Calculus
FOR Engineers
%
Tr.i.cJu o^^jr r.i^
LIBRARY
OF THE
UNIVERSITY OF CALIFORNIA.
aass
CALCULUS FOE ENGINEEES,
QtambxiUQt:
PRINTED BY J. AND C. F. CLAY,
AT THE UNIVERSITY PRESS.
THE
CALCULUS FOE ENGINEEES
BY
JOHN PERRY, M.E, RSc, F.R.S.
WH. SCH., ASSOC. MEMB. INST. C.E.,
PROFESSOR OF MECHANICS AND MATHEMATICS IN THE ROYAL COLLEGE OF SCIENCE,
LONDON: PRESIDENT OF "IBan^Xtsiil^^O^ OF ELECTRICAL ENGINEERS.
FIFTH IMPRESSION.
Honlron :
EDWARD ARNOLD,
37, BEDFORD STREET.
lAll Rights reserved,]
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PEBFACB.
This book describes what has for many years been the
most important part of the regular course in the Calculus
for Mechanical and Electrical Engineering students at the
Finsbury Technical College. It was supplemented by easy
work involving Fourier, Spherical Harmonic, and Bessel
Functions which I have been afraid to describe here because
the book is already much larger than I thought it would
become.
The students in October knew only the most elementary
mathematics, many of them did not know the Binomial
Theorem, or the definition of the sine of an angle. In July
they had not only done the work of this book, but their
knowledge was of a practical kind, ready for use in any
such engineering problems as I give here.
One such student, Mr Norman Endacott, has corrected
the manuscript and proofs. He has worked out many of
the exercises in the third chapter twice over. I thank him
here for the care he has taken, and I take leave also to
say that a system which has, year by year, produced many
men with his kind of knowledge of mathematics has a
good deal to recommend it. I say this through no vanity
but because I wish to encourage the earnest student. Besides
I cannot claim more than a portion of the credit, for I
do not think that there ever before was such a complete
J 2/ 12b
VI PREFACE.
harmony in the working of all the departments of an
educational institution in lectures and in tutorial, labora-
tory, drawing office and other practical work as exists in
the Finsbury Technical College, all tending to the same
end ; to give an engineer such a perfect acquaintance with
his mental tools that he actually uses these tools in his
business.
Professor Willis has been kind enough to read through
the proofs and I therefore feel doubly sure that no important
mistake has been made anywhere.
An experienced friend thinks that I might with advantage
have given many more illustrations of the use of squared
paper just at the beginning. This is quite possible, but if
a student follows my instructions he will furnish all this sort
of illustration very much better for himself Again I might
have inserted many easy illustrations of integration by
numerical work such as the exercises on the Bull Engine
and on Beams and Arches which are to be found in my book
on Applied Mechanics. I can only say that I encourage
students to find illustrations of this kind for themselves;
and surely there must be some limit to spoon feeding.
JOHN PERRY.
EoYAL College of Science,
London,
16th March, 1897.
TABLE OF CONTENTS.
PAGE
Introductory Remarks ..,.,,.. 1
Chap. I. The Study of ^** . 6
Chap. II. The Compound Interest Law and the Harmonic
Function 161
Chap. III. General Differentiation and Integration . 267
CALCULUS FOE ENGINEERS.
INTRODUCTORY.
1. The Engineer has usually no time for a general mathe-
matical training — mores the pity — and those young engineers
who have had such a training do not always find their mathe-
matics helpful in their profession. Such men will, I hope,
find this book useful, if they can only get over the notion
that because it is elementary, they know already all that it
can teach.
But I write more particularly for readers who have had
very little mathematical training and who are willing to work
very hard to find out how the calculus is applied in Engineer-
ing problems. I assume that a good engineer needs to know
only fundamental principles, but that he needs to know these
very well indeed.
2. My reader is supposed to have an elementary know-
ledge of Mechanics, and if he means to take up the Electrical
problems he is supposed to have an elementary knowledge of
Electrical matters. A common-sense knowledge of the few
fundamental facts is what is required; this knowledge is
seldom acquired by mere reading or listening to lectures;
one needs to make simple experiments and to work easy
numerical exercises.
In Mechanics, I should like to think that the mechanical
engineers who read this book know what is given in the
elementary parts of my books on Applied Mechanics and the
Steam and Gas Engine. That is, I assume that they know
P 1
2 CALCULUS FOR ENGINEERS.
the elementary facts about Bending Moment in beams, Work
done by forces and the Efficiency of heat engines. Possibly
the book may cause them to seek for such knowledge. I take
almost all my examples from Engineering, and a man who
works these easy examples will find that he knows most of
what is called the theory of engineering.
3. I know men who have passed advanced examinations
in Mathematics who are very shy, in practical work, of the
common formulae used in Engineers' pocket-books. How-
ever good a mathematician a student thinks himself to be,
he ought to practise working out numerical values, to find
for example the value of a^ by means of a table of logarithms,
when a and h are any numbers whatsoever. Thus to find
V'014, to find 2*36o~<''2«, &c., to take any formula from a
pocket-book and use it. He must not only think he knows ;
he must really do the numerical work. He must know that
if a distance 2 4 5 4 has been measured and if one is not sure
about the last figure, it is rather stupid in multiplying or
dividing by this number to get out an answer with many
significant figures, or to say that the indicated power of an
engine is 324"65 Horse power, when the indicator may be in
error 5 per cent, or more. He must know the quick way of
finding 3'216 x 4571 to four significant figures without using
logarithms. He ought to test the approximate rule
(1 +a)"=l + ?ia,
or (1 + a)" (1 + ^)'« = 1 + 7ia + m^,
if a and /S are small, and see for himself when a = '01 or
— -01, or yS = ± -025 and ti = 2 or ^ or — 1^, and m = 4 or 2
or — 2 or J or any other numbers, what errors are involved in
the assumption.
f As to Trigonometry, the definitions must be known. For
example, Draw BA C an angle of, say, 35°. Take any point B
and drop the perpendicular. Measure AB and ^0 and AC
as accurately as possible. Is AC^ -{- BC^ = AB^ 1 Work
BC AG
this out numerically. Now -r^ = sin 35°, -rv^ = cos 35°,
BC ^^ ^^
-j-^ = tan 35°. Try if the answers are those given in the
tables. Learn how we calculate the other sides of the
INTRODUCTORY. 3
triangle ABC when we know one side and one of the acute
angles. Learn also that the sine of 130° is positive, and the
cosine of 130° is negative. Also try with the book of tables if
sin {A-\- B) — sin A , cos B + cos A . sin B,
where A and B are any two angles you choose to take.
There are three other rules like this. In like manner the
four which we obtain by adding these formulae and subtract-
ing them, of which this is one,
2 sin a . cos /S = sin (a + y8) + sin (a — /8) ;
also cos 2^ = 1 - 2 sin^ A =2 cos^ ^ - 1.
Before readers have gone far in this book I hope they will
be induced to take up the useful (that is, the elementary
and interesting) part of trigonometry, and prove all rules for
themselves, if they haven't done so already.
Calculate an angle of 1*6 degrees in radians (1 radian is
equal to 57'296 degrees) ; see how much the sine and tangent
of this angle differ from the angle itself Remember that
when in mathematics we say sin x, x is supposed to be in
radians.
I do not expect a man to know much about advanced
algebra, but he is supposed to be able to give the factors of
a?^ + 7^ 4- 12 or of x"^ — o? for example ; to be able to simplify
expressions. It is not a knowledge of permutations or com-
binations or of the theory of equations, of Geometrical Conies
or tangent planes to quadrics, that the Engineer wants.
Happy is the Engineer who is also a mathematician, but
it is given to only a few men to have the two so very different
powers.
A prolonged experience of workshops, engineers and
students has convinced me that although a Civil Engineer
for the purposes of surveying may need to understand the
solution of triangles, this and many other parts of the
Engineer's usual mathematical training are really useless to
the mechanical or electrical engineer. This sounds un-
orthodox, but I venture to emphasise it. The young engi-
neer cannot be drilled too much in the mere simplification
of algebraic and trigonometrical expressions, including ex-
pressions involving J —1, and the best service done by
1—2
4 CALCULUS FOR ENGINEERS.
elementary calculus work is in inducing students to again
undergo this drilling.
But the engineer needs no artificial mental gymnastics
such as is furnished by Geometrical Conies, or the usual
examination-paper puzzles, or by evasions of the Calculus
through infinite worry with elementary Mathematics. The
result of a false system of training is seen in this, that not
one good engineer in a hundred believes in what is usually
called theory.
4. I assume that every one of my readers is thoroughly
well acquainted already with the fundamental notion of the
Calculus, only he doesn't know it in the algebraic form. He
has a perfect knowledge of a rate, but he has never been
accustomed to write -^ ; he has a perfect knowledge of an
area, but he has not yet learnt the symbol used by us,
lf(x).dos. He has the idea, but he does not express his
idea in this form.
1 assume that some of my readers have passed difficult
examinations in the Calculus, that they can differentiate any
function of x and integrate many ; that they know how 'to
work all sorts of difficult exercises about Pedal Curves and
Roulettes and Elliptic Integrals, and to them also I hope to
be of use. Their difficulty is this, their mathematical know-
ledge seems to be of no use to them in practical engineering
problems. Give to their afs and y's a physical meaning,
or use p's and vs instead, and what was the easiest book
exercise becomes a difficult problem. I know such men
who hurriedly skip in reading a book when they see a
~ y or a sign of integration.
5. When I started to write this book I thought to put
the subject before my readers as I have been able, I think —
I have been told — very successfully, to bring it before some
classes of evening students; but much may be done in
lectures which one is unable to do in a cold-blooded fashion
sitting at a table. One misses the intelligent eyes of an
audience, warning one that a little more explanation is needed
INTRODUCTORY. 5
or that an important idea has already been grasped. An
idea could be given in the mere drawing of a curve and
illustrations chosen from objects around the lecture-room.
Let the reader skip judiciously; let him work up no
problem here in which he has no professional interest. The
problems are many, and the best training comes from the
careful study of only a few of them.
The reader is expected to turn back often to read again
the early parts.
The book would be unwieldy if I included any but the
more interesting and illustrative of engineering problems. I
put off for a future occasion what would perhaps to many
students be a more interesting part of my subject, namely,
illustrations from Engineering (sometimes called Applied
Physics) of the solution of Partial Differential Equations.
Many people think the subject one which cannot be taught
in an elementary fashion, but Lord Kelvin showed me long
ago that there is no useful mathematical weapon which an
engineer may not learn to use. A man learns to use the
Calculus as he learns to use the chisel or the file on actual
concrete bits of work, and it is on this idea that I act in
teaching the use of the Calculus to Engineers.
This book is not meant to supersede the more orthodox
treatises, it is rather an introduction to them. In the
first chapter of 160 pages, I do not attempt to differentiate
or integrate any function of x, except x^. In the second
chapter I deal with e^^, and sin (aw + c). The third chapter
is more difficult.
For the sake of the training in elementary Algebraic
work, as much as for use in Engineering problems, I have
included a set of exercises on general dinerentiation and
integration.
Parts in smaller type, and the notes, may be found too
difficult by some students in a first reading of the book. An
occasional exercise may need a little more knowledge than
the student already possesses. His remedy is to skip.
CHAPTER I.
x».
6. Everybody has already the notions of Co-ordinate
Geometry and uses squared paper. Squared paper may
be bought at sevenpence a quire : people who arc ignorant
of this fact and who pay sevenpence or fourteen pence a
sheet for it must have too great an idea of its value to use
it properly.
When a merchant has in his office a sheet of squared
paper with points lying in a curve which he adds to day by
day, each point showing the price of iron, or copper, or cotton
yarn or silk, at any date, he is using Co-ordinate Geometry.
Now to what uses does he put such a curve ? 1. At any
date he sees what the price was. 2. He sees by the slope of
his curve the rate of increase or fall of the price. 3. If he
plots other things on the same sheet of paper at the same
dates he will note what effect their rise and fall have upon
the price of his material, and this may enable him to pro-
phesy and so make money. 4. Examination of his curve for
the past will enable him to prophesy with more certainty
than a man can do who has no records.
Observe that any point represents two things; its
horizontal distance from some standard line or axis is called
one co-ordinate, we generally call it the x co-ordinate and it
is measured horizontally to the right of the axis of y ; some
people call it the abscissa ; this represents time in his case.
The other co-ordinate (we usually call it the 3/ co-ordinate or
tlie ordinate, simply), the vertical distance of the point above
some standard line or axis; this represents his price. In the
newspaper you will find curves showing how the thermometer
and barometer are rising and falling. I once read a clever
article upon the way in which the English population and
wealth and taxes were increasing; the reasoning was very
SQUARED PAPER. 7
difficult to follow. On taking the author s figures however
and plotting them on squared paper, every result which he
had laboured so much to bring out was plain upon the
curves, so that a boy could understand them. Possibly this
is the reason why sDme writers do not publish curves: if they
did, there would be little need for writing.
7. A man making experiments is usually finding out
how one thing which I shall call y depends upon some other
thing which I shall call x. Thus the pressure p of saturated
steam (water and steam present in a vessel but no air or
other fluid) is always the same for the same temperature.
A curve drawn on squared paper enables us for any given
temperature to find the pressure or vice versa, but it shows
the rate at which one increases relatively to the increase of
the other and much else. I do not say that the curve is always
better than the table of values for giving information ; some
information is better given by the curve, some by the table.
Observe that when we represent any quantity by the length
of a line we represent it to some scale or other ; 1 inch
represents 10 lbs. per square inch or 20 degrees centigrade
or something else ; it is always to scale and according to a
convention of some kind, for of course a distance 1 inch is a
very different thing from 20 degrees centigrade.
When one has two columns of observed numbers to plot
on squared paper one does it, 1. To see if the points lie in
any regular curve. If so, the simpler the curve the simpler
is the law that we are likely to find. 2. To correct errors of
observation. For if the points lie nearly in a simple regular
curve, if we draw the curve that lies most evenly among the
points, using thin battens of w^ood, say, then it may be taken
as probable that if there were no errors of observation the
points would lie exactly in such a curve. Note that when
a point is — 5 feet to the right of a line, we mean that it is
5 feet to the left of the line. I have learnt by long ex-
perience that it is worth while to spend a good deal of time
subtracting from and multiplying one's quantities to fit the
numbers of squares (so that the whole of a sheet of paper is
needed for the points) before beginning to plot.
Now let the reader buy some squared paper and without
asking help from anyone let him plot the results of some
8 CALCULUS FOR ENGINEERS.
observations. Let him take for example a Whitaker's
Almanack and plot from it some sets of numbers ; the average
temperature of every month last year; the National Debt
since 1688; the present value of a lease at 4 per cent, for
any number of years ; the capital invested in Railways since
1849 ; anything will do, but he had better take things in
which he is interested. If he has made laboratory observa-
tions he will have an absorbing interest in seeing what sort
of law the squared paper gives him.
8. As the observations may be on pressure jj and tem-
perature t, or p and volume v, or v and t, or Indicated Horse
Power and Useful Horse Power of a steam or gas engine, or
amperes and volts in electiicity, and we want to talk generally
of any such pair of quantities, I shall use x and y instead
of the p'q and v^ and ^ s and all sorts of letters. The short
way of saying that there is some law connecting two variable
quantities x and y is F (x, y) =0...(1), or in words "there
is some equation connecting x and y." Any expression
which contains x and y (it may contain many other letters
and numbers also) is said to be a function of x and y and we
use such symbols as F(x, y),f{x, y), Q(x, y) etc. to represent
functions in general when we don't know what the expres-
sions really are, and often when we do know, but want to
write things shortly. Again we use F{x) or f{x) or any
other convenient symbol to mean "any mathematical ex-
pression containing x" and we say " let f(x) be any function
of x!* Thus y =f(x) ... (2) stands for any equation which
would enable us when given x to calculate y.
X' ifi
The law ^ + ^t^ = 1 comes under the form (1) given
above, whereas if we calculate y in terms of x and get
y = 4- 1 \/25 — x'^ we have the form (2). But in either case
we have the same law connecting y and x. In pure mathe-
matics X and y are actual distances ; in applied mathematics
X and y stand for the quantities which we are comparing and
which are represented to scale.
9. ' Graph ' Exercises.
I. Draw the curve y = 2 + -^j-^x-.
Take ^ = and we find 2/ =2; take x = l, then 7/ = 20333 ;
'graph' exercises. 9
take X = 2, then y = 2 + IS^o = 2-133; and so on. Now plot
these values of x and y on your sheet of squared paper. The
curve is a parabola.
II. Draw the curve y = 1 — ^cc-\- -J^x^ which is also a
parabola, in the same way, on the same sheet of paper.
III. Draw the curve xy = 120. Now if x = l, y=120;
if X = 2, y = 60 ; if x = S, y = 4<0; if a; = 4, y = 30 and so on :
this curve is a rectangular hyperbola.
IV. Draw yx'-'^' = 100 or y = 100^-i-«^ If the student
cannot calculate y for any value of x, he does not know how
to use logarithms and the sooner he does know how to use
logarithms the better.
V. Draw y = ax''^ where a is any convenient number. I
advise the student to spend a lot of time in drawing members
of this great family of useful curves. Let him try ?i = — 1
(he drew this in III. above), 7i=— 2, ?i=— 1-J, n=—^, ?i=— 0*1,
n = 0, n = "I, ri = f , n = l, n=l\,n = '2 (this is No. I. above),
?i = 3, w = 4 &c.
YI. Draw y = a sin (bx + c) taking any convenient
numbers for a, h and c.
Advice. As hx-{-c is in radians (one radian is 57'2958
degrees) and the books of tables usually give augles in
degrees, choose numbers for h and c which will make the
arithmetical work easy. Thus take 6 = 1-7- 114"6, take c the
number of radians which correspond to say 30°
(
this is ^ or '5236
Let a = 5 say. Now let x — 0, 10, 20, &;c., and calculate y.
Thus when «?= 6, 3/ = 5 sin [yrxTp + "5236) ; but if the
angle is converted into degrees we have
2/ = 5 sin (16 + 30 degrees) = b sin 33° = 2723.
Having drawn the above curve, notice what change would
occur if c were changed to or ^ or ^ or ^ . Again, if a
were changed. More than a week may be spent on this curve,
very profitably.
10 CALCULUS FOR ENGINEERS.
VII. Draw y = ae***. Try 6 = 1 and a = 1 ; try other
values of a and h ; take at least two cases of negative values
for h.
In the above work, get as little help from teachers as
possible, but help from fellow students will be very useful
especially if it leads to wrangling about the subject.
The reason why I have dwelt upon the above seven cases
is this: — Students learn usually to differentiate and integrate
the most complicated expressions: but when the very simplest
of these expressions comes before them in a real engineering
problem they fight shy of it. Now it is very seldom that an
engineer ever has to face a problem, even in the most intri-
cate part of his theoretical work, which involves a knowledge
of more functions than these three
y = ax^y y = a sin {hx + c), y = ad^,
but these three must be thoroughly well understood and the
engineering student must look upon the study of them as his
most important theoretical work.
Attending to the above three kinds of expression is a
student's real business. I see no reason, however, for his not
having a little amusement also, so he may draw the curves
X' 4- y^ = 25 (Circle), ^ + f^ = 1 (Ellipse),
|g-^ = l (Hyperbola),
and some others mentioned in Chapter, III., but from the
engineer's point of view these curves are comparatively un-
interesting.
10. Having studied y = e~^^ and y = 6 sin (ex + ^) a
student will find that he can now easily understand one of
the most important curves in engineering, viz :
y = be~** sin (ex + g).
He ought first to take such a curve as has already been
studied by him, y = h sin (ex + g) ; plot on the same sheet of
paper y = e~^^ ; and multiply together the ordinates of the
two curves at many values of x to find the ordinate of the
EQUATIONS TO LOCI.
11
new curve. The curve is evidently wavy, y reaching maximum
and minimum values; y represents the displacement of
a' pendulum bob or pointer of some measuring instrument
whose motion is damped by fluid or other such friction, x being
the time, and a student will understand the curve much better
if he makes observations of such a motion, for example with
a disc of lead immersed in oil vibrating so slowly under
the action of torsional forces in a wire that many ob-
servations of its angular position (using pointer and scale
of degrees) which is called y, x being the time, may be
made in one swing. The distance or angle from an extreme
position on one side of the zero to the next extreme position
on the other side is called the length of one swing. The
Napierian logarithm of the ratio of the length of one swing to
the next or one tenth of the logarithm of the ratio of the
first swing to the eleventh is evidently a multiplied by half
the periodic time, or it is a multiplied by the time occupied
in one swing. This logarithmic decrement as it is called,
is rather important in some kinds of measurement.
11. When by means of a drawing or a model we are able to find the
path of any point and where it is in its path when we know the
position of some other point, we are always able to get the same
information algebraically.
Example (1). A point jPand a straight line DD being given ; what
is the path of a point P when it moves so that its distance from the
point F is always in the same ratio to its distance from the straight line?
Thus in the figure let FF=exPI> ... (I), where e is a constant.
Draw FFX at right angles to DD. y
If the distance FD is called a; and
the perpendicular FG is y ; our
problem is this ; — What is the equa-
tion connecting x and y? Now all
we have to do is to express (1) in
terms of x and ^. Let FF be called a.
Thus
FF= \/FG^+FG^= Vy^ + (^'-a)2
so that, squaring (1) we have
This is the answer. If e is 1 the
curve is called a parabola. If e is
greater than 1, the curve is called an
Fig. 1.
hyperbola. If e is less than I, the curve is called an ellipse.
12
CALCULUS FOR ENGINEERS.
Example (2). The circle APQ rolls on the straight line OX,
What is the Path of any point P on. the circumference? If whenP
touched the line it was at 0, let OX and OF bo the axes, and let SP
be X and PT be y. Let the radius of the circle be a. Let the angle
PCQ be called <^. Draw CB, peq^endicular to PT. Observe that
PB=a . sin PCB=a sin (</> - 90) = - a cos (/>,
BC= a cos PCB = a sin (/>.
Now the arc QP= a .cf) = OQ. Hence SiS x=OQ — BC, and
/c have
y = BT^-PB,
x=axf)- a sin 0|
y = a —acoscfij
(3).
If from (3) we eliminate we get one equation connecting x and y.
I^ut it is better to retain </> and to have two equations because of the
greater simplicity of calculation. In fact the two equations (3) may
be called the equation to the curve. The curve is called the oycloid
as all my readers know already.
Example (3). A crank and connecting rod work a slider
in a straight path. Where is the slider for any position of the crank ?
Let the path be in the direction of the centre of the crank shaft.
P.^-
Fig. 3.
If A is the end of the path, evidently ^0 is equal to l-\-r, r being
length of crank.
MECHANISMS.
13
It is well to remember in all such problems that if we project all
the sides of a closed figure upon any two straight lines, we get two in-
dependent equations. Projecting on the horizontal we see that
.(1).
s+l cos (f)+r cos 6=1 + r\
Projecting on the vertical ^ sin ^ = r sin ^ J
If we eliminate from these equations we can calculate s for any
value of 6. The student ought to do this for himself, but I am weak
enough to do it here. We see that from the second equation of (1)
COS0
so that the first becomes
-V^-i
sin2^,
s=l\l-^l-^^^sm^0J+r{l-cosd)..* (2).
Students ought to work a few exercises, such as; — 1. The ends
A and B of a rod are guided by two straight slots OA and OB which
are at right angles to one another ; find the equation to the path of
any point P in the rod. 2. In Watt's parallel motion there is
a point which moves nearly in a straight path. Find the equation to
its complete path.
In fig. 4 the Mean Position is shown as OABC. The best place
for P is such that BP/PA = OAICB. Draw the links in any other
Fig. 4.
* Note that if as is usual, -j^isa. small fraction, then slnca ,Jl-a=l-\a
■when a is small, we can get an approximation to the value of s, which can
be expressed in terms of d and 2d. This is of far more importance than it
here seems to be. When the straight path of Q makes an angle a with the
Hne joining its middle point and O, if a is not large, it is evident that s is
much the same as before, only divided by cos a. When a is large, the
algebraic expression for s is rather complicated, but good approximations
may always be found which will save trouble in calculation.
14 CALCULUS FOR ENGINEERS.
position. The complete path of P would be a figure of 8. 3. Find
the equation to the path of a point in the middle of an ordinaiy con-
necting rod. 4. A^ the end of a link, moves in a straight path COC\
being the middle of the path, with a simple haraaonic motion
OA=asmpti where t is time ; the other end B moves in a straight
path OBD which is in a direction at right angles to COC ; what is ^'s
motion ? Show that it is approximately a simple harmonic motion of
twice the frequency of ^. 5. In any slide valve gear, in which there
are several links, &c. driven from a uniformly rotating crank ; note
this fact, that the motion of any point of any link in any particular
direction consists of a fundamental simple harmonic motion of the
same frequency as the crank, together with an octave. The proper
study of Link Motions and Radial valve gears from this
point of view is' worth months of one's life, for this contains the secret
of why one valve motion gives a better diagram than another.
Consider for example the Hackworth gear with a curved and with a
straight slot. What is the difference ? See Art. 122.
12. Plotted points l3ring in a straight line. Proofs
will come later; at first the student ought to get well ac-
quainted with the thing to be proved. I have known boys
able to 'prove mathematical propositions who did not really
know what they had proved till years afterwards.
Take any expression like y = a + hx, where a and h are
numbers. Thus let y = 2 + l^x. Now take ^7 = 0, a?=l,
a; = 2, a? = 3, <Src. and in each case calculate the corresponding
value of y. Plot the corresponding values of a; and y as the
co-ordinates of points on squared paper. You will find that
they lie exactly in a straight line. Now take say y = 2-\-Sx
or 2 + ^a; or 2 — ^a; or 2 — Sx and you will find in every case
a straight line. Men who think they know a little about
this subject already will not care to take the trouble and if
you do not find yourselves interested, I advise you not to
take the trouble either; yet I know that it is worth your
while to take the trouble. Just notice that in every case I
have given you the same value of a and consequently all
your lines have some one thing in common. What is it?
Take this hint, a is the value of y when a? = 0.
Again, try y=:2 + l^x, y = 1 +l^x, y = 0-i-lix,
y = ^l + lix, 2/ = -2 + l^a?,
and so see what it means when h is the same in every case.
You will find that all the lines with the same b have the
SLOPE OF A LINE. 16
same slope and indeed I am in the habit of calling h the
slope of the line.
If y = a 4- 6^, when oc = x-^, find y and call it y^y
when a; = a?! + 1, find y and call it y^.
It is easy to show that y.^—y^ = h. So that what I mean by
the slope of a straight line is its rise for a horizontal distance
1. (Note that when we say that a road rises ^ or 1 in 20,
we mean 1 foot rise for 20 feet along the sloping road. Thus
■jjj is the sine of the angle of inclination of the road to the
horizontal; whereas our slope is measured in a different way).
Our slope is evidently the tangent of the inclination of the
line to the horizontal. Looking upon y as a quantity whose
value depends upon that of x, observe that the rate of in-
crease of y relatively to the increase of x is constant, being
indeed h, the slope of the line. The symbol used for this
rate is -p. Observe that it is one symbol; it does not mean
/Y V 7/
-^ — ~ . Try to recollect the statement that if y= a + hx,
a ^ X
-~—h, and that if -,— = h, then it follows that y:=A-\-hx,
dx dx
where A is some constant or other.
Any equation of the first degree connecting x and y
such as Ax-\-By = G where A, B and C are constants, can
G A
be put into the shape y = ^ — ^x, so that it is the equation
A
to a straight line whose slope is — ^ and which passes
C
through the point whose x = 0, whose y = -n, called point
( 0, p ) . Thus 4^ + 2^/ = 5 passes through the point x = 0,
y = 2^ and its slope is — 2. That is, y diminishes as x in-
creases. You are expected to draw this line y = 2|^ — 2^7 and
distinguish the difference between it and the line 3/ = 2 J + 2x.
Note what is meant by positive and what by negative slope.
Draw a few curves and judge approximately by eye of the
slope at a number of places.
13. Problems on the straight line.
1. Given the slope of a straight line; if you are also
16
CALCULUS FOR ENGINEERS.
told that it passes through the point whose x = 3, and whose
y = 2, what is the equation to the line ? Let the slope be
0-35.
The equation is y = a + 0'35.'^^^. where a is not known.
But (3, 2) is a point on the line, so that 2 = a + 035 x 3,
or a = 0"95 and hence the line is y = 0'95 + 35a?.
2. What is the slope of any line at right angles to
y=za-\-bxl Let -45 be the given line, cutting OX in C. Then
h = tan BOX. If DE is any line at right angles to the first,
its slope is tan DEX or — tan DEC or - cot BGE or — ^ .
So that y = A—j-a)is typical of all lines at right angles
to y = a-{-h.v] A being any constant.
3. Where do the two straight lines Ax -\-By-\-C—0
and Mx + Ny -{■8 = meet ? Answer, In the point whose x
and y satisfy both the equations. We have therefore to do
what is done in Elementary Algebra, solve simultaneous
equations.
4. When tan a and tan ^ are known, it is easy to find
tan (a — /S), and hence when the straight lines y=a + hx
and y = m-\-nx are given, it is easy to find the angle between
them.
5. The line y = a-\-bx passes through the points a?=l,
y = 2, and a? = 3, y = 1, find a and h.
EMPIRICAL FORMULAE. 17
6. A line y = a-{-bx is at right angles to y = 2 + Sa) and
passes through the point x = l, y = l. Find a and h.
14. Obtaining Empirical Formulae.
When in the laboratory we have made measurements of two
quantities which depend upon one another, we have a table showing
corresponding values of the two, and we wish to see if there is a simple
relation between them, we plot the values to convenient scales as the
co-ordinates of points on squared paper. If some regular curve (a
cm*ve without singular points as I shall afterwards call it) seems as if
it might pass through all the points, save for possible errors of
measurement, we try to obtain a formula y = f(x), which we may call
the law or rule connecting the quantities called y and x.
If the points appear as if they might lie on a straight line, a
stretched thread may be used to help in finding its most probable
position. There is a tedious algebraic method of finding the straight
line which represents the positions of the points with least error, but
for most engineering purposes the stretched string method is suffi-
ciently accurate.
If the ciu-ve seems to follow such a law as y~a-\-hx^, plot y and
the square of the observed measurement, which we call x^ as the co-or-
dinates of points, and see if they lie on a straight line. If the curve
ax
seems to follow such a law as y= j ■ (1), which is the same as
^.\.hy = a, divide each of the quantities which you call y by the corre-
sponding quantity x ; call the ratio X. Now plot the values of X and
of y on squared paper ; if a straight line passes through the plotted
points, then we have such a law as X=A-{-By, or - — A-\-By^ or
, so that (1) is true.
^ l-Bx'
Usually we can apply the stretched thread method to find the
probability of truth of any law containing only two constants.
Thus, suppose measurements to be taken from the expansion part
of a gas engine indicator diagram. It is im]3ortant for many purposes
to obtain an empirical formula connecting p and v, the pressure and
volume. I always find that the following rule holds with a fair amount
of accuracy pv" = C where s and C are two constants. We do not
much care to know C, but if there is such a rule, the value of s is very
important*. To test if this rule holds, plot log jo and log v as the co-
* There is no known physical reason for expecting such a rule to hold.
At first I thought that perhaps most curves drawn at random approximately
like hyperbolas would approximately submit to such a law as yx^= C, but I
found that this was by no means the case. The following fact is. worth
mentioning. When my students find, in carrying out the above rule that
log p and log v do not lie in a straight line, I find that they have
P. 2
18 CALCULUS FOR ENGINEERS.
ordinates of points on squared pa^^er (common logarithms will do).
If they lie approximately in a straight line, we see that
\ogp+s\ogv=c
a constant, and therefore the rule holds.
When we wish to test with a formula containing three independent
constants we can often i*educe it to such a shape as
Av+Bw+Cz=l (2),
where y, w, z contain x and y in some shajje. Thus to test if y=—- — ,
we have y + cxy —a-\-hx^ ov- + -xy — x=\. Here y itself is the old v,
xy is the old w>, and x itself is the old z.
If (2) holds, and if t?, \o and z were plotted as the three co-ordinatea
of a ix)int in space, all the points ought to lie in a plane. By means
of three sides of a wooden box and a number of beads on the ends of
pointed wires this may be tried directly ; immersion in a tank of water
to try whether one can get the beads to lie in the plane of the siuface
of the water, being used to find the plane. I have also used a descrip-
tive geometry method to find the plane, but there is no method yet
used bv me which compares for simplicity with the stretched thread
method in the other case.
But no hard and fast rules can be given for trying all sorts of em-
pirical formulae upon one's observed numbers. The student is warned
that his formula is an empirical one, and that he must not deal with it
as if he had discovered a natural law of infinite exactness.
AVhen other formulae fail, we try
y—a-\-hx-\-cx'*--\-^-\'^^")
because we know that with sufficient terms this will satisfy any curve.
When there are more than two constants, we often find them by a
patient application of what is called the method of least squares. To
test if the pressure and temperature of saturated steam follow the
rule jo = a(^+/3)"... (3), where B is temperature, Centigrade, say, three
constants have to be found. The only successful plan tried by me is
one in which I guess at /3. I know that /3 is nearly 40. I ask one
student to try ^ = 40, another to try i3 = 41, another ^=39 and so on;
made a mistake in the amount of clearance. Too much clearance and too
little clearance give results which depart in opposite ways from the straight
hne. It is convenient in many calculations, if there is such an empirical
formula, to use it. If not, one has to work with rules which instruct us to
draw tangents to the curve. Now it is an excellent exercise to let a number
of students trace the same curve with two points marked upon it and to let
them all independently draw tangents at those points to their curve, and
measure the angle between them. It is extraordinary what very different
lines they will draw and what different angles they may obtain. Let them
all measure by trial the radius of curvature of the curve at a point ; in this
the discrepancies are greater than before.
SLOPE OF A CURVE.
19
they are asked to liiid the rule (3) which most accurately represents
p and 6 between, say jo = 71b. per sq. inch, and p = 150. He who gets
a straight Hne lying most evenly (judging by the eye) among the points,
when \ogp and log(^-f /3) are used as co-ordinates, has used the best
value of 4. The method may be refined upon by ingenious students.
(See end of Chap. I.)
15. We have now to remember that if y = a-^hx, then
-^ — h, and if -^ = h, then y = A -{■ hx, Avhere A is some
ax ax ^
constant.
Let us prove this algebraically.
\iy = a-\-hx. Take a particular value of x and calculate
y. Now take a new value of x, call it ^ + hx, and calculate
the new y, call it y -t- Sy,
y ^^y — a-\-h{x-\- hx).
Subtract y—a-\-hx and we get
hy = hhx, or -^ = 6,
ox
and, however small hx or Zy may become their ratio is 6, we
therefore say ~ = h.
Fig. 6.
16. In the curve of fig. 6 there is positive slope
{^ increases as x increases) in the parts AB, DF and HI and
2—2
20
CALCULUS FOR ENGINEERS.
Fig. 7.
negative slope {y diminishes as x increases) in the parts
BD and FH. The slope is at 5 and F which are called
points of maximum or points where i/ is a maximum; and it
is also at D and H which are points of minimum. The
point E is one in which the slope ceases to increase and
begins to diminish : it is a point of inflexion.
Notice that if we want to know the slope at the point P
we first choose a point F which
is near to P. (Imagine that
in fig. 6 the little portion of
the curve at P is magnified a
thousand times.) Call PS=x,
PQ=:y-NF=x+Bx,FL=ij+ht/,
sotha,tPM=Bx,F3I==Si/. Now
FM/PM or By/Bx is the average
slope between P and F. It is
tan FPM. Imagine the same
sort of figure drawn but for a
point F' nearer to P. Again,
another, still nearer P. Ob-
serve that the straight line FP or F'P or F"P gets gi-adually
more and more nearly what we mean by the tangent to the
curve at P. In every case hyjhx is the tangent of the angle
which the line FP or F'P or F'P makes with the horizontal,
and so we see that in the limit the slope of the line or dy/dx
at P is the tangent of the angle which the tangent at P
makes with the axis of X. If then, instead of judging
roughly by the eye as we did just now in discussing fig. 6, we
wish to measure very accurately the slope at the point P ; —
Note that the slope is independent of where the axis of X is,
so long as it is a horizontal line, and I take care in using my
rule here given, to draw OX below the part of the curve
where I am studying the slope. Draw a tangent PR to
the curve, cutting OX in P. Then the slope is tan PRX.
If drawn and lettered according to my instructions, observe
that PRX is always an acute angle when the slope is
positive and is always an obtuse angle when the slope is
negative.
Do not forget that the slope of the curve at any point
means the rate of increase of y there with regard to a?, and
WHAT IS SPEED? 21
that we may call it the slope of the curve or tan PRQ or by
the symbol -^- or "the differential coefficient of y with
regard to x," and all these mean the same thing.
Every one knows what is meant when on going up a
hill one says that the slope is changing, the slope is diminish-
ing, the slope is increasing ; and in this knowledge he already
possesses the fundamental idea of the calculus.
17. We all know what is meant when in a railway train we
say ^'we are going at 30 miles per hour.^' Do we mean
that we have gone 30 miles in the last hour or that we are
really going 30 miles in the next hour ? Certainly not. We
may have only left the terminus 10 minutes ago ; there may be
an accident in the next second. What we mean is merely this,
that the last distance of 3 miles was traversed in the tenth
of an hour, or rather, the last distance of 0*0003 miles was
traversed in 0*00001 hour. This is not exactly right ; it is
not till we take still shorter and shorter distances and divide
by the times occupied that we approach the true value of
the speed. Thus it is known that a body falls freely
vertically through the following distances in the following
intervals of time after two seconds from rest, at London.
That is between 2 seconds from rest and 2*1 or 2*01 or 2001,
the distances fallen through are given. Each of these
divided by the interval of time gives the average velocity
during the interval.
Intervals of time in seconds
Distances in feet fallen through
Average velocities
We see that as the interval of time after 2 seconds is
taken less and less, the average velocity during the interval
approaches more and more the true value of the velocity at
2 seconds from rest which is exactly 64*4 feet per second.
We may find the true velocity at any time when we know
the law connecting s and t as follows.
Let s = 16'lt2, the well known law for bodies falling
freely at London. If t is given of any value we can calculate
•1
•01
•001
6*601
•6456
•064416
6601
64*56
64416
22 CALCULUS FOR ENGINEERS.
s. If t has a slightly greater value called t-\- St (here St
is a symbol for a small portion of time, it is not S "Xt, but a
very different thing), and if we call the calculated space
s + Ss, then s -^ hs = U'l (t + Btf or l&l {f + 2t. St + (Sty].
Hence, subtracting, Ss = 16*1 {2t . St 4- (Sty], and this formula
will enable us to calculate accurately the space Ss passed
through between the time t and the time t+St. The average
velocity during this interval of time is Ss -^ St or
^ = 32-2^+ 10- 1 8/.
Please notice that this is absolutely correct ; there is no
vagueness about it.
Now I come to the important idea; as St gets smaller
and smaller, ^ approaches more and more nearly 32*2^, the
other term 16'1S^ becoming smaller and smaller, and hence
we say that in the limit, Ss/St is truly 32-2^. The limiting
value of ^ as St gets smaller and smaller is called , or the
rate of change of s as t increases, or the differential coefficient
of s with regard to t, or it is called the velocity at the time t
Now surely there is no such great difficulty in catching
the idea of a limiting value. Some people have the notion
that we are stating something that is only approximately
true ; it is often because their teacher will say such things as
"reject 16'1S^ because it is small," or "let dt be an infinitely
small amount of time" and they proceed to divide something
by it, showing that although they may reach the age of
Methuselah they will never have the common sense of an
engineer.
Another trouble is introduced by people saying " let
Ss ds
St — and kt or -y- is so and so." The true statement is, " as
of at . . Ss
St gets smaller and smaller without limit, ^ approaches more
and more nearly the finite value 322^," and as I have already
said, everybody uses the important idea of a limit every day
of his life.
SLOPE AND SPEED.
23
From the law connecting s and t, if we find -r- or the
velocity, we are said to differentiate s with regard to the
time t. When we are given ds/dt and we reverse the above
process we are said to integrate.
If I were lecturing I might dwell longer upon the correct-
ness of the notion of a rate that one already has, and by
making many sketch'es illustrate my meaning. But one may
listen intently to a lecture which seems dull enough in a
book. I will, therefore, make a virtue of necessity and say
that my readers can illustrate my meaning perfectly well to
themselves if they do a little thinking about it. After all
my great aim is to make them less afraid than they used to be
of such symbols as dy/dx and fy . dx.
18. Given s and t in any kind of motion, as a set of num-
bers. How do we study the motion ? For example, imagine
a Bradshaw's Railway Guide which not merely gave a few
stations, but some hundred places between Euston and Rugby.
The entries might be like this: s would be in miles, t in
hours and minutes. .9 = would mean Euston.
10 o'clock
3
10. .10
5
10. .15
7
10.. 20
7
10. .23
9
10.. 28
12
10. .33
&c.
One method is this : plot t (take times after 10 o'clock)
horizontally and s vertically on a sheet of squared paper and
draw a curve through the points.
The slope of this curve at any place represents the velocity
of the train to some scale which depends upon the scales for
s and t.
24 CALCULUS FOR ENGINEERS.
Observe places where the velocity is great or small.
Between ^=10.20 and i = 10.23 observe that the velocity
is 0. Indeed the train has probably stopped altogether.
To be absolutely certain, it would be necessary to give s
for every value of t, and not merely for a few values. A
curve alone can show every value. I do not say that the
table may not be more valuable than the curve for a great
many purposes.
If the train stopped at any place and travelled towards
Euston again, we should have negative slope to our curve
and negative velocity.
Note that acceleration being rate of change of velocity
with time, is indicated by the rate of change of the slope of
the curve. Why not on the same sheet of paper draw a curve
which shows at every instant the velocity of the train ?
The slope of this new curve would evidently be the accelera-
tion. I am glad to think that nobody has yet given a name
to the rate of change of the acceleration.
The symbols in use are
s and t for space and time ;
ds
velocity i; or -^ , or Newton's symbol s ;
acceleration -j- or -yz , or Newton's s.
at dv
Rate of change of acceleration would be -r- .
Note that ^ is one symbol, it has nothing whatever to do
d^ X s
with such an algebraic expression as , — -— . The symbol is
Gt X I
supposed merely to indicate that we have differentiated s
twice mth regard to the time.
I have stated that the slope of a curve may be found by
drawing a tangent to the curve, and hence it is easy to find
the acceleration from the velocity curve.
ACCELERATION.
26
19. Another way, better than by drawing tangents, is
illustrated in this Table :
-.
acceleration
t
seconds
s
feet
V
feet per
second
in feet per
second per
second or
or 8sJ8t
8vldt
•06
•0880
14-74
•07
•2354
13-49
-125
•08
•3703
1222
-127
•09
•4925
10-95
-127
•10
•6020
9-66
-129
•11
•6986
8-35
-131
•12
•7821
704
-131
•13
•8525
In a new mechanism it was necessary for a certain
purpose to know in every position of a point A what its
acceleration was, and to do this I usually find its velocity
first. A skeleton drawing was made and the positions of
A marked at the intervals of time t from a time taken
as 0. In the table I give at each instant the distance
of A from a fixed point of measurement, and I call it s.
If I gave the table for all the positions of A till it gets
back again to its first position, it would be more instructive,
but any student can make out such a table for himself
for some particular mechanism. Thus for example, let s
be the distance of a piston from the end of its stroke.
Of course the all-accomplished mathematical engineer will
scorn to take the trouble. He knows a graphical rule
for doing this in the case of the piston of a steam
engine. Yes, but does he know such a rule for every
26 CALCULUS FOR ENGINEERS.
possible mechanism ? Would i^t be worth while to seek
for such a graphical nile for every possible mechanism ?
Here is the straightforward Engineers' common-sense way
of finding the acceleration at any point of any mechanism,
and although it has not yet been tried except by myself
and my pupils, I venture to think that it will commend
itself to practical men. For beginners it is invaluable.
Now the mass of the body whose centre moves like the
point A, being m (the weight of the body in pounds at
London^ divided by 32*2)*, multiply the acceleration in
feet per second per second which you find, by m, and you
have the force which is acting on the body increasing the
velocity. The force will be in pounds.
* I have given elsewhere my reasons for using in books intended for
engineers, the units of force employed by all practical engineers. I have
used this system (which is, after all, a so-called absolute system, just as
much as the c. a. s. system or the Foundal system of many text books) for
twenty years, with students, and this is why their knowledge of mechanics
is not a mere book knowledge, something apart from their practical work,
but fitting their practical work as a hand does a glove. One might as well
talk Choctaw in the shops as speak about what some people call the
English system, as if a system can be English which speaks of so many
poundals of force and so many foot-poundals of work. And yet these same
philosophers are astonished that practical engineers should have a contempt
for book theory. I venture to say that there is not one practical engineer
in this country, who thinks in Poundals, although all the books have used
these units for 30 years.
In Practical Dynamics one second is the unit of time, one foot is the
unit of space, one ijoiind (what is called the weight of 1 lb. in London) is the
unit of force. To satisfy the College men who teach Engineers, I would say
that "The unit of Mass is that mass on which the force of 1 lb. produces
an acceleration of 1 ft. per sec. per see."
We have no name for unit of mass, the Engineer never has to speak
of the inertia of a body by itself. His instructions are "In all Dynamical
calculations, divide the weight of a body in lbs. by 32'2 and you have its
mass in Engineer's units — in those units which will give all your answers in
the units in which an Engineer talks." If you do not use this system evei-y
answer you get out will need to be divided or multiplied by something before
it is the language of the practical man.
Force in pounds is the space-roXe at which work in foot-pounds is done,
it is also the f/we-rate at which momentum is produced or destroyed.
Example 1. A Kaxnxner head of 2^ lbs. moving with a velocity of 40 ft.
per sec. is stopped in '001 sec. What is the average force of the blow?
Here the mass being 2^-^32•2, or -0776, the momentum (momentum is
mass X velocity) destroyed is 3-104. Now force is momentum per sec. and
hence the average force is 3•104-^•001 or 3104 lbs.
Example 2. Water in a jet flows with the linear velocity of 20 ft. per sec.
DIFFERENTIATION OF ax^. 27
20. We considered tlic case of falling bodies in which
space and time are connected by the law 6' = iu^^? where cf
the acceleration due to gravity is 32 "2 feet per second per
second at London. But many other pairs of things are con-
nected by similar laws and I will indicate them generally by
y = ax2.
Let a particular value of a be taken, say a = -^. Now take
a) = 0, x = l, x==2, 00 = 3, &c. and in every case calculate y.
Plot the corresponding points on squared paper. They
lie on a parabolic curve. At any point on the curve, say
where x = S, find the slope of the curve ( I call it ~j^] , do
the same at a; = 4, ^ = 2, &c. Draw a new curve, now, with the
same values of x but with ~- as the ordinate. This curve
ax
shows at a glance (by the height of its ordinate) what is
the slope of the first curve. If you ink these curves, let
the y curve be black and the -~ curve be red. Notice that
the slope or -~ at any point, is 2a multiplied by the x of the
point.
We can investigate this algebraically. As before, for
any value of x calculate y. Now take a greater value of x
which I shall call x + Bx and calculate the new y, calling it
y + 8y. We have then
y -\- By = a (x -{- Bxf
= a[x^^+2x.Bx + (8xy}.
Subtracting ; By = a [2x . Bx + (Bx)-].
By
Divide by Bx, -— = 2ax + a . Bx.
(relatively to the vessel from which it flows), the jet being 0*1 sq. ft. in cross
section ; what force acts upon the vessel ?
Here we have 20 x -1 cu. ft. or 20 x -1 x 62-3 lbs. of water per sec. or
a mass per second in Engineers' units of 20 x -1 x 62*3-1- 32-2. This mass is
3-^7, its momentum is 77 '4, and as this momentum is lost by the vessel
every second, it is the force acting on the vessel.
A student who thinks for himself will see tliat this force is the same
whether a vessel is or is not in motion itself.
28 CALCULUS FOR ENGINEERS.
Imagine Bw to get smaller and smaller without limit and
use the symbol —^ for the limiting value of -A > ^^^ we have
-p = 2aic, a fact which is known to us already from our
CLX
squared paper*.
21. Note that when we repeat the process of differentiation
we state the result as -^ and tho answer is 2a. You must
become familiar with these symbols. If y is a function of
X, ,- is the rate of change of y with regard to a?; t4 is the
rate of change of -^ with regard to x.
Or, shortly; ^^ is the differential coefficient of ^
with regard to a? ; ;p is the differential coefficient of y with
regard to x.
Or, again ; integrate ~; and our answer is ~- ; inte-
grate ~ and our answer is y.
You will, I hope, get quite familiar with these symbols
and ideas. I am only afraid that when we use other letters
than x& and y^ you may lose your familiarity.
* Symbolically. Let 2/=/(a;)..,(l), where /(a;) stands for any expression
containing x. Take any value of x and calculate y. Now take a slightly
greater value of x say x + 8x and calculate the new y ; call it y + 8y
then y + dy^f{x + 8x) (2).
Subtract (1) from (2) and divide by dx.
Sy_ f{x + dx)-f{ x)
5^- 5i ^^^•
,^ , dw . f(x + 8x)-f(x)
What we mean by ^~ is the limiting value of —— — ^-^-^ as dx is
ax ox
made smaller and smaller without limit. This is the exact definition of -^ .
dx
It is quite easy to remember and to write, and the most ignorant person may
get full marks for an answer at an examination. It is easy to see that the
differential coefficient of of {x) is a times the differential coefficient of / (x)
and also that the differential coefficient of/ {x) + F.{x) is the sum of the two
differential coefficients.
MOTION. 29
The differential coefficient of
where a, b and c are constants, is
cix
The integral of + b-\-kx with regard to w is A-\-bx + ^kaf,
where A is any constant whatsoever.
Similarly, the integral of 6 + A;^ with regard to z is
A+bz-i- ^kz\
The integral oib-^kv with regard to v/i& A + bv-\- ^kv"^.
It is quite easy to work out as an exercise that if y=aa^^ then
~-=Mx^, and again that if y = ax^^ then -p=4a^. All these are
examples of the fact that if y = ax\ then -^—naod^'K
In working out any of these examples we take it that —■ becomes -—
OX OiX
dii
or that ^y=8xx-^ more and more nearly as dx gets smaller and
smaller without limit.
This is sometimes written y-\-by=y-{-bx . -f ,ov
f{x+bx)^f{x) + bx'^.. (I).
22. Uniformly accelerated motion.
dt
d^s
If acceleration, ^2~^ (!)•
Integrate and we have ~z=b-\-at— velocity v. Observe
that we have added a constant 6, because if we differentiate
a constant the answer is 0. There must be some information
given us which will enable us to find what the value of the
constant b is. Let the information be v = Vo, when i = 0.
Then b is evidently v^.
ds
So that velocity v— ^- =Vo + at (2).
30 CALCULUS FOR ENGINEERS.
Again integrate and 5 = c + ?;„< + ^at-. Again you will notice
that we add an unknown constant, when wc integrate. Some
information must be given us to find the value of the constant
c. Thus if s = So when ^ = 0, this 5o is the value of c, and so
we have the most complete statement of the motion
s = So + Vot + ^at^ (3).
If (3) is differentiated, we obtain (2) and if (2) is differen-
tiated we obtain (1).
23. We see here, then, that as soon as the student is
able to differentiate and integrate he can work the fol-
lowing kinds of problem.
I. If s is given as any function of the time, differentiate
and the velocity at any instant is found ; differentiate again
and the acceleration is found.
II. If the acceleration is given as some function of the
time, integrate and we find the velocity; integrate again
and we find the space passed through.
Observe that s instead of being mere distance may be
the angle described, the motion being angular or rotational.
Better then call it 0. Then 6 or -^- is the angular velocity
and S or J— is the angular acceleration.
24. Exercises on Motion with constant Accele-
ration.
1. The acceleration due to gravity is doiunwards and is
usually called g, g being 32*2 feet per second per second at
London. If a body at time is thrown vertically upwards
with a velocity of F© feet per second; where is it at the end
of t seconds ? If s is measured upwards, the acceleration is
— g and s^V^Jt — \gt^. (We assume that there is no resist-
ance of the atmosphere and that the true acceleration is g
downwards and constant.)
Observe that v=Vo—gt and that v=0 when Vo—gt =
or t = ~- . When this is the case find s. This eives the
highest point and the time taken to reach it.
When is s = again ? What is the velocity then ?
KINETIC ENERGY. 31
2. The body of Exercise 1 has been given, in addition
to its vertical velocity, a horizontal velocity Uo which keeps
constant. If oo is the horizontal distance of it away from
(I?X (17'
the origin at the time ^, — = and -7- = Uq, x — ujt. If we
call s by the new name ?/, we have at any time t,
y = V,t - \gt\
iXj ^^ UQVy
V of
and if we eliminate t, we find y=—^x — ^g-- which is a
Parabola.
3. If the body had been given a velocity V in the
dii'ection a above the horizontal, we may use V sin a for Vq
and V cos a for Uf^ in the above expressions, and from them
we can make all sorts of useful calculations concerning pro-
jectiles.
Plot the curve when V= 1000 feet per second and a = 45°.
Again plot with same V when a = 60°, and again when
a = 30°.
25. Kinetic Energy. A small body of mass vi is at
5 = when ^ = and its velocity is Vq, and a force F acts
upon it causing an acceleration F/ni. As in the last case
at any future time
v^Vf, + - t...(l), and s = -^Vot + I ~ t" (2),
(2) maybe written s^^tilv^A — t\ and it is easy to see
from (1) that this is s = \t{vQ-\-v), and that the average
velocity in any interval is half the sum of the velocities at
the beginning and end of the interval. Now the work done
by the force F in the distance s is Fs. Calculating F from
till
(1), F={v — Vq)— and multiplying upon s we find that the
t
work is ^m (v^ — Vq^) which expresses the work stored up in a
moving body in terms of its velocity. In fact the work
done causes ^mvo^ to increase to imv^ and this is the reason
why ^mv'^ is called the kinetic energy of a body.
32 CALCULUS FOR ENGINEERS.
Otherwise. Let a small body of mass iii and velocity v
pass through the very small space hs in the time ht gaining
velocity hv and let a force F be acting upon it. Now
ov
F= m X acceleration or ^= 77i -^ and Bs=^v , Bt so that
hv
F,hs=^ nivBt -^ =1^1' vBv = BE,
ot
if BE stands for the increase in the kinetic energy of the body
oE
-K- = m . -y. But our equations are only entirely true when
Bs, Bt, &c., are made smaller and smaller without limit :
Hence as -Y- = mv, or in words, "the differential coefficient of
av
E with regard to v is mv" if we integrate with regard to v,
E = \mv^ + c where c is some constant. Let E — when
v==0 m that c = and we have E — |mt;l
Practise differentiation and integration using other letters
than X and y. In this case -^ stands for our old -^^ If we
•^ dv dx
had had —■ = mx it might have been seen more easily thaf
y = Jmar* + c, but you must escape from .the swaddling bands
of X and y.
26. Exercise. If x is the elon/^ation of a spring
F
when a force F is applied and if a; = - , a representing the
stiffness of the spring ; F .Bx'is the work done in elongating
the spring through the small distance Bx. If F is gradually
increased from to i^ and the elongation from to x, what
strain energy is stored in the spring ?
The gain of energy from x to x-\-Bx is BE = F . Bx, or
iilF
rather ~=— = F = ax, hence E = ^ax^ + c. Now if ^ = when
x=^0, we see that c = 0, so that the energy E stored is
E=iax' = iFx..,{l).
It is worth noting that when a mass M is vibrating at the
end of a spiral spring ; when it is at the distance x from its
ELEOTRTO CTT1CUI^,„„,^ 33
"... ^< ■ '"'' ■ ^".
position of equilibrium, the potential energy is ^ax^ and the
kinetic energy is Jilfy- or the ^/tot^l, energy is ^Mv^ + ^ax"^....
Note that when a forc0 F is required to pi-oduce an
elongation or compression ^ in a rod, or a deflexion ^r in a beam,
and ii F=^ ax where a is some constant, the energy stored up
as strain energy or potential energy is ^ax^ or ^Fx.
Also if a Torque T is required to produce a turning
through the angle ^ in a shaft or spring or other structure,
and if r = aO, the energy stored up as strain energy or
potential energy is |a^ or ^T6. If T is in pound-feet and
is in radians, the answer is in foot-pounds.
Work done = Force x distance, or Torque x angle.
27. If the student knows anything about electricity let
him translate into ordinary language the improved Ohm's law
V = RC + L.dC/dt (1).
Observe that if R (Ohms) and L (Henries) remain con-
dC
stant, if C and -^ are known to us, we know F, and if the
law of F, a changing voltage, is known you may see that there
must surely be some means of finding C the changing current.
Think of L as the back electromotive force in volts when the
current increases at the rate of 1 ampere per second.
If the current in the primary of a transformer, and there-
fore the induction in the iron, did not alter, there would be
no electro-motive force in the secondary. In fact the E.M.F.
in the secondary is, at any instant, the number of turns of
the secondary multiplied by the rate at which the induction
changes per second. Rate of increase of / per second is what
we now call the differential coefficient of / with regard to
time. Although L is constant only when there is no iron or
else because the induction is small, the correct formula being
V= BC-h /V ,- (2), it is found that, practically, (1) with
L constant is of nearly universal application. See Art. 183.
28. If y = ax" and you wish to find -,- , I am afraid that
34 CALCULUS FOR ENGINEERS.
I must assume that you know the Binomial Theorem
which is : —
(x + by = a?« + nbo)''-^ + ^ ^^^ " ^ 62a;«-2
+ -^ ^ ^ h'x''-^ + &c.
It is easy to show by multiplication that the Binomial
Theorem is true when n = 2 or 3 or 4 or 5, but when n = ^ or
J or any other fraction, and again, when n is negative, you
had better perhaps have faith in my assertion that the
Binomial Theorem can be proved.
It is however well that you should see what it means by
working out a few examples. Illustrate it with n = 2, then
?i = 3, 71 = 4, &c., and verify by multiplication. Again try
n = — l, and if you want to see whether your series is correct,
just recollect that (x + h)~^ is j and divide 1 by a? + 6 in
the regular way by long division.
Let us do with our new function of co as we did with aao^.
Here y = aoo'\ y-\-hy = a{x + hxY =a[x^-\- n.hx . x^-^
Tl ill/ ~~ W
H — ^-^ — - {hx)' x^~^ + terms involving higher powers of hx].
Now subtract and divide by hx and you will find
!=•{
n . ^"-1 + ^^\ ^^ {hx) X''-' + &c.
We see now that as 8x is made smaller and smaller, in
the limit we have only the first term left, all the others
having in them Bx or (8xy or higher powers of Bx, and they
must all disappear in the limit, and hence,
^ = nax"-i. (See Notes p. 159.)
Thus the differential coefficient of a^ is 6x^, of x^^ it is
2^x^^, and of x~^ it is — |^~i
When we find the value of the differential coefficient of
any given function we are said to differentiate it. When
given -^ to find y we are said to integrate. The origin of
/"
. dec. 35
the words differential and integral need not be considered.
They are now technical terms.
Differentiate ax^ and we find nax'^~^.
Integrate naaf^^ and we find ax^ -\- c. We always add a
constant when we integrate.
Sometimes we write these, ^- (ax'^) = nax'^~^ and
dx^ ^
\nax'^~^ . dx = aaf\
Obsei've that we write / before and dx after a function
when we wish to say that it is to be integrated with regard
to X. Both the symbols are needed. At present you ought
not to trouble your head as to why these particular sorts of
symbol are used*.
You will find presently that it is not difficult to learn
how to differentiate any known mathematical function. You
will learn the process easily ; but integration is a process of
guessing, and however much practice we may have, ex-
perience only guides us in a process of guessing. To some
extent one may say that differentiation is like multiplica-
tion or raising a number to the 5t]i power. Integration is
like division, or extracting the 5th root. Happily for the
engineer he only needs a very few integi-als and these are
* When a great number of things have to be added together in an
engineer's office — as when a clerk calculates the weight of each little bit of a
casting and adds them all up, if the letter w indicates generally any of the
little weights, we often use the symbol I,io to mean the sum of them all.
When we indicate the sum of an infinite number of little quantities we
replace the Greek letter s or S by the long English s or / . It will be seen
presently that Integration may be regarded as finding a sum of this kind.
Thus if y is the ordinate of a curve ; a strip of area is y . dx and jy . dx
means the sum of all such strips, or the whole area. Again, if 8m stands for
a small portion of the mass of a body and r is its distance from an axis, then
r^ . 57/1 is called the moment of inertia of dm about the axis, and Sr^ . dm or
/
r^ . dm indicates the moment of inertia of the whole body about the axis.
Or if 5 F is a small element of the volume of a body and m is its mass per
unit volume, then ir^m . dV is the body's moment of inertia.
3—2
36 CALCULUS FOR ENGINEERS.
well known. As for the rest, lie can keep a good long list of
them ready to refer to, but he had better practise working
them out for himself.
Now one is not often asked to integrate n(Mf~^. It is too
nicely arranged for one beforehand. One is usually asked to
inteoi-ate 6a?"*...(l). I know that the answer is -,...(2).
How do I prove this ? By differentiating (2) I obtain (1), there-
fore I know that (2) is the integral of (1). Only I ought to
add a constant in (2), any constant whatever, an arbitrary
constant as it is called, because the differential coefficient of
a constant is 0. Students ought to work out several
examples, integrating, say, a*^, hsc^, hx^, ax~^, ci^, auo^. When
one has a list of differential coefficients it is not wise to use
them in the reversed way as if it were a list of integrals, for
things are seldom given so nicely arranged.
For instance J4>a^ .da; = a^. But one seldom is asked to
integrate ^oc^, more likely it will be Sx^ or 5./"^, that is given.
We now have a number of interesting results, but this
last one includes the others. Thus if
y = (f? or y — x^ or y = a?^ or 3/= a;"*,
we only have examples of y — x*^, and it is good for the
student to work them out as examples. Thus
ax
If 7i = 1 this becomes \x^ or 1. If ?i = it becomes ()x~'^
or 0. But we hardly need a new way of seeing that if y is a
constant, its differential coefficient is 0. We know that if
y = a-^hx + cx^ + ex^-\- &c. + (/x^\
f = + h + 2cx-^ Sex" + &c. + ng
with this knowledge we have the means of working quite
* I suppose a student to know that anj^thing to the power is unit}-. It
is instructive to actually calculate by logarithms a high i-oot of any number
to see how close to 1 the answer comes. A high root means a small power,
the higher the root the more nearly does the power approach 0.
Then ^ = + 6 + 2c^ + Sex" + &c. + ngx''-^
INTEGRATION OF *'»'. 37
half the problems supposed to be difficult, that come before
the engineer. ,
The two important things to remember now, are : If
y = ax'K then -~ = nax'^~^ ; and if -:,*- = bx^, then
-^ ' dx ' dx '
y = i^''"+i + c,
^ m + 1
where c is some constant, or
/
bx'^dx = *•'«+! + c.
m + 1
I must ask students to try to discover for themselves
illustrations of the fact that if y = x'\ then -i^ = ??a'"-^ I do
-^ dx
not give here such illustrations as happened to suit myself;
they suited me because they were my own discovery. I
would suggest this, however :
Take y = af\ Let x = l'()2, calculate y by logarithms.
Now let X = 1*03 and calculate y. Now divide the increment
of y by *01, which is the increment of x.
Let the second x be 1"021, and repeat the process.
Let the second x be 1*0201, and repeat the process.
It will be found that ^^ is approaching the true value of
^^ which is 5 (l-02y.
Do this again when y = x^"^ for example. A student need
not think that he is likely to waste time if he works for
weeks in manufacturing numerical and graphical illustrations
for himself. Get really familiar with the simple idea that
if V = j;" then -,- = nx'^-^ :
^ dx '
that \ax^ . dx — — — - ip*+i + constant :
J 5 + 1
that \av^ . dv = \f^^ + constant.
j sH-1
Practise this with s= '7 or "8 or I'l or —5 or —"8, and use
other letters than x or v.
/
38 CALCULUS FOR ENGINEERS.
29. Exercises. Find the following Integrals. The
constants are not added.
Ix". dx. Answer, \x^. \v^ . dv. Answer, ^if.
/v~*. dv. The answer i.> - — v^~^-
J 1-5
j\^V' .dv or jv^ , dv. Answer, ^v^,
r^.dt. Answer, 2e^
j-dx or lx~^.dx. Here the rule fails to help us for
of . .
we get -zr which is x , and as we can always subtract an in-
finite constant our answer is really indeterminate. In our
work for some time to come we need this integral in only
one case. Later, we shall prove that
I— dx = log X, and 1 ^— dx = log (x + a),
and if y = log a?, -— = - and l-dv = log v.
If j9 = av^, then ~ = Sav\
If t; = mrK then ^ = _ ^^^rl
at
30. If pv = Rt, where i^ is a constant. Work the
following exercises. Find -— , if v is constant. Answer, — .
Find ,- , if « is constant. Answer, — .
at ^ p
The student knows already that the three variables p, v
and t are the pressure volume and absolute temperature of a
''dp
gas. It is too long to Avrite -^ when v is constant." We
ctt
1.
PARTIAL DIFFERENTIATION. 39
shall use for this the synibol [-4i) , the brackets indicating that
the variable not there mentioned, is constant.
Find (;/ ) • Answer, As p=Rt . v~^ we have [-T-) = — Rtv~\
and this simplifies to —pv~^.
Find f ^' ] . Answer, As v = Rt . p~^ we have (;t- ) = — Rtp~^,
and this simplifies to — vp''\
Find (I) . Answer, As i = | . /, we have Q = ^ .
Find the continued product of the second, fifth, and third of
the above answers and meditate upon the fact that
fdv\ fdt\ (dp\___
[dtj [dp) [dv)"
Generally we may say that if u is a function of two
variables x and y, or as we say
'dv>
\dxj
efficient of u with regard to x when y is considered to
be constant.
These are said to be partial differential coefiKcients.
31. Here is an excellent exercise for students : —
Write out any function of x and y ; call it u.
Find ( ^7- ) • Now differentiate this with regard to y,
assuming that x is constant. The symbol for the result is
dhi
dy .dx'
It will always be found that one gets the same answer if
one differentiates in the other order, that is
d^u d^u
.(3).
then we shall use the symbol ( t- ) to mean the differential co-
dy . dx dx . dy
40 CALCULUS FOR ENGINEERS.
Thus iry 11 = 0,^ + if+ a(i?ij + hx]f.
Again, \-^\ = + 3?/= 4- ax- + ^hxy,
and r/""V = ^ + ^ + 2aa; + 2bij,
which is the same as before.
A student ought not to get tired of doing this. Use other
letters than w and y, and work many examples. The fact
stated in (3) is of enormous importance in Thermodynamics
and other applications of Mathematics to engineering. A
proof of it will be given later. The student ought here to
get familiar with the importance of what will then be
proved.
32. One other thing may be mentioned. Suppose we
have given us that ii is a function of x and y, and that
(du
^ =aa? + hf-\-co(?y-\-gxy\
Then the integral of this is
11 = lax* + by^x + Jc^y + ^gxy -\-f( y),
where f{y) is some arbitrary function of y. This is added
because we always add a constant in integration, and as ?/ is
regarded as a constant in finding (-7- J we add f(y), which
may contain the constant y in all sorts of forms multiplied
by constants.
33. To illustrate the fact, still unproved, that if y = log x,
then ^- = - . A student ouo^ht to take such values of x as 3,
dv X ^
3001, 3*002, 3"003 &c., find y in every case, divide increments
of y by the corresponding increments of x, and see if our rule
holds good.
STEAM AND POWER. 41
Note that when a mathematician writes logo) he always
means the Napierian logarithm of x.
34. Example of / — =log « + constant.
It is proved in Thermodynamics that if in a heat engine the work-
ing stuff' receives heat // at temperature t, and if t^ is the temperature
of the refrigerator, then the work done by a perfect heat engine would
be ir.'-Jo,orH(l-f).
If one pound of water at Iq is heated to ^j, and wc assume that
the heat received per degree is constant, being 1400 foot-lbs. ; what is
the work which a j^erfect heat engine would give out in equivalence
for the totiil heat ? Heat energy is to be expressed in foot-pounds.
To raise the temperature from t to t-\-dt the heat is 14005^ in foot-lb.
This stands for JI in the above expression. Hence, for this heat we
have the equivalent work 8 W= l-iOOdt ( 1 - -^ j , or, rather,
"^^ =1400-1400^.
at t
Hence W= \AQOt - UOOIq log t + constant.
Now W=0 when t=tQ,
= 1400^0 - 1400^0 log ^0 + constant,
therefore the constant is known. Using this value we find equivalent
work for the heat given from t^ to ^i= 1400 (t^ - to) - 1400^0 ^^S j" •
If now the pound of water at t^ receives the heat X^ foot-lb. (usually
called Latent Heat) and is all converted into steam at the constant
temperature t^ , the work which is thermodynamically equivalent to this
is L^ll--]. We see then that the work which a perfect steam
engine would give out as equivalent to the heat received, in raising
the pound of water from t^ to t^ and then evaporating it, is
1400 {t, - 1,) - urn, loge ^^ + X, (^1 - ^f^ .
Exercise. What work would a perfect steam engine perform per
pound of steam at ^^ = 439 (or 102 lb. per sq. inch), or 165^0, if ^q = 374
or 100' C. Here L^ = 681 ,456 foot-pounds.
The work is found to be 107,990 ft.-lb. per lb. of steam. Engineers
usually wish to know how many pounds of steam are used per hour
per Indicated Horse Power, w lb. per hour, means pr! ^^ ft.-lb. per
minute. Putting this equal to 3.3(KM) we find 2V to be 18-35 lb. of steam
per hour per Indicated Horse Power, as the requirement of a perfect
steam engine working between the temperatm'es of 165° C and 100° C.
42
CALCULUS FOR ENGINEERS.
35. Exercises. It is proved iii Thermodynamics when
ice and water or water and steam are together at the same
temperature, if Si is the volume of unit mass of stuff in the
higher state and Sq is the volume of unit mass of stuff in the
lower state. Then
where t is the absolute temperature, being 274 + 6" C, L
being the latent heat in unit mass in foot-pounds. If we
take L as the latent heat of 1 lb. of stuff, and Si and Sq are
the volumes in cubic feet of 1 lb. of stuff, the formula is
still correct, p being in lb. per sq. foot.
I. In Ice-water, So = '01747, 8^ = '01602 at ^=274 (cor-
responding to 0° C), p being 2116 lb. per sq. foot, and
X = 79 X 1400. Hence ^ = - 278100.
at
And hence the temperature of melting ice is less as the
pressure increases ; or pressure lowers the melting point of
ice ; that is, induces towards melting the ice. Observe the
quantitative meaning of -^ ; the melting point lowers at the
rate of '001 of a degree for an increased pressure of 278 lb.
per sq. foot or nearly 2 lb. per sq. inch.
II. Water Steam. It seems almost impossible to
measure accurately by experiment, Si the volume in cubic
feet of one pound of steam at any temperature. So for water
is known. Calculate s^ — Sq from the above formula, at a few
temperatures having from Regnault's experiments the follow-
ing table. I think that the figures explain themselves.
e^c
t
absolute
pressure in
lb. per sq.
inch
P
lb. per
sq. foot
Sp
5t
assumed
dp
dt
L
in foot-
pounds
Sj-So
100
374
14-70
211G-4
81-5
105
379
17-r)3
2524
94
87-8
740,710
22-20
110
384
20-80
2994
LATENT HEAT. 43
It is here assumed that the vakie of dpjdt for IDS'" C. is
half the sum of 81 "5 and 94. The more correct way of
proceeding would be to plot a great number of values of
hplht on squared paper and get dpjdt for 105° C. more
accurately by means of a curve, f
s, - So for 105° C. = 740710 -- (379 x 87*8) = 22-26. Now
So = "OIG for cold water and it is not worth while making any
correction for its warmth. Hence we may take s^ = 22*28
which is sufficiently nearly the correct answer for the present
pui-pose.
Example. Find Si for 275° F. from the following, L being
t°F. 248°
4152
257° 266° I 275'
284° ; 293°
302°
4854 I 5652 6551
7563 8698
9966
Example. If the formula for steam pressure, p=a6^ where a and b
are known numbers, and 6 is the temperature measured from a certain
zero which is known, is found to be a useful but incorrect formula
ibr re])reseiiting Reguault's experimental results ; deduce a fornuila for
the volume s^ of one p<nnid of steam. We have also the well known
formula for latent heat L—c-et^ where t is the absolute temperature
and c and e are known numbers. Hence, as i? which is the same as
dv • ' de
^ is haB^-\ s^-SQ=(c-et)^tbae^-\
After subjecting an empirical formula to mathematical operations
it is wise to test the accuracy of the result on actual experimental
numbers, as the formula represents facts only approximately, and the
small and apparently insignificant terms in which it difl:ers from fact,
may become greatly magnified in the mathematical operations.
36. Study of Curves. When the equation to a new
curve is given, the practical man ought to rely first upon his
power of plotting it upon squared paper.
Very often, if we find ~ or the slope, everywhere, it
gives us a good deal of information.
If we are told that a.\, y^ is a point on a curve, and we are
asked to find the equation to the tangent there, we have
simply to find the straight line which lias the same slope
as the curve there and which passes through x^, y^. The
normal is the straight line which passes through x^, y^ and
whose slope is minus the reciprocal of the slope of the curve
there. See Art. 13.
CALCULUS FOR ENGINEERS.
P (fig. 8) is a point in a curve APB at which the
tangent FS and the normal FQ are drawn. OX and Y are
the axes. OR — x, RF — y, tan FSR = ;,- •, the distance SR
is called the subtangent ; prove that it is equal to ?/ -i- -^ .
The distance RQ is called the subnormal; it is evidently
dii
equal to y -^ . The length of the tangent FS will be found
to be y\J\-\-{,\ , the length of the normal FQ is
y W^^ ^£i • ^^^ Intercept OS is ^- - y '^ .
Example 1. Find the length of the sub tangent and sub-
normal of the Parabola y = mx'^,
- dy
dx
2nix.
Hence
Subtangent = mx- -=- 2?/iif or ^x.
Subnormal = y x 2nix or 2m^x^.
Example 2. Find the length of the subtangent of y = Wio;",
dy_
dx
mnx'
Subtangent = nix'^ -i- vinx'^~^ — xju.
CURVES. 45
Example 3. Find of Avhat curve the subnormal is constant
in length,
dy d^ 1
y-f- = a or -r = ~ n.
^ dx dy a-^
The integral of - ?/ Avith regard to y is a* = ~ 3/^ + a con-
Cl Act
stant h, and this is the equation to the curve, where h may
have any value. It is evidently one of a family of parabolas.
(See Art. 9 where xs and y's are merely interchanged.)
Example 4. The point ^ = 4, ?/ = 3 is a point in the para-
bola y = '^a^. Find the equation to the tangent there. The
dy
slope is -^ = 1- X f A'"^ or, as ;r = 4 there, the slope is f x ^ or -| .
The tangent is then, y = m+ g^. To find m we have ^ = 3
when a? = 4 as this point is in the tangent, or 3 = 9/1 + | x 4,
so that ??i is 1 J and the tangent is 3/ = li|- + %x.
Example 5. The point ^ = 32, 1/ = 3 is evidently a point in
the curve y = 2-{-^x-'. Find the equation to the normal there.
dij
The slope of the curve there is ^ = -^x' * = yj^ and the
slope of the normal is minus the reciprocal of this or — 160.
Hence the normal is y = m — \ Q>Ox. But it passes through
the point x=S2, y = S and hence S = m— KiO x 32.
Hence m = 5123 and the normal is y = 5123 — 160;r.
Example 6. At what point in the curve y = aa?""" is there
the slope 6? dy
-f = - nax-'^-K
ax
The point is such that its x satisfies — nax~^^~'^ = h or,
1
— j-j . Knowing its x we know its y from the
equation to the curve. It is easy to see and well to remem-
ber that if ^1, 3/1 is a point in a straight line, and if the slope
of the line is 6, then the equation to the line most quickly
written is
x — x^
46
CALCULUS FOR ENGINEERS.
Hence the equation to the tangent to a curve at the
point a?i, 2/1 on the curve is
•^^ y^ = the -^ at the point.
X — Xx dx
And the equation to the normal is
x — Xi aij
- = — the ^ at the point.
Exercise 1. Find the tangent to the curve x^y^^ = a at
the point x^, y^ on the curve. Answer, — X'\- - y=m-\- n.
Exercise 2. Find the normal to the same curve.
Answer, — (a? — a?,)
m
Xi
(!/ - y.) = 0.
Exercise 3. Find the tangent and normal to the parabola
y" = ^ax at the point where x = a.
Answer, y = x -\- a, y = Sa — x.
Exercise 4. Find the tangent to the curve
y =^ a -{-hx ■\- ca^ + ex^
at a point on the curve x^, y^.
Answer, ^ ~^-^ = 6 + 2cXi -f ^ex^^
X — Xi
37. When y increases to a certain value and then
diminishes, this is said to be a maximum value of y;
when y diminishes to a certain value and then increases,
this is said to be a minimum value of y. It is evident
that for either case -^ = 0. See Art. 16 and fief. 6.
dx ^
Example 1. Divide 12 into two parts such that the
product is a maximum. The practical man tries and easily
finds the answer. He tries in this sort of way. Let x be one
part and 12— x the other. He tries x = 0, x = l, x = 2, &c.,
in every case finding the product. Thus
a^|0|l|2i3 4 5 6 7 8 9
Product I i 11 i 20 27 32 35 36 35 32 27
MAXIMA AND MINIMA. 47
It seems as if ^' = 6, giving the product 36, were the correct
answer. But if we want to be more exact, it is good to get
a sheet of squared paper; call the product y and plot the
corresponding values of x and y. The student ought to do
this himself
Now it is readily seen that where y has a maximum or a
minimum value, in all cases the slope of a curve is 0. Find
then the point or points where dy\dx is 0.
Thus if a number a is divided into two parts, one of them
X and the other a — x, the product is y = ^ (a — a;) or ax — x^,
and £ =a — 2x. Find where this is 0. Evidently where
2x — a or x= \a.
The practical man has no great difficulty in any of his
problems in finding whether it is a maximum or a minimum
which he has found. In this case, let a = 12. Then x = Q
gives a product 36. Now if a? = 5 999, the other part is 6*001
and the product is 35-999999, so that a? = 6 gives a greater
product than x = 5*999 or x= 6*001, and hence it is a maxi-
mum and not a minimum value which we have found. This
is the only method that the student will be given of dis-
tinguishing a maximum from a minimum at so early a period
of his work.
Example 2. Divide a number a into two parts such
that the sum of their squares is a minimum. If x is one
part, a—x is the other. The question is then, if
y = ar' + (a — xf, when is y a minimum ?
2/ = 2^-2 + a^ - 2ax,
-^= 4^ — 2a, and this is when x — ^a.
dx 2
Examjile 3. When is the sum of a number and its
reciprocal a minimum ? Let x be the number and y =x + -.
When is 2/ a minimum ?
The differential coefficient of - or x~'^ being — x~^, we have
dy -, 1 11..^. ^ .
-IT" = 1 -, and this is when a? = 1.
48
CALCULUS FOR ENGINEEKS.
The student ought to take numbers and a sheet of squared
paper and try. Trying a;= 100, 10, 4 &c. we have
100
2/
100-01
10
4
2
1
i
i
i
101
4-25
2-5
2
2-5
^
4i
Now let him plot x and i/ and he will see that y is a
minimum when a; = 1.
Example 4. The strength of a rectangular beam of
given length, loaded and supported in any 2:)articular way, is
proportional to the breadth of the section multiplied by the
square of the depth. If the diameter a is given of a cylindric
tree, what is the strongest beam which may be cut from it ?
Let X be its breadth. Then if you draw the rectangle inside
the circle, you will see that the depth is Ja'^ — x-. Hence
the strength is a maximum when y is a maximum if
y = x((r--x-),
or 3/ = a\x — x-\
j^ = a- — 3a", and this is when x—~=^,
dx V3
In the same way find the siiffest beam which may be cut
from the tree by making the breadth x the cube of the depth
a maximum.
This, however, mav wait till the student has read
Chap. III.
Example 5. Experiments on the explosion of mixtures
(at atmospheric pressure) lead to a roughly correct rule
p = 83-3-2a?,t
where p is the highest pressure produced in the explosion,
and X is the volume of air together with products of previous
combustions, added to one cubic foot of coal gas before ex-
plosion. Taking px as roughly proportional to the work done
in a gas engine during explosion and expansion ; what value
of X will make this a maximum ?
FUEL ON A VOYAGE. 49
That is, when is 83^3 — S'2x^ a maximum ? Answer, When
83 — 6'4fX= 0, or x is about 13 cubic feet.
I am afraid to make Mr Grover responsible for the above
result which I have drawn from his experiments. His most
interesting result was, that of the above 13 cubic feet it is
very much better that only 9 or 10 should be air than that
it should all be air.
Example 6. Prove that ax — x^ is a maximum when
x — \a.
Example 7. Prove that x — x^ is a maximum when
^=:^V3.
, Example 8. The volume of a circular cylindric cistern
being given (no cover) when is its surface a minimum ?
Let X be the radius and y the length ; the volume is
irafy = a, say (1).
The surface is ttx^ + 27rxi/ (2).
When is this a minimum ?
From (1), y is — ; using this in (2) we see that we
TTX"
must make
, 2a . .
TTX^ H a mmimum,
X
27rx r=0 or x^ = - .
a? TT
„ iTX^y
x^= — — or x=^y.
TT
The radius of the base is equal to the height of the cistern.
Example 9. Let the cistern of Ex. 8 be closed top and
bottom, find it of minimum surface and given volume.
The surface is 2irx- + ^irxy, and proceeding as before we
find that the diameter of the cistern is equal to its height.
Example 10. If v is the velocity of water in a river and
X is the velocity against stream of a steamer relatively to the
luater, and if the fliel burnt per hour is a + bx^; find the
p. 4
60 CALCULUS FOR ENGINEERS.
velocity a? so as to make the consumption of fuel a minimum
for a given distance m. The velocity of the ship relatively
to the bank of the river is ac—v, the time of the passage is
, and therefore the fuel burnt during the passage is
OS —~ V
m(a-\- ha?)
x — v
Observe that a -\- hx^ with proper values given to a and h
may represent the total cost per hour of the steamer, in-
cluding interest and depreciation on the cost of the vessel,
besides wages and provisions.
You cannot yet differentiate a quotient, so I will assume
a?
a — 0, and the question reduces to this: when is a
x — v
minimum ? Now this is the same question as : — when is
X "^ V
— r— a maximum ? or when is ar^ — vx~^ a maximum ? The
a?
differential coefficient is — 2ar^ + ^vx~\ Putting this equal
to we find x = '^v, or that the speed of the ship relatively to
the water is half as great again as that of the current.
Notice here as in all other cases of maximum and mini-
mum that the engineer ought not to be satisfied merely with
such an answer. x='^v is undoubtedly the best velocity, it
makes a?j{x — v) a minimum. But suppose one runs at less
or more speed than this, does it make much difference ? Let
V = 6, the best a? is 9,
— x-=243 if «; = 9.
.-r — 6
= 250 if a; = 10.
= 256 if a;=8;
and these figures tell us the nature of the extra expense in
case the theoretically correct velocity is not adhered to*.
* Assuming that you know the rule for the differentiation of a quotient
— usually learnt at the very beginning of one's work in the Calculus, and
without assuming a to be as above, we have
(a;-i?)36a;*=a + 6x',
26x3_3tv«2_rt (1).
CUllRENT FROM CELLS. 51
Example 11. The sum of the squares of two factors of a
is a minimum, find them. If x is one of them, - is the
X
Qj^ (it/ ZiO'
other, and y'=x^-{-— is to be a minimum, -^ = 2x , and
x^ ax x^
this is when x^ = ct- ^r x — \'a.
Example 12. To arrange n voltaic cells so as to obtain
the maximum current through a resistance R. Let the E.M.F.
of each cell be e and its internal resistance r. If the cells
are arranged as x in series, n-jx in parallel, the E.M.F. of the
battery is xe, and its internal resistance is - - . Hence the
current G == xe-^ I \- R]
a
As the student cannot yet differentiate a quotient, we
shall say that (7 is a maximum when its reciprocal is a
... fx'r . „\ xr R
minimum, so we ask when is h R] -^^e or 1 —
\ n J n X
Given the values of a, 6 and v the proper value of x can be found by trial.
Thus let the cost per day in pounds be 30 + ^V^^ so that a = 30, 6 = ^ and
let r = 6. Find x from (1) which becomes
.r3_ 9^2- 300 = (2).
I find that a; = 11-3 is about the best answer.
Tliis is a cubic equation and so has three roots. But the engineer needs
only one root, he knows about how much it ought to be and he only wants
it approximately. He solves any equation whatsoever in the following
sort of way.
Let x^-9x--S00 be called f{x). The question is, what value of x
makes this 0? Try a;= 10, f{x) turns out to be - 200,
X
10
j
8
12
11
11-3 1
/(^)
-200 I
-360
+ 176
-57
-6
whereas we want it to be 0. Now I try a; = 8, this gives -360 which is
further wrong. Now I try 12 and I get 176 so that x evidently lies between
10 and 12. Now I try 11 and find - 57. It is now worth while to use
squared paper and plot the curve y=f{x) between a; = 10 and a;=12. One
can find the true answer to any number of places of decimals by repeating
this process. In the present case no great accm*acy is wanted and I take
a: = 11*3 as the best answer. Note that the old answer obtained by assuming
a = is only 9. A practical man will find much food for thought in thinking
of these two answers. Note that the captain of a river steamer must always
be making this sort of calculation although he may not put it down on
paper.
4—2
52 CALCULUS FOR ENGINEERS.
. . ?' jR
minimum ? Its differential coefficient is and this is
when R = — , which is the internal resistance of the battery.
Hence we have the rule : Arrange the battery so that its
internal resistance shall be as nearly as possible equal to the
external resistance.
Example 13. Voltaic cell of E.M.F. = e and internal re-
sistance r; external resistance i?. The current is (7= ^z.
The power given out is P = RC\ What value of R will make
P a maximum ? P = R
{r + Rf
To make this suit such work as we have already done we may
(?• + Ry
say, what value of R will make ^^ — j~^ a minimum, or
7^ + 2Rr + jR^
D ~ <^r ')'^R~^ + 2r + P a minimum ?
Putting its diiferential coefficient with regard to R equal
to we have - i-^R-^ -t- 1 = so that R = r, or the external
resistance ought to be equal to the internal resistance.
Example 14. What is the volume of the greatest box
which may be sent by Parcels post ? Let x be the length,
y and z the breadth and thickness. The P. 0. regulation is
that the length plus girth must not be greater than 6 feet.
That is, we want v = xyz to be a maximum, subject to the
condition that x-\-2{y + z) = Q. It is evident that y and z
enter into our expressions in the same way, and hence y = z.
So that a; + 4i/ = 6 and v = xy^ is to be a maximum. Here as
a; = 6 — 4^ we have -y = (6 ■- 4^) y^ or 6^* — 4^' to be a maxi-
fj'i)
mum. Putting -T- = we have 12^ — 123/^ = 0. Rejecting
v = for an obvious reason, y = 1, and hence our box is 2 feet
long, 1 foot broad, 1 foot thick, containing 2 cubic feet.
Find the volume of the greatest cylindric parcel which may
be sent by Post. Length being I and diameter d,l + ird — Q
IT . . 4
and -r Idr is to be a maximum. Answer, I = 2 feet, d = — feet,
4 TT
Volume = 8 -^ TT or 2'55 cubic feet.
WATER IN STEAM CYLINDER. 58
Example 15. Ayrton-Perry Spring. Prof. Ayrton and
the present writer noticed that in a spiral spring fastened at
one end, subjected to axial force F, the free end tended to
rotate. Now it was easy to get the general formula for the
elongation and rotation of a spring of given dimensions, and
by nothing more than the above principle we found what these
dimensions ought to be for the rotation to be great.
Thus for example, the angle of the spiral being a the
rotation was proportional to sin a. cos a. It at once followed
that a ought to be 45°.
Again, the wire being of elliptic section, x and y being
the principal radii of the ellipse, we found that the rotation
was proportional to
To make this a maximum, the section (which is propor-
tional to xy) being given. Let xy = 5, a constant, then the
above expression becomes
"-— ^ — r, and this is to be a maximum.
s^ o sy^
Here we see that there is no true maximum. The larger
we make y or the smaller we make y (for small values of y
the rotation is negative but we did not care about the direc-
tion of our rotation, that is, whether it was with or against
the usual direction of winding up of the coils) the greater is
the rotation. This is how we were led to make springs of
thin strips of metal wound in spirals of 45°. The amount
of rotation obtained for quite small forces and small axial
elongations is quite extraordinary. The discovery of these very
useful springs was complete as soon as we observed that any
spring rotated when an axial force was applied. Students
who are interested in the practical application of mathematics
ought to refer to the complete calculations in our paper
published in the Proceedings of the Royal Society of 1884.
Example 16. From a Hypothetical Indicator Diagram
the indicated work done per cubic foot of steam is
w — 144^1 (1 -f- log r) — 144?"j[)3 — x.
54 CALCULUS FOR engineehs.
where jh and jXi arc the initial and back pressures of the
steam ; r is the ratio of cut off (that is, cut off is at -th of
the stroke) and a? is a loss due to condensation in the
cylinder, w depends upon r.
1st. If X were 0, what value of r would give most
indicated work per cubic foot of steam ?
We must make .— = 0, and we find — =^ — 144«.i =
or ?' = — . If it is brake energy which is to be a maximum
per cubic foot of steam, we must add to p^ a term represent-
ing engine friction.
2nd. Mr Willans found by experiment in non-condensing
engines that ?• = — ^— gave maximum indicated w. Now
if we put in the above -,— = we have ^ — 144^^ — i =0.
So that ~~ = ^^^ (p, + 10) - 144^3 or ^^^ = 1440. So
that .r = 1440r + constant. Hence Mr Willans' practical rule
leads us to the notion that the work lacking per cubic foot of
steam is a linear function of 7\
This is given here merely as a pretty exercise in maxima
and*minima. As to the practical engineering value of the
result, much might be said for and against. It really is as if
there was an extra back pressure of 10 lb. per sq. inch which
represented the effect of condensation.
Mr Willans found experimentally in a non-condensing
engine that the missing water per Indicated Horee Power
hour is a linear function of ?• using the same steam in the
boiler, but this is not the same as our .^. We sometimes
assume the ratio of condensed steam to indicated steam to
be proportional to log r, but a linear function of r will agree
just as well with such experimental results as exist.
Example 17. The weight of gas which will flow
per second through an orifice from a vessel where it is at
ECONOMY IN CONDUCTORS. 55
pressure />o into another vessel where it is at the pressure jfj
is proportional to «> V 1 — « ^ ; where a is pjpf^ and 7 is a
known constant, when is this a maximum ? That is, when is
ay — a y a. maximum ? See Art. 74, where this example
is repeated.
Differentiating with regard to a and equating to we
find j^
In the case of air 7 = 1*41 and we find p = "527^0 > that is,
there is a maximum quantity leaving a vessel per second
when the outside pressure is a little greater than half the
inside pressure.
Example 18. Taking the waste going on in an electric
conductor as consisting of (1) the ohmic loss ; the value of
G^r watts, where r is the resistance in ohms of a mile of going
and coming conductor and C is the current in amperes;
(2) the loss due to interest and depreciation on the cost of
the conductor. I have taken the price lists of manufacturers
of cables, and contractors' prices for laying cables, and I find
that in every case of similar cables, similarly laid, or suspended
if overhead, the cost of a mile of conductor is practically
proportional to the weight of copper in it, that is, inversely
proportional to the resistance, plus a constant. The cost of
it per year will depend upon the cost of copper per ton,
multiplied by the number taken as representing rate per
cent, per annum of interest and depreciation. We can state
this loss per year or per second, in money per year or per
second and the ohmic loss is in watts. We cannot add them
together until we know the money value per year or per
second of 1 watt. There are three things then that decide
the value of the quantity which we call t^. I prefer to
express the total waste going on in watts rather than in
pounds sterling per annum and I find it to hey— C^r -\ f- 6,
where h is some constant. The value of t may be taken as
anything from 17 to 40 for the working of exercises, but
66 CALCULUS FOR ENGINEERS.
students had better take figures of their own for the cost ot
power, copper and interest*.
For a given current (7, when is the y, the total waste, a
minimum ? that is, what is the most economical conductor for
a given current ? ^~^^~';^ ^^^ *^^^ ^^ ^ when r — -^.
Thus if ^ = 40, r = ^.
Now if a is the cross section of the conductor in square
•04
inches, r = — nearly, so that (7= 1000a, or it is most econom-
ical to provide one square inch of copper for every 1000
amperes of current.
When w is a function of more than one independent
* The weight of a mile of copper, a square inches in cross section, is to
be figured out. Call it ma tons. If p is the price in pounds sterling of a
ton of copper, the price of the cable may be taken as, nearly, pma + some
constant. If R is the rate per cent, per annum of interest and depreciation,
then the loss per annum due to cost of cable may be expressed in pounds as
R
-r^l'wa + some constant. If £1 per annum is the value of %v watts, (ob-
100
serve that this figure w must be evaluated with care. If the cable is to
have a constaiat current for 24 hours a day, every day, w is easily evaluated),
then the cost of the cable leads to a perpetual loss of zrj-r. wpma + some
100
constant. Now taking a= — , we see that our t^ is -^7=7^77 .
** r 2000
Men take the answer to this problem as if it gave them the most econom-
ical current for any conductor under all circumstances. But although the
above items of cost are most important, perhaps, in long cables, there are
other Items of cost which are not here included. The cost of nerves and
eyesight and comfort if a light blinks ; the cost in the armature of a dynamo
of the valuable space in which the current has to be carried.
If a man will only write down as a mathematical expression the total
cost of any engineering contrivance as a function of the size of one or more
variable parts, it is quite easy to find the best size or sizes; but it is not
always easy to write down such a function. And yet this is the sort of
problem that every clever engineer'is always working in his head; increasing
something has bad and good effects; what one ought to do is a question in
' maxima and minima.
Notice also this. Suppose we find a value of x which makes y a maxi-
mum ; it may be, that quite different values of x from this, give values of
y which are not very different from the maximum value. The good practical
engineer will attend to matters of this kind and in such cases he will not
insist too strongly upon the use of a particular value of x.
ELECTRIC CONDUCTORS. 57
variable^ say x and y. Then | — - J = 0, y being, considered
constant during the differentiation, and f — j = 0, x being
considered constant during the differentiation, give two equa-
tions which enable the values of x and y to be found which
will make u a maximum or a minimum. Here, however,
there is more to be said about whether it is a minimum or
a maximum, or a maximum as to x and a minimum as to y,
which one has found, and we cannot here enter into it.
Sometimes in the above case although it is a function of
X and y, there may be a law connecting x and y, and a little
exercise of common sense will enable an engineer to deal
with the case. All through our work, that is what is wanted,
no mere following of custom; a man's own thought about his
own problems will enable him to solve very difficult ones
with very little mathematics.
Thus for example, if we do not want to find the best conductor
for a given current of Electricity: if it is the Power to be
delivered at the distant place that is fixed. If the distance is n miles,
and the conductors have a resistance of r ohms per mile (go and return),
if \\ is the potential, given, at the Generating end, and C is the
current. Then the potential at the receiving end being T' ; F^ - T^= Cnr.
CV=PiH fixed, and the cost per mile is y = C^r + -...{l), where t^ is
known. When is y a minimum?
Here both C and r may vary, but not independently. V= V^ - Cnr
and P=(7F, — (7%r ... (2). One simple plan is to state ?/ in terms of r
CV —P
alone or of C alone. Thus r from (2) is r= — ^ — (3).
Substituting for this jn (1), we get
CV,-P . f-a^n
y=-ir-^cv;^p ^^^-
Here everything is constant except (7, so we can find the value of C
to make y a minimum, and when we know C we also know r from (3).
At present the student is supposed to be able to difibrentiate only
^", so he need not proceed with the problem until he has worked a few
exercises in Chap. III.*
* To differentiate (4) is a very easy exercise in Chap. III. and leads to
j>, = — i + - — - — -/xr^ — 5svo > and on putting this equal to we obtain
d(J n {CVi~Py JT o
the required value of C. It would not be of much use to proceed further
58 CALCULUS FOR ENGINEERS.
In ray Cantor lectures on Hydraulic Machinery, I wrote out an
expression for the total loss in pounds per annum in Hydraulic
transmission of povirer by a piiwj. I gave it in terms of the
maximum pressure, the power sent in, and the diameter d of the pipe.
It was easy to choose d to make the total cost a minimum. If how-
ever I had chosen jo, the pressure at the receiving end as fixed, and the
power delivei-ed as fixed, and therefore Q, the cubic feet of water per
second, and if I had added the cost of laying as proportional to the
square of the diameter, I should have had an expression for the total
cost like i(^ pqi
when the values of a, h and c depend upon the cost of power, the
interest on the cost of iron, &c. This is a minimum when its differen-
tial coefficient with regard too? is zero or ^cld—hal(^d'^-\-Zh^^M~^^
and d can be obtained by trial. The letters 6 and c also involv^ the
strength of the material, so that it wjxs possible to say whether wrought
iron or ca»t iron was on the whole the cheaper. But even here a term
is neglected, the cost of the Engine and Pumps.
The following example comes in conveniently here, although it is
not an example of Maximum or Minimum.
An Electric Conductor gives out continuously a amperes
of current in every mile of its length. Let x be the distance of any
jKiint in miles from the end of the line remote from the generator, let
C be the current there and V the voltage. Let r ohms per mile be the
resistance of the conductor (that is, of one mile of going and one mile
of coming conductor). The current given out in a distance hx is dC,
or rather hx -j- , and the power is bx . V '-f- y «o that if P is the
power per mile (observe the meaning of per),
''='-f :•••(')•
Also if V is voltage at x and V+bVat x+8x] —
As the i-esistance is i' . 8x the current is dV-rr . dx^ or rather, since
these expressions are not correct until 8x is supposed smaller and
smaller witliout limit,
^-l^ (^)-
unless we had numerical values given us,
91 = 10 miles, P= 20000 watts, f^=H)00, find G and then r.
Consult a Paper in the Joimial of the Institution of tJie Society of Tele-
graph Engineers, p. 120, Vol. xv. 1886, if there is any further difficulty.
It has not yet been sufficiently noted that if V^ and P and r are given,
there is a limiting length of line
n=JV/4rP,
and when this is the case P is exactly equal to the ohmic loss in the con-
ductor.
ELECTRIC TRACTION.
59
As -T- = ^h C=t<.<,- if C is when .r = 0.
Hence if r is constant (2) becomes
dV
rax = , so that T^ = Fq + i ^^-^-^
.(3),
^o being the voltage at the extremity of the line. .
(1) becomes P^aV^+^ra'^x'^ (4).
Taking ro = 200 volts, a = 25 amperes per mile, r=l ohm per mile,
it is easy to see by a numerical example, how the power dispensed per
mile, and the voltage, diminish as we go away from the generator.
X
''
P
200
5000
1
212-5
5312
2
250
6250
3
312-5
7812
4
400
10,000
If Fi is the voltage at the Dynamo and the line is n miles long
T\=F^-|-iam2from(4).
The power per mile at the extremity being l\ = aVQ, if we are given
V^ and 1\ to find T'q, we shall find that n cannot be greater than
l\-V2^o,
and this gives the limiting length of the line.
If we wish, as in Electric traction to get a nearer approach
to uniform P, let us try
C=ax-hx^ (5),
where a. h, c are constants.
1 ^y 7^
- - - = ax - oyf\
r ax
\=V,+\rax-^-^^x<^^
(6).
dC
As P=V ,- , or V {a - cbx^-'^), we can easily detennine the three
constants a, b, c so that P shall be the same at any three points of
thehne. Thus let r=l ohm, Fo=100 volts, and let P= 10000 watts,
where .^;=0, x=l mile, .r=l| miles.
We find by trial that
(7=100.r-14-75.r2-iu
and from this it is eas}'^ to calculate Cat any point of the line
60
CALCULUS FOR ENGINEERS.
Example 19. A machine costs ax-\-hy/\t'& value to mc is
proportional to xy, find the best values of x and y if the cost
is fixed. Here xy is to be a maximum. Let c = cw; + 63/, so
that V = J -~ j.^> ^^d ^y is r ^ — T ^^. This is a maximum
Hence ax = hy = 0/2 makes xy a
when j = 2 r X or ax — -^ .
b 2
maximum.
Examj)le 20. The electric time constant of a cylindric
coil of wire is approximately
II = mxyzl{ax +hy + cz),
where x is the mean radius, y is the difference between the
internal and external radii, z is the axial length and m, a, b, c
are known constants.
The volume of the coil is 27rxyz.
Find the values of x, y, z to make u a maximum if the
volume of the coil is fixed. Let then 27r . xyz = g ; when is
— I a minimum ? That is, substituting for z, when
yz xz
xy
90
is ax + bi/ 4- ^ — = v, say, a minimum ? As x and y are p(ir-
fectly independent we put f ^ J = and ( ;t~ ) = 0,
or
ft +
o-.^^=o.
ITT^/.^-
Fig. 9.
and + 6 - s^^^, = 0,
2'7rxy^
so that A'^y = ^^^ ,
a^r
627r
and
ax
- or y — T, .1 - -7-
a27r
x^ = -^ or
a'^27r
^ V a=27r
W ¥2'ir'
and 2^
'lirxy
or -?
7«¥
V C»27i^
HANGING CHAIN.
61
38. The chain of a suspension bridge supports a load
by means of detached rods ; the loads are about equal and
equally spaced. Suppose a chain to be really continuously
loaded, the load being w per unit length horizontally. Any
veiy flat uniform chain or telegraph wire is nearly in this
condition. What is its shape ? Let be the lowest point.
OX is tangential to the chain
and horizontal at 0. OF is
vertical. Let P be any point
in the chain, its co-ordinates
being x and y. Consider the
equilibrium of the portion
OP. OP is in equilibrium,
under the action of Tq the
horizontal tensile force at 0,
Tthe inclined tangential force
at P and wx the resultant
load upon OP acting ver- -p. ^^
tically. We employ the laws
of forces acting upon rigid bodies. A rigid body is a body
which is acted on by forces and is no longer altering its
shape.
If we draw a triangle whose sides are parallel to these
forces they are proportional to the forces,
and if 6 is the inclination of T to the c. 5^ ^a
Y
\ I ^M \ ,
horizontal
and
T
wx
cos 0.
•(1).
wr
tan^.
.(2),
but tan ^ is j^ , so that ^ = 7^ ;«;... (3);
dx dx Tq
Pig. 11.
1 w
hence, integrating, y = ^7rr^ + constant.
- -to
Now we see that y is when x is 0, so that the constant
is 0. Hence the equation to the curve is
1 ^ ^
(*).
62 CALCULUS FOR ENGINEERS.
and it is a parabola. Now tan ^ is ^ o^, so that sec- 6 is
1-f^,^^ Andasr = rosec^, r=ToA/l+^U-^...(5).
From this, all sorts of calculations may be made. Thus
if I is the span and D the dip of a telegraph wire, if the whole
curve be drawn it will be seen that we have only to put in
(4) the information that when x—\l,y = D,
and the greater tension elsewhere is easy to find.
In the problem of the shape of any uniform chain, loaded
only with its own weight, the integi'ation is not so easy. I
give it in a note*. When it is so flat that we may take the
* The integration in this note requires a knowledge of Chapter in.
If the weight of the portion of chain OP, instead of being wx is %cs,
where s is the length of the curve from to P, the curve y is called the
Catenary, Equation (3) above becomes
| = ^^.or,e«ingr,=.c.^V.i (X).
If bs is the length of an elementary bit of chain, we. see that in the limit
{bsY={8xf^{byf
60 that "i- = A / 3- + If
and hence -^ = . — . This being integrated gives y + c=v^c2 + «-...( 2),
"« 'slc' + s^
the constant added in integration being such that « = when y = 0. From (2)
we find 6-2=^2^ 2j/c... (3), and using this in (1), we have
dx _ c
the integral of which is
as when y=0, a?=0, if is the origin, no constant is to be added. Putting
this in the exponential form
c^lc _ y ^ ^. ^. JyT. ^ 2yc,
transposing and squaring we find
HEATING SURFACE OF BOILER
a3
load on any piece of it as proportional to the horizontal pro-
jection of it, we have the parabolic shape. |
39. Efficiency of Heating Surface of Boiler.
If 1 lb. of gases in a boiler flue would give out the heat
in cooling to the temperature of the water (0 maybe taken
as proportional to the difference of temperature between
gases and water, but this is not quite correct), we find from
Peclet's experiments that the heat per hour that flows through
a square foot of flue surface is, roughly, md\ Let ^ = ^i at
the furnace end of a flue and 6 = 0^ at the chimney end.
Let us study what occurs at a place in the flue.
The gases having passed the area S in coming from the
furnace to a certain place where the temperature is 6, pro-
Or changing the origin to a point at the distance c below O, as at O in fig. 12
>Yhere SP is y' and RP is x, we have
This is sometimes called
Using (1) we find
sometimes called
y' = c cosh xjc.
, I xlc -x!c\
s = c sinh xjc.
.(4).
Nv R
A
ly
O
»
Fig. 12.
Note that tables of the values of sinh u and cosh u have been publighed.
Returning to the original figure, the tension at P being T,
T
W8
AB
BG
-p , and from (3),
dy
■y+c,
so that
W8 S
Hence T=zio{y + c) or T=wy'
64 CALCULUS FOR ENGINEERS.
ceed further on to a place where S has become aS' 4- ^S and
has become 6 + B0 (really SO is negative as will be seen).
A steady state is maintained and during one hour the gase i
lose the heat 7n0^ . BS through the area hS. If during the
hour W lb. of gases lost at the place the amount of heat
-W.Bd, then
orrather ^ = -^-g. «•
That is, integrating with regard to 0,
where c is some constant.
Putting in 6 = ^i the temperature at the furnace end
when S=0, we hav©'
' Wl Wl
0= —-pf-^c orc = ^,
711 Ui m Oi •
80 that (2) becomes
^•=^^U^J (3).
This shows how 6 diminishes as S increases from the
furnace end, and it is worth a student's while to plot the
curve connecting S and 6. If now S is the whole area of
heating surface and ^ = ^2 at the smoke-box end,
»-^a-i) <•>
The heat which one pound of gases has at the furnace
end is 6^, it gives up to the water the amount 61—62.
Therefore the efficiency of the heating surface may be
taken as
. j^t_ ^i- 62 . .
^'^~dr ^^^'
and it follows from (4) that
1
^ 6,mS
HEATING SURFACE. 65
Now if W is the weight of coals burnt j^er hour ;
W=13W' if air is admitted just sufficient for complete
combustion ; W = about 20 W in the case of ordinary
forced draught ; W = about 26 W in the case of chimney
draught. In these cases 0^ does not seem to alter inversely
as W, as might at first siglit appear : but we do not know
exactly how 0^ depends upon the amount of excess of air ad-
mitted. We can only say that if W'-^S is the weight of coal per
hour per square foot of heating surface and we call it tu, there
seems to be some such law as B = z, , where a clepends
1 -h aw ^
upon the amount of air admitted. In practice it is found
that a = 0*5 for chimney draught and 0'3 for forced draught,
give fairly correct results. Also the numerator may be taken
as greater than 1 when there are special means of heating
the feed water.
Instead of the law given above (the loss of heat by
gases in a flue oc 6^^), if we take what is probably more likely,
that the loss is proportional to 0,
Then (1) above becomes
de mO ^ ^'
W
or S= loe: ^ + constant (2).
Let 0= 6j at furnace end or when S=0 so that our con-
. W
stant IS — log Oi and (2) becomes
m
S
'i'-^ai) »■
If S is the area of the whole flue and 6^ is the temperature
at the smoke-box end, then
^ = Zlog^i (4),
Sm ^
66 CALCULUS FOR ENGINEERS.
The efficiency E J-^^ (5)
becomes E=^\-~^ = l-e ^^ (6).
Or if w is the weight of fuel per square foot of heating
surface as above (6) becomes
j5'=l-e"a«' (7).
40. Work done by Expanding Fluid*. If p is the
Eressure and v the volume at any instant, of a fluid which
as already done work W in expanding, one good definition of
pressure is p—'-r- •..(!)) or in words, pressure is the rate at which
work is done per unit change of volume. Another way of putting this
is: if the fluid expands through the volume hv there is an increment
8 IF of work done so that 'p .hv=hW^ or jo=-— , but this is only
strictly tiiie when hv is made smaller and smaller without li«iit, and so
(1) is absolutely true. Now if the fluid expands according to the law
p%}'=c. a constant ... (2) : p = cv~'j and this is the differential coefficient
dW
of W with regai*d to v or, as we had better write it down, -j— = cv'.
We therefore integrate it according to our rule and we have
T^=Z7^i^-'-*-^+<^' ^^^'
where C is some constant. To find C\ let us say that we shall only
begin to count W from v=Vi. That is, W=0 when v^Vi, Then
0= -^ v.^-' + C, so that C= - ^ v.^-'.
Insert this value of G in (3) and we have
F=^^(^i--V-) (4), ■
which is the work done in expanding from v^ to r.
Now if we want to know W when ^ = ^2) we have
^^'i2=rr-/^2'"'-^^'"'^ ^^^-
* Observe that if for p and v we write y and x this work becomes very
easy.
WORK BY EXPANDING FLUID.
67
This answer may be put in other shapes. Thus from (2) we know that
c^p^v^' or pav/,
so that
iir —P_^.u, 1-8.
a formula much used in gas engine and steam engine calculations.
There is one case in which this answer turns out to be useless ; try
it when s — \. That is, find what work is done from Vy to Vg by ^ fluid
expanding according to the law (it would be the isothermal law if the
fluid were a gas)
•pv—c.
If you have noticed how it fails, go back to the statement
dW
dv
(7).
You will find that when you integrate x"^ with regard to m, the
general answer has no meaning, cannot be evaluated, if m= - 1. But
I have already said, and I mean to prove presently that the integral of
x"^ is log^. So the integral of (7) is
W=c\ogv-{-C.
Proceeding as before we find that, in this particular case.
^1
(8).
41. Hypothetical Steam Engine Diagram.
Let steam be admitted to a cylinder at the constant pressure jt?,.
the volume increasing from to
Vi in the cylinder. The work done
is VjjOj. Let the steam expand
to the volume Vg according to the
law 'pijP^c. The work done is
given by (6) or (8). Let the back
pressure bejOg, then the work done
in driving out the steam in the
back stroke is p^^. We neglect
cushioning in this hypothetical
diagram. Let v^-~v^ be called r
the ratio of cut-off. Then the
nett work done altogether is
Fig. 13.
5-^2
68
CALCULUS FOR ENGINEERS.
If pe is the effective i)re.s.sure so that ^e^^ ^^ equal to the above iiett
work Ipe is measured from actual indicator diagrarus, as the average
pressure) ; putting it equal and dividing by v^ we have on simplifying
■M-)
■Vz-
In the special case of s = l we find pe=Pi p^ in the same
way.
42. Definite Integral.
Definition. The symbol I /(a?) . dx
tells us: — " Find the general integral of/ (a?); insert in it the
value a for ic, insert in it the value 6 for a? ; subtract the
latter from the former value*." This is said to be the
(1), tells us to integrate u (which is a
function of x and ?/), with regard
to y, as if x were constant ; then
insert F{x) for y and also f(x) for
y and subtract. This result is to
be integrated with regard to x,
and in the answer a and b are in-
serted for X and the results sub-
tracted.
I. If w = 1, dx . dy evidently
means an element of area, a little
rectangle. The result of the first
process leaves
Fig. 14.
/:
{F{x)-f{x)}dx (2)
still to be done. Evidently we
have found the area included between the curves y = F{x) and y~f{x) and
two ordinates at a: = a and x = b. Beginners had better always use form (2)
in finding areas, see fig. 14.
II. If u is, saA', the weight of gold per unit area upon the above men-
tioned area, then u . dx . dy is the weight upon the little elementary area
dx .dy, and our integral means the weight of all the gold upon the area I
have mentioned.
When writers of books wish to indicate generally that they desire to
integrate some jwoperty u (which at any place is a function of x, y, z),
throughout some volume, they will write it with a triple integral,
///
u .dx.dy. dz,
and summation over a surface by
I j V . dx . dy.
AREAS OF CURVES.
69
integral of f{x) between the limits a and h. Observe now
that any constant which may be found in the general
integral simply disappears in the subtraction.
Iq integrating between limits we shall find it convenient
to work in the following fashion.
Example, to find I x^ . dx. The general integral is ^xf^
J h
and we write I x"^ .dx—\ ^a^ — ^d^ — ^bK
Symbolically. If F{x) is the general integral of /{x)
then rf(x) . dx = P F(x)] =F(a) - F(b),
J b U J
Note as evidently true from our definition, that
I f(x),dx = - I f(x).dx,
J b J a
and also that
T/w • ^^' = f Vw • ^^' + f "/(^) • ^^•
43. Area of a curve.
some function of x and
let PS be the curve. It
is required to find the
area MPQT.
Nowiftheareai¥PQr
be called A and OT=x,
QT=y, OW = X -ir Sx,
WR = y-\-8y, and the area
MPRWheA + 8A then
BA = siTesiTQIiW.
Let y of the curve be known as
/
Fig
TW
, 15.
Indeed some writers use
V over an area, and
jto .
Ih
dS to mean generally the summation of
els to mean the summation of 2a along a line or
what is often called "the line integral of ic. The line integral of the pull
exerted on a tram car means the work done. The surface integral of the
normal velocity of a fluid over an area is the total volume flowing per
second. Engineers are continually finding line, surface and volume integrals
in their practical work and there is nothing in these symbols which is not
already perfectly well known to them.
70
CALCULUS FOR ENGINEERS.
.(1).
If the short distance QR were straight,
Therefore ^ = y + i^y» as ha) gets smaller and smaller
and in the limit -y— =?/
Hence A is such a function of x that y is its differential
coefficient, or A is the integral of y.
In fig. 16 CQD is the curve y=a-\- hx^ and EMGF is the
curve showing
so that A is the integral of
y. In what sense does A
represent the area of the
curve CD ? The ordinate
of the A curve, GT^ repre-
sents to some scale or other,
the area of the y curve
MPQT from some standard
Fig. 16. ordinate MP.
The ordinate TQ represents to scale^ the slope of
EF at G. Observe, however, that if we diminish or increase
all the ordinates of the A curve by the same amount, we do not
change its slope anywhere, and y, which is given us, only tells
us the slope of A. Given the y curve we can therefore find
any number of A curves ; we settle the one wanted when we
state that we shall reckon area from a particular ordinate
such as MP. Thus, in fig. 16 if the general integral of y is
F(x)-{- c. If we use the value x = OM we have, area up to
MP from some unknown standard ordinate =F(OM)-i-c.
Taking x = ON, we have area up to NR from some
unknown standard ordinate =F{ON)-\-c. And the area
between MP and NR is simply the difference of these
F(ON) — F{OM), the constant disappearing.
Now the symbol I y .dx tells us to follow these instruc-
J OM
tions : — integrate y ; insert ON for x in the integral ; insert
OM for ^ in it ; then subtract the latter. We see therefore
AREA OF PARABOLA.
71
that the result of such an operation is the area of the curve
between the ordinate at OM and the ordinate at ON.
If y and a; represent any quantities whatsoever, and a
curve be drawn with y as ordinate and x as abscissa, then the
integral j ;
y .dx is represented by the area of the curve, and
we now know how to proceed when we desire to find the sum
of all such terms rs y . 800 between the limits x=b and a)=a
when Bw is supposed to get smaller and smaller without
limit.
Example. Find the area enclosed between the parabolic
curve OA, the ordinate AB
and the axis OB. Let the
equation to the curve be
y = ax^ (1),
where PQ = y and OQ = x.
Let QR = Bx.
The area of the strip PQRS
is more and more nearly
ax^ . Sx,
as Bx is made smaller and
smaller; or rather the whole
rOB
area is I aa^ . dx, whic'h is
^0
Y
M
A
X
p
c
o|
' G
I
k B
Fig. 17.
OB -\
OSn
Now what is a in terms o{ AB and
x==OB. Hence by (1)
AB = a.OB^, so that a =
OB^
AB
OB^'
(2).
When y^AB,
Therefore the area = 5 =^, OJ^ ^AB . OB ;
that is, frds of the area of the rectangle OMAB.
Observe that the area of a very flat segment of a circle is
like that of a parabola when OB is very small compared
with BA,
72
CALCULUS FOR ENGINEERS.
Exercise 1. Find the area between the curve y = mx "
and the two ordinates at a? = a and x — h.
The answer is
'h
I
irix"''^ . dx =
1-71
nt
1-n
(^,i-»_ai-n).
Observe (as in Art. 40) that this fails when n — 1; that
is, in the rectangular hyperbola.
In this case the answer is
JaX
m
\ogx
= m log - .
The equation to any curve being
y = a -\- hx -{- ca^ -{■ ex^ 4-/^,
the area is A = ax f ^bx^ + ^cx^ 4- \ex* + i/ar*.
Here the area up to an ordinate at x is really measured from
the ordinate where x = 0, because A =0 Avhen x=0. We
can at once find the area between any two given ordinates.
Exercise 2. Find the area of the curve y — a^x between
the ordinates at *• = a and x — ^.
aTx^ .dx = a\ ix^'] = ^ (y8^ - a^).
Exercise 3. Find the area of the curve yx" = a between
the ordinates at a; = a and x — fi.
Answer : a \ x~'- .dx = a
— x~
= a(a-'-^-').
44. Work done by Expanding Fluid. When we use
definite integrals the work is somewhat shorter than it was in Art. 40.
For i{p = cv~', the work done from volume v^ to volume V2 is
/V2 Fv-i 1 "1 c
cy~*. dv or c\ y^"* or _- — -(^'2^"'- V')
The method fails when s = l and then the integral is
c\ "logeV =cl0ge— .
L»i J ^1
CENTKE OF GRAVITY.
73
45. Centre of Gravity, Only a few bodies have
centres of gravity. We usually
mean the centre of mass of a
body or the centre of an area.
Fig. 18.
If each little portion of a
mass be multiplied by its dis-
tance from any plane, and the
results added together, they are
equal to the whole mass multi-
plied by the distance of its centre,
X, from the same plane. Expressed algebraically this is
If each little portion of a plane area, as in fig. 18, be
multiplied by its distance from any line in its plane and the
results added together, they are equal to the whole area
multiplied by the distance of its centre x from the same line.
Expressed algebraically this is Saaf = xl^a.
Example. Find the centre of mass of a right cone. It
is evidently in the axis OB of the
cone. Let the line OA rotate
about OX, it will generate a
cone. Consider the circular slice
PQR of thickness hx. Let
OQ = x, then PQ or
AB
y^OB'^-
The mass of PR multiplied
by the distance from to its
centre is equal to the sum of the
masses of all its parts each mul-
tiplied by its distance from the
plane YOY^. The volume of the slice PR being its area iry'^
multiplied by its thickness hx ; multiply this by rn the mass
per unit volume and we have its mass miry- . hx. As the
slice gets thinner and thinner, the distance of its centre from
gets more and more nearly x. Hence we have to find the
sum of all such terms as mirxy- . hx, and put it equal to the
whole mass {^irm . ABr . OB) multiplied by x, the distance of
Fig. 19.
74
CALCULUS FOR ENGINEERS.
its centre of gravity from 0. Putting in the value of y^ in
terms of x we have
roB fAB\^
I rmr ( -^f^ ] ^'^ • dx equated to ^TrmAB^ . OB . x.
rOB FOB -]
Now af,dx== i^ = iOB\
and hence '"^'^ (Tfn] iOB^^^ir^ii . AB^.OB.x.
Hence x = J OB. That is, the centre of mass is f of the
way along the axis from the vertex towards the base.
46. It was assumed that students knew how to find the
volume of a cone. We shall now prove the rule.
The volume of the slice PR is it .y"^ ^ hx and the whole
volume is
£ V • d^ = \y (oi)' *■= • '^- = - i^) [p]
or J of the volume of a cylinder on the same base AG and of
the same height OB, If we had taken y — ax all the work
would have looked simpler.
Example. Find the volume and centre of mass of
uniform material (of mass m per
unit volume) bounded by a para-
boloid of revolution.
Let PQ = y, OQ=^x, QS=Bx.
Let the equation to the curve
OFA hey=aa^ (1).
The volume of the slice PSR
is Try^ . Bx; so that the whole
rOB
volume is 1 tt . a^x . dx or
^0
Y
n
■^'^
A
O^
[4
S
B ^
Fig. 20.
l-iraKOB^ (2).
SURFACE OF REVOLUTION. "75
Now what is a ? When
7/ = AB, a)=: OB,
so that from (1), AB = a . OB^ and a is ^-^ . Hence the
volume is ^ir -7y^ OB'^ or
^TT.A&.OB (3).
That is, half the area of the circle AC multiplied by the
height OB. Hence the volume of the paraboloid is half the
volume of a cylinder on the same base and of the same height.
(The volumes of Cylinder, Paraboloid of revolution, and Cone
of same bases and heights are as 1 : i : J.)
Now as to the centre of mass of the Paraboloid. It is
rOB
evidently on the axis. We must find I jriir . y^x . da), or
jmira) . a^js . da;, or 7/t7ra- Ix^ . dx,
-OB
or mTTcv^ • i^M ^^^^ *his is ^mira^ . 0B\ Inserting as before
the value of «- or -jTW ^^ have the integral equal to
IniTT.OB^.AB^. This is equal to the whole mass multiplied
by the x of the centre of mass, x, or m^Tr . AB^ . OB . x, so that
x = ^OB. The centre of mass of a paraboloid of revolution
is |rds of the way along the axis towards the base from the
vertex.
Example. The curve y = ax^ revolves about the axis
of X, find the volume enclosed by the surface of revolution
between x = and x = h.
The volume of any surface of revolution is obtained by
integrating iry"^ . dx. Hence our answer is
TT
{'a'x-^ . dx = J^ Vx'^-^A = ^-^ 6^+\
J 2n 4- 1 Lo J 2/1 + 1
Find its centre of mass if in is its mass per unit volume.
For any solid of revolution we integrate m . xiry^ . dx and
76 ^ CALCULUS FOR ENGINEERS.
divide by the whole mass which is the integral of mir'fdx.
If m is constant we have
rb rb
niTT I a;a"iv-"' . clx = niirtv^ I a;2"+i dx
Jo Jo
~2n + 2Lo J~2n + 2^ '
and the whole mass is _ 6-"+\ so that Jj = _ ^ b.
2/1 +1 2?i -I- 2
Suppose m is not constant but follows the law
'ill = //i„ 4- caf.
To find the mass and centre of mass of the above solid. Our
first integral is
TT |(?/toA' + caf+') ttU-'^ . dw, or (/ V [(vi^''+' + cx''''+'+^) clx,
|_o 2n + 2 2/1 + 5 + 2 J ^
ass is a^TT I (?yio + co."*) x^ . c?a;
aV \^, af''^' +-« ^ ^ ar^+'-^»] (2).
L2m+1 2n + 5+l J ^'^
Substituting h for ^' iu both of these and dividing (1) by (2),
we find x.
An ingenious student can manufacture for himself many
exercises of this kind which only involve the integration
ofd?".
An arc of the ellipse ~ + ^^ = l revolves about the axis of
X, find the volume of the portion of the ellipsoid of revolu-
tion between the two planes where x = and where x = c.
Here y^ = — (a^ — x-). The integral of tti/^ is
or
or
|[k-K]=-^;(a=c-|c').
LENGTHS OF CURVES
'^
The volume of the whole ellipsoid is -t- iira; aud of ^
it IS -^ a^ ^^
o
77
ihere
47. Lengths of Curves. In fig. 21 the co-ordinates of
P are x and y and of Q they
are a; -\- Sx and 3/ + S^. If we
call the length of the curve
from some fixed place to P by
the name s and the length PQ,
8s, then (8sy=(Bxy-\-(8i/y more
and more nearly as 8x gets
smaller, so that
8s
Boo
-^A-(I)■
or rather, in the limit
Fig. 21.
CD
dx
V'-©"-
To find 5 then, we have only to integi'ate
^/'-(i)'
It is unfortunate that we are only supposed to know as
yet j x^ . dx, because this does not lend itself much to exer-
cises on the lengths of curves.
Example. Find the length of the curve y — a-\-hx (a
straight line) between the limits x = Q and x = c.
t^ ds [^ — ~
5=1 j^.dx=\ ^/l+6^(^a; =
x\/l+b'
^cVl+6^
78
CALCULUS FOR ENGINEERS.
Exercise. There is a curve whose slope is »Ja^x^—l,
find an expression for its length. Answer: s= -«;'**+->/-.
Other exercises on lengths of curves will be given later.
48. Areas of Surfaces of Revolution. When
the curve APB revolving
about the axis OX de-
scribes a surface of revo-
lution, we have seen that
the volume between the
ends AC A' and BDB' is
the integral of iry^ with
regard to x between the
limits OG and OD.
Again the elementary
area of the surface is what
is traced out by the ele-
mentary length PQ or hs
and is in the limit 2iry . ds.
Hence we have to integrate
ds
dx
Fig. 22.
roD ^
I liTV .-T-.dx, and as the law of the curve is known, y .
Joe dx
00
or
can be expressed in terms of x.
Example. The line y = a-\-bx revolves about the axis of
X ; find the surface of the cone between the limits x=0 and
x=c.
-^ = 6, so that the area is 27r
dx JO
/.■^■-©■'
= 2'7r*Jl+b^r{a-\-bx)dx = 2'ir's/l -^¥ I ax + ^boA
= 27rVITF(ac + Jtc^.
The problem of finding the area of a spherical surface is
here given in small printing because the beginner is supposed
to know only how to differentiate x'^ and this problem
requires him to know that the differential coefficient of y^
AREAS OF SURFACES.
79
with regard to x is the differential coefficient with regard to
y multiplied hy -~ , or 2y .-^ . As a matter of fact this is
not a real difficulty to a thinking student. The student can
however find the area in the following way. Let V be the
47r
volume of the sphere of radius r, F= -^ r^, Art. 46. Let
o
V-{- SVhe the volume of a sphere of radius r*+ 8r, then
which is only true when Sr is supposed to be smaller and
smaller without limit. Now if S is the surface of the
spherical shell of thickness Br, its volume is Bi^ . S. Hence
Br .S = Br . ^irr^ and hence the area of a sphere is ^iri^.
Example. Find the area of the surface of a sphere. That
is, imagine the quadrant of a circle AB oi radius a, fig. 23, to
A
Fig. 23.
revolve about OX and take double the area generated. We
have as the area, 47r j y a / \ -\- {-^\ dx.
In the circle x'^ +y^ = <^% or y = \/«^ - x-,
X
V'
..^..,.|=0.or|,= -^-
80
CALCULUS FOR ENGINEERS.
Hence as
l + m^^l+^
\dx')
r y
Guldinus's Theorems.
PCD
49. If each elementary portion hs of the length of a
curve be multiplied by x its distance from a plane (if the
curve is all in one plane, x may be the distance to a line in
the plane) and the sum be divided by the whole length of
the curve, we get the x of the centre of the curve, or as it is
sometimes called, the centre of gravity of the curve. Observe
that the centre of gravity of an area is not necessarily the
same as the centre of gravity of the curved boundary.
Volume of a Ring^.
BC, fig. 24, is any plane
area ; if it revolves about
an axis 00 lying in its
own plane it will generate
a ring. The volume of
this ring is equal to the
area of BG multiplied by
the circumference of the
circle passed through by
the centre of area of BC.
Imagine an exceed-
Fig' 24. ingly small portion of the
area a at a place P at the distance r from the axis, the
volume of the elementary ring generated by this is a . 27rr
and the volume of the whole ring is the sum of all such
terms or F= 27r2ar. But ^ar = rA, if A is the whole area
of BG. The student must put this in words for himself; r
means the r of the centre of the area. Hence V= 2irf x A
and this proves the proposition.
II. Area of a Ring. The area of the ring surface is
the length of the Perimeter or boundary of BG multiplied
by the circumference of the circle passed through by the
centre of gravity of the boundary.
Imagine a very short length of the boundary, say hs^ at
MOMENT OF INERTIA. 81
the distance r from the axis ; this generates a strip of area
of the amount hs x 27rr. Hence the whole area is ^irZZs . r.
But 2^5 . r = r X 6- if r is the distance of the centre of gi-avity
of the boundary from the axis and s is the whole length
of the boundary. Hence the whole area of the ring is
27rr X 8.
Example. Find the area of an anchor ring whose sec-
tion is a circle of radius a, the centre of this circle being
at the distance JR, from the axis. Answer : — the perimeter
of the section is 27ra and the circumference of the circle-
described by its centre is 27rR, hence the area is 47r-ajR.
Exercise. Find the volume and area of the rim of a fly-
wheel, its mean radius being 10 feet, its section being a
square whose side is 1'3 feet. Answer:
Volume = (1-3)2 x 27r x 10 ; Area = 4 x 1-3 x 27r x 10.
50. If every little portion of a mass be. multiplied by the
square of its distance from an axis, the sum is called the
moment of inertia of the whole mass about the axis.
It is easy to prove that the moment of inertia about any
axis is equal to the moment of inertia about a parallel axis
through the centre of gravity together with the whole mass
multiplied by the square of the distance between the two
axes. Thus, let the plane of the paper be at right angles
to the axes. Let there be a little mass
m at '^P in the plane of the paper.
Let be the axis through the centre of
gravity and 0' be the other axi-s. We want
the sum of all such terms as m . (O'Py.
Now (O'py = (ovy + OP' +2. 00'. oq,
where Q is the foot of a perpendicular from
P upon 00\ the plane containing the two
axes. Then calling Xm. (O'Py by the name
/, calling Xm . OP^ by the name /q, the
moment of inertia about the axis through
the centre of gravity of the whole mass,
then, / = (0'0)^Sm + /o + 2.00'.2:m.OQ. But tm.OQ
means that each portion of mass m is multiplied by its
distance from a plane at right angles to the paper through
P. 6
S2
CALCULUS FOR ENGINEERS.
the centre of gravity, and this must be by Art. 45. So
that the proposition is proved. Or letting 2m be called M
the whole mass
I = Io + M.(0'0)2.
Find the moment of inertia of a circular cylinder of
length I about its axis.
Let fig. 26 be a section,
I the axis being 00. Con-
— o sider an elementary ring
shown in section at TQPR
of inside radius r, its out-
side radius being r + hr.
Its sectional area is I .hr
T
R
O— -
Fig. 26.
SO that the volume of the ring is 2irr . 1 . 8r and its mass is
m^irrl . h\ Its moment of inertia about 00 is ^irml .t-^.dr
and this must be integrated between the limits r — R
the outside radius and ?• = to give the moment of inertia
of the whole cylinder. The answer is /q = lirmlRK The
MR^
whole mass M=ml7rR\ So that 7o = -«— • If we define the
radius of gyration as Jc, which is such that Mk^ = Io, we
have here k^ = ii2^ or A; = -p^ R.
The moment of inertia about
the axis NS is
I=Io + M.R' = ^3IR%
so that the radius of gyration about
jsrs is R Vf .
Moment of inertia of a circle
about its centre. Fig. 27. Con-
sider the ring of area between the
Fig. 27. circles of radii r and r + Br, its area
is 27rr . Br, more and more nearly
as Br is smaller and smaller. Its moment of inertia is 27rr^ . dr
and the integral of this between and R is ^ttR* where R
is the radius of the circle. The square of the radius of
R^
gyration is JttjR'^-t- the area =-^
MOMENTS OF INERTIA.
83
At any point in an area, fig. 28, draw two lines OX
and OF at right angles to one
another. Let an elementary area
a be at a distance x from one of
the lines and at a distance i/ from
the other and at a distance r from
0. Observe that ax^ 4- ay- = «/'^
so that if the moments of inertia
of the whole area about the two
lines be added together the sum
is the moment of inertia about the
point 0. Hence the moment of
inertia of a circle about a diameter is half the above, or
JttjR*. The square of its radius of gyration is ^R-.
The moment of inertia of an ellipse about a principal
diameter AOA. Let OA = a, OB = b.
.y^^
B
y f
\ >
/ /
\ \
aI
B
Fig. 29.
The moment of inertia of each strip of length BT is t
times the monaent of inertia of each strip PQ of the circle,
MT a
because it is at the same distance from A OA and ytfi = y •
MQ h
This is a property of ellipse and circle well known to all
engineers. But the moment of inertia of the circle of radius
h about AOA is \it¥, so that the moment of inertia of the
ellipse about AOA is \'Tr¥a. Similarly its moment of inertia
about BOB is \'iTa%.
The above is a mathematical device requiring thought,
not practical enough perhaps for the engineer's every-day
work; it is given because we have not yet reached the inte-
6—2
84 CALCULUS FOR ENGINEERS.
gral which is needed in the straightforward working. The
integral is evidently this. The area of the strip of length
ST and breadth % is 2x . Sy, the equation to the ellipse
beins: -„ + '7. = 1, so that ic=^T "Jh^ — y^
Then 2J if.2x.dy or 4^[ ^fs/b^ -if .dy=^ I.
The student ought to return to this as an example in
Chap. III.
51. Moment of Inertia of Rim of Fly-wheel. If
the rim of a fly-wheel is like a hollow cylinder of breadth
I, the inside and outside radii being R^ and R.,, the moment
of inertia is ^irml I 1-^ .dr or ^irml — =i7rml(R/—Ri*).
J jR, \_Ri ^ J
The mass is 7r(jR2'-i^f)^//i'=^irsay,_so that I=\{R.^-\-R^^M.
The radius of gyration is V^ (^R^ + Ri^). It is usual to
calculate the moment of inertia of the rim of a fly-wheel as
if all its mass resided at the mean radius of the rim or
Oil?
--^ — ^ . The moment of inertia calculated in this way is
to the true moment of inertia as ^y^- — ^- . Thus if
R.2 = R + a, Ri = R — a, the pretended I divided by the true
/ is 1 -^ (1 -f 0:2) ^^^ i^ ^ is small, this is 1 — -^^ nearly. If
the whole mass of a fly-wheel, including arms and central
boss, be M, there is usually no very great error in assuming
that its moment of inertia is / = R'^M.
52. A rod so thin that its thickness may be neglected
is of length I, its mass being m per unit length, what is
its moment of inertia about an axis at right angles to it,
through one end ? Let w be the distance of a point
from one end. An elementary portion of length Sw of mass
in . Boo has a moment of inertia x^ ,m .dx and the integral of
this from ^ = to x = l \^ i^nP, which is the answer. As 7nl
^ .X £83s>Q is the whole mass, the
^U ' --1 ^^^^ square of the radius of
Fig. 30. gyration is ^l^ /© the
CENTRE OF GRAVITY.
85
moment of inertia about a parallel axis through the middle
of the rod, at right angles to its length, is
^ml^ — tnl.liz) or ml^ (^ — {) or -^mlK
So that the square of this radius of gyration is -^P.
We shall now see what error is involved in neglecting the
thickness of a cylindric rod.
If 00 is an axis in the plane of the paper at right angles to the
axis of a circular cylinder,
through one end, and OP is <-»
a; and R is the radius of the
cylinder, its length being I ;
if p is the mass of the cylin-
der per unit volume ; the
moment of inertia about 00
of the disc of radius It and
thickness 8.v is irR^pbx . x*- +
the moment of inertia of the disc about its own diameter. Now we
saw that the radius of gyration of a circle about its diameter was
— , and the radius of gyration of the disc is evidently the same. Hence
its moment of inertia about its diameter is \ WirR^ . hx . p, or | irpTt^ . 8x.
Hence the moment of inertia of the disc about is
7rR^p{xK8x+im.dx).
If OAj the length of the rod, is I, we must integrate between and I,
and so we find
--<>-
-A
Fig. 31.
'-nR^p(^+k
mi\
The mass m per unit length is irR'^pj so that
I—m
3 IP
?)■
This is the moment of inertia about an axis through the centre of
gravity parallel to 00.
53. Example. Where is the Centre of Area of the
parabolic segment shown in fig. 20 ? The whole area is
lAGy<OB.
The centre of area is evidently in the axis.
86
CALCULUS FOR ENGINEERS.
The area of a strip PSR is 2i/ . Bx and we musb
integrate 2xy . Bx. Now y — ax^ where a — AB ^ OBK
rOB ^^
Hence 2 1 x vym^* . dx = ^AO . OB . x. The integral is
2 g^^ Ux^l or f -^j 0J5t, so that iAB . OB' = ^AB.OB.x
or x = ^OB,
Find the centre of area of the segment of the sym-
metrical area bounded by + ?/ = ax'^ between x = b and x = c.
i:
We must divide the integral 2 1 x . ax"^ . dx by the
area
2 I ax'' . dx.
Or
2a r"-^""
X =
C«+i _ hn+i ' n + 2'
Many interesting cases may be taken. Observe that if
the dimensions of the figure be given, as in fig. 20 : thus if
AB and PQ and BQ are given, we may find the position of the
centre of the area in terms of these magnitudes.
54. Moment of Inertia of a Rectangle.
The moment of inertia of a rectangle about the line 00
ATTRACTION. 87
through its centre, parallel to one side.
Consider the strip of area between
OP = y and OQ = y + hj. Its area is b . By
and its moment of inertia about 00 is
b .y^ . By, so that the moment of inertia of
the whole rectangle is
bd'
rhd rid-]
b f.dyorb\if\
J -id L-idA
"' li
This is the moment of inertia which is
so important in calculations on beams. Fig. 33.
55. Force of Gravity. A uniform spherical shell of
attracting matter exercises no force upon a body inside it.
On unit mass outside, it acts as if all its mass were gathered
at its centre.
The earth then exercises a force upon unit mass at any
point P outside it which is inversely proportional to the
square of r the distance of P from the centre. But if P is
inside the earth, the attraction there upon unit mass is the
mass of the sphere inside P divided by the square of r.
1. If the earth were homogeneous. If m is the mass per
unit volume and R is the radius of the earth, the attraction
on any outside point is — mR^ -r- r\
The attraction on any inside point is -^ mi^ -h t^ or
— mr. The attraction then at the surface being called 1, at
o
any outside point it is R^ -r- r- and at any inside point it is
r -f- R. Students ought to illustrate this by a diagram,
2. If m is greater towards the centre, say m — a — br,
then as the area of a shell of radius r is 4<TTf\ its mass is
4mr^ . Ill . 5r, so that the whole mass of a sphere of radius r
rr ^^
47r r^ (a — br) dr, or -^ ar^ — irbr^. Hence on any
Jo ^
4i7r
inside point the attraction is -;r- a?' — irbr^ and on any
o
outside point it is f ^ aR^ — irbR^j r^.
is
88
CALCULUS FOR ENGINEERS.
Dividing the whole mass of the earth by its volume -^ jK^
o
we find its mean density to be a — ^hR, and the ratio of its
mean density to the density at the surface is
(4a-36i^)/(4a-46E).
56. Strength of thick Cylinders.
The first part of the following is one way of putting the
well known theory of what goes on in a thin cylindric shell
of a boiler. It prevents trouble with + and — signs after-
wards, to imagine the fluid pressure to be greater outside
than inside and the material to be in compression.
Consider the elementary thin cylinder of radius r and of
thickness 8?\ Let the pressure
inside be p and outside p + Bp
and let the crushing stress at
right angles to the radii in the
material be q. Consider the
portion of a ring PQSR which
is of imit length at right angles
to the paper.
Radially we have p + 8p
from outside acting on the area
RS or (r + Br) BO if QOP = BO,
because the arc RS is equal to
radius multiplied by angle; and
2:> .r. BO from inside or
{p + Bp){r-\-B7^)Bd-pr.Be
the whole the radial force from the outside more
and more nearly as B6 is smaller and smaller. This is
balanced by two forces each q . Br inclined at the angle B6,
and just as in page 165 if we
draw a triangle, each of whose
sides GA and ^J5 is pamllel
to q . Br, the angle BAG being
Bd, and BG representing the
radial force, we see that this radial force is q . Br . Bd, and
and
Fig. 34.
is on
Fig. 35.
this
expression is more
smaller and smaller. Hence
more nearly true as BO is
(p + Bp) (r + Br) B0-pr.Be = q.Br, B0,
HYDRAULIC PRESS OR GUN. 89
or p .^7^ -\- r.Bp+ 8p.Sr = q.Sr,
or rather p-\- r -¥- = <! (1),
since the term Bp is in the limit.
When material is subjected to crushing stresses p and q
in two directions at right angles to one another in the plane
of the paper, the dimensions at right angles to the paper
elongate by an amount which is proportional to p + q.
We must imagine the elongation to be independent of r
if a plane cross section is to remain a plane cross section, and
this reasonable assumption we make. Hence (1) has to be
combined with p-\-q = 2A (2)
where 2A is a constant.
Substituting the value of q from (2) in (1) we have
dp 2A 2p
or -f- = ^ .
a?' r r
Now it will be found on trial that this is satisfied by
P = ^+J (3).
and hence from (2), q = A — -; (4).
To find these constants A and B. In the case of a gun
or hydraulic press, subjected to pressure po inside where
r = ro and pressure outside where r = 7\. Inserting these
values of p in (3) we have
sothat ^=i^o.(i-l) = Po,;g^,
90 CALCULUS FOR ENGINEERS.
The compressive stress — q may be called a tensile stress f,
/-^V,'-V r^ ^^^.
/ is greatest at r = 7\ and is then
fo = Po-^—', (6).
This is the law of strength for a cylinder which is initially
unstrained. Note that po can never be equal to the tensile
strength of the material. We see from (5) that as r increases,
/ diminishes in proportion to the invei-se square of the
radius, so that it is easy to show its value in a curve. Thus
a student ought to take 7\ = 1% r^ = 0'8, p^ — 1500 lb. per sq.
inch, and graph / from inside to outside. / will be in the same
units as p. (5) may be taken as giving the tensile stress
in a thick cylinder to resist bui^ting pressure if it is initially
unstrained. If when p^ — there are already strains in the
material, the strains produced by (5) are algebraically added
to those already existing at any place. Hence in casting
a hydraulic press we chill it internally, and in making a gun,
we build it of tubes, each of which squeezes those inside it,
and we try to produce such initial compressive strain at
r = ro and such initial tensile strain at r = ri, that when the
tensile strains due to Pq come on the material and the cylinder
is about to burst there shall be much the same strain in the
material from r^ to i\*
* In the case of a eylindric body rotating with angular velocity a, if p is
the mass per unit volume ; taking into account the centrifugal force on the
element whose equilibrium is considered, above the equation (1) becomes
p + r ^ - r^pa? = q and the solution of this is found iohep = A+ Br~^ + ^pa^r^
and by inserting the values of p for two values of 7- we find the constants A
and B ; q is therefore known. If we take p~0 when r=rQ and also when
This is greatest when r = r^.
If the cylinder extends to its centre we must write out the condition that
the displacement is where r=0, and it is necessary to write out the valuer
GAS ENGINE. 91
TJim cylinder. Take ro = R and 7\=Il-\- 1, where t is very
small compared with R,
. 2Rr' + 2Rt + t' PoR /^ t t-\l(^ t\
Fig. 86.
Now -jz and ^r-^^ and ^r^ become all smaller and smaller
li ZH- ZK
as t is thought to be smaller
and smaller. We may take f=V3 . .. (7)
(7) as a formula to be used ' t ' "
when the shell is exceedingly ^j^ ^
thin and (8) as a closer ap- ^~^"^2 ^^^'
proximation, which is the
same as if we used the average radius in (7). In actual
boiler and pipe work, there is so much uncertainty as to the
proper value of/ for ultimate strength, that we may neglect
the con-ection of the usual formula (7).
57. Gas Engine Indicator Diagram. It can be
proved that when a perfect gas (whose law is pv = Rt for a
pound of gas, R being a constant and equal to K—k the
difference of the important specific heats; 7 is used to
denote Kjk) changes in its volume and pressure in any
of the strains, Eadial strain =pa - g/3 if a is the reciprocal of Young's
Modulus and ^/a is Poisson's ratio, generally of the value 0-25.
In this way we find the strains and stresses in a rotating solid cylinder,
but on applying our results to the case of a thin disc we see that equation
(2) above is not correct. That is, the solution is less and less correct as the
disc is thinner. Dr Chree's more correct solution is not difficult.
&2 CALCULUS FOR ENGINEERS.
way, the rate of reception of heat by it per unit change
of volume, which we call h (in work units) or -^— , is
{jpv) + 'p (1),
rf—ldv
'-~i[t^A ^'^-
Students ouorht to note that this -~ is a very different
thing from [-i] , because we may give to it any value we
please.
We always assume Heat to be expressed in work units so as
to avoid the unnecessary introduction of J^ for Joule's equiva-
lent.
Exercise 1. When gas expands according to the law
ptf = c...{^) a constant, find li.
Answer : // = ^^— :. p (4).
y- 1
Evidently when 5 = 7, k = 0, and hence we have pyy = con-
stant as the adiabatic law of expansion of a perfect gas.
7 is 1*41 for air and 1'37 for the stuff inside a gas or oil
engine cylinder. When s= 1, so that the law of expansion is
pv constant, we have the isothermal expansion of a gas, and
we notice that here h=p, or the rate of reception of heat
energy is equal to the rate of the doing of mechanical energy.
Notice that in any case where the law of change is given by
(3), h is exactly proportional to p. If .9 is greater than 7 the
stuff is having heat withdrawn from it.
If the equation (1) be integrated with regard to v we
have J^oi = — 3T (piVi - p^Vo) + T^oi- • -(5). Here H^^ is the heat
given to a pound of perfect gas between the states po, Vq, t^
and jOi, -yi, ti, and TFoi is the work done by it in expanding
from the first to the second state.
This expression may be put in other forms because we
have the connection pv = Rt...(6). It is very useful in cal-
culations upon gas engines. Thus, if the volume keeps
ELASTICITY. 93
constant Wqi is and the change of pressure due to ignition
and the gift of a known amount of heat may be found. If
the pressure keeps constant, TFoi is p(vi — Vo) and the change
of volume due to the reception of heat is easily found.
/J JT rit
Another useful expression is -p =1^~r +'P"'0) where k
is a constant, being the specific heat at constant volume.
Integrating this with regard to v we find
Ho, = k(t,-to)+W,, (8).
This gives us exactly the same answer as the last method,
and may at once be derived from (5) by (6). In this form
one sees that if no work is done, the heat given is k (ti — to)
and also that if there is no change of temperature the heat
given is equal to the work done.
58. Elasticity is defined as increase of stress -r increase
of strain. Thus, Young's modulus of elasticity is tensile or
compressive stress (or load per unit of cross section of a tie
bar or strut) divided by the strain or fractional change of
length. Modulus of rigidity or shearing elasticity is shear
stress divided by shear strain. Volumetric elasticity e is
fluid stress or increase of pressure divided by the fractional
diminution of volume produced. Thus if fluid at p and v,
changes to p-\-Bp, v-\-Sv: then the volumetric stress is
Bp and the volumetric compressive strain is — Sv/v, so
that by definition e = — Bp-. — , or e = — v J- . The
•^ ^ V ov
definition really assumes that the stress and strain are
smaller and smaller without limit and hence e — — v -/-. . .(1).
dv
Now observe that this may have any value whatsoever.
Thus the elasticity at constant pressure is 0. The elasticity
at constant volume is — oo . To find the elasticity at con-
stant temperature, we must find ( i^ ) • see Art. 30. As
pv = Rt, p^ Rtv-\ Here Rt is to be constant, so that
^^- Rtv-^ and e=Rtv^^^p.
®-
94 CALCULUS FOR ENGINEERS.
It is convenient to write this et and we see that Kt^ the
elasticity at constant temperature, is p. This was the value
of the elasticity taken by Newton ; by using it in his calcu-
lation of the velocity of sound he obtained an answer which
was very different from the experimentally determined velo-
city of sound, because the temperature does not remain
constant during quick changes of pressure.
Exercise. Find the elasticity of a perfect gas when the
gas follows the law pvy = c, some constant. This is the
adiabatic law which we found Art. 57, the law connecting j9 and
V when there is no time for the stuff to lose or gain heat by
conduction, p = cv~y, so that -t- = — 7C^;~>'~^ and
6? = + vycv~y~^ or ycv~y, or yp.
It is convenient to write this e^, and we see that in a
perfect gas en = yet. When this value of the elasticity of
air is taken in Newton's calculation, the answer agrees with
the experimentally found velocity of sound.
59. Friction at a Plat Pivot. If we have a pivot of
radius R carrying a load W and the load is uniformly
distributed over the surface, the load per unit area is
w=^ W-T-irR^ Let the angular velocity be a radians per
second. On a ring of area between the radii r and r-\-8r
the load is w27rr . 8r, and the friction is fjLiv27rr . Sr, where
fjb is the coefficient of friction. The velocity is v = otr, so that
the work wasted per second in overcoming friction at this
elementary area is ^^irwoLr^ . Br, so that the total energy
wasted per second is
^irwoLfi \ ?'^ . dr = ^irwafxR^ = |« ^ WR,
Jo
On a collar of internal radius Ri and external R^ we have
27rwafi r'.dr^ ^irwdfi (R^^ - R^% W^irw {R^^ - R^) and
J Ri Jl3 _ J^ 3
hence, the energy wasted per second is '{afiW j^\ _ pV
60. Exercises in the Bending of Beams. When
the Bending moment ilf at a section of a beam is known,
we can calculate the curvature there, if the beam was
CURVATURE. 95
straight when unloaded, or the change of curvature if the
unloaded beam was originally curved. This is usually written
1 1 1 if
- or = -^^ ,
where / is the moment of inertia of the cross section
about a line through its centre of gravity, perpendicular
to the plane of bending, and E is Young's modulus for the
material. Thus, if the beam has a rectangular section of
breadth h and depth d, then I=^hd^ (see Art. 54); if the
beam is circular in section, I=-:R\ if J2 is the radius of
tbe section (see Art. 50). If the beam is elliptic in section,
I — T ^^^> if ^ aiid h are the radii of the section in and at
4
right angles to the plane of bending (see Art. 50). t
Curvature. The curvature of a circle is the reciprocal
of its radius, and of any curve it is the curvature of the circle
which best agrees with the curve. The curvature of a curve
is also " the angular change (in radians) of the direction of
the curve per unit length." Now draw a very flat curve, with
dii
very little slope. Observe that the change in -^ in going
from a point P to a point Q is almost exactly a change of
angle change in -^ is really a change in the tangent of
an angle, but when an angle is very small, the angle, its sine
dii
and its tangent are all equal . Hence, the increase in -^
from P to Q divided by the length of the curve PQ is the
average curvature from P to Q, and as PQ is less and less
we get more and more nearly the curvature at P. But the
curve being very flat, the length of the arc PQ is really hx^
and the change in -/ divided by hx, as hx gets less and less,
dv
is the rate of change of -f- with regard to x, and the symbol
^2y ° ax ^2y
for this is -—^ . Hence we may take -y^ as the curvature of
dx^ '' ^ dx^
a curve at any place, when its slope is everywhere small.
96 CALCULUS FOR ENGINEERS.
If the beam was not straight originally and if y' was its small
deflection from straightness at any point, then ,^ was its original
curvature. We may generalize the following work by using ^^ (y -y')
instead of --j— everywhere.
It is easy to show, that a beam of uniform strength, that is a beam
in which the maximum stress / (if compressive ; positive, if tensile,
negative), in every section is the same, has the same curvature every-
where if its depth is constant.
If d is the depth, the condition for constant strength is that
-J '\d=±f a constant. But -y=^x curvature, hence curvature
-E.d'
Exercise. In a beam of constant strength if c^= , .
d^V '2f
Then ^-^ = -^ {a-\-hx). Integrating we find
ET 1 TTI
2?' ^=<' + ^+i^'^^ ■^^.y = e^rCX-\-\ax''--V\.hx^,
where e and c must be determined by some given condition. Thus
if the beam is fixed at the end, where ^=0, and -^- = there, and
CLX
also y = there, then c = and e — 0.
In a beam originally straight we know now that, if
X is distance measured from any place along the beam to
a section, and if y is the deflection of the beam at the
section, and / is the moment of inertia of the section, then
^ = ^ a)
dx2 EI ^ ^'
where M is the bending moment at the section, and E is
Young's modulus for the material.
We give to -^ the sign which will make it positive if
M is positive. If M would make a beam convex upwards
and y is measured downwards then (1) is correct. Again,
(1) would be right if M would make a beam concave up-
wards and y is measured upwards.
BENDING.
97
Example /. Uniform beam of length 1 fixed at one
end, loaded with weight W at the other. Let x be the
distance of a section from the fixed end of the beam. Then
M= W{l — x), so that (1) becomes
.(2).
w
<3L
X—
■l-x— ^
Fig. 37.
Integrating, we have, as E and / are constants,
EI dy J . ,
From this we can calculate the slope everywhere.
To find c, we must know the slope at some one place.
Now we know that there is no slope at the fixed end, and
0, hence c = 0. Integrating again,
EI
To find G, we know that ^ = when x = 0, and hence
(7=0, so that we have for the shape of the beam, that is, the
equation giving us y for any point of the beam,
hence -y^ = where x
dx
W
la-)
.(3).
We usually want to know y when x = l, and this value of y
is called D, the deflection of the beam, so that
B
Wl'
SEI
.(4).
Example II. A beam of length I loaded with W at
the middle and supported at the ends. Observe that if
half of this beam in its loaded condition has a casting of
P. 7
98
CALCULUS FOR ENGINEERS.
cement made round it so that it is rigidly held ; the other
half is simply a beam of length ^l, fixed at one end and
W\
W
Fig. 38.
loaded at the other with ^W, and, according to the last
example, its deflection is
ZEI ^ 4:8. EI
(5).
The student ought to make a sketch to illustrate this method
of solving the problem.
Example III. Beam fixed at one end with load w per
unit length spread over it uniformly.
The load on the part FQ is wx PQ or w{l — x).
P Q
— X >* ^'1-X ■*■
Fig. 39.
The resultant of the load acts at midway between P and Q,
so, multiplying hy ^{1 — x), we find M at P, or
M=\w{l-xy (6).
Using this in (1), we have
w dx^
or.
Integrating, we have
2EJ
w
This gives us the slope everywhere
^0 ax
BEAMS. 99
Now -r =0 where ^ = 0, because the beam is fixed
ax
there. Hence c = 0.
Again integrating,
2EI
w
y = ^px" - ^laf + yijja;* + G,
and as ?/ = where a; = 0, = 0, and hence the shape of the
beam is
y=^((ilV-ila^ + a,-) (7).
1/ is greatest at the end where a; — I, so that the deflection is
^=M&7^' "'^ = 8 EI <«>•
if Tr='Z(;^, the whole load on the beam.
Example IV. Beam of length I loaded uniformly with
lu per unit length, supported at the ends.
Each of the supporting forces is half the total load. The
moment about P of ^zvl, p q
at the distance PQ, is --------
-<,— X — *^-il-x-^
iwl
against the hands of a >
watch, and I call this i^i
direction positive; the
moment of the load Fig. 40.
'^^ (2 ^ ~ ^) ^t the average distance ^PQ is therefore negative,
and hence the bending moment at P is
^lul (hi — x) — ^w ( JZ — xy, or ^^Z- — ^wx'^ . . .(9),
so that, from (1), EI ,^ = ^wP - \wx'^,
y being the vertical height of the point P above the middle
of the beam, see Art. 60. Integrating we have
dy
dx
EI -^ = \wl~x — \wa? 4- c,
a formula which enables us to find the slope everywhere.
7-2
100 CALCULUS FOR ENGINEERS.
c is determined by our knowledge that ^ = where a; = 0,
and hence c = 0. Integrating again,
and C=0, because y = 0, where x = 0. Hence the shape of
the beam is y = aaW'T^^^^^ " 2^)-"(l^X V is greatest where
X = i?, and is what is usually called the deflection D of the
beam, or ^= oQ±pf if" ^= ^^^ the total load.
61. Beams Fixed at the Ends. Torques applied
at the ends of a beam to fix them (that is, to keep the end
sections in vertical planes) are equal and opposite if the
loading is symmetrical on the two sides of the centre of the
beam. The torques being equal, the supporting forces are
the same as before. Now if 7ti is the bending moment
(positive if the beam tends to get concave upwards) which
the loads and supporting forces would produce if the ends
were not fixed, the bending moment is now ni — c because
the end torques c are equal and opposite, and the supporting
forces are unaltered by fixing.
^^"^ d^^=^r w-
If the beam is uniform and we integrate, we find
EI .-— = ini.dx — ex -\- const (2).
Take x as measured from one end. We have the Uvo
conditions : / = where x = 0, and ,- = where x = lAf I
ax dx
is the length of the beam. Hence if we subtract the value
of (2) when ic = from what it is when x=l, we have
n I ri
0=1 7n . dx — cl, or c = J j m . dx,
that is, c is the average value of m all over the beam.
The rule is then (for symmetric loads): — Draw the diagram
of bending moment m as if the beam were merely suppoi^ted
BEAMS FIXED AT THE ENDS. 101
at the ends. Find the average height of the diagram and
loiuer the curved outline of the diagram by that amount.
The resulting diagram, which will be negative at the ends, is
the true diagram of bending moment. The beam is concave
upwards where the bending moment is positive, and it is
convex upwards where the bending moment is negative, and
there are points of inflexion, or places of no curvature, where
there is no bending moment.
Example. Thus it is well known that if a beam of length
I is supported at the ends and loaded in the middle
with a load W, the bending moment is \Wl at the middle
and is at the ends, the diagram being formed of two
straight lines. The student is supposed to draw this diagram
(see also Example II.). The average height of it is half the
middle height or ^Wl, and this is c the torque which must
be applied at each end to fix it if the ends are fixed.
The whole diagram being lowered by this amount it is
evident that the true bending moment of such a beam if its
ends are fixed, is ^Wl at the middle, half-way to each end
from the middle so that there are points of inflexion there,
and — iTT^ at each end. A rectangular beam or a beam of
rolled girder section, or any other section symmetrical above
and below the neutral line, is equally ready to break at the
ends or at the middle.
Example. A uniform beam loaded uniformly with
load w per unit length, supported at the ends ; the diagram
for 711 is a parabola (see Example IV., where ilf = ^tul^ — ^wx^);
the greatest value of m is at the middle and it is ^tul^ ; m is
at the ends. Now the average value of m is | of its middle
value (see Art. 43, area of a parabola). Hence c = -^wP.
This average value of m is to be subtracted from every value
and we have the value of the real bending moment every-
where for a beam fixed at the ends.
Hence in such a beam fixed at the ends the bending
moment in the middle is i^^ivl-y at the ends — -^i^wP, and the
diagram is parabolic, being in fact the diagram for a beam
supported at the ends, lowered by the amount -^wP every-
where. The points of inflexion are nearer the ends than in
the last case. The beam is most likely to break at the ends.
Students ought to make diagrams for various examples of
102 CALCULUS FOR ENGINEERS.
symmetrical loading. Find m by the ordinary graphical
method and lower the diagram by its average height.
When the beam symmetrically loaded and fixed
at the ends is not uniform in section^ the integral
of (1) is
^t-i>-^lT (^).
and as before this is between the limits and I, and hence
to find c it is necessary to draw a diagram showing the value
111/
of -J everywhere and to find its area. Divide this by the
area of a diagram which shows the value of y ever3rwhere,
or the average height of the Mjl diagram is to be
divided by the average height of the 1// diagram and
we have c. Subtract this value of c from every value of
m, and we have the true diagram of bending moment of the
beam. Graphical exercises are much more varied and interest-
ing than algebraic ones, as it is so easy, graphically, to draw
diagrams of m when the loading is known.
The solution just given is applicable to a beam of which
the / of every cross section is settled beforehand in any
arbitrary manner, so long as / and the loading are symmetrical
on the two sides of the middle. Let us give to / such a
value that the beam shall be of uniform strength every-
M
where ; that is, that -jz=fc or ft . . .(4), where z is the greatest
distance of any point in the section from the neutral line on
the compression or tension side and/, and/^ are the constant
maximum stresses in compression or tension to which the
material is subjected in every section. Taking /<. as
numerically equal to ft and z = ^d, where d is the depth
M
of the beam, (4) becomes Trc?= ± 2/"... (5), the + sign being
taken over parts of the beam where if is positive, the — sign
where 31 is negative. As I -y c^^ = 0, or, using (5-),
/iff^^^O (6),
BEAMS FIXED AT THE ENDS.
103
the negative sign being taken from the ends of the beam to
the points of inflexion, and the positive sign being taken
between the two points of inflexion. We see then that to
satisfy (6) we have only to solve the following problem.
In the figure, EATUGQE is a diagram whose ordinates
represent the values of -^ or the reciprocal of the depth of
the beam which may be arbitrarily fixed, care being taken,
however, that d is the same at points which are at the same
distance from the centre. EFGE is a diagram of the values
of m easily drawn when the loading is known. We are re-
quired to find a point P, such that the area of EPTA = area
Fig. 41.
of POO'Tj where is in the middle of the beam. When
found, this point P is a point of inflexion and PR is
what we have called c. That is, m — PR is the real
bending moment M at every place, or the diagram EFG
must be lowered vertically till R is at P to obtain the
diagram of M. Knowing M and d it is easy to find /
through (5).
It is evident that if such a beam of uniform strength
is also of uniform depth, the points of inflexion are half-
way between the middle and the fixed ends. Beams of
uniform strength and depth are of the same curvature
ever3rvvhere except that it suddenly changes sign at the
points of inflexion.
61. In the most general way of loading, the bending
moments required at the ends to fix them are different
flrom one another, and if m^^ is the torque against the hands of a
watch apphed at the end A, and ?«._, is the torque with the hands of a
104 CALCULUS FOR ENGINEERS.
watch at the end /?, and if the bending moment in case the beam were
merely supported is m : —
Consider a weightless unloaded beam of the same length with the
torques oui and wio applied to its ends; to keep it in equilibrium it is
necessary to introciuce equal and opposite supporting forces P at the
m,
c
m,
^A
--T--
4--— •
X-
— ^
B '
P
>
Fig. 42.
ends as shown in the figure. Then Fl + m2 = m^f the forces &c. being
as drawn in fig. 42, so that P=— i-^ — ^.
If then these torques m.^ and wij are exerted they must be balanced
by the forces P shown ; that is, at -B a downward force must be exerted ;
this means that the beam at B tends to rise, and hence the ordinary
supporting force at B must be diminished by amount P. At any place
C the bending moment will be m (what it would be if the beam were
merely supported at the ends) — m^-P. BC. . .(1). If one does not care
to think much, it is sufficient to say : — The beam was in equilibrium
being loaded and merely supported at the ends ; the bending moment
at any place was m ; we have introduced now a new set of forces which
balance, the bending moment at C due to these new forces is
-(ma + P.iiC).
So that the tnie bending moment at C is m - m^ - P. BC.
9)1
Suppose r/i2 = 0, then P= j-, and the bending moment at C is
I
m-^.BCovm-P.BC.
62. Beam fixed at the end A^ merely supported at
B which is exactly on the same level as A. As Wg^^ ^"d letting
BC^x, we have the very case just mentioned, and
El'^^^^m-Px (2).
We will first consider a uniform beam uniformly loaded
as in Example IV., Art. 60. It will be found that when x is measured
from the end of the beam, the bending moment in = \wlx -\wx'^-^ if the
beam is merely supported at its ends and \o is the load per unit length.
Hence (2) is
EI^ = ^wlx-\wx''--Px (3),
El-^=\wlx'^-\wx^-\Px^\c (4).
BENDING. 105
We have also the condition that -i\=0 where ^=^...(5), for it is to be
observed that we measure ,v from the unfixed end.
Again integrating,
£;ii/=^wla;^-^wa)*-^Pa^ + CA' (6).
We need noi add a constant because y is when ^ is 0.
We also have y=0 when x=l. Using this condition and also (5)
we find
0=j\v>l^-^PP + c (7),
0=^^wl^-^PP+d (8),
and these enable us to determine P and c.
Divide (8) by I and subtract from (7) and we have
(d=^wP - ^PP or P= Iwl,
hence from (7 ), 0'=^wP — ^wP+CfC—- -^g wP.
We have the true bending moment,
^wlx - J wx^ - \ ivlxj
and (6) gives us the shape of the beam.
63. If the loading is of any kind whatsoever and
if the section varies in any way a graphic metliod of inte-
gration must be used in working the above example. Now if the value
of an ordinate z which is a function of ^ be shown on a curve, we have
no instrument which can be relied upon for showing in a new curve
/
z . dx, that is, the ordinate of the new curve representing the area of
the z curve up to that value of .r from any fixed ordinate. I have
sometimes used squared paper and counted the number of the squares.
I have sometimes used a planimeter to find the areas up to certain
values of .v, raised ordinates at those places representing the areas to
scale, and drawn a curve by hand through the ten or twelve or more
points so found. There are integrators to be bought ; I have not cared
to use any of them, and perhaps it is hardly fair to say that I do not
believe in the accuracy of such of them as I have seen.
A cheap and accurate form of integrator would not only be very
useful in the solution of graphical problems ; it would, if it were used,
give great aid in enabling men to understand the calculus.
Let us suppose that the student has some method of showing the
value of I z. dx in a new curve ; the loading being of any kind what-
y
soever and / varying, since
dx^
106 CALCULUS FOn ENGINEERS,
we have on integrating,
"£-} i'^-'-^l~r-' w.
We see that it ia necessary to make a diagram whose ordinate every-
where is -j and we mast integrate it. Let I -^ o?^ be culled fi ; when
w.
x=lf IX becomes the whole area of the -j diagram and we will call
this ^j.
It is also necessary to make a diagram whose ordinate everywhere
is -J and integrate it. Let J -.dxhQ called X, When x=l^ X becomes
X
the whole area of the j diagram and we will call this X^.
Then as in (9), d * when x=lj
o=f,,-p. x,+c ao).
Integrating (9) again, we have
Ey=\ii. dx - P \X . dx + ex + C.
In this if we use y — O when x=0, we shall find C'=0, and again if
y=0 when x=l, and if we use Mj and Xj as the total areas of the /x
and X curves we have
= Mi-P.Xi + e;* (11),
from (10) and (11) i^ and c may be found, and of course P enables us
to state the bending moment everywhere, -c is the slope when .'t-
is 0.
64. Example. Beam of any changing section fixed
* Without using the letters fi, X, yUj , X^ &c. the above investigation is : —
e:i]--/:t--/:^^ <-'•
Integrating again between the limits and I and recollecting that y is
the same at both limits
f4]=0=f7'"i..-p/"'f'^%.Z (11).
a;=0 J J J- J J J-
The integrations in (10) and (11) being performed, the unknowns P and c
can be calculated ; the true bending moment everywhere is what we started
with,
m - Px,
BENDING. 107
at the ends^ any kind of loading. Measuring x from one
end where there is the fixing couple ?«2)
M=m-m,-rj; (1),
^'g = ?-'?-''7 (^).
^j-,= I 7 • dx-m^ I ~r -P \ -y-+constant (3),
Let fi = I — 'y-^ and fi^ the whole area of the -^ curve;
(dx 1
Let F= / -y and Fj the whole area of the y curve; Let
foe dx 1'
X=l ' and Xi the whole area of the ~ curve; then
= /Xi-m2Fi-PXi (4).
Again integrating
y=j/jL.dx — 111.2 1 Y ,dx — P \X .dx-\- const.
Calling the integi^als from to I of the //,, F and X curves
Mj, Yj, and X^, we have
= Mi-7?i.,Yi-PXi* (5),
and as rrio and P are easily found from (4) and (5), (1) is
known.
* We have used the Bymbols /a, A', Y, /j.^, X^, Y^, M, Y, Mi, Xi, Yi fearing
that students are still a little unfamiliar with the symbols of the calculus ;
perhaps it would have been better to put the investigation in its proper
form and to ask the student to make himself familiar with the usual symbol
instead of dragging in eleven fresh symbols.
After (3) above, write as follows ; —
U d}='=} 07"^- '"■'). T-n.-r w-
Again integrating between Umits
l-U=N"^a.-.J'l'^^-r['f'i^ (5,
x=o J J J- J J ■'■ yoyo-'
The integrations indicated in (4) and (5) being performed, the unknowns m^
and P can be calculated and used in (1). The student must settle for
himself which is the better course to take; to use the formidable looking
but really easily understood symbols of this note or to introduce the eleven
letters whose meaning one is always forgetting. See also the previous note.
108
CALCULUS FOR ENGINEERS.
65. In Graphical work. Let A GB (fig. 43) represent m,
the bending moment, if the beam were merely supported at
the ends; let AD represent nv^ and let BE represent Wg.
Join DE. Then the difference between the ordinates of
Fig. 43.
ACB and of ADEB represents the actual bending moment;
that is the vertical ordinates of the space between the straight
line DE and the curve AFCGB. It is negative from A to
H and from / to B, and positive from H to I. F and G are
points of inflexion.
Useftil Analogies in Beam Problems. If io
is the load per unit length
on a beam and M is the bend-
ing moment at a section
(positive when it tends to
make the beam convex up-
wards*), a; being horizontal
distance, to prove that
dF = ^ ^^>-
If at the section at P
fig. 44, whose distance to the
right of some origin is oc there
is a bending moment M in-
dicated by the two equal and
p. ^, opposite arrow heads and a
shearing force S as shown,
being positive if the material to the right of the section is
* This convention is necessary only in the following generalization.
BENDING. 109
acted on by downward force, and if PQ is hx so that the load
on this piece of beam between the sections at P and Q is
w .hx\ if the bending moment on the Q section is M + hM
and the shearing force S + hS, then the forces acting on this
piece of beam are shown in the figure and from their equili-
brium we know that
hS — io.hx or — - = w (2),
and taking moments about Q,
3I+S.Sx + iw (8xy=^M+ MI,
or -tT- = o + 1^ . ox,
ox ^
and in the limit as hx is made smaller and smaller
^ = S (3),
and hence (1) is true.
Now it is well known that in beams if y is the deflection
d2y M
d^"EI
.(4).
If we have a diagram which shows at every place the
value of Wy called usually a diagram of loading, it is an
exercise known to all students that we can draw at once by
graphical statics a diagram showing the value of M at every
place to scale ; that is we can solve (1) very easily graphi-
cally*. We can see from (4) that if we get a diagram
M
showing yn- at every place, we can use exactly the same
method (and we have exactly the same rule as to scale) to
find the value of y ; that is, to draw the shape of the
beam. Many of these exercises ought to be worked by all
engineers.
* We find M usually for a beam merely supported at the ends. Let it
be AGB, fig. 43. If instead, there are bending moments at the ends we
let AD and BE represent these and join DE. Then the algebraic sum of
the ordinates of the two diagrams is the real diagram of bending moment.
110 CALCULUS FOR ENGINEERS.
Example. In any beam whether supported at the ends
or not : if -m; is constant, integrating (1) we find
— - — h-\-iux and M = a-\-hx-\- ^tuaf (5).
In any problem we have data to determine a and b.
Take the case of a uniform beam uniformly loaded and
merely supported at the ends.
Measure y upwards from the middle and x from the
middle. Then i/= where x = ^l and — ^l,
and = tt — ^hl + ^wl-. I
Hence 6 = 0, ti = — Itvl^ and (5) becomes
M = - iwl" ■}- ^wx"" (6),
which is exactly what we used in Example IV. (Art. 60)
where we afterwards divided M by EI and integrated twice
to find y.
Let i be -^- or the slope of the beam.
e,. dy . di M dM ^ dS
dx ' dx EI dx dx
we have a succession of curves which may be obtained
from knowing the shape of the beam y by differentiation, or
which may be obtained from knowing w, the loading of the
beam, by integration. Knowing w there is an easy graphical
rule for finding MjEI, knowing MjEI we have the same
graphical rule for finding y. Some rules that are obviously
true in the w to MjEI construction and need no mathematical
proof, may at once be used without mathematical proof in
applying the analogous rule from MjEI to y. Thus the area
of the MjEI curve between the ordinates x^ and x^ is the
increase of i from x^ to x.2y and tangents to the curve showing
the shape of the beam at x^ and x^ meet at a point which
is vertically in a line with the centre of gravity of the
portion of area of the MjEI curve in question. Thus the
whole area of the M/EI curve in a span IIJ is equal to the
di/
increase in -~ from one end of the span to the other, and
CONTINUOUS GIRDERS. Ill
the tangents to the beam at its ends H, J meet in a point
P which is in the same vertical as the centre of gravity of
the whole M/EI curve. These two rules may be taken as
the starting point for a complete treatment of the subject
of beams by graphical methods.
If the vertical from this centre of gravity is at the
horizontal distance HG from H and GJ from «/, then P is
higher than H by the amount HG x ij^, the symbol % being
used to mean the slope at ^ ; J" is higher than P by the
amount GJ x { at J. Hence J is higher than H by the
amount
HG.i^+GJ.ij,
a relation which may be useful when conditions as to the
relative heights of the supports are given, as in continuous
beam problems.
67. Theorem of Three Moments. For some time,
Railway Engineers, instead of using separate girders for
the spans of a bridge, fastened together contiguous ends
to prevent their tilting up and so made use of what
are called continuous girders. It is easy to show that
if we can be absolutely certain of the positions of the
points of support, continuous girders are much cheaper than
separate girders. Unfortunately a comparatively small
settlement of one of the supports alters completely the
condition of things. In many other parts of Applied
Mechanics we have the same difficulty in deciding between
cheapness with some uncertainty and a gi-eater expense with
certainty. Thus there is much greater uncertainty as to the
nature of the forces acting at riveted joints than at hinged
joints and therefore a structure with hinged joints is pre-
ferred to the other, although, if we could be absolutely
certain of our conditions an equally strong riveted structure
might be made which would be much cheaper.
Students interested in the theory of continuous girders
will do well to read a paper published in the Proceedings of
the Royal Society/, 199, 1879, where they will find a graphical
method of solving the most general problems.t I will take
here as a good example of the use of the calculus, a uniform
girder resting on supports at the same level, with a uniform
112
CALCULUS FOR ENGINEERS.
load distribution on each span. Let ABC be the centre line
of two spans, the girder originally straight, supported at
A, B and G. The distance from J. to .S is ^i and from B to G
20£5S'
-^^^^^^^^mxww^mm
Fig. 45.
is I2 and there are any kinds of loading in the two spans.
Let A, B and G be the bending moments at A, B and G
respectively, counted positive if the beam is concave upwards.
At the section at P at the distance x from A let m be
what the bending moment would have been if the girder
on each span were quite separate from the rest. We have
already seen that by introducing couples 77I2 and mi at A and
B (tending to make the beam convex upwards at A and B)
we made the bending moment at P really become what
is given in Art. 61. Our m^ — — A^m^^ — B, and hence the
bending moment at P is
B-A_ d'j
m-i- A -^ X
EI
■(1),
where m would be the bending moment if the beam were
merely supported at the ends, and the supporting force at A
is lessened by the amount
^ A-B
.(2).
I.
Assume EI constant and integrate with regard to x and
we have
/
B — A
m .dx -\- Ax + ^x^ — J 1- Ci
^'■i (^^)-
Using the sign jltn. dx . dx to mean the integration of
the curve representing J7a.dx we have
r r T) A
\\m.dx.dx'^\Ax^+^x' — ^ + Ci^p -f e = ^Y . 3/ . . . ( 4).
THEOREM OF THREE MOMENTS. 113
As 7/ is when ic=0 and it is evident that llm.dx.da; =
when cc = 0, e is 0. Again y = when a;=lj^. Using the
symbol //^ to indicate the sum jjm.dx.dic over the whole
span,
fi, + iAk'-{-ili'(B-A) + cA = (5).
From (3) let us calculate the value of EI -—- at the point
B, and let us use the letter a^ to mean the area of the m
curve over the span, or I m . dx, so that EI ~ at 5 is
a^-\-Ak + \k{B-A) + c, (6).
But at any point Q of the second span, if we had let BQ = x
we should have had the same equations as (1), (3) and (4)
using the letters B for A and G for B and the constant Ca.
Hence making this change in (3) and finding EI -^- at the
point B where a: = 0, we have (6) equal to Co or '^
C2-Ci = ai + ^^i4-i^i(5-^) (7),
and instead of (5) we have
/i, + \BU 4- ^U {C-B)-¥ cA = (8).
Subtracting (5) from (8) after dividing by Zj and l.j we
have
c,-c, = ^-f^ + iAh-iBk + lh{B-A)-ih{C-B)...(9).
The equality of (7) and (9) is
an equation connecting A, B and C, the bending moments
at three consecutive supports. If we have any number of
supports and at the end ones we have the bending moments
because the girder is merely supported there, or if we have
two conditions given which will enable us to find them in
case the girder is fixed or partly fixed, note that by writing
P. 8
114 CALCULUS FOR ENGINEERS.
down (10) for every three consecutive supports we have a
sufficient number of equations to determine all the bending
moments at the supports.
Example. Let the loads be w^ and w^ per unit length
over two consecutive spans of lengths l^ and l^. Then
m = \wlx — \wa^, jm.dx= \wla^ — \wa?,
Hence rti = T^^'j ^^^ \\m.dx ,dx = -^wlaf^ — -^wa^.
Hence /Aj = -^w^k\ fi^ = i^wj^i*.
Hence ^ + tti — y^ becomes -^-fW.}.^ + r-^ Zi» — ^w^li^,
or ^\ (ukI/ 4- w^li"),
and hence the theorem becomes in this case
Al, + 2B (I, + k) + Gl, + i (wj2^ + w^k') = (10).
If the spans are similar and similarly loaded then
A-h4^B-\-C+^wP=^0 (11).
Case 1. A uniform and uniformly loaded beam rests on
three equidistant supports. Here A = C = and B = — ^wl\
m = \w {Ix -- aP), and hence the bending moment at a point
P distant x from A is
\w {Ix - a^) + --^wl\
The supporting force at A is lessened from what it would be
if the part of the beam AB were distinct by the amount
shewn in (2), — ^ — or ^wl. It would have been ^l, so now
it is really ^wl at each of the end supports, and as the total
load is 2wl, there remains ^^-wl for the middle support.
Case 2. A uniform and uniformly loaded beam rests on
four equidistant supports, and the bending moments at
these supports are A, B, G, D. Now A = D=0 and from
symmetry B = G. Thus (11) gives us
+ 5J5+irf = or B=^G=^--^^wl\
SHEAR STRESS IN BEAMS.
115
If the span AB had been distinct, the first support would
have had the load ^wl, it now has ^wl — -^^wl or -^qwI. The
supporting force at D is also -^-^wl. The other two supports
divide between them the remainder of the total load which
is altogether ^wl and so each receives \^wl. The supporting
forces are then -^^wl, \^wl, \^wl and -^-^wl.
68. Shear Stress in Beams. Let the distance
measured from any section of a beam, say at 0, fig. 46,
to the section at A be x, and let OB — x-\-hx. Let the
bending moment at G'AG be M and at B'BD be M + hM,
C D
H
E— IF
C D
—
E
Fig. 46.
Fig. 47.
Fig. 48.
OAB (fig. 46) and A A (fig. 47) represent the neutral
surface. We want to know the tangential or shear stress /
at E on tlie plane GAC. Now it is known that this is
the same as the tangential stress in the direction EF on
the plane EF which is at right angles to the paper and
parallel to the neutral surface at AB. Consider the equi-
librium of the piece of beam ECDF, shown in fig. 47 as ECE,
and shown magnified in fig. 48. We have indicated only
the forces which are parallel to the neutral surface or at
right angles to the sections. The total pushing forces on DF
are greater than the total pushing forces on CE, the tangential
forces mi EF making up for the difference. We have only to
state this mathematically and we have solved our problem.
At a place like ^ in the plane CAC at a distance y from
the neutral surface the compressive stress is known to be
M . ,
p = Y y^ ^^^ ^^ ^ ^® *^^ breadth of the section there, shown
as IIH (fig. 47), the total pushing force on the area ECE is
rAC \f Jlf rAC
P= b'^y.dy or P^'y- hy.dy (1).
J AE J- -^ J AE
8—2
116 CALCULUS FOR ENGINEERS.
Observe that if b varies, we must know it as a function of
y before we can integrate in (1). Suppose we call this total
pushing force on EC by the name P, then the total push-
ing force on DF will be P + 8^ . -r* . The tangential force on
EF is fx area of EFovf.hx. EE, and hence
f.Bx.EE = Bx.^~ovf= J:^ . i? (2).
*^ dx -' EE dx ^ '
Example. Beam of uniform rectangular section^ of
constant breadth 6 and constant depth d. Then
p 12m p^ , i2iifr^ 1
F^^-^{\d^-AE%
and hence /= ^ |(t^'3_^/;.)^^^ (8);
so that /is known as soon as M is known.
As to M, let us choose a case, say the case of a beam
supported at the ends and loaded uniformly with iv lb.
per unit length of the beam. We saw that in this case, x
being distance from the middle
M — \wl^ — \wx^.
Hence - 7- = — wx, so that (3) is
dx
f=~{^l''-AE^)wx (4).
If we like we may now use the letter y for the distance -4 -£^,
and we see that at any point of this beam, x inches measured
horizontally from the middle, and y inches above the neutral
line the shear stress is
/=-5«*-2'')'^ ^■'^-
The — sign means that the material below EF acts on
the material above EF in the opposite sense to that of the
arrow heads shown at EF, fig. 48.
SHEAR STRESS IN BEAMS. 117
Observe that where 2/ = the shear stress is greater
than at any other point of the section, that is, at points
in the neutral line. The shear stress is at C. Again, the
end sections of the beam have greatest shear. A student
has much food for thought in this result (5). It is interest-
ing to find the directions and amounts of the principal
stresses at every point of the beam, that is, the interfaces at
right angles to one another at any point, across which there
is only compression or only tension without tangential stress.
We have been considering a rectangular section. The
student ought to work exercises on other sections as soon as
he is able to integrate hy with regard to 3/ in (1) where h is
rAG
any function of y. He will notice that I hy.dy is equal to
J AE
the area of EHGHE, fig. 47, multiplied by the distance of its
centre of gravity from A A.
Taking a flanged section the student will find that / is
small in the flanges and gets greater in the web. Even in a
rectangular section /' became rapidly smaller further out
from the neutral line, but now to obtain it we must divide by
the breadth of the section and this breadth is comparatively
so great in the flanges that there is practically no shearing
there, the shear being confined to the web ; whereas in the
web itself / does not vary very much. The student already
knows that it is our usual custom to calculate the areas of
the flanges or top and bottom booms of a girder as if they
merely resisted compressive and tensile forces, and the web
or the diagonal bracing as if it merely resisted shearing.
He will note that the shear in a section is great only where
, or rather -^\-f] is great. But inasmuch as in Art. QQ
we saw that -j- — S, the total shearing force at the section,
there is nothing very extraordinary in finding that the actual
shear stress anywhere in the section depends upon -r- . In
a uniformly loaded beam -7- is greatest at the ends and gets
less and less towards the middle and then changes sign,
hence the bracing of a girder loaded mainly with its own
weight is much slighter in the middle than at the ends.
118 CALCULUS FOR ENGINEERS.
Deflection of Beams. If a bending moment M acts at a
section of a beam, the part of length bx gets the strain-energy \ — '' ,
because M . dx/FI is the angular change (see Art. 26), and therefore
the whole strain-energy in a beam due to bending moment is
m-'^ <«\
If / is a shear stress, the shear strain-energy \}er unit volume is
lf2/2iV...(7), and by adding we can therefore find its total amount for
the whole beam.
By equating tlic strain-energy to the loads multiplied by half the
displacements produced by them we obtain interesting relations. Thus
in the case of a beam of length /, of rectangular section, fixed at one
end and loaded at the other with a load W; at the distance x from the
end, M= Wx and the energy due to bending is
172^.2
dx=WH^lQEI (8).
1 p TF2;
^Ej^ I
The above expression (5) gives for the shearing stress
f=\^iid?-f)W (9).
The shear strain-energy in the elementary volume h .bx . by is
b.bx . by.f'^/2K Integrating this with regard to y from - ^c^ to -{-^d
we find the energy in the slice between two sections to be
SWH.bx/biYbd,
so that the shear strain-energy in the beam is 3 WH/dNbd... (10).
If now the load W produces the deflection z at the end of the beam
the work done is ^ Wz,..{\l).
Equating (11) to the sum of (8) and (10) we find
.= i^ + iL ^^ (12)
Note that the first part of this due to bending is the deflection as
calculated in Art. 60, Example I. We believe that the other part due
to shearing has never before been calculated.
If the deflection due to bending is z^ and to shearing is z^^
zjz.^=iomy3Ed^.
Taking A^= § ^ as being fairly correct, then zjs.^ = Al^j^d^. If a beam
is 10 inches deep, when its length is 8-6 inches the deflections due to
bending and shear are equal ; when its length is 86 inches, the deflection
due to bending is 100 times that due to shear ; when its length is 0-79
inch, the deflection due to bending is only 1 /100th of that due to
shear. Probably however our assumed laws of bending do not apply
to so short a beam.
BENDING OF STRIPS.
119
69. Springs which Bend. Let fig. 49 show the centre
line of a sirring fixed at A,
loaded at B with a small
load W in the direction
shown. To find the
amount of yielding at B.
The load and the deflec-
tion are supposed to be
very small. Consider
the piece of spring
bounded by cross sec-
tions at P and Q. Let FQ = 8s, the length of the spring
between B and P being called s.
Fig. 49.
The bending moment at P is W . PR or W.w if x is, the
length of the perpendicular from P upon the direction of W.
Let BR be called y. Consider first that part of the motion
of B which is due to the change of shape of QP alone ; that
is, imagine AQ to be perfectly rigid and PB a rigid pointer.
The section at Q being fixed, the section at P gets an angular
M
change equal to Ss x the change of curvature there, or Bs -^rt-
8s Wx . .
or — '^p^ ...(1), where E is Young's modulus and / is the
moment of inertia of the cross section. The motion of B due
to this is just the same as if PB were a straight pointer ;
in fact the pointer PB gets this angular motion and the
motion of B is this angle, multiplied by the straight
distance PB or
Ss. Wx
EI
PB
.(2).
Now how much of B's motion is in the direction of IF ?
PR X
It is its whole motion x -p^ or x p-^ and hence B'a
motion in the direction of W is
hs . Wx"
EI
.(3).
120 CALCULUS FOR ENGINEERS.
Similarly B'a motion at right angles to the direction of
W is ^-i^ (4).
In the most general cases, it is easy to work out the
integrals of (3) and (4) graphically.
We usually divide the whole length of the spring from B
to A into a large number of equal parts so as to have all the
values of Bs the same, and then we may say (s being the
s W
whole length of the spring) that we have to multiply —w—
upon the average values of -^ and -j- for each part. In a well
made spring if b is the breadth of a strip at right angles to
the paper and t its thickness so that / = -f^bt^ we usually
have the spring equally ready to break everywhere or
~r42~~f> ''^ constant. When this is the case (3) and (4)
become
2/.^^ ^ and ^-^'^^ y
And if the strip is constant in thickness, varying in
breadth in proportion to x, then
If X and y are the x and y of the centre of gravity of
the curve (see Art. 48)
^- is the total yielding parallel to W,
li/t
•^-^ is the total yielding at right angles to 'W.
70. Exercises. The curvature of a curve is
When the equation to a curve is given it is easy to find
STATIC PRESSURE OF FLUIDS. 121
-^ and v^and calculate - where r is the radius of curvature.
ax da? r
This is mere exercise work and it is not necessary to prove
beforehand that the formula for the curvature is correct.
1. Find the curvature of the parabola y — ax^ at the
point d? = 0, 3/ = 0.
2. The equation to the shape of a beam, loaded uni-
w
formly and supported at the ends is 3/ = aqtpt {^^^^^ ~ 2^'*),
see Art. 60, where the origin is at the middle of the beam ;
I is the whole length of the beam, lu is the load per unit
length, E is Young's modulus for the material and / is the
moment of inertia of the cross section. Take I = 200, w = 5,
E = 29x 10«, / = 80, find the curvature where x = 0. Show
that in this case f-^) may be neglected, in comparison with
1, and that really the curvature is represented by -vt,.
Show that the bending moment of the above beam is
^^RFT ^^^ " ^'^•^' ^^^^ *^^^ ^^^^ ^^ greatest at the middle
of the beam.
3. Find the curvature of the curve i/= a log x +bx-\-c
at the point where x = Xi.
71. Force due to Pressure of Fluids. Exercise 1.
Prove that if p, the pressure of a fluid, is constant, the
resultant of all the pressure forces on the plane area A is Ap
and acts through the
centre of the area. -"t^j ;
2. The pressure in ''\
a liquid at the depth h \
being wh, where w is the ^^\^
weight of unit volume,
what is the total force due
to pressure on any im-
mei*sed plane area ?- Let
DE be the surface from
which the depth h is
measured and where the ^^S- ^0.
122 CALCULUS FOR ENGINEERS.
pressure is 0. Let BG be an edge view of the area ; imagine
its plane produced to cut the level surface of the liquid DE
in D. Let the angle EDO be called a. Let the distance
DP be called x and let DQ be called x + hx, and let the
breadth of the area at right angles to the paper at P be
called z. On the strip of area z . hx there is the pressure
wh if h is PH the depth of P, and h^x sin a, so that the
pressure force on the strip is
wx . sin OL . z . hx,
rue
and the whole force is F=wsuyol\ x .z .dx (1 ).
J 1)B
Also if this resultant acts at a point in the area at a distance
X from D, taking moments about P,
rDC
FX=tu sina x- .z.dx (2).
J DB
rDC
Observe in (1) that I x ,z ,dx — Ax,
J DB
if A is the whole area and x is the distance of its centre of
gravity from P. Hence, the average pressure over the
area is the pressure at the centre of gravity of the
area.
CDC
Observe in (2) that I x^z .dx — I the moment of inertia
J DB
of the area about P. Letting I — k^A , where k is called the
radius of gyration of the area about P, we see that
F=iu sin a . Ax, FX = w sin a . Ah^.
Hence X = — . . .(3), the distance from D at which the
x
resultant force acts.
Example, If DB = and the area is rectangular, of
constant breadth h ; then
CDC h
I=h x^.dx = ^DC\
Jo «5
and ^ = 6 . PC so that k^ = ^DC\ Also x = J PC. Hence
X = fPC, that is, the resultant force acts at | of the way
CENTRIFUGAL FORCE IN FLUIDS.
123
down the rectangle from i) to (7 and the average pressure is
the pressure at a point half way down.
It is an easily remembered relation that we find in (8).
For if we have a compoimd pendulum, whose radius of
gyration is k and if x is the distance from the point of
support to its centre of gravity and if X is the distance to
its point of percussion, we have the very same equation (3).
Again, if X is the length of the simple pendulum which
oscillates in exactly the same time as the compound one,
we have again this same relation (3). These are merely
mathematical helps to the memory, for the three physical
phenomena have no other relation to one another than a
mathematical one.
Whirling Fluid.
72. Suppose a mass of fluid to rotate like a rigid body about
an axis with the angular velo-
city of a radians per second.
Let 00 be the axis. Let P
be a particle weighing w lbs.
Let OP
o<>
X.
The centrifugal force in
pounds of any mass is the
mass multiplied by the square
of its angular velocity, multi-
plied by X. Here the mass
w
is - and the centrifugal force
• "U) „
IS - o?x. Fig. 51.
9
Make PR represent this to scale and let PS represent w
the weight, to the same scale, then the resultant force, repre-
sented by PT, is easily found and the angle RPT which PT
w
makes with the horizontal. Thus tan RPT — w^— ot^x or
9
g — a^x, being independent oi w: we can therefore apply our
results to heterogeneous fluid. Now if y is the distance of
the point P above some datum level, and we imagine a curve
drawn through P to which PT is (at P) tangential, and if at
124
CALCULUS FOR ENGINEERS.
every point of the curve its direction (or the direction of its
tangent) represents the direction of the resultant force; if
such a curve were drawn its slope ~ is evidently — y- and
its equation is y = — -2 log x + constant (1 ).
The constant depends upon the datum level from which y is
measured. This curve is called a line of force. Its direction
at any place shows the direction of the total force there. We
see that it is a logarithmic curve.
Level Surfaces. If there is a curve to which PT is a
Fig. 52.
FLUID MOTION.
125
normal at the point P, it is evident that its slope is positive
and in fact
dx
— —X,
so that the curve is
y — ^x^-\- constant (2),
the constant depending upon the datum level from which y
is measured. This is a parabola, and if it revolves about the
axis we have a paraboloid of revolution. Any surface which
is everywhere at right angles to the force at every point is
called a level surface and we see that the level surfaces in
this case are paraboloids of revolution. These level surfaces
are sometimes called equi-potential surfaces. It is easy to
prove that the pressure is constant everywhere in such a
surface and that it is a surface of equal density, so that if
mercury, oil, water and air are in a whirling vessel, their
surfaces of separation are paraboloids of revolution.
The student ought to draw one of the lines of force and
cut out a template of it in thin zinc, 00 being another edge.
By sliding along 00 he can draw many lines of Force. Now
cut out a template for one of the parabolas and with it draw
many level surfaces. The two sets of curves cut each other
everywhere orthogonally. Fig. 52 shows the sort of result
obtainable where aa^ hb, cc are the logarithmic lines of force
and A A, BB, GO are the level paraboloidal surfaces.
73. Motion of Fluid. If AB is si stream tube, in the
126 CALCULUS FOR ENGINEERS.
vertical plane of the paper, consider the mass of fluid between
sections at P and Q of length Bs feet along the stream, and
cross-section a square feet, where a and Bs are in the limit
supposed to be infinitely small. Let the pressure at F be
p lbs. per square foot, the velocity v feet per second, and let
F be at the vertical height h feet above some datum level.
At Q let these quantities be ^ + Sp, v + Bv and h + Bh.
Let the fluid weigh w lbs. per cubic foot.
Find the forces urging PQ along the stream, that is,
forces parallel to the stream direction at FQ.
pa acts on one end F in the direction of motion, and
(p-\-Bp)a acts at Q retarding the motion. The weight of
the portion between F and Q is a . Bs . w and, as if on an
inclined plane, its retarding component is
. , ^ height of plane ^ Bh
weight X -, — ^Tu— r n — or a. Bs .w ^,
° length ot plane Bs
Hence we have altogether, accelerating the motion from F
towards Q,
pa — (p + Bp) a — a . Bs . w . ^— .
But the mass is — — , and -7- is its acceleration, and we
have merely to put the force equal to — ^ — '- — . -=- . We have
then, dividing by a,
5. ^ Bh Bs .w dv
-bp-8s.w-^= -J- .
^ Bs g dt
Now if Bt be the time taken by a particle in going from
Bs
F to Q, v — j- with greater and greater accuracy as Bs is
. dv .
shorter and shorter. Also, the acceleration , is more aud
Bv . ^^
more nearly kt . (It is more important to think this matter
out carefully than the student may at first suppose.)
Hence if Bs is very small, Bs .-j- — -^.Bv — v ,Bv, so that
at ot
we have Bp + w. Bh-^-v .Bv = (1),
FLOW OF FLUID. 127
or as we wish to accentuate the fact that this is more and
more nearly true as Ss is smaller and smaller, we may
write it as
^ + dh-\--.dv==0 (2)*,
w g
or integrating, ^ + 9- + I = constant (2).
dp
We leave the sign of integration on the -^ because w may
vary. In a liquid where w is constant,
/* + ^ + --= constant (3).
2g w ^ -^
74. In a gas, we have lu ocpif the temperature could be
kept constant, or we have the rule for adiabatic flow w x py,
where 7 is the well-known ratio of the specific heats. In either
of these cases it is easy to find I — and write out the law. This
law is of universal use in all cases where viscosity may be
neglected and is a great guide to the Hydi^aulic Engineer-
Thus in the case of adiabatic flow w=cpy , the inteOTal of -^ is
f dp 1 /" _i 1 V 1 1
we have h-{--;r--\ — ©"^constant (4).
In a great many problems, changes of level are insignificant and we
* After a little experience with quantities like 5p &c., knowing as we do
that the equations are not true unless 5/), &c. are supposed to be smaller
and smaller without limit and then we write their ratios as ^f , &c., we
an
get into the way of writing dp, &c. instead of 5p, &c.
Again, if /(x) . dx-\-F{y) dy + <p [z) dz = (1),
then 1/H 'dx+ I F {y) . dy + I <p{z) . d2 = a constant (2).
There is no harm in getting accustomed to the integration of such an
equation as (1), all across.
128 CALCULUS FOR ENGINEERS.
often use i?^^.,^^*^ constant (4) for gases. Thas, if «« is the
cs
pressure and zvq the weight of a cubic foot of gas inside a vessel at
places where there is no velocity and if, outside an orifice, the pressure
isp; the constant in (4) is evidently + — jOq«, and hence, outside the
C8
surface, v^=-^{pQ'-p') (5), and as c is Wq-^PqV it is easy to make
all sorts of calculations on the quantity of gas flowing per second.
Observe that if p is very little less than joq, if we use the approxi-
mation (1 +«)" = 1 -\-naf when a is small, we find
v'=^^{Po-p) (6),
a simple rule which it is well to remember in fan and windmill problems.
In a Thomson Water Turbine the velocity of the rim of the wheel is
the velocity due to half the total available pressure ; so in an air turbine
when there is no great difference of pressure, the velocity of the rim of
the wheel is the velocity due to half the pressure difference.
Thus if Pq of the supply is 7000 lbs. per square foot and if p of the
exhaust is 6800 lbs. per square foot and if we take ?^'q=0*28 lb. per
cubic foot, the velocity of the rim V is, since the difference of pressure
is 200 lbs. per square foot,
y
% (100) = 151 feet per second.
Returning to (5) ; neglecting friction, if there is an orifice of area
A to which the flow is guided so that the streams of air are parallel, Q
the volume flowing per second is Q=vA and if the pressure is p, the
weight of stuff flowing per second is
W==vAw,
1 1
or since w = cp^ ^ and iOQ=cp^ ,
If the student will now substitute the value of v from (5) and put
a for p/pQ he will obtain
W
=^"'-N/r^^:(--^^^) ^^)-
Problem. Find p the outside pressure so that for a given inside
pressure there may be a maximum flow.
FLOW OF GASES. 129
It is obvious that as p is diminished more and more, v the velocity
mcreases more and more and so does Q. But a large Q does not
necessarily mean a large quantity of gas. We want W^ to be large.
When is W a maximum 1 That is, what value of a in (7) will make
a^l-a-^ )ova^-a ^
a maximum 1 Differentiating with regard to a and equating to
2 , / -.v 1
?„;— _(i+i)„9=o
dividing by ay we find a = ( ^ )
JUL
In the case of air 7/= 1-41 and we find p= •527po-
That is, there is a maximum quantity leaving the vessel per
second when the outside pressure is a little greater than half the inside
pressure. »
Problem. When p is indefinitely diminished what is v ?
Answer -. v= j,./ -^ — .
This is greater than the velocity of sound in the ratio a/ — — t- ,
being 2 -21 for air. That is, the limiting velocity in the case of air is
2413 feet per second x */ — - , where t is the absolute temperature
inside the vessel and there is a vacuum outside.
Students ought to work out as an example, the velocity of flow
into the atmosphere.
Returning to equations (2) and (4), we assumed h to be of little
importance in many gaseous problems of the mechanical engineer. But
there are many physical problems in which it is necessary to take
account of changes in level. For example if (2) is integrated on the
assumption of constant temperature and we assume v to keep constant,
we find that p diminishes as h increases according to the compound
interest law considered in Chap. 11. Again under the same condition
as to V, but with the adiabatic law for w we find that p diminishes with
h according to a law which may be stated as " the rate of diminution
of temperature with A, is constant." These two propositions seem to
belong more naturally to the subject matter of Chapter II.
130 CALCULUS FOR ENGINEERS.
75. A great number of interesting examples of the use of
(2) might be given. It enables us to understand the flow of
fluid from orifices, the action of jet pumps, the attraction of
light bodies caused by vibrating tuning-forks, why some
valves are actually sucked up more against their seats
instead of being forced away by the issuing stream of fluid,
and many other phenomena which are thought to be very
curious.
Example 1. Particles of water in a basin, flowing very
slowly towards a hole in the centre, move in nearly circular
paths so that the velocity v is inversely proportional to the
distance from the centre^ Take v = - where a is some con-
x
stant and x is the radius or distance from the axis. Then (13)
(Art. 73) becomes
k^f-, + P = f;.
2gx- w
Now at the surface of the water, p is constant, being the
pressure of the atmosphere, so that, there
and this gives us the shape of the curved surface. Assume
c and a, any values, and it is easy to calculate h for any value
of X and so plot the curve. This curve rotated about the axis
gives the shape of* the surface which is a surface of revo-
lution.
Example 2. Water flowing spirally in a horizontal plane
follows the law v = - if ^ is distance from a central point.
w h^
Note that p=:G^—h- - .
^ g X''
The ingenious student ought to study how p and v vary
at right angles to stream lines. He has ohly to consider
the equilibrium of an elementary portion of fluid PQ, fig. 53,
subjected to pressures, centrifugal force and its own weight
in a direction normal to the stream.
He will find that if ~- means the rate at which p
FLOW OF LIQUID. 131
varies in a direction of the radius of curvature away from
the centre of curvature and if a is the angle QPR, fig. 53, the
stream being in the plane of the paper, which is vertical,
dp wv^ . ,^,
-T = lusma (1).
ar g r
If the stream lines are all in horizontal planes
f = ^^' (2).
ar g r ^
Example 3. Stream lines all circular and in horizontal
planes in a liquid, so that h is constant.
If -y = -, where 6 is a constant,
r
dp _ lu h-
dr g ' T^'
p:=— I , + constant (3).
^ ^ g r-
We see therefore that the fall of pressure as we go out-
ward is exactly the same as in the last example. Show that
this law, V = hjr, must be true if there is no ' rotation ' (See
Example 5).
Example 4. Liquid rotates about an axis as if it were a
rigid body, so that v = 6/-, then
dr g '
This shows the law of increase of pressure in the wheel of a
centrifugal pump when full, but when delivering no water.
Exercise. The pressure at the inside of the wheel of
a centrifugal pump is 2116 lbs. per sq. foot, the inside radius
is 0*5 foot, the outside radius 1 foot. The angular velocity
of the wheel is 6 = 30 radians per second ; draw a curve show-
ing the law of p and r from inside to outside when very
little water is being delivered. If the water leaves the wheel
by a spiral path, the velocity everywhere outside being
9—2
132 CALCULUS FOR ENGINEERS.
invei-sely proportional to r, draw also the curve showing
the law of p in the whirlpool chamber outside.
Example 5. The expression
ti^ 1
which remains constant all along a stream line, may be
called the total store of energy of 1 lb. of water in the
stream if the motion is steady.
^^ dE 1 dv 1 dp dh , « ^. ,.,.
Now -r- =~ V j--\ r-4-T- becomes irom equation (1),
d?' g or w dr ar .
r ~ g 2\r dr)
This expression ^ \- + -r) is called the "average angular
velocity" or "the rotation" or the 'spin' of the liquid.
Hence
dE 2v
^j- = — X rotation.
dr g
When liquid flows by gravity from a small orifice in a
large vessel where, at a distance inside the orifice, the liquid
may be supposed at rest, it is obvious the E is the same in
d IR
all stream lines, so that -T7 is 0, and there is no ' rotation '
anywhere. ^
If when water is flowing from an orifice in a vessel we
can say that across some section of the stream the velocity is
everywhere normal to the section and that the pressure is
everywhere atmospheric, we can calculate the rate of flow.
It is as well to say at once that we know of no natural
foundation for these assumptions. However wrong the
assumptions may be, there is no harm in using them in mere
exercises on Integration. There being atmospheric pressure
at the still water level, if v is the velocity at a point at the
depth A, if a is an element of area of the section, Q= 2a 'J2gh
the summation being effected over the whole section, Q being
the volume flowing. Thus if the section is a vertical plane
and if at the depth h it is of horizontal breadth z, through
EXERCISES ON INTEGKATlON. ISS
the area z . Sh water is flowing with the velocity '^2gh, so
that ^2gh . z .Bhis the elementary volume flowing per second,
and if hi and h^ are the depths of the highest and lowest
points of the orifice, the total flow is Q = V2^ ( zh^ . dh.
J hi
Example 6. Rectangular section, horizontal breadth 6,
Q = ^¥gh{\^.dh = |6\/2^ M 1 = lb \f2g (h.^ - h^^).
Example 7. Triangular section, angle at depth h^, base
horizontal of length b at depth h... Then within the limits
of integration it will be found that z= j v (— /<i 4- h).
fl2 — ill
Hence Q = |^^ f(- hih^^ + A?) dh = ^-^-^f W-^hJi^ + W-)\
«2— "W h'i — hi L/t^ J
If the ratio hJjhi be called r, it will be found that
(^ = i^^f J6H-10ri+25l.
When the student has practised integration in Chap. III., he
may in the same way find the hypothetical flow through
circular, elliptic and other sections.
Keturning to the rectangular section, there is no case
practically possible in which h^ is 0, but as this is a mere
mathematical exercise let us assume hi = 0, and we have
Q= |6 s^'lgh^. Now further assume that if there is a rect-
angular sharp-edged notch through which water flows, its
edge or sill being of breadth b and at the depth h^, the
flow through it is in some occult way represented by the
above answer, multiplied by a fraction called a coefficient of
contraction, then Q = cb \l2gh^. Such is the so-called theory
of the flow through a rectangular gauge notch. A true
theory was based by Prof. James Thomson on his law of flow
from similar orifices, one of the very few laws which the
hydraulic engineer has to depend upon. We are sorry to
think that nearly all the mathematics to be found in standard
treatises on Hydraulics is of the above character, that is, it
has only an occult connection with natural phenomena.
134 tJALClTLtJS FOR ENGINEERS.
76. Magnetic Field about a straight round wire.
There are two great laws in Electrical Science. They concern
the two circuits, the magnetic circuit and the electric circuit,
which are always linked through one another.
I. The line integral (called the Gaussage whatever the
unit may be) of Magnetic Force round any closed
curve, is equal to the current [multiplied by 47r if the
current is in what is called absolute C.G.S. units (curious
kind of absolute unit that needs a multiplier in the most
important of all laws); multiplied by 47r/10 if the current
is in commercial units called Amperes].
II. The line integral (called the Voltage whatever
the unit may be) of Electromotive Force round any-
closed curve is equal to the magnetic current (really,
rate of change of induction) which is enclosed. [If the in-
duction is in absolute C.G.S. units, we have absolute Voltage
in C.G.S.; if the induction is in Webers the Voltage is
in Volts.
We are to remember that in a non-conducting medium
the voltage in any circuit produces electric displacement, and
the rate of change of this is cun-ent, and we deal with this
exactly as we deal with currents in conducting material.
When we deal with the phenomena in very small portions
of space we speak of electric and magnetic currents per unit
area, in which case the line integrals are called ' curls/
Leaving out the annoying 47r or 47r/10, we say, with Mr
Heaviside, " The electric current is the curl of the magnetic
force and the magnetic current is the negative curl of the
electric force." When we write out these two statements in
mathematical language, we have the two great Differential
Equations of Electrical Analysis.
The Electrical Engineer is continually using these two
laws. Many examples will be given, later, of the use of the
second law. We find it convenient to give here the following
easy example of the first law.
Field about a round wire. A straight round wire of
or A amperes, so
that G
10
on
r
°%10
MAGNETIC FIELD. loO
If H is the magnetic force at a distance r from the centre
of the wire, the Gaussage round the circle of radius r is
Hx27rr, because H is evidently, from symmetry, the same
all round. Hence, as Gaussage = 47rC,
2 Al
Inside the wire, a circle of radius r encloses the total
current — G, and hence H inside the wire at a distance r
from the axis is
2rG r 2r A
^ ['' To ^
If BG is a cross section of the round wire of radius a,
and if OD is any plane n
through the axis of the f\ p Q p /^"^
wire, and \oJ y \0^
OP==r,OQ = ri-Br: ^ f". 54.
then through the strip of area FQ, which is / centimetres
long at right angles to the paper, and Sr wide, area I.Br,
there is the induction If per sq. cm. [We take the perme-
ability as 1. If /A is the magnetic permeability of the me-
dium, the induction is yS = fiH per sq. cm.], or H .I.Br
through the strip of area in question. If there are two
parallel wires with opposite currents, and if OB is the plane
through the axes of the two wires, the fields due to the two
currents add themselves together. If 0' is the centre of the
other wire, the total i^ at P is 2C ( ^p + yyp
77. Self-induction of two parallel wires. Let the
radius of each wire be a, and the distance between their centres
h, the length of each being I between two planes at right angles
to both. The wires are supposed to be parts of two infinite
wires, to get rid of difficulties in imagining the circuit com-
pleted at the ends.
The total induction from axis to axis is the sum of the
[^ G dr
two amounts, 4>l - - — from the outside of each wire to the
1
J a
136 CALCULUS FOR ENGINEERS.
fa rC
axis of the other and 41 \ —- dr from the axis of each wire
Jo Cb'
to its own surface. This is
21G j 2 log - + 1 1 , or - . j 2 log - + 1 [ in absolute units.
Dividing by 10^ we have it in commercial units.
This total field when the current is 1, is the self-induction
L of the circuit (we imagine current to be uniformly distri-
buted over the section of the wire), and
y = ^ U^g~i + 1^ in c.G.s. units,
in Henries per centimetre length of the two circuits.
78. Function of Two Independent Variables.
Hitherto wo have been studying a function of one variable,
which we have generally called x. In trying to under-
stand Natural Phenomena we endeavour to make one
thing only vary. Thus in observing the laws of gases, we
measure the change of pressure, letting the volume only
change, that is, keeping the temperature constant, and we
find p cc - , Then we keep v constant and let the tempera-
ture alter, and we find p cc t (where ^ = ^"^ C. + 274). After
* Notice that one Henry is lO' absolute units of self-induction ; our
commercial unit of Induction called the Weber is 10* absolute units of
Induction.
dA
The Henry suits the law : \o\iB= RA+L — ,
The Weber suits Volts = J?^ + iV . ^ ,
where R is in ohms, A amperes, L Henries, N the number of turns in a
circuit, I Weber's of Induction.
In Elementary Work such as is dealt with in this book, I submit to the
use of 47r and the difficulties introduced by the unscientific system now in
use. In all my higher work with students, such as may be dealt with in a
succeeding volume, I always use now the rational units of Heaviside and I
feel sure that they must come into general use.
TWO INDEPENDENT VARIABLES. 137
much trial we find, for one pound of a particular gas, the law
pv = Rt to be very nearly true, R being a known constant.
Now observe that any one of the three, p, v or t, is a
function of the other two; and in fact any values what-
soever may be given to two, and the other can then be found.
Thus p = R^ (1),
we can say that p is a, function of the two independent vari-
ables t and V.
If any particular values whatsoever of t and v be taken
in (1) we may calculate p. Now take new values, say t + Bt
and v + Sv, where St and 8v are perfectly independent of one
another, then
. s^ 7? ^ + ^^ J s> n t + Bt j^t
p-\-6p = R ^ and 6/; = it s R - .
V -\-ov ^ V ■\-ov V
We see therefore that the change Sp can be calculated if the
independent changes Et and Bv are known.
When all the changes are considered to be smaller and
smaller without limit, we have an easy way of expressing Bp
in terms of Bt and Bv. It is
'^-m^^-o (^>-
This will be proved presently, but the student ought fii'st
to get acquainted with it. Let him put it in words and
compare his own words with these : " The whole change in ^j
is made up of two parts, 1st the change which would occur
inp if V did not alter, and 2nd the change in p if ^ did not
alter." The first of these is St x the rate of increase of p
with t when v is constant, or as we write it ( -^ ) Bt, and the
second of these is Bv x the rate of increase of p with v if t
is constant.
This idea is constantly in use by every practical
man. It is only the algebraic way of stating it that is
unfamiliar, and a student who is anxious to understand the
subject w^ill manufacture many familiar examples of it for
himself.
138 CALCULUS FOR ENGINEERS.
Thus when one pound of stuff which is defined by its p^ v and t^
changes in state, the change is completely defined by any two of the
changes bp and hi\ or bv and bt, or bp and bt^ because we are supposed
to know the characteristic of the stuff, that is, the law connecting p, v
and t.
Now the heat bH given to the stuff in any small change of state
can be calculated from any two of bv, bt and bp, and all the answers
ought to agree. As we wish to accentuate the fact that the changes
are supposed to be exceedingly small we say
dH=k . dt+ l.dv\
= K. dt-\-L.dp> (3),
= r .dp+V.dv)
where the coefficients k, I, A', A, P and V are 'all functions of the state
of the stuff, that is of any two of v, t and p. Notice that k . dt is the
heat required for a small change of state, defined by its change of
temperature, if the volume is kept constant : hence k is called the
specific heat at constant volume. In the same way K is called the
specific heat at constant pressure. As for I and L perhaps they may
be regarded as some kinds of latent hent, as the temperature is supposed
to be constant.
These coefficients are not usually constant, they depend upon the
state of the body. The mathematical proof that if bH can be calcu-
lated from bt and bv^ then dH=k . dt + l. dv, where k and I are some
numl;ers which depend upon the state of the stuff, is this : — If bH can
be calculated, then bH=L bt-^l.bv-\-a(bty + b{bv)^ + c{bt.bv)+e{btf +
terms of the third and higher degrees in bt and bv, where k, I, a, b, c, e
&c. are coefficients depending upon the state of the body. Dividing by
either bt or bv all across, and assuming bt and bv to diminish without
limit, the proposition is proved.
Illustration. Take it that for one pound of Air, (1) is
true and R is, say, 96, p being in lb. per sq. foot and v in
cubic feet.
As « = 96 - ,
Hence, from (2), hp = ^^^ .M — - .hv (4)
Example. Let t = 300, p = 2000, t; = 1 44.
If t becomes 301 and v becomes 14*5 it is easy to show
that p will became 199283. But we want to find the change
in pressure, using (2) or rather (4),
^ 96 , 2000 , ^^^„
^i^= 14^ >< 1 - 14:4 >< •!=- 7-22 lb. per sq.ft.,
whereas the answer ought to be — 7*17.
'dp\ 96 fdp\
.dt)~ V ' \dvj~
V '
Bp = ^l,Bt-^.Sv..
^ V V
SPECIFIC HEATS. 139
Now try Bt = 1 and Bv = '01 and test the rule. Again, try
^^ = •01 and Bv = "001, or take any other very small changes.
In this way the student will get to know for himself what
the rule (1) really means. It is only true when the changes
are supposed to be smaller and smaller without limit.
Here is an exceedingly interesting exercise : — Suppose
we put hp = in (2). We see then a connection between Bt
and 8v when these changes occur at constant pressure. Divide
one of them by the other ; we have ^ when p is constant,
or rather
fdp\
dv\ \dtj
(dv\ \dtj ,^s
\Tt)-'W{ '^''^•
\dv)
At first sight this minus sign will astonish the student
and give him food for thought, and he will do well to manu-
facture for himself illustrations of (5). Thus to illustrate
it with pv = Rt Here
'dv\ _ R (dp\ _ R (dp\ _ Rt p
/dv\ _ K /dp\ _ R
\dtl ~ ~p ' \dt) ~ V '
Ivj t^ V
and (5) states the truth that
^^_^^( _P
The student cannot have better exercises than those
which he will obtain by expressing hv in terms of ht and Sp,
or ht in terms of Bp and hv for any substance, and illustrating
his deductions by the stuff for which pv = Rt t
79. Further Illustrations. In (.3) we have the same answer
whether we calculate from dt and dv, or from dt and dp, or from dp
and c?y. Thus for example,
Jt . dt+l .dv = K . de + L . dp (6).
We saw that ^P={-j^) (^i+{-r-] dv, and hence substituting this
for dp in (6) we have
k.dt + Ldv = K.dC + Li'^^]dt-{-L{^^)dv:
(1)"-^:
140 CALCULUS FOR ENGINEERS.
This is true for any independent changes dt jind do\ let do=0^ and
again let dt=0^ and we have
^-A-+^g) (7),
^At) («>•
Again, in (6) substitute ^^=(j:)^^+(j~) ^^P^ ^^^^ we have
k.dt + l(^dt-\-l(^\dp = K.dt + L.dp.
Equating coefficients of dt and of dp as before we have
^•+'60=''^' '-'^
<©=^ (^°)-
Again, putting h .dt-\-l .dv = P .dp+ V . doy and substituting
we have /: . <lt + Ld,' = p(^£\ dt+p{^ dv+ V. dv,
-d *=^(|) (").
al»o ;=/>(*)+ F (12).
Again, putting K . dt + L . dp^ P . dp+ V . dv, and substituting
we have K (j-j dp+K l-r-j dv + L . dp = P . dp+ V . dt\
""■> ^(1)+^=^' c-^)'
^(1)='^ (»>•
The relations (7), (8), (9), (lO), (ii), (12), (13) and (14) which
are not really all iudepeiulcnt of one another (and indeed we may get
others in the same way) are obtained merely mathematically
and without assuming any laws of Thermodynamics. We have called
THE TWO ELASTICITIES. 141
ZT, heat; t temperature &c., but we need not, unless we please, attach
any physical meaning to th^ettei-s.
The relations are true for any substance. Find what
they become in the case of the stuff for which pv = Rt (the mathe-
matical abstraction called a perfect gas). We know that
-^ 1 = — , so that (7) becomes k=K-{-L— (7)*,
(s)
/dp\ ^_P ^^ ^j^3^^ (8) becomes l=-L^
(dv\_ R
\dt)~ p
(dv\ _ V
\dp)~~p
\dt)~ V
(dp\_'P
\dv) V
V
(8)*,
SO that (9) becomes k + l — = K (9)*
so that (10) becomes -Z- = X (10)*,
so that (11) becomes ]c=P- (11)*
so that (12) becomes l=-r(-+y (12)^
It is evident that these are not all independent ; thus using (10)*
in (9)* we obtain (7)*.
80. Another Illustration. The Elasticity of our stuff is
defined, see Art. 58, as
dp
dv
Now if t is constant, we shall write this e^ = — v ( -, - j , or the
elasticity when the temperature remains constant.
If it is the adiahatic elasticity e„ which we require, we want to
know the value of -^ when the stufl neither loses nor gains heat. In
the last expression of (3) put dH=Oj and the ratio of our dp and our
dv will then be just what is wanted or l-f) = — p> *^® ^ being
affixed to indicate that ff is constant or that the stuff neither loses
Y
nor gains heat. Hence eij=v ^.
Taking Y from (14) Art. 79 and P from (11),
€n_ \dvj \dtj
ei J 7T%
142 CALCULUS FOR ENGINEERS.
but we have already seen as in (5) that* f ^^ ) -i- ( -^ ) = — ( -r-; ] and
hence for any substance -5 = ^- (15).
This ratio of the two specific Heats is usually denoted by the
letter y. Note that neither of the two laws of Thermodynamics nor a
Scale of temperature is referred to in this proof.
81. General Proof. If u is a function of a- and y, we may
write the statement in the form u=f(x, y). Take particular values
of X and y and calculate u. Now take the values x-\-bx and y + by,
where bx and by are perfectly independent of one another, and calculate
the new u^ call it u + bu. Now subtract and we can only indicate our
result by
bu=f{x+bx, y+by)-f{x, y).
Adding And subtracting the same thing f{x^ y + by) we have
bic=f{x+bx, y+by)-fix, y+by)+f{x, y+by)-f{x, y).
This is the same as
^^_ /(^+&r,y+dy)-/(a;,y+&y ) g^, ^f{^,y+b y)-f{x,y) ^^^^^^^^
Now if bx and by be supposed to get smaller and smaller without
limit, the coefficient of by
or/(^+MzZ(^y) becomes ^^^^ or (f),
by dy \dy)
the X being constant. In fact this is our definition of a differential
coefficient (see Art. 20, Note). Again, the coefficient of bx becomes
the limiting value of — > ^/ J \ ^ if) ^ i^ecause by is evanescent.
Writing then u instead of /(^, y) we have
«»"=(i)<^-+(g)'»y (")•
Thus if w = ax^ + 6y^ + cxy, du = i2ax + cy) dx-\-{2by + ex) dy.
82. Notice that although we may have
dz=M. dx-\-JSr. dy (18),
where M and iV are functions of x and y ; it does not follow that ^ is a
function of x and y. For example, we had in (3)
dll'=k.dt+l.dv^
where k and I are functions of t and v. Now H the total heat
which has been given to a pound of stuff is not a function of v and t ;
it is not a flinction of the state of the stuff. Stufi' ma^
COMPLETE DIFFERENTIAL. 143
receive enormous quantities of heat energy, being brought back to its
original state again, and yet not giving out the same amounts of he^at as
it received. The first law of Thermodynamics states however chat if
dE = dH — p . dv, where p . dv is the mechanical work done, we can
give to E the name Intrinsic Energy because it is something
which is a function of the state of the stuff. It always comes back to
the same value when the stuff returns to the same state.
Our E is then some function of t and v, or of t and jo, or of p and v,
but H is not !
The second law of Thermodynamics is this : — If dH be divided by
t where t is ^°C. + 274, 6°Q. being measured on the perfect gas thermo-
JTT
meter, and if — be called c?0, then ^ is called the Entropy of the
stuff, and </> is a function of the state of the stuff.
83. It is very important, if
dz=M.dx-\-N .d)/ (18),
where M and N are functions of x and y, to know when z is a function
of X and y. If this is the case, then (18) is really
that is, M is ( ;t- ) and iV is (-r-)^
^"^ i^^"^« (^) = ( as^) (1^)'
because it is known that -. — — ,- = -^ ^ .
dy .dx dx . a//
d2u d2u
* Proof that - — —. = -— T= •
dy . dx dx . dy
We gave some illustrations of this in Art. 31, and if the student is not yet
familiar with what is to be proved, he had better work more examples, or
work the old ones over again.
Let w=/(a?, y);
(^) is the limiting value of f^'^ + ^'^^y)-f('^'y^ as 8x gets smaller and
smaller. Now this is a function of y, so — ( ^ j or ^^ is, by our defini-
tion of a differential coefiicient, the limiting value of
1 l f{x + 8x,y + dy)-f{x,y + Sy) f(x + 8x,y)-f{x,y) ]
by \ 8x $x \
as 8y and 8x get smaller and smaller.
144 CALCULUS FOR ENGINEERS.
Here we have an exceedingly important rule: — If
dz=M.dx + N.dy (18),
and if 2 is a function of x and v (another way of saying that z is a
function of x and y is to say that dz=M ,da;-\-N .dy is a complete
differentiaVjy then
"£f (-)•
(dM\ _ (o
Working the reverse way, we find that ^ — -p is the limiting value of
1^ { f{x+Sx, y + S y )-f{x-\-dx, y ) f{x, y + Sy)-f{x ,y)\
8x I Sy dy )
as Sy and 5x get smaller and smaller. Now it is obvious that these two are
the same for all values of 5x and %, and we assume that they remain the
same in the limit.
* M .dx + N.dy (1),
where M and N are functions of x and y, can always be multiplied by
some fonction of x and y which will make it a complete differential.
This multiplier is usually called an integrating factor. For, whatever
functions of x and y, M and N may be, we can write
dx-~N ^^'r
and this means that there is some law connecting x and y. Call it
^,.„.c.t.en(f).(^-)| = ,3,
and as -^ from (3) is the same as in (2) it follows that (— )"^(*7— ) = xr»
and hence (—\= /xM, f — j = fiN, where /* is a function of x and y or
else a constant.
Multiplying (1) by /* we evidently get
(^h-i^^y w.
and this is a complete differential. It is easy to show that not only is
there an integrating factor /x but that there are an infinite number of them.
As containing one illustration of the importance of this proposition I will
state the ^teps in the proof which we have of the 2nd law of Thermo-
dirnamictt.
1. We have shown that for any substance, of which the state is defined
by its t and v,
dH=k.dt+l.dv (5),
where k and I are functions of t and v.
Observe that t may be measured on any curiously varying scale of tempera-
ture whatsoever. We have just proved that there is some function /u. of t and
THERMODYNAMICS.
145
84. The First Law of Thermodynamics is this :
If dE= dH-p . dv, or dE=- k.dt + {l -p) dv, then dE
is a complete differential that is, ^returns to its old value when
V by which if we multiply (5) all across we obtain a complete differential ;
indeed there are an infinite number of such functions. Then calling the
result d0, d(f> = fi.dH=fik.dt + fxl.dv (6).
Let us see if it is possible to find such a value of n that it is a function
of t only. If so, as the differential coefficient of fik with regard to v {t being
supposed constant) is equal to the differential coefficient of fil with regard to
t {v being supposed constant),
(dk\ _ dti (dX\
\di)t~ Tt'^^Utjv
fdk\ _fdl\
or I — J =( ^1 +-. -^
» /* at
But the first law of Thermodynamics (see Art. 84) gives us
.(7).
and hence
/dk\ _ /dl\ _(dp\
\dv)t~\dt)v \dt)
l/dp\ _ 1 djx
T\dtJ "■ " |I ■ dt
.(8),
(9).
This then is the condition that /j.. dH is & complete differential, fi being a
function of temperature only. Obviously for any given substance (9) will
give us a value of /x which will answer ; but what we really want to know is
whether there is a value of /* which will be the same for all substances.
2. Here is the proof that there is such a value. I need not here give to
students the usual and well-
known proof that all rever-
sible heat engines working
between the temperatures t
and t-5t have the same
efficiency. Now let ABCD
be a figure showing with in-
finite magnification an ele-
mentary Caru ot cycle. Stuff
at A at the temperature
t-8t\ A I shows the volume
and AK the pressure. Let
^D be the isothermal for
t - dt and BC the isothermal
for t, AB and CD being
adiabatics.
Notice carefully that the
distance AG or WB {JV is in
DA produced to meet the
ordinate at B) is {dpjdt) dt.
Now the area of the
parallelogram ABCD which
represents the work done, is
BW X XZ (if parallelograms on the same base and between the same parallels
be drawn, this will become clear). Call XZ by the symbol 5v (the increase
of volume in going along the isothermal from B to C), and we see that
p
\
R
\
\
b
%
A^
L
.
w V
V^
1
Q
A
\
M e
N
^^^8
A^ .
•-^
t>
O
>
: \
( 2
Fig. 55
P.
10
146 CALCULUS FOR ENGINEERS.
the nett work done in the Carnot cycle is (dpldt)5t. 5v. Now the Heat
t and V return to their old values, (or another way of putting it is that
dE for a complete cycle is 0).
/
We have seen that the differential coefficient of k with regard to v,
t being constant, is equal to the differential coefficient of l—p with
regard to ty v being constant, or
(a-(a-(s «
This statement, which is true for any kind of stuff, is itself sometimes
called the first law of Thermodynamics.
The Second Law of Thermodynamics is this ; -— or
t
taken in at the higher temperature is, from (3), Art. 78, equal to l.dv and
hence — =j— - — = efficiency = -- ( - . j 8t... (10), and this is the same for all
substances.
As it is the same for all substances, let us try to find its value for any
one substance. A famous experiment of Joule (two vessels, one with gas at
high pressure, the other at low pressure witli stopcock between, immersed
in a bath all at same temperature ; after equalization of pressure in the
vessels, the temperature of the bath keeps its old value) showed that in
gases, the intrinsic energy is very nearly constant at constant temperature,
or what is the same thing, that / in gases is very nearly equal to p, and it is
also well known that in gases at constant volume, p is & linear function of
the temperature. Whether there really is an actual substance possible for
which this is absolutely true, is a question which must now be left to the
higher mathematicians, but we assume that there is such a substance and
in it
I \dty " p \dtj " + 274 ^ '*
if is the Centigrade reading on the Air Thermometer. If then we take
t = d + 2l4: as our scale of temperature and (11) as the universal value of
- ( '--; I , then, from (9), -=--.-r-,or — = — ^or log t + log^= a constant,
I \dt J ^ t fi at t /Jt. o or-
or Ai=r-, where c is any constant. This being an integrating factor for (5),
1
we usually take unity as the value of c or ^=- as Camot's function.
It is not probable that, even if there is one which is independent of j? or
V, there really is so simple a multiplier as ^— o;^ (where is the Centigrade
temperature on the air thermometer) or that there is such a substance as we
have postulated above. Calling our divisor t the absolute temperature, we
believe that for ordinary values of 0, t is ^ + 274, and the greater is, the
more correctly is t represented by ^ + 274; but when is very small, in all
probability the absolute temperature is a much more complicated function
of 6. The great discoverers of the laws of Thermodynamics never spoke of
- 274° C. as the absolute zero of temperature.
PERFECT GAS. 147
d<f> = -. dt-\-- .dv ...{21), is a complete diftereutial, and hence the
differential coefficient of - with regard to v, t being considered
t
constant, is equal to the differential coefficient of - with regard to t,
V being constant t.
1 /^'\ ^\dt)~^
Hence
or
(ii-Q.-i »
This statement, which is true for any kind of stuff, is itself some-
times called the second law of Thermodynamics.
Combining (20) and (22), we have for any stufl'
(t)A (->-
a most important law, T
Applying these to the case of a perfect gas we find
that (23) becomes - = -, or 1= — , or l=^p (24).
Hence (20) is (-^ ) =0. It is not of much importance perhaps,
practically, but a student ought to study this last statement as an
exercise, k is, for any substance, a function of v and t, and here we are
told that for a perfect gas, however k may behave as to temperature, it
does not change with change of volume. Combining (24) with (9)* &c.
(p. 141), already found, we have K-k—R, and as Regnault found that
K is constant for air and other gases, k is also constant, so tliat
y-i- y.
We can now make exact calculations on the Thermodynamics of a
perfect gas if we know K and R.
l=p^ L= —V, P= , ]'= ^^-Aj-» where y=x-
85. The statements of (3) Art. 78 become for a pound of perfect
gas
dH = k.dt + p.dv >|
= K . dt - V . dp
V— 1 -Y— 1
dv.
.(1).
+ The rule for finding the differential coefficient of a quotient is given in
Art. 197.
10—2
148 CALCULUS FOR ENGINEERS.
I often write this la«t in the shape d{pv)-{-p . dv (2),
also d£J=k .dty or ^=X'^ + constant (3).
It is easy to obtain from this other forms of E in terms of p and v.
To the end of this article, I consider the stuff to be a perfect gas.
Example 1. d(h = k . - +^ . dv. or as - = —
. t t ' t V
d<b = i- — + — . do.
^ t V
Hence, integrating,
<f)=k log t-\-R log V + constant, or (f> = log ^v" + constant (4).
Again d<^ = --.dt-. dpy but - = — .
t t P
Hence d<f)= -dt dp.
Integrating
<f) =>: K log t-Ii log p + constant, or </> = log t^p " " + constant ... (5).
Substituting for t its value -., we have (5) becoming
(/) = log jo^y*'+ constant (6).
The adiabatic law^ or constant, may be written down at
once. Keducing from the above forms we find
or t^-yp = constant,
■ or pv^ = constant.
Students ni^,y manufacture other interesting exercises of this kind
for themselves.
Example 2. A pound of gas in the state p^^ v^^^ t^ receives the
amount of heat ^^j^, what change of state occurs? We get our informa-
tion from (1).
I. Let the volume % keep constant. Then dH=k.dt
from (I).
The integral of this between ;„ and t^ is II^^^ = Jc{t^ — t^^ and we may
calculate the rise of temperature to t^ .
Or again, dH= - - dp.
GAS ENGINE.
149
Hence, the integral, or H^^ = -^ {p^ - p^), and we may calculate
the rise of pressure.
II. Let Pq the pressure, keep constant.
dH= K . dt, hence ^^i = ^^ (^i - ^o)-
Again dH= -^ dv, hence H^^ = -^"^^ (v^ - v^).
III. At constant temperature.
dH=p.dv or Hq]^= \ p.dv=W, the work done by the gas in ex-
J Vo
panding.
IV. Under any conditions of changing pressure and
volume.
Jf^^ = h (t^ — tf^ -\- work done.
Also from (2), Hqi = r(i^i^i— iOo^*o) + work done.
If ir=0, the work done=X*(^Q — ^i)
We often write the last equation of (1) in the convenient shape
^^=7^f^^-W
If in this we have no reception of heat,
dff ^ ,. dp
or _ = 0, then v^ + y.^ = 0,
or -" + y — = or, integrating, log p-\-y log v = constant,
or j02;'>'=: constant.
This is the adiabatic law again.
Example 3. In a well known gas or oil engine cycle of
p
.(7).
Fig. 56.
150 CALCULUS FOR ENGINEEIIS.
operations, a pouiul of gas at p.j,, v-,, t.,, iudicatcd by the point A is
compressed adiabatically to B, wnere we have p^, v,, ^j. The work
done upon the gas is evidently (from iv.) Jc{t^ — t^, being indeed the
gain of intrinsic energy.
Heat given at constant volume from B to C where we have
Work done in adiabatic expansion CJ)=k(t^-t^).
Nett work done = work in Ci>-work in AB=
-^ = efficiency eJ'-^l' -^^^=1- ^^ (8).
But we saw that along an adiabatic W~'^ is constant, and hence
From this it follows that -t = t — \-] > ^"d ^^^^"^ of these
t —t h h \^h)
= * — ^ . Using this value in (8) we have
efficiency = 1 - f^j (9),
a formula which is useful in showing the gain of efficiency produced by
diminishing the clearance v^.
Students will find other good exercises in other cycles of gas engines.
Change of State.
86. Instead of using equations (3) Art. V8, let us get out equations
specially suited to change of state. Let us consider one pound of
substance, m being vapour, 1 - m being liquid (or, if the change is from
solid to liquid, m liquid, l-m solid), and let
^2 = cubic feet of one pound of vapour,
*i = » )) of one pound of liquid,
jt>= pressure, t temperature, p is a function of t only.
Tf V is the volume of stuff in the mixed condition,
v=ms2 + {l —m)sj
= (*2 — «i) wi + Si, or v = mu-{-Xi (1),
if we write u for ,% — Si.
AVhen heat dJJ is given to the mixture, consider that t and m
alter. In fact, take t and m as independent variables, noting that t
and VI define the state. If 0-2 and o-^ be specific heats of vapour and
li(iuitl, when in the saturated condition (for example, 0-2 is the heat given
to one pound of vapour to raise it one degree, its pressure rising at
LATENT HEAT. 151
the same time according to the proper law), then the 7>i lb. of vapour
needs the heat ma^.dt^ and the 1— m of liquid needs the heat
{\ — m)<j^.dt and also if dm of liquid becomes vapour, the heat L . dm
is needed, if L is latent heat. Hence
dII={(a-2-o-j)m + ai]dt+L.dm (2).
If E is the Intrinsic Energy, the first law of Thermodynamics gives
dE=^dlI-p,dv (3).
Now if m and t define the state, v must be a function of m and t^ or
Using this in (3) and (2) we find
dE = |(cr2 - o-i) m + o-i -p {^j^ dt-\-\L-p {~y^ dm ... (4).
Stating that this is a complete differential, or
^{(..,-,.,)m + .,-^(|)}=||i;-^(^)}.
we have, noting from (1), that ( -i— ) = ?/,
dL , dp fdv\ dp ,_.
-^ + 0-1 — (To = -& . -1— I , or w -4- (5).
dt^ ^ '" dt \dm)' dt ^ '
Now divide (2) by t and state that d(^ = — is a complete dif-
ferential,
*
(6),
dm \ ~^t ]~dt \i)
dm
dL L
or -n-^f^\-^^2 = -7 ('^)-
Hence, with (5) we have T = U -s^ (8),
87. To arrive at the fundamental Equation (8) more
rapidly. In fig. 57 we have an elementary Carnot cycle for one pound
* w7 ( T ) ~ g as will be seen later on when we have the rule for
differentiating a quotient. But indeerl we may as well confess that to
understand this article on change of state, students must be able to perform
' differentiation on a product or a quotient,
152
CALCULUS FOR ENGINEERS.
of stuff. The co-ordinates of the point B are FB=s^ the volume, and
BG the pressure jt? of 1 pound of liquid. At constant tem^^ratrire t, and
Fig. 57.
also constant pressure, the stuff expands until it is all vapour at FC^s^ ;
CD is adiabatic expansion to the temperature t-bt at J). J) A is iso-
thermal compression at t-bt and AB is the final adiabatic operation.
The vertical height of the parallelogram is bt ~ , and its area, repre-
dv
senting the nett work, is bt . -£ (s.^- s^). The heat taken in, in the
: dp
operation BO is X, and the efficiency is bt -^ («2 ""*i) "^'^- ^^^^ ^^ i^ i'^
a Carnot cycle this is equal to — and so we obtain (8). t
88. The Entropy. From (6) we find o-^,-<ri = ^
we can write (2) as
dt\t)'
and
or
dH=tr^dt-^t.d(^^y\
Hence, the entropy d(l> = - — = ^^ dt + di-^\^
t t \ /
jut
.(9).
<(- = - +
c?^-f constant
.(10).
In the case of water, o-j is nearly constant, being Joule's equivalent.
(We have already stated that all our heat is in work units), and
= — +0-1 log f-j-f constant (11).
Hence the adiabatic law for water-steam is
TnT t
-T- + o"! log £- = constant (12).
It is an excellent exercise for students to take a numerical example. .
EXAMPLES. TWO VARIABLES. 153
Let steam at 165°C. (or ^ = 439) expand adiabatically to 85°C. (ori = 359).
Take a-i = 14(X) and L in work units, or take (t^ = \ and take L in heat
imits. In any case, use a table of values of t and L.
1. At the higher ^2=4^9 1^^ '^2='"- (This is chosen at random.)
Calculate ii\ at, say ^^ = 394, and also m^ at ^^=359.
Perhaps we had better take L in heat units as the formula
Z = 796 --695^
is easily remembered.
Then (12) becomes
r,r, (^ - -695) +log^^= m^ (^ - '695) +log^^ ,
logr+wi2(^-'695
^ /2/
695
h
If we want m^ we vise t^ instead of t^.
Having done this, find the coiTesponding values of v. Now try if
there is any law like
pi;* = constant,
which may be approximately true as the adiabatic of this stuff.
Rej^eat this, starting with m^=-% say, instead of '7.
The t, <t> diagram method is better for bringing these matters
most clearly before students, but one or two examples like the above
ought to be worked.
89. When a complete differential dn, is zero, to solve
the equation du = 0. We see that in the case,
{x" - ^xy - t\f) dx + {if - ^xy - 'Ice") dy = 0,
we have a complete differential, because
-J- (xf^ — 4fxy — 2y^) = — 4;x — ^y,
^ if - 4^y - 2^0 = - 4y - 4a-,
so that they are equal. Hence it is of the form
Integrating a?^ — ^xy — 2?/^, since it is ( y j , Avith i-egard to
154 CALCULUS FOR ENGINEERS.
X assuming y constant, and adding, instead of a constant, an
arbitrary function of y, we get ^ as
u = :^a^-2afy-2ifx + <i>{y).
To find </) {y\ we know that \~j~]—y^~ ^^V — ^^'
Hence -2a^- ^yx + ^ <f> {y) = y"" - ^.x-y - 2a^,
ay
Hence ^ </, (y) = y^ or </> {y) = \f.
Hence xi — \a? — 2x^y — ^xy"^ -\- ^y^ =zc.
We have therefore solved the given differential equation when
we put this expression equal to an arbitrary constant.
Solve in the same manner,
1 + ^ j (/^ - 2 ^ c?y = 0. Answer x^-y^= ex.
., , 2x.dx n Saf\ ,
holve f- ( -; —jdy=0. Answer x^ — y^ — cy^.
J ^J J '
-Solve (3^- + 3y - ^ dx + {^x - 1 + 3/) dy = 0.
Answer x^y"^ + ot^y^ + 4^^ ^ ^y"^ ^ ^^yz _ ^y%^
90. In the general proof of (17) given in Art. 81, we assumed that
X and y were perfectly independent. We may now if we
please make them depend either upon one another or
any third variable z. Thus if when any inde^Hindent quantity z
becomes z + bz, x becomes x + bx and y becomes y+by, of course u
becomes u + bu. Let (IG) Art. 81 be divided all across by bzy and let
bz be diminished without limit, then (17) becomes
du _ /o?w\ dx /du\ dy
di ~ \dx) dz "^ XTy) ^
.(1).
Thus let u = ax^-\-hy^+cxy^
and let x = ez^, y = yz'^.
and consequently
-T- = {2ax + cy) nez:^ ~^-\- {2by + ex) mgz^ ~ ^
EXAMPLES. 155
In this we may, if we please, substitute for x and y in terms of z,
and so get our answer all in terms of z.
This sort of example is rather interesting because it can be worked
out in oiu* earlier way. In the expression for w, substitute for x and y
in terms of z, and we find ii^aeh'^^+hgh^'^-'rcegz:^'^^^ and
-^ = 2?iae%2n - 1 + 2 w6^%2m - 1 -I- (,i + m) cegz"" + "*-!.
It will be found that this is exactly the same as what was obtained
by the newer method. The student can easily manufacture examples
of this kind for himself.
For instance, let y=uv where u and v are functions of ^, then (1)
tells us that
dy dv . du
a5 = "S+^d5E'
a formula wliich is usually worked out in a very different fashion.
See Art. 196.
In (1) if y is really a constant, the formula becomes
du _ du dx
dz " dx ' Hz '
which again is a formula which is usually worked out in a very different
fashion. See Art. 198.
In (1) assume that z=x and that y is a function of x, then
dx \dx) "^ \dy) dx ^'^^^
The student need not now be told that ;,- is a very different thing
from
(du\
\dx) •
Example. Let u = ax^ + hy'^ + cxy^
and let y=gx'^.
Hence (2) is, j- = {2ax + cy) + (26y + ex) mga/^-K
More directly, substituting for y in u, we have
u = ax^ + 6^2^2m ^ ^^^» + 1^
J- = 2ax + 2?»6^2^'2'« - 1 + (771 + 1 ) cgx^,
and this will be found to be the same as the other answer.
156 CALCULUS FOR ENGINEERS.
If M is a function of three independent variables it is easy to prove,
as in Art. 81, that
,MJ'-\a.+m,,U^-)^, (3).
91. Example. When a mass m is vibrating with one degree
of freedom under the control of a spring of stiffness a, so that if x is
the displacement of the mass from its position of equiHbrium, then ax
is the force with which the spring acts upon the mass ; we know that
the potential energy is ^ ax^ (see Art. 26), and if v is the velocity of the
mass at the same time t, the kinetic energy is ^tnv^, and we neglect the
mass of the spring, then the total store of energy is
When X is 0, r is at its greatest; when v is 0, x is at its greatest.
1. Suppose this store E to be constant and differentiate with
regard to ^, then
= mv^^ + ax--^^ (1)'
. dx ... d^x « dv ,
or as V IS ~j- , writnig -y-^ for -j- we have
^1-^^-0 •••■ ^^)-
which is (see Art. 119) tlie well known law of simple harmonic motion.
2. If the total store of energy is not constant but diminishes at
a rate which is proportional to the square of the velocity, as in the case
rf If
of Fluid or Electromagnetic friction, that is, if -r = - Ev^ then (1)
becomes - Fv^^mv y-4- ax—j , or (2) becomes
d2x F dx a ^^
Compare (1) of Art. 142.
92. Similarly in a circuit with self induction L and resistance R,
joining the coatings of a condenser of capacity K, if the current is C\ and
if the quantity of electricity in the condenser at time t is K V so that
dV
C= -K --J-J ^LC'^ is called the kinetic energy of the system, and \KV^
is the potential energy, and the loss of energy by the system per second
is RC'^. So that if E is the store of energy at any instant •
E=^LC'' + \KV\
EXAMPLES. 157
or LC"^- V. C+EO^=0,
at
or L^-V+RC=0,
at
d2V . R dV^ 1 Tr /%
aF + L^+EK'^=^ ^^^-
Differentiating this all across and replacing K -r- with C we have a
similar equation in C. Compare (4) of § 145.
93. A mass m moving with velocity v has kinetic energy \mv^.
If this is its total store E^
If E diminishes at a rate proportional to the square of its velocity
as in fluid friction at slow speeds,
dE jy ., dv
dv F ,^,
or -j-= V (5).
dt m ^ '
We have a similar equation for the dying out of current in an
electric conductor, \LC'*- being its kinetic energy, and RC^- being the
rate of loss of energy per second.
94. In (2) of Art. 90, assume that u is a constant and we find for
example that if u=f{x, y) = G
(dn^\_y)\ . ( df{x, y) \ di_
\ dx )^\ dy ) dx~ '
so that if /(^-j y)=o' or =0, we easily obtain g^.
1. Thus if x^^f = c, 2..+2y . ^^=0, or ^= -^'.
2. Alsoif^^+f-!-l=0,?|' + |^/ = 0,orf'=-^;i;,
a^ W- ' a^ h^ dx ' dx a^ ij
3. Again if u = A a;"' + %",
Hence i{ u = 0, or a constant, we have
dy _ mAx'^~^
dx~ nBy'^~'^
158 CALCULUS FOR ENGINEERS.
4. If u==Zi + 1:..
a,'" y-
2a; , 2y ,
5. If ^-3 +3,3 _ 3«.^^ = i^ find ^ . Answer : ^, = ''''' — ^ .
6. If .rlogy— ylogA'=0,
t/y^y /^log.
t/y_y ^^logy-y\
.v — xj
95. Example. Find the equations to the tangent and normal
to the ellipse ' ., + t^ = l, at the point A'l, yi on the curve.
•^ 4. 2^ ^^-0 or at the tx)int ^^- -?^' ^i
Hence the equation to the tangent is
y-^yi^_62.ri
a;—a;i ^ Vi
''' -^+-52-;^ + ^'
and as x^ and y^ are the co-ordinates of a point in the curve, this is 1.
Hence the tangent is — / + '^ = 1.
a^ b^
The slope of the normal is v^ — , and hence the equation to the
normal is ti^i = " ^J ^
x-Xi b^ x^
APPENDIX TO CHAP. I.
Page 19. In an engineering investigation if one arrives at mathe-
matical expressions which cannot really be thought about because
they are too complicated, one can often get a simple empirical formula
to replace them with small error within the limits between which they
have to be used. Sometimes even such a simple expression as a-|-6.r,
or x^ will replace a complicated portion of an expression with small
error. Expertness in such substitution is easily attained, especially in
calculations where some of the terms can be expressed numerically or
when one makes numerical experiments.
MAXIMA AND MINIMA.
159
Exercise 1. The following observed numbers are known to follow
a law like y = a-\-bx, but there are errors of observation. Find by the
use of squared paper the most probable values of a and h.
X
2 3
4i
6
7
9 12
13
y
5-6 6-85
9-27
11-65
12-75
16-32 20-25
22-33
Ans. y = 2*5 + 1 'hx.
Exercise 2. The following numbers are thought to follow a law like
y=zax\{\-^sx). Find by plotting the values of y\x and y on squared
paper that these follow a law ylx + sy=a and so find the most probable
values of a and s.
X
•5
1 2
0-3
1-4
2-5
y
•78
•97 r22
-55
1-1
1-24
Ans. y = 3^/(1 + 2.r).
Exercise 3. If p is the pressiu-e in poimds per square inch and if v
the volimie in cubic feet of 1 lb. of saturated steam,
p 6-86
14-70
28-83
60-40
101-9
163-3 250-3
V 53-92
2636
14-00
6992
4-28
2-748 1-853
Plotting the common logarithms of p and v on squared paper test
the truth oi pv^'^^^^Ti^.
Exercise 4. The following are results of experiments each lasting for
four hours ; / the indicated horse-power of an engine, transmitting
B horse-power to Dynamo Machines which gave out E horse-power
(Electrically), the weight of steam used per hour being TFlb., the weight
of coal us^ per hour being G lb. (the regulation of the engine was by
changing the pressure of the steam). Show that, approximately,
fr=800+21/, i5=-95/-18, ^=-935-10, C=4-2/-62.
1
B
^
^y
C
190
163
143
4800
730
142
115
96
3770
544
108
86
69
3080
387
65
43
29
2155
218
19
1220
—
Page 34. It has been suggested to me by many persons that I
ought to have given a proof without assuming the Binomial Theorem,
and then the Binomial becomes only an example of Taylor's. In spite
160 CALCULUS FOR ENGINEERS.
of the eminence and experience of my critics, I believe that my method
is the better — to tell a student that although I know he has not proved
the Binomial, yet it is well to assume that he knows the theorem to be
correct. The following seems to me the simplest proof which does not
assume the Binomial.
Let y—^^i x+bx=x\^ y + ^^=yi«
(1) Supjjose n a positive integer ; then
In the limit, when bx is made smaller and smaller, until ultimately
Xi=x\ the left-hand side is -^ and the right-hand side is x^~'^+x'^~^
-f... to n terms : so that j-=7iar»~i.
dx
(2) Suppose n a positive fraction, and put n= - where I and m are
positive integers. We have — — ^- = -^ = ~i,i — ;^ where a;'" =2,
Xi — X X% — X Zt — *
Xi=Zj^^j and so on.
^-linut of (h-^)ih'-'+^'h'~'+ +^'-')
dx~ {zi-z){z{^-i + z.Zi''^-'^ + -|-2'»-i)
(3) Suppose n any negative number = —m say, where vi is positive,
then noticing that Xi~^-x~'^= *'
we have
■x-'" 1 x\
Xy—X
/It wi ^^"^
Now the limit of ^ =mx'^-'^ by cases (1) and (2) whether 111
Xi — x •'
be integral or fractional.
.'. -r-=— -^ ' mx'^ - 1 = — 771^- '»- 1 = MA*"" 1.
Thus we have shown that -j- (^*) = 7w?»»-i, where ?i is any constant,
positive or negative, integral or fractional.
COMPOUND INTEREST LAW. 161
CHAPTER II.
e* and sinx.
97. The Compound Interest Law. The solutions
of an enormous number of engineering problems depend
only upon our being able to differentiate x'\ I have given
a few examples. Surely it is better to remember that
the differential coefficient of x^^ is nx^~^, than to write
hundreds of pages evading the necessity for this little bib
of knowledge.
We come now to a very different kind of function, e*,
where it is a constant quantity e (e is the base of the
Napierian system of logarithms and is 2*7 183) which is
raised to a variable power. We calculate logarithms and
exponential functions from series, and it is proved in Algebra
that
The continuous product 1 . 2 . 3 . 4 or 24 is denoted by .4
or sometimes by 4 !
Now if we differentiate e* term by term, we evidently
obtain
SO that the differential coefficient of ^ is itself e*. Simi-
larly we can prove that the differential coefficient of e^^ is
a^". This is the only function known to us whose rate of
increase is proportional to itself; but there are a great many
phenomena in nature which have this property. Lord
Kelvin's way of putting it is that "they follow the compound
interest law."
P. 11
162 CALCULUS FOR ENGINEERS.
dy
Notice that if -i = ay (1),
that is, the rate of increase of y is proportional to y itself,
then
y = be" (2),
where h is any constant whatsoever ; h evidently represents
the value of y when a; = 0.
Here again, it will be well for a student to illustrate his
proved rule by means of graphical and numerical illustrations.
Draw the curve y = d^, and show that its slope is equal to its
ordinate. Or take values of a?, say 2, 2-001, 2002, 2003, &c.,
and calculate the corresponding values of y using a table of
Logarithms. (This is not a bad exercise in itself, for practical
men are not always quick enough in their use of logarithms.)
Now divide the increments of y by the corresponding incre-
ments of X. An ingenious student will find other and
probably more complex ways of getting familiar with the
idea. However complex his method may be it will be
valuable to him, so long as it is his own discovery, but let him
beware of irritating other men by trying to teach them
through his complex discoveries.
98. It will perhaps lighten our study if we work out a
few examples of tbe Compound Interest Law.
Our readers are either Electrical or Mechanical En-
gineers. If Electrical they must also be Mechanical. The
Mechanical Engineers who know nothing about electricity
may skip the electrical problems, but they are advised to
study them ; at the same time it is well to remember that
one problem thoroughly studied is more instructive than
thirty carelessly studied.
Example 1. An electric condenser of constant capacity
K, fig. 58, discharging through great resistance R. If v is the
potential difference (at a particular instant) between the
condenser coatings, mark one coating as v and the other as
on your sketch, fig. 58. Draw an arrow-head representing
the current G in the conductor ; then C =v-t-R.
But q the quantity of electricity in the condenser is Kv
LEAKAGE RESTSTANCE. 163
and the rate oi diminution oiq per second or — ^ or —K-r-. is
the very same current. Hence ,^
-^
.dv
di
V
~R'
dv
di'-
= —
kr'-
EX
— _ _ ^
^'' ~di~~KR'"' Fig. 58.
That is, the rate of diminution of v per second, is propor-
tional to V, and whether it is a diminution or an increase we
call this the compound interest law. We guess therefore that
we are dealing with the exponential function, and after a little
experience we see that any such example as this is a case of
(1), and hence by (2)
v=^he~^' (8).
It is because of this that we have the rule for finding the
leakage resistance of a cable or condenser.
For (log 6 -log 2;) = ^^.
So that if Vi is the potential at time ti and if v^ is the
potential at time fg
KR{logb-\ogv,) = t^,
KR (log h — log V2) = ^2.
Subtracting, KR (log v^ — log Vo) — U — ti
So that R = {U - t,)/K log ^ .
It is hardly necessary to say that the Napierian logarithm
of a number n, \ogn, is equal to the common logarithm
logio?! multiplied by 2-3026.
Such an example as this, studied carefully step by step by
an engineer, is worth as much as the careless study of twenty
such problems.
Example 2. Newton's law of cooling. Imagine a
body all at the temperature v (above the temperature of sur-
11-2
164 CALCULUS FOR ENGINEEKS.
rounding bodies) to lose heat at a rate which is proportional
to v.
Thus let -T. = — «^j
at
where t is time. Then by (2)
V = 6e-«« (4),
or log h — log V = at.
Thus let the temperature be Vi at the time t^ and v^ at
the time t^, then log Vi — log I'g = a (^2 — ^i), so that a can be
measured experimentally as being equal to
Example 3. A rod (like a tapering winding rope or like
a pump rod of iron, but it may be like a tie rod made of
stone to carry the weight of a lamp in a church) tapers
gradually because of its own weight, so that it may have
everywhere in it exactly the same tensile stress / lbs. per
square inch. If 3/ is the cross section at the distance
X from its lower end, and if y + 5y is its cross section
at the distance oc-{- Sx from its lower end, then /. By is
evidently equal to the weight of the little portion between
OS and w + 8x. This portion is of volume BiV x y, and if w is
the weight per unit volume
f.Bi/ = tv.y. B.€ or rather ;r; = rpl/-
—
Hence as before, y=ibe^ (5).
If when a7 = 0, y = yo, the cross section just sufficient to
support a weight W hung on at the bottom (evidently
fy^zrzW), then yQ=b because e°=l.
It is however unnecessary to say more than that (5) is
the law according to which the rod tapers.
Example 4. Compound Interest. £100 lent at
3 per cent, per annum becomes £103 at the end of a year.
The interest during the second year being charged on the
increased capital, the increase is greater the second year, and
is greater and greater every year. Here the addition of
interest due is made every twelve months ; it might be
SLIPPING OF A BELT.
165
made every six or three months, or weekly or daily or every
second. Nature's processes are, however, usually more
continuous even than this.
Let us imagine compound interest to be added on to the
piincipal continually, and not by jerks every year, at the rate
of r per cent, per annum. Let P be the principal at the
end of t years. Then hP for the time ht is -— ^ P .ht or
dP r 1^^
-zr- — -—■ P, and hence by (2) we have
dt 100 ^ ^ ^
P^he^\
where 6 = Po the principal at the time t = 0.
Example 5. Slipping of a Belt on a Pulley. When
students make experiments on this slipping phenomenon,
they ought to cause the pulley to be fixed so that they may
see the slipping when it occurs.
The pull on a belt at W is Ti, and this overcomes not only
the pull To but also the
friction between the belt
and the pulley. Consider
the tension I' in the belt
at P, fig. 59, the angle
QOP being 6] also the
tension T+BT at >Sf, the
angle QO>Sf being (9 + 8(9.
Fig. 60 shows part of
OPS greatly magnified,
SO being very small. In
calculating the force pres-
sing the small portion of
belt P>Si against the pulley
rim, as we think of PS as a shorter and shorter length, we see
that the resultant pressing force is T. hd*, so that /ju.T.BOis
* When two equal forces T make a small angle dd with one another,
find their equilibrant or resultant. The three forces are parallel to the sides
of an isosceles triangle like fig. 01, where AB=CA represents 2', where
Fig. 59.
Fig. 61.
1G6 CALCULUS FOR ENGINEERS.
the friction, if /x, is the coefficient
of friction. It is this that BT
is required to overcome. When
fi.T .BO is exactly equalled by
Fig. 60. BT sliding is about to begin.
Then fi . T , B6 = BT or, ^ = fiT, the compound interest law.
Hence T=^he^^. Insert now T^T, when ^ = 0, and T = T^
when e^QOW or 6, , and we have T, = h, T, = ToCf'K
In calculating the horse-power H given by a belt to a
pulley, we must remember that H={1\ - T,) F-=- 33000, if
Ti and T^ are in pounds and V is the velocity of the belt in
feet per minute. Again, whether a belt will or will not tear
depends upon T^ ; from these considerations wo have the
well-known rule for belting.
Example 6. Atmospheric Pressure. At a place
which is h feet above datum level, let the atmospheric
pressure be p lbs. per sq. foot ; at h + Bh let the pressure
be p-\-Bp {Bp is negative, as will be seen). The pressure
at h is really greater than the pressure at h+Bh by the
weight of air filling the volume Bh cubic feet. If w is the
weight in lbs. of a cubic foot of air, —Bp = w. Bh. But lu = cp,
where c is some constant if the temperature is constant.
Hence — Bp= cp. Bh...(l), or, rather Jr = — cp. Hence, as
before, we have the compound interest law; the rate of fall of
pressure as we go up or the rate of increase of pressure as we
come down being proportional to the pressure itself. Hence
p = ae~'^^y where a is some constant. If ^; = po> when /i = 0,
then a = j^oj so that the law is
p=Poe-'^ (2).
an
As for c we easily find it to be -- , w^ being the weight
Po
of a cubic foot of air at the pressure po. If t is the constant
(absolute) temperature, and Wq is now the weight of a cubic
w 274
foot of air at 0° C. or 274° absolute, then c is — — .
Po t
BAG =56 and BC represents the equilibrant. Now it is evident that as
5^ is less and less, BG-^AB is more and more nearly 5^, so that the
equilibrant is more and more nearly T . 50.
ATMOSPHERIC PRESSURE. 167
If w follows the adiabatic law, so that pw-y is
constant or w=cp^^y where 7 = 1*414 for air. Then (1) be-
comes — Bp = cp^ly Bh or f~ = cBh or rather — I ^ = ch or
7
1 , 1
^1 P ^ =oh+G. If ^ =po where A = 0, we can find G, and
we have p y=po y— ch (3),
7
as the more usually correct law for pressure diminishing
upwards in the atmosphere.
Observe that when we have the adiabatic law pyy = 6, a
constant, smd pv=Rt; it follows that the absolute temperature
1— -
is proportional to p y ,
So that (3) becomes
1 7-1
li 7
So that the rate of diminution of temperature is
constant per foot upwards in such a mass of gas.
Compare Art. 74, (4), if v is 0.
Example 7. Fly-wheel stopped by a Fluid Frictional
Resistance.
Let a be its velocity in radians per second, / its moment of
inertia. Let the resistance to motion be a torque proportional
to the velocity, say Fa, then
Fa = — Ix angular acceleration (1),
/ J + i^a=0 (2),
da F
* = -/"•
Here rate of diminution of angular velocity a, is proportional
to a, so that we have the compound interest law or
oL — a^e ^ (3),
where a^ is the angular velocity at time 0.
168 CALCULUS FOR ENGINEERS.
Compare this with the case of a fly-wheel stopped by
solid fHction. Let a be the constant solid-frictional torque.
(1) becomes a = — Ida/dt,
or da/dt + a/I = 0,
or a = — at/T -f a constant,
or a = ao — a^// (4),
where Oo is the angular velocity when t = 0.
Returning to the case of fluid frictional resistance, if
M is a varying driving torque applied to a fly-wheel,
we have
^=^»+4; <«>•
Notice the analogy here with the following electric circuit
law.
Example 8. Electric Conductor left to itself.
Ohm's law is for constant currents and is V= RC, where
R is the resistance of a circuit, C is the current flowing in it ;
V the voltage. We usually have R in ohms, C in amperes,
V in Volts. When the current is not constant, the law
becomes
V = RC + I.g (1).
dC
where -7- is the rate of increase of amperes per second, and
L is called the self-induction of the circuit in Henries. It is
evident that L is the voltage retarding the current when the
current increases at the rate of one ampere per second.
1. IfF=Oin(l)
dC__R^
dt~ L '
which is the compound interest law.
-St
Consequently C = CoC ^ (2).
EXERCISES IN CURVES. 169
so that from (1) V==Ra-\-(Rh- Lgh) e-^K
Now let R = Lg or ff—Y and we have V= Ra, so that the
voltage may keep constant although the current alters.
Putting in the values we have found, and using Vq for the
constant voltage so that a= Fq -4- R, we find
G^'^+be'L' (3).
If we let C=0 when ^ = 0, then 6 = — ^, and hence we may
write G= -^{l-e~L^) (4).
The curve showing how G increases when a constant voltage
is applied to a cii'cuit, ought to be plotted from ^ = for some
particular case. Thus plot when Fo = 100, -R = l, L—01.
What is the current finally reached ?
99. Easy Exercises in the Differentiation and Integration
of e**^.
1. Using the formula of Art. 70, find the radius of
curvature of the curve y = eF, where a; = 0. Answer : r = \/8.
.(
2. A point ;ri, y^ is in the curve y = hef^, find the
equation to the tangent through this point.
Answer: y^Zyi — Vl^
X — Xi a
Find the equation to the normal through this point.
Answer : - — ^ = .
X - Xy^ ?/i
Find the length of the Subnormal. Answer : y -- or y^-ja.
Find the length of the Subtangent. Answer \y-\--^- or a.
170 CALCULUS FOR ENGINEERS
3. Find the radius of curvature of the catenary
y = - (e^'c + g-*/''), at any place. Answer : r — y^jc.
At the vertex when a? = 0, r = c.
4. If y = Ae''^^ where i stands for V — 1. Show that if
i behaves as an algebraic quantity so that i^ = — 1, i' = — i,
{* = !, i^ = I, &c. then ^ = - a^y.
5. Find a so that y = Ael^ may be true when
Show that there are two values of a and that
y = Ae-"^ + 5e-3«.
6. Find the subtangent and subnormal to the
/J
Catenary y = ^ {^"^ + e"*'"), or, as it is sometimes written,
y = c cosh xjc.
Answer :
X
the subtangent is c coth - or c (e^^" + e~^l*')l{d^'*' — e~^l*^),
c
C 2ul7 c
the subnormal is ^ sinh — or 2 (e^^^ — g-^^/c).
7. The distance PS, fig. 8, being called the length of the
tangent, the length of the tangent of the above catenary is
2z cosh^ - / sinh - .
2 cj c
The length of PQ may be called the length of the normal,
rn
and for the catenary it is c cosh^ - or y^/c.
c
8. Find the length of an arc of the catenary
c - —
y— (e" -{-e « ). The rule is given in Art. 38. Fig. 62
shows the shape of the curve, being the origin, the distance
A being c. The point P has for its co-ordinates x and y.
CURVES.
171
Now -f-= ^(e'' —e ^). Squaring this and adding to 1
and extracting the square root gives us ^(e^ + e ^). The
integral of this is ^ (e ^ — e ^ ) which is the length of the
arc AP, as it is when x = 0. We may write it, s = c sinh w/c.
\. R
pA
A
O
s
Fig. 62.
9. Find the area of the catenary between OA and
SP, fig. 62.
Area =1 -^ (e^ ■{■ e *" ) dw^
OS
OS
c2 ros 1 _«~|
Or the area up to any ordinate at x is & sinh x\c.
or ^ (e '^ -e " ).
The Catenary i/ = -{e^"^ + e~^"') revoh^es about the axis of x^ find
the area of the hour-glass-shaped surface generated. See Art. 48.
Area = - \ {e^ "^ -\- e~="''f . dx
%
ire-
=-^(^
IC ^ ff-'lx IC'
)+'ncx
between the ordinates at x = x and .^'=0.
It is curious that the forms of some volcanoes are as if
their own sections obeyed the compound interest law like an
inverted pump rod. The radii of the top and base of such a
172 CALCULUS FOR ENGINEERS.
volcano being a and h respectively and the vertical height h,
find the volume. See Art. 46. Taking the axis of the
volcano as the axis of x, the curve ij = be^^ revolving round this
axis will produce the outline of the mountain if c = -r log ^ .
The volume is ttJ 6- . e"'"" . da)=^ ^- e-'^^
= 2^(e--l).
IT
Now a — be^^, so that our answer is ^ {a^ — 6-).
100. Harmonic Functions.
Students ought to liavc already plotted sine curves like
y = asm{bx + e)...(l)on squared paper and to have figured out
for themselves the signification of a, b and e. It ought to be
unnecessary here to speak of them. Draw the curve again.
Why is it sometimes called a cosine curve ? [Suppose e to
be ^ or 90°.] Note that however great a; may be, the sine of
{bx-\-e) can never exceed 1 and never be less than — 1. The
student knows of course that sin = 0, sin j (or 45°) = '707,
sin ^ (or 90°) = 1, sin -J (or 135°) = '707, sin tt (or 180°) = 0,
sin ^ (or 225°) = - '707, sin '-^ (or 270°) = - 1, sin ^~
(or 315°) = --707, sin 27r (or 360') = again and, thereafter,
sin ^ = sin (^ — 27r). Even these numbers ought almost to
be enough to let the wavy nature of the curve be seen.
Now as a sine can never exceed 1, the greatest and least
values of y are a and — a. Hence a is called the amplitude
of the curve or of the function.
When A" = 0, y — a sin e. This gives us the signification
of e. Another way of putting this is to say that when ba;
was = — e or A" = — J , y was 0. When x indicates time or
when bx is the angle passed through by a crank or an
eccentric, e gets several names ; Valve-motion engineers call
SINE FUNCTIONS.
173
it the advance of the valve ; Electrical engineers call it the
lead or (if it is negative) the lag.
Observe that when bx — 27r we have everything exactly
the same as when x was 0, so that we are in the habit of
calling -J the periodic value of (i\
Besides the method given in Art. 9, I advise the student
to draw the curve by the following method. A little know-
ledge of elementary trigonometry will show that it must
give exactly the same result. It is just what is done in
drawing the elevation of a spiral line (as of a screw thread)
in the drawing office. Draw a straight line OM. Describe
Fig. 63.
a circle about with a as radius. Set off the angle BOG
equal to e. Divide the circumference of the circle into any
mimber of equal parts numbering the points of division
0, 1, % 3, &c. We may call the points 16, 17, 18, &c., or
32, 33, 34, &c., when we have gone once, twice or more times
round. Set off any equal distances from B towards M on
the straight line, and number the points 0, 1, 2, &c. Now
project vertically and horizontally and so get points on the
curve. The distance BM represents to some scale or other
the periodic value of x or 27r/6.
If OG is imagined to be a crank rotating uniformly
against the hands of a watch in the vertical plane of the
paper, y in (1) means the distance of G above OM, hx means
the angle that OG makes at any time with the position OM,
and if x means time, then h is the angular velocity of the
crank and 27r/6 means the time of one revolution of the crank
174 CALCULUS FOR ENGINEERS.
or the periodic time of the motion, y is the displacement at
any instant, from its mid position, of a slider worked vertically
from G by an infinitely long connecting rod.
A simple harmonic motion may be defined as one which
is represented by s = a sin (bt + e), where s is the distance
from a mid position, a is the amplitude, e the lead or lag or
advance, and 6 is 27r/T or ^irf where T is the periodic time or
/is the frequency. Or it may be defined as the motion of a
point rotating uniformly in a circle, projected upon a diameter
of the circle (much as we see the orbits of Jupiter's satellites,
edge on to us), or the motion of a slider worked from a
uniformly rotating crank-pin by means of an infinitely long
connecting rod. And it will be seen later, that it is the sort
of motion which a body gets when the force acting upon it is
proportional to the distance of the body from a position of
equilibrium, as in the up and down motion of a mass hanging
at the end of a spring, or the bob of a pendulum when its
swings are small. It is the simplest kind of vibrational
motion of bodies. Many pairs of quantities are connected by
such a sine law, as well as space and time, and we discuss
simple harmonic motion less, I think, for its own sake, than
because it is analogous to so many other phenomena. Now
let it be well remembered although not yet proved that if
dy
y = a sin (bx + c) then -?i = ab cos (bx + c)
and ly . dx = -- cos(bx + c).
101. When c = and t = 1 and a = 1 ; that is when
y — ^mx (1),
let us find the differential coefficient.
As before, let a; be increased to x-\-hx and find y + By,
y-^hy — sin {x + hx) (2).
Subtract (1) from (2) and we find
hy = sin {x + hx) — sin x,
or 2 cos (x + \Bx) sin \hx. (See Art. 3.)
Hence ^ = cos (a? + JS^) ^^^^ (3).
a sin (Kv -{• c). 175
It is easy to see by drawing a small angle a and recollect-
ing what sin a and a are, to find the value of sin a -r a as a
gets smaller and smaller. Thus in the figure, let POA be
the angle. The arc PA
divided by OP is a, the
angle in radians. And
the perpendicular PB
divided by OP is the pj g^
sine of the angle. Hence
= -pr-r and it is evident that this is more and more
a PA
nearly 1 as a gets smaller and smaller. In fact we may
take the ratios of a, sin a and tan a to one another to
be 1, more and more nearly^ as a gets smaller and
smaller. If we look upon ^Bx as a in the above expression,
we see that in the limit, (3) becomes
dy r • *
-r^ = cos X if y = sin x.*
* The proof of the more general case is of exactly the same kind.
Here it is :
i/ = actln(bx + c),
y + Sy = a sin {b{x + Sx) + c},
Sy = 2a COB {bx+c + ^b, Sx) sin (^ 6 . dx). See Art. 3.
Sx ^ ^ ^ ^b.Sx
Now make 5a; smaller and smaller and we have
— = ab cos (bx 4. c) and hence I a cos (bx + c) . dx = - sin (bx + c).
Again, to take another case : —
If ysa cos (bx + e), this is the same as
y = asin( 6x+e + - J=a 8in(6aT + c), say.
Hence -=^ = a6 cos (6a; + c)
dx ^
= a&C08 ( 6a: + e + - j
^ = — ab sin (bx + e).
Hence / a sin (bx + c) . dx =— - cos (bx + c)«
176
CALCULUS FOR ENGINEERS.
And hence
I cos A'
dx
sm a%
102. Now it is not enough to prove a thing like this, it
must be known. Therefore the student ought to take a book of
Mathematical Tables and illustrate it. It is unfortunate that
such books are arranged either for the use of very ignorant
or else for very learned persons and so it is not quite easy to
convert radians into degrees or vice versa. Do not forget
that in sin x or cos x we mean x to be in radians. Make out
such a lit
tie bit ot tab
le as this, w
hich is taken
L at ranc
iom.
Angle in
degrees
X
or angle in
radians
y = sina;
5y
dy~$x
Average
'6y_
dx
40
•6981
•6427876
•0132714
•7583
41
•7156
•6560590
•0130716
•7512
•7547
42
•7330
•6691306
If it is remembered that Bi/ -^ Bx in each case is really the
average value of -^^ for one degree or ^01745 of a radian, it
will be seen why it is not exactly equal to the cosine of x.
Has the student looked for himself, to see if "7547 is really
nearly equal to cos 41° ?
103. It is easy to show in exactly the same way that if
dy F
y = cos X, ji- = - sin x and Isin x . dx = - cos x. The —
sign is troublesome to remember. Here is an illustration :
Angle in
degrees
X
or angle in
radians
i/ = C08a;
negative
8x
Average
5y
5x
20
21
22
•3491
•3665
•3840
•9396926
•9335804
•9271839
•0061122
•0063965
-•3513
-•3666
- -3584
sin d.
vn
Notice that y diminishes as x increases. Notice that
sin 21° or sin (-3665) = '3584.
104. Here is another illustration of the fact that the
differential coefficient of sin x is cos x. Let AOP, fig. 65, be
6. Let AOQhe -h SO. Let PQ be a short arc drawn with
as centre. Let OP = 0Q= 1. PR is perpendicular to QB.
Fig, 65.
Then AP ^y = sin e, BQ = sm(0 -^ Be) = i/ + By, QP = B0
and EQ = 8y. Now the length of the arc PQ becomes
more and more neai-ly the length of a straight line between
P and Q as B6 is made smaller and smaller.
Thus Yyp 01^ ]Si is more and more nearly equal to cos PQR
or cos 0.
imit*' -J2J = cos d if y = sin 0.
In the limit
d9
Similarly if z = cos = OA, Bz = — BA = — RP and
dz RP . >,
d9=-QP = """^-
Illustrations like these are however of most value when a
student invents them for himself. Any way of making the
fundamental ideas familiar to oneself is valuable. But it is
a great mistake for the author of a book to give too many
illustrations. He is apt to give prominence to those illustra-
tions which he himself discovered and which were therefore
invaluable in his own education.
105. Observe that ii ^=A sin ajc 4- B cos ax;
<^% and ^=«V
Compare this with the fact that if y = e«*, ;7-^=aV) ;7^=^V ^^
12
p.
178 CALCULUS FOR ENGINEERS.
the higher applications of Mathematics to Engineering this resemblance
and difference between the two functions ^ and sin ax become im-
portant. Note that if i stands for s] -\ so that i^= -\^ {*=1, &c.
Then if y=e^ 1^~ ~^V> ;7:^=^"V j^^^ ^^ ^i*^ *^® s^^^® function.
Comi)are Art. 99.
106. Exercise. Men >yho have proved Demoivre's theorem
in Trigonometry (the proof is easy ; the proofs of all mathe-
matical rules which are of use to the engineer are easy ;
difficult proofs are only useful in academic exercise work) say
that for all algebraic purposes, co^ax=\{e^^-\-e~^^)^ and
sin CW7 = ^ (e^*** — e~^^). If this is so, prove our fundamental
propositions.
107. Example. A plane electric circuit of area A sq.
cm. closed on itself, can rotate with uniform angular velocity
about an axis which is at right angles to the field, in a uni-
form magnetic field H. H is supposed given in c.G.s. units;
measuring the angle ^ as the angle passed through from the
position when there is maximum induction HA through the
circuit ; in the position 6, the induction through the circuit is
evidently A.H. cos 6. If the angle 6 has been turned
through in the time t with the angular velocity q radians
per second, then 6 = qt. So that the induction /= AH cos qt.
The rate of increase of this per second is —AqH ^in qt, and
this is the electromotive force in each turn of wire. If there
are n turns, the total voltage is — nAqH^m qt in C.G.S. units;
if we want it in commercial units the voltage is
- nAqH 10"8 sin qt volts,
being a simple harmonic function of the time. Note that the
term voltage is now being employed for the line integral of
electromotive force even when the volt is not the unit used.
Example. The coil of an alternator passes through a
field such that the induction through the coil is
/= ^0 + A^ sin {6 + fi) + Ar sin {rO + Cr),
where 6 is the angle passed through by the coil. If q is the
ILLUSTRATIONS. 179
relative angular velocity of the coil and field, 6 = qt. If there
are n turns of wire on the coil, then the voltage is w -7- , or
nq {Ai cos (qt + ej) + ^,.r cos (rqt + e^)].
So we see that irregularities of ?• times the frequency in
the field are relatively multiplied or magnified in the
electromotive force.
108. In Bifilar Suspension, if W is the weight of
the suspended mass, a and b the distances between the
threads below and above, h the vertical height of the threads;
if the difference in vertical component of tension is n times
the total weight W, and 6 is the angle turned through in
azimuth, the momental resistance offered to further turning is
{{l-n')Wjsmd (1).
Note that to make the arrangement more ' sensitive ' it is
only necessary to let more of the weight be carried by one of
the threads than the other.
The momental resistance offered to turning by a body
which is 6 fi-om its position of equilibrium, is often propor-
tional to sin 6. Thus if W is the weight of a compound
pendulum and OG is the distance from the point of support
to its centre of gravity, W . OG . sin 6 is the moment with
which the body tends to return to its position of equilibrium.
If M is the magnetic moment of a magnet displaced
from equilibrium in a field of strength H, then HM sin 6 is
the moment with which it tends to return to its position of
equilibrium. A body constrained by the torsion of a wire or
a strip has a return moment proportional to 0. When
angular changes are small we often treat sin as if it were
equal to 6. Sometimes a body may have various kinds of
constraint at the same time. Thus the needle of a quadrant
electrometer has bifilar suspension, and there is also an elec-
trical constraint introduced by bad design and construction
which may perhaps be like a6 + bO^. If the threads are stiff,
their own torsional stiffness introduces a term proportional to
6 which we did not include in (1). Sometimes the constraint
is introduced by connecting a little magnetic needle rigidly
12—2
180 CALCULUS FOR ENGINEERS.
with the electrometer needle, and this introduces a term
proportional to sin 6, In some instruments where the
moving body is soft iron the constraint is nearly propor-
tional to sin 26, Now if the resisting moment is M
and a body is turned through the angle hO, the work
done is M. Bd, Hence the work done in turning a
body from the position 6^ to the position ^.^, where S.^ is
greater than ^i, is / M .dd.
. 6,
Example. The momental resistance offered by a body
to turning is asin^ where 6 is the angle turned through,
what work is done in turning the body from 6^ to 6.J
Answer, / a . sin 6 .d6 = — a (cos ^2 — cos d^=^a (cos 6^ — cos 6.).
J e,
Example. The resistance of a body to turning is partly
a constant torque a due to friction, partly a term b6-{-c&-,
partly a term esinO; what is the work done in turning from
^ = to any angle ?
M the torque = a-{-bd + cd" + e sin =f(0) say,
V the work done
= aO + ibO' + icd'-\-e(l- cos 6) = F(e) say.
This is called the potential energy of the body in the
position 6.
-^-j where
/ is the body's moment of inertia about its axis. When a
body is at 6 its total energy E is \I i~\ + F {$).
If the total energy remains constant and a body in the
position 6 is moving in the direction in which 6 increases,
and no force acts upon it except its constraint, it will con-
tinue to move to the position 6^ such that
So that when the form oi F {6) is known, 61 can be calculated,
if we know the kinetic energy at 6.
ILLUSTRATIONS. 181
Thus let M=e sin 0, so that F(0) = e(l- cos 0), then
1/ (jY + e (1 - cos (9) = e (1 - cos e,),
from which ^i the extreme swing can be calculated.
Exercise. Show that if the righting moment of a ship
is proportional to sin 4)6 where is the heeling angle, and
if a wind whose momental effect would maintain a steady
inclination of llf degrees suddenly sends the ship from rest
at ^ = and remains acting, and if we may neglect friction,
the ship will heel beyond 33J degrees and will go right over.
Discuss the effect of friction.
A body is in the extreme position 0^, what will be
its kinetic energy when passing through the position
of equilibrium ? Answer, F (Oi).
Thus let M^hO + ce^-^e sin 6.
Calculate a, the angular velocity at = 0, if its extreme
swing is 45°. Here
6, = ^ and F(e) = ^6^ -f ^c6' + e(l- cos 6),
from which we may calculate a.
Problem. Suppose we desire to have the potential
energy following the law
V=F(e)= a6^ + hO^ + ce'""^ + h sin 26,
and we wish to know the necessary law of constraint, we see
dV
at once that as Jllf = -y^ ,
M = f al9'' -f 3&^ + mc€'''^ + 2/i cos W.
Problem. A body in the position 6^, moving with the
angular velocity a, in the direction of increasing 6, has a
momental impulse m in the direction of increasing 6
suddenly given to it ; how far will it swing ?
The moment of momentum was /a, it is now loi + m and
if a' is its new angular velocity ot' = a + y .
182 CAI.CULUS FOR ENGINEERS.
The body is then in the position 0^ with the kinetic
energy |/ ( a + y ) and the potential energy F{Oq), and the
sum of these equated to F{6^ enables 6i to be calculated.
The student will easily see that the general equation
of angular motion of the constrained body is
fid) may include a term involving friction.
109. Every one of the following exercises must be worked
carefully by students ; the answers are of great practical use
but more particularly to Electrical Engineers. In working
them out it is necessary to recollect the trigonometrical
relations
cos 2(9= 2 cos2^ - 1 = 1-2 sin2^,
2 sin ^ . cos </) = sin (^ + </>) + sin {6 - ^),
2 cos 6 . cos <fi = cos (^ + </)) + cos {$ — (j)),
2 sin ^ . sin <^ = cos (0 — (f)) — cos (6 + <^).
Such exercises are not merely valuable in illustrating the
calculus; they give an acquaintance with trigonometrical
expressions which is of great genei-al importance to the
engineer.
The average value of /(.r) from ii' = ooi to .T = a?.^ is
evidently the area I f(x) . dx divided by .r., — a'i.
Every exercise from 6 to 20 and also 23 ought to be
illustrated graphically by students. Good hand sketches of
the curves whose ordinates are multiplied together and of the
resulting curves will give sufficiently accurate illustrations.
X sin 2ax
sin- (Lxdx
o r . , , cos (a-\-h)a; cos (a — h)x
3. sm ax , cos bx . dx = ~ ~ ^ rf-
J 2(a + 6) 2(a-6)
2 4a
X . sin 2ax
4>a
EXERCISES. 183
f . . , , sm(a~b)x sin (a + h) x
k sm ax . sin bx . d^• = —^ r-r ^, , . (—
; 2 (a — 6) 2 (« + b)
5. I cos a^ . cos bx.cix — — ^, ~ h
sin {a + b) x sin (a — b) x
2(a + 6) "^ ~27a^;^6)~
-I'
6. The area of a sine curve for a whole period is 0.
r2.1T r2jr "I
I sinfl7.fZic=— cosiT = — (1 — 1) = 0.
7. Find the area of the positive part of a sine curve,
that is
I sin ^ . c^a; = — cos x = — (— 1 — 1) = 2.
Since the length of base of this part of the curve is tt,
2
the average height of it is - . Its greatest height, or ampli-
tude, is 1.
8. The area of y — a-\-b sin x from to 27r is 27ra and
the average height of the curve is a.
9. Find the average value of sin- x from a; = to ^ = 27r.
As cos 2^' = 1 - 2 sin^ x, sin^ ^ = | (1 - cos 2x). The integral
of this is |.2J - J sin 2x, and putting in the limits, the area is
(^27r-isin47r-04-isinO) = 7r. The average height is
the area -j- 27r, and hence it is ^.
10. The average value of cos- x from ^; = to iz? = 27r is \.
In the following exercises s and r are supposed to be
whole numbers and unequal :
11. The average value of a niri^ {sqt ■\- e) from ^=0 to
^ = T is g , s being a whole number and q = ^irlT. T is the
periodic time.
12. The average value of aQ,o^^isqt-\-e) from i = to
13. I cos sqt . sin sqt ,dt=^0.
184 CALCULUS FOR ENGINEERS.
[^
14. I sin sqt . sin rqt ,dt — 0.
Jo
rT
15. I cos sqt . cos rqt .dt — 0.
Jo
16. I sin sqt . cos rqt .dt = 0.
Jo
17. The average value of sin^ sqt from to ^T is ^.
18. The average value of cos^ sqt from to ^T is J.
19. I sm 55^ . sin rqt . a^ = 0.
rhT
20. I cos 55^ . cos rqt ,dt = 0.
21. Find /sin a; . sin (x + e) . dx.
Here, sin {x-\- e) = sin a? cos e -\- cos a; . sin e.
Hence we must integrate sin'* x . cos e-^-smx . cos a; . sin e,
a? sin 2a;
/■
sin'' X .dx =
(sin X . cos X .dx = \ (sin 2a? . (^a; = — J cos 2a;,
and hence our integral is
fx sin 2a;\ , ^
( ^ J— J cos e — % cos 2a7 . sin e.
22. Prove that (sin qt . sin (^-^ + e) . f/^
/^ sin2gA 1 ^ , .
23. Prove that the average value of sin qt . sin {qt ± e) or
of sin (qt + a) sin (5'^ + a ± e) for the whole periodic time T
(if q = y) ^s i cos e.
USE OF IM AGIN ARIES. 185
This becomes evident when we notice (calling qt-\- a = (p),
sin (ft . sin ((f) ±e) = sin (ft (sin (j) cos e ± cos (f) . sin e)
= sin^ <l> . cos e ± sin (f> . cos </> . sin e.
Now the average value of sin- (/> for a whole period is |^,
and the average value of sin <^ . cos <^ is 0.
IT
By making a = ^ in the above we see that the average
value of cos qt . cos {qt ± e) is ^ cos e, or the averages value
for a whole period of the product of two sine flinctions
of the time, of the same period, each of amplitude 1, is
half the cosine of the angular lag of either behind the
other.
24. Referring to Art. 106, take
cosa(9 = J(e'^ + e"'^),
or take e^^ = cos ad -f i sin aO,
e"^ = cos ad — i sin aO,
and find L^ cos ad . dO.
This becomes \j{e^''-''^'-Ve^'-''^')dd
(b+ai)0 . 1 ib~ai)e)
e + V . e y
b — ai
* (6 + ai
1 W f 1 aid 1
i^ "^r-; — ^-^ +1 —
6 ~ ai
and on substituting the above values it becomes
I e cos a^ .
Similarly we have
e cos a0.dd= - — ^ e^ (b cos a^ + a sin a^) . . .(1).
fe^« sin adds = -^^-^ e^« (b sin a(9 - a cos a(9) (2).
186 CALCULTTS FOn ENGINEERS.
110. Notes on Harmonic Functions. In the fol-
lowing collection of notes the student will find a certain
amount of repetition of statements already made.
111. A function ^ = tt sin qt is analogous to the straight
line motion of a slider driven from a crank of length a
(rotating with the angular velocity q radians per second)
by an infinitely long connecting rod. x is the distance of the
slider from the middle of its path at the time t At the zero
of time, ^ = and the crank is at right angles to its position
of dead point, q = 27rf = -=- , if T is the periodic time, or
if/ is the frequency or number of revolutions of the crank
per second, taking 1 second as the unit of time.
112. A function x = a sin {qt -}- e) is j ust the same, except
that the crank is the angle € radians (one radian is 57 '2957
degrees) in advance of the former position; that is, at
time the slider is the distance a sin e past its mid-position*.
* The student is here again referred to § 10, and it is assumed that he
lias drawn a curve to represent
x=6c""'sin {qt + i) (1).
Imagine a crank to rotate uniformly with the angular velocity q, and to
drive a slider, but imagine the crank to get shorter as time goes on, its
length at any time being fl€~ .
Another way of thinking of this motion is: —
Imagine a point P to move with constant aiujnlar velocity round O,
keeping in the equiangular spiral path APBCDEF; the motion in question
is the motion of P projected upon the straight line MON and what we have
called the logarithmic decrement is tt cot a if a is the angle of the spiral.
SIMPLE HARMONIC MOTION.
187
113. A fiuictiou X = a sin (ryi + e) + o! sin {(jt + e') is the
same as X = A sin (5'^ + ^); that is, the sum of two crank
motions can be given by a single crank of proper
length and proper advance. Show on a drawing the
positions of the first two when ^ = 0, that is, set off
YOF = e and OF = a,
YOQ = e' and OQ = a/.
Complete the parallelogram OPRQ and draw the diagonal
OR, then the single crank OR = A, with angle of advance
YOR = E, would give to a slider the sum of the motions
which OP and OQ would separately give. The geometric
proof of this is very easy.
Imagine the slider to have
a vertical motion. Draw
OQ, OR and OP in their
relative positions at any
time, then project P, R
and Q upon OX. The
crank OP would cause
the slider to be OP' above
its mid-position at this
instant, the crank OQ
would cause the slider to
be OQ' above its mid-po-
^^g- ^'^' sition,the crank OR would
cause the slider to be OR' above its mid-position at the same
instant; observe that OR' is always equal to the algebraic
sum of OP' and OQ.
We may put it thus: — "The s.H.M. which the crank OP
would give, + the s.H.M. which OQ would give, is equal to
the S.H.M. which OR would give." Similarly "the s.H.M.
which OR would give, - the s.H.M. which OP would give, is
equal to the s.H.M. which OQ would give." We sometimes
say: — the crank OR is the sum of the two cranks OP and OQ.
Cranks are added therefore and subtracted just like vectors.
that is, the constant acute angle which OP everywhere makes with the curve,
or Tfcot a = aT/2 and g = 27r/r, so that cot a = alq. If fig. 66 is to agree with
fig. 67 in all respects NM being vertical and P is the position at time 0, then
€= angle iVOP -7r/2.
188 CALCULUS FOR ENGINEERS.
114. These propositions are of great importance in dealing
with valve motions and other mechanisms. They are of so
much importance to electrical engineers, that many practical
men say, "let the crank OP represent the current." They
mean, " there is a current which alters with time according
to the law (7= a sin (5^ 4- e), its magnitude is analogous to
the displacement of a slider worked vertically by the crank
OP whose length is a and whose angular velocity is q and
OP is its position when t = 0." t
115. Inasmuch as the function w = a cos qt is just the
same as asinf^^ + ^j, it represents the motion due to a
crank of length a whose angle of advance is 90°. At any
time t the velocity of a slider whose motion is
{V = a sin (qt + e),
due
is v^aq cos (qt 4- f) = -^ or ^r
= aq sin
(,* + e + |).
that is, it can be represented by the actual position at any
instant of a slider worked by a crank of length representing
aq, this new crank being 90° in advance of the old one.
cl?x dv
The acceleration or -^^ or -7- or v is shown at any instant
by a crank of length aq^ placed 90° in advance of the v
crank, or 180° in advance of the x crank, for
Accel. = - aq^ sin {qt + e)
= aq^ sin {qt-\-e + ir).
The characteristic property of S.H. motion is that, numerically,
the acceleration is q^ or 47r-/- times the displacement, /being
the frequency.
If anything follows the law a sin {qt + e), it is analogous
to the motion of a slider, and we often say that it is repre-
sented by the crank OP ; its rate of increase with time
is analogous to the velocity of the slider, and we say that it
is represented by a crank of length aq placed 90° in advance
of the first. In fact, on a S.H. function, the operator djdt
multiplies by q and gives an advance of a right angle.
ELECTRIC CIRCUIT.
189
116. Sometimes instead of stating that a function is
A sin (qt + e) we state that it is a sin qt + b cos qt.
Evidently this is the same statement, if a? + h^ = A^ and if
h
tan e = -.
a
It is easy to prove this
trigonometrically, and gra-
phically in fig. 68. Let
The crank OP is the sum
of 08 and OQ, and tan e or
tan FOP = -.
a
Fig. 68.
117. We have already in Art. 100 indicated an easy
graphical method of drawing the curve
x = asm (qt + e),
where a; and t are the ordinate and abscissa.
Much information is to be gained by drawing the two of
the same periodic time,
a; = a sin (qt + e) and x — a sin (qt + e),
and adding their ordinates together. This will illustrate 113.
118. If the voltage in an Electric Circuit is V volts, the
current G amperes, the resistance R ohms, the self-induction
L Henries, then if t is time in seconds,
V==RC+L
dC
dt'
.(1).
Now if = Co sin qt,
T- = G^q cos qt, •
so that Y= RGq sin qt + LG^q . cos qt,
and by Art. 116 this is
. V=Go^/W+LY^^^(qi + ^) (2);
190 CALCULUS FOR ENGINEERS.
nR^ 4- L^q^ is called the impedance ;
tan e = -^* = — — ^ , if/ is frequency ;
6 is lag of current behind voltage.
Hence again if V = VoSin qt (3),
then C = ;7=^Bin(qt-tan-^)l (4).
Notice that if V is given as in (3) the complete answer
for G includes an evanescent term due to the starting
conditions see Arts. 98, 147, but (4) assumes that the simple
harmonic V has been established for a long time. In practical
electrical working, a small fraction of a second is long enough
to destroy the evanescent term.
119. We may wiite the characteristic property of a simple
harmonic motion as
S+<»'*=o W-
(compare Arts. 26 and 108) and if (1) is given us we know
that it means
x = a8mqt+hQ0>iqt or x =^ A ^m {qt + e) (2),
where A and e, or a and h are any arbitrary constants.
Example. A body whose weight is W lb. has a simple
harmonic motion of amplitude a feet (that is, the stroke is
2a feet) and has a frequency / per second, what forces give
to the mass this motion ?
If X feet is the displacement of the body from mid-
position at any instant, we may take the motion to be
x= a sin qt or a sin 2irf. t,
and the numerical value of the acceleration at any instant is
^ir^px and the force drawing the body to its mid -position is
in pounds ^ir^f'^xW-^ 322, as mass in engineer's units is weight
in pounds in London -r- 32 "2, and force is acceleration x mass.
CONNECTING KOD. 191
120. If the connecting rod of a steam or gas engine were
long enough, and we take W to be the weight of piston and
rod, the above is nearly the force which must be exerted
by the cross-head when the atmosphere is admitted to both
sides of the piston. Observe that it is when a? is and is
proportional to o), being greatest at the ends of the stroke.
Make a diagram showing how much this force is at every
point of the stroke, and carefully note that it is always act-
ing towards the middle point.
Now if the student has the indicator diagrams of an
engine (both sides of piston), he can first draw a diagram show-
ing at every point of the stroke the force of the steam on
the piston, and he can combine this with the above diagram
to show the actual force on the cross-head. Note that steam
pressure is so much per square inch, whereas the other is the
total force. If the student carries out this work by himself
it is ten times better than having it explained.
Since the acceleration is proportional to the square of the
frequency, vibrations of engines are much more serious than
they used to be, when speeds were slower.
121. As we have been considering the motion of the piston
of a steam engine on the assumption that the connecting rod is
infinitely long, we shall now study the effect of shortness of
connecting rod.
In Art. 11, we found 6" the distance of the piston from the
end of its stroke when the crank made an angle 6 with its dead
point. Now let x be the distance of the piston to the right
of the middle of its stroke in fig. 3, so that our x is the old s
minus r, where r is the length of the crank.
Let the crank go round uniformly at q radians per second.
Again, let t be the time since the crank was at right
angles to its dead point position, so that 0— - = qt, and we
find ^
or
x z= - r cos -^ I \l - \/ 1 -^ sin2 oi ,
x= r iiin qt + I \i — \/ 1 — j~ cos^ qt [•
192 CALCULUS FOR ENGINEERS.
Using the approximation that Vl — a = 1 — ^a if a is
small enough we have
X =^ r sm qt -\- ^. cos- qt
But we know that 2 cos- qt-'l= cos 2qL (See Art. 109.)
Hence x = r sin qt + ^i ^^os (2qt) - — (1).
We see that there is a fundamental simple harmonic motion,
and its octave of much smaller amplitude.
Find -jr and also -f- . This latter is
at dt^
acceleration = — rq- sin qt y cos '2.qt.
It will be seen that the relative importance of the octave
term is four times as great in the acceleration as it was in
the actual motion. We may, if we please, write 6 again for
2^ + 2 and get
d\
r-if'
— - = r^' cos 6 + ~ cos
2(9.
.2^2
r^q
When ^ = 0, the acceleration is rif- H — ? .
When B — 90^ the acceleration is — -
V
When = 180°, the acceleration is — rq^ + —j- ,
(q is 27rf, where / is the frequency or number of revolutions
of the crank per second).
If three points be plotted showing displacement x and
acceleration at these places, it is not difficult by drawing a
curve through the three points to get a sufficiently accurate
idea of the whole diagram. Perhaps, as to a point near the
middle, it might be better to notice that when the angle
OPQ is 90°, as P is moving uniformly and the rate of change
of the angle Q is zero, there is no acceleration of Q just then.
This position of Q is easily found by construction.
VALVE GEARS. 193
The most important things to recollect are (1) that
accelerations, and therefore the forces necessary to cause
motion, are four times as great if the frequency is doubled,
and nine times as great if the frequency is trebled ; (2) that
the relative importance of an overtone in the motion is
greatly exaggerated in the acceleration.
122. Take any particular form of link motion or radial
valve gear and show that the motion of the valve is always
very nearly {t being time from beginning of piston stroke or
qt being angle p?^sed through by crank from a dead point),
x = a^ sin {qt + 61)4-02 sin {^qt + 63) (1).
(There is a very simple method of obtaining the terms Oj
and a^ by inspection of the gear.) When the overtone is
neglected, a^ is the half travel of the valve and ei is the angle
of advance. In a great number of radial valve gears we find
that 6-2 = 90°. The best way of studying the effect produced
by the octave or overtone is to draw the curve for each term
of (1) on paper by the method of Fig. 63, and then to add
the ordinates together. If we subtract the outside lap L
from cc it is easy to see where the point of cut-off is, and how
much earlier and quicker the cut-off is on account of this
octave or kick in the motion of the valve.
In an example take «i = 1, e^ = 40°, a^ = '2, e^ — 90"^.
The practical engineer will notice that although the
octave is good for one end of the cylinder it is not good for the
other, so that it is not advisable to have it too great. We
may utilize this fact in obtaining more admission in the up
stroke of modern vertical engines ; we may cause it to correct
the inequality due to shortness of connecting rod.
Links and rods never give an important overtone of
frequency 3 to 1. It is always 2 to 1 .
In Sir F. Bramwell's gear the motion of the valve is,
by the agency of spur wheels, caused to be
a? = «! sin {qt -f Cj) + Oo sin {^qt -t- e,) (2).
Draw a curve showing this motion when
a, = 1-15 inch, e^ = 47", a._ = '435, e, = 62°.
P. 13
194 CALCULUS FOR ENGINEERS. >A^
If the outside lap is 1 inch and there is no inside lap, find
the positions of the main crank when cut-off, release, cushion-
ing and admission occur. Show that this gear and any gear
giving an overtone with an odd number of times the funda-
mental frequency, acts in the same way on both ends of the
cylinder.
123. If a? = Oi cos (^1^ + €i) -H 02 cos {q4 ■\- e^ (3),
where q^^^irfi and q^—'^irf^,
there being two frequencies; this is not equivalent to one
S. H. motion. Suppose a^ to be the greater. The graphical
method of study is best. We have two cranks of lengths a^
and tta rotating with different angular velocities, so that the
effect is as if we had a crank A rotating with the average
angular velocity of a^, but alternating between the lengths
tti + tta and ai — tta ; always nearer a^s> position than Og's ; in
fact, oscillating on the two sides of Oi's position. If q^ is
nearly the same as q^ we have the interesting effect like
beats in music*.
Thus tones of pitches 100 and 101 produce 1 beat per
second. The analogous beats are very visible on an in-
candescent lamp when two alternating dynamo-electrical
machines are about to be coupled up together. Again, tides
of the sea^ except in long channels and bays, follow nearly
the s. H. law ; a^ is produced by the moon and a^ by the sun
if ai=2'la2, so that the height of a spring tide is to the
height of a neap tide as 3*1 to 1*1. The times oi full are
times of lunar full. The actual tide phase never differs more
than 0*95 lunar hour from lunar tide ; 095 lunar hour = 098
solar.
124. A Periodic Function of the time is one which
becomes the same in every particular (its actual value, its
rate of increase, &c.) after a time T. This T is called the
* Analytically. Take cos (^vf^t + eg) =co8 {^irf-J, - 2t (/^ -/2)f + e^},
therefore a; = r cos (2t/i« + 6) ,
where r^ = a-^ + a^ + 2a^a^ cos {27r (/^ -/g) t + Cj - Cg} ,
and the value of tan 6 is easily vrritten out.
ROTATING MAONETTO FIELD. 195
periodic time and its reciprocal is called f the frequency.
Algebraically the definition of a periodic function is
where n is any positive or negative integer.
125. Fourier's Theorem can be proved to be true. It
states that any periodic function whose coraplete period is T
(and g is ^tt\T or 27r/) is really equivalent to the sum of a
constant term and certain sine functions of the time
fif) = ^0 + ^1 sin {qt + E^ + A^ sin {%qt + E^ +
^3 sin (3^'^ + £'3) 4- &:c (1).
In the same way, the note of any organ pipe or fiddle
string or other musical instrument consists of a fundamental
tone and its overtones. (1) is really the same as
/(^) = ^0 + «i sin 5^ + 61 cos qt + ag sin ^qt -f 6., cos Iqt + &c.,
if ar + &r = A^ and tan JS^^ = — , &c. t
126. A varying magnetic field in the direction x follows
the law X = a sin qt where t is time. Another in the
direction y, which is at right angles to x, follows the law
Y—a cos qt.
At any instant the resultant field is
R = VZmT^ = a = a constant
making with y the angle 6, where tan = F/X, or ^ = qt
Hence the effect produced is that we have a con-
stant field R rotating with angular velocity q.
When the fields are
X = Oi sin (qt + Cj) and F = aa sin (qt + e^),
it is better to follow a graphical method of study. The
resultant field is represented in amount and direction by the
radius vector of an ellipse, describing equal areas in equal
times.
Let OX and OY, fig. 69, be the two directions mentioned.
Let OAi in the direction OX = a^ . With 0-4 1 as radius describe
13—2
196
CALCULUS FOR ENGINEERS.
a circle. Let YOO be the angle ei. Divide the circle into
many equal parts starting at and naming the points of
division 0, 1, 2, 3, &c. Draw lines from these points parallel
to OY, Let 0A» in the direction OF be ag. Describe a
circle with OA2 as radius. Set off the angle X'OO'as e^ and
divide this circle at 0', 1, 2, 3, 4, &c., into the same number
of parts as before. Let lines be drawn from these points
parallel to OX, and where each meets the corresponding line
from the other circle we have a point whose radius vector
at any instant represents, in direction and magnitude, the
resultant magnetic field.
If OX and OY are not at right angles to one another,
the above instructions have still to be followed.
If we divide the circle OA2 into only half the number of
parts of OJ-i we have the combination of X = Oj sin (qt + fi)
and F= ttg sin (2qt 4- 62).
If we wish to see the combination X = any periodic
function and Y any other periodic function, let the curve
SINE CURVE.
197
from Mo to N^ show F, M^N^ being the whole periodic time ; and
let the curve from M^ to N^ show X, the vertical distance M^Ni
Fig. 70.
being the whole periodic time. If Pi and Pj are points on the
two curves at identical times, let the horizontal line from P^
meet the vertical line from Pj in P. Then at that instant
OP represents the resultant field in direction and
magnitude.
Carry out this construction carefully. It has a bearing
on all sorts of problems besides problems on rotating mag-
netic fields.
127. The area of a sine curve for a whole period or for
any number of whole periods is zero. This will be evident if
one draws the curve. By actual calculation; let s be an
integer and g= -— -,
•2. 1 r^ '
sin sqt.dt = cos sat
because cos5 -^ Tor cos s2it = 1 and cos = 1.
/;
.-/cos.^P
cos j
0,
.J,
Again, I
^0
cos sqt .dt = —
27r,,,
■^. 1
sm sqt
.0 J
1 / . 27r ^
sin
)=..
because sin s~^ T= sin s27r = and sin 0=0.
198 CALCULUS FOR ENGINEERS.
128. If the ordinates of two sine curves be multiplied
together to obtain the ordinate of a new curve : the area of
it is for any period which is a multiple of each of their
periods. Thus if s and r are any integers
T
sin sqt . cos rqt . dt = (1),
o
sin sqt . sin rqt. dt = (2),
/,
I cos sqt . cos rqt . dt = . .(3).
These ought to be tried carefully. '1st as Exercises in
Integration. 2nd Graphically. The student cannot spend
too much time on looking at these propositions from many
points of view. He ought to see very clearly why the
answers are 0. The functions in (1) and (2) and (3) really
split up into single sine functions and the integral of each
such function is 0. Thus
2 sin sqt . cos rqt = sin (s + r) ot 4- sin {s — r) qt,
and by Art. 127, each of these has an area 0.
The physical importance of the proposition is enormous.
Now if 5 = r the statements (2) and (3) are untrue, but (1)
continues true. For
•(4),
/•T rT
I sin^ sqt . dt = / cos^ sqt . dt = ^T
.'o Jo
whereas (1) becomes the integral of ^ sin 2sqt which is 0. (4)
ought to be worked at graphically as well as by mere inte-
gration. Recollecting the trigonometrical fact that
cos2^ = 2cos2^-l or 1 - 2 sin^ ^,
and therefore that
cos^ qt=^ cos 2qt + i, sin^ g'^ = J — J cos 2qt,
the integration is easy and the student ought to use this
method as well as the graphical method.
129. To illustrate the work graphically. Let 0C\ fig. 71,
be T. Taking s = 2, the curve OPQRSC represents sin sqt
ELECTRICAL ILLUSTRATION.
199
Its maximum and minimum heights are 1. Now note that
sin^^^^ is always + and it is shown in OP'Q^R^Sfi. It
fluctuates between and 1 and its average height is \ or
Fig. 71.
the area of the whole curve from to (7 is ^T. The fact
that the average value of sinsqt. x sinrqt is 0, but
that the average value of sin sqt x sin sqt is \, is one of
the most important in practical engineering "work.
130. Illustration in Electricity. An electric dynamo-
meter has two coils ; one fixed, through which, let us suppose,
a current G flows ; the other moveable, with a current c. At
any instant the resultant force or couple is proportional to
Go and enables us to measure Gc. But if G and c vary
rapidly we get the average value of Gc. Prof. Ayrton and
the author have carried out the following beautifully illus-
trative experiment. They sent a current through the fixed
coil which was approximately, G=Gq sin 27rft. This was
supplied by an alternating dynamo machine. Through the
other coil they sent a current, c = Co sin 2'nf't whose frequency
could be increased or diminished. It was very interesting to
note (to the average practical engineer it was uncanny,
unbelievable almost) that although great currents were
passing through the two coils, there was no average force —
in fact there was no reading as one calls it in the laboratory.
Suppose / was 100 per second, /' was gradually increased
from say 10 to 20, to 30 to 40 to 49. Possibly about 49 to
200 CALCULUS FOR ENGINEERS.
51 a vague and uncertain sort of action of one coil un the
other became visible, a thing not to be measured, but asf
increased the action ceased. No action whatever as f
became 60, 70, 80, 90, 97, 98, 99, but as / approached 100
there was no doubt whatever of the large average force ;
a reading could be taken and it represented according to the
usual scale of the instrument ^CqCo; when/' increased beyond
100 the force suddenly ceased and remained steadily until
/' became 200 when there was a small force to be measured ;
again it ceased suddenly until /' became 300, and so on. We
know that if C and c had been true sine functions there
would have been absolutely no force except when the
frequencies were exactly equal. In truth, however, the
octaves and higher harmonics were present and so there were
slight actions when /and/' were as 2 to 1 or 1 : 2 or 1 : 3, &c.
This is an extremely important illustration for all electrical
engineers who have to deal with alternating currents of
electricity.
131. Exercise in Integration. C and c being alternating
currents of electricity. When G = Co sin qt and c = cv sin(qt ± e)
and these two currents flow through the two coils of an
electro-dynamometer, the instrument records ^ CoCo . cos e as
this is the average value of the product Cc.
When G and c are the same, that is, when the same
current G = G^ sin (qt + e) passes through both coils, the
instrument records the average value of G^ . dt, or
^j G,'8m'{qt + e).(lt (1),
which we know to be ^Gq^ The square root of any such
reading is usually called the effective current, so that
~r^ Gq is what is known as the effective value of C^o sin qt
V^
Effective current is defined as the square root of
mean square of the current. Thus when an electrical
engineer speaks of an alternating current of 100 amperes he
means that the effective current is 100 amperes or that
G= 141*4 sin {qt + a). Or the voltage 1000 means
t; = 1414 sin (</^ + /3).
EFFECTIVE CUKRENTS.
201
Exercise. What is the effective value of
tto 4- ^1 sin {qt + ti) + A.2 sin {^qt + 63) + &;c. ?
Notice that only the squares of terms have an average
value, the integral of any other product being during a
complete period. Answer: \/ a^ -^ ^{A-^ -{-A.^ -{■ kc).
Observe the small importance of small overtones.
If V = — (sin qt + J sin ^qt + i sin aqt &c.).
(2),
we shall see from Art. 135 that this is the Fourier expression
for what is shown in the curve (fig. 72) the distance OM being
called Vq and the distance OQ being the periodic time T, where
27r
q= rp } s-i^d V is measured upwards from the line OQ.
M
Fig. 72.
The effective « = ^^ Vl + * + ^V + ,V + &<=.
(3).
Again in fig. 73, where PM = Vq and OQ = T,
IT'
(sin qt — ^ sin Sqt + ^ sin 5qt — &c.) . . .(4).
Fig. 73.
The effective ?; = —-";. VI +-J^ + ^ + &c (5).
Again note the small importance of everything except
the fundamental term.
202 CALCULUS FOR ENGINEERS.
Exercise. If C=Go-{-A^smqt-\- B^ cos qt
4- ^2 sin 2qt + B^ cos 2qt + <fec ((3),
and if
c = Co + «! sin qt 4- 6i cos g'< + aa sin 2^'^ 4- h^ cos 2^^ + &c. . . .(7).
Average Cc = G^c^ + J (^itti 4- BJ)^ 4- ^./ta + BJ)^ + &c.). . .(8).
It will be seen that there are no terms like A.]),^ or A.^^,
132. Let AB and BG be parts of an electric circuit. In
yj^^ AB let the resistance be R
^- --^- -■-- „„ ->, ^j^^ j^^ there be no self-in-
^•vsAAAAAAAAA/Tfnrinnpnnr-*^ duction. In BG let the re-
^ ^ T,- ^ i ' ^ sistance be r and let there be
^'^' ^^- self-induction Z. If 0= Co sin ^i
is the current passing. Let Vab &c. represent the voltage
between the points A and B, &c. Let F^ mean the effective
voltage between A and B.
Vab = -^C'o sin qt,
Vbc ■■= Go Vr« 4- ly sin ^^^ + tan-^ -^') , see Art. 118,
Vac = Co \/(i24-r)2 4-?Y sin (^« 4- tan"^ ^^-) ,
F..= -^i2C„, F..= i^^^V^+>'
F^o=-^(i2 + r)Ooyi4- ^'^'
Observe that V^c is alTvays less than Vj^^ 4- Vbc^ or
the effective voltage between A and C is always less
than the sum of the effective voltages between A and
B and between B and C.
Thus take Co =141*4, jR=l, ?'=1, lq=l, and illustrate
a fact that sometimes puzzles electrical engineers.
133. Rule for developing any arbitrary function
in a Fourier Series.
-|*The function may be represented as in fig. 75, P^ repre-
sents the value of y at the time t which is represented by
FOURIER ANALYSIS.
203
OE, 00 represents the whole periodic time T. At G the
curve is about to repeat itself. (Instead of using the letter
F
^^""^^^
/
/
|\
G
' E >
C
Fig. 75.
t we may use x or any other. We have functions which are
periodic with respect to space for example.) Assume that y
can be developed as
y^a^^-ai sin qt + hi cos qt -}- a.^ sin 2qt + h.^ cos 2qt
+ as sin 3^^ + 63 cos ^qt 4- &c. . . .(1),
where
27r
It is evident from the results given in Art. 127 that a^
is the average height of the curve, or the average
value of y. This can be found as one finds the average
height of an indicator diagram. Carry a planimeter point
from to FPHGCO, and divide the whole area thus found by
OC. If we have not drawn the curve ; if we have been given
say 36 equidistant values of y, add up and divide by 36.
The reason is this; the area of the whole curve, or the
integral of y between the limits and T, is a^T, because
the integi'al of any other term such as Wi sin qt or 63 cos Sqt
is 0. In fact
rT rT
sin sqt .dt or I cos sqt . dt is 0,
J .'0
if s is an integer.
tti is twice the average height of the curve which results
from multiplying the ordi nates y by the corresponding
204 CALCULUS FOR ENGINEERS.
ordinate of sin^^; for, multiply (1) all across by sinqt,
and integrate from to T, and we have by Art. 128
rT rr
I y.sinqt.dt==0-\-aij ain' qt,dt -^-O -\-0 + &i}c,=^aiT...{l);
Jo Jo
dividing by T gives the average value, and twice this is
evidently Oi. Similarly
rT
y.coaqt,dt-=^b/r (2).
.0
In fact, by the principles of Ai't. 128, a.^ and bs are twice
the average values of y sin sqt and y . cos sqt^ or
2 f^
^^»=^t/ y. sin sqt, dt
i .'
(3).
dt
134. In the Electrician newspaper of Feb. 5th, 1892, the
author gave clear instructions for carrying out this process
numerically Avhen 36 numbers are given as equidistant
values of y.
In the same paper of June 28th, 1895, the author de-
scribed a graphical method of finding the coefficients.
The graphical method is particularly recommended for de-
veloping any arbitrary function.
Students who refer to the original paper will notice
that the abscissae are very quickly obtained and the curves
drawn.
In this particular case we consider the original curve
showing y and time, to be wrapped round a circular cylinder
whose circumference is the periodic time. The curve is pro-
jected upon a diametral plane passing through ^ = 0. Twice
the area of the projection divided by the circumference of
the cylinder is di. Projected upon a plane at right angles to
the first, we get bi in the same way. When the curve is
wrapped round s times instead of once, and projected on
the two diametral planes, twice the areas of each of the
FOURIER ANALYSIS.
205
two projections divided by s times the circumference of the
cylinder give cig and hg*
Prof Hem'ici's Analyzers, described in the Proceedings
of the Physical Society, give the coefficients rapidly and
accurately. The method of Mr Wedmore, published in the
Journal of the Institution of Electrical Engineers, March 1896,
seems to me very rapid when a column of numbers is given as
equidistant values of y,
135. When a periodic function is graphically represented
by straight lines like fig. 72 or fig. 73 we may obtain the
development by direct integration. Thus in fig. 76, the
Electrician's Make and Break Curve:
w
P Q
Fig. 76.
y=OA, or 2^0 say, from t- to t = OP =^iT;
y = from t = iT or OP, to t==T or OQ.
277
Evidently ao = Vo, 5' = -^;
2 ri^
/ b
M
2 fi^ 2 r*^
as = 7w I 2i'o . sin sqt . dt, h = jp\ 2vq cos sqt . dt.
4i;o T [■
hT
27r
"^=-T-25;^Lr'- r^
, 4t;o T V^T 27r 1
* The method is based upon this, that
a, = ^\y.m.nsqt.dt = -—^\y.d (cos sqt) = - — / 1/ • ^ (cos sqt).
Drawing a complete curve of which y (at the time t) is the ordinate and
COB sqt is the abscissa, we see that its area as taken by a Planimeter
divided by sir gives a^ . This graphical method of working is made use of
in developing arbitrary functions in series of other normal forms than sines
and cosines, such as Spherical Zonal Harmonics and Bessels.
By the above method
' ^'=hfy'
d (sin sqt).
206
CALCULUS FOR ENGINEERS.
2^0 , ... 2vo
aii= (cos SIT — cos 0) =
/ if « ii
\- 2 if s ij
if « is even\
IS odd /
= — if .9 is odd,
Sir
2v,
ha=—^ (sin sir - sin 0) = 0.
STT
Hence the function shown in fig. 76 becomes
y = Vf,+ -- (sin at + \ sin Zqt + J sin ^qt + &c.) (1).
IT
M
Fig. 77.
If the origin is half-way between and A (fig. 76), as in
fig. 77, so that instead of what the electricians call a make
and break we have Vq constant for half a period, then — Vq
for the next half period, that is, reversals of y every half
period, we merely subtract v^, then
y = — ^ (sin qt-\-l sin ^qt + i sin oqt + &c.) . . .(2).
Let the origin be half-way between and P, fig. 70; the
^ of (1) being put equal to a new t-\-\T,
sin sqt where s is odd, becomes sin sq {t -f \T),
sin s -^ (t-\-^T) or sin Uqti-s'^],
where 5=1, 5, 9, 13 &c. this becomes cos sqt,
„ 5 = 3, 7, 11, 15 &c. „ „ — cos sqty
and consequently with the origin at a point half-Avay between
and P,
or
y = vo + — (cos qt-^ cos Sqt + i cos Sqt - 1 cos Iqt + &c.).
FOURIER ANALYSIS. 207
136. To represent a periodic function of x for all values
of X it is necessary to have series of terms each of Tvhich
is itself a periodic function. The Fourier series is the
simplest of these.
137. If the values of y, a function of x, be given for all
values of x between a; = and x — c\y can be expanded in a
series of sines only or a series of cosines only. Here
we regard the given part as only half of a complete periodic
function and we are not concerned with what the series
represents when x is less than or greater than c. In
the previous case y was completely represented for all
values of the variable.
I. Assume y = ai sin qx + a^ sin 2qx + &c. where q = tt/c.
Multiply by ^insqx and integrate between the limits
and c. It will be found that all the terms disappear except
as sin'^ sqx . dx which is Ja«c, so that a^ is twice the average
value oiy,^ixisqx.
Thus let y be a constant m, then
/,
2 /•'
C /a
, 2m p
m sm sqx .dx= cos sox
esq Lo ^ _
= (cos 57r — 1 ) = — II s IS odd,
csq^ ' SIT
= if s is even.
Hence m=^ — (sin qx-\-\ sin Zqx + \ sin hqx + &c.)*.
II. Assume y = 6o + &i cos 5-5; + &2 cos 2^-^ + &c. Here 60
is evidently the mean value of y from a? = to a; = c. In the
* Exercise. Develope y—mx from a;=0 to .r=c in a series of sines.
mx — a^ sin qx + a^ sin ^qx + Ac, where g- = - ,
«8 IS ~ I mx . sm sg^a; . «« = -g^i" ^^^ ^^^ ~ *5^ • ^^^ -""/^ •
For this integral refer to (70) page 365.
Hence
2mc f IT 1 . 27r 1 . Stt „ \
'= 1 sin-a;--8in — a;+-sin — x- &c. j.
TT \ c 2 c 6 c J
208 CALCULUS FOR ENGINEERS.
same way as before we can prove that hg is twice the average
vakie of t/ cos sqx*.
138. In Art. 118 we gave the equation for an electric
circuit. The evanescent term comes in as before but we shall
neglect it. Observe that if V is not a simple sine function of
t, but a complicated periodic function, each term of it gives
rise to a term in the current, of the same period. Thus if
F=Fo + SF,8in(5^^ + 6'«)'^ (1),
R \IB? + X V^2 V ^ R ] ^ ^
If Lq is very large compared with R we may take
^■^i-^Isq''''^^''^^^''^ •••^^^•
Thus, taking the make and break curve for V, fig. 76,
4F
F=F„+ — \Hmqt-\■\sm^t^&c.) (4),
TT
V 2VT
^=^- ^«^ (^os qt-hi cos 3^^ + ^ cos 5qt + &c.) . . .(5),
T
which is shown by the curve of fig. 73, being at j .
139. When electric power is supplied to a house or
contrivance, the power in watts is the average value of CV
where G is current in amperes and V the voltage.
* Thus let y=ma? between x=0 and x = c. Evidently bQ=^mc, and we
^ , tmr 4wt / 1 „ 1m „ \
find y~~o ( cos9a; + -co8 3gx + 2^cos5gar + <S:c. J .
There are many other normal forms in which an arbitrary function of
X may be developed. Again, even of sines or cosines there are other forma
than those given above. For example, if we wish generally to develope y a
function of x between and c as y = Sa^ sin a„^x by the Fourier method, the
essential principle of which is I sin a,j.T . sina^-r . rfa;=0, where m and n
are different ; we must have o„ and o-, roots of — ^.— - =a. In the ordinary
Fourier series s is oo .
i — =: COS e — tan^ —
v/r2 4- pQ"^ V rj
ALTERNATING CURRENT POWER. 209
Let F= Fo sin qt and 0—0^ sin (g'^ — e).
Then P = \GqVq cos ef, or half the product of the ampli-
tudes multiplied by the cosine of the lag. When the power
is measured by passing G through one coil of a dynamometer
and allowing F to send a current c through the other coil, if
this coil's resistance is r and self-induction I
c= ,^' - sinfg^-tan-^^) (6).
What is really measured therefore is the average value
of Gc, or
i
Usually in these special instruments, large non-inductive
resistances are included in the fine wire circuit and we may
take it that Iq is so small in comparison with r that its square
may be neglected. If so, then
cos le — tan~^~)
apparent power V rJ
true power cos e
Observe that tan~^ — is a very small angle, call it a,
apparent power cos e cos a + sin e sin a
-E-f = = cos a -H sm a . tan e.
true power cos e
Now cos a is practically 1, and sin a is small, and at first
sight it might seem that we might take the answer as
nearly 1.
But if e is nearly 90° its tangent may be exceedingly
large and the apparent power may be much greater
than the true po'virer.
It is seldom however that e approaches 90° unless in coils
of great diameter with no iron present, and precautions taken
to avoid eddy currents. Even when giving power to a
choking coil or unloaded transformer, the effect of hysteresis
is to cause e not to exceed 74°.
140. True Power Meter. Let EG and GD be coils
wound together as the fixed part of a dynamometer, and let
P. 14
210
CALCULUS FOR ENGINEERS.
<c
^
<#^c _.^
DB be the moveable coil. The current C + o passes from E
to G. Part of it c goes along the
non-inductive resistance GF which.
has a resistance R. The part G
flows from G to D and D to B and
through the house or contrivance.
The instantaneous value of Rc.C
is the instantaneous power.
The coils EG and GD are care-
fully adjusted so that when c =
and the currents are continuous
currents, there shall be no deflec-
tion of the moveable coil DB. Hence the combined action
of C + c in EG and of C in GD upon C in DB is force
or torque proportional to cG, and hence the reading of the
instrument is proportional to the power. With varying
currents also there will be no deflection if there is no metal
near capable of forming induced currents.
Fig. 78.
-rB
o ^ o
C—
-0
141. The student ought to get accustomed to translating
into ordinary language such a statement as
(1) of Art. 119. Having done so, consider a
mass of W lb.* hanging from a spring whose
stiffness is such that a force of 1 lb. elongates
it h feet. If there is vibration ; when W is at
the level CO, fig. 79, a? feet below (we imagine
it moving downwards) its position of equili-
brium 00, the force urging it to the position
of equilibrium is x-^h pounds, and as the
W
moving mass is — (neglect the mass of the
spring itself or consider one-third of it as being added to the
moving body),
W . X
— X the acceleration = r .
9 h
xo
The acceleration = -t4t • The acceleration is then pro-
* The name W lb. is the weight of a certain quantity of stuff ; the inertia
of it in Engineers' units is W-i-S2'2.
Fig. 79.
MECHANICAL VIBRATION. 211
portional to x, and our ^ stands for g-- in (1) of Art. 119, and
(2) shows the law connecting a) and t
Notice carefully that the + sign in (1) is correct. The
body is moving downwards and x is increasing, so that doo/dt
is positive. But -j— is negative, the body getting slower in
its motion as a) increases.
142. Imagine the body to be retarded by a force which
die
is proportional to its velocity, or b -r. • Observe that this acts
as J acts, that is upwards, towards the position of equi-
librium.
Hence we may write
Wd^x dx X
Jdi^^^di-^h'^^ (^>-
We shall presently see what law now connects x and t in
this damped vibration.
143. Suppose that in the last exercise, when the body is
displaced x feet downwards, its point of support B is also y feet
below its old position. The spring is really only elongated
by the amount x—y, and the restoring force is —7-^ • Con-
sequently (1) ought to be
Wd'x dx x_y
J dt'^^ drh'h ^^^'
Now imagine that the motion y is given as a function of
the time, and we are asked to find a; as a function of the time.
y gives rise to what we call a forced vibration. If 2/ =
we have the natural vibrations only.
We give this, not for the purpose of solving it just now,
although it is not difficult, but for the purpose of familiar-
izing the student with differential equations and inducing
him to translate them into ordinary language.
14—2
212 CALCULUS FOR ENGINEERS.
144. Notice that if the angular distance of a rigid body
from its position of equilibrium is 0, if / is its moment of
inertia about an axis through the centre of gravity, if H6 is
the sum of the moments of the forces of control about the
•in
same axis, and if jP vr is the moment of frictional forces
at
which are proportional to velocity,
4'+^!+^^=^^' <'^)'
if B' is the forced angular displacement of the case to which
the springs or other controlling devices are attached.
145. The following is a specially good example. Referring
back to Example 1 of Art. 98, we hud CR, the voltage in
the circuit, connecting the coatings of the condenser. If we
take into account self-induction L in this circuit, then the
voltage V is ♦
^C' + ii = '' w-
We may even go further and say that if there is an
alternator in the circuit, whose electromotive force is e at any
instant (e, if a constant electromotive force would oppose G
as shown in the figure)
RG + L^ = v-e (5).
But we saw that the current G =^ — K-jj (6).
Using this value of G in (5) we get
LK^ + RK^+v = e (7).
Kow imagine that e is given as a function of the time and
we are asked to find t; as a function of the time.
e gives rise to what we call a forced vibratory current
in the system. If e = we have the natural vibrations only
of the system. Having v, (6) gives us (7.
FORCED VIBRATIONS. 218
146. If (7) is compared with (2) or (3) we see at once
the analogy between a vibrating mechanical system
and an electrical one.
They may be put
W d^x ..dx X y ^f , . -
^S+^S+^=i'E^-*'^-i (^)-
W
The mass — corresponds with self-induction L.
The friction per foot per second &, corresponds with the
resistance R.
The displacement x, corresponds with voltage v, or to be
seemingly more accurate, v is Q the electric displacement
divided by K.
The want of stiffness of the spring h con-esponds with
capacity of condenser K.
The forced displacement y corresponds with the forced
E.M.F. of an alternator.!
147. The complete solution of (8) or (9), that is, the
expression of x or i; as a function of t, will be found to
include: —
(1) The solution if i/ or e were 0.
This is the natural vibration of the system, which dies
away at a rate which depends upon the mechanical friction
in the one case and the electrical friction or resistance in the
other case. We shall take up, later, the study of this vibra-
tion. It ought to be evident without explanation, that if y
or e is 0, we have a statement of what occurs when the
system is left to itself
(2) The solution which gives the forced vibrations
only.
The sum of these two is evidently the complete answer.!
148. Forced Vibration. As the Mechanical and
Electrical cases are analogous, let us study that one about
214 CALCULUS FOn ENGINEERS.
which it is most easy to make a mental picture, the mechanical
case. We shall in the first place assume no friction and neglect
the natural vibrations, which are however only negligible
when there is some friction. Then (8) becomes
dt?^wi.'—my <i*^>-
Let y = a sin qt be a motion given to the point of support
of the spiral spring which carries W ; y may be any compli-
cated periodic function, we consider one term of it.
We know tha t if y were 0, the natural vibration would
be a; = 6 sin f i 4/ ^-jj + ^^^ ) , where h and m might have any
values whatsoever. It is simpler to use n"^ for glWh as
we have to extract its square root, n is 27r times the
frequency of the natural vibrations of W. We had better
write the equation as
— -f i^^x = n^y = n-a sin qt (11).
Now try if there is a solution, 00 = A r,in qt + B cos qt If so,
since ,- = — Aq- sin qt — Bq^ cos qt ; equating the coefficients
of sinqt and also those of cosqt, — Aq- -[■ n^A = n^a, so that
A =— — ^ , and — Bq^ + ii^B = 0, so that B = unless n = q,
n^ — q^ ^ *
We see that we have the solution
^^n^Zrjf^'^'i^ (12)-
This shows that there is a forced vibration of W which is
synchronous with the motion of the point of support ; its
amplitude being , times that of the point of support.
Now take a few numbers to illustrate this answer. Let a = 1,
let " be great or small. Thus ^ = -X means that the forced
n ^ n ^^
frequency is one tenth of the natural frequency.
FORCED VIBRATIONS.
215
?
Amplitude of j
'/
Amplitude
n
W 's motion j
n
JF's motion
•1
1
101 1
1
X
•5
1^33:3
1^01
-50
•8
2^778
1-03
-16^4
•9
5^263
1^1
- 4-762
•95
1026 ;
15
- 0-8
•97
1692
20
- 0-333
•98
2525
50
- 0042
•99
50^25
100
- 0010
Note that when the forced frequency is a small fraction
of the natural frequency, the forced vibration of TT is a
faithful copy of the motion of the point of support B ; the
spring and W move like a rigid body. When the forced is
increased in frequency the motion of TT is a faithflil magni-
fication of B's motion. As the forced gets nearly equal to the
natural, the motion of W is an enormous magnification
of B's motion. There is always some friction and hence the
amplitude of the, vibration cannot become infinite. When
the forced frequency is greater than the natural, W is always
a half-period behind B, being at the top of its path when
B is at the bottom. When the forced is many times the
natural, the motion of W gets to be very small ; it is nearly
at rest.
Men who design Earthquake recorders try to find a
steady point which does not move when everything else is
moving. For up and down motion, observe that in the last
case just mentioned, W is like a steady point.
When the forced and natural frequencies are nearly equal,
we have the state of things which gives rise to resonance
in acoustic instruments ; which causes us to fear for suspen-
sion bridges or rolling ships. We could easily give twenty
interesting examples of important ways in which the above
principle enters into engineering problems. The student
may now work out the electrical analogue for himself and
study Hertz' vibrations.
149. Steam engine Indicator vibration. The
motion of the pencil is to faithfully record the force of
216 CALCULUS FOR ENGINEERS.
the steam on the piston at every instant ; this means that
the natural vibrations of the instrument shall be very quickly
destroyed by friction. Any friction as of solids on solids will
cause errors. Indeed it is easy to see that solid friction
causes diagrams to be always larger than they ought to be.
Practically we find that if the natural frequency of the
instrument is about 20 times that of the engine, the diagram
shows few ripples due to the natural vibrations of the indi-
cator. If the natural frequency is only 10 times that of the
engine, the diagram is so ' upset ' as to be useless.
F
The frequency of a mass — at the end of a spring whose
if
yieldingness is hy see Art. 141, is ^ ^gj Wh, neglecting friction.
We shall consider friction in Art. 1 GO. What is the frequency
of a mechanism like what we have in an indicator, controlled
by a spring ? Answer : If at any point of the indicator
w
mechanism there is a mass - , and if the displacement of
this point is .9, when the displacement of the end of the spring
(really the piston, in any ordinary indicator) is 1 ; imagine
w w
that instead of — we have a mass s^ — at the end of
9
hXsHv'
the spring. Thus the frequency is ^ a/ ^
To illustrate this, take the case shown in fig. 80 ; GAB is
a massless lever, hinged at 0, with
S- the weight W at B. The massless
e spring is applied at A.
I When A is displaced downwards
from equilibrium through the dis-
tance x, the extra pull in the spring
€)
X
is -r . The angular displacement of
^ig- 80. the lever, clockwise, is j-j . Mo-
UA
ment of Inertia x angular acceleration, is numerically equal
W
to moment of force. The Moment of Inertia is — OB^,
9
VIBRATION INDICATOR. 217
The angular acceleration is yr-j , where x stands for -^ , so
that -0i?^;^ + 7. 0^ = 0,
g OA h
^+^.^.^ = 0.
OR
And yya ^^ what we called 5, so that s^W takes the place
of our old W when W was hung directly fronri the spring.
150. Vibration Indicator. Fig. 81 shows an in-
strument which has been used for indicating quick vertical
vibration of the ground.
A
c o a o
Fig. 81.
The mass GPQ is supported at F by a knife edge, or by
friction wheels. The centre of gravity G is in a horizontal
line with P and Q. Let FG=a, GQ=:h, PQ = a+b=l
The vertical spring AR and thread BQ support the body at
Q. As a matter of fact AR is an Ayrton-Perry spring, which
shows by the rotation of the pointer R, the relative motion of
A and Q; let us neglect its inertia now, and consider that
the pointer faithfully records relative motion of A and Q.
It would shorten the work to only consider the forces at P
and Q in excess of what they are when in equilibrium, but for
clearness we shall take the total forces.
When a body gets motion in any direction parallel to
the plane of the paper, we get one equation by stating
that the resultant force is equal (numerically)
to the mass multiplied by the linear acceleration
of the centre of gravity in the direction of the
resultant force. We get another equation by stating
218 CALCULUS FOR ENGINEERS.
that the resultant moment of force about an axis
at right angles to the paper through the centre of
gravity is equal to the angular acceleration, multi-
plied by moment of inertia about this axis through
the centre of gravity. I shall use x, x and ic to mean
fj nf* Cm IT
displacement, velocity and acceleration, or x, -j- and -v- .
Let P and A get a displacement x^ downward. Let Q
be displaced x downward. Let the pull in the spring be
Q = Qq-\- c{x — Xi) where c is a known constant (c is the
reciprocal of the h used in Art. 141). Let W be the weight
of the body. Then if Pq and Qo be the upward forces at the
points marked P and Q when in the position of equilibrium,
Q„(a + Z>)= Wa and Po+Qo= W.
J, bW ■ aW
Hence ^o = ^^^, Qo = ^-7-^ (1),
Q=Qo + c{x-x,).
Now G is displaced downwards =- Xi H ? x, so that
^ a+6 a+b
W-P-Q='^{b-£, + aB]^^ (2).
The body has an angular displacement clockwise about its
centre of mass, of the am<
moment of inertia about G
X -— X
centre of mass, of the amount r« So that if / is its
a-{-b
--Qb + Pa = ^^^^^(x-x,) (3).
W
Hence (2) and (3) give us, if M stands for — , and if
/ = Mk^ where k is the radius of gyration about G,
VIBRATION INDICATOR. 219
If ki is the radius of gyration about P, we find that (4)
simplifies to
I [^
if n stands for tta/ ir?= ^tt x natural frequency, and e^ stands
for 1 — r-2 • Call X — Xihy the letter y because it is really y
that an observer will note, if the framework and room and
observer have the motion x^. Then 2^ y = x — XiOv x = y-\-Xx
y + Xi-\- n^ (y + ^'i) = ^Xi 4- n^Xi .
So that y + n^y = (^-l)xj_ (5),
or y + w'2/ + p^i = ^ (6)-
Let Xj^ = A sin qt.
We are neglecting friction for ease in understanding our
results, and yet we are assuming that there is enough
friction to destroy the natural vibration of the body.
We find that if we assume y = a sin qt, then
_ al ([- .
That is, the apparent motion y (and this is what the
pointer of an Ayrton- Perry spring will show ; or a light
mirror may be used to throw a spot of light upon a screen),
is T-« -r^ — ^ times the actual motion of the framework and
A-j* n^ — q^
room and observer. If q is large compared with n, for
example if q is always more than five times n, we may take
it that the apparent motion is y-^ times the real motion and
is independent of frequency. Hence any periodic motion
whatever (whose periodic time is less say than Jth of the
periodic time of the apparatus) will be faithfUlly indi-
cated.
Note that if al ~ Tc^ so that Q is what is called the point
of percussion, Q is a motionless or * steady ' point. But in
practice, the instrument is very much like what is shown in
220 CALCULUS FOR ENGINEERS.
the figure, and Q is by no means a steady point. Apparatus of
the same kind may be used for East and West and also for
North and South motions.
151. Any equation containing ■— or -^ or any other
differential coefficients is said to be a " Difierential
Equation/' It will be found that differential equations
contain laws in their most general form.
Thus if o) is linear space and t time, the statement -f^ =
d^x . ^^
means that ,- , (the acceleration), does not alter. It is the
most general expression of uniformly accelerated motion.
When we integrate and get -7^= ci, we have introduced
the more definite statement that the constant accelei-ation is
known to be a. When we integrate again and get
dx
^ = at + h,
we are more definite still, for we say that h is the velocity
when ^ = 0.
When we integrate again and get
X = ^af -i-bt-^-c,
we state that x = c when ^ = 0.
Later on, it will be better seen, that many of our great
general laws are wrapped up in a simple looking expression
in the shape of a differential equation, and it is of enormous
importance that when the student sees such an equation he
should translate it into ordinary language.
152. An equation like
g-^S-«2-^l-^=^ «.
if Py Q, R, S and X are functions of x only, or constants, is
said to be a linear differential equation.
Most of our work in mechanical and electrical engineering
leads to linear equations in which P, Q, &c., are all constant
with the exception of X. Thus note (8) and (9) of Art. 146.
DIFFERENTIAL EQUATIONS. 221
Later, we shall see that in certain cases we can find the
complete solution of (1) when X is ; that is, that the solu-
tion found will include every possible answer. Now suppose
this to be y =/(*'). We shall see that it will include four
arbitrary constants, because ^^ is the highest differential
coefficient in (1), and we shall prove that if, when X is not 0,
we can guess at one solution, and we call it y= F (os), then
j,=f{x) + F(x)
is a solution of (1). We shall find in Chap. III. that this is
the complete solution of (1).
In the remainder of this chapter we shall only consider
P, Q, &c., as constant ; let us say
da^^^d^^^d^^^dx^^y-^ ^^^'
where A, B, G, E are constants and Xis a function o{ a;.
We often write (2) in the form
153. Taking the very simplest equation like (3). Let
J-«y = o • W'
it is obvious (see Art. 97) that
y = Me^'' (5)
is the solution, where M is any constant whatsoever.
154. Now taking ^-«'.y = (6),
we see by actual trial that
2/ = if €«* + i\r6-«^ (7)
is the solution, where M and N are any constants whatsoever.
But if we take ^^n^y = (8),
222 CALCULUS FOR ENGINEERS.
we see that as the a of (6) is like ni in (8) if i means V— 1,
then
y =: Me''^ -\- Ne-""^ (9)
is the sohition of (8). If we try whether this is the case, by
difFerentiat.Ton, assuming that i behaves like a real quantity
and of course i^ = — 1, i^ = — i, i* = l, {^ = i, &c., we find that
it is so. But what meaning are we to attach to such an
answer as (9) ? By guessing and probably also through re-
collection of curious analogies such as we describe in Art. 106,
and by trial, we find that this is the complete solution also,
y — Mi sin tix + N^ cos nx (10).
As (10) and (9) are both complete solutions (Ai-t. 152) be-
cause they both contain two arbitrary constants which may be
unreal or not, we always consider an answer like (9) to be the
same as (10), and the student will find it an excellent exercise
to convert the form (10) into the form (9) by the exponential
forms of sinew; and coscu?, Art. 106, recollecting that the
arbitrary constants may be real or unreal. Besides, it is im-
portant for the engineer to make a practical use of those
quantities which the mathematicians have called unreal,
155. Going back now to the more general form (3) when
X = O, we try if 2/ = Me'^ is a solution, and we see that it is
so if
, m^ + Am^ -\- Bni" -{■ Gm-\- E = ^ (1).
This is usually called the auxiliary equation. Find the
four roots of it, that is, the four values of m which satisfy it,
and if these are called 7?ii, "nfi^y Wg, m^, we have
as the complete solution of (3) when X = 0; M^, &c., being
any arbitrary constants whatsoever.
156. Thus to solve
if we assume y = e*"*, we find that m must satisfy
?M^ + oiit^ -f om^ — om — 6 = 0.
DIFFERENTIAL EQUATIONS. - 223
By guessing we find that m = l is a root; dividing by
m — 1 and again guessing, we find that m= — l is a root ;
again dividing by m + 1 we are left with a quadratic expres-
sion, and we soon see that m= — 2 and m = — 3 are the
remaining roots. Hence
is the complete solution, il/i, i/s) &c., being any constants
whatsoever.
157. Now an equation like (1) may have an unreal root like
7n + 7ii, where t is written for J— 1, and if so, we know from
algebra, that these unreal roots go in pairs ; when there is
one like m + ni there is another like ni — ni. The corre-
sponding answers for y are
y — M € ('"~"^' ^ 4- N 6 <"*+*"^ *
or e"*^ {i¥i€-^'^ + iVi€+"'*),
and we see from (10) that this may be written
y = e"** {31 sin nx + N cos nx],
where M and iV" are any constants whatsoever.
158. Suppose that two roots m of the auxiliary equation,
happen to be equal, there is no use in writing
because this only amounts to (M^ + M,^) e*"^ or ife**^ where M
is an arbitrary constant, whereas the general answer must
have two arbitrary constants. In this case we adopt an
artifice ; we assume that the two roots are m and 7n + h and
we imagine h to get smaller and smaller without limit :
= e"'^ (Ml -H if^e^^^),
but by Art. 97, e^ = 1 + /^A- + ^ + j^3 4- &c.,
therefore y = e"^^ (m^ + M^ + MJiw + M^ ^ + ^c) -
Now let i/2^ be called N and imagine h to get smaller
224 CALCULUS FOR ENGINEERS.
and smaller, and M2 to get larger and larger, so that MJi may
be of any required value we please, say N, and also
as h gets smaller and smaller without limit we find
If this reasoning does not satisfy the reader, he is to
remember that we can test our answer and we always find it
to be correct.
159. It is in this way that we are led to the fol-
lowing general rule for the solving of a linear differential
equation with constant coefficients. Let the equation be
fl + A^^+Bf;^, + S^. +G^^ + % = 0...(1).
Form the auxiliary equation
m** + ^m^-i + ^m'*-^ + &c. + Gm + iT = 0.
The complete value of y will be expressed by a series of
terms : — For each real distinct value of m, call it a, , there will
exist a term Mie*»'*'; for each pair of imaginary values a^ ± ySgi,
a term
6»*^ (ilfa sin ^^ + iVs cos ^.^) ;
each of the coefficients M^y M^, N2 being an arbitrary
constant if the corresponding root occurs only once, but a
polynomial of the r— 1th degree with arbitrary constant
coefficients if the root occur r times.
Ea^rcue. g+ 12^ + 66 f|+ 206 g
aa:^ dx* dx^ do(?
+ 345^ + 234y = 0.
Forming the auxiliary equation, I find by guessing and
trying, that the five roots are
- 3, - 3, - 2, - 2 + 3i, - 2 - 3i.
Consequently the answer is
y = (ifi + N^x) €-^ + M^e-"^ + €-^ (i/g sin Sx + N, cos 3^).
NATURAL VIBRATIONS. 225
Exercise. 1. Integrate ^t^ - 4 ^ 4- 3y = 0.
Answer : y= A^+ Bg^.
% lutegrate g-loJ+34y = 0.
Answer : y = e^ {A sin ^x + B cos 3^j.
3. Integrate g + 6| + 9, = 0.
Answer : y = (^ + 5i») e~^.
4 Integrate g-l^g + esg- lo6|+ 169, = 0.
Here vv' - 12m^ + 627?t2 - Ib^m + 169 = 0, and this will be
found to be a perfect square. The roots of the auxiliary
equation will be found to be
3 + 2i, 8 + 2'i, 3 - 1i, 3 - 2i
Hence the solution is
y = 6^* {(^1 + B^x) sin 2*- + (^2 + BrO)) cos 2^j.
We shall now take an example which has an important
physical meaning.
Natural Vibrations. Example.
160. We had in Art. 146, a mechanical system vibrating
with one degree of fi-eedom, and we saw that it was analogous
with the surging going on in an Electric system consisting of a
condenser, and a coil with resistance and self-induction. We
neglected the friction in the mechanical, and the resistance
in the electric problem. We shall now study their natural
vibrations, and we choose the mechanical problem as before.
If a weight of TTlb. hung at the end of a spring which
elongates x feet for a force of x -^Ji lb., is resisted in its
motion by friction equal to 6 x velocity, then wc had (8) of
Art. 146, or IL^ h^ --0
d^x hq dx xq ^ ,^.
-rfF + F-rf*+m = *^ (!>•
i\ 15
^26 CALCULUS FOR ENGINEERS.
Let ^^ be called 2/ and let -^ = ?i= ; (1) becomes
5+2/.J + n'. = (2).
Forming the auxiliary equation we find the roots to be
We have different kinds of answers depending upon the
values of / and n. We must be given sufficient information
about the motion to be able to calculate the arbitrary con-
stants. I will assume that when t ia the body is at w =
and is moving with the velocity Vq.
I. Let / be greater than n, and let the roots be — a
and - 13.
IL Let /be equal to n, the roots are —/and — /
IIL Let /be less than n, and let the roots be — a ± bi.
IV. Let/= 0, the roots are ± ni.
Then according to our rule of Art. 159,
In Case I, our answer is
dos
H
^ = 0, we can calculate A and B and so find x exactly in
terms of ^ ;
In Case II, our answer is
In Case III, our answer is
X = €-«* {A sin U + B cos U] ;
In Case IV, our answer is
X — A^mnt'\- B cos nt.
161. We had better take a numerical example and we assure
the student that he need not grudge any time spent upon it
and others like it. Let n = 3 and take various values of /.
For the purpose of comparison we shall in all cases let a? =
"*" dx
when ^ = 0; and ^. = 20 feet per second, when i = 0.
and if we are told that x = when < = and -y; = Vo when
NATUKAIi VIBRATIONS.
Case IV. Let/= 0, then a;== A sin nt + 5 cos nt,
= AxO + Bxl, so that 5 = 0,
da)
227
dt
= nA cos ?«^ — iiB sin n^,
20 = SA, so that ^ = ^-.
Plot therefore *•= 6*667 sin 3^.
This is shown in curve 4, fig. 82. It is of course the
ordinary curve of sines: undamped S.H. motion.
Fig. 82.
Case III. Lety = "S. The auxiliary equation gives
771 = - -3 ± \/-09 - 9 = - -3 ± 2-985t.
Here a = '3 and 6 = 2-985 in
x^e-^'^lAsinU-tBcosbt} (1).
You may not be able to differentiate a product yet, although
we gave the rule in Art. 90. We give many exercises in
Chap. III. and we shall here assume that
dx
— = - ae-"* (A sin ht + B cos bt)
+ 6e-««(^cos6e-5sin60 (2).
15—2
228 CALCULUS FOR ENGINEERS.
Put .£=0 when t = and -r: = 20 when t = 0. Then
at
B = from (1) and
and hence os = 6-7€-*«* sin 2-985^.*
This is shown in curve 3 of fig. 82. Notice that the period
has altered because of friction.
Case II. Let/= 8. The roots of the auxiliary equation
are ?/i = — 3 and — 3, equal roots. Hence
^• = (^+^0^"'' (!)•
Here again we have to differentiate a product and
~ = B€-''-S{A+Bt)e-'^' (2).
Putting in a; = when ^ = and -17 = 20 when ^ = 0,
^ = from (1) and J5 = 20 from (2).
Hence o) = 20^ . e-'K
This is shown in curve 2 of fig. 82.
Case I. Let f=o. The roots of the auxiliary equation
are — 9 and -- 1,
0! = Ae-''^ + Be-\
^=-9^6-««-l?6-'.
at
Putting in the initial conditions we have
= A-hB, 20^-9A-B.
Hence J. = - 21, 5 = 2^,
«<'=2i (€-*-€-»«)•
This is shown in curve 1 of fig. 82.
Students ought to take these initial conditions
OS =10 when t^O and -77 = when t = 0.
dt
VIBRATIONS. 229
This would represent the case of a body let go at time
or, in the electrical case, a charged condenser begins to be
discharged at time 0.
Notice that if we differentiate (1), Art. 160, all across we
have f using v for -j-) ,
d^v hq dv a ^
— ^ — P— ?) =
dt^^ Wdt^ Wh
We have therefore exactly the same law for velocity or
acceleration that we have for x itself.
Again, in the electrical case os K -r: represents current,
if we differentiate all across we find exactly the same
law for current as for voltage. Of course differences are
produced in the solutions of the equations by the initial
conditions.
162. When the right-hand side of such a linear differential
equation as (2) Art. 152 is not zero and our solution will give the
forced motion of a system as well as the natural vibrations, it
is worth while to consider the problem from a point of view
which will be illustrated in the following simple example.
To solve (11) Art. 148, which is
-J- -\- ii^x — n^a sin qt (1),
the equation of motion of a system with one degree of
freedom and without friction.
Differentiate twice and we find
d*w ^d^x , , . .
Hence from (1), ^+(712 + 52)^+^^^ = (2).
To solve (2), the auxiliary equation is
m^ + (ii' + 50m2 + g'2|i2^0 (3),
and we know that + ni are two roots and + qi are the other
two roots. Hence we have the complete solution
«? = ^ sin nt + B cos nt -^-G sinqt-^ D cos qt (4).
2S0
CALCULUS FOR ENGINEERS.
© @ ©
Now it was by differentiating (1) that we introduced
the possibility of having the two extra arbitrary constants C
and D, and evidently by inserting (4) in the original equation,
we shall find the proper values of C and D, as they are really
not arbiti-ary. It will be noticed that by differentiating
(1) and obtaining (2) we made the system more complex,
gave it another degree of freedom, or rather we made it
part of a larger system, a system whose natural vibrations
are given in (4). When we let a mass vibrate at the end of
a spring, it is to be remembered that the centre of gravity of
the mass and the frame which supports it and the room,
remain unaltered. Hence vibrations occur in the supporting
frame, and there is friction tending to still the vibrations.
If there is another mass also vibrating, this effect may be
lessened. For example in fig. 83, if M vibrates at the end of
the strip MA, clamped in
the vice A, any motion
of M to the right must be
accompanied by motion of
A and the support, to
the left. But if we have
two masses Mi and il/g (as
in a tuning fork), moving
in opposite directions at
each instant there need be
no motion of the supports,
consequently the system
MjMo vibrates as if there
were less friction, and this
principle is utilized in
tuning forks. Should a
motion be started, different
from this, it will quickly
become like this, as any part of the motion which
necessitates a motion of the centre of gravity of the
supports, is very quickly damped out of existence.
The makers of steam engines and the persons who use
them in cities where vibration of the ground is objected to,
find it important to take matters like this into account.
163. If y is a known function of oc, we are instructed by
Fig. 83.
SYMBOLS OF OPERATION. 231
(3), Art. 152, to perform a complicated operation upon it.
Sometimes we use such a symbol as
{e'-\-Ae'-{-Bd^+ce+E)y = x,
to mean exactly the same thing ; 6y meaning that we differ-
entiate y with regard to x, 6^y meaning that we differentiate
y twice, and so on.
6, 6^, &c., are symbols of operation easy enough to under-
stand. We need hardly say that 6'^y does not mean that
there is a quantity 6 which is squared and multiplied upon
y: it is merely a convenient way of saying that y is to be
differentiated twice. 66y would mean the same thing.
On this same system, what does (6 -\-a)y mean ? It means
^+ay. What does (6'' + A0 + B)y mean? It means
(Py dy
~T^ -}- -4 1^ -h By. (6-{-a)y instructs us to differentiate y and
add a times y, for a is a mere multiplier although is not so,
and yet, note that (6 -\-a)y= 6y + ay.
In fact we find that 6 enters into these operational
expressions as if it were an algebraic quantity, although it is
not one.
If u and V are functions of x we know that
e {u + v)= 6u + Ov.
This is what is called the distributive law.
Again, if a is a constant, 6au = a6u, or the operation 6a is
equivalent to the operation aO. This is called the com-
mutative law.
Again 6^6^ — 6^+^.^ ^\^\^ jg ^j^^ index law. When these
three laws are satisfied we know that 6 will enter into
ordinary algebraic expressions as if it were a quantity. 6
follows all these laws when combined with constants ; but note
that if u and v are functions of x, vdu meaning v -j- , is a
very different thing from 6 , uv. When we are confining our
232 CALCULUS FOR ENGINEERS.
attention to linear operations we are not likely to make
mistakes.
Thus operate with 6-{-b upon (0 -f a) y. Now
{e-\-a)y=-ey-\-ay or ^ + ay.
Operating with 6 ■\-h means "diflFerentiate (this gives us
/7/S"^^^) ^^^ ^^^ ^ times -^- + ay" Consequently it gives
We see, therefore, that the double operation
{e + h){d + a)
gives the same result as
[&'-\-{a-\-h)e^-ah].
In this and other ways it is easy to show that although 6 ia
a symbol of operation and not a quantity, yet it enters into
combinations as if it were an algebraic quantity, so long as all
the quantities a, h, &c. are constants. Note also that
is the same as {0 + h){d -\- a).
The student ought to practise and see that this is so and
get familiar with this way of writing. He will find that it
saves an enormous amount of unnecessary trouble. Thus
compare such expressions as
(ae-hh)(ae-h^)y
with jaa^-^ + (a^ +ah)e + h^} y,
164. Suppose that Dy is used as a symbol for some curious
operation to be performed upon y, and we say that Dy = X ;
does this not mean that if we only knew how to reverse the
SYMBOLS OF OPERATION. 233
operation, and we indicate the reverse operation by D~^ or p. ,
then y = I)~^ X or yy ? We evidently mean that if we operate
with D upon D~^X, we annul the effect of the D~^ operation.
Now if -^ + ay = X , or l-j- -\- a] y — X , or {6 ■{■ a) y =X , let us
indicate the reverse operation b}^
y-^^x-^^y^^^i^+^rx (1),
X X ,^^
or , or ^ (2).
ax
Keeping to the last of these; at present ^ is a mere
symbol for an inverse operation, but v-\-a
y=-^a (^>
submits to the usual rules of multiplication, because (3) is
the same as (6 + a)y= X (4);
and yet (4) is derived from (3) as if by the multiplication of
both sides of the equation by {6 + a),
Again,take^ + (a + 6)^+a62/ = Z (5),
or {e^+{a-{-h)e + ab]y=^X (6),
or (e + a)(d + h)y = X (7).
Here the direct operation 6 + a performed upon (0 -{•h)y
gives us X ; hence by the above definition
(^+6)y=^„ (8).
and repeating, we have
2/
~(e-hh)\d-\-aj "^^^'
234 CALCULUS FOR ENGINEERS.
But it is consistent with our way of writing inverse operations
to write (6) as
^^e'-^(a+b)e+ab ^^^^'
and so we see that there is nothing inconsistent in our treating
the 6 -{-h and ^ + a of (9) as if ^ were an algebraic quantity.
165. We know now that the inverse operation
{^+(a+6)(9 + a61-^ (1),
may be effected in two steps ; first operate with (6 -\- h)~^ and
then operate with (6 + a)~\
Here is a most interesting question. We know that if 6
were really an algebraic quantity,
1^1 1 ^ .(2).
^ + (a + 6) ^ + a6 b-a\d-\-a O-^b,
And it is important to know if the operation
irrs(5n"ff+b) ^^^'
is exactly the inverse of ^ + (a + b) d + ab?...(4). Our
only test is this ; it is so, if the direct operation (4) com-
pletely annuls (3). Apply (3) to X and now apply (4) to
the result; if we apply (4) to ^ X, we evidently obtain
dX tf + ct 1
(d-{-b)X or -T- + 6X ; if we apply (4) to j^v X we evi-
dently obtain (^ + a) X or -^ + aX, and
1
b — a
We see therefore that (3) is the inverse of (4), and that we have
the right to split up an inverse operation like the left-hand
side of (2) into partial operations like the right-hand side of
(2). We have already had a number of illustrations of this
when the operand was 0. For it is obvious that if a^, oui, &c.
are the roots of the auxiliary equation of Art. 159, it really
means that
e^'^A e>'-' + J5l9"-=^ + ka. ■\- GO ■\- H
splits up into the factors {6 — ttj) {6 — a^), &c,
f + 6X-(f + «X)^=X.
SYMBOLS OF OPERATION. 235
Observe that if -J^ = X, or 6y = X, or y — -^, or y = 6~^X,
the inverse operation ^^ simply means that X is to be inte-
grated. Again, 0~^ means integrate twice, and so on*.
* Suppose in our operations we ever meet with the symbols 6^ or d~^
or 6"^ &c., what interpretations are we to put upon them ? It is not very
necessary to consider them now. Whatever interpretations we may put
upon them must be consistent with everything we have ah"eady done. For
example 6^ will be the same as 60^ and d~ -^ will be the same as d^ d~^
or 6~^6^. We have to recollect that all this work is integration and we use
symbols to help us to find answers ; we are employing a scientific method of
guessing, and our great test of the legitimacy of a method is to try if our
answer is right; this can always be done. Most of the functions on
which we shall be operating are either of the shape ^e"* or B sin bx or sums
of such functions. Observe that
^^e«^=^a»
if n is an integer either positive or negative. There is therefore, a likelihood
that it will help in the solution of problems to assume that
or that e^A6*^ = Aa^6^ (1).
Again 0B sin bx = Bb cos bx = Bb sin. (bx + ^\ ,
e^B sin bx= - Bb^ sin 6a; = Bb^ sin (&a; + tt) ,
and ^ 0^B sin bx=Bb'' Bin (bx + n'^j (2).
Evidently this is true when n is a positive or negative integer ; assume it
true when n is a positive or negative fraction, so that
623 sin bx = Bb2 sin f bx + ^^ j
.(3).
There are certain other useful functions as well as e*"* and si7i bx such
that we are able to give a meaning to the effect of operating with 6^ upon
them. It will, for example, be found, if we pursue our subject, that we shall
make use of a function which is for all negative values of x and which is a
constant a for all positive values of x. It will be found that if this function
is called /(.r) then
0if(x)=a4-x-^ (4),
and the meaning of d^ or ^^ or ^~i or 9~^ &c. is easily obtained by
differentiation or integration. The Mnemonic for this, we need not call it
bn
proof or reason, is d^x'^ =~=^~ x"^'^. Let w=i, m=0 and we have
*^ \m-n ^
286 CALCxn.us for engineers.
166. Electrical Problems. Circuit with resistance
R and self-induction L,
let -^ be indicated by 0, then
V
V=(R + Le)G or =
R + Ld'
In fact in all our algebraic work wo treat R + "LO as if it
were a resistance.
Condenser of capacity K farads. Let V volts, be voltage
between coatings. Let G be current in amperes into the
condenser, that is, the rate at which Q, its charge in coulombs,
is increasing. Or ^ = 7^ = "7^ (-^^^ ^^' ^^ ^ ^ usually
dV
assumed to be constant, (7 = TT—r- .
at
The conductance of a condenser is KO, therefore
Hence the current into a condenser is as if the condenser
had a resistance — ^.
Fig. 84.
Circuit with resistance, self-induction and capacity, fig. 84.
All problems are worked out as if we had a total resistance
R + I-^ + xFZJ (!)•
kS
dhx^=-. — ^x 2. But |- i has no meaning. Give it a meaning by assuming
|_IL2 ~
that whatjs true of integers, is true of all numbers, and use gamma function
of I or ! ^ which is Jir instead of l~^' It is found that the solutions
effected by means of this are correct.
EASY RULE FOR ELECTRICAL PROBLEMS. 287
167. In any network of conductors we can say exactly
what is the actual resistance (for steady currents) between
any point A and another point B if we know all the resist-
ances 7\, r^, &c. of all the branches. Now if each of these
branches has self-induction ^i, &c. and capacity Ki, &c.
what we have to do is to substitute r^ + l^d -f t^-t, instead of
Ti in the mathematical expressions, and we have the resistance
right for currents that are not steady.
How are we to understand our results ? However com-
plicated an operation we may be led to, when cleared of
fractions, &c. it simplifies to this; that an operation like
a-\-b0-\- cB' 4- dO' + ed'+fd' + &c
a' + b'd + c'0'' + d'e' + e'e'~tf'e'-{-&}c. ^ ^'
has to be performed upon some voltage which is a function
of the time. On some functions of the time which we have
studied we know the sort of answer which we shall obtain.
Thus notice that if we perform (a -}- bd -\- &}c.) upon e** we
obtain
{a + b2 + col' + da^ + ea.^ +/a^ + &c.) e*^ (2).
Consequently the complicated operation (1) comes to be a
mere midtiplication by A and division by A', where A is the
number a -f 6a + ca^ + &c. and A' is the number
a + Vol + c V -|- &c.
Again, if wc operate upon in sin {nt + e), observe that
6"- would give — mn^ sin {nt + e).
and
^ „ „ +mn^mn{nt-\-e\
and so on ;
whereas
would give inn cos {nt + e).
6^ „ „ — mu^ cos {nt + e).
6^ „ „ + mn^ cos {nt + e).
And hence the complicated operation (1) produces the same
effect as -^ — ^^ , where
p — a — cn^-hen* — &c., q=^b — dn- -^fn^ — &c.
a = a' - c'n2 -\-e'n^ - &c., ^ = b' - dn" +fn' - &c.
238 CALCULUS FOR ENGINEERS.
Observe Art. 118, that p + qO operating upon m sin (tit 4- e)
multiplies the amplitude by Vp^ + q^ii^ and causes an advance
of tan~^— . The student ought to try this again for himself,
although he has already done it in another way. Show that
(p + qO) sin nt = V^T^ sin (nt + tan-^ 2!!:^ .
Similarly, the invei-se operation l/(a + ^6) divides the
amplitude by Va" + yS^/i^ ^j^^j produces a lag of tan~^ — , and
hence
^-j-|gmsm(7i«+e)
a labour-saving rule of enormous importance.
168. In all this we are thinking only of the forced
vibrations of a system. We have already noticed that
when we have an equation like (1) or (2) Art. 152, the
solution consists of two parts, say y —f(x)-{- F{x)\ where
f{x) is the answer if X of (2) is 0, the natural action of
the system left to itself, and F {x) is the forced action.
If in (2) we indicate the operation
{ d' . d^ ^ d'' ^ d „\ , ^
then D (y) = X gives us
Where I)-^{0) gives /(^) and B-^(X) gives F(x).
Thus if^+ay^O, or fj- + a]y = 0, or (^ + a)y = 0, we
know Art. 97, that y = ^.e-**^.
O
a
so that if ^■-\-ciy'= X, the complete solution is
Hence we see that x is not nothing^ but is Ae'
ELECTRICAL EXAM^lS^^^V , ^^ 239
"We are now studying this latter paiX't^the forced part, only.
In most practical engineering problems the exponential terms
rapidly disappear.
169. Thus in an electric circuit where V=(R + Ld) C, if
V = Vq sin qt,
we have already found the forced value of G,
^_ Fosin^
R+Ld'
and according to our new rule, or according to Art. 118, this
becomes
0= . ZV s^^ f^^-tan-^^) ....(1).
But besides this term we have one
-R-^Ld'^'^TT^'
and according to the above rule (Art. 168) this gives a term
A,e ^' (2).
Or we may get this term as in Art. 97,
iJ(7+X^ = 0,or ^=-j;0.
This is the compound interest law and give's us the answer (2),
and the sum of (2) and (1) is the complete answer. If we
know the value of G when ^ = 0, we can find the value of the
constant Ai ; (2) is obviously an evanescent term.
Thus again, suppose V to be constant = Vq,
=
R + ie-
V
It is evident that G = ~ is the forced current, for if we
Y ^
operate on G=-~ with R + LOwe obtain Fo, and the e vanes-
240 CALCULUS FOR ENGINEERS.
cent current is always the same with the same R and L
whatever Fmay be, namely Ai€~l^,
The complete answer is then
^t
n
G^A,e-L^+^^ (2).
Let, for example, C = when t — Oy then
V V
and (2) becomes (7= — " (l - e"!') (3).
The student ought to take Fo=100, jK=1, L=1 and
show how C increases. We have had this law before.
170.. Example. A condenser of capacity K and a
non-inductive resistance r in parallel; voltage V at
their terminals, fig. 85. The two currents are c— V/r,
C=KOV, and their sum is C+c=v(^^ + Ke\ or v(^-'~^^\
V
so that the two in parallel act like a resistance .. ,T^n *
o g^c>
C
K
Fig. 85.
UV^Vosinnt,
r ''
K
C + c = — Vl + r'^K'^n'', sin ini + tan-^ rKn\
r
c = — sin nt, C = V^n sin ( n^ 4- ^ ) •
ELECTRICAL EXAMPLES. 241
171. A circuit with resistance, self-induction and
capacity (fig. 86) has the alternating voltage V= Fosin z*^
established at its ends ; what is the current ?
V
Answer, G= -—, and by Art. 167
R + Ld + ^^
r- ^'^'^ ^^ V *
l + RK.e + LK.e^ {l-LK7i') + RKe
C = : Sin nt +K — tan ^ - — ^ rr » •
^{l-LKnJ + R^KHi-' V 2 l^-LKnV
The earnest student will take numbers and find out by
much numerical trial what this means. If he were only to
work this one example, he would discover that he now has a
weapon to solve a problem in a few lines which some writers
solve in a great many pages, using the most involved mathe-
matical expressions, very troublesome, if not impossible, to
follow in their physical meaning. Here the physical mean-
ing of every step will soon become easy to understand.
' Numerical Exercise. Take Fo = 1414 volts, K=l micro-
farad or 10-«, R = 100 ohms and n = 1000, and we find the
following effects produced by altering L. We give the
following table and the curves in fig. 87 :
Fig. 86.
A BCD shows how the current increases slowly at first from
A where Z = as X is increased, and then it increases more
rapidly, reaching a maximum when Z= 1 Henry and diminish-
ing again exactly in the way in which it increased. EFG
shows the lead which at X = 1 changes rather rapidly to a
lag. The maximum current (when LKn^=l) is the same as
if we had no condenser and no self-induction, as if we had a
mere non-inductive resistance R. It is interesting to note
in the electric analogue of Art. 160 that this LKn^=l is the
relation which would hold between L, K and n (neglecting the
P. 16
242
CALCULUS FOR ENGINEERS.
small resistance term) if the condenser were sending surging
currents through the circuit R, L, connecting its two coatings.
/y, ill
Henries.
0-1
0-2
0-8
0-4
0-5
0-6
0-7
0-8
0-9
0-95
0-975
100
1025
Effective
Lead of
Henries.
Effective
ciuTent, in
current, in
current, in
amperes.
degrees.
amperes.
0-995
84-28
1-05
8-944
1110
83-67
11
7071
1-240
82-87
1-2
4-472
1-414
81-87
1-3
3-162
1-644
80-53
1-4
2425 1
1-961
78-67
1-5
1-961 1
2-425
75-97
1-6
1-644 1
3-162
71-57
1-7
1-414 !
4-472
63-43
1-8
1-240 !
7071
450
1-9
1-110
8-944
26-57
20
0-995
9-701
1403
2-5
0-665
1000
30
0499
9-701
-14-03
Lead of
current, in
degrees.
-26-57
-45-0
-63-43
-71-57
-75-97
-78-67
-80-53
-81-87
-82-87
-83-67
-84-28
-86-18
-87-13
10
T~
~~~
— 1
-f
■"s.
\
'»J
\
s'
A
\
K
\
„5
8 ^
B
i ^
p
/
V
\
iu 3
/
f
V
\
v^
/
^^.^
V
Q-
'*"
—
-«
-=
-2 '4
100 o
•6 -8 1-0 1-2 1-4 1-6 1-8 2-0 22 2-4 2-6 2-8 S'O
8ELF-1NDUCI10N IN HENRIES.
1003
Fig. 87.
Experimenting with numbers as we have done in this example
is much cheaper and much more conclusive in preliminary
work on a new problem, than experimenting with alternators,
coils and condensers.
IDLE CURRENTS OF TRANSFORMERS. 243
172. Even if a transformer has its secondary open there
is power being wasted in hysteresis and eddy currents, and the
effect is not very different from what we should have if there
was no such internal loss, but if there was a small load on.
Assume, however, no load. Find the effect of a con-
denser shunt in supplying the ''Idle Current."
The current to an unloaded transformer, consists of the
fundamental term of the same
frequency as the primary voltage,
and other terms of three and five
times the frequency, manufactured
by the iron in a curious way.
With these "other terms" the con-
denser has nothing to do ; it cannot
disguise them in any way ; the total
current always contains them. We shall not speak of them,
as they may be imagined added on, and this saves trouble,
for if the fundamental term only is considered we may
imagine the permeability constant ; that is, that the primary
circuit of an unloaded transformer has simply a constant
self-induction.
In fact between the ends of a coil (fig. 88) which has
resistance R and self-induction L, place a condenser of capacity
K. Let the voltage between the terminals, be F= Fo sin nt.
Let G be the instantaneous current through the coil and
let c be the current through the condenser, then + c is the
current supplied to the system.
Fo sin nt
^C*c
V
{
p
o
C-.c'-
Fig. 88.
Now G:
R+LO
and c = Fo sin ??^ -7- -^^ , or c = Kn Fo cos nt,
0^o=[^^^Ke)v.sinnt
_ 1 + RKd -H LKd' ^_ l-LK n' + RK.e y
R + LO R + Ld
by our rule of Art. 167.
It is quite easy to write out by Art. 167 the full value of
16—2
244
CALCULUS FOR ENGINEERS.
+ c, but as we are not concerned now with the lag or lead,
we shall only state the amplitude. It is evidently
"V R' + JJn^
and the effective value of (7 + c (what an ammeter would
give as the measure of the cuiTent), is this divided by \/2.
Observe that C + c is least when
(Note that if L is in Henries and n = 27r x frequency (so
that in practice n = about GOO), K is in farads. Now even a
condenser of J microfarad or J x 10"" farad costs a number of
pounds sterling. We have known an unpractical man to
suggest the practical use of a condenser that would have cost
millions of pounds sterling.)
When this is the case, the effective current C + c, is
R/JR^ + L'^n- times the effective value of G.
The student ought to take a numerical case. Thus in an
actual Hedgehog Transformer we have found iJ = 24 ohms,
L = 6'23 Henries 7i = 509, corresponding to a frequency o^
about 81*1 per second. The effective voltage, or Fo-rV2
is 2400 volts. In fig. 89 we show the effective current cal-
culated for various values of K. The current curve ABCD is
a hyperbola which is undistinguishable except just at the
•9
2-8
i:
h
1
■""■
—
-%
^^
~
/
r-
vo
'8
\
/
•6
1
s.
1
/
D
s
A
i
/
\
',*
i/
\
%
4
r
\
ll
/
i
-6
s.
1
/
B^
s.
;
/
J
\
A
...
...
—'
—
._
...
-10
•* -5 -6 -7 -8 -9 10 M
CAPACiTV in MirROfARAOS
•2 1-3 1-4 \-h
Fig. 89.
vertex, from two straight lines. The total current is a
minimum when K — Lj(R? -^ LSi^)^ in this case 618 micro-
ELECTRICAL EXAMI»LES. 245
farad ; and the effect of the condenser has been to diminish
the total cun-ent in the ratio of the resistance to the im-
pedance. It is interesting on the curve to note how the
great lag changes very suddenly into a great lead.
173. If currents are steady and if points A and £ are
connected by parallel resistances 7\, 7\, r^y if V is the
voltage between A and B, and if the three currents are
Ci, C2, C3, and if the whole current is (7; then
_F _7 _F
In fact the three parallel conductors act like a conductance
1 1 IN
- + -■ + -
Also if G is known, then
C
111,
i\ r^ Vs
Now let there be a self-induction I and a condenser of
capacity k in each branch, and we have exactly the same
instantaneous formulae if, for any value of r, we insert
The algebraic expressions are unwieldy, and hence nu-
merical examples ought to be taken up by students.
174. Two circuits in parallel. They have resistances
Vi and ?-2 and self-inductions l^
and I2 . how does a total current ASL^i^i&SliSiSi
C divide itself between them ?
"^ISlSUUlSlSlMSi
If the current were a con-
tinuous current, Cj (fig. 90) in
the branch 7\ would be " p- ' qq
246 CALCULUS FOR ENGINEERS.
TT V • r.+ LO ,^
Jtieiice it IS now Ci = -. rz~n ^.
If a= Oo sin nt, then by Art. 167
'^'"''"Vcn + ny + i/i + Lyrr
sin «^ + tan^ — tan^ -^^
In the last case suppose that for some instrumental
purpose we wish to use a branch part of C, but with a
lead. We arrange that tan~^ tan~^ — ^^ shall be
7*2 ?•! + ?•,
equal to the required lead, and we use the current in the
branch r^ for our purpose.
175. Condenser annulling effects of self-induction.
When the voltage between points A and B follows
any law w^hatever^ and we wish the current flowing into
Y
A and out at B to be exactly -^ , whatever V may be, and
when we have already between A and B a coil of resistance
R and self-induction L, show how to arrange a condenser
shunt to effect our object.
Connect A and 5 by a circuit containing a resistance ?•,
self-induction I and condenser of capacity K, as in fig. 91.
The total current is evidently
V V
r 1 k '^ I ^ .
M-hLd^ ^j. 1 '
r^ld + j^^
AJUMilfiJUMiU
R L or, bringing all to a common
Fig. 91. denominator and arranging
terms, it is
1 + e (vK + RK) -^e^-{lK + LK) y
R+e (RrK -\-L)+&' {RIK -I- LrK) + LIKO'
Observe that as V may be any function whatsoever of the
time Ave camiot simplify this operator as we did those of
ELECTRICAL EXAMPLES. 247
Art. 167. Now we wish the effect of the operation to be the
same as p F! Equating and clearing of fractions we see that
R + e (BrK+ B?K) + &" (IilK+ RLK)
must be identical with
jR + (9 {RrK + X) + (9^ {RIK + LrK) + LIKO'.
As V may be any function whatsoever of the time, the
operations are not equivalent unless LIK = ; that is, l—O)
so there must be no self-induction in the condenser circuit,
RrK + R^K = RrK + L ; that is, i(^= 4 ;
RIK + RLK = RIK + LrK ; that is, R = r;
so the resistance in the condenser circuit must be equal to
that in the other.
In fact we must shunt the circuit R + L^ by a
1 Ii
condenser circuit R + ==3 where K = =^ .
176. If in the last case F= Fq sin nt, the operator may
be simplified into
l-K(l-\-L)n'-^K(r+R)e
R-K(Rl + rL) n^ + (9 {RrK + L - LlKn'} '
, . R-K(Rl-\- rL) n^ RrK + L- LlKn'
^^^'* l-K{l-\-L)n' ~ K(r + R)
then although the adjustment alters when frequency alters,
we have for a fixed value of n the current flowing in at A and
out at B proportional to V and without any lag. If R — r
Y
the current is equal to ^ .
177. To explain why the effective voltage is some-
times less between the mains at a place D, fig. 92, than
at a place B further away from the generator. This
is usually due to a distributed capacity (about J microfarad
per mile is usual) in the mains. We may consider a dis-
tributed capacity later ; at present assume one condenser of
248 CALCULUS FOR ENGINEERS.
capacity K between the mains at B. Let the non-inductive
resistance, say of lamps, between
000000/ ? *^® mains at B be r. Let the
jn resistance and self-induction of
Kfrnlr the mains between D and B be
If R and L. Let v be the voltage
— X ^^ ^» ^^^ ^ the current from D
Fig. 92. to B.
The current into the condenser is v -^ ifn^^ ^^^'•
V
The current through ?- is - , so that
O=(if0 + i). (1).
The cZro/) of voltage between D and 5 is
{R + X^) or (jK + LO) (kB + ^) t;,
or 1^ 4- (rK + ^) (9 + LKe\ V.
Now if V = t;o sin nt, the fZrop is
The voltage at D is the drop plus v, or
1 b A-f ir7 square of effective voltage at D
square of effective voltage at B
= {l + ^-LKn^\{RK + ^^'n\
and there are values of the constants for which this is less
than 1. As a numerical example take
r= 10, R = '\, K=lx 10-^ w = 1000,
and let L change from to 05, "01, '02, '03 &c.
TWO CIRCUITS. 249
The student will find no difficulty in considering this
problem when r + ^^ is used instead of r in (1) ; that is, when
not merely lamps are being fed beyond B, but also coils
having self-induction.
Most general case of Two Coils.
178. Let there be a coil, fig. 93, with electromotive force
E, resistance i2, self-induction L,
capacity K\ and another with e, r, RLC m ric
I, h Let the mutual induction f § § I
be m. Em § fe |o«
Using then R for R -\- Ld ■\- -^^, I |
I r for r -{-
the equations are
and r for r -1-/(9 4--^^^. ^^^- '^^•
E = RG + m6c,]
e = meC + rc j ^^'
Notice hoAV important it is for a student not to trouble
himself about the signs of and c &c. until he obtains his
answers.
From these we find
Re — mOE ,^.
'=w^.^^ ^^>'
^ rE-mOe ,„.
^^Rr-m^e' ^^^-
We can now substitute for i2, r, E and e their values and
obtain the currents.
Observe that E may be a voltage established at the
terminals of part of a circuit, and then R is only between
these terminals.
The following exercises are examples of this general case.
There are a gi-eat. many other examples in which mutual
induction comes in.
179. Let two circuits (fig. 94), with self-inductions,
250 CALCULUS FOR ENGINEERS.
be in parallel^ with mutual induction m between
them.
umsmsmsu
1 r c
Fig. 94.
(1) At their terminals let v = Vo sin nt ; (2) of last exercise
becomes c = fs iTTv. v- Or, chanerinof R into R-\-L6 and r
into r 4- 16,
R + {L-m)e
^""[Rr- (LI - m'') n^) + (Zr + IR) 6 ^*
(2) How does a current A sin nt divide itself between two
such circuits ? Since -p^ = ^ we can find at once -^
G r-mO G+c
A ^ A R + {L-m)e
and >. -— . Answer : c = Tr, x ^Vr r—7i — r^ operatmsf
G + c {R + r)-{-{L+l-2m)0 ^ ^
on -4 sin nt.
We think it is hardly necessary to work such examples
out more fully for students, as, to complete the answers they
have the rule in Art. 167.
180. In the above example, imagine each of the
circuits to have also a mutual induction with the com-
pound circuit. We shall use new letters as shown in fig. 95.
^
r
?
c,lir,
viznnnfg W ^ c+Co=C
c,Lr„
Fig. 95.
If V is the potential difference between the ends of the
two circuits which are in parallel. Using r, /jl and m to stand
for r 4- W, fiO and mO,
V = ?-iCi + fiCi + nh (ci + Co).
ROTATING FIELD. , 251
Hence the equations are
V = (7\ + 71h) Ci + (/X + nil) Ca,
V = (ft + m^) Ci + (r., + mg) c.,,
^. _ (r, + m,)-(/^ + mO ^
' (ri + mi)(r2 + 7?i2)-(/A + mi)(/A + m.2) *"^ ^'
with a similar expression for Cg.
Also a total current C divides itself in the following way
^ 7-1+ ro — 2//- ^ '^'
If we write these out in full, we have exceedingly pretty
problems to study, and our study might perhaps be helped by
taking numerical values for some of the quantities. If we care
to introduce condensers, we need only write ^ + ^^ + r^ with
proper affixes, instead of each ?•; fi becomes fiO and m
becomes mO.
To what extent may we make some of the ma negative?
I have not considered this fully, but some student ought to
try various values and afterwards verify his results with
actual coils. Taking (2) without condensers
_ f\+ (h+ fn-2 — 1^ — '^^h) p
181. Rotating Field. Current passes through a coil
wound on a non-conducting bobbin ; the same current
passes through a coil wound on a conducting bobbin. The
coils are at right angles and have no mutual induction ;
find the nature of the fields which are at right angles at the
centre of the two bobbins. Let the numbers of turns be rii
and 71.2 • Instead of a conducting bobbin imagine a coil closed
on itself of resistance 7\, and ng turns and current c. For
simplicity, suppose all three mean radii the same, and the
coils 7i2 and % well intermingled. One field F^ is propor-
tional to iiiG per square centimetre, call it iiiC The other F,
is proportional, or let us say equal to,??2(7 + ^gC per square cm.
Take the total induction / through each of the coils as
252 CALCULUS FOR ENGINEERS.
proportional to the intensity of field at its centre, say b times.
Then for the third coil, we have
= nc + risdl or = rgC + bn-iO {n.C -\- n^c),
bi^n-iOC
so that — c =
and hence F., = v..G
rs-hbn,^e'
\-^bn^^e n + bn^^e'
If then G= Cq sin qty
Fi = 71jGq sin qt,
■^■2= 2* = — , sin 7^-tan-i q):
Art. 126, shows the nature of the rotating field. We
can assure the student that he may obtain an excellent
j-otating field in this way.
It is evident that bn-i really means the self-induction of the
bn^ . .
third coil, and — - means its time constant. A coil of one
turn, — that is, a conducting bobbin, will have a greater time
constant than any coil of more than one turn wound in the
same volume. It is evident that if the bobbin is made large
enough in dimensions, we can for a given frequency have an
almost uniform and uniformly rotating field by making
n^^b— q=nj.
This is one of a great number of examples which we
might give to illustrate the usefulness of our sign of opera-
tion 6.
182. In Art. 1 78 let E= V the primary voltage of a trans-
former, the primary circuit having internal resistance B and
self-induction L ; let the secondary have no independent
E.M.F. in it; let its internal resistance be 7\ and self-induction
I and let it have an outside non-inductive resistance p, of
lamps. Let the voltage at the secondary terminals be i^ = cp.
TRANSFORMERS.
253
Then in (1), (2) and (3) Art. 178, let F^V, e=0; instead of R
use R-\-Ld. Instead of r use r + Id which is really i\-{- p + W,
....(1).
(7 =
Br + (Rl -trL)e + {LI - m"") ^
(r-^W)V
(2).
Rr + (Rl + rL)e+(Ll-m^)d' ""
Note that the second equation of (1) Art. 178 is
0==meG + (r + W)c (2)*
I. From (2)*, if C = Co6**, c =
r + la
— c
1)
ma
r + la
If r is small compared with la
— c
u
m
I '
c
1
"o
1
Rss
r
V
•o-
I
1
F^
Fi^. 96.
II. If C = Cq sin qt, again using (2)*
Wr^-\'Pq'' V 2 rJ
1
Hence
effective c
effective G
V^ + £'
Except when the load on the secondary is less than it
ever is usually in practice, r is insignificant compared with
Iq (a practical example ought to be tried to test this) and we
may take
C=Gq sin qt.
c= Gg-j- sin (qt — tt),
or
-c
C
m
T
.(3>
254 CALCULUS FOR ENGINEERS.
It may become important in some application to re-
member that the ratio of the instantaneous values of — c and
G is that of
r sin qt + Iq cos qt to mq cos qt,
and this sometimes is oo . t
Returning to (1). Let LI — m^ (this is the condition
called no magnetic leakage) and let Rr be negligible. In
any practical case, Ri' is found to be negligible even when r
is so great as to be several times the resistance of only one
lamp, t
'f'^- -^ = B^Z W'
SO that — c is a faithful copy of F as a function of the time.
C is so also.
If N and n are the numbei*s of windings of the two coils
on the same iron,
m : L : l^Nn : N^ : 71^ (5),
N
so that — c = ^ (6);
that is, the secondary current is the same as if the trans-
formed voltage ( — 7^ ^ ) acted in the secondary circuit, but
as if an extra resistance were introduced which I call the
transformed primary resistance IR-^).
If the volumes of the two coils were equal, and if the
volumes of their insulations were equal, R ^^ would be equal
to Vy the internal resistance of the secondary. Assume it
so and then
-^=2;tt7 ^'^^
DROP IN VOLTAGE. 255
JSf
also /)c or v = — (8).
P
As Vi is usually small compared with p,
and — * is called the drop in the secondary voltage
due to load.
^2 IP
As — = P, the power given to lamps ; - = -^ and the
2r
fractional drop is — ^ P and is proportional to the Power, or
to the number of lamps which are in circuit.
183. The above results may be obtained in another way.
Let / be the induction, and let it be the same in both
coils. Here again we assume no magnetic leakage,
V^RC-^NOI (1),
O^rc + ndl (2).
Multiplying each equation by its N or n and dividing by its
^ or r and adding
NV . (m n\ ^^
-^ = ^ + U+-)^^ (•^)'
where A — NC + nc, and is called the current turns.
Now when we know the nature of the magnetic circuit,
that is, the nature of the iron and its section, a square centi-
metres, and the average length X centimetres of the magnetic
circuit, we know the relationship between A and /. I
have gone carefully into this matter and find that whatever
be the nature of the periodic law for A, so long as the
frequency and sizes of iron &c. are what they usually are in
practice, the term A is utterly insignificant in (3). Reject-
ing it we find
'^ SZhk '^^--^^^ '"' "^'"^ ""^^' •••^*^-
256 CALCULUS FOR ENGINEERS.
Thus, in a certain 15 00- watt transformer, J? =27 ohms,
N = 460 turns, internal part of r = '067 ohms, ii = 24 turns,
effective V is 2000 volts or F= 2828 sin qt where q = 600 say,
a= 360, \ = 31. When there is no load r = x ; on full load
r = nearly 7 ohms.
We have called R -^^ the transformed resistance of the
. / 24 \2
primary. It is in this case 27 ( .^^1 or '073 ohms.
If the primary and secondary volumes of copperthad been
equal, no doubt this would have been more nearly identical
with '067, the internal resistance of the secondary.
^r,.— or — '- is the fractional drop in / from what it is at
no load. When at full load r* = 7 ohms the fractional drop is
greatest, and it is only 1 per cent, in this case. Because of
its smallness we took a fractional increase of the denominator
as the same fractional diminution of the numerator of (4).
Consider / at its greatest, that is, at no load ; -^V is the
2828
d
integral of F or — -^^^ cos 600^. So that the amplitude
. 2828 ^^^
° ^^600x460 •
Multiply this, the maximum value of / in Webers, by 10*
to obtain c. a s. units, and divide by a = 360, and we find
2856 c. G. s. units of induction per sq. cm. in the iron, as the
maximum in this transformer every cycle.
61 being ^ F / f 1 + vfa — ) > we have from (2) the same
value of — re that we had before in (6) of Art. 183.
184. Returning to (7) of Art. 182. Let us suppose that
there is magnetic leakage and that Vx is really r^ + VO.
If one really goes into the matter it will be seen that this
is what we mean by magnetic leakage. Then we must
divide by
p + 2n + 21' By
instead of /> -f 2ri. In fact our old answer must be divided by
1+ 2r .
MAGNETIC LEAKAGE. 257
or neglecting 2?'i as not very important in this connection ;
2^
our old answer must be divided by 1 -\ 6. This means
. . P
that the old amplitude of v must be divided by
/-
1 H ^ or 1 H ^,— nearly,
^ P"
if the leakage is small, and there is a lag produced of the amount
tan-i — ^ . We must remember that a is 27r/* if f is the
frequency. We saw that P, the power given to the lamps, is
inversely proportional to p, so we see that the Aractional
2r P
drop due to mere resistances is — \- , the fractional drop
due to magnetic leakage is la^pF", and the lag due to
magnetic leakage is an angle of afP radians where a is a
constant which depends upon the amount of leakage, and /
is the frequency.
185. Only one thing need now be commented upon in
regard to Transformers. If Fis known, it has only to be inte-
grated and divided by N to get I. Multiply by 10^ and divide
by the cross-section of the iron in square centimetres, and we
know how /3, the induction per sq. cm. in the iron, alters with
the time. The experimentally obtained /3, H curve for the
iron enables us to find for every value of ^ the corresponding
value of H,w[\d H multiplied by the length of the magnetic
cuTCUit in the iron gives the gaussage, or y^ x the ampere
turns A. Hence the law of variation of A is known, and if
there is no secondary current, we have the law of the
primary current in an unloaded transformer or choking
coil. This last statement is, however, inaccurate, as one
never has a truly unloaded transformer, even when what is
usually called the secondary, has an infinite resistance.
186. Sir W. Grove's Problem ; the effect of a condenser
in the primary of an induction coil when using alternating
currents.
ABB, fig. 97, is the primary with electromotive force
E— Ed sin nt, resistance R and self-induction L. BA is h
P. IT
25S CALCULUS FOR ENGINEERS.
condenser of capacity K, and ?' is a non-inductive resistance
in parallel with the condenser. G
qq^q'q^j^ -? the current in the primary, has an
S JTI amplitude Co, say.
® Kli1j|r rpjjg condenser has the resist-
*^" It is quite easy to write out
the value of Cq when r and K have any finite values*.
But for our problem we suppose r=0 or else r=oo . When
r = 0, the resistance \^R-\-L6 and the current is EI{R + Z^),
n^-—^ (1).
When r=oc , the resistance is
R + Le+-^^ or j^ ,
or ^^ j^ , by Art. 167,
A rr.- £Jo'K'n^ Ej .
\Kn J
Now (2) is greatei' than (1) if 2KLn^ is greater than 1,
so that the primary current is increased by a condenser of
capacity greater than o;-— g. Again, there is a maximum
current if iT = y— ^ ; in this case the condenser completely
destroys the self-induction of the primary.
* When both r and K have finite values, the parallel resistances between
2S and A, together form a resistance rl{l + rKd), and the whole resistance of
the circuit for C is jR + L^ + - tptSO that
1 + rKd
(1 + rKd)EQ sin lit
{R + r- LrKii') + {RrK+L) d
02 = -
" ~ (E + r - LrKn^f + (RrK-^LYii^ '
and the lag of C is easily written.
ALTERNATORS IN SERIES. 259
187. Alternators in series. Let their e.:m.f. be ei and
e.2 and let C be the current through both. The powers exerted
are efi and 020. Now if
ei=E sin (nt+a) and e.2=Esm (nt — a),t ei +e.j=2j&cos a . sin?^^.
If I is the self-induction of each machine, r its internal
resistance, and if 2R is the outside resistance and if Pi and
P2 are the average powers developed in the two machines,
^ 2^cosa. sin?i^ Ecosa • / . , , In \
^ = -zTTTi — ^ ^tt = ,-' sm nt — tan~^ v,
2;^ + 2r-f-2P V(P4-r)2 + ^'^'' V i^ + r/
= ilf cos a sin {nt — e) say,
Pi = p/^ cos a . cos (a + e),
Po = ^Jlfii' cos a . cos (a — e).
Hence P2 is greater than Pi, and machine 2 is retarded
whilst machine 1 is accelerated ; hence a increases until
a = - , and when this is the case, cos a = 0, so that Pi = 0,
P2 = and the machines neutralize each other, producing no
current in the circuit. Alternators cannot therefore be
used in series unless their shafts are fastened together.
188. As we very often have to deal with circuits in
parallel we give the following general formula ; if the electro-
motive forces ^1, ^2 and e^, fig. 98, are constant,
v^e^ — c^Vi = e.2 - 0.^2 ^e-i-c^n ( 1 ),
and Ci + C2 + C3 = (2).
Given the values of e^, e._,, e^ and 7\, i\, r^ we easily find
the currents, because
, = (?i + ?. + ^V(i + i + l) (3).
Fig. 98.
17—2
200 CALCULUS FOR ENGINEERS.
Now if the es are not constant, we must use ?'i + liO, &c.,
instead of mere resistances.
189. Alternators in Parallel. Let two alternatoi-s, each
of resistance r and self-induction I, and with electromotive
forces, ei = E sin (iit + a), and e^ = E sin (nt — a), be coupled
up in parallel to a non-inductive circuit of resistance R.
What average electrical power will each of them create,
and will they tend to synchronism ? If Ci and e.2 were con-
stant or if I were 0, then v = ei — CiV = e., — cr = (ci + c^) R.
And hence
^^ = 2r^H^+3-^-^
fcH^+S"^^
^'~'2rR-¥r'
Now alter r to ?• 4- 16, because the e's are alternating.
The student will see that we may write
62 =61 (a — bO),
e^ = 62 (a -H hO),
where a^ -h 6W =1, a = cos 2a, 6/1 = sin 2a. Then
I
hi-^)-'Ki^')
with a similar expression for Ca in terms of e, except that 6
is made negative. If we write out (1) by the rule of Art. 167,
there is some such simplification as this : —
Let tan <f) = — ^— - — =^ and tan ^^ = ^— ^
^ 2Rr + r^ — Pn^ ^ R+r — aR
tan ylr^ = t,-; d •
^ R + r-aR
Then c, = if sin (ni -f a — <^ -}- i|ri),
C2 = il/ sin (nt — a — <j) -{- yfr^,
the angles <^, yjry, s^^ being all supposed to be between
and + 90^
ALTERNATORS IN PARALLEL. 261
The average powers are
P„_ = ME cos ((f>-^|r.,),
where M'= ^ .^^.. l^^""^ , ^J' E^.
Po COS (<l>-ylr^y
It' 11= oo we see that
, , In ^ . sin a , ,
tan 6 = — , tan -v/r, = , = — tan vr...
^ r ^ 1 — cos a ^ "
Hence P,^ cos(^-^ )
P. COS(«/)+>^i)
In this case it is obvious that Pi is greater than P2.
P
The author has not examined the general expression for p-
with great care, himself, but men who have studied it say
that it shows P, to be always greater than Po. Students
would do well to take values for i\ I, R and a and try for
themselves. If Pi is always greater, it means that the
leading alternator has more work to do, and it will tend to go
slower, and the lagging one tends to go more quickly, so that
there is a tendency to synchronism and hence alternators
will work in Parallel.
190. Struts. Consider a strut perfectly prismatic, of
homogeneous material, its own weight neglected, the resultant
force F at each end passing through the centre of each end.
Let ACB, fig. 99, show the centre line of the bent strut. Let
FQ = yhe the deflection at P where OQ = x. Let OA=OB = I.
y is supposed everywhere small in comparison with the length
21 of the strut.
Fy
Fy is the bending moment at P, and ^ is the curva-
ture there, if E is Young s modulus for the material and / is
262 CALCULUS FOR ENGINEERS.
the least moment of inertia of the cross section everywhere,
P, about a line through the centre of area of the sec-
tion. Then as in Art. 60 the curvature bemg - t~*
we have
II--^^ (1)
Now if the student tries he will find that, as in
the many cases where we have had and again shall
have this equation, (see Art. 119)
7/ = a cos X
\lm <^)
satisfies (1) whatever value a may have. When
a: = we see that y — a, so that the meaning of a is
known to us ; it is the deflection of the strut in the
middle. The student is instructed to follow carefully
99. the next step in our argument.
When A" = /, 2/ = 0. Hence
FY
a cos ^ a/ VTr = (3).
* Notice that when we choose to call -~ the curvature of a curve, if the
expression to which we put it equal is essentially positive, we must give such
a sign to -r^ as will make it also positive. Now if the slope of the curve of
fig. 99 be studied as we studied the curve of fig. 6, we shall find that
-r-^ is negative from a; = to a: = O^ , and as y is positive so that ^ is positive,
«re must use - —-^ on the right-hand side.
It will be found that the comjilete (see Arts, loi and 159) solution of any
such equation as (1) which may be written
ia y = ^ ^^^ nx + B sin nx
where A and B are arbitrary constants. A and B are chosen to suit the
particular problem which is being solved. In the present case it is evident
that, as ?/ = when x = l and also when x= - I,
0=iA cos 111 + B sin 111,
= A cos 111 - B sin 7il, so that B ia 0.
STRENGTH OF STRUTS. 263
Now how can this be true ? Either a = 0, or the cosine
is 0. Hence, if hending occurs, so that a has some value,
the cosine must he 0. Now if the cosine of an angle is the
angle must be ^ or -^ or -^ , &c. It is easy to see why we
confine our attention to » *.
Hence the condition that bending occurs is
is the load which will produce bending. This is called
Euler's law of strength. The load given by (4) will produce
either very little or very much bending equally well. It
is very easy to extend the theory to struts fixed at both ends
or fixed at one end and hinged at the other.
For equilibrium under exceedingly great bending, the
equation (1) is not correct, as -r~ is not equal to the curva-
ture when the curvature is great, but for all engineering
purposes it may be taken as correct.
191. We may take it that F given by (4), is the load
which will break a strut if it breaks by bending. If / is
the compressive stress which will produce rupture and
A is the area of cross section, the load /J. will break the
strut by direct crushing, and we must take the smaller
of the two answers. In fact we see that /A is to be
taken for short struts or for struts which are artificially*!"
protected from bending, and (4) is to be taken for long struts.
Now, even when great care is taken, we find that struts are
neither quite straight nor homogeneous, nor is it easy to
load them in the specified manner. Consequently when
loaded, they deflect with even small loads^ and the}^ break
with loads less than either /A or that given by (4).
* This gives the least value of W. The meaning of the other cases is
that y is assumed to be one or more times between x = and x = l, so that
the strut has points of inflexion.
t This casual remark contains the whole theory of struts such as are used
in the Forth Bridge.
264 CALCULUS FOR ENGINEERS.
Curiously enough, however, when struts of the same section
but of different lengths are tested, their breaking loads
follow, with a rough approximation to accuracy, some rule as
to length. Let us assume that as F=fA for short struts,
and what is given in (4) for long struts, then the formula
F = -a
.(5)
may be taken to be true for struts of all lengths, because it is
true both for short and for long ones. For if I is gieat
we may neglect 1 in the denominator, and our (5) is really (4) ;
again, when / is small, we may regard the denominator as
only 1 and so we have W=fA. We get in this way an
empirical formula which is found to be fairly right for all
struts. To put it in its usual form, let I — Ak^, k being the
least radius of gyration of the section about a line through
its centre of gravity, then
I±
I-
•(6),
where a is ^//Ett^ or rather / and a are numbers best
determined from actual experiments on struts.
If F does not act truly at the centre of each end, but at
the distance h from it, our end condition is that y = h when
00=^1. This will be found to explain why struts not perfectly
ti-uly loaded, break with a load less than what is given in (4).
Students who wish to pursue the subject are referred to
pages 464 and 513 of the Engineer for 1886, where initial
want of straightness of struts is also taken account of
192. Struts with Lateral Loads. We had better confine our
attention to a strut with hinged ends. If the lateral loads are such
that by themselves and the necessary lateral supporting forces, they
produce a bending moment which we shall call ^ (.r), then (1) Art. 190
becomes
Thus let a strut be uniformly loaded laterally, as by centrifugal force
or its own weight, and then <j> {a:)~^ w' {l — x)^ if w' is the lateral load
per unit length.
STRUTS LATERALLY LOADED. 265
We find it slightly more convenient to take <f) (x) = ^Wl cos ^x
where W is the total lateral load ; this is not a very different law.
Hence
dh/ F , Wl n ^
d^^+m^-^^W'''2i''=^ (1)-
We find here that
y="-r3 cos- r (2).
Observe that when i^=0 this gives the shape of the beam.
The deflexion in the middle is
.. = -i?^ (3),
and the greatest bending moment fi is
M = ^yi+iTf/,or
-i--^7(^^--) (^)-
If TF=0 and if fx has any value whatever, the denominator of (4)
must be 0. Putting it equal to 0, we have Euler's law for the strength
of struts which are so long that they bend before breaking. If Euler's
value of F be called U, or U=ETii^/4l^y (4) becomes
>^=i^''^T&F • '^)-
If Zc is the greatest distance of a point in the section from the
neutral line on the compressive side, or if I-^Zc=^Z, the least strength
modulus of the section, and A is the area of cross section, and if / is
the maximum compressive stress to which any part of the strut is
subjected, n F
Z'^ A-
Using this expression, if /3 stands for -r (that is Euler's Breaking
load per square inch of section), and if w stands for -r (the true break-
A
ing load per inch of section), then
(^-?)(^-f)=:^ («)•
This formula is not difficult to remember. From it lo may be found.
Example. Every point in an iron or steel coupling rod^ of length
26 inches, moves aljout a radius of r inches. Its section is rectangular,
266
CALCULUS FOR ENGINEERS.
d inches in the plane of the motion and b at right angles to this. We
may take W=lbdrn^'7-62940y in pounds, where 7i = number of revolu-
tions per minute. Take it as a strut hinged at both ends, for both
directions in which it may break.
1st. For bending in the direction in which there is no centrifugal
force where/ is ^^^
12'
Euler's rule gives
48^2
.(7).
Now we shall take this as the endlong load which will cause the
strut to break in the other way of bending also, so as to have it equally
ready to break both ways.
2nd. Bending in the direction in which bending is helped by
centrifugal force. Our w of (6) is the above quantity of (7) divided by
bdf or taking
iS'=3xlO",
b^
'?t'=6'17x^xlO«.
Taking the proof stress / for the steel used, as 20000 lb. per sq. inch
(remember to keep^ low, because of reversals of stress), and recollecting
the fact that / in this other direction is r-s^ , we have (6) becoming
b'
8-4 X 108 1 - 308
12
62
d^
)-
nH^r-T-d
(8).
Thus for example, if 6= 1, Z= 30, r= 12, the following depths c? inches,
ai'e right for the following speeds. It is well to assume d and calculate
n from (8).
<^ 1 I 1-5 2 2-5 3 4 6
0|205 277 327 368 437 545
Exercise. A round bar of steel, 1 inch in diameter, 8 feet long, or
Z = 48 inches. Take i^=1500 lb. Show that an endlong load only
sufficient of itself to produce a stress of 1910 lb. per sq. in., and a
bending moment which by itself would only produce a stress of 816 lb.
per sq. inch ; if both act together, produce a stress of 23190 lb. per
sq. inch.
For other interesting examples the student is refeiTed to The
Philosophical Magazine for March, 1892. t
CHAPTEE III
ACADEMIC EXERCISES.
193. In Chapter I. we dealt only with the differentia-
tion and integration of ic" and in Chapter II. with €«^ and
sin aw, and unless one is really intending to make a rather
complete study of the Calculus, nothing further is needed.
Our knowledge of those three functions is sufficient for
nearly every practical engineering purpose. It will be found,
indeed, that many of the examples given in this chapter might
have been given in Chapters I. and II. For the differen-
tiation and integration of functions in general, we should
have preferred to ask students to read the regular treatises,
skipping difficult parts in a first reading and afterwards
returning to these parts when there is the knowledge
which it is necessary to have before one can understand
them. If a student has no tutor to mark these difficult parts
for him, he will find them out for himself by trial.
By means of a few rules it is easy to become able to
differentiate any algebraic function of cc, and in spite of our
wish that students should read the regular treatises we are
weak enough to give these rules here. They are mainly used
to enable schoolboys to prepare for examinations and attain
facility in differentiation. These boys so seldom learn more
of this wonderful subject, and so rapidly lose the facility in
question, because they never have learnt really what -^
means, that we are apt with beginners to discourage much
practice in differentiation, and so err, possibly, as much as the
older teachers, but in another way. If, however, a man sees
clearly the object of his work, he ought to try to gain this
facility in differentiation and to retain it. The knack is
easily learnt, and in working the examples he will, at all
268 CALCULUS FOR ENGINEERS.
events, become more expert in manipulating algebraic and
trigonometric expressions, and such expertness is all-important
to the practical man.
In Chapters I. and II. we thought it very important
that students should graph several illustrations of
y = cwJ", y = ae^, y = a sin {hcc + c).
So also they ought to graph any new function which comes
before them. But we would again warn them that it is better
to have graphed a few very thoroughly, than to have a hazy
belief that one has graphed a great number.
The engineer discovers himself and his own powers in
the first problem of any kind that he is allowed to work out
completely by himself. The nature of the problem does not
matter; what does matter is the thoroughness with which
he works it out.
Graph y = tan ax. We assume that the student has
already graphed y = «€** sin nx.
194. If y =f(x), so that when a particular value of on is
chosen, y may be calculated ; let a new value of x be taken,
x+Bx, this enables us to calculate the corresponding value
ofy,
or y + By^f(x-{-Bx).
Now subtract and divide by Bx, and we find
By_ f{x-^Bx)--f(x )
Bx~ Bx ' ^^^•
We are here indicating, generally, what we must do with
any function, and what we have already done with our famous
three, and we see that our definition of dy/dx is, the
limiting value reached by (1) as Bx is juade smaller and
smaller without limit.
195. It is evident from this definition that the differential
coefficient of af(x), is a multiplied by the differential co-
efficient of fix), and it is easy to show that the differential
coefficient of a sum of functions is equal to the sum of the
differential coefficients of each. In some of the examples of
Chapter I. we have assumed this without proof
DIFFERENTIATION OF A PRODUCT. 269
We may put the proof in this form : —
Let y=u + v + w, the sum of three given functions of x.
Let X become os-{- hx,f\et u become ii + Su, v become v-{-Sv,
and w become w -f ^iv. It results that if ij becomes y + 5y, then
8y = Bit -^ hv + Biu,
- By Bit Bv Bw
and 5^^ = ^ + ?- + F~ .
ox ox ox ox
, . ,, ,. . dy du dv dw
and m the limit -r^ = ^- + t" + -? •
ax ax ax ax
196. Differential Coefficient of a Product of two Func-
tions.
Let y = uv where u and v are functions of x. When x
becomes x + Bx, let
y-\-By = {ii + Bi()(v + Bv) — uv •{- ^l . Bv + v . Bu ■{■ Bu . Bv.
Subtracting we find
By = a . Bv -^^ v . Bu -{■ Bu . Bv>
, By B^ Bu Bu ^
and f-^U'^-hv K- + ^'Bv.
ox ox ox ox
We now imagine Bx, and in consequence (for this is
always assumed in our work) Bti^ Bv and By to get smaller
du
and smaller without limit. Consequently, whatever -r*- may
be, -r- . Bv must in the limit become 0, and hence
ax
dy_ dv da
dx dx dx'
The student must translate this for himself into ordinary
language. It is in the same way easy to show, by writyig
uvto as uv X w, that if t/ = uvw then
dy dw , du dv
-r- = wv -7— + viv -,- + wit -J- "
ax ax ax ax
Illustrations. If y = 10a;'' then, directly, ^ = lOa^. But
we may write it y = 5^' x 2x^.
270 CALCULUS FOR ENGINEERS.
Our new rule gives
^ = 00^ (8^) + 20,-^ (15*'0 = ^O*-" + SOaf = 70af.
The student ought to manufacture other examples for
himself.
197. Differential Coefficient of a Quotient.
Let ?/ = - when u and v are functions of x.
Then y + oy= — —^ .
Subtract and we find
^ __ w 4- Sw u _v .8u — to.^v
"^ " v+Bv V v'^+v .Sv '
Bu Bv
hi/ __ 8x Bx
Bx" v'^ + v .Sv
Letting Bx get smaller and smaller without limit, v . Bv
becomes 0, and we have
du dv
dy dx dx
dx v^
Here again the student must translate the rule into
ordinary language, and he must get very well used indeed to
du
the idea that it is ^ -v" which comes first : —
ax
Denominator into differential coefficient of nume-
rator^ minus numerator into differential coefficient
of- denominator, divided by denominator squared.
A few illustrations ought to be manufactured. Thus
24a;^ . ,i ^ . t dy .^ ^
y = -^— IS really Sa?^, and -j- — ¥)oi^.
ox ax
By our rule, ■£= ^^ ^—^ = 40^.
d^^dy dz^ 271
dx dz ' dx'
The student ouffht to work a few like y = -^r— - = 5x~^ or
again y = — ^ = — 1^~^, and verify for himself.
198. If y is given as a function of z, and ^ is given as a
function of x, then it is easy to express y as a function of oj.
Thus if 2/ = 6 log (az^ 4- g) and ^ = c + c?^ + sin ex, then
2/ = 6 log {a (c-\- dx-\- sin eaj)^ + ^}.
Now under such circumstances, that is, y =/{z) and
2 = F (x), if for X we take x + 5^, and so calculate z + S^, and
with this same z -\-Bz we calculate y + %, then we can say
that our 8y is in consequence of our Bx, and
Bx Sz 8x ^ ^'
This is evidently true because we have taken care that the
two things written as Bz shall be the same thing. On this
supposition, that the two things written as Sz remain the
same however small they become, we see that the rule (1) is
true even when Sx is made smaller and smaller without
limit, and as we suppose that Bz also gets smaller and
smaller without limit,
dx"dz ' dx ^ ^*
This is such an enormously important proposition that
a student ought not to rest satisfied until he sees very
clearly that it is the case. For we must observe that the
symbol dz cannot stand by itself ; we know nothing of dz by
itself; we only know of the complete symbols dy/'dz or dzjdx.
We are very unwilling to plague a beginner, but it would
be fatal to his progress to pass over this matter too easily.
Therefore he ought to illustrate the law by a few examples.
Thus let y = az^ and z=hx\ As -^ = Saz^, -y- = 2hx, we have
^ dz dx
-r ' -T- = Qabz^x or Qab^x^. But by substitution, y = ab^af,
and if we differentiate directly we get the same answer. A
student ought to manufacture many examples for himsel£
.(3).
272 CALCULUS FOR ENGINEERS.
An ingenious student might illustrate (2) by means of three
curves, one connecting z and x, the other connecting z and y
and a third produced by measurements from the other two,
and by means of them show that for any value of x the slope
of the y, X curve is equal to the product of the slopes of the
other two. But in truth the method is too complex to be
instructive. By an extension of our reasoning we see that
dy _ dy dw du dv
dx dw ' du ' dv ' dx
199. It is a much easier matter to prove that
dv dx
:t^x-- = i (4),
dx dy ^ ^
by drawing a curve, because it is easy to see that -y- is the
cotangent of the angle of which -~ is the tangent.
Otherwise: — if by increasing i» by &c we obtain the
increment By of y, and if we take this same By, so foimd, we
ought to be able to find by calculation the very same Bx with
which we started. Hence
By Bx ^ _.
On this proviso, however small Bx may become, (5) is
true and therefore (4) is true.
200. To illustrate (2). If a gas engine indicator
diagram is taken, it is easy to find from it by applying
Art. 57, a diagram for h, the rate at which the stuff shows
that it is receiving heat in foot-pounds per unit change of
volume, on the assumption that it is a perfect gas receiving
heat from some furnace. (In truth it is its own furnace;
the heat comes from its own chemical energy.) Just as
dW
pressure is -j- , the rate at which work is done per unit
change of volume : so ^ is -=- . Observe that h is in the
dv
same units as p, and to draw the curve for h it is not necessary
to pay any attention to the scales for either p or v. They
ILLUSTRATIONS. 273
may be measured as inches on the diagram. We know of
no better exercise to bring home to a student the meaning
of a differential coefficient, than to take the indicator
diagram, enlarge it greatly, make out a table of many values
of j> and V, and find approximately ^ for each value of v.
This is better than by drawing tangents to the curve. Using
rlTT
these values, and having found the values of A or -p at
every place, suppose we want to find the rate per second at
which the stuff is receiving heat. If t represents time,
-rr — —r- . -77 , and hence it is only necessary to multiply h
, dv
dv
As -7T is represented by the velocity of the piston, and as
the motion of the piston is, as a first approximation, simple
harmonic, we describe a semicircle upon the distance on the
diagram which represents the stroke, and the ordinates of the
di)
semicircle represent -j- . We have therefore to multiply
every value of h by the corresponding ordinate of the semi-
circle, and we obtain, to a scale easily determined, the
diagram which shows at every instant -^ .
Havins: seen that -^^ = -^ . -^ and that — = 1^-7-, we
° dx dv dx dx dy
shall often treat dx or dy as if it were a real algebraic
quantity, recollecting however that although dy or dx may
appear by itself in an expression, it is usually only for
facility in writing that it so appears ; thus the expression
M.dx + N.dy = (1),
may appear, where M and N are functions of x and y ; but
this really stands for M+N-^ = (2).
Again, if y = ax^, we may write
dy = 2ax . dx (3),
P. 18
274 CALCULUS FOR ENGINEERS.
but this only stands for ~~ = 2ax (4).
Our main reason for doing it is this, that if we wish to
integrate (3) we have only to write in the symbol /, whereas,
if we wish to integrate (4) we must describe the process in
words, and yet the two processes are really the same. We have
already used dx and dy in this way in Chap. I.
Mere mathematical illustrations of Art. 198 may be manu-
factured in plenty. But satisfying food for thought on the
subject, is not so easy to find. The law is true; it is not
difficult to prove it ; but the student needs to make the law
part of his mental machinery, and this needs more than
academic 'proof.'
Let us now use these principles.
201. Let y = log x ; this statement is exactly the same as
X = e'^ Hence -,- — e^ — x and -f- = - . We used the idea
ay ax x
that the integral of x~^ is logd?, in Chap. I., without proof.
It is the exceptional case of the integration of «".
202. If the differential coefficient of sin a? is known to
be cos X, find the differential coefficient of sin ax,
y = sin ax = sin u\iu = ax,
dy J du
-Y- = cos u and -r = a,
du dx
xi X ^y dii du
so that -!i r= -^ . — = cos uxa — a cos ax.
dx du dx
Find the differential coefficient of 3/ = cos ax, knowing
that the differential coefficient of sin x is cos a?,
y = cos ax = Bm.lax + -^\ — sin u say, where ;7- = «,
dy dy du / 7r\ .
^ — ~l '~^ — ^^s u X a = a cos {ax-{--^j — — asinaa.
FUNDAMENTAL CASES. 275
203. Let 2/ = log(x+a).
Assume x-\-a= ti, or y = loer u, then -,- = 1 and -j^ = - ,
dy _dy dii _\ _ 1
rfa; dii ' dx u x+a'
204. y = tan x. Treat this as a quotient, y = ,
cos X
dy __ cos a? . cos a; — sin a; (— sin x) _ 1
(ia? ~ cos'^ X cos** a^ '
The student ought to work this example in a direct
manner also.
205. y = cot X. We now have choice of many methods.
cos X
Treat this as a quotient, y = -. ,
^ ^ sm X
dy _ sin x (— sin x) — cos x (cos x) _ 1
c^a; ~ sin^ x sin^ a? '
or we might have treated it in this way,
y = u~^ if ii = taniJ7,
ir'' X -T- = — i^~- X
COS^'fl?
1 1
cosec^ ic.
dx dx cos* fl?
11 1
tan^a; cos'^a? sin^a?
206. Let y = sin aar^, say y = sin t^, and it = aa-^.
Then -r- = 2aa7,
and -=^ = cos u,
du
so that ~ = cos ^6 X 2ax = 2aa7 cos ax^,
dx
Let 3/ = 6*""^*, say y = e", and w = a sin a?, so that
dy „ c?u
^ = e", J- = a cos a?,
so that ~- = e^a cos x, or a cos a? . e**'^"*
18—2
276 CALCULUS FOR ENGINEERS.-
207. y — sec x. We may either treat this as a quotient,
or as follows ; y = (cos ijc)~^ = u~^ if u = cos x.
du . dy dy du ,/ • x
-Y- == — sin a?, j^ = -/ . J— = — u-^ (— sm a?)
ax ax du ax ^ '
sin 33
sec X . tan x.
cos'*;??
208. In Art. 1 1 the equation to the cycloid was given in
terms of an auxiliary angle <!>; x = a<t> — a sin (f), y = a — a cos (f).
Find -— and ~ at any point.
Here ^^^ J±^^^^
dx d(j) ' dx d(f> ' d(j)
. ,,. ,. sin</>
= a sm d>(a — a cos 0) = = ^~ .
^'^ ^^ l--cos<^
Also '^ = - (^] = — (^A X ^
dx^ dx \dx) d<t> \dx) dx
(1 — cos <f)) cos <f) — sin d) (sin 6) .
= ^^ ^^ — ^ JLX2 ^ -^ (a - a cos </))
(1 — cos <^)'* ^^
-1 ^_a
'"a(l-cos</))2'~ 2/'*-
209. If a^ + 2/'*=a'* (1),
If we want -,- in terms of x only we must find y from (1)
and use it in (2). But for a great many purposes (2) is
useful as it stands.
x^ ifi
In the same way, if — + ^ = 1,
?f 4.?^^=0 or ^--^^
a?'^ V" dx dx~ a'y'
Agam, if --2-n=l» :r ="~2--
• FUNDAMENTAL CASES. 2V7
Also if u;^ 4- 2/^ = a^j
If y = i^-^i sin 2a; + yg sin 4.x;, -j- = cos* a).
If 2/ = i t^^^^ *' + t^^ ^' 1^>~ ^^^*' ^'
Let // = V^2 + a^ = u^ if m = a-^ + ct^, ^— = 2a;, so that
— - = 4?4-i X 2x or V
210. Let y = sin~^ x. In words, y is the angle whose
sine is x. Hence x = sin y,
dx
dy
cos y — 's/i— sin'-^ y = Vl — a;^.
Hence
dy
dx ^/l-.a^'
We have extracted a square root, and our answer may
be + or — . We must give to -^ the sign of cos y.
211. Similarly if y = cos~^ x,
dy 1
dx Vl-x^'
212. Let y — tan~^ x, so that x = tan y,
~ = —= 1 + tan^y = 1 +x\
dy cos^ y '^
213. Similarly if 2/ = cot"^ x, then -f-= — ^ — -5 ■
278 CALCULUS FOR ENGINEERS.
214. It will be seen that (2) and (4) of Arts. 198 and
199 give us power to differentiate any ordinary expression,
and students ought to work many examples. They ought
to verify the list of integrals given at the end of the book.
A student ought to keep by him a very complete list of
integrals. He cannot hope to remember them all. Some-
times it is advisable to take logarithms of both sides before
differentiating, as in the following case :
y = x^. Here logy = ^ log ;r,
1 dy 1 ,
y ' dx X
| = .^(l + log,.).t
215. In the following examples, letters like x, y, z^ v, w,
0, &c. are used for the variables ; letters like a, b, c, m, n, &c.
are supposed constant. A student gets too familiar with x
and y. Let him occasionally change x into ^ or ^ or v, and
change ?/ also, before beginning to differentiate. Ho ought
to test the answer of every integral by differentiation.
List of Fundamental Cases.
-7- x'' = nx''-\ L'"» . dx = *•»«+! ;
dx J m 4- 1
^(log^) = i, jl.dx^logx;
d , . . [ , 1 .
^r-(sm mx)= m cos mx, j cos 'tnx .dx— — sm mx ;
dx J m
d . . . r 7 1
-J- ( cos mx)= — m sin mx, \ sm mx ,dx — cos nix ;
dx^ ^ J m '
., . a [ dx 1
(tan ax) = — ^ — , — - — = ~ tan ax ;
cos^a./; j cos'' fr.r a
dx
d , . . a [ dx 1 ^
-^ (cot ax) = — -^— — , -^-T— - = cot ax \
dx^ ^vo^ax J^m^ax a
ADVICE TO STUDENTS. 279
Sin"
d , . , . 1 { dx
-j- (sin ^x) = . , , =
d .. , . 1 f dx 1 ^ , ^
Many integrals that at first sight look different are really
those given above. Even the use of \/~ or ^~ instead of
the numerical symbol of power or root, disguises a function
to a beginner. Thus
1 . 1 _,
3 ■- IS -X*,
ci\/x (I
and its integral is
1 / x-^+' \ 3 3
or z:r-X\
a V- i + 1/ 2a
216. In some of the following integrals certain substitu-
tions are suggested. The student must not be discouraged if
he cannot see why these are suggested ; these suggestions are
the outcome of, perhaps, weeks of mental effort by some
dead and gone mathematician. Indeed, some of them are no
better than this, that we are told the answer and are merely
asked to test if it is right by differentiation.
Just here, in learning the knack of differentiation and
integi'ation, the student who has a tutor for a few lessons has
a gi'eat advantage over a student who works by himself from
a book. Nevertheless the hardworking student who has no
tutorial help has some advantages ; what he learns he learns
well and does not forget. The man who walks through
England has some advantages over the man who only takes
railway journeys. In learning to bicycle, I think that on
the whole, it is better to be held on for the first few days ;
learning the knack of differentiation and integration is not
unlike learning to bicycle.
Exercises and Examples.
1. i/ = xhgx, ^=l+%^'-
280 CALCULUS FOR ENGINEERS.
-^ ' dx 2^Jx
3. y = losf (tan a?), -t-= - — ^r- -
'^ ^^ ^' dx sin 2a;
, 1 — tan X du ... .
4. y — , -T-— — (sin X 4- cos x\
•^ sec a; eta; ^
1/1 \ ^y 1
o. y = log(log<.), ^ = ^1^.
tto; — ^— —
6. a; = 6«< sin 6^, "^ = ^/«' + ^' • ^* «" (6^ + c),
where tan c = - .
a
We here use the simplification of Art. 116. The student will
note that by page 235, Q (standing for djdt), operating n times
upon sin ht^ multiplies its amplitude by 6" and gives a lead
of n right angles. He now sees that if operates n times
upon e"* sin ht, it multiplies by {it? 4- 6^)"^^ and produces a
lead nc.
Thus ^1 = (a= + I/) €«« sin (6^ + 2c) ;
and ^^ = (a-^ + 62)^€«' sin {ht + 3c).
^, fl-e dp 1
i^ = 2tan-yj-^^, J = -^7j^
8. 2/ = log (6^ + 6-0, :£
c?i/ €^ — e"
cZ^ e^ + e-^ '
9- y = ^^^, t=t^/^.
c?a;
10. y = aa^+hx + c, -^^ = 2aa; + &.
12. p^cv-''\ ^=-l-37ct;-'^'
13
EXERCISES. 281
dv
14. \av'''dv^--^v-''\
lav-'-
15. Haf + bt + c)dt=: ^at^ + ^bf + ct + g.
16. I V^ . (^« = \x^ . c?a; = §a-^
17. /f i«/r3.<^e=JJ^ = -i.
19 ! dx 1 r dx 1 1 _, a; ^
"^' 2r^
V m
1
20. I v^oT+l; . dv. Here let a + 1; = 2/ so that dv = c?y, and
we have jy^ . dy = ly^ = ^ (a + v)\
= f Sa f— + 3a^ I a^ /— ,
4 — m 3 — m 2 — m 1— m
and in this it is easy to substitute t i- a for y.
282 CALCULUS FOR ENGINEERS.
22. I '- -T . Let a + 6^ = y so that h .dx = dy,
J (a + bx)^
23
24
3. I -7^=r_ .dt=^-\/a^- 1\ evidently.
. I . Let x — a — y,dx = dy
25. Since ~i-, = ^(-^ --^-l
g^—d? zaxx — a x-^ a/
J a?-a? 2a ^ ^^ ^^ ^^ 2a ^^ + a
^. ., , r dx \ , x — a
r fix
26. If x- + 2Ax + B has real factors, then - — ^— j ^
j x^-{-2Ax-\-B
is of the form just given.
But if there are no real factors, then the integral may
r (2x
^^ ''"'^^'' j a? + 2Ax + A^ TB^rAi ^"^ if y = *' + ^ and
a^ = B — A^ we have | ~ ^-„ which is - tan~* - .
J y^ + a^ a a
27. I tan X. dx=— I dx. This is our first example
J J cosx ^
of a great class of integrals, where the numerator of a fraction
is seen to be the differential coefficient of the denominator.
Lot ?/ := cos X, then dy = — sin x . dx, so that the above integral
is — I -^ , or — log y, or — log (cos x).
EXAMPLES. 283
28. Let/' (x) stand for the differential coefficient oif{x),
and Ave are asked to find I — \r7-: — ■ Let f{x) = y, then
f {oc) . dx = dy, so that the integral becomes
/--=logy=log/(^).
Hence, if the numerator of a fraction is seen to be the differ-
ential coefficient of the denominator, the answer is
log (denominator).
_^ f X . dx If 2bx .dx 1 , , , „.
31. Reduce I ^ ^ — - to a simpler form. If the
J Ct "T" OX "T" cx
numerator were 2cx + b, the integral would come under our
rule in Ex. 28. Now the numerator can be put in the shape
n ,^ 7x ^^
-(2c^ + 6) + m- — ,
so we may write the integral as
r 2cx + h J f nh\ C dx
^j a + bx + cx^^'^'''^V''~2c)j a + bx + cx'
The latter integral is given in Example 26.
na f x + b , . f2x .dx C b. dx
= vUog(a2 + .7'^) + -tan-i-.
„_ r sin X .dx 1 f—b sin x .dx 1 i . 7 v
33. — — T = -,- -— , = -rlog(a + £*cos*').
Ja + b cos i« b J CI + 6 cos a; 6 °
284 CALCULUS Foil ENGINEERS.
34 14^ n_+jog^-log^^
J xAogx J xiogx
_ /' (I + \ogx)dx Cdx
J a; log a; J x
= log (x log x) — log X
= log X 4- log (log ;r) — log X
= log (log a;).
When expressions involve a"* and (a + 6^)**, try substi-
tuting y = a + bx or y=^ - + b.
X
35. Thus f .--^^-.- = -^-Jl_-.
*17 f <^^' __Jl_lA] (f' + bx
^^'"^ J (a + 6^+1 " 2^;^ (a + ba^y-
2m - 1
2ma J (a-hbx-y
and so we have a formula of reduction.
C da
J (a-hh
39
When expressions involve Va 4- bx try y^ = a-{- bx.
rp,, C x.dx 2 (2a — 6^) / =—
Thus -— ..^=_-A_^^ — ^s/a + bx.
J Wa + bx ob^
40. I - Vl 4- log ic . dx. Try ?/ = 1 -f logo;.
Answer : f (1 + log a;)^.
41. I VT~^=^- Try€^ = i/. Answer: tan-^e*.
217. Integration by Parts. Since, if u and v are
functions of x,
d . ._ dv du
dx^ ^^ dx dx'
uv= j u.dv-\- I V . da,
EXAMPLES. 286
or lu . dv = uv — I V . du (1).
We may write (1) as ju '-r . dx=uv — Iv .-j-.doc.
By means of this formula, the integral I u . dv may be
made to depend upon j v .du.
r rl
42. Thus to find p**. log a?, c?^. Let 2^ = log a? and -t-=^",
so that V = r . Formula (1) gives us :r loe: x — \ ^ dx,
a;«+i /, 1 \
or log X .
43. fx . 6" , dx.
Let it — x;--j- = e^^, so that v==-€^"'; then formula (1) gives
us \ x.e"^^ .dx = - .re"* I e"* . dr = - a?€"* ^e"*
J a aj a a^
= - €«* (a? ) .
a \ aJ
44. I e** . sin bx . dx. Call the answer A.
Let u = sin 6^, v = - e^^, then formula (1) gives us
^ _ _ gaa; gj^ hx \ e^^ . COS bx ,dx = - e"^ sin 6^ 5.
a a J a a
But similarly 1 e^* . cos 6a; . dx, which we have called B,
may be converted, if we take u = cos bx and v = - e"*;
^ _ _ guas cos hx-\- - I 6*** sin bx.dx — - e"* . cos bx-\-~A,
a a J a a
286 CALCULUS FOR ENGINEERS.
Hence A = - e"* sin h.v ( - e"^ cos hx-\- -A] ,so that
a a \a a )
f . , , €"* (a sin bx—b cos bx)
A = l e^"" sin bx .dx= - -^ ;— -y^ .
€°^(a cos bx-\-b sin bx)
€** COS bx . dx = —
Similarly i^ = I
a- + b"-
218. By means of Formulae of Reduction we reduce
integrals by successive steps to forms which are known to us.
They are always deduced by the method of integi'ation by
parts. Thus
I ar«e"* . dx = - A-«6«^ - - I x''-' . €«^ . dx.
j a aj
If then we have to integrate ar*e"*, we make it depend
upon x^e"^', again using this formula of reduction we make
^a^ax depend upon ar^e"*, and so on, till we reduce to x^e"^ or t*"*,
whose integral we know.
Thus L-^e^ . (^ = o^-^t* - 3 U-^e* dx
= x^e^-^\x'e'-2 jxe^ . dxl
= (x^-Sx'-^Qx-(j)6\
Some General Exercises.
45. 2/ = « sin^ bx, -^^ab sin 2bx.
46. y=^b sin ax^, -~ = bnax^~^ cos ax^.
^ dx
47. y = (a + 6a;")»", -^ = nbx^-^m {a + 6a?»*)»*-^
48. y = (a + 6^') e'^*, -^ = €'^(b + ac + bcx).
EXERCISES. 287
49. y — a"^, -^=^a^ . log a.
50. y=\ogaiCy -r-=—\ .
^ ^ dx xioga
-- a — tdv a
dv
52. v=^^a^-t\ — =
53. u =
v^ du Sv""
(l_V2)f' dv (1-V2)f
_ ^/a + t dv _ Vg ( V^- Va)
57. 2/ = log (sin a?), -7^ = cot x.
58. „ = logya-^sJ=-^,.
1 /l — cost dy 1
61. ^ = tan"^ / , -TT = / .
Jl+P-Jl-t" dt Ji-ti
62. a; = sec-i^, ^= "^
dt tjf-l'
63. 3/ = sin (log v)y-r = ■ cos (log v).
288 CALCULUS FOR ENGINEERS,
^- _l-\-x dy'^\ — 2x — ob^
66. « = log (cot t;), -^ = - -^-s" .
■^ *=* ^ ^ dv sin 2v
67. 5=e'(l-f»X^=€*(l-3f''-^).
69. ^ =
'^- /^-fd-r rf(9~ (€^-1)'^ ■
hr-. Ti- X Zl . Zl 4.1, 4. ^^^ COS ^
71. If a? = tan 6 + sec ^, prove that -j^ =
c?^ (1 - sin ey
d^x 2
72. If a; = ^ log ^, prove that 3^ = g •
78. If 2/ = 6~^ cos a?, prove that -^^ + 4?/ = 0.
.T>. T^ ^ 4.1,4. d*y 24
74. If2,= ^__, prove that ^ = ^-^-^.
75. L"»-^ (a 4- &a:") ^/« dx.
(1) If ^/5' be a positive integer, expand, multiply, and
integrate each term.
(2) Assume a + hx^ — y^ ; and if this fails,
(3) Assume aaj~" +h — y^: this also may fail.
76. \ af^ (a + x)^ . dx. Let a + x — y"^, then dx—2y. dy, and
co^y^—a, so that we have 2 {(y*— 2ay^-h a^)y^ ,dy, or
2 I (3/^ — 2a3/* 4- ay) c?y, which is easy.
EXERCISES. 289
— 2x~^ .dx = 2i/ .dy so that we have
78. [ — ^^ . Try a'^a?"* + 1 = ^^ ^nd we find
1 rc^^_ 1 _ 0)
cPx
79. If a? = -4 sin nt +B cos w<, prove that -j-^ + ?i2a; = 0.
80. If 1* = xy, prove that ^^, = ^ ^ + ^-^^i •
81. Illustrate the fact that -^ — i- = ^ 7- (see Art. 83)
ay . ax ax . ay ^
in the following cases :
X
w = tan~^ - , ^* = sin (aa;** + 63/**),
w = sin {oi^y), u = a; sin y + y sin a?,
« = ba^ log ai/, li = log (tan ^ ) ,
82. y = e**^ sin"* 6a;, -^ = e*'* sin"*-^ 6a? (a sin hx + m6 cos hx).
83. ic = €-«'cos6^, ^ = (a« + 6=')~2 e""' cos(6^ -n^) where
tan d = - .
a
^1 A 1.2.3.4
84. 2/ = a;^loga?,^- ^—
19
290 CALCULUS FOR ENGINEERS.
,,.. d^y 2 COS 00
85. y = log(8m^),J=-^j^.
86. Uv = Aiof'^y^^ + A^y^^ + &;c., where
aj + 6i = Oa + 62 = &c. = n,
V is called a homogeneous function of x and y of n dimensions.
Show that fl7(-7^)+v(-T-)= ^^- Illustrate this when 1; = — ^
and V = V ^* + 3/"-
87. In general if u =/ (1/ 4- a^) + i" (y — ax), where / and
-P are any functions whatsoever, prove that
dhi,_^(Pu
dx^^'' dy''
the differentiation of course being partial.
88. If v = (x^ + f + ^)-i, prove that g + + g = 0.
89. If 5 = ae--* sin y8^ satisfies ^ + 2/ j, + w^s = 0, find /
and n^ in terms of a and /3, or find a and ^ in terms of /
and n^,
90. If y = €** is a solution of
find a. As an example take
^__2^^-^^ + 2^ =
da^ da? dx^ dx '
and find its solution.
Answer : y — ae^-\' be~^ + ce^ + e, where a, 6, c, e are any
constants whatsoever.
219. To integrate any fraction of the form
Ax"^ + Bx"^-"- + Cx"^-^ + &c.
ax"" + 6a?^-i + ca;"-2 + &c.
where ?^ and n are positive integers.
(1).
PARTIAL FRACTIONS. 291
If m is greater than or equal to n, divide, and we have a
quotient together with a remainder. The quotient is at once
integrable and we have left a fraction of the form (1) in which
m is less than n. Now the factors of the denominator can
always be found and the fraction split up into partial
fractions.
For every factor of the denominator of the shape x — a
assume that we have a partial fraction ^ ; for every
factor of the shape x^ -\- oix-\- ^ assume that we have a partial
of the shape — 77; if there are n equal factors each of
them being x— a assume that we have the corresponding
partial fractions
Thus for example, suppose we have to deal with a fraction
which we shall calK^^^ and that F{pii) splits up into factors
a? — a, X — y8, a?- + ax + h, {x — 7)** ; we write
/(^__4_ ^ Gx^B E
F(x) " x-a'^ x-^'^ af + ax-{-b'^ (^-7)"
+ 7— ^^-T^+&c (2).
Now multiply by F{x) all across and we can either
follow certain rules or we can exercise a certain amount of
mother wit in finding A, B, C, D, E, F, G, &c.
Notice that as we have an identity, that is, an equation
which is true for any value of x, it is true if we put a; = a or
X— ^ or a? = 7 or x'^ + ax-hb—0. Do all these things and
we find that we have obtained A, B, E, C and D. To
find G we may have first to differentiate our identity and
then put ^ = 7 and so on. You will have found it more
difficult to understand this description than to actually carry
out the process.
Having split our given fraction into partials the integra-
tion is easy.
19—2
292 CALCULUS FOR ENGINEERS.
^^ a? A B Gx-\-D
Q1 ^^ _______ _| J
Hence x'' = ZX^^ -f I) + 5 (^ - 1) (^=^ + 1) + {Cx -{-D)(x- If.
Let dr^+l = 0, and we have with not much difficulty
(7= - J. i) = 0. Put a? = 1, and we have ^ = J. To find B,
make x = 0, and we find B=^. Hence we have to inte-
grate
11 11 1 X
2(07-1)2 2^-1 2l+af'
and the answer is
-| -^ + ilog(«-l)-Jlog(^= + l).
When there are r equal quadratic factors, we assume the
partials
(x' + cuc + ^y {x' + axi-^y-^'^ °*
It is not difficult to see how all the constants are deter-
mined. We seldom, however, have complicated cases in our
practical work.
92. Integrate ^-^^-^ or -^^^^-^—^^;
assume it to be equal to
M F P
X "^a;+3"^a;-2'
so that x^-\-x-l^M{x-\-^){x-'2,)-^Nx (a;- 2)+ P^ {x^- 3).
As this is true for all values of x, put x = and find M,
put ^ = — 3 and find N, put x—l and find P. Thus we find
that the given fraction splits up into
Qx 3a7+3 2a;-2'
so that the integral is
Jloga?-i-ilog(a^ + 3) + ilog(^-2).
EXERCISES. 298
5^2
= — + 15a; - 6 log (a; - 1) + 41 log {x - 2).
94. 1^-7^3-6'^
/■/^u.^^1 1 32 1 ^243 1 \,
95 [ ^-^^
= J tan-1^4- i log(H-a^')- i log(l 4-^).
^3-5a;24-3^ + 9 a; + l"^(a;-3)2 ^--3'
and we find ^ = - 8, A = - 5, ^3== 17 ;
so that the integral is
-81og(^4-l)+-^ + l71og(a;~3).
^^^- /^ + i- 3 = ^ l"g (^' + 3) + i log i^ - !)■
294 CALCULUS FOR ENGINEERS.
107 r (2a; - 5) dx __ 7 ^ + 1
j(a; + 3)(a; + i> 2(a; + l)'^^^''^^+3*
108. - = -7=tan^ — 7=—.
a;+4
220. Maxima and Minima. If we draw any curve
with maxima and minima points, and also draw the curve
showing the value of -~ in the first curve, we notice that ; —
where y is a maximum, -f- = and -~ is negative; whereas,
1 • • • ^V /x J d^y • :• T<. •
^vnere y is a mmimum, t =^ and -^^ is positive, it in
any practical example we can find no easier way of discrimi-
nating, Ave use this way.
Notice, however, that what is here called a maximum,
value, means that y has gradually increased to that value
and begins to diminish, y may have many maximum and
MAXIMA AND MINIMA. 295
minimum values, the curve beinsr wavy. Notice that -^
(Py . . ^^
may be and ~- = so that there is neither a maximum
a 3c
nor a minimum value, y ceasing to increase and then begin-
ning" to increase again. See M, fig. 6.
1. Find the maximum and minimum values of .
Answer : ^ and — J.
2. Find the createst value of .— : t-tt-^ r .
Answer :
{a + hy
3. Prove that a sec 6-\-h cosec ^ is a minimum when
3
tan 6
= V5
4. When is , a maximum ? Answer : x = ^.
5. When is x'^ {a — xy^ a maximum or minimum ?
. ma
Answer : x = , a maximum.
m + n
6. Given the angle (7 of a triangle, prove that
sin^ A + sin^ 5 is a maximum and cos'^ A + cos^ J5 is a mini-
mum when A=B.
7. y = asmx-\-h cos x. What are the maximum and
tninimum values of y ?
Answer : maximum is y = \/aFTb^, minimum is - Va^ + ¥.
8. Find the least value of a ton + bcot 6.
Answer : 2 ^ah.
9. Find the maximum and minimum values of
^^^ + 2^ + 11
i»2 4- 4^ + 10 '
Answer : 2 a maximum and f a minimum.
Students ought to plot the function as a curve on squared
paper.
296 CALCULUS FOR ENGINEERS.
10. Find the maximum and minimum values of
a^-x + 1
x^ + os-l'
Answer : maximum, — 1.
11. Find the values of x which make y = ztr —
^ a;- 10
a maximum and a minimum.
Answer : x=A gives a maximum, a;= 16 a minimum.
12. What value of c will make v a maximum if ?; = - logo?
Answer : c = e.
13. If^ = ^^ -^ yt^Nah gives a mmimum value
t
1 A sin" 6 ^ IT . . ,
14. X — , ^ , u —- £(ives a maximum value to x.
1 - cos ^ 3 °
15. What value of c will make v a minimum if
V = q , ? Answer: c = i.
16. When is 4^— 15a^4- 12a; — 1 a maximum or minimum ?
Answer : a; = J a maximum ; a; = 2 a minimum.
17. tan'^ a? tan** (a — a;) is a maximum when
tan (a — 2a;) = tan a.
n-\-m
1 ft _ ^^ ^ = 3 a maximum,
9 + ^v ^ = — 3 a minimum.
19. Given the vertical angle of a triangle and its area,
find when its base is a minimum.
20. The characteristic of a series Dynamo is
^=T^a W.
MAXIMA AND MINIMA. 297
where a is a number proportional to the angular velocity of
the armature, and a and s depend upon the size of the iron,
number of turns &c., E is the E.M.F. of the armature in volts
and (7 the current in amperes. If r is the internal resistance
of the machine in ohms and i^ is an outside resistance,
the current
^'=4^ (2),
and the power given out by the machine is
P:=OR (3).
What value of R will make P a maximum ?
Here (2) and (1) give --^ -^ = (7.
So that 1+5(7=-^, C=if-^-lV
r-\-R' s \r -\-R J
'-7(;:fji-')'.-<"'S-».
we have {-^ - iT + 2i2 {-^- \\ {- ,— ^o,.! = 0-
Rejecting ^— 1 = because it gives (7=0, we have
— r-15 — 1 = 7 7^ > and from this R may be found if r and
r-\-R (r + Ry ^
a are given. Take a = l'2, 5 = 0'03, r = *05 and illustrate
with curves.
21. A man is at sea 4 miles distant from the nearest
point of a straight shore, and he wishes to get to a place 10
miles distant from this nearest point, the road lying along
the shore. He can row and walk. Find at what point he
ought to land, to get to this place in the minimum time, if he
rows at 3 miles per hour and walks at 4 miles per hour.
Assume that he can equally well leave his boat at one place
as at another.
298 CALCULUS FOR ENGINEERS.
Fig. 100, ^C=4, CT=10 . Let him land at D where
CD = X. Then AD = Vl6+a-^ and D^ = 10 - a;.
Hence the total time in hours = r 1 -. — .
3 4
This is a minimum when ^^(16 +^)~*=J, or J^'=16+a;*,
or a; = 4*535 miles.
22. The candle power c of a certain kind of incandescent
lamp X its probable life I in hours, was found experimentally
to approximate on the average to
Iq _ ]^Qll-897— 00764et>
where v is the potential difference in volts. The watts w
expended per candle power were found to be
tt; = 3-7 + 108 O07-07667W
The price of a lamp being 2^., the lamps being lighted
for 560 hours per year, and one electrical horse-power (or
746 watts) costing £2 for this year of 560 hours, find the
most economical v for these lamps, so that the total cost in
lamps and power may be a minimum.
—J- lamps are needed per year, each costing £01. Cost
. 56 .
per year is then -y- in pounds, and this is for c candles, so that
. 56
cost per year in pounds per candle is y- . Now £1 per year
746
means —^ watts, so that the cost per year per candle is
56 746 ,,
_x— watts.
This added to w gives total cost in watts.
INDETERMINATE FORMS. 299
We have Ic and lu as functions of v. Hence
R« y 74fi
<j\j /^ -x\j -|^Q_ii.697+o 07545W i g-^ _I_ ]^Q8 007— 07667V
It
is to be made a minimum.
Answer : i; = 101*15 volts.
221. Sometimes when a particular value is given to ^ a
function takes an indeterminate form. Thus for example
in Art. 43, the area of the curve y=mx~^^ between the ordinates
/* TYl
mx~^^ . dx was z (6^~"^ — a}~^).
Now when «,= 1 the area becomes ^ (1 — 1) or ^, and
this may obviously have any value whatsoever.
fix)
In any such case, say 4r/—'J > '^^ fipd — ^ and F {a) = 0, we
£ {X)
proceed as follows. We take a value of x very near to a and
find the limiting value of our expression as x is made nearer
and nearer to a in value. Thus \q\} x — a-\- hx.
Now as hx is made smaller and smaller it is evident
dfix)^
that f{x + hx) is more and more nearly f{x) + hx . — — . If
in this we put x=a, f{x) or f(ci) disappears, and conse-
quently our fraction*— ^ becomes more and more nearly
The rule then adopted is this : — Differentiate the numerator
oidy and call it a new numerator; differentiate the denomi-
nator only and call it a new denominator; now insert the
critical value of x, and we obtain the critical value of our
fraction. The process may need repetition.
loP" X
Example 1. Find the value of ° when x = l,
X ~~' L
300 CALCULUS FOR EiNGINEERS.
First try, and we see that we have 0/0. Now follow the
1
oc
above rule, and we have - , and inserting in this a; = 1 we get
1 as our answer.
2. rmd ^—„ — ^, r— when a? = c.
First try a? = c, and we get 0/0.
Now try ^, _^ , , and again we get 0/0.
Now repeating our process we get r -
sc—l 1
3. Find ^- when x=l. Answer: -.
a?" - 1 n
4. Find when x = 0. Answer : log j- .
5. Try the example referred to above. The area of a
curve is ^j (6^~" — a^"") = A. If m, b, a are constants,
what is the value of A when n = 1? Writing it as
m — = ,
1—71
differentiate both numerator and denominator with regard
to n, and we have, since
A(5i-»)=6^-«.log6x(-l),
6^~" log b - a^~" log a
7)1 - ,
and if we insert n = 1 in this, we get
m (log b — log a) or m log - ,
CL
which is indeed the answer we should have obtained if
instead of taking our integral p~^ . dx as following the rule
i
a;~" . dx — + c,
— 7l-\-l
GLOSSARY. 301
we had remembered that in this special case
]a;~^ . dx—logiv.
I'
222 J Glossary and Exercises.
Asymptote. A straight line which gets closer and closer
to a curve sls x ov y gets greater and greater without limit.
Thus y = - Jx^ — a^ is a Hyperbola. Now as x gets greater
^ a .
and greater, so that - is less and less important, the equation
X T
approaches more and more y=- x, which is the asymptote.
a
The test for an asymptote is that -^ has a limiting value
for points further and further from the origin, and the inter-
dx
cept of a tangent on the axis of x, x — y -^ , has a limiting
value, or the intercept on the axis oiy.y — x -j-, has a limit-
ing value. ^^
Point of Inflection. A point where -^^ changes sign.
Point of Osculation. A point where there are two or
more equal values of — .
Cusp. Where two branches of a curve meet at a common
tangent.
Coi^Jugate Point. An isolated point, the coordinates
of which satisfy the equation to the curve.
Point d' AiTet. A point at which a single* branch of a
curve suddenly stops. Example, the origin in y = a? log x.
The Companion to the Cycloid. x-a{\ -cos (/>),
The Epitrochoid. x={a + h) cos <^ - m6 cos f ^ + 1 J </>,
y = (a -f- 6) sin — m6 sin ( T- -f 1 J (f>,
302 CALCULUS FOR ENGINEERS.
where h is the radius of the rolling circle, a is the radius of
the fixed circle, and mb = distance of tracing point along
radius from centre of rolling circle. Make m=l, and this is
the Epicycloid.
The H3rpotrochoid. ^ = (a— fe)cos<^+m6cos f ^-1 J<^,
y= (rt— 6)sin<^— ??i6sin(7- — 1 ]</).
Make ??i = 1, and we have the Hypocycloid.
Take a = 46, and obtain a Hypocycloid in the form
a;* + y* = ai
Take a = 26, and obtain the Hypocycloid which is a
straight line.
In obtaining the Cycloid, Art. 11, let the tracing point be
anywhere on a radius of the rolling circle or the radius pro-
duced and obtain a? = a (1 — m cos <^), y = a (<^ + m sin <^).
If wi > 1, or < 1, we have a prolate or a curtate Cycloid.
The Lemniscata (a^ + y^y= a^ (jx^ ^y^) becomes in polar
coordinates r* = a* cos 26,
and taking successively ^ = 0, ^ = '1, &c., we calculate r and
graph the curve easily.
The Spiral of Archimedes, r = ad.
The Logarithmic or Equiangular Spiral, r = ae*^
The Logarithmic Curve. y=a\ogba)-{- c.
The Conchoid a^ = (a + xy (¥ - x') becomes
r = a + 6 sec 6.
The CiBBoid y^ = a^/{2a — x) becomes r = 2a tan 6 . sin d.
The Cardioide. r = a (1 — cos 0).
The Hyperbolic Spiral. r6=a.
The I.ituusis?^2^ = a2.
The Trisectrix. r = a (2 cos ^ + 1).
1. In the curve y=—^ ^, show that there are points of
a ~i~ X
inflexion where a? is and a VS; the axis of x is an asymptote
on both sides ; there are points of maxima where x = a and
— a ; the curve cuts the axis of ac at 45°.
EXAMPLES. 303
2. In aj^y = 36^?^ — a^ show that there is a point of inflexion
where x = b, y= —r -
3. If y^x = 4a'^ (2a — a?), show that there are two points of
, „ . , 3a ^ 2a
inflexion when oo= -zr- , y=^±-j=-.
^ V3
4. If 2/^ {x^ — a^) = ^, show that the equations to the
asymptotes are y=^-\-x and y=- w.
5. The curve a^ — y^ — a^ cuts the axis of x at right angles
at a? = a where there is a point of inflexion.
6. Show that y = a^xjiah + ^) has three points of in-
flexion.
7. Prove again the statements of Exercise 2, Art. 99,
and work the exercises there.
8. Find the subtangent and subnormal to the curve y = e***
Answer : subtangent - , subnormal ae^^^.
9. Find the subnormal and subtangent to the catenary.
=!{-■■•
c
i
subtangent = c
Answer : subnormal = -J6<'— e "l
c - c
10. Find the subtangent of the curve
x^ — Saycc + y* = 0,
Hence ^ ^ay-j^
dx y^ — ax
Subtangent at point x, y is y-^ = y ^ — .
(ty cty — ocr
304 CALCULUS FOR ENGINEERS.
11. In the curve 2/^ = find the equation to the
asymptote. Here i/^ = a?^ I — \=ia^ f 1 -\ 1- --- + &c. j
by division. As x is greater and greater, — &c. get smaller
and smaller and in the limit (see Art. 3)
3/=±^(l+^).
y=±{x^-a\
So we have a pair of asymptotes y = x-\-a, y = — x — a.
Again, the straight line x = a,a> line parallel to the axis of
y, is also an asymptote, y becoming greater and greater as
X gets nearer and nearer to a in value.
12. Find the tangent to y* + a^ = a*.
Hence at the point x^, y^ the tangent is ~ — ^ = — a/^\
13. In the curve 3/ — 2 = (a? — 1) Va; — 2, where is -^ = x ?
At what angle does the curve cut the axis ?
This is infinity where x — 2 and then 3/ = 2 ; that is, the
tangent at (2, 2) is at right angles to the axis of x.
Where y=0/\t will be found that a? = 3 and -t- = 2.
14. In the curve y^ — ax^ ■\- a? , find the intercept by the
tangent on the axis of 3/, that is, find y — cc-^.
Sy^ -^ = 2ax + Sx\
JSXAMPLES AND EXERCISES. 305
bo tnat we want y — x — ^r— — or -^ r- and
3i/2 %f
this will be found to be
3 \a +xj '
15. The length of the subnormal at x, y, is 20^^, what is
the curve ?
Here 2/-,— = 2a V. Hence ^y^ = ^a^x* or y = ax^ is the
dx
equation to the curve, a parabola. The subtangent is ^ y ,
^^^•2f^^3>or2^,oria;.
16. Show that the length of the normal to the catenary^
is - y\
17. Show that y^ — x^-\- 2hx^y = has the two asymptotes
b , b
y^x-^andy=-x--^.
18. Show that the subtangent and subnormal to the cii'cle
y^==2ax — x^, are — — — and a — x respectively, and to the
IT 2 ^\^ 2X4.1. 2ax-x' ,b\ .
ellipse y^ = —Azax — x^) they are and -- (a — x\
a?
19. Find the tangent to the cissoid y* =
2a — X
Answer : y = j ^^^^^y l {(3a -x)x^- ax\,
20. What curve has a constant subtangent ?
dx ^ dy , ^?
2/ -J- = a or a^ = a -^ , or x= a log y + c or y = Ce^y
the logarithmic curve.
21. Show that a? — y^-\- ax^ = has the asymptote
P. 20
306 CALCULUS FOR ENGINEERS.
22. Show that a curve is convex or concave to the
axis of X as y and -— have the same or opposite signs. See
Art. 60.
223. The circle which passes through a point in a curve,
which has the same slope there as the curve, and which has
also the same rate of change of slope, is said to be the
circle of curvature there. If the centre of a circle has
a and h for its co-ordinates, and if the radius is r, it is easy
to see that its equation is
(a;-a)» + (i/-6)» = ?-« (1).
Differentiating (1) (and dividing by 2) and again dif-
ferentiating we have
a.-a + (y-6)J=0 (2),
and i+(,_,)g+(|J = (3).
writing p for -~ and q for j^ we have from (3)
y-^ = — ^ (4);
using this in (2) we have
X'-a=^—^p (5).
Now p and q and x and y at any point of the curve being
known, we know that these are the same for the circle of
curvature there, and so a and h can be found and also r. If
the subject of evolutes were of any interest to engineers, this
would be the place to speak of finding an equation connecting
a and 6, for this would be the equation of the evolute of the
curve. The curve itself would then be called the involute to
the evolute. Any practical man can work out this matter
for himself. It is of more interest to find r the radius of
curvature. Inserting (4) and (5) in (1) we find the curvature
1 = ^_ (6)
.(8>
CURVATURE. 307
A better way of putting the matter is this: — A curve
turns through the angle Bd in the length Bs, and curvature
is defined as the limiting value of
hd 1 dO
B^'^^'r^dS (^)-
Now tan = -~ = p, say, so that 6 = tan~^ j^. Hence
dd _ 1 f^
ds ~ 1 -i-p- ' ds '
Now ^ = ViT^^=-.'^ = -.^.
dw ^ dp dx dp dx"
H-~ \-f.-%i[Htr\'
Exercises. 1. The equation to a curve is
ic^ - 1500a;2 + 30000^ - 3000000y = 0.
Show that the denominator of - in (8) is practically 1
from ;r = to iz; = 100. Find the curvature where x — 0.
2. In the curve y = ar* — 4a'3 — ISar^ find the curvature
at the origin. Answer : 36.
3. Show that the radius of curvature at x=^a,y = Ooi
the ellipse — + ^r = 1, is — .
4. Find the radius of curvature where a; = 0, of the
parabola, y^ = 4aa;. Answer : 7' — 2a.
5. Find the radius of curvature of y = 6e"^.
(1 + a^h-6^^)^
Answer : q = a^be^"^, p = ci&e«^, r = -'^ ,, „^ , so that
where a; = 0, r = ^ -j — ^ .
a-'o
6. Find the radius of curvature of y = a sin bx.
. (l + a262cos2 6a?)5 .,,, .
Answer : /• = ^^ j-—. — , — — . Where x = 0, r=QC, or
— ab^ sin ox
curvature is : where 6a; = - , ?■ = r, ♦
2 — ao^
20—2
308 CALCULUS FOR ENGINEERS.
7. Find the radius of curvature of the catenary
Answer -. r=.^ . At the vertex where a; = 0, 1/ = c, r = c.
8. Show that the radius of curvature of
7/^ {x — 4m) = 7nx {x — Sm),
at one of the points where y = 0, is -^ , and at the other, ^r- .
9. Find the equation of the circle of curvature of the
curve y* = 4m'^ — a^, where x=0, 3/ = 0.
10. The radius of curvature of Sa^y = x^, is ?' = ^ . .
11. In the ellipse show that the radius of curvature is
(a' — e-x"^)^ -^ ah, where e^=\ ^ , e being the eccentricity.
12. Find the radius of curvature oi xy = a.
Answer: (a;^ 4- 2/")* -i- 2a.
13. Find the radius of curvature of the hyperbola
Answer: (eV — a^)^ -r- ab, where e- = l-\ — .
a^
14. In the catenary the radius of curvature is equal and
opposite to the length of the normal.
15. Find the radius of curvature of the tractrix, the
dx
equation to which is y -7- = Ja^ — y^.
224. Let f(x,y,a) = (1)
be the equation to a family of curves, a being a constant for
each curve, but called a variable parameter for the family,
ENVELOPES. 309
as it is by taking different values for a that one obtains
different members of the famil}^ Thus
f(a;,y,a + Ba) = (2)
is the next member of the family as Ba is made smaller and
smaller. Now (2) may be written (see (1) Art. 21)
/(^,2/,a)+Sa.^/(^,2/,a) = (3),
and the point of intersection of (2)f and (1) is obtainable by
solving them as simultaneous equations in oo and y; or again,
if we eliminate a from (1) and
^J(^.y,a) = (4).
we obtain a relation which must hold for the values of x and
y, of the points of ultimate intersection of the curves formed
by varying a continuously ; this is said to be the equation of
the envelope of the family of curves (1) and it can be
proved that it is touched by every curve of the family.
Example. If by taking various values of a in
m ,
^ a
we have a family of straight lines, find the envelope. Here
f(x, y, a) — is represented by
y ow? = (1)*,
and differentiating with regard to a we have
+ ^■-^^ = (4)*,
1 a" , m
or - == — or a^ = _ ,
X m X
Using this in(l)*we have
y — ^mx — X A / — = 0,
or y — 2 \/nix = 0, or y^ = 4>mx,
a parabola.
310
CALCULUS FOR ENGINEERS.
Example. In Ex. 3, Art. 24, if projectiles are all sent out
with the same velocity F, at different angular elevations a,
their paths form the family of curves,
Fsin a
2/ =
\9
Fcosa ^^F^cos^a'
or . y — xa-^ma^{a^'\-\) = 0,
where a stands for tan a and is a variable parameter, and
1 9
Differentiating with regard to a,
— x-\- 2ma^a = or a = + - — :
2mx
1 ,/ 1
,4fm^x^
is the equation to the envelope, or
1
+ 1
)..
-y == — rtiOi^ -f
4>m
This is the equation to a parabola whose vertex is j — or
F-
— above the poii^t of projection.
225."^ Polar Co-ordinates. If instead of giving the
position of a point P in .r and y co-ordinates, we give it in
terms of the distance OP called
r, the radius vector, and the angle
QOP (fig. 101) called (9, so that
what we used to call x is r cos
and what we used to call y is
r sin 6, the equations of some
curves, such as spirals, become
simpler. If the co-ordinates of
P' are r + Sr and 6 + hO, then
^ig- 101- in the limit PSP' may be looked
upon as a little right-angled triangle in which PS = r . BO,
SP' is Sr, PP' or 8.9 = Jr'(8dy + (8ry so that
dd y ' ^ \de)
POLAR CO-ORDINATES. 311
Also the elementary area POP' is in the limit ^r"^ .hd, and
the area enclosed between a radius vector at ^j and another
at ^2 is i I r"^ ' dO, so that if ?' can be stated in terms of 6 it
is easy to find the area of the sector. Also the angle </>
between the tangent at P and ?' is evidently such that
dO
tan cj) = PS/P'S or, t -r- — tan <f). This method of dealing
with curves is interesting to students who are studying
astronomy.
If r = a*** (the equiangular spiral)
— = ha^^ log a, and so, ?' -7-; , or ?'-f- -^ = 1/6 log a,
so that tan <^ is a constant ; that is the curve everywhere
makes the same angle with the radius vector.
Let ^ = rcos 6 so that x is always the projection of the
radius vector on a line, x — a^^ cos 6. Now imagine the radius
vector to rotate with uniform angular velocity of — ^ radians
per second starting with ^ = when ^ = 0, so that 6 — — qt,
then X = a~^9^ cos qt.
Thus we see that if simple harmonic motion is the pro-
jection of uniform angular motion in a circle; damped
simple harmonic motion is the projection of uniform
angular motion in an equiangular spiral. See Note, Art. 112.
Ex. 1. Find the area of the curve r = a (1 + cos 0). Draw
the curve and note that the whole area is I r^ . dO, or ^iral
Ex.2. Find the area of ?^= a (cos 2^+ sin 2^). Answer: Tral
Ex. 3. Find the area between the conchoid and two radii
vectores. Answer :
b^ (tan e^ - tan d^) + 2ab log (tan (7r/4 - i^a) -f- tan (7r/4 - ^6,)},
226. Exercises. 1. Find the area of the surface gener-
ated by the revolution of the catenary (Art. 38) round the
axis of y.
2. Prove that the equation to the cycloid, the vertex
being the origin, is
x=:a(0 + sinO) y = a{l- cos 6),
if (fig. 102) PB = x, PA= y, OCQ = 6,
312
CALCULUS FOR ENGINEERS.
Show that when the cycloid revolves about OF it generates
(0_.2 Q\
•~9 — q) > and when it revolves about OX it
Y
F
-)V
y.^^
o
A X
Fig. 102.
generates the volume ttW If it revolves about EF it
generates the volume bir^a^.
3. Find the length of the curve ^ay^ — 4a^.
Answer, . = /J^ 1 + f . <te = fa j(l + ^^ - 1
4. Find the length of the curve y^ = 2ax — x^.
Answer : s = a vers~^ x/a.
5. Find the length of the cycloid. See Art. 47.
Answer : 5 = 8a (1 — cos ^(f>) = 8a — 4 V4a^ — 2ay.
parabola y —
^x + Va + X
6. Find the length of the parabola y — ^4iax, from the
vertex.
Answer : 5 = Nax -^ x^ -{■ a log
Va
7. Show that the whole area of the companion to the
cycloid is twice that of the generating circle.
8. Find the area of r = he^l'^ between the radii n and ra,
•^2
r«2
using A — \ ^r^ . d6.
J 0,
Answer: -r{r.^ — r^).
EXERCISES. SltS
y
9. Show that in the logarithmic curve x — ae'^,
5 = clog J + Vc^ + a?-^ + C^.
c + Vc^ + a-^
10. Show that in the curve r = a (1 + cos ^), using
s = 4a sm ^ .
11. Show that in the curve r = ^e*/*,
5 = rVl + c2+(7.
12. Show that in the cycloid,
ds \l 2a'
and consequently s = 4 Va^ — ^ay — 2a.
13. Show that in the curve x^ + 2/* = a*,
s = ^ciSx^.
x" f
14. The ellipse, — + r^ = 1 revolves about the axis of x.
^ a- 0^
Prove that the area of the surface generated is
where e^=l -.
x^ 1/^
15. Show that the whole area of the curve, -^ + fi = 1
IS fTrao.
16. Find the area of the loop of the curve, y=xtJ —33- .
Answer : 2a=^ ( 1 — -7 j .
314 CALCULUS FOR ENGINEERS.
17. Find the whole area of y = a; + Va^ — {jc\
Answer: Tral
18. Find the area of a loop of the curve r^- = a^ coa 20.
Answer : Jal
19. Find the area of the ellipse — „ + r;: = 1 ; that is, find
four times the value of the integral
•«6
/,
«
20. Find the area of the cycloid in terms of the angle <p
(Art. 11).
Answer : a^ (|</) — 2 sin </> + J sin 2<f)) ; and if the limits are
(f> = and </) = 27r we have the whole area equal to 3 times
that of the rolling circle.
227. A body of weight W acted upon by gi-avity,
moves in a medium in which the resistance = a?^", where
V is the velocity and a and n are constants.
9 dt
What is the velocity when acceleration ceases? Let Vi
be this terminal velocity, av^^ = W, or our a = Wvr\
dt I 1
^di^
so that
Thus let n= 2, t=^ log ^-^^ ,
2g ^ Vi — v
^ , at dx
or v=^Vi tann -- = ^rr •
V. dt
EXAMPLES. 315
If a; is the depth fallen through,
Vi^. 1 Qt
X — — losf . cosh ^- .
228. Our old Example of Art. 24.
A point moves so that it has no acceleration horizontally
and its acceleration downward is ^ a constant. Let y be
measured upwards and x horizontally, then
d'x
dt^
"' dt'
ff-
dx
dt~
--C,
dt
dy dx
'~di ' It''
' dx
)
d'y
dt'~
d^y dx
^'^dx'' ~dt
= c2
d^
dhj
d^~
.-9
dy
dx
Hence
y-^-lS'^^ + o.c+h... (I),
which is a parabola. Compare Art. 24.
If Ave take y— when .r=0, 6=0. Also we see that a is the
tangent of the angle which the path makes with the hori-
zontal at d? = and c is the constant horizontal velocity. If a
projectile has the initial velocity V with the upward inclina-
tion a, then c = Fcos a, and tan a = a, so that (1) becomes
1 gx^
y= — ^ TF^ — ;; — f- oc tan a.
229. Exercises on Fourier.
1. A periodic function of x has the value /(^) = mx, from
iv — to a? = c where c is the period, suddenly becoming
316 CALCULUS FOR ENGINEERS.
and increasing to mc in the same way in the next period.
27r
Here, see Art. 133, 7 is — ,
c
mju = cto -\- ttj sin qx + ^1 cos qx + &e.
+ ttg sin sqx + hg cos sqx + &c.
tto is ^mc,
Gg = - I mx . sin sqx . dx, hg=- I mx . cos sqx . dx.
c J c J
Answer :
7}hC
mx = ^mc (sin qx +i sin 2qx + J sin Sqx -f i sin iiqx -f&c.).
2. Expand a? in a series of sines and also in a series of
cosines.
Answer : x = 2 (sin ii" — J^ sin 2a; + J sin 3a; — &c.) from — tt
to tt;
also a* = — (sin a? — ^ sin Sx -f ^ sin 5x — &c.) from to ^ ,
TT
2
TT 4
and ''^" = 9 (<^os A' 4-^ cos 3.r -i-^ cos 5a; + &c.).
TT
3. Prove t = sin x -\- 1 sin 3./; + i sin 5a^ + &c.
4
4. Show that
5. Integrate each of the above expansions.
230. 1. The radius of gyration of a sphere about a
diameter being k and the radius a, prove that
Here, since a;' + 3/^ = a", and the moment of inertia of a
circular slice of radius y and thickness hx about its centre, is
|y.Sa;.
Taylor's theorem. 317
The moment of inertia is
I iriny'^ .da)Xi/'=l irmy^ . dx = m^Tra*,
Jo Jo
and the mass is m^TraK
2. In a paraboloid of height h and radius of base a, about
the axis, P = Ja^.
About the diameter of the base k^ — ^ (a^ 4- h^).
3. In a triangle of height h, about a line through the
vertex parallel to the base, k^ = ^h\
About a line through centre of triangle parallel to base
k^=^^h\
231. Taylor^ s Theorem.
If a function of « + h, be differentiated with regard to ic,
h being supposed constant, we get the same answer as if we
differentiate with regard to h, x being supposed constant.
This is evident. Call the function f{u) where u = x-\-h.
Then -y- f(u) = y- fiu) x ^- = -j- fCu) as -y- is 1, and this is
dx-^ ^ ^^ dir ^ ' dx du-^ ^ ^ dx
the same as ifrfi}^) because
Assume that f(x-\-h) may be expanded in a series of
ascending powers of h.
f(x + h) = Xo + XJi + X,h'^+XJi'-^&c (1),
where Xq, X^, X^ Sac. do not contain k
^^!i^tA) = o + X,+ 2Z^ + 3X3/1^+ &c (2),
an,
§f<^)=i^+^\h + ^.h' + &c (3).
dx dx dx ax
As (2) and (3) are identical
^ _dXo ldX,_ 1 d'Xo
'" dx ' ^"^2 rf;^; "1.2 dx' '
Z.=
1 dX^ 1 d'Xo
' S dx 1.2,2 dx" '
318 CALCULUS FOR ENGINEERS.
A.lso if h = in (1) we find that Xo =f(x). If we indicate
-r-^f(^) ^y f'{^)y <ihen Taylors Theorem is
f (X + h) = f (X) + hf (x) + ^ i" (x)
^' r'(x) + &c (4).
1.2.3
After having differentiated f{x) twice, if we substitute for x,
let us call the result /" (0) ; if we imagine substituted for
X in (4) we have
/W=/(O) + A/(0) + j^/"(0)
1.2.3
/"(0) + &c (5).
Observe that we have no longer anything to do with the
quantity which we call x. We may if we please use any
other letter than h in (5) ; let us use the new letter x, and
(5) becomes
/(^•)=/(0) + ^/(0) + j^/'(0)
r'(0) + &c (6);
1.2.-3
which is called Maclaurin's Theorem.
The proof here given of Taylor's theorem is incomplete,
as we have used an infinite series without proving it con-
vergent. More exact proofs will be found in the regular
treatises. Note that if x is time and s =f(t) means distance
of a body from some invariable plane in space ; then if at the
present time, which we shall call to, we know s and the
velocity and the acceleration and -5-, &c.; that is, if we know
all the circumstances of the motion absolutely correctly at
the present time, then we can predict where the body will be
at any future time, and we can say where the body was at
any past time. It is a very far-reaching theorem and gives
food for much speculation.
TAYLOR AXD MACLAURIN. 319
232. Exercises on Taylor. 1. Expand (^4-/^)'^ iu
powers of h.
Here f{x) = x^, f {x) = nx''-\ f {x) = n {n - V^x'^'S &c,
and hence
{x 4- A)" = X'' + n/t^"-i + ^^ Y ~ -^ fe"-2 + &c.
This is the Binomial Theorem, which is an example of
Taylor.
2. ExjDand log (.r + h) in powers of A.
Here /(..) = log (..), /(.;) = 1
X
/'(4 = -*-=,/"(.■) = + 2*-',
1 A/^ 1 /i^ 1
and hence log (a; + A) = log ^ + A -^ -3 + -^ -5 — &c;
^ X A X o Qj
If we put a; = 1 ,we have the useful formula
log(l+A) = + A-| + ^-&c.
8. Show that
sin (x + h) = smx-\- h cos ^ — y— « sin os — ■= — ^— s cos a} + &c.
4. Show that
cos (a? + A) = cosa; — Asini»— = — jrcos^+ z — ^— ^ sina? + &c.
6. What do 3 and 4 become when ^ = ?
233. Exercises on Maclaurin. 1. Expand sin x in
powers of x,
fix) ^ sin X, -/(0) = 0,
/(^) = cos^, /(0)=1.
/'(^) = -sina;, /'(0)=0,
/» = -cos^, r{0)=-l,
/X^) = sin^j, /v(0)=0,
&c. /'(0)=1.
/>io /^>5 /j,-»7
Hence sin a; = i^* — -r^ + — — -c + &c.
|3 • 1 5 \7_
320 CALCULUS FOR ENGINEERS.
».2
or
2. Similarly cos ^ = 1 — 'r^ -I- 77 — 77, + &;c.
\A ]i E.
Calculate from the above series the values of the sine and
cosine of any angle, say 0*2 radians, and compare with what
is given in books of mathematical tables.
3. Expand tan~^ w. Another method is adopted.
The differential coefficient oHair^xisr—, — , , and by actual
division this is 1 — a;^ + a;* — a.* -{- x* — &c.
Integrating this, term by term, we find
tsin-^a; = x- J^ + ^a* - -f a;' + J^ - &c.
We do not add a constant because when a? = 0, tan~* ic =« 0.
4. Expand tan (1 — x) directly by Maclaurin.
5. Show that
a*= 1 +0; loga+ 1^ (logay + ^ (log ay+ &c
6. Show that tan a; = a? + 7r4-T^ + &c.
6 15
234. Expand e'^, compare with the expansions of sin Q
and cos Qy and show that
e-** = cos — i sin 6,
cos (9 = J (e^'^ + e-'^),
sin^ = -i(e^'^-e-^).
Evidently (cos Q ±i sin ^)" = cos nd + % sin nB^
which is Demoivre's Theorem.
In solving cubic equations when there are three real roots,
we find it necessary to extract the roots of unreal quantities
by Demoivre. To find the gth root of a + hi where a and
h are given numerically. First write
a + ln = r (cos -\-i sin 0)»
DIFFERENTIAL EQUATIONS. 321
Then r cos = a, r sin. 6 = b, r = Va* + b^, tan 6 = ^- Cal-
6
culate r and 6 therefore.
- / 1 1 \
Now the qth roots are, r^ f cos - ^ + 1 sin - ^ J ,
r9 jcos - (27r + (9) + i sin - (27r + 0)] ,
r^ ]cos - (47r + ^) + t sin - (4>7r + ^) [ (Sec.
We easily see that there are only q, qth. roots.
Exercise. Find the three cube roots of 8.
Write it 8 (cos 0-\-i sin 0), 8 (cos 27r + i sin 27r),
8 (cos 47r + { sin 47r) and proceed as directed.
235. The expansion of e^^ is
1+he + ^h'0' + j-|-^A3<9» + &c.
Now let 6 stand for the operation -7- , and we see that
h —
f(w +h)—e^^f(ai) ; or e ^f{x), symbolically represents Taylor's
Theorem.
236. An equation which connects x, y and the differ-
ential coefficients is called a Differential Equation. We
have already solved some of these equations.
The order is that of the highest differential coefficient.
The degree is the power of the highest differential coeffi-
cient. A differential equation is said to be linear, when it
would be of the first degree, if y (the dependent variable), and
all the differential coefficients, were regarded as unknown
quantities. It will be found that if several solutions of a
linear equation are obtained, their sum is also a solution.
Given any equation connecting x and y, containing
constants; by differentiating one or more times we obtain
sufficient equations to enable" us to eliminate the constants.
P. 21
322 CALCULUS FOR ENGINEERS.
Thus we produce a differential equation. Its primitive
evidently contains n arbitrary constants if the equation is of
the nth. order.
Exercise. Eliminate a and b from
y^ax^+bx (1),
Hence (1) becomes
^'i^-^4x^^y-' (2>-
If we solve (2) we find y = Aa^ + Bx, where A and B are
any arbitrary constants.
237. In the solution of DifiPerential Equations we begin
with equations of the First Order and the First Degree.
These are all of the type M^-N-^ —0, where M and N
are functions of x and y. We usually write this in the shape
M. dx + N.dy=0.
Examples.
1. (a + x) (b + y) dx -i-dy=0 or (a + x) dx -\- r dy — 0.
Integrating we have the general solution
a^ + i;z;2 4-log(6 + 2/) = (7,
where C is an arbitrary constant.
It is to be noticed here, as in any case when we can
separate the variables^ the solution is easy.
Thus if /(^) F{y).dx-^^ {x) .'f{y).dy = Oy^Q have
f{x).dx ^{y).dy
<^(.') ^ F{y) -"'
and this can be at once integrated.
FIRST ORDER AND FIRST DEGREE. 323
{l+x)i/.dx-\-(l -y)x.dy={),
\ogx + x + \ogy-y=G,
log xy — G+y — x.
Integrating, we have sin"^ x + sin~^ y — c.
This may be put in other shapes. Thus taking the sines
of the two sides of the equation we have
X Vl-y2 ^ y Vl_^ = (7.
'^^ (^ " ^ • iTv ^^^^"^^^ (2/ + 2/') dy = (x + ^) (^^.
Answer : ^y^ ^ j^^s _ |^ ^_ ^^ _j_ constant.
5. ^±5") = ^. Answer: (1+^) (1+2/^) = c^.
l-¥y^ ay ^ f \ a f
6. sin ^ . cos y . c^o; — cos a? . sin y . (ii/ = 0.
Answer: cos y = c cos ij?.
7. {f ■\- xy'')dx -^ {cc" - yo(P')dy =^.
Answer : loff - = c + ^ .
y ^y
^' -^. + a/t-^=^- Answer: VlT^^ + VIT? == C.
238. Sometimes we guess and find a substitution
which answers our purpose. Thus to solve
dy _y^ — x
dx Ixy '
. dx
we try y — Nxv, and we find \-dv=^0, leading to
log a? + — = c.
Solve (y - x) Vr+^ ^ = ?i ( 1 + 7/2)1
21—2
324 CALCULUS FOR ENGINEERS.
239. If M and N are homog^eneous functions of ic
and y of the same degree : assume y = vx and the equation
reduces to the form of Art. 237.
Example 1. ydx + (2 ^fxy — a;) dy — 0. Assume y — vx,
dy==v.dx-{-x.dv,
vx.dx-^ {2x s/v — x)(v.dx + x. dv) = 0,
(2^^) dx + (2.X-2 VS - x") dv = 0,
2dx 2 Vti - 1 , ^
X \v V^J
2 log .r 4- 2 log^; + 2v-* = 0,
log XV + v~* = (?,
Answer: y = C€ V »,
where c is an arbitrary constant.
Let y — vx and we find the answer
x^ — y^
+ log((;^-yi)(.^-y)i)-=0.
Answer: a^ ^'^.Ay ^-A^. '
Remember that two answers may really be the same
although they may seem to be altogether different.
4. Solve (x^ + Sxy"") dx + (y^ -f 3^y) dy = 0.
Answer : ^-^ 4- 6.r'y^ + y*=C.
V
HOMOGENEOUS EQUATIONS. 325
5. Solve 3^24.(^^4.^)^ = 0.
Answer : Zxy + x" + 2^^ ^ q^ (2y + xf.
6. Solve [x — y cos ^ J c?^' 4- a; cos - .dy =0.
Answer : x = ce *
_x
7. (y/ — x)dy-\-y. dx = 0. Answer : 1y = ce K
8. j;c?2/ — y-dx — ^x^ 4- y^ . cZ.:^ = 0. Answer : x^ = c^ + 2cy.
9. x-\-y~ ^%j. Answer : (./; - y ) e^ " ^ = (7.
QjX
240. Of the form {ax + by-{-c) dx + (a'i^- + b'y + c') (/y = 0.
Assume x = w + a, y = v + y^, and choose a and /3 so
that the constant terms disappear.
Thus if (3^' - 2y + 4>) dx -{■ (2x - y -^l)dy ==0 ', •a.sdx^dw
and dy = dv, we have
(Sw-{-S^ji-2v-2^-\-4>)dw + (2w-\-2oL-v~^ + l)dv = 0.
Now choose a and /9 so that
3a- 2^ + 4 = and 2a-/34-l = 0,
or -a4-2 = 0, or a = 2, /3=5.
Therefore the substitution ought to be x=tu-^2, y = v-\-b,
and the equation becomes a homogeneous one.
Exercise. (3y - 1 x -{■ 1) dx -^ {1y - Sx + 3) dy=^0. ^ 'n.'-
Answer : (y — x + 1)^ (y + x — ly — c.
r, • dy 2x — y-\-l _
Exercise. -f + ^ ^ — ^ = 0.
ax zy — x—1
Answer : x^ — xy-{-y'^-\-x —y=c.
241. Exact Differential Equations are those which
have been derived by the differentiation of a function of x
and y, not being afterwards multiplied or divided by any
function of x and y. Consult Art. 83.
.326 CALCULUS FOR ENGINEERS.
Mdx 4- Ndy = is an exact differential equation if
.dM\ (dN\, ,, df{x,y) ,^ d., .
the primitive being /(^, y)=c. It will be found that
(x" '-Safy)dx-\-(y^- a-') dy = 0,
is an exact differential equation.
Then x'-Sx^y=^ ^-^'^f''^\
OjCO
Integrating therefore, as if y were constant, and adding Y
an unknown function of y, instead of a constant,
f{x,y) = la^-x^y^Y.
Differentiating as if /c were constant, and equating to Ny
we have
dV
-J- = y^ and hence Y=^y^ + c.
Hence ^x" - afy + ^y^ + c = 0,
where c is any arbitrary constant.
242. Any equation M . dx-\- N .dy = may be made
exact by multiplying by some function of x called an Inte-
grating Factor. See Art. 83. For the finding of such
factors, students are referred to the standard works on
differential equations.
243. Linear equations of the first order.
These are of the type
s+^y=« W'
where P and Q are functions of x.
The general solution is this. Let IP.c^a* be called X,
then
y=e^{/e^.Q.dx+cJ (2),
where C is an arbitrary constant.
LINEAR EQUATIONS. 327
No proof of this need be given, other than that if the
vakie .of y is tried, it will be found to satisfy the equation.
Here is the trial : —
(2) is the same as ye^ = I e^Qdx-\- G (3).
and
dy
J XT
Diiferentiating, and recollecting that -y— = P, (3) becomes
dx
e^ + ye^P=e^Q (4),
ov -~- + Py — Q, the original equation.
To obtain the answer (2) from (1), multiply (1) by e^ and
we get (4) ; integi-ate (4) and we have (3) ; divide by e^ and
we have (2).
We have, before, put (1) in the form
(e + P)y=Q or y^ie + PrQ.
and now we see the general meaning of the inverse operation
(d + P)-\ In fact if I P.dwhe called X,
{d + P)-i Q means, e^^ || e=^ . Q . dx + c| . . .(5).
Thus if Q is 0, (0 + P)-i = Ce-^. Again, if P is a con-
stant a, and if Q is 0, then (6 + a)~^ = Ce~"^, where C is an
arbitrary constant. We had this in Art. 168.
Again, if Q is also a constant, say n,
{6 4- a)-' n = e""^ I ( He'*^ . dx -[- C
{!■
= a€-«* + - (6),
ft
where G is an arbitrary constant. See Art. 169. "
Again, if Q = e^^,
((9 + a)-' e^ = e-^'-^ | j €'«+^) "^ . dx ■{■ g\
1
.(7).
328 CALCULUS FOR ENGINEERS.
It is easy to show that when a = — h,y = (G-i-a;) €~"*
li Q — h sin (ex + e),
(6 + a)-' h sin {ex -\-e) = e-«* j h | €*»* sin {ex + e) . dx + C
= Ce-«^ + , sin fco; + e - tan-^ -^ (8).
244. Example. In an electric circuit let the voltage
at the time t be F, and let (7 be the current, the resistance
being R and the self-induction L. We have the well-known
equation
F=ie(7 + X^,
or ^^^G=\rV.
Now
dt L L
--t (1 f-t )
and hence C=6 ^ ]j j^^ F . cZ^ 4- constant -4 [ (1).
Of this we may have many cases.
1st. Let F at time 0, suddenly change from having been
a constant Fj, to another constant V^. Put F=F2 therefore
in the above answer, and we have
0=r^'UF,e^ + 4|
V -it
y
To determine A we know that = ^ when^=0; there-
in!
V V V — V
fore -^^ = ^^ 4- ^ so that A = — -„ — ^ and hence
±1 li K
c=Yi-Y^^I-^r^' (2)
ELECTRICAL EXAMPLE. 329
K/ --t\
Thus if Fi was 0, ^=^(1--^^) (3)>
showing how a current rises when a circuit is closed.
V -^t
Again if F2 is 0, ^=:^^^ ^^)'
showing how a current falls when an electromotive force is
destroyed.
Students ought to plot these values of C with time.
Take as an example, F^=100 volts in (3), i2 = 1 ohm,
L = -01 Henry.
Again take Vi= 100 volts in (4). Compare Art. 169.
2nd. Let V at time suddenly become
Vq sin qt,
a=e ^
^(F
V f -t
~ le^ .smqt.dt-\-A
itfR .
R \ ^r ^^ f -f sin qt — q cos qt )
-^t V
This becomes C = Ae ^ + , " sin (qt -e).. .(5),
where tan e = ~ .
M
The constant JL, of the evanescent term Ae ^ , depends
upon the initial conditions ; thus if (7 = when ^ = 0,
= ^.-^=^ sing,
or A = VoLql(R' + Df).
Students ought to plot curves of several examples, taking
other initial conditions.
330 CALCULUS FOR ENGINEERS.
245. Example. A body of mass M, moving with velocity
V, in a fluid which exerts a resistance to its motion, of the
amount fv, is acted upon by a force whose amount is F at
the time t. The equation is
M%+fv = F.
Notice that this is exactly the electrical case, if M stands
for Z,/for R, F for Volts V, v for C; and we have exactly the
same solutions if we take it that F is constant, or that F
alters from one constant value Fi to another F.., or that F
follows a law like Fo sin qt
This analogy might be made much use of by lecturers on
electricity. A mechanical model to illustrate how electric
currents are created or destroyed could easily be made.
The solutions of Linear Differential Equations with
constant coefficients have such practical uses in engineering
calculations that we took up the subject and gave many
examples in Chap. II. Possibly the student may do well
now to read Art. 151 over again at this place.
246. Example, x ^ = ay-^x + \,
dy a ^ 1
OC X X
I-
— . dx=X= — a loff a-.
X ^
Observe that e-'*^^'^'" = x'"', e^^^K'^ = x^.
Hence y = x^\ L"" ( 1 + -\ dx -\- c\ y
y = x^ ||(^-« + x-^-^) dx-^c\,
" \ — a a
the answer, where C is an arbitrary constant.
LUBRICATION OF JOURNAL.
331
247. There being continuous lubricating liquid between the surfaces
AB and EF as of a brass and a journal.
OC=/iQ the nearest distance between them.
At the distance ^, measured along the arc
OAj let the thickness be h. Anywhere in.
the normal line there, representing the
thickness, let there be a point in the liquid
at the distance y from the journal, and
let the velocity of the liquid there be 2t.
Then if p be the pressure it can be shown
'^- 541 (^)'
if /u, is the coefficient of viscosity of the
lubricant, and Wq the linear velocity of
the journal, and ic is the velocity of the
liquid at any place; we have no space for the reasoning from (1)
G^Uq dh
h^ duo
Fig. 103.
leading to
d'^p 3 dh dp
dx^ h ' dx dx
.(2).
Let
dp _
dx
0,
then
d4 3
dx h
dh 6/iWg ^^^ _ /^
dx ' ^ A3 dx''
This is of the shape (1) Art. 243, h in terms of x being given.
Let X= JP . dx= f f . ^ . dx=3\ogh, e^=h\ Hence
(t> = h
\l
-A3
QjiUq dh
h^ dx
dx + &
}.-<^=-|=
A-'(6;.«„A + = ^?+f3.
The solution depends upon the law of variation of A. The real case
is most simply approximated to by h = hQ + ax^: using this we find
p = C'
GixUq
~2h
V
C'-C
If students were to spend a few weeks on this example they might
be induced to consult the original paper by Prof. 0. Reynolds in the
Phil. Trans, vol. 177, in which he first explained to engineers the theory
of lubrication, t
Numerical Exercise. Let OB = 2-59, OA = 11 -09 centimetres.
fi = 2-l6, Aoor 0C=0-001135, a- -0000082, W(, = 80cm. per second.
332 CALCULUS FOR ENGINEERS.
Calculate C and C\ assuming jo = at B and at A.
Now calculate the pressure for various values of x and graph it on
squared paper. The friction per square cm. being /ij- at y=0, the
total friction F will be found to be
11
between the limits A and B. The total load on the bearing is \p.dx
bearing is / ^ . (
between the limits, if AB covers only a small part of the journal, and
may be calculated easily in any case.
The bearing is supposed to be infinitely long at right angles to the
pai)er in fig. 103, but forces are reckoned i)er cm. of length.
248. Example. Solve
Writing this in the form (^ - 4^ + 3) 7/ = 2e^,
Now (e^ - 4^ + 3)- = i (^-^^ - ^ :^ i) •
Indeed we need not have been so careful about the J as
it is obvious that the general solution is the sum of the
two (^— 3)~^ and (^ — 1)~^ each multiplied by an arbitrary
constant.
Anyhow, y = ^__^_^,
and this by (7) Art. 243 is
or y=(C^+a')^-hG.je\
249. Equations like ^ 4- Py = Qy", where P and Q,
as before, are functions of oo only.
Divide all across by y* and substitute z = y^~'\ and the
equation becomes linear.
Example. (1 — w"^) -^ — cci/ = ax\f.
DIFFERENTIAL EQUATIONS. S38
Substituting z = y~^ we find
dz xz ax
dx \—x^ \-x'^'
Answer :
2/-1 = (1 _ ^)*{_ a(l -j;2)-^ + 0} = - a + OVr^T^^.
Exercise. ^~j -^V =y^ log iz*.
Answer : - = 1 + Oa; + los: a?.
2/ ^
250
. 1. Given C^Y-ay = 0.
This is an equation of the first order and second degree.
Solve for -^ and we find two results,
dx
dt/
-^ — ai/ = 0, so that log y—ax—A^ = 0,
-^ 4- a^ = 0, so that log y + ax— A2 = 0.
Hence the solution is
(log y - ax - A,) (log y + ax-A^)= 0.
It will be found that each value of y only involves one
arbitrary constant, although two are shown in the equation.
2. Given i + (^^Y = ^,
This is an equation of the first order and third degree.
Hence 2 = (.^-l)*,
and y = |(.r-l)HG
334 CALCULUS FOR ENGINEERS.
3. Given (^y_7^ + 12 = 0.
\dxj ax
This is an equation of the first order and second degree.
(2-*)g-»)=»^
(3/ - 4^ + Ci) (y - S.'Zr + Co) = 0.
251. Clairaut's equation is of the first order and of any
degree
y^xp-Vf{p) (1),
where p is -^ and f{p) is any function of ~ .
Differentiate with regard to x, and we find
^■^iMt=' <^)-
So, either ^ = (3)
^ + ^fiP) = 0.... (4)
will satisfy the equation.
1=0., =0.
Substituting this in (1) we have
y = CX+f{c) (5),
which is the complete solution.
Eliminating^ now between (1) and (4) we obtain another
solution which contains no arbitrary constant. Much may
be said about this Singular Solution as it is called. It is
the result of eliminating c from the family of curves (5), and
is, therefore, their Envelope. See Art. 224.
Example of Clairaut's equation.
m
We have the general solution (5) ... y — cxA — , a family
c
HIGHER ORDER THAN FIRST. 385
of straight lines the members of which differ in the values of
their c.
Hence y = 2 Jmx or y^ = 4ma;, a parabola which we found
to be the envelope of the family in Art. 224.
This curve satisfies the original equation, because in any
infinitesimal length, the values of x, y and -^ are the same
for it as for a member of the family of straight lines.
252. If a differential equation is of the form
it can be at once solved by successive integration. We
have had many examples of this in our work already.
253. Equations of the form -^^=f{y); multiply by 2 ^
cix dx
and integrate and we have
Extracting the square root, the equation may be solved,
as the variables are separated.
Thus let g = ay
Proceeding as above,
y -=^dx.
Integrating we find
x=^^\og{ay + JaY-^C}-\-C' (1).
336 CALCULUS FOR ENGINEERS.
If this equation (1) is put in the shape
ce'^^ = ay-\- Ja^ + C,
it becomes c^e^'' - ^ayce*"^ = C,
^ la
or ?/=^6«^ + ^e-«^ (2),
which looks different from (1) but is really the same. (2) is
what we obtain at once if we solve according to the rule for
linear equations, Art. 159.
254. Solve -7^= a -s~, an equation of the third order
and first degree.
Let, g = ?, theng = a5,
9 = 6e«* so that t^ = - «"^ + G,
^ a?
or y == Ae"^ •\- Coo+ C\ where A, 0, C are arbitrary constants.
This also might have been solved by the rule for linear
equations, Art. 159.
255. Solv(
a. dp
^^=:dx SO that -=log {p+ V«2+ 1) +C",
or (7e<»-jp = Vp2 + l.
1 -^ d.
20 ~dx
- 1 -- dti
Squaring, we find p = ^Ce** — ^r^ e "
EXERCISES. 337
Integrating this we have
X X
where G and c are arbitrary constants.
256. General Exercises on Differential Equations.
(1) {a? + y-) dx + Ja^^^ .dy = 0.
x X y
Answer : sm"^ - + — tan"^ ^ = c.
a a- a
/^\ dy X 1
^^ dx^l-\-x^'^ 2x(l+x')'
. , _ 1 + JU^']
Answer
1 f ,, l-fVl+^^'l
(3) _ + ycOS*--^.
Answer : i/ = sin i» — 1 + Ce" ^^"'^.
Answer : (2y -x'^-c) (log (a; + y - 1) + ^ - c} = 0.
(6) J+2^y = 2a^y.
Answer : y = (Ce^^' + ^a {2x^ + l)}"*.
(6) 1 = f ^V + ?^ ^ . Answer : y' = 2c^ + c\
^ ^ \dxj y dx
(7) x-^ + y = y^logx. Answer: y == {ex + log x + 1)-\
(8) y = 2x^- + y^ C^Y . Answer : y^ = 2c^ + c\
(9) Solve (l + 5)c^^-2|c^y=.0.
1st, after the manner of the Exercise of Art. 241.
2nd, as a homogeneous equation.
Answer: af — y"==cx,
P 22
338 CALCULUS FOR ENGINEERS.
(10) Solve — '- f- 1 -^ J c?y = after the manner of
Art. 241. Answer : x^ — y^ = cif.
Answer : y = {x-[-\y [^ {x + 1)^ + C).
(12) (a; + 2/)2 ^ = a\ Answer : y-a tan-^ ^^ = c.
(13) xy (1 4- a??/2) ^ = 1. Answer : - = 2 - 2/^ + ce-^J/'.
Answer : (y —a log a; — c) (?/ + a log a; — c) = 0.
Answer : y = (ao + aio; + aoa?'' + a^) e~^.
Answer : a? = (ai + dj^) e-^' + /-^r— Tvj •
Answer : 3/ = (a? + 1)" (e* + c).
(19) (1 —x^)-f- — xy = axy\ Answer : - = c Jl — a^ - a.
(1^) yi = sin (^ - 6). Answer : cot |j - ^-^ I = <^ + c.
Answer: 3/ = C^ie'^^ + e^ Jf Og-^j cos a; + fOg + ^jsinajl .
(21) Change the independent variable from a; to i in
(1 — a?) -~ — a? -^ = 0, if a; = sin f, and solve the equation.
ELLIPTIC INTEGRALS. 339
(22) Also in {a- +x'')-j^^-\-2x -^ = 0, ii x=a tan t, and
solve the equation. %
«■ '■ '-■• S=-£/©'.
, „ d's [dt dH „fdHV\ fdty
*""* ^- -dt= = -\ds-d^-^W)\^\ds)-
3. Prove as i^ = -^ / tt, so we have also
ax dt I dt
d^y _ fdx (Py _ d'x dy\ ^ /dxV
dx^ ~ [dt df dt^ ' dt) ' \dt) '
and find the equivalent expression for -^^ .
4. If X = e* show that b,sx^ — x-^-~-t: this equals -^ .
dx dt dt ^ dt
Also a^^^Jl-^^^-fl-l]^
^^^"^ "^ da?~ dP dt~\dt V dt'
and a^^-(l-2\(^-l]^
^"""^ "^da--\dt V\dt ^Jdt'
5. Change the independent variable from x to t if x = €*,
-^--\-x^
dx^ dx
in x^ ;r 2 + ^' ;/ + ^^^y = ^> ^^^ ^^^^^ *^® equation.
258. If we try to find by the method of Art. 47 the length of the arc
of an elhpse, we encounter the second class Elliptic Integral which is
called E (k, x). It may be evaluated in an infinite series. Its value
has been calculated for values of k and x and tabulated in Mathemati-
cal tables.
When the angle through which a pendulum swings is not small, and
we try to find the periodic time, we encounter the first class Elliptic
Integral which is called F (k, x). It can be shown that the integral of
any algebraic expression involving the square root of a polynomial of
the third or fourth degree may be made to depend on one or more of
the three integrals
22—2
340 CALCULUS FOR ENGINEERS.
rx /i _ z.2«.2 re .
f 1 \ i"" ^
7r(w, h. 6)=\
;o(l+7ism2^)
do
\Jl-k^Hm^6'
k, which is always positiA e and less than 1, is called the tnodulus.
n, which is any real number, is called the parameter.
The change from the x form to the 6 form is effected by the substi-
tution, :r = sin 6. When the limits of F and E are 1 and in the x case,
or - and in the B case, the integrals are called complete, and the letters
K and E m erely are used for them. B is called the amplitude and
\/\ — ^2sin2^ is called by the name A^.
If u = F{k, x) = F{ky B\ then in dealing with fimctions which have
the same k if we use the names
^ = amw,
x=Bnu (in words, x is the sine of the amplitude of w),
*^l — x^=cmi (or Vl — -^ is the cosine of the amplitude of u),
Vl - k^x^=dnu (or Vl - k'^^^ is the delta of the amplitude of u),
it is found that
sn^w-f cn2?< = l, dn'^u+k^.Bn^u=l, -=-(amw)=dnw, &c.
Also am { — u)=— am . ti, &c.
., / . \ sn w. en V. dn v + cni/. snv.dn . -M
Also sn {u±v) = Yx-== r, ,
and similar relations for en {u±v) and dn {u±v).
Expressions for sn (tc + v)+sn(u- v), &c. follow. Also for
sn2u, cn2w, dn2u.
So that there is as complete a set of formulae connecting these
elliptic functions, as connect the Trigonometrical functions, and there are
series by means of which tables of them may be calculated. Legendre
published tables of the first and second class integrals, and as they have
known relations with those of the third class, special values of these
and of the various elliptic functions may be worked out. If complete
tables of them existed, it is possible that these functions might be
familiar to practical men.
259. To return to our dififerentiation of IHinctions
of two or more variables.
HEAT CONDUCTION EQUATION. 341
1. If u = z'^ + 7/^ -\- zy and ^ = sin a; and y = e^,
du
and hence ^ = (Si/^ + ^) e* + (2^ + y) cos a;. If this is expressed
all in terms of x we have the same answer that we should
have had, if we had substituted for y and z in terms of x in
u originally, and differentiated directly.
2. If u= a/ — — „ , where v and w are functions of «?,
V v^ + w-
nnd -Y .
ax
3. If sin (xy) = ma?, find -^ .
4. If %i — sin~^ - , where z and y are functions of x, find
da?*
5. If \i = tan~^ - , show that du, = ^-^^ — -^ .
y f + 2^
260. Exercise. Try if the equation
dx^~Kdt ^ ■
has a solution like v = e*^ sin (qt + yx), and if so, find a and y,
and make it fit the case in which v=0 when x = oo , and
v = asm qt where x = 0. We leave out the brackets of
'dv""
C)
&c.
dv
-7- = «€"* sin (qt + yx) + e*^ 7 cos (qt + ^cij),
— — = a^e*^ sin (g^ + yx) + a7€*^ cos (qt + 7.<c)
+ a7e** cos (qt + yx) — 6**7^ sin (qt + 70;).
dv
Also -^ = ge*^ cos (qt + 7^),
342
CALCULUS FOR ENGINEERS.
SO that to satisfy (1) for all values of t and oc
a"-rf=:0 or a= ±y,
and
ay + ay
_2
As - is not zero, a = + 7 only,
2a=» = ^, a= +
K
Hence we have
v= A e"'* sin (qt -f cue) -f i^e"** sin (qt — ax)
where A and B are any constants, and ol= /u -x— or a/
if ^ = 27r/2. Now if i; = when x= oc , obviousl}^ A=0. If
V = a sin qt where x=0, obviously B = a.
Hence the answer is
Trn
V — ae
V " sin (27r?^^ - .'?? W
K )
.(2).
261. Let a point P be moving in a curved path
^PQ; let^P = a^,j5P = y,
Cm *ir (ill
-^ and -~ being the velocities in the directions OX and OY^
-j~ and -7^ being the accelerations in the directions OX
and OY.
Let OP=r, BOP —6,x = r cos ^, ?/ = r sin ^. The acceler-
ation or velocity of P in any
direction is to be obtained just
as we resolve forces. Thus the
velocity in the direction r is
Jcos^4-Jsin^...(l),
and in the direction PT which
is at right angles to r, the
velocity is Pig^ 104
-t«in^ + ^!cos^ (2).
Y
A
t\
\
;
p
o\
I
3
dt
dt
POLAR CO-ORDINATES. 343
Now diiferentiating a; and y, as functions of the variables
r and 0, since ( ^— j = cos and f -^ J = — ?* sin 6,
dx dr ^ . ^ dO ,.-,.
5«=*'=°^^-'-^'°^rfi ^^^'
dy dr . ^ ^dO ,^.
Solving (3) and (4) for j- and r -j- , we find
dr dx ^ dy . ^
= ;i- cos ^ +-^ sm d (o),
(^^ dt dt
dO dx . ^ a
dt dt dt
dO dx . ^ dy ^ ,_.
From (1) and (2) we see therefore that -j- is the velocity
de
dt
in the direction OP and that r -^ is the velocity in the
direction PT. Some readers may think this obvious.
Now if we resolve the x and y accelerations in the
direction of OP and PT, as we did the velocities, and if we
again differentiate (3) and (4) with regard to t, we find
d^x d^y
Acceleration in direction OP = -i- cos ^ + -i^ sin ^ . . .(7),
d^x d^y
Acceleration in direction PT— — ^ sin ^ + -j^ cos ^...(8).
And
S={£-'©"}-*-('Sf-S) '"•■■<'>.
S=(S-(S)l"»»-(4JS-S)~«-»
And hence, the acceleration in the direction r is (and
this is not very obvious without our proof),
dt' '\dt)
(iiX
344 CALCULUS FOR ENGINEERS.
and the acceleration in the direction PT is
^dtdt^"^ dt^'°' rdtK dt) ^^^^-
r^ -J- is usually called A. It is evidently twice the area per
second swept over by the radius vector, and (12) is - -^ .
262. If the force causing motion is a central Force,
an attraction in the direction PO, which ia a function of
r per unit mass of P, say /(r) ; or mf{r) on the mass m at
P; then (12) is 0, or r^ -^- = constant, or h constant. Hence
under the influence of a central force, the radius vector
sweeps out equal areas in equal times.
Equating mf{r) to the mass multiplied by the accelera-
tion in the direction PO we have
/<')"©-!?■ "»>
But r"^ -^ = h a. constant. As r is a function of 6
dt
dr _dr dd _dr h
di~dd'di~der^'
^ ^\^h _ h (dA^^)h
df'~\d6'7'' 7^\dd)]r''
and we can use these values in (13) to eliminate t.
If we use - for r, (13) simplifies into
f(r)=h2u2(^+u) (14).
If f(r) = a?'~" or a^*", an attraction varying inversely as
the ?ith power of the distance, -^ + ^^ = I^ '^^~^ — ^w"~^ say.
Multiplying by 2 -^ and integrating, we have
(sr— «-?!— <->•
CENTRAL FORCE. 345
Thus let the law be that of the inverse square;
f{r) — ar~^ or ait^ ; (14) becomes
(Pu a ,
Let lu — iL — h, then
The solution of this is,
-m; = ^ cos (^ + B),
and it may be written
i,= l = ^{H-ecos(^-a)l (IG).
This is known as the polar equation to a conic section,
the focus being the pole. The nature of the conic section
depends upon the initial conditions.
(15) enables us, when given the shape of path, to
find the law of central force which produces it. Thus if a
particle describes an ellipse under an attraction always
directed towards the centre, it will be found that the force of
attraction is proportional to distance. It is easier when given
this law to find the path. For if the force is proportional
to PO, the X component of it is proportional to oo, and the y
component to y. If the accelerations in these directions are
written down, we find that simple harmonic motions of the
same period are executed in these two directions and the
composition of such motions is well known to give an
elliptic path. If the law of attraction is the inverse cube
or f{r) = (7?'~^ = au^, (14) becomes -j^- + ?^ = r^ ^^•
If ^-l = a-, w = ^6*« + 56-*^
h
If l-~ = ^2 u = Asm0 + Bcos0e,
fi-
giving curiously different answers according to the initial
conditions of the motion.
346 CALCULUS FOR ENGINEERS.
263. If a? = r cos 0,y = r sin 6, so that if y is a function
of x, r must be a function of ^ ; if i* is any function of x and y,
it is also a function of r and 6.
Express f -7- ) and f -^ j in terms of the polar co-ordinates
r and 6.
\drJ~\dx)dJi^^\d^)d^ ^^^'
6 being supposed constant, and
fdu\ _ fdu\ dx /dic\ dy
\de)~\Tx)de^\d^jlde ^^^'
r being supposed constant.
dx
Now -77; if r is constant = — ?- sin 0,
au
-M. if r is constant = r cos ^,
-7- if ^ is constant = cos ^,
— ^ if ^ is constant = sin ^.
ar
Treating (-7- J and (-7-) in (1) and (2) as unkno\vn, and
finding them, we have
(|) = .„..g).Jeos..(|) (4).
Notice that in [-,— ], the bracket means that y is supposed
to be constant in the differentiation.
In ( -7- j , it is ^ that is supposed to be constant.
r, ^, <^ CO-ORDINATES.
347
In (o) or (4) treat ( i-l or ( -, ) as it is treated, and find
^i— and Ti^ . However carefully one works, mistakes are
da^ dy^ *^
likely to occur, and this practice is excellent as one must
think very carefully at every step. Prove that
d2u d2u
d^
di^
d^'^dy2"dr^ "^^r dr'*"?d9^'
1 du
r dr
1 d2u
.(5).
264. Sometimes instead of x, y and z, we use r, 6, (f)
co-ordinates for a point in space. Imagine that from the
centre of the earth (Fig. 105), we have OZ the axis of the
earth, OX a line at right angles to OZ, the plane ZOX being
through Greenwich ; OF a line at right angles to the other
two. The position of a point P is defined by x its distance
from the plane ZOY, y its distance from the plane ZOX, z its
distance from the equatorial plane YOX. Let r be OP the
distance of the point from 0. Let <^ be the west longitude
or the angle between the planes POZ and XOZ\ or if Q be
the foot of the perpendicular from P upoii XOY^ the angle
QOX is <^. Let Q be the co-latitude or the angle POZ.
Then it is easy for anyone who has done practical geometry
to see that, drawing the lines in the figure, QfiO is a
348 CALCULUS FOR ENGINEERS.
right angle and OR = x, QR — y, also PQO is a right angle,
a; = rsm6. cos </>, y = r sin 6 .sincf), z = r cos 6. If u is a
given function of x, y and z, it can be expressed in terms of
r, 6 and ^, by making substitutions. It is an excellent
exercise to prove
du . ^ , du cos 6 . cos <f) cZm sin ci c??^
-y- = sin ^ cos </> ;t- H ^ -ja ^n -ji ,
dx ^ dr r do r sm 6 d<f>
du . ^ . , du cos 6 ,sin<f> du cos <f> du
dy ^ dr r dd r sm 6 d<f>
du _ ^ cZ?^ sin ^ du
dz dr r dO'
It will be noticed that we easily slip into the habit of
leaving out the brackets indicating partial differentiation.
The average student will not have the patience, possibly
he may not be able to work sufficiently accurately, to prove
that
d2u d^ d2u_d2u l^d^
33? + dy2 ■*■ dF " dp" "*" r2 d(92
1 d2u 2du cotg du
■^r2sin2^'d^ + 7d?"^"J2-d9 ^ ^•
This relation is of very great practical importance.
265. The foundation of much practical work consists in
understanding the equation
d^u dhc d^u__ldu
dx^'^df'^'d?~'K~dt ^ ^'
where t is time. For example, we must solve (1) in Heat
du
Conduction Problems if u is temperature, or in case ;t- =
and u is electric or magnetic potential, or velocity potential,
in Hydrodynamics.
(1) is usually written
^^-4S (^>-
ZONAL HARMONICS. 349
We see then in (A) the form that V"u takes, in terms of r,
and </> co-ordinates.
We know that if u is symmetrical about the axis of z,
that is, if u is independent of </>, the above expression
becomes
_, dhi 1 d-ii 2 da cot 6 du ,^.
^'''=d;^-^?w^-^rd^+-^d0 <^>-
266. Students are asked to work out every step of the
following long example with great care. The more time
taken, the better. This example contains all the essential
part of the theory of Zonal Spherical Harmonics^ so
very useful in Practical Problems in Heat, Magnetism,
Electricity, Hydrodynamics and Gravitation. When to is
independent of </> we sometimes write (2) in the form
du K id f „du\ 1 d / . ^du
'hMei^-'U ^^>'
dr
u being a function of time t, r and 6, The student had
better see if it is correct according to (3).
dxi
If -^ = 0, show that the equation becomes
„d^u - du ^ ndu d^u _ ,^.
,._ + 2r^^ + cot^^^ + ^, = (2).
Try if there is a solution of the form u = RP where R is
a function of r only, and P is a function of 6 only, and show
that we have
T^d^R 2rdR ,AdP 1 d'P ,^,
R^^^Rd^=-''''^^p~de-pd¥ (^^-
Now the left-hand side contains only r and no 0, the
right-hand side contains only 6 and no r. Consequently each
of them must be a constant. Let this constant be called C
and we have
drR ldR_RG_
dr' +r dr r^ ~" ^^^'
%^.oie%^PG = (5).
360 CALCULUS FOR ENGINEERS.
There is no restriction as to the value of 0, and it must
be the same in (4) and (5), and then the product of the two
answers is a value of u which will satisfy (2). The solutions
of many linear Partial Differential Equations are obtained
in the form of a product in this way. There are numberless
other solutions but we can make good practical use of these.
We have then reduced our solution of the Partial Differ-
ential Equation (1), to the solution of a pair of ordinary
differential equations (4) and (5). Now a solution of (4) may
be found by trial to be r*", and when this is the case we have
a method (see Art. 268) of proving the general solution to be
ii = ^r»« + 5r-(»«+i' (6),
where G is m (m + 1); anyhow (6) will be found by trial to
answer. Using this way of writing G in (5) and letting
cos = fjb, we find that we have an equation called Legendre's
Equation, an ordinary linear equation of the 2nd order
We now find it convenient to restrict m. Let m be a
positive integer, and try if there is a solution of (7) in the
form
P = 1 + Aj/M + A^^ + A^^ + &c.
Calling it P^ (/it) or P,^ (0), the answers are foimd to be
Pq (0) = 1, if m is put 0, Pi (0) = ^, if m is put 1,
Pg (0) = 3^2 _ ^, if ^ is put 2, Ps (6) = f /!« - f /x, if m is put 3,
P, (0) = ^fi* - ^fi^ + 3, if m is put 4.
A student will find it a good exercise to work out these
to Pg. My pupils have worked out tables of values of Po, Pi,
P2, &c., to F7 for every degree from ^ = to ^=180°. See
the Proceedings of the Physical Society, London, Nov. 14,
1890, where clear instructions are given as to the use of
Zonal Harmonics in solving practical problems.
We see then that
(Ar- + j^,)p„(<)) (8)
BESSEL FUNCTION. 351
is a solution of (1). A practical problem usually consists in
this : — Find u to satisfy (1) and also to satisfy certain limit-
ing conditions. In a great number of cases terms like (8)
have only to be added together to give the complete solution
wanted.
In the present book I think that it would be unwise to do
more in this subject than to set the above very beautiful
exercise as an example of easy differentiation.
(8) is usually called The Solid Zonal Harmonic of
the mth degree^ P^ (6) is called the Surface Zonal
Harmonic of the mth degree.
267. In many axial problems, u is a function only of
time and of r the distance of a point from an axis, and we
require solutions of (1) which in this case becomes
d^u \ du ^\ du
dr^ r dr k di *
Let us, as before, look for a solution in the form
u = BT (2),
where i? is a function of r only and J' is a function of t only.
(1) becomes
rpd^_^lrpdE^lj^dT
dr^ r dr k dt'
Dividing by RT
Rdr'^r Rdr~ kT dt~ ^"^ ^^^'
where /^^ is a constant.
Jim
Then -m = - /cfi^dt or log T= — k/jlH + c, or
T=Ge-'^f^'i (3),
where G is an arbitrary constant. We must now solve
d'R IdR ,^ ^ ,,,
^+-r;^+^^^=^ (^)-
Let r = - and (4) becomes
d2R 1 dR . „ ^ ,^^
dF + xdJ + ^ = ^ (^>-
352 CALCULUS FOR ENGINEERS.
Assume now that there is a solution of (5) of the shape
we find that A = G = E= G=0 and in fact that
x^ X^ x^ x^
This is an important series first used by Fourier, although
it has Bessels name. It is called the Zeroth Bessel and
the symbol Jo (*") is used for it. Tables are published which
enable us for any value of x to find Jo(x). Thus then
R = Jo {^r) is a solution of (4), and hence
u=Ge-'^|'''J,{^lr) (7)
is a solution of (1). Any solution of (1) needed in a practical
problem is usually built up of the sum of terms like (7), where
different values of iju and different values of G are selected to
suit the given conditions.
268. In the linear differential equation
when P and Q are functions of x, if we know a particular
solution^ say y=^v, we can find the general solution.
Substitute y = vii, and we get
"d^+^id-. + ^Vd^-"' (2).
Calling -p= u\ (2) becomes
U V
or
log u' + log v^+ I P .dx = constant.
LINEAR EQUATIONS. S53
Let I F .dx = X then u' or -^ = A ~ e-^,
J ax v^
u^B + Al^^e-'^.dx (3).
Thus we lind the general solution
y = Bv-\- Av I — €~^ .dx .(4),
where A and B are arbitrary constants. Even if the
right-hand side is not zero, the above substitution will
enable the solution to be found, if v is a solution when the
right-hand side is 0.
Easy Example. One solution of
-r^ + a^x = is ?/ = cos ax.
dx^ "^
Find the general solution.
Here P = so that jP.dx^X^ 0.
f dx
Hence y = B cos ax + A cos ax I — -i, — ,
•^ J cos^ax
and as I — - — = - tan ax,
J cos^ ax a
we have as the general solution
y = B cos ax+G sin ax.
Exercise. We find by trial that y = x^"' is a solution of
ic* ^ + 2a; ^ - m (m + 1) y = 0, see Art. 266.
Show that the general solution is y=Ax^-^ — ;^^.
X
Exercise. We find by trial that y — e^^ is a solution of
d^v
-T^ = a^y, show that the general solution is y = Ae^^ + i?e~^*.
P. 23
S54
CALCULUS t'OR I:NGINEERS.
E.rercise. We saw that u — P^n{t^) is a solution of
Legcndre's equation Art. 2GG, prove that xl =APin(fi)-\-BQmi^)
is the general solution, where
Q,„W=P,„(^)/^j-^^-
Wl"
Qm(H') or Qui(0) is called the Surface Zonal Harmonic
of the second kind.
Exercise. We saw that Jo(x) was a solution of the
Bessel equation (5), show that the general solution is
AJq {x) -f BKq {x)y where
Kq (x) is called the Zeroth Bessel of the second kind.
269. Conduction of Heat. If material supposed to
be homogeneous has a plane face AB,
If at the point P which is at the dis-
tance X from AB, the temperature is v,
and we imagine the temperature the same
at all points in the same plane as P
parallel to -4^ (that is, we are only con-
sidering flow of heat at right angles
dv
to the plane AB), and if t- is the rate of
rise of temperature per centimetre at P,
then — A; -7- is the amount of heat flowing
per second through a square centimetre of
area like PQ, in the direction of increas-
Fig. 106. -^g ^ rpj^-g -g ^^^ definition of k, the
conductivity. We shall imagine k constant.
k is the heat that flows per second per square centimetre,
when the temperature gradient is 1. Let us imagine PQ
exactly a square centimetre in area. Now what is the flow
dv
across TS, or what is the value of — A; -j- at the new place,
dx
dv
which is x + hx from the plane ABl Observe that —k-j-
is a function of x\ call it f{x) for a moment, then the
HEAT CONDUCTIVITY. 355
space PQTS receives heat f(x) per second, and gives out
heat f(x + Bx) per second.
Now f(x-\- Bw) -f{x) = Bx i^^ .
These expressions are of course absolutely true only
when Bx is supposed to be smaller and smaller without limit.
We have then come to the conclusion that —Bx-j- f(x)
is being added to the space PQTS every second : this is
— Bx^ri— f^' 1-] or +k.Bx-j—.
ax \ ax J oar
But the volume is 1 x Bx^ and if w is the weight per cubic
centimetre, and if s is the specific heat or the heat required
to raise unit weight one degree in temperature, then if t is
time in seconds,
5j dv
W .ox .S-Ti
dt
also measures the rate per second at which the space
receives heat. Hence
1 ^ d'^v ^ dv
tc, ox. -j-- — W.OX.S.-Tly
da? at
dH ws dv /,v
d^^T-dt ^^)-
This is the fundamental equation in conduction of heat
problems. Weeks of study would not be thrown away upon
it. It is in exactly this same way that we arrive at the
fundamental equations in Electricity and Hydrodynamics.
If flow were not confined to one direction w^e should have
k
the equation (1) of Art. 265. — is often called the diffusivity
^ ws
for heat of a material, and is indicated by the Greek letter
K', wsis, the capacity for heat of unit volume of the material.
Let us write (1) as
dh^ __ 1 ^ /.^K
dx^'Kdi ^"^'
23—2
356 CALCULUS FOR ENGINEERS.
270. It will be found that there are innumerable solutions
of this equation, but there is only one that suits a particular
problem. Let us imagine the average temperature every-
where to be (it is of no consequence what zero of tempera-
ture is taken, as only differences enter into our calculations),
and that
F= a sin 27r;i^, or a sin ^'^ (3),
is the law according to which the temperature changes at
the skin where a; is ; w or ^ means the number of com-
plete periodic changes per second. Now we have carefully
examined the cycle of temperature of steam in the clearance
space of a steam-cylinder, and it follows sufficiently closely
a simple harmonic law for us to take this as a basis of
calculation. Take any periodic law one pleases, it consists
of terms like this, and any complicated case is easily studied.
Considering the great complexity of the phenomena occurring
in a steam-cylinder, we think that this idea of simple
harmonic variation at the surface of the metal, is a good
enough hypothesis for our guidance. Now we take it that
although the range of temperature of the actual skin of the
metal is much less than that of the steam, it is probably
roughly proportional to it, so we take a to be proportional
to the range of temperature of the steam. We are not
now considering the water in the cylinder, on the skin and
in pockets, as requiring itself to be heated and cooled ; this
heating and cooling occurs with enormous rapidity, and the
less there is of such water the better, so it ought to be
drained away rapidly. But besides this function of the
water, the layer on the skin acts as creating in the actual
metal, a range of temperature which approaches that in the
steam itself, keeping a large. Our n means the number of
revolutions of t lie engine per second.
To suit this problem we find the value of v everywhere
and at all times to be what is given in (2) of Art. 260,
V = ae
"V^sin (2',mt-x^^'j (4).
This is the answer for an infinite mass of material with
one plane face. It is approximately true in the wall of a thick
PENETRATION OF HEAT
357
cylinder, if the outside is at temperature 0. If the outside
is at temperature v and the thickness of the metal is h
1)'
we have only to add a term y ^ to the expression (4).
This shows the effect of a steam-jacket as far as mere con-
ductivity is concerned. The steam-jacket diminishes the
value of a also. Taking (4) as it stands, the result ought
to be very carefully studied. At any point at the depth x
there is a simple harmonic rise and fall of temperature every
revolution of the engine ; but the range gets less and less as
the depth is greater and greater. Note also that the changes
lag more as we go deeper. This is exactly the sort of
thing observed in the buried thermometers at Craigleith
Quarry, Edinburgh. The changes in temperature were 1st
of 24 hours period, 2nd of 1 year period ; we give the
yearly periodic changes, the average results of eighteen
years* observations.
Depth in feet
below surface
Yearly range
of temperature
Fahrenheit
Time of highest
temperature
3 feet
6 feet
12 feet
24 feet
16138
12-296
8-432
3-672
August 14
„ 26
Sept. 17
Nov. 7
Observations at 24 feet below the surface at Calton Hill,
Edinburgh, showed highest temperature on January 6th.
Now let us from (4) find the rate at which heat is
flowing through a square centimetre; that is, calculate —kj
for any instant ; calling a/ — = a.
dv
dx
— aae~*^ sin {2irnt — ax) — aae~'^^ cos {27rnt — ax),
where x = 0, that is, at the skin, it becomes when multiplied
358 CALCULUS FOR ENGINEERS.
hy —k; l—k -J-] = + kaa {sin 27mt + cos 27mt]
= kaa V2 sin (2'Trnt + ^] , by Art. 116.
This is + for half a revolution when heat is flowing into
the metal, and it is — for the other half revolution when heat
is flowing out of the metal. Let us find how much flows
in ; it will be equal to the amount flowing out. It is really
the same as
kaa V2 | sin 27rnt . dt = kaa^/2 . - , where t = - ,
Jo IT n
= «\/
2kws
nir
That is, it is inversely proportional to the square root of
the speed and is proportional to the range of temperature.
We have here a certain simple exact mathematical result ;
students must see in what way it can be applied in an
engineering problem when the phenomena are very compli-
cated. We may take it as furnishing us with a roughly
correct notion of what happens. That is, we may take it
that the latent heat lost by steam in one operation is less
with steam jacketing, and with drying of the skin ; is pro-
portional to the range of temperature of the steam, to the
surface exposed at cut off, and inversely proportional to the
square root of the speed. Probably what would diminish it
more than anything else, would be the admixture with the
steam of some air, or an injection of flaming gas, or some
vapour less readily condensed than steam. When we use
many terms of a Fourier development instead of merely
one, we are led to the result that the heat lost in a steam
cylinder in one stroke is
{0,-0,){h + ^^AI>Jn,
where 6^ is the initial temperature and 6^ the back pressure
temperature, r the ratio of cut ofi", n the number of revolu-
tions per minute, A the area of the piston h and c constants
LIST OF INTEGRALS. 359
whose values depend upon the type of engine and the
arrangements as to drainage and jacketing.
271. Students will find it convenient to keep by them
a good list of integrals. It is most important that they
collect such a list for their own use, but we have always
found that it gets mislaid unless bound up in some book of
reference. We therefore print such a list here. Repetition
was unavoidable.
Fundamental cases :
1. Ix"',dx = V x'"+\
J m + 1
2. j - .dx= log X.
4. la^.dcc = -, a*.
J log a
5. /cos mo) .dx= — sin mx.
J m
sin mx .dx = cos mx.
m
6. /si
7. I cot x.dx = log (sin x).
8. I tan X .dx= - log (cos x),
9. I tan X . sec x.dx = sec x.
10. I sec^ x.dx — tan x.
cosec^ X .dx = — cot x.
I
„. /,
, dx 1 .
12. I = - tan ax.
cos^ ax a
360 CALCULUS FOR ENGINEERS.
13. / -^—z — = — cot ax.
J sm^ax a
1 ^ f dx • , ^
14. I — . = sin~* - .
^ V a^ - ar" «
f dx 1 , X
J a'^ + x^ a a
li
10. / — _ =-sec'
I
dx 1 X
—- zzi SGC~~ ~
X ^a?-a? a a
17*. I cosh ax .dx — - sinh ax.
J a
* From 17 to 23 we have used the symbols (called Hyperbolic sines,
cosines &c. )
Binhx=i (c* - e-*), and cosech x— . . ,
cosha' = i(c*+6~*),8echa:= — ^^ ,
^^ ' coah«
^ , sinh a; e^-1 ^, 1
tanh x= — i— = -iiT, — - , coth a; =
cosha: e'^ + l' tanhx'
Also if 2/ = sinh x, a: = sinh"^ »/.
It is easy to prove that
sinh (a + h) = sinh a . cosh h + cosh a . sinh h,
cosh (a + &) = cosh a cosh h + sinh a . sinh t,
sinh {a-b) = sinh a cosh & - cosh a sinh Z>,
cosh (a - ?^) = cosh a cosh t - sinh a sinh t,
sinh 2a = 2 sinh a . cosh a,
cosh 2a = cosh^ a + sinh^ a = 2 cosh'^ a - 1 ,
= 2sinh2a + l.
If we assume that aJ -1, or i as we call it, submits to algebraic rules and
t2= - 1, t^= -t, 1*^1, 1^ = 1 &c. we can write a + bi as r (cos 6 + i sin ^), where
r^=a'^+h^, and tan ^=- . It is easy to extract the 7tth root of a + bi: being
r^ " I cos - + 1 sin - j , and by adding on 27r to 6 as many times as we please,
we get n, 7ith roots. ■
We also find that c** = cos a + i sin a,
— ia . .
e = cos a - 1 sm a.
If z = a + bi-r (cos + i sin ^) = r . e**,
log 2 = log r + id = ii log {a^ + b"^) + z tan"' - .
HYPERBOLIC FUNCTIONS. 361
18. I sinh ax.dx = - cosh ax,
J a
19. I sech^ ax .dx = - tanh ax.
J a
20. I cosech^ ax.dx = — coth ax.
J a
21. [ --M^ = sinh-^ - = log [x + V^M^^K
22. 1 -^^— = cosh-^ - = log {x + ^/^^~^'l •
23. f,^,==ltanh-f=llog
J a^ — x- a a za °
a-\-x
a —X
This is indeterminate because tan~i - may have any number of times 2ir
in it, and indeed the indeterminateness might have been expected as e '"=1.
Evidently cosh a;= cos ix,
sinh x= -i sin ix.
sinh is usually pronounced shin,
tanh is usually pronounced tank.
Prove that if if = sinh~ia; or a; = |(e" - e~«), only positive values of u being
taken, then e" = x + ^/l + x^, and therefore u = sinh"^ x = log {x + >^1 + x"^).
Similarly cosh-ix = log {x+ Jx- - 1),
tanh-ia; = ^log^— |,
sech-ia: = logQ+yi-l),
cosech-i a; = log (^ + ^^^ + 1) •
Now compare
/ -7==5= 8in"^a;, / , = sinh-i rr = log (a; + Vl+^).
y aJi-x^ j ^1+x^
f dx I
I - .- = cosh-^a; = log {x + ^Jx"^ - 1).
j-j^,= tan i.r. j j--,= tanh-»a:=ilog ^~^.
362 CALCULUS FOR ENGINEERS,
dx 1 , a; — a
24 ( ^ = -i- log
25. f^^ = ver«-.^, f- , _...
dx ,x[ dx i . ,x — a
— - SIQ"
27. I \/^Ta* . rfa? = ia; VaM^ + ia' log {a; -f Vi»Ta«}.
28. I Var' - a^ . da? = Ja; v/^t^ _ ^2 _ |^2 i^g [_^ _^ \/.x"2 - a^}.
c?a; 1 . , 0. 1 , a
■ - = — sm~^ - or = -
Na?- — a? ^ ^ ^
29. I — ^'^^ = - - sin"^ - or = - cos~* - .
±0^
a + Vo^T^
30. f ,j^-=^iog ;
J x\/a^±a? a a + V «' ± a^
31. |-\/aM^.d'a; = \/anr^-a
32. I - si a? -a? . rfa; = Va;^ - a^ - a cos"^ - .
J a? X
r x.dx ,
r x.dx ,
35. f a? V^To^ . rfa; = J ^{¥Tc^y.
36. l*a;\/a2-a;2.(Za; = -iV(a2-a^)».
37. I V2aa; - a?^ . rfa? = ^^ sJ2ax-a^ + ^sin-*^^
] 2 2 a
33 r ^ ^ fx-\
i(a; + l)V^-l Va; + X*
LIST OK INTEGRALS.
40. I A / , . dx = sin"^ x — s/\— a?.
J V 1 — «
^1- \sJ"'-^^'dx=^J{x^a){x^h)
363
4- (a — 6) log {sjx 4- a + -s/a; + 6).
42. I a;"^-i (a + 6a;")9 . c^a?.
1st. If ^/g' be a positive integer, expand, multiply and
integrate each term.
2nd. Assume a + hx^ — y^, and if this fails,
3rd. Assume ax~'^ ^-h — y^. This also may fail to give
an easy answer.
43. I sin~^ X .dx — x sin~^ x-{- Jl— x"^.
44. I X log x.dx= — (log a? — 1^).
45. I xe*^"^ dx = - 6«* ( ic j .
46. I a?"6«^ . c^a; = - aj'^e"^ - - / a;«-i6«* . c^a--.
J a a]
Observe this first example of a formula of reduction to
reduce n by successive steps.
^^' j^'^^~"^^r:n.^— l■^7;^^ja;--^'^''•
^ 1 1 r e"^
48. I 6**^ log a? . (^a? = - ^'^ log a; I — dx.
49. rJ^=jiogl±£!^=loglcot(^-f)l.
j cos a; ^ ^ 1 - sm a; ^ ( V4 2/J
1.
364 CALCULUS FOR ENGINEERS.
51- — 7T = / r *^^ 7 ; it a > 6,
^ Jb—a tan ^ + Jb+a
log if a < 6.
Jb^ - «=» /j; ^ ^ /,— -
Jb — a tan ^ — V^ + fi
^^ r . , e*''^ (c sin aa? — a cos cw;)
o2. e*'* sm ax .ax= — ^ r- — ^ .
J a^ ^ c-
f _ , ^^ (c Go^ ax ■\- a sin aa;)
53. I e<^ cos aa; . rfa; = — ^^ :— — r ^ .
f . „ , cos X . sin"~^ a; w — 1 f . „__ ,
o4. sin** a? . c?a? = + sin"-^ x . dx.
J n n J
[ dx _ cos a? 71 — 2 f dx
J sin** x~ (w — 1) sin**"* a; ri. — 1 j sin""*-^ x '
r 6^* . .
56. I €"* sin" x.dx= ~- sin**~^ x (a sin x — n cos a?)
J a^ + n^ ^
n(n-l) r .^^,^^^^
a2 + ?i=' j
-K,* r . . , sin (7?t — w) •''■ sin(m + 7i)a?
57*. I sin 7«a; . sm ?ja; . aa; = — ^r-^ r ^ ^ .
J 2(m-n) 2(m + n)
-o f 7 sin (?/i — ?i) a; sin(m + w)a?
58. I cos 7?ia7 . cos 7ix . aa? = -^) f- H — ~ ^- .
J 2(m-n) 2(m + ?i)
-^ r . 7 cos(m+?i)a; cos(7ti—n)x
59. I sin mx . cos ??a; . ax = -^ ^ — -^^ f- .
; z (??i + n) 2 (m — n)
60. I sin- na? . dx =:la; — -— sin 2na;.
j 4fn
61. I cos^ ?ia; .dx — — sin 2wa; + Jo;.
* In integrating any of these products 57 to 61 we must recollect the
following formulae :
2 sin vix . sin nx = cos (m -n) x- cos (w + 7j) a;.
2 cos wa; cos 7/a; = cos (m-w)a: + cos (m + n) x.
2 sin wx cos jja: = sin (m + n) a: + sin (m - n) a;,
cos 2«x = 2 cos^ ;ia; - 1 = 1 - 2 siu^ nx.
FORMULAE OF REDUCTION. 365
In the following examples 62 to 67, m and n are supposed
to be unequal integers.
rrr or 2n-
62. I sin 1710) . sin nx .dx= 0.
Tir or 27r
63. I COS ??!« . COS iix . dx = 0.
Jo
64. j sin^ nx,dx = - , I cos^ nx. dx = ^ , if /j is an
Jo -^ Jo ^
I si]
Jo
integer
•2jr
65. I sin mx . cos nx. dx=
66. / sin ?7ZA' . cos nx. dx=0 if 7?i — n is even.
Jo
67. I sin 771^ . cos nx .dx— „ if vi — n is odd.
17V — ?r
Jo
'-/
68. I sin"* ic . cos x ,dx =
sin"*+i a;
m+1
^r. r «. • 7 cos'"-^"a;
69. I cos"* a; . sin a; . ttic = =—
J m + 1
Hence any odd power of cos x or sin x may be integrated,
because we may write it in the form (1 — sin'^a^)"'cos x or
(1 — cos^ xj^ sin a?, and if we develope we have terms of the
above shapes. Similarly sin^ x . cos^ x may be integrated
when either p or q is an odd integer.
70. I a?"* sin x.dx = — x^ cos x-^-mi x^"'~^ cos x . c?a7.
71. I ic*" co^ x.dx = x"^ sin x —m j x'^~^ sin a; . c?a;.
H-^ fsina; , 1 sin a; 1 fcosx ,
* 2, I — — - . dx = = — -— H :, I — — -; dx.
»_-, fcosx y 1 cos a; 1 Tsina;,
J a;"* m-la;"*-^ m-lja;^"~^
74. [tan** x.dx = (^^^y~' ^ |'(tan x^-' . dx
366 CALCULUS FOR ENGINEERS.
75. A'" Sin-l X.dx = = :r I ,- .
J 71+1 n+lj l-^-a?
77. — -r = , tan-i ,.==^ , if 4ac > 6^
ja-\-bx+ cx^ V4ac - 6" v 4ac - 6^
1 , 2cx + h- ^J¥ - Aiac ... ,,
log . , it 4ac < o-'.
Vft^ - 4ac 2ca; + 6 + Vt" - 4ac
2cx + 6
If X = a + bx + cx^ and 5' = 4<ac — 6^, then
^^ fc^a; 2cx-{-h 2c fdx
„ tdx _ 2cx + b ( 1 3c\ ic^ ("da;
'^- JX^- q [iX'-'^qX)'^ q'jX-
Q^ r^' . dx bx+2a b [dx
01 fdx 1 . a^ b [dx
^^' J^ = 2-J''Sx-Tajx'
^^ fdx b , X 1 ^(¥ c\ fdx
83. f__^__ = 2sin-A/^.
N{x-a){b-x) V b-a
84. f_=iL==_=-Asin-.y^pp
Js/{a + bx){oL-px) V6^ V c^^ + 6a
85. I , = — sec^ I — .
Jx\lx''-a'' «^ \^/
ob.
r rig? _ j_ , vv_+^^
a
87.
LIST OF INTEGRALS. 367
f dx
^ \/a-\-hx-\- caf
= -p log f c^ + 2 + Vc (a^bx + cx'^yj .
„„ f dx 1 • -1 ^cx-b
J Na-\-hx — ca? Vc v 4ac + 6^
89. i£±g^ may be altered to
•^ (^ + 2ca?) 2pc-gh J^
2c y/a + bx + ex"" 2c V^ 4- hx + c«2 '
and so integrated.
90. Any integral of the form 1 ^ — ^ — T^r^^> where
P, Q, ii and S are rational integral functions of x, can be
rationalized by the substitution oi ax-\-b = v^.
91. Any integral of the form 1 ,_ dx, where U is
a+bx-\- cx^, can be rationalized, (1) when b^ — 4>ac is positive
and c negative, by the substitution , ^ = - — L.
(2) When ¥ — 4iac is positive and c positive by
Vc"VF _ 2y
\/62-4ac~l~2/'*
(3) When 6- — 4ac is negative and a positive by
VcVF _ 1 +y '
A/4ac-62 ~ 1-2/' '
4c
If ?7 = a 4- 6a; 4- ca;^, g = 4ac — ¥, S = — ,
92 f cZa;_^ 2(2ca;_j-6)
[_dx__ 2(2cx-{b)\/ U 2S(n-l) f dx
JU^s/U (2n-l)qU'' "^ 2n-l JU^-i^'
368 CALCULUS FOR ENGINEERS.
^. r, (2ca;+6)\/|7 1 [dx
nr r /- (2c^ + 6)VF/.. 3\ 3 [dx
ra;rfa; _ J^ _ ^ [ dx
J W^ s/a "^ I X 2ja)
1 . _ / bx + 2a \ .. ^„
V - a \x V 6^ - 4ac/
-2\fU ., .
= — = if a = 0.
f dx ^ 'JU b C dx
J x-^U~~ d^ 2aj^VF"
272. The solutions of many Physical Problems are given
in terms of certain well-known definite integrals some of
which have been tabulated. The study of these is beyond
the scope of this book. I say a few words about The
Gamma Function which is defined as
[ e-^.x^'-Kdx^Tin) (1).
By parts, I e~^ . a" ,dx=- e'^x^ +n\e-^. a;"-^ . dx.
Putting these between limits it is easy to prove that
— e~^x'^ vanishes when x = and a; = x .
And therefore / e-^x'^ ,dx = nj e~^ . x^~^ .dx .. .(2).
THE GAMMA FUNCTION. 369
Hence T (?i4-l) = n r(w) (3),
so that if n is an integer
r(n + l) = 1.2.3.4, &c. n=\n (4).
Notice that j/i has a meaning only when n is an integer,
whereas F (ii) is a function of any value of n.
Tables of the values of F (n) have been calculated, we need
not here describe how. The proof that
ra) = V^ (5),
as given in most books, is very pretty. The result enables
us through (3) to write out r(f) or F(— f), &;c.
A very great number of useful definite integrals can be
evaluated in terms of the Gamma Function.
V IT
Thus 1. [' sin« e,dd=j cos*^ d.dO
o r «.-./i ^«-lJ r '»'"+' (fo T(m)V(n)
4. j a;«e-«* dx = a-<'*+^) F (?i + 1 ).
Jo 2a Vy 2a ^^'
1. i £ i/»'-> (1 - y)""-' dy = r (I) r (m) - 2r (I + m) .
2r (£+? + !)
24
fsi
sinP e cos9 e,d0 =
370
APPENDIX.
The following notes are intended to be rccod in connection with the
text on the page whose number appears before the note. The exact
position on the page is indicated by a t.
Page 3. The ordinary propositions in Geometry ought to be
illustrated by actual drawing. The best sign of the health of our race
is shown in this, that for two generations the average British boy has
been taught Euclid the mind destroying, and he has not deteriorated.
Euclid's proofs are seemingly logical; advanced students know that
this is only in appearance. Even if they were logical the Euclidian
Philosophy ought only to be taught to men who have been Senior
Wranglei-s. 95 per cent, of the schoolboys, whose lives it makes
miserable, arc as little capable of taking an interest in abstract reason-
ing as the other five per cent, are of original thought.
Page 43. There is a much more accurate method for finding ^
described in my book on the Steam Engine.
Page 48. This rule is not to be used for values of .r greater
than 16.
Page 63. In Art. 39 I have given the flue investigations as usually
given, but prefer the following method, for I have for some time liad
reason to believe that in a tube of section A, X)erimeter P, if irib. of
gases flow per second, the loss of heat per second per unit area of tube
is proportional to
vejt,
if t is the absolute temperature of the gases and v the velocity of the
gases. Now v oc Wt-^A^ so that the loss of heat per second is pro-
portional to
weiA,
Proceeding as before
-w,de=cwe.dsiA,
or -A^=G.dS,
a
or -A\oge-\-c=CS,
where c and G are constants. As before, this leads to
c=^log^i, >^=^log|.
A 6
. Whole >S'=7j log ^-, so that the efficiency becomes
E^\-e-^^l^.
APPENDIX. 371
Now if it ia a tube of perimeter F and length ^, JS/A becomes PI I A
or l/m where m is known as the hydraulic mean depth of the flue, so
that
This makes the efl&ciency to be independent of the quantity of stuff
flowing, and within reasonable limits I believe that this is true on the
assumption of extremely good circulation on the water side. This
notion, and experiments illustrating it, were pubhshed in 1874 by
Prof. Osborne Eeynolds before the Manchester Philosophical Society.
Mr Stanton has recently {Phil. Trans. 1897) published experiments
which show that we have in this a principle which ought to lead to
remarkable reductions in the weights of boilers and surface condensers;
using extremely rapid circulation and fine tubes.
Page 95. It ought not to be necessary to say here that the
compressive stress at any point in the section of a beam such as
AC AC, fig. 47, is -jz, if z is the distance (say JB) of a point on the
compression side of the neutral line A A from the neutral line. The
neutral line passes through the centre of the section. If z is negative
the stress is a tensile stress. The greatest stresses occur where z is
greatest. Beams of uniform strength are those in which the same
greatest stress occurs in every section.
Page 111. Mr George Wilson (Proc. Royal Soc., 1897) describes a
method of solving the most general problems in continuous beams
which is simpler than any other. Let there be supports at ^, ^, C, Dj E.
(1) Imagine no supports except A and E, and find the deflections at
B, C and D. Now assume only an upward load of any amount at B,
and find the upward deflections at B, (7, and D. Do the same for C
and B. These answers enable us to calculate the required upward
loads at J5, C and B which will just bring these points to their proper
levels.
Page 139. For beginners this is the end of Chap. I.
Page 146. In all cases then,
dll=l'.dt + t(^\dv (23*).
Exercise 1. By means of (23) express K, I, Z, P and V in terms
of k and write out the most general form of equation (3) in terms of k.
Show that among many other interesting statements we have what
Maxwell calls the four Thermodynamical relations,
U/ \dpJt' WJp \dp)i' \dt) \dv)t' \d(t>), [dvU
Exercise 2. Prove that in fig. 55,
Area ABCB=AE. AF, =A U. AJ==AO.AM=AQ. AB,
and show that these are the above four relations.
24—2
372 APPENDIX.
Exercise 3. Show by using (23) with (7), (8), (11) and (14) that for
any substance
and that (20) becomes
(dk\ d^
so that k =1-0 -{-t j( ^ j dv,
where I'q ls a function of temperature only.
Bividing dR=K.dt+L.('
differential, show that we are
Dividing dH=K .dt+L.dp by t and stating that it is a complete
led to
K=K,-tj
where Kq is a function of temperature only.
Page 152. Another way is merely to recognize (8) as being the
same as
bH=^k.ht + l.bv,
for l — t(~\, and when a pound of stuff of volume s^ receives bH=L
at constant temperature (or bt = 0) in increasing to the volume
«2 (so that dv = «2-«i) we have, since dpjdt is independent of v,
8Zr=Z = + i.^.(52-«i).
Page 188. Sine functions of the time related to one another by
linear operators such as a-{-h6-\-c€^-\-etc. + e6~^+fB~^-\-Qic.^ where
6 means -y- , are represented by and dealt with as vectors in the manner
here described. Eepresenting an electromotive force and a current in
this way, the scalar product means Power. Dr Sumpner has shown
{Proc. Roy. Soc, May 1897) that in many important practical problems
more complicated kinds of periodic functions may be dealt with by the
Vector Method.
Page 190. The symbol tan~i means " an angle whose tangent is."
Page 195. To understand how we develope a given function in a
Fourier Series, it is necessary to notice some of the results of Art. 109,
very important for other reasons ; indeed, I may say, all-important to
electrical engineers.
Page 195. Article 126 should be considered as displaced so as to
come immediately before Art. 141, p. 210.
Page 202. The problem of Art. 125 is here continued.
APPENDIX. 373
Page 208. The beginner is informed that 2 means "the sum of all
such terms as may be written out, writing 1 for s, 2 for 5, 3 for 5, and
so on."
Page 209. See Ex. 23, p. 184.
Page 213. The student ought to alter from v to (7 or to ^ in (9)
as an exercise.
Page 213. See Art. 152.
Page Ml. Remember that the effective value of as.\n{nt-\-e) is
a^sj2.
Page 254. Here again a student needs a numerical Example.
Page 256. After copper read " and their insulations."
Page 259. Then
e-^-\-e.^=E [sin {nt + a) + sin (nt — a)] = 2i7 cos a sin w^.
{See Art. 109.)
Page 266. All other statements about this subject that I have seen
are of infantine simplicity, but utterly wrong.
Page 269. Insert " and in consequence."
Page 278. In the same manner show that if
y=«^ ^=«*ioga.
Page 281. See the eighth fundamental case, Art. 215.
Page 299. See (1) Note to Art. 21.
Page 301. Article 225 should be considered as displaced so as to
precede Art. 222.
Page 305. See Ex. 8, Art. 99.
Page 309. Or (3).
Page 310. Art. 225 should be read before Art. 222.
Page 331. I have taken an approximate law for h and so greatly
shortened the work.
Page 359. Viscosity. All the fluid in one plane layer moves with
the velocity v\ the fluid in a parallel plane layer at the distance bx
moves with the velocity v-\-bv in the same direction; the tangential
force per unit area necessary to maintain the motion is /* ^ or /* ^- ,
where /x is the viscosity.
Example 1. A circular tube is filled with fluid, the velocity v at
any point whose distance from the axis is r being parallel to the axis.
Consider the equilibrium of the stuff contained between the cylindi-ic
spaces of radii r and r+5r of unit length parallel to the axis. The
374 APPENDIX.
tangential force on the inner surface is 2wrfi. ,~ and on the outer
surface it is what this becomes when r is changed to r-\-hr or
^TTfi -J- (i'-f) 5^1 the difference of j^ressure between the ends gives us
a force - 2wr ~ dr if x is measured parallel to the axis. The mass of
the stuflF is 27rr . 6r . m if dr is very small and if m is the mass per unit
volume ; its acceleration is -^ if ^ is time ; and hence, equating force to
mass X acceleration and dividing by 27r/x8r,
d / dv\ r dp _r.m dv
dr \ dr) fi dx~ fi dt '
Example 2. Let -^ be constant; say that we have a change of
pressure P in the length I so that t^ = y • Let a state of steady flow
have been reached so that -f=Oy then
dr\ dry fi I
If r -7- be called u and if i'/^u be called 2a. then -^ + 2ar = 0, so that
dr '^ dr '
du 4- 2ar . dr = 0, or u + ar^ = constant c.
rj^ + ar^=^c, or ^ + ar=- (1),
dv-\-
{ar-i^dr=0,
c
vi-j^ar' + ^-, = C (2).
Evidently as there is no tangential force where r=0, -f- = thei-e,
c must be 0. Hence
v + ^ar^ = C (2).
Now v=0 where r^r^, the outer radius of the fluid, and hence
(2) becomes v=^a {r^^-r^).
The volume of fluid per second passing any section is
27r / rv .dr=^ - aro* = nrQ*P/8lfx.
This enables us to calculate the viscosity of a fluid passing through a
cylindrical tube if we know the rate of flow for a given difference of
pressure.
INDEX.
The References are to pages.
Academic exercises, Chap. iii.
Acceleration, 24, 30, 188, 220
Adiabatic law, 92, 148, 167
Advance, 186
Air in furnaces, 65
Air turbine, 128
Alternating current formulae, 183 — 5,
189, 209, 239
Alternator, 178
Alternators in parallel and series, 259
Amplitude, 172
Analogies in beam problems, 108
— in mechanical and electri-
cal systems, 213
Angle between two straight lines, 16
Angular displacement, 33
Angular vibrations, 212
Apparent power, 209
Approximate calculations, 2
Arc, length of, 170
Archimedes, spiral of, 302
Area, centre of, 85
— of catenary, 171
— of curves, 69
— of parabola, 71
— of ring, 80
— of sine curve, 197
— of surfaces of revolution, 75, 78
Asymptote, 301
Atmospheric pressure, 166
Attraction, 87, 344
Average value of product of sine
functions, 185
Ayrton, 53, 199
Ballistic effects, 181
Basin, water in, 130
Beams, 48
— fixed at the ends, 100—108
— of uniform strength, 102
— shear stress in, 115
— standard cases, 97 — 99
Beats in music, 194
Belt, slipping of, 165
Bending, 94—121
— in struts, 262
Bessels, 205, 352, 354
Bifilar suspension, 179
Binomial Theorem, 34
Boiler, heating surface, 63
Bramwell's valve gear, 193
Bridge, suspension, 61
Calton HUl, 357
Cardioide, 302
Capacity of condensers, 162, 236, 240
Carnot cycle, 145, 152
Carnot's function, 145
Catenary, 62, 170
Central force, 344
Centre of gravity, 73, 85
Centrifugal force, 123
Centrifugal pump, 131
Chain, hanging, 61
Change of state, 150
— of variable, 339
Characteristic of dynamo, 296
Circle, 10
— moment of inertia of, 82
Circuitation in electricity, 134
Cissoid, 302
Cisterns, maximum volume of, 49
376
INDEX.
The References are to pages.
Clairaut's equation, 334
Clearance in gas engines, 150
— in pumps, 131
Commutative law, 231
Companion to cycloid, 301
Complete differential, 143—145, 153
Compound interest law, 161, 164
Concavity, 306
Conchoid, 302
Condensation in steam cylinders,
54, 153, 358
Condenser, electric, 162, 212, 236
— annulling self-induction,
247
— with induction coil, 257
Conductivity of heat, 341, 354
Conductors, network of, 237
Cone, 73, 78
Conjugate point, 301
Connecting rod, 191
Constraint, 179
Continuous beams, 111
Convexity, 306
Cooling, Newton's law of, 163
Co-ordinate geometry, 6
Co-ordinates, r, 0, <p, 347
— polar, 310, 342
Cosines, development in, 207
Cos-la;, 277
Cot-la;, 277
Cot X, 275
Coupling rod, 265
Craigleith quarry, 367
Crank, 173
Crank and connecting rod, 12, 191
Curl, 134
Current, effective, 200—202
— electric, 33, 168, 189, 239
Curvature, 96, 120, 169, 306
— of beams, 96—121
— of struts, 262
Curves, 43
— ' area of, 69
— lengths of, 77, 312
Cusp, 301
Cycloid, 12, 276, 302, 312
— companion to, 301
Cylinder, heat conduction in sides of
steam-engine, 356
— moment of inertia of, 82,
85
Cylinders, strength of thick, 88
Cylindric body rotating, 90
Damped vibrations, 10, 211, 225—
228
Definite integral, 68
Deflection of beams, 96, 118
Demoivre, 320
Development, Fourier, 202
— in cosines, 207
Differential coefficient, 21, 28, 268
— complete, 143, 153
— equation, 220-225
— equations, general exer-
cises on, 337
— equations, partial, 341,
346—351
— partial, 139
Differentiation of function of more
than one varia-
ble, 340
— of product, 269
— of quotient, 270
Diffusivity for heat, 355
Discharge of condenser, 156
Displacement, angular, 33
Distributive law, 231
Drop in transformers, 255, 257
Dynamics of a particle, 344
Dynamo, series, 296
df>'
dx dy
^( = ^.^,155.271
dx dz dx
e<^, 10
e<" sin bx, 10, 285
e'^cosbx, 286
Earth, attraction of, 87
Earthquake recorder, 215
Economy in electric conductors, 65,
57—59
— — lamps, 294
— hydraulic mains, 58
Eddy currents, 209
Effective current, 200—202
Efficiency of gas-engine, 150
— of heat-engine, 41
— of heating surface, 64
Elasticity, 93, 141
Electric alternator, 178
— circuit, 33, 168, 178, 189,
208, 212, 236, 239, 247
INDEX.
377
The References are to pages.
Electric condenser, 162, 236, 240—245
— — and Ruhmkorf
coil, 257
— — as shunt, 243,
246
— conductor, 168
— — economy in, 55,
57—59
— current, efifective, 200—202
— illustration, alternating cur-
rents, 199
— lamps, economy in, 294
— make and break curve, 201,
205
— power meter, 209
— time constant of coil, 60, 160
— traction, 59
— transformer, 33, 249, 253
— vibrations, 156, 212, 213, 225
— voltage, effective, 202, 247
— voltaic cells, 51, 52
Electricity, partial differential equa-
tions in, 349
— problems in, 33, 51, 52,
55, 57—59, 60, 134—
136, 156, 157, 162, 168,
178,182—186,189,195,
202,205,208—210,212,
225, 236, 239—261
— gelf-induction in, 135
Electrodynamometer, 200
Electromagnetic theory, fundamental
laws of, 134
Electromotive force, 134
— — in moving coil,
178
Ellipse, 8, 10, 11, 83, 158, 276
Ellipsoid of revolution, 76
Elliptic functions, 339
— integrals, 339
Empirical formulae, 17
Energy, intrinsic, 143
— kinetic, 31, 180
— of moving body, 156
— potential, 180
— stored in compressed spring,
33
Engineer, 1
Entropy, 143, 152
Envelopes, 309
Epicycloid, 302
Epitrochoid, 301
Equality of forces, 217
Equations, differential, 220—225
— partial differential, 341,
346—351
— solving, 51
Equiangular spiral, 186
Euler's law for struts, 265
Evanescent term, 169, 190, 208, 240
Exact differential equation, 339
Exercises, 38
— general, on differential
equations, 33/
— — on integration
and differen-
tiation, 279
— maxima and minima, 295
— on curves, 169, 311
— on integration of sin x and
cos^, 182
— on Maclaurin, 319
Expansion of functions, 317 — 320
— of gas, 17
Experiments, 7
Explosions, 48
Exponential and trigonometrical for-
mulae, 177, 185, 190,
222, 320
— theorem, 161
e«, 161
Factor, integrating, 328
Factorial fractions, 235
Factorials and gamma function, 369
Falling body, 21, 30
Feedwater missing in steam cylinder,
358
Ferranti effect, 247
Field, magnetic, about straight wire,
134
— — rotating, 195, 251
Flow, maximum, of gas, 128
— of gas, 54
— of liquid, hypothetical, 133
Fluid friction, 167
— motion, 125
— pressure, 121
— whirhng, 123
— work done by, 66
Flywheel, 84
— stopped by friction, 167
Force, central, 344
— due to jet, 27
— — pressure of fluids, 121
— lines of, 124
378
INDEX.
T/te References are to pages.
Force of blow, 26
— of gravity, 87
— unit of, 26
Forces on moving bodies, 180
Forms, indeterminate, 299
Formulae of reduction, 284, 286
Fourier development, 208
— exercises on, 315
— proof, 183, 184, 197, 201
— rule, 202, 204
— theorem, 195
Fractions, partial, 291
Frequency, 186
Friction at pivot, 94
— fluid, 167
— solid, 168
Frustrum of cone, 78
Fuel on voyage, 49
Function, 8
— average value of sine, 185
— Bessel, 352
— elliptic, 339
— gamma, 369
— hyperbolic, 172, 360
— of more than one variable,
137, 341
Fundamental integrals, 278
— rules on electricity, 134
Furnaces, air in, 65
«7, 26
Gamma functions, 235, 369
Gas, 38
— elasticity of, 94
— engine, 91
— — diagram, heat in, 272
— — formulae, 17, 91, 147,
150, 272
— flow of, 55, 127
— perfect, 136, 150
— work done by, 66, 72
Gauge notches, 133
Gaussage, 134
Gear, valve, 193
General case of two coils, 249
— exercises in differential equa-
tions, 337
— — in differentiation
and integration,
279
— rule, 271
— — for differential equa-
tions, 224
General rule for operators, 237
Girders, continuous, 111
— shear stress in, 117
Glossary, 301, 302
Gordon's rule for struts, 264
Gradient, temperature, 354
Graph exercises, 8 — 10
Graphical Fourier development, 204
— work in beams, 108
Gravity, 87
— motion of centre of, 230
Groves' problem, 257
Guldinus' theorems, 80
Guns, 90
Hammer, 26
Hanging chain, 61
Harmonic functions, 172, 186
Harmonics, zonal, 205, 354
— — spherical, 349
Heat conductivity, 354
— — equations, 341,
347—352
— equations, 138
— experiments at Edinburgh, 357
— latent, 41, 43
— lost in steam cylinders, 356
— reception in gas-engine, 272
— specific, 93, 141
Heating surface, 63
Hedgehog transformer, 244
Henrici, Professor, 205
Henry, 136
Hertz, 215
Horse-power and steam, 41
Hydraulic jet, 26
— press, strength of, 89
— transmission of power, 58
Hydraulics, 133
Hyperbola, 10, 11
Hyperbolic functions, 360
— spiral, 302
Hypocycloid, 277, 302
Hypotrochoid, 302
Hysteresis, 209, 255
Idle current in transformers, 243
Imaginaries, 185
Incandescent lamp, economy in, 298
Independent variables, change of, 338
— — more than
one, 56, 136,
154
INDEX.
379
Ttie References are to payes.
Indeterminate forms, 299, 300
— multipliers, 158
Index law, 231
Indicator diagram, 53, 67
— — gas-engine, 91, 27*2
— vibration, 215
Induction, 136
— coil and condenser, 257
— in transformers, 256
— mutual, 249
— self-, 33, 60
— self-, and capacity, 240
Inertia, moment of, 81
— — of cylinder, 81,
85
— — of ellipse, 83
— — of flywheel, 84
— — of rectangle, 86
— — of rod, 84
Illustrations of meaning of differen-
tiation, 37, 40, 42, 162, 176, 177
Inflexion, point of, 20, 301
Instruments, measuring, 179
Integral, definite, 68
— double, 68
— line, 69
— surface, 69
Integrals, elliptic, 339
— list of, 359—369
Integrating factor, 144, 327
Integration, 23, 35
— by parts, 285
— exercises on, 183
— of fractions, 282
Interest law, compound, 161
Intersection of two straight lines, 16
Intrinsic energy, 143
Isothermal expansion, 92
Joule's equivalent, 43
Journal, lubrication of, 231
Kelvin, 161
Kinetic energy, 31
Labour-saving rule, 237
Lag, 209
Latent heat, 15, 42
Lateral loads, struts with, 264
Lamp, incandescent, 298
Law, adiabatic, 148
— commutative, 231
— of flow of heat, Peclet's, 63
Law of cooling, Newton's, 163
— compound interest, 161
— distributive, 231
— of Entropy, 148, 152
— Euler's, for struts, 265
— of expansion of steam, 17
— of falling bodies, 21
— Index, 231
— of loss of heat in steam cylin-
der, 356
— of ^ and t, 18
— of Thermodynamics, first, 142
— — — second, 146
— of vibratory systems, 225 — 230
Lead in branch electric circuit, 247
Leakage of condenser, 162
Legendre's equation, 350
Lemniscata, 302
Length of arc, 170
— of curve, 77, 312
Level surface, 124
Limit, meaning of, 22
Line integral, 69, 134
Linear equations, 220, 326, 353
Link motions, 14, 193
Liquid, flow of, 130
List of Integrals, 359—369
Lituus, 302
Loci, 11
Logarithmic curve, 10, 302
— decrement, 11
— function, 40
— spiral, 302
Logarithms, 2, 161
Log a;, 274
Lubrication of journal
Maclaurin, 319
Magnet suspended, 179
Magnetic field about straight wire,
134, 195
— — rotating, 195, 251
— force, 134
— leakage, drop due to, 257
Make and break curve, 201, 205
Mass, 26
— energy of moving, 157
— of body, variable, 76
— vibrating at end of spring, 156
Maxima and minima, 20
— — exercises on,
46, 47, 60,
294
380
INDEX.
The References are to pages.
Maximum current from voltaic cells,
62
— flow of air, 128
— parcel by post, 53
— volume of cistern, 49
— power from dynamo, 297
Measuring instruments, 179
Merchant and squared paper, 6
Meter, electric power, 209
Minimum, 20
Modulus, 340
Moment of inertia, 80
Momentum, 26
Motion, 30, 157
— angular, 33, 212
— of fluids, 125
— of translation, 344
— in resisting medium, 314
Multipliers, indeterminate, 158
Mutual induction, 250
Natural vibrations, 225, 220
Negative and positive slope, I'J
Network of conductors, 237
Newton's law of cooling, 163
Normals, 15, 43, 169
Notches, gauge, 133
Numerical calculations, 2
Octave, 192
Ohm's law, 33
— — modified, 136, 168, 189,
208, 236
Operation, symbols of, 231
Orifice, flow of gas through, 54
Oscillation, 123, 156, 190, 210, 211,
225—30
Otto cycle, 149
Parabola, 8, 11, 27, 31, 61, 71
Paraboloid of revolution, 74
Parallel motion, 13
— alternators in, 261
Parameter, 340
~ variable, 308
Partial differential equations, 341,
347—352
— differentiation, 39, 137, 155,
269, 341
— fractions, 224, 234, 291, 294
Particle, dynamics of, 344
Parts, integration by, 284
Peclet's law of flow of heat, 63
Pendulum, 179
Percussion, point of, 123
Perfect gas, 38
— — thermodynamics of, 147
— steam-engine, 41
Periodic functions, 194, 203
— motions in two directions,
196
— time, 186
Perpendicular lines, equations to,
16
Pivot, friction at, 94
Point, conjugate, 301
— d'arr^t, 301
— of inflection, 301
— of osculation, 301
— moving in curved path, 342
Polar expressions, 342
Pound, unit of force, 26
Poundal, 26
^n»(^).350
Positive and negative slope, 19
Potential energy, 32
Power, apparent, 209
— electric, 208
— — transmission of, 58
— meter, electric, 209
— true, 209
Press, hydraulic, 90
Pressure, 136, 273
— atmospheric, 166
— fluid, 121
Primary, transformed resistance of,
254
Product, differentiation of, 155, 269
Projectile, 310, 315
Pulley, slipping of belt on, 165
Pump rod, 164
Quotient, differentiation of, 270
Radial valve gears, 14, 193
Badian, 9
Radius of curvature, 169, 306
— of gyration, 82
Radius vector, 342
Rate of reception of heat, 92
Ratio of spcific heats, 93, 139, 141
Ratios, trigonometrical, 182
Reduction, formulae of, 284, 286
Recorder, earthquake, 215
Rectangle, moment of inertia of, 86
Resistance, electric, 33
INDEX.
381
The References are to pages.
Resistance, leakage, 163
— operational, 236
Besistances in parallel, 245
Resisting medium, motion in, 314
Resonance, 215
Resultant of any periodic functions,
197
Revolution, surfaces of, 78
— volume of solids of, 75
Rigid body, 61
— — motion of, 217
Ring, volume and area of, 80
Rod, moment of inertia of, 85
Roots of equations, 224, 321
Rotating field, 195, 251
Rotation in fluids, 132
Ruhmkorff coil, 257
Rule, 4
r and d co-ordinates, 342
r, df <t> co-ordinates, 347
Secant x, 276
Secohm, 136
Self-induction, annulled by con-
denser, 247
— — of parallel wires,
135
Series, alternators in, 257
— development in, 207
— dynamo, 296
Shape of beams, 109
Shear stress in beams, 115
Shearing force in beams, 108
Simple harmonic motion, 173
— — — damped, 311
Sina;, 9, 161, 274
Sine curve, area of, 173
— functions, 172
Sines, curve of, 9, 173
— development in, 207
Singular solution, 334
SHpping of belt, 165
Slope of curve, 15, 19, 70
Solution of forced vibration equa-
tions, 214
— of linear differential equa-
tions, 229
Sound, 94
Specific heat, 139
— heats, ratio of, 93, 141
— volume of steam, 42
Speed, 21
Spherical Harmonics, 205, 349, 354
Spin, 132
Spiral flow of water, 130
— hyperboUc, 302
— of Archimedes (or equi-
angular), 302
— line, 173
— logarithmic, 302, 311
Spring with mass, vibrating, 156
Springs, 21, 32, 53
— which bend, 119
Square root of mean square, 200, 202
Squared paper, 6, 7
State, change of, 150
Steady point, 219
Steam, 41
— work per pound of, 53
— -engines, 41
— -engine indicator, vibrations
of, 215
— — piston, motion of, 191
Stiffness of beams, 48
Straight line, 14
Strains in rotating cylinder, 90
Stream lines, 126
Strength of beams, 48
— of thick cylinders, 88
— of thin cyhnders, 91
Struts, 261
— with lateral loads, 264
Subnormal, 44, 169
Substitution, 279
Subtangent, 44, 169, 305
Successive integration, 335
Sum, differentiation of a, 268
Surface heating, 63
— integral, 69
— of revolution, 75, 78
— level, 124
Suspension, 179
— bridge, 61
Swinging bodies, 179, 182
Symbols of operation, 233
— — — simplification
of, 237
Synchronism, 215
Table of Fundamental Forms, 278
Tangents, 43, 158, 169
Tana;, 275
Tan-la:, 277
Taylor's theorem, 317
Temperature, 136
882
INDEX.
The Refererwes are to pages.
Temperature, absolate, 145
— gradient, 354
— in rocks, <feo. , 357
Terminal velocity, 314
Theorem, Binomial, 34
Theorem of three moments, 111
— Guldinus', 80
— Demoivre's, 320
— Maclaurin's, 318
— Taylor's, 317
Thermodynamics, 42, 138, 153
Thomson, Professor James, 133
Three moments, 111
Tides, 194
Time constant of coil, 60, 160
t, diagram, 153
Torque, 33
Torsion, 179
Traction, electric, 59
Transformer, 33
Transformers, 252, 257
— condenser shunt, 243
— idle current of, 243
Triangle of forces, 61
Trigonometry, 2
Trigonometric and Exponential
Functions, 177, 185, 222, 234
Trigonometric Formulae, 182
Trisectrix, 302
Tuning-forks, 230
Turbine, air, 128
Two circuits, 249
Valve gears, 14, 193
Variables, independent, 57, 136, 341
Variable mass of body, 76
— parameter, 308
Velocity, 21, 30, 188
Vibration, 156, 190, 210, 212, 216,
225, 230, 238
— (electrical), 156, 212, 213
— indicator, 217, 219
— of indicator, 215
Volcanoes, 172
Voltage in moving coil, 178
Voltaic cells, 51, 52
Volume of cone, 74
— of ellipsoid of revolution, 76
— of paraboloid of revolution,
74
— of ring, 80
— of solid of revolution, 76
Voyage, fuel consumed on, 49
Water in steam cylinder, 358
Watt's parallel motion, 13
Wedmore, 205
Weight, 26
Whirhng fluid, 123
Willans, 54
Work, 31—32
— in angular displacement, 35
— per pound of steam, 31, 53
— done by expanding fluid, 66,
72
— — gases, 149
Uniform strength in beams, 102, 103
Uniformly accelerated motion, 29
Unreal quantities, 3, 177, 185, 222,
320
X . e«*, 285
x-K- 6—160
Zonal Harmonics, 205, 349, 354
CAMBRIDGE : PRINTED BY J. & C. F, CliAY, AT THE UNIVERSITY PRESS.
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-EW
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