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Graphical calculus,
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GRAPHICAL CALCULUS
Graphical Calculus
ARTHUR H. BARKER, B.A., B.Sc.
SENIOR WHITWOETH SCHOLAR 1895
LATE ASSISTANT TO THE PROFESSOR OF ENGINEERING, YORKSHIRE COLLEGE, LEEDS
WITH AN INTRODUCTION BY
JOHN GOODMAN, A.M.I.C.E.
PROFESSOR OF ENGINEERING AT THE YORKSHIRE COLLEGE, VICTORIA UNIVERSITY
LONGMANS, GREEN, & CO.
LONDON, NEW YORK, AND BOMBAY
1896
Ail rights reserved
M-'S-o
INTRODUCTION
All •■ up-to-date " teachers of engineering and applied
sciences generally now recognize the vast superiority
of graphical over purely mathematical methods of im-
parting instruction of almost every description. The
former are much more convincing to the student,
because they appeal to the eye, the training of which
is one of the chief objects to be aimed at in the
education of an engineer. There is no doubt that this
method is capable of great extension with advantage.
In this little book, for instance, we see graphical con-
structions of a very simple character employed to teach
what, to the beginner, are somewhat abstruse mathe-
matical principles.
The attempt to employ purely mathematical, in
preference to graphical methods, seems to me quite as
absurd as attempting to teach geography by giving the
position of towns in terms of their latitude and longi-
tude, and explaining the shape of a country by giving
the equation to the coast-line instead of by employing the
graphical method, i.e. exhibiting a map. The teacher
vi Introduction.
who attempted the former method would indeed be
considered unpractical, and would, I fear, meet with
but a very limited share of success ; yet, strange to
say, such a method is precisely that which teachers of
mathematics are trying to employ with a much more
subtle subject than geography — the Calculus. Is it,
then, to be wondered at that many technical students
shudder at the bare sign of integration, " the long S,"
as they are wont to call it ? Not because they cannot
manipulate the symbols — far from it — but because they
have not the faintest notion of the physical meaning of
the processes.
I have frequently had students come under my
notice who, although fairly good mathematicians as far
as bookwprk is concerned, yet, through not having had
the advantage oi 2. practical mathematical training such
as we give at the Yorkshire College, were utterly at sea
when they came to apply their mathematics to such
a simple engineering problem as, for instance, finding
the quantity of water flowing over a V notch ! It is
primarily to help such to acquire an intelligent working
knowledge of the Calculus that Mr. Barker has written
this little book. Even if it have only a tithe of the
success that its author has had in teaching mathematics
by this method, it will still be eminently successful. I
can unreservedly say that this is exactly the style of
book that I have been wanting to see for years, and
I believe it will prove to be of very real value to those
■ Introduction. vii
students of engineering who wish to get a stage beyond
the barest elements of the subject.
Some of those who know my propensity to scoff at
the mathematics so commonly drummed into technical
students, may, on seeing this introduction, exclaim, " Is
Saul also among the prophets ? " To such I would
say that if as a student it had fallen to my lot to be
under such a teacher as Mr. Barker, I should always
have been numbered among the prophets, though
possibly the minor ones.
JOHN GOODMAN.
The Yorkshire College, Leeds,
April, 1896.
AUTHOR'S NOTE
The author takes this opportunity of expressing his
thanks to Professor Goodman and Mr. Frederick
Grover, A.M.I. C.E., of the Yorkshire College, for their
kind assistance ; and also to Mr. P. Nicholls, B.Sc,
Wh.Sc, for the great care with which he has read and
revised the proof-sheets.
CONTENTS
CHAPTER PAGE
I. Introductory.— Curves and their Equations . . i
II. Graphical Differentiation and Integration . . 12
III. Nomenclature and General Principles .... 31
IV. General Principles . .46
V. Genebal Principles — {CoKtinued) . ... 63
VI. Differential Coefficients of Trigonometrical
Functions . . ..... 79
VII. Differential Coefficients of Logarithmic Func-
tions . . . .96
VIII. Differentiation of a Function ok a Function of
a Variable with Respect to that Variable . 105
IX. Integration .... . .... 122
X. Methods of Integration 136
XI. Miscellaneous Applications of Differentiation . 148
XII. Miscellaneous Applications of Integration . . 166
Appendix— Barker's Planimeter 187
GRAPHICAL CALCULUS
CHAPTER I.
INTRODUCTORY. — CURVES AND THEIR EQUATIONS.
§ I. Co-ordinates of a Point.
The exact position of a point in a plane is completely known
if its perpendicular distances from two intersecting lines in that
plane are known. Thus, suppose the lines OX, OY (Fig. i)
represent to some scale two hedges of a field meeting at right
angles, and we are told that an article is buried in the field at
a given depth at a point A, whose perpendicular distance from
the hedge OY is 20 yards, and from OX 30 yards.
It is clear that the position of the article could be im-
mediately found by measuring 20 yards from O along OX to
L, and 30 yards along LA perpendicular to OX.
Definitions. — The lines OL, LA, would be called the co-
ordinates of the point A, with reference to the axes OX, OY ;
OL is the abscissa, and LA the ordinate ; and the point A
would be described in mathematical language as the point
whose abscissa is 20, and whose ordinate 30, or shortly as
" the point (20, 30)."
§ 2. Equation to a Line.
Suppose, however, we are told that the distance of the
point A from OX is half (its distance from OY) + 20 yards.
From this condition alone we could not find the exact
B
2 Graphical Calculus.
position of the point, for there are many points in the field,
in addition to the point A, of which the statement would be
equally true. Thus if we take OM = 50 yards along OX,
and MB at right angles = (^ x 50 + 20) yards = 45 yards,
we should find a point B which would also " satisfy the con-
dition " that its distance from OX was half its distance from
OY + 20 yards. Or we might have taken ON = 80 and
NC = 60 ; or, indeed, any arbitrary (or, as it is called, " in-
dependent ") distance along OX, and calculated and measured
the corresponding distance perpendicular to OX. We could
thus find any number of points " satisfying the given condition."
All these points would be found to lie on a certain straight
line in the field, and no point which is not on the straight
line would be found to satisfy the condition.^ And, further,
any point which is on the line will satisfy it.
We should then be sure of finding the buried article, if we
were to dig a trench of the given depth along a line repre-
sented by AC.
Now let us attempt to discover the position of the article
by an algebraical process.
• This statement should be tested by plotting the points to scale on a
plan of the field.
Introductory. — Curves and their Equations. 3
Let X be the perpendicular distance of the buried article
from OY ; and y the perpendicular distance from OX.
Then we have —
jl/ = - + 20
2
This is the only equation we can obtain from the data,
and we are here met with the same difficulty as before,
namely, that there are an infinite number of possible solutions
to the equation, each solution corresponding to one particular
point on the plan. Now, just as {x = 20, y - 30) may be
taken to represent the point A, so the equation y = --\- 20
may be taken to represent the line AC. In other words, the
line AC is a picture or geometrical representation of the
equation _y = - + 20. To put it in still another way, the line
AC shows the relation between the value of x and that of
/ — \. 20), or y, corresponding to all values of x or y. Thus
suppose we wish to find from the diagram what is the value
of jc or (^ - + 20 ) when x is 59*2, say, or any other arbitrary
value, we measure off to scale 59-2 along OX, and erect a per-
pendicular to OX from the point so found. The length of this
perpendicular cut off by the line AC gives the required value.
The algebraical counterpart of this process is as follows :
In the equation y = --^20, find the value of y when x is
59-2. To solve this we have merely to substitute 59-2 for x in
the equation, and solve the resulting simple equation in ;- ; we
thus find the value of y corresponding to .» = 59-2. This is
easier and more accurate than the graphical process. Another
example of the same thing is, " Find where the line y=--\-2o
cuts the axis of y." At the required point it is obvious that
4 Graphical Calculus.
X = o. Hence substituting this value of x in the equation,
we obtain a simple equation in y, viz. y = 20, which gives
the distance of the point from OX.
This is expressed by saying that —
y = — 1-20
2
is the " equation to the line AC."
A little reflection will show the student that j' = - + 20
2
is also the equation to the continuation of the straight line AC
in both directions anywhere in its length, and not only of that
part of it which is to the right of OY and above OX.
Convention of Signs. — In this connection it must be
noticed that if distances to the right of OY are called positive,
those measured to the left are to be called negative. Thus
if a distance = 10 yards be measured to the left of OY, its
distance from OY is said to be — 10. The reason for this may
be gathered from consideration of the following case : —
Suppose a man starts from P to walk to Q, a distance of 4 miles.
After walking to S, he turns back and walks in the other direction for
zi miles to R. The total distance,
H~ ^ *F "■ 'I' '^ H irrespective of direction, which he
" *" S % has now walked is ij + 2| = 4
^'°- 2- miles ; but, considered with respect
to his original destination, the effective distance he has walked is — i mile,
i.e. he has i mile to make up before he begins to be any nearer to Q
than he Was when he started. This may be conveniently expressed by
prefixing the negative sign to distances walked towards the left ; thus,
ii + (-2i)=-i, which represents the total distance he has walked
towards Q.
Suppose, then, we take a distance along OX = — 10. The
corresponding value of y will be ^ X (-10) + 20 = +15,
The point D (^ 10, +15) is also on the line AC.
In the same way distances below OX are reckoned as
minus quantities. Thus when x= -50, jf = -J x (-50) + 20
= -5, the point E (-50, -5) being also on the line AC.
.Sjcaw//,?.— Find where the given line cuts OX.
Introductory. — Curves and their Equations. 5
§ 3. Equation to a Curve of the Second Degree.
Suppose now that, instead of the condition j = ^ + 20, we
2
had had the condition that the distance from OX = -^ of (the
square of the distance from OY) + 20 yards.
Dimensions of Quantitiss. — It may be noted, in passing, that, strictly
speaking, it is as absurd to speak of one distance being = the square of
another distance, as this expression is usually understood {i.e. that a line
is equal to an area), as it would be to
say, for instance, that a square foot is
equal to 6'2 gallons. It is an absurdity
of the same kind as is often committed
in mechanics, when speaking of an' ac-
celeration of so many feet per second,
instead of so many feet-per-second per
second. Conventionally, however, it is
to be understood in the following
sense : We know by geometry that MP^ (Fig. 3) = AM. MB. Now,
when MB = 1", there are as many square inches in the square on MP as
there are linear inches in AM. And so, disregarding dimensions, we say
that AM = MP". For instance, if MP were = 3", AM would = 9".
This idea may also be expressed by saying that we are to regard a line
as a geometrical method of representing a number, and not necessarily a
number of inches. Thus a line 3" long represents essentially on the simple
inch scale the number 3, and may at pleasure stand for 3 seconds, 3
degrees, 3 feet per second, 3 square inches, or 3 units of any kind what-
soever.
Our relation expressed algebraically is —
^ = ^ -f 20
40
Exactly as before, any value may be arbitrarily assigned to
X, and the corresponding value of _y or ( f- 20 J calculated.
Thus, if ^ = 10 —
100 ,
y = 1- 20 = 22'i;
40 ^
Or, ^ = 20 gives J/ = 30
X = 30 gives 7 = 42 '5
and so on.
6 Graphical Calculus.
All these and similarly calculated points will be found to
lie on a certain curve (Fig. 4), instead of, as before, on a
straight line, and no point
which is not on the curve
will satisfy the equation,
and any point which is on
the curve will satisfy it.
This equation, since it
contains the second power
of one of its "variables,"
is said to be of the second
degree, and the curve is, as
before, called "the curve
Fig. 4.
y = h 20," and the diagram of the curve exhibits graphi-
40
cally the relation between x and y, or, in other words, between ■
x^
X and h 20, for all values of x.
40
The equation to any curve, then, gives a relation which
must be " satisfied " by the co-ordinates of any point on the
curve. We have " plotted " or " traced " these curves by
arbitrarily assigning a series of values to x {i.e. treating x as
an independent variable quantity which can assume any value
we please), and calculating the value which y (the " dependent
variable ") assumes in consequence of x having that particular
value we have assigned to it.
In this book the values of the independent variable are,
to avoid confusion, always measured in a horizontal direction,
and those of the dependent variable vertically.
§ 4. Experimental Curves.
Curves may also be obtained by other means than trans-
lating an algebraical equation into geometry, which is practically
the way in which we have obtained the preceding curves.
Introductory. — Curves and their Equations. y
The results of series of experiments in physical or engineer-
ing science are, whenever possible, plotted on paper.'
This method exhibits relations between mutually dependent
and therefore simultaneously varying quantities far more
clearly than rows of figures or pages of symbols can possibly do.
The method may be best explained by taking a simple and
familiar example. Suppose a kettle of cold water is set on a
fire, and the temperature of the water observed at intervals
of, say, one minute by means of a thermometer, a note being
taken of the time and simultaneous value of the temperature.
Suppose the initial temperature of the water is 60° Fahr.
Our readings are booked as follows : —
Time (minutes).
After
Temperjit
ure (degrees).
60
I
95
2
126
3
152
4
..
174
5
193
6
209
■» T _ ._!_ _ C
7
r -1 /-xxr /rM
„ ^\ 1 j-^._
212
Mark off along OX (Fig. 5) equal distances i, 2, 3, to repre-
sent minutes, and along OY (to scale) temperatures beginning
at any convenient temperature. Then find a series of points
on the paper whose abscissae represent the observed times,
and whose ordinates represent the corresponding temperature to
any assumed scale. Thus the point (i min. 95°) will be found
by the intersection of a vertical through i min. and a horizontal
through 95°. All the points are to be found in the same
way, and a freehand curve drawn through them. This will be
a time-temperature curve, exhibiting the result of our experi-
ment very clearly j for not only does the height of the curve
' The paper most convenient for this purpose is what is called "squared
paper," which is ruled in small squares of o'l inch side, and can be bought
at any stationer's.
8 Graphical Calculus.
at any point show at once the value of the temperature at the
corresponding time, but the shape of the curve conveys at a
glance a general idea
how the rate of rise
of temperature varies
throughout the experi-
ment. It is clear that
when the temperature is
rising rapidly (as it will
do at first), the curve is
steeper than when it is
rising less rapidly later
on in the experiment.
Thus the amount of rise
of temperature between
say o and i min. is greater than the amount of rise between 3
and 4 min., and therefore the average upward slope of the
curve between and i min. is greater than the upward slope
between 3 and 4 min.
Curves of this kind are quite familiar to us. Thus we
have in the daily papers curves representing time variations
in the height of the barometric column or of the thermometer.
The differential calculus is chiefly concerned with the slope
of curves, and it is therefore important that we should get
accurate ideas of how this slope is to be measured.
FiQ. 5-
§ 5. Methods of measuring the Slope of a Line.
Suppose we have a sloping line AB, and we wish to
determine exactly how much it slopes with respect to a
horizontal AC. We can do this in several ways, any of which
we can use when convenient.
(i.) We can find the number of degrees in the angle BAG
by a protractor.
(ii.) We can measure the length of the arc CD by a steel
Introductory. —Curves and their Equations. 9
tape, and also find the length of AC. By these two measure-
ments we could easily reproduce the angle on paper.
(iii.) The method we shall
always use in the differential
calculus is as follows : From
any point B in AB drop a
perpendicular BC to AC.
Measure CB and AC. Divide
the length of CB by that of
AC. We thus obtain the length
of a line MP where AM = i."
Thus, suppose CB = 2", AC
CB 2
= 3". Then -— = - = 0-667. It is clear, if AC is three
AU 3
times AM, that therefore CB is three times MP, and therefore —
MP = ^=^^
3 AC
(Read again note on p. 5.)
This is, of course, true wherever we take B on the line AB.
If AC, instead of being 3", had been o'o42" suppose, we should
have found CB = o*o28, and therefore MP = = - as
0'042 3
before; so that wherever B may be on AB we can always
CB
represent the ratio — by the line MP, where AM = i, even
though B is quite close to A. The student should convince
himself of the reality of this result by trial and accurate measure-
ment. To obtain the length of MP with some accuracy, it is
convenient to make AC = 10". Thus in this case CB would
be 6-67. Hence MP = 0-667.
§ 6. Trigonometrical Ratios.
, . . , ^ , . perpendicular.
It IS convenient to have a name for the ratio ;
base.
It is called the "tangent of the angle of slope," and the
relation is expressed thus in Fig. 6 —
lO Graphical Calcubts.
Tan BAG = •— = MP
There are other ratios of an angle which must be perfectly
familiar to the student before he can make any considerable
progress in this subject.^ Thus -- is called the " sine of the
AB
angle BAG," written thus —
OR
Sin BAG = •— r
AB
AC
also 7— is called the " cosine of BAG," written thus—
A.ij
AG
Gos BAG = — r
AB
AG
also " cotangent of BAG," written "cot BAG," = ^
AB
" secant of BAG," written "sec BAG," = -^
"cosecant BAG," written "cosec BAG," = r—
GB
also —
length of any arc GD . , ^ , , „ . ^
; ; — — -— = cn-cular measure of the angle BAG ;
length of Its radius AG
or, in other words, the number of " radians " in that angle.
It is clear that when the arc GD, measured with a flexible
steel tape along the circumference, = radius AG, the ratio
arc GD ^ , , , „.„
— —— =1. In that case the angle BAG = i radian =
radius AG
57'03° about.
It is clear that, provided GB is always perpendicular to AG,
all these ratios are quite independent of the position of G on
the line AG, for as AC increases, BC and AB and the arc GD
' It is highly desirable that a student should have a knowledge of
elementary trigonometry before commencing this subject.
Introductory. — Curves and their Equations. ii
also increase in the same ratio, if the angle BAG remains
unaltered.
EXAMPLES,
1. Draw the following curves on both sides of the axes : —
(i.) y = x.
(ii.) y = 2.x,
(iii.) J/ = 2x + 3.
(iv.) 37 = X - 6.
(v.) y = x\
(vi.) ><' = j; + 4. (Notice the double sign for / : ^ = ± >J x + 4).
(vii.) j' = A~'-4.
(viii.) xy = 4.
(These curves may be obtained by giving arbitrary values to either x ory.)
2. What are the meanings of m and c in the line jy = mx + c ?
{A»s. m is the tangent of the angle of slope of the line to OX. c is
the distance from O, where the line cuts OY.)
3. Find where the curves
(i.) ay = j; + 3
(ii.) y = x^ — 2
(iii.) y'i = x -I
cut the axes of X and Y.
(Find the value oi x and_)/ successively when the other variable = o.)
4. Find the equation to a line cutting OY at a distance = 3 below O
and inclined to OY at 60°. (See Example 2.)
5. Why is no part of the curve y = x^ below OX ?
(Ans. Because, whether the value of x be positive or negative, the square
of it must essentially be positive, i.e. the ordinate must be ahoveO'%.. Sup-
pose any point of this curve were 2 inches below OX ; we should have —
— 2 = 3(?
ox X = kJ — 2
And since the square ,root of a negative quantity is essentially imaginary,
being neither + nor — , it is evident that we can have no real point of the
curve below OX.)
6. Devise geometrical constructions for determining the value of the
sine, cosine, circular measure of any angle ; also for -: — , — ; — , etc.
■" " sine cosme
(See iii. p. 9.)
CHAPTER II.
graphical differentiation and integration.
§ 7. Illustration of Graphical Differentiation.
Every straight line, however long or short, must have a
definite inclination to every other line in its plane. In this
Fig. 7.
work we are chiefly concerned with that function' of the
' Any variable quantity p whose value depends on the value of another
variable quantity q, is said to be a " function " of that quantity. Thus the
sine of an angle is a function of that angle, because the value of the sine
depends on that of the angle.
Graphical Di^erentiation and Integration. 13
inclination of lines to the horizontal which we have called
the tangent of the angle of slope (§ 6).
If we have a figure made up of straight lines, it is quite
easy to determine the slope of each rectilinear part of it to a
horizontal line. Suppose, for instance, we have given an
elevation (drawn to scale) of a certain section of railway, such
as Fig. 7, in which vertical heights are much exaggerated for
the sake of clearness.
On most railways what are called the "gradients" are
indicated on boards, such as that illustrated in Fig. 8, placed
by the side of the line. The
meaning of this is that while
the line is level on the right of
the board, for every 100 feet
measured horizontally, towards
the left the line falls i foot
in 100; or, in other words, the
LEVEL
tangent of the angle of slope ^^^^^^^^^^^^:^^^^:^^
is 1^. The direction in which p^^ 3
the line slopes is shown by the
obvious slope of the board. If the line is level, or of no
slope, it might in the same way be indicated o in 100, but the
word " level " is used instead.
Now, we may very conveniently draw underneath the actual
elevation of the railway in Fig. 7, another curve showing the
slope at each point; or, choosing the scale of i inch = i in roo,
the rate at which the line is rising just at that point for every
100 feet horizontal. It is of the highest importance that the
student should thoroughly grasp the exact meaning of this
lower curve, for in it is contained the very kernel of the whole
subject. Suppose we take any point A on the upper curve
(which we may call a curve in spite of the fact that it consists
wholly of straight lines).
Draw a vertical from A to cut the lower curve in A\ Then
the height of the lower curve d'PC shows the amount (i foot)
14 Graphical Calculus.
which the line would rise if it continued with the same
slope as it has at A for loo feet horizontally. The fact
that the slope may change just after the point A is passed
does not in the least affect the height just at the point
A\ for this is only dependent on the tangent of the angle
of slope just at the point A. Neither is the height of the
lower curve at A^ any guarantee that the line will actually rise
I foot in the next loo feet horizontal, any more than the fact
that a train may be travelling at the rate of 60 miles per hour
is any guarantee that in the next hour it will actually travel
60 miles, or any other distance. It possibly may entirely
change its velocity in the next half-second, as in the case of a
collision, but this does not alter the fact that at the point in
question it was travelling at 60 miles per hour. In exactly the
same way, at the point B, the line is rising at the rate of i foot
per 100 feet, and this is not affected by the fact that from a
couple of feet to the right the line runs perfectly level. The
important point to observe is that the height of the lower curve
represents a rate of rise, and not necessarily an actual rise. It
^ ■ c ^ ^^ .- corresponding small vertical rise.
represents, in fact, the ratio — - — ° :
small horizontal distances.
The fact that both the numerator and denominator of this
fraction may be indefinitely small does not affect the value of
the ratio, as explained in § 5.
Now, just to the right of point B the upper curve suddenly
changes its slope from i in 100 to o in 100 ; and, conse-
quently, the lower curve drops suddenly from i to o. The
line has no slope from near B to near C, and therefore the
lower curve has no height between the same two points.
Near C the rise changes into a fall, i.e. the rate of rise
becomes negative, and so the height of the lower curve becomes
negative, i.e. is below the axis of X (see § 2). At D the negative
slope suddenly changes to a positive one, and so the lower
curve suddenly jumps up above the axis.
If the student has thoroughly mastered the preceding
Graphical Differentiation and Integration. 15
explanation, he will have little difficulty with the rest of the
subjects treated of in this book, and we have, therefore, given
it a very full explanation — fuller, perhaps, than many students
will find necessary. The principle here explained is exactly
that running through the whole of the subject, and the student
cannot be too familiar with it. The lower curve is called the
" derived curve " of the upper one, and the height of the lower
curve at any point gives the value of the " differential co-
efficient " of the upper one at the corresponding point. The
student will understand these expressions better a little later
on. We shall then also see that the railway companies, by
means of boards, such as Fig. 8, really " differentiate " the
curve of the railway for the information of the engine-drivers.
§ 8. Example of Graphical Integration.
Hereafter, in all cases, points on the derived curves are in'
dicated by dashes, thus : F^ on the first daived and P" on the
second derived correspond to P on the primary, etc.
Let us now look at the converse process. Suppose we are
given this derived curve or curve of slopes, and are required
to deduce from it the actual elevation of the railway. The
student will find this easy if he has mastered the principle on
which the derived curve was obtained. We see, in the first
place, that along the line PB there must be an even uphill
slope of I in 100; for the fact that P^B^ is parallel to O'X^
shows that the slope is constant. Hence PB must be
perfectly straight as far as B. A question that meets us at
the outset is a very suggestive one. Where are we to start to
draw the curve ? The bearing of this question on the subject
of integration will be fully explained later on. At present it
is sufficient to notice that there is nothing whatever in our
derived curve to tell us where the point P is to be taken in
the vertical 0^0 ; that is to say, our curve of slopes does not
give us the height of the point P, or any other point, above
1 6 Graphical Calculus.
datum level. Taking, then, any arbitrary point P as starting-
point, we see that the line slopes upwards i in loo as far
as a point corresponding to B\ after which it is level as far
as a point vertically above C\ afterwards sloping downward
\ in 100 as far as D, then upwards \ in loo. So that we
can draw the actual shape of the upper curve, having given its
curve of slopes.
Meaning of the Arbitrary " Constant " in Integration.—
If we had taken any other point, Pj, as starting-point in the
same vertical line, we should have obtained the dotted curve
which is precisely similar to the one we have obtained, but shifted
higher or lower, according as Pj is higher or lower than P, and
the distance between the two curves, measured perpendicularly
to OX, is "constant" all along the curve. It is clear that
while we cannot find the absolute height of any point on
the curve so obtained, yet we can ascertain definitely the
difference of height of two points on it, for this difference is
the same wherever the curve, as a whole, may be shifted to.
§ 9. Differentiation of Continuously Varying
Outline.
It is not necessary, for our process, that the real elevation
of the curve should consist of straight lines. We might take
any continuously curving outline, such as the hill illustrated
in Fig. 9, and determine the curve of slopes for it. Now, in
a rounded outline like this, the slope does not change suddenly,
as it did in our assumed case of the railway, but it gradually
changes from point to point. We say familiarly that the hill
is steeper at one part than at another, or that as we go higher
up the hill it gets steeper and steeper, or the steepness
increases every step we take, and so on. Now, we have
already explained how we are going to measure this steepness,
viz. by the tangent of the angle of slope (§ 5). But we come
again to a question which must be carefully considered in all
practical applications, viz. what scale are we to use ? Observe
Graphical Differentiation and Integration. ly
that the horizontal scale is the same for both curves, whereas
a vertical height or ordinate on the upper curve has an entirely
different meaning from an ordinate on the lower ; the former
represents an actual height, the latter a ratio. For con-
venience and clearness, we shall at present adopt as a unit
whatever is represented by i" on the upper figure, "■ which is
Fig. 9.
here 100'. The vertical scale of the derived curve is here
full size where PM is unit length, i.e. the height in the given
units represents the tangent of slope at the corresponding
point of the upper curve.
To find the derived curve, draw a number of ordinates
• In the original drawing, of which Fig. 9 is a reduced copy, PM was
made = i"- This remark applies to all the figures.
C
1 8 Graphical Calculus.
to the upper curve and produce them downwards. We are
about to determine the actual slope of the hill at each of the
points where these ordinates cut the outline of the hill. For
the sake of avoiding confusion in the figure, we shall do this
for two points only, viz. P and Q ; all the others are to be
treated in exactly the same way. At the point P draw a line
PT, touching the curve. The slope of the hill at the point
P is evidently exactly the same as that of this line, for a little
to the left of P the hill slopes more, and a little to the right
less ; so that at the point P the slope is the same. Through
P draw PM horizontal = i" (representing loo'), and draw
MT vertical. This line measures, say, 1*2" = 120 feet.
Then, just at the point P the hill is sloping 120 feet in 100.
Take any convenient base-line, 0^X\ on which to draw the
derived curve, and make /T* = MT ; then /T' represents the
slope at the point P. At Q the hill is sloping downwards.
Draw the tangent as before, and make QN = i", and draw
NS vertical, and make ^'Q^ on the lower curve = NS, and
measured in the same direction, i.e. downwards. Repeat this
process for all the points where the ordinates cut the outline.
It is convenient, before commencing, to ink in the original
curve, so that any line may be removed from the figure after it
is done with. It is usually more accurate to draw the tangent
first, and the ordinate afterwards ; the point of contact can
then be more accurately found. The drawing should be made
with a hard pencil, sharpened to a fine point. Unless the
original curve is very accurately drawn, it is impossible, even
with great care, to obtain very accurate results, owing to the
difficulty of drawing in the tangents correctly. When the
points are all obtained, draw an even curve through them all.
This is the derived curve, or curve of slopes of the upper
curve. The height of it represents, at any point, the corre-
sponding value of the differential coefiScient of the function
which represents the height of the upper curve.
Graphical Differentiation and Integration. 19
§ 10. Remarks on Derived Curves.
There are several important facts to be observed about
two curves standing in this relation to one another.
1. At the highest point of the hill the slope is, of course,
nothing, the tangent being horizontal ; for suppose the slope
were slightly downwards, then a point on the left of the
highest point would be slightly higher than the highest point,
which is absurd, (This assumes that the curve is continuous,
i.e. that there are no angles or sharp points in it.) The height
of the derived curve is therefore nothing.
2. This is also the case at the very bottom of the valley,
for a similar reason.
3. Conversely, at the point corresponding to that at which
the derived curve cuts the axis of X\ the height of the
priiliary curve is either a maximum or minimum.
4. In the case of a maximum, the derived curve slopes
downwards from left to right, i.e. has a negative slope. In the
case of a minimum, the derived curve slopes upwards from
left to right, i.e. has a positive slope.
By a "maximum" or "minimum" we do not mean absolutely the
highest or lowest point on the curve, but merely a point to the left of
which the slope is in a different direction from what it is on the right.
Fig. I
Thus in Fig. ID the points AAA are all maxima, and BBB are
all minima. For each of these points the derived curve crosses its own
base-line.
20 Graphical Calculus.
5. At the point where the curve of the hill changes from
convex to concave, or vice versA, with respect to the axis OX
its slope is greatest, and consequently the derived curve highest
or lowest, as the case may be ; i.e. at these points the derived
curve has no slope (see observation i). Such a point is called
a " point of inflection " on the original curve.
§ II. Meaning of a Tangent to a Curve.
We have spoken of " drawing a line touching a curve," but,
as much of what follows depends on the relation between
a curve and its tangent, it is important to get clear ideas as
to the precise meaning of the expression, "a line touching
a curve." Euclid's definition applied to a circle is that the
line meets the curve, but, being produced, does not cut it.
This definition, although applicable to a circle, would not be
applicable, for instance, to such a curve as that illustrated in
Fig. 10, where the tangent at P cuts the curve again at S.
The following is a more modern conception of a tangent to
a curve. Consider a fixed point P and a movable point Qi
on a continuous curve of any shape. Join PQi, and produce
it in both directions. Conceive that Q, with the line PQ
always passing through it, moves gradually towards P, occu-
pying successively such positions as PQj, PQ3. " In the limit "
when Q is just on the point of coinciding with P (being, in
fact, "infinitely near" to P), the line PQ is a tangent to the
curve at P. In this position the infinitely small portion of
the curve PQ may be regarded as coinciding with the portion
PQ of the touching line. This shows that a curve may be
regarded as composed of an infinite number of indefinitely
short straight lines joined end to end. Bearing this explana-
tion in mind when we have to do with an extremely short
bit of a curve, we shall treat it as a straight line, as the
various explanations are thereby rendered much simpler and
clearer.
Graphical Differentiation and Integration. 21
§ 12.
We have also a remark to make with respect to the modern mathe-
matical conception of the meaning of such expressions as "zero,"
"infinitely small," "infinitely great," "absolute equality," and so
on. It may be stated at once that the human mind is incapable of
conceiving any reality corresponding to any of these expressions. The
modern conceptions of them may be briefly summed up thus : zero or o
means " something 5ra.z!i\s.x than anything" Thus, take a quantity whose
weight is the thousand-millionth part of the weight of a hydrogen atom ;
zero weight means some weight smaller than this — smaller, indeed, than
anything that can be named. The same idea is contained in the words
" infinitely small." In the same way, " infinitely great" means "some-
thing greater than anything." " Absolute equality " between two quantities
exists when the difference between them is what we have defined above
as zero.
§ 13-
Let us now cease to regard the upper curve of Fig. 9 as the
elevation of a hillside, and look at it simply as a curve traced
on paper. The lower curve may still be called the curve of
slopes of the upper one. The length of the ordinate of the
lower curve, corresponding to a point P on the upper, now
simply shows the instantaneous rate at which the upper ordinate
is increasing per inch increase of the abscissa, i.e. the rate of
increase during the instant in which the tracing point is
passing through P ; of course, after the point has passed
through P the rate is no longer the same.
Comparing together § S, § 9, and § 12, we see that if, in
Fig. II (which represents a curve PB and its " first derived"),
P and Q are two points on the primary curve close together,
KO
then = /T^ (This is only an absolute equality when Q
PK
is infinitely near to P.) Therefore, since pq is small, KQ
= /P^ X PK =/T^ X/y, nearly. That is to say, the
number of linear inches in the short line KQ is very nearly
equal to the number of square inches in the thin rectangle
P^Ky/\ In the same way, let pq = qr = rs, etc. ;
22 Graphical Calculus.
Then LR = rectangle Q'LVy
MS = R^VIV/
and so on. Suppose this process continued as far as B, and the
results added together; then clearly to the same approximation —
KQ + LR + . . . 4- HB in inches = FKV/ + Q^LVy + . . .
+ U^HV^«^ in sq. inches (a)
Now, however long or short pq,qr, etc., are, the left-hand
side of this equation always = CB. Imagine, then, that instead
of pq, qr, etc., being finite, as
in the figure, we had been able
to draw the ordinates /P, ^Q,
etc., so close together that pq,
qr, etc., are infinitely small, so
small that pb contains an in-
finitely great number of small
parts, each = pq. This will
X not in the least alter the total
value of the left-hand side of
equation («), for this must of
necessity be = CB. But when
pq, qr, etc., are infinitely small,
the right-hand side is equal to
the area under the curve P^B\
, bounded on each side by the
ordinates /^P^ and ^^B^ ; for it
is clear that the sum of these
rectangles only differs from this
area by the sum of the little triangles P*K'Q\ Q'L'R\ etc., and
these all added together are evidently considerably less than
the rectangle K^D, which is infinitely small compared to the
area P^B^^^\ when /V^ becomes infinitely small (see § 32 on
" Orders of Infinitesimals ").
So we see that the dwindling of pq to an infinitely small
quantity produces two effects simultaneously —
Fig. II.
Graphical Differentiation and Integration. 23
(i) Makes the equations such as
PK ^
absolute equalities instead of only approximations.
(2) Makes the equation —
Sum of rectangles P^Ky/^ + etc. = area of curve
an absolute equality instead of an approximation.
On the other hand, it does not interfere with the absolute
equality KQ + LR + . . . = CB. It therefore makes equation
(o) equivalent to the assertion —
CB in inches = area P^B'ijy in sq. inches . (/3)
(see note to § 3 and § 1 2).
This result must not be regarded as an approximation. It
is an absolute and complete equality.
Our proof of the equality (jS) would have been only an approximate
one if we had imagined the strips as of finite thickness ; for in that case
the two equations on which (j8) depends would be only approximately true.
They are only absolutely true (§ 12) "in the limit " when pq, etc., are
infinitely small.
Here, then, we have a most convenient way of measuring
any irregular area with a curvilinear outline. Suppose, for
instance, it is required to find the area of the lower figure
between any two ordinates, we must regard it as the curve of
slopes of some other curve, which we must proceed to draw.
We shall presently explain the practical method of drawing
this curve. Assuming that it has been drawn, and that the
upper curve in Fig. ii has been so obtained, then, in order
to find the area of the lower curve between any two ordinates
taken at random, we have only to find the difference between
the two corresponding ordinates of the upper curve, and we
have at once the required area in square inches.
24
Graphical Calculus.
§ 14. Practical Integration.
The method of finding the upper curve, having given the
lower one, is the same in principle as that by which, in § 8,
we deduced the curve of the railway from its curve of gradients.
Take a curve, such as the lower one in Fig. 12, of which
we are desired to find the area between any two ordinates.
Then we know that the length of any ordinate of this curve
in inches must = tangent of slope of required curve at the
point where continuation of that ordinate cuts it. Draw a
number of continued ordinates, and figure them as shown.
Graphical Differentiation and Integration. 25
Set off on the axis of X, OS = i", and set up SK vertical.
Cut off on SK, Si, S2, etc., etc., equal to the ordinates o,
I, 2, 3, 4, respectively. Join Oo, Ox, etc., as shown. Now
consider any one of these ordinates, say 3. If we had had
the upper curve given, and had been required to deduce the
height of the lower on the ordinate 3 by the process explained
in § 9, it is evident that the triangle we should have drawn
would have been exactly equal and similarly situated to the
triangle OS3. Hence O3 must be parallel to the tangent to
the upper curve at the point where ordinate 3 cuts it.
Arguing in a similar way about the other points, it is clear
that what we have to do is to draw a curve of which the
slopes at the points where the lines o, i, 2, etc., cut it, are the
same as the slopes of the lines Oo, Oi, O2, etc. On pro-
ceeding to draw the curve, we take an arbitrary point T from
which to start (re-read carefully § 8 on the " Constant ").
The next question is as to how we are to obtain the form
of the curve between two ordinates. This we shall decide as
follows. Any two neighbouring ordinates will be assumed
so close together that the part of the
curve between them is indistinguishable
from a circular arc which has the same
slopes at its ends as the actual curve
would have. It will not be necessary
for our purpose to find either the ''
actual position of the centre or the
length of the radius. The method of
obviating this necessity is seen from
Fig. 13, which is an enlarged view
of two neighbouring ordinates. P is
the assumed starting-point. Draw
through P, PR, and PS parallel to the
directions of the curve on the ordinates P and Q. These
directions are obtained from Oi, O2, O3, etc., in Fig. 12.
Bisect RP3 by PQ. It is evident, from the geometry of the
Fig. 13.
26
Graphical Calculus.
figure, that the circular arc when drawn will pass through Q.
Q may next be taken as a fresh starting-point and the process
repeated, and so on for the next ordinate. Thus a series of
points are found through which the curve may be drawn.
The difference between any two ordinates of the upper curve
will, as already proved, give the area of the lower curve
between the same two ordinates.
Z1Z
Fig. 14.
It is to be noticed that when the lower curve dips below the base, the
corresponding portion of area is to be reckoned negative.
This is the process of "integration," and corresponds exactly
to the algebraical process known by that name in mathematics.
It is necessary, for the above process, to use a hard sharp-
pointed pencil, otherwise great inaccuracies may creep in.
If carefully performed, the process is at least as accurate as
Graphical Differentiation and Integration. 27
any of the ordinary processes for finding of areas. The author
of this work has devised a mechanical integrator, described in
the Appendix, whereby the integral curve may be automatically
drawn.
§ 15. Differential Coefficient considered as a Rate
OF Increase.
In cases where the curve which we are differentiating is one
representing the results of a series of experiments, the derived
curve is often of great importance. As an illustration of
this, we will take the case of the experiment described in
§ 4. Suppose we had found, in that experiment, that the
temperature had risen uniformly up to 212° — that is to say,
that, during the first half-minute, if the temperature had
risen 9^° (suppose), then during
the fifth, or any other half-minute,
it would also have risen 9^°.
Suppose the total time occupied
= 8 min. Now, if we divide the
total increase of temperature, viz.
152° (represented by O212, Fig.
14), by the time occupied, viz. 8
min., represented by ON, we shall
obtain the amount of rise of tem-
perature in I min. We have seen
from § s that we shall also obtain
the hne MP. It is also otherwise
obvious that, since OM = i min.,
MP = amount of rise of tempera-
ture in I min. Thus the tangent
of the inclination of OP to OX
represents the rate at which the temperature is rising. Our
derived curve in this case would be a line parallel to O^X^
at a height = MP. This would indicate that the rate of rise
of temperature was constant all along the curve.
tf" P' X
Fig. 15.
28
Graphical Calculus.
Although the actual curve was not a straight line, it may
easily be seen that the tangent of inclination at a point P
(Fig. 15) still represents the rate of increase at the point P,
for the small " element " of curve at the point P is also part of
the tangent to the curve (§ 11), and the rate of increase is
therefore the same as the rate of increase along the tangent,
i.e. = height of derived curve at the point.
§ 16. Other Applications.
Innumerable other applications of the same principle may
be found. In almost every case of a curve derived from
experiment, a distinct and tangible
meaning may be ascribed to the
height of the derived curve. One
of the most important applications
is the case where a curve is drawn
Q
/
F
X
!
y
secoiiUs
P
'I
1
s
P
q'
&
^
secoiuU 1
0'
'P
Fig. 16.
to represent the motion of a moving body. Take the case
of a man walking at a uniform rate along a road. Suppose
Graphical Differentiation and Integration. 29
we plot vertically his distance from the starting-point, and
horizontally the corresponding time. Thus after i sec. (repre-
sented by O/) (Fig. 16) he has walked a distance repre-
p? oQ
sented by/P j V- = ^ = /P^ = his rate of walking. The
(jp \jq
derived curve is here a horizontal line, as in the last section.
Successive Differentiation. — But suppose he walked a
greater distance in the second second than in the first, and
a greater still in the third, and so on, the height of the derived
curve would still represent his instantaneous velocity at the
corresponding point ; for just at the point P, for instance, he
is increasing his distance from any fixed starting-point at the
rate of MS =/T^ yards per second, so the middle curve is
a curve of velocities. Now, if we differentiate the derived curve,
we shall obtain a curve showing the time-rate at which his
velocity is varying at each point, for at P^ his velocity is
increasing by IVrS^ yards-per-second every second = /"P".
Hence /"P" represents the numerical value of his acceleration
at the point P. This curve is called the second derived
curve of the time-distance curve. If we differentiate again,
we shall obtain a curve showing the rate at which his accele-
ration is changing. The dimensions of these latter units
would be yards-per-second-per-second per second. This
curve is called the third derived of the time-distance curve,
and so on. It is easy to see that if the velocity had increased
uniformly (or the time velocity curve had been an inclined
straight line), the acceleration curve would have been a
horizontal line, or the acceleration would have been constant
or uniform, as in the case of a falling body. Thus we see
that velocity = time rate of change of distance, acceleration =
time rate of change of velocity, etc.
The student should think very carefully over this argu-
ment, because, in addition to its intrinsic importance, it forms,
perhaps, the most perfect illustration of the application of
the calculus to science that could be found. Curves may
30 Graphical Calculus.
sometimes be differentiated by special constructions not in-
volving the drawing of tangents. Several cases of this will
be given later on.
Examples.
1. The space passed over by a body falling from rest is given by —
;' = 16^"
Draw this curve, selecting suitable scale, and differentiate it twice
graphically. Compare the curves obtained with the curves (i.) y = y.t,
(ii.) y = 32. What are the meanings of j/, y, y ?
2. Draw also the curve —
y=loi+ \(>e
giving the space described by a body thrown downwards with velocity
ft.
10 — ; ' differentiate it twice, and compare with curves derived from (i).
sec*
Show that the first derived differs by a constant from the corresponding
curve of (i), and the second derived is the same as the second derived of
(l). What would have been the difference introduced if the body had
been thrown upwards ?
3. Draw the curve xy = 12. Integrate it, and find, by means of the
integrated curve, the area of the given curve between the ordinates —
jc = 3 and X = 12. (Ans. — 16'63 sq. inches.)
4. Draw the curves (in the first quadrant only) —
_ I _
y--x'&ndy= v'^
and find their areas between the ordinates x = 3 and x = S. Am. i6'l7
and 1 1*6.
5. Find between the same ordinates (by reducing the scale of the
integrated curve) the area of—
y = x^ i.e. y = xi. Atts. 89 sq. inches.
' This is a very convenient and suggestive notation for the unit of
velocity, i.e. foot-per-second. It is clear that a velocity of say 6 feet per
second is the same velocity as 12 feet per two seconds, or 18 feet per three
seconds. This notation brings this out very clearly for obviously
sec.
= '- = It will be found on examination that anv quantitv
2 sec. 3 sec, ^ ^ ■'
preceded by the word " per " is invariably a denominator. A logical
extension of the same notation is used to denote unit of acceleration. An
increase of velocity of s feet-per-second every second is denoted —
ft.
J. —
sec. ft,
sec. "~ sec'
All physical units are treated in the same way.
CHAPTER III.
nomenclature and general principles.
§ 17. to show that the height of the derived
Curve may reasonably be denoted by — .
dx
We have shown, in the last chapter, the geometrical meaning
of the processes of differentiation and integration. We now
proceed to explain the system of symbols that accompanies
it. Suppose we are required to find the height of the derived
curve of the curve y = 0?, corresponding to the point (2, 4).
Now, it would obviously be very inconvenient to be com-
pelled to draw the curve y = 0? \.o scale, and differentiate it
graphically in order to find the height of the derived curve
at one point. An algebraical method of calculating it would
be much more convenient, and this is what we are about to
explain. The method is as follows : —
1. Calculate the height of the primary at a point Q (Fig. 18)'
whose abscissa is slightly greater than that of P, the given point.
2. Find the difference MQ between the ordinates, and
divide it by PM, the difference between the abscissae.
3. Investigate what the result would be if PM were to
gradually diminish until it became infinitely small. Now, if
PM is a considerable size, the value of — — , viz. the tangent
PM'
of the angle of slope of PQ, will differ considerably from the
' Fig. 18 is not drawn to scale.
32
Graphical Calculus.
tangent of angle of slope of the tangent to the curve at P.
When PM is infinitely small, there will be no such difference.
Our algebraical method of limiting values is equivalent to
taking Q infinitely near to P. The algebraical result of the
process must not, therefore, be regarded in the light of an
approximation. It is an absolute exact truth (see § 12, p. 20).
Fig. j8.
The difficulty in understanding this (if there be one) is
due to an imperfect appreciation of the meaning of the
expression, "when Q is infinitely near to P." Now, PM is
called the " increment of x." It is written A.x, and implies
the amount by which x is supposed, to increase from an
by 2t,
Nomenclature and General Principles. 33
assumed particular value x. Similarly, MQ is called the
corresponding " increment oiy" and is written Aj/ ;
Tan QPM therefore =^
^x
Now, the point P being (2, 4), let the abscissae of Q be
2^, i.e. A^ = \. Its ordinate then = (2^)2 = 6^.
Hence t^y = {(2^)^ - (2)^} = 2^
Thus when x increases' by ^ from the value 2, y increases
1
A^
Ax
Similarly, if we take Ax = I —
^y = ItV
Av
and therefore — = 4x when Ax = x
Ax *
Similarly, if A^ - o'oooooi
1 Ay
then — = 4 '00000 1
Ax
Again, if we take Ax = —J —
Then A;/ = {(i|)2 _ (2)2} ^ _i.|
Ay -iS
Therefore — = 4i when Ax = \
A^- -i ^*
If we plot a curve (Fig. 19) showing the relation between
Av
Ax (abscissa) and — (ordinate) for the point (2,4), we shall
find it is in this case a straight line. Where this line cuts
OY, the value of Ax is obviously = o ; i.e. the increment of a-
(PM in Fig. 18) is " indefinitely small" or " vanishes," and in
Ay
this case clearly -— = 4 exactly.
A^
When this is the case, i.e. when Ax, and therefore Ay,
D
34
Graphical Calculus,
diminish indefinitely, we write dx and dy instead of Aar and
Ay, and the ratio (as indicated in Fig. 19)
dy
dx
„ dy .
But — is not
dx
like ^
in being an actual
Aa;
fraction with numerator and
denominator — that is to say,
the dy and dx are not separate
quantities which have actual
numerical values : -- must be
dx
taken as a single symbol repre-
senting a definite finite quan-
tity, although dy and dx are
each infinitely small (see
bottom of p. g).
Of course, it is only for the
point (2, 4) that the value of
Values of t^a: dy
_ -7- = 4. If we had chosen the
Fig. ig. ^^
point (3, 9) instead, we should, in exactly the same way, have
Indeed, if we had taken any point {a, a?) on
dy
found -,-= 6.
dx
dy
the curve, the value of -7- would have been za.
dx
The student
should work out a few cases of this in the same manner as
shown above. If he does it thoughtfully, he will probably
be able to see the reason of it.
The algebraical process corresponding to that explained
above is as follows —
Letjv = x^ (i.)
Let X increase by Ajk, and in consequence j' by A_y.
[In the process explained above, we took particular values J, i, — i> etc.,
for Lx, and calculated the corresponding numerical value for tyy. Here
we calculate the general algebraical value for Aj in terms of i^x as
follows.]
Nomenclature and General Principles. 35
Then, since the point (a; + Aa-, y + A)*) is by supposition
on the curve J = s^, we have (see p. 6)—
y ■\- t^y = (x ■\- Aa;)2 . . (ii.)
Subtracting (i.) from (ii.) in order to find the difference
between the ordinates, we have —
Aj = (« + t^xf - x^ = 2xl^x + {Lxf
Dividing by A* the difference between the abscissa
(PM in Fig. 18)—
Aji*
-— = 2X -Y t^X
which, when Aa; is indefinitely diminished, becomes —
dy
~r = 2X
ax
because Lx (or, as it would then be written, dx) becomes
an infinitely small quantity. Therefore, as in § iz, the dif-
dy
ference between — and 2x being infinitely small, we say that
ax
dy
— = 2x absolutely.
(The student should compare this process with that ex-
plained above at every step. Only thus can he fully realize
its meaning.)
Therefore the ordinate of the first derived curve, of the
curve y = x^, is always twice the abscissa, i.e. the derived
curve is a straight line whose equation is y'^ = 2x.
Exercise. — In exactly the same way, the student can
calculate the height of the derived curve for y = x^. He will
find it to be / = 3^^ ; iox y = «*, it is y = /^x\
It will now be evident that the process of finding the algebraical value
of — is that of obtaining the equation to the curve which shows the
dx
relation between ^ (ordinate) and i^x (abscissa), and finding, as at p. 4,
^x
36
Graphical Calculus.
where this curve cuts the vertical axis. This, it is clear, gives the value
-^ when Ajc = o.
Ax
We can here prove part of the general proposition that
when y = x", ~- = nofi" ' ". This is true, as a matter of fact,
ax
whatever n may be, positive or negative, integral or fractional.
The reader is, however, not yet in a position to understand
the complete proof, so we shall confine ourselves here to
the limited case of positive integral indices, which will be
necessary for purposes of
illustration. The rest of
the proof will be given
as the reader is ready
for it (§§ 24, 34).
All complete curves of the
form y = x'^ where n is posi-
tive and > I, are in general
j( shape similar to two of the
four branches of the curves
shown in Fig. 20.
Where n is an even posi-
tive integer, the curve j/ = jr»
resembles the curve PiOP,
and when n is odd, P^OP. The
reason of this is that, whether x
is positive or negative, y or
jc" is always positive when n
is positive and even (thus-
(-2)' = -t- 16), so that the'
ordinate y always lies above'
XOX. When n is odd, j/ or x^ is always negative when x is negative, so that
in such curves as jc = ;i:' the curve on the left of OY always lies below XOX.
For instance, in the curve j/ = «', if ar = - 2, j/ = ( — 2)' = - 8. The branch
3
OP3 is included in such cases asj/' = x^, i.e. y = ±x'^. In cases where n
is < I, but > o, the curves resemble Fig. 20 turned through a right angle,
i.e. looked at with OY horizontal. When « is < o, the curves resemble'
hyperbolas, of which OY, OX are asymptotes.
It is a most interesting exercise to trace the variations of the curve^
Fro. 20.
Nomenclature and General Principles. 37
y = x^, as « varies between +00 and -00. It afifords, among other
things, a beautiful illustration of the meaning of the statement that
x°= I.
Let 7 = a;" (i.)
If X increase by Aa;, and y in consequence by t^y, we have,
as before —
y -{.^y = {x-\- A^)" . . (ii.)
Subtracting (i.) from (ii.), we have —
^y = {x-\- Aa-)" — .-v;"
Ay _ {x + A^)" — a:"
A.T Aa;
Expanding by the binomial theorem —
x^ + nx^'-^l^x + ~ ^' x^'-Hi^xY + . . . - ^"
A_v 1^2 ^
Aa' A^
= «^"~^ +Aa; X some other quantity
which equation, when A jc vanishes, becomes f- = «;</"~"j or, as
it may be written —
4^")
= nx^
dx
§ 18. Meaning of dy and </:« when used alone.
In some methods of treating and writing the calculus the
expressions dy and dx are used apparently alone. This seems
to cause great difficulty to students, because of a sort of
indefiniteness in the actual values to be assigned to the
quantities denoted by dy and dx. It will be found on ex-
amination, however, that in such cases there is always an
dy
implicit reference to the ratio — . It is merely a somewhat
more convenient way of referring to the ratio, and is introduced
for the purpose of saving space and for convenience of printing.
38 Graphical Calculus.
Thus though dy, standing by itself and considered apart from
anything else, is numerically absolutely meaningless, yet when
we write, as in the curve above, dy = 2xdx, we really mean
that -^ = 2x, which indicates that if x increases by a small
dx
quantity, y increases by a quantity lx times as great.
So in general we may write —
ly = ^y,lx . . . (a)
dx
where hy, tx are corresponding small increments of y and x,
which may be of definite magnitude. If, however, 'hx and Sy
are of small but finite magnitude, the equation (a) becomes an
approximation, though usually an extremely close one, and
not an absolute equality.
dy
8* is here called a " differential," and — therefore a
dx
" differential coefiScient " (see illustration, § 25).
The fact that two quantities of indefinite magnitude can have
a definite ratio sometimes causes students trouble. This may
be got over by reflecting that the quantities 2« and 3« have
always the ratio f whatever the absolute value of 2 « or 3«.
§ 19-
It is easy to see that when y = constant, i.e. when the
primary curve is a straight line parallel to OX, since the
dy
slope is at all pomts of this line nothing, — = o. Or, regarded
dx
algebraically, the statement y = constant is equivalent to the
statement that y does not vary, and therefore dy (which means
the amount of variation of y corresponding to a variation dx
in x) = o, and therefore —
dy
dx ~
Nomenclature and General Principles. 39
§ 20. Successive Differentiation.
At § 17 it was explained how it was sometimes necessary
to differentiate a derived curve. The primary curve in that
case was a time-distance curve, the first derived a time-velocity,
and the second derived a time-acceleration curve. Now, the
height of the primary being denoted by y, and that of the first
dy
derived by y or — , the height of the second derived may,
dx
d-^
on the same principle, be denoted by y or -—-. The latter,
dx
however, is very inconvenient to write and print. It is
therefore shortened by treating it as a simple fraction in which
d stands for some definite algebraical quantity. (In reality,
of course, it does not mean anything of the kind.) Thus we
have —
'^KdiJ d-'y
dx {dxY
d^y
or, omitting the bracket, —
The student must be careful to notice that this quasi-
dy
dx
fraction is nothing but a shortened form of — -, that the
expression has absolutely nothing to do with x\ and that
at present he may regard ^ as merely symbolical. In the
same way, the height of the third derived is —
(Py_
da?
dx ~ dx^
^ do? cPy
d'y
that of the fourth, — , and so on.
40 Graphical Calculus.
§ 21. Notation of Integration.
It will be seen from § 13 that the process of graphical
integration consists of a construction whereby a curve is
obtained of which the tangent of angle of slope is at all points
= ordinate of curve we wish to integrate. The algebraical
process is the exact counterpart of this, and consists in
obtaihing an expression which, when differentiated algebraically,
will give as a result the expression which we wish to integrate.
There is no general method of performing this reverse operation.
Indeed, in a great number of cases it cannot be performed
at all except by the aid of an infinite series. We are in all
cases obliged to rely on our previous experience of differentia-
tion. If the expression is of a type of which we have had no
previous experience, we cannot do anything with it until we
have twisted it into a shape which we do recognize as the
result of some differentiation with which we are already
acquainted.
Suppose, for instance, we wish to integrate ■^x^. This
means thaty = 3^^ is to be the first derived of the curve we
wish to find. The problem is stated thus for an " indefinite "
integral—
f2,x'''dx
or, in the case of a definite integral —
) a
■^o?dx
a
These expressions will be presently explained.
This symbol f may be regarded in two ways : (i) It may
be taken simply as a question mark. The meaning then is —
J
3^:2
? expression will give 2>x'^ when differentiated with respect to x
(2) It may be taken to be the letter s, the first letter of the
word " sum," thus —
/
Nomenclature and General Principles.
b
41
3^V^
Fig. 21.
The sum between the ordinates [x=i) and [x=a) of all such rectangles as ■^xMx
For it is evident that the area in square inches of a very thin
vertical strip of the curve such as that y
shown in P'ig. 21 is ^x^dx (see §§ 13
and 17), and the sum of all the thin
strips into which the area between b and
a may be divided = whole area of curve
between these two ordinates = difference
between corresponding ordinates of upper
curve, as already explained (§ 13).
Now, let us consider what expression
will give 3^;^ when differentiating with respect to x as
independent variable. (The student will understand the
last expression better after reading the next chapter.)
Consider what is the rule just proved (§ 17) for differentiating
^jc". We have found the differential coefficient to be ^a:'""^'.
Hence the answer to the question _/"«.*;'""■"(& is jc". It is
easily seen that the given expression 3^;^ is of the form «.»<"""
where « = 3, hence —
J^x'^dx = a?
A more complete solution, as will be presently explained, is
" st^ + some constant."
Hence we see that J'^x'dx = a;* is exactly the same
d(xP)
equation as = 3^S but put into another form. In just
the same way as — = 5 is the same thing as 5 x 4 = 20.
4
This is sometimes symbolically expressed by saying that
y and d " cancel one another.'' Thus multiplying both sides
d(ii?)
of the equation — — - = ^x'' by dx we have —
dx
42 Graphical Calciibis.
Now multiply hy f. We obtain —
fd(x^) =fix'^dx
or, sinceyand (/cancel —
a;' =J^x'^dx
This is not altogether a happy analogy, for f and d
do not cancel on the right-hand side. The idea is that if
any quantity, A (represented here by the ordinate of the
upper curve), be divided into a large number of parts, and
then all the parts be added together, the quantity A is re-
produced.
§ 2 2. The "Constant" in Integration.
The expression " jc^ + constant " is known as the "in-
definite integral " of 3^;^. It is a general expression for the
height of every possible primary which has y = 3^^ for its first
derived. We have already seen (§§ 8, 13, etc.) that there are
an infinite number of such curves corresponding to different
starting-points on the line OY, If a value K (suppose) be
assigned to the " constant," the value of x" + K at any point
also represents absolutely the area of the curve y = 3^;^ between
that ordinate y and the ordinate corresponding to the point
where the curve j* = ^ -f K cuts the axis of x. This point
may be found by putting y = o m the equation and solving
for X. Thus here —
^ = -^K
This may be generally explained as follows. Suppose we
have any curve P^Q^ (Fig. 22) of which the equation is
y =f^{x) (where f\x) is a shorthand symbol for " any ex-
pression containing x "), and suppose, having integrated it, we
obtain a curve PQ or TK, or some other parallel curve of which
Nomenclature and General Principles. 43
the equation is j/ = f(x) + c, where f{x) is some different ex-
pression containing x. Then, in accordance with the notation
already explained, we have ff\x)dx = f{x) + constant.
Then the expression f{x) + ^ is called the indefinite integral
Fig. 22.
of f\x). Let X = Oq, Then for some particular value of
^1 f{^) + ^ = ?Q- For some other value of c,f{x) + f = ^K.
Since the slope at K is the same as the slope at Q, the height
^'Q^ is the same in each case. Now, ^Q = area of lower
curve between the ordinate /P^ (corresponding to the point
44 Graphical Calculus.
where the upper curve cuts OX) and ^^Q^; also ^K = area
between /T' and ^^Q\ and so on.
If we wish to find the (shaded) area of the lower curve
between two definite ordinates, * = i and x = 2 (suppose),
the expression for the area \sf\f\x)dx.
This is called a " definite integral," and we have already
seen (§ 14) that the area is found by taking the difference in
length between the corresponding ordinates of the upper curve.
These ordinates are found by substituting i and 2 in turn for
X in the expression /(:\;) + ,;. The notation for this is/(i) + c
and/(2) + c, denoting respectively the ordinates aK and ^B.
Hence clearly —
i:^
f\x)dx = {f{2)+c}-{/{l)+c}
I
/:
= /(2)-/(l)
Here again we see that the actual value of c is unimportant,
since it disappears in the final result.
The following notation is usual as a shortened form of the
expression on the right-hand side of the above equation : —
'/'{x)dxJ\f{x)-\-?\
iL J
A further exposition of these facts, with actual numerical
examples, will be found in Chapter IX.
The. nomenclature adopted when it is necessary to
differentiate any expression twice has been explained in § 20.
It is also often necessary to integrate an expression twice.
The notation then adopted is for an indefinite integral —
ff{f"{x)}dxdx
which means —
Suppose the quantity inside the [ ] brackets, viz. —
f{fV)}dx = {f{x)+c}
Nomenclature and General Principles. 45
the above expression obviously means —
f{f(x)^c\dx
It is necessary to notice that the constant c, which is re-
quired (as already explained) for the first integral, must be
included under the sign of integration for the second operation.
The geometrical meaning of this statement will be apparent to
any one who difterentiates a curve twice and then integrates
the result twice. Obviously, the final result will be very much
affected by the (arbitrary) height of the starting-point for the
first integrated curve above 0\ for the height of the curve
obtained by the first integration determines the slope of the
curve obtained by the second integration.
For a definite integral the notation is —
{f"\x)\dxdx
na nc
J bJ a
bJ d
which, as before, means —
Further explanation is not possible at this stage.
Examples.
1. Differentiate ^^ x', x\ x\ jr'", «"», x^, ^, wiih respect to x.
2. Differentiate {a'f, (i^f"; (^)", with respect to—
(i.) a^, b", c*, respectively.
(ii.) a, b, c, respectively.
(iii.) dP, b", c^, respectively.
(Substitute in each case x for the quantity with respect to which the
function is to be differentiated. Then differentiate with respect to x, and
substitute again. Thus, in case iii. —
^dKx") J2C (f-O I2C C^) _12C (13,-,)]
and = — X = — x- J
dx q q 9
J. Integrate 7^", im", (w"-', i.
CHAPTER IV.
general principles.
§ 23. Changing the Independent Variable.
We have hitherto regarded the value of the quantity denoted
by y as absolutely dependent on the value we give to the
independent variable denoted by x. Thus in our illustrative
curve y = %'■ we calculated the points on it by giving arbi-
trary values to x, and calculating the corresponding value of
y. We have also (and we shall continue to do so) always
plotted the independent variable horizontally.
dy I
To show -- = ~. — Now, we might write the relation
dx dx
dy
y = x'^ m the form x -± >Jy, and, in calculating points on
the curve, give arbitrary values to y, and calculate x by
finding the square roots of these values. Now, if we con-
tinued to plot y vertically and x horizontally, we should, by
this process, obtain exactly the same curve as before. But '
by our convention we are to plot the values of the inde-
pendent variable (which is now y) horizontally.
What, then, is the relation between the two curves thus
obtained ?
Let OPQ (Fig. 23) represent the curve y = x\ Let
the curve be drawn on a piece of tracing-paper held over the
original curve. Holding the point O fixed, turn the tracing'
paper through a right angle in the direction of the arrow ; we
shall thus obtain the curve OPiQi, which is clearly the same
General Principles. 47
curve as before, but in which the original positive values of
y, such as ^ Q, are plotted horizontally in the negative direction,
i.e. towards the left, as at Oq^. If we "reflect" the curve
OPiQi along OY, i.e. imagine the curve turned about OY as
axis into the position OP2Q2, this defect is remedied. This
latter curve is the same as would have been obtained by
plotting the curve x = ^^ va. the ordinary way, but with y
Y
fi! qip (f
Fig. 23.
horizontally and x vertically. If we now differentiate this
curve graphically in the ordinary way, what we shall obtain
will be a curve showing the values of -— for all values of v
ay ■'
Now, what we have to prove is that, taking any point P on
the first curve, and the same point P2 on the other curve —
Tangent of slope of (OP) x tangent of slope of (OP2) = i
It will, perhaps, appear that a. simpler and more direct method of
exhibiting this relation would have been to find the slope of the original
curve relative to the axis of y, and to treat the axis of Y exactly as we
dx
have previously treated the axis of X, plotting values of - horizontally
dy
along a base on the left of the original axis of Y. This would, however,
have involved temporarily reversing our ideas of positive and negative direc-
tions. It is easy to see that the present construction is in reality the same
as this would have been if we had also rectified the signs. The additional
utility of the present construction will be more apparent at a later stage.
Taking an adjacent point Q, it is clear that wherever Q is —
PM = MiQi = M2Q2 and MQ = PiM, = P^M^
48 Graphical Calculus.
Hence —
MQ M^, ^ ^
PM P2M2
And as this is true wherever Q is, it is true in the limit, when
dv doc
it moves up to and ultimately coincides with P, i.e. ^ X "^~ = i>'
a result which, though it was certainly to be expected, we
should not have been justified in assuming merely because
— and — look like fractions. Thus we see that, whenever
dx dy
dy . . •, ■ ,
we wish to find a value for — m any given case, if it happens
doc
to be easier to find -- from something we already know, we
ay
dy
are at liberty to do so, and thence deduce -r- by inverting the
value so found.
Some students find a difficulty here which is not easy to express in words.
It is as follows. This proof, as it stands, only holds when Q is at the same
distance from P as Qj is from Pj. When this is the case, it is clear that
— X — = 1. But the values o!dy and i/a; are not definite. We only know
dx dy
that they are indefinitely small, and, provided they are indefinitely small,
they can have any order of smallness. How, then, are we to know that the
dy and dx in the first factor are the same dy and dx as in the second factor?
The answer is, that, provided dy and dx are indefinitely small, the value of
the ratio — is constant for any particular point, no matter what the order
dx
of smallness of dy and dx. The dy and dx of the first factor must be taken
together ; likewise the dy and dx of the second factor. The difficulty arises
dy
from thinking of — as a variable fraction instead of a fixed ratio of two
dx
variable but mutually dependent quantities.
' It is easy to see, from the note on p. 47, that this result is merely
another form of the trigonometrical relation tan 9 X tanf fl) = I, where
fl is the angle of slope of a curve at any point.
General Principles. 49
Thus we see that if we differentiate curve OPQ and
OP2Q2 with OX as base, and- take any two corresponding
ordinates of the two derived curves {e.g. pV and /2P2 are cor-
responding ordinates), the rectangle formed with these two
derived ordinates as sides will contain exactly one square
inch.
Now, we have hitherto, for the sake of clearness, regarded the value of
y as being dependent on the value of x. For the purposes of the calculus
this is not in the least necessary. Our results are just as good if, as a
matter of fact, the value of x depends on that of y, or if the value of
both X and y depend on that of another variable .i, which, although its
variations may be the prime cause of the simultaneous variations of x and
y, does not appear in the equations at all. What the calculus is really
concerned with is the fact that x and y do actually vary together in such a
way that every definite value of x corresponds to a definite value or values
of y, and it is not concerned with what may or may not have been the
ultimate cause of those simultaneous variations.
§ 24. Differentiation of x'^
We can utilize this principle at once to further our proof
dy
dx
dy
of the general proposition that if ^ = x", -- = w^v'"-".
Let n be of form — where m is an integer.
m
1 i_ _
Let _j/ = a:™ where *"' means '^x (see chapter on indices
and surds in any algebra).
As far as we have hitherto gone, we cannot differentiate
this.
\_
It is easy to see that the curve j = a;™ is the same as the
curve X = j™, for the same values of x and y satisfy both
equations.
If we differentiate x = y" with respect to j, we get—
^ = »?y-"
dy -^
50 Graphical Calculus. ■
This operation corresponds exactly to turning our curve through a
right angle, reiiecting along OY, and differentiating graphically.
Hence we have —
dy I
dx »2y™-w
But we require the value of — in terms of x, and not of j.
Substituting a;" for 7, we get —
dy \ \
dx
I I
m' "^
X "
=
m
which
is of the required form
. Thus the proposition holds good
when :
n is of the form --.
m
§ 25. Illustrative Example,
The following example will give an idea of the practical
meaning of this principle.
Suppose we take a barometer up a mountain-side from sea-
level, and note the height of the barometer at short intervals of
vertical rise. (The vertical heights must, of course, be known, in-
dependently of the barometer.) Let a height-barometer reading
curve be plotted from these observations. Let 100 feet = i unit
on the diagram, and barometer-reading scale be full size. The
characteristics of the curve that would be obtained by such
a proceeding are shown exaggerated in Fig. 24. On differen-
tiating this curve graphically, we shall obtain a first derived,
which, since the primary always slopes downwards, lies
entirely below 0^X\ Thus /^P^ shows the rafe of rise, or,
General Principles,
Si
in other words, Py shows the rate of fall, of the barometer
at P per loo feet lift.
Notice carefully the significance of the signs here. A fall is a negative
rise. If the barometer falls + 0'5 inch, it may be said to rise — o'5 inch.
Thus, suppose /^P^= 0-5 inch. The meaning of this is that
, . _, , , - ,, . o.i; mch ,
at the pomt P the rate of fall is — "^ — r (see note on p. •?o),
100 feet lift ^ f o /
feetO
Fig. 24.
Now, we may, as at § 18, p. 38, conveniently write Sy = —
dx
X ix ; e.g. assume that we lift the instrument through 6- inches
(8,*:), when at an altitude given by Op. The instrument will
rise (see note above) by an amount 8y =— x Ix, where
dx dx
=/^P^ is the current rate of rise per 100 feet lift. Hence —
o'S inch
Sy = - — ^-i — X 0-5 foot
100 foot
52 Graphical Calculus.
The dimension " foot " cancels out
_ 0-5 inch _ ^ .
TiTir ■
that is to say, the barometer falls -^\^ of an inch.
If the student be not well versed in the method of
dimensions, he should carefully note the illustration given here.
Of course, this only holds where ^x is so small that the point
on the curve whose abscissa is (x + 8a;) is not any appreciable
distance from the tangent to the curve at the point P. It
%y dy hy dy
implies that -^ = —-. If - differs sensibly from ~, there will
ox ax ox dx
be an error introduced. This point has been already explained
several times (§§ 7, 12, 13, 15, etc.) in various aspects. If the
student does not understand it, he is referred to the sections
quoted.
Now, suppose we turn our primary curve through a right
angle and reflect it on OY. We shall obtain the upper curve
in Fig. 25, which is exactly the same curve as Fig. 24 viewed
under another aspect. The height of the derived curve now
represents the vertical distance through which we must lift
the barometer in order that it may fall i inch, assuming that the
rate of vertical lift per inch fall of the barometer remains
constant ; or, in other words, /T* represents the instantaneous
rate of lifting per inch rise of barometer. Now, what we have
just proved amounts in this case to this —
Rate of vertical lift (inhundreds of feet)per inch fall of barometer
rate of fall of barometer in inches per 100 feet lift
In the particular case considered above, rate of fall of
barometer per 100 feet lifted through was 0*5 inch per 100
feet at a certain point. Hence, at that point rate of lifting
per I inch fall of barometer = — = 2 ; i.e. the instrument
0-5
General Principles. 53
must be lifted 2 units or 200 feet if the barometer is to fall
I inch (assuming constant rate of falling), which is, of course,
otherwise obvious.
This illustration may probably present some difficulty to
the student, partly owing to the essential difficulty (to a
beginner) of viewing the same ratio under two aspects, and
partly from the confusion introduced by the practically
o'
Fig. 25.
necessary difference of scale in the vertical and horizontal
directions. We have already had several instances in which
the scale was intentionally arranged so as to be as simple
as possible. This is intended as an exercise in variation of
scale. The best way to understand confusing examples of
this kind is to keep the mind fixed on the diagram rather than
on the form of the words.
54 Graphical Calculus.
Exercise. — Draw a curve of any shape and differentiate it.
Turn it through a right angle, as explained in § 23. Reflect
on OY, and differentiate again. Mark corresponding points
on the two curves, and show by the method explained in § 2 ,
Fig. 3, that the mean proportional between the heights of the
derived curves is always i inch. (This is, of course, merely an
application of the principle that tan 6 x cot ^ = i.)
§ 26. Differentiation of Sum and Difference
OF Functions.
Suppose we have given two elementary curves; for
example, \J\ and (?) in Fig. 26, which represent y = x^ and
y = fJx- Draw another curve whose ordinates are = the
sums of the corresponding ordinates of the given curves.
This may easily be done graphically. In Fig. 26, all pairs of
corresponding ordinates (such as P/, ^ Q) of Q and (7)
are together equal to the ordinate (such as rR) of G).
Differentiate Q and (7), and place the derived curves on the
immediate right of the corresponding primaries ; thus Q
is the first derived of 0, Q^ of 0, and (e) of 0. We
shall now show that there is the same relation between the
ordinates of 0, 0, and as there is between those of
0, and e.g. that/P' + ^^Q^ = ^^R\
Proof. — By construction —
/P + ?R = rR
that is, f L + M =r «N ;
also by construction —
jS + ^T = «U
Hence by subtraction —
LS + MT - NU
General Principles.
Di\dding through by PL = QM = RN, we have—
LS MT _ NU
PL ■*" QM ~ RN
55
Fig. 26.
that is, when S is close to P, and therefore T and U close
to Q and R respectively —
/P^ + /Q' = ^'R'
(see note on p. 48).
56
Graphical Calculus.
Roughly speaking, the meaning of this is, that if we have two slopes
(which we may imagine as wedges cut out of a pack of cards) of the same
length piled on top of one another in the way shown in Fig. 27, the
resulting slope (tangent of angle) is the same as that of the other two added
together. This can be easily seen.
Now, the equation to curve (T) in Fig. 26 is evidently —
Ord. of(3).
y
Ord. of (i).
Ord. of (2).
Our result tells us that its derived equation is —
Slope of (3).
Slope of (i).
2X
+
Slope of (2).
(I)
as already shown in §§ 17 and 24.
Exactly similar rleasoning applies to the differential
co-efficient of differences of func-
tions. In this case the ordinates
of curve (3) are to be made = the
differences of the ordinates of (7)
and (T). The figure can be easily
made from Fig. 26, by exchanging
the places occupied by curves Cy)
and (3), and also those of (T) and
(6^ It is then easily seen that the
former proof applies also to this.
Indeed, the proof given above for
the sum will hold throughout in-
dependently if we change the -i- sign
into — . If the wedges (i) and (3)
in Fig. 27 also change places, the result may be seen to be
the same as the -f result viewed under another aspect.
Generally speaking, our result may be written thus :
\iy = u-\-v — 'w-\-t— r, etc., where u, v, etc., are any
functions of x, we have —
Fig. 27.
General Principles. 57
dy^ _du dv dw dt dr
dx dx dx ctx dx dx
There is no difficulty in extending the proof in this
manner. It is left as an exercise for the student.
§ 27. Illustrative Example.
A simple practical example of this principle, which, though
not scientifically quite accurate, as will presently be explained,
is instructive and easy to understand, is as follows.
A man holds two appointments, in one of which his salary
is ;^i8o, with an annual increase of ;^2o per annum. In
the other his salary is ;^85, with an annual increase of ;£is
per annum. His income tax at this time is ^7 per annum,
and is increasing at the rate of £^\ per year per year. It
is required to find what is the total net rate of increase of
income.
Call the salaries jTi, ja, the income tax y.^, and total
income y. Let x represent the number of years reckoned
from this time. We have J = jVi + J2 — y%- It is clear that,
if we take A^ to be any integral number of years —
— = rate of increase on first appointment
A^
-/?= „ „ second „
Ax
Av-)
^^ = ,, ,, mcome tax
Ax
, A,)- Aj/j Aja
In this simple case it is easy to see that T~ = ^ + ^
_ ^^ which is another way of saying that total net increase
Ax
of income = rate of increase of salary on first appointment +
rate of increase of salary on second appointment — rate of
increase of income tax.
This notation is not strictly applicable to the case in point,
58
Graphical Calculus.
^
because the rate of pay does not increase every instant, but
step by step, each step being one year broad, and we have,
P therefore, to assume such
particular values for Aa; as
will make the expressions
^, etc., give a correct idea
of the rate of increase. If
the salary or ralte of pay
increased every instant, this
stepped figure (28) would
merge into a straight line
as shown, and the notation
Years I 2
Fig. 28.
dy
dx
could then have been applied to it.
Exercise. — The line AB, Fig. 28, is itself the derived curve
of another curve. What do the ordinates of this other curve
represent ? Ans. The aggregate earnings of the man.
Another illustration of a more scientific character will be
found in the following. A man in a corridor train commences
to walk along the corridor in the same direction as the train
is moving. Suppose the ordinate of curve (i\, Fig. 26, repre-
sents the distance travelled by the train in a time (after the
moment of starting) represented by the abscissa. Let curve
^2^ represent in the same way the distance walked by the
man along the corridor (or, as it is expressed in kinematics,
" relative to the train "). Then curve (7) represents the total
distance moved by the man through space (relatively to the
ground) in time represented by the corresponding abscissa.
Our principle states that ordinate of derived curve of (V)
+ ordinate of derived curve of (T) = ordinate of derived curve
of (3)' which, as we see from § 17, is equivalent to stating
that at any instant velocity of train + velocity of man along
General Principles. 59
corridor = total velocity of man relative to rails. Differentiating
again, we have acceleration of train + acceleration of man
along corridor = acceleration of man relative to rails.
Let the student follow out the case where the man walks
in the opposite direction along the corridor.
§ 28. To Differentiate nf(x).
An important analytical principle can be deduced from
a special case of this result.
Suppose each of the two curves (T) and ^2^ in Fig. 26
had been exactly alike ; then curve \Zj would be twice as
high as either Q or (^, and (^ would be twice as high as
or 0. Thus—
if jy = 2U
dy du
then — = 2 ,
dx ax
dy dv
or if V = ■») = 3 ,
•^ ^ ' dx dx
This law evidently holds for any integer whatsoever. It
also clearly holds for any fraction. For in this case curve
Q is half as high as curve (j^, so that the derived of (i\
is half as high as the derived of (T)- Proceeding in this way,
we can prove the principle for any positive or negative integer
or fractio;n.
HencCj when n is any quantity —
dy du
,iy = nu, ^^ = n-
Exercise. —Prove the result when n is a negative quantity.
6o Graphical Calculus.
§ 29. Integrals of Sums and Differences of
Functions.
It is obvious that the principle proved in § 26 applies
equally well to integrationSj for in Fig. 26 the curves Q,
\2^, and (3^ are respectively the integrals of ^4^, ^s), and @.
Now, ^6J is the sum of (\\ and (7), and the ordinate of Q
represents its area. The ordinate of ^2^ represents the area
of (Y) (reckoned as explained at p. 42), and that of Q the
area of (4). Hence since (T) + ^2^ = ^3^, it is clear that —
Area of Q + area of Q = area of Q
This proof assumes as self-evident the fact that a curve cannot have
more than one first derived curve.
This may easily be proved independently, for if we take a
corresponding " element of area " of each curve (as an
infinitely thin strip is called), such as that shaded in the
figure, it is clear that since —
area of strip of (T) + area of strip of \s) = area of strip of (ej
The same relation holds for all the common strips into
which each curve may be divided. It therefore holds for
the sum of all the strips, i.e. for the whole areas of the curves.
The same is obviously true mutatis mutandis for the difference
of two curves.
As a special case of this, we see that —
fnf{x)dx = nff{x)dx
where n is any quantity whatever, and f{x) has the meaning
already explained on p. 42.
General Principles. 6i
For since ii^P^x) =f{x) +f{x) + ...\.on terms, we have
as a.hoyefnf{x)dx=ff{x)dx+ff{x)dx ... to « terms
= nff{x)dx.
We can now easily find the integral of any multiple of any
power of X. Suppose we require f loifdx. The power of
the integral must clearly be o^. Let us differentiate o^, and
compare the result with the proposed expression. We obtain
ds?. This is clearly twice too great, so the desired integral
is \o^.
P
Similarly, required f ■j^x'dx (where /, q, r represent
either numerical quantities or expressions whose values
do not depend on that of oc), we find that the d.c. of
0^^'^ is Ir + x)3f. This must be multiplied by —, — —
'Jq{r + i)
to obtain the desired result ; hence the required integral is
'Jq{r + i)
Many other expressions can be reduced by simple alge-
braical or trigonometrical operations to forms which can be
differentiated or integrated by means of these rules.
For instance, to fmdif{x + aydx, we have —
{x — ay = A-^ — 2ax + a^, and therefore
f {x — aydx —fx^dx — 2af xdx -\- f a^dx
x"
= ax- + ax
3
Examples.
1. Differentiate (by expanding) {x + 3)^ (x + a)\ (i + \(>x^ + 64^*)^,
(x + d)(x - 6). Ans. 2{x + 3), 4[x + a)', i6j:, 2x + a-l>-
2. Integrate —
62 Graphical Calculus.
(ii.) {x + 3)=. Ans. - + 3*" + '^x^ + 2>]x.
(iii.) (^-3)'. ^«j. — — 3^' + ??jK'-27;r.
4 2
(iv.) {x-\-d)[x-a). Ans. d'x.
(v.) V'*Ml + 2ffi* + a»). ^«j. i^t_i*;.r».
CHAPTER V.i
GENERAL PRINCIPLES {continued).
§ 30. Products of Functions.
It must be carefully noticed that the principles explained in
the last chapter cannot be extended by analogy to multi-
plication and division of functions.
For instance, if —
y = uv
where uv stand for any expressions involving x, it by no means
follows that —
dy du dv
dx
or that if —
that therefore-
—
dx dx
u
y
~ V
du
dy
_dx
Tx
dv
dx
' In this and the succeeding chapters almost all the diagrams are
reduced copies of drawings made to scale. In many cases the scale with
which the curves are to be measured is given. The student should always
measure the curves, aiid make as large accurate drawings to scale for
himself as possible. Three times full size is a convenient scale for a half-
imperial sheet.
64 Graphical Calculus.
The student must be very much on his guard against
assuming results from analogies of this kind. He should in
all cases return to the curves, and think each principle out
on its own merits.
diuii)
The graphical proof of the formula for — — will be best
understood if we first give a brief algebraical one.
Suppose we have three variable and mutually related
quantities, which we shall denote (i), (2), (3) respectively,
of which the values of (i) and (2) both depend directly on
the value of an independent variable x, so that curves can be
obtained which show the relations between (i) and x and
between (2) and x. When x has any value we like to give it,
(i) and (2) each assume definite corresponding values.
Now (3) is to vary in such a way that, whatever the value
of X, its value is always equal to the product of the corre-
sponding values of (i) and (2). Thus it is clear that the value
of (3) must also depend entirely on the value of x, and the
problem is — if the value of x changes slightly at a time when
the values of (i), (2), and (3) are respectively u, v, and y — to
find what is the relative magnitude of the consequent change
in the value of (3).
Obviously, from the data —
y = uv . . . . (i.)
Now, if X changes from the value it now has, it is clear
that the values of (i), (2), and (3) will also change from the
values u, v, and y respectively.
Suppose a change in the value of x of the magnitude A*
causes the three quantities to become {u + A«), (z; + Lv\
(y + Aj;), respectively. Then, since (3) always = (i) x (2), we
must have —
y -\- Ay = (u + Au)(v + Az/)
On multiplying out, this becomes —
y + Ay = uv + uAv + vAu + AuAv
General Principles. 65
but since J)/ = uv, we have, on subtraction—
^y = ti\v + vt^u 4- AkAj; . . (ii.)
and therefore-
Av Az; , t^u , Aw
tt
+ ^a.+ a:;^^ • • ('"•)
t^x Ajc Ax ' Ax'
If this change in x had been infinitely small, all the
quantities, Ajy, Aw, Lv, would have been infinitely small too,
and the equation (iii.) would have been —
^ _ dv du dv
dx dx dx dx ' '
dy dv , du
^^' ~dx T' ^ T ^""^ ''"''■^ magnitude, although
dy, du, dv, dx are infinitely small (as already explained in
§§ 5, 17, 18, etc.). Hence the last term in (iv.), being the
product of a finite quantity with an infinitely small one, does
not afifect the equation at all, as it is infinitely small compared
to the other terms (see § 12). Therefore we have —
dy _ dv du
dx dx dx
That is to say, when the variable quantities (i), (2), (3)
have the values u, v, y, the rate of change of (3) per unit
increase of x = « x rate of change of (2) -\-v y, rate of change
of(i)._
This result may be very clearly demonstrated graphically.
The ordinates of curves (i), (2), (3) (Fig. 29) represent the
values of the three variables for all values of x. The length
of ordinate of (3) always = product of corresponding ordinates
of (i) and (2). Curves (i) and (2) being given, and the scale
with which they are to be measured, (3) can always be found.
Thus if /P = o"4, and ^Q = 2-o, then ^R = o"8, and
so on.
66
Graphical Calculus.
Let/P, ^Q, ;-R be the definite values u, v,y respectively,
and let PL = QM = RN be the value Aat.
Fig. 29.
Then LS = ^u, MT = ^.v, NU = Ay
Draw a rectangle ACDB, of which AB = /P, AC = ?Q.
General Principles. 67
It is clear that.the number of square inches (see note on p. 17)
in this rectangle = number of inches in rR. Produce AB,
AC, so that AE = jS, AF = fT. Then-
Rectangle AFGE = u\]
hence gnomon EDF = NU = ^y
But gnomon EDF = rect. ED + rect.. DF + rect. DG
= BD . BE + CD . CF + CF . BE
i.e. NU = A7 ^y.Ayi +yAy% + ^yAy-,
All these increments have been produced by an increment
PL = Aa; in x.
This is true however near Q is to P, or however far off
it is. It is therefore true when PL is infinitely small. But
when PL is infinitely small, the small rectangle DG is infinitely
small compared to the rectangles ED, DF; for, comparing
the area of rectangle DG with that of, say, FD, when PL and
therefore also CF and BE have become infinitely small, we
see that, although each of these rectangles has the same
breadth, CF, yet the length of DF, viz. CD, being of finite
magnitude, contains an infinite number of lines = EB, which
is the length of the rectangle DG, and therefore rectangle DG
is infinitely smaller than DF.
Thus Ajj/jAj's vanishes in the equation for Aj, being infinitely
small compared to the other terms, and our equation may be
written dy = udv + vdu (see pp. 37 and 21), remembering that
all these increments have been produced by an increment dx
in X. This may be signified to the eye by writing equation
in the form —
dy dv du
— = u — + V—-
dx dx dx
To illustrate this, differentiate curves (i) and (2), placing
each derived curve on the immediate right of its primary at (4)
and (s). Then multiply curves (i) and (5) together, just as
(i) and (2) were multiplied together to produce (3), and place
the result at (7). Do the same with (2) and (4), placing
68 Graphical Calculus.
the resulting curve at (6). Now, on adding together (6) and
(7), we shall, as the proof shows, obtain (8), which is the first
derived of (3), as may be found by trial.
The actual equations to the curves are shown in the
diagram.
§ 31. Illustration of D.C. of Product.
Before proceeding to a more general case of products,
we shall consider an example of the application of this
principle. Given that at the present time the total number
of poor-houses in the country is 7251, and the average number
of paupers in each 112, the rate of increase in the number of
poor-houses in the country is 8 per year, and the annual
decrease of the average number of paupers in each is 17.
Find how the total number of paupers in the country is
varying.
It is clear that —
Total number of paupers = total number of poor-houses X
average number of persons in
each.
\jt\. y = total number of paupers.
j'l = total number of poor-houses.
y^ = average number of paupers in each.
Thenjf =jCi Xjt'a
Let X be the number of years reckoned from the present
time.
Now —
— = rate of increase of poor-houses
dx
= +
i
dx
-"^ — rate of increase of average persons in each
= -r?
General Principles. 69
dy
- = required rate of increase of total number of
paupers.
Hence we have, from what we have just proved —
dy
— = -7251 X 17 + 112 X 8
ax
= -ii,4307-
Hence, from these data, the total number of paupers is
decreasing at the rate of 11,4307 annually.
Exercise. — Why is this number not exactly the same as
that which would have been obtained by finding the value of
(7251 X 112 — 7259 X 110-3)?
dy
Ans. Because the rate of increase — is not constant.
dx
The value here found for it is only the momentary rate of
increase (see remark on p. 14).
§ 32. D.C. OF A Continued Product.
We can easily proceed from this result to a more geneial
expression for the continued product of a number of functions.
Suppose, for instance, y = uvw, where u, v, and w are, as before,
shorthand symbols for " any expressions involving x."
Here our primary curves are jCi = «, J2 = »> Jfs = ^«', etc.
The above equation may be written as a product of two
quantities, thus —
y = (uv) X w
In this form we can differentiate it by the previous section,
thus —
dy d(mi) , dw
— = w^ — ' + uv —
dx dx dx
We can differentiate (uv) as before, so as to get the first term
of the right-hand side in a simpler form —
70 Graphical Calcultts.
dy r du , dv\ ,
■^ = W I V 1- u — I +
dx \ dx dx'
dw
uv —
dx
Removing the bracket, we obtain —
dy dw , du , dv
-^ = uv + vw 1- 'i^u —
dx dx dx dx
In the same way we can proceed to the differentiation of
the continued product of four functions, the result being
as follows —
If jy = rstu
dy du dr ds dt
— = rst— + stu h tur 1- urs —
dx dx dx dx dx
These results are often more conveniently and sym-
metrically written —
\ dy T. du 1 dv i ds
_^= - +- +_ etc.
y dx u dx V dx s dx
the result being obtained by dividing each side of the upper
equation by the corresponding side of the equation _;' = ursf.
Exercise. — Draw three curves at random, and a fourth
showing the value of the product of the three. Differentiate
them graphically, and exhibit the truth of the above result as
accurately as possible. Prove the result independently of the
proof in § 30.
The chief difficulty in understanding this result is due to the multitude
of different symbols which are often (as the student is prone to think)
needlessly introduced into the proof. He is apt, for instance, to stumble
over the symbols «, v, etc., and to ask himself, in the case of such an
equation as j/ = «, " What is the use of introducing u at all, if we are
already dealing with a quantity _j/, which denotes exactly the same thing?"
The answer to this is that, whereas y stands for the ordinate to the curve,
u is used for brevity, instead of (fx), and means " some expression in-
volving X," and may stand for any such expression ; and the equation
y = ti means " there are, for any value of x, one or more definite values
for y ; " i.e. " x andy are dependent on one another." If the student finds
other difficulties of this kind, the best plan is to express his difficulty in
General Principles. 7^
words and write it out. It will, in most cases, be found that the veiy
exercise of explaining accurately to himself what his difficulty is (besides
being of high educational value in itself), will enable him to explain
the difficulty away. To obtain a clear understanding of any point,
there is nothing like seeking for a geometrical explanation by assuming
curves about which to reason instead of symbols. It is much easier
to reason about the curves themselves than about the symbols denoting
them.
§ 33. Orders of Infinitesimals.
The case of the d.c. of the continued product of three
functions may be proved independently of § 30, and by the
same method as was adopted in that section. This proof
illustrates very well the principle of what are called " orders
of infinitesimals." While leaving the working out of the com-
plete proof as an easy exercise for the student, we shall give
as much of it as will enable us to show the meaning of this
expression.
Suppose the ordinates of curve (Oj), in Fig. 30, represent
the products of the corresponding ordinates of (Oi), (O2), (O3).
I imagine a rectangular block made, the lengths of whose
edges are equal to a particular set of corresponding ordinates
of three given curves.
Let 0-i,p = Qi^_q=, etc., be the current value of the
independent variable.
Let ab = /P, be = ^Q, mn = rK.
Then the number of cubic inches in the white block (of
which only one face, abed, can be seen in this view) = the
number of inches in j-S.
Now, if X increases by A.a; = // =, etc., then a conse-
quent simultaneous increase, Ay^, Ay„, A73, and Ay, will take
place in y^, y.,, y^, and y. These increments are respectively
KT, LU, MV, NW. The block, therefore, increases to
amrjhe. The number of cubic inches in this increment =
NW. This increment consists of all the shaded parts of the
block, together with a slab at the back, which in this view
72 Graphical Calctdus.
is entirely concealed. This cubical increment is made of
several parts.
(i.) Three large flat plates (shaded light in Fig. 30), degfc,
Fig. 30.
cbmnlk, and the hidden plate at the back. These are repre-
sented byjCsjViAj/j, ji/2j)/3Ayi,j)'ijC2A;;3, respectively.
(ii.) Three rectangular four-sided prisms (shaded darker
in the figure), ghi, Inz, fckl; the magnitudes of these are
respectively, jiA/^Aj-j, yj^y^ Aj/j, y-iC^y^^y^.
General Principles. 73
(iii.) The small rectangular piece ijl (black), whose
magnitude is ^.y-^ti.y^b^y^.
Now, suppose Ajc, and consequently also A^i, Aj'., Aj'a, and
Aji, to dwindle indefinitely till they are infinitely small = dx,
dy\ dy.2, dy^, and dy. Then —
(i) The flat plates, such as gedcf. become indefinitely thin
in one direction {de) ; and although the other edges, such as
eg, ef, are the same size as the original block, yet these flat
plates are clearly, as regards their cubical contents, infinitely
small compared to the block abed.
(2) In the same way, the prismatic pieces (such as fckl)
though in two directions {cf, kl) they are just the same size
as the plates {gedcf), nevertheless become cubically infinitely
small compared to the plates.
(3) Again, the small piece (black), though in every two
directions it is the same size as one of the darkly shaded
prismatic pieces, is nevertheless infinitely small compared to
any of them.
Hence we see that though we may have any number of
different quantities of the same kind, and all infinitely small,
yet they may have " orders '' of smallness among themselves,
i.e. one quantity (3) may be infinitely small compared to
another quantity (2), which is itself infinitely small compared
to another quantity (i), which in turn may be infinitely small
compared to a quantity y, and so on. This would be expressed
by saying that the plates are an " infinitesimal of the first order,''
the prismatic pieces "an infinitesimal of the second order,"
and the small cubical piece " of the third order," and so on.
We have already had several instances of infinitesimals of different
orders. Thus in § 13, Fig. 11, what we showed with respect to each of
the infinitely small "elementary" vertical rectangles of which the lower
curve was composed was in reality, that each rectangle differed from the
corresponding strip of the curve by a small triangle, which was infinitely
small compared to the infinitely thin rectangle ; in other words, that the
difference was an "infinitesimal of the second order," and the sum of an
infinite number of these infinitesimals of the second order was, comparable
ill size with the rectangle K'D, an infinitesimal of the first order.
74 Graphical Calculus.
§ 34. D.C. OF x"".
We are now, for the first time, in a position to complete
the proof of the formula for d.c. of ^"-
We have not hitherto proved that it is true either for such
expressions as sn or *'" "'.
E
If J = X'i
111
y = Xi y, ofl Y. ^ . . . to/ factors
We can differentiate it in this form from the rule for
1
continued products, for we have seen (§ 24) that whenjji = ofi
the rule holds good.
Hence we have, from § 32 —
dy I ^- ,1 (i-i)
-^ ■= xf Y. X'' v. ■ . .to/— I factors X -x^''
dx q
+ the same quantity {p — 1) times repeated
since all the factors are alike
^ f ( iV-i I (l.i) 1
= / X j V^c'/ X -x^i ' \
= -I X
j j- 1 i-g
« X :« if
which is of the required form.
Now, let us suppose that —
y = x-^
where nt itself, apart from its sign, is any integral or fractional
positive quantity.
Another way of writing the equation is —
_ I
■^ ~ x»
dy .
Now, we shall find — indirectly thus. Fmd an expression
dx
for the differential coefficient of a:'" X — , which we know
x'"
General Principles. 75
from other sources = o (for a;"' x — = i, and from 5 ig
we know that when y ~ \ — = o),
dx '
Having found this expression, if we equate it to zero we
shall have a simple equation involving — , which by
(IX
solution will give us what we require in terms of the d.c. of x^,
which we know :
dfx'-X—^ d(l-^
\ x^J Kx'^J I , ,,
dx dx ^"
Hence —
d
therefore-
Kx"' y I
dx "^^
\ x'" J X
.•V'" _ 1 w?^!™-" = o
dx X-"-
dx
^(m - 1)
= -m — r—
i-m-Vi
= — mx
As this proof is usually given, it involves a difficulty to the beginner
which he often finds difficult to express in words. It is usual to write —
then j/x'" = I
Differentiating both sides of this eqiiation, we have, etc. Though the
student cannot find anything to actively object to in the words in italics,
and though he may understand the process of differentiating a product,
yet, because he does not understand the meaning of the reasoning, the
proof fails to convince him. If he compares the reasoning given above
with what is usually given, i.e. if he substitutes for y in terms of x in
the product to be differentiated, he will find it easier to understand.
76 Graphical Calculus.
Taking the meaning of the italicized words, literally they may be
assmned to mean that if two curves, « = yoi^ and 2=1, are the
same, then their derived curves are also the same with respect to any
variable whatever. The fact that one of the factors of the product yx^,
viz. y, does not contain x, need not trouble us, for we know that although
y does not appear to contain x at all, yet it does so in reality, for the
value oiy may be expressed, if we please, in terms of x. Indeed, if it did
not depend on x the expression -j- would be utterly meaningless.
§ 35. D.C. OF A Quotient.
We can easily find the d.c. of a quotient of two functions
It
of X by an application of this principle ; for suppose y = -,
where u and v are any functions oi x —
Then yv = it
Differentiating both sides of this equation by the product-rule
on p. 65—
dv dy du
y + v-^ = —
dx dx dx
dy
This IS a simple equation to find - -, giving —
dx
du dv
dx
dy dx dx
, . . It .
or, substituting - for y —
du dv
V u —
dx dx
If
If this is not clear to the student, let him substitute, as
an example, say, {x^ for u, and {x^) for v, and f — j or (x-h)
for J, throughout the proof here given.
dy
The expression for — should be committed to memory.
General Principles. yj
The same thing could be proved directly from the curves
Fig. 29. The bracketed numbers refer to the ordinates of
the corresponding curves, and ^(i) means the " first derived
of(i)."
(i)
Given (2) = — , required the equation of (5). We have —
,,, (7) (8) - (6)
'"-{^)-
(0
43)-
-!!*'
(I)
(i)43) - (3)4i)
(^r
Or we
might prove the same
thing thus :
y =
U f I \
- = t([ - j
__
UV~'^
dy _
d{v-'} .
u— + 1
,-.^
dx
dx
dx
=
u x|(-i)»"
■'?} + '
dx^ V
du
dx
(The student will not understand this last step till Chapter
VIII. is reached.)
du dv
V « —
dx dx
sin X dy
As an example, we might have y = , to find -— . We
log X dx
can write at once —
d{sm. x) . d{\oB x)
J log^-^^ ^— sm^-'^-^ — -
dy dx dx
'dx ^ (log xf
which we cannot as yet further simplify.
78 Graphical Calculus.
Every practical example of the product-rule furnishes also
an example of this rule.
It is to be noticed that, since in the plate ordinate of
(6) + ordinate of (7) = ordinate of (8), therefore —
area of (6) + area of (7) = area of (8)
i.e. area of (6) = area of (8) — area of (7)
i.e. area of (6) = ordinate of (3) — area of (7)
all areas being taken between corresponding ordinates, or
reckoned as explained at p. 42. The bearing of this on the
integration of expressions will be explained later on.
Examples.
1. From the illustrative example given in § 31, find, by inversion of
this, the rate at which the poor-houses in the country are increasing, given
total number of paupers (= 112 X 7251) and the average in each poor-
house (112) and the rates of increase of these ( — 11,430 and— 17
respectively).
2. Find from the rule for the d.c. of products the result for ^^, '^
dx ' dx '
dy
etc. (Thus y-xy.xy.x, therefore —=, etc.) Prove the rule for
dx
positive integers in this way by induction.
3. The length, width, and height of a cubical block of crystal are given
respectively by the equations —
L, = /i (I -|-(r, t)
L2 = 4 (l + a^ t)
L3 = 4 (l + as t)
where /„ 4, 4 are the length, width, and height at a temperature at 0° C,
«i) "it "z are constants, and t the temperature centigrade.
Find (i) the rate of cubical expansion of the whole crystal per degree
rise of temperature. (2) The rate of expansion of a block which is i cubic
inch at 0° C. (3) The rate of expansion of a block which is I cubic inch
at a temperature t.
Is the rate of (i) cubical expansion, (2) linear expansion, constant at
all temperatures ?
CHAPTER VI.
DIFFERENTIAL COEFFICIENTS OF TRIGONOMETRICAL
FUNCTIONS.
§ 36. D.C. OF Sin x.
We have proved that for all values of n = «:t'"~^', and
ax
we now proceed to deduce expressions and derived curves
for other functions of x.
Let J' = sin x.
When sin x, sin 6, and similar trigonometrical expressions
are used in abstract mathematics, the quantities x, 6, etc.,
invariably refer to an angle oi x or 6 radians, and not degrees.
When degrees are meant, the symbol ° is never omitted.
Thus sin 2° means the sine of 2 degrees; but sin 2 would
mean the sine of 2 radians, or 2 x 5 7 '2 95°. The reason for
this will appear as we proceed.
The following practical process will give a tangible con-
ception of the meaning of the curve y = sin x.
Draw a circle (Fig. 31) with i unit radius,^ and divide and
number the circumference, starting at A counter-clockwise,
into, say, 32 equal parts. Draw a horizontal line OX
through the centre of the circle. Drop perpendiculars from
each of the points of division to BOA. Take OX = circum-
ference of circle = 2 x 3*142 = 6'284, and divide it into
32 equal parts. Set up perpendiculars and number them
' The unit may conveniently be 3 inches long.
8o
Graphical Calculus.
corresponding to the numbers on the circle, through each
of the points of division, equal in length to the corresponding
perpendicular to BOA (thus ii = ii, 22 = 22, etc. ; these
numbers are not shown on Fig. 31), and in the same direction
as these are drawn. Draw, with great care, a smooth curve
through all the points thus found. This is the curve y = sin x,
for the number of units in the abscissa (e.g. Op) = number
of radians in the corresponding angle (AOP ), and the ordinate
fP represents to scale the numerical value of the sine of that
angle.
Fig. 31.
It should be noticed that when we speak of the angle AORj, we refer
to the whole amount of angle (in this case greater than two right angles)
included by the arc APiRj, and not the smaller angle included by the other
part of the circumference.
Now differentiate this curve graphically. If the work is
accurately done, a curve will be obtained precisely similar to
the original curve, but moved to the left by a distance = 1*57,
which is half the length of one of the loops.
(Considerable accuracy may be obtained if a large number of points be
taken,'and the scale of the drawing increased.)
Differential Coefficients of Trigonometrical Functions. 8i
The reason for this peculiarity will now be shown.
Take two points PiQi on the circle (Fig. 31), and find the
corresponding points PQ on the curve. Draw through ordi-
nates P/\ Q/. We have then —
0/ = arc APi ; /P = MPi, etc.
O^ ^ arc AQi ; qCl = NQi
therefore ?«Q = LQi ; Vm = PjQi
therefore - = —
But when Q moves up to P in the limit, the figure PiQiL
becomes a small right-angled triangle, similar and similarly
situated to the triangle RiKO, where ORi is perpendicular to
OPj. Also — — becomes — , or the tangent of the angle of
P;« dx
slope of the curve at the point P.
dy KR,
Hence ^ = /F^^=KR.
since ORj = i unit
Hence we have/^P^ = rR
Now, pr is evidently = PiRi = i"57 = -. Hence the
2
height of the derived. curve at any point Pj is the same as the
height of the original curve at a point i'S7 to the right of
P ; in other words, the derived curve is exactly similar to the
original curve moved i"S7 to the left. Its equation must
therefore be —
/ = sin(.. + ^)
Now, if we turn the triangle ORjK round the point O as
centre, through a right angle into the position OR2K2, each
side becomes parallel to a side of the triangle PjOM, and
since OR2 = PiO, we have' —
RaK., = OM = cos X (see § 6)
82 Graphical Calculus.
Hence we have —
(f(sin sc)
dx
= cos X
For instance, the tangent of the angle of slope of the curve
y = shi X where x = vt, units, suppose, = cos i'3 radian
= cos 74'5° = o'2 67.
If we differentiate again, we shall have the original curve
moved 3*14 to the right; we have, therefore —
— = sin (x + TT) = — sin X
^(cos x)
or — ^ = — sin ^
dx
which result may also be proved independently in the same
way. This should on no account be omitted by the student.
Again —
rt?^(cos x) d^(sm x)
dx'' dx''
d^(sin x)
-cos X
dx'
= sm ^
and so on.
The meaning of the negative sign in the result —
= — sin ^ is to be carefully noted. It affords a most instruc-
tive example of the meaning of derived functions.
Consider the angle AOPi as being " generated " by the
line OPi turning round O in the " counter-clockwise " direc-
tion. As X increases in value, cos x or OM decreases, i.e. the
increment of cos x (corresponding to a positive increment of
x) is negative. Thus — ■
d{co% x) _ increment of cos x _ negative quantity
dx increment of x positive quantity
= negative quantity
Differential Coefficients of Trigonometrical Functions. 83
The negative sign indicates that when x increases cos x
diminishes in the first quadrant. Now, in the second quadrant
the arithmetical magnitude of cos x increases ; but as its
sign is negative, since M is then on the left of O (see § 2), we
could no more say that the absolute value of cos x is in-
creasing under these circumstances, than we could say that
the value of a man's estate is increasing because his debts are
increasing. So that, in the second quadrant, ( — sin x) is still
negative, i.e. cos x diminishes, while x increases. In the third
quadrant, sin x being negative (since P is below BOA) — sin x,
dy
or — , is positive, as it should be, because in this quadrant cos x
dx
increases algebraically along with x, just as the value of a
man's estate increases when his debts decrease. In the fourth
quadrant cos x increases as x increases, because ( — sin oi) is
positive.
§ 37. Motion of Mechanism of Direct -acting Engine.
An example of the use of these results is found in the
investigation of the motion of the mechanism of an ordinary
steam-engine. Neglect, for the sake of simplicity, the effect
of the obliquity of the connecting-rod, or assume that the
crank-pin works in a slot perpendicular to the stroke of the
piston.
Let OPi (Fig. 32) represent the crank of an engine of 2-
feet stroke working at 60 revolutions per minute. Required
the piston velocity when the crank is inclined at 30°, suppose.
Let a curve be plotted to scale, showing the distance of
the cross-head from its central position, corresponding to the
total distance travelled by the crank-pin, starting at far dead
centre.
The horizontal scale is to be the same as the vertical.
Thus at any point P, corresponding to Pi on the circle, the
crank-pin has moved through a distance APj = O/, and its
displacement from the central position is clearly OMj = pY,
84
Graphical Calculus.
When the crank-pin has reached Qi the piston displacement is
ON = ^Q on the other side of centre, and distance moved by
crank-pin is APiQi = Oq, and so on. Plotting all such values
on the curve, we clearly get a curve of cosines to a certain
scale. Now, if we diiferentiate the curve graphically, the
meaning of our derived curve will depend on the length we
take lines such as PL. (The crank-pin is assumed to have a
constant velocity.)
Fig. 33.
Fig. 33. Fig. 34.
(i) If we take them equal to i inch, we shall get a curve
the length in inches of whose ordinates show the values of the
ratio —
small displacement of piston _ piston velocity
displacement of crank-pin in same time crank-pin velocity
(2) If we take PL to scale = distance travelled by crank-
pin in I second, the ordinates in inches will show the absolute
velocity of the piston in feet per second.
Differential Coefficients of Trigonometrical Functions. 85
(3) If PL is taken = crank radius, LK will give us the
velocity of the piston on the same scale as OPi, represents the
crank-pin velocity. In any case, whatever the length of PL, we
shall always get a negative curve of sines to some scale or other.
The height of the derived curve also represents the velocity
of the piston to some scale, which it is necessary to determine
from common-sense principles. Thus, suppose the linear scale
is a quarter full size, and we take PH any arbitrary distance, say
10 inches. Then HT represents, on the given linear scale, the
distance that the piston would have moved during the time
taken by the crank-pin to describe ro inches x 4 = 40 inches
if the piston speed had remained the same as it is at the point
P. Thus HT represents the velocity of the piston at the point
P to the same scale as PH represents the velocity of the
crank-pin. From this we can easily find the scale of velocities.
The crank-pin moves at the rate of2XiX3'i4 feet per
second = 6'28 feet per second, and the scale is such that PH
represents this velocity. The method, therefore, of construct-
ing the scale is as follows —
Make P,H, (Fig. 33) = PH, and describe an arc of circle
with radius PL = 6'28 units, cutting H,L in L.
Mark oif the points of the scale as shown, and project
to PH.
We thus get a scale of piston velocities applicable to the
derived curve, which would be obtained by making all lines
such as PH of the given arbitrary length.'
Now, it is clear, from what has been said, that distances
along the line OX may be taken to represent to some scale
either (i) displacement of crank-pin, or (2) time occupied in
making that displacement ; for since the velocity of the crank-
pin is constant = in this case 6-28 feet in i second, the same
distance which represents 6-28 feet on the linear scale along
■ The ordinates of the curves of velocity and acceleration have been
reduced in the figure owing to want of space. The student should draw
a larger figure for himself.
86 Graphical Calculus.
OX will also represent i second. Thus, if we differentiate the
derived curve, bearing this in mind,' we shall be able easily to
find the scale of accelerations applicable to the curve thus
produced (§ i6).
Thus, suppose P^H^ represents 0*25 second. It is clear that
H^T^ = /"P" represents a rate of change of velocity of H^T^
(measured by the velocity scale, already constructed) in o'2 5
second, i.e. four times that change of velocity in i second.
Thus, suppose H^T\ when measured by the velocity scale, to
represent 4*52 feet-per-second. Then the acceleration scale is
such that /"P" represents a rate of change of velocity of 4*52
feet-per-second per i of a second, i.e. i8'o8 feet-per-second
per second, and we can proceed as before to construct a
complete scale with which to measure accelerations on the
second derived curve.
The student who desires to understand the subject thoroughly should
on no account omit to perform the complete process himself, and think
out himself ab initio all the principles involved in the construction of his
own scales. It need not be pointed out to him that the whole process is
utterly useless unless he can construct exact scales for himself by which to
measure his curves. He should be able to alter his scale at pleasure, in
case he has not room enough to adhere to one.
The derivation of the second derived curve is of the
highest importance in calculations respecting the inertia of
moving parts in high-speed engines. It will be found, in the
process of graphical differentiation, unless very great care be
taken in the exact determination of a large number of points
on the curves, that great and cumulative errors may be made
in the drawing of the tangents to the curves. For this reason,
other and more accurate methods are preferable where it is
possible to find them. In particular, very simple and accurate
methods are known for determining the curves here found by
the process just explained. Those processes also take account
of the varying obliquity of the connecting-rod. We might
' It is best to mark off O^X* in seconds or half or quarter seconds.
Diffei-ential Coefficients of Trigonometrical Functions. 87
have done the same by a modification of the construction for
determining the curve of displacements ; but if we had done so,
the process would not have corresponded with the algebraical
investigation to be given shortly. A proof is here given which
shows the real though obscure connection of the following
process with that of graphical differentiation. The student
should not fail to perform the operation by both processes
and compare them.
Divide the circle representing (to scale) the path of the
crank-pin into a number (32) of equal parts. Draw the line
of centres, and put in the centre lines of the various positions
of the connecting-rod with one end on the line of centres and
the other end on the circle at the points of division. Produce
Fig. 35-
the connecting-rod, if necessary, till it cuts the vertical
through O in H ; then OH represents the velocity of the
piston to the same scale as OP represents the velocity of the
crank-pin. For consider two infinitely near positions of
the crank-pin P and Pj. Draw in the two positions of the
connecting-rod QP and QiPi. Draw a horizontal PjN, and
with centre Q and radius QP describe a small arc of a circle
PN. Then NQQjPi is a parallelogram, for Q^Pj = QN, and
NPi is parallel to QQi. Therefore NPj = QQi- Thus, while
the crank-pin describes PPi, the piston describes NPi. Now,
in the limit when PPi is indefinitely small, the small figure
88 Graphical Calculus.
NPPi becomes a triangle of which PPi is perpendicular to
OP, PN to PH, and NPj to OH. Hence if we turn the small
triangle NPPi through a right angle round the point P in the
right-handed direction, each of its sides will be parallel to one
of the sides of the triangle POH. Hence we have by similar
triangles —
PjN _ OH
PiP " OP
. velocity of piston at point P OH
velocity of crank-pin OP
i.e. on the same scale as that on which OP represents the
velocity of the crank -pin, OH represents that of the piston.
Hence, plotting the values of OH on a base representing the
path of the crank-pin unrolled, we get a curve of velocities
of the piston which is in practice more accurate than that
obtained by direct differentiation.
In the same way it may be proved that if HM be drawn
horizontal to meet OP produced, and ML vertical to cut QP
in L, and LK perpendicular to the connecting-rod, then OK
represents the acceleration of the piston to the same scale as
OP represents the radial acceleration of the crank-pin, viz.
toV, or — , where u represents the angular velocity of the crank
in radians per second, and v the linear velocity of 1;he pin. A
curve plotted on a similar base to the preceding and vertically
underneath it, shows the value of the piston acceleration.
The student should not fail to draw this curve, and to demonstrate to
himself that it is the same curve as would have been derived by graphical
differentiation from the curve of velocities, as obtained by the method of
Fig. 34-
It is impossible for any one to properly appreciate the extremely
instructive points involved in these constructions without thoughtfully
drawing the curves to scale.
Now, the algebraical investigation of the same thing is the
exact counterpart of the process first described, neglecting
the effect of obliquity of the rod.
Differential Coefficients of Trigonometrical Functions. 89
Let y represent the displacement of the piston from its
central position ;
X the angular displacement of the crank, starting from A
(Fig. 32) in radians;
A-nd r the radius of the crank-pin circle.
Tatx\.y = T cos x
dy
-=- =r — r sm a;
ax
dy
= — sm ^
rax
Now, rdx = distance moved by crank-pin, while the crank
describes the angle dx radians (for, as in § 6, since angle
arc , . ....
= — ^r: — , therefore arc = angle X radius) ; dy = distance
moved by piston in same time.
dy . .
Hence — — ratio of velocities of piston and crank-pin.
rdx
IT
Take any particular value for the angle, say 30° = -7
= AOP.
Velocity ratio at P = — sin ^ = — 5
i.e. the piston is moving backwards half as fast as the crank-
pin is moving.
§ 38. D.C OY y= Sin-' x.
From the result already obtained —
(/(sin x)
dx
cos A-
combined with the principle deduced in § 23, we can at once
find an expression giving the height of the first derived curve
of the curve/ = sin"^ x.
90 Graphical Calculus.
As already shown, this means " y is equal to the angle
(in radians) whose sine is x."
Let the curve OQiPj in Fig. 36 be Y = sin X. Rotate
this curve through a right angle into position dotted, and
reflect it, and we obtain the curve x = s'my, or, as it may be
written, y = sin~^ x.
X is thus geometrically substituted for Y, and y for X. x and X are
both plotted horizontally.
This is the curve OPQP2. Consider the point P.
— = limit of r-7 when Q is infinitely near to P,
ax PN
= hmitof —
d^i I I
(Pi cos X cos y
for every value of X = corresponding value of j'. This result
is perfectly satisfactory, and is all we require if y is to be the
independent variable ; but if x is, as usual, the independent
variable, we can immediately find the value of this in terms
of X by substituting ; thus —
— — = + - = +
cosj' Vi -sin^jc Vi -0^
For the derived curve, therefore, we get—
I
y=±
V7^
which is usually written without the double sign, because
writing a double sign may be considered as part of the prqcess
of finding a square root.
The shape of this curve is shown in the figure. It consists
of two infinite branches as shown. At P the slope is /^P^ ; but
at P2, of which the abscissa is the same as that of P, the slope
Differential Coefficients of Trigonometrical Functions. 91
Fig. 36.
92 Graphical Calculus.
is f''^i. The meaning of the double sign is thus rendered
evident. Each of these two branches approach the lines SA,
but it is clear they never actually touch it in finite space ; for
dy
at the point S, -4- is infinitely great, and though by taking
the point P near to S we can make the distance of the point
P^ from the line SA as small as we please, yet the ordinate
of P^ becomes enormously great, and the actual co-ordinates
of the point S^ would be (i, oo .) When a line and a curve
have this relation to one another, i.e. the curve continually
approaches as near as we please to the line, but never
actually meets it in finite space, the line is said to be an
" asymptote " ^ to the curve. These asymptotes are of great
importance in the general tracing of curves. In general, both
co-ordinates of the point of contact are infinite.
D.C. of Cos"^ X. — It is clear that by moving the vertical
curve downwards through a distance = -, so that the point
2
S is on the line OX, we shall obtain the curve y = cos"'^,
since we obtain the curve Y = cos X by moving the horizontal
curve to the left through the same distance. Now, it is clear
that this does not in the least alter the derived curve, so that —
(f(cos ''^ oc) _ (f(sin ~'^ x) i
dx dx ~ v'l — ^2
as may be proved independently, thus —
y = cos"' X
X = cos y
dx . , ,
— = — sm jc = + V I —X-
dy
dy _ ^ I
dx 'J-i.-y?
the same expression as before.
' From three Greek words, signifying " not falling together."
Differential Coefficients of Trigonometrical Functions. 93
It will, of course, be seen that neither of the curves y =
sin"' X nor y = cos"' x can have an abscissa > i or < — i ;
for there is no possible angle which has a sine or cosine
> I or < — I. The same thing may be seen in the equation
to the derived curve ; for if x becomes greater than i , say 2,
we have —
I
an imaginary expression, for it is impossible that a negative
quantity should have a real square root, since the square of
any real quantity, positive or negative, has a positive sign.
§ 39-
In a similar way the d.c's of tan x, cot x, sec x, and cosec
X can be obtained. The principles involved have in previous
Q, hcctx P,
Fig. 37.
sections been fully explained, and as these can also be easily
obtained from the dc.'s of sin x and cos x, we shall merely
give brief geometrical proofs. The student should in all cases
draw the actual curves.
Make OA = i inch (Fig. 37). Draw tangents at A and D.
Consider the point B on the circle.
Let AB, or the angle AOB in radians, = x.
Then AP = tan x, PQ = A(tan x\ BC = A^
94 Graphical Calculus.
PQ PN
" PNBC
PQ OP
" PN ' OB
PQ OP
" pn'oa
= sec^ X in limit
since angle QPN = angle AOP
Also DPi = cot X, PiQi = A cot x.
dx EC
QiPi
BC
= limit of -Qi^^^^
MQi BC
QiPi OQi
MQi ■ OD
= — cosec^ X
Again, sec x — OP, A(sec oc) = NQ.
4sec ;t) ,. . .NQ
-=^- — - = limit of —■
dx BC
NQ PN
" pn'bc
NQ OP
" pn'oa
= tan X . sec x
Again, cosec x = OPi, A(cosec x) = — MPj.
4cosec x) .... MP,
-^ — ; = limit of
dx BC
Differential Coefficients of Trigonometrical Functions. g5
_ MP MQi
MQi' BC
MPi OQj
MQi"OD
= —cot X . cosec X
Exercises. — (These exercises are of the highest importance. )
Prove each of these results from the d. c.'s of sin x and cos x
on the principles explained in Chapter V. in the following
manner : —
^(tan x^ _ d f s\n x\ _ cos^ x + sin^ x
dx dx \ cos X ' cos^ x
I
= sec X
cos" X
Prove also the following results by the same method as
that explained for y = sin'^.r, drawing the curve in each
case.
^(tan~'jc)_ I
dx
i+x''
d{coi-'
■x)^
I
dx
i+a?
(/(sec-'
x)^
I
dx
X'Jx] — I
d(cosec-'^
x)_
I
dx
X^K? — I
Thus, \iy = tan"' x, then x — tan y.
— = sec^ V = I + tan^ y = i + x^
dy
dy _ I
dx I + X-
Prove these results graphically by tracing the curves and
inverting them.
CHAPTER VII.
DIFFERENTIAL COEFFICIENTS OF LOGARITHMIC FUNCTIONS.
§ .40. D.C. OF Log x.
We will now consider the curve y = log x. A remark
similar to the one we made in defining the meaning of such
expressions as sin x applies here, viz. that in abstract mathe-
matics log X with no suffix signifies, not the ordinary logarithm
as found in log tables, but the '' natural " logarithm to base
" e " where e is the value of the infinite series —
I +£+ ^+ |_+ . . • = 27167 . . ., etc.
which is the value which the expression ( i + - j assumes
when n is infinitely great. The student cannot hope to
understand this fully unless he be acquainted with the
algebraical theory of logarithms, which is found in any fairly
advanced book on algebra, such as Hall and Knights' "Higher
Algebra.'' He may, nevertheless, obtain approximate values
of the natural logarithm of a number by multiplying its
ordinary logarithm (to base 10) by the log of 10 to base e,
viz. 2 "303 about.
Calculate in this way the natural logarithm of o'25, o"5, 075,
1-25, i'5o, 2'o, 3'o, 4'o, 5'o, 7"5, 10. Plot points whose
abscissse are the numbers here given in inches or other units, and
ordinates the calculated logarithms. Carefully draw a smooth
curve through these points. This curve crosses the line OX at
Differential Coefficients of Logarithmic Functions. 97
a point whose abscissa is i ; for with any base whatever
log I = o.
On the left of point (1,0) care must be taken : thus from
the tables we can find logu, 0*5 = i '69897, which for our
X
/•
/
/
Fig. 38.
g8 Graphical Calculus.
purpose is practically equivalent to 0700, since we cannot
plot correct to -nnnr ii^ch.
Hence we have —
logic 0-5 = - 1 + 070 = -0-3
hence hyp. log 0-5 = -0-3 X 2'303 = — 0*69
and similarly for the other points.
Draw a curve through the points. This curve is shown in
a full line in Fig. 38.
Now differentiate this curve graphically. The general
shape of the curve obtained will be as shown in the lower
part of Fig. 38. Take a number of points such as P' on
the curve, measure with a decimal scale /T^ and 0]^\ multiply
their lengths together, and the result will be found to
be always i if the work is accurately done. Its equation
must therefore be xy'' = i, ory = -.
X
Another way of exhibiting this fact very clearly is to take
a number of points P on the primary, through which erect
/T perpendicular = i inch. Join OT. Then OT will be found
parallel to the tangent PR at T.
Exercise. — The whole curve may therefore be drawn by
the method explained' in § 14. Draw it in this way, and
compare it with the curve just plotted.
These exceedingly important facts may be proved alge-
braically as follows. Consider another ordinate ^Q near p?,
distance h from it.
Let Op = x; pV = log x.
Then Oq =^ x ^ h
qQ, = log {x + h)
Therefore, with the usual notation —
A>' _ log {x + h) —log X
IS.X h
Differential Coefficients of Logarithmic Fjinctions. 99
which, from the nature of logarithms
_ I ( x-\-h\
X h \ X y
X \ X /
AT
Write n instead of -. This becomes —
h
ilog(i+i)"
X ^ nJ
When h or Aa- diminishes indefinitely, it is clear that
n increases indefinitely. When, therefore, this takes place,
log f I + - ) becomes, from the definition above, log i? = i.
At the same time, when t^x or h dwindles indefinitely, — be-
before,
comes -f^, or height of derived curve : hence — = v' = i, as
dx dx ^ X
§ 41. Illustrations.
Some interesting and instructive results may be derived
from these equations. On the same base as before, plot the
logarithms as found in ordinary tables. A curve will be
obtained similar to the other in general character, but flatter.
As we have seen, it is —^ as high at all points. It is the lower
2*3
dotted curve in the figure. Its equation \% y = ft. log^ x.
It crosses the line OX at the same point as the other. Its
slope at this point may be approximately found from the
tables, for we have —
logio I = o'ooo,oooo
logio I'oooi = o"ooo,o434
lOO Graphical Calculus.
Hence at this point Ay = o"ooo,o434
^x = 0*0001
Ay
— , which, when. Aa; is so small as o'oooi, will be very
nearly = -£,= o-434-
This is also evident from the equation —
if jF = /t lege X
J = ^ (see § 28)
ax X
which (when x = 1) = /j..
It is interesting to notice that, in an ordinary book of logarithms, the
height of the derived curve of the curve of ordinary logarithms is given
by the side of the tables, so as to enable any one using the tables to
"interpolate." This height is called "difference" in the tables. The
principle made use of in the calculation of intermediate logarithms is
Sy = — • Sx. The value of — is given as a "difference."
dx ax
We can obtain another curve of the same character by
plotting the lengths taken from a slide rule on the same
base. This is the upper dotted curve in Fig. 38. The
graduations of a slide rule are ruled proportional to the
logarithms of the numbers engraved on the rule, so that
addition and subtraction on the rule, which are easily per-
formed mechanically by sliding one scale over the other, are
equivalent to multiplication and division respectively. On
the ordinary small "Gravet" rule, log 10 is represented by
i2'5 cm. = 4'92i inches.
Hence this latter curve is = 2*135 times as high
2*302
as the e curve, and its slope at the point (1,0) is 2*135 (§ 28).
The result we have obtained may also be written
— =\og x-\-c. In this form it is extremely useful to the
X
/
Di^erential Coefficients of Logarithmic Functions, ibi
engineer in enabling him to find the work done by a gas
(such as air) in expanding isothermally or at constant tem-
peratures. This will be fully considered in the next chapters.
§ 42. D.C. OF €'■.
By inverting the curve of logarithms, as explained in § 23,
we can prove a result of great importance.
The curve Pi (Fig. 39) is the curve Y = log X. Rotate
it about point O into the position dotted, and reflect on OY,
and we get a curve whose equation might be written con-
formably with those of sin x and cos x, etc., y = log~^ x, or
y is the number whose logarithm is x.
It is usual, however, to write the equation y = e', for e"" is
obviously, from the definition, the number whose logarithm
to base e is x.
If the student cannot understand this, he is referred to any book on
algebra which contains a chapter on logarithms.
If this curve is differentiated graphically, the result will
be a curve which is exactly in every respect like the primary
curve. In other words, the peculiarity of this curve is that
if the tangent at a point P be produced so as to meet OX
in S, then, wherever P is on the curve, S/ will be exactly
I inch, for the triangle S/P is evidently exactly equal and
similarly situated to the triangle we should have drawn for the
point P in differentiating the curve in the ordinary way. This
is expressed by saying that the " subtangent " is constant, Sp
being the subtangent.
This result may be proved as follows : —
Ify = e^
then X = log y
_ dx _ I
"dy y
102
Graphical Calculus.
Aa interesting property of this curve is that, if a series
of abscissae are taken in arithmetical progression, the corre-
sponding ordinates are in geometrical progression. The
Differential Coefficients of Logarithmic Functions. 103
student should prove this algebraically from the equation to
the curve.
The whole curve should be obtained by this method,
by taking ordinates 0-25 inch apart. The ratio of the
ordinates will be i"-^, the value of which must be cal-
culated, and the successive ordinates found geometrically by
a construction similar to that of Fig. 3. All logarithms can
be graphically obtained from this curve by measuring the
abscissae corresponding to an ordinate whose length = number
whose log is required.
The whole of the results in this chapter and the last must
be thoroughly learnt off by heart. The student who wishes
to proceed with the subject will save himself much time and
annoyance by making himself perfectly familiar with them at
the outset. It is not too much to say that one-half of the
difficulty usually met by elementary students of the integral
calculus is due to an imperfect knowledge of these few simple
results. The student can best learn them by deducing them
for himself once every day, and constantly picturing to him-
self the curves representing the functions and their differential
coefficients. He thus obtains a practical and real familiarity
with the functions, such as he could not get by studying the
symbols only. Unless he is gifted with an exceptional
memory, he will find even the few here collected difficult to
remember otherwise than by understanding what they mean.
The results should be as familiar forwards as they are back-
wards ; 1?.^. he should know that | — ^ = sec"^ x just as
di^tc'^ X) _
r dx
J x^ x^ — :
well as that
dx xslx' - 1
104
Graphical Calcuhcs.
Direct.
Inverse.
Function.
Diff. coefficient.
Function.
Diff. coefficient.
xf"
sin X
nx" - 1
cos A-
— sin X
sec^ ;ir
— cosec'^ X
sec ;ir tan x
— cosec ;f cot .^■
I
X
sin"' X
COS"' ,r
tan-' ;tr
cot-i X
sec-' ;ir
cosec-'.r
I
-1 ;
cos X
Vi-.r^
I
+
tan X
cot X
Vl -x^ -
I
1+^2
— I
I +;r2
1
cosec X
.rv'^^ —I
— I
log X
a-V-i^ — I
Examples.
1. What is tlie equation to llie inverse curve of the lower dotted curve
in Fig. 38? Is the subtangent constant in this curve? Is the first derived
curve lil<e the primary curve ? Prove your answer graphically and
analytically.
2. Assuming the result for the d.c. of e', prove algebraically by
inversion the result for^ = log j..
3. Differentiate y = a?-.
ofU^^^"^'"*
'^^:si^^^^^'^''
CHAPTER VIII.
DIFFERENTIATION OF A FUNCTION OF A FUNCTION OF A
VARIABLE WITH RESPECT TO THAT VARIABLE.
§ 43-
We have considered, in the preceding chapters, the process of
differentiation of simple functions of a variable x (such as
sin X, log X, etc.) with respect to that variable — that is, the
relative magnitude of the change produced in the value of the
function by a small change in the value of the variable.
Now, this small change in the value of the variable may
have been itself produced by a change in some other variable
{z, suppose), on the value of which x depends, and it is often
necessary to know the ratio between a change in the value of
the given function of x and a small change in the value of z
(which latter produces a certain change in x, and in conse-
quence a change in /(x), the function to be differentiated).
In other words, we have to differentiate some function of x
(say log x) with respect, not to x, but to a, i.e. to find the
value of — . Of course, this would not be possible
az
unless there were some relation subsisting between x and z,
such that x takes up a definite value corresponding to any
given value of z (see the note at the end of § 23, on p. 49).
As the meaning of this process is usually very confusing to
the beginner, and as it is important that he gets clear ideas on
it, we shall illustrate it by an everyday example.
io6 Graphical Calculus.
Suppose a tradesman starts in business for himself at the
beginning of the year 1870. At the beginning of that year he
earns profit at the rate of ;^2oo per annum, or about lu. per
day, or \s. if\d. an hour.
Suppose this rate of profit gi-adually and regularly increases
by ;£^2o per annum every year, so that, for instance, in the
middle of 1870 he is earning lu. 6^. per day, or £,2\o per
year; and at the beginning of 187 1 he is earning £,120 per
year, or about 125. a day. It is clear that his average rate
of profit throughout 1870 has been .^^210 per annum, which
sum also represents his total earnings for the year. At the
beginning of 1872 he is earning ^^240 a year, and so on (i.).
Let the current rate of profit at any time be denoted
by £,z per annum, and suppose his current rate of living
expenditure at the same time is given by £,zi per annum,
denoted hy y (ii.).
It is required to find the rate per annum at which his
rate of living expenditure is increasing.
This example can most easily be understood by following the curves on
Fig. 41. The " dimensions " of this rate of increase will be " pounds-per-
annum every year," in the same way as the dimensions of an acceleration
are " feet-per-second every second." It would be incorrect to measure
this rate of increase in " pounds-per-annum," because " pounds-per-
annum " are the dimensions of an income or annual expenditure, and not
a rate of annual increase of income or of annual expenditure (cf. Fig. 17).
Let y be his rate of living per year, and z his rate of earning
profit (both in pounds per annum), at a time represented by x
years counted from the beginning of 1870.
It is evident that the relation between y and z is —
y = ^ . . . . («)
This being the algebraical expression of supposition (ii.)
above.
Also we have —
^ = £^°° + £2CX . . . {b)
which expresses supposition (L).
D.C. of a Function of a Function. 107
dv
We have then to find the value of — . Now, from equation
{a) we can (§ 34) easily determine — , or his rate of increase
dz
of expenditure per ;^i increase of income;^ but this is not
dz
what we want. Also from \b) we can find --,-, or his rate of
increase of income per year (§ 16) j but neither is this what
we require.
Now, from these two equations, {a) and {b), we can obtain
another involving only y and x, for we can substitute £,^00
+ £iox instead of « in the equation —
y = zi
This process is called "eliminating z between (a) and {b)."
We thus obtain —
y = {£^°'=> + ^2o^)s . . .(c)
Here we are fixed, for we have hitherto proved no rule which
will enable us to differentiate this expression with respect to
X. We have, in fact, come to a point where we must
differentiate a function (viz. the power %) of a function (viz.
200 + 20JI:) of a variable {x) with respect to that variable.
If the student has followed the previous reasoning carefully, he will
probably suspect that we shall find what we require by multiplying together
the two d.c.'s already found ; that is —
dy _dy dz
dx dz dx
but he must be very careful to notice that he has no right whatever to
take this result as proved merely because , and — look like fractions.
dz dx
He should know already that dx, dy, and dz are not quantities to which
^ If we take /'lOO as the unit, -y represents the amount by which his
dz
rate of expenditure increases per ^100 increase of income. In this case,
however, z would represent the profit in hundreds of pounds per annum,
and we should take account of this algebraically- by modifying equations
(«) and ifi) according to the units we are working in.
io8
Graphical Calculus.
definite values can be assigned, and tlierefore to c ancel out one dz with
another without inquiring into the meaning of the process is an operation
dy
which is quite as illegitimate as it would be to cancel out the rf in , and
to put -7- = -. In certain cases the latter might be true, but in the great
dx X
majority of cases it would not be. It would signify that the tangent at a
point P of a curve passes through the origin O, which is obviously generally
untrue.
The student's aim should be to grasp the meaning underlying all these
symbols. He should never perform algebraical operations of this kind in
a haphazard fashion without making himself acquainted with the principle
dy dy dz .
involved. In this case it is perfectly true that — = — X — , but it requires
dx dz dx
proof before it can be accepted, and it is only to be taken as another
analogy between the laws relating to differential coefficients and those
relating to fractions.
Draw the two curves representing relations {a) and {b)
as shown in Fig. 40. (The curves in the figure are not
drawn to scale.) Notice that curve (a) does not involve the
idea of time, but simply shows the expenditure corresponding
D.C. of a Function of a Function. 109
to any income. Also that, since, during the period under
consideration, the rate of profit is always greater than ;£'2oo
per annum, we have nothing to do with the dotted part of
the curve.
Curve (^) shows, in the length of its ordinate, the income
corresponding to a time given by the abscissa. To obtain a
curve showing time — annual expenditure — we must combine
the abscissae of (3) with the corresponding ordinates of (a).
Thus, consider a time two years after January, 1870, i.e.
January, 1872. The income is given by /jFj. Transfer this
to O2/2 as shown. Then p^^ gives the living expenditure at
this date. Take a base, O3X, divided exactly like OiX, and
transfer the ordinates such as/aPg to/gPs, where O3/3 = Oi/i.
This curve, when drawn, is the result of graphically eliminating
z between {a) and (p).
Consider corresponding ordinates, Qi, Q^, Qj, adjacent to
Pi, P2, P3, where Q2, Q3 are obtained from Qi, exactly as P2, P3
were obtained from P,.
Then clearly —
LQi = P2M
MQ2 = NQ3
P3N = PjL
Hence we have —
NQ3 ^ MQ2 ^ MQ2 LQi
P3N PiL P2M'PiL
From the way in which the curves were constructed, this
is true wherever Q may be. Now, when Qj approaches Pi, so
that PiL dwindles indefinitely, it is clear that all the other
quantities in the above equation do the same ; and when this
is the case, the equation becomes —
dy _dy dz^
dx dz dx
for the three ratios which are contained in the equation
I lo Graphical Calculus.
become respectively v-, ^, and — , whatever actual values
dx dz dx
the quantities denoted by NQ3, P3N, P2M may have, pro-
vided always these are infinitely small (since the part of the
curve along which Q may move consistently with this con-
dition is an infinitely short straight line, as already explained
in § 13 ; see also note on p. 48).
Differentiating all three curves, then we see that any
ordinate of {V) X corresponding ordinate of (d) - correspond-
ing ordinate of (d). The criterion of correspondence is, of
course, not the same as that in the case of the curves multi-
plied together in the ordinary sense. Thus piVl, pi'^l, A'Ps',
are corresponding ordinates, although O/// is not = Olpl.
It is clear that the curve ((!) in this case is represented
by-
y = l^rS-i X 20
= y(200 + zojf)-^ X 20 =
V200 -J- 20J£;
§ 44-
(i.) On the same principle, we can differentiate such ex-
pressions as (sin x)^.
The curves in this case are —
y = z^_. .
z — sin X
By
He
eliminating z,
re—
we
obtain —
y = (sin xy
dy
dz='' •
dz
— = cos ^ .
dx
ia)
■ W
D.C. of a Function of a Function. ill
Hence —
dy dy dz
-f^ = — •-; — 2z cos X
dx dz dx
= 2 sin X cos X . . . (c')
The letters denoting the equation correspond to the same
letters in the illustration.
(ii.) A frequent application of the same principle occurs
X
m the differentiation of such expressions as sin -. Here we
X
might be tempted to think that the d.c. was cos -. But it
a
must be carefully noticed that this would be the d.c. with
X
respect to -, and not to x (see Examples II. at end of Chapter
III.).
Here0 = - . . . . (i)
a
y = s'm z («)
dy dy dz i x
— = ^ — = - cos-
dx dz dx a a
X .
The mistake m this case arises from the fact that - is, for
a
convenience, not usually enclosed in a bracket, although it
might be if desired.
(iii.) Take another case : j' = log (sin xy
Let z = sin X . . . . (J>)
then J = log . . . . (a)
dy _ dy dz
dx dz ' dx
I I
= - • cos X = -; COS X
z sm X
= cot X
(iv.) Take a more complicated case : y = {f cos xf.
112 Graphical Calculus.
Let z = e' cos x . . . (a)
y = z'' . . . . li)
dy _dy dz
dx dz dx
dy
-f = nz"^-'
dz
dz
-J- must be found by the rule for products of functions
given in § 30, thus- —
dz d(cos x) d(e')
— = e'-^ ' + cos X -^-i
dx dx dx
= — if" sin X -\- e" cos x
= f'(cos X — sin x)
Hence —
dy , X , /
-r- = Me" cos X)"'^ X t'ICOS X — Sin x)
dx
After a certain amount of practice, the student will find
that he is able to dispense with the z substitution, and to
write down the result without any intermediate step.
(v.) A difficulty arises to beginners when they have to
differentiate such an expression as, say, ^ with respect to x.
They are tempted to write down as the result 3^^, forgetting
that this is the d.c. with respect to ^, and not to x. They
are often unable to trace the meaning of differentiating ^,
which does not appear to contain x, with respect to x. If so,
they should read again the note at the end of § 23, and
remember that there could not be such a thing as a d.c. of
^^ with respect to x unless a relation such as is there de-
scribed subsisted between ^ and x.
-^ is therefore 3^^ . —
dx dx
ie^ =^ ^
' ' dx dq' dx
D.C. of a Function of a Function. 113
The RULE, therefore, is as follows : —
To differentiate any function of a quantity enclosed in a
bracket with respect to a variable x (e.g. cos (log x)) —
(i.) Differentiate the expression, treating the whole quantity
in the bracket as an independent variable. This would give
us — sin (log x).
(ii.) Multiply this by the differential coefficient with respect
to the variable of the quantity enclosed in the bracket. Here
the d.c. of (log x^ is -. Hence —
X
rfjcos (log x)\ _ —sin (log oc)
dx X
If there are two or more brackets enclosed one within the
other, it is easy to see by induction that we must first treat
the whole of the outside bracket as an independent variable,
and proceed inwards, treating each bracket in turn as the
independent variable, multiplying all the successive results
together. Careful attention to the following example will
enable the student to understand this.
Letjc = [log {log (sin «*)}]".
(i) Differentiate as though the quantity contained in the
[ ] brackets were an independent variable. This gives us —
« [log {log (sin^)}]"-i
(2) From our rule, it is clear that this must be multiplied
by the d.c. of the quantity contained in the [ ] brackets.
Hence we have, as it were, to start the same process over
again, absolutely neglecting everything outside the [ J brackets.
This, according to our rule, will involve treating the quantity
in the { } brackets as an independent variable. Thus far we
have —
«[log {log (sin e^)}]"-^ X
{log (sin e-'^)}
I
114 Graphical Calculus.
Now, in order to obtain the d.c. mentioned in (2) above,
we must multiply \ r; by the d.c. of the quantity in. the .
{logCsme')}
I } brackets, which in turn involves treating the expression
in the ( ) brackets as independent variable. This gives—
wflog {log (sin f^jH"-' X 7- — :7 X 7^— T,
^ ^ ^ ^^ ''J J {log(sm«^)} (smd^)
which we must then multiply by the d.c. of the quantity in
the ( ) brackets. This involves treating if" as an independent
variable. (The student is apt to stumble at the last step,
because e^ is not enclosed in visible brackets.) Finally, the
whole expression must be multiplied by the d.c. of e' with
respect to x. The whole expression is then
«[log {log (sin «^)}1"-i X X -—■ — Jx X cos ^^ X f
■^ { log (sin f"^)} (ime)
The student should not be satisfied till he can write out
any complicated result like this at sight, without any sub-
stitutions. He must learn to fix his attention on each bracket
in turn, treating it quite apart from anything else, and regard-
ing the next bracket proceeding inwards as the independent
variable. If he finds it impossible at first to avoid getting
the thread of his thoughts entangled among the brackets, he
should get a separate piece of paper and cross each bracket
out as it is done with. He will thus find an apparently
extremely complicated expression quite simple to differentiate.
§ 45. Applications.
The application of this rule is the source of much of the
difficulty which the student meets with in applying elementary
calculus to science. Differential coefficients of quantities are
sometimes treated of with respect to variables, with which the
D.C. of a Function of a Function. \\%
quantities have no apparent connection. New variables are
often arbitrarily introduced, and d.c's assumed with respect
to them ; so the student is quite bewildered by the multiplicity
of symbols. He is again reminded that the very existence of
a d.c. of any quantity with respect to a variable involves the
existence of a definite relation such that, other variables being
constant, the assumption of a particular value by one fixes
the value of the other.
For instance, suppose that each of the following variables,
(a), (p), {c), {d), (e), etc., are exclusively dependent on (A), the
temperature during the winter : —
(a) The number of unemployed workmen.
(d) The demand for overcoats.
(c) The amount of railway traffic.
(rf) The sale of skates,
(i?) The death rate, etc.
We are assuming that we have curves given, representing
the value of each of these variables, corresponding to values
of (A). Derived curves could be obtained representing their
rates of increase or decrease per degree-rise of the thermo-
meter. From these curves we could find a relation such as
AJ , for -^ = —^ ■ -Vt ) although the demand for overcoats
4^)' d(e) d(A) d{e)' ^
might have no apparent connection with the death rate.
Or, again, we might introduce the arbitrary variable time,
although in our original curves the idea of time did not
enter ; but, in order to make such a relation as — have any
' ' at
determinable value, we must have given a curve showing the
relation between any one of these variables and the time.
Suppose the primary and first derived of the time-temperature
curve had been given. Then we have, say —
d{i) ^d{d) dA
dt ~ dA ■ dt
■ and so on.
Ii6 Graphical Calculus.
Direct-acting Engine. — We have already had a disguised
example of the application of this principle in the case of
the engine in § 37, which we now proceed to explain more
fully. What we actually wish to find in the problem is the
velocity of the piston, and this, as we have seen in § 16, is
the first derived function of the time-displacement relation,
dp
or ~.
dt
Now, the geometrical relation between crank angle B and
piston position / furnishes us with the means of finding the
dp
value of — for any value of 0. This quantity (neglecting
obliquity) we have seen to be —r sin 0. Hence we have
only to multiply by the corresponding value of the relation
— (i.e. the height of the first derived of the time-angle curve)
dt
^ , dp ^ dp dp dd
m order to find-, for - = -.--.
Now, we know from the data of the problem that the time-
angle curve is a sloping straight line, since the motion of the
crank is a uniform rotation, i.e. the amount of angle described
is proportional to the time ; hence the first derived is a hori-
zontal line, or the " angular velocity is constant." The height
of this first derived is given in the problem, for we are told
the crarik turns at 60 revolutions a minute, or ztt radians per
second. Hence for the time-piston displacement first derived
curve we have —
dp dp dO . „
dt do dt
Turning back to § 37, we find the assumption that-
small displacement of piston
corresponding small displacement of crank-pin
_ velocity of pistoti
velocity of crank-pin
D.C. of a Function of a Function.
It is easily seen that this is the same thing as —
dp
dp ~dt
117
or that-
rdQ d6
r —
di
dp _dp dO
Tt~ Je'Jt
In that section we avoided the general assumption by
showing, from other considerations, that in that particular
case the result held good.
It is now quite easy to correct this investigation for obliquity
of the connecting rod. It is clear that, corresponding to the
position C of the crank-pin, the actual displacement of the
Fig. 41.
piston / is not OM, but ON, where CN is a circle with P
as centre.
Let angle MOC = 6.
MFC = ^.
PC = /.
OC = ;-.
Then/ = ON = OM + MN
oi p = r cos 6 + {I —I cos <l>) . . (i.)
Now CM = r sm 6 = I sin <^
r
Therefore sin <^ = -sin 6 (ii.)
1 1 8 Graphical Calmlus.
Differentiating (i.) with respect to t (see note at end of
§ 34), we have (see p. 112 (v.)) —
dp ■ . d6,, . ^ dcl>
-T, = - ^ Sin 6 . — + / Sin <i . -;^
d^ dt ^ dt
because the d.c. of / = o (see § 19). But from (ii.) this
We also have from (ii.) —
f cos 0— = I cos i> —
dt ^ dt
d& rcosO d6
or ^ = • —
df /cos^ dt
_ r cos 6 d6
Vp-r'sm'e- dt
since /cos <f> = /v'l-sin^ = / \/ 1 _^ sin' 6 = ^i--7''?,m^6
Substituting this value of -j in equation (iii.) above, we
obtain —
dp ■ t,d6 ( r cos e \
-- = ;• sin ^ — — - 1
dt « ^ V/2_r''sin'6l ^
which is the exact value of the velocity of the piston. If the
connecting rod = n X length of crank, this becomes —
dp . .dOf cos (9 \
-£- = — rsin^— I— '
dt dt\ 'Jn'^-sm^ 0''
This expression is rather complicated. It is simplified
as follows : Sin' 6 can never be > i, whereas «' is always
comparatively large, usually about 25. Hence 'Jn^ — sin' 6,
D.C. of a Function of a Function. 119
being very nearly ^ >Jn-, is put =«. The maximum error
in doing this is very small, for V'24 = 4-9, and ^25 = 5 ;
but when sin^ ^ = i, cos ^ = o, so that even this estimated
error of 2 per cent, in the value of the fraction appears to
have a much greater effect than it actually has. Making this
approximation we have, since 2 sin Q cos B — cos 16 (see any
book on trigonometry).
dp . V sin 2
-^ = —V sm '
dt 2n
where V = r— = velocity of crank-pin
= a constant
Differentiating again with respect to the tim e, we obtain
the acceleration —
-i-r = — V cos 0— + — cos 26 -r (see p. in (11.))
dt^ dt 211 dt ^ "
It is well to test the accuracy of equations of this kind by a process
known as "taking dimensions." It is clear that in any equation what-
ever all the terms must be of the same kind. It would, for instance, be
absurd to have an equation such as the following : —
ft. ft.
2.'Zr+3,Tri= 5 seconds
for a velocity can by no conceivable process be added numerically to an
acceleration, much less can the sum of the two be equated to a time.
Similarly, if our equation is correct, it is certain that all its terms, however
obtained, must be of the- same kind. This we can test by finding what are
the dimensions of each of the terms (see note on p. 30, also p. 51). If
these are not alike, it is very certain our equation must be wrong. Now,
the left-hand side of the equation is obviously an acceleration, being the
time-rate of variation of a velocity. This is also suggested by the form
in which it is written, since d^p would naturally suggest a length. If this
had been written ((^^) we should have expected it to represent a small
area. dC' represents, naturally, the square of a small element of time.
I20 Graphical Calculus .
Hence -77 has for its dimensions — '-., i.e. tlie dimensions of an accelera-
te/' sec'
tion. Now consider the right-hand side. V is a velocity having for its
J. . ft. . length I
dimensions . Cos 6 has for its dimensions , = - ; or, m other
sec. length I
words, has zero dimensions, or the dimensions of a simple number or
"numeric." Now, — has for its dimensions . , but an angle has no
at time
dimensions, for the same reason that a cosine has none. Hence the total
dimensions of the first term are — '— X - X = — -; = an acceleration.
sec. I sec. sec'
Let the student work out for himself the dimensions of the second term
on the right-hand side. In an ordinary algebraic equation, such as
^ + Z^ + ■ . • = o, each of the letters must be assumed to be of zero
dimensions, i.e. to represent numerics (see note on p. 5).
Both the expressions for the velocity and the acceleration
de
are given m terms of 6 and -— , and can therefore be found
at
numerically by substitution.
Examples.
1. By the method of Fig. 40 obtain the curves —
(i.) _j/ = sin (x').
(ii.) y = sin (log x).
(iii.) y = log (sin x).
(iv.) y = \o% (log^).
(v.) ^ = log (cos ey.
(vi.) y - log {log (a -f to')}.
Differentiate them by (i.) multiplying together corresponding ordinates
(as explained on p. no) of the respective derived curves; (ii.) by the
method of Fig. 9 ; (iii.) algebraically, and plot the curve by calculation.
Compare the results.
2. Differentiate —
(i.) log(V'jr-a+ ^ x - b"). Ans. . =r
Z'J (x—a){x—b)
(ii.) /<™ sin™ rx. Ans. /"« sin™-' rx (a sin rx + pir cos rx).
' For this example and the next, the process must be a compound one.
Thus : Find;)/ = cos «" by the method of Fig. 40, and y =log cosr" by a
repetition of the process. Find by induction which derived curves must be
taken for the factors of the result.
D.C. of a Function of a Function. 121
(iii.) x". Take logs thus —
Let y=x''
therefore log j/ = x log x
I dy X
Differentiating - — = log x-\--
y ax X
g=>-(logx+.)
= ::C (log ^ + I)
,. s ' ~ ^^^ -^
(iv.) , (Ans. — (cos X + sm a^).)
sec :f
(v.) A . Ans. l^ X a^ log / X (i + log *•). Take logs twice in
succession.
(vi.) [log {log (log ^)}]. Ans. — — ■
X log X • log (log x)
(vii.) ^z ax —x". Ans. ( . )
X'J 2ax —XV
Ix 'y
(viii.) tan-i ■. Ans.
I -x^ I + x'^
3. Find the exact velocity and acceleration of a piston of an engine,
given crank 8 inches, connecting rod 30 inches, revolutions 95 per minute,
at angles 30°, 45°, 60°, 90°, 120°, and 150°.
CHAPTER IX.
integration.
§ 46. Examples of Integration.
We have already explained, in Chapters II. and III., the real
nature and nomenclature of the process called integration,
and have obtained the integral of one function of x, viz.
x", which is —
•^ n-f- I
It has also been pointed out that, to effect any proposed
integration, it is essential that we have a previous knowledge
of the process of differentiation ; and it is only by working
backwards from this knowledge that we can obtain an expres-
sion for an integrated curve, though we can graphically find
the curve itself independently of its equation.
There are many simple integrals which we can write down
at once if we know the corresponding proposition of the dif-
ferential calculus; but it should be clearly understood that
there is no general method by which we can deduce the
integral of a function from first principles in the same way
ais we have deduced the d..c. of various functions. The process
of integration is essentially a tentative process depending on
a previous knowledge of the differential calculus, just as
the process of division in arithmetic is a tentative process
depending on a previous knowledge of multiplication. It is
impossible, therefore, to attain, proficiency, or even facility, in
Jntegration. 123
integration without a previous familiarity with the differential
calculus. The expressions the integrals of which we can write
down at once are the results of the differentiations explained
in Chapters VI. and VII. Thus we have —
y cos X dx = %va X ■{■ c
which means precisely the same thing as —
(/(sin X A- c)
dx
= cos X
I 2
in much the same way as — = 3 means precisely the same
thing as 3 X 4 = 12.
Again —
/"( — sin x)dx = cos x + c
which means the same thing as —
d(cos X + 1:)
— > ' = — sm X
dx
This may also be written —
ysin X dx = —cos x — c
d( — cos — c)
for -'^ = sm X
dx
AlsoyjTsin x dxdx, which means, as already pointed out — •
/(/sin X dx)dx =f{ — co5 x ■\- c)dx
= -sin X -\- ex + e (see § 36 and p. 45).
Also —
Jff^in X dxdxdx =f\_f{A^^'^ xdx)dx}']dx
-/[/ {(-COS X + c)dx}yx
= /" ( — sin X -\- ex -]- e)dx
- cos X + -.t" + ex -\-/
124 Graphical Calculus.
Also we have I dx, which is usually shortened
J \ l—X^
into —
dx
J
; = sin~^ ^ or = cos"^ x
1/ i—x^
corresponding to —
</(sin~' x) _ I _ d{cos-^ x)
dx ~ Vi -x"- dx
This result is sometimes confusing to the student. How
is it that the same expression can have two different integrals ?
To answer this we must refer to §§ 8, 16, 22, in which it was
shown that in graphically integrating a curve we have to
assume some arbitrary point to commence from, which point
may be at any height above or below the base-line OX. At
whatever point we start from in the same vertical line, we shall
obtain the same shape of curve. If we draw two such curves
starting at different points, any ordinate of one is greater than
the corresponding ordinate of the other by a definite and con-
stant amount.
Now, bearing this in mind, let us look at Fig. 35,
which shows the curve y = sin"' x and its derived curve
/ = ± . If we integrate the latter graphically, start-
V I— ^^
ing at O, we shall, of course, obtain the curve y = sin" x;
but if we start at a point P at a distance - = i-57 units lower
down, we shall obtain the precisely similar curve _y = cos"' x
as shown. In fact, the angle whose sine is x is greater by
exactly -, or 90° than the angle whose cosine is x, for all
2
values of x. It will be evident from this example that a
complete solution of the integral — -^=n is not sin"' x or
J VI— ^^
Integration. 125
cos~^ X, but it must be some function of x which will include
both these functions and an infinite number of other similar
ones, for we may start to draw our curve from any one of an
infinite number of points on the vertical OY. Now, any curve
dx
fJ i—x^
which would answer to the description j* = + -7 ~' ^ 1 i-e-
t/
which might be obtained from the curve / = + — ^=^=r by
fJ i—x"
graphical integration would be included in the equation
y = sin~' x -\- some constant.
For different values of the constant we should get different
curves, but all of exactly the same shape. If the constant were
f - - J we should obtain j/ = sin-^:« — -, which we have shown
above to be the same thing zs, y = cos"'' x, whereas if the
constant were o we should have y = sin"' x.
Now, in every case of an " indefinite integral," i.e. without
any limits specified (see § 2 2), this unknown constant must be re-
presented by a letter, though it is often omitted for convenience,
unless more than one successive integration is required, when
it must never be omitted (see Examples i and 2, p. 30 ; also
p. 4s), It will also be evident that we can in general find
the exact value of this constant, if we know one point through
which the integrated curve passes. But in the above case, if
we know that ^ = o when _j' = o, we know that the constant
must be either o or iitt where n is an integer. If we know,
dy
in addition, that -7- at O = +1, we know that the constant
dx
is either o or 2mr, i.e. it X an even number, either of which
dy
would give us exactly the same result. If — at O is — i, then
the constant must be {211 + i)t, i.e. tt x any odd number.
This is perfectly definite, for all odd numbers would give the
same result.
126 Graphical Calculus.
§ 47. Example of Quadrature of Area.
It has been already shown, both graphically (§ 10) and
algebraically (§ 22), how the constant disappears if the
integral is taken between definite limits.
As an illustration, let us find the area of the curve y ■=%va.x
(Fig. 42) between the limits x = \ and x = 2. It has been
already shown (§ 2 1) that this area would be represented by—
I sin xdx = [— cos a; + constant!
This is the solution usually employed, the constant being
represented by " C." For the sake of definiteness, let us
assume a particular value for this constant, say \\ units. Now,
if this constant had been o, the integrated curve j/ = — cos«
would have cut OY at a point where _v = — cos o = — i,
as shown in the dotted line in Fig. 42 ; but since we have
arbitrarily added i^ to this value, the curve lies as shown,
where OA = o"S. Draw in the limiting ordinates /'P and
^'Q at distances of i and 2 units from OY. The upper
curve is _j? = —cos x + i^, and we know (§ 13) that i/Q — p?
in inches = number of square inches in area P'/V'Q'- Now — •
^Q = —cos 2 radians + 1^
= -cos 114° 39' + li
= 0-401 + i"5 = I '9°
/P = —cos I radian + i'5
= -cos 57° 19'+ 1-5
= -0-540+ 1-5 = 0-96
Hence —
?Q-/P = (o'4oi + i'5)-(-o-54o + I'S)
= 0-401 + 0-540 + I '5 - 1-5
= 0-941 sq. units
The constant, it will be seen, disappears entirely in the-
Integration.
127
result, so that its absolute magnitude is a matter of no
importance.
We shall often find that the adoption of two different
methods of integration will give us a different result for the
Fig. 42.
same function. If, however, the work has been correct, it will
invariably be found, on examination, that the two curves re-
presented by the two results are of exactly the same shape
and are exactly alike in every particular, except that one is
higher than the other by a definite and constant quantity all
128 Graphical Calculus.
along its length, and that the one algebraical expression for
value of y can be expressed as the sum of the other and a
constant.
§ 48. Work done by Expanding Gas.
Another function for which we found the d.c. was log x.
The result was - (see § 40).
Hence-
J- = log^
or, as we have seea the general integral to be, log x ■\- c.
This result is of great importance. It is constantly occur-
ring in engineering problems. It furnishes, for instance, a
solution of the question as to the amount of work done by
compressed air or steam in expanding from one pressure or
volume to another.
Take the case of air. Boyle's law tells us that if air
expands or contracts at a constant temperature, the pressure
varies inversely as the volume, or, in other words, pv = con-
stant. This constant can easily be calculated from the mass
of air and the temperature. For i lb. of air at 32° Fahr. the
value of the constant is 26,214, when the pressure is measured
in pounds weight per square foot, and the volume in cubic
feet. For half this quantity of air the constant is, of course,
13,107 ; for at the same pressure the volume is half what it
was before, and therefore the product pv has half its previous
value. In the same way, since the volume varies directly as
the absolute temperature {i.e. temperature Fahr. -f- 461° nearly),
pressure being constant, this product must vary according to
the same law, as may easily be seen by imagining the pressure
kept constant, while the temperature, and therefore the volume,
varies. The constant may in all cases be calculated by finding
the value of the expression, 53-2 X '^ X t, where w = mass of
gas in pounds, t = absolute temperature.
Integration. 129
Now, if all these values of/ and v for a given mass of gas
at a given constant temperature be plotted on a curve (pres-
sures-vertical), the resulting curve will be a rectangular hyper-
, , , . . constant ^ . , „ . ,.
bola, whose equation is / = It is the " indicator
v
card " of the expansion, and it is shown in all works on the
steam-engine, in a similar way to that adopted in § 14, that
the area under the curve between any two ordinates repre-
sents the amount of work done during the expansion between
the corresponding volumes.
The chief difficulty in understanding the working of
these problems is that of units, which will continually harass
the student till he masters it once for all. He must here
imagine the curve drawn to full inch scale, i.e. i inch vertical
= I lb. per sq. foot, i inch horizontal = i cub. foot. Under
these circumstances, i sq. inch on the diagram represents
-nr X ft.^ = I ft. -lb. If the scale had been i inch vertical
ft.^
= 1000 lb., and i inch horizontal = 10 ft.', an area of i sq.
inch on the diagram would have represented 1000 — ^ X 10 ft.°
= 10,000 ft. -lbs.
This may be taken as an example of the general method of finding the
scale in which an area, such as an indicator diagram, measures a quantity,
whose value we require. The rule is, consider what quantity would be
represented by a square figure one inch long and one inch high.
Taking, then, the full-size diagram, we have its equation—
26214
V
for I lb. of air at 32° Fahr.
Its integrated curve, as we have seen, is a curve of loga-
rithms, each of whose ordinates is X 26214 (see § 40), whose
equation is therefore —
y = 26,214 log V -^ c
K
130 Graphical Calculus.
The area, then, of the lower curve between any two ordinates
(say, where the volumes are 5 cubic feet and 9 cubic feet) is
the difference between the two corresponding ordinates of the
upper curve —
= 26,2 14 log 9 -j- f — 26,2 14 log 5—1?
= 26,2 14 (log 9 - log s)
= 26,214 log T = 26,214 X o"588
' = IS.4I3'8
This is the area of the curve, and since each square inch
represents i ft.-lb., the total work done is i5,4i3'8 ft.-lbs. If
any constant other than 26,214 had been given with the same
ratio of expansion, this constant, instead of 26,214, would
have been multiplied by log f.
Thus, suppose in an air-compressor, diameter of cylinder
= 10 inches, stroke = 2 feet; required the work done per
stroke in compressing air isothermally up to 6 atmospheres.
Here volume of air compressed per stroke = 10^ x 0-7854 X 24
= i884'96cub. in.
The corresponding pressure is that of the atmosphere, viz.
147 lbs. per square inch;
The constant therefore = 1885 X 147 = 27709*5
Notice very carefully the effect of altering the units from
ft.' and -r-^ to in.^ and .-\. If we plot this expansion curve
ft.^ m.^
in these units, one square inch of the diagram will represent
I — ^ X I in.'* = I in.-lb. Therefore we must divide the area
in.^
by 12 to get foot-pounds. The result is —
27709-5 log f in.-lbs. = ^7709'5^X ^19^\ ,_,^^,.
' Part of this work, viz. 2_Z 11 ft.-lbs., has been done by the
12
atmosphere which presses on the suetion side of the piston.
Integration. 131
The student should in every case pay great attention to
the units in which he is working, otherwise he will find himself
hopelessly confused. For instance, if in the above case he
lbs.
had taken pressures in -^ and volumes in ft.^, then one square
inch of diagram would have represented ^-r^ x ft.^ = 1728
in .-lbs.
It may be noted that in an actual air-compressor the work
would have been greatly in excess of this, because a large
amount of heat is developed in the air by the process of
compression, which increases the pressure, and therefore also
the work done.
If the compression is effected without any heat being lost,
as will very nearly be the case if it is done very rapidly, it may
be shown that —
/z;i'*°8 = constant
The constant here, also, will have to be calculated either
from known simultaneous values of the pressure and volume,
with the help of a table of logarithms, or from the temperature
and volume and mass. In this latter case the constant = my,
53'2 X T X z/"'*"* = c, suppose. Here, as before, the work done
is —
/:
but/ = ~
hence integral = J^^^^ = ^""^^cv^dv = ^J%Vz/
where « = — i '408
' Here it will be noticed that it is impossible to integrate this as it
stands, because the expression to be integrated, viz. /, does not contain v
at all, and dv tells us that the expression has to be integrated with respect
to V. Hence our object is to change p into an expression containing no
o'Ca&x variable except zi. This we must do by "eliminating"^ between
the two equations, or, in other words, substituting for p an equal value in
c
terms of v, t.e. -^,_
132 Graphical Calculus.
Hence the work done is —
« + I
y" + i 1
or, substituting /iZ/i^*"^ for c and — 1-408 for «-
■"■^ r I
A»:
%,^j_- 0-408
V~
Hence, substituting % and v^ in turn for v, and subtracting,
we obtain —
L_(^^^,^l-408 X z,^-»-™ _ ^^^,^1-403 -^ ^^-0-408)
o'4o8
but since we know \hz!ip-ffli^^ = p-pi'^'^, we can write this—
which represents the work done.
§ 49. Integrals to be Lear^jt.
The following integrals must be learnt by heart. The
corresponding differentials are also given in § 42.
i
^v^ =
'dx
ft + I
= log„ *
X
«V« =
log. «
sin xdx = — cos x
cos a;i& = sin x
dx
= cot X
sin' X
dx
) COS'' X
h
J:
I
k
Integration. 133
tan X
\/d- —£■ ^
dx I , a:
= -tan"^
c^ ■\- or a a
dx I - X
= - sec" ^ -
X aJx'' — d^'^ '^
These are the most important elementary integrals. The
integration of any expression which can be integrated is
effected by transforming it by processes which will be shortly
explained, into forms of which the integral is known.
Many simple cases can be so transformed at once. For
instance, required —
J rx
This can be written —
f x~^dx
This is evidently an example of the x"" integral given above,
where n = —\. Since we know that x^dx = — ; — ^"+\we
have evidently —
/•
X~idx = A--5 + 1 :
= 2\/x
To test the accuracy of this, let us differentiate the latter
expression —
d(2 \fx) _ d(2xl}
= 2 X i X x(^-'^)
= X-i = —^
dx dx
I
^ X
134 Graphical Calculus.
or again, required —
C //-r /■ T
C dx fi -i
J (M' iP"
Here — - = n
r
I I 1— i
hence required integral = — x — ^ Y. x "
pr I_£
r
Again, requiredy cos mxdx —
Here, if we try sin mx as the resulting integral, we shall find,
on differentiating it —
^(sin mx)
; — m cos mx
ax
It is evident, therefore, that sin mx is m times too great;
therefore, instead of taking sin mx, we evidently ought to have
had — sin mx, which on differentiating gives cos mx. This is,
111
therefore, the correct result.
It is to be noticed that functions of {x + a), where a is
any constant, can usually be treated exactly like the same func-
^ . ^ sin (.r + «) / . X
tion of X. For instance, ; = cos {x + a), and
dx
d\o%(x-\-a) I ^ , , r , , ~. ■ ,
— 2 ' _ ^ for the d.c. of {x + «) with respect to
doc oc ^* Or
X = X (see § 43). This principle does not, of course, extend
to such expressions as log {0^ + <^), whose d.c. is evidently
I
OC + a''
Similarly, we are continually dependent on our previous
experience of the differential calculus to enable us to effect
any proposed integration, and it is thus evident that we must
have- the d.c.'s of the elementary functions at our finger-ends
before we can hope to attain facility in integration.
Integration. 1 3 5
/dx
. Even
d'-\- x'
in simple case of this kind we are at once hopelessly lost, unless
we happen to know that -y ( tan"' - ) = -5— — =
Now, if we tried tan-' -, to see if it would produce
«' '"" "" «' + x'-
on differentiation we should find it was a times too large ; hence
the correct result is -tan"' -.
a a
Such a procedure is, no doubt, extremely unsatisfying to the
student; at the same time it is the only one that is open
to him, and he must be content to make the best of it by
continued practice. He will find even the elementary integrals
and a few easy applications to be of great service to him in
practical work.
Examples.
Integrate a"*, (^'^ X c, sin x cos x, ( — ; — '. + - \ I
° \ I + sin ^ I — sin ;i; /j
/ I I \ / I \ f (^ - p\- + 2j»>x •,
K^V^T^'^b ' ^a^^^b} \x - sin x} \ '.x'' + ff ]'
,3, i ; I •*' W.fZ^I
CHAPTER X.
METHODS OF INTEGRATION.
§ SO. Integration by Expansion.
Many expressions can be integrated by expanding them into
separate terms by some algebraical or trigonometrical process.
This method should always be tried before any other.
Take, for instance, f (c^ + o^^dx. On expansion, this
becomes —
/(«" + 3aV + SfflV + s^)dx
This expression, as we have shown (§ 29) —
= f(fdx +fzc^o^dx +fza^x^dx +fx^dx
= a'x + aV + ffl'^' + \x'
Again —
dx
f-lf^ '—)dx
\ 2a\x-a X + a /
dx I [ dx
2a I x—a 2a \ x-\- a
= Liog{x-a)-^^\og{x + a)
I , x — a
= —log— —
2a x-\-a
This is a very important result.
Again —
y sin^ xdx = -I y 2 sin^ a:
Methods of Integration. 137
= iy (i —cos 2x)dx = \fdx—\fco^ 2xdx
_ i
~ 2 ■
Examples.— \xi\.^gxzX& {x -{■ af, (px + qf, {x + a){x-a),
|5f , (cos X + sin xY, ^qr^g, ^. (Split this deno-
minator into {x + \f6){x— V^).)
It is not necessary or desirable to give the answers. It is
always possible to find whether the result is correct by dif-
ferentiating the result obtained. The d.c. of the answer should,
of course, be the same as the function to be integrated.
§ 51.
If it is found impossible to reduce the proposed expres-
sion into a series of simple integrable forms, the next thing to
be done is to try whether it is possible to write it in the form
of two factors, of which one is the d.c. of the other, or of some
power or root of the other. If so, the expression can be
immediately integrated.
Thus to find y (a:' -1- aV)ii/x Here we see that we can
write this in the form —
fx'/x' + a^dx = A/2X X {0^ + a')VA-
Now, 2x is the differential coefficient of 0^ -f- (fi.
Now, obviously, if we differentiate (x- -t- a^)i + 1, we shall
obtain (§ 43)—
|(a;^ -f- (f)l X2X = zx'Jx'' + a'
This is three times too large. Hence the required function
is lix" + ay..
Again, required f {px^ -\- zqx -)- rf-{px + q)dx. This
becomes —
\f{px^ + ■^qx -f r)\(2px -f 2q)dx
138 Graphical Calcidus.
By a similar process, we obtain —
\ X f (/^' + 2qx + ;f
A similar case is —
r «:»; + ^ , , / zax + 25
J ax' + 2bx ■{- c J ax + 20x + f
Here the numerator is the d.c. of the denominator. When
this is the case, we at once write down the integral
log {ax^ + 2bx + c). The student will see the reason for this
if he differentiates this expression.
losf X ^ I \
Examples. — Integrate — ^— ( this expression = (log x)x- ) ,
I X X ( %\n x\
~', > — T"; — ;> , tan * I = I sec ^ tan x
xXogx px'-^-g^ 'Jpx^ + q^ ^ cos a;/'
V COS^ X J '
sin 2x
§ 52. Integration by Substitution.
The next method to be tried is more difficult to under-
stand. It consists in changing the variable from x to some
other, usually z, by substituting z for some function of x con-
tained in the expression to be integrated. Ey this means, as
will presently be shown, we can often, by judicious substitution,
reduce a complicated expression in a: to a simple one in z.
This is really treating the expression exactly as in the last
case, but we shall be able to deal by this method with more
difficult examples.
Take a simple case for the purposes of illustration. Find
f x\^
the area of the curve y = \i ■\-- \ between the ordinates
X = 0-5 and x= 2*0. As we have seen, this area is represented
by [\-\--\dx. Now, suppose we substitute z instead
Methods of Integration. 139
JX = 2'0
zMx.
X = 0-5
At first sight, the meaning of this is far from clear. The
student will have seen that before we attempt to integrate any
expression we must first of all get it into some form in which
functions of one variable only are present, otherwise we
cannot be sure of what we are doing. Since we cannot
integrate this expression with respect to x, we are going to
make zor(i-\--\ the independent variable, to see whether
it is any easier to integrate in that way. To do this we must
completely change every x in the expression into the corre-
sponding value in terms of z by means of the known relation
between z and x. In doing this we must not omit to change
the dx into some multiple of dz. In order to explain the
(x\^
I + - 1 . The
easiest way of doing this is as follows : —
Draw a curve (Fig, 43) showing the relation between x and z,
or I + -, for all values of x between the given limits. For
2
convenience of reference, call the ordinate of this curve z.
This curve, which is shown at {a) in the diagram, has for its
equation 2 = x +-• Now the ordinates of our given curve
ji; = r I + f j are the f power of the ordinates of curve {a).
Transfer the ordinates of curve (a) to another horizontal base
OjXi, as shown, and on this base draw the curve y = zi by
calculating the ordinates with a table of logs or a slide rule,
and erect ordinates to it from all the points /a, ^'a etc. This
curve consists of two branches (as shown) with a "cusp" at
the origin (see Fig. 20), i.e. the tangent at the origin touches
two branches of the curve at the origin. Next draw another
base OcX as shown, and divide it exactly as 0„X„ is divided, and
I40
Graphical Calculus.
erect ordinates from each of the points of division. Transfer
each of the ordinates of curve {b) to the corresponding mdivazXt of
this new curve as shown. Draw a smooth curve through each
of the points so found (much of the actual construction is left
out in the figure for the sake of clearness). This curve is easily
seen to be the given curve y = j i + - j . Draw in the
limiting ordinates a; = 0*5 and ;*: = 2*0, and the corresponding
ordinates of curve iV). It is clear that, although we cannot
at once find the area of curve {c), yet we can at once find
i -It -1 -i a
Fig. 43.
that of curve {b) between the corresponding ordinates, for it is
fiddz, taken between proper limits, these limits being the
X
values of I + when x is 0-5 and 2'o respectively, which values
are I'as and 2'o.
Our object, then, is to find a relation between the area of
curve {c) and that of {b). Consider any shaded element of
area of {c), and the corresponding elements of (b) and {a).
These elements of (b) and {c) are the same height, and their
J- , , • , , , , , , PaMo MiQi
areas are directly as their breadths, and clearly -
PM
P,M,
dz
dx
dz
Now -f- = h always, that is to say, dz = idx.
Hence the
Methods of Integration. 141
area of curve {b) between the ordinates 1*25 and 2"o = J area
of given curve if) between the ordinates 0*5 and 2'o for each
element of (^) = ^ corresponding element of {c). The reason-
dz
ing would have been exactly the same if — had not been
constant, as in the next example, for instance. The above
shows the geometrical meaning of the following reasoning,
which is that given in most text-books : —
To find I f 1 + - ) dx.
X
Let I + - :
2
Thsxi.^dx = dz
dx = 2dz
Hence f i + - j ^(/* =f2zldz
= 2 X \zl = \^
Take another example (the student should not fail to
r xdx
draw the curves in this example and the next) : I j ^ ^
Put /a' + x'' = z ;
Then a^ + x" = z^
And therefore 2xdx = 2zdz, or dx = giz
z J
. rx-- -dx
Hence ( / '^ = \ —^ — =fdz = z
]>Ja^ + x' J z
= Vfl^ + x""
Again, if we desire to find
r dx
142 Graphical Calculus.
=^^
(^ = z — x.
Th
en X' -\- c? =
Z^—2ZX + x'^
.-.z'-a^^
: 2ZX
dz
.: 2z— =
dx
, dz
2Z + 2X-r
dx
dz
or "T =
dx
z
Z — x
dz
dx
z — x
f dx
C dx
jz—x
logz =
fdz
~ }'
log {x +
Vx'- + a"
•Jx^
+
a-)
The student can only hope to learn what substitution will
be required in any given case by continued practice. He is
referred to a larger book for further examples.
Examples. — Integrate , ; — ; — r,, —,
x — a ax -\- b i i
'J{x-cf + b'^ 'Jax^ + 2bx + c 'J x^ + 4' 'J\x + af + 16
§ 53. Integration by Parts.
Another method of great importance is known as " inte-
gration by parts," which presents considerable difficulty to the
beginner, because of the large number — eight — of different
functions which have to be simultaneously borne in mind.
The process is, in reality, the opposite of that explained in
§ 30 for differentiating a product.
Before commencing the following explanation, the student
should carefully read § 30. That article showed how to find
the d.c. of an expression represented by curve (3), which
was the product of two other expressions represented by (i)
and (2). It was there seen that if^
Methods of Integration. 143
(4) be the first derived of (i)
(5) be the first derived of (2)
(6) be the product of (2) and (4)
(7) be the product of (i) and (5)
(8) be the sum of (6) and (7)
Then (8) is the first derived of (3)
But suppose we had been given curves (6), (2), and (4)
in their correct places, and had been required to complete
Fig. 44, we should evidently have proceeded as follows —
{a) integrated (4), thereby producing (i)
(b) multiplied together (i) and (2), producing (3)
{c) differentiated (2), producing (5)
id) multiplied together (i) and (5), producing (7)
{e) addedtogether (6) and (7), producing (8)
Now, it was shown on p. 75, that since ordinate of (6)
4- ordinate of (7) = ordinate of (8), therefore area of (6)
+ area of (7) = area of (8) (§ 29, p. 59), all areas being
taken between corresponding ordinates.
But area of (8) is represented by the difference of corre-
sponding ordinates of (3), since (8) is the first derived of (3)
(§ 13).
Hence we have —
(/) area of (6) = ordinate of (3) — area of (7)
Now, suppose that, instead of the curves (6), (2), and (4),
we had given their equations, and had been required to find the
area of (6) (or, in other words, its integral), we might proceed
exactly as at {a), {b), {c), {d) above, by writing down the equa-
tions to the curves instead of drawing the curves themselves, and
by the same processes as there described finding the equations
(3) and (7). Then by the use of relation (/) we can make
the integral of (6) depend on the integral of (7). The use of
the process consists in this, that sometimes (7) is an easier
expression to integrate than (6). If, in any given case, it is
144
Graphical Calculus.
not so, then the process is of no assistance, and some other
must be tried.
Fig. 44.
An example will make the meaning of this clear.
Required the integral of a;" log x, an expression which we
cannot integrate immediately.
Methods of Integration. 145
Here equation (6) is j' = x" log x
equation (4) is j = x"
equation (2) isjji = log x
The product of any ordinate of (4) with the corresponding
one of (2) is then equal to that of (6). Now, obviously, as at
above, (i), being the integral of (4), must be j; = o<^'^*^\
(l>) . . . (3), being (i) X (2), is^ = ^^^ x x'"*^\
{c) . . . (5) being the first derived of (2) is jK = -
X
^n +1) J
(</)... (7),being(i)x(s), isj)/=-^^j X '^
a;"
(« + i)
Area of (6). Ordinate of (3). Area of (7).
I X'^ log X ,
loff X t x'^dx
if) We have I x" log x dx = —^^^ ^'"+1' '
+ 1 j (« + i)
A-'""'^'loga: 0:'"+^'
(« + i) (ft + ly
Exactly the same method may be applied to findy :t: sin x,
which is curve (6) in Fig. 44. This is left as an example for
the student.
In working examples, the beginner will save himself a great
deal of confusion if he writes the symbols down in the same
relative positions as the corresponding curves in Fig. 44, till he
has become thoroughly familiar with the process.
Thus - cos X
X
-~X COS X
sm X
I
jusm X
— cos x
{x sin X— cos x)
L
146
Graphical Calculus.
The symbols corresponding to the curves in the diagram,
the order of writing down is as follows : (6), (4), (2), (i),
(S), (3), (7). (8) may be left out,
as it is not needed.
Another geometrical illustration which
the student should completely analyze
for himself is as follows. In Fig. 45,
PABQ represents the area I udv, and the
] a
I
f du-
rh
u dv
'a.
A
Fig. 45.
6
area CPQD represents the area I vdu
\'
when c and d are respectively the values
which V assumes when a has the values a and b. Here it is clear that
area PABQ = area CPABQDC - area CPQD, or fudv = uv — fvdu,
all taken between corresponding limits, u and v are both supposed to
be dependent on the independent variable x, which does not appear in
the diagram (see § 45 and note to § 23 on p. 49). Hence the above equa-
tion is really a shortened form of
— i.
dx
dx
Of course, it is only by trial that the student can discover
whether or not the process is of any use to him ; that is to say,
whether or not equation (7) is any simpler to integrate than
curve (6) ; if not, of course the process is useless. It is to be
noticed that integration by parts can almost always be tried in
two ways, viz. by putting each factor in turn in position (4).
Thus, in this instance, if we had written the process thus —
2-^
sin X
X' sm X
X
cos X
X sm X
\c^ cos X
1 ^2
5'*
{x sin X -\- \x^ cos x)
cos X is no easier to integrate than
it is evident that
X sin X.
Take another case, f log xdx. Here there is, apparently,
only one factor, but we can use i for the other factor, and
proceed thus —
Methods of Integration.
147
log X
X log ;
log X
Hence y log xdx = x log x — fidx
= X log X — X
It is sometimes possible to integrate an expression by in-
tegrating by parts twice in succession. Care must be taken
which expression is placed in position (4) in the second opera-
tion, otherwise the only result will be to reproduce the original
expression. Several of the following examples must be treated
in this way.
Examples.
Integrate x cos px, xex, x'^e", x sin x cos x, xn log x, xii(log x)^,
«"* sin dx. (Integrate by parts twice, and reduce the required integral by
the principles of simultaneous equations from the equations so obtained.)
Sin /^x (put J a: = 2, as in § 52 ; then integrate by parts), (log xY,
x' log x'.
CHAPTER XI.
miscellaneous applications of differentiation.
§ 54. Maxima and Minima.
In § 13 (i.), (ii.), (iii.), which should be re-read, it was shown
that the height of a derived curve corresponding to a maxi-
mum or minimum on the primary (which terms were there
explained) was o, or at that point the primary curve was of no
slope. Conversely, if at any point the first derived cuts OX,
then the corresponding point of the primary is a maximum or
minimum. It was also shown in (iv.) that we can distinguish
between a maximum and a minimum by the direction of slojie
of the first derived curve. In other words, if the second
derived curve is at that point above the axis of X, then the
point on the primary is a minimum. If at that point the
height of the second derived curve is negative, then the point
on the primary is a maximum.
These principles are of great use in practice from the ease
with which we can find algebraical expressions for the height of
a derived curve. If we have an expression involving x which,
as X increases, first increases and then diminishes, we can
find that value of x at which the function has attained its
maximum value, by finding that value of x which makes the
first derived function = o. In other words, differentiate the
function and equate the d.c. to zero, thus giving an equation
to find X (see Example in § 2). Differentiating the d.c. thus
found with respect to the same variable, we evidently
Miscellaneous Applications of Differentiation. 149
obtain an expression for the height of the second derived
curve at any point. Substitute in this expression the value
of X found by equating the first derived function to zero. If
the result is negative, the height of the primary is at the corre-
sponding point a maximum ; and if positive, a minimum.
The successive derived curves of sin x form a very in-
telligible illustration of this. Suppose we wish to find at what
points the value of sin ^ is a maximum or minimum. The
height of the first derived curve is clearly (Fig. 46) given
by cos X for all values of x. Now, equating this to 2ero, we
Fig. 46.
find a value such as O^S^ of x for which the first derived curve
cuts the axis of x.
These values are found from the equation cos x = o.
Hence x—-— — , etc., each of which corresponds to
222
either a maximum or a minimum.
To find which is which, differentiate cos x, giving — sin x.
Now substitute the above values of x in the expression
— sin X, and see whether the result is a positive or negative
value.
ISO
Graphical Calculus.
— sin - = — I, showing that at a point distant - from OY
2 2
the primary curve attains a maximum.
„ „ minimum.
■ ^1"
- sm — = + I
2
— sm ^— = — I
maximum,
etc., etc.
These can easily be verified in the diagram.
The way in which this principle can be applied in practical
case will be seen by considering an example. Suppose a man
running along a footpath AO (Fig. 47) wishes to reach a house
at H.'in the middle of a ploughed field, in the shortest possible
time. Suppose that he can run on the footpath at the rate of
15 feet per second, but only at 10 feet per second in the
ploughed field. What will be his quickest route across the
field ? If he ran across AH, he would have the shortest possible
distance to go, but his speed
would be only 10 feet per second
instead of 15 along the footpath,
whereas if he went to O along
the footpath he would have a
greater distance to go, but his
average speed would be greater
than in the other case. It is
clear that the shortest time
would be occupied by taking
some intermediate route, A/H,
where he would partly utilize his
superior speed along the footpath, and partly the advantage of
cutting off the corner /OH. It is our object to determine
the position of the point/.
Suppose the distances are OA = 500 feet, OH = 150 feet.
Let 0/ (the distance to be found) = x.
Then distance to be run in the field = ^ x'- + (150)^
Fig. 47.
Miscellaneous Applications of Differentiation. 151
Time occupied = ^ + (^5°) seconds
10
Distance to be run along road = 500 — ^
Time occupied = ^ ° ~ '^ seconds
Total time occupied = y = "^^' + ^^5°)' + 5°°-^
10 15
We have now to find the minimum value of y.
Plot various values oiy along OA as base, and draw a curve
through the points so found. Thus ^Q represents the time
occupied if the man leaves the path at g. Now, this curve is
obviously parallel to OA at the point where the derived curve
(a line not shown in the figure) cuts 0^X\ This value oix will
be found by equating the d.c. of y to zero (see example in § 2).
Simplifying the equation, we have —
3^^;^ + (150)'' -(- 1000 —2X
y
30
Now, since we only wish to find for what value of x this is a
minimum, it is clear that we may simply consider the numerator,
for if the numerator is a minimum, the whole fraction will be a
minimum. (If the denominator contained x, or anything de-
pendent on X, we could not legitimately disregard it.) Similarly,
we may disregard the 1000, since it is always the same, what-
ever be the value of x, and our problem becomes to find for
what value of*, 3^^;^ + (150)^ — 2* is a minimum.
Exercise. — Let the student determine for himself the
graphical interpretation of the process of disregarding these
constants.
Differentiating this, we obtain (§ 43) —
X 2* -2
2 V*2+ (150)2
This represents a multiple of the tangent of slope of the upper
152 Graphical Calculus.
curve, and must therefore = o at a point corresponding to P.
Therefore, to find x, we have the equation —
3^
'Jx''^{x^of
or, simplifying-
• s^s^ = 4 X (150)'
whence ^'^ = -f 150''
therefore x = 134 yards about
the negative value being obviously inadmissible. |
Another familiar example shortly worked oiit may serve
to further illustrate the process.
According to the post-office regulations for the size of
parcels which may be sent by parcels post, tlje length of
parcel + girth must not be greater than 6 feet, Required
the greatest volume which can be sent.
Let X feet be the girth ; then (> — x = length.
Now, with any given perimeter, the figure containing the
greatest area is well known to be a circle. 1
The area of a circle of girth x - —
47r
.*. volume of a parcel of girth x and length (f^—x) = — {6 — x)
47r
Therefore the question becomes for what value of x has
6x^ — x^ a maximum value.
Differentiating and equating to zero, we have —
12;* — ^x^ = o
Neglecting the solution x — o, which obviously gives a
minimum, we have —
X = 4
The volume is therefore - = 2-55 cubic feet
Miscellaneous Applications of Differentiation. 153
Examining these three examples, the student will see that
the rule is — Express the quantity of which we have to find
the maximum or minimum value in terms of one variable;
differentiate the expression with respect to that variable;
equate to zero, and solve the resulting equation. The solu-
tion gives the value of the variable for which the expression
has either a maximum or a minimum value. If it is not
apparent on inspection whether the value so found gives a
maximum or minimum, differentiate the first differential
coefficient, and substitute in the expression thus found the
value in question. If this gives a negative result, the value
gives a maximum; and if a positive result, a minimum (see
§ 10, 4)-
Example i. — A man weighs 160 lbs.; he attaches a rope
to tlie top of a post 20 feet high, with the object of pulling it
over : what is the greatest bending moment he can produce
at the base of the post, assuming that the pull he can exert
on the rope varies as the sine of the angle which the rope
makes with the ground ? Ans. 1600 lbs. feet.
Example 2. — Find the minimum weight of a cylindrical
boiler made of ^" plate necessary to hold 200 cubic feet of
water. Neglect overlap of plates. One cubic inch of iron
weighs 0-28 lb. Ans. 3850 lbs. (about).
Example 3.— One leg of a pair of compasses is held
vertical with its point stuck in a board, and the compasses
are rotated about this leg as axis : find what angle the other
leg must make with the vertical, in order that the bending
moment tending to open the compasses may be a maximum,
given that the legs are uniform, each 5" long and each
\/'~
o-S6°-± V 7 + "■5'-*
weighing \ oz. Ans. cos 6 = — ^ where
g = 32-2, and o) = angular velocity.
154 Graphical Calculus.
§ 55. Indeterminate Forms.
It sometimes happens that we have to find the value of
some function of x which cannot be obtained by simple
substitution, because on giving x the required value the
function assumes the form -, 0°, o x oc , or some such form.
o
In these cases the application of the principles of the
differential calculus enables us to effect a very simple solution.
For instance, we might have to find the value of —
X^ — \2X ■\- Q ,
when X — ■^
o(? — d^''' — ^x -\- 24
It is easy to see that this fraction assumes the form -
o
when ^ = 3, and we can therefore not determine its value
by simple substitution. Assume that the two cm'ves APB,
CPD represent the values of the numerator and denominator
respectively of any fraction I -j)-\ ) , which assumes the form -
\f4^x) J o
when X = OP. Now suppose, for the sake of definiteness,
dy
that we have found — for the curve APB to have the value
dx
i"S at the point P, and for the curve CPD the value 2-5 at
the same point. Draw the tangents ST and RQ at the point
P. Take two verticals TQM and SRN near to P, but on
- . , MT .
opposite sides of it. It is clear that r-r— is constant wherever
we take T on the line ST. Now, when very near the point
P, the points T and Q (as has been frequently explained)
are respectively on the curves representing the value of the
numerator and denominator of the given fraction. Now, when
T Ms travelled through P to an extremely small distance
the other side of P, it is clear that the value of the rafio
has not altered either in sign or magnitude, because the
Miscellaneous Applications of Differentiation. 155
numerator and denominator have both changed sign, and there-
fore their ratio has the same sign as before. Thus we have
MT NS , , , .
rr^ = rr^ = constant whatever be the sign and magnitude
of the values PM, PN. It is therefore true when they are
"infinitely small." Now, it is impossible to imagine what
happens at the point P in just the same way as it is impos-
sible to imagine an infinitely great or an infinitely small
Fig. 48.
Fig. 49.
quantity. Assuming the curve representing the value of the
ratio at all points along OX is continuous, i.e. does not make
a sudden vertical jump at the point P — and there is no reason
for supposing it would — we say that at the point P it has the
same value as it has at a point infinitely near to P. We
therefore express the fact that, however small PM is, the
dJM
MQ KF _ dx
value of the fraction
MT KE dfjx)
dx
, by saying that the
iS6 Graphical Calculus.
value of the fraction at the point P = ratio of the values of
the d.c.'s at that point.
If the curves AB, CD are of the form shown at APB,
df(x) df.(x)
CPD in Fig. 49, it will happen that -^^^ and ~^ are
GrjC (IOC
both equal to zero, in which case, by the same reasoning as
before, the ratio will be that of the values of the second derived
functions at the point. Again, one curve may be of the form
APB, and the other of the form EPF, in which case the ratio
will be = o or oc . This will be shown by one derived function
vanishing, while the other is finite.
Again, if the curves, in addition to having no slope at P,
have a point of inflection at that point, the second derived
functions will vanish at that point, and the ratio required will
be that of the heights of the third derived curves at the point.
In general the required ratio will be that of the first pair of
derived functions which do not both vanish at the point.
Examples. — Find the value of —
tan X — sec x + i , , , ,
(i.) when .« = o. \Ans. i.)
tan a: — sec -r + i
... , log «
'■• ^ when a; = I. (Ans. i.)
(iii.) ^ — when x = o. (Ans. \.)
(iv.) , when .a; = A (.Ans. 00.)
^ ' (x-pf ^ ^ '
§ 56. Equation to Tangent to a Curve.
It is clear that, given any equation to a curve, and any point
on it, we can at once write down the equation to the tangent
and normal at that point; for we know that the equation
to any line passing through a point (ab) whose tangent
of slope is m, is (y — b) = m(x — a) ; for this equation, being
Miscellaneous Applications of Differentiation. 157
of the first degree, clearly represents a straight line. It is
also satisfied by the point (a, b^, and since it is of the form
y = mx-{-c when simplified, it represents a line inclined at
dy
tan"^ m. Hence, substituting — for m, we have —
(y — h) - —{x — a) for the tangent
and {y — b) = — — - for the normal
dy
dx
Example i. — Find the equation to a tangent and normal of
the curves = — , at a point on it whose abscissa is 5.
Here it is clear ^ = 2-5 ; -y- = i- Hence equation to
ax
tangent is 0-2-5) = (a- -5), and to normal j- 2-5 = 5-x.
2. — Prove the subnormal in the curve y^= 2mx'is equal
to m.
Take any point (x^yi) on the curve. Since the point is on
the curve we have —
hence ji = 'J 2mxi
hence the point {x^, >J 2mx^ is on the curve.
Find where the normal at the point (x^, iJ zmx-^^ cuts OX by
putting J = ova. its equation, and show that this point is at a
constant distance from the point whose abscissa is x-^.
§ 57. Radius of Curvature.
Consider any curve APQ (Fig. 49). It is clear that y,
— , etc., are not the only quantities in connection with the
dx
curve which assume definite values for any assumed value
158 Graphical Calculus.
of X. Other such quantities are S, the length of the curve
reckoned from A; ^, the angle of slope in radians, etc.
We are, therefore, quite within our rights in speaking of
such differential coefficients as — , -j- (see § 45). It would
be easy to plot, for instance, a curve showing in its ordinate
the length of the curve reckoned from A corresponding, to
each value of x. This should be done by measurement.
The first derived of this curve would represent the value
ds
of — . But we can see that another geometrical function
Fig. 50. Fig. 51.
would represent this value independently of such a curve.
. . PQ
It is clearly the limit of -— , or the secant of the slope at P,
which function will vary when the slope of the curve varies.
ds
Consider what is the geometrical meaning of — . It is
the hmit of the ratio of the length PQ (Fig. 50) to the dif-
ference (measured in radians) between the angle of slope of the
curve at P and at Q. Consider first what this ratio means
for the circle APQ (Fig. 51).
We have by definition —
OP^^ = ds, where OR = i inch
Miscellaneous Applications of Differentiation. 159
This holds good just as well for the circle in Fig. 50, for
it is clear that the angle d<^ between PT and QS = the angle
POQ. OP is here clearly the radius of curvature of a circle,
which most nearly coincides at P with the curve. Hence, to
find the value of the radius of curvature at any point on the
curve, we have to direct our attention to finding the value of
ds
-y. at that point : that is, the height of the first derived curve
d<p
of the curve representing the values of <f) (abscissa) and i
(ordinate).
This curve can easily be drawn by measurement. For the
co-ordinates of any point corresponding to P, Fig. 50, we must
take AP (the arc) for ordinate, and TK (arc) for abscissa
where PT = i inch. Differentiating this curve graphically, we
obtain a curve showing in its ordinate the length of the
radius of curvature.
The algebraical process is directed towards obtaining the
ds
value of — from the successive derived coefficients of the
d<l>
primary curve with respect to x.
We have —
ds _ i_
d<ji dcji
■ ■ (§23)
ds
I
dcl> dx
dx ' ds
■ • (§43)
dy
Now, tan <^ = ^•
Diff'erentiating this
with respect to
X (§§ 43> 44), we
obtain —
d
tan (b \dx/
dx dx
i6o Graphical Calculus.
or <ec^ <i— = —
dx dx^
d4> d^y ^,
^'''''dx^d^^'°''^
Also -- = cos <^
ds
_^ I _ sec° cj)
~ dd> dx" dy , , dy
Txds d?'"'"^ ^
Now, we know that sec^ (f) = i + tan^ <^.
Hence sec <^ = \/ i 4. ( — )
Hence r = -^^-^— ^^—
which gives the value of r in terms of the corresponding height
of the first derived and seqond derived curve.
This process seems confusing at first. The student should bear in mind
that the object is to express -r- in terms of -~- and -r^. In order to do
this, we must first separate ds and d<j> by means of § 43. By that section
we Icnow that —
^(/) d(j} dx
ds dx ds
Now, we can find — by expressing the relation between ^ and x,
and differentiating it with respect to x. This relation is —
dv
tan * = -T-
dx
1 dy
or <* = tan"' -f^
dx
On differentiating either of these, we obtain -^ in terms of -j-, and
Miscellaneous Applications of Differentiation. i6i
-T^^. We also know that -y- = cos ip. These equations are combined, as
shown in the above article.
There are several other expressions for this most important
function, but the use of them involves ideas beyond the scope
of the present work.
Example i. — Find the smallest radius of curvature of the
curve y — e\
Example 2. — Prove that, when a ball is projected obliquely
upwards, the centrifugal force due to the curvature of the path
at the highest point just balances the weight of the ball.
Fig, 52.
At its highest point the ball is moving horizontally (§ 10)
with velocity v — '-, suppose. Now, after / seconds the co-ordi-
sec.
nates of the position of the ball are x = vt, and;/ = igfi
The equation of the path of the ball, therefore, is—
-y =
— Lcr~ = ^ "''■
OCT
If 2ir
This is found by eliminating t between x = vt (i.) and
X
V
dy g
dx v'
(Py ^_^
dx- v^
y = ig^'^ (ii-)' ^•^- substituting - for t in (ii.).
Hence
l62 Graphical Calculus.
Hence radius of curvature at the point (o, o,)
(i +o^)i v"
i.e. a distance — vertically downwards.
S
Let in = mass of ball.
Centrifugal force on ball due to curvilinear path
imt' _ mv"
~ r 1?
mg = weight of ball
§ 58. Illustration of Taylor's Theorem.
It may not be out of place in the present work, without
going into the proof of what is called "Taylor's theorem"
(which will be found in any book on the differential calculus),
to give an illustration of the meaning of that very compre-
hensive proposition, which will, perhaps, enable the student
to grasp its meaning better.
The proposition is as follows : If any curve is represented
by the equation y =/{x), and if we know the height of the
curve and all its derived curves corresponding to one value :*;
of the variable, then assuming that the function is " continuous,"
i.e. that neither the function nor its derived functions become
infinite for any of the values of x under consideration, the
height of the curve at a point whose abscissa is (x + h) is —
Ax)^hf\x) + ^/\x)^^^f"x^. . . (i.)
where /'(*), f"{x), etc., are the heights of the first, jsecond,
etc., derived curves at the point whose abscissa is x.
Miscellaneous Applications of Differentiation. 163
The function we shall take for illustration will be 0^.
The curve is ^ = jc*, and the height which we shall calcu-
late corresponds to an abscissa
(x -f h).
Assume that the curve (Fig. 53)
represents the distance travelled
by a particle, as in Fig. 17.
Let Op = X, pq = h.
Then if OPQ represent the
curve, y=/(x), it is clear that-
^Q represents /(:« + h).
Of course, we could in this
case arrive at the height ^Q by
cubing {x ■\- A); but we can also
arrive at it by another process,
which has the advantage that it
is applicable to all other func-'
tions of .V. Differentiate the
primary function. Then it is clear
that the area P>VQ' = MQ.
c
1
/a
T
' '/
/
M
y.
'/
//
1
1
/
v
s
p'
A
/
N
0'
f
' '1
f'
Fig. 53-
^Q = ^M + MQ =/(x) + area P'/V'Q'
=/(x)+ P'/VN + SP'N +Q'P'S
=/{x)+ hf(x) -\-\hy.h tan SP'N + area Q'P'S
= /(*)+ hfx + iH hr{x) + area Q'P'S'
Now, this small area Q'P'S' is called " the remainder after
three terms of the series." An expression is found for it in
all books on the calculus. Its actual value in this case is h^.
Working the above formula (i.) out, we find —
f(x) ^ X'
f{x) = sx"-
f'{x) = 3.2^^-
r{x) = 3-^.1
/""(x) = o
164 Graphical Calculus.
fix -\-h) = x^+ xo^h + ^^h^ + ^'^^^ + o
^ ' ^ 1.2 1.2.3
= 0? ■}- ^x"p + 3^^ + ff
which agrees with the ordinary formula.
Again, suppose we are given sin 30° = o'5, and are required
to find, say, sin 35°. We have —
X = 2,0° = — radians
6
^ t8o
= —T radians
36
Here/(^) = sin .v = J
^r
f (x) = cos X = — -
2
/"(x) = -sin A- = -i ^
f"(x) =-CQ5X ^--^
2
f"'{x) = sin X, etc.
fix + h) =f{x) + hf{x) + ^J\x) + ^ /'"{«) ■ • .
IT V3 ^2 J JT^
X r^ X -f^ + 77i X — — - X h etc.
I V ^ TT* I
X -^ + -7i X
1.2.3 2 36* 1.2.3.4
= o'S + o"o75 — o*oo22
i.e. sin 35° = 0-573. . .
The process being carried to any desired degree of accuracy.
Again, given logio 2 = 0-301, required logi, 3.
Here (x) = jx log *
= 0-434 log X
h = I
Miscellaneous Applications of Differentiation. 165
2jU,
/""(^) = -^,etc.
Now, .T = 2, ^ = I J and logio 2 = o'3or.
, . ,^ o'4^4 o*4'?4 , o"434 X 2 ^
Hence logj„(.v+/5) = 0-301 + -==^ - ^^-- + ^^^ „ -, etc.
°^ ' "^ 2 1.2.4 1.2.3.8
= 0-536 - 0-061 = 0-475 = log 3
The process being, as before, carried to any desired degree
of accuracy.
Examples.
(1) Given log 1=0, find the distance between marks I and 2 on a slide
rule where the distance between i and 10 is 12-5 cm. (The equation to
the curve isy — p log x, find the value of/.) <
(2) Find by calculation /lyp log 3"2S, ,^^/ log 4'2i, /y/ log 7, given
log 1=0.
(3) Find by calculation cos 4°, cos 66°, sin 72°, etc.
(4) Find the value of S''*, 3^", etc., by calculation from the values of
5=, 3*, etc. (The equations are here 7 = ^"'»ssl, etc.)
CHAPTER XII.
MISCELLANEOUS APPLICATIONS OF INTEGRATION.
Fig. 54.
§ 59. The Cubature of Solids.
We have already shown the application of integration to the
finding of areas.
It was shown in Chapter I., § 3, that a line may represent
an area. If a line in one direction represents an area, and
if a line at right angles to it repre-
sents a linear distance, then it is
clear that the area of the rectangle
formed on these two lines as sides
will represent a volume. Thus, if
the number of inches in AB (Fig.
54) represents the sectional area of
a prism in square inches, and AD = length of prism, it is
clear that the number of square inches in ABCD represents
the number of cubic inches in the substance of the prism.
If the sectional area of the prism is not constant all along
the length, but varies from point to point, then if a curve BEG
be drawn so that the length of the ordinate FE at any point
F represents the value of the sectional area at that point, then
it is easy to show by splitting the area up into vertical
elements, exactly as explained in § 13, that the area of the
figure BECDA still represents the volume of the irregular
solid.
Exercise. — Draw any irregular curve about 10 inches long.
Miscellaneous Applications of Integration. 167
and a straight line of similar length. Imagine a solid
generated by the curve revolving about the line as axis.
Find graphically the whole volume generated by the method
of sectional areas.
Now, the algebraical method of obtaining the volume of a
solid is the counterpart of this process. It consists in obtain-
ing the equation to the line of sectional areas BEC (Fig. 54),
and integrating it with respect to x between the ordinates AB
Fig. 55.
and DC. Let it be required to obtain the volume of a cone
of the dimensions shown in Fig. 55.
To obtain the equation of the line of sectional areas, con-
sider what will be the height of the curve at a distance x from
O. Now, the radius of the circular section of the cone at that
00
point will clearly be ^ tan ^ = ^ X x = -, and the area of
TTX
this circle — = height of curve /P at that pomt.
9
In other
1 68 Graphical Calculus.
words, the equation to the curve of sectional areas isj; = -x^.
Hence we require the area of this curve between the limits o
and 15. This is given by —
— ax = - I x'^dx
o 9 9 o
IT IS
9 o
V1
9^ 3 3^
IS'^r 5 X 5 X 57r .
= ^^ — =- =i25TrcuD.m.
27 I ~'
It is easy to show that the formula - X — gives exactly
the same result as that derived from the common rule, " \ of
volum.e of cylinder on same base ; " for, taking r = radius of
base, k = height, we have, as a result —
■Cl)^'^-^^^i^'=i-^^
The formula for the volume of a sphere of radius a is
obtained in the same way.
Taking the origin at the centre of the elevation of the
sphere, the equation to the curve of sectional areas is clearly
■y - Tr(a^ — x"-). The integral is therefore —
a
« r ^3
■iz(c^ — s^)dx = -K
Xd'x--
— a
-a L 3.
|Tra°
The same result may also be obtained by imagining the
sphere split up into concentric spherical shells, remembering
that area of surface of sphere is 4ir;l
Miscellaneous Applications of Integration. 169
Example i. — Obtain the volume of a cone of height h,
the base being an ellipse whose semi-axes are a, h (area of an
ellipse = irab). Ans. \ -Kabh.
2.— Obtain the volume of a solid paraboloid generated by
the revolution of the parabola, f- = 4«a-, round its axis,
between the planes x - ^ and jc = 9. Ans. 112 tta.
§60.
By a slight extension of this principle, we can obtain such
results as the following.
Find the total mass of a sphere of radius 10 inches, whose
density varies as the square of the distance from the centre,
the density (mass per unit volume) at the surface being 0*25 lb.
per inch^ Consider an elementary spherical shell of infinitely
small thickness dx, and of radius x. The surface of this shell
is ifTTO^'. The volume of it is \tTO^d.x. The quantity of matter
in it is clearly /^irs^pdx, where p is the density or quantity of
matter per unit volume at a distance x from the centre. Now
0-2S lb. . , , .
we have 10- : x' :: —.--.,- : p, smce the density vanes as the
square of the distance from the centre.
mu <■ °'25 X X^ lbs.
Therefore p = - , — . . — -
10'- m.^
Hence, substituting this value of p in the above expression,
we see that required total mass is the result of adding together
all such small masses as f h^ttx*' x — | J dx between the given
limits ; that is —
/r(4'^"'^T^^)''"=(T^''*^o
500
10
o
X lo"' = 2007r lbs.
170 Graphical Calculus.
§ 61. Graphical Solution of Differential Equations.
Problems sometimes arise in which we are given a relation
subsisting between two or more of the primary or successive
derived functions of a quantity, and we are required to find
either the primary or some other function connected with it.
These problems are very confusing to the beginner, and we
shall show in what way many of them can be attacked graphically
by the careful application of the principles already explained.
For example, a train weighs 50 tons exclusive of the engine-
The resistance to motion due to mechanical friction alone is
constant at all velocities, and is of the magnitude of say 8 lbs.
per ton. The resistance due to other causes (such as that due
to the atmosphere) varies directly as the i*7th, power of the
velocity in — '— , being, let us say, = "0025 X v^"'- Suppose
sec.
we are also given a curve showing the magnitude of the
pull in the drawbar as the speed varies, and are required
to find —
(i.) The maximum velocity attainable on the level.
(ii.) The time occupied in attaining it.
(iii.) Distance travelled in that time.
A method similar to the following may be used in attacking
problems of this kind. Suppose AHB (Fig. 56) is the given
velocity-pull curve, of which both scales must be given (the
figure is not drawn to scale) where the drawbar pull is plotted
• ft-
vertically in tons suppose, and velocity horizontally in — .
We have now to draw on the same base the curve of
resistances. This resistance consists of two parts —
(a) Frictional resistance, which is constant whatever the
velocity.
(^) Other resistances, which vary as the 17th power of the
velocity.
Find the total value of {a) for the whole train.
Miscellaneous Applications of Integration. 171
This is —
8 lbs.
ton
X 50 tons = 400 lbs.
Set this value off at OjD to the same scale as the drawbar
pull is plotted in, and draw a horizontal line DE. This is the
curve of frictional resistances.
On DE as base plot a curve DB, whose ordinates show
172 Graphical Calculus.
the corresponding values of ■00252;^'' to the given scale. Then
it is clear that the height of the curve DB above OiXj at any
point shows the total resistance to the motion of the train
at a constant velocity represented by the abscissa.
Now, of the total force in the drawbar pulling the train
only part is required to overcome the actual constant-velocity
resistance. The whole force over and above this part is
employed in increasing the velocity of the train, i.e. in produc-
ing acceleration. This latter surplus force is clearly given by
the length of the ordinates between the two curves AB and
DB. At the point B this surplus vanishes ; .the maximum
velocity is therefore given by 0-J), for here the total force in
the drawbar is absorbed in overcoming constant-velocity
resistances, and there is none left to increase that velocity.
Transfer these lengths of ordinate between the two curves to
corresponding positions on the base O2X2, thus GH = IJi, etc.
This gives a curve ahb.a which shows the net force producing
acceleration at all velocities.
Now, from this curve we can easily deduce another showing
the actual value of the acceleration produced. We have from
Dynamics —
, . . ft.
mass m tons X acceleration m ;
sec."
Force in tons =
g
where g is the acceleration due to gravity, = 32 — '— about.
sec.^
Hence —
ft. -12.
Acceleration in = force in tons x ^—
sec 50
It is obvious that we need not consider the mass of the engine in this
equation, because any force necessary to accelerate the engine does not
appear in the drawbar at all, being absorbed in increasing the velocity of
the engine. It is only the surplus force not absorbed in the engine itself
that appears as a pull in the drawbar.
Miscellaneous Applications of Integration. 173
Hence if we reduce all ordinates of curve ahb in the ratio
If we shall obtain a curve KjLaMa^a, which gives the actual
acceleration in — '-„ corresponding to any velocity.
Now, we know that if the velocity and the acceleration were
each plotted separately on the same time base, the former would
be the integral of the latter (§ 16), and the problem resolves
itself into the finding of the time base, i.e. from known simul-
taneous values of the velocity and acceleration to deduce a
time velocity and a time acceleration curve. This we may do
in the following manner : Divide the curve KjLjMo into small
parts K2L2, L2M2, etc., such that each part is nearly straight,
and draw ordinates at all these points. Take a base O^X^ as
shown coUinear with O2X2. Let the time from the instant of
starting be reckoned from 0\ It is obvious that the point
O^ is on the curve of velocities. The height of the acceleration
curve at this point is clearly O2K2 = 0"N, which lines also
represent the tangent of slope of the velocity curve. Take
O^S to represent i second, and set up ST vertical. Transfer
the ordinate O2K2 to this line as shown ; then, as in § 14, O^K
must be tangential to the curve of velocities at the point 0\
Next consider the point L2. Here the acceleration is /2L2 and
the velocity 04i- Hence wherever the ordinate of the time-
velocity curve corresponding to the point L2 may be, the slope
of that curve at the point where that ordinate cuts it, must be
the slope of the line O^L (where L is proj ected from Lj). Hence,
as shown in § 14, the point on the time-velocity curve corre-
sponding to Lj must lie on the line bisecting KO^L, and since
the height of the point above O^X' is given by O2/2, we can
find the point P by projecting as shown. This gives another
point on the velocity curve, and we can proceed exactly as
before to find a third point. Thus the whole curve may be
drawn and the acceleration curve plotted as we go along.
The distance from O" of the point where the acceleration
curve cuts 0"X", gives the required time. The space passed
174 Graphical Calculus.
over in this time can be found by integrating the time-velocity
curve in the usual manner.
Similar problems can often be solved algebraically if we
can find the equations to the curves involved, as in the follow-
ing example : —
A smooth tube 15 feet long turns round a vertical axis
at 2 revolutions per second, so as to be always horizontal. A
smooth marble is placed in the tube at the vertical axis of
rotation : find its velocity when it is swirled out at the other
end.
It is clear that if a piece of string of length x were tied to
the marble, the radial acceleration would be w^je, and the
tension in the string mii?x, so that when the marble is free
in the tube, at a distance x from the axis it has an accelera-
tion i^x along the tube. Now, if we plot values of <^x along
a line representing the tube we obtain a curve of accelerations,
but not a curve of time-acceleration, so that the area of this
curve does not represent the velocity, as we see to be the case
from §§ 14-17 together. If, however, we could by any means
transform this curve, as in § 43, to a time acceleration curve
by transferring the ordinates to a time base, then we could
integrate it graphically.
We are, in fact, required to integrate u^x with respect to t,
the time, in order to find the velocity. Hence the problem
is to find some function which, when differentiated with respect
to t, will produce u^*.
Now, we have that —
Jdx\
d?x \df)
"^ = ^=~^r • • • ^'^
that is, f iiP'xdt = ^
Consider what is the relation of the element of area u^xdt
to the element of area n^xdx, which would be the correspond-
ing element on an x base, i.e. on the length of the tube.
Miscellaneous Applications of Integration. 175
It is clear from § 5 1 that —
uy'xdx = ui^x—dt
at
and therefore f ofixdx = f ui^x ~dt . . . i[\.)
•^ ^ dt ^ '
dec
Now, if we multiply each side of (i.) above by — , we shall
dt
find that we are able to integrate both sides with respect to t,
and thereby obtain an equation for — '-.
dt
Thus / 0)=X-7- -dt = \—;:--r-dt
■' dt jdt^ dt
which we see, from (ii.), becomesy (o'^jc^x = I -— — dt.
d'^sc dx
It is easy to see that the expression — • —- is the differential
coefficient with respect to «■ of ^ f — j , for, as in § 50, we
have split the expression -7:^ . v iiito two factors, one of which
I —J is the d.c. of the other [-r:)-
Therefore we have —
2
*.■«■ = i(f)
dt
There is no constant required, since, as explained in § 22,
the line of velocities cuts OX when x = o.
Hence — = wx
dt
The velocity is therefore —
4X'rXis = 6o7r feet per second
176 Graphical Calculus.
Dividing by x, we obtain —
r dx
— — - = O)
X 'at
Integrating again with respect to t —
log X = tot
or ic - e"" -f- const.
which constant in this case is — i, as may be seen by con-
sidering the instant of starting.
If this latter equation be differentiated twice with respect
to t, the original equations (i.) will be reproduced.
A bird's-eye view of the whole of this problem may be
, . 1 . d'X
obtamed by considering that — - = ui^x, i.e. the height of the
second derived of the time-distance curve is a positive con-
stant multiple of the height of the time-distance curve itself.
The only curve that satisfies this condition is a; = e'\ where c is
a constant.
Examples. — (i) Given -j-^ = \ ( y^ j^ find y in- terms
doc ^ (tec ^
of X, (i.) graphically and (ii.) analytically. (For the latter,
dy
divide both sides of the equation by -- and integrate. Obtain
resulting equation for — in the e form. Invert it, and integrate
uCC
with respect Xa.y.) Ans. x = — 4«'~V.
(2) Solve the tube problem graphically.
§ 62. Rectification of Curves.
It is sometimes very useful to be able to find the length
of curves.
We saw in the last chapter, § 57, that we might have a
curve plotted on the de base showing in its ordinate the length
Miscellaneous Applications of Integration. 177
of the curve measured from a fixed point on it. The tangent
of slope of this curve will clearly be the secant of the angle of
slope of the original curve.
The student will understand this without difficulty if he
works the following exercise. Draw a curve (a) of any shape,
and take a point P on it ; make another curve (3) on an x
base, by measurement from A, whose ordinates represent the
corresponding length of curve (a) measured from P. Differen-
tiate it graphically, and show that the derived curve so obtained
is the same as would have been obtained by plotting values of
the secant of the angle of slope of {a) on an x base. Thus it
is clear that the integrated curve of a curve showing the values
of the secant of the angle of slope of the original curve is a
curve showing the length of the original curve.
Now, if <^ be the angle of slope of the original curve, we
have, from trigonometry —
sec'' 1^ = I + tan^ ^.
that is, sec ^ = \/ i 4. ( -^ )
\dx /
ds
Sec 1^ we have called — . Hence —
length of curve = S= I V i + (
"LYdx
dx'
the limits being taken as required.
Thus, find the length of wire rope required to hang between
two pillars 120 yards apart, assuming the curve of the rope is
given by the equation —
ml". . \
- \e"' + e «)
2
This is the actual curve in which a rope hangs, and is
called the " catenary curve." The axis of ^ is a horizontal
line at a depth m below the lowest point of the curve; m also
represents the length of the same kind of rope which weighs as
much as the tension at the lowest point of the rope.
178 Graphical Calculus.
Assume m = 100.
The sag of the rope is then = y — 100, where j; =■• greatest
ordinate.
Sag of rope = so^^ia + e'wj — 100
= 50(1-82 + ^-:|j) -100
= 5° X o"37 = 18-5 yards
tis
dx
H -(I)"}' = «'= + -)
Therefore the whole length, being twice the half-length,
is —
since s is measured from the vertex,
= 100 I I"82 5- )
V I'82/
= 127 yards nearly
Example. — Plot a catenary from the given equation.^
Draw a curve on an x base representing the secant of its angle
of slope, and obtain the above result by integrating this
curve.
' This may best be done by first obtaining the curve y = «"' (i.) by the
method described on pp. 102, 103, i.e. find geometrically the lengths of a
series of equidistant ordinates in geometrical progression, i.e. with a con-
stant common ratio which must be calculated from the equation. The
curve ji = e " (ii.) can then be found by combining this curve with the
curve ^ = — by the method of Figs. 40, 43. These two curves (i.) and
X
(ii.) should then be added together (Fig, 26), and the result divided
by-.
Miscellaneous Applications of Integration. 179
§ 63. Centres of Gravity.
The finding of centres of gravity and moments of inertia
is an application of the integral calculus of very great service
to engineers. The principle of these methods is exactly the
same as that adopted in elementary mechanics, but the proofs
are very much simplified by the application of the calculus.
The proposition relied on in all these methods is that the
resultant of a system of forces acting on a body has the same
tendency to twist it about any arbitrary point or line as the
sum of the twisting tendencies of each of the forces taken
separately. This proposition, as applied to the special purpose
before us, is embodied in the following rule. Find the mass
of each of the separate parts of the object, and multiply each
by the algebraical distance (+ or — ) of the centre of gravity of
that part from any convenient straight line. Add all these
products together, and divide the sum by the sum of the masses.
The result is the distance of the centre of gravity from the line.^
The following graphical process embodies this rule, and
applies the principles already explained.
Let it be required to find the centre of gravity of a piece
of plate cut into the shape of a curve of sines between the
ordinates .» = o and a- = 2 (Fig. 58). It will be seen that, for
the graphical method, any arbitrary curve of any shape what-
ever might be used ; but as we shall also give the correspond-
ing analytical process, it will be convenient to consider a
curve of which the equation is known.
Draw a number of ordinates to the curve at convenient
and well-defined distances from O, such as 0-2 inch, 0-4 inch,
o'6 inch, etc. Measure the length of each ordinate, such as
yR, with a decimal scale, and multiply it by the scale length of
the corresponding abscissa Or.
' A very lucid explanation of this and similar propositions will be
found in Professor Goodman's treatise on " Applied Mechanics."
i8o
Graphical Calculus.
There is no difficulty in devising a method whereby this may be done
graphically, if desired. Thus to multiply together the lines AB, AC (Fig.
57), complete the rectangle, and mark
off DE = I inch, and complete as
shown ; then AF represents the pro-
duct required. The actual ordinate,
however, may be obtained much more
rapidly and accurately with a slide
rule, an instrument with which every
engineer or scientist should be familiar.
Fig. 57-
Plot the value thus found
along the same ordinate as at
rRi and carefully draw a curve, ORiPj, through all the points
so obtained. Now, the area of this curve will be the moment
of the whole area about O in inch units to the same scale as
the area of ORQ represents the actual area. That is to say,
suppose OX, OY are both held horizontally, then the tendency
in inch units which the force exerted by gravity on the plate
Fig. 58.
ORQ has to twist the plate about the line OY (supposed held
fixed) is represented by the area of ORiPjQiY in square inches,
to the same scale as the area of ORQ represents the weight
of the plate.
Thus if the area of ORQ is = (a) sq. inches, and ORiQi
= ip) sq. inches, and weight of the plate = 2 ounces ; then,
Miscellaneous Applications of Integration. i8i
since {a) sq. inches represents 2 ounces, i sq. inch = - ounces,
2
and the moment of the weight about OY is therefore b -K -
ounce-inch units. For consider an element of area /Pi of
breadth dx. It is clear that the moment of the element pY
about OY is /i X /P X (/« X O/, where /a represents the mass
of a square inch of the plate. Now, p? x Op = pF^, and /Pi
X dx = area of element /Pi ; hence the area of the element
/Pj X /A = moment of element /P about OY.
- - , . , weight of plate
Now, the quantity «, as we have seen, = — ° *
area of plate
2 oz. .
= -•-■ — :, m above example. Hence, to the same scale as the
area of /P represents the weight of the corresponding strip
of plate, /Pi represents the moment of that strip about OY.
The same may be said of all corresponding strips, and is
therefore true of them all taken together.
Now, let X be the distance of centre of, gravity from O Y.
Then we have area of ORjPiQi = area of OPQ X X.
_ area of ORiPiQ,
Hence X = ^^^ —
area OPQ
Integrate both the curves graphically, as already explained,
or find their areas by the planimeter or otherwise ; set off a line
representing the area of the curve OPiQi vertically, as at CB
(Fig. 6) ; set off a line representing the area OPQ to the same
scale horizontally, as at AC. Join AB, maiie AM = i inch,
and draw MP vertical; then MP represents the distance of the
centre of gravity from OY.
The process must be repeated with OX vertical to get the
distance of the centre of gravity from OX.
We thus get two intersecting lines, each of which contains
the centre of gravity, which point is therefore found at the
point of intersection of the lines.
N 3
i82 Graphical Calculus.
The student will now have no difficulty in understanding
the following process, which is inserted without explanation ;
as it is the same step by step as the graphical process just
described.
J^jx sin xdx
X =
/"^ sin xdx
^[sin X —X cos x] , a •,
: °i — — J (see § 53)
l[-cosx]
sin 2 — 2 cos 2 —
— cos 2 4- cos o
sin ii4'59°— 2 cos ii4'59°
I —cos ii4'S9°
sin 65'4i° + 2 cos 6s'4i°
I + cos 65 '41
sin 65° 24' 4- 2 cos 65° 24'
I + cos 65° 24'
o'9092i + o'83252
nearly
i"4i626
= I ■23" nearly
Examples. — Find the centre of gravity of a triangle, a cone,
a frustum of a cone, an arc of a circle, a slice of a sphere, a
rod whose density varies as the «th power of its distance from
one end, any section of a parabolic plate, a theoretical indicator
card (no compression).
§ 64. Moments of Inertia.
Moments of inertia may be found in a similar way to that
employed for centres of gravity. The moment of inertia of
an area about an axis in its plane is analogous to what we
have already described to be the moment of an area about an
axis; but whereas each ordinate in Fig. 57 = ordinate of
area x distance of ordinate from axis of Y, each ordinate in
Miscellaneous Applications of Integration. 183
the corresponding Fig. 59 for the moment of inertia = ordinate
SiSo X (Oj)Ii
Let PQRS be any area of which it is desired to find the
moment of inertia about the axis OY. Set up ordinates
' There is a good deal of confusion in the minds of students as to the exact
connection between the moment of inertia of an area about a line in its plane,
and what is called the moment of inertia of a solid body imagined spinning
about an axis. In the discussion on centres of gravity, the same point
arose in connection with the relation between the geometrical first moment
of an area about a line- and its mechanical analogue. Without' going fully
into a question which has more to do with rigid dynamics than calculus,
we may point out that a clear conception of the meaning of the moment of
inertia of a body may be obtained by considering it as the angular mass
of the body and a couple as angular force. The meaning of this will be
clear from the following analogy. If a force acts on a mass perfectly free
to move^ —
Force in poundals = mass in pounds X acceleration in feet per second per
second *
Similarly, if a couple acts on a mass perfectly free to turn round —
Couple in ft.-poundals = moment of inertia X angular acceleration in
radians per second per second
or angular force = angular mass X angular acceleration
Again —
Momentum = mass X velocity
angular momentum = moment of inertia x angular velocity
... (\ mass X (velocity)'
kmetic energy = { \ ■, ^, , , 1 -i >»
"' WW. (angular velocity)
A conception of the magnitude of unit moment of inertia in foot and
pound units may be derived from the consideration that if a body has unit
moment of inertia round an axis, and is rotating at unit angular velocity,
it will do \ ft.-poundal of vifork before being brought to rest. The con-
nection between the geometrical moment of inertia and the mechanical
one consists simply in the introduction of the factor p, or the mass per
unit area, into the expression for the element. The mechanical significance
of this is easily seen from the above remarks. The geometrical moment
of inertia is, as it were, a, skeleton which we may endow with life either
by multiplying by pounds mass per square inch ; it then comes in for
calculating kinetic energies, etc. ; or if used for such purposes as the
determination of stresses in beams, we multiply it by pounds weight per
square inch (tension or compression), and in other ways which need not
be here mentioned.
1 84
Graphical Cakuhis.
parallel to the given axis, as in § 63, and multiply the length
S1S2 of each by the square of its distance from the given axis
(O^)^, and set up the length so oTjlained on each of the ordi-
nates; thus jS = S1S2 X OSl The area of the curve so obtained,
found in any miinner, is the moment of inertia of the area
about that line in inch units. The proof of this is almost
identical with that given for the similar method used in con-
nection with finding centres of gravity in § 63, and may easily
be completed by the student himself.
Just as before, if jj' = f{x) be the equation to the boundary
line of the curve, all we have to do is to find the value of
yx^dx, which gives us at once the value of the moment of
Fig. 39.
inertia. The value of this area or definite integral, divided by
the greatest distance from the axis of the boundary of the
figure, gives us what is known as the modulus of the section.
Miscellaneous Applications of Integration. 185
It may be found graphically, in the manner explained in § 63
as the length of a line. Also the value of this area or integral
divided by the area of the iigure gives the value of the square
of the radius of gyration.
Thus, required the dynamical moment of inertia of a flat
circular plate of mass m lbs., whose radius = r, about an axis
passing through its centre perpendicular to its plane. Con-
, , sider a circular element of the plate, radius x, breadth dx;
Its area is iirxdx
its mass is 2-n-xdx X a = 2-Kxdx x — -„
where fi is mass of unit area of plate —
its moment nl inertia about O obviously = ^ —
Fig. 60.
Hence, whole moment of inertia—
= — f^o^dx
*2 «>' "
r
2 m
= ^(i^*):
r-
The geometrical moment of inertia is \ar^ or -^^ which,
multiplying by the mass per unit area, becomes ^mt^ as above.
1 86 Graphical Calculus.
Consider the case of a cylindrical shaft under torsional
stress. Suppose /is the shearing stress per square inch at a
distance = i inch from the axis. Clearly the stress at any
distance x from the centre is fx. Since the stress varies as
the strain, and the strain varies as the distance from the axis.
Area of an elementary annular ring of radius x x breadth dx
= 2Trx X dx
total stress on this layer zirxdx xfx= 2-irfx^dx
moment of this stress about axis = 2irfx^dx x x = 2-nfx'dx
Plot values of this along the horizontal radius, and find the
area of the curve so obtained, and compare result with the
result of the integral of 2-nfsi^dx between limits r and o.
Examples.
I. Find the moment of inertia of a fly-wheel, outside diameter 6 feet,
sectional area of rim 4x5, inside diameter of rim 5 feet 4 inches, six
arms each of sectional area an ellipse of axes 3J and 2. Boss, a cylinder
8 inches diameter x 8 inches long, with a 3-inch hole for the shaft. Mass
of cubic inch of iron, o"26 lb. = p. (Method. — Plot a curve on the
horizontal radius of the wheel as base, showing the value of pax^, where a
is the area of metal cut through by an imaginary cylinder o£ radius x,
concentric and coaxal with the wheel.) Find the scale on which the area
of this curve represents the moment of inertia. Find the radius of gyration
by a graphical process (find the weight of the wheel graphically as the
area of a curve, showing the values of pa). Then —
I =MR'
M
R = . /"^
M
Find this graphically by the process of Fig. 3.
2. Find the moment of inertia, by graphical method, of a rectangular
section, a box section, a triangular section, and a circular section about
axes in their planes and passing through their centres of gravity, and
compare your results with that given by the formula 7iah^ —
where « = r! f°'' rectangular section.
T^g for triangular section.
t'j for circular section.
a = area of section.
h = height in plane of bending.
APPENDIX.
Barker's Planimeter.
This instrument was devised by the author for the purpose of
mechanically drawing the integral curves, on the principle explained
in § 14. . It is here described for the first time. It consists of a
horizontal slide AB, carrying a slider DF, to which is rigidly
attached the vertical slide CE, which is fitted accurately perpen-
dicular to AB. The vertical sUde CE carries a long slider GH, to
which is fitted the tracer P, and to which is pivoted at L the rod
KL, which slides through the piece M. M is itself pivoted on a
clamp as shown, and the clamp can be secured to any part of the
piece D, which is graduated. By means of a double parallelogram
of jointed rods, KL is kept parallel to the piece N, one point of
which is pivoted on a slider Q, on which a vernier is engraved ; at
the same time Q is allowed to assume any position on the vertical
slide. This piece N carries a wheel with points on its periphery,
and the bearings of the wheel are so attached to the piece N that
its plane is always parallel to the axis of KL. It is clear that when
the pointer P traces out a:ny curve, the wheel will roll out its
integral, for the tangent of angle of slope of the upper curve is
clearly proportional to the ordinate of the curve traced out by P.
The adjustments required are, in the case of a curve on a base, that
the instrument must be so placed that AB is parallel to OX, the
given base ; also KL must be parallel to AB when the tracer P is
on the base OX. However, it may be used for finding areas inde-
pendently of this adjustment, for if the pointer P be placed on the
curve whose area is required — such as an indicator diagram — and
the reading of Q taken, and the tracer be then carried round the
curve to the starting-point, and Q read again, the difference of the
i88
Graphical Calculus.
readings gives the area required, in units which depend on the
position of M on D. This may be adjusted to read in any units
within the working limits of the instruments.
The instrument may also be used for differentiating a curve, by
tracing out a curve with the wheel ; but it is rather difficult to hold
the wheel sufficiently steady.
Fig. 6i.
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