


9845 


i 223 


80 


9848 


9851 


9854 


9857 


9860 


9863 


9866 


9869 


9871 


9874 


122 


81 


9877 


9880 


9882 


9885 


9888 


9890 


9893 


9895 


9898 


9900 


122 


82 


9903 


9905 


9907 


9910 


9912 


9914 


9917 


9919 


9921 


9923 


122 


83 


992.5 


9928 


9930 


9932 


9934 


9936 


9938 


9940 


9942 


9943 


112 


84 


9945 


9947 


9949 


995i 


9952 


9954 


9956 


9957 


9959 


9960 


I I I 2 


85 


9962 


9963 


9965 


9966 


9968 


9969 


997 1 


9972 


9973 


9974 


I I I 


86 


9976 


9977 


9978 


9979 


9980 


9981 


9982 


9983 


9984 


9985 


I I I 


87 


9986 


9987 


9988 


9989 


9990 


9990 


9991 


9992 


9993 


9993 


O O O I I 


88 


9994 


9995 


9995 


9996 


9996 


9997 


9997 


9997 


9998 


9998 


O O O O 


89 


9998 


9999 


9999 


9999 


9999 


I'OOO 


I'OOO 


i-ooo 


i-ooo 


i-ooo 


O O O O O 


90 


i-ooo 























406 



MATHEMATICAL TABLES 



TABLE VI. NATURAL COSINES 



1 

B, 


0' 


6' 


12' 


18' 


24' 


30' 


36' 


42' 


48' 


54' 


Mezn Differenci's. 


3 


00 


01 


02 


03 


04 


0-5 


06 


0-7 


0-8 


09 


1' 2' 3' 4' 5' 





I'OOO 


ooo 


ooo 


ooo 


I'OOO 


I'OOO 


9999 


9999 


9999 


9999 


o o o o o 


1 


9998 


9998 


9998 


9997 


9997 


9997 


9996 


9996 


9995 


9995 


O O O O 


2 


9994 


9993 


9993 


9992 


9991 


9990 


9990 


9989 


9988 


9987 


I I 


3 


9986 


9985 


9984 


9983 


9982 


9981 


9980 


9979 


9978 


9977 


O O II 


4 


9976 


9974 


9973 


9972 


997 1 


9969 


9968 


9966 


9965 


9963 


O O II 


5 


9962 


9960 


9959 


9957 


9956 


9954 


9952 


9951 


9949 


9947 


01 12 


6 


9945 


9943 


9942 


994 


9938 


9936 


9934 


9932 


9930 


9928 


01 12 


7 


9925 


9923 


9921 


9919 


9917 


9914 


9912 


9910 


9907 


9905 


O I 22 


8 


9903 


9900 


9898 


9895 


9893 


9890 


9888 


9885 


9882 


9880 


01 22 


9 


9877 


9874 


9871 


9869 


9866 


9863 


9860 


9857 


9854 


9851 


O 22 


10 


9848 


9845 


9842 


9839 


9836 


9833 


9829 


9826 


9823 


9820 


I 223 


11 


9816 


9813 


9810 


9806 


9803 


9799 


9796 


9792 


9789 


9785 


I 223 


12 


9781 


9778 


9774 


977 


9767 


9763 


9759 


9755 


9751 


974 s 


I 233 


13 


9744 


974 


9736 


9732 


9728 


9724 


9720 


97*5 


9711 


9707 


I 233 


14 


9703 


9699 


9694 


9690 


9686 


9681 


9677 


9673 


9668 


9664 


1 234 


15 


9659 


9655 


9650 


9646 


9641 


9636 


9632 


9627 


9622 


9617 


12234 


16 


9613 


9608 


9603 


9598 


9593 


9588 


9583 


9578 


9573 


9568 


12234 


17 


9563 


9558 


9553 


9548 


9542 


9537 


9532 


9527 


9521 


95i6 


12334 


18 


95" 


955 


95oo 


9494 


9489 


94 8 3 


9478 


9472 


9466 


9461 


12345 


19 


9455 


9449 


9444 


9438 


9432 


9426 


9421 


9415 


9409 


943 


12345 


20 


9397 


9391 


9385 


9379 


9373 


9367 


9361 


9354 


9348 


9342 


12345 


21 


9336 


9330 


93^3 


9317 


93" 


9304 


9298 


9291 


9285 


9278 


12345 


22 


9272 


9265 


9259 


9252 


9245 


9239 


9232 


9225 


9219 


9212 


12346 


23 


9205 


9198 


9191 


9184 


9178 


9171 


9164 


9157 


9150 


9M3 


12356 


24 


9135 


9128 


9121 


9114 


9107 


9100 


9092 


9085 


9078 


9070 


2456 


25 


9063 


9056 


9048 


9041 


9033 


9026 


9018 


9011 


9003 


8996 


3456 


26 


8988 


8980 


8973 


8965 


8957 


8949 


8942 


8934 


8926 


8918 


3456 


27 


8910 


8902 


8894 


8886 


8878 


8870 


8862 


8854 


9846 


8838 


3457 


28 


8829 


8821 


8813 


8805 


8796 


8788 


8780 


8771 


8763 


8755 


3467 


29 


8746 


8738 


8729 


8721 


8712 


8704 


8695 


8686 


8678 


8669 


1 3 4 6 7 


30 


8660 


8652 


8643 


8634 


8625 


8616 


8607 


8599 


8590 


8581 


13467 


31 


8572 


8563 


8554 


8545 


8536 


8526 


8517 


8508 


8499 


8490 


23568 


32 


8480 


8471 


8462 


8453 


8443 


8434 


8425 


8415 


8406 


8396 


23568 


33 


8387 


8377 


8368 


8358 


8348 


8339 


8329 


8320 


8310 


8300 


23568 


34 


8290 


8281 


8271 


8261 


8251 


8241 


8231 


8221 


8211 


8202 


23578 


35 


8192 


8181 


8171 


8161 


8151 


8141 


8131 


8121 


8111 


8100 


23578 


36 


8090 


8080 


8070 


8059 


8049 


8039 


8028 


8018 


8007 


7997 


23579 


37 


7986 


7976 


7965 


7955 


7944 


7934 


7923 


7912 


7902 


7891 


24579 


38 


7880 


7869 


7859 


7848 


7837 


7826 


7815 


7804 


7793 


7782 


24579 


39 


7771 


7760 


7749 


7738 


7727 


7716 


7705 


7694 


7683 


7672 


24679 


40 


7660 


7649 


7638 


7627 


7615 


7604 


7593 


758i 


7570 


7559 


24689 


41 


7547 


7536 


7524 


7513 


75i 


7490 


7478 


7466 


7455 


7443 


2 4 6 8 10 


42 


7431 


7420 


7408 


7396 


7385 


7373 


736i 


7349 


7337 


7325 


246 8 10 


43 


73H 


7302 


7290 


7278 


7266 


7254 


7242 


7230 


7218 


7206 


246 8 10 


44 


.7193 


7181 


7169 


7157 


7145 


7133 


7120 


7108 


7096 


7083 


246 8 10 


45 


.7071 


7059 


7046 


7034 


7022 


7009 


6997 


6984 


6972 


6959 


2 4 6 8 10 






1 





















MATHEMATICAL TABLES 
TABLE VI (contd.) 



407 



1 

Q 


0' 
00 


6' 
01 


12' 
02 


18' 
03 


24' 
0-4 


30' 

0-5 


36' 
06 


42' 
0-7 


48' 
08 


54' 
0-9 


Mean Differences. 


1' 2' 3' 4' 5' 


45 


7071 


759 


7046 


7034 


7022 


7009 


6997 


6984 


6972 


6959 


246 8 10 


46 


6947 


6934 


6921 


6909 


6896 


6884 


6871 


6858 


6845 


6833 


246 8 ii 


47 


6820 


6807 


6794 


6782 


6769 


6756 


6743 


6730 


6717 


6704 


246 9 ii 


48 


6691 


6678 


6665 


6652 


6639 


6626 


6613 


6600 


6587 


6574 


2 4 7 9 ii 


49 


6561 


6547 


6534 


6521 


6508 


6494 


6481 


6468 


6455 


6441 


247 9 II 


50 


6428 


6414 


6401 


6388 


6374 


6361 


6347 


6334 


6320 


6307 


247 9 II 


51 


6293 


6280 


6266 


6252 


6239 


6225 


6211 


6198 


6184 


6170 


257 9 II 


52 


6157 


6i43 


6129 


6115 


6101 


6088 


6074 


6060 


6046 


6032 


257 9 12 


53 


6018 


6004 


5990 


5976 


5962 


5948 


5934 


5920 


5906 


5892 


2 5 7 9 12 


54 


5878 


5864 


5850 


5835 


5821 


5807 


5793 


5779 


5764 


5750 


2 5 7 9 12 


55 


5736 


5721 


577 


5693 


5678 


5664 


5650 


5635 


5621 


5606 


2 5 7 10 12 


56 


5592 


5577 


5563 


5548 


5534 


55i9 


5505 


5490 


5476 


5461 


2 5 7 10 12 


57 


5446 


5432 


5417 


542 


5388 


5373 


5358 


5344 


5329 


53i4 


2 5 7 10 12 


58 


5299 


5284 


527 


5255 


524 


5225 


5210 


5195 


5180 


5165 


2 5 7 10 12 


59 


515 


5135 


5120 


5105 5090 


575 


5060 


5045 5030 


5015 


3 5 8 10 13 


60 


5000 


4985 


4970 


4955 


4939 


4924 


4909 


4894 4879 


4863 


3 5 8 10 13 


61 


4848 


4833 


4818 


4802 


4787 


4772 


4756 


4741 4726 


4710 


3 5 8 10 13 


62 


4695 


4679 


4664 


4648 4633 


4617 


4602 


4586 4571 


4555 


3 5 8 10 13 


63 


454 


4524 


4509 


4493 447 s 


4462 


4446 


4431 4415 


4399 


3 5 8 10 13 


64 


4384 


4368 


4352 


4337 4321 


4305 


4289 


4274 4258 


4242 


3 5 8 ii 13 


65 


4226 


4210 


4195 


4179 


4163 


4M7 


4131 


4115 4099 


4083 


3 5 8 ii 13 


66 


4067 4051 


435 


4019 


4003 


3987 


3971 


3955 ' 3939 


3923 


3 5 8 ii 14 


67 


3907 


3891 


3875 


3 8 59 3843 


3827 


3811 


3795 3778 


3762 


3 5 3 ii 14 


68 


3746 


3730 


37M 


3697 3681 


3665 


3649 


3633 , 3616 


3600 


3 5 8 ii 14 


69 


3584 


3567 


3551 


3535 35i8 


3502 


3486 


3469 3453 


3437 


3 5 3 ii 14 


70 


3420 


344 


3387 


3371 3355 


3338 


3322 


3305 


3289 


3272 


3 5 8 ii 14 


71 


3256 


3239 


3223 


3206 3190 


3173 


3156 


3140 


3123 


3107 


3 6 8 li 14 


72 


390 


3074 


3057 


3040 3024 


3007 


2990 


2974 2957 


2940 


3 6 8 ii 14 


73 


2924 


2907 


2890 


2874 


2857 


2840 


2823 


2807 2790 


2773 


3 6 8 ii 14 


74 


2756 


2740 


2723 


2706 


2689 


2672 


2656 


2639 


2622 


2605 


3 6 8 ii 14 


75 


2588 


257 1 


2554 


2538 


2521 


2504 


2487 


2470 


2453 


2436 


3 6 8 ii 14 


76 


2419 


2402 


2385 


2368 


2351 


2334 


2317 


2300 


2284 


2267 


3 6 8 ii 14 


77 


2250 


2233 


2215 


2198 


2181 


2164 


2147 


2130 


2113 


2096 


3 6 9 ii 14 


78 


2079 


2062 


2045 


2028 


201 1 


1994 


1977 


1959 


1942 


1925 


3 6 9 ii 14 


79 


1908 


1891 


1874 


1857 


1840 


1822 


1805 


1788 


1771 


1754 


3 6 9 ii 14 


80 


1736 


1719 


1702 


1685 


1668 


1650 


1633 


1616 


1599 


1582 


3 6 9 12 14 


81 


1564 


1547 


1530 


1513 


1495 


1478 


1461 


1444 


1426 


1409 


3 6 9 12 14 


82 


1392 


1374 


1357 


134 


1323 


1305 


1288 


1271 


1253 


1236 


3 6 9 12 14 


83 


1219 


1201 


1184 


1167 


1149 


1132 


i"5 


1097 


1080 


1063 


3 6 9 12 14 


84 


1045 


1028 


IOII 


0993 


0976 


0958 


0941 


0924 


0906 


0889 


3 6 9 12 14 


85 


0872 


0854 


0837 


0819 


O8O2 


0785 


0767 


0750 


0732 


0715 


3 6 9 12 14 


86 


0698 


0680 


0663 


0645 


0628 


0610 


0593 


0576 


0558 


0541 


3 6 9 12 15 


87 


0523 


0506 


0488 


0471 


0454 


0436 


0419 


0401 


0384 


0366 


3 6 9 12 15 


88 


0349 


0332 


0314 


0297 


0279 


0262 


0244 


0227 


0209 


0192 


3 6 9 12 15 


89 


0175 


0157 


0140 


OI22 


OIO5 


0087 


0070 


0052 


0035 


0017 


3 6 9 12 15 


90 


oooo 























4S 



MATHEMATICAL TABLES 



TABLE VII. NATURAL TANGENTS. 




& 

ID 

c 


0' 
00 


6' 
01 


12' 
02 


18' 
03 


24' 
0-4 


30' 
05 


36' 
0-6 


r 

42' 
0-7 


48' 
08 


54' 
09 


Mean Differences. 


1' 2' 3' 4' 5' 





oooo 


0017 


0035 


0052 


0070 


0087 


0105 


0122 


0140 


oi57 


3 6 9 12 15 


1 


0175 


0192 


0209 


0227 


0244 


0262 


0279 


0297 


0314 


0332 


3 6 9 12 15 


2 


0349 


0367 


0384 


0402 


0419 


437 


454 


0472 


0489 


0507 


3 6 9 12 15 


3 


0524 


0542 


0559 


0577 


594 


0612 


0629 


0647 


0664 


0682 


3 6 9 12 15 


4 


0699 


0717 


0734 


0752 


0769 


0787 


0805 


0822 


0840 


0857 


3 6 9 12 15 


5 


0875 


0892 


0910 


0928 


0945 


0963 


0981 


0998 


1016 


1033 


3 6 9 12 15 


6 


1051 


1069 


1086 


1104 


1122 


H39 


"57 


"75 


1192 


I2IO 


3 6 9 12 15 


7 


1228 


1246 


1263 


1281 


1299 


1317 


J 334 


*352 


137 


1388 


3 6 9 12 15 


8 


1405 


1423 


1441 


1459 


H77 


1495 


1512 


1530 


1548 


1566 


3 6 9 12 15 


9 


1584 


1602 


1620 


16^8 


1655 


1673 


1691 


1709 


1727 


1745 


3 6 9 12 15 


10 


1763 


1781 


1799 


1817 


1835 


1853 


1871 


1890 


1908 


1926 


3 6 9 12 15 


11 


1944 


1962 


1980 


1998 


2016 


2035 


2053 


2071 


2089 


2IO7 


3 6 9 12 15 


12 


2126 


2144 


2162 


2180 


2199 


2217 


2235 


2254 


2272 


229O 


3 6 9 12 15 


13 


2309 


2327 


2345 


2364 


2382 


2401 


2419 


2438 


2456 


2475 


3 9 12 15 


14 


2493 


2512 


2530 


2549 


2568 


2586 


2605 


2623 


2642 


2661 


3 6 9 12 16 


15 


2679 


2698 


2717 


2736 


2754 


2 773 


2792 


2811 


2830 


2849 


3 6 9 13 16 


16 


2867 


2886 


2905 


2924 


2943 


2962 


2981 


3000 


3019 


3038 


3 6 9 13 16 


17 


3057 


3076 


3096 


3H5 


3134 


3153 


3172 


3I9T 


3211 


3230 


3 6 10 13 16 


18 


3249 


3269 


3288 


3307 


3327 


3346 


3365 


3385 


344 


3424 


3 6 10 13 16 


19 


3443 


3463 


3482 


3502 


3522 


3541 


356i 


358i 


3600 


3620 


3 7 10 13 16 


20 


3640 


3659 


3679 


3699 


3719 


3739 


3759 


3779 


3799 


3819 


3 7 I0 J 3 17 


21 


3839 


3859 


3879 


3899 


3919 


3939 


3959 


3979 


4000 


4O2O 


3 7 I0 T 3 17 


22 


4040 


4061 


4081 


4101 


4122 


4142 


4i63 


4183 


4204 


4224 


3 7 10 14 17 


23 


4245 


4265 


4286 


4307 


4327 


4348 


4369 


4390 


4411 


4431 


3 7 10 14 17 


24 


4452 


4473 


4494 


45i5 


4536 


4557 


4578 


4599 


4621 


4642 


4 7 ii 14 18 


25 


4663 


4684 


4706 


4727 


4748 


4770 


4791 


48i3 


4834 


4856 


4 7 ii 14 18 


26 


4877 


4899 


4921 


4942 


4964 


4986 


5008 


5029 


5051 


5073 


4 7 ii 15 18 


27 


5095 


5"7 


5139 


5161 


5184 


5206 


5228 


5250 


5272 '5295 


4 7 ii 15 18 


28 


5317 


5340 


5362 


5384 


5407 


5430 


5452 


5473 


5498 


5520 


4 8 ii 15 19 


29 


5543 


5566 


5589 


5612 


5635 


5658 


5681 


5704 


5727 


5750 


4 8 12 15 19 


30 


5774 


5797 


5820 


5844 


5867 


5890 


5914 


5938 


596i 


5985 


4 8 12 16 20 


31 


6009 


6032 


6056 


6080 


6104 


6128 


6152 


6176 


6200 


6224 


4 8 12 16 20 


32 


6249 


6273 


6297 


6322 


6346 


6371 


6395 


6420 


6 445 


6469 


4 8 12 16 20 


33 


6494 


6519 


6 544 


6569 


6594 


6619 


6644 


6669 


6694 


6720 


4 8 13 17 21 


34 


6745 


6771 


6796 


6822 


6847 


6873 


6899 


6924 


6950 


6976 


4 9 13 17 21 


35 


7002 


7028 


754 


7080 


7107 


7 J 33 


7*59 


7186 


7212 


7239 


4 9 13 18 22 


36 


7265 


7292 


7319 


7346 


7373 


7400 


7427 


7454 


748i 


7508 


5 9 14 18 23 


37 


7536 


7563 


7590 


7618 


7646 


7673 


7701 


7729 


7757 


7785 


5 9 14 18 23 


38 


7813 


7841 


7869 


7898 


7926 


7954 


7983 


8012 


8040 


8069 


5 9 14 19 24 


39 


8098 


8127 


8156 


8185 


8214 


8243 


8273 


8302 


8332 


8361 


5 10 15 20 24 


40 


8391 


8421 


8451 


8481 


8511 


8541 


8571 


8601 


8632 


8662 


5 10 15 20 25 


41 


8693 


8724 


8754 


8785 


8816 


8847 


8878 


8910 


8941 


8972 


5 10 16 21 26 


42 


9004 


9036 


9067 


9099 


9131 


9163 


9195 


9228 


9260 


9293 


5 ii 16 21 27 


43 


9325 


9358 


9391 


9424 


9457 


9490 


9523 


9556 


9590 


9623 


6 ii 17 22 28 


44 


9657 


9691 


9725 


9759 


9793 


9827 


9861 


9896 


9930 


9965 


6 ii 17 23 29 


45 


i-oooo 


0035 


0070 


0105 


0141 


0176 


O2 1 2 


0247 


0283 


0319 


6 12 18 24 30 



MATHEMATICAL TABLES 



409 



TABLE VII. (contd.) 



1 

I 


0' 
00 


6' 
01 


12' 
0-2 


18' 
03 


24' 
04 


30' 
0-5 


36' 
0-6 


42' 
07 


48' 
0-8 


54' 
09 


Mean Differences. 


1' 2' 3' 4' 5' 


45 


I-OOOO 


0035 


0070 


0105 0141 


0176 


O2I2 


0247 


0283 


0319 


6 12 18 24 30 


46 


1-0355 


0392 


0428 


0464 


0501 


0538 


0575 


0612 


0649 


0686 


6 12 18 25 31 


47 


1-0724 


0761 


0799 


0837 


0875 


0913 


0951 


0990 


1028 


1067 


6 13 19 25 32 


48 


1-1106 


H45 


1184 


1224 


1263 


1303 


1343 


1383 


1423 


1463 


7 13 20 27 33 


49 


1-1504 


1544 


1585 


1626 1667 


1708 


1750 


1792 


1833 


1875 


7 14 21 28 34 


50 


1-1918 


1960 


2OO2 


2045 


2088 


2131 


2174 


2218 


2261 


2305 


7 14 22 29 36 


51 


1-2349 


2393 


2437 


2482 


2527 


2572 


2617 


2662 


2708 


2753 


8 15 23 30 38 


52 


1-2799 


2846 


2892 


2938 


2985 


3032 


3079 


3127 


3i75 


3222 


8 16 24 31 39 


53 


1-3270 


33i9 


3367 


34i6 


3465 


35H 


3564 


3613 


3663 


3713 


8 16 25 33 '41 


54 


1-3764 


3814 


386 5 


39i6 


3968 


4019 


4071 


4124 


4176 


4229 


9 17 26 34 43 


55 


1-4281 


4335 


43 88 


4442 


4496 


4550 


4605 


4659 


47i5 


4770 


9 18 27 36 45 


56 


1-4826 


4882 


4938 


4994 


5051 


5108 


5166 


5224 


5282 


5340 


10 19 29 38 48 


57 


1-5399 


5458 


5517 


5577 


5637 


5697 


5757 


5818 


5880 


594i 


10 20 30 40 50 


58 


1-6003 


6066 


6128 


6191 


6255 


6319 


6383 


6447 


6512 


6577 


ii 21 32 43 53 


59 


1-6643 


6709 


6775 


6842 


6909 


6977 


745 


7H3 


7182 


7251 


II 23 34 45 56 


60 


1-7321 


739i 


7461 


7532 


7603 


7675 


7747 


7820 


7893 


7966 


12 24 36 48 60 


61 


1-8040 


8115 


8190 


8265 


8341 


8418 


8495 


8572 


8650 


8728 


13 26 38 51 64 


62 


1-8807 


8887 


8967 


9047 


9128 


9210 


9292 


9375 


9458 


9542 


14 27 41 55 68 


63 


1-9626 


9711 


9797 


9883 


997 


0057 


oi45 


0233 


0323 


0413 


15 29 44 58 73 


64 


2-0503 


0594 


0686 


0778 


0872 


0965 


1060 


"55 


1251 


1348 


16 31 47 63 78 


65 


2-1445 


1543 


1642 


1742 


1842 


1943 


2045 


2148 


2251 


2355 


17 34 51 68 85 


66 


2-2460 


2566 


2673 


2781 


2889 


2998 


3109 


3220 


3332 


3445 


18 37 55 73 92 


67 


2-3559 


3673 


3789 


3906 


4023 


4142 


4262 


4383 


4504 


4627 


20 40 60 79 99 


68 


2-4751 


4876 


5002 


5129 


5257 


5386 


5517 


5649 


5782 


5916 


22 43 65 87 108 


69 


2-6051 


6187 


6325 


6464 


6605 


6746 


6889 


734 


7179 


7326 


24 47 71 95 119 


70 


2-7475 


7625 


7776 


7929 


8083 


8239 


8397 


8556 


8716 


8878 


26 52 78 104 131 


71 


2-9042 


9208 


9375 


9544 


97 J 4 


9887 


0061 


0237 


0415 


0595 


29 58 87 116 145 


72 


3-0777 


0961 


1146 


1334 


1524 1716 


1910 


2106 


2305 


2506 


32 64 96 129 161 


73 3-2709 


2914 


3122 


3332 


3544 ! 3759 


3977 


4197 4420 


4646 


36 72 108 144 1 80 


74 


3-4 8 74 


5105 


5339 


5576 


5816 6059 


6305 


6554 


6806 


7062 


41 81 122 163 204 


75 


3-7321 


7583 


7848 


8118 


8391 8667 


8947 


9232 


9520 


9812 


46 93 139 1 86 232 


76 


4-0108 


0408 


0713 


IO22 


1335 1653 


1976 


2303 


2635 


2972 




77 


4-33I5 


3662 


4 OI 5 


4374 


4737 5107 


5483 


5864 6252 


6646 




78 


4-7046 


7453 


7867 


8288 


8716 9152 


9594 


0045 6504 


0970 




79 


5-I446 


1929 


2422 


2924 


3435 


3955 


4486 


5026 5578 


6140 




80 


5-67I3 


7297 


7894 


8502 


9124 


9758 0405 


1066 


1742 


2432 




81 


6-3138 


3859 


4596 


535 


6122 


6912 


7720 


8548 


9395 


6264 


Mean differences are 


82 


7' IJ 54 


2066 3002 


3962 


4947 


5958 


6996 


8062 


9158 


0285 


no longer suffici- 


83 


8-1443 


2636 


3863 


5126 


6427 


7769 


9152 


0579 


2052 


3572 


since the differ- 


84 


9-5I4 


9-677 


9-845 


IO-O2 


IO-2O 


10-39 


10-58 


10-78 


10-99 


II-2O 


ences vary con- 


85 


"43 


n-66 


11-91 


12-16 


12-43 


12-71 


13-00 


13-3 


13-62 


13-95 


siderably along 
each line. 


86 


14-30 


14-67 


15-06 


I5-46 


15-89 


16-35 


16-83 


17-34 


17-89 


18-46 




87 


19-08 19-74 20-45 


21-20 


22-02 22-90 


23-86 


24-90 


26-03 


27-27 




88 


28-64 30-14 31-82 


33-69 


35-8o 


38-19 


40-92 


44-07 


47-74 


52-08 




89 


57-29 


63-66 


71-62 


81-85 


95-49 


114-6 


143-2 


191-0 


286-5 


573-0 




90 


00 























MATHEMATICAL TABLES 



TABLE VIII. LOGARITHMIC SINES. 



Q 


0' 
00 


6' 
0-1 


12' 
02 


18' 24' 
03 0-4 


30' 
05 


36' 
06 


42' 
0-7 


48' 
0-8 


54' 
0-9 


Mean Differences. 


1' 2' 3' 4' 5' 





00 


3-2419 


54 2 9 


7190 


8439 


9408 O2OO 


6870 


1450 


1961 




1 


2-2419 


2832 


3210 


3558 


3880 


4 J 79 4459 


4723 


497 1 


5206 




2 


5428 


5640 


5842 


6035 


6220 


6397 


6567 


6731 


6889 


7041 




3 


7188 


7330 


7468 


7602 


773i 


7857 


7979 


8098 


8213 


8326 




4 


8436 


8543 


86 47 


8749 


8849 


8946 


9042 


9i35 


9226 


93i5 


16 32 48 64 80 


5 


9403 


9489 


9573 


9655 


9736 


9816 


9894 


997 


0046 


OI2O 


13 26 39 52 65 


6 1-0192 


0264 


0334 


0403 


0472 


0539 


0605 


0670 


0734 


0797 


ii 22 33 44 55 


7 


0859 


0920 


0981 


1040 


1099 


H57 


1214 


1271 


1326 


I38l 


10 19 29 38 48 


8 


1436 


1489 


1542 


1594 


1646 


1697 


1747 


1797 


1847 


1895 


8 17 25 34 4 2 


9 


1943 


1991 


2038 


2085 


2131 


2176 


2221 


2266 


2310 


2353 


8 15 23 30 38 


10 


2397 


2439 


2482 


2524 


2565 


2606 


26 47 


2687 


2727 


2767 


7 M 20 27 34 


11 


2806 


2845 


2883 


2921 


2959 


2997 


3034 


3070 


3107 


3143 


6 12 19 25 31 


12 


3179 


3214 


3251 


3284 


3319 


3353 


3387 


3421 


3455 


3488 


6 ii 17 23 28 


13 


3521 


3554 


3586 


3618 


3650 


3682 


3713 


3745 


3775 


3806 


5 ii 16 21 26 


14 


3837 


3867 


3897 


3927 


3957 


3986 


4015 


4044 


473 


4IO2 


5 10 15 20 24 


15 


4130 


4158 


4186 


4214 


4242 


4269 


4296 


4323 


4350 


4377 


5 9 14 18 23 


16 


4403 


4430 


4456 


4482 


4508 


4533 


4559 


4584 


4609 


4 6 34 


4 9 13 17 21 


17 


4659 4684 


4709 


4733 


4757 


4781 


4805 


4829 


4853 


4876 


4 8 12 16 20 


18 


4900 


4923 


4946 


4969 


4992 


5015 


5037 


5060 


5082 


5104 


4 8 ii 15 19 


19 


5126 


5H8 


5I7 


5192 


5213 


5235 


5256 


5278 


5299 


5320 


4 7 ii 14 18 


20 


5341 


536i 


5382 


5402 


5423 


5443 


5463 


5484 


5504 


5523 


3 7 10 14 17 


21 


5543 


5563 


5583 


5602 


5621 


5641 


5660 


5679 


5698 


5717 


3 6 10 13 16 


22 


5736 


5754 


5773 


5792 


5810 


5828 


5847 


5865 


5883 


59oi 


3 6 9 12 15 


23 -5919 


5937 


5954 


5972 


5990 


6007 


6024 


6042 


6059 


6076 


3 6 9 12 15 


24 -6093 6110 


6127 


6144 


6161 


6177 


6194 


6210 


6227 


6243 


3 6 8 ii 14 


25 


6259 6276 


6292 


6308 


6324 


6340 


6356 


6371 


6387 


6403 


3 5 8 ii 13 


26 


6418 6434 


6449 


6465 


6480 


6495 


6510 


6526 


6541 


6556 


3 5 8 10 13 


27 


6570 6585 6600 


6615 


6629 


6644 


6659 


6673 


6687 


6702 


2 5 7 10 12 


28 


6716 6730 ! 6744 


6759 


6 773 


6787 


6801 


6814 


6828 


6842 


2 5 7 9 12 


29 


6856 6869 6883 


6896 


6910 


6923 


6937 


6950 


6963 


6977 


2 4 7 9 ii 


30 


6990 7003 


7016 


7029 


7042 


755 


7068 


7080 


793 


7106 


2 4 6 9 ii 


31 


7118' 7131 


7M4 


7156 


7168 


7181 


7193 


7205 


7218 


7230 


2 4 6 8 10 


32 


7242 7254 


7266 


7278 


7290 


7302 


7314 


7326 


7338 


7349 


2 4 6 8 10 


33 


736i ' 7373 


7384 


7396 


7407 


7419 


7430 


7442 


7453 


7464 


2 4 6 8 10 


34; -7476 | 7487 


7498 


7509 


7520 


753i 


7542 


7553 


7564 


7575 


24679 


35 


7586 


7597 


7607 


7618 


7629 


7640 


7650 


7661 


7671 


7682 


24579 


36 


7692 


773 


7713 


7723 


7734 


7744 


7754 


7764 


7774 


7785 


23579 


37 


'7795 


7805 


7815 


7825 


7835 


7844 


7854 


7864 


7874 


7884 


23578 


38 


7893 


7903 


7913 


7922 


7932 


794i 


7951 


7960 


797 


7979 


23568 


39 


7989 


7998 


8007 


8017 


8026 


8035 


8044 


8053 


8063 


8072 


23568 


40 


8081 


8090 


8099 


8108 


8117 


8125 


8i34 


8i43 


8152 


8161 


13467 


41 


8169 


8178 


8187 


8i95 


8204 


8213 


8221 


8230 


8238 


8247 


13467 


42 


8255 


8264 


8272 


8280 


8289 


8297 


8305 


8313 


8322 


8330 


13467 


43 -8338 


8346 


8354 


8362 


8370 


8378 


8386 


8394 


8402 


8410 


13457 


44 -8418 


8426 


8433 


8441 


8449 


8457 


8464 


8472 


8480 


8487 


13456 


45 -8495 


8502 


8510 


8517 


8525 


8532 


8540 


8547 


8555 


8562 


12456 


1 





















MATHEMATICAL TABLES 



411 



TABLE VIII. (contd.) 



E 


0' 


6' 


12' 


18' 


24' 


30' 


36' 


42' 


48' 


54' 


Mean Differences. 


t 
Q 


00 


0-1 


02 


03 


04 


05 


06 


07 


08 


09 


1' 2' 3' 4' 5' 


45 


^8495 


8502 


8510 


8517 


8525 


8532 


8540 


8547 


8555 


8562 


12456 


46 


8569 


8577 


8584 


8591 


8598 


8606 


9613 


8620 


8627 


8634 


12456 


47 


8641 


8648 


8655 


8662 


8669 


8676 


8683 


8690 


8697 


8704 


1 2 3 5 6 


48 


8711 


8718 


8724 


8731 


8738 


8745 


8751 


8758 


8765 


8771 


2 346 


49 


8778 


8784 


8791 


8797 


8804 


8810 


8817 


8823 


8830 


8836 


2 345 


50 


8843 


8849 


8855 


8862 


8868 


8874 


8880 


8887 


8893 


8899 


2345 


51 


8905 


8911 


8917 


8923 


8929 


8935 


8941 


8947 


8953 


8959 


2345 


52 


8965 


8971 


8977 


8983 


8989 


8995 


9000 


9006 


9012 


9018 


2345 


53 


9023 


9029 


9035 


9041 


9046 


9052 


9057 


9063 


9069 


9074 


2345 


54 


9080 


9085 


9091 


9096 


9101 


9107 


9112 


9118 


9123 


9128 


2345 


55 


9134 


9139 


9144 


9149 


9155 


9160 


9165 


9170 


9175 


9181 


2334 


56 


9186 


9191 


9196 


9201 


9206 


9211 


9216 


9221 


9226 


9231 


2334 


57 


9236 


9241 


9246 


9251 


9255 


9260 


9265 


9270 


9275 


9279 


2234 


58 


9284 


9289 


9294 


9-298 


9303 


9308 


9312 


9317 


9322 


9326 


2234 


5 S 


933i 


9335 


9340 


9344 


9349 


9353 


9358 


9362 


9367 


937 1 


1234 


60 


'9375 


9380 


9384 


9388 


9393 


9397 


9401 


9406 


9410 


9414 


1234 


61 


9418 


9422 


9427 


9431 


9435 


9439 


9443 


9447 


945i 


9455 


1233 


62 


'9459 


9463 


9467 


9471 


9475 


9479 


9483 


9487 


9491 


9495 


233 


63 


'9499 


9503 


9507 


95io 


95H 


95i8 


9522 


9525 


9529 


9533 


233 


64 


'9537 


9540 


9544 


9548 


955i 


9555 


9558 


9562 


9566 


9569 


223 


65 


"9573 


9576 


958o 


9583 


9587 


9590 


9594 


9597 


9601 


9604 


223 


66 


9607 


9611 


9614 


9617 


9621 


9624 


9627 


9631 


9634 


9637 


i 223 


67 


9640 


9643 


9647 


9650 


9653 


9656 


9659 


9662 


9666 


9669 


i 223 


68 


9672 


9675 


9678 


9681 


9684 


9687 


9690 


9693 


9696 


9699 


I I 2 2 


69 


9702 


9704 


9707 


9710 


97*3 


9716 


9719 


9722 


9724 


9727 


O I I 2 2 


70 


'973 


9733 


9735 


9738 


9741 


9743 


9746 


9749 


975i 


9754 


O I I 2 2 


71 


9757 


9759 


9762 


9764 


9767 


977 


9772 


9775 


9777 


9780 


I I 2 2 


72 


9782 


9785 


9787 


9789 


9792 


9794 


9797 


9799 


9801 


9804 


I I 2 2 


73 


9806 


9808 


9811 


9813 


9815 


9817 


9820 


9822 


9824 


9826 


O I I 2 2 


74 


9828 


9831 


9833 


9835 


9837 


9839 


9841 


9843 


9845 


9847 


I I I 2 


75 


'9849 


9851 


9853 


9855 


9857 


9859 


9861 


9863 


9865 


9867 


I I I 2 


76 


9869 


9871 


9873 


9875 


9876 


9878 


9880 


9882 


9884 


9885 


O I I I 2 


77 


9887 


9889 


9891 


9892 


9894 


9896 


9897 


9899 


9901 


9902 


I I I I 


78 


9904 


9906 


9907 


9909 


9910 


9912 


9913 


9915 


9916 


9918 


O I I I I 


79 


9919 


9921 


9922 


9924 


9925 


9927 


9928 


9929 


993i 


9932 


O I I I 


80 


'9934 


9935 


9936 


9937 


9939 


994 


9941 


9943 


9944 


9945 


I I I 


81 


9946 


9947 


9949 


9950 


9951 


9952 


9953 


9954 


9955 


9956 


O O I I I 


82 


9958 


9959 


9960 


9961 


9962 


9963 


9964 


9965 


9966 


9967 


O I I I 


83 


9968 


9968 


9969 


997 


9971 


9972 


9973 


9974 


9975 


9975 


O O O I I 


84 


9976 


9977 


997 s 


997 s 


9979 


9980 


9981 


998i 


9982 


9983 


O O O I 


85 


9983 


9984 


9985 


9985 


9986 


9987 


9987 


9988 


9988 


9989 


O O O O 


86 


9989 


9990 


9990 


9991 


9991 


9992 


9992 


9993 


9993 


9994 


O O O O O 


87 


'9994 


9994 


9995 


9995 


9996 


9996 


9996 


9996 


9997 


9997 


O O O O O 


88 


'9997 


9998 


9998 


9998 


9998 


9999 


9999 


9999 


9999 


9999 


O O O O O 


89 


9999 


9999 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


O O O O O 


90 


O'OOOO 























4T2 



MATHEMATICAL TABLES 



TABLE IX. LOGARITHMIC COSINES 



8 

& 


0' 


6' 


12' 


18' 


24' 


30' 

j 


36' 


42' 


48' 


54' 


Mean Differences. 


of 



00 


0-1 


0-2 


03 


04 


05 


0-6 


0-7 


08 


09 


1' 2' 3' 4' 5' 





o-oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


oooo 


9999 


O O O O 


1 


1-9999 


9999 


9999 


9999 


9999 


9999 


9998 


9998 


9998 


9998 


o o o o o 


2 


9997 


9997 


9997 


9996 


9996 


9996 


9996 


9995 


9995 


9994 


O O O O O 


3 


9994 


9994 


9993 


9993 


9992 


9992 


9991 


9991 


9990 


9990 


O O O 


4 


9989 


9989 


9988 


9988 


9987 


9987 


9986 


9985 


9985 


9984 


o o o o o 


5 


9983 


9983 


9982 


9981 


9981 


9980 


9979 


9978 


997 s 


9977 


O O O I 


6 


9976 


9975 


9975 


9974 


9973 


9972 


9971 


9970 


9969 


9968 


O O I I 


7 


9968 


9967 


9966 


9965 


9964 


9963 


9962 


9961 


9960 


9959 


O I I I 


8 


995 8 


9956 


9955 


9954 


9953 


9952 


9951 


9950 


9949 


9947 


I I I 


9 


9946 


9945 


9944 


9943 


9941 


9940 


9939 


9937 


9936 


9935 


O O I I I 


10 


9934 


9932 


9931 


9929 


9928 


9927 


9925 


9924 


9922 


9921 


I I I 


11 


9919 


9918 


9916 


9915 


9913 


9912 


9910 


9909 


9907 


9906 


O I I I I 


12 


9904 


9902 


9901 


9899 


9897 


9896 


9894 


9892 


9891 


g88q 


I I I I 


13 


9887 


9885 


9884 


9882 


9880 


9878 


9876 


9875 


9873 


9871 


I I I 2 


14 


9869 


9867 


9865 


9863 


9861 


9859 


9857 


9855 


9853 


9851 


I I I 2 


15 


9849 


9847 


9845 


9843 


9841 


9839 


9837 


9835 


9833 


9831 


I I I 2 


16 


9828 


9826 


9824 


9822 


9820 


9817 


9815 


9813 


9811 


9808 


I I 2 2 


17 


9806 


9804 


9801 


9799 


9797 


9794 


9792 


9789 


9787 


9785 


O I I 2 2 


18 


9782 


9780 


9777 


9775 


9772 


9770 


9767 


9764 


9762 


9759 


I I 2 2 


19 


9757 


9754 


975i 


9749 


9746 


9745 


974 1 


9738 


9735 


9733 


I I 2 2 


20 


9730 


9727 


9724 


9722 


9719 


9716 


9713 


9710 


9707 


9704 


I I 2 2 


21 


9702 


9699 


9696 


9693 


9690 


9687 


9684 


9681 


9678 


9675 


I I 2 2 


22 


9672 


9669 


9666 


9662 


9659 


9656 


9653 


9650 


9647 


9643 


II223 


23 


9640 


9637 


9634 


9631 


9627 


9624 


9621 


9617 


9614 


9611 


II223 


24 


9607 


9604 


9601 


9597 


9594 


9590 


9587 


9583 


958o 


9576 


II223 


25 


9573 


9569 


9566 


9562 


9558 


9555 


955i 


9548 


9544 


9540 


II223 


26 


9537 


9533 


9529 


9525 


9522 


95i8 


9514 


95io 


9507 


9503 


1 i 2 3 3 


27 


9499 


9495 


9491 


9487 


9483 


9479 


9475 


9471 


9467 


94 6 3 


II233 


28 


9459 


9455 


945i 


9447 


9443 


9439 


9435 


9431 


9427 


9422 


1 i 2 3 3 


29 


9-1 1 8 


9414 


9410 


9406 


9401 


9397 


9393 


9388 


9384 


9380 


11.234 


30 


9375 


9371 


9367 


9362 


9358 


9353 


9349 


9344 


934 


9335 


II234 


31 


9331 


9326 


9322 


9317 


9312 


9308 


9303 


9298 


9294 


9289 


12234 


32 


9284 


0279 


9275 


9270 


9265 


9260 


9255 


9251 


9246 


9241 


12234 


33 


9236 


9231 


9226 


9221 


9216 


9211 


9206 


9201 


9196 


9191 


2 334 


34 


9186 


9181 


9175 


9170 


9165 


9160 


9155 


9149 


9144 


9139 


2 334 


35 


9134 


9128 


9123 


9118 


9112 


9107 


9101 


9096 


9091 


9085 


2345 


36 


9080 


9074 


9069 


9063 


9057 


9052 


9046 


9041 


9035 


9029 


2 345 


37 


9023 


9018 


9012 


9006 


9000 


8995 


8989 


8983 


8977 


8971 


2 345 


38 


8965 


8959 


8953 


8947 


8941 


8935 


8929 


8923 


8917 


8911 


2345 


39 


8905 


8899 


8893 


8887 


8880 


8874 


8868 


8862 


8855 


8849 


1 2 3 4 5 


40 


8843 


8836 


8830 


8823 


8817 


8810 


8804 


8797| 


8791 


8784 


1 2 3 4 5 


41 


8778 


8771 


8765 


8758 


8751 


8745 


8738 


8731 


8724 


8718 


12356 


42 


8711 


8704 


8697 


8690 


8683 


8676 


8669 


8662 


8655 


8648 


1 2 3 5 6 


43 


8641 


8634 


8627 


8620 


8613 


8606 


8598 


8591 


.8584 


8577 


12456 


44 


8569 


8562 


8555 


8547 


8540 


8532 


8525 


8517 


8510 


8502 


12456 


45 


8495 


8487 


8480 


8472 


8464 


8457 


8449 


8441 


8433 


8426 


13456 



























MATHEMATICAL TABLES 



413 



TABLE IX (contd.) 



V 

1 

Q 


0' 
00 


6' 
01 


12' 18' 
0-2 03 


24' 
0-4 


30' 
05 


36' 42' 
06 0-7 


48' 
08 


54' 
0-9 


Mean Differences. 


1' 2' 3' 4' 5' 


45 


1-8495 8487 


8480 8472 8464 


8457 


8449 8441 


8433 


8426 


13456 


46 


8418 


8410 


8402 


8394 9386 


8378 


8370 8362 


8354 


8346 


13457 


47 -8338 


8330 


8322 


8313 8305 


8297 


8289 8280 


8272 


8264 


1 3 4 6 7 


48 -8255 8247 


8238 


8230 8221 


8213 


8204 8195 


8187 


8178 


1 3 4 6 7 


49 -8169 8161 


8152 


8143 8134 8125 


8117 


8108 


8099 


8090 


13467 


50 


8081 


8072 


8063 


8053 


8044 


8035 


8026 


8017 


8007 


7998 


23568 


51 


7989 


7979 


7970 


7960 


795i 


794i 


7932 


7922 


79i3 


7903 


23568 


52 


7893 


7884 


7874 


7864 


7854 


7844 


7835 


7825 


7815 


7805 


23578 


53 


7795 


7785 


7774 


7764 


7754 


7744 


7734 


7723 


7713 : 7703 


23579 


54 


7692 


7682 


7671 


7661 


7650 


7640 


7629 


7618 


7607 7597 


24579 


55 


7586 


7575 


7564 


7553 


7542 


7531 


7520 


7509 


7498 


7487 


24679 


56 


7476 


7464 


7453 


7442 


7430 


7419 


7407 


7396 


7384 


7373 


246 8 10 


57 


7361 


7349 


7338 


7326 


7314 


7302 


7290 


7278 


7266 


7254 


246 8 10 


58 


7242 


7230 


7218 


7205 


7*93 


7181 


7168 


7156 


7H4 7i3i 


246 8 10 


59 


7118 


7106 


793 


7080 


7068 


7055 


7042 


7029 


7016 7003 


2 4 6 9 II 


60 


6990 


6977 


6963 


6950 


6937 


6923 


6910 


6896 


6883 ' 6869 


247 9 II 


61 


6856 6842 i 6828 


6814 


6801 


6787 


6773 


6759 


6744 


6730 


2 5 7 9 12 


62 


6716 6702 


6687 


6673 


6659 


6644 


6629 


6615 


6600 


6585 


2 5 7 10 21 


63 


6570 6556 


6541 


6526 


6510 


6495 


6480 


6465 6449 


6434 


3 5 8 10 13 


64 


6418 6403 


6387 


637 1 


6356 


6340 


6324 


6308 


6292 


6276 


3 5 8 it 13 


65 


6259 6243 


6227 


6210 


6194 


6177 


6161 


6144 


6127 


6110 


3 6 8 ii 14 


66 


6093 6076 


6059 


6042 


6024 


6007 


5990 


597 2 


5954 


5937 


3 6 9 12 15 


67 


5919 5901 


5883 


5865 


5847 


5828 


5810 


5792 


5773 


5754 


3 6 9 12 15 


68 


5736 5717 


5698 


5679 


5660 5641 


5621 


5602 


5583 


5563 


3 6 10 13 16 


69 


5543 


5523 


5504 


5484 


5463 5443 


54 2 3 


5402 


5382 


536i 


3 7 10 14 17 


70 


534 1 


5320 


5299 


5278 


5256 


5235 


5213 


5192 


5170 


5M8 


4 7 ii 14 18 


71 


5162 


5 I0 4 


5082 


5060 


5037 


5015 


4992 


4969 


4946 


4923 


4 8 ii 15 19 


72 


4900 


4876 


4853 


4829 


4805 4781 


4757 


4733 


4709 


4684 


4 8 12 16 20 


73 


4659 


4 6 34 


4609 


4584 


4559 4533 


4508 


4482 


4456 


4430 


4 9 13 17 21 


74 


443 


4377 


4350 


4323 


4296 


4269 


4242 


4214 


4186 


4158 


5 9 14 18 23 


75 


4130 


4102 


473 


4044 


4i5 


3986 


3957 


3927 


3897 


3867 


5 10 15 20 24 


76 


3837 


3806 


3775 


3745 


3713 3682 


3650 


3618 


3586 


3554 


5 ii 16 21 26 


77 


3521 


3488 


3455 


3421 


3387 3353 


3319 


3284 


3250 


3214 


6 it 17 23 28 


78 


3179 


3143 


3io/ 


3070 


3034 2997 


2959 


2921 


2883 


2845 


6 12 19 25 31 


79 


2806 


2767 


2727 


2687 


2647 


2606 


2565 


2524 


2482 


2439 


7 14 20 27 34 


80 


2397 


2353 


2310 


2266 


2221 


2176 


2131 


2085 


2038 


1991 


8 15 23 30 38 


81 


1943 


1895 


1847 


1797 


J 747 


1697 


1646 


1594 


1542 


1489 


8 i? 25 34 42 


82 


1436 


1381 


1326 


1271 


1214 


"57 


1099 


1040 


0981 


0920 


10 19 29 38 48 


83 


0859 


0797 


734 


0670 


0605 


539 


0472 


0403 


0334 


0264 


ii 22 33 44 55 


84 


0192 


OI2O 


0046 


9970 


9894 


9816 


9736 


9655 


9573 


9489 


13 26 39 52. 65 


85 


2-9403 


9315 


9226 


9135 


9042 


8946 


8849 


8749 


8647 


8543 


16 32 48 64 80 


86 


8436 


8326 


8213 


8098 


7979 


7857 


7731 


7602 


7468 


733 




87 -7188 


7041 


6889 


6731 


6567 


6397 


6220 


6035 


5842 


5640 




88 -5428 


52O6 


4971 


4723 


4459 


4*79 


3880 


3558 


3210 


2832 




89 


2419 


I96l 


M50 


0870 


O200 


9408 


8439 


7190 


5429 


2419 




90 


CO 























414 



MATHEMATICAL TABLES 
TABLE X. LOGARITHMIC TANGENTS. 



o 


0' 
00 


6' 
01 


12' 
02 


18' 
0-3 


24' 
04 


30' 
05 


38' 
06 


42' 
0-7 


48' 
08 


54' 
0-9 


Mean Differences. 


1' 2' 3' 4' 5' 





00 


3-2419 


5429 


7190 


8439 9409 


02 oo 


0870 


M5o 


1962 




1 


2-2419 


2833 


3211 


3559 


3881 4181 


4461 


4725 


4973 


5208 




2 


5431 


5643 


5845 


6038 


6223 6401 


657 1 


6736 


6894 


7046 




3 


7194 


7337 


7475 


7609 


7739 


7865 


7988 


8107 


8223 


8336 




4 


8446 


8554 


8659 


8762 


8862 


8960 


9056 


915 


9241 


933i 


16 32 48 64 81 


5 


9420 


9506 


959i 


9674 


9756 


9836 


9915 


9992 


0068 


0143 


13 26 40 53 66 


6 


1-0216 


0289 


0360 


0430 


0499 


0567 


0633 


0699 


0764 


0828 


ii 22 34 45 56 


7 


0891 


0954 


1015 


1076 


H35 


1194 


1252 


1310 


1367 


1423 


10 20 29 39 49 


8 


1478 


1533 


1587 


1640 


1693 


1745 


1797 


1848 


1898 


1948 


9 17 26 35 43 


9 


1997 


2046 


2094 


2142 


2189 


2236 


2282 


2328 


2 374 


2419 


8 16 23 31 39 


10 


2463 


2507 


2551 


2594 


2637 


2680 


2722 


2764 


2805 


2846 


7 14 21 28 35 


11 


2887 


2927 


2967 


3006 


3046 


3085 


3123 


3162 


3200 


3237 


6 13 19 26 32 


12 


3275 


3312 


3349 


3385 


3422 


3458 


3493 


3529 


3564 


3599 


6 12 18 24 30 


13 


3634 


3668 


3702 


3736 


3770 


3804 


3837 


3870 


3903 


3935 


6 ii 17 22 28 


14 


3968 


4000 


4032 


4064 


495 


4127 


4158 


4189 


4220 


4250 


5 10 16 21 26 


15 


4281 


43ii 


4341 


4371 


4400 


4430 


4459 


4488 


4517 


4546 


5 10 15 20 25 


16 


4575 


4603 


4632 


4660 


4688 


4716 


4744 


477i 


4799 


4826 


5 9 14 19 23 


17 


4853 4880 


4907 


4934 


4961 


4987 


5014 


5040 


5066 


5092 


4 9 13 18 22 


18 


5118 5143 


5169 


5195 


5220 


5245 


5270 


5295 


5320 


5345 


4 8 13 17 21 


19 


537 5394 


5419 


5443 


5467 


549i 


55i6 


5539 


5563 


5587 


4 8 12 16 20 


20 


5611 


564 


5658 


5681 


5704 


5727 


5750 


5773 


5796 


5819 


4 8 12 15 19 


21 


5842 


5864 


5887 


5909 


5932 


5954 


5976 


5998 


6020 


6042 


4 7 ii 15 19 


22 


6064 


6086 


6108 


6129 


6151 


6172 


6194 


6215 


6236 


6257 


4 7 ii 14 18 


23 


6279 


6300 


6321 


6341 


6362 


6383 


6404 


6424 


6 445 


6465 


3 7 10 14 17 


24 


6486 


6506 


6527 


6547 


6567 


6587 


6607 


6627 


6647 


6667 


3 7 10 13 17 


25 


6687 


6706 


6726 


6746 


6765 


6785 


6804 


6824 


6843 


6863 


3 7 10 13 16 


26 


6882 


6901 


6920 


6939 


6958 


6977 


6996 


7i5 


734 


753 


3 6 9 13 16 


27 


7072 


7090 


7109 


7128 


7146 


7^5 


7183 


7202 


7220 


7238 


3 6 9 12 15 


28 


7257 


7275 


7293 


73H 


7330 


7348 


7366 


7384 


7402 


7420 


3 6 9 12 15 


29 


7438 


7455 


7473 


749i 


7509 


7526 


7544 


7562 


7579 


7597 


3 6 9 12 15 


30 


7614 


7632 


7649 


7667 


7684 


7701 


7719 


7736 


7753 


7771 


3 6 9 12 14 


31 


7788 


7805 


7822 


7839 


7856 


7873 


7890 


7907 


7924 


794i 


3 6 9 ii 14 


32 


7958 


7975 


7992 


8008 


8025 


8042 


8059 


8075 


8092 


8109 


3 6 8 ii 14 


33 


8125 


8142 


8158 


8i?5 


8191 


8208 


8224 


8241 


8257 


8274 


3 5 8 n 14 


84 


8290 


8306 


8323 


8339 


8355 


8371 


8388 


8404 


8420 


8436 


3 5 8 n 14 


35 


8452 


8468 


8484 


8501 


8517 


8533 


8549 


8565 


8581 


8597 


3 5 8 ii 13 


36 


8613 


8629 


8644 


8660 


8676 


8692 


8708 


8724 


8740 


8755 


3 5 8 ii 13 


37 


8771 


8787 


8803 


8818 


8834 


8850 


8865 


8881 


8897 


8912 


3 5 8 10 13 


38 


8928 


8944 


8959 


8975 


8990 


9006 


9022 


9037 


9053 


9068 


3 5 8 10 13 


39 


9084 


9099 


9H5 


9130 


9146 


9161 


9176 


9192 


9207 


9223 


3 5 8 10 13 


40 


9238 


9254 


9269 


9284 


9300 


9315 


9330 


9346 


936i 


937 


3 5 8 10 13 


41 


9392 


9407 


9422 


9438 


9453 


9468 


9483 


9499 


9514 


9529 


3 5 8 10 13 


42 


9544 


9560 


9575 


9590 


9605 


9621 


9636 


9651 


9666 


9681 


3 5 8 10 13 


43 


9697 


9712 


9727 


9742 


9757 


9773 


9788 


9803 


9818 


9833 


3 5 8 10 13 


44 


9848 


9864 


9879 


9894 


9909 


9924 


9939 


9955 


997 


9985 


3 5 8 10 13 


45 


o-oooo 


0015 


0030 


0045 


0061 


0076 


0091 


0106 


0121 


0136 


3 5 8 10 13 



MATHEMATICAL TABLES 



TABLE X. (contd.) 



g> 
Q 


0' 
00 


6' 
01 


12' 18' 
02 03 


24' 30' 
04 05 


36' 
06 


42' 
0-7 


48' 
08 


54' 
09 


Mean D;ffer;uces. 


1' 2' 3' 4' 5' 


45 


oooo 0015 0030 0045 0061 


0076 


0091 


OIO6 OI2I OI36 


35 8 10 13 


46 


0152 


0167 


Ol82 0197 O2I2 


0228 


0243 


0258 


0273 0288 


35 8 10 13 


47 


0303 


0319 


0334 0349 0364 


379 


395 


0410 


0425 0440 


35 8 10 13 


48 


0456 0471 


0486 0501 0517 


0532 


0547 


0562 


0578 0593 


35 8 10 13 


49 


0608 0624 


0639 0654 0670 


0685 


0700 


0716 


0731 0746 


35 8 10 13 


50 


0762 0777 


0793 0808 ; 0824 


0839 


0854 


0870 


0885 0901 


35 8 10 13 


51 


0916 


0932 


0947 0963 0978 


0994 


IOIO 


IO25 


1041 1056 


35 8 10 13 


52 


1072 


1088 


1103 nig 1135 


1150 


1166 


1182 


1197 1213 


35 8 10 13 


53 


1229 


1245 


1260 1276 1292 


1308 


1324 


1340 


1356 I37 1 


35 8 ii 13 


54 


1387 


1403 


1419 


H35 


1451 


1467 


1483 


1490 


1516 1532 


35 8 ii 13 


55 


1548 


1564 


1580 


1596 


1612 


1629 


1645 


1661 


1677 1694 


35 8 ii 14 


56 


1710 


1726 


1743 ! 1759 


1776 


1792 


1809 


1825 


1842 1858 


35 8 ii 14 


57 


1875 


1891 


1908 1925 


1941 


1958 


1975 


1992 


2008 2025 


36 8 ii 14 


58 


2042 


2059 


2076 2093 


2110 


2127 


2144 


2161 


2178 


2195 


3 6 9 ii 14 


59 


2212 


2229 


2247 


2264 


228l 


2299 


2316 


2 333 


2351 


2368 


3 6 9 12 14 


60 


2386 


2403 


2421 


2438 


2456 


2474 


2491 


2509 


2527 


2545 


3 6 9 12 15 


61 


2562 


2580 


2598 


2616 


2634 


2652 


2670 


2689 2707 


2725 


3 6 9 12 15 


62 


2743 


2762 


2780 


2798 


2817 


2835 


2854 


2872 


2891 


2910 


3 6 9 12 15 


63 


2928 


2947 


2966 


2985 


3004 


3023 


3042 


3061 


3080 3099 


36 9 13 16 


64 


3"8 


3i37 3i57 


3176 3196 


3215 


3235 


3254 


3274 3294 


3 6 10 13 16 


65 


3313 


3333 3353 


3373 3393 


3413 


3433 


3453 


3473 


3494 


3 7 I0 13 17 


66 


35H 


3535 3555 


357 6 


3596 


3617 


3638 


3659 


3679 


3700 


3 7 10 14 17 


67 


3721 


3743 3764 


3785 


3806 


3828 


3849 


3871 


3892 


3914 


4 7 ii 14 18 


68 


3936 


3958 398o 


4002 


4024 


4046 


4068 4091 


4"3 


4136 


4 7 ii 15 19 


69 


4158 


4181 4204 


4227 


4250 


4273 


4296 4319 


4342 


4366 


4 8 12 15 19 


70 


4389 


44U ' 4437 


4461 


4484 


4509 


4533 4557 


458i 


4606 


4 8 12 16 20 


71 


4630 


4 6 55 


4680 


475 


4730 


4755 


4780 


4805 


4831 


4857 


4 8 13 17 21 


72 


4882 


4908 


4934 


4960 


4986 


5013 


5039 5066 


5093 


5120 


4 9 13 18 22 


73 


5M7 


5174 


5201 


5229 


5256 


5284 


5312 5340 


5368 


5397 


5 9 14 19 23 


74 


54 2 5 


5454 


5483 


5512 


5541 


5570 


5600 5629 


5659 


5689 


5 10 15 20 25 


75 


5719 


5750 


578o 


5811 


5842 


5873 


5905 


5936 


5968 


6000 


5 10 16 21 26 


76 


6032 


6065 


6097 


6130 


6l6 3 


6196 


6230 


6264 


6298 


6332 


6 II 17 22 28 


77 


6366 


6401 


6436 


6471 


6507 


6542 


6578 


6615 


6651 


6688 


6 12 18 24 30 


78 


6725 


6763 


6800 


6838 


6877 


6915 


6954 


6994 


7033 


773 


6 13 19 26 32 


79 


7"3 


7154 


7*95 


7236 


7278 


7320 


7363 


7406 


7449 


7493 


7 14 21 28 35 


80 


7537 


758i 


7626 


7672 


77 l8 


7764 


7811 


7858 


7906 


7954 


8 16 23 31 39 


81 


8003 


8052 


8102 


8152 


8203 


8255 


8307 


8360 


8413 


8467 


9 17 26 35 43 


82 


8522 


8577 


8633 


8690 


8748 


8806 


8865 


8924 


8985 


9046 


10 20 29 39 49 


83 


9109 


9172 


9236 


9301 


9367 


9433 


9501 


9570 


9640 


97" 


ii 22 34 45 56 


84 


9784 


9857 


9932 


0008 


0085 


0164 


0244 


0326 


0409 


0494 


13 26 40 53 66 


85 


1-0580 


0669 


0759 


0850 


0944 


1040 


1138 


1238 


1341 


1446 


16 32 48 64 81 


86 


I-I554 


1664 


1777 


1893 


2OI2 


2135 


2261 


2391 


2525 


2663 




87 


1-2806 


2954 


3106 


3264 


3429 


3599 


3777 


3962 


4155 


4357 




88 


1-4569 


4792 


5027 5275 


5539 


5819 


6119 


6441 


6789 


7167 




89 


1-7581 


8038 


8550 


9130 


9800 


0591 


1561 


2810 


457 1 


758i 




90 


-00 























4i 6 



MATHEMATICAL TABLES 



TABLE XI. EXPONENTIAL AND HYPERBOLIC FUNCTIONS 



X 


e x 


6-x 


cosh x 
e x +e-* 


sink x 
e x e~ x 


tanh x 
_e*-e~* 


2 


2 


e x +e~* 


1 


1-1052 


9048 


1-0050 


i 002 


0997 


2 


1-2214 


8187 


I-020I 


2013 


1974 


3 


1-3499 


7408 


1-0453 


3045 


2913 


4 


1-4918 


6703 


1-0811 


4108 


3799 


5 


1-6487 


6065 


1-1276 


5211 


4621 


6 


1-8221 


5488 


1-1855 


6367 


537 


7 


2-0138 


4966 


1-2552 


7586 


6044 


8 


2-2255 


4493 


1-3374 


8881 


6640 


9 


2-4596 


4066 


I-433I 


1-0265 


7163 


10 


2-7183 


3679 


I-543I 


1-1752 


7616 


1-1 


3-0042 


3329 


1-6685 


1-3357 


8005 


12 


3-3201 


3012 


1-8107 


1-5095 


8337 


13 


3-6693 


2725 


1-9709 


1-6984 


8617 


1-4 


4-0552 


2466 


2-1509 


1-9043 


8854 


15 


4-4817 


2231 


2-3524 


2-1293 


9051 


1-6 


4-953 


2019 


2-5775 


2-3756 


9217 


17 


5-4739 


1827 


2-8283 


2-6456 


9354 


1-8 


6-0496 


i653 


3-1075 


2-9422 


9468 


19 


6-6859 


1496 


3-4I77 


3-2682 


9563 


20 


7-3891 


1353 


3-7622 


3-6269 


9640 


2-1 


8-1662 


1225 


4 -I 443 


4-0219 


9704 


22 


9-0251 


1108 


4-5679 


4-457 1 


9758 


2-3 


9-9742 


1003 


5-037 2 


4-937 


9801 


2-4 


11-0232 


0907 


5-557 


5-4662 


9837 


25 


12-1825 


0821 


6-1323 


6-0502 


9866 


26 


13-4638 


0743 


6-7690 


6-6947 


9890 


2-7 


14-8797 


0672 


7-4735 


7-4063 


9910 


28 


16-4446 


0608 


8-2527 


8-1919 


9926 


29 


18-1741 


0550 


9-1146 


9-0596 


9940 


30 


20-0855 


0498 


10-068 


10-018 


9951 


3-1 


22-1980 


0450 


11-122 


11-076 


9959 


32 


24-5325 


0408 


12-287 


12-246 


9967 


33 


27-1126 


0369 


13-575 


I3-538 


9973 


34 


29-9641 


0334 


I4-999 


14-965 


9978 


35 


33-II55 


0302 


16-573 


16-543 


9982 


36 


36-5982 


0273 


I8-3I3 


18-285 


9985 


3-7 


4 -4473 


0247 


20-236 


2O-2II 


9988 


3-8 


44-7012 


0224 


22-362 


22-339 


9990 


39 


49-4024 


O2O2 


24-711 


24-691 


9992 


40 


54-5982 


0183 


27-308 


27-290 


9993 


4-1 


60-3403 


0166 


30-178 


30-162 


9995 


4-2 


66-6863 


0150 


33-351 


33-336 


9996 


4-3 


73-6998 


0136 


36-857 


36-843 


9996 


4-4 


81-4509 


OI23 


40-732 


40-719 


9997 


4-5 


90-0171 


01 1 1 


45-014 


45-003 


9997 


4-6 


99-4843 


OIOI 


49-747 


49-737 


9998 


4-7 


109-9472 


0091 


54-978 


54-969 


9998 


4-8 


121-5104 


0082 


60-759 


60-751 


9999 


4-9 


134-2898 


0074 


67-149 


67-141 


9999 


5-0 


148-4132 


0067 


74-210 


74-203 


9999 



INDEX 



Abbreviations, i 

Altitude, 366 

" Ambiguous " case in the solution 

of spherical triangles, 360 
Amsler planimeter, theory of, 266 
Analysis, harmonic, 342 
Anchor ring, moment of inertia of, 256 
Applications of the Calculus, 300 

et seq. 

Applications of Differentiation, 88 
' et seq. 

Applied electricity, examples in, 317 
Arc, length of, 201 
Archimedean spiral, 258 
Area of cardioid, 262 
Areas by polar co-ordinates, 261 
Areas by sum-curve method, 118 
Arithmetic mean, probable error of, 

378 

Arrangement of electric cells, 317 
Azimuth, 367 



Contraflexure, point of, 93 
Cooling curves, 21 
Coradi integraph, 126 
Curvature of an arc, 308 
Cycloid, equation of, 69 



D, the operator, 26, 286 

d?s 

d p , meaning of, 9 

Declination, 367 
Definite integral, 118, 137 
Deflection of muzzle of a gun, 316 
Derivative, 9 
Derived curve, 12 
Differential coefficient, 9 
Differential equations, exact, 279 
-- , homogeneous, 281 
-- , solution of, 270 et seq. 

- of type - = /(*), 271 



B 



Beam problems, 38, 93, 123, 307 
Belt round pulley, tension in, 330 
Bending moment on ship, 126 
Buoyancy, curve of, 125 



Cable, approximate length of, 202 

Calculation of small corrections, 107 

Cardioid, area of, 262 

Catenary, 321 

Centre of Gravity, 211 et seq. 

of irregular solids, 225 

of solids of revolution, 228 
Centre of Pressure, 232, 336 
Centroid, 211 et seq. 

of sections by calculation, 220 

by drawing, 251 

Centroid vertical, determination of, 
by double sum curve method, 
218 

Circular parts, Napier's rules of, 359 
Complementary function, 289 
Compound pendulum, time of oscilla- 
tion of, 324 



dv 



=b > 2 75 



of the second degree, 294 
Differential, total, 82 
Differentiation, applications of, 88 
et seq. 

, graphic, 12 

, logarithmic, 85 
Differentiation of ax n , 26 

exponential functions, 47 

- function of a function, 63 

- hyperbolic functions, 54 

- inverse trigonometric functions, 

76 

- log x, 51 

- product, 70 
quotient, 73 

- trigonometric functions, 56 
Differentiation, partial, 79 
Double integral, 123 

Double sum curve method for fixing 
the centroid vertical, 218 



MATHS. FOR ENG. 
E E 



417 



4i8 



INDEX 



E 



Efficiency of engine working on the 
Rankine cycle, 304 

Electric condenser, time of discharge 
of, 318 

Ellipse, perimeter of, 205 

Elliptic integral, 205 

Entropy of water, 304 

Equations, differential, 270 

Euler's formula for struts, 328 

Exact differential equations, 279 

Expansion in series, 108 

Exponential functions, differentia- 
tion of, 47 

, integration of, 129 



Integration by partial fractions, 146 

by parts, 155 

by substitution, 148 

Integration, graphic, 118 
Integration, meaning of, 115 
Integration of exponential functions 

129 

powers of x, 127 

trigonometric functions, 134 

Interpolation, using Taylor's theorem, 

112 



K 



k, symbol for swing radius, 240 
Kinetic energy, 240 



First moment, 211 

Fixed beams, deflection of, 309 

Fleming's graphic method of finding 

R.M.S. values, 264 
'Footstep bearing, friction in, 331 
Forced vortex, 338 
Fourier's theorem, 342 
Friction on wheel disc in fluid, 335 



Least squares, theorem of, 374 
Leibnitz, 3 
Length of arc, 201 
Length of cable, approximate, 202 
List of integrals, 175 
Logarithmic differentiation, 85 
Logarithmic functions, differentia- 
tion of, 51 
Logarithmic spiral, 258 



Gamma function, 173 
Gauss's error curve, 374 
Goodman scheme for fixed beams, 313 
Governor, problem on, 75 
Graphic differentiation, 12 et seq. 
Graphic integration, 1 1 8 et seq. 
Graphic solution of spherical triangles, 
368 



H 



Harmonic analysis, 342 et seq. 
Harrison's method of harmonic 

analysis, 346 
Homogeneous differential equations, 

281 

Hour angle, 367 
Hydraulics, examples on, 334 et seq. 



Inertia, moment of, 237 et seq. 
Inflexion, point of, 93 
Integral, definite, 118, 137 

, double, 123 

, indefinite, 118, 137 

Integrals, list of, 175 
Integraph, the Coradi, 126 



M 



Maclaurin's theorem, 108 
Maximum and minimum values, 88 
Maximum intensity of shear stress, 

3H 

Maxwell's needle, 250 
Mean spherical candle-power, 262 
Mean values, 180 et seq. 
Modulus of rigidity, determination of, 

325 

Moment of inertia, 237 et seq. 
Moment of inertia of anchor ring, 256 

circle, 246 

compound vibrator, 249 

cylinder, 247 

pulley wheel, 248 

rectangle, 244 

sphere, 250 

Tee section, 245 

Moments, ist and 2nd, by construc- 
tion, 251 
Muzzle of gun, deflection of, 316 

N 

Napier's rules of circular parts, 359 

Neutral axis, 238 

Newton, 3 

Notch, triangular, 334 



INDEX 



419 







Oblate spheroid, volume of, 200 



Parallel axis theorem, 241 

Partial differentiation, 79 

Pendulum, time of swing of, 324 

Perimeter of ellipse, 205 

Perpendicular axes theorem, 243 

Planimeter, theory of, 266 

Point of inflexion, 93 

Polar co-ordinates, 257 et seq. 

Pressure, centre of, 232, 336 

Probability, 370 

Probability of error, 372 

Probable error of arithmetic mean, 

378 
Prolate spheroid, volume of, 200 



K 



Radius of gyration, 240 

Rankine cycle, efficiency of, 304 

Reduction formulae, 163 

Right-angled spherical triangles, solu- 
tion of, 358 

Root mean square values, 188 et seq. 

, Fleming's gra- 
phic method for, 264 

Rousseau diagram, 262 



Strength of materials, examples on, 

321 et seq. 
Stresses in thick cylinders, 325 

spherical shells, 327 

Struts, formulae for, 328 
Sub-normal, length of, 42 
Sub-tangent, length of, 42 
Sum curve, 119 

Surface of solid of revolution, 208 
Swing radius, 240 



Taylor's theorem, 108 

Tee section, centroid of, 220 

, moment of inertia of, 245 

Thermodynamics, examples in, 300 
et seq. 

Thick cylinders, stresses in, 325 

Time to empty a tank, 334 

Total differential, 82 

Tractrix, 333 

Transition curve, 338 

Triangle, spherical, 355 

, solution of, 357 

Triangular notch, measurement of 
flow by, 324 

Trigonometric functions, differentia- 
tion of, 56 

, integration of, 134 

Trigonometry, spherical, 355 



Schiele pivot, 333 

Second moment, 211 

Shear stress in beams, 314 

Simple harmonic motion, 60 

Simpson's rule, proof of, 141 

" Sine " rule for spherical triangles, 

. 357 
Solid of revolution, centre of gravity 

of, 228 

Solid of revolution, volume of, 195 
Solution of right-angled spherical 

triangles, 358 

Solution of spherical triangles, 357 
Sphere, moment of inertia of, 250 

, volume of zone of, 200 

Spherical excess, 356 
Spherical triangle, 355 
Spherical trigonometry, 355 et seq. 
Spheroid, volume of, 200 
Spiral, Archimedean, 258 
, logarithmic, 258 



Values, mean, 180 

, root mean square, 188 

Velocity of piston, 65 
Volume of oblate spheroid, 200 

prolate spheroid, 200 

solid of revolution, 195 

Vortex, forced, 338 



W 



Wattless current, 187 

Wheel disc in fluid, friction on, 335 

Work done in complete theoretical 

cycle, 303 
expansion of a gas, 302 



Zenith, 366 

Zone of sphere, volume of, 200 







s 

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