TIGHT BINDING BOOK
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OSMANIA UNIVERSITY LIBRARY
Call No. 5~f T Accession No. %- 1 ^4-^ ST
Author
Title
This book should be returned on or before the date
last marked below.
'IFFERENTIAL AND
[NTEGRAL CALCULUS
by John Haven Neellev.PLD.
PROFESSOR OP MATHEMATICS IN
THE CARNEGIE INSTITUTE OF TECHNOLOGY
and Joshua IrvinoePh. D
ASSOCIATE PROFESSOR OF MATHEMATICS
IN YALE UNIVERSITY
SECOND EDITION
HE MACMILLAN COMPANY
New York 1939
Second Edition Copyrighted, 1989,
BY THE MACMILLAN COMPANY
All rights reserved no part of this book may be
reproduced in any form without permission in
writing from the publisher, except by a reviewer
who wishes to quote brief passages in connection
with a review written for inclusion in magazine
or newspaper.
->nted in the United States of America
Published July, 1989
First edition copyrighted and published, 1932,
By The Macmillan
PREFACE TO FIRST EDITION
This texjt on the Differential and Integral Calculus is presented
with the Belief that it is well adapted for use both in academic
colleges and in engineering schools.
The student who studies this subject because of his attraction,
to mathematics is not well equipped if he lacks a fair appreciation
of the wide applications of the Calculus in modern science and in
engineering On the other hand, the student who is required to
use the Calculus in some chosen field of science can make more
intelligent and extensive applications if he understands the under-
lying principles of the subject. Hence, whether mathematics is
to be regarded as the jjueen of thejscjences or as the tool of the
scientist, the study of the^Calculus for the futu! ^teacher of mathe-
matics and for the future engineer should differ omy in the degree
of emphasis placed on the theory and the applications.
As is well known, a complete rigorous proof of some of the
theorems ; n the Calculus is out of the question for the beginning
student, whereas the applications are easily made and are of
extreme importance. With this in mind, the authors have en-
deavored to use only proofs which are valid but which may involve
certain assumptions, the proof of which belongs properly in an
advanced course. These assumptions are pointed out to the
student; for at this stage he should be instructed to examine
proofs more critically, in order that he may realize some of the
difficulties to be encountered, and also that he may avoid the
common pitfalls.
A comprehensive review of as much analytic geometry as is
required in the Calculus has been included in Chapters I, II, and
VIII. These may be omitted, or they may be used only for
reference, at the discretion of the teacher.
The authors are sincerely grateful to Professor W. A. Wilson,
^ho has kindly made many pertinent suggestions; to Professor
vi PREFACE TO FIRST EDITION
O. T. Geckeler, who has contributed many problems and valuable
suggestions, and who has prepared the material for several impor-
tant parts of the text; and to the Macmillan Company, the
publishers, who have been most considerate in cooperating with
the authors and the editor in the publication of this text.
J. H. NEELLEY
J. I. TRACEY
August, 1932
PREFACE TO SECOND EDITION
In revising this book, special attention has been given to the
selection and arrangement of problems. To increase classroom
utility many additional carefully chosen problems have been
interspersed throughout the text ; especially those of a less involved
nature, for purposes of drill, and on the other hand, a number of
highly challenging problems.
Although the general plan and organization of the book remains
unchanged, new material has been added in Chapters VII, VIII,
and IX and some other sections have been rewritten.
The authors are grateful for helpful suggestions from the many
teachers familiar with the earlier edition, and to those who have
manifested an interest in the revision by constructive criticism.
J. H. N.
J. I. T.
June, 1939
vu
CONTENTS
CHAPTER PAGE!
I. COORDINATE SYSTEMS. GRAPHS 1
II. EQUATIONS OF DEFINED CURVES. EMPIRICAL EQUATIONS 31
III. THE DERIVATIVE . . . . > 70
IV. DIFFERENTIATION OF ALGEBRAIC, LOGARITHMIC, AND
EXPONENTIAL FUNCTIONS 88
V. SOME APPLICATIONS OF THE DERIVATIVE 103
VI. DIFFERENTIALS. THEOREM pf r "MEAN VALUE^^T? ... 131
TRIGONOMETRIC FUNCTIONS CURVATURE 147
SOLID ANALYTIC GEOMETRY . . . 777T 1 186
PARTIAL DIFFERENTIATION, DIRECTIONAL DERIVATIVES. \
ENVELOPES . j /.... -rTy^ 205
/ X- INTEGRATION . ^ 229
fXI. METHODS OF INTEGRATION. DEFINITE INTEGRALS.... 239
XII. APPLICATIONS AND INTERPRETATIONS OF INTEGRALS 271
'XIII. THE DEFINITE INTEGRAL AS THE LIMIT OF A SUM 295
XIV. MULTIPLE INTEGRATION 316
XV. ADDITIONAL APPLICATIONS OF THE FUNDAMENTAL
^ THEOREM 341
XVI. INFINITE SERIES WITH CONSTANT TERMS 366
XVII. POWER SERIES AND SOME APPLICATIONS 380
XVIII. HYPERBOLIC FUNCTIONS 410
XIX. EXACT DIFFERENTIALS AND LINE INTEGRALS 421
XX. SOME DIFFERENTIAL EQUATIONS OF THE FIRST ORDER. . 428
XXI. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 445
XXII. APPLICATIONS OF DIFFERENTIAL EQUATIONS. STATICS,
DYNAMICS, ELECTRICITY 469
STANDARD INTEGRALS 483
INDEX 491
ix
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER I
COORDINATE SYSTEMS GRAPHS
1. Coordinate Systems. A system of coordinates is a method
c representing the position of points by means of numbers. In
ohe elementary geometry of the plane two systems are in general
e, rectangular coordinates and polar coordinates.
2. Rectangular Coordinates. To establish a system of rectan-
gular coordinates in a plane, it is necessary to draw a pair of per-
pendicular lines, called axes, and to have an appropriate unit of
length.*
The perpendicular lines are the x axis and the y axis and their
^oint of intersection is the origin. It is customary to draw the
x axis horizontally and the y axis vertically; the four quadrants
into which they divide the plane are numbered as in trigonometry.
The position of any point in the plane is designated by two real
numbers, called coordinates, which represent the respective distances
Jrom the axes to the point as measured in terms of the given unit.
These coordinates are called ab-
scissa and ordinate.
The abscissa of a point is the dis-
mcefrom the y axis to the point. j> 2
The ordinate of a point is the dis-
to. ^e from the x axis to the point. _
L the coordinates of a point are
x and y, respectively, they are writ- p p
ten in the form (x, y), the first * * * 4
number always being the abscissa.
In any system of coordinates it is F i
Essential that a given pair of num-
ffers shall designate one and only one point. Hence it is nec-
to distinguish between the coordinates of such points as
* There are coordinate systems in which the axes are not perpendicular, but little
use is made of oblique axes in elementary geometer.
1
2 DIFFERENTIAL AND INTEGRAL CALCULUS [On. I
represented in the figure; this is done by making abscissas and
ordinates directed line-segments.
If AB is a directed line-segment, then BA is AB, that is,
reading a segment in the opposite direction changes its sign. Thus,
if the length of the segment BA is considered as + 6 units, then
AB is 6 units. The signs for directed segments in rectangular
coordinates are usually as follows:
A horizontal segment when read from left to right is positive, if
read from right to left it is negative.
A vertical segment when read upward is positive , if read downward
it is negative.
If an oblique segment is directed, it is positive when read upward
and negative when read downward.
However, we shall use an arrow head to designate the positive
direction along each axis, as it is sometimes more convenient to
reverse the positive direction along one of the coordinate axes.
To plot a point, when its rectangular coordinates are given, is to
mark its position in the plane with reference to the coordinate axes.
Thus the point (4, 3) is the point 4 units from the y axis in the
positive direction of the x axis and 3 units from the x axis in the
negative direction of the y axis.
I In practice, it is customary to
start at the origin and measure
a distance along the x axis, in this
case 4 units to the right, then meas-
2L. ure from this point along a perpen-
dicular to the x axis, in this case 3
units downward, and mark the
point.
We may then think of the point
(4, 3) as being the intersection
of two lines, one parallel to the y
axis and 4 units to the right of it, the other parallel to the x axis
and 3 units below it.
The axes should always be marked so as to show the scale of
units used.
3. Theorem for Directed Lines. // 0, any point on the line
A iAz, is taken as an origin, then the directed line-segment A\A*
expressed in terms of OA\ and OA% is always equal to OA Z minus
OAi.
i 5]
COORDINATE SYSTEMS GRAPHS
Three cases are to be distinguished, one when is between AI
and Az, the others when is outside the segment AiA%. Let be
between A\ and A 2 ; then
Aj
But AiO =
hence
FIG. 3
as was to be proved. The other two cases are left as exercises.
4. Horizontal and Vertical Projections of a Line-Segment.
Let PiP 2 be any line-segment whose extremities have the coordi-
nates (xi, y\) and (x 2 , 2/2) respectively. Through each extremity
of the segment draw a horizontal
and a vertical line. The distance
between the vertical lines, meas-
ured along a horizontal', that is,
APz or MiM 2 , is known as the
horizontal distance from Pi to P 2 ,
or the horizontal projection of
PiP 2 . Similarly the distance be-
tween the horizontal lines as meas-
ured along a vertical, namely,
PiA or NiN 2 is known as the ver-
tical distance from PI to P 2 , or the
vertical projection of PiP 2 .
To express these horizontal and vertical projections in terms of
the coordinates of PI and P 2 , we have
MiM 2 = MiO + OM 2 = - xi + x 2 .
Hence the horizontal projection of PiP 2 equals x 2 XL Similarly,
N,N 2 = N& + ON* = - yi + 7/ 2 .
Hence the vertical projection of PiP 2 equals y% T/I. That is, the
horizontal projection of PiP 2 is the abscissa of the last-named
point minus the abscissa of the first-named point; the vertical
projection of PiP 2 is the ordinate of the last-named point minus
the ordinate of the first-named point.
5. Length of a Line-Segment. From Fig. 4, we observe
that the horizontal and vertical projections of PiP 2 , namely,
FIG. 4
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
and Pi A respectively, are the two legs of the right triangle
. Hence, letting PiP<* be represented by d, we have
But
and
Hence
(D
d 2 = Pi
PiA = PiMi + MiA = -
AN*
+ AP 2 \
= 2/2- 2A>
EXAMPLES
1. Given A(- 7, 3) and B(- 1, - 5), find
(a) the vertical projection of AB;
(b) the horizontal projection of BA ;
(c) the length of the segment AB.
SOLUTION.
u
r
[N
B(e,-c)
FIG. 5
(a) The vertical projec-
tion of AB is (- 5) - (3) = - 8 units.
(6) The horizontal projection of BA
is (- 7) - (- 1) = - 6 units.
(c) The length of A B is
d = V(-!-6) 2 -h (~8) 2 = 10 units.
2. Given A(c, d) and B(e, - c) find
(a) the horizontal projection of
AB]
(6) the vertical projection of BA ;
(c) the length of the segment AB.
SOLUTION, (a) The horizontal pro-
jection of A B is
DB = DM + MB = - c + e
c c.
(6) The vertical projection of BA is
ZX1 = DAT + NA - (- C )
(c) The length of AB is
X/QAJV + ND)* + (DM -f
4. fg _
COORDINATE SYSTEMS GRAPHS
PROBLEMS
1. Prove the theorem for directed lines when
(a) the point O is on AB extended;
(b) the point is on BA extended.
2. Find the horizontal and the vertical projections of the line-eegment
PiP 2 ; of P 2 Pi; also find the length of the segment for PI( 4, 4) and
P 2 (l, 3). Ans. 5, 7; - 5, - 7; V74 units.
3. The same as Problem 2 for Pi(0, 1), P 2 (- 8, 0).
4. The same as Problem 2 for P^- 3, 3), P 2 (3, - 3).
5. The same as Problem 2 for Pi(\/3, 2), P 2 (3, - V3).
6. The same as Problem 2 for PI(O, 6), P 2 (c, d).
7. The same as Problem 2 for PI(WI, - n), P 2 (n, - m).
Ans. n m, n m; m n, m n', (m n) V2 units.
8. The same as Problem 2 for Pi(- 0, A), P 2 (/&, - g).
6/ Inclination. Slope.
hich its positive direction
inclination of a line is the angle
makes with the positive direction of the
iv
x axis.
The slope of a line is the tangent of its inclination.
M X
FIG. 6
From either figure we have
tan a
FIG. 7
AP*
BM
But
and
AP 2 = AM + MP 2 = -
PiA = PiR + RA = Xi + x z
- 2/i>
6 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
Hence, letting the slope of FiP 2 be m, we have
(II) ta
That is, the slope of the line joining two given points is the differ-
ence of the ordinates of the two points divided by the difference of
their abscissas taken in the same order.
7. Conditions for Parallelism and Perpendicularity. If two
lines LI and L 2 are parallel, they have the same inclination and
their slopes are equal, that is, the condition for two lines with slopes
mi and w 2 respectively to be parallel is
/TPTT \ ,. .
(111,; /Mi == 772 2 .
Conversely, if their slopes are equal, tan ai = tan a 2; hence the
inclinations are equal and the lines are parallel.
If LI and L 2 are perpendicular, their inclinations must differ by
90. If LI has the greater incli-
nation, then
= <* 2 + 90.
= tan ( 2 + 90)
= ctn a 2
1
tan a 2
But
FIG. 8
tan ai = mi, and tan 2 = ra 2 .
Therefore
(IV)
= , or m\m<2, = 1.
Conversely, if Wira 2 = 1, then tan on = I/tan <* 2 = ctn <* 2 .
Hence the inclinations of the lines must differ by 90 and the lines
are perpendicular. Therefore:
A necessary and sufficient condition for two lines to be perpendicu-
lar is that their slopes be negative reciprocals.
8. The Angle between Two Lines. By the angle between
two lines is meant the angle formed by the positive directions from
9]
COORDINATE SYSTEMS GRAPHS
their intersection, the angle being measured counter-clockwise.
This angle is the one which lies entirely above the intersection of
the lines.
In Fig. 9, ft is the angle between
L 2 and LI, and therefore
(V)
or
P = 1 ~ 2,
tan /3 = tan (i a 2 )
tan a\ tan
1 + tan ai tan a 2
Hence
FIG. 9
Therefore the angle between two lines is the greater inclination
minus the lesser, or the tangent of the angle between two lines in terms
of their slopes is (nil ra 2 )/(l + rai??? 2 ), where^nti is the slope of
the line with the greater inclination.
9. Mid-Point of a Line-Segment. The coordinates of the
mid-point P of the line-segment joining the points PI and P 2 may
be expressed in terms of the coordinates of PI and P 2 as follows.
Draw the horizontal and vertical lines through PI, P, and
P 2 , as shown in Fig. 10. Since PiP = PP 2 , we have at once
BP = CP 2 . But
Y BP = BN + NP = - zr+ x,
and
= - x + x 2 .
Similarly, the vertical projec-
FlG - 10 tion of PiP is equal to the verti-
cal projection of PP 2 ; and, by similar reasoning, we find
_
y
+
8
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
Therefore, the coordinates of the mid-point of the line-segment
PiP 2 are
(VII)
y
EXAMPLES
1. Find the angles of the triangle whose vertices are A( 7, 6), B(2, 2),
andC(-3, -4).
SOLUTION. From the triangle ASC y we have the slope of CA given by
SA SM -f MA 4+6 _ 5
CS CN + NS 3 - 7 ~ 2 *
Similarly, or from the formula for
the slope of a line segment, we find
that the slope of BA is 8/9 and that
of CB is 2/5. Hence, CA and CB are
perpendicular, their slopes being nega-
tive reciprocals. That is, Z BCA is a
right angle. The angle between CA
and BA is equal to Z CAB] therefore
tan Z CAB =
(-8/9) - (-5/2) 1
2'
Hence Z CAB = 26 33.9'.
The remaining angle of the triangle is the supplement of the angle between
the lines BA and CB. Hence formula (VI) gives
(-8/9) - (2/5)
tan Z DBA =
__ o
(-8/9) (2/5)
Therefore
tan Z ABC == tan (180 - Z DBA) = - tan Z DBA = 2,
or
Z ABC -63 26.1'.
2. Given A (1, 4) and B(5, - 2). (a) If B is the mid-point of AC, find the
coordinates of C. (b) If AD is perpendicular to AB and AD = AB, find the
coordinates of D.
SOLUTIONS, (a) Let the coordinates of C be (xi, 2/1), then, since B is the
mid-point of AC, we have from the mid-point formulas
K xi + l n 2/i+4
whence (x\ t yi) is the point (9, 8).
Using slopes, since the slope of A B is 3/2, we see that B is 6 units below
and 4 units to the right of A. Accordingly, C is 6 units below and 4 units to
the right of B, which locates it at (9, 8).
9]
COORDINATE SYSTEMS GRAPHS
9
(ft) The slope of AD is 2/3, since it is perpendicular to AB. Then since
AD = AB, D is either 6 units to the right and 4 units above A, or 6 units to
the left and 4 units below A. These locate D at either (7, 8) or (- 5, 0).
These results may also be obtained
algebraically as follows. Let D have
the coordinates (x, y). Then, using
the slope of AD, we have
y -4 2
x - I 3*
Also,
we have
AD = AB 2yl3,
- I) 2
- 4)2
Solving these two equations simul-
taneously, we have
(1) (x - l)^ + (y - 4)2 = 52,
but from the first equation
(2) y _ 4 - | (X - 1).
\B(5 t -*)
C(* ltVt )
FIG. 12
Substituting this value for (y 4) in (1), and solving for (x 1), we get
a; 1 = =fc 6, z = 7 or - 5;
whence, by (2), y 8 or 0.
PROBLEMS
1. Given the triangle with vertices (6, 8), (- 4, - 2), (8, 4)
(a) Find the length of the sides. Ans. 6\/5, 2\/5, 10V2 units.
(b) Find the length of the medians.
(c) Find the slope and inclination of each side.
(d) Find the angles of the triangle.
Ans. (c) 1/2, 26 33.9'; - 2, 116 33.9'; 1,45.
Ans. (d) 71 33.9'; 18 26.1'; 90.
2. The same as Problem 1 for (- 6, - 1), (- 2, - 4), (4, 3).
3. Find the distance from ( 4, 6) to the mid-point of the segment joining
(7, 1) and (- 3, - 9). Ans. 2\/34 units.
4. Do the following sets of three points lie on a line? Prove your answers,
(a) (- 9, 2), (- 2, - 1), (11, - 7); (6) (3, - 1), (23, 14), (15, 8).
5. Construct a line through (4, 4) with a slope of 7/5. Do the points
(- 6, 10) and (10, - 12) lie on this line? Find the points on this line with
integral coordinates which are nearest to the point (4, 4).
6. What equation must the coordinates of P(x, y) satisfy if P is 11 units
from the point (7, 2)?
10 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
7. If the point (a, 6) is in the second quadrant, where is the point (6, a)?
What are the coordinates of the mid-point of the line joining these points, and
in what quadrants may it lie?
8. Prove analytically that the lines joining the mid-points of the sides of
any triangle are each equal to half the opposite side and parallel to it. [Take
the vertices at points (2a, 0), (26, 0), and (0, 2c).]
9. The mid-points of the sides of a given triangle are (2, 3), (4, 1), and
( 1, 2), respectively. Find the vertices.
Ana. (5, -4), (-7,0), (3,6).
10. Find the inclination of each of the following lines. Construct each line
by using only one point and its slope:
Through (a) (- 2, 0) and (5, - 3); (6) (1, 3) and (- 2, 7); (c) (4, - 5)
and (- 3, - 6).
11. One end of a line-segment 13 units long is ( 4, 8) and the ordinate of
the other end is 3. What is the abscissa of that end? Ans. 8, or 16.
12. The vertices of any quadrilateral are taken at (2a, 0), (& 26), (2c, 0)
and (2c/, 2e). Prove that the lines joining the mid-points of the' sides taken
in order form a parallelogram. What are the coordinates of the intersection
of the diagonals of the parallelogram?
13. Two of the vertices of an equilateral triangle are at (2, 2\/3) and
( 2, 2\/3). Find the coordinates of the third vertex.
Ans. (- 6, 2\/3) or (6, - 2\/3).
14. The extremities of a diagonal of a square are at ( 5, 2) and (3, 6).
Find the coordinates of the other vertices.
15. The same as Problem 14, for the points (- 6, 2) and (2, - 4).
Ans. (1, 3) or (- 5, - 5).
16. The point P(x, y) is as far from the origin as it is from the point (4, 6).
What equation must its coordinates satisfy?
17. A given line has a slope of 2/3. Find the slope of a line which makes
with the given line: (a) an angle of 45; (6) an angle of 135. Ans. (a) 5.
18. Find the slope of a line which makes an angle of 60 with a line whose
slope is 2v / 3- How many solutions are there?
10. Graphs. The locus of all points whose coordinates x
and y satisfy^a given equation is called the curve or graph of the
equation. 'if the equation is algebraic, the corresponding graph is
an algebraic curve. Other equations and curves, such as exponen-
tial and trigonometric, are called transcendental. The graph of
an equation can be approximated by plotting a series of points
whose coordinates satisfy the equation, and then drawing a smooth
curve through them.
10] COORDINATE SYSTEMS GRAPHS 11
The following procedure will facilitate the work involved:
(a) Solve the equation for y in terms of x.
(6) Make out a table of values for x and y by assigning positive
and negative values for x and calculating for each the corresponding
value or values of y.
(c) Plot the points designated by x and y as abscissa and ordi-
nate respectively. Then draw a smooth curve through the points.
If it is inconvenient or impossible to solve the equation for y, it
may be solved for x in terms of ?/, in which case arbitrary values
are to be assigned to y and corresponding values for x are calculated
to make out a table of values.
Certain information about the graph of a given equation can be
found which will enable the student to sketch the curve by plotting
only a few points rather than by making out a lengthy table of
values of x and y. The topics to be considered include intercepts,
symmetry, extent of curve, horizontal and vertical asymptotes.
INTERCEPTS. The x intercepts are the abscissas of the inter-
sections of the curve with the x axis. They are found by setting y
equal to zero in the given equation and solving the result for x.
Similarly, the y intercepts are the ordinates of the intersections of
the curve with the y axis. They are found by setting x equal to
zero in the given equation and solving the result for y.
EXAMPLE
Find the intercepts of x z - 2 y z - 4 x - 5 = 0.
SOLUTION. Setting y = in the equation we have
x 2 4 x 5 = 0, x = 5 or 1.
Similarly for x = 0, we get y = V 2.5. Hence the x intercepts are 5
and 1; the ^-intercepts are imaginary, in other words the curve does not
cross the y axis.
SYMMETRY. Two points are symmetric with respect to the
x axis when they have the same abscissa and their ordinates differ
only in sign. If they are symmetric with respect to the y axis,
they have the same ordinate and their abscissas differ only in sign.
Two points are symmetric with respect to the origin as a center
when their respective coordinates are numerically equal but of
opposite signs. *"
If a curve is symmetric with respect to the x axis, each point
on the curve has the point symmetric to it with respect to the
x axis also on the curve. Then if (x y y) represents any point on
12
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
the curve, the point (x, y) must also be on the curve. That is,
if x and y represent any pair of numbers which satisfy the equation
of the curve, that equation will be satisfied by x and y also.
We have then the following test :
// in the given equation y can be substituted for + y without
changing the equation, then the graph is symmetric with respect to
the x axis.
Obviously this condition is satisfied if y is involved in the
equation only with even powers.
Similarly, if in a given equation x can be substituted for + x
without changing the equation, then the graph of the equation is
symmetric with respect to the y axis.
If the substitution of both x for -f- x and y for -f y in a
given equation does not change the equation, then its graph is sym-
metric with respect to the origin.*/
EXTENT OF THE CURVE. In general, it is easy to determine
whether the graph of a given equation is a closed or an open curve,
and whether there is any region of the plane between two horizontal
lines, or between two vertical lines, in which the curve does not
exist. To do so, solve the equation for y in terms of x; if then this
value of y involves a square root or an even root such that all
values of x between, say, x = a and x = b make y imaginary, then
no part of the curve can be between the vertical lines x = a and
x = 6. On the other hand, if values of x between a and b are the
only ones which make y real, then the curve lies wholly between
these two vertical lines. If y is real for all values of x, the curve is
unlimited in its extent along the
x axis.
Next solve the equation for x
in terms of y and find in a similar
manner whether or not any val-
ues of y make x imaginary, and
whether or not the curve is re-
stricted or unlimited in extent in
the direction of the y axis.
FIG. 13
EXAMPLE
Consider the equation
mentioned above. Its graph is symmetric with respect to the x axis, but not
with respect to the y axis, nor the origin. Why?
,11] COORDINATE SYSTEMS GRAPHS 13
SOLUTION. Solving the equation for each variable we have
y - =fcV(l/2)(x - 5)(x + 1), x - 2
From the first of these we observe that all values of x between x = 1
and x = + 5 make y imaginary, while all values of x greater than or equal to 5
and less than or equal to 1 make y real. From the second, all values of y
give two real values of x. Also as x increases beyond 5 or decreases beyond
1, y 2 increases and the curve extends indefinitely, as indicated in Fig. 13.
PROBLEMS
Discuss and plot the graph of each of the following equations. (Nos, 1-20.)
1. x 2 = 4y. 11. 9x 2 - 4t/ 2 = 0.
2. y* + 4 x + 12 = 0. 12. x 2 + 3 y* - 15 y = 0.
3. 3 x = y* - 6. 13. XT/ = 12.
4. x 2 + 6 y - 15 = 0. 14. xy* = 12.
5. x 2 -f 2/ 2 ~ 6 x + 4 1/ = 0. 15. x 8 + 4 |/ 2 = 0.
6. x 2 4- y 2 - 6 2/ = 16. 16. y = 4 x - x 8 .
7. 4 x 2 + 2/ 2 = 24. 17. 2/ 2 - x 8 - 3 x 2 .
8. x 2 + 4 1/ 2 = 16. 18. y = x - 1/x.
9. 3 x 2 - 2/ 2 = 12. 10. y(x* + 4) = 8.
10. x 2 - 3 y* + 12 = 0. 20. y = x 2 (x - 2).
Discuss the graph of each of the following equations, and plot by giving
appropriate lengths to the literal coefficients. (Nos. 21-28.)
21. 2/ 2 = 2 px. 25. x 1 / 2 -f i/ 1 / 2 - a 1 / 2 .
22. xy = 2 a. 26. x 2 / 3 + y 2 / 3 = a 2 / 3 .
23. x 2 /a 2 -f # 2 /& 2 = 1- 27. |/ = 8 o 3 /(x 2 + 4 a 2 ).
24. x 2 /a 2 - # 2 /& 2 + 1=0. 28. 2/ 2 = ax 3 .
11. Asymptotes. An asymptote of a curve is a straight line
which the curve approaches continuously in such a way that the
distance between the line and the curve approaches zero as they are
indefinitely extended. If a curve has either horizontal or vertical
asymptotes they are easily found when the equation is solved for x
and y. Thus if y in terms of x is an algebraic fraction and any
value of x such as x a makes the denominator zero and the
numerator different from zero, then y increases indefinitely as x
approaches a and x = a is a vertical asymptote. Similarly, to
find horizontal asymptotes, solve for x. If the result is a fraction
14
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
whose denominator becomes zero for some value of y such as
y = b when the numerator differs from zero, then y = 6 is a
horizontal asymptote.
EXAMPLE
Trace the curve x*y x z 2y-}-4Q.
SOLUTION. The intercepts on the x
axis are -f 2, 2, on the y axis 2.
The graph is symmetric with respect to
the y axis. Solving for x and y, we have
x 2 - 4
-5
Fia. 14
If y is between 1 and 2, is imaginary;
hence no part of the curve is between the
lines y 1 and y = 2. The horizon-
tal asymptote is y = 1 ; the vertical are
x = \/2. This information with a few
points on each branclj give the curve as
shown in Fig. 14.
12. Intersections of Curves. From the definition of the graph
of an equation, it follows that if two curves intersect, the coordi-
nates of each point of intersection must satisfy both equations.
Hence, to find the coordinates of all points of intersection of two
curves, solve the corresponding equations simultaneously.
In solving two equations simultaneously only real values for
both x and y will give a real point of intersection, so imaginary or
complex values may be disregarded.
EXAMPLE
Find the intersections of the curves
whose equations are 9 x 2 -f- 4 y 2 37,
y = x 2 - 1.
SOLUTION. Eliminating x 2 between the
two equations, we have
4 y 2 + 9(2, + 1) = 37,
or
4 y 2 + 9 y - 28 == 0,
whence
(y + 4)(4y - 7) - 0.
Therefore
y
For y = 4, x
FIG. 15
3 and there are no intersections; for y = 7/4,
13]
COORDINATE SYSTEMS GRAPHS
15
x = Vll/4 = 1.66. Hence the intersections are (1.66, 1.75) and
( 1.66, 1.75). The solution should be checked by drawing the graphs.
PROBLEMS
Discuss and draw the asymptotes, then plot the graph of each of the follow-
ing equations. (Nos. 1-16.)
xy* - 2 y = 4.
= 3 -f 3 y.
1. xy + 2 x = 4.
2. y = 2 x + xy.
3. y = 2 x -f xy + 2.
7. xy* = 4 - x.
8. zi/ 2 -4x = 12.
17. Find the intersection of 3 x 8 y
9.
10.
11. x*y + 3 z 2 = 6.
12. x*y - 2 y = x 2 + 4.
13. z?/ 2 = x -f 2 ?/ 2 .
0. 14. (y -f 4)(z I) 2 = 1.
15. xy* + 2/ 2 - 4 x - 2 y = 0.
16. z?/ 2 - 2 ?/ 2 - 4 z = 4.
20 and 2x -7 y = 10.
Ans. (12, 2).
18. Find the intersections of (x + I) 2 -f (*/ + I) 2 = 13 with each of
the coordinate axes.
19. Find the intersection of 13 x -f 3 y = 9 and 14 x - 4 y = 35.
Find the intersections of each of the following pairs of curves and check
by drawing their graphs. (Nos. 20-26.)
20. 2 x + !/ = 1, i/ 2 -f 4 x = 17.
21. 4 x 2 4- ?/ 2 = 25, 8 x + 3 y -f 25 = 0. Ans. (- 2, - 3).
22: 2 y = 12 + x, x 2 + 4 y = 19.
23. x 2 - !/ 2 + 4 = 0, x 2 - y = 8. Ans. ( \A2 4), ( v^, - 3).
24. xy -f 8 = 0, y z = 4 x.
25. x 2 = 4 y, y = 8/(x 2 + 4). Ans. ( 2, 1).
26. x 2 + 4 y 2 + 6 x = 0, 2 x 2 - ^ 2 = 12.
13. Trigonometric Functions, Circular
Measure. If the independent variable x
is an angle, it is desirable to have a natu-
ral unit of measure to represent its mag-
nitude. This is obtained by measuring
the angle in radians. Describe an arc of
a circle with its center at the vertex of
the angle. The central angle whose inter-
cepted arc is equal in length to the radius
is one radian. Hence for any central angle
the length of the intercepted arc divided by the radius gives the
radian measure of the angle.
FIG. 16
16 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
That is, if s is the length of the arc and r the radius, then
o
= - radians.
Since the circumference is 2 wr,
180 = TT radians, 90 = ?r/2 radians,
1 = 7T/180 radians = 0.01745 radians.
1 radian 180/Tr degrees = 57 17.7'.
To plot the graph of trigonometric functions, make out a table
of values of the angle in radians and of the required function.
Thus:
X
sin x
tan x
sec x
0.00
0.00
1.00
7T/6 = 0.52
0.50
0.58
1.15
7T/3 = 1.05
0.87
1.73
2.00
7T/2 = 1.57
1.00
00
oo
27T/3 =2.09
0.87
-1.73
-2.00
5 ir/6 = 2.62
0.50
-0.58
-1.15
7T 3.14
0.00
0.00
-1.00
.."V
>^^Vr
1
2 5\.
X
FIG. 17. y = sin x
The graph of
y = sin x
will cross the x axis for each
value of x equal to an integral
multiple of TT. Moreover, since
sin ( x) = sin x = y the
graph of y = sin x is symmetric
with respect to the origin. The
same is true for y tan x. But
since sec ( x) = sec x the graph
of y = sec x is symmetric with
respect to the y axis. These curves are given below. Likewise,
asymptotes of the graphs of tan x and sec x are shown.
A function whose values are repeated after a definite interval p
of the variable, so that
14]
COORDINATE SYSTEMS GRAPHS
17
is called a periodic function. The interval p is called the period.
The maximum absolute value of the function is called the ampli-
tude. Thus, in the function sin x, the period is 2 TT and the ampli-
tude is 1. The period of tan x is p = TT and it does not have a
finite amplitude. The period of the secant is 2 ?r.
y
FIG. 18. y = tan x FIG. 19. y = sec x
If we consider the function
y = k sin nx y
it is evident that sin nx will be zero when nx is zero or any integral
multiple of TT, that is, when x is 0, ir/n, 2 ir/n, etc. Again, the
period of this function is 2 ir/n. The maximum value of sin nx is
+ 1 when nx is ir/2 or differs from w/2 by a multiple of 2 TT, that
is to say, when x is ir/2 n or differs from
7T/2 n by an integral multiple of the period
2 ir/n. Hence the amplitude of the func-
tion k sin nx is k.
14. Inverse Trigonometric Functions.
The function
y = sin" 1 #
means in direct notation
x = sin y.
The function sin" 1 x is called the inverse of
the function sin x y and is read the angle
whose sine is x, or arc sine x. Hence the
graph of y = sin" 1 x differs from that of y sin x only by having
the coordinate axes interchanged. The same is true for the other
inverse trigonometric functions.
FIG. 20. y = sin- 1 x
18
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
15. The Algebraic Sum of Functions. If the given function
consists of a sum of functions, such as
the following alternative method may be used to find the graph of
the function. Plot on the same axes the graphs of y = fi(x) and
y = / 2 (x). Along each vertical line of the coordinate paper mark
the point whose ordinate is the sum of the ordinates of fi(x) and
/ 2 (;r). The desired graph is the smooth curve through the points
located in this manner. Of course, this method can only be used
for values of x which make both/i(x) and/ 2 (a;) real.
EXAMPLE
Draw the graph of y = (3/2) cos x -f sin (2 z/3).
SOLUTION. Plot with the same axes the graph of y = (3/2) cos x and the
graph of y = sin (2 x/3). The addition of corresponding ordinates gives the
required graph, as shown in Fig. 21.
FIG. 21
PROBLEMS
Plot the graph of each of the following and find its period. (Nos. 1-9.)
1. y = cos x. 6. y = tan (x + ?r/4).
2. y = ctn x. 7. y = sin x -f cos x.
3. y esc x. 8. y = cos x sin (x/2).
4. y = 2 cos (z/2). 9. y = (1/2) ctn (7rz/4).
5. y = 2 sin (irz/3).
Plot the graph of each of the following and tell which are periodic.
10. y = (1/2) sin- 1 x. 13. x = 2 ctn~ l 2 i/.
11. y = 2 cos^ 1 2 Z. 14. y = 3 sin 2 rr.
12. 2 y = tan (x/2 - */4). 15. y = 3 sin- 1 2 S.
16. y =1 +cos(irz/2).
16]
COORDINATE SYSTEMS GRAPHS
19
17. y = x -f 2 sin x, - 2 w < x i 2 IT.
18. y = x sin x, 27r^a;^2gr.
19. z = 2 sin- 1 (i/ - 2).
20. y = c 08 "" 1 ( + !) ~ " ~ y = *"
21. i/ = 2 cos a; - cos 2 a;, = x < 2 w.
22. t/ = cos z - 2 sin (x/2), < x i 4 TT.
23. s = 7r/2 + tan- 1 (y/2) .
24. ?/ = sin (irx/2) - cos (2 vx/3), for 1 period.
25. y x cos (wx/2).
26. y = x 2 cos TTX.
16. Exponential and Logarithmic Curves. The equation
T/ = a x , in which a is any constant, is an exponential equation.
If a is positive, y is real for all real values of x and a table of values
can easily be obtained from a table of logarithms by the relation
log y = x log a. If a is negative, y is not real for all values of x
between any two given values. Thus, if x is given a fractional value
whose denominator is an even number, y would be an even root of
a negative number and hence imaginary. Hence the graph of
y = a* is a smooth unbroken curve only if a is positive. We shall
consider only such equations. Since (1/6) x = (b)~ x , if the base
is less than one, the equation can be written with the reciprocal
base. Hence we assume throughout that a is positive and greater
than unity.
The number represented by e (= 2.71828 )> the base of
natural logarithms, is of great importance in exponential equations.
We shall use the symbol log instead of log e hereafter. Thus if
y = e x , log y = x.
EXAMPLE
Draw to the same set of axes the graph of y
values 1.5, e, 4, and l/e.
SOLUTION. Form the table shown below.
a* when a has each of the
X
(1.5)*
e x
4*
e~ x
-2
0.44
0.135
0.06
7.39
-1
0.67
0.37
0.25
2.72
1.00
1.00
1.00
1.00
1
1.50
2.72
4.00
0.37
2
2.25
7.39
16.00
0.135
20
DIFFERENTIAL AND INTEGRAL CALCULUS [On. I
* 9
FIG. 22
The student should observe carefully the nature of the graph of y = a*
as a takes values that increase from a = 1. Also note that the graphs of
y = (1/4)* = 4-*, y = (l/e) x = e~*, y = (2/3)* = (1.5)-* can be obtained
from the graphs of y = 4*, y == e x , y = (1.5)* respectively by a rotation of
these graphs through 180 about the y axis. Or, we can say that y a x
and y = a-* are each symmetric to the
other with respect to the y axis; as either
can be obtained from the other by chang-
ing x to x.
MULTIPLICATION OF ORDINATES.
The graphs of certain equations
may be obtained by the multiplica-
tion of the corresponding ordinates
of auxiliary graphs. Thus if the
graphs of y = fi(x) and y = / 2 (a?)
be drawn to the same set of axes,
then the graph of y = /i(#) -fz(x) can
be obtained for all values of x which
make both /i(x) and / 2 (x) real by
multiplying the corresponding ordinates of the two auxiliary graphs.
The equation
y = ae~* sin kx
plays a very important role in physics. Its graph is known as the
curve of damped vibration and can be drawn by multiplying the
ordinates of the graphs of y = aer x and y = sin kx.
EXAMPLE
Draw an accurate graph of \
y (3/2)<r sin (irx/2). e \
SOLUTION. Draw the graph * v
of y = (3/2)e~ x and also of **
y = sin (irx/2) to the same
axes. Along each ordinate mark
the ordinate corresponding to ^
the product of the two given
ordinates and draw a smooth
curve through the points lo-
cated in this way. The dotted
curves of the figure are called
auxiliary graphs and the other
curve is the desired graph.
Note also that the two curves
y = db (3/2)e~* are boundary curves which the desired curve touches but
never crosses-
V
tf5
FIG. 23
16]
COORDINATE SYSTEMS GRAPHS
21
The logarithmic equation is called the inverse of the exponential
since y = a x may be written x = log a y. Therefore, if we write
y = loga x, where a is any positive number except unity, its graph
is the same as that of x = a v .
EXAMPLES
1. Draw the graph of y - log x and also of y = log x to the same set of
axes.
SOLUTION. Form the table shown below.
X
0.5
1
2
3
4
5
logio x
00
-0 30
0.30
0.48
0.60
0.70
log x
00
-0 69
0.69
1.10
1.39
1.61
Notice that the y axis is an asymptote of each curve.
FIG. 24
FIG. 25
2. Draw the graph of y = log Vx z 4.
SOLUTION. The equation may be written
y - (1/2) log (* - 4).
The lines x - 2 are asymptotes and the curve crosses the x axis at
x = \/5. If x* < 4, then y is the logarithm of a negative number, which is
complex. Hence no part of the curve lies between the asymptotes. A few
points with the information above give us the graph of Fig. 25.
This equation may be written
y - (1/2) [log (x + 2) + log (x - 2)].
However, the student cannot find the graph by the method of the addition
of ordinates. If we plot y = (1/2) log (x + 2) and y = (1/2) log (x - 2),
only the values of x which make both x + 2 and x 2 positive can be used
in this way. That is, log (x 2) is complex between x = 2 and x 2
and hence the sum of the ordinates is complex; then for x < 2 both
22
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
log (x -f 2) and log (x 2) are complex while the sum of these complex num-
bers is real. Hence this method would give only the part of the curve for
which x > 2. The auxiliary curves are shown in Fig. 26.
Y
\
1
7~~
1 2
FIG. 26
PROBLEMS
Plot the graph of each of the following equations.
1. y = (1/2) e 2 *.
2. y = (1/2)-.
3. y == (1/2) e*, -
4. y = *-.
5. y = e l/(l -*>.
6. y - 2'/< 1 -*>.
7. y = logic (3 - x)
15. y = (e* + e-')/2,
16. y = (e* - e~*)/2,
17. y = (1/3) 1 /*, - 5 < x ^ 5.
18. u = e"* 2 2 x 2
19. y = ze*/5, - 2 i x ^ 2.
20. y = 4 e~* sin x.
21. y = or 1 e*/ 2 , - 1 < x < 3.
22. y = e~* 2 cos x.
23. y = e-*/ 2 sin irx, -2 < x < 2.
8. y = log (3 - a;), - 3 :
9. y = log (3 -h x).
< a; ^ 3. 10. y = log V9 - a: 2 .
11. y = log Vx 2 - 9.
12. y = log sin 2 x.
13. y = log (a: 2 + 4).
14. y = log [l/(a: 2 -f 4)].
(Hyperbolic cosine.)
(Hyperbolic sine.)
24. y = e- z2 sin x, - 3 < a; < 3.
25. y = 2 X/3 cos x, ^ a: < 4.
26. y = 2*/ 2 sin 3 x, -2<x^2.
27. y = e-*/ 3 sin 2 x, - 2 < a: < TT.
28. y = e-*/ 4 sin TTX, - 3 < a; < 2.
29. y = e*/ 2 cos 2 irx, 1 < a; < 2.
30. y = - 3 e-* sin TTX, < x < 4.
31. y = 2 3-*/ 2 sin Zvx, - l<x^2.
17. Polar Coordinates. In a system of ^o/ar coordinates
the position of a point is determined by measuring a distance and a
direction instead of measuring two distances as in rectangular
17]
COORDINATE SYSTEMS GRAPHS
23
FIG. 27
coordinates. The system consists of a fixed point 0, called the
pole, and a fixed line through called the polar axis or initial line.
The polar coordinates of any point P in the plane are two
numbers r and 6; r represents the
distance OP and is called the radius
vector of P, and 6 is the angle which
OP makes with the polar axis and is
called the vectorial angle. Each co-
ordinate may be positive or negative;
6 is positive when measured from
the polar axis in a counter-clockwise
direction, negative if clockwise; r is
positive if OP is alongfr the terminal
side of 6, and negative if OP is taken
in the opposite direction, along the terminal side produced through
the pole.
Any pair of real numbers (r, 0) will determine uniquely a point
P. However, the point P may be designated by more than one
pair of coordinates. Thus the pairs (2, 3 ?r/4), (2, 5 Tr/4), and
( 2, 7T/4) each designate the same point.
To plot points in polar coordinates, the paper should be ruled
with concentric circles about the pole and radial lines through the
pole, as shown in Fig. 28.
If an equation is given in r and 0, it can frequently be solved for
r in terms of 6. Then by making out a table of corresponding
values the graph of the equation is obtained by plotting these
points and drawing a smooth curve through them. The student
should not assume that the complete graph is obtained when the
curve crosses itself. If in doubt, he should continue the table of
values until successive points are repeated.
EXAMPLES
1. Plot the graph of r = 2 3 sin 0.
SOLUTION. Giving values varying at intervals of 30, we have:
e
30
60
90
120
150
180
210
240
270
300
330
360
r
2
0.5
-0.6
-1
-0.6
5
2
3.5
4.6
5.0
4.6
3.5
2
Point
A
B
C
D
E
F
G
//
/
/
K
L
A
24
DIFFERENTIAL AND INTEGRAL CALCULUS [CH. I
60
soo*
The complete graph is given by an
interval of variation of 6 from to
360, since 6 and (0 + 2?r) give the same
value of r and the same vectorial angle.
Between 6 = sin- 1 (2/3) in the first and
second quadrants r is negative. This
curve is one form of the limagon.
2. Plot the graph of
r 2 = a 2 cos 2 6.
SOLUTION. Give 6 values at inter-
vals of 15. If B is between 45 and
270 135, cos 2 B is negative and r is imagi-
Fia. 28 nary.
e
15
30
45
135
150
165
180
r
Point
A
0.93a
B
71 a
C
71 a
D
93 a
E
a
A
This curve is called thelemniscate. It is completely given by an interval of
variation of from to 180. This is true since and (0 + TT) each give the
same value for r 2 , and also since ( r, 0) and (i r, -f- TT) locate the same two
points. (See Fig. 29.)
185<
225
FIG. 29
3. Plot the graph of r = e<*.
SOLUTION. Let a = 0.2 units and express ad in radians.
-27T
IT
-7T/2
1
D
IT/4
7T/2
37T/4
TT
37T/2
2TT
00
Point
-1.26
0.3
A
-0.63
0.5
B
-0.31
0.7
C
0.16
1.2
0.31
1.4
0.47
1.6
G
0.63
1.9
H
0.94
2 6
7
1.26
3 5
/
This curve is called the logarithmic spiral. (See Fig. 30.)
118]
COORDINATE SYSTEMS GRAPHS
25
18. Relations between the Coordinate Systems. At times it is
convenient to transform an equation given in rectangular coordi-
nates into a corresponding polar form, and conversely. The usual
arrangement is to take the pole at
the origin of the rectangular coor-
dinates and the polar axis along the
positive half of the x axis. It is
apparent that for any point P in
the plane the following relations
will then exist between its two sets
of coordinates (#, y) and (r, 8) : ,
(VIII) x = r cos 6, y = r sin 6;
(IX)
r 2 = x 2 + y\
e = tm->*
X
- COS -'VPT;
FIG. 31
EXAMPLE
Transform into rectangular coordinates the equation of the lemniscate
r 2 = a 2 cos 2 0.
SOLUTION. Write cos 2 = cos 2 sin 2 0, then the equation becomes
r 2 = a 2 (cos 2 6 sin 2 0). Therefore, if the axes are as shown above,
or (x* + y 2 ) 2 = a 2 (z 2 -
PROBLEMS
Plot the graph of each of the following equations. (Nos. 1-27.)
1. r cos = 3.
2. r sin = 2.
3. r = 4 sec 0.
4. r = 5 esc 0.
5. r + 2 sin = 0.
6. r = 4 cos 0.
7. r = 1/(1 -f cos0).
8. r - 4/(3 - 2 cos 0).
9. r = 6/(2 - 3 sin 0).
10. 2 r cos + r sin =
3.
11. r = 2(1 - sin 0).
12. r = 3(1 + cos 0).
13. r = 2 + 3 cos 0.
14. r = 3 - 2 sin 0.
15. r = 1 -f cos 2 6.
16. r = 1 - 2 sin 0.
17. r 2 = 4 cos 20.
18. r = 1 -f 2 cos (0/2).
19. r = 1 -f sin (0/2).
20. r 2 = a 2 sin 2 0.
26
DIFFERENTIAL AND INTEGRAL CALCULUS [CH. I
21. r = o/(l - e cos 0), (e < 1, = 1, > 1).
22. r = a. 25. r = a sec 2 (0/2).
23. r = a cos 2 0. 26. r = 2*.
24. r = a sin 30. 27. r = 2 cos 3 sin 0.
28. Transform into rectangular coordinates the equations of Problems 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 16.
20. Transform the following equations into polar representation.
(a) xy = 2 a, (c) z 2 + r/ 2 + 2 ay = 0, (e) * 2 (2/ 2 - a 2 ) + t/ 4 = 0,
(6) z 2 = 2 py, (d) x 2 + az + by = 0, (/) z 3 + 2/ 3 - 3 a*?/ - 0.
19. Parametric Equations. We have seen that the equation
of a curve may be given by a single relation connecting the coordi-
nates x and y of any point of the curve. Also x and y may each be
expressed in terms of a third variable, which is called a parameter,
and the two equations together are called the parametric equations
of the curve. The parameter may or may not have geometric
significance. Each value arbitrarily assigned to the parameter
fixes one or more values for x and y } giving thereby corresponding
points on the curve. If the parametric equations of a curve are
given, the equation in x and y may be obtained provided the pa-
rameter can be eliminated between the given equations.
EXAMPLES
1. Given x = at*, y = at 2 . Plot the curve and find its equation in terms of
x and y.
SOLUTION. Form the table below and plot the points. A smooth curve is
obtained by joining the points in order, as shown in Fig. 32.
t
x
y
Point
1/2
dra/8
a/4
B
dbl
a
C
db2
db8a
4a
D
Eliminating t t we have t = Vy/a; then x = db a(i//a) 3/2 or y 8 = ax*.
2. Given x a cos 0, y = 6 sin 9, obtain the graph and the equation in
rectangular coordinates.
20]
COORDINATE SYSTEMS GRAPHS
27
SOLUTION. In this example the parameter is the angle 0. The table below
gives the coordinates of points on the curve in Fig. 33.
.Y
FIG. 32
x
y
Point
a
A
30
a\/3/2
b/2
B
60
a/2
6\/3/2
C
90
D
To eliminate we have cos x/a, sin
sin 2 -|- cos 2 1 , we find the equation to be
= y/b. Using the relation
PROBLEMS
Make a table of values and plot the graph of each of the following pairs of
equations. (Nos. 1-10.)
6. x = 4 cos 0, y = 3 sin 0.
7. x = 4 sec 0, y 6 tan 0.
8. x a cos 3 0, y = a sin 3 0.
9. z = 3/(* - 1), */ = 2/(* 4- 1).
10. a; = 3 */(J 3 + 1), y = 3 * 2 /(* 8 -f- 1).
11. Derive the rectangular equations of Problems 1 to 10 inclusive by
eliminating the parameter.
Ans. (1) 4 x + y = 4; (3) xy = 6; (6) z 2 /16 -f t/ 2 /9 1; (10) x 3 + y* = 3 xy.
20. Transformation of Coordinates. It is sometimes desirable
to simplify the equation of a given locus by referring it to a new
set of coordinate axes, since the form of the equation depends on
the position of the axes with respect to the curve. This is known
1. x = 1 - t, y = 4 1.
2. x = 1 - < 2 , y = 4 t.
3. x = 2/t, y = 3 t.
4. s = l-f2*, t/ = 6/.
5. # = a sin i, y a cos tf.
28
DIFFERENTIAL AND INTEGRAL CALCULUS [On. I
as transformation of coordinates. If the new axes are parallel to
the old axes, respectively, the transformation is called transforma-
tion by translation. If the new axes have the same origin as the
old axes but are oblique to them, the transformation is called
transformation by rotation.
If the coordinates of any point P are (x, y) when referred to one
set of axes, and (x', y'} when referred to the other set, then either
set of coordinates can be expressed in terms of the other. Like-
wise, the equation of a given locus referred to one set of axes can
be transformed into the equation
of the same locus referred to the
other set of axes.
M
Y'
FIG. 34
AP =
BP =
21. Formulas of Translation.
In Fig. 34 let the x and y axes be
parallel to the x 1 and y f axes, re-
spectively. Let the coordinates of
0' be (h, k) when referred to the
former set; that is, OC = h and
CO' k. Let P be any point in
the plane with coordinates (x, y}
and (#', y'), respectively; that is,
MP = y,
NP = y'.
Now AP = AB + BP and MP = MN + NP; hence
(X) x = x' + ft, y = y' + k.
22. Formulas of Rotation. Let
the x' axis make an angle with
the x axis. Let P, any point in
the plane, have coordinates (x, y)
and (x', t/')> respectively. Then
OM = x, MP = y, ON = x', and
NP y'. Now
OM = OT - SN,
MP=TN + SP. FIG. 35
But OT x' cos 0; likewise SN = y' sin 0, TN = x' sin 6, and
SP = y' cos 0. Hence
(XI) x = x' cos 6 - y' sin 0, y = x' sin 6 -f y' cos 9.
22] COORDINATE SYSTEMS GRAPHS 29
EXAMPLES
1. Change the axes by translation so that the equation 2 x 2 -f 3 -h 7 = y
transforms into an equation without a term in x' and without a constant term.
SOLUTION. Translating the origin to (h, k), we have
2(z' + /i) 2 + 3(*' + h) + 7 = y' + k.
Collecting terms, we find
2 z' 2 -f (4 h + 3)z' + (2 /i 2 + 3 h + 7 - fc) = y'.
By the conditions of the problem we must have
Hence /& = 3/4, k = -f 47/8. That is, by translating the origin to
( 3/4, 47/8), the transformed equation is
2 z' 2 = y'.
2. Find the transformed equation for x 2 y 2 = a 2 when the axes are
rotated through 45.
SOLUTION. From formulas (XI) we have
&' 2/0, y = o^fa' + y')
Hence
or
2 x'y' + a 2 = 0.
3. By completing the squares, find the translation which will remove the
first degree terms from the equation 2x 2 3 y 2 4 x 2y = 13.
SOLUTION. Completing the squares, we have
The first degree terms will be eliminated if we make
x - 1 - x', y +
that is, if
y f -
Hence, if the origin is translated to the point h =* 1, k = 1/3, the trans-
formed equation will be
2x /2 -S?/' 2 = ^, or 6z' 2 -9y /2 ~ 44.
30 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. I
PROBLEMS
Simplify each of the following equations by translation. (Nos. 1-4.)
1. x* - & x + 4 y = 10. Ans. x' 2 + 4 y' = 0.
2. 3 y = x 2 H- 4 x - 8.
3. x 2 + 3 ?/ 2 - 4 x + 6 y = 6, to (2, - 1). 4ns. x' 2 + 3 2/' 2 = 13.
4. 3 x 2 - 2 ?/ 2 + 6 a: -f 8 y = 10, to (- 1, 2).
Simplify each of the following equations by rotation. (Nos. 5-8.)
5. xy = 12, 6 = - 7T/4. Ans. y' 2 - x' 2 = 24.
6. x 2 - y 2 + 20 = 0, 6 = 7T/4.
7. 2 x 2 -f xT/v/3 + y 2 = 7, - 7T/6. Ans. 5 x' 2 + y' 2 = 14.
8. 2 x 2 - xy V3 + 2/ 2 = 10, = 7T/3.
Simplify each of the following equations by completing the squares. (Nos.
0-11.)
9. x 2 -f y 2 4- x - T/ = 6. Ans. 2 x /2 + 2 !/' 2 = 13.
10. 2z 2 + 3s/ 2 -8x-f 6t/ = 11.
11. 6 x 2 - 2 t/ 2 - 9 x -f 8 y = 10. Ans. 48 x' 2 - 16 y' z = 43.
Apply translation and then rotation to simplify the equation. (No. 12.)
12. x 2 -f xy + !/ 2 - 2 x -f 2 y = 2, to (2, - 2) and then = 7r/4.
ADDITIONAL PROBLEMS
Draw the graph of each of the following equations. These curves are im-
portant and many are of historic interest.
1. r = a(l + 2 cos 0), Trisectrix.
2. r = a 6 sin 9, Lima$ons (if a = b, cardi&ids).
3. r = a cos 3 6, Three-leafed rose.
4. r = a sin 2 0, Four-leafed rose.
5. x 1 -f xy 2 = 2 ay 2 , Cissoid.
6. |/ = 8 a 3 /(z 2 -f 4 a 2 ), TFifc/i.
7. yJ/ajJ = (a + x)/(a - x), Strophoid.
8. x = 3 a*/(l -f * 3 ), 2/ == 3 a* 2 /(l + * 3 ), Fofcwm.
9. y = (a/2)(e* /0 + e-*/), Catenary.
10. y = a sin fcx -f 6 cos kx t Simple harmonic motion.
11. y = ae" 62 **, Probability curve.
12. r = ad, Spiral of Archimedes.
13. log r = ad, Logarithmic spiral.
CHAPTER II
EQUATIONS OF DEFINED CURVES
EMPIRICAL EQUATIONS
23. The Straight Line. A straight line is determined by two
fixed points PI and P 2 on it, or by one point PI and its slope m.
To find the equation of the line, we must obtain a relation
connecting the given values x\ } y\, m } and the coordinates of a
point P(x, y), which is assumed to be any point on the line.
Draw the line through PI
with the given slope m. We
know that the slope of a line
may be obtained from the
coordinates of the extremi-
ties of any segment of the line
as in 6. Hence, if the slope
is m, we have the relation
(1) y
= m (x -
FIG. 36
which is called the point-slope form of the equation of the line.
We may also derive equation (1) from Fig. 36 as follows:
SP SM + MP
tan a m =
PiS
+ RS '
OT
for any position of P(x, y) on the line.
Let the line be located by any two fixed points PI and Pz and
let P(x, y) be any other point on the line. Since the slope of the
segment PiP equals the slope of the segment PiP2, we have
(2)
which is called the two-point form of the equation of the line.
This may also be derived from Fig. 37.
31
32
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
It is, however, preferable to use equation (1) even when two
points are given. The right-hand member of (2) is ra, therefore
use this number m and either of the given points with the point
P(XJ y) to write the point-slope
form.
Let the given point Pi be
the point where the line crosses
the y axis. If the y intercept is
b y then the point Pi is (0, b).
Substituting these values for x\
and y\ in (1), we have
(3)
y = mx + b,
F IG . 37 called the slope-intercept form
of the equation of the line.
If the line is given by its intersections with the axes, and if the
x intercept is a and the y intercept 6, these points are (a, 0)
and (0, b) and the slope of the
line is m = OM/NO = b/(- a).
Substituting this value of m in (3),
we have
(4) jj + i-1.
which is known as the intercept
form of the equation of the line.
It is important to realize that
these are four forms of the same
equation, and from the given data
regarding the line we determine which form is the more conven-
ient to use.
24. The Linear Equation. We shall now show that the general
equation of the first degree in x and y, that is, the linear equation
(1) Ax + By + C = 0,
where A, 5, C are any constants, but A and B not both zero, always
represents a straight line.
First, assume B j 0. Solving (1) for y y we have
/o\ ... A C
(2) y = ^ x p ;
FIG. 38
24]
DEFINED CURVES
33
which is in the form of equation (3) of the preceding article, where
m = A/B and b = C/B. That is, when (1) is solved for y
the coefficient of x is then the slope of the line and the constant term
is the y intercept.
Second, assume B = 0. Then (1) reduces to x = C/A,
which is the equation of a line parallel to the y axis whose x inter-
cept is C/A. Since any line either crosses the y axis and has a
slope and y intercept, or else is parallel to the y axis, it follows that
every linear equation can be written in the form y = mx + b or
in the form x = a. Hence the equation Ax + By + C =
represents a straight line.
It is important to recognize the slope of a line from its equation.
Since m = A/B, we can say that if B ^ 0, and if the terms in x
and y are on the same side of the equality sign, the slope of the line is
the quotient of the coefficient of x by the coefficient of y with its
sign changed.
EXAMPLES
1. Find the equation of the line through the point ( 3, 2) which makes
an angle of 120 with the x axis.
SOLUTIONS, (a) The slope of the line is m = tan 120 = V3. If
P(x, y) is any point on the line, we have from the point-slope form of the
equation of the line
- y - (- 2)
- =
xVZ -f y + 2 4- 3\/3 = 0.
(b) Again, from Fig. 39 we have
tan 120 = - \/3 = TP/P,T, or
TM 4- MP
- V3
- V3
+ NT'
y
+ y + 2 -f 3\/3 = 0.
FIG. 39
2. Find the equation of the line through the points (2, 3) and ( 3, 1).
SOLUTIONS, (a) The slope of the line ism = [3 - (- l)]/[2 - (- 3)] =4/5.
Then, from the point-slope form, if we use the point (2, 3),
x -2
34 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
Y Simplifying, we have
+ 4x 5 y -f 7 = 0.
(6) By line-segments, from Fig. 40,
TP, _ NP
Pi*
TR + RP-
'NS + SP
+ MT ''
1 +3
FIG. 41
FlG - 40 4 _ *; 4. 7 - n
3. Find the equation of the line crossing the y axis 4 units above the
origin and making with the y axis an angle of 60.
SOLUTION. From Fig. 41, we see
that the inclination of the line is 150
or 30. Hence m = tan 150 = - A/3/3,
or m = tan 30 = V3/3. From either
the slope-intercept form or the sides of
the triangles, we have the equations of
the lines as
x db 2/V3 =F 4V3 = 0.
4. Find the equation of the line
through ( 2, 3) which is perpendicu-
lar to the line 2 x 3y 3=0.
SOLUTION. The slope of the given line is m = - [2/(- 3)] = 2/3. Hence
the slope of the required line is 3/2. The equation of the line by either
method outlined in Example 1 is
3 x + 2 y = 0.
25. Parallel and Perpendicular Lines. Given the two lines,
LI with equation A\x + B\y + Ci = 0, and L 2 with equation
AIX + B 2 y + C 2 = 0. Call the slopes of these two lines mi
and m 2 , respectively. It follows then that mi = Ai/Bi and
w 2 = A 2 /B 2 , where B\ and B 2 are both different from zero.
The condition that L\ and L 2 be parallel is (7) that the slopes
be equal. Hence mi = ra 2 , that is, A\/B\ = A 2 /# 2 , or
\*"J ~A~ IT" *
A 2 02
Therefore, if two lines are parallel the coefficients of x and y in the
equations of the lines are proportional, and conversely.
26] DEFINED CURVES 35
If LI and 1/2 are perpendicular, raiW 2 = 1 (7). Therefore
(-4i/fii)(-A,/ft) = - l,or
(2) A^t + B,B 2 = 0.
Conversely, if this relation exists, the lines LI and L 2 are
perpendicular.
26. Special Cases. If, in the equation Ax + By + C = 0,
we have A = 0, the equation becomes y = C/B, which is in
the form y b and is a line parallel to the x axis. If B = 0,
the line is x = C/A, parallel to the y axis. If C = 0, the line
is Ax + By = or y = ( A/B)x. This equation is satisfied
by the point (0, 0); hence the line passes through the origin.
The converse is also true; that is, if the line passes through the
origin, the constant term in its equation is zero.
PROBLEMS
Find the equation of each of the following lines. (Nos. 1-15.)
1. Through (- 3, 4) and (4, - 5). Ans. 9 x + 7 y = 1.
2. Through (2, - 4) with slope - 6/5.
3. With x intercept 4 and inclination of Tr/4. Ans. x y -j- 4 = 0.
4. With y intercept 3 and slope 2/3.
5. Through ( 2, 4) and parallel to the line joining the origin to (3, 2).
Ans. 2 x - 3 y = 8.
6. Through ( 3, 5) and perpendicular to line joining ( 3, 5) and
(-5,4).
7. With x and y intercepts 3/7 and 5/2, respectively.
Ans. 35 x - 6 y + 15 = 0.
8. Through (5, - 3) with ot = 135.
9. Through (- 4, 3) with a = 120. Ans. V3 x + y + 4V3 -3 = 0.
10. Through (6, - 1) with an inclination of 150.
11. Perpendicular to the line 3 x 4 y = 5 and through ( 5, 6).
Ans. 4z + 32/-f-2 = 0.
12. Parallel to the line 7 x -f 6 y = 12 and through ( 7, 3).
13. Through (-2, -3) and making an angle of 45 with the line
y = 7 x + 11. Ans. 3 x - 4 y = 6, and 4 x + 3 y + 17 = 0.
14. Through (5, 3) and the intersection of the lines 3 x 4 y = 7 and
2x + y =1.
15. Through ( 3, 4) and the mid-point of the segment joining (4, 2) and
(- 1, -7). Ans.
36 DIFFERENTIAL AND INTEGRAL CALCULUS [CH. II
16. Find the equation of the perpendicular bisector of the segment joining
(-2, l)and (-6, -5).
17. A diagonal of a square has its extremities at ( 1, 1) and (3, 5);
find the equations of its sides.
Ans. x - 5y = 4, 5z + ?/ + 6=0, x - 5y +22 = Q, 5 a; + y = 20.
18. The vertex of the right angle of an isosceles right triangle is at (3, 1)
and another vertex is at (5, 4). Find the equations of the three sides.
19. The sides of a square have equations x + 3y 10, x -{-By = 20,
3z i/ + 5 = 0, and 3 x y = 5. Find the equations of the diagonals and
find their point of intersection.
f O~_y*<,,_L_1K A
(3/2, 9/2).
x + 2 y = 15,
20. The vertices of a triangle are at (3, 3), (- 1, 5), and (6, 0). Find
the equation of the line through each vertex parallel to the opposite side.
21. Find the perpendicular distance between the parallel lines 3 x = y 2
and 6 x - 2 y - 1 = 0. Ans. VTU/4 units.
22. Find the length of the perpendicular from the intersection of the two
lines x - 6 y = 1 and 10 y - x + 5 = to the line z + 2 ?/ + 1 =0.
23. Find the equations of the locus of points equidistant from the lines
x = a and y - b. Ans. x y = a 6, x + y = a -{- b.
24. Given the triangle A(- 4, - 2), 5(6, 4), C(- 2, 8).
(a) Find the equations of its sides.
(b) Find the equations of the medians.
(c) Find the equations of the altitudes.
(d) Find the equations of the perpendicular bisectors of the sides.
(e) Find the intersection of the medians (centroid).
(f) Find the intersection of the altitudes (orthocenter) .
(g) Find the intersection of the perpendicular bisectors (circumcenter).
(h) Prove that the points (e), (/), and (g) are collinear (Eider line).
27. The Circle. A circle is the locus of a point at a given distance
from a fixed point.
Let (/i, fc) be the fixed point called the center, and let r be the
given distance which is the radius. From the definition any
point P(x, y) on the locus must satisfy the relation CP = r, where
C is the center (h, k). From the right triangle in Fig. 42, we have
(CN + NRY + (RM +
This relation, which may be obtained from the distance formula,
gives the equation of the circle as
27]
DEFINED CURVES
37
(1) (x - ft) 2 + (y - fe) 2 = r\
If we expand this equation and collect terms, we have
x 2 + y 2 - 2 te - 2 % + h 2 + & 2 - r 2 = 0,
which is of the form - ,
(2) x 2 + y 2 + Dx + Ey + F = 0. v
These are the two standard forms of the equation of a circle,
and since (h, k) may be any point, and r any distance, this equation
may represent any circle in the plane. Since any circle can be
written in either of these forms,
it follows that an equation of the
second degree represents a circle
only when the coefficients of x 2 and
y 2 are the same and the equation
has no term in xy.
We note the following special
positions of the circle with respect
to the coordinate axes :
If the equation has no term
in x, then D = 2 h = 0, or
h = 0, which means that the
center lies on the y axis. Similarly, if there is no term in y, then
k = and the center is on the x axis. If the terms in x and y
are both missing, the equation reduces to
(3)
N
FIG. 42
and the center is the origin. Conversely, if the center is at the
origin the equation of the circle will be in form (3).
If F = 0, then /i 2 + fc 2 = r 2 , and the circle will pass through the
origin.
If the equation is given in form' (2), it can be reduced to form (1)
by completing the squares in x and y.
Since equations (1) and (2) each have three constants, we see
that three conditions, which may be expressed as algebraic rela-
tions connecting these constants, are necessary and sufficient to
determine the equation of the circle.
EXAMPLES '
1. Find the center and radius of the circle whose equation is
Gy = 23.
38
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
SOLUTION. Divide each member of the equation by 2 and complete the
squares. Then
23 .25 9
or
(,-9-
Hence the center is (5/2, 3/2) and the radius is V20 units.
2. Find the equation of the circle through the points (0, 0), ( 1, 3),
and (5, - 3).
SOLUTION. Since nothing is known of the center or radius, we shall use the
form (2) of the equation of the circle. Substituting the coordinates of each
point in that equation successively, we have
[34 -h 5 D - 3 E + F = 0.
Solving these, we find that D 11 and E = 7. Hence the desired
equation is
x 2 -f */ 2 - 11 a; - 7y = 0.
3. Find the equation of the circle through the point ( 6, 2) which is tan-
gent to the y axis and has its center on the line x + y 1.
SOLUTION. Since the circle passes through (6, 2), it must lie to the left
of the y axis; hence h = r. Then its equation has the form
(x -{- r) 2 + (y - fc) 2 = r 2 .
Using the point ( 6, 2), we may
write
(-6 +r) 2 -f (2 - fc) 2 = r 2 .
Also, since the center is on the
given line, we have
h + k -f 1 = 0,
-15
-10 -5"
From these relations we have
A: 2 - 16 k -f 28 = 0,
X
FlG - 43 whence
k = 2, 14; ft = - 3, - 15; r = 3, 15.
Hence there are two circles satisfying the given conditions; and they are
\x + 3) 2 -f (y - 2) 2 = 9, and (x + 15) 2 + (y - 14) 2 = 225.
4. Prove that an angle inscribed in a semicircle is a right angle.
(28]
DEFINED CURVES
39
SOLUTION. Take the origin at the center and the x axis along the given
diameter LM. Let A (a, b) be any
point on the circumference with mi jy
and m-i 'the slopes of AL and AM
respectively. Then mi = b/(a 4- r),
m 2 = b/(a r); whence we have
rmmz = 6 2 /(a 2 r 2 ). But we must
use the fact that A is on the circle,
which gives a 2 -f- 6 2 = r 2 . Sub-
stituting - 6 2 for a 2 - r 2 we get \L(-r,0) \M(r.O)
mim 2 = 1. Hence AL and AM
are perpendicular (7).
Q(r,e)
P(r,0)
FIG. 44
28. The Line and the Circle in
Polar Coordinates. The equation
of any line perpendicular to the
polar axis and at a distance a from
the pole is
(1)
r cos 8 = a.
FIG. 45
If the line is parallel to the polar
axis and at a distance b from it, the equation is
(2) r sin 6 = b.
These equations are the transformations from rectangular to
polar coordinates of x = a and
y = 6 [ 18, (VIII)], where the
pole is the origin and the polar
axis is the x axis.
If the center of a circle of ra-
dius a is at the pole, its equation
is obviously
(3) r = a.
If the center is on the polar axis
with the pole at one extremity
of a diameter, then any point
P(r, 6) on the circle will satisfy the relation
f(r,e)
\M(2a,0)
FIG. 46
(4)
r = 2 a cos 0.
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
Similarly, if the center is on the 90 vector with the pole on the
circle we have
(5) r = 2 a sin 6.
If the pole is on the circle and the intercepts on the polar axis
and 90 vector are c and d re-
spectively, the polar equation is
readily found by transforming the
corresponding rectangular equa-
tion x 2 + y 2 ex dy = 0. Thus
I " \ I (Fig. 47)
(6) r = c cos 9 + d sin 6.
29. The Line and the Circle in
Parametric Form. If x and y are
FIG. 47 each given in terms of a third
variable t and if each is of the first degree in the variable, the rec-
tangular relation connecting x and y is linear. Thus if x = at + b
and y = ct + d, solving each for t and equating the results, we
have
ex ay = be ad.
Each value of t assigns one value to x and one to y. These values
are the coordinates of a point on the
line corresponding to t.
Given a circle whose center is the
origin, draw the radius to any point
P(x, y) on the circumference. De-
noting by B the angle which the
radius OP makes with the x axis, we
have (Fig. 48)
(1) jc= rcos6, y = rsinO,
as parametric equations of the circle.
The elimination of 0, as explained
. .
in 19, gives the equation
29] DEFINED CURVES 41
PROBLEMS
Find the center and radius of each of the following circles. (Nos. 1-6.)
1. z 2 -f 2/ 2 - 10 y = 0. Ans. (0, 5), 5 units.
2. x 2 -f 2/ 2 + 6 x - 4 y = 7.
3. x 2 + 2/2 - 16 x + 12 y -f 1 = 0. Ans. (8, - 6), 3vTI units.
4. x z + 2/ 2 - 5 x + 4 y = 2.
5. 4 z 2 + 4 z/ 2 - 16 z + 24 y + 51 = 0. Ans. (2, - 3), 1/2 unit.
6. 36 x 2 -f 36 t/ 2 - 180 x + 48 ^ + 97 = 0.
Find the polar equation of the circle in the following two problems. (Nos. 7-8.)
7. With center at (2, ir/2) and radius 2 units. Ans. r = 4 sin 0.
8. Passing through the origin with center at (4, 0).
Find the equation of the circle satisfying each of the following sets of con-
ditions and draw its graph. (Nos. 9-16.)
9. With center at ( 3, 4) and passing through (1, 1).
Ans. x z + 2/ 2 -f- 6 x - 8 y = 0.
10. With center at ( 2, 4) and passing through the origin.
11. With center on the line 2x 3^ + 2=0 and passing through the
points (4, - 2) and (11, 3). Ans. x* + y* - 10 x - 8 y + 4 = 0.
12. Tangent to the x axis, with one extremity of a horizontal diameter at
the point (2, - 2). (Two cases.)
13. Concentric with the circle x 2 + y 2 5x -f- 4t/ = 1, and passing
through ( - 2, 4). Ans. x* + 2/ 2 - 5 x + 4 y = 46.
14. Through the points (0, 3), (- 3, 0) and (0, 0).
15. Through the points (1, 0), (2, 0) and (0, 3).
Ans. 3x 2 + 32/ 2 -9x~ll2/ + 6=0.
16. Which has the segment of2z 3?/4-6=0 cut off by the reference
lines~as a diameter.
17. Find the locus of the vertex of a right angle if its sides pass through the
points (1,1) and (1, 7), respectively. Ans. x 2 + y 2 2x 82/4-8 = 0.
18. Find the locus of points four times as far from ( 2, 1) as from (1, 3).
19. Transform into polar coordinates the equation of the circle through the
origin with center (4, 2). Ans. r -f 8 cos 6 4 sin = 0.
20. Find the locus of a point if the square of its distance from the point
(1, 3) is six times its distance from the y axis.
21. Find the points at which the y axis cuts the circle having the segment
from (2, 1) to (- 4, - 3) as a diameter. Ans. (0, - 1 d= 2\/3).
22. Find the circles passing through ( 3, 4) and tangent to both axes.
42
DIFFERENTIAL AND INTEGRAL CALCULUS [On. II
23. Find the circle with center at (5, 1) and tangent to the line
4 x + 3 y = 3. Am. x 2 + y 2 + 10 x - 2 y + 10 = 0.
24. Find the circle through the origin and tangent to the line 5 x 2 y = 16
at the point (2, 3).
25. Chords of the circle r + 10 cos 6 =* are drawn through the pole.
Find the locus of their mid-points. Ans. r -h 5 cos 0.
26. Chords of the circle r = 6 sin & are drawn through the pole and each
produced its own length. Find the locus of the extremities.
27. A chord of the circle x 2 -\- y 2 -f 2 x ~f 4 y = 44 has its mid-point at
(3, 1). Find the equation of the chord and its length.
Ans. 2 x - 3 y -f 9 = 0; 12 units.
28. The equations of two circles are x 2 -f y 2 + A -f- #i2/ + FI = Q and
2 _j_ 2/2 _|_ ) 2X _j_ E 2 y _j_ /? 2 _ o. Find the relation connecting these coeffi-
cients if the circles are (o) equal; (6) concentric; (c) tangent to each other.
30. The Parabola. A parabola is the locus of points equi-
distant from a fixed point and a fixed line. The fixed point is called
the focus and the fixed line the
directrix.
Let F be the focus and D'D the
directrix, with p the distance be-
tween the focus and the directrix.
Through F draw AF perpendicular
*X to the directrix.
The x and y axes may be taken in
any convenient position with respect
to the focus and directrix, and the
corresponding equation of the curve
obtained. However, the simplest
and standard form of the equation
is found by taking the origin midway between the focus and di-
rectrix with the line AF as one of the axes, say the x axis. If
P(x, y) is any point on the parabola, from the definition we have
EP = FP. But
A(-f,0)
D'
N
[P(x,y)
I
M F(
FIG. 49
EP = EN + NP = + x,
and
FP =
MP 2 =\](OF - OAf) 2 +
31] DEFINED CURVES 43
Hence
VHF-i-
Squaring and collecting terms, we get the form
Since we see that the curve is symmetric to the x axis, the line
AF is called the axis of the parabola. The intersection of the
curve and its axis is the vertex. The curve is open and extends
indefinitely to the right of the y axis.
Any chord of the parabola through F is called a focal chord.
The focal chord perpendicular to the axis of the parabola is called
the latus rectum; its length is 2 p.
Equation (1) may be written in the form
(2) x = ky\
where k is any constant not zero. The graph extends to the right
or to the left from the origin according as k is positive or negative.
In a similar manner the equation
(3) y = kx\
where k is any constant not zero, is a parabola with vertex at the
origin and with its axis along the y axis. It is concave upward or
concave downward according as k is positive or negative.
31. Other Forms of the Equation of the Parabola. Two
important equations of the parabola are
and
(2) x = ay 2 + by + c.
The former equation defines a parabola with its axis parallel to
the y axis. For, on completing the square, we have
_ |
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
or
From this equation, we see that a translation of the origin to
the point ( 6/2 a, c 6 2 /4 a) will reduce equation (1) to the
form (see 21)
(3) y f = ax' 2 , I
Likewise equation (2) represents a parabola Avith axis parallel
to the x axis. '
If we substitute for x and y in either equation (1) or (2) the
coordinates of any three distinct points not on the same straight
line, we obtain three corresponding relations connecting the co-
efficients a, 6, and c. The solution of these relations will deter-
mine a unique set of values for a, 6, and c. Hence :
Through any three distinct points not on the same straight line,
one and only one parabola can be drawn with a vertical axis, and one
and only one parabola can be drawn with a horizontal axis.
32. Construction of the Parabola. When the focus and direc-
trix are given, points on the parab-
ola may be found readily by means
of a ruler and compasses. Draw
any line parallel to the directrix
and on the same side as the focus.
Let its distance from the directrix
be a. With F as a center and a
radius a, mark the points on the
given line at a distance a from F.
These are points on the parabola.
If the directrix is taken parallel to
either set of ruled lines of the co-
ordinate paper, points on the curve
FIG. 50 can be marked rapidly with the
compasses, and an accurate sketch of the curve can be drawn.
33. Parabolic Segment A segment cut from a parabola by a
chord perpendicular to the axis is known as a parabolic segment.
If we take the curve y = kx* and draw through two points PI
and P 2 the chords perpendicular to the axis (Fig. 51), we have
D
Pe
P,
PS
f 3
R
r>
tf
a
7
\a
F
r
/i
p*:
P'
D
P*'
P*
<1
p;
p*
2/2 =
33]
DEFINED CURVES
whence
or
2/2
(2
But 2 #i and 2 z 2 are the lengths of the two chords and yi and y*
are their respective distances from the vertex. Hence without
reference to the coordinate
axes we have the theorem:
In any parabola the squares
of any two chords which are
perpendicular to its axis are
to each other as their distances
from the vertex of the parabola.
In the segment QiOPi, QiP l
is called the base and ONi the
altitude of the segment. In
practical problems the base of
the parabolic arch is usually
called the span.
EXAMPLES
1. Find the equation of the parabola with y = 6 as directrix and ( 4, 2)
as focus. Find the vertex and extremities of the latus rectum.
SOLUTION. Assume P(x, y) any
the parabola. Draw PF
and draw PD perpendicular to the
directrix. From the definition, we have
But
PD = PS + SD = -2/4-6,
D
Y
point
and d
directr
PD =
P
and
FP =
Hence
(x-
or
r
P<T r
L
V
/\M F(~4,2)
i! "
f ' ' 's ' o
FIG. 52
\
FP = V(FL + LMY + (MS + SP)*
(-2
(y -
The vertex is evidently (4, 4) and the ends of the latus rectum are
(-8, 2) and (0,2).
46
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
2. In the parabolic segment shown in Fig. 53, find x.
SOLUTION. It is evident that
the two chords are, respectively,
15 + x and 15 x feet. Hence
(15 +
(15 -
10
4 '
or
15 + x
15 - x ''
VTo
2
15'
FIG. 53
Solving for x, we find
s = 5(7 - 2VlO) = 3.38 feet.
PROBLEMS
Ans. (- 9/4, 0), 4 3 = 9, (- 9/4, =fc 9/2).
Ans. (0, - 9/16), 16 y = 9, (=fc 9/8, - 9/16).
Find the focus, directrix, and extremities of the latus rectum of the follow-
ing parabolas. (Nos. 1-8.)
1. x 2 - 2 y = 0. Am. (0, 1/2), 2 y + 1 = 0, (dbl, 1/2).
2. x* + 12 y = 0.
3. y 2 + 9 x = 0.
4. y 2 - 6 a? = 0.
5. 4 x* + 9 y = 0.
6. 5 t/ 2 + 16 z = 0.
7. z 2 + 2 py = 0.
8. 2/ 2 = kx.
Derive the equation of the parabola which has the following given parts.
(Nos. 9-16.)
9. Vertex at the origin, focus at (0, - 5/2). Ans. x* + 10 y = 0.
10. Focus at (- 9/4, 0), directrix x = 9/4.
11. Vertex at (5, 1) and directrix y = - 3.
Ans. x 2 - 10 x - 16 y + 41 = 0.
12. Vertex at (2, - 1) and focus at (- 1, - 1).
13. Focus on the x axis, vertex at the origin, passes through (4, 3).
Ans. 4 2/ 2 =*= 9 x.
14. Vertex at the origin, directrix parallel to the x axis, passes through
the point (3, 1).
15. Focus at ( 2, 4), one extremity of the latus rectum at (3, 4).
Ans. x z + 4 x + 10 y = 61, or x 2 + 4 x - 10 y + 19 = 0.
16. Extremities of the latus rectum at (1, 6) and (1, 4). (Two cases.)
17. Find the locus of a point whose distance from (2, 3) is two units more
than its distance from the line x + 5 = 0. Ans. y 2 6 y 18 x = 36.
33]
DEFINED CURVES
47
18. Find the locus of a point whose distance from the point ( 2, 8) is
three units less than its distance from the x axis.
19. Find x in the parabolic segment, Fig. 54. Ans. 2.252 ft.
20. Find AC in the parabolic segment, Fig. 55.
21. Find h in the parabolic segment, Fig. 56. Ans. 10.8 ft.
100'
30'
180'
200'
FIG. 56
22. A parabolic arch of height h spans a stream of width w. (a) What part
of the width of the stream has a clearance of at least (3/4) hi (b) What part
has a clearance less than (h/2)?
(c) What do these results be-
come if h = 24 feet and w = 150
feet?
23. Find the height of a para-
bolic segment if A O = OB = 60',
CB = 10', and CD = 12'. (See
FIG. 57
C B
Fig. 57). Ans. 39 ? T ft.
24. Find the height of a parabolic arch if its base is 24 feet and if there is a
clearance of 10 feet at a distance of 4 feet from the end of the base line.
25. Find the equation of the parabola with a vertical axis which passes
through the points (- 5, 1), (3, - 1), and (0, - 4).
Ans. 4 y = x 2 + x - 16.
26. Find the equation of the parabola with a horizontal axis which passes
through the three points in Problem 25.
27. Find the equation of the parabola with a horizontal axis which passes
through the points (- 3, 0), (1, 4), and (6, - 4).
Ans. 32 x = 13 t/ 2 - 20 y - 96.
28. Find the equation of the parabola with a vertical axis which passes
through the three points in Problem 27.
Find the vertex of each of the following parabolas. Translate the origin to
the vertex and find the transformed equation.
29. y* - 8 * -f 6 y + 17 = 0. Ans. (1, - 3); y'* = 8 x'.
30. 2 x = 8 + 4 y - y 2 .
31. y = 6 - 2 x ~ x*/3. Ans. (- 3, 9); x' z + 3 y' = 0.
32. y = 2 x 2 - 5 x - 3.
48
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
34. The Ellipse. An ellipse is the locus of a point the sum of
whose distances from two fixed points is a constant.
Let F' and F be the two fixed points at a distance of 2 c from each
other. These points are called the
Y foci. Let the sum of the distances
of any point P (x, y) on the ellipse
from F' and F be 2 a. Take 0,
the mid-point of F'F, as the origin
and F'F along the x axis. The
coordinates of the foci are then
X (db c, 0). Any point P on the
ellipse must satisfy the definition
F'P + FP = 2 a.
Then, using the distance formula,
FlG. 58 r TV ro u
or from Ing. 58, we have
V(x + c)* + y* + V(z-c) 2 + 7/ 2 = 2 a.
Transposing one radical, say the second, squaring both sides, and
collecting terms, we find
4 ex = 4 a 2 4 av(x
Solving for the radical, we have
y
V(x -
= a
ex
Squaring again, we obtain, after collecting terms,
or
(a 2 - c 2 ) x 2
+-
2 r*
a 2 - c 2
1.
From Fig. 58 it is obvious that 2a>2cora>c; then if we
set a 2 c 2 = b 2 it follows that 6 2 is positive and hence b is real for
every ellipse. The equation takes the form
If*
~
This equation shows that the ellipse is symmetric with respect
35]
DEFINED CURVES
49
to both axes and to the origin. The x intercepts are db a, the
y intercepts b. Solving for y and x in turn, we get
Fio. 59
These forms show that no real point of the curve can have an
abscissa numerically greater than
a, nor an ordinate numerically
greater than b. The ellipse is a
closed curve and is inscribed in the
rectangle whose sides are x = dz a,
y = db b. The segment V'V is
called the major axis, its extremi-
ties V and V being the vertices.
The segment B 'B is the minor
axis. The length of the major axis
is 2 a, and the length of the minor
axis is 2 6.
The chord through either focus perpendicular to the major axis
is called a latus rectum. To find its length, set x = db c in equation
(1), then y = =t b 2 /a. Hence the length of the latus rectum is
2 b 2 /a.
If the foci are on the y axis, the equation of the ellipse is
< 2 > +='
35. Limiting Forms. Eccentricity. If c = 0, since we have
a 2 c 2 = 6 2 , it follows that a = 6. That is, if F' and F are made
to coincide at the center, the major and minor axes become equal,
and the equation becomes
i2 I ,,/2 - /v2
*/ ^ 1^ vv
Hence the circle is the limiting form of the ellipse as the foci
approach coincidence.
On the other hand, if F' and F approach V and V, respectively,
when they coincide c = a and 6 = 0. Hence, as c approaches
a and 6 approaches 0, the ellipse flattens and approaches as a
limiting form the line-segment V'V.
The shape of the ellipse depends on the relative lengths of
50
DIFFERENTIAL AND INTEGRAL CALCULUS [On. II
c and a. The ratio c/a is called the eccentricity of the ellipse. It
is denoted by e and may have any value between and 1, these
two extremes being the eccentricities for the limiting forms of the
circle and the line-segment, respectively.
36. More General Forms of the Equation of an Ellipse. It is
important to recognize that the equation
(1) Ax* + By* = C -
represents an ellipse provided A, B, and C have the same sign.
Writing this in the form
CIA ^ C/B '
we see that the major axis is along the x axis or the y axis accord-
ing as A < B or A > B.
Completing the squares, we may write the equation
(2) Ax 2 + By* + Dx + Ey + F = 0,
where A and B have like signs, in the form
where C" = D 2 /(4 A) + E 2 /(4 B) - F. Hence, translating the
origin to the center (- D/2 A, E/2 B), we obtain the trans-
formed equation
Ax' 2 + By' 2 = C",
which is like equation (1) if C' has the same sign as A and B.
EXAMPLES
1. Given the ellipse 25 x 2 -f 9 y 2 = 196. Find
the coordinates of the vertices and foci, and find
the eccentricity.
SOLUTION. Dividing both sides by 196, we
have
/14V T /14V
\JJ W
FIG. 60
Hence the major axis is along the y axis, and
a = 14/3, b = 14/5. Then c = 56/15. The ver-
tices are (0, it 14/3); the foci are (0, d= 56/15);
and e = c/a = 4/5.
2. Derive the equation of the ellipse if its foci are at ( 2, 3) and (6, 3)
with one vertex at ( 4, 3).
36]
DEFINED CURVES
51
SOLUTION. The center of the ellipse is the mid-point between the foci or
(2, 3). Hence the semi-major axis a is 6, so the other vertex is (8, 3).
If a = 6, and c = 4, then b = 2 VS. The extremities of the minor axis
are (2, 3 -f 2 V5) and (2, 3 - 2 V). By the definition of the ellipse, we see
that the sum of the distances from any
point P(x, y) on the curve to (2, 3)
and (6, 3) is 2 a = 12. Hence L we
find FT + FP = 12. That is,
V(x + 2)2 + (y - 3)2 +
V(x - 6) 2
- 3) 2 = 12.
FIG. 61
Rationalizing, we have
5 x* -f 9 y 2 - 20 x - 54 y - 79 = 0.
3. Change the equation of Example
2 to the form of (1) by completing
squares.
SOLUTION. Write the equation in the form
5(s - 2) 2 + Q(y - 3) 2 = 79 -f 20 + 81 = 180.
Then a translation of the origin to the center (2, 3) gives
5 x' 2 + 9 y" 2 - = 180.
PROBLEMS
Find the coordinates of the vertices and foci, the eccentricity, the lengths
of the major and minor axes, and sketch each of the ellipses whose equations
are given below. (Nos. 1-6.)
1. 4 x* + 2/ 2 = 12.
2. x* + 4 y* = 16.
3. 2 x 2 + 5 y 2 = 40.
4. 7 x 2 + 5 y 2 = 70.
5. 4 x 2 + 9 ?/ 2 = 72.
6. 8 z 2 + 3 y 2 = 48.
Find the remaining values a, 6, c, e; then write the equation of each ellipse
when the given parts are as follows. (Nos. 7-10.)
7. Foci at ( 3, 0), e = 1/2. Ans. a = 6, b = 3\/3, z 2 /36 + 2/ 2 /27 =1.
8. Center at the origin, minor axis 4 units along the y axis, semi-major
axis V5 units.
Ans. (0, db 2V3), (0, 3), V3/2, 4V3, 2V3.
Ans. ( 2V5, 0), (=h 2\/3, 0), V3A 4\/5, 4\/2.
Ans. ( 3V2, 0), ( VTO, 0), VS/3, 6\/2, 4\/2.
9. Center at the origin, extremities of a latus rectum ( 10/3, 4).
Ans. a = 6, b - 2V5, c = 4, e = 2/3, x 2 /20 + y 2 /36
1.
52 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
10. Extremities of minor axes ( 8, 0), e = 3/5.
From the definition of an ellipse derive its equation if it has the following
parts. (Nos. 11-15.)
11. Foci at (- 2, 3) and (6, 3), e = 2/3.
Ans. 5(x - 2) 2 + 9Q/ - 3) 2 = 180.
12. Vertices at (- 1, 8) and (-1, - 4), e = 2/3.
13. Extremities of the minor axis ( 1,0) and ( 1, 4), one focus
(2, - 2). Ans. 4(s + I) 2 + 13(y + 2) 2 = 52.
14. One vertex at (3, 4), nearer focus at (1, 4), e = 1/2.
15. Center at ( 2, 3), one focus at ( 2, 1), the major axis is twice as
long as the minor axis. Ans. 12 (x + 2) 2 -f- 3(y 3) 2 = 64.
16. Find the equations of the two ellipses with axes parallel to the coordi-
nate axes, respectively, which have the foci and the extremities of the minor
axes on the circle x z + y 2 - 2 x + 4 y - 20 = 0.
Locate the center, and translate the origin to the center. Find the trans-
formed equation and sketch. (Nos. 17-21.)
17. 4 x 2 + y z = 8 x - 4 y. Ans. (1, - 2), 4 x'* + 2/' 2 - 8.
18. z 2 + 4 t/ 2 - 2 x - 16 y + 1 = 0.
19. 2 z 2 + 3 y 2 -f- 8 x - 6 y - 1. Ans. (- 2, 1), 2 x' 2 -f- 3 y'* = 12.
20. 7 x 2 + 2 y 2 + 7 x + 6 y + 25/4 = 0.
21. x = 3 + 2 sin 0, */ = 1 + 3 cos 0. Ans. (3, 1), 9 x' 2 + 4 ?/' 2 = 36.
22. A line of constant length a -f 6 units has its extremities on the coordi-
nate axes. Find the locus of a point P on this line which is a units from one
extremity and b units from the other.
23. The axes of an ellipse coincide with the coordinate axes. Find its
equation if it contains the points (2, 4) and (6, 2).
Ans. 3 x* + 8 y 2 = 140.
24. The same as Problem 23, if it contains the points ( 3, 5) and (4, 1).
25. A semi-elliptic arch spans a roadway 150 feet wide. If the center of
the arch is 30 feet above the road, what width of the road will have a clearance
of at least 20 feet? Ans. 50\/5 = 111.8 ft.
26. A semi-elliptic arch is to be built over a four-lane highway. It is
required that the arch shall be at least 20 feet above the two central sections,
each of which is 15 feet wide, and at least 15 feet above the two outside sec-
tions, each of which is 10 feet wide. Find the necessary height and span of
the arch.
37. The Hyperbola. The hyperbola is the locus of a point the
difference of whose distances from two fixed points is a constant. Let
the two fixed points which are called the foci be taken on the x axis
at F' and F with the origin at the mid-point. Call 2 c the length
37}
DEFINED CURVES
53
of F'F. Assume P(x, y) to be any point on the hyperbola; then
we have, by the definition above,
F'P - FP = 2 a,
or
FP - F'P = 2 a,
where 2 a is the constant differ-
ence. Corresponding to these rela-
tions, we have the equation
P(*,V)
F\-c,0)
F(c,0)
D X
FIG. 62
V(x - c) 2 + y 2 = 2 a.
The rationalized form of this
expression is identically the same
as the form obtained for the ellipse in which case the sum of the
same two radicals is 2 a.* That is,
~9 ^.2
= 1.
a'
But in the triangle F f PF the difference of FT and FP is always
less than F'F, that is, 2 a < 2 c or a < c. Hence, by writing
6 2 for a 2 c 2 , so that b 2 is positive and b is real, the equation
becomes
"
Equation (1) shows that the
hyperbola is symmetric with
respect to both the axes and the
origin. The x intercepts are a.
There are no y intercepts.
If x 2 < a 2 , y is imaginary,
which means that no part of the
curve can lie between the lines
x = a. The segment V'V is
called the transverse axis, V
and V being the vertices. The
FlG - 63 segment B'B is called the con-
jugate axis. is the center of the hyperbola. The ratio c/a
* This is due to the fact that the rationalized form of each of the four radical
equations V a "N/6 = d= Vc is obtained by equating to zero the product
- Vb - Vc)(- Va + Vb - Vc)(- Vi - Vfc + Vc).
54
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
is called the eccentricity of the hyperbola; it is denoted by e,
and is always greater than 1. The chord through either focus
perpendicular to the transverse axis is called a latus rectum.
If the foci are on the y axis the "equation is
(2} -^-i
(2 > a 2 b 2 ~~ *'
ASYMPTOTES. The hyperbola has two asymptotes which pass
through the center of the curve. These are lines which the hyper-
bola approaches so that its distance from them approaches zero as it
is indefinitely extended. The equation of any line through the
center is y = mx, and its intersections with the hyperbola are
found by solving this equation simultaneously with (1 ) . This gives
ab abm
Vb 2 - a 2 ra 2 '
y =
- a' 2 m 2
The intersections will be real or imaginary according as the
expression under the radical, 6 2 a 2 m 2 , is positive or negative.
However, if 6 2 a 2 w 2 = 0, that is, if m = db b/a, the curve
approaches these lines but has no finite intersection with them.
To prove this statement, let (x\, y\) be a point on either line
y = (b/a)Xj and let (x\, 2/2) be a point on the hyperbola with the
same abscissa. Then
b f
a*^
The difference of these ordinates is
ab
FIQ. 64
Xi + >Ai 2 - a 2
Hence, as Xi increases indefinitely, y%
approaches 2/1, since yi y 2 approaches
0, and the hyperbola approaches the lines
y = zh (b/a)x as asymptotes.
The two curves
r 2 >/ 2
/o\ * y , t
( 3 ) ^ - w = =*= l ;
are called conjugate hyperbolas. They have the transverse and
39]
DEFINED CURVES
55
conjugate axes interchanged, have the same asymptotes, a common
center, and foci at the same distance from the center. The
eccentricities are e\. = c/a and e z = c/6, respectively.
Draw the rectangle x = a, y = it b. Its diagonals are the
asymptotes to each of the con-
jugate hyperbolas (3). The circle jy
circumscribing this rectangle inter-
sects the axes in the foci for each
curve.
38. Equilateral Hyperbolas. If
a = by the two conjugate hyper-
bolas have the same shape and
eccentricity. The asymptotes are
the lines x db y = and the equa-
tions of the two curves become
(1) jc 2 - y 2 = a 2 .
These hyperbolas are called equi- FIG. 65
lateral or rectangular hyperbolas.
A more important form of the equation of the equilateral hyper-
bola is obtained by taking the asymptotes as the coordinate axes.
This equation may be obtained from (1) by using the equations of
transformation by rotation ( 22), for = 45. These equations
are
x = (x/2/2)(*' - ?/), y = (x/2/2)(z' + y').
Substituting in (1) and simplifying, we find the result
(2)
x'y' =
39. Other Forms of the Equation of the Hyperbola. The
following forms of the equation of the hyperbola are important :
(1)
Ax"
where A and B are of unlike sign. By completing the squares,
and translating the origin to the center, (1 ) becomes
(2)
Ax' 2 + By' 2 = C',
where A and B are the same as in (1). This hyperbola has its
transverse axis along the x' axis or y' axis according as C" has the
56
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
same sign as A or as B. If C' = 0, then (2) is the product of
linear factors, and the graph is a pair of straight lines.
The equation
(3 ) * = $!' y
where a, &, c, d have any values, is an equilateral hyperbola. It
is evident that ex + d = or x = d/c is a vertical asymptote.
Solving (3) for x, we get
x - ^dy-b
*L> " .
cy - a'
from which we see that y = a/c is a horizontal asymptote. If we
translate the origin to the intersection of these asymptotes
( d/c, a/c) by writing (x' d/c) and (y' + a/c) for x and y,
respectively, equation (3) reduces to
(4) x'y' = k,
where fc = (be ad)/c 2 .
It can be shown that the asymptotes of the hyperbola in equation
(2) are the linear factors of Ax' 2 + By' 2 0. This brings us to a
useful converse theorem which is stated without proof.
// LI and L 2 are any two linear expressions in x and y, and k has
any value except zero } then L\ - L 2 = k is some hyperbola whose
asymptotes are LI = and L 2 = 0.
EXAMPLES
1. Given the hyperbola 9 x 2 - 16 y* + 100 = 0, find the coordinates of
the foci and vertices, the eccentricity,
and the equations of the asymptotes
and of the conjugate hyperbola.
SOLUTION. Write the equation in the
form
FIG. 66
hence the equations of the asymptotes are
Here the transverse axis along the
y axis is of length 5, the vertices being
(0, 5/2); c = Va* + 6" = 25/6, and
the foci are (0, =fc 25/6); e = 5/3. The
slopes of the asymptotes are 3/4;
y
39]
DEFINED CURVES
57
The conjugate hyperbola is
ioy /5
/ V2
2. Find the equation of the hyperbola
with foci at (2, - 1) and (- 6, - 1) with
e = 2.
SOLUTION. The center is (2, 1).
The transverse axis is along y 1;
e = c/a = 2 and c = 4. Hence a = 2 and
the vertices are (0, 1), and ( 4, 1).
Let P(x, y) be any point on the hyper- F'(-6,
bola. Then we must have, from the defini-
tion, F'P - FP = 2 a, or
6) 2
I) 2 -
2)2
4.
Rationalizing this equation, we obtain the
equation
FIG. 67
0.
PROBLEMS
Find the vertices and foci, the eccentricity, the equations of the asymptotes
and sketch each of the following hyperbolas. (Nos. 1-8.)
1. 3z 2 -22/ 2 -{-'24 = 0. _
Arts. (0, db 2 V3), (0, =fc 2\/5), V5/3, x V3 y V2 = 0.
2. 4 z 2 ~ ?/ 2 = 16.
3. 9 x 2 - 16 ?/ 2 = 144. Ans. (db 4, 0), ( 5, 0), 5/4, 3 x =fc 4 2/ = 0.
4. 3 x 2 - 4 2/ 2 + 60 = 0.
5. 5 y 2 - 3 x* = 30.
Ans. (0, V6), (0, =fc 4), 2\/6/3, a; V3 ?/ >/5 = 0.
6. 9 x 2 - 4 2/ 2 - 49.
7. 3 t/ 2 - 7 x 2 24.
db 2/V3 = 0.
Ans. (0, 2v^2), (0, db 4V5/7),
8. 7 z 8 ~ 3 y 2 84.
Find the remaining values a, 6, c, e; then write the equation of the hyperbola
with the given parts below. (Nos. 9-12.)
9. Center at the origin, a focus at (5, 0), e = 2.
Ans. a 5/2, 6 = 5 V5/2, 12 x 2 - 4 s/ 2 = 75.
10. Extremities of the conjugate axis ( 5, 0), e = 3/2.
58 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
11. Foci at (0, 6), slope of asymptotes 4/3.
Ans. a = 24/5, 6 = 18/5, e = 5/4, (5 y/4) 2 - (5 z/3) 2 = 36.
12. Center at the origin, extremities of a latus rectum at (rfc 14/3, 8).
Derive the equation of each hyperbola in the following cases. (Nos. 13-16.)
13. Center at (- 2, 2), one focus (- 2, 7), one vertex (- 2, - 1).
Ans. (x + 2) 2 /16 - (y - 2) 2 /9 + 1=0.
14. Foci at (3, 5) and (13, 5), e = 5/4.
15. Extremities of the conjugate axis ( 1, 2) and (1, 6), e = V2.
Ans. x 2 - y z -f- 2 x - 4 y = 19.
16. Vertices at (3, 2) and (3, - 4), length of latus rectum 32/3 units.
17. Find the equation of the hyperbola with center at the origin, foci on the
x axis, and passing through the points (5, 9/4) and (4V2, 3).
Ans. z 2 /16 - 2/2/9 = L
18. Two vertices of a triangle are fixed at ( a, 0). Find the equation of
the locus of the third vertex if the product of the slopes of the variable sides
is fe 2 /a 2 .
19. Find the locus of a point whose distance from (3, 4) is 3/2 its distance
from the line x = 1. Ans. 5 x* - 4 ?/ -f 6 x -f 32 y = 91.
20. Find the locus of a point whose distance from ( 2, 4) is twice as
great as its distance from the line y + 1 =0.
Find the center of each of the following hyperbolas. Translate the origin
to the center and find the eccentricity. (Nos. 21-24.)
21. 5 x 2 - 2 ?/ 2 -f 20 x - 4 y - 18 = 0. Ans. (- 2, - 1), V772.
22. 3 x* - 3 y 2 - 4 x -f 8 y = 31.
23. 3 z 2 - 2 y 2 + 12 x + 4 y + 20 - 0. Ans. (- 2, 1), Vo73.
24. 4 x z - 8 y 2 + 4 x + 32 t/ -f 1 = 0.
Translate the origin to the center and find the transformed equation.
(Nos. 25-26.)
25. x = (5 y + 8)/(4 y - 1). Ans. (5/4, 1/4), 16 x'y' = 37.
26. y = (3x - 5)/(2s + 7).
27. One vertex of a hyperbola is at (1, 3), the corresponding focus at (1, 6).
If the slope of one asymptote is 3/4, find the center and eccentricity.
Ans. (1, - iy 2 ), 5/3.
28. Given the equilateral hyperbola a? 2 y 2 = a 2 . Prove that the dis-
tance of any point on the curve from the center is the mean proportional
between its distances from the foci.
29. Change 9 x 2 4 y 2 -f- 36 = to parametric form if y = 3 sec 6 is to
be one equation.
30. Sketch and name x = 3 tan t, y 2 = 4 sec t. Transform to its
rectangular equation.
40]
DEFINED CURVES
59
P(r,0)
40. Conies. The parabola, ellipse, and hyperbola are included
in a class of curves called conies or conic sections. A conic may be
defined as the locus of a point such that its distance from a fixed
point is in a constant ratio to its
distance from a fixed line. The
constant ratio is the eccentricity e,
and the curve is an ellipse, parab-
ola, or hyperbola according as e
is less than, equal to, or greater
than 1. The fixed point is a focus
and the fixed line a directrix. To
derive the equation of the conic in
polar coordinates, let the given focus
be the pole and let the directrix be per-
pendicular to the polar axis, and
at a distance p from the focus. Then if P(r, 0) is any point
on the locus, by the definition we have OP/NP = e, that is,
OP = e-NP. Hence r = e(p + r cos 6), from which we find
\o~
M
D'
FIG. 68
(D
T *~~
ep
1 e cos 8
In general the equations
r= =*=**
1 =t e cos 9 '
1 d= e sin
D'
represent conies. The latter equa-
tions are the forms obtained when
the directrix is parallel to the polar
axis.
EXAMPLES
1. Plot the graph of r(l + cos 6) = 4.
SOLUTION. This is a parabola since e = 1. Then p = 4. Some pairs of
values are:
e
60
90
120
180
240
270
300
360
r
2
8/3
4
8
oo
8
4
8/3
2
from which the curve can be drawn (Fig. 69).
60
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
2. Draw the graph of r(3 - 2 sin 6) + 6 = 0.
SOLUTION. This equation may be written
-2 - (2/3)3
1 - sin
o
1 5 sin
o
Hence e = 2/3, p 3, and the locus is an ellipse (Fig. 70).
6 30 90 150 180 210 270 330
_o -3 -6 -3
360
-2 - 3/2 - 6/5 - 3/2 - 2
Let the student show that the
graph of r(3 + 2 sin 6) =6 gives the
same ellipse. Explain
(-2,0
\(-6,90)
J
D'
D
FIG. 70
FIG. 71
3. Draw the graph of r (4 5 cos 0) = 9.
SOLUTION. Write the equation
9
4
1 T cos 1 "~ 7 cos ^
Hence c = 5/4, p = 9/5. The curve is a hyperbola. We observe that the
values of which make 1 (5/4) cos = will make r become infinite. These
values of give the lines through the pole which are parallel, respectively,
to the asymptotes.
60 90 120 180 240 270 300 360
r -9 6 9/4 18/13 1 18/13 9/4 6 -9
Note that the transverse axis between ( 9, 0) and (1,180) is of length 8.
PROBLEMS
Plot the graph of each of the following equations. (Nos. 1-8.)
1. 2 r 4 -f r sin 0. 3. r = 2 + r cos 0.
2. 3 r + 4 = 5 r sin 0. 4. r = 3 - 2 r sin 0.
41]
DEFINED CURVES
61
5. r(4 - 4 sin 0) = 9. 7. r(2 cos 6 - 5) + 6 - 0.
6. r(4 cos - 3) = 9. 8. r(l -f cos 6) = 5.
Derive the equation of each of the following curves. (Nos. 9-12.)
9. The circle of radius 5 units tangent to = at the pole.
Ans. r = db 10 sin 6.
10. The ellipse with e = 1/2, r sin $ -f 2 = as directrix, and focus at the
pole.
11. The hyperbola with e = 3/2 and a line 10 units from the pole per-
pendicular to the polar axis as directrix, and focus at the pole. (Two cases.)
Ans. 2 r = 3(10 dt r cos 0).
12. The same as Problem 11 except the directrix is parallel to the polar axis.
13. Transform x 2 y(2 a y) to polar coordinates. Name and draw its
graph. Ans. r 2 a sin 6.
14. Discuss r(l e cos 6) - ep with reference to symmetry, asymptotes,
closed or open form, for e 1, 2, 0.2, 3, 0.3.
15. The same as Problem 14 for r(l -f- e sin 0) = ep, e = 0.5, 1, 2.
16. Find the locus of P if its distance from the pole is 2/3 of its distance
from r cos 9 -f 3 = 0.
41. Cycloid. Involute
of a Circle. These curves
have some important ap-
plications and their equa-
tions are usually given
in parametric form. The
cycloid is the path traced
by a point on the circum-
ference of a circle as it
rolls along a straight line. FlG 72
Let C be the center of a
circle of radius a and let P(x } y) be any point on the circumfer-
ence.
To find the locus of P as the circle rolls along the x axis, let 0,
the point where P is in contact with the line, be the origin. We
can express the coordinates of P in terms of the angle 6 which the
radius CP makes with the vertical line CN. Thus
x = OM = ON - MN = arc NP - PS = ad - a sin 0,
and
y = MP = NC - SC = a - a cos 0.
Hence
(1)
x = a(6 - sin 8), y = a(l - cos G).
62
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
Each value for in (1) fixes a value for x and a value for y and
hence a point on the cycloid.
// a string is wound about a fixed
circle, the path traced by any point of
the string as it is kept taut and un-
wound from the circle is called an invo-
lute of the circle.
Let be the center of the fixed circle
and P be the point of the string which
meets the circle at R on the x axis.
Let T be the point of tangency corre-
sponding to the point P and call 6 the
angle between the radii OT and OR.
FIG. 73 Then
TP = arc RT = aB.
If we draw TS perpendicular to OR and NP perpendicular to ST,
we find
x = OM = OS + NP = a cos 6 + a6 sin 6,
and
y = MP = ST - NT = a sin - ad cos 6.
Therefore, the involute of the circle has the equations
fjt = a(cos6 -f 9 sin 6),
[y = a(sin 6 - 6 cos 6).
(2)
PROBLEMS
Draw the graph of each of the following pairs of parametric equations.
(Nos. 1-3.)
1. x = 2(0 - sin 0), y = 2(1 - cos 0).
2. = 1 cos 0, y = sin 0.
3. z = 3(0 + sin 0), y = 3(1 - cos 0).
4. Sketch a section of an involute of a circle with a radius of 3 units.
42. Empirical Equations. Often the exact form of an equation
is not known, the only information obtainable being a table of
corresponding values of the related variables. Even then, the
values of the variables are inexact if they are obtained by measure-
ment. In that case, the problem is to find a formula or relation
between the variables which the given values satisfy approxi-
mately. There are six such formulas, with two constants, generally
43] DEFINED CURVES 63
used for fitting a curve to the points determined by the given data.
By fitting we mean the constants may be determined so that the
graph of the formula will come as near the plotted points of the
given data as the accuracy of the observations demands.
43. Two-Constant Formulas. The formulas with two constants
commonly used are:
(1) y = mx + b (straight-line formula),
(2) y = ax 2 + b (parabolic formula) f
(3) y = ax n (power formula),
(4) y = ae bx , or y = ab x (exponential formula),
(5) xy = ax + b (hyperbolic formula) ,
(6) xy = ax + by (hyperbolic formula).
We determine which formula to use as follows:
(a) If the table of data is plotted and the curve suggested by the
points is a straight line, assume formula (1).
(b) Let x 2 = z and (2) becomes a straight-line formula with
each point located at (z, y).
(c) Taking the logarithm of each member of (3), we have
log y = log a + n log x,
which is linear in log y and log x. Hence, if the points plotted
from the logarithms of the given data suggest a straight line,
assume formula (3). Either natural or common logarithms may
be used.
(d) Treating formula (4) in like manner, we have
log y = log a + bx log e, or log y = log a + x log 6,
which are linear in x and log y. Hence, if the points plotted with
the given values of one variable as abscissas and the logarithms
of the values of the other as ordinates suggest a straight line,
assume formula (4).
(e) Setting xy z, we see that formula (5) takes the form
z == ax + b and that it is linear in x and z. Here we plot the
given data with the ordinates replaced by the products of corre-
sponding variable values and, if the points suggest a straight line,
we assume formula (5).
64 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
(/) Formula (6) is usually written
a a- b - i
y + x~ l >
and may be used if l/x and l/y suggest a straight-line formula.
44. Methods of Determining Constants, (a) GRAPHICAL
METHOD. Plot the points determined by the given data. In
so doing, choose as coordinates either (x, y) or (x 2 , y) or
(log x, log y), or (x, log t/), or 0, xy), or (l/x, I/?/), where x and y
are corresponding values of the given data.
If the formula is one of the two-constant formulas of 43, one
set of points will suggest a straight line. Draw a straight line
fitting the points and substitute the coordinates of two points on
the line in the formula chosen to get two equations involving the
unknown constants. These are solved simultaneously for the
constants.
In constructing the straight line, it is desirable to have as many
of the plotted points as near the line as possible, as well as the same
number on each side of it. Of course, the two points used to
determine the constants have their coordinates read from the
graph and consequently are inaccurate. However, better results
are obtained if the two points used to determine the constants
are chosen as far apart as possible in the group of plotted points.
(6) METHOD OF AVERAGES. A method which takes into account
all the data, and not merely two selected points which may not
even occur in the given observations, is that of averages. In
using this method, we must determine the formula to be used as in
the previous method, but, having fixed on the formula, the prob-
lem is arithmetical. Proceed as follows: Substitute each pair of
values used in plotting the points in the linear formula assumed.
That gives as many equations as there are pairs of corresponding
values. Then divide these equations into two groups as nearly
equal in number as possible. Add corresponding members of the
equations of each group, thus obtaining two equations to deter-
mine the two constants. Solve these equations simultaneously
and make the proper substitutions in the original formula.
If the form of the desired formula is known from the nature of
the problem or is given together with the data, the method of
averages does not depend upon any graphical observations. In
any case, this method is generally the better to apply because it
145]
DEFINED CURVES
65
takes into account all given data and does not depend upon read-
ing fractional values of coordinates. However, if the data are not
necessarily of equal validity, the graph will suggest the pairs of
values which are questionable and probably negligible.
45. A More General Parabolic Formula. The parabolic
formula
(1) y = ax* + bx + c
occurs occasionally. The method of averages may be used at
once. Merely divide the observational equations into three
groups, add corresponding members of the equations of each
group, and solve the three equations for a, 6, and c.
EXAMPLE
Find a two-constant formula for the following data:
X
2.1
5.6
9 3
11.5
y
20
18.92
17.34
15.8
14.96
SOLUTION. We find from trial that points with coordinates x and logic y
fit a straight line. The table of values for those points and the corresponding
graph are given below.
X
logic y
1 3010
2.1
1.2769
5 6
1.2390
9.3
1 . 1987
11.5
1 . 1749
1.35
1.30
1.25
1.20
1.15
1.10
1.05
1.00
<*
"
**.
^
**,
*,
^.
<.
) 1 2 3 4 5 6 7 8 9 101112 X
FIG. 74
(a) GRAPHICAL METHOD. Draw a
straight line fitting the points plotted.
The straight line seems to pass through
the points (0, 1.3) and (11.5, 1.17).
One recognizes the absurdity of attempting to read or plot the second point
chosen for the assumed scale. However, using the selected points, we have the
two equations from
logio y = logio a -f bx logio e
as follows:
1.3 = logio a, 1.17 = logio a + b (11. 5) (0.4343).
Solving these, we find
logio a = 1.3, b = 0.0260,
66
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. II
whence a = 19.95, and the desired formula is
y = 19.95 e-.o26o*.
(b) METHOD OF AVERAGES. Substitute in
logio y = logio a -f bx logic e.
Then
1.3010 = logio a -f 0,
1.2769 = logio a -f 2.1(0.4343)6.
Adding, we find
Also
Adding, we have
2.5779 = 2 logio a + 0.91206.
1.2390 = logio a -f 5.6(0.4343)6,
1.1987 - logio a + 9.3(0.4343)6,
1.1749 = logio a -I- 11.5(0.4343)6.
3.6126 = 3 logio a + 11.4655 6.
Solving the equations derived by additions, we find
logio a = 1.3004, 6 = - 0.0252,
whence a = 19.97, and the desired formula is
y = 19.97 e--2*.
PROBLEMS
The following sets of data satisfy approximately the given formula. Find
the laws.
X
0.5
1
1.5 2
2.5
3
y
31
0.81
1.29 1
85 2 51
3.02
for y = ax -f 6.
Ans. y =
1.1 x - 0.30.
x
10
20
30 ^
10 50
60
2.
y
3.26
4.73
6.24 7.
49 9.01
10.51
for y = TWO; -f 6.
x
6.0
6.9
7.5
8.7
a
y
7.5
11.5
13.8
20.5
f or y * a -f fo 2 *
Ans. y = 0.5 x 2 - 10.6.
X
2.0
4.7
7.1
8.4
y
75.6
55.7
24.5
2.4
f or y a + 6x 2 or y = 0-6*.
45]
DEFINED CURVES
67
5.
for y =
6.
for y =
7.
for y =
8.
for y =
9.
for y =
10.
for pu n
11.
for y =
12.
for y =
13.
for y =
14.
forB =
15.
for xy -
X
3.0
4.1
5.3
7.0
y
1 90
5 75
11 80
24.10
ax 2 -f- bx.
Ans. y = 0.70 a: 2 - 1.47 a;.
5
10
15
20
30
5.0
6.8
7.4
8.0
8.7
y
ax 2 -f- 6 or x = axy -\- by.
10
y 20.0 24 3
ax 2 + bx + c.
15
20
25
30
28 3 32.1 35 6 39.0
Ans. y = - 0.004 x 2 + 0.9 x + 15.5.
X
0.5
2
3 5
4.5
y
0.1
9
2 2
3.3
ax n .
1
1 5
2
2 5
3.0
3.5
1
4 1
8 5
14 1
21
29 1
y
ax 2 -f b.
P
= c.
Ans. a = 2.5, b = -1.5.
26.4
22.4
19 1
16.3
14
14 7
17 5
20 8
24 5
28.8
x
1.22
426
0.047
0.005
y
676
074
0.004
ax n .
Ans. a = 0.40, n =* 1.78.
X
1.124
0.342
0.511
0.730
y
1.002
0.604
0.494
0.414
a(y/x)\
X
5
10
15
20
25
30
y
6.1
6.8
7.4
8
8.5
9.2
x/(ax + b).
Ans. a = 0.1, b = 0.5.
d
1.5
3
4.0
6.0
B
13.43
75.13
152 51
409.54
X
1
0.50
0.25
0.17
0.10
y
ax
0.77
+ by.
0.45
0.34
J:
0.20
ins. xy = 1.4
0.16
9 x - 0.09 y.
68
DIFFERENTIAL AND INTEGRAL CALCULUS [CH. II
X
1.3
3.4
6.2
8.3
y
56
91
1.11
1 18
16.
for xy
17. An iron plate with one straight edge has its width given by d at inter-
vals x along the straight edge. Find a law for the other edge if we have
X
4
8
12
16
20
d
2
4
5
6
9
12
Arcs, d = 0.006 (x - 8) 3 + 4.81.
ADDITIONAL PROBLEMS
1. Find the locus of a point P if the difference of its distances from the
fixed points (0, i c) is a constant 2 a. Ans. A hyperbola.
2. Write the general equation of a line through PI so that it will have only
one arbitrary constant.
3. Write the general equation of the line through PI and parallel to
ax -f by + c = 0. Ans. a(x Xi) + b(y yj = 0.
4. The same as Problem 3 except perpendicular to the given line.
In Problems 5-8, find the equation of each of the lines described below.
5. Through the origin and making an angle of 45 with x = 2 y -f- 3.
Ans. y = 3 x.
6. Through (4, 6) and perpendicular to 60; 7 y + 1 =0.
7. Through (1, 1) and perpendicular to the line through that point
and (2, 3). Ans. 3x + 4i/ + 7 = 0.
8. Through (3, 5) and parallel to the line through (2, 5) and (-5, - 2).
9. Find the distance from x + 2 y = 5 to (2, 6). Ans. 3\/5 units.
10. Given A (2, 5) and PI. Find the equation of APi, its length, its
slope, and its mid-point.
11. Find the equation of the circle with its center at the point (2, 2)
which is tangent to x + y = 6. Ans. (x + 2) 2 + (y + 2) 2 = 50.
12. Find the equation of the circle through (3, 4) and tangent to the x
axis at ( 1, 0).
13. The channel of a river for some distance remains equidistant from a
rock in the stream and a straight shore line. If the rock is 600 feet from the
shore, derive an equation for this part of the channel. Ans. y 2 1200 x.
14. Find the circle through the vertex and focus of x 2 = 8 y with its center
on x y 2 0.
15. Find the circle whose diameter is the chord of y* I x cut off on
x - y + 1 = 0. Ans. (2 x -f 3) 2 + (2 y + I) 2 = 18.
45]
DEFINED CURVES
69
16. Find the equation of the parabola with its focus at the center of the
circle 3 z 2 4- 3 ,y 2 -f- 12 x 6 y = 10 if its directrix is parallel to the x axis
and tangent to the circle. (Two cases.)
17. Find the locus of a point whose distance from (0, 9) is twice its distance
from y = 4. Arts, x 2 - 3 ?/ 2 -f- 14 y + 17 = 0.
18. A parabolic arch of 100 ft. base has 12 ft. clearance 6 ft. from one end of
the base. What clearance has the arch at its top?
19. A parabolic arch is 10 ft. high and 15 ft. wide at its base. How far from
the end of the base is the clearance 6 ft.? Ans. 4.74 ft.
20. Find the equation of the circle which passes through the vertex and the
extremities of the latus rectum of the parabola x 2 -p- 10 y =0.
21. If two vertices of a triangle are fixed at ( a, 0), find the locus of the
third vertex if the product of tangents of the base angles is 6 2 /a 2 .
Ans. Ellipse.
22. Using the definition of the ellipse, derive its equation if the foci are
(2, 2), (6, - 3) and it passes through (2, 0). What is the value of e?
23. Name each curve, sketch, and transform to rectangular representation:
(a) r(l - sin 0) = 3; (6) r(3 + 2 sin 6) = 5; (c) r(2 - 3 cos 0) = 5.
Ans. (a) Parabola, x 2 6 y -\- 9.
24. Two lighthouses are 5 miles apart. If a distress signal is heard at one
lighthouse 15 seconds before it is heard at the other, what path should a vessel
take to locate the signal? (Assume sound travels 1 mile in 5 seconds.)
25. Which law, xy = ax -j- b, y 2 = ax -j- 6, or xy = ax -f- by, is best adapted
for the following data?
8.05 7 54 5 16 3 22 1.57
1.20
1 41
2 15
2.91
10.3 - 1.1 x.
2 59
Ans. y 2
26. The no-load magnetization curve of a direct-current generator taken at
1200 r.p.m. was found by test to include the following points:
7
1
0.2
3
0.4
0.5
0.6
E
4.4
29.6
52.6
73.1
91.5
105.3
115.1
I
8
1
1.2
1.4
1.6
2.0
E
126 4
132 1
134.7
135.9
136.4
137.1
where 7 is field current in amperes and E is no-load voltage. Find the em-
pirical equation for this curve:
(a) Using Froelich's equation E al/(b -f- 7).
(6) Using a modification of Froelich's equation E = al/(b + /) 4- c.
(c) Using a power series 7 = a + bE -f- cE 2
27. Derive equation (6), 28, directly from a figure.
CHAPTER III
THE DERIVATIVE
46. Constants. Variables. A quantity which has a fixed
value is called a constant. These are of two kinds. An absolute
constant is a fixed number, as 2, 3/2, v% TT, logio 17, sin 24 15'.
An arbitrary constant is one which is represented by some letter, as
a, c, fc, m. Such a constant is assumed to have a definite value
which it retains throughout a particular problem.
A variable is a quantity which may have different values in
a given problem. Thus the temperature of a certain object is a
variable quantity. The length of a chord of a circle of radius 10
is a variable which may have any value between and 20. A
variable is usually represented by some letter in the latter part of
the alphabet, as #, y, u y v.
47. Functions. A function of a given variable x is another
variable quantity which has one or more definite values corresponding
to each value assigned to the variable x. To illustrate, 2\/25 x*
is a function of the variable x since it is also a variable quantity,
and depends for its value on the value assigned to x. A function
of x is commonly represented by the symbol /(#), which is read
"/ of x," or by a single letter, as y. Then x is called the inde-
pendent variable and/(x), or y, is called the dependent variable or
function of x. In general we speak of x simply as the variable and
of /(x), or t/, as the function. We may refer to the function men-
tioned above either by f(x) = 2^25 x 2 or y = 2\/25 x 2 ,
where f(x) or y are merely other symbols for the expression
2\/25 x 2 . This function has the following geometric interpre-
tation: In a circle of radius 5 units, if any chord is drawn at a
distance of x from the center, then the length of the chord, y, is
2\/25 x z . Hence we can say that this function expresses the
length of a chord of the given circle in terms of its distance from the
center of the circle.
A function of a variable may be represented (a) by an equation
connecting the variable and the function, and (6) by a graph in
which the corresponding values of the variable and function are
70
49] THE DERIVATIVE 71
the abscissas and ordinates, respectively. If a function is given
by a formula or equation, the corresponding pairs of values may be
obtained from it and the graph constructed. However, two physi-
cal quantities may be related without being connected by a known
formula. Thus, the temperature at a given place may be a func-
tion of the time. A table of values of time and temperature can
be recorded by observation and the results plotted to form a graph,
but it is impossible to write a formula expressing the temperature
in terms of the time, or to calculate the temperature at some future
time.
Since y = f(x) means that y is a function of x, in a similar
manner z f(u, v) means that z is a function of the two vari-
ables u and v. To distinguish between different functions in the
same discussion, different letters are used, as/(x), g(x\ </>(x), $(x).
Throughout the same discussion, however, the same symbol
refers always to the same function. Thus if we have given
V f( x ) = 2 x \/9 + x 2 , then /(4) means the value of f(x)
when 4 is substituted for x. That is, /(4) = 3 ; similarly
/(O) = - 3, /(- 2) = 4 - \/l3, /(a) = 2 a - ~~
48. Inverse Functions. If y = /(x), then any value assigned
to y will determine one or more corresponding values of x. That
is, x is also a function of y, or x = 0(y); then </>(?/) is called the
inverse function of /(#). To get </>(?/), solve the equation
y = f( x ) f r x - There are many cases where this cannot be
done by means of elementary algebra, but the relation x = </>(y)
in general exists. Examples of inverse functions are
y = x z + 2x-l and x = - 1 \/2 + y\
y = 3 log a 0/2) and x = 2 a v/3 ;
y = (1/3) s i n (2 x) and x = (sin- 1 3 y)/2.
49. Explicit and Implicit Functions. In the form y f(x) we
call y an explicit function of x since y is given explicitly in terms of x.
However, if two variables x and y are connected by a relation
of the type <p(x, y) = 0, then y is called an implicit function of x,
since the existence of this relation implies that y is a function of x.
Likewise, x is an implicit function of y. Solving </>(x, y) = for
either variable gives an explicit function in terms of the other.
For example,
<0, y) = x 2 + 2 xy - y 2 + 4 =
72 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. Ill
gives x and y implicitly. Solving for each variable in turn, we
obtain the explicit functions y = x rfc \/2 x 2 + 4, and its inverse,
x = - t/ V2 y 2 - 4.
SO. Types of Functions. The functions which are used in an
elementary course of the calculus are algebraic functions, and
certain transcendental functions including logarithmic functions,
inverse logarithmic or exponential functions, trigonometric functions,
and inverse trigonometric functions. These are the functions which
are of fundamental importance also in a study of the physical and
engineering sciences.
PROBLEMS
1. Express the area of a square inscribed in a circle as a function of its
radius. Ans. A 2 r 2 .
2. Express the volume of a right circular cone whose altitude is one-half
the radius of its base as a function of the altitude.
3. Express the volume of a right circular cylinder whose altitude is equal
to its diameter of a base as a function of the radius of the base. Ans. V 2 xr 3 .
4. Express the surface of a cylinder of volume V cubic units in terms of the
radius of the base.
5. Express the volume of a vessel made of a cylinder with hemispheres on
each end as a function of the length of the vessel, if the length of the cylindrical
part is three times the radius of the ends. Ans. V - 13 irl z /375.
6. Express the volume of a sphere as a function of its surface. Express the
surface as a function of the volume.
7. A right triangle has a hypotenuse 10 units long. From the vertex of
the right angle the altitude and median are drawn to the hypotenuse. Express
the area of the triangle as a function of the segment of the hypotenuse between
the median and the altitude. Ans. A - 5V25 z 2 .
8. The velocity of a falling body varies as the square root of the distance
it has fallen. If the velocity is 32 ft. per second when it has fallen 16 ft.,
express the distance as a function of the velocity.
Write the inverse of each of the following functions. (Nos. 9-16.)
9. y = 1 # 2 . Ans. x db Vl y.
10. z = 2 - 3 ?/ -f 2 y*.
11. y = 3 sin (x + /2). Ans. x = cos~ l (^/3).
12. x = 2 cos- 1 (2 y/3).
13. y sin x cos x. Ans. x = (1/2) sin" 1 2 y.
14. y = e **.
51] THE DERIVATIVE 73
15. y = log Vo-HTo*. Ans. x - =b Ve^ - a*.
16. * (1/2) (e* + e-).
Express explicitly each variable as a function of the other in the following
cases. (Nos. 17-22.)
17. Z 1 / 2 -f 7/ 1 / 2 - 4. Ans. y = (4 - z 1 / 2 ) 2 , x = (4 - 2/ 1 / 2 ) 2 .
18. a; 2 / 3 - 2/ 2 / 3 = 8.
19. x 2 + 2xy + 4y 2 = 5.
Arcs. ?/ = (1/4) (- 3 =t V20 - 3 a:' 2 ), a; = - y V5 - 3 ^ 2 .
20. 3 sin a;?/ = 2.
21. 2 log x - 3 log y = 4. Ans. ?/ = (z/e 2 ) 2 / 3 , x = e 2 2/V?/.
22. 2 log sy = 5.
51. Limits. An idea of fundamental importance in the calculus
is that of the limiting value of a function of a variable.
Let f(x) be any function of the variable x. Then, as x approaches
any constant a, if the corresponding values of f(x) approach a definite
constant I in such manner that the numerical value of the difference
I f(x) becomes and remains less than any preassigned positive
number, however small, then I is said to be the limit of /(x).
The notation
lim/(x) = J
x *-a
is read " the limit of f(x), as x approaches a, is I." In other words,
this statement means that/(x) is as near as we like to Z if x is near
enough to a.
The student is already familiar with illustrations of a function
approaching a constant as a limit. Thus the sum of the first
n terms in the geometric progression
(1) S n = 1 + 1/2 + 1/4 + 1/8 + - - - + 1/2"- 1
is evidently a function of n, the number of terms involved. The
limit which S n approaches as n increases is 2, and the expression
lim .S n = 2
n *
is read " the limit of 8 n as n increases without limit is 2."
Again, the area C of a circle is defined as the limit approached
by the area A of a regular inscribed or circumscribed polygon as
the number, n, of its sides increases without limit, or
lim A = C.
74 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. Ill
A function may or may not attain its limit. In the series (1),
the addition of successive terms will never add up to 2 ; nevertheless
the limit exists.
52. Theorems on Limits. Certain theorems on limits are
stated without proof. In each one the existence of the limit is
implied.
THEOREM I. The limit of an algebraic sum of any finite number
of variables is equal to the same algebraic sum of their respective
limits.
THEOREM II. The limit of the product of any finite number of
variables is equal to the product of their respective limits.
THEOREM III. The limit of the quotient of two variables is equal
to the quotient of their respective limits, provided the limit of the
denominator is not zero.
THEOREM IV. // / ^ g ^ h and if Urn f = Urn h, then we have
lim g = Urn h.
53. Infinitesimals. An infinitesimal is a variable whose limit
is zero. From the definition of the limit of a function it follows
that the difference between a function and its limit is an infinitesi-
mal. Hence if
= I
then I f(x) is an infinitesimal whenever a - x is an infinitesi-
mal. As other examples, we may state that if v is an infinitesimal,
so are also kv (where k is any constant), sin v, and (1 cos v).
54. Continuous Functions. A function f(x) is said to be con-
tinuous for x = a if
lim/(oO =
A function is continuous in the interval x = Xi to x = x 2 if it is
continuous for all values of x in this interval.
A function is said to be discontinuous for x = a if the condition
for continuity is not satisfied. The only functions which we shall
consider are those which are in general continuous, but which may
have a discontinuity for some value or values of the variable. As
examples of discontinuity we mention the following:
54]
THE DERIVATIVE
75
(a) When the function /(#) increases without limit or decreases
without limit as x approaches a, that is, when
lim J(x) = oo .
at *a
Thus the function y = 1/z 2
(Fig. 75) is continuous for every
value of x excepting x = 0, for
which it is not defined.
(b) The function f(x) may ap-
proach different values as a limit
according as the variable x ap-
proaches a from a value greater
than a, or less than a. Thus the
function y = 1/(1 x) decreases
indefinitely as x approaches 1
from a greater value, but in-
creases indefinitely if x ap-
proaches 1 from a smaller value. This function is not defined for
x = 1 but is continuous for all other values. (Fig. 76.)
X
FIG. 75
X
FIG. 76 FIG. 77
(c) Another illustration is shown in the graph of .
y = 2 1 /*.
As x approaches from positive values y increases without limit
but as x approaches from negative values y approaches the
limit 0. This graph possesses what is known a an end point at the
origin. The function is not defined for x = 0. (Fig 77.)
76 DIFFERENTIAL AND INTEGRAL CALCULUS [On. Ill
55. Evaluation of Limits. It is necessary to know the values of
some important limits used in the calculus. Among these are:
3in6"
(a)
lim
en
,
J
if is measured in radians.
In Fig. 78, OAB is a circular sector with AC & tangent at A, and
c central angle 6. We have
A OAB < sector OAB < A OAC.
That is,
or, on dividing by (r 2 /2) sin 6,
A . e
1
FIG. 78 ' sin 6
But lim [sec 0] = 1 ; therefore, by Theorem IV, 52,
o [sin J
Hence the reciprocal ratio will have the limit 1, or
(1) limp~-_ 1 = 1.
cos 6
< sec 0.
(6)
Since
we have
lim
']
1 cos i
sin 6
1 cos sin
tan 2'
Hence, by Theorem II, 52,
,. fl - cos 01 r fsin 0~| r P. 0*1 , n
j im s- i im __ . i im tan jr = 1 0,
^oL * J e ^ Q L * J *-*L J J
55]
THE DERIVATIVE
77
or
(2)
f . fl - cos 61 _
lim a = 0.
0-0 L e J
(c) lim (1 + x) 1 /*. The proof that lim (1 + z) 1/z exists is
x -0 x K)
beyond the scope of this book, but it is given in advanced courses*
and will be assumed when needed.
Since this limit does exist, we can approximate its value by the
following logarithmic calculation.
Let
then
y =
logio y = logio (1 +
logio (1 + x).
By referring to a more extensive table of logarithms (an 8- or
10-place table), we obtain the values below.
X
(I/a;) log,o (1 + x)
y = (1 -f- x)U*
IO- 1
0.41393
2.5938
io- 2
43214
2.7048
10~ 3
0.43408
2.7169
10~ 4
43427
2.7181
-io- 1
45757
2.8679
-10~ 2
0.43648
2 7320
-io- 3
0.43451
2 7196
-io- 4
0.43432
2.7184
It is known that, as x approaches zero either from positive or
negative values, (1 + x) 1/z approaches a definite limit known as e,
which is one of the most important constants in mathematics. To
eight significant figures, it is
lim (1 + x) 1 / 2 = e = 2.7182818,
and to the same number of significant figures,
logio e = 0.43429448.
* Pierpont, The Theory of Functions of Real Variables, Vol. 1, 306-308.
78 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. Ill
PROBLEMS
1. Discuss e llx between x = 1 and x = 1.
2. Discuss e 1 ^ 1 -*) between x 1 and x 3.
3. Discuss and draw the graph of y \/(x 2) from x 3 to x = 4.
4. Discuss and draw the graph of y = 2/(x I) 2 from x = 3 to x = 4.
5. Find the nature of the function sin (1/x) as x approaches zero.
6. Draw a careful graph of y = (1 -f- x) l/x from x = 0.9 to x = 4.
7. Evaluate (a) lim x ~~ , (6) lim -?^. ^ins. (a) 4; (6) 1.
~ ~ 4 *o sin
o T 1 x / N V Sln 30 /, s |. 1 ~f COS 4
8. Evaluate (a) hm : . (b) lim ^
^.^0 sin V- ir/4 cos 2
^ Tl 1 A / \ 1- Sm3 ^ /tN 1-
9. Evaluate (a) lim . . . . (6) lim
v - - "\n 3 20' v - -
(a) 1/8; (6) 1/2.
56. Increments. If a variable changes from one value to
another, the difference of the two values, obtained by subtracting
the first value from the second, is called the increment of the
variable. Thus if x changes from x\ to o: 2 , the increment of x is
rr 2 x\. This increment is represented by the symbol Ax (read
" delta x ") so that Ax = rr 2 x\. When the variable x is given
an increment in a discussion or a problem, it is customary to
assume the values x and x + Ax rather than x\ and x 2 , respectively.
Now consider any function of x, such as f(x) = x 2 6 x + 7.
It is evident that a change in the value of the variable x will pro-
duce in general a change in the function. Thus, when x = 2,
/(2) = - 1, and if Ax = 4, that is, x + Ax = 6, then /(6) = 7.
As x changes to x + Ax the function changes from f(x) to
f(x + Ax) ; the difference f(x + Arc) f(x) is called Af(x), the
increment of the function. Or, if we call the function y, an
increment Ax assigned to the variable produces a corresponding
increment Ay in the function so that y + Ay f(x + Ax) and
Ay = f(x + Ax) f(x). Hence the increment of the function,
Ay t is expressed in terms of both the variables x and Ax.
EXAMPLE
Given the function y = x 2 6 x -\- 7. Calculate the increment of the
function.
SOLUTION. First assign the variable x an increment Ax. Then
y + Ay = f(x + Ax) = (x -f Az) 2 - 6(z + Ax) + 7.
But
^ = f( x ) = a; 2 - 6 a: +7.
57]
THE DERIVATIVE
79
Subtracting, we find
Ay = f(x -h Ax) - f(x) = 2 Z- Az -|- A? - 6- As,
which is a function of x and As.
57. Average Rate of Change. Consider the function y /(x).
A change of Ax in the variable produces a corresponding change
of Ay in the function, the ratio At// Ax is defined as the average rate
of change of the function with respect to the variable in the given
interval x to x -f- Ax.
Let the graph of the function be drawn and mark on it any gen-
eral point P(x, y). Give x an increment Ax and mark the point Q
on the graph whose abscissa is (x + Ax). Its ordinate is (y + Ay).
Figure 79 is the graph of the
function AY
y
- 6 x + 7.
X
FIG. 79
The points P and Q of the
graph are taken at (2, 1) and
(6, 7), respectively, which cor-
respond to the values of x and
x + Ax arbitrarily selected in
the preceding article. Here a
change in the variable from 2
to 6, making Ax = 4, produced
a change in the function from
1 to 7, making Ay = 8.
Hence in this interval the average rate of change of the func-
tion with respect to the variable, that is, Ay/Ax, is 2.*
From the graph above it is evident that Ay/Ax is the difference
of the ordinates of P and Q divided by the difference of the abscis-
sas taken in the same order and is, therefore, the slope of the secant
line PQ. The same statement would hold true for the graph of
any other function. Hence we have the following important
relation.
The average rate of change of a function with respect to a
variable in a given interval is equal to the slope of the secant line
* This does not mean that the function is changing twice as fast as the variable
throughout the interval. An examination of the graph reveals that in the first
part of this interval, as the variable changes from 2 to 3, the function is actually
decreasing; while in the latter part of the interval, as the variable changes from
5 to 6, the function increases from 2 to 7, which is five times the corresponding
change in the variable.
80 DIFFERENTIAL AND INTEGRAL CALCULUS [Cit. Ill
joining the two points on the graph of the function corresponding to
the extremities of the interval.
PROBLEMS
Express &y as a function of x and Ax arid find its value corresponding to
the values given in each of the following cases. (Nos. 1-8.)
1. y = 2x 2 , x = 2, Ax = 0.5. Ans. 4 x-Ax + 2 Ax 2 , 4.5.
2. y = x - 3 x 2 , x = - 1, Ax = 1.
3. y = x 3 - a; -f 4, a: = 2, Ax = 0.25. * _
4ns. 3 z 2 - Ax -f 3 a:- A? -f Ax 3 - Ax, 3*V
4. y = 5 x - x 3 .
5. j/ = x + 1 + 1/x 2 , x = 3, Ax = 2.
s. Ax - (2 x- Ax -f- Ax 2 )/X 2 (x + Ax) 2 , 434/225.
6. ?/ = x 1/x, x = 4, Ax = 2.
7. T/ = x 2 - 2/x 2 , x = - 3, Ax = 2.
Ans. 2 x- Ax + Ax 2 + (4 x- Ax -f 2 A?)/x 2 (x + Ax) 2 , 88/9.
8. y = x 3 - 2/x, x = 3, Ax = 0.01
Find the average rate of change of each of the following functions and
evaluate for the given interval. (Nos. 9-18.)
9. s = (1 + 0/(1 - for A*. Ans. 2/1(1 - t) (1 - t - A*)].
10. s = 100 t - 16 t\ t = 3, AZ = 2.
11. /(x) = 2 - x/(x - 1), x = 0, Ax = 0.5. 4ns. 2.
12. /(y/) = 2 y/(2 - 3 7/ 2 ), T/ = 2, AT/ = 1.
13. /(O = ^ 2 - -f 1A, < = 2, A< = - 0.5. Ans. 2|.
14. /(x) = x/(x 2 - 1) + 3, x = 2.5, Ax = - 0.5.
15. y = V2x + l, x 4, Ax = 0.3. 4ns. 0.328.
16. i/ = 1/V3 + , a; = 3, Ax = 2.
17. J(y) = l/y - 2/ 3 - 4, y = 3, Ay = 1. 4ns. - 37 A-
18. /(x) = V4 - x 2 , x = 1, Ax = - 0.25.
58. The Derivative. Many important properties of a function
of a variable are found with the aid of a related function called
the derivative of the function with respect to the variable. Let the
given function be y = f(x). The derivative is obtained as
follows :
First, assign to the variable x an increment Arc and calculate AT/,
the corresponding increment of the function.
58] THE DERIVATIVE 81
Next, divide Ay by Ax and evaluate the limit of this quotient as
Ax approaches zero. That is, find
Km -. fan /(
Ax
This limit is the derivative of y with respect to x.
We have seen ( 56) that Ay is a function of x and Ax. The
same is true of the quotient Ay/ Ax, but the limit of this quotient as
Ax approaches zero, namely, the derivative, is a function of x
alone.
The symbol most frequently used to represent the derivative
with respect to the independent variable x is d/dx. Thus the
derivative of y with respect to x is dy/dx or (d/dx)f(x). The
primed symbols y' or /'(x) are also used. Some texts use the
symbols D x and D x y in place of d/dx and dy/dx, respectively.
It is important to remember that whatever symbol is used it
represents the result obtained by performing the operations above
on the given function. Thus,
(1) -**
dx &c-+oAx
Similarly, if u is a function of t, then du/dt ~ lim (Au/At), or,
if s = f(v), then s f or/' (v) = lim (As/Av).
EXAMPLES
1. Find dy/dx if y = x 2 - 3 x.
SOLUTION. Give x an increment Ax and calculate A?/. Thus
y + &y = (x + Ax) 2 - 3(x 4- Ax),
?/ = x 2 3 x.
A?/ = 2 x- Ax + Ax 2 3 Ax.
Dividing both sides by Ax, and calculating the lim (At/ /Ax), we find
^ = 2 x + Ax - 3,
whence
lim ^ = lim (2 x -f Ax - 3) = 2 x - 3,
or
82 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. Ill
2. Given /(O = (2/0 + 5, find /'(/).
SOLUTION. Assign the independent variable t an increment At, then
whence
AffO = 2
A* f (J -f
and
3. Find d/dx of ax 2 -f 2 fez -f c.
SOLUTION. Let y - ax z + 2 bx -}- c. Then we have
y + Ay = a(x -h Ax) 2 -f 2 6(x + A*) -f c,
hence _
Ay = 2 ox- Ax + a- Ax 2 -f 26-Ax,
and
Ay
= 2 ax -f a- Ax 4-26.
Ax
lim ^ = lim (2 ax + a- Ax -f 2 6) =2 ax -f 2 6;
Ax *-o "^
Therefore
that is,
-T- (ax 2 + 2 bx -f c) = 2(ax -f- b).
The student should now be able to comprehend the following
definition.
DEFINITION OF A DERIVATIVE. Given a continuous function of a
variable, if the increment of the function is divided by the increment
of the variable, the limit of the quotient, as the increment of the variable
approaches zero, is called the derivative of the function with respect
to the variable.
All functions which we shall consider will be differ entiable,
that is, the derivative of the function is in general another con-
tinuous function which may become discontinuous only for
particular values of the variable. For any value of x for which
the limit of ky/kx exists, the derivative is said to exist.
PROBLEMS
of each
4.
2. y = x 3 - 5 x.
PROBLEMS
Find the derivative of each of the following functions. (Nos. 1-17.)
1. y = x 2 - x 4- 4. Ans. 2 x - 1.
I 59]
THE DERIVATIVE
83
Ans. 3 x 2 + 8 x - 5.
Am. - l/(x - 1)2.
Ans. 100 - 32 1.
Ans. 2 y - 1 -h 4/2/2.
Ans. 4/(2 - 3 x) 2 .
3. y = x 3 -f 4 x 2 - 5 x + 7.
4. i/ = (x 2 - x - I) 2 .
5. y = l/(x - 1).
6. /(x) = l/(3 -2x).
7. /(O = 32 -f 100 * - 16 **.
8. /(t/) = 2 -f 2//(i/ - 1).
9- /(t/) = t/ 2 - y - 4/y.
10. /(x) = x 2 - 1/x 2 at x = 1.
11. /(x) = 2x/(2 - 3x).
12. /(O = t* - (t - 2)/(t + 2).
13. /(s) = 2 a/(4 - 3 s 2 ). Ans. (84-6 s 2 )/(4 - ;
14. /(x) = Vl x. (HINT: Rationalize numerator of AI//AZ.)
15. /(x) = Vx + 1 - V5. An. l/(2Vx~TT) -
16. !/ = ox/(x a).
17. /(w) = a/(o 2 w 2 ). ATW. 2 aw/(a 2 w 2 ) 2 .
18. If ?/ has a derivative with respect to x for a given value of x, what con-
dition must be satisfied by Ay as Ax -* 0?
19. Find/'(0 if f(t) = (t - 3)- 1 / 2 .
Ans. - (1/2) (t - 3)->/ 2 .
20. Find the derivative of any of the
functions given in the problems on p. 80.
59. Geometric Interpretation of
the Derivative. Let us assume Fig.
80 to be the graph of y = /(x) with
P(x,y) any point on it. Give x an _-
increment Ax, then Ay/Ax is the
tangent of Z. RPQ, or the slope of
the secant line joining P(x, y) and
Q(x + Ax, y + At/) (see 57).
Let Ax approach zero, then, since /(x) is a continuous function,
A?/ approaches zero. That is, the point Q moves along the graph
and approaches P as a limit. Hence if At// Ax has a limit, it is
the slope of the limiting position of the secant line. But, by defi-
nition, we have
The tangent to any curve at a point P is the limiting position
of the secant joining P and another point Q on the curve as Q
approaches P.
FIG. 80
84 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. Ill
Therefore, lim (Aiy/Az) is the slope of the tangent line to the
Ax +Q
graph of y = /(x) at the point P(x, y}. Or, we can say:
The numerical value of the derivative dy/dx for any given value
x\ assigned to the variable is the slope of the tangent line at the
point on the graph ofy = f(x) whose abscissa is x\.
EXAMPLE
Draw the graph of 3 y = x* 2 x~ 4 x by making a table of values and
showing the slope at, each point marked.
SOLUTION. Finding the derivative, we obtain
4
4
*
\
Make a table of values of #, ?/, and
y'. Through each point (x, y) draw
an arrow in the direction indicated by
ij' y the slope of the tangent.
X
y
y'
-2
-8/3
16/3
-1
1/3
1
-4/3
1
-5/3
-5/3
2
-8/3
3
-1
11/3
FIG. 81
60. Physical Interpretation of the Derivative. Let two physi-
cal quantities be connected by a functional relation. Calling them
u and v we have, u = f(v). Then any change Af in the variable v
produces a corresponding change Aw in u. The ratio of these incre-
ments, Aw/Ay is the average rate of change of u with respect to v
in the interval Ay. Now let Av approach zero; then AM approaches
the limit zero, since u is assumed to be a continuous function of v.
Hence if Au/Av has a limit, it is defined to be the rate of change of u
with respect to v at the beginning of the interval. That is:
The lim (Aw/Ay), or du/dv is the exact rate of change of the
function u with respect to the variable v and is measured in units
of u per unit v.
Thus the distance s, in feet, of an object falling by the influence
of gravity is a function of the time t t in seconds. An increment of
time A changes the distance by As. Then As/A is the average
rate of change of s per unit t in the interval. This we call the
60]
THE DERIVATIVE
85
average velocity during the interval and is measured in feet per
second. The value of ds/dt for any given t is the exact velocity at
that instant in feet per second.
Also the temperature T at which water boils is a function of
h, the altitude. If T is measured in degrees C, and h in meters,
then dT/dh is the rate of change of degrees C per unit A, that is,
degrees per meter.
Again, the pressure p and the volume v of a gas in a container are
connected by a functional relation. If p is measured in pounds
per unit area and v in cubic inches, then dp/dv for any value of v
is the rate of change of p with respect to v, which is measured in
(Ibs. per unit area) per cu. in.
EXAMPLE
If a body falls from rest under the influence of gravity, the relation between
the velocity v in feet per second, and the distance fallen s in feet, is approxi-
mately v =* 8V 7 *. (a) At what rate is
v changing with respect to s when s is
4 feet? 36 feet? (b) For what value of s
are v and s changing at the same rate?
SOLUTION. To find dv/ds, give s an
increment A.s- and calculate Av. Then
v -f Av = 8 + As,
Av = 8(vY-f As - Vs),
Aw/As = S(Vs + As - Vs)/As.
Rationalizing the numerator, we have
Av/As = 8 As/As (Vs~+ As -f Vs)
whence
S2
= S/(Vs -f An -f
dv ,. Av
= hm -- =
S As->0 AS
32
FIG. 82
(a) When s = 4 ft., dv/ds = 2, that is, v is changing at the rate of 2 units
of v per unit change in s. When s = 36 ft., dv/dx = 2/3, and therefore v is
changing at the rate of 2/3 units v per unit s, or 2/3 (ft. /sec.) per ft.
(b) The velocity v is changing at the same rate as s when dv/ds = 1. That
is
4
^P = 1, s = 16 ft.
This does not mean that v and s are the same, for when s - 16 ft., v is
32 ft. per sec., but v is then changing at the same rate as s is changing.
Draw the graph of v &Vs, taking s as abscissa and v as ordinate. By
86 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. Ill
using dv/ds, we find the slopes of the tangents at s = 4 and s = 36 to be 2 and
2/3, respectively. Also the slope of the tangent is 1 at the point where s = 16.
We observe that * and v are numerically the same in their respective units
when s = 64, but then dv/ds = 1/2; that is, v is then changing 1/2 as fast as
s is changing.
PROBLEMS
Form a table for x, y, and dy/dx for each of the following equations and plot
points showing the direction of the curve at each point. (Nos. 1-10.)
1. y = x* + 4 x. 6. (y - 3)(x - 2) = 6.
2. x 2 =* 6 y. 7. y x 3 from x 2 to x = 2.
3. 2 xij = 9. 8. y = 3 x - x\
4. x = 4 y - y 2 . 9. xy - 5 x* -= 4.
^5. 3 x = (y - I) 2 . 10. 3 y = x* - 4 x.
11. Find the rate of change of the area of a sector of a circle of radius a
units with respect to the angle. Ans. a 2 /2 sq. units per unit of angle.
12. The distance a body falls from rest under the force of gravity is
s = if, /2 Find its velocity at any time. Find its acceleration. What is its
velocity arid the distance fallen after 3 seconds? What is its velocity after it
has fallen 48 ft.?
13. The law connecting the pressure and the volume of a fixed quantity of
gas at constant temperature is /;/; = c. What is the rate of change of p with
respect to v when v is 4 cubic units? What is the rate of change of v with re-
spout to /; when i> is 4 cubic units? How are these two rates related?
Ans. c/16 units p per unit v, 16/c units v per unit p] reciprocals.
14. A ball thrown vertically upward has its distance from the starting
point given by .s* = 100 / 16 < 2 . When does it stop rising? What is its
velocity at the end of 4 seconds?
15. Find the values of x for which the tangents to y = 3 x 2 and y x 3
are parallel. Ans. 0, 2.
16. Find the direction a particle is moving at the point determined by
x 2 if it follows the graph of y 2 x x 2 .
17. Find the rate of change of the volume of a sphere with respect to its
radius. Evaluate the rate when r 3 units. Ans. 36 TT cubic units per unit r.
18. Find the rate of change of the volume of a sphere with respect to its
diameter. Evaluate for r 2 units.
19. Set up the volume of a solid made up of a right circular cylinder with a
hemisphere on each end. If the length of the cylinder is twice the radius of an
end, find the rate of change of the volume of the solid with respect to the
radius of one end. Ans. 10 TIT* cubic units per unit r.
20. In Problem 19, find the rate of change of the solid with respect to its
total length.
60] THE DERIVATIVE 87
ADDITIONAL PROBLEMS
Find the average and exact rate of change of the following functions.
(Nos. 1-8.)
1. 3 x - x 2 . Ans. 3 - 2 x - Az, 3 - 2 x.
2. 4 - 2 y - y*.
3. 1/w - w 2 . 4ns. - l/w(u -f Aw) - 2 w - Aw, - 1/w 2 - 2u.
4. u/(u - 1).
5. v 3 - l/v. Ans. 3 v* + 3 v Ay + A? -f 1 M" + Au), 3 y 2 + 1/0*.
6. 2 a; 2 - 3/s.
7. 2/Vz - 3.
Ans. - 2/Vx - 3Vz -f As - 3(Vz -f 3 -f Vx + Ax - 3),
- (x - 3)~ 3 / 2 .
8. V.c -f i/j.
Draw the graph of each of the following equations with the help of the
derivative. (Nos. 9-14.)
9. y = 3 x + x 2 . 12. 3 a? = y 2 -f 2 y - 5.
10. ay = - 8. 13. 6 y = 4 x - x 4 .
11. 2 y = 3 - x\ 14. xy -2x 2 - 4 = 0.
15. Find points on xy 5 x 2 = 4 where the slope of the tangent is 1.
Ans. (1,9), (- 1, -9).
Express each of the following quantities as a function of the suggested vari-
able. Find the rate of change of each function with respect to its variable.
(Nos. 16-24.)
16. The area of a circle in terms of its circumference.
17. The volume of a box with square base as a function of its altitude if
h is 3 times a side of its base. Ans.
18. The surface of a box with square base as a function of a side of the base
if its volume is constant.
19. The total surface of a circular cylinder as a function of the radius of one
end if its volume is constant. Ans. 2 irr 2 + 2 F/r, 4 irr 2 F/r 2 .
20. The volume of a box made by cutting squares from the corners of a
rectangular sheet 12" by 6" as a function of the side of the squares.
21. The total surface of a cone as a function of its altitude if r = 2 h.
Ans. 2 7r/i 2 (2 + V5), 4 wh(2 + V6).
22. The volume of a cone as a function of its altitude if r = (h/2) .
23. The volume of a sphere as a function of the area of ajajreat circle.
Ans. (A/Gw)V^A, (1/4
24. The volume of a sphere in terms of its surface.
CHAPTER IV
DIFFERENTIATION OF ALGEBRAIC, LOGARITHMIC,
AND EXPONENTIAL FUNCTIONS
61. Derivation of Formulas. The method of forming the
derivative of a function, as explained in the last chapter, is per-
fectly general and can be applied to all differentiate functions.
However, by differentiating a special type of function we obtain a
formula which, when memorized, may be used to write down the
derivative of any function belonging to that type.
A thorough knowledge of the formulas derived in the following
articles is essential.
62. Derivative of a Constant. Let the function be a con-
tant c. Call it y and write y = c. Then any increment Ax
assigned to the independent variable x will not affect the function,
since it is constant. Hence we have
y + AT/ = c, and Ay = 0,
whence
Ax ~~ '
and
.. Ay dy
hm ~ = ~- = 0.
Ax >0 AX (IX
Therefore
> *-
The derivative of a constant with respect to any variable is zero.
This result is evident if we consider the graph of y c, which is
a straight line parallel to the x axis. For any two points (x, c)
and (x + Ax, c) on the line, the rate of change of y with respect to x
is zero. In other words, the slope of the graph is always zero.
63. Derivative of the Independent Variable. Let the function
be x. Then an increment Ax will produce the same increment in
the function y. That is, Ay = Ax.
Then
Ax = lf
65] DIFFERENTIATION 89
and
$H - 1
dx ~ '
or
<n>
derivative of a variable with respect to itself is unity.
Is this result apparent from the graph of y = x? Explain.
64. Derivative of a Constant Times a Function. Given the
function cu y where c is a constant and u is any differentiate function
of x. Give x an increment Ax; this will change u to u + Aw,
and the function y = CM to
j/ + A?/ = c(w + Aw),
then
AT/ = c-Aw,
A?/ Aw
_ !1 == __ J
Ax Ax
.. A?/ ,. Aw ,. Aw rfw
lim -~- = hm c = c Inn - = c -r- *
Ax AX-^O Ax AX K) Ax ax
AX+Q x AX-^O x AX K)
since w is a differen liable function of x. Hence
(Ill)
I4.A, W.A.
The derivative of a constant times a function is equal to the constant
times the derivative of the function.
65. Derivative of an Algebraic Sum of Functions. Let w, v }
and w be any differentiate functions of x, and consider the
function
y = u + v w.
Give x an increment Ax. This will cause w, y, and w each to
assume a corresponding increment. Then
y + At/ = w + Aw + v + Ay (w + Aw),
Ay = Aw + Ay Aw,
Ay _ Aw Ay __ Aw
Ax Ax Ax Ax '
dy ,. fAw . Ay Awl dw . efo dw
2. llTYl I -f- - I TT I- >
dx ^->oL^ x ^^ A^J ax dx dx
90 DIFFERENTIAL AND INTEGRAL CALCULUS [On. IV
by Theorem I, 52. Therefore
: + v w) du . dv dw
The derivative of an algebraic sum of functions is the corresponding
algebraic sum of the derivatives of the functions. A finite num-
ber of functions is assumed.
66. Derivative of a Power of a Function. Let y = u n , u being
any differentiable function of x. Give x an increment Ax, then
y + Ay = (u + Au) n .
Assuming n to be a positive integer and expanding by the bi-
nomial theorem, we have
y + A?/ = u n + nu n ~^Au + -( U ~ l ^u n -*-Au 2 H ----- h At?,
AT/ - nu n ~ l -Au
-&
Ay n Au n(n 1) Aw A?i T n-i
~ = m^" 1 - -- -- 10 u -^ Au + -+-- -Au .
Ax Ax 1-2 A.r Ao;
Taking the limit of both sides as Ax approaches zero and keeping
in mind that lim An = 0, since u is a continuous function of x, and
Ax
that du/dx exists, then
dy _ , du
~ = nu n ~ l -r- 9
dx dx
or*
/TTN
(V)
Since Formula V is known to be perfectly general, we shall use it
for all values of n.
67. Derivative of the Product of Two Functions. Let u and v
be any differentiable functions of x. We wish to find dy/dx when
* Formula V is derived on the assumption that n is a positive integer. We shall
prove (70) that the formula is true for n any rational number, and we shall discuss
the general case in 73.
68] DIFFERENTIATION 91
Assign to x an increment Ax. Then
y + Ay = (u + Au)(v + Ay),
Ay ~ U'Av + V'Au + AW'Ay,
and
Ay Av , Au , AU'Av
,_. Y_ ft j _ I at _ I _ f
Ax Ax Ax Ax
Then
* dy ,. Av . ,. Au , ,. Aii'Av
~~ u lim - -- h v lim + hm -
dx A*-MAz AX-+&X AX-+ Ax
The last term may be written either lim Au lim (Av/ Ax), or
Az-*0 Ax *0
lim (Au/Ax) - lim Au, either of which is zero since du/dx and
As K) Ax *0
dv/dx exist and lim Aw = lim Av = 0. Hence
Ax >0 Aar^0
cfa/ f?y , du
-j- = W-7- + V-j-9
ax dx dx
or
rtrr\ d(u-v) dv . du
(VI) -^ + '
derivative of the product of two functions is the first function
times the derivative of the second plus the second function times the
derivative of the first.
68. Derivative of the Quotient of Two Functions. Let u
and y be any differentiable functions of x. We wish to find dy/dx
where y is u/v, and v is not zero. Proceed in the usual way to assign
an increment Ax to x. Then
. . u + Au
v + *y-7+-&.
u + Au u v*Au U'Av
&y = ,
y + Ay y y(y + Ay)
Au Av
A|/ _ Ax Ax
Ax y(y + Ay)
Taking the limit of both sides as Ax approaches zero and remem-
bering that du/dx and dv/dx exist and that lim Ay = 0, we find
Ax >0
du dv
y ~~5i u ~^
ay _ dx ax
dx ^ '
92 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IV
or
du dv
The derivative of the quotient of two functions is the denominator
times the derivative of the numerator minus the numerator times the
derivative of the denominator all divided by the square of the denomina-
tor.
SPECIAL CASES, (a) If u c ; y = c/v, and
(Vila) --- -.-
v / dx\yj u 2 dx
This is the same result we would obtain by differentiating y = cv~ l
by Formula V, The student can now prove the validity of Formula
V for n any negative integer, by differentiating the function
l/v n by Formula VII a.
(6) If v = c, y = u/c ~ (l/c)Uj and
(VII 6)
d^ /u\ ^l du
dx \c) ~~ c ' dx '
This is just a repetition of Formula III, for if c is a constant so
is 1/c. This case should always be recognized as a constant
(1/c) times a function and differentiated accordingly.
EXAMPLES
1. Differentiate x 3 - 3 x 2 /2 + 5 x - 7 with respect to x.
SOLUTION. Let y x 3 3 x 2 /2 -\-5x-7, whence y is an algebraic sum
of functions. Then d(x*)/dx = 3 z 2 , by V; d(ll x*/2)/dx = 3 Z, by III and
V; d(5 x)/dx = 5, by III arid II; and d(7)/dx = 0, by I. Hence
^ = 3z 2 -3z + 5.
dx
2. Find dy/dx if y = (x 3 - 3 z 4 ) 5 .
SOLUTION. Here y i^ where u x* 3 # 4 . But, by V,
and, as in Example 1,
^ = A ( X * _ 3 3.4) 3 x * _ 12 x \
dx dx
Substituting for u and du/dx their respective values, we have
^ = 5(x 8 - 3 x 4 ) 4 (3 x 2 - 12 x 3 ) = 15 x 14 (l - 3 x) 4 (l - 4 x).
ax
68]
DIFFERENTIATION
93
3. Find dy/dx if y = (x 2 - 2)(x - 3 x 3 ).
SOLUTION. This function is a product u v where u is (x 2 2) and v is
(x 3 x 3 ). Then, by VI, we have
= (a* - 2)(x - 3*') + (u -
)~ (x 2 - 2)
(x 2 - 2)(1 -- 9 x 2 ) -Mx - 3 x 3 )(2 x)
- 15 x 4 + 21 x 2 - 2.
4. Differentiate (x 3 x 2 )/V2x 5 x 3 with respect to x.
SOLUTION. This is a quotient u/v, and hence by VII, calling y the function,
we have
dy = V2 x - 5 x*(l - 6 x) - (x - 3 x 2 ) - (1/2) (2 x - 5 x 3 )-" 2 - (2 - 15 x 2 )
ax 2i x ox 1
2(2 x - 5 x 3 )(l - 6 x) - (x - 3 x 2 )(2 - 15 x 2 )
2(2 x - 5x 3 ) 3 / 2
= 2x - 18x 2 4 5x 3 4- 15 x 4
2(2x-5x 3 ) 3 ' 2
PROBLEMS
Differentiate each of the following functions with respect to its variable.
1. 3 x 2 - 4 x 3 - 7.
2. 3 x 3 - 3/x 3 + x 3 /3.
3. ax 2 bx 4 c 4- dx~ l .
4. (3x - l) 2 (x - I) 3 .
5. (x 2 -2x) 3 .
6. (x 4 4- 4) (4 -x) 2 .
7. (a - x 2 )/(a 4 x 2 ).
8. 7/(# 3 4 8).
9. Vl 4 y 4- Vl - y. Ans. (V\
10. a/ 2 - 2) 3 / 2 .
Ans. 3 x(x 2 4 4) 1 / 2 - 3(4 - x)- 5 / 2 /2.
Ans. 6x 12 x 2 .
A ns. 2 ax b dx~*.
Ans. 6(x 2 - 2x) 2 (x - 1).
Ans. - 4 ax/ (a + x 2 ) 2 .
11. (x 2 -f 4) 3 / 2 - (4 - x)~ 3 / 2 .
- 2) 3 (5 2/ 3 -f 12 y 4- 2)/(i/ 2 4 2) 2 .
. (4 4-2x - 3x 2 )/2(l -x) 3 / 2 .
. (1 - 2 x 3 )/3 x 2 / 3 (x 3 4- 1) 4/3 .
94 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IV
20. xVx 2 a 2 a z x/Vx 2 a 2 .
21. [(a + bx*)/(a - bx z )J'*. Ans. (abx)/(a -f 6z 2 ) 3/4 (a - kr 2 ) 5 / 4 .
22. (a 2 -
23. (Vx* + 1 -f Vz* - l)/(Vs* -f 1 - Vz 2 - 1).
.4ns. 2 (a; -f ,
24. VaJ 2 2 &J/V2 c d in three ways, using V, VI, and VII.
25. Suppose y = au -f &tf 2 > where w and w are differentiable functions of x.
What formulas are used when you obtain dy/dxl
26. Suppose y = (au - v 2 )/w l(2 , where u, v, and w are differentiable functions
of t. What formulas are used in obtaining dy/dt?
69. Derivative of a Function of a Function. If y is a function
of u and u in turn is a continuous function of x, then an increment
Ax assigned to x will produce a corresponding increment Aw in u,
and AT/ in T/. For any value of these increments; provided Au is
not zero,* w r e have
Ay _ Ay Au
Ax Au Ax
whence, taking the limit of both sides, we find
Au
dy ,. FA?/ Awl ,. Ay .. A
= lim ~ - = hm - lim
since Au approaches the limit zero as Ax approaches zero. Then
9
)Ax
..
dx "" C/l/ (/JC
// ?/ /5 a function of u and u is a function of x, the derivative of y
with respect to x is the product of the derivatives of y with respect to u,
and of u with respect to x.
To express y directly as a function of x, we must eliminate u
between the two given functions.
From Formula VIII we have at once:
dy
(Villa) % = **
du du
dx
* For a proof of VIII when AM = 0, see Pierpont, Theory of Functions of Real
Variables, Vol. I, p. 234.
71] DIFFERENTIATION 95
70. Derivative of Inverse Functions. If the given function is
expressed as the inverse of y = f(x), namely, x = <p(y) (see 48),
then for corresponding increments Ax and A?/, provided Ax ^ 0,
*y =sJL.
Ax "" Ax
A^
Taking the limits of both sides as Ax approaches zero, we have
dy
That is, the derivative of y with respect to x is the reciprocal of the
derivative of x with respect to y.*
In this case dx/dy, that is <}>' (y), is expressed in terms of y, and
its reciprocal dy/dx will be given also in terms of y.
71. Parametric Equations. Suppose x and y are both given in
terms of a parameter , that is,
x = 0(0, y =/(0.
From 69 it follows at once that
dy
dy _ dy dt __ dt
dx dt dx dx
lit
* Since Formula V has been proven valid for n any negative integer ( 68) as
well as positive, now we can show it is valid for n, any rational number. If p and q
are any integers, then consider the function,
y = u l/t * t where u = v p .
Then
yq = u t and y = v p ^.
Since u is an integral power of both y and v,
dy di)
But by IX,
du q q
Then, since dy/du exists, we have, from VIII,
Hence for v and y, differentiable functions of x, from VIII we have
^ = P rP /
dx Q
tvhere n is any rational number.
= rP fl-i, or = m>-,
dx Q dx dx dx
96 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IV
Therefore
dx g'(t)
This formula will give dy/dx in terms of t.
72. Differentiation of Implicit Functions. Given a function of
x and y, say f(x, y) = 0. It is possible to find dy/dx without
.solving the equation f(x, y) = for either variable explicitly in
terms of the other.
If each term in /(z, y) = be differentiated with respect to x,
the resulting expression df/dx = will contain terms in x, y, and
dy/dx. Solving this equation for dy/dx, we have the desired result
in terms of x and y.
EXAMPLES
1. Find ds/dt if s = (1 + r)/(l - r) and r - V2 * - J 2 .
SOLUTION.
tte = (1 -r)(l) -ft + r)(-~ 1) ^ 2
dr (1 -r) 2 (1 -r) 2 '
Also
Hence, by VHI,
da _ 2(1 -
dt "~
2. Find rfy/da; when 3 = Vl + if/(\ - ?/).
SOLUTION. First find dx/dy.
4- ?y 2
dy (1 - 2
-f ?/ 2 ) 1/2 +
Therefore, by IX,
dx 1 +y
3. Find c/y/c/x if x = 3 a/(l -f ^ 3 ) and y =, (
SOLUTION.
^ = (1 -f ^ 3 )(6aQ - (3a**)(3* 2 ) = 3 at(2 - t 3 )
dt ~ (1+ 1*)* (1 -f * 3 ) 2
72]
DIFFERENTIATION
97
Also
dx ^ (1 -M 3 )(3a) - (3a<)(3< 2 ) 3 o(l -
df
(1 +
(1 + < 3 ) 2
Then, by X, we have
dy
1(2 -
3a(l - 2
1 -
4. Find dy/dx if z 3 + ?/ 3 - 3 azr/ -f a 3 = 0.
SOLUTION. Differentiating implicitly with respect to z, we have,
Solving for dy/dx, we get
Find dy/dx if:
4-
d|/ __ ay x 2 1
dx ^ 2 ax
PROBLEMS
4 y + 1).
. 1/[2(1 -
- 3 z + x - 1)].
Find dx/dy if:
3. y = (x 2 -
4. y = (x 2 - 2)/(l - x 2 ).
Find dy/dx and dx/dy if:
5. x = J - < 2 , y = < + < 2 . Ans. (1 -f 2 /)/(! - 2 /), (1 - 2 /)/(! + 2 0-
6. x = I 3 - 3 , y = 1 - 2 t*.
7. x = 3 - 0\ y = 2 3 . 4na. - 3 (9, - 1/3 0.
8. x = < 3 , y = (1 - * 2 )3/2 at t = 1.
9. x = 2 J, y = 2Vt*~-t at ^ = 2. Ans. 3/2V2, 2>/2/3.
10. x = a t,y-=bt- gt*/2 t and find * if cfy/da; = 0.
11. x 2 + y 2 = a 2 . Ans. z/#, y/3.
12. x 2 / 3 + y 213 = a 278 .
13. x* + xy + y* = a\
Ans. - (2 x -f 2/)/(2 y + *), - (2 y + *)/(2 a; + y).
14. t/ 2 (2 a x) = x 3 at x = o.
15. a: 2 ?/ -f 4 a 2 t/ = 8 a 3 at (2 a, a). Ans. 1/2, 2.
16. z 2 - 4 xy -f x + y + 3 = 0.
17. y\x + 2 y) = x - 2 y. Ans. dy/d* = (1 - y 2 )/2(l + asy + 3 t/ 2 ).
98 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IV
18. x - Vxy + y = 4.
19. *V - x/y = 7.
4rw. (y - 2 xy*)/(x + 2 x 2 */ 3 ), (x +
20. Vx + y + Vz y = 1.
21. a; 2 4- (a 2 ?/) 2 / 3 = a 2 . Ans. dy/ds = - 3
73. Derivative of the Logarithm of a Function. Let the func-
tion be y = loga u and assign to u an increment Aw, then >
2/ + AT/ = log a (w + Aw),
ky = loga (W + AW) -
A?/ 1 , / A?A
-~- = log ( 1 H )
Aw Aw y w f
To evaluate the limit of the right-hand member as Aw approaches
zero, it is written in the following form:
A?/ w . / t . A?A 1, / Aw\ tt/AM
Ji lO tT I 1 ~T~ I \r\rr I I _1_ 1
Aw w Aw na
Now
A?A 1, / Aw\
I = - lo S [ ! H -- I
w y w rt y w y
'?/ i ,. r /, . A?A W / AM "| i. r r / t . AW\ M/AM I
r- = - lim loga [ 1 H -- ) = - loga hm ( 1 H -- )
w w A u_>oL V u ) J u L A -^o\ u) J
du
But, by 55 (c), we have
(A?/\ u/Au
1 + J = Hm (1 + x)v* = e.
^ /
^ X-+Q
Therefore*
,.
du u
Hence, if w is a differentiate function of x, we have, by VIII,
. M
v x dx u 6 c/jc
* Certain assumptions are involved in this proof. One is that u is definitely not
zero, and must be positive if the function log a u is real. Another is that the limit
of the logarithm of a function is the logarithm of the limit of the function. This is
true under the existing conditions but the proof belongs in a more advanced course.
Still another is the actual existence of the lim (1 -f~ Au/u) M /A which has already
been pointed out in 55 (c). It will be understood that the use of formulas XI
and XIa involves these assumptions.
74] DIFFEkENTIATION 99
If y = log Uj this becomes
, YT x d(\og u) _ 1 du
(XIa) dx -n'dT
NOTE: The proof that Formula V is valid for any value of n,
irrational as well as rational , depends on Formula XI a, provided
the function y u n satisfies the restrictions imposed by the use of
that formula. (See footnote on p. 98.)
That is, taking the logarithm of both sides, we have
log y = n log u.
Since u is a differentiate function of x, so is log u, also n log u or
log y; hence, differentiating implicitly, we have
1 dy _ I du
n * "~j .
y dx u dx
or
dy ydu . du
" /y\ x. . - / fil/
dx u dx dx
The formulas for differentiating the product of functions, and
the quotient of two functions, are readily obtained in a similar manner.
EXAMPLE
Find dy/dx if y = log [vV -f x 2 /(2 ax - x 2 )].
SOLUTION. From the fundamental laws of logarithms we can write y as
follows:
y = (1/2) log (a 2 -f x 2 ) - log (2 ax - x 2 ).
Then, by XIa, we have
fy. = I 2x _ 2a - 2x = x _ 2(o - a?)
dx " 2 ' o 2 -f x 2 2ax - x* ~ a 2 + x* 2 ax - x* '
or
<fy = a 8 -f- 2 a 2 x - 2 a 8
da? ~ (a 2 + z 2 )(2az - x 2 )*
74. Derivative of the Exponential Function. Assume the
function to be y = a u , where u is any differentiate function of x.
Take the natural logarithm of both sides, then
log y = u log a.
Since u is log y times a constant, du/dy exists, likewise its reciprocal
100 DIFFERENTIAL AND INTEGRAL CALCULUS [On. IV
dy/du, and therefore dy/dx. Then, differentiating with respect to
x, we get
1 dy du ,
- -~ = -r- log a,
y dx dx &
and
dy , cu
p = y log a -7- ,
dx dx
or
,,___. d(c u ) n , cfi/
(xii) ^r = alo & a d~x'
If a = e, this becomes
du
If ?y = e x , we have
(XII ft)
Hence the exponential function e x has the interesting property
that its rate of change with respect to the variable is always equal
to the value of the function. Likewise the graph of the function
e x has the slope at every point equal to the ordinate of that point.
EXAMPLES
1. -^ (3* 8 / 2 ) = log 3 3' 8 / 2 3 x 2 /2 - (3/2)z 2 3 *'/ 2 log 3.
2. Y (e 1 -* 2 ) = e l ~**( - 2 x) - - 2 .re 1 -* 2 .
PROBLEMS
Find the derivative of each of the following functions with respect to its
variable. (Nos. 1-12.)
L 3 log (x* + 4). Am. 6 x/(x* + 4).
2. logVr=V-
3. log V(T~i:"^)/(l -f
4. log v 7 ^ 3 - 1)(J - i 2 ).
5. 3 t-*. .Arts. - 6 fcr".
6. 4* 2 - 1 .
7. e s * - 2 c ll *. Ans. 3 e 3 ' + 6 e s "/^
: 74]
DIFFERENTIATION
101
8. e 2 *(9x 2 -fGx -2).
9. Iog 10 (2x-f VI^Tl). Ans. 2 Iogi c/V4x 2 -f 1.
10. (log a; 2 ) 2.
11. ?/ = x r . (HINT: Take log of both sides.) Ans. x x (l -f log x).
12. c 2 = (I -f c 2 *)/(l - e 2 *).
Find dy/dx and dx/dy in each of the following cases (Nos. 13-15.)
13. x = c 2t , y = te l , evaluate at t = 1. Ans. 1/c, e.
14. x = 2%'', ?/ Z log /, evaluate at = 1.
15. x = 2^ 2 , y = te- 1 . Ans. (I - l)/(3c t Vt) t (3
Find dy/dx in each of the following cases. (Nos. 16-25.)
16. y = log V(x -h l)/(x - 1).
17. y = log lV2x -3/(^ 2 - a; 8 ) 2 !
18. x = c
19. y = x log Vl - x.
20. y = log 3 (xe 2x ).
21. y = e x * log x 2 .
22. xy = 4 log (xy).
23. e*" - 4 x?/ - 2.
24. x log y 2 + y log x 2 = a.
25. If x = Va 2 - ?/ 2 - a log [(a 4- Va' 2 - y
\x = (11 x 2 - 25 x 4- 12)/x(l - x)(2 x - 3).
Ans. dy/dx = x/2(x - 1) 4- log Vl - x.
Ans. dy/dx = (2 e* 2 /x)(2 x 2 log x 4- 1).
'. - yjx.
find dy/dx.
Ans. 7//Va 2 y 2 .
ADDITIONAL PROBLEMS
Find the derivative of each of the following functions. (Nos. 1-7.)
1. y = ax 3 - bx 2 + c/x*. Ans. 3 ax 2 - 2 bx - 3 c/x 4 .
2. x - (3y -4)(27/ - t/ 2 ) 2 .
3. 5 = (3 - 2 < 4- ^ 2 ). -Ans. ds/d* = 3 - 4 t -f 3 / 2 .
4. y = x/(x 2 -f 4).
5. y - x/(x - Vx 2 + 4).
= (x -f- Vx 2 + 4)/(x - Vx 2 -f- 4) Vx 2 + 4.
6. y = log (x - Vx 2 -f 4).
7. # = log V(e x e~ x )/(e x 4- e~ z ). Ans. dy/dx 2/(c 2x e~ 2x ).
Find the derivative of each variable with respect to the other in Nos. 8-11.
8. xy = (x 4- ?/) 2 .
9. xy = x 4- e~ v . Ans. dx/dy = (x 4- <
102 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IV
10. xy -f 2 x 4- 3 y = 6, and find what value of x makes dy/dx = 0.
11. x = e-<, y e* t+l . Ans. dy/dx = - 2 e ai+1 , dx/dy = - e~ 3t - l /2.
12. Find the slope of y = ze* at the point where x = 0. Also find x where
the slope is zero.
13. Find the slope of x* + 3 xy + y 2 f= 5 at (1, 1). Ans. - 1.
14. Draw the curve y = e~* 2 using the derivative at each point plotted.
15. The name as Problem 14 for y = log x.
16. The same as Problem 14 for y e l/x .
17. The same as Problem 14 for y = xc~ z .
18. Show that y = (a/2) (e zfa 4- e~ x/n ) and y = a -f 2 /2 a have the same
slope at their intersection (0, a). This means the curves are tangent to each
other.
Find the derivative of each of the following cases. (Nos. 19-29.)
19. y = a 1 *- 1 . Ans. 2 xa**- 1 log a.
20. y = log </(x - 3)/(x + 3).
21. y = 3 c*- 1 /* 2 . Ans. 3(1 -f 2/z 3 )e*- l / a:3 .
22. y = cu; 2 a* 3 .
23. = a"*. 4ns. a a +1 log a.
24. 5* 2 * - log 2 (xy).
25. s = 2~" - (2 O 2 . 4ns. - 2 l ~ 2t log 2 - 8 *.
26. w = e 1 -' log V/72^~9.
27. x 2 -f x?/ - ?/ 2 = 0. Ans. dy/t/x = (2 x -f y)/(2 y + x).
28. x 2 - 2 x Vx 7 ?/ + 2 2/Vxy - ?/ 2 = 0.
29. a; = log 2 , ?/ = log 2 ^ at t = e. Ans. dy/dx = dx/dy = 1.
30. Find the slope of x 3 // 2 + 1 = x -f 2 y at (1, 2).
31. Find the slope and the rate of change of the slope with respect to x for
xy = x-y. Ans. (1 - y)/(l + x), 2(y - !)/(! -f *) 2 .
32. Find the slope and the rate of change of the slope with respect to x for
x = 3 t 2 , y = 2 t z and evaluate each at t 2.
CHAPTER V v-
SOME APPLICATIONS OF THE DERIVATIVE
75. Tangents. Normals. We have seen (59) that at any
point PI(#I, ?/i) on the graph of y f(x) the slope of the tangent
is the value of the derivative for x x\. If the derivative has
been obtained implicitly, and is expressed in terms of both x and ?/,
then the substitution of x\ and y\ for x and y, respectively, will give
the slope of the tangent at PI. That is,
dyi , , ,.
= = w i ^ constant).
Then the equation of the tangent at P\ is ( 23),
(1) y y\ = mi(x xi).
The normal to a curve at a point PI is the line perpendicular
to the tangent at PI. Hence the slope of the normal is (7)
__ L = _ *:! = _ JL
dy\ dyi mi '
dx\
and the equation of the normal at PI is
(2) y - yi = - ^ (x - *).
76. Angle of Intersection of Two Curves. By the angle of
intersection of two curves is meant the angle between the tangents
to the respective curves at a point of intersection. Let PI be a
point of intersection of the curves whose equations are y f\(x)
and y = fz(x). Then, if the slopes of the tangents at PI are Wi
and ra 2 , the angle of intersection ft is (8) such that
(3) m m2
v ' r 1
where m\ is the slope of the tangent with the greater inclination. Or,
103
104
DIFFERENCIAL AND INTEGRAL CALCULUS [CH. V
if the inclination of each tangent line is found after the slope is
obtained, then ( 8)
(3 cr) P = i a 2 .
EXAMPLES
1. Find the equations of the tangent and the normal to the curve
y = x 2 + 4 x + 2 at the point where the tangent is perpendicular to the
line 2 x - 4 y + 5 == 0.
SOLUTION. The slope of the given line is m = 1 /2, hence the slope of the
required tangent is 2, and that of the normal is 1/2. From the equation
y = x* + 4 x + 2
we have, on differentiating,
The point on the curve which has its
slope 2 is located then by
2 x + 4 = - 2.
Hence the point of contact desired is
(-3, - 1), and from (1) and (2) the
tangent and normal are, respectively,
2z+7/ + 7=0,
FKJ. 83 and
x 2y + 1 =0.
2. Find the angle between the curves
X 2 - 3 y = 3 and 2 z 2 + 3 ?/ 2 = 30,
at their intersection in the second quadrant.
SOLUTION. Solving the equations simultaneously, we find the real inter-
sections are (3, 2) and ( 3, 2). Differentiating, we obtain, from the first
equation,
d Ji = ?_?
dx 3
and, from the second equation,
dy = __2x.
dx 3y'
Hence, at the point ( 3, 2), these slopes are
2x 9
-- - - 2 = mi,
76] SOME APPLICATIONS OF THE DERIVATIVE 105
and
~~ 3]y "" ~ m2 '
respectively. Hence, by (3),
tan = ~ 2 _"^ 1 =3, = 71 33.9'.
PROBLEMS
Find the equations of the tangent and the normal at the required point on
each of the following curves. (Nos. 1-11.) Draw the graph in each case.
1. xy = 8, at (2, 4). Ans. 2x-ft/-8=0, x-22/ + 6 = 0.
2. y x 2 -f- 2 x -f- 3, at the point where the tangent is perpendicular to
x - 2 y = 2.
3. y 2 3 4- 4 x 2 x 3 , where the tangent has the inclination 45.
Ans. x y -f 2 = 0, 27 x - 27 y -f 22 = 0, x -f 2/ - 6 = 0,
27 x + 27 y - 58 = 0.
4. y = x 3 3 x 2 6 x -f 12, at the point where x = 2; at the points
where the slope is 3.
5. y = x 3 + 4x 2 , at (- 1,3).
6. x 2 2 xy -f 4 y = 0, where the slope is 3/2.
7. x 2 -f- 4 ?/ 2 = 8, at the point in the first quadrant where the tangent is
parallel to the line through the positive ends of the major and minor axes.
8. y = log x 2 , where the tangent is parallel tox 2 ?/ -f 6 = 0; perpen-
dicular to x -}- y = 1.
9. y = x log x, where the tangent has slope 3/2.
Ans. 3 x 2 # 2 Ve = 0, 4 x -f- 6 y 7VJ = 0.
10. y = log (2 x e), at x = e.
11. y = 2e~ x / 3 , at the crossing of the y axis.
Ans. 2x-f3y =6, 3x 22/-J-4 = 0.
12. For what values of x are tangents on 3 y = 4 x 2 and y ~ x 8 perpen-
dicular?
Find the angle of intersection between each of the following pairs of curves.
(Nos. 13-21.)
13. x + y + 2 = 0, x 2 + y* - 10 y = 0. Ans. tan-*(l/7).
14. xy = 2, x 2 -f 4 y = 0.
15. y 2 = 4 x, x 2 + 2/ 2 = 5. Ana. tan^- 3).
16. x 2 + 3 y = 3, x s - j/ a -f 25 - 0.
106 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
17. x 2 = 2(y + 1), y - 8/(o; 2 + 4). Ans. r/2.
18. 2 y 2 9 x =* 0, 3 s 2 + 4 y 0, in the fourth quadrant.
19. ?/ 2 = 8 x, 4 x 2 + 7/ 2 = 32. 4ns. tan~*(3).
20. x 2 // = 4, r/(z 2 + 4) = 8.
21. 2/ = (l/2)6'- x/2 and the ?/ axis. Ans. tan~ 1 (4).
22. Docs y = 3 x bisect the angle between y = 2 x and y = 4 x? Prove
your answer.
23. Show that a tangent to a parabola makes equal angles with its axis and
with the line from the focus to the point of contact.
24. Show that the tangent to Vx + v^ = \Ta at PI is yyr 1 '* + xxr 1 '* = a 1 / 2 .
25. A tangent to the curve xy c forms a right triangle with the coordi-
nate axes. Show that the area of this right triangle always has the constant
value 2 c sq. units.
77. Increasing and Decreasing Functions. A function of a
variable is said to be an increasing function if it increases as the
variable increases. It is a decreasing function if it decreases as the
variable increases.
Consider the graph of the function, say y = f(x), and trace the
curve from left to right so that the variable (or abscissa) is increas-
ing. Then the function (or ordinate) is increasing if the curve is
rising; it is decreasing if the curve is falling.
Since the derivative gives the rate of change of the function
with respect to the variable, if the derivative is positive, the junction
is increasing; if the derivative is
negative y the function is decreasing.
Recalling the definition of the
derivative, we see immediately
I \ that the sign of Ai//Ax will be
1 (C>0 1 positive or negative according as
(p) ^ TY ^h e infinitesimals AT/ and Az have
the same or opposite sign. In
Fig. 85, any value of x in the in-
terval from x = a to x = b will
FlG 85 make the derivative positive and
throughout this interval, y in-
creases as x increases. Again, in the interval x = b to x c, the
derivative will be negative and throughout this interval, y decreases
as x increases. For any value of x for which the function changes
from an increasing to a decreasing function, or conversely, the de-
77] SOME APPLICATIONS OF THE DERIVATIVE 107
rivative must change sign; hence it must be zero, if it exists at
all, for that value of x. At the corresponding point on the graph
of the function, the slope being zero, the tangent is horizontal.
EXAMPLES
1. Find the intervals of the variable x in which the following function
y = f(x) 3 -f~ 12 x Q x 2 -{- 2 x 3 increases and decreases, respectively.
SOLUTION. Find the values of x for which the derivative is zero, that is, for
which the function may change from increasing to decreasing, or conversely.
Now
^ = 12 - 18 x -f 6 x z = 6(2 - \)(x - 2).
When dy/dx 0, x = 1, 2. Since dy/dx is continuous, the sign of the
derivative can change only at x 1 and x 2. Hence it has one sign in each
of the intervals z < 1, I < x < 2, and x > 2. Try values of x in each interval.
Thus for x = 0, dy/dx = 12 and hence is positive for x < 1. Similarly,
x = 1.5 makes dy/dx = - 3/2 or dy/dx < for 1 < x < 2. Also x = 3
makes dy/dx 12 and so dy/dx > for x > 2. We may represent this by
the following diagram.
x = 1
x = 2
x < 1
dy/dx positive
y increasing
1 <x < 2
dy/dx negative
y decreasing
x> 2
dy/dx positive
y increasing
2. Find the interval of time t in which a body moves in the direction
in which its distance s from a fixed point is measured positively, if
s - - 20 - 24 t + 9 t 2 - t\
SOLUTION. As in Example 1, set the derivative equal to zero. Then
J<?
~ = - 24 -f 18 t - 3 t* = - 3(* - 2)(t - 4) = 0,
whence
t = 2, 4.
Therefore, using t =* 0, 3, and 5, we get the information shown below.
t = 2 t = 4
t <2
ds/dt negative
s decreasing
2 <t < 4
ds/dt positive
s increasing
t> 4
ds/dt negative
s decreasing
Hence the body moves in the direction in which s is measured positively
in the interval 2 < t < 4.
108 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
PROBLEMS
Find the intervals of the variable in which each of the following functions
ncreases and those in which each decreases. (Nos. 1-7.)
1. y = x 3 3 x 2 . Ans. Increases x < 0, x > 2; decreases < x < 2.
2. y = x*(x - 2) 2 .
3. y = x 3 - 3 x* - 6 x -f 12.
4ns. Increases 3 < 1 Vs, s > 1 -f V3;
decreases l-V3<a;<l + V3.
4. s = 2 t* - 2U 2 -f 60 t + 5.
5. s = t 4 /4 - 7 *'/3 - 4 t* - 2.
Arcs. Increases 1 < # < 0, x > 8',
decreases x < 1, < x < 8.
6. y = ze*.
7. s = log t/t. Ans. Increases < t < e] decreases t > e.
8. If p = 3 v z 7 v 2 4- 4, when is dp/dv increasing?
9. If 6 ,s = 2 2 3 3 2 -h 12 t 4, where s represents the distance of a
particle from a fixed origin, for what interval of t is the particle moving in the
iirection opposite to which s is measured positively? Ans. 1 < t < 2.
10. Tho height of a ball is given by h = 120 t 16 t 2 . How long and how
high will it rise?
11. The position of a point on a straight line as given by its distance ,s from
some starting point is represented by s l*/4 8 / 3 /3 -f- 10 t* 16 t + 7.
When is the motion opposite to the positive direction for 5? Ans. t < 4.
12. Whore does the slope of y x 4 /12 -f # 3 A> ~ # 2 -f 3 z decrease?
13. A particle moves along a line with a velocity given by v = 1 -}- 3 2 2 t*.
When is its velocity increasing? When is its acceleration decreasing?
Ans. < t < 1; t > 0.5.
14. A variable rectangle is inscribed in the area bounded by the parabola
r 2 = 8 y and its latus rectum. One side of the rectangle lies along the latus
rectum. As one vertex of the rectangle moves along the curve so that y
increases from y to y = 2, for what values of y will the area of the rec-
tangle increase?
15. A variable rectangle is inscribed in a circle of radius a units with sides
parallel to the reference lines. As a vertex P(x, y) moves along the circle from
a. position where x = to x = a, for what values of x will the area of the
rectangle decrease? Ans. x > a V2/2 units.
78. Second and Higher Derivatives. We have found that
the derivative of a function of a variable, as f(x), is in general
another function of that variable, f'(x). This new function can
be differentiated with respect to the variable giving what is known
79] SOME APPLICATIONS OF THE DERIVATIVE 109
as the second derivative of the original function. This we repre-
sent by /"(#). Similarly, the derivative of the second derivative,
if it exists, is/'"(z), the third derivative of the original function.
Calling the original function y we have the corresponding symbols
for the higher derivatives.
Second derivative, /"(*) = = y" = d = Dly.
Third derivative, J'"(x) = = y 1 " = ~ = Dly.
From the meaning of the derivative, it is evident that dy'/dx
is the rate of change of y f with respect to x. Interpreted on the
graph of y = /(x), this means the rate of change of the slope of the
tangent with respect to the abscissa of its point of contact. Also
( 77) y' is an increasing or decreasing function according as y" is
positive or negative. Then as x increases, if y" is positive the
slope of the tangent is increas-
ing; that is, in moving along
the curve to the right the tan-
gent will continually turn in a
counter-clockwise direction so
that the curve will be concave
upward. This is illustrated in
Fig. 86 along the arcs A to /?,
and C to D. If y" is negative,
y' is decreasing; that is, in mov-
ing along the curve to the right, FlG
the tangent will turn in a clock-
wise direction so that the curve is concave downward. This is il-
lustrated along the arcs B to C, and D to E.
79. Points of Inflection. If the derivative y f changes from an
increasing to a decreasing function as x increases, that is, if the
curve changes from concave upward to concave downward, as at
B or D in Fig. 86, then the second derivative, y", changes from a
positive to a negative value. Likewise, if y' changes from a
decreasing to an increasing function, as at C, then y" changes
from a negative to a positive value. At such points of the curve,
B, C, and D, the second derivative changes sign and becomes zero if
it exists; otherwise it may become infinite. These points on the curve
110 DIFFERENTIAL AND INTEGRAL CALCULUS [On. V
FIG. 87
at which the direction of concavity changes are called points of
inflection. At a point of inflection the tangent crosses the curve,
since an arc which is concave upward is above the tangent at any
point of the arc, whereas if the arc is concave downward it is below
the tangent at any point of the arc.
EXAMPLE
1. If the equation of a given curve is
y = 2 x 3 - 9 x* + 12 x - 3, for what values
of x will it be concave downward and for
what values will it be concave upward?
SOLUTION. This is the same function
which is given in Example 1, 77. Then
y' = G(x 2 -3x-f 2), y" = 6 (2 x - 3).
Hence y" is positive or negative ac-
cording as x > 3/2 or < 3/2. Hence the
graph is concave downward to the left of
x 3/2 and concave upward to the right
of x = 3/2.
PROBLEMS
Find the second and the third deriva-
tives of eiich of the following functions. (Nos. 1-6.)
1. y = x</2 - 3 x z . Ans. 6 x 2 - 6, 12 x.
2. y - :rV12 +93+4.
3. y = e -*\ Ans. - 2 e~ x * (1-2 x 2 ), 4 xe~ x * (3-2 x 2 ).
4. y = xe*.
5. y = log 2 x. Ans. 2(1 - log x)/x\ 2(2 log x - 3)/z 3 .
6. s = Va 2 - t 2 .
Find d*y/dx* and d*y/dx*. (Nos. 7-9.)
7. x = t - t 2 , y = t + t 2 . Ans. 4/(l - 2 *) 3 , 24/(l - 2 O 5 .
8. x = * + 1/J, y =* t - l/t.
9. Find ete/df, c/ 2 s/r// 2 if s 2 - s + * 2 = 0.
Ans. 2 t/(l - 2 s), 8 * 2 /(l - 2 s) 3 + 2/(l - 2 s).
Find the points of inflection of each of the following curves and observe their
intervals of concavity. (Nos. 10-19.)
10. y = z 3 - 3 x 2 .
11. 12 y = x 4 - 4 x 3 - 18 x 2 + 26 x + 51.
Ans. (1, 1), (3, 5); upward for x < 1, x > 3, downward for
- 1 < x < 3.
12. y = x* 4 x 2 . Find the direction of the tangent at the inflection.
80] SOME APPLICATIONS OF THE DERIVATIVE 111
13. y = e-* 2 . Ans. ( A/2/2, e- 1 ' 2 ) ; upward x < - A/2/2, x > V2/2,
downward - V2/2 < x < V2/2.
14. ?/ = 4 x 3 6 x 2 + 3. Find the slope of the inflectional tangent.
15. y = log (x 2 - 2x + 3).
Ans. At x = 1 =b V2; upward 1 A/2 < x < 1 -f- A/2, downward
x < 1 - A/2, x > 1 + A/2.
16. y = xe~ x .
17. y = x/(a 2 + x 2 ).
ylns. At x = 0, d= a A/3; upward a A/3 < x < 0, x > a A/3,
downward x < a A/3, < x < a A/3.
18. y = 6x/(x 2 + l).
19. y = (1/x) log x.
Ans. (e 3 ' 2 , 3e~ 3 / 2 /2); downward < x < e 3 ' 2 , upward x > e 3 ' 2 .
20. Is the curve y x 4 /2 3 x 2 ever concave upward?
21. Discuss the concavity of y = x 3 3 x 2 -f 4 at x == 2, 1, 1, 2.
Ans. Down, down, neither, upward.
22. Is the curve y x* 3 x 2 20 x -f 40 concave upward or downward
at x = 1? Is ?/ increasing or decreasing at x = 1?
23. What are the signs of dy/dx and d 2 y/dx 2 for each of the following cases?
The curve is (a) concave down but rising; (6) concave up and rising; (c) con-
cave up but falling; (d) concave down and falling.
24. Test the curve ?/ = x 4 4 x 3 -f- 6 x 2 -f 21 x 7 for points of inflection.
How do you explain the result?
80. Maxima and Minima. Let f(x) be a continuous single-
valued function of the variable x. By single -valued we mean
that the function has one and only one real value for each value of
the variable. If, as x increases, the function y first increases, then
decreases, as in Fig. 88 from A to C, there will be one value of the
variable, say x = b, for which the function is greater than it is for
any value of x either a little greater, or a little less than 6. Then
f(b) is called a maximum value of the function; that is, MB in
Fig. 88 represents a maximum value of /(#).
Similarly, as x increases, if the function decreases, then increases,
as from C to E, for some value of the variable, as x = d, the func-
tion will be less than it is for any value of x either a little greater, or
a little less than d. Then/(d) is a minimum value of the function;
that is, ND represents a minimum value of /(x).
If y changes from an increasing to a decreasing function, as
when x increases through the value x = 6, then the derivative
112 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
dy/dx must change from a positive to a negative value. But if y
changes from a decreasing to an increasing function, as when x
increases through x d, the deriv-
ative dy/dx must change from a
negative to a positive value. At
B and at D, where the function
assumes its extreme values, the
derivative, if continuous, must
be zero. We have then the follow-
ing theorem.
Given the function f(x), if, as x
increases through the value x a,
the derivative changes from a posi-
tive to a negative value, the function /(#) is a maximum at x = a; if
the derivative changes from a negative to a positive value, the function
f(x) is a minimum at x = a.
The derivative may change sign by becoming zero, or by becom-
ing infinite. These two cases are illustrated in Fig. 89. The
function is a maximum at A where y r = 0, and a minimum at B
where y' becomes infinite.
\
y'-o
FIG. 89
FIG. 90
As x increases, the derivative may become either zero, or infinite,
without changing sign. Thus if the graph of the function has a
point of inflection with a horizontal tangent, as at A in Fig. 90,
both y f and y" become zero. Here y' decreases until it becomes
zero at A, then increases without changing sign. B is a point of
inflection with a vertical tangent, that is, y' is infinite, but the sign
of y' on either side of B is the same. At such points the function is
neither a maximum nor a minimum.
81] SOME APPLICATIONS OF THE DERIVATIVE 113
81. Critical Values. Tests for Maxima and Minima. The
values of the variable which make the derivative of the function
zero, or make the derivative infinite, are called critical values of the
variable. Each value of the variable for which the function is a
maximum or a minimum is a critical value; but from 80 it is
evident that the converse is not true. Hence to find the maximum
and the minimum values of a function, obtain the critical values
of the variable, that is, the values for which f'(x) = 0, and those
for which !//'() = 0, and apply to each value one of the follow-
ing tests :
FIRST TEST. Let x a be a critical value. Substitute in
the original function f(x) a value of a: a little less than a, then again
substitute a value a little greater than a. If /Or) in both cases is
less than /(a), then /(a) is a maximum. If /(x) in both cases is
greater than /(a), then /(a) is a minimum.
SECOND TEST. Substitute a value of x, first a little less than a,
and then a little greater than a in the derived function /'(a;), and
observe the sign of the derivative in each case. If the sign of the
derivative changes from positive to negative, the function is a
maximum for x = a. If the sign of the derivative changes from
negative to positive, the function is a minimum for x = a.
In applying either of the tests above for a given critical value, care
must be taken that no other critical value lies in the interval between
the two values selected for the test.
THIRD TEST. If /'(#) is continuous for x = a, when a is a
critical value, then substitute a for x in the second derivative,
/"(:r). If /"(a) is negative, then f'(x) is a decreasing function;
but a is a critical value, hence /'(x) decreases through the value
zero for x = a, and the graph is concave downward. Hence if
/"(a) is negative, /(a) is a maximum.
Similarly, if /"(a) is positive when a is a critical value, then
f'(x) is increasing through the value zero, the curve is concave
upward, and /(a) is a minimum. If /"(a) is zero, the curve
usually has a point of inflection at x = a, but not always.
The first and second tests may be applied to all critical values of
the variable; the third test only to those critical values for which
the first derivative is zero. To say that the derivative becomes
infinite for some value of the variable, as x = a, is merely to say
that it increases or decreases without limit and is not defined for
that particular value.
114 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
EXAMPLES
1. locate the maximum and minimum points, and points of inflection of
y = 3 3 - 3 3 2 - 9 x - 3. Trace the graph.
SOLUTION. Differentiating, we have
= 3(x - 3) (3 -f- 1),
and ?/" = 6(3 - 1).
Set y' = to find critical values,
x = - 1, 3.
Set ?/" = to locate possible points of in-
flection. This gives
3 = 1.
Here the third test is the simplest.
If x = 1, y f = 0, y" is negative,
Fl - 91 hence ?/=/(- 1) = 2 is a maximum. If
x = 3, y' = 0, y" is positive, hence y = /(3) = - 30 is a minimum.
Since y" changes sign at x = 1, the point (1, 14) is a point of inflection,
with a slope of curve at that point of 12. Note that different scales are
used for abscissas and ordinates in Fig. 91 and this must be taken into con-
sideration in estimating the slope at any point.
2. Examine (x a) 1/3 (2 x a) 2/3 for
maxima and minima.
SOLUTION. Denoting the function by
?/, we have?/ = (x a) 1/3 (23 a) 2/s , and
-504-
' = . -
V 3(3 - a) 2 / 3 ~*~3(23 - a) 1 / 3
= 6 3 - 5 a
3(3 - a) 2 / 3 (2z - a) 1 / 3 '
From y' 0, and \/y' 0, we have the
critical values
6
a.
Use the second test on these values.
Setting x = a/3 and 2 a/3 in turn, we have y' positive and negative respec-
tively. Hence x a/2 makes y a maximum.
Test 3 = 5 a/6, using the values 2 a/3 and 9 a/ 10. These show y' to be
successively negative and positive, so the function has a minimum value at
3 = 5 a/6.
Apply the test to x = a, with the values 9 a/10 and 2 a. These show y'
positive in both cases and hence at x a, \/y' = 0, and the graph of the
function has a vertical tangent with neither a maximum nor a minimum.
At 3 = a there is a point of inflection, as shown. The maximum, at the
point (a/2, 0), is called a cusp.
81] SOME APPLICATIONS OF THE DERIVATIVE 115
PROBLEMS
Find the critical values of the variables in each of the following functions.
(Nos. 1-16.) Test each and find the maximum and minimum values of the
functions.
1. x 9 - 3 x*.
Arcs. Max. (0) at x 0. Min. (- 4) at x 2.
2. 2 z 3 - 3 x 1 - 12 x -f- 2.
3. z 3 (z - 2) 2 .
Ans. Max. (3456/3125) at x 6/5. Min. (0) at z = 2. Inflection
at x = critical value.
4. * 2 /2 - z 3 A
5. x 4 - 2 z + 3.
Ans. Min. (21/16) at x = 3/2. Inflection at a; = 0.
6. 6z/(z 2 + l).
7. 3e-* s . Arcs. Max. (3) at z = 0. (NOTE: & * 0.)
8. ze x .
9. 3 xe~*. Ans. Max. (3 e" 1 ) at a: = 1.
10. (x + 2) 2 /3 (3 - 5) 2 .
11. 3z 5 - 65 x 3 + 540 3.
Arcs. Max. at a? = 3, 2. Min. at z = 2, 3.
12. (2 a; - a) 1 "/ (3 * -f 2 a) 1 / 3 .
13. z/log z. -Aws. Min. (e) a,t x = e. What about a? = 1?
14. x l >*.
15. ae cx -f fo- cx , (a, 6 positive). Ans. 2 Va/> (min.).
16. 3 e 21 + 5 e- 21 .
17. If t/ = S(o; a^t) 2 ** where x t are constants, i = 1,2, n, what values
of ^ make y a minimum? Ans. x = x t .
18. The equation of the curve described by a jet of water projected from a
hose may be represented by y = kx - (1 -f /c 2 )z 2 /100. What does k repre-
sent? What value of k will make the water reach the greatest height on a
wall (a) 25 feet from the nozzle? (b) 45 feet?
19. The total waste per mile in an electric conductor is w cV -f k*/r.
What resistance r will make the waste a minimum if the current c is kept con-
stant? Ans. k/c units.
20. The work done by a voltaic cell of constant E.M.F. and constant
internal resistance r in sending a steady current through an exterior circuit of
resistance R is kE*R/(r -f- R)* in a given time. What value of R makes the
work a maximum? Ans. R = r.
* y here means the sum of n terms formed by giving i consecutive values from
1 to n inclusive.
116 DIFFERENTIAL AND INTEGRAL CALCULUS [CH. V
82. Applications of Maxima and Minima. Many important
problems require for their solution that the maximum or minimum
value of some quantity be found. Suppose the problem be to
find the maximum area of a geometric figure which satisfies given
conditions, or to find the minimum amount of material required to
build a tank of given capacity, or to find when two moving objects
will be nearest together; in all such problems there is a definite
method of procedure for finding a solution.
First, the quantity which is to be a maximum or a minimum is
always the function to be examined. Hence express the function
in terms of the variable or variables which occur in the problem.
If the function is obtained in terms of a single variable, the critical
values of the variable and the maximum or minimum values of
the function can then be found as in the preceding article.
If the function is expressed in terms of two variables, then an
additional relation connecting those two variables must be found.
Using this relation, one of the variables can be eliminated so that
the function will be expressed in terms of a single variable.
Similarly, if the function is given in terms of three variables,
two additional relations connecting the three variables must be
found. With these relations, two of the variables can be elimi-
nated and the function obtained in terms of a single variable.
After finding the necessary additional relations connecting the
variables, the remainder of the solution may be varied as shown in
Example 2 below.
Find
EXAMPLES
1. A right circular cone is circumscribed about a sphere of radius a.
the dimensions of the cone if its volume is a minimum.
SOLUTION. Since the volume V is to be a minimum,
it is the function in the problem. Hence the function is
To express the function in terms of a single variable,
we must find a relation connecting h and r. Using
similar triangles in Fig. 93,
Fio. 93
(2)
AE
BE
h-a
AC
82] SOME APPLICATIONS OF THE DERIVATIVE 117
Squaring both sides and solving for r 2 , we find
-rS-.-
Substituting (3) in (1), we get
Hence
dV 7ra*h r h -4a
v ' dh 3 L(^-2a) 2 J
Placing the derivative equal to zero, we find
h = 0, h = 4 a.
Both the function and its derivative become infinite if h = 2 a and hence
this value of h cannot be considered. The value h = is extraneous since it
does not satisfy (2). By applying the second test to the critical value h = 4 a,
we observe that dV /dh is negative for h < 4 a, and positive for h > 4 a.
Hence h = 4 a makes V a minimum. That is, the altitude of the cone is twice
the diameter of the sphere and the dimensions of the cone are h 4 a, r = a V2.
2. What are the most economical proportions for an open cylindrical can
of given capacity, if no allowance is made for waste of material?
SOLUTION (a). Obviously this means that the volume V of the cylinder is a
constant, and that the amount of material M in the can (which forms the
lateral surface and one base) is to be a minimum. Then M is the function.
Calling the radius r, and the altitude h,
(1) M = wr* + 2 wrh t
where r and h are connected by the relation,
(2) 7rr 2 /i = V 9 a constant.
Here it is easier to eliminate A, hence from (2) we use
(3) *-
Substituting (3) in (1), we find M in terms of r, or
9 V
(4) af-- + =jf.
Then
dM n 2V
(5) -^-=2^-
Using the value of V from (2), we have
(6)
Hence
(6) ~ 2 TTT - 2 irh - 2 ir(r - h).
dM n u i,
-T = when r = h.
dr
118 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
Differentiating (5), we see that d*M/dr 2 > for all possible r. Therefore M is a
minimum for an open cylinder with fixed volume if its altitude is equal to its
radius.
SOLUTION (6). Differentiate implicitly, say with respect to r, both relations
(1) and (2); then
-, + ,
and
(8)
Substituting the value of dh/dr from (8) in equation (7), we find
(9) = 2 7r(r + h - 2 h) = 2 ir(r - A).
Hence
-T = when r = h.
dr
If the independent variable r < h, dM /dr < 0, and if r > h> dM /dr > 0,
therefore M is a minimum.
3. The amount of fuel consumed per hour by a certain steamer varies as the
cube of its speed. When the speed is 15 mi./hr., the fuel consumed is 4j/ tons
of coal per hour at $4 por ton. The other expenses total $100 per hour.
Find the most economical speed, and the cost of a voyage of 1980 miles.
SOLUTION. The cost of the voyage C is the function. The cost per hour
is (kv* -f- 100) dollars where
k ** 15~ 3 ^ 375 '
The time of the trip is s/v where s is the distance. Then
dC / v _ 25
dv ~ 46> V375 v*
Equating this to zero, we find the critical value
v s = (375) (25) = (125) (75).
Whence
v = 5V/75 = 21.086 mi./hr.
C" = 4 s(l/375 + 50/t> 3 ) which is positive for all positive values of v, and
hence, by the third test, C is a minimum. dC '/dv is infinite when v = 0, but this
makes C infinite. The cost of the trip for 5 = 1980 miles is
r - f 2 "' + i
C - + 1
21.086
82] SOME APPLICATIONS OF THE DERIVATIVE 119
PROBLEMS
1. The sum of two positive numbers is 10. Find the numbers if their
product is a maximum. Ans. 5, 5.
2. The sum of two positive numbers is 12. Find the numbers if the sum
of their squares is a minimum.
3. A page of a book must have 18 sq. in. of printed matter and must have
2 in. margins at top and bottom and 1 in. margins on each side. What dimen-
sions will require the least amount of paper? Ans. 5 in. by 10 in.
4. What number exceeds its square by the greatest amount?
5. (a) A man has 100 rods of fencing and wishes to erect it along three
sides of a rectangular field which borders on a straight shore line. What
dimensions will give the maximum area? Ans. 25 by f)0 rods.
(6) For any given length of fence, what is the shape of the rectangle?
6. What is the area of the largest isosceles triangle which may be inscribed
in the parabolic segment bounded by y 2 = 8 x and x 8, if its vertex is at
the point (8, 0)?
7. The strength of a rectangular beam varies as the product of its breadth
6 and the square of its depth h. What is the relation between b and h for the
strongest beam which may be cut from a log of radius a units?
Ans. h = 6\/2.
8. A power house on a river bank supplies power to a plant on the other
side and 3 mi. down stream. If the river is 2 mi. wide and the power line costs
4/5 as much per mile on land as under water, what line would be cheapest?
Would the line under water be changed if the plant were farther down stream?
9. Find the area of the largest rectangle which may be inscribed in a
parabolic segment of 30 unit base and 20 unit altitude.
Ans. 400>/3 sq. units.
10. Same as Problem 9 for a semicircle of radius a units.
11. Find the dimensions of the maximum rectangle inscribed in the ellipse
-f 2/V& 2 = 1. Ans. aV2, bV2 units.
12. A rectangular box with square base and cover is to contain 800 cu. ft.
If material for the bottom costs 15 cts., for the top 25 cts., and for the
sides 10 cts. per square foot, what is the least possible cost of the box?
13. Find the shortest distance from the line 2 x -f y ~ 3 to the point
(- 6, 0). Ans. 3V5 units.
14. Find the shortest distance from( 6,0) to the hyperbola z 2 i/ 2 -fl6 =0.
15. A telephone company finds that it makes a net profit of $15 per phone
for 1000 phones or less in a given period. If the profit decreases 1 ct. per phone
over 1000, what number of phones would yield the greatest profit?
Ans. 1250 phones.
16. What are the dimensions of the largest right circular cylinder which
may be inscribed in a sphere of radius a units?
120 DIFFERENTIAL AND INTEGRAL CALCULUS [On. V
17. (a) What is the least material needed to make an open circular cylindri-
cal can of volume 8 TT cu. in.?
(6) For any fixed volume what must be the relation between the radius and
altitude of such a can? What if closed at both ends?
Ans. 12 TT sq. in.; r = h, h = 2r.
18. The distance of a body from a fixed point is given by the relation
s = Z 4 /12 - 5 t*/6 + 2 t* + 3 t + 1.
If the body moves along a straight line, when is it moving most slowly?
19. (a) An ellipse of 6 and 8 unit axes is revolved about its major axis.
What is the volume of the largest right circular cone which may be inscribed in
the solid if its vertex is at an end of the major axis? Ans. 128 vr/9 cu. units.
(b) Prove that such an inscribed cone always has its altitude equal to two-
thirds of the corresponding axis of the ellipse.
20. (a) Find the equation of the line through (4, 3) which cuts off the
triangle of least area in the first quadrant.
(b) What are the intercepts of such a line if it goes through (a, b)?
21. For a given hypotenuse of 2 k units, what is the area of the largest right
triangle? Ans. Area = k' 1 sq. units.
22. A ship sails south 6 mi./hr. and another east 8 mi./hr. At 4 P.M. the
second ship crosses the path of the first at the point whore the first was at
2 P.M. When are the ships closest to each other?
23. Given an amount of lumber to make a rectangular box of largest
volume. What dimensions should be used if there is no top and if the base
dimensions are in the ratio 2:1? Ans. 6:3:2.
24. A right triangle with hypotenuse 3 in. is revolved about one leg. What
dimensions will the triangle have if the volume generated is a maximum?
25. The intensity of light varies inversely as the square of the distance from
its source. If two lights are 300 yds. apart, and one light is H times as strong
as the other, where should an object be placed between the lights to have the
least illumination? Ans. 200 yds. from the stronger light.
26. Find the dimensions and volume of the right circular cylinder of largest
surface which may be inscribed in a right circular cono of dimensions r and h.
27. What percent of a precious stone, spherical in shape, may be saved if it is
cut in the shape of a right circular cone? Ans. 29%$%.
28. Same as Problem 27 except that the shape is a regular pyramid with
square base.
29. A silo is to be built in the form of a cylinder with a hemispherical roof.
The floor and wall are of the same material but the roof costs 2 l / 2 times as
much per sq. unit as the floor. Find the most economical shape.
Ans. h = 4r for cylinder.
30. Prove that the maximum and minimum line segments from the point
(h, k) to the curve whose equation is y = f(x) meet the curve at points where
the tangent is perpendicular to the segment.
83] SOME APPLICATIONS OF THE DERIVATIVE 121
83. Rates. The essential meaning of the derivative has been
shown to be the rate of change of the function with respect to the
variable. If the function in question is a linear function, that is,
involves the variable to the first degree, as y = ax + 6, then
dy/dx = a, or y changes at a constant rate with respect to x.
The student is already familiar with the fact that the graph of
such a function is a straight line and a, the coefficient of x, is the
slope of the line. Thus if r and s are connected by the relation
5 = 2 r + 5, then s is decreasing at twice the rate r increases
since ds/dr = 2, and 5 is the value of the function s when
r = 0.
If the function y = f(x) is not linear, then the derivative /' (x)
is also a variable quantity and will depend for its definite values
upon particular values assigned to x. That is, the rate of change
of the function with respect to the variable will depend on the
variable.
A case of extreme importance is that in which the independent
variable is the time 2, or else the independent variable is itself a
function of the time. If y = f(t), then dy/dt is the rate of change
of y per unit time. In many physical problems the time rate of
change dx/dt of the independent variable x is given, or can be
found, and the time rate of change dy/dt of a related variable y is
desired. To solve such a problem, the relation y = f(x) connect-
ing the variables must be obtained; then the
desired rate is given at once by the formula
(ty = d\i t dx m
dt ~ dx' dt
EXAMPLES
1. A balloon in the form of a right circular cone
vsurmounted by a hemisphere and having its diameter
equal to the height of the cone is being inflated. How
fast is its volume V changing with respect to its total FIG.
height /i? What is the result when h = 9 units?
SOLUTION. Given h 3 r, to find dV/dh. Hence we must express V in
terms of h. The volume of the cone is irr 2 2 r/3, and that of the hemisphere is
2 7rr 3 /3. Adding these, we have
(IT
81
122 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
Then
dV
dh
27
Hence V is changing 4 irh 2 /27 times as fast as h. When h = 9, dV/dh = 12 ?r,
that is, V is then changing 12 TT times as fast as h.
N
FIG. 95
mi./hr., and y is increasing 10 mi./hr., we have
2. A ship is 41 miles due north of a
second ship. The first sails south at the
rate of 9 miles per hour, the second sails
west 10 miles per hour, (a) How rapidly
are they approaching each other l*/2 hours
later? (b) How long will they continue
mi ' to approach each other?
SOLUTION. After t hours of sailing let
x be the distance of the first ship from
the intersection of the courses, y the dis-
tance of the second ship, and z the distance
between them. Since x is decreasing 9
dx .
- = - 9 mi./hr.,
= 10 mi./hr.
We require in (a) to find dz/dt when = 1^; in (b) to find what value of t
makes z a minimum.
(a) To find dz/dt express z in terms of x and ?/, the variables whose rates are
given. Evidently
(1) z 2 = z 2 + y 2 .
Now 2, T/, and hence 2 are each functions of t y since
(2) x = 41 - 9 t, y = 10 *.
Differentiating (1) with respect to t, we find
(3)
When t =
Hence
that is, in
mi./hr.
-93
ctt ~ z
i, from (2) and (1) we have
x = 27.5, y = 15,
dz - 247.5 + 150
di~ -
2 = 31.325.
-3.11 mi./hr.,
31.325
i hours the ships are approaching each other at the rate of 3.11
83] SOME APPLICATIONS OF THE DERIVATIVE 123
An alternative method of solution is to express z directly in terms of t by
substituting equations (2) in (1).
(6) The ships will continue to approach until z is a minimum, or until
dz _ - 9x -f 10y _ n
dt ~ I '
that is, when
- 9 x + 10 y 0.
Substituting values of (2), we have,
- 369 -f- 8U + 100 t = 0, 181* 369,
or t 2.04 hrs. nearly.
PROBLEMS
1. A spherical balloon is inflated so its volume increases 12 cu. ft./min.
How fast is the radius changing when it is 6 ft.? Compare the rates of change
of the volume and surface when the radius is 8 ft.
Ans. 1/(12 IT) ft./min.; 4 : 1 numerically.
2. If y x 3 6 z 2 + 3 x + 5, at what points are the ordinates and the
slopes of the curve changing at the same rate with respect to x?
3. If the amount of wood in a tree is proportional to the cube of its
diameter, compare the rates of the growths of two trees of 3 ft. and 6 ft. diame-
ters. Ans. 1:4.
4. Two ships are at the same point. One leaves at 10 A.M. sailing east
9 mi./hr.; the other at 11 A.M. sailing south 12 mi./hr. How fast are they
separating at noon?
5. A point moves along the curve y = 4 2 x 2 so that its abscissa is
decreasing 5 units per second. How fast is its ordinate changing as the point
passes through (1, 2)? Ans. 20 units/sec.
6. The height of a ball thrown upward is given by h = 120 t 16 Z 2 .
How fast is the ball rising or falling at t 3 sec., 4 sec.? How long does it
rise?
7. A man 6 ft. tall walks at 2 ft. /sec. toward a light 10 ft. above the ground.
How fast is the length of his shadow decreasing? How fast is the end of his
shadow moving? Ans. 3 ft. /sec.; 5 ft. /sec.
8. A barge, whose deck is 10 ft. below the level of a wharf, is drawn in by
a cable through a ring in the floor of the wharf. A windlass at the level of the
deck hauls the cable in 5 ft. /sec. How fast is the barge moving toward the
wharf when it is 20 ft. away? Is there a maximum velocity for the barge?
9. A conical funnel of height and radius each 6 units contains a liquid
which escapes at the rate of 1 cu. unit/min. How fast is the surface falling
when it is 4 units from the top of the funnel? Ans. l/(4 TT) units/min.
10. In Problem 9, how fast is the inner surface of the funnel being exposed
above the liquid?
124 DIFFERENTIAL AND INTEGRAL CALCULUS [On. V
11. In Problem 9, how fast is the area of the exposed surface of the liquid
changing? Ans. 1 sq. unit/mm.
12. A body is being raised by a rope over a pulley 25 ft. above the body.
A man's hand holding the end of the rope is 5 ft. above the body. If the rope is
50 ft. long, at what rate will the body start to rise if the man walks away from
under the pulley at 10 ft./min.?
13. In a right triangle the legs are increasing 1 unit/sec, and 2 units/sec,
respect! voly. At what rate is the hypotenuse changing at the time the legs
are 3 and 4 units in length? Ans. 2% units /sec.
14. Solve Problem 13 if the first leg increases 1 unit/sec, and the other
decreases 3 units /sec.
15. An isosceles triangle has its vertex at P(x t y) a point of the curve
y = e x . Its base is along the x axis with one extremity fixed at the origin.
If P is moving along the curve so that its ordinate is increasing 5 units /sec.,
how fast is the area of the triangle changing? Ans. 5(z -f- 1) sq. units/sec.
16. In Problem 15 let the vertex P be on the curve y = x log x. Find the
rate of change of the area of the triangle and evaluate this for the point (e, e).
17. A man walks across a bridge at the rate of 5 ft. /sec. and a boat beneath
him passes down stream 12 ft. /sec. If the bridge is 30 ft. above the water, how
fast are man and boat separating 4 sec. later? What does this rate approach
as I increases without limit?
Ans. 338/V901 ft./sec. s 11.30 ft./sec.; 13 ft./sec.
18. A train is moving along an elevated track 20 ft. high at 30 ft./sec.
Immediately below it, a truck is going in the same direction 10 ft./sec. How
fast are the train and the truck separating one minute after the train is above
the truck?
19. The lower end of a ladder 26 ft. long is being pulled away from a vertical
wall at 3 ft./sec. How fast is the upper end, resting against the wall, descend-
ing when the lower end is 10 ft. from the wall; 24 ft. from the wall? When are
both ends moving at the same rate? Ans. \ l /i ft./sec.; 7K ft./sec.
20. A trough 8 ft. long has for a cross-section an isosceles trapezoid of
altitude 1 ft., upper base 4 ft., lower base 2 ft. If water is poured into the
trough at the rate of 2 cu. ft./min., how fast is the depth increasing when it is
6 inches?
21. Water is being poured into a 10 ft. trough at the rate of 25 cu. in. /sec.
If the ends are isosceles triangles with altitude equal to one half of the base,
find the rate of rise of the level of the water when it is 10 inches deep.
Ans. 1/96 in. /sec.
22. A balloon is rising vertically 10 ft./sec. from a point on the ground.
After 1 min., how fast is it receding from an observer 800 ft. from the point?
23. Water is running out of a horizontal cylindrical tank 9 ft. long and 3 ft.
in diameter. When the water is 1 ft. deep, the surface area of the water is de-
creasing 2 sq. ft./min. At what rate is the depth decreasing?
Ans. 2V2/9 ft./min.
84] SOME APPLICATIONS OF THE DERIVATIVE 125
24. If the circumference of a great circle on a sphere is decreasing 2 in. /sec.,
show that the rate at which the volume of the sphere decreases is numerically
equal to the area of the square circumscribing the great circle.
25. A spherical tank of 10 ft. diameter is receiving water at 12 cu. ft./min.
At what rate is the depth of the water increasing at 8 ft.? Am. 3/(4 TT) ft./min.
26. An arc light is 24 ft. above one side of a street which is 30 ft. wide. A
man 6 ft. tall walks along the opposite side at the rate of 5 ft. /sec. When he is
40 ft. from the point opposite the light, how fast is the tip of his shadow mov-
ing? How fast is his shadow lengthening?
27. The adiabatic law for the expansion of air is pv n = c where n = 1.41,
approximately. If the air has a volume of 600 cu. in. at 40 Ib. pressure per
sq. in., what is the rate of change of the volume with respect to the pressure
when p = 40 Ibs./sq. in.? Approximately how much will the volume be
changed due to an increase of 2/3 Ib./sq. in. in the pressure?
Ans. 10.64 cu. in./unit p\ 7.09 cu. in. decrease.
28. A point moves along the curve y x log x so that y decreases at the
rate of 2 units/sec, (a) How fast is x changing when the point crosses the
line x = 3 //? (6) Find how fast the slope of the graph is changing, (c)
Show that the abscissa, ordinate, and slope are changing at the same rate when
the point crosses the x axis.
84. Rectilinear Motion. One kind of time-rate problem which
deserves special mention is rectilinear motion.
Let a particle move along a straight line so that its distance 5
from a fixed origin on the line is a function of the time t. We have
seen ( 60) that the rate of change of s with respect to t, the time
derivative of s, is the velocity v of the particle. Hence if 5 = /(<),
by differentiating we have
Similarly, the rate of change of v with respect to t, the time
derivative of v, is the acceleration a of the particle. Hence
dv d 2 s -//x.x
If the velocity is constant, the motion is called uniform. If the
acceleration is constant the motion is called uniformly accelerated
motion. Thus, a particle near the earth's surface, which is subject
to the force of gravity only, moves with an acceleration of
g = 32 ft./sec. 2 , approximately.
126 DIFFERENTIAL AND INTEGRAL CALCULUS [On. V
EXAMPLE
A ball thrown vertically upward has its distance in feet from the starting
point given by s = 104 t 16 t 2 , where t is measured in seconds. Find its
velocity, acceleration, and the height the ball will rise. How high is the ball
after 3 seconds; after 4 seconds? What distance does the ball pass over
during the fourth second?
SOLUTION. The velocity and acceleration arc given at once by
v = ^ I = (104 - 320ft./sec.;
a = ~ = - 32 ft./sec. 2
The ball will rise until s is a maximum, that is, when
<fc n
V = dt = "
Henco 104 32 t = 0, t = 3J4 sec. The height it will rise is the value of
for t = 3J4, or
s = 169 feet.
For t = 3 sec., A = 168 ft.; and for t = 4 sec., s = 160 ft. The distance
the ball moves during the fourth second is not As for t = 3 and At = 1, since
the velocity changes sign during the fourth second and hence As changes sign.
Since we have found the maximum value of .s to be 169 ft., we see that during
the fourth second the ball rises 1 ft. and falls 9 ft. Therefore the distance
traveled during the fourth second is 10 ft.
PROBLEMS
The following laws refer to straight-line motion in each case.
1. A body moves so that s = t 2 8 t -f 7. When will its velocity be
positive? Ans. t > 4.
2. A body moves with v = 1 -f 3 t 2 2 t*. When is its acceleration
decreasing?
3. If ,s = 100 t 16 2 , when is (a) s increasing; (6) v decreasing; (c) a
increasing? Ans. t <3 l /%\ aU t] constant.
4. If s = t 3 2 P, when is v increasing? Is s increasing ordecreasing at
t = 1?
5. What is the direction of motion of a body if its distance s from a fixed
point is given by s = 2 < 3 - 2U 2 + 60 t + 5?
Ans. s increasing 2 < t < 5, decreasing t < 2, t> 5.
6. The same as Problem 5 if (a) s = 6 -f 24 t - 15 t 2 - 2 t\ (6) s = t 2 log t.
7. The same as Problem 5 if 5 = t* - 3 t 2 + 3 t -f 4.
Am. Always forward except v = at t = 1.
84] SOME APPLICATIONS OF THE DERIVATIVE 127
8. The motion of a point is determined by s = t* 8 t + 7. When is
it speeding up?
9. The velocity of a car after t min. is given by v = t* 21 t 2 -f- 80 t.
When is it in reverse? Ans. 5 < t < 16.
10. If v = t 3 5 t 2 -h 7 t 3, when is the distance s increasing? How
fast does the point move when its acceleration is a maximum or a minimum?
11. If s = t 4 - 2 t 3 - 12 t 2 -f 36 t - 10, when is s increasing? When is
the particle not in motion? When is its velocity decreasing? When is v
constant? Has its acceleration an extreme value?
Ans. - VQ < t < 3/2, t > V6; t = \/6, 3/2; - 1 < t < 2;
t = - 1, 2; min. at t = 1/2.
12. A particle moves according to each of the following laws. Graph each
of the functions s, v, and a against the variable t on the same set of reference
lines. Study graphs to obtain data about increasing, decreasing, extreme, and
stationary values of s, v, and a.
(a) 8 = I 3 / 3 - t*. (c) s = 4(4 - e-</ 2 ).
(b) s = log V2 - 3 t. (d) s = (log t)/t.
ADDITIONAL PROBLEMS
1. Find the tangent and the normal to xy = 6 at (2, 3).
Ana. 3 z + 2 ?/ - 12 = 0, 2z-3i/-f5=0.
2. Find the tangent and the normal to t/ 3 = 2 z 2 -f 3 ?/z 2 at ( 1, 2).
3. Show that the tangent to y 2 = 2 px at (xi yi) is yiy - p(x -f z t ).
4. (a) The line 4 # 3 y 55 is tangent to the curve whose equation is
3 y = X 3 _ 3 X 2 _ 20 x + 25. Find the point of contact.
(6) Find the point at which the line 3 x + 4 y = 54 is normal to the same
curve.
5. Find the line through (4, 3) which is parallel to the tangent to
y = x 4 - 3 x 2 -f 23 x at x = - 2. Ans. 3&-p + 9 = 0.
6. Find the line through ( 2, 1) which is perpendicular to the tangent
to y = x s 2 z at x = 1.
7. Show that the tangent to the ellipse x z /a 2 -f- y z /b 2 = 1 at the point
(#1, t/i) is Ziz/a 2 -f 2/i2//& 2 = 1.
8. Find the area of the triangle formed by the tangent to y 2 9 x at
(4, 6), the normal at the same point, and the x axis.
9. Find the slope of the tangent to x = at, y - bt (1/2) gt 2 at any point.
Find the value of t which makes the slope zero. Interpret this value.
Ans. (b - gt)/a, b/g.
Find the angles between the following pairs of curves. (Nos. 10-12.)
10. xy = 4, x 2 - y* = 6.
128 DIFFERENTIAL AND INTEGRAL CALCULUS [Ca. V
11. 4 x* -f y* = 52, 3 2/ 2 = 16 x. Ans. tan- 1 (54/23).
12. x 2 = 4 ay, y 2 = 8 a 3 /(* 2 + 4 a 2 ).
13. Find the angle between the tangent to x 2 = 4 ?/ -f- 4 at ( 2, 0) and
the line through (3, - 2) and (- 2, 0). Ans. tan" 1 (3/7).
14. Find the tangents to y = 2 x 2 2 and y x 2 + I = at their inter-
sections and find the angle between them.
15. The curves y = 2 e~ I/2 and y = log (x + e) 2 meet on the y axis. Find
the angle of intersection. Ans. tan~ x [(2 + e)/(2 e)].
16. If P(x, y) is on y x 3 and the tangent at P cuts the x axis and the
y axis at (? and R, respectively, show that PR = 3 PQ.
Find the intervals in which each of the following functions increase and those
in which they decrease. (Nos. 17-19.)
17. x 3 + 4 x 2 .
Ans. Increases x < 8/3, x > 0; decreases 8/3 < x < 0.
18. x 4 /2 - 3 x 2 .
19. 12 /* - 2t* - t*.
Ans. Increases t < - (3 + vT05)/4, < t < - (3 - vT()5)/4;
" decreases - (3 -f VlOf>)/4 < t < 0, t > - (3 - Vl05)/4.
20. If the position of a particle is given by s = t 3 6 2 + 9 Z 12, when
is y increasing, when decreasing?
21. If a point on a line is s units from a starting point, find when its veloc-
ity is increasing if:
(a) 8 - 3 + 4 t -f 30 i 2 -f 8 t 3 - t*',
(b) s = 5 4 - 14 J 3 -f 300 J 2 H- 360 .
ylrcs. (a) - 1 < t < 5; (6) For all t.
Examine each of the following curves for inflections and types of concavity.
(Nos. 22-26.)
22. y = x 3 - 3 x* - 6 x + 12.
23. ?y = x 3 - x 2 -f 6 x - 1.
Aws. (1/3, 25/27); upward x > 1/3, downward x < 1/3.
24. y = 2<r 2 * 2 .
25. y = 3 x 4 - 4 x 3 - 6 x 2 + 4.
Ans. (- 1/3, 95/27), (1, - 3); upward if x < - 1/3, z > 1,
downward 1/3 < x < 1.
26. y = log(x 2 + 1).
Locate maximum and minimum values of each of the following functions.
(Nos. 27-29.)
27. 4 x 3 - 6 x 2 -f 3. Ans. At x - 0, max. (3); at x - 1, min. (1).
28. x 4 - 12 x' + 36 x 2 - 50.
29. 3 x 4 - 4 x 3 .
s. Min. ( 1) at x = 1; inflection at critical value # = 0.
84] SOME APPLICATIONS OF THE DERIVATIVE 129
30. Draw the graphs of ?/ = x log x, and y /log x to the same set of
axes. Show that each curve passes through the point of the other at which y
is a minimum.
31. Draw a careful graph of y = x lf * (x -f 4) between x - 4 and x 3.
Find high or low points and points of inflection.
Ans. (- 1, - 3) low point; (0, 0) and (2, 6^2) inflections.
32. A potato crop is now 120 bushels and worth $1 per bushel. If the crop
would grow 20 bushels per week and lose 10 cts. per bushel in price, when
should they be dug to get the best value? ,
33. An isosceles triangle with its vertex at (0, 0) and with a horizontal base
above the vertex has the ends of its base on x z + 2 y =4. Find the area of
the largest such triangle. Ans. 8/3 V3 sq. units.
34. An open box with a square base is to be made with a given inner sur-
face. For a maximum volume, find the relation between the height of the box
and the side of the base.
35. A covered box whose base has sides 2 : 1 is to contain 360 cu. ft. If the
bottom costs 4 cts., lid cts., and sides 3 cts. per sq. ft., what are the dimen-
sions for a minimum cost? Ans. 6^3 X 3^X3 X 20^/3/3 ft.
36. A closed cylindrical vessel is to contain a fixed volume V. (a) Find the
relation of the radius and the height of the most economical vessel. (6) If
the curved surface and the ends of this vessel each have a thickness of a units,
show that the shape of the vessel should remain unaltered for different values
of V.
37. A fixed quantity of metal is to be divided between two molds, one a
sphere of radius r, and the other a cube of side s. When will the total surface
of these solids have an extreme value?
Ans. Max. if s = 2 r; min. if s = or if r = 0.
38. Find the maximum trapezoid which can be inscribed in the ellipse
8 x 2 -f- 9 ?/ 2 = 72 if one base of the trapezoid coincides with the major axis of
the ellipse.
39. A body of weight w is dragged along a horizontal plane by a force F
whose line of action makes an angle 6 with the plane. Find when F is least if
F mw/(m sin -j- cos 0), where m is the coefficient of friction.
Ans. For B = tan^m.
40. What point on 4 y = x 2 is: (a) nearest the point (0, 4)? (6) nearest
the line x y = 5?
41. The perimeter of a sector of a circle is 50 units. What radius will make
the area of the sector a maximum? Ans. r =* 1/4 of the perimeter.
42. Water escapes from a conical vessel at the rate of 2 cu. units/sec, and
is poured in at the rate of 5 cu. units/sec. The altitude of the vessel is 10 in.
and the diameter at the top is 15 in.
(a) At what rate is the depth of the water increasing when 4 in. deep?
(b) At what rate is the top surface of the water increasing?
(c) At what rate is the conical surface being inundated?
130 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. V
43. Find the dimensions of the largest circular cylinder which may be cut
from a right circular cone of height h, and radius of hase r.
Ans. Alt. of cylinder = h/3.
44. At a certain time, the radius of a cylinder is 2 ft. and is increasing at the
rate of 1 ft./hr., and the altitude is 4 ft. and is decreasing at the rate of
1 ft./hr. When will the cylinder have a maximum volume?
45. A ship is sailing north at the rate of 10 mi./hr. Another ship 190 miles
north of the first ship sails S 60 E at the rate of 15 mi./hr. When are they
nearest each other? Ans. In 7 hrs.
46. A car running 60 ft. /sec. passes directly beneath a balloon at the
instant a bomb is released. The height of the bomb after t seconds is
800 16 t 2 ', at what rate is the distance between the car and the bomb chang-
ing at the end of 5 seconds?
47. A conical funnel loses water so that its depth decreases 2 in. /sec. when
the water is 6 in. deep. If the funnel is 9 in. deep and 6 in. across the top,
how fast is the wet surface of the funnel decreasing at that insta-nt?
Ans. (8/3) TrVlO sq. in. /sec.
48. Find the rate of change of the total surface of a cylinder if h 4 r
at the instant r is 4 units and if the volume is decreasing 10 cu. units/min.
49. At a given instant the legs of a right triangle are 4 and 9 units respec-
tively. Assume that the shorter leg is caused to decrease 2 units/sec., and the
area to increase 5 sq. units/sec. Plow is the longer leg changing?
Ans. Increasing 7 units/sec.
50. The velocity of a stream of water issuing from the nozzle of a fountain
is given by the formula v 2 = 2 gh, where g is the acceleration of gravity,
32 ft. /sec. 2 , and h is the height of the surface of the water above the nozzle.
If the surface of the water is falling at the rate of 6 inches per hour, at what
rate is v changing when h 25 ft.?
51. Given a semicircle lying above its horizontal diameter. Chords are
drawn parallel to the diameter and on each chord as a diameter a circle is
drawn. What chord will have the highest point of its circle at a maximum
distance from the diameter of the semicircle? Ans. r\/2 long.
52. A rhombus A BCD is made by fastening together with hinges 4 rods
of length 13 inches each. If A and C are drawn together at the rate of
5 in. /sec., at what rate is the area of the rhombus changing when AC = 10
inches?
53. A tank standing on level ground is kept full of water to the depth of
a ft. ; water issues horizontally from a small hole, at a distance of h ft. below the
surface, with the velocity V2 gh ft. /sec. What value of h will make the water
strike the ground at the greatest possible distance from the tank?
Ans. h = a/2.
54. A cone-shaped container of dimensions a and r units is filled with water.
Find the radius of the solid sphere which when placed in the container will
displace the greatest amount of water.
Ans. ar/(Va 2 + r 2 r) units.
CHAPTER VI
DIFFERENTIALS. THEOREM OF MEAN VALUE
85. Order of Infinitesimals. We have seen that the incre-
ments of the independent variable and of the function, when
used in the process of differentiation, are infinitesimals. An
important idea in the use of infinitesimals is that of their order.
Given two infinitesimals u and v such that v is a function of u
These infinitesimals are said to be of the same order provided
(1) limj = k,
u- *0 u
where k has a finite value not zero.
If the lim (v/u) = 0, then v is said to be of a higher order than u.
In general, if
(2) lim ~ = k * 0,
U *.Q U
where fc is finite, then v is said to be an infinitesimal of the nth
order with respect to u.
If u and v are of the same order, then from relation (1) we may
write
(3) v = ku + 5,
where 5 is a function of u that approaches zero as u approaches
zero.
Dividing both sides of (3) by u and taking limits, we have
lim - = 0.
Thus 5 is an infinitesimal of higher order than u.
86. Differential of a Function. In any given function, as
y = /W, if 1 (Ay/As) = f'(x) ^ for a given value of x, then
Ax -0
At/ and Ax are infinitesimals of the same order for that value of x.
From the reasoning of the preceding article we can write
(1) A-4f(a?) =/'(*)-Az + S,
131
132 DIFFERENTIAL AND INTEGRAL CALCULUS^ [Cn. VI
where 5, a function of Ax, is an infinitesimal of higher order than
Ax, since
lim =().
The expression /'(x) Ax in (1) is known as the principal part of
the increment of the function and is called the differential of the
function. Hence:
The differential of a function is the product of the derivative of
the function by the increment of the independent variable.
The symbol for the differential of a function y is dy, so that
(2) dy = df(x) = /'(x)-Ax.
If the function equals the independent variable, that is, if
/(x) = x, then/'(x) = 1, and we have
(3) dx = Ax.
Hence, the differential of the independent variable is its increment.
However, the differential of any
function, other than a linear func-
tion, will differ from the increment
of the function.
Let Fig. 96 represent the graph
of the function y = /(x). Let
P(x, y) be any point of the graph
and Q any other point of the graph
located by giving x an increment,
Draw the tangent PT at P and
let a be its inclination. Now
the value of the derivative /'(x)
at P is the slope of PT, which is the tangent of its inclination.
Hence
N
FIG. 96
or
. // / \
tana = /'(x) = ,
RS = /'(x) -Ax = dy,
since this is the definition of the differential of the function.
However, the increment AT/ of the function is RQ.
Obviously, if P is taken at a point of the graph such that the arc
PQ is concave downward, then dy is greater than Ay. It is impor-
86] DIFFERENTIALS. THEOREM OF MEAN VALUE 133
tant to keep in mind that for any continuous function the difference
between Ay and dy approaches zero as Ax approaches zero; in
other words, the difference of Ay and dy is an infinitesimal of higher
order than Ax.
Since the differential of a function is its derivative multiplied
by the differential of the independent variable, all formulas for
differentiation become differential formulas when multiplied by
the differential of the independent variable. Thus from
we have
but by definition
d , . dv , du
-7- (UV) U-r- + V-j- ,
dx dx dx
j/ \ dv , . du ,
d(uv) = u-j-dx + v-j-dx,
dx dx
dv 7 7 du , ,
-7- dx = dv : ~r dx = du,
dx dx
hence
d(uv) = u dv + v - du,
and similarly for each of the formulas for differentiation.
EXAMPLES
1. Find dy if y = (x - z 2 )(2 - 2 x - z 2 ) 1 / 2 .
SOLUTION. Since dy f(x)dx, we have
* - fe ( -~2 f ^ + - ' - ^a - a ')]
= (2-7x + 3a: 2 -f-3 x 3 )(2 - 2 x - x 2 )- 1 / 2 cte.
2. If xy + a; 3 = y - 3 z/V evaluate dy for x = 1, and dx = 0.03.
SOLUTION. The differential of such an implicit function can be written
down at once as follows :
xdy + ydx + 3 x*dx = dy - 3 y dx ~
y
When x l f y = 3. Substituting these values, and dx = 0.03, in the
equation above, we get
dy - 3(0.03) + 3(0.03) - dy - 3[(- 3) (0.03) -
or
dy = 0.09 unit.
134 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VI
PROBLEMS
Find the differential of each of the following functions. (Nos. 1-6.)
1. y = (1 + 2 x)/(l - 2 x). Ans. 4 dx/(l - 2 x) 2 .
2. y = 2 xV4 x*.
3. /(s) = (a + s) Va - s. Ans. i[(a - 3 s)/Va - s]ds.
4. /( s ) = V(a~
5. /(O = </log t. Ans. df(t) = [(log t - l)d*]/log 2 1.
Determine dy in terms of x, y, and dx from each of the following equations.
(Nos. 7-13.)
8. y log x = x 2 .
9. x = log t, y = e~ 2 '. Ans. dy = 2
10. x 2 /a 2 + y 2/3/^2/3 = i.
11. (2 x + I) 2 / 3 (2 y - I) 2 / 8 = 5.
12. log (y/x) - xy = 7.
13. e z 6 W = xy. Ans. dy = [_(c x
Evaluate the differential of one variable for the given data in each of the
following cases. (Nos. 14-19.)
14. y = x 3 - x, x = 2, dx = - 0.002.
15. 2 x 2 + 3 y 2 = 7, x = 1, dx = 0.15. Ans. dy = =t V3/(10V5).
16. Vx + Vy = 3, y = 2, dy = - 0.2.
17. s = Hog <, t = e, ds = 0.03. Ans. dJ = 0.015.
18. x = a~", y = a, dy = 0.01 a.
19. x = log J 2 , y = log 2 1, t = e, dt = 0.02 e. Ans. dx = dy = 0.04.
20. If is an infinitesimal, is each of the following an infinitesimal?
(a) 3 + 2 e, (b) e + 2 e 2 , (c) sin c, (d) cos , (e) /(! + c 2 ), (/) * 2 (1 + e 2 ),
(0) lg > W e< - Which are of the same order as e? Which of higher order?
21. Given s = 80 t 16 t 2 , calculate the difference between As and ds
when (a) t = 2, dt = 0.1; (6) t = 4, dt = - 0.2. Ans. - 0.16; - 0.64.
22. Given y = x 3 x and x = 3, find the difference between Ay and dy if
Ax = dx = 0.02.
23. If y = x 2 - 2/x 2 find Ay - dy for x = 2, dx = 0.03. Ans. 0.115.
24. If s = e - log t find As - ds for t = 3, dt = 0.1.
87] DIFFERENTIALS. THEOREM OF MEAN VALUE 135
87. Approximations. Errors. The differential of a function
affords a simple method of approximating the change in a given
function due to a small change in the independent variable. Sup-
pose, for instance, it is desired to find the change in the area of a
circular metal plate due to expansion caused by a rise in tempera-
ture. Simple measurements give the diameter or radius of the
plate before and after the change in temperature. Now if A is the
area and r the radius of the plate, the problem is to approximate AA
by finding dA for a given r and Ar. To solve, merely express A as a
function of r and find its differential. Then substitute in the
expression for dA the values of r and dr, respectively, and obtain the
desired approximation.
In many such problems as the one above, it would be absurd
to find A^4 by the more cumbersome method of increments, since
any values obtained by measurement, as r and dr, are in themselves
merely approximations.
All calculations which are based on measurements involve errors
and these may be approximated by differentials. Thus if it is
assumed that the error in measuring the radius of a circle does not
exceed 0.1 inch, then the possible error in the calculation of the
area can be approximated by finding dA when d r 0.1 inch.
If x is given an increment, A# = dx, in the function y f(x), the
function becomes y + A?/. However, an approximation of this
value by differentials is y + dy = f(x) + }'(x)dx.
EXAMPLES
1. Heat applied to a metal plate expands its diameter from 15 inches to
15.14 inches. Approximate the change in area.
SOLUTION. The function is A = Trr 2 . To find dA when r = 7.5 in. and
dr = 0.07 in.,
dA = 2irr-dr
= 15 7r(0.07) == 1.05 TT sq. in. approximately.
We note in this example that dA may be interpreted as a rectangular strip of
length 2 Trr and width dr, whereas AA is a circular ring of thickness dr bordering
i circle of radius r.
2. Find by differentials the reciprocal of 5.03.
SOLUTION. This example involves the reciprocal function,
Convenient values for x and dx may be chosen provided dx is relatively
small. Since the reciprocal of 5 is 0.2, let x = 5 and dx = 0.03. By differen-
136 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VI
tials we can find the approximate change in the function due to a change of
0.03 in the variable. Thus
1
dy = - ~-dX
= _ 0.0012.
That is, the reciprocal of 5.03 is approximately 0.0012 less than the reciprocal
of 5. Hence
y -f dy = 0.2 - 0.0012 = 0.1988.
3. Approximate by differentials a root of the equation x 2 -f 3 x 6 = 0.
SOLUTION. Substituting x = 0, 1, 2, we find there is a root such that
1 < x < 2.
Let y = x 2 + 3 x - 6. Then
dy -= (2 x + 3) dx.
For x 1, y = 2. But for the root, y should equal 0, whence we let the
change in y, namely, dy = + 2 and we have
- 2*4-3 ~ 2T + 3 ~ '
The new value of x is then
x + dx = l + 0.4 = 1.4.
But x = 1.4 makes y = 0.16, so taking dy = 0.16,
The desired approximation is then
x + dx = 1.4 - 0.028 = 1.372,
which should be compared with the solution by the formula.
88. Relative Error. When errors of measurement are involved
in a problem, the ratio of the magnitude of the error to the magni-
tude of the quantity is usually more significant than the magnitude
of the error itself. Evidently, the same actual error made in
measuring both a large quantity and a small quantity may be
negligible in the former case but not in the latter. If an error Ao;
is made in the measurement of a given quantity x, then Ax/x is
called the relative error. If any function y = f(x) is computed
from data which are in error, the ratio dy/y is an approximation for
the relative error in the function and 100 (dy/y) is the approximate
percentage error.
88] DIFFERENTIALS. THEOREM OF MEAN VALUE 137
The relative error may be approximated directly by logarithmic
differentiation since the differential of log y is dy/y.
The applications of the differential may be extended to func-
tions of two or more variables. This, proved in Chapter IX, will
be assumed here.
EXAMPLE
If the radius of a right circular cone is measured as 5 in. with a possible
error of 0.02 in , and the altitude as 8 in. with a possible error of 0.025 in.,
what are approximations for the possible relative error, and the possible per-
centage error in the volume as computed from these measurements?
SOLUTION. We have the values r 5 in., dr =t 0.02 in., h 8 in.,
dh db 0.025 in. The double sign must be used since the error may be
positive or negative. The function is
V =\r*h,
o
whence, taking its logarithm and writing the differential, we have
dV _ ~dr dh
T~ r +7T
Using the positive values for dh and dr, we may write
dV
- = 0.008 + 0.0031 = 0.0111,
which is the approximate relative error in F; or
dV
100^- = 1.11%,
which is the approximate percentage error in V.
If negative values are taken for dr and dh, the results are numerically the
same as those above. However, if the values of dr and dh differ in sign, the
results are numerically less than those given above. Hence only one set of
calculations is necessary to determine the possible errors.
PROBLEMS
1. The radius of a circular plate increases by heating from r<> to r -f Ar.
Find an expression in terms of r and Ar for (a) the increase in the area ; (6) an
approximation for the increase in the area; (c)^he error if the result of (6) is
assumed correct. Ans. w(2 r Ar -|- Ar 2 ); 2 7rr Ar; TrA^sq. units.
2. The cost of painting a hemispherical dome is 20 cts. per sq. ft.
Approximate the error in the estimated cost due to a 3 in. error in measuring
the radius as 50 ft. How accurately must the radius be measured if the
possible error in the estimated cost must not exceed $10?
3. Approximate the volume of a right circular cone with a vertical angle
of 7T/2, if the diameter of its base is 2.9997 units. Ans. 1.1247 -K cu. units.
138 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VI
4. A sphere is to be cut from a cube of edge 2 x. If the diameter of the
sphere is to be 2 x and an error of 1 % is made in measuring x, approximate the
error in the amount of material to be cut from the cube.
5. Approximate the volume of a right circular cylinder of radius 4.97 in. if
its altitude is three times the radius of one end. Ans. 368.25 TT cu. in.
6. How accurately must the diameter of a sphere be measured if it is
necessary that the error in the calculated area shall not exceed 0.1%?
7. The edge of a cube is near 6 units. How accurately must it be measured
to give an error not to exceed 1 cu. unit in the volume? Ans. 1/108 unit.
8. Derive a formula for an approximation of the volume of a thin spherical
shell of thickness t. How much is the error in your formula?
9. The hypotenuse and a leg of a right triangle are measured as 5 and 4
inches, respectively. If there is a possible error of 0.01 in. in each measure-
ment, approximate the possible error in the other leg if it is computed from
these data. Ans. 0.03 in.
10. A central angle is computed from the measurements of the radius and
the arc. If 2% errors are possible in each, approximate the possible error in
the angle.
11. Approximate the change in the total surface of a right circular cylinder
of altitude 10 ft. and radius 4 ft. if its volume is changed 1/2 cu. ft. and the
altitude is kept constant. Ans. 9/40 sq. ft.
12. How accurately must the altitude of a right circular cone with
r = (4/3) h be measured if it is necessary that the percentage error in the
calculated volume shall not exceed 3%?
13. If s = Jcpv 2 and p is changed by + 2% and s by 3%, find an approxi-
mation for the change in v. Also the relative and percentage changes in v.
Ans. - 0.025 v, - 0.025, - 2.5%.
14. If the edge of a cube is near 5 in. and there is a possible error of 0.02 in.,
approximate the resulting relative and percentage errors in the volume and in
the surface.
15. A sphere's mass is determined as 1 oz. with a possible error of 0.05 oz.
and its diameter as 2 in. with a possible error of 0.02 in. Approximate the
possible error in its computed density. Ans. 3/5 w units.
Use differentials to approximate the values of each of the following expres-
sions. (Nos. 16-26.)
16. v/35.
17. ^26. Ans. 2%$.
18. 1/V50.
19. (123.5) 4 / 3 . 4ns. 615.
20. e 2 - 01 , if e* = 7.389.
21. log 10.2, if log 10 - 2.303. Ans. 2.323.
90] DIFFERENTIALS. THEOREM OF MEAN VALUE 139
22. 7 1 - 98 , if log 7 = 1.946.
23. 2/[l + (2.001) 2 ].
24. x 9 - x, if x = 2.002.
25. x 4 -f 4 z 2 + 1, if x = 1.997.
26. x 4 - 2 or 3 f 2 x\ if a; is 2 =b 0.015.
Ans. 0.3999.
Ans. 32.856.
Approximate irrational roots of each of the following equations. (Nos.
27-30.)
27. x 3 - x - 3 = 0.
28. x 3 - 3 x + 5 = 0.
29. x 3 + 3 x - 10 = 0.
30. x 3 - x 2 - 5 = 0.
Ans. 1.674.
Ans. 1.699.
89. Rollers Theorem. Let F(x) be a continuous, single-valued
function which vanishes for x = a and x = 6. As a; varies from
a to b, it is evident that F(x)
either increases then decreases, or
decreases then increases, so that
F(x) will have at least one maxi-
mum value or one minimum value
between x a and x = b. Now
if we add the further condition that
F'(x) is continuous in the interval
x = a to x = b, then F f (x) must
become zero at least once in this
interval. This gives rise to the
following theorem:
ROLLERS THEOREM. // F(x) is a single-valued function which
vanishes for x = a and x = b, and if both F(x) and F'(x) are
continuous* in the interval x a to x = 6, then F f (x) will vanish for
at least one value of x in this interval.
90. Theorem of Mean Value. Let Fig. 98 represent the graph
of y = F(x) from x = a to x = b. Let R be [a, F(a)] and S be
[6, F(b)]. Then the slope of the secant RS, which is the average
rate of change of the function F(x) from x = a to x = b, is
F(b) - F(a)
b a
FIG. 97
* It is in fact sufficient to assume that F'(x) exists everywhere, but the theorem
as stated is all that we shall need.
140 'DIFFERENTIAL AND INTEGRAL CALCULUS [CH. vi
The equation of the secant line RS may be written in the form
(a,0)
For any value of x in the given
interval, the difference of the two
ordinates represented by the y
of equation (I), and by F(x) is
another function of x, say <t>(x).
Then
(2) <I>(x)=F(b)-F(x)
b a
-p
We see by inspection that </>(x)
vanishes for x = a, and for x = b.
Hence, by Rollers Theorem, <'(x) will vanish for some value of
x as x\ between a and b. That is,
(3)
or
(4)
= 0, for
x =
But F'(XI) is the slope of the tangent to y = F(x) at the point
located by x = Xi and the right-hand member of (4) is the slope
of the secant RS. We have then the following theorem :
THEOREM OF MEAN VALUE. // F(x) is a single-valued function
and Twth F(x) ancTf^x) are continuous in the interval x a to
x = 6, then
F(b) - F(a) = *'(*) (6- a),
where a < JCi < b.
This is also called the law of the mean. There are two other
forms in which this law has important applications. One is
obtained by substituting x for 6, thus making a variable interval.
Then
(5) F(x) = F(a) + F'(xJ(x - a}, a < Xl < x.
The other is obtained by letting a = x, b = x + Ax, then the
interval is Ax, and,
(6) F(x + &x) = F(x)+Ax-F'(x + k-Ax) 9 < k < 1.
91] DIFFERENTIALS. THEOREM OF MEAN VALUE 141
y =
91. Extended Theorem of Mean Value.
be given parametrically by the equations
x = g(t),
where g(t), /(), g'(t) and /'(*)
are continuous in the interval
t = a to t = b. Furthermore,
we must assume g'(t) ?* 0, so
that dy/dx = f'(t)/g'(f) shall
exist and be continuous in the
interval. The coordinates of R
and 8 are then [g(a), f(a)] and
[0(6), /(>)] respectively. Follow-
ing the method of the preceding
section, we find the equation of
the secant RS to be
Now let the function
FIG. 99
(1) y
For any value of t in the interval, the difference of the correspond-
ing ordinates to the curve and to the secant line is the difference of
f(t) and the right-hand member of (1), namely,
- /()
(2)
/(O-
0(6) - g(a)
Call this function if>(t). It vanishes for t = a and for t = b;
hence, by Rolle's Theorem, if>'(t) will vanish for some value of t,
as t = ti, between a and 6, that is
(3)
or
(4)
J w g(W - /()
/(&) -/(a) _ /'(M
=Q for =
a < ti < b.
This is the extended theorem of mean value or extended law
of the mean.* If we make the interval variable by letting 6 = t,
it takes the form
/(O - /(fl) /U)
(5)
a <ti<t.
* This theorem is due to the French mathematician Cauchy and (4) is frequently
called Cauchy's formula. The geometric proof above waa suggested by Professor
A. A. Bennett in the American Mathematical Monthly, 1924, Vol. 31, p. 41.
142 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VI
The geometric interpretation of the law of the mean or the
extended law of the mean is obvious from Figs. 98 and 99, namely,
that the slope of the tangent at Q is equal to the slope of the secant
RS. Or we can say if a function and its derivative are continuous
in any interval, for some value of the variable within that interval
the rate of change of the function will be the same as the average rate of
change of the function throughout the interval.
92. The Indeterminate Form 0/0. If two functions f(x) and
g(x) both vanish for x = a, their quotient f(a)/g (a) has no mean-
ing. However, the lim (f(x)/g(x)] may be perfectly definite.
x *-a
We have already seen this in the case of (sin x)/x for x = 0.
Also in the derivative of the function y = f(x), both numerator
and denominator of (Ay) /(Ax) approach zero as a limit, yet the
limit of the quotient exists. Such a limit, if it exists, may be
found for special cases as in 55, but a more general method can be
obtained from the extended law of the mean. From (5), 91, we
have
/(*)-/() _ /'(*.) a <x <x
g(x)-g(a) ~ ^0 ' <*'<*
But /(a) = g(d) - 0; and, since lim Xi = a, it follows that
g'(a]
if g'(a) 7* 0. If both f'(x) and g'(x) are zero for x = a, this
method may be extended to the form
r ,* r
hm t = hm ~~ === hm
x -+a (j(x) x -+ a g (X)
where all derivatives of f(x) and g(x) up to the nth derivative are
zero for x = a. If either nth derivative is not zero for x = a the
limit of the quotient can be evaluated.
EXAMPLES
1. Find lim .
x >3 % $
SOLUTION. This quotient may be found by writing
** - x - 6 _ (x + 2)(x - 3)
a-3 - ^=~3 ~ * + 2 '
93] DIFFERENTIALS. THEOREM OF MEAN VALUE
only if x 7* 3. Using the method of the last section we have
~2 /y f{ O /y 1
|. J-> Jj U .. A JO 1 .,
lim - = km = 5.
2. Evaluate lim
*-Ci I "&
log a
SOLUTION.
,. a x ,. a x ,. 1
lim - = lim ; - : - = lim = a.
x-+a log Z - log a
3. Evaluate lim
. ,
-K X 3 3 Z H- 2
SOLUTION. Call the numerator /(x) and the denominator 0(z), then
v /(*) . , 4
lim -7 = - , indeterminate.
0'
,. r x A *
lim ^77-r = lim ^7-^ - -r = - f indeterminate.
x-+i9 (*) x-+i3(x 2 -1) 0'
f"(x] .
lim f~f~t r == lim
f~f~t ~ ~~ ^j
Q (x) x _+ l bx 6
93. Other Types of Indeterminate Forms. If the functions
/(x) and g(x) are such that lim f(x) = ^o and lim g(x) = oo,
x -a a; >-a
then the fraction f(x)/g(x) assumes the indeterminate form
oo / oo . The same method is then used as in the case for the indetermi-
nate form 0/0. A rigorous demonstration of this fact belongs more
properly in an advanced course and will be omitted here.
If lim /(#) = 0, and lim g(x) = <*> y then f(x) - g(x) becomes
x >a x >a
oo for x = a, which is indeterminate. In this case we write the
function
or
or
1/f(x y
so that the transformed expression takes one of the forms 0/0 or
oo / oo 9 and its limit, if it exists, can be evaluated by the method
explained in 92.
An exponential function may assume one of the indeterminate
forms 0, 00, or 1 for some value of the variable. In this case the
144 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VI
logarithm of the function may assume the form > and its limit,
if it exists, can be evaluated. (See footnote, 73.)
The difference of two functions for some value of the variable
may assume the indeterminate form oo oo. In this case a
transformation may be found which will reduce the function to one
of the other types of indeterminate forms which have been men-
tioned.
EXAMPLES
1. Find the lim
-
log x
SOLUTION. This function has the indeterminate form oo/oo.
,. x logx .. 1 -f logz
lim , , = lim - r ry =
a^o, x 4- log x x _+a> 1 -f l/x
since the denominator has the limit 1, while the numerator increases without
limit.
2. Evaluate lim x l l^~'\
X+1
SOLUTION. Call this function /(x), then log f(x) =[!/(! x)~\ log x.
Fie re f(x) has the indeterminate form 1, for x = 1 ; and log f(x) has the form
oo 0. Writing log/(x) as (log x)/(\ x), we have
1
JL
i 1 - x " ^\ - 1
.. log x ,. x .
lim - = lim - = 1.
Stance
lim log/(z) = 1, or lim x lf(l ~ x) = e~ l - l/e.
3. Find lim ( l
- 1 x)
SOLUTION. This function has the indeterminate form oo oo for x = 0.
Vrite it in the form (x - e x 4- l)/x(e x - 1). Then
r x - e x 4- 1 ,. 1 - e x
hm n x(e * _ n = lim n ^x . ex _ i =n'
jp ^.Q Js\V L) % ^.Q U^ T^ o JL U
Differentiating again, we get
J gZ gX 1 1
lim ; lim ^ = lim
^ x + 2 2
PROBLEMS
Evaluate the limit, if it exists, of each of the following cases. (Nos. 1-19.)
1. lim [(x 2 - 2 x)/(x* - 4)]. Ans. 1/2.
oj-^2
2. lim [(x - 2)/(x 2 + 4)].
af---oo
3. lim [(5 x - 4 z 2 )/(2 z 2 - x 4*4)1 Ans. - 2.
93] DIFFERENTIALS. THEOREM OF MEAN VALUE 145
4. lim [x/(l - *)].
z-*o
5. lim [(3 x - 4 z 2 )/(2 re 2 + e* - 1)]. Arts. 3.
a>-M)
6. lim [x log a?].
x-*o
7. lim l(x m - l)/(z n - 1)]. Ans. m/n.
x-+l
8. lim [(log x) /z 2 ].
a>-*.oo
9. lim [(log z 2 )/(z - 1)]. Ans. 2.
z-*l
10. lim[{log(3-a)f/(* -a)].
X*<x>
11. lim [(e-i - a-i)/(z - 1)]. Ans. (1 - logo)/2.
x *!
12. lim
13. lim [(1 + /err) 1 /*]. Att$. c*.
15. lim ~ _^JL. Ans. 1/2V6.
x-^2 3 x - 2v 15 - 3 x
16. lim [_x n log z 2n ].
17. lim [(1 -I- az) >-**>/*;]. ATIS. e' b .
X-+Q
18. lim [(log Z)*/CI-IK *>].
X >
19. lim [(e a * + 6x) c / x ]. X"' Ans. e c <+ 6 >.
x-K) V^
20. Given the curve whose equation is y = (1 + x) lfx . Prove that the
limiting value of the slope of the tangent as the point of contact approaches
the y axis is e/2.
ADDITIONAL PROBLEMS
1. If the edge of a cube is 3 in. and is expanded to 3.002 in., approximate
the volume. Ans. 27.054 cu. in.
2. Approximate the volume of a right circular cone of h == 2 r if ^ r is
measured as 4.98 in.
3. A sphere of radius 2 ft. has its radius reduced 3 in. Approximate the
resulting change in its surface. What is the exact change?
Ans. 4 TT sq. ft. ; 3J * sq. ft.
4. A spherical casting of r =* 1 ft. is smoothed down so that the radius is
decreased 0.1 in. Approximate the xpj^me removed. What is the error in
the approximation?
146 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VI
5. Show that (x + dx) 2 is approximately equal to x 2 -f- 2 x dx.
Derive formulas similar to the one in Problem 5 to approximate each of the
following expressions. (Nos. 6-10.)
*. \/(x + dx).
7. Vx + dx. Ans. Vx + dx/(2Vx).
8. $(x + dx)*.
9. log(x -f- dx). Ans. log x + dx/x.
10. a<*+ d *>.
Approximate the irrational roots of each of the following equations, (Nos.
11-16.)
11. x* - x* + 5 = 0. Ans. - 1.43.
12. x 3 - 4 x 2 - 2 x + 8 = 0.
13. z 3 - 3 x + 1 0. Ans. - 1.88, 0.42, 1.59.
14. z 3 - 8 x + 2 = 0.
15. z 3 + z - 4 = 0. 4ns. 1.38.
16. x* - 11 z 2 + 24 = 0.
17. Two iron spheres are each approximately 10 inches in diameter. When
immersed in a pail of water it is found that one sphere displaces 20 cu. in. more
of water than the other. Approximate the difference in their radii
Ans. l/(5 ?r) in.
18. What is the percentage error in u if there is an error of 1 in the fourth
decimal of logio ul
19. What error in the common logarithm of a number will be produced
by an error of 1% in the number? Ans. 0.0043.
20. The motion of an object is given by s = t* + 4 t 2 3 t -f 5. Approxi-
mate the distance moved from t = 2 to t - 2.02 sec.
21. The same as Problem 20 except from t = 1 to t = 2.998 sec.
Ans. 51.904 units.
22. A point moves along the curve x = 3^, y = 1 2v7. Find the
rectangular equation of its path and approximate the change in y when x
changes from 9 to 8.97.
23. Rectangles with sides parallel to the two axes are inscribed in the
area bounded by y 2 - Qx and its latus rectum. The value of y for one of these
rectangles is measured as 1.5 units with a possible error of 0.04 unit, approxi-
mate the possible error resulting in its area. Ans. 0.03 sq. units.
24. If s = kp l{2 v 2 j approximate the relative change and the percentage
change in s for 3% change in v and 1J^% change in p.
CHAPTER VII
TRIGONOMETRIC FUNCTIONS CURVATURE
94. Derivative of sin u. Let y = sin u. Give the independent
variable u an increment; then
y + Ay = sin (u + Au),
and
A?/ = sin (u + Aw) sin u.
Using the formula
.
sm
we get
. w o [A + B-] . [A - BI
sin B = 2 cos = sin ^ ,
. / . Aw\ . Aw
A?/ = 2 cos ( u + -^ \ sin -^
Then
, Aw\ .
Aw Aw
. Aw
Au
~2
Let Aw -> ; then
. Aw
/ Aw\ sm "2"
lim ( w + - ] = w, and, by 55, lim z_
Au *-0 \ ^ / Aw0 AW
T
Hence
dy
~- = cos w.
aw
If w is a differentiable function of x, since
of?/ _ r^i/ du
dx du dx '
then
(XIH) ^ (sin u) = cos u ^
147
148 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
95. Derivative of cos u. The derivative of cosu may be
obtained by using either the relation cos u = sin (ir/2 u) or
cosV = sin (7T/2 + u). Using the former, we have
^ (cos u) = ^ [sin (ir/2 - u)]
= COS (7T/2 ~U)~ (T/2 - li)
"COB Or/2 -
Therefore
(XIV) - (cos u) = - sin u -
96. Derivative of tan u, ctn i/. To differentiate the tangent
function, write tan u = sin u/cos u. Then using the formula for
differentiating a quotient, we have
d , . . . d , ,
7 cos u - -r- (sin u ) sin u -7- (cos w)
d /A x dx dx
(tan
7 IfM/li W/J rt
dx cos 2 u
_ cos 2 u + sin 2 u rfw
cos 2 u dx
I du
cos 2 u dx
Hence
(XV) 4- (tan u) = sec 2 u ~
v ' dx v ' dx
In a similar manner we derive the formula
(XVI) 4~ ( ctn u ) ^ - cs 2 u ' T^ '
97. Derivative of sec u, esc u. Write sec u = I/cos u. Then
d , ,
, 3- (cos w) . ,
d f , __ qj; sm^ atz
(sec w )
~T ~~ ^ n ~i
dx cos 2 w cos 2 w dx
That is,
(XVII) (sec u) = sec u - tan u
97] TRIGONOMETRIC FUNCTIONS CURVATURE 149
In like manner we find
(XVIII) ~T- (esc u) = esc u ctn u -T-
EXAMPLES
1. Find dy/dx if y = (1/3) sin 2 z - tan 2 3 x.
SOLUTION.
^ = I (cos 2 a) (2) - 2 (tan 3 or) (sec 2 3 a?) (3)
2
= -- cos 2 x 6 tan 3 a: sec 2 3 x.
o
2. If r - [sec< (5/2) - cos 2 (0/2)] ctn (5/2), find Jr/rfo.
SOLUTION. This function is of the form u - v. Hence,
-f- ctn - ( 4 sec 3 - sec - tan s ) I s I ~~ 2 cos ^ ( sin -
1 f 2 ^ _ 1 4 ?. 2^10 4 ^ 1 2 ^
"22 2 2 2 "^ C 2 "*" C S 2 '
PROBLEMS
Differentiate each of the following functions with respect to its variable.
(Nos. 1-16.)
1. y - ctn (z/3). Ans. - (1/3) esc 2 (z/3).
2. y cos 3 x 2 .
3. y = sin 2 5 /. Ans. 5 sin 10 t.
4. s = tan 2 2 J.
5. x = sin (2 - 3 y) cos (2 </ - 1).
Ans. - 2 sin (2 3 y) sin (2 y - 1) - 3 cos (2 - 3 ?/) cos (2 y - 1).
6. ?/ = sin 3 2 x cos 3 x.
7. y = sin 2 x e cos 2 *. Ans. (1 /2) sin 4 x - e coa 2 .
8. !/ = log sin 2 (3x/2).
9. s = sin (log 0- Ans. (1/0 cos (log t).
10. 2/ = log (sec 3 x -f tan 3 x).
11. y = log {[1 - sin (x/2)]/[l + sin (x/2)] }. Ans. - sec (x/2).
12. y = (tan x)/x 2 .
13. y = 4 sin (x/2) - 2 x cos (x/2). Ans. x sin (x/2).
lec x -f tan x
tan x
HINT: (sec x + tan x) and (sec x tan x) are reciprocals. Why?
150 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VII
15. y = 2 sin 2 (z/2) cos (x/2).
Ans. sin (z/2) [2 cos 2 (x/2) - sin 2 (z/2)].
16. y = esc 2 (1 - 2 x) ctn (1-2 z).
Find the derivative of one variable with respect to the other in each of the
following cases. (Nos. 17-26.)
17. xy = sin 3 x. Ans. dy/dx = (\/x) (3 cos 3 x y).
18. xy = e cos2x .
19. cos (x + 2/) + cos (x /) = 0. ylns. dy/dx tan x ctn y.
20. a:// -f- ctn xy = 0.
21. sin 2: cos ?/ + cos 2 :c sin 2 y = 1.
/I/is, dy/dx (2 sin 2 a; sin 2 y cos x cos ?/)/(2 cos 2 x cos 2y sin x sin y).
22. r 2 sin 3 = a cos 0. Find dr/dO.
23 . y=log J=^2-
\ 1 + cos
24. r sec = sin 2 0.
25. x = a(0 -f sin 0), y = a(l cos 0).
^4 /is. dy/dx = (sin 0)/(l 4- cos 0) = tan (0/2).
26. x = a cos + aO sin 0, ?/ = a sin aO cos 0.
Find dy/dx and d 2 y/dx 2 for each of the following cases. (Nos. 27-31.)
27. x 2 sin t, y = 2 cos t. Ans. tan Z, (1/2) sec 3 t.
28. x = 2 sin -+- 3 cos t, y sin t.
29. x = a tan 0, // = b sin 0. ATW. (6/a) cos 3 0, (3 6/a 2 ) cos 4 sin 0.
,30. x - a sin 3 0, y = a cos 3 0.
31. a: = e 2 ', ?/ = 2 + cos 4 .
^4ns. (2 sin 4 )/e 2 *, 4(sin 4 < 2 cos 4 OA 2 *.
32. Derive the formula for differentiating cos u with respect to x by incre-
ments.
33. The same as Problem 32 for tan u.
98. Derivatives of the Inverse Trigonometric Functions.
Let
y = sin" 1 u
where u is a differentiable function of x, then
u = sin y.
98] TRIGONOMETRIC FUNCTIONS CURVATCRE 151
Differentiating u with respect to y, we have
du
Hence
dz
Observe that the double sign enters in the result. This
is because y is a many- valued function of u in y sin" 1 u,
whereas u is a single- valued function of y in u = sin y. Since
du/dy cos y, and the smallest values of y for which cos y is
positive lie between ir/2 and ?r/2, we shall ordinarily restrict
the values of y to satisfy 7r/2 = y ^ 7r/2, so that the derivative
will not be negative. Then
du
(XIX) -^-(sin- 1 ^ JL-.
Differentiating y = cos" 1 u in a similar manner, we find
du
-y- = sin i/.
%
Here again y is a many-valued function of u and if we restrict
y to an interval for which sin y is positive or zero, namely,
< 2/ = ^ the derivative will then be negative or zero, and
<Jy = _ J__ =
dw sin !/
or
du
(XX) (cos- lu )=--
If y = tan" 1 u, u = tan y, and du/dy = sec 2 y, then
dy ^ 1 = 1 = 1
dw sec 2 y 1 + tan 2 y 1 + tt 2
Hence
du
(XXI)
152 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
For y = ctn" 1 u, du/dy = esc 2 y. Hence
dy _ 1 1 _
du
esc 2 y
Therefore
(XXII)
1 + ctn 2 y
du
f -L i \
-j- (ctar 1 u) = - r ~. 7,
dx ^ ' 1 + u 2
1 +
In like manner we obtain the additional formulas
du
d *lv
(XXIII) -f
uVu 2 - 1
where ri u < ir/2, or TT ^u < ir/2, and
(XXIV)
-r- (CSC- 1 U) =
- 1
where < u ^ ir/2, or TT < T/ ^ w/2.
PROBLEMS
Differentiate each of the following functions with respect to its variable
(Nos. 1-16.)
1. y = tan- 1 x 2 . Ans. 2x/(l+ a; 4 )
2. y = ctn' 1 (1/x 2 ).
- 3. y = sin" 1 6 .
4. y = cos^Vl x 2 .
5. y = x sec" 1 x.
6. y x 2 tan" 1 (a/x).
7. s = ctn- 1 [</(! - OJ-
i)
8. y = x csc^Vl -f x 2 .
9. ?/ = x 2 sin" 1 (2/z).
10. y = xVl x 2 + cos"" 1 a:.
11. s = e co * '-sin- 1 (1 - 0-
Ans. - e oos < [I/ V2 < - t* + sin i sin-^l - *)].
12. y = x sin^Vl x 2 Vl x 2 .
13. y = x tan" 1 ax (I/a) logVl -f a 2 x 2 . Ans. tan"" 1 ax.
*. 6/ Vl - 36 i 2 .
Ans. 1/Vx 2 1 + sec" 1 x.
Ans. - 1/(1 -2t + 2 P).
Ans. 2 x sin- 1 (2/x) - 2 x/Vx 2 - 4.
99] TRIGONOMETRIC FUNCTIONS CURVATURE 153
14. y = tan^Vx 2 - 1 + cos- 1 (1/x).
15. y = 3 x ctn~ J 3 x -f logVl -f 9 x\ Ans. 3 ctn" 1 3 x.
16. y = (tan- 1 a; 2 ) 3 .
Find the derivative of one variable with respect to the other in the following.
(Nos. 17-26.)
17. x sin' 1 x -f cos 2 x sin 2 y = 1.
ylns. dy/dx = [2 sin 2 x sin 2 ?/ sin' 1 x x/Vl # 2 ]/(2 cos 2 # cos 2 ?/).
18. 2/Vz 2 + 2 a - tan- 1 Vs* + 2 3 = 0.
19. y 3 sin x -}- y tan" 1 re.
Arw. dy/dz = [l/(z 2 + 1) - y 3 cosx]/(l + 3 y* sin x).
20. z 2 ?/ 2 = x sin" 1 2 .
21. z = sin-M, T/ = cos~ l . Ans. 1.
22. x = sin- 1 2 , y = tan" 1 2 ^.
23. log(o: 2 + 2/ 2 ) ~ tan-^/z) - 0. Ans. dy/dx (2 a; + y)/(a; - 2 y).
24. log sin 2 ?/ -f sec" 1 4 x = 0.
25. tan- 1 2 a; + 7/2.^* = 7>
s. dy/dx - [y 2 sin 35 e ax - 2/(l -j- 4 o: 2 )]/(2 ?/e co8a: ).
26. a: + V2 a?/ ?/ 2 = a cos" 1 [(a
27. If an angle x is measured in degrees, how may you differentiate a
function of it with respect to x?
28. Answer Problem 27 for tan" 1 x log cos x.
99. Applications of Trigonometric Differentiation. Many im-
portant problems involving trigonometric functions are included
under the various topics discussed in Chapter V. In fact, many
problems in that chapter may be solved readily by using trigono-
metric relations to express a required function.
Whenever the differentiation of trigonometric or inverse trigono-
metric functions is involved it is essential to remember that the
formulas were derived on the assumption that the angle is expressed
in radian measure. All these formulas depend on the derivative
of sin u, containing lim [(sin0)/0] which is unity only when is
0-*0
expressed in radians.
EXAMPLES
1. A wall is to be braced by means of a beam which must pass over a
lower wall 6 feet high and standing a feet from the first wall. What is the
shortest beam that can be used?
154 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
SOLUTION. Let I be the length of the beam and let it make an angle 6 with
the horizontal. Then
I (a -\- x) sec and x = b ctn 6.
Hence
I a sec + 6 esc 0]
then
dl
do
= a soc 6 tan 6 b esc ctn = 0.
^ Solving, we have
a sin b cos
COS 2
sin 3 _ b
cos 3 a '
sm 2 0~'
tan =
36
FIG. 100
Applying the third test, we have
dH
-j~ = a sec (sec 2 -f tan 2 0) -f- 6 esc (esc 2 -f ctn 2 0).
Hence dH/dO 2 is positive and I is a minimum. The length of the beam is
I
- (a 2 / 3 -f b 2 / 3 ) 3 / 2 .
2. The hands of a tower clock are 4]^ ft. and 6 ft. long, respectively. How
fast arc the ends approaching at four o'clock?
SOLUTION. Let s be the distance between the ends, and the angle between
them. Then using the law of cosines, we have
6 .2 = 6 2 + (4.5) 2 - 12(4.5) cos
= 56.25 - 54 cos 0.
Hence
o ds K . . d9 , d s 27 sin (70
2 s -7 = 54 sm - -r , and -r = ---- ~
dt dt dt s dt
At four o'clock = 2 ir/3, s 2 = 56.25 -f- 27 = 83.25. Since is decreasing
at the rate of (2 TT ir/Q) =11 TT/G radians per hr., we have
dO = 11
dt ~
Therefore
360
radians per minute.
360
dt \/83.25
= - 0.246 ft./min.
3. An acute angle A of a right triangle is computed from measurement of
the sides a and b. If 1 % error is assumed in each measurement, approximate
the possible error in the calculated value of A.
99] TRIGONOMETRIC FUNCTIONS CURVATURE 155
SOLUTION. Since tan A = a/b, then A = tan~ l (a/6) and
b da adb
dA =
a 2 + 6 2
To approximate the largest possible error in -A, assume that da =h 0.01 a
anddb =* =F 0.01 6; then,
PROBLEMS
1. Find the slope of the normal to the curve x = 2 cos 3 1, y =*= 2 sin 3 / at
t = 7T/6. ylns. V3.
2. Find the angle between the curves y = cos x and y = cos 2 # between
x = 7T/2 and x = TT.
3. Find the maximum vertical width of the cardioid r = 1 -f- cos 0.
Ana. 3 A/3/2 units.
4. Has lim (sin 0/0) 1 any importance in the development of the
o*o
calculus? Mention instances in which it is used.
5. Find the equation of the tangent and the normal to x == a cos 0,
y b sin at the point determined by 7r/4.
Am. bx 4- ay = abV2, (ax - by)V2 = a 2 - b 2 .
6. Find maximum and minimum values of x -f- 2 sin x.
7. Find 6 if tan + ctn is a minimum. ^4ws. ?r/4.
8. An isosceles triangle has equal sides of 6 inches which include the
angle 0. If increases l/sec., how fast is the area of the triangle changing?
9. In Problem 8 find the intervals of in which the area of the triangle
increases and decreases. Ans. < < ir/2, ir/2 < < ir.
10. Examine y sin 2 x 1 for inflections. Evaluate maximum and
minimum values for y.
11. Find maximum and minimum values of sin 2 x by calculus and check.
Ans. 1, 0.
12. Find the slope of a cycloid at any point.
13. Show that y e~ 2x cos 2 x and y = e~ 2x have a common tangent at
any common point.
14. Derive our known formulas for cos 2 from sin 2 =2 sin cos 0.
15. A plane is moving horizontally 200 mi./hr. at an elevation of 4200 ft.
How fast will an angle of elevation from the ground change if it is near 30?
^4.W5. l/sec.
16. Find the minimum area cut from the first quadrant by a line through
(1,3).
156 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
17. Approximate the change in tan if 6 decreases 15' from T/4.
Ans. - 0.0087 unit.
18. Find maximum and minimum values for logVl -{- x 2 + ctn" 1 x.
19. If y sin 2 3 , is y increasing at x = 7r/4; is the curve concave up at
x 7T/6; what is an approximate value of y if x = 45 10'?
Ans. No; no; 0.4913.
20. Find an approximation for the change in cos if 6 decreases from Tr/6
by 0.01 radian.
21. Show that a line through any point of a rolling wheel tangent to the
generated cycloid cuts the wheel at its highest point.
22. A roofer wishes to make an open gutter of maximum capacity, its cross-
section being an isosceles trapezoid with lower base and equal sides 10 in. each.
What should be the width of the top?
23. Approximate logio sin 33 if logio sin 30 = 0.301 and logio e = 0.434.
Ans. - 0.262.
24. A man on a wharf 20 ft. above the water pulls in the rope of a boat so
that the boat approaches the wharf at the rate of 3 ft. /sec., when it is 15 ft.
distant. At what rate is the rope being drawn in?
25. A man slides a 10 ft. ladder up a wall, so that the foot of the ladder
approaches the wall at the rate of 3 ft. /sec. How fast is the top of the ladder
going up the wall when the bottom is 4 ft. distant? Ans. 6/V21" ft. sec.
26. Given y = x 2 sin^ 1 x. Find (a) the approximate change in x for
0.02 unit change in y } (b) the approximate relative and percentage errors
in y for 0.02 unit error in x.
27. Evaluate the following limits.
, x ,. sin 2 6 tan
(a) hm - - -- Ans. 2.
x-*o sec x - cos x
, N ,. e* cos x
(c) hm - r- - Ans. 1.
x *o sin x
. j. ,. tan x sin re 1
--
28. Find a point of inflection on y = (x -f 1) tan" 1 x.
29. From the point A(r, 0) on the circle x 2 + y 2 = r 2 the arc AB and the
tangent AT are drawn so that AT - AB. Find the limiting intersections of
TB with the diameter through A as B approaches A. Ans. ( 2 r, 0).
30. AB is a diameter and the center of a circle of radius r. The chord
AC makes an angle with AB. A ray of light from A is reflected by the circle
at C and the reflected ray meets AB at P. Find the limiting position of P
as C approaches B. Ans. OP * r/3.
31. Solve Problem 30 if the source of light is (a) at M the mid-point of OA ;
(fc) at N 9 such that A is the mid-point of ON.
Ans. (a) OP r/4; (b) OP = 2 r/6.
100] TRIGONOMETRIC FUNCTIONS CURY AT U RE 157
32. Two diametral paths cross a circular courtyard at right angles. A man
walks along one path at a uniform rate, and a lamp at one extremity of the
other path ca^sts his shadow on the circular wall surrounding the court. At
what rate is his shadow moving when the man is midway between the center
and the extremity of the path?
33. A beam 30 feet long is to be carried horizontally around the corner of
two passageways which intersect at right angles. One passageway is 10 feet
wide ; how narrow may the other one be to permit the beam to pass, if we as-
sume that no allowance is necessary for its thickness? Ans. 11.225 ft.
100. Angle between the Radius Vector and Curve. Let the
equation of the curve in polar coordinates be
(1) r=/(0),
and let \l/ be the angle between
the vector OP and the tangent
P7 7 to the curve at P (r, 0). Then
give 6 an increment A0, and let
Q(r + Ar, 6 + A0) be the corre-
sponding point on the curve.
Draw OQ, PQ, and draw PR
perpendicular to OQ. Call <j> the
FIG. 101
angle between OQ and the secant PQ.
OPR and PQ/J,
RP = RP
RQ OQ - OR
r sin A0
tan =
Then in the right triangles
r sin A0
A0
r + Ar r cos A0
4-
A0" 1 "
A0
Let A0 approach zero, then Q approaches P as its limit and the
secant PQ approaches the tangent TP as its limit. Hence angle <f>
approaches \l/ as its limit and we have
r sin A0
lim tan 4> = tan ^ = lim
lim
_ 4- -
A0^
r sin A0
cos A0)
A0
r F Ar
lim Af)
^^ ^0| ZAt/
r(l cos A0)
A0
158 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VII
But, by 55,
,. sin A0 . , ,. 1 cos A0 n
hm r- = 1, and hm - - = 0,
&" A6+0 &v
therefore
, . r dB
tan \1/ = -7- = r -r
t/r ar
Hence the angle ^ between the radius vector OP and the curve
r = /(0) at P is given by
(2) tanv|, = r^.
The inclination a of the tangent TP is given by
(3) a = 6 + i|r.
To find the angle ft between two curves whose equations are
given in polar coordinates, the angles \l/\ and ^ 2 , which they make
respectively with the radius vector to their intersection, must be
found, then
(4) P = *i - *2, v|>i ^ v|/ 2 .
If a polar curve passes through the pole, the radius vector
corresponding to the value of for which r = is tangent to the
curve. Hence, at the pole, \[/ = 0. If two curves intersect at the
pole, = 0i 2 , where 61 and 2 are the values of in the respec-
tive equations for which r = 0.
EXAMPLES
It Find the angle which the curve r = 3 + 2 cos makes with the vector
e = 45.
a r 3 + 2 cos
SOLUTION. tan \f/ = = - - -
dr 2 sin
</0
Hence if 6 = 45, this becomes
3 4- V/2
tan ^ = ^ ~ - - 3.1213,
or
^ = 180 - 72 14.1' - 107 45.9'.
2. What angle does the radius vector to any point of the cardioid
r = a(l cos 0) make with the curve?
101] TRIGONOMETRIC FUNCTIONS CURVATURE 159
SOLUTION.
r a(l cos 6) . 9
tan ^ = = : ^ - = tan =
dr a sin 2
dB
Hence the angle is one-half that which the radius vector makes with the
polar axis.
PROBLEMS
Find the angle of intersection of each of the following pairs of curves.
(Nos. 1-14.)
1. r = a0, = 7T/2. Ans. 57 31.1'.
2. r = a sin 3 (0/3), <9 = ir/6.
3. r 2 = a 2 sin 2 0, = *-/6. Ans. 7r/3.
4. r = 2 - 3 cos 0, = 7r/3.
5. r = a(l - cos 0), = Tr/6. Ans. *-/12.
6. r 2 = a 2 sin (0/2), = ir/3.
7. r = a[2 - (2/3) cos 0], = T/3. Ans. 70 53.6'.
8. r = 2, r = 2 -f- 3 sin 0.
9. r = 1/2, r = cos 0. Ans. Tr/3.
10. r = sin 2 0, r = cos 2 0.
11. r = 4 cos 0, r = 4 sin 2 0. Ans. 0, *-/2, tan" 1 3V3.
12. r = a cos 0, r = a(l cos 0) in the first quadrant.
13. r = 2 + cos 0, 4 r cos = 5. Ans. tan~K~ 5/V3) - ir/6.
14. r = o sin 0, r 2 = a 2 cos 2 0.
15. Show that the spiral r = e ad cuts each radius vector at the same angle.
16. Find the inclination of r = a(l 4- cos 0) where it crosses = ir/2.
(The angle the tangent to the curve makes with the polar axis.)
17. Find the inclination of each curve of Problem 12 at that intersection.
Ans. 7T/6, 7T/2.
18. The same as Problem 17 for Problem 8.
101. Curvilinear Motion. If the path of a moving point P is a
plane curve, the coordinates of P are functions of the time t, as
(i) * = 0(0, y=/(0-
These are called the equations of motion, and are the parametric
equations of the path.
We have already considered motion in a straight line in 84.
In curvilinear motion, the distance s measured along the arc of the
160 DIFFERENTIAL AND INTEGRAL CALCULUS [Ca. VII
curve is also a function of the time, and the velocity, as in recti-
linear motion, is v = lim (As/ At) = ds/dt. The velocity is repre-
sented by a directed line-segment or vector whose length is the
value of ds/dt. It is determined from equations (1) as follows:
In time At the point P is displaced
a distance Ax in the direction of the
x axis, and a distance Ay in the di-
rection of the y axis. The limit of
the ratios Ax/ At and Ay /At as At ap-
proaches zero, namely, dx/dt and
dy/dt, are the components of the
velocity v in the directions of the
coordinate axes. In other words,
they are the projections of the vec-
tor v on lines through P parallel to
the coordinate axes. These vector
f and v y respectively. Let v make
FIG. 102
components are designated v.
an angle 6 with v x , then
(2)
= = v cos 6,
= ^ = v sin 8.
From these we have
(3)
v = Vv* 2 +
dy
dt
,
dx
which give the magnitude and direction of the velocity. Hence
the direction of the velocity at any instant is along the tangent
to the curve at P.
Writing the positive value of v in the form
ds
dt
we have at once an expression for the differential of arc ds since
s, x, and y are functions of t, namely,
102] TRIGONOMETRIC FUNCTIONS CURVATURE 161
whence
(4) ds = \/dx 2 + dy 2
EXAMPLE
A particle is moving on the curve x ad 6 sin 0, y = a b cos 0. Find
the rectangular equation of its path, the velocity in its path, and the point
where the direction of motion has the greatest inclination.
SOLUTION. The equation of the path is found by eliminating the param-
eter 0. From the second equation,
Substituting this in the first equation, we get
, a
x = a cos" 1 p^ V6 2 (a y) 2 .
Since x, y, and must be functions of t,
dx , , ^ de
Vv =dy =
Hence
That inclination will be greatest which has the greatest absolute value for its
slope. The slope is
6 sin
dx v x a 6 cos
dy^ __ ab cos 6 2 cos 2 6 2 sin 2 _ 6(a cos 6)
d0 "~ (a 6 cos 0) 2 ~~ (a 6 cos 0) a
Then
-~ = when = cos" 1 - ,
de a
= oo when = cos" 1 T
If 6 < a, = cos" 1 (6 fa) gives the maximum slope.
If a < 6, the direction of motion is vertical when = cos" 1 (a/6).
102. Acceleration in Curvilinear Motion. Let the change in
velocity along the x axis in time A be &v x and the corresponding
162 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
change along the y axis be Ay y . Then
,. Ai> x dv x d 2 x
and
dt*
= a v
are the components of the acceleration in the direction of the
coordinate axes, respectively. These are likewise represented by
vectors, and the total acceleration a is
(1) a = Vfe'W+t/'W = Va, 2 + a, 2 .
It is very important to observe here that the acceleration of a
moving point in curvilinear mo-
tion is not given by the value
dv/dt as is the case in rectilin-
ear motion.
The velocity is directed
along the tangent to the curve,
and dv/dt is the component of
the acceleration in the direc-
tion of motion. It is called the
tangential acceleration a r .
The component of the accel-
eration at right angles to the
direction of motion is called
the normal acceleration a N .
FIG. 103
Differentiating v = \/v x 2 + v v 2 with respect to t, we have
(2)
dv _ v x a x + v y a v __ x'x" + y'y
+ y' 2 ) 1 / 2
Let the vectors PL and PM represent the horizontal and verti-
cal components of the acceleration of a point P moving along a
given curve. Then the vector PQ represents the total accelera-
tion. The projections of PQ on the tangent and normal at P are
PH, the tangential acceleration, and PK, the normal accelera-
tion. Let the angle between the vectors a x and a T be 6; then
PH = KQ = EM + FQ,
or
a T = a v sin + a x cos 6.
102] TRIGONOMETRIC FUNCTIONS CURVATURE 163
Substituting for sin 6 and cos 6 the values from (2), 101, we obtain
the result given above in (2).
Similarly, to find the normal acceleration, we have
PK = PE - FM = a y cos - a x sin 6,
and substituting as before, we obtain
/,\ v x a y v y a x x'y" y'x"
(6) QN = = / TJ p 1
where the primes, as in (2), indicate differentiation with respect to t.
The normal component of the acceleration produces no effect
on the speed of the moving point but it is the centrifugal force
which causes the change in the direction of motion.
PROBLEMS
Find the equation of the path in rectangular coordinates and the velocity
of a point along the path if its equations of motion are as stated below. (Nos.
1-9.)
1. x = 2t, y x=2/(* 2 + l), t = 1.
Ans. y(x z -+- 4) =8, V units per unit time.
2. x = 2 - 3 cos t, y = 3 -f 2 sin t. Also find a.
3. x = i V* 2 + 9, y = t. Ans. x 2 - y z = 9, V(2J 2 + 9)/(J 2 + 9).
4. x = 2 t, y = 2V t - t 2 . Also find a.
5. x = 4 cos 3 2 t, y = 4 sin 3 2 /. Find the smallest positive value of t for
which v is a maximum. Ans. x 2/3 -f ?/ 2/3 = 4 2/3 ; 12 sin 4 J; t = -r/S.
6. x = log *, y = 2VT. Also find a.
7. x = # 2 ', y = log c~*. Ans. x = e~ 2v , \/4 e 4 * -f- 1.
8. x = at b sin , y = a 6 cos . Also find a.
9. x = e~ 2 ' cos 2 , y = e~ 2t sin 2 t.
Ans. x = Vx 2 + i/ 2 cos' 1 [log (x 2 + 2/ 2 )~ 1/2 ], 2 e- 2t V2.
10. If v x = 5 ft./sec. on x 2 -f y 2 = 10, find v at x = 2.
11. If v* = 15 ft./sec. along x 2 - y 2 = 144, find v at (13, 5).
Ans. 41.8 ft./sec.
12. A point moves along x 2 y 2 = 25. Where is v* = v v f
13. A body moves along y 2 x 2 + 4 = with v x = 4 units /sec. Find
v tf and a v . Ans. 4 x/y, 64/y 3 .
14. A body moves along x 2 3 y z - 6 with v x = 2 units/sec. Find
v at (3, 1).
15. A particle moves along y = cosx at 1 unit /sec. Find v x , v v at
x = 7T/6. Ans. 2/V5 units /sec.; 1/V5 units /sec.
164 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
16. Motion along y = 2 sin (2 x 3) has v x = 5 units /sec. Where is the
motion fastest; slowest?
17. A projectile followed the curve whose equation may be taken as
y = x (z 2 /l(H)), where x and y are measured in feet. If the horizontal
velocity was 20 ft. /sec., how fast was the projectile rising at x = 0? What
was its velocity when x = 100 ft.? Ans. 20 ft. /sec., 20 V2 ft. /sec.
18. A particle moves along y = sin x. Find a point where v x = 2 v v .
19. The laws of motion of a particle in a plane are x a cos t -\- fo,
y = a sin / -f c. Find the rectangular equation of its path and show that its
velocity is constant. Ans. (x 6) 2 + (y c) 2 = a 2 .
20. A man walks along the 200 ft. diameter of a semicircular courtyard at
5 ft. /sec. How fast is his shadow moving along the wall if the sun's rays are
perpendicular to the diameter?
103. Angular Velocity. If a wheel turns, the angle which is
generated is dependent upon the time and hence is a function of
the time. The rate of change of 9 with respect to the time is
called the angular velocity and is generally denoted by co. That is,
P(x,V)
FIG. 104
Likewise the angular acceleration is
(2)
a dt
terms of x and y as follows :
For a point P(x t y) we can find
the expression for the w of OP in
Differentiating with respect to t, we get
dx
or
(3)
dO ^
dt 1 +
CO =
- yv x
Hence o> depends on the position of P and on its horizontal and
vertical component velocities.
103] TRIGONOMETRIC FUNCTIONS CURVATURE 165
If the point P is at a fixed distance from 0, we have the special
case of the angular velocity of a point on a circle. Then the
distance traversed by P is a function of 6, namely rd, and its veloc-
ity is
ds dd
dt
= rco.
Therefore co = v/r, where v is the linear velocity of the point on
the circle. Also
__ do) 1 dv _ a
a ~ ~dt ~ rdt ~ r'
where a denotes the linear acceleration of the point along the cir-
cumference of the circle.
Angular velocity is measured in radians per unit of time, or in
revolutions per unit of time.
EXAMPLES
1. A point on the rim of a flywheel of
radius 5 ft., which is 4 ft. above the level
of the center of the wheel, has a vertical
velocity of 50 ft. /sec. Find the angular
velocity of the wheel.
SOLUTION. The point P is located by
the equations
x 5 cos 0, y = 5 sin 6.
Since v v is given, we have
* = r )0 FIG. 105
radians/sec.
When the ordinate of P is 4, cos = 3/5 and
d$ = 50
Y dt " 3
-I
2. Find the expression for the angular
velocity of the line joining the origin to a
point on the line 2 x 3 y = 4. Eval-
uate this when the point is at (5, 2)
on the line, and is moving along it at the
rate of 5 ft. /sec.
SOLUTION. 6 = tan" 1 - ,
x
then
FIG. 106
xv v yv x
166 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
and 2 v x 3 v v = 0, or v x = (3/2) v v \ also x = (3 2/ + 4)/2. Hence
Now t) = Vi^ 2 + (;* = V13 ,'/* ( 101), and for v = 5 ft./sec.,
^ = 25, v ,-^=;
then at the point (5, 2),
20
co = ,= = d= 0.1913 radian/sec.
29v / i3
The double sign permits motion up or down the line.
PROBLEMS
1. A wheel 4 feet in diameter revolves on a fixed axis at the rate of 20
r.p.m. Find v x of a point of the rim 1 ft. above the center.
An*. 27T/3 ft./sec.
2. In Problem 1, find v x and v v of a point on the rim 1 ft. below the axis.
3. For the wheel of Problem 1 suppose v x = 2 ft./sec. at a point on the
rim 1 ft. above the axis and to the left of the center. Find the direction the
wheel revolves and its co in r.p.m. Ans. Clockwise, 60/?r r.p.m.
4. A wheel 10 feet in diameter makes 3 r.p.m. on its fixed axis. What are
co, v x , v v for a point on the rim 3 ft. above the axis?
5. A flywheel of fixed center revolves in a counter-clockwise direction
20 r.p.m. If its radius is 5 feet, at what points of the rim is v x equal to
160 TT ft./min.? Am. 3 ft. to left or right of center.
6. A line joins the origin and a point of x -f 3 y = 4. Find the co of this
line in terms of x and v x as the point moves along x -f 3 y = 4. Evaluate
when the point is at (7, 1) and the motion is 3 units /sec.
7. A point moves along the parabola r = 6/(l sin 0) so that the angular
velocity of the radius vector to the point is 1.25 radians per second. If r is
measured in inches, how fast is r changing at = 30? Ans. 15V3 in. /sec.
8. A point moves along the hyperbola r = 4/(l -f 2 cos 6) so that r is
increasing 3 in. /sec. Find co of the radius vector at ir/3.
9. A point P moves along the parabola 4 y 2 = 5 x so that its ordinate
increases 2 units per second. Find co of the line joining P to the origin.
Ans. -8/(4z + 5).
10. A hoop of radius 2 ft. rolls so that co of a radius is 4 radians per second.
Find v x and v v of a point on the rim.
11. A wheel 4 ft. in diameter rolls along a straight horizontal road at 20
r.p.m. Find v and a, and the direction of motion of a point on the rim when it
is 1 ft. below the center of the wheel and rising.
Ans. 4 7T/3 ft./sec.; 8 *- 2 /9 ft./sec, 2 ; *-/3.
104] TRIGONOMETRIC FUNCTIONS CURVATURE 167
12. A 6-foot wheel rolls along a horizontal straight road 30 mi./hr. What
are v x , v v of a point on the rim? Evaluate at high and low and mid-point.
13. A variable point P on a rod OA which rotates with uniform angular
velocity about describes the curve whose equation, with as the pole, is
r = a(l + cos 0). What is the velocity of P at any time?
Ans. kV2 ar.
14. A rod OA revolves in a vertical plane with uniform w about 0.
A movable point P on the rod is constrained by a cam to describe
r a (2 cos 0), where OP = r, and is the angle between OP and the
horizontal. Find the component of v along the rod, and the total v.
104. Simple Harmonic Motion. If a point P oscillates in a
straight line through a fixed point as center, so that its accelera-
tion is always proportional to the directed distance from P to 0,
then P is said to describe simple harmonic motion.
Let be an origin and let the distance of P from be 5. By
the definition, the acceleration of P must satisfy the relation
(1)
(i)
a =
a
R(o,r)
While the general solution of (1) for s is obtained in a later chapter,
it is obvious that the equation is satisfied if s is proportional to
either sin kt or cos kt.
To illustrate this motion,
let a point Q move along
the curve y = rsinx so
that the horizontal com-
ponents of its velocity is a
constant k. Then, if t =
when a: = 0, the coordinates
of Q are:
x = kt, y = r sin kt. t FIG. 107
Let the projection of Q on the y axis be P(0, y). Then, since
d z y/dt 2 = fcV sin kt = - fc 2 ?/, it is evident that P describes
simple harmonic motion as it oscillates between R and S. If the
time is t Q when x = 0, then OM = fc( fo) and
(2) y = r sin &(* / ) = r sin (W P),
where = kt<>.
The period of P is the time of one complete oscillation and is
the difference of the two values of t when the angle kt ft varies
168 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
from to 2 TT. Hence the period of t is 2 v/k. When P is at R or
at 8, OP has a maximum absolute value r, which is called the
amplitude of the motion. The phase angle is ft or kt Q) and fo is the
phase expressed in time.
Another illustration is as follows: Let Q move with constant
speed v in a circular path. Then
its projection P upon any line in the
plane of its path describes simple
harmonic motion. Take a diameter
as the line and call it the x axis.
If the time is / when the point is
at /?, and t when at Q, then
6 = k(t to), where the constant
k is v/r, the angular velocity.
Then for any position of P
FIG. 108
(3) x = r cos k (t - t )
= r cos (kt - p).
The velocity of the point P is
(4) ^ = - kr sin (kt - p),
and its acceleration
(5) ^ = - & 2 r cos (to - p) = - k' 2 x.
d1
When P is at R or at S, the distance x has an extreme value and
dx/dt is zero; that is, when sin (kt ft) = or = (n?r + /3)/fc,
where n is an integer. The velocity of P has an extreme value
when the acceleration is zero, that is, when P is at 0, or when
cos (kt - 0) = 0. This makes t = [(2 n + I)TT + 2 0]/2 fc, where
n is any integer.
Equations (2) and (3) may be written in the form
(6)
= A sin kt + B cos
where s represents the distance in each formula.
The importance of simple harmonic motion is due to its repre-
sentation of the motion of vibrating bodies such as that of a weight
suspended by a spring, or a vibrating string or wire. Since the
104] TRIGONOMETRIC FUNCTIONS CURVATURE 169
acceleration of any particle is proportional to the force acting on it,
we have
d*s
F = pin -Tg = pmrk 2 cos (kt 0) = pmk 2 s.
Hence the force which tends to restore the vibrating particle to its
central position is directly proportional to its distance from that
central position and is always directed toward the center.
EXAMPLE
1. Write the law of motion for a particle which moves with simple harmonic
motion of period 5 seconds and amplitude 4 feet.
SOLUTION. Take the motion along the x axis. Then
x = r cos (kt (3),
where r, the amplitude, is 4, and the period 2 ir/k = 5. Hence k 2 7r/5.
Then
x = 4 cos
PROBLEMS
1. Find the velocity, the acceleration, the period, and the amplitude of the
particle which moves according to the law s 2 c sin (kt -f- 0).
Am. 2 ck cos (kt -f 0); - /c 2 s; 2 TT//C; 2 c.
Discuss the motion defined by each of the following equations. (Nos. 2-8.)
Compare the acceleration and the distance.
2. x = cos 3 t.
3. s = 2 sin (t/2 + Tr/4).
4ns. cos (/2 + 7T/4); - s/4; 4 TT; 2; S. H. M.
4. s = a cos bt + c sin &.
5. x = sin * - (1/2) cos 2 .
Ans. cos + sin 2 ; sin Z + 2 cos 2 ; not S. H. M.
6. s = sin t cos t 2 cos 2 t.
7. s=2-4: sin 2 -f (3/2) sin 2 *.
Ans. 3 cos 2 4 sin 2 ; 4 s; TT; 5/2.
8. s = (1/2) cos 2 - (1/3) cos 3 t.
9. An alternating electric current varies in intensity according to the
law c = a sin kt. What is the maximum current, and what is the frequency?
Ans. o; ir/k is time between c = o, c = a.
10. A particle moves according to the law s = de kt -f c 2 e~ kt . Find its
velocity and its acceleration. Is this motion simple harmonic? Explain.
170 .FFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
11. >\ flywheel 5 ft. in diameter revolves with a uniform velocity of 40
revoluti >ns per min. What law defines the motion of the projection of a point
on the rim on a plane parallel to the shaft of the wheel? Discuss.
Am. s = (5/2) cos (4 ^/3 + k); a = - 167r 2 s/9; p = l l / 2 sec.; amp. = 5/2.
12. If .s = 3 6 sin 2 2 t expresses the motion of a body in a straight line
show that the motion is simple harmonic and discuss it.
13. If a particle moves in simple harmonic motion with v = 3 ft. /sec.
when s = 3 ft. and v = 4 ft. /sec. when s = 2 ft., find the law of motion and
discuss it. Ans. s = $VW/7 sin $V7/5; P = 2 7rV5/7; amp. = 6^3/7.
14. The amplitude of a simple harmonic motion is 6 ft., and when the body
is midway between the mean and extreme positions, the numerical value of
its velocity is 3 ft. /sec. Find the period and discuss the motion.
105. Curvature.
In a given circle of radius R the length of arc
As between two points P and Q is
proportional to the angle A a between
their tangents; that is,
As = R Aa,
where A a is expressed in radians.
Then
\ x ~\~
is the constant rate of change of the
length of arc with respect to its in-
clination. Since s is measured in linear units and a in radians, R
is the rate of change of s per radian. The reciprocal ratio,
FIG. 109
Aa
As
_
R'
which is the rate of change of a per unit length of arc, is called
the curvature of the circle. Hence a given circle of radius R has
a curvature l/R.
If, however, we consider any other curve, the rate of change of
the length of arc with respect to its inclination is in general a
variable quantity. Let P and Q be any two points on a given
curve separated by a length of arc As. Let the inclinations of
the tangents at P and Q be a and a + Aa respectively. The ratio
Aa/As is the average rate of change of a with respect to the arc s.
106] TRIGONOMETRIC FUNCTIONS CURVATURE 171
It is called the average curvature of the arc PQ. Now let Q
approach P; then
(1) Km ^ = g = K>
is defined as the curvature at P.
That is, the curvature of a curve at a
given point is the rate of change of its
inclination with respect to its length of
arc.
106. Formulas for Curvature. Let
y = /(#) be the equation of a given
curve. Since the slope of the curve
at Pis
(1) y' = tan a
the inclination is therefore
(2) a = tan" 1 y f .
Then
ds dx ds 1 + y' 2 ds
But, from the differential of arc ds in (4) 101, we have
d s
Substituting this in (3), we have
/ 4 \ K _. rf = y"
w ds (1+y' 2 ) 3 / 2 '
If y is considered the independent variable in the
may write
equation, we
tan a. = -7- . a. = tan" 1
dx'
dy
dx
dy
Differentiating with respect to y and using the relation
172 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
we obtain
x"
(5) K= "(I +%")""
where the primes indicate differentiation with respect to y.
If the curve is defined by the parametric equations
* = 0(0, y = m,
we have
dy dt d 2 y \dxj di dt d& dt dt 2
dx 2 ~~ dt
AteV
(dt)
Substituting these in (4), we obtain
*V z/
\ rr _ A y _ y
where the sign is that of #', and where the primes indicate differen-
tiation with respect to t. If we take the sign of the radical in (4)
as positive, then the sign of K will depend on that of y" y that is,
it will be positive or negative according as the curve is concave upward
or concave downward.
EXAMPLE
Find the curvature of the parabola x 1/2 -f y 1/2 = a 1/2 , at any point (x, y)
of the curve.
SOLUTION. Differentiating implicitly, we obtain
-y'V. Vy l y
- Vy V^ _ + >x
y " 2x 2x
then
_ 2x 3 / 2 a 1 / 2
J\.
2(x -f
107. Curvature in Polar Coordinates. If the equation of the
curve given in polar coordinates is
(1) r -/().
107] TRIGONOMETRIC FUNCTIONS CURVATURE 173
then
(2) J
Now a = + \f/ and, by 100,
da _ da dB ^
ds ~~ d6 ds
tan \f/ = = ,
50
= tan" 1
/
Then
(3)
da
2 r /2 -
To express ds in polar coordi-
nates, use the transformation
FIG. Ill
x = r cos 0, t/ = r sin ( 18, VIII),
dx = cos 6 dr r sin rf0, d?/ = sin dr + r cos
then
or
(4)
ds =
Substituting (3) and (4) in (2), we get
(5)
2 r' 2 - rr
(r 2 + r' 2 ) 3 / 2 '
the independent variable being 0.
EXAMPLE
Find the curvature of the cardioid r a(l cos 0) at any point.
SOLUTION. Here r' = a sin 0, r" - a cos 0.
Substituting in the formula, we obtain
K
a 2 [(l - cos 0) 2 + 2 sin 2 (9 - cos 0(1 - cos 0)]
a 3 [(l - cos0) 2 +sin 2 0] 3 / 2
3(1 cos0) 3 3
a[2(l - cos 0)] 3 / 2 2 aV2(l - cos 0) 2V2ar
174 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
PROBLEMS
Find the curvature of each of the following curves at any point and cal-
culate the numerical value when definite points are indicated.
1. y = xe~* at x 1. Ans. 1/e.
2. 9 x 2 + 4 y* = 36 at (2, 0).
3. z 2 /9 - 2/ 2 /4 = 1 at (3, 0). Ans. 3/4.
4. y = log sec x.
5. xy 2 = a 2 (a x) at x = a. Ans. 2/a.
6. y = x 3 (x - 2) 2 at (2, 0).
7. y - e x at the point where the curvature is a maximum.
Ans. (- log V2, VI72), # = 2/(3V3).
8. Vx - Vy = V^".
9. y = e~ 2x sin 3 x at the origin. ylns. G/CSv/lO).
10. ?/ = (x z log z)/e at a; = e.
11. x = t + * 2 , y = - * 2 at ^ = 1. Ans. - 2/(5v / H)).
12. # = a cos 0, y = 6 sin ^ at ^ = 7r/2.
13. a: = 2 cos <, ?/ = sin 2 <t> at = TT. ^Iws. - 1/(4V^),
14. x = 2 , y = 2 *V at < = 1.
15. x cos t, y t sin t at = 7r/3 .
^ns. - 18(4 TT - 3V3)/(ir 2 -f 6 7iV3 -f 54) 3 / 2 .
16. x = log sin 2 0, y = lg cos 2 at = ir/8.
17. x = e* cos f, ?/ = e' sin < at t = ir/2. Ans. - l/(e*'*V2).
18. x = e-** cos 2t, y = e~ 2 sin 2 /.
19. r = eO. Ans. (0 2 + 2)/e(0 2 + I) 3 / 2 .
20. r 2 = o 2 sin 2 0.
21. r = o(l + cos 0) at = 7r/4 . Ans. 3/(2 aV2 + \/2).
22. r = a sec 2 (0/2).
108. Circle of Curvature. Center of Curvature. Let P be
any point on the curve y = /(#). TVie ctrc/e i^/i^cA ^'s tangent to
the curve at P on the concave side, and which has the same curvature
as the curve at P, is called the circle of curvature at P.
Since the circle is tangent to the curve at P, its center must lie
on the normal at P. The curvature of the circle is K = 1/7?.
Hence to construct the circle draw the normal at P, and on the
concave side of the curve mark a point C such that PC = R = l/K.
With this point as center, describe a circle with radius R.
109] TRIGONOMETRIC , UNCTIONS CURVATURE 175
The center C of this circle is called the center of curvature at P,
and the radius R is called the radius of curvature at P. Hence, at
any point of the curve, the radius of curvature is the reciprocal of
the curvature, that is,
1 (\ _L ./'2N3/2
(1} /?= = ' y '
To find the coordinates (h, K)
of C we use the relations, distance
PC = R and slope PC = - l/y 1 .
These give
(2) jR 2 = (h x) 2 + (k i/) 2
and
(3) (h-x)+y'(k-y) = 0.
Eliminating (h x) between these, we have
FIG. 112
(4)
fc 2/ =
y r
where the positive sign in (4) must be used, since (k y) will
have the same sign as y" ', as C is on the concave side of the curve.
Then
(5)
/z = x -
109. Osculating Circle. TAe
osculating circle of a given curve
at a point P is the limiting position
of the circle through P and two other
points P 2 and P 4 of the curve as P 2
and P 4 approach P.
Let y = f(x) be the equation of
a given curve and let the circle
through the points P, P 2 , P 4 on the
curve have center at C'(A', k')
and radius R f . Its equation is
(1) (x - h'Y + (y - fc') 2 - R' 2 = 0.
Since P, P 2 , and P 4 are on both the curve and the circle, if we
176 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
denote by $(x) the left-hand member of (1) when y is replaced by
/(a;), then <j>(x) will vanish for the abscissas of P, P*, and P 4 , that
is,
(*) = 0,
(2)
= 0,
= 0.
Hence, by Rollers Theorem, <j> r (x) must vanish for at least two values
of Xj one at PI for x\ between x and 2 , the other at Pa for # 3 between
x z and #4, that is,
(3) 4/fe) = 0, 0'(s,) = 0.
Applying the same theorem again, <t>"(x) must vanish for some
value of the variable between x\ and Xa, say #o, then
(4)
= 0.
Now let P 2 and P 4 approach P; then P ,Pi, Pa will each approach
P as a limit, and h 1 ', ft', and 72' will approach A, fc, and 72 of the
osculating circle at P. Then, corresponding to (2), (3), and (4),
we have the limiting values of 0, 0', and </>", namely,
+ (0 - fc) 2 - fl 2 = 0,
= 0,
= 0,
These relations enable us to find
A, ft, and R in terms of x, y, y'
and y". Solving these, we have
(8)
h = x-
FIG. 114
But this is just the circle of curva-
ture at P.
Hence, the circle of curvature at any point of a curve is the osculating
circle at that point.
110] TRIGONOMETRIC FUNCTIONS CURVATURE 177
This means that the circle of curvature has three coincident
intersections with the curve; hence it will in general cross the
curve at P. This circle " fits " the curve more closely at P than
any other circle that can be drawn. To show the direction of the
curve at P, we draw the tangent to the circle, since both have the
same slope y'. In exactly the same way, the circle of curvature
shows the curvature at P, since both circle and curve are readily
shown to have the same value of y".
110. Limiting Position of Intersection of Adjacent Normals.
Let P and PI be two points on the curve y = f(x). At each point
draw the normal to the curve; the slopes will be \/y r and
l/i/i 7 , respectively.
Let (h 1 , k 1 ) be the intersection of these normals; then
(x - 7i 7 ) + y'(y - k') = 0,
(xi-V) + lfc'(yi- fc') = 0.
Let
(x - h 1 ) + y'(y - fc') = *(x),
where y = f(x). Then,' by (1),
(2) iKs) = 0, $(xi) = 0.
Hence, by Rollers Theorem, we have
(3) f&Q) 0,
where XQ is some value of the vari-
able between x and x\. Now let
PI approach P along the curve;
then PO will approach P as a limit
and the point (ft 7 , 7c 7 ) will approach some limiting inter-
section, say (h, k). The limiting values of \l/(x) and $'(x) will be
one-half of <t>'(x) and <t>"(x], respectively, in (6) and (7) of the pre-
ceding article. That is, h and k must satisfy the relations
\
FIG. 115
(4)
(5)
= (x-h) 4-y'fo-fc') =0,
= l + t/" + (y-k}y" = 0.
Solving these for h and fc, we get the center of curvature at P.
Therefore, the center of curvature at P is the limiting position of the
intersection of the normals at P and PI on the curve as PI approaches P.
178 DIFFERENTIAL AND INTEGRAL CALCULUS [CH. VII
EXAMPLES
1. Find the radius of curvature of the cycloid
x = a(0 sin 0), y = a(l cos 0).
SOLUTION. Since the equation is given parametrically use (6), 106.
Differentiating with respect to 0, we have
x' a(\ cos 0), y' a sin 0,
x" a sin 0, y" a cos 6.
Then
, a 2 (cos e cos 2 9 sin 2 0} __ 1 _ _ __ 1
a 3 (2 - 2 cos 0) 3 / 2
2 aV2(l - cos 0) 4 a sin 0V 2 )
since 1 cos = 2 sin 2 (0/2). Hence
R 4 a sin -
2. Find the radius of curvature
and center of curvature of the curve
y xe~* at its maximum point.
SOLUTION. Differentiating, we have
y
x), y" = e-*(* - 2);
FIG. 116
R =
so that ?/' = and y" = - l/e <
when x = 1. Hence (1, l/e) is the
maximum point.
[1 -f e-**(\ - rr) 2 ] 3 '
QX( X 2)
T^he sign of R shows that the curve is concave downward at the point. We
have also
As a check, the distance P(l, l/e) to C(l, l/e e) must equal 7?.
PROBLEMS
Find R for each of the following curves. (Nos. 1-21.)
1. 2xy = a 2 . 4ns. (l/a 2 )Cc 2 -f 2/ 2 ) 8 / 2 .
2. 2/ = tan z at x = ir/4.
3. 2xy + x + y = a?.
Ans. [(2 x + I) 2 + (1+ 2 2/) 2 ] 3 /Y[4(2 x + !)(!+ 2 T/)].
4. Ellipse, s 2 /a 2 + y*/b* = 1 at an extremity of the minor axis; at a
vertex.
5. Hyperbola, z 2 /a 2 j/ 2 /6 2 = 1 at a vertex; at an extremity of the latus
rectum. Ans. b*/a; (6 2 /o 4 )(a 2 + c 2 ) 8 / 2 .
6. y = log a; at (1, 0).
110] TRIGONOMETRIC FUNCTIONS CURVATURE
7. y = x sin (I/a;) at x = 2/T. Ans. - 16-S/2A 3 .
8. Catenary, y = (a/2)(e x ' a -f e~ xla ).
9. Hypocycloid, z 2/3 -f 2/ 2/3 = a 2 / 3 . Ans. 3*^1^.
10. Cycloid, x a cos" 1 [(a y)/a} V2 ay ?/ 2 . Ans. 2V/2 ay.
11. x = 2 cos , y = 3 sin t. Ans. - (4 + 9 ctn 2 3/ V(6 esc 3 0-
12. x = e~~* cos t, y e~* sin at t = 0.
13. x = 4 cos 3 0, y = 4 sin 3 where 7? is a maximum. Ans. 6.
14. x = t -{- cos , y = 4 sin t at = 0.
15. Involute, x = a (cos -\~ 6 sin 6), y a (sin 6 cos 0). Ans. a0.
16. r = a sin 2 (9.
17. r = a cos 3 (0/3). Ans. (3 a/4) cos 2 (0/3).
18. r = a(l 2 cos0).
19. r = 4(1 - sin 0) at = ir/6. Ans. 8/3.
20. r 2 = a 2 cos 2 0.
21. r 2 = a 2 sec 2 0. Ans. - r 3 /a 2 .
Evaluate the radius of curvature at the point indicated for each curve;
draw the curve and the corresponding circle of curvature. (Nos. 2229.)
22. z 2 = 4 y at (2, 1).
23. x = 4 + 2 y - y 2 at (1, - 1). Ans. 17VJ7/2.
24. x 2 +47/ 2 = 12 at (2, V2).
25. z = cos 2 f, i/ = cos 4 / at * - TT. Ans. 17 V 17/4.
26. x = t* + 3 t, y = 3 t 2 at t = 1.
27. z = a sin 3 0, 2/ = a cos 3 at = 7r/4. Ans. 3 a/2.
28. At the point (10, 5) on the parabola whose vertex is at the origin and
whose axis lies along the x axis.
29. At the point where y x cuts the parabola Vx -f- Vy Va. Is
there an extreme value for R, at this point? Ans. Yes, a/V2.
30. Find the point of the cycloid which has the largest radius of curvature.
Is there a point of least R? Explain the test for extreme value.
31. Prove that the normal acceleration of a point moving along the curve
% 0(0> y = /(O is a N = v*/R, where v is the velocity of the moving point, and
R is the radius of curvature of the curve.
32. A point is moving along the arc of the curve 2 y - x 3 at the rate of
8 ft. /sec. Approximate its change in direction during the half second after it
passed the point (2, 4).
.80 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
33. Approximate by differentials the change in the inclination of the curve
4 y - x 2 by moving along the arc of the curve from the point (2, 1) a distance
As =0.1 unit. ** 10.9'.
111. Evolutes. The locus of the centers of curvature of a given
curve is called the evolute of that curve.
Let the equation of the given curve be
(1) =/<*).
We have found ( 108) the coordinates of the center of curvature
C corresponding to a point P(x, y) on the curve. As P moves
along the curve, the point C varies and traces some locus. Call
its coordinates (x,y); then
y'(l+y /2 ) - l+l/' 2
(2) *=* -77 , y = JM -77 ^
may be considered as parametric equations of the evolute, each being
expressed in terms of the parameter x. For, since the coordinates
of P must satisfy (1), y and its
derivatives y 1 and y" are all func-
tions of x, and each value assigned
to x determines one or more points
P by (1), and corresponding points
Cby(2).
The rectangular equation of the
evolute is found by eliminating^
and y between equations (2) and
(1). In some problems this is readi-
ly done by solving equations (2)
for x and y, respectively, in terms of
randy, and then substituting the resulting expressions in (1).
The slope of the evolute at any point is
dy
dy ___ dx
fa
Differentiating equations (2) with respect to x, we find
FIG. 117
dx
dx
i
y" z + 3
/2 -
V" -
111] TRIGONOMETRIC FUNCTIONS CURVATURE 181
and
dy _ . , 2y'y"*-y'"-y'*y'"
dx ~ y "*" y" 2
3 y'y"* - y'" - y'*y'"
hence
y" 2
But y' is the slope of the tangent to the given curve (1) at the
point P. Therefore we have the following theorem.
The tangent to the evolute at any point C is the normal to the given
curve at the corresponding point P. That is, the radius of curva-
ture PC at any point is tangent to the evolute at C.
When referred to the evolute, the given curve y = f(x) is called
an involute.
EXAMPLE
Find the equation of the evolute of the parabola y 2 2 px.
SOLUTION, y' = ,
Then
Substituting these values in y z = 2 px, we obtain
27 pj/ 2 = 8(5 - p) 3 .
PROBLEMS
Find the center of curvature for any point of the following curves. (Nos.
1-12.)
1. 1/2 = 4 Xf Ans. (3x4-2, - i/ 3 /4). %
2. y xe~* at x = 1.
3. 2 zy = a*. Ans. [(3 z 2 + y*)/2 x, (x 2 + 3 2/ 2 )/2 ?/].
4. t/ = x cos x at x = x.
5. 2/ = log x for minimum #. Ans. [2V2, - (1/2) (3 4- log 2)].
182 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VII
6. x = t - t\ y = t + t* at t = - 1.
7. re = a cos 0, y = b sin 3 at = *-/2. Ans. [0, 6 - a 2 /(3 6)].
8. x = sin 2 J, ?/ = cos 2 2.
9. z = e~ 2 < cos 2 J, i/ = e~ 2 < sin 2 J aU = 0. Ans. (0, - 1).
10. r 1 - 2 sin 2 0.
11. r 2 = a 2 sec 20.
Ans. For coincident reference systems (2 x 3 /a 2 , 2 2/ 3 /a 2 ).
12. r = 1/(1 + cos 0), B = 7T/2.
Find the equation of the evolute of each of the following curves. (Nos.
13-19.)
13. y z =* 8 x. Ans. 3(- 16 ) 2 / 3 = 8(2 - 4).
14. x 2 H- 4 y 2 = 4.
15. 2 zy - a 2 . ^ns. (5 + 2 / 3 - (2 - ) 2 ' 3 = 2 a 2 / 3 .
16. x = 3 , y = * 2 + 1.
17. x = ^ sin t, y = 1 cos .
^ns. x - cos- 1 (1 + y) + V- 2y - y*.
18. x sin , ?/ = cos 2 .
19. z 2 / 3 -|- ^ 2 / 3 = 4. Ans. (x + ^) 2 / 3 -f (5 - i/) 2 ' 3 = 8.
ADDITIONAL PROBLEMS
1. Given the curve y = x/2 -f sin 2 a:.
(a) What are the values of x for maximum and minimum values of i/?
Ans. ( T 7 2 -f n)7r, (H -|- n)7r.
(6) Find /^ and K and evaluate for the first positive values of x giving a
maximum and a minimum y. Ans. 1/V3, V3; 1/V3, V3.
(c) What are the values of x for inflections? Ans. x/4 -|- mr/2.
(d) Approximate the change in the direction of the curve from (TT, w/2)
for a change in x of 0.01. Ans. da = 0.016 radian.
2. Transform ds 2 = dx z -f- rf?/ 2 to polar representation.
3. Find the acute angle for which 3 tan -f- ctn is a minimum.
Ans. tan
4. If d?s/dfi = & 2 s, is s = a sin Atf -h b cos & a possible value for s?
5. Where between B = and = Tr/2 does sin increase half as fast as
the arc in the unit circle? Ans. 7r/3.
6. A barge is being drawn toward a wharf by means of a cable attached to
a ring in the floor of the wharf 10 ft. above the windlass on the deck of the barge.
(a) Express the length of the cable and the distance of the barge from
the wharf as functions of the angle at the windlass between the cable and the
horizontal.
111] TRIGONOMETRIC FUNCTIONS CURVATURE 183
(b) If the cable is being wound in at the rate of 4 ft. /sec., how fast is the
angle of (a) changing?
(c) How fast is the barge approaching the wharf?
(d) Evaluate the results of (b) and (c) when the barge is 24 ft. from the
wharf.
(e) Does the rate of change of the distance from the barge to the floor of
the wharf have a maximum rate of change? If so, where?
(f) Does the velocity of the barge have a maximum value? If so, where?
(g) Approximate the change in the length of the cable and the distance
of the barge from the wharf if the angle changes 0.1 after the distance is 24 ft.
7. Given the curve y = x sin" 1 x.
(a) Find v along the curve at x = 1/2 if v v ~ 2 units/sec.
Ans. 2V(io - 4\/3)/(7 - 4\/3) units/sec.
(6) Approximate the change in x for 0.02 unit change in y at x 1/2.
Ans. dx = (0.02 V3)/ (2 - V3) units.
(c) Approximate the relative and percentage changes in y for a change of x
from 0.5 to 0.52. Ans. (2 - V3)/(1.18V3), (2 - V3)/(0.0118V3).
8. A hoop of radius 2 ft. rolls so that co = 4 radians per second. An insect
at t = is at the bottom of the hoop and is crawling along the hoop at the rate of
8 in. /sec. Find the path followed by the insect and its velocity along the path.
9. A wheel of radius a rolls along a straight track with constant angular
velocity. Find the maximum and minimum velocity of a point on the rim
and describe its position for each of these extremes.
Ans. 0, v = 0, point at rest on track; 6 = TT, v 2 ow, point
moving at top of wheel twice as fast as the hub.
10. A flywheel of radius 5 ft. and fixed center has an angular velocity of
16 radians per second. Find:
(a) The parametric equations of a point on the rim.
(b) The values of v x , v v , and v of a point on the rim.
(c) The position of a point if its v x is a maximum. Find also the value
of that maximum.
(d) The same as part (c) for v v ,
11. A wheel rolls along a straight horizontal road with constant velocity.
Show that the highest point of the wheel moves twice as fast as the two points
on the rim which are half the radius distant from the road.
12. The equations of motion of a point are
x = log sin t, y = 2Vt.
(a) Find v along the path.
(6) For what values of t, v, and a does the point cross the y axis?
13. At what point on 16 x 2 H- 9 y* 400 does y decrease at the same rate
as x increases as a particle moves around the curve clockwise? What is the
velocity of the particle at that point? Ans. (3, 16/3); v x V2 units/sec.
184 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VII
14. Find the velocity of a point moving along the curve:
(a) x 2 + 2 y* = 12 with v v = 2 units/sec, at (2, - 2).
(6) y = x*(x - 2) 2 with v x * 2 units/sec, at (1, 1).
(c) ?/ = sin x, with v a = 4/ V5 units/sec, at a: = 7r/3.
(d) ?/ = 4 x 2 , with t> x = 5 units /sec. at (1, 3). Also find .
15. A particle moves around a circle of radius a units at b revolutions per
second. Find how fast it is moving away from a point c units from the circle.
What is the speed when the motion is parallel to the line from the center of the
circle to the fixed point?
Ans. [2 -n-ab(a -f c) sin 0]/V2 a (a + c)(l cos 0) + c 2 units /sec.;
[2 wab(a + c)]/ V2 a(a + c) + c 2 units/sec.
16. A statue 20 ft. high stands on a pedestal 20 ft. high. If the statue
subtends an angle at some point in the plane of the base of the pedestal:
(a) At what distance from the pedestal is a maximum?
(6) Approximate the change in if the distance changes from 10 ft. to 11 ft.
Find K and R for each of the following cases. (Nos. 17-28.)
17. y 2 = 8 x at y = 3. Ans. R = - 7}|.
18. y = e~ 3x cos 3 x at the point (0, 1).
19. y = x/2 sin 2 x.
(5 -4 sin 2 3 + 4 sin 2 2 x) 3 / 2
t =
16 cos 2 a:
20. x = a cos , ?/ = b sin at t = (l/2)7r.
21. x = sin /, y = cos 2 * at J = (1/2)*-. Ans. 72 = - 17VT7/4.
22. Z = te*', y = t*e*< at t = 0.
23. The cycloid at any point. Ans. R = 4 a sin (0/2).
24. x a0 b sin 0, y = a b cos 0. Find the center of curvature at
any point.
25. x = sin 2 0, y = log cos at = (1/4)*-. Ans. R = V2.
26. a: = 2(cos -f sin 0), 2/ 2(sin cos 0).
27. r = e*. Ans. R = rVl + a 2 .
28. r 2 = a 2 esc 2 0.
29. Find the second and third derivatives of y with respect to x in each
of the following cases.
(a) x = P, y = 1 - 2 < 3 . Ans. - 3/(2 0, 3/(4 * 3 ).
(6) x = sin , i/ = cos t.
(c) x = t\ y = e 3 '. Ans. e 3 '(3 * - 2)/J 5 , e 3 <(9 J 2 - 18 * + 10) /3 t 9 .
(d) x = /(*), y = flr(0-
30. Derive formula (5) of 106 directly from formula (4).
111] TRIGONOMETRIC FUNCTIONS CURVATURE 185
31. Express d*x/dy* in terms of derivatives of y.
Ans.
Vy\ __(dy\(d*y
dx*f \dx) \dx*
(dy\
\dx)
32. Light travels from a point A in one medium to a point B in another.
A plane surface separates the media. If light has the velocities Vi and t> 2 in the
respective media, find the path so the time of passage is a minimum.
FIG. 118
33. A section of the earth through the poles is an ellipse, the polar radius
being 3949.9 mi. and the equatorial radius 3963.3 mi. Using the results for
the radius of curvature of an ellipse in Problem 4, 110, calculate the length
of a meridian arc of 1 at the equator, and at the pole.
A ns. 68.708 mi.; 69.405 mi.
CHAPTER VIII
SOLID ANALYTIC GEOMETRY
112. Introduction. The ability to write simple equations for
certain surfaces is very important; therefore it seems advisable
to insert at this point a short chapter on solid analytic geom-
etry. Its contents may be used as a review of the material which
will be used in the following chapters, as it contains sufficient
information to permit students to understand the remaining chap-
ters on the calculus.
113. Rectangular Coordinates in Space. A point in space may
be located by a set of three real numbers. For rectangular
coordinates these numbers repre-
sent the distances of the point
from three mutually perpendicular
planes which meet in three mutu-
ally perpendicular lines called the
x axis, y axis, and z axis.
If x, y, and z represent these dis-
tances for the point P of Fig. 119,
then 0,1 = DP = x, AB = FP = y,
and OE = BP = z. The signs of
these coordinates are chosen so as to
represent points in all eight octants.
Thus, if we consider any point B of the xy plane, it has positive
and negative coordinates, as explained for plane cartesian co-
ordinates. Then P has the same x and y coordinates as B', its z
coordinate is positive if the point is on one side of the xy plane
and negative if on the other. Obviously, this rectangular system
of coordinates is a very simple extension of the cartesian coordi-
nates in the plane.
114. Cylindrical Coordinates, Three real numbers locate a
point in space just as readily if the points of the plane are located
by means of polar coordinates. Thus, we locate a point B(r, 6)
in the horizontal plane and then measure a distance z perpendicu-
lar to that plane. This fixes a point P(r 9 6, 2), and the system of
186
FIG. 119
(116)
SOLID ANALYTIC GEOMETRY
187
coordinates is called cylindrical. This system is very useful for
expressing the locus of points which lie on surfaces of revolution
about one of the coordinate axes.
If we take the two sets of axes coin-
cident, so that the positive x axis falls
along the polar axis OA and the positive
z axis remains the same, we have
(1) x = r cos 6, y = r sin 6, z = z,
and
P(r,e,z)
(2)
r =
9 = tan~
B(r,e)
and z = z as the relations between the
two systems of coordinates. They are
used to transform equations from one FIG. 120
system of coordinates to the other.
115. Spherical Coordinates. Let P be any point in space.
Then its spherical coordinates are p, Q, and <, where these quanti-
ties are defined as shown in the figure.
The relations giving the rectangular co-
ordinates in terms of the spherical coor-
dinates are
Ix = p sin (j> cos 8,
y = p sin <j> sin 6,
z = p cos 4>.
We observe that the p of spherical coor-
dinates is not the r of cylindrical, but
that p = r esc </>.
116. Direction Cosines. If a direct-
ed line through the origin makes the
angles a, 0, and y with the coordinate axes, these angles are called
the direction angles of the line, and their cosines are called the
direction cosines of the line. Any line parallel to such a line and
having the same direction has the same direction angles and the
samp/airection cosines.
THEOREM. // a, P, and y are the direction angles of a line, then
FIG. 121
/
'(I)
cos 2 a + cos 2 p + cos 2 Y = 1.
PROOF. Let MN represent a line with direction angles a, 0, 7.
188 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VIII
Draw OP through the origin parallel to MN. Let r represent OP.
Evidently
(1)
FIG. 122
x = r cos a, y r cos 0, z = r cos 7.
N But
(2) x 2 + y 2 + z 2 = r 2 .
Substituting for x, y, z, we have
r 2 cos 2 a + r 2 cos 2 /3 + r 2 cos 2 7 = r 2 ,
or, dividing by r 2 , we obtain (I).
Y Any three numbers proportional to
the direction cosines of a line are
called direction numbers of the line.
If a, b, c are three such numbers,
(3)
cos a cos f3 cos 7
= *,
and substitution of the values of the direction cosines from relations
(3) in (I) gives
Therefore the proportionality factor is
(4)
Jfe - X
dz Va 2 + 6 2 + c 2
Hence
a
-k- ^2 _|_ 2 _|_ C 2
/TT\ A
rn*R 6
(II) <
Va 2 + b 2 + c 2
Va 2 + b 2 + c 2
FIG. 123
When a line is directed, the proper choice of sign for k can be made,
since its direction angles are fixed.
The numerical values of the direction cosines of the line which
passes through the two points P\(XI, yi, Zi) and P<z(x%, y^ z 2 ) are
proportional to the projections #2 x\ 9 y% y\, z 2 Zi of the
117]
SOLID ANALYTIC GEOMETRY
189
line-segment P\Pi upon the coordinate axes. This is evident
from Fig. 123, as
cos a =
PiQ
cos ft =
PiR
cos 7 =
Pig
PiP.'
PiP,'
where
PiQ = x z Xi, PiR = 2/2 - 2/i, PiS = 2 2
arc the projections of PiPg on the coordinate axes.
THEOREM. The length of PiPz is given by the formula
(III) L = V(x 2 - Xi) 2 + (y, - y,) + (z 2 - Zl ) 2 .
L = V(x 2 - Xi) 2 + (y,
PROOF. From Fig. 123, we have
L 2 = P&* + P
Pis',
and substitution of the values of the line-segments gives the
theorem.
117. The Angle between Two Lines. Suppose two lines
through the origin are li and Z 2 with direction angles ai, ft, 71,
and a 2 , 02, 72, respectively. Take P
and Q as any pair of points, one on
each line. We define the angle be-
tween two lines as that angle between
the positive directions of the lines.
Then
(1) cos e =
Since
op 2
2 OP OQ
OP* = x 2 + 7/ 2 + z\
/2
and
FIG. 124
+ (t/ -
+ (z -
cos v =
xx
yy
zz
+ 2/ 2 + z 2 ' \/a: /2 + i/' 2 + z' 2
or
(IV) cos 8 = cos ai cos a 2 + cos pi cos p 2 + cos Vi cos \ 2 .
190 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VIII
If li and 1 2 are perpendicular, cos 6 = cos 90 = 0, whence, for
two perpendicular lines,
(2) cos ai cos 2 + cos & cos /3 2 + cos 71 cos 72 = 0.
However, x, y, z and x', y' y z' are direction numbers of the two
lines since they are drawn through the origin. Hence
(3) xx' + yy' + zz f = 0.
Therefore the sum of the products of corresponding direction numbers
of two perpendicular lines is zero.
Evidently the ratios of correspond-
ing direction numbers of two paral-
lel lines are equal.
EXAMPLE
What kind of quadrilateral has the
vertices (2, 4, - 1), (4, 4, 0), (3, 2, 2),
(1, 2, 1) in the order given?
SOLUTION. The direction numbers of
the sides of the quadrilateral are: of AB,
2, 0, 1; of BC, 1, 2, - 2; of CD, 2, 0, 1;
of DA, 1, 2, - 2.
FIG. 125 Since AB and CD have direction
numbers which are proportional, these
lines are parallel. The same is true about BC and DA. Since the sum of the
products of corresponding direction numbers of A B and BC is zero, these lines
are perpendicular. Hence A BCD is a rectangle.
PROBLEMS
1. Find the direction angles of the line which makes equal angles with the
coordinate axes. Ans. 54 44. 1 7 .
2. Find the direction cosines of a line with 4, 3, 1 as direction numbers.
3. If a = 45, = 75, what is the value of 7? Ans. 48 51', 131 9'.
4. What kind of a triangle has the vertices (7, 3, 4), (1, 0, 6), (4, 5, - 2)?
5. Find the angle between the lines with direction cosines l/Vl4, 2/Vl4,
3/VIi and 2/V30, - 1/V30, 5/V30. Ans. 42 57'.
6. Find the angle between the lines with 2, 1, 2, and 1, 1, 1 as direc-
tion numbers.
7. Find the length of the sides of the triangle with vertices (3, 1, 3),
(1, - 2, 1), and (1, 0, - 3). Ans. \/29, 2\/5, \/5 units.
8. Are the points (2, - 4, 3), (- 6, 12, - 9), (8, - 16, 12) on a straight
line?
119] SOLID ANALYTIC GEOMETRY 191
9. Find the coordinates of a point 8 units from the origin if a. = 60
and 7 = 45. Ans. (4, 4, 4\/2).
10. Find the coordinates of a point 4 units from the origin if a = 120,
/3 == 45.
11. Find the angles of the triangle with vertices ( - 1, 3, 5), ( - 2, - 1, - 1),
(2, 3, - 4). Ans. 42 24', 87 32.5', 50 3.5'.
12. Find the angles of the triangle with vertices at (1, - 3, 2), (0, 2, -1),
and (- 1, 4, 1).
118. Surfaces. Space Curves. If we write an equation con-
necting the three variables of either system of coordinates defined
previously, points whose coordinates satisfy the equation lie on a
surface. For, let an equation, in rectangular coordinates for
instance, be solved for 2, so that it takes the form
(1) z = /(*, y).
Points of the locus defined by (1) may be located by giving
x and y special values and then computing the resulting values
for z. If f(x, y) is an algebraic function, the values of z are finite
in number and in general are distinct. The fact that they are
either distinct or equal shows that the lines through the points on
the xy plane perpendicular to that plane meet the locus in points
and hence the locus has no thickness. This type of locus is called
a surface.
Of course equations may be formed which are not satisfied by
the coordinates of any point in space, as was the case in the plane.
We are not interested in any such at present.
A space curve may be regarded as the intersection of any two
surfaces which pass through it. The equations z = fi(x, y) and
z = fi(x, y) of the surfaces considered as simultaneous are equa-
tions of the curve of intersection.
119. Cylinders. A surface generated by a straight line which
moves parallel to a fixed line and which intersects a given curve is
called a cylinder. The straight line is called the generatrix and
the curve is the directrix. If the directrix is taken as f(x, y) = 0,
say, and if the generatrix is perpendicular to the xy plane, the
cylinder is represented merely by the equation
(1) /(*, y) =
When an equation contains two variables only, we plot the locus
192 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VIII
as though it represented points in the given plane. Then the locus
is formed of all the straight lines through the points of the plane
locus parallel to the axis of the missing variable. Such special
equations represent cylinders. If two variables are missing, the
locus represents planes parallel to the plane of the missing variables
and through the points of the axis corresponding to the roots of
the given equation.
120. Surfaces of Revolution. If a plane curve is revolved about
a faxed line in its plane, the curve generates a surface of revolution
whose axis is the fixed line. The following examples illustrate a
very easy method of deriving the equation of such a surface.
EXAMPLES
1. Find the equation of the surface generated by revolving the ellipse
z 2 /a 2 -f z 2 /6 2 = 1 about its minor axis.
SOLUTION. The equation of the ellipse means that for any point P(z, z) on
the curve we have
RP MP
1. - S + g-1.
Now suppose the first quadrant of the el-
lipse revolved about the z axis to some posi-
tion ST, so that P takes the position Q(x, y, z).
Then RP moves to RQ and MP to NQ. Hence
equation (1) becomes
FIG. 126
(2)
RQ
o 2
But RQ = $ 2 = x 2 -f y 2 and NQ = z 2 = z 2 , where x, y, z are the coordinates
of Q, any point of the surface generated by the ellipse.
Substituting in (2), we find the desired equation to be
x* + y* , z 2 ,
~~
2. What is the equation of the surface generated by revolving the parabola
z 2 - 4 x about the line x = 2 in the xz plane?
SOLUTION. From the equation of the parabola and Fig. 127, we have, for
any point P(x, 5),
(3)
MP =4SP = 4(2 -PR).
121]
SOLID ANALYTIC GEOMETRY
193
Suppose the parabola revolved about the line x = 2 until P moves to the
position Q(x, y, z). Then MP = 2 becomes NQ, and PR becomes QR. Draw
QT perpendicular to PR. Now NQ = z, and, from the right triangle QRT, we
have
~QR* = TR Z + TO 2 = (2 - x) 2 + y\
where x, y, z are the coordinates of Q, any
point of the surface generated by the
parabola.
Substituting these values in (3), we find
Z 2 = 4[2 _ V(2 - x) 2 + 2/ 2 ],
which, when rationalized, becomes
(z 2 - 8) 2 = 16[(2 - x) 2 + 2/ 2
FIG. 127
PROBLEMS
Find the equation of the surface generated by revolving
1. y 2 - 2 px about its axis. Ans. y 2 + z 2 = 2 px.
2. x 2 /a 2 y z /b 2 = 1 about its transverse axis.
3. x 2 /4 -f- 2/ 2 /9 = 1 about its major axis.
Ans.
z 2 )/4
4. x 2 = Sy about y 2 in the xy plane.
5. 2 x 3 ?/ = 6 about x = 5 in the xy plane.
Ans. (3 y - 4) 2 = 4(x 2 + 2 2 - 10 x + 25).
6. 3z = 2?/ + l about x = 1 in the xy plane.
7. 3 y = 2 x 2 + 6 about x = 3 in the ajy plane.
Ans. 216(y - 2) = (2 z 2 + 2 2 2 - 12 x - 3 y -f 6) 2 .
8. x 2 + 2/ 2 = 4 about x = 4 in the x?/ plane.
9. x = 2/ 2 2t/-f3 about x = 2 in the x?/ plane.
Ans. (y 2 - 2 y + 5) 2 = x 2 + 2 2 + 4 x + 4.
10. x 2 = 4 z about z = 2 in the X2 plane.
11. x -f- z = 5 about x = 6 in the xz plane.
Ans. (1 + z) 2 = (6 - x) 2 + 2/ 2 .
12. 3 z 2 = 2 y + 1 about z = 1 in the 2/2 plane.
121. The Plane. The Line. In the plane ABC, let P(x, y, z)
be any point, and let ON be the perpendicular upon the plane
from the origin.
Suppose ON is of length p and has the direction angles a, , y.
194 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VIII
Then the coordinates of N are p cos a, p cos /3, p cos 7; hence the
direction numbers of NP are
x p cos a,
But the lines ON and
hence by (3), 117,
cosa(x pcosa) + cos/3(?/ pcos/3)
This, when simplified, becomes
i/ ~~ P cos /3, z p cos 7.
are perpendicular for all positions of P;
cos 7(2 pcosy) = 0.
(V)
x cos a + y cos p + 2 cos "Y =
Since this equation is satisfied by
the coordinates of all points of the
plane and those points only, we have
the following theorem.
THEOREM. The equation of a plane
is always of the first degree in x> y,
and z.
Any equation of the first degree in
x, y, and z may be changed into the
form of (V) by dividing by the square
root of the sum of the squares of the
coefficients of the variables. Since p,
in (V), is positive, choose the sign for the radical which will make
the constant term positive when it is transposed to the right-hand
member of the equation.
The equations of any line in space are those of any pair of planes
through it. If the line passes through PI with direction numbers
a, 6, and c, its equations may be written in the symmetric form
FIG. 128
(VI)
x -
_ y yi _ z -
b c
122. The Sphere. A sphere is a surface each point of which is
at a fixed distance Rfrom its center (h } k y I). Its equation is readily
seen to be
(VII) (x - hY + (y - *) + (z - 0* = #2.
If the center of the sphere is the origin, its equation reduces to
(VIII) X 2 + y2 +z t^ R 2
123]
SOLID ANALYTIC GEOMETRY
195
If the sphere is tangent to the xy plane at the origin, its center is
on the z axis and has the coordinates (0, 0, R). Whence its
equation is
or
(IX)
x* + tf + (z
+ y 2 + 2 2 db 2 Rz = 0,
where the positive sign makes the sphere lie below, and the negative
sign above, the xy plane. Similar equations may be written for
spheres tangent to other coordinate planes.
Equations (VIII) and (IX) become in cylindrical coordinates
(X)
r 2 + z 2 = R 2 , r 2 +
2 Rz = 0.
These same equations in spherical coordinates are
(XI) p = R, p 2 R cos <|> = 0.
123. The Right Circular Cone. The surface generated by a
straight line turning about a fixed point on itself and passing through
a given curve is called a cone.
Suppose the fixed point of the line
to be on the z axis and the given curve
to be a circle in the rd plane with its
center at the origin. Then for any
point P of the cone, we have, from
Fig. 129,
Q p
tana= VQ>
FIG. 129
where a is the angle any position of the
generating line makes with the z axis,
the axis of the cone. But QP = r
and VQ = z k. Hence the equation of the right circular cone
of vertical angle 2 a. which has its axis along the z axis and its ver-
tex at (0, 0, fc) is
(XII) (z - ft) tan a = r.
When the vertex is at the origin, this equation reduces to
(XIH) r = z ten a.
196 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VIII
These two equations in rectangular coordinates are
(XIV) jc 2 + y 2 = (z- kY tan 2 a, x 2 + y* = z 2 tan 2 ,
respectively.
In spherical coordinates, the same equations are
(XV)
(p cos <(> ft) tan a = p sin <(>, <j> = a.
It is worth while to observe that equations (XIV) are composed
of terms which are squares of linear expressions in x, y, and z, respec-
tively, and one of the terms is of different sign from the other two
when all are written in the same member of the equation. These
two characteristics are sufficient for an equation to represent a
cone with its axis parallel to one of the coordinate axes.
124. The Ellipsoid. The sur-
face given by
(XVI) g + g + 1* i
is called an ellipsoid. It has the
following properties :
(a) The intercepts on the coordi-
nate axes are x = a, y = dz 6,
z = d= c.
(6) The coordinate planes cut
the surface in curves called traces.
FIG. 130
They are the ellipses
a 2 + b 2 *' b 2 + c 2 lj a 2 + c 2 lm
(c) The equation of any section of the surface made by a plane
parallel to one of the coordinate planes is that of an ellipse. Thus,
the section in the plane y = k is
which may be written
(6* -
125]
SOLID ANALYTIC GEOMETRY
197
As fc changes from to b or to 6, the axes of the ellipses cut out
by the plane y = fc decrease and at fc = 6 they become points.
These plane sections of a surface, parallel to one of the coordinate
planes, afford the best method for sketching a surface. Several
such sections are shown in Fig. 130, which represents one-eighth
of the ellipsoid.
If any two of the constants a, b, c are equal, the surface is an
ellipsoid of revolution, or a spheroid.
z
125. The Hyperboloid of One
Sheet. The surface represented by
is a hyperboloid of one sheet and has
the following properties :
(a) The intercepts on the x axis
are x = a, and y = db b on the y
axis; but the surface does not meet
the z axis.
(6) The traces in the coordinate planes are
FIG. 131
+ .*7 -i y ** -j ^ ^
T o ~"~" -* y 70 9 ~~ J- , o ~~ o ~~~ J-
cr o 2 o 2 c 2 a 2 c 2
The first of these is an ellipse and the other two are hyperbolas.
(c) The equation of the section in the plane z k is
-2 (< 2 + fc 2 ) 2 (< 2 + fc 2 )
= 1,
which is an ellipse for all values of fc; the axes of the ellipses cut
out by such planes increase indefinitely as fc <*> .
In sketching these surfaces it is best to draw the traces in the
coordinate planes; then draw plane sections parallel to one of
these traces, choosing sections which are closed curves (ellipses or
circles) if such exist. Figure 131 shows one-eighth of the surface
with several elliptic sections.
If a = 6, the surface is a hyperboloid of revolution of one sheet
and the closed sections are circles.
198 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VIII
This surface has an interesting property which makes it useful
for gears. By writing (XVII) in the form
x 2 z 2 _ y 2
a 2 ~~ c 2 ~ b 2 '
both members can be factored into linear factors. These factors
permit us to consider the planes
/i "\ *^ ^ //i y \ x . z i
a c y bj a c k
If a point PI(XI, 2/1, 21) is on the intersection of these two planes,
its coordinates satisfy their equations; hence
b J a c k\ b
Multiplying the corresponding members of these two relations and
transposing, we have
5i! j. y - ?i! = 1
a 2 ^ b 2 c 2
This means that the point PI lies on the surface represented by
equation (XVII). Therefore, since PI is any point of the line of
intersection of the two planes, their line of intersection lies on the
surface. With k as a parameter, the equations (1) represent a
system of lines on the hyperboloid of one sheet. This same con-
clusion can be reached by means of the system of lines
hence the hyperboloid of one sheet may be regarded as a ruled
surface with two sets of rulings. No two lines of the same system
meet, but each line of either system meets all of the other.
126. The Hyperboloid of Two Sheets. The surface repre-
sented by the equation
(XVIII) - - + j - ^ = 1
is called a hyperboloid of two sheets. Evidently the intercepts
on the y axis are y = b and the surface does not meet the x or
z axes.
127] SOLID ANALYTIC GEOMETRY 199
The traces in the xy and yz planes are hyperbolas with equations
respectively. There is no trace in
the xz plane.
The equation of a section in a
plane parallel to the xz plane, say
y = k, is
+ ^
b 2
which is an ellipse if k > b or
if k < b. The axes of these
sections increase indefinitely as
fc oo . For fc = 6, the section has the equation
FIG. 132
which is a />om. There is no section corresponding to y = k
for 6 < k < b.
One-eighth of the surface is shown in Fig. 132, drawn by means
of sections made by planes y & t .
If a = c, the surface is a hyperboloid
of revolution of two sheets.
127. The Elliptic Paraboloid. If
taken as shown in Fig. 133, this sur-
face has the equation
(XIX)
4- sL - C7
^ + 52 - cz -
The intercepts are all zero. The traces
*Y in the coordinate planes are
X 2 ?V 2 X 2 t/ 2
tf + 6* " ' a* = CZ) 6~ 2 = CZ '
The first of these is a point and the other two are parabolas. Sec-
tions parallel to the xy plane are
^4-^.rf:
200 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VIII
which are ellipses for all values of k such that ck > 0. The axes
of these sections increase indefinitely as ck > + oo .
Figure 133 shows one-fourth of the surface for the case of
c > 0, drawn by means of the closed parallel sections lying in the
planes z k % .
If a = b y the surface is a paraboloid of revolution.
128. The Hyperbolic Paraboloid. The surface whose equation
is
y/ 2 r 2
TO &-*-
is called a hyperbolic paraboloid. The surface has all of its inter-
cepts zero.
Its traces in the coordinate planes
are
= cz
The first of these may be written
FIG. 134
and therefore represents two straight lines intersecting at the
origin. The other two traces are parabolas. All sections in the
planes z = fc, are hyperbolas and their axes increase indefinitely
as ki > OQ .
If c and ki have the same sign, the transverse axes of these
hyperbolic sections are parallel to the y axis, and when c and fc
have opposite signs the transverse axes are parallel to the x axis.
Figure 134 shows this surface for c > 0. Since it does not have
closed sections, we should sketch it, as shown, by means of hyper-
bolas.
129. The Elliptic Cone. If the vertex is placed at the origin
and its axis along the z axis, the elliptic cone has the equation
(XXI)
130J
SOLID ANALYTIC GEOMETRY
201
All intercepts are zero and the traces in the coordinate planes are
~
The first of these is a point, and each of the others is a pair of
straight lines meeting at the origin. z
Sections in the planes z = fc t , that
is, perpendicular to the axis of the
cone t are
a;* y 2 = kj z
a* + &2 ~ C 2 *
These are ellipses which increase in
dimensions as k l > =t < . Figure
135 shows one-fourth of the surface,
drawn by means of sections parallel
to the xy plane, and the traces in the
other planes. FIG. 135
If a = b, the cone is circular.
The equations of the surfaces discussed in this chapter have
been given their simplest forms by the proper choice of the coordi-
nate planes.
130. Translation of the Axes. If the axes are translated to
the new origin O'(h, fc, Z)> as referred to the z, y, z axes, we have for
any point P,
pc = x' + h,
(XXII) \y = y' + k,
(z = z' + I
Suppose the equation of a sur-
face is given in terms of x', y',
and z' and it is desired to find the
equation of the surface as referred
to the x, y, z axes. It is merely
necessary to replace the x f , y', and
z' by their values as given by
(XXII).
Thus, in the equations of the preceding articles, we must replace
x > V> z by x h, y k, z I if the point we choose for the origin
is at the point (A, k, I). However, the content of these articles
FIG. 13d
202 DIFFERENTIAL AND INTEGRAL CALCULUS [On. VIII
still holds if we consider the planes x = h, y = k, z las reference
planes when we find the intercepts, traces, and sections of any
surface.
EXAMPLE
What surface has the equation
Sketch the surface.
SOLUTION. Since first-degree terms are present in the equation of the
surface, it is necessary to complete the squares of each pair of terms involving
one variable. This gives
(z -h I) 2 -f 3(y 2)2 4(2 I) 2 = 12.
Z'
This equation can be written in the form
(z ~ I) 2
(x + I) 2 (y - 2) 2
12 "*" 4
= 1.
FIG. 137
In this form' we recognize the equation
as that of a hyperboloid of one sheet, as the
equation is similar to equation (XVII),
125.
The fact that x -f 1, y - 2, and z - 1
replace the x, ?/, and z of (XVII) merely
means that the center of the hyperboloid
is at the point ( 1, 2, 1).
Draw lines through this point which are parallel to the x, y, and z axes
respectively. Then sketch as shown in Fig. 137. Here we have used the
traces of the surface in the planes x = !,?/ = 2, z = 1 and sections in the
planes z = k v . One-eighth of the surface is shown.
PROBLEMS
1. Derive the equation of the plane whose intercepts on the axes are re-
spectively a, 6, and c units.
Discuss, sketch, and name the surfaces represented by the following equa-
tions. (Nos. 2-33.)
10. xy -f 3 x - 4 y = 3.
= 0. 11. r = 2 cos0.
12. r = 3 sin 0.
0. 13. r = a cos 3 0.
14. z -f- r = 0.
15. t/ 2 -h z 2 = 2 x.
16. 4 z 2 - y 2 -f 2 2 = 0.
17. r 2 = 1 + z.
2. x -f- y = 4.
3. 3z-3t/-2z + 6
4. x 2 + 7/2 + z 2 = 9.
5. x* -f y 2 + * 2 - 4 z
6. r 2 + z 2 = 16.
7. 2/ 2 = 4 - z.
8. y 2 = x - 2.
9. x 2 -f z 2 - 2 02 = 0.
130] SOLID ANALYTIC GEOMETRY 203
18. 4 x - y 2 - 4 z 2 = 4. 26. 6 x 2 - 3(y - 4) 2 + 4 z 2 = 0.
19. z = 6 - 2 x 2 - ?/ 2 . 27. r 2 = 4 - z.
20. (x - I) 2 4- 4 I/ 2 = 4 z 2 . 28. 4 x 2 4- 3 2/ 2 + 4 z 2 = 36.
21. x 2 + y 2 - z 2 = 16. 29. 4 x 2 4- 3 */ 2 - 4 z 2 = 0.
22. r\/3 = z. 30. x 2 /9 - ?/ 2 /25 - z 2 /25 = - 1.
23. x 2 4- 4 ?/ 4- z 2 = 4. 31. r = o(l + cos 0).
24. x 2 - 4 ?/ 2 4- z 2 = 4. 32. r 2 = a 2 cos 2 0.
25. x 2 - 4 ?/ 2 - z 2 = 4. 33. r = z 2 .
Draw each of the following lines and find its direction numbers. (Nos.
34-35.)
34. (a) 2 x - 3 y = 0, 5 x + 16 y = 0.
(6) x 4- y 4- z = 3, x 4- y = 6.
35. (a) x 4- 3 ?/ - 3 z = 4, 3 x 4- 4 y 4- 6 z = 6.
(6) (x - 4)/6 = (y + 3)/3 = - (z 4- 2)/2. Ans. (6) 6, 3, - 2.
36. Find the equation of the plane through (1, 2, 3) and perpendicular to
the line 3x + 2i/4-l =0, 5x-2z-3 =0.
Sketch each of the following pairs of surfaces on one reference scheme and
show the curve of intersection. (Nos. 3743.)
37. x 2 4- 2/ 2 = 4 z, x 2 4- 2/ 2 4- z 2 = 9-
38. r = 4 cos 0, r z.
39. r = 2 sin 0, r 2 -f z 2 = 4.
40. 3 x 2 4- 4 7/ 2 = - 2 z, x 2 4- y 2 ~ 2(z 4- 2) = 0.
41. 2/ 2 = 4 z, x 2 = 4 z.
42. x 2 4- z 2 = 9, ?/ 2 4- z 2 = 9.
43. r 2 = 3 z, r cos + z = 2.
44. Write the equation of the right circular cone with a vertical angle of 120
at the origin if its axis is along the z axis.
45. Write the relation connecting the three sides of a right triangle. Sketch
the relation. Ans. z 2 = x 2 4- 2/ 2 -
46. Write the equation of the right circular cone of vertical angle 7r/2 at
(0, 0, 2) and axis along the z axis.
47. Write the equation of the right circular cylinder which has the x axis
as one element and has (0, 0, 1) as the center of its yz trace.
Ans. y* 4- (z - I) 2 = 1.
48. Write the equation of an ellipsoid whose center is the origin if its semi-
axes are 2, 3, 4 units, respectively, and lie along the reference lines.
204 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. VIII
49. Write the equation of the paraboloid of revolution whose vertex is
(0, 0, 2) and which cuts the xy plane in a circle of radius 2 units.
Ans. 4 2 z = x 2 -f- y*.
50. A paraboloid of revolution is formed by revolving about its axis a
parabolic segment whose base and altitude are each 10 units. Choose a refer-
ence scheme and derive the equation of the surface.
Find the direction numbers of each of the following lines. (Nos. 51-52.)
51. (a) 2 x + y - 3 z - 7, x-4y-f3s+9=0.
(b) 3 x - 2 y + z = 12, x - 5 y - 2 z = 10.
Ans. (a) 1, 1,1; (6)9,7, - 13.
52. (a) 2 x - y + z = 7, x + y + 2 2 = 11.
(6) x + 5 y - 3 z = 10, 2x-3y-z = 10.
Find the equation of the plane determined by each of the following sets of
conditions. (Nos. 53-54.)
53. (a) Passing through the points (8, - 2, 6), (3, 4, - 3), (2, 2, - 2).
(6) Passing through the point (1, 1, 2) and the line x 3y 2z = t
x + y+z-2=Q.
Ans. (a) 6 x - 7 y - 8 z - 14 = 0; (6) 4 x + z - 6 == 0.
54. (a) Passing through the points (4, 2, 1), (-1, - 2, 2), (0, 4, - 5).
(6) Passing through the point (0, 2, 3) and the line 3 x z -f 12 = 0,
x + 3 y + 17 = 0.
Find the equation of the plane through each of the following pairs of inter-
secting lines. (Nos. 55-56.)
3 - y - z - l X + 2 - y ~ 3 - * ~ X
~ ~~ ~~ - ~~ - '
' 3 ~ 5 ~ ~2~ ~^2~ - ~T~ - 3
Ans. 23z-M3!/-22-f9=0.
56. [3 x - 2 y = 0, x -f 3 y - 2 2 = 3] and [> + y - z = 0, 5 y - 3 z = 0].
CHAPTER IX
PARTIAL DIFFERENTIATION DIRECTIONAL
DERIVATIVES ENVELOPES
131. Partial Derivatives. A function of two or more inde-
pendent variables can >e differentiated with respect to any one
of the variables if the others are considered as constants during
the operation. Such a derivative is called a partial derivative
of the function. j
If we consider z = f(x, ?/), the first partial derivative of z with
respect to x is defined as follows: I
(1) lim /(* + A*, y) - f(x, y)
and we represent the same by one of the symbols
dz _ df(x, y) _ f( ,
to to f : (x> y) -
The student should recognize (1) as the definition of the deriva-
tive of a function of one variable x] that is what it really means,
since y is assumed to remain constant.
Similarly we have
(2) = Km f( x > y + A ^ ""/fa y) .
<ty Ay-*) Ay
If we have a function of more than two independent variables
the first partial derivatives of the function with respect to each
variable are defined and found just as in the case of f(x, y). All
variables are treated as constants except the one with respect to
which the derivative is taken.
We are familiar with functions that depend upon two or more
variables; for example: the volume or surface of a cone depends
upon the radius of its base and its altitude; the volume of a gas
depends upon its temperature and the pressure to which it "is
subjected; the motion of a body is dependent on all the forces
acting upon it. Also, in filling a cylindrical can with water we
have the volume of the water changing in depth only; in rolling
205
206 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
paper into rolls for printing, the volume changes due to a change
in the radius of the roll alone. These are illustrations of functions
of more than one dimension, however, only one dimension is
allowed to change. Hence the rate of change of the volume in
each case is a partial derivative.
To illustrate this, let the rectangle
OABC of Fig. 138 be a plate heated in
such a manner that the temperature at
I any point P(x, y) is given by some
y*Ai/) function of x and y. If T P /(x, y)
I then T Q = /(x + Ax, y) as P and Q
Q(x*&x,y) have the same ordinate. Hence the
4"^ temperature changes in going from P
to Q by the amount
, y) -f(x,y).
As this change is due to a change in x alone, the temperature along
the line PQ has as its rate of change at P the following limit :
AX
which is exactly an illustration of a first partial derivative of a
function of two variables arid is accordingly dT/dx.
Similarly, the rate of change of T in the direction PR is dT/dy.
However, the change of T in any other direction PS will depend
upon both Ax and A?/ and hence the rate of change is not a partial
derivative,
EXAMPLES
1. If T = e* sin y, find dT/dx, dT/dy at the point (1,2).
SOLUTION. T x = e* sin y and at (1, 2) is e sin 2 or 2.4717. Also
T v = e* cos y and at (1, 2) this is - 1.1312.
2. Find the rate of change of the volume of a right circular cone with respect
to the radius of its base when the radius is 4 inches, if its altitude remains
6 inches.
SOLUTION. The volume of the cone ie V (l/3)7rr 2 /i and we want
dV/dr evaluated for r = 4" and h = 6". Since
dV 2 .
d7 = 3^
the desired quantity is (2/3)ir-4-6 = 16 ?r cu. in. per in.
133] PARTIAL DIFFERENTIATION 207
132. Higher Partial Derivatives. If z = f(x, y) is differenti-
ated twice with respect to x or y, or once with respect to each
x and y, we have second order partial derivatives. These second
partial derivatives are represented by the symbols
d^ _ 3 fdz\ d*Z
~ dx* - }xx(X > y) > d7j\ ~~ Jxv(X '
_ / / >> d (
~ Jvx(X > y) >
dx\du ~ dxdy
If f(Xj y) is a continuous function with continuous derivatives,
dxdy dljdx
A similar notation is used for higher derivatives than the second
of functions of two or more independent variables. Thus
lL(?-*\ = ^-f ( \ 1Y 2 *\ _ &* _ / / N
dx\dx*J dx* Jxxx(X> y)) dy \fyte) ~ d^di " J - XVU , (X > y) >
dy\dx*J dydx 2 '""'<"> dx \ d
/( T ii\
wv\ ) y/t
dx \dxdyj dx 2 dy Jvxx ^' ! "'dy\dy*J~ dy
and similarly for the fourth and higher derivatives.
EXAMPLE
If z = x z y + 2 xe lf , show that z xy - z yx .
SOLUTION. The first partial derivatives of z are z x = 2 xy -f- 2 c }f * t and
z v = x 2 2 xe l/ v/y 2 . Differentiating the first of these with respect to y
and the second with respect to x, we have
Z x y Z 3J
133. Geometric Representation of z x and z y . Tangent Plane.
Since z = /(x, y) may be considered to represent a surface, a very
simple geometric representation of the partial derivatives dz/dx
and dz/dy can be given.
Suppose a portion of the surface z f(x, y) is represented by
ABCD of Fig. 139.
The plane x = Xi cuts the surface in the curve QN and LM is
* For a proof of this see Goursat-Hedrick, Mathematical Analysis, Vol. I, 11.
208 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
cut out by the plane y = 2/1. Then WPi is the value of z at the
intersection of the curves QN and LM . If we let WE = PiK = Ax
arid W// = T\F = A?/, then Wl\ = /On, 2/0, #^ = /(^i v + Ax, 2/0,
and HN = f(x\, 2/1 + AT/). Whence
A'M = /(a?i + Az, j/i) - /(xi, 2/0,
and
*W = /(a?i, 2/i + Ay) ~ ,
This means that
is the slope of the curve LAf at the
point PI. And similarly
= lim
FN
is the slope of QN at PI.
Hence the first partial derivatives of z f(x, y) are the slopes of
the curves cut from the surface by sets of planes which are parallel to
the yz and xz planes, respectively.
It is now easy to find the equation of the tangent plane to the
surface at PI. This plane intersects the planes x = x\ and y = y\
in lines tangent .to QN and ML at PI.
These tangent lines are
(1)
and
(2)
dz l ,
- Zl = (y -
z - Zl =
to!
a: Xi = 0,
0.
The equation of any plane through PI is
(3) A(x- x,) + B(y - i/O + CO - zi) = 0.
If this plane contains the tangent line (1), whose direction numbers
are 0, 1, dzi/diji, we must have
Similarly, if (3) contains the line (2), we must have
d"i
A + C = 0.
133] PARTIAL DIFFERENTIATION 209
Hence*
dzi _ A , d2! _ B
foi~ c> ana a^~ c*
Solving (3) for 2 Zi and substituting these values, the equation of
the tangent plane becomes
EXAMPLES
1. If z = 3 z 2 - 2 i/ 2 + 2 sy, find ds/Ox and
SOLUTION. Differentiating z as though x were the only variable in it, we
find dz/dx = 6 x + 2 y. Similarly for y, dz/dy = 2 z 4 ?/.
2. Find the equation of the tangent plane to the paraboloid z = 2 x 2 + 4 t/ f
at the point (J, 1, 6).
SOLUTION. At the point dz/dx = 4 and dz/cty = 8. Hence the tangent
plane is z 6 = 4(x 1) + S^y 1) or
4z-f8i/-z--6=0.
3. If x and i/ are given paramctrically by the two independent variables
r and 6 such that x e zr cos 6, y = e r sin 0, find x r , 0*0, ?/ r , y Q .
SOLUTION. Evidently x r = 2 e 2r cos 0, y r = e r sin ^. Also we have
X Q e 2r sin B, y Q e r cos 0.
4. Suppose in Example 3 above that x and y are the independent variables.
Find r X9 r v , O x , V .
SOLUTIONS. It is essential that the student realize that r and are now
the dependent variables and each is a function of x and y. Therefore, r and d
have partial derivatives with respect to each x and y. Hence we may solve
the given equations for r and in terms of x and y and find these partial
derivatives just as in the preceding examples. However, such a solution is
often inconvenient or more difficult than the method given below. That is,
find the partial derivatives of both sides of each of the two equations
x = e 2r cos 0, y e r sin
with respect to x, as is done in implicit differentiation. Recalling that y is
assumed constant, we have
1 = 2e 2 'cos0- ^-e^sintf. ,
ox ox
\ \-
= e r sin ^- -f e r cos -^-
dx dx
* These results can be obtained also as follows. The section of (3) by the plane
x - xi is obtained by setting x xi - in (3) . Solve the resulting equation for
z zi and compare with (1). Proceed similarly for (2).
210 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
Since such equations are always linear in the partial derivatives, they are
very readily solved. Thus we get
dr cos e dB sin
dx e zr (l + cos 2 6} ' dx e 2r (l + cos 2 0)
These results may be expressed in terms of x and y with the help of the original
equations. They become
dr
2x
dx (3,2 H
h Vz/^ 4- 4 X 2) v?/ 4
+ 4x 2 '
do
7/V2
VX A A*2 J_ \/,.4 _1_ A >.2 \A/4 _i_ A ^2
Similarly, 6V/% and 00/d?/ may be found.
PROBLEMS
Find the partial derivatives specified below. (Nos. 1-5.)
1. z = 6 2 x 2 i/ 2 , find z x , z yy z xy . Am. 4 x, 2 y, 0.
2. u - x 2 + 5 x?/ 2 , find xw xa! + (l/2)yu yx .
3. 4 x 2 - 2/ 2 -f z 2 = 0, find z xy z v , z xx , z vx at (2, 5, 3).
Arw. - 8/3, 5/3, - 100/27, 40/27.
4. u = x 2 2 z?/ + 3 ?/ 2 , find u yx , u xv .
5. a: 8 ^ 4 - 2 ?/ 2 2 3 -f- a; 2 ?/ 2 +2 = 0, find z x , z tf .
Ana. (2 xy + 3 x 2 z 4 )/(6 ?/z 2 - 4 x 3 z 3 ),
(2 x*y -f x 3 z 4 - 4 ?
6. u - (x 2 -f 2/ 2 )/(2/ 2 x 2 ), show that w, v = u yx .
7. u = x 3 ?/ 2 2 x?/ 4 + 3 xV, show that xw* + yu v - 5 w.
8. w = (x 2 -f 2/ 2 ) l/3 , show that 3 xu vx -f 3 ?/w vv + u v = 0.
9. w = (ax -f- 6y + c ^) n , show that xu x 4- yw v -{- zw, = nw.
10. z = logVx 2 -f J/ 2 , find z JX + 2 Vtf .
11. z x 2 sin 2 y, show that z xv = z yx -
12. w = sin x 2 z/, is w xy = w vx ?
13. x = r cos 0, y = r sin d, find x r , t/^. Ans. cos 0, r cos 0.
14. x = e r sin 0, t/ == e r cos 0, find x r , t/ d .
15. x r cos 0, y = r sin 0, find r z , tf . Ans. x/r or cos 0; x/r 2 or (cos 0)/r.
16. x = e r sin &, y = e r cos 0, find r y , X .
17. z ~ e"^ 8 cos x, show that *,, 2 z xx =
18. = 2at (cos 2 y - sin 2 y\ find z x + z vv .
133] PARTIAL DIFFERENTIATION 211
19. z = (z 2 + y 2 ) tan- l (*/2/), find z X9 z y , z vx .
Ans. y + 2x tan" 1 ^/*/), 2 y tan-'Oc/lf) - , (i/ 2 - z 2 )/(x 2 -f ^ 2 ).
20. 2 = e v/x sin (y/x), find 2/2 v + 22,.
21. 2 = x sin ?/ ye* t find z,,*. ylns. cos y e*.
22. z = x 2 e~ 2 " -f y log [3 esc (z/2)] -f sin ?/ tan" 1 a;, find z x , z v , z vy .
23. Given the surface z = 2 x 2 -f 3 ?/ 2 4, find the slopes of the curves cut
from this surface by planes through (1, 1, 1) parallel to the xz and yz planes.
Ans. 4, - 6.
24. Given the surface z = x* y 2 -f 3 x 7, find the slopes of the curves
cut from this surface by the planes x = 3, y = 2.
25. Use the area of a triangle in terms of two sides and the included angle
to find the rate of change of the area with respect to the angle; with respect to
either side. Ans. (l/2)xy cos 0, (l/2)y sin 0, (l/2)x sin 0.
26. Using the law of cosines, derive the rate of change of one side with
respect to the opposite angle.
27. If the resistance of the air is not neglected in considering vibrations
of strings, the equation y lt -f 2ky t = a 2 y xx occurs. Is the following function,
y = e~ kt sin ax cos (tVa 2 a 2 /c 2 ), a solution of this equation?
28. The equation z ti = c 2 (z xx + z vu ) occurs in the theory of the vibration
of stretched membranes. Show that z = sin otx sin (3y sin (ctVa 2 -\- ft 2 ) is a
function satisfying this equation.
29. A function defining the potential of a point of a thin sheet due to
current flow must satisfy the relation v xx -f v vv = 0. Are the functions
v = 1 - [ta,n~ l (y/x)']/Tr and v = - [log (x 2 + ?/ 2 )]/(2 TT) solutions of this
equation?
Find the equation of the tangent plane to each of the following surfaces at
the point indicated. (Nos. 30-39.)
30. xz + 2 x -f 4 z = 5, (2, 5, 1/6).
31. x 2 - y 2 + z 2 = 6, (T, 2, - 3). Ans. x-2y-3z = Q.
32. x 2 + y 2 /* - z 2 /V = I, (- 1, 2, 3).
33. 2x + y 2 /2 + z 2 /4 = 0, (- 3/8, 1, 1). Ans. 8x + 4y + 2z = 3.
34. x 2 - 4z + 2y = 0, (4,0, 5).
35. xy + yz + zx = 0, (2, - 3, - 6). Ans. 9x + 4y+z = Q.
36. x 2 4- y 2 + z 1 - - 6 z = 0, (2, 2, 4).
37. x 2 -f y z - 3 z = 2, (- 2, - 4, 6). Ans. 4z-f-8t/-f3z + 22=0.
38. 2 x 2 - y 2 - 3 z 2 - 4 y + 6 z + 2 = 0, (3, - 2, 4).
39. x 2 + 3 ?/ 2 - 4 z 2 + 2 x - 12 y + 8 z - 7 = 0, (1, - 2, 4).
-Gi/ Gz-f-H =0.
212 DIFFERENTIAL AND INTEGRAL CALCULUS [Ca. IX
134. The Increment and Differential of z = /(*, y). If x and y
are given increments Ax and A?/, the resulting increment of z is
(1)
Az = /(x +
+ Ay) - /(x,
Since TFP = z and JC = z + Az, we have Az - JC - W P = JBC.
Now, assuming that z and its derivatives are continuous func-
tions, the point C will approach P as a limit as PK and PF approach
zero as a limit in any manner whatsoever.
The line-segment RC is made up of RT = KM and TC. We
have shown in the preceding article that
(2)
r
hm
KM
dx'
whence (85)
(3) KM
= 3T + e i>
where ei > as
(3), we have
. From
FIG. 140
lim
TC
(4) KM = f^ + djA*.
yc/x y
Also we see that TC is such that
= slope of MC.
However, as Ax > 0, the curve MC approaches coincidence with
PAT, and TC with FN. Hence if we take the limit of TC/MT
as both MT = PF = At/ and Ax approach zero as a limit we have
r TC r FN dz
(5) lim TTT^ = hm
whence
(6)
where e 2 > as Ax > 0, Ay > 0. This gives
(7)
re
134] PARTIAL DIFFERENTIATION 213
Substituting the values of the line-segments KM and TC in Az,
we get
(8) Az = J A* + g Ay + * A* + * 2 - Ay.
This formula for the change in z due to changes in x and y is of
no value for finding the change of a given function. Such a prob-
lem is merely an arithmetical calculation of the difference of the
two values of the function.
However, if we observe that e\ and e 2 depend upon Ax and Ai/,
we see that for sufficiently small values of Ax and Ay the two
terms e\ Ax + e^ Ay become very small. Hence the remaining
two terms of Az may be used to approximate the change in z.
These two terms are defined as the differential of z, and represent
the approximate change or error in z for given changes or errors in
x and y. That is,
/*\ i 9z f , 9z ,
(9) dz = ~d X + ~dy,
where we write dx and dy for Ax and Ay as both are independent
increments. This is a useful formula, as the partial derivatives
and increments are readily combined to give approximate changes
in the function of two variables.
As dx and dy are independent increments,- we have
(10) d x z = I? dx, d v z = I dy,
where d^ and d y z represent the partial differentials of z with
respect to x and y respectively. We notice that dz becomes
dxZ when dy = 0, and d y z when dx 0. Therefore the differential
of z is the sum of the partial differentials of z with respect to each x
and y.
EXAMPLES
1. If z = x* - 2 xy + y\ find dz, d x z, d y z.
SOLUTION. Since dz/dx = 2 x 2 y and dz/dy 2 x -\- 2 y, we
have d x z 2(x y)dx and d y z 2(y x)dy. From these we obtain
dz = d x z -f c^z = 2(x - y)dx + 2(y - z)di/.
2. In measuring two sides of a triangle which include an angle of 30
one side is found to be 27 inches with a possible error of 0.10 inch, and the other
214 DIFFERENTIAL AND INTEGRAL CALCULUS [On. IX
13 inches with a possible error of 0.05 inch. What is an approximate value
for the largest possible error in the area of the triangle due to the errors in
measuring the sides?
SOLUTION. The area A of the triangle may be given by the relation
A = (l/2)xy sin 30 = (l/4)xy. An approximate error in A due to errors in
x and y is
d A = -r (y dx -f x dy).
We have given x = 27 in., y = 13 in., dx = 0.10 in., and dy = rfc 0.05 in.
Since the largest possible value for dA is desired, we assume dx and dy
to have the same sign since their coefficients have the same sign. Sub-
stitution gives dA = (1/4) 13 (1/10) + (1/4) 27 (1/20) = 53/80 sq. in.
The actual error is A A = A>i A\, whore A 2 is computed by using
x 27.1 in. and y = 13.05 in., and A\ by using x 27 in., y = 13 in. Such
computation being usually laborious, the value of dA is used as an approxi-
mation of the error A A.
The relative error is defined as &A/A and again the approximate relative
error dA/A is generally requested. The percentage error is of course 100 times
the relative error. It is convenient to take the logarithm of a function if the
approximate relative error is desired. Thus for this example, we have
log A = log x -f log y log 4.
Then the differential gives
dA dx dy 0.10 0.05
^ ^'
- 0075 -
This shows that an approximation for the relative error of a function is the
sum of the approximations of the relative errors of each factor. This expressed
as a percentage is evidently (3/4)%.
The expressions for Az and dz are readily extended to the cases
of three or more variables. Thus, if u = f(x, y, z), we have
A du . . du . . du A , A . . ,
AM = Ax + Ay + Az + 61 Ao; + e 2 A?/ + e z Az,
by a line of reasoning that is essentially that given for two variables
except for the geometric interpretation.
The differential of u is defined as
, du 7 , du j , du 7
du = dx + dy + dz,
dx dy dz
and is used to approximate the value of An just as dz is used to
approximate Az.
Exactly similar relations hold for four or more independent
variables as long as the number of independent variables is finite.
134] PARTIAL DIFFERENTIATION 215
PROBLEMS
Use the differential in evaluating the following unless exact results are
called for.
1. If x differs from 4 units by at most 0.02 unit and y from 3 by 0.01
unit, approximate the difference between x 2 + y 2 and 25 sq. units.
Ans. 0.22 sq. unit.
2. A right circular cylinder of altitude 10 ft. and radius 4 ft. has its
altitude increased 0.1 ft. and its radius decreased 0.01 ft. Approximate the
change resulting in the volume.
3. Approximate the possible error in the area of an ellipse (irab sq. units)
due to a 1% error in a and a 2% error in 6. Ans. 3%.
4. If z(y z x 2 ) = xy and x = 2, y = 3, Ax = 0.0010, A?/ = 0.0123,
find (a) Az; (6) dz; (c) the approximate relative change in z; (d) the per-
centage change in z.
5. Approximate the volume of a circular cylinder with radius 2.997 in.
and altitude 10.02 in. Ans. 90 TT cu. in.
6. The hypotenuse and a leg of a right triangle are 5 in. and 4 in., respec-
tively. If the hypotenuse is decreased 0.01 in. and the leg increased 0.01 in.,
approximate the change in the third side, the triangle being kept a right
triangle.
7. The power consumed in an electrical resistor is given by P E 2 /R
watts. If E 100 volts and R 5 ohms, how does the power change if E is
decreased 2 volts and R is decreased 0.3 ohm? Ans. dP = 40 watts.
8. If the length and width of a rectangle are measured as 3.5 and 2.3 ft.
with 0.01 ft. possible error in each, approximate the possible error in the
computed area.
9. Find the exact and approximate error in the calculated volume of a
right circular cone if r = 4.95 in. and h = 4.1 in. but 5 in. and 4 in. are used.
Ans. 0.1534 cu. in.; x/6 cu. in.
10. If 8 = LH^/Vo and L, H, D are measured as 4, 3, and 25 in. respec-
tively with a possible error of 1% in each, approximate the possible percentage
error in S.
11. Approximate the relative error in Problem 9. Ans. 1/200.
12. If z = x 3 + i/ 3 - 3 x 2 t/, and x = 2. y = 3, Ax = 0.01, Ay = 0.001,
find A0 and dz.
13. If x and y are measured as 5 and 3 units with possible errors of 0.02
unit in x and 0.01 in T/, approximate the possible resulting error in
M = xVx 2 - y\ Ans. db 97/400.
14. The radius of a right circular cone is measured as 7 ft. and its slant
height as 25 ft. with a possible error of 1% in each. What possible error
results in the computed volume?
216 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
15. Two sides of a triangle are 6 in. and 8 in. long, respectively, and they
include an angle of 30. Approximate the change in the area if the shorter
side is decreased 1/25 in. and the longer is increased by the same amount.
Ans. 1/50 sq. in.
16. Approximate the possible error in the triangle of Problem 15 if the sides
were measured accurately to within 0.1 in.
17. Find the maximum percentage error allowable in measuring the
dimensions of a circular cylinder so that the computed volume will be correct
to within 1%. Ans. If dr = dh then 100(r/r/r) = h/(2 h r).
18. Find the maximum percentage error allowable in measuring the diam-
eter of a sphere if the computed volume must be correct to within 1%.
19. The length L and the period P of a simple pendulum are connected by
the law 4 7T 2 /, = P*g. If L is calculated for P = lsec.,0 = 32 ft. /sec. 2 , approx-
imate the error in L if P is really 1.02 sec., g 32.01 ft. /sec. 2 Find the approx-
imate percentage error also. Ans. 1.29/4 ?r 2 units, 129/32%.
20. Using the pendulum law of Problem 19 for possible errors of 0.1%
and 0.2%, respectively, in L and g, approximate the possible percentage error
in P.
21. Suppose L in Problem 19 is measured as 100 units with a possible error
of 1/2 unit, arid P as 2 units with a possible error of 0.01 unit. Approximate
the possible percentage error in g. Ans. 13^%.
22. If R = (V 2 sin 2 0)/32, approximate the possible error in computing R
for V 5000 ft. /sec. with a possible error of 10 ft. /sec. and = 7r/3 with a
possible error of 3 rnin.
23. The altitude and diameter of a circular cylinder arc measured as 10
in. and 6 in., respectively. If a 4% error is made in the diameter, what error
in the altitude will counteract this error in the computed volume?
Ans. 0.8 in.
24. Approximate the resulting change in each of the following cases.
(a) z = log(x 2 -f 7/ 2 ) for x = y = 1 and dx = =b dy = 0.01.
(6) q = e */2 sm 2 y for x = 2, y = ir/3 and dx = it 0.01, dy = db 0.01.
(c) T = (tan y) /(log x), for x = 2, y = x/4, and dx = db 0.1, dy = 3'.
25. The density D of a body is Wi/(Wi - W 2 ) where Wi is its weight in
air and Wz its weight in water. If W\ 1200 gr. and Wi 10 gr., what is
approximately the greatest percentage error in D as computed from these
weights if there is a possible error of 0.50 gr. in Wi and 0.01 gr. in TF 2 ?
Ans. 0.00119%.
135. The Derivative of z = /Or, y) with Respect to Some
Other Variable. If it is desired to find the rate of change of z
with respect to some other variable t, it is obviously necessary that
x and y be functions of t. This makes z a function of t and the
135]
PARTIAL DIFFERENTIATION
217
derivative of z with respect to t is defined as in the case of a function
of one variable.* Thus, if x = </>(t) and y = \l/(t) t we have
dz . Az /dz Ax . dz Ay Ax Ay
or
(2)
Az r /dz Ax
= um V" ' TT
dx
y\
7 I >
2 /
d* "" dx
where Ax and Ay approach zero as a limit as At > 0, and whence
ei and e 2 > 0. It is assumed, of course, that cte/eft and dy/dt exist.
Let A BCD represent a portion of
the surface z = f(x, y).
Then the equations x = </>(t),
y ^(0 define a cylindrical surface
NM. This cylinder intersects the
surface in the curve NK and the xy
plane in the curve LM. Hence
dz/dt represents the rate of change
of QP with respect to t as Q moves
along the curve LM.
A function u of three variables x, y,
and z, each of which depends upon
another variable t, has as its deriva-
tive the expression
,~, du _ du dx du d\
( ' Tt ~ d~x "di + ~dii ' di
FIG. 141
du
where u, x, y, and z are differentiate functions of t.
It z = f(x, y) and y = F(x), so that is a function of x alone, we
may find the rate of change of z with respect to x as in (2) . That is,
dz .. Az r /dz Ax . dz Ay .
T- = hm = hm ( + --^ + ei
dx Ax-^oAa; AX^O\^ ^ ty Ax
Then if Ay, e^ and e 2 > when Ax > we have
Ax
Ax
.
+
Ay
T^
Ax
\
r
J
dz _ dz dz_ dy
~dx ~ ~dx ~dy* dx'
* Of course x and y may not be functions of t alone, as x = <f>(t, u), y = $(t, it),
and still the derivative of z with respect to t have the same derivation, provided
z = /( 5 ; t j/) is still a function of t only.
(4)
if lim Ay /Ax exists.
218 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
Similarly, by considering z a function of y alone, we get
(5)
dz
dy
dz dx . dz
dx dy dy'
K
FIG. 142
the relation
These formulas (4) and (5) are
evidently the same as (2) if a; re-
places t or y replaces t, respectively.
The plane z = c intersects the sur-
face z f(x y y) along the curve LK.
If we let z be constant in relations
(4) and (5), the left-hand member of
each is zero and these become, re-
spectively,
L 4_ = n
dx'*' dy' dx '
dz_
dx
dy ' dy "'
Solving either of these, we obtain
(6)
dx
This gives the slope of the curve RS, which is the projection of
the~curve LK upon the xy plane.
If z = f(x, y) is written as F(x, y } z) = 0, where z is given
implicitly as a function of x and y, we find expressions for dz/dx
and dz/dy by assuming first y constant and then x constant.
Thus, if y is constant, differentiation of F(x, y, z) = with
respect to x gives
dF dF dz =
dx + dz ' dx " J
where dz/dx means dz/dx for y constant. Solving this for dz/dx,
we find
(7)
9F
9z 9x ., 9F
= - lf
dz
135] PARTIAL DIFFERENTIATION 219
Similarly,
BF
S--S- E-*
dz
EXAMPLES
1. If the radius of a right circular cone is increasing 3 inches per second
and its height is decreasing 4 inches per second, how fast is its volume changing
when r = 6 inches and h = 12 inches?
SOLUTION. The volume of the cone is V (1/3) irr z h. Therefore
dV dV dr , dV dh 2 , dr , 1 , dh
_ _ , . _ _i_ _ . __ . - ir<rn -4- irr*
dt dr dt ^ dh dt 3 dt ^ 3 dt
But r = 6 in., h = 12 in., dr/dt = 3 in. /sec., and dh/dt = 4 in. /sec., from
the conditions given. Hence, substituting, we have
~ = ?7r-6-12-3 +ir-36(- 4) = 967TCU. in./sec.
(it o o
Since dV /dt is positive, the volume is increasing 96 TT cu. in./sec. at that instant.
2. If z = sin" 1 ^! + x)/(l -f 2/)D and x sin t y y cos , find the rate of
change of z with respect to t when = 0.
SOLUTION. By differentiation, we find
dz ^ _ _ 1 _ 1 dx __ 1 _ _ 1 -fa; dy
dt r /T + x \ ' 1 + y ' dt /!+ x ' (l+y)*'dt'
But dx/dt cos t, dy/dt sin t', so for t we have x 0, y 1,
dx/dt - 1, cfa//(& = 0. Substituting these values in the expression above,
we obtain
dz
I F 1 _
I / 1 vLl + 1
V 1 ~ vmy
(14- I) 2
PROBLEMS
1. The volume and radius of a cylindrical boiler are expanding at the rate
of 0.8 TT cu. ft./min. and 0.002 ft./min., respectively. How fast is the length
of the boiler changing when the volume is 20 ir cu. ft. and the radius is 2 ft.?
Ans. 0.19 ft./min.
2. The radius of a right circular cone is decreasing 2 in./sec. and its slant
height is increasing 3 in./sec. How fast is the volume changing when r 4 in.,
h = 3 in.?
220 DIFFERENTIAL AND INTEGRAL CALCULUS [On. IX
3. Suppose that the x and y coordinates of a point on the surface
x 2 -f 2 if 3 -h z = are changing 3 units/sec, and 4 units /sec., respec-
tively. How is z changing at (2, 3, 19)? ylrts. Decreasing 60 units /sec.
4. A particle at A on a line AB which is 28 ft. long starts toward B at
4 ft. /rniri. and at the same time another particle leaves B in a direction which
makes 60 with BA at 4 ft./min. How fast are the two particles approaching
each other after 1 min.?
5. The top of a 20 ft. ladder leaning against a vertical wall slides down at
the rat/? of 3 ft. /sec. (a) How fast is the foot moving along the ground, which
makes an angle of 2 ?r/3 with the wall, when the ladder makes an angle of 7r/6
with the ground? (6) At the same position how fast is the area of the triangle
formed by ladder, wall, and ground changing?
Ans. 3 ft. /sec.; no rate of change.
6. The legs of a right triangle at a given time are 2 ft. and 4 ft. and are
increasing 1 ft./min. How fast is the area of the triangle changing? How
fast is the perimeter changing?
7. In a right circular cone of altitude 4 ft. and radius 3 ft., the slant height
is increasing 0.1 ft. /sec., and the radius is decreasing 0.2 ft. /sec. How fast is
the volume changing? Ans. Decreasing 0.775 TT cu. ft. /sec.
8. Find the rate of change of the area of a triangle if its shorter side is
decreasing 0.05 in. /sec. and its longer side is increasing 1/30 in. /sec., when
they are 6 in. and 8 in. respectively! The included angle is ?r/6.
9. A point on the surface z = log xy at the point (2, 3, log 6) is moving so
that the x coordinate increases 1 unit/sec., and the y coordinate decreases
4 units/sec. How is z changing? Ans. 5/6 unit/sec.
10. Given z = tan~ 1 [x/(l + 2/)]> # = cos t, y sin t. Find (a) dz/dt at
t = 7T/4; (b) dz/dx at t = ir/6.
11. The radius and altitude of a right circular cone at the instant when
r = 12 ft., h 5 ft., are changing so that r is increasing 0.2 ft. /sec. and the
slant height decreasing 0.1 ft. /sec. How fast is the total surface changing?
Ans. 6.2 TT sq. ft. /sec.
12. Two sides and the included angle of a triangle are such that the shorter
side is decreasing 0.1 ft. /sec. and the longer side is increasing 0.15 ft. /sec., the
angle is decreasing 0.1 radian /sec. How fast is the area changing for the
respective values 10 ft., 15 ft., ?r/3?
13. We move along a hill, a part of which may be represented by
z = 4 x 2 2 ?/ 2 , in such a direction that x increases 2 units /sec., arid y
decreases 3 units/sec. How is z (the height) changing?
Ans. dz/dt = 4(3 y x) units /sec.
14. If e = (V~a 2 &*) /a, find the rate of change of e when a = 8 in.,
6 = 6 in. and each is changing at 0.5 in. /sec.
136J PARTIAL DIFFERENTIATION 221
15. Find the rate of change of an acute angle of a right triangle if the
opposite and adjacent legs are 10 ft. and 12 ft., respectively, and are changing
at the rate of 1 ft./min. and 2 ft./min. Ans. 8/61 rad./min.
16. Given any continuous function of three variables, F(x, y, z) = 0, show
that
dx dy dz _
dy dz dx
17. Given Van der Waal's formula (p -f a/v 2 ) (v b) ct, where a, 6,
and c are constants. Find dp/do, dv/dt, and dt/dp. Check your results by
using Problem 16.
136. The Directional Derivative of z=/(x,y). Since the
value of z is determined when x and y are known, z is fixed for each
point of the xy plane. Then if the
point P(x, y) moves, in general,
z changes value and therefore has
a rate of change at each point of
the plane.
Suppose we desire the rate of
change of z for the point P(x, y)
if P moves in the direction PR. /'^
Let PR make an angle B with the
x axis. If we assume that P has ^ IG 143
moved to an adjacent point Q on
PR, the change in z is its increment due to the corresponding incre-
ments Ax and Ay.
The derivative of z for the point P in the direction PR is defined as
the limit of the change of z divided by the distance Ar which P moves
in the given direction as Ar 0. That is,
/1N dz ,. Az ,. /dz Ax . dz Ay . Ax , Ay\
(1) T = hm = lim ( +- - + ei - + e 2 -~- }
dr Ar-*o Ar Ar-*o \d# Ar dy Ar Ar Ar J
Since Ax and Ay > as Ar > we have e\ and e 2 approaching
zero with Ar and therefore
( . dz _ dz dx dz dy
dv dy dr du dr
But dx/dr = lim (Ax/Ar), and Ax/Ar = cos 0, from triangle PQS.
Ar *0
Also Q*P along the straight line QP and therefore the
ratio Ax/Ar is constant. Whence dx/dr = cos 8. Similarly,
222 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
dy/dr = sin 0. These values substituted in relation (2) give the
directional derivative of z as
(3)
dz
dr
The student must realize that this expression for dz/dr is entirely
a function of 6 when the point P has been chosen. That is,
dz/dx and dz/dy are constants evaluated by means of the coordi-
nates of P. Then, for a given 0,
the direction PR is fixed and dz/dr
represents the rate of change of PH
as H moves along the curve IIT.
(Fig. 144.)
Since dz/dr is a function of 6
alone, we can find the 6 which makes
the directional derivative a maximum
or minimum for any given point.
Notice that 6 is between and 360
and is measured from the direction of
the positive x axis.
EXAMPLES
1. If z = # 2 -f 2/ 2 , find the rate* of change of z for the point (3, 4) in the
direction of the point (2, 6). What direction makes the rate of change a
maximum?
SOLUTION, dz/dr = 2 x cos -f 2 y sin 0, and at (3, 4) this becomes
V- = 6 cos -f 8 sin 0.
dr
Now 0, as shown in Fig. 145, is the second-quadrant angle which has its cosine
equal to 1/V5 and its sine equal to 2/V5. It follows that we have
dz/dr = - 6/V5 + 16/V5 = 2V5.
To find the which makes dz/dr a maximum, we have, as the first necessary
condition,
FIG. 144
~(~) = - 6 sin + 8 cos = 0,
de \dr) '
whence tan = 4/3. This means that is either an acute angle or a third-
quadrant angle. To determine which, we need the second derivative of dz/dr
with respect to 0. This is
( \ = -
d8* \dr)
6 cos & 8 sin 0.
136]
PARTIAL DIFFERENTIATION
223
This second derivative is negative for positive sin 6 and cos and therefore
dz/dr is a maximum if = tan" 1 (4/3) in the first quadrant. Since sin and
cos 6 are negative for in the third quadrant, we have
dr
for the tan" 1 (4/3) in the third quadrant and hence dz/dr is a minimum in that
direction.
If = tan" 1 (4/3), then cos = =b 3/5, sin 6 = 4/5. Hence for the
acute angle dz/dr = 6(3/5) + 8(4/5) = 10,
and for the third-quadrant angle dz/dr is
10. We point out that the numerical
values of dz/dr arc the same in the two di-
rections. This means that the extreme
values of dz/dr are the same except that for
one direction z is increasing and for the
opposite direction z is decreasing.
2. If the temperature T at any point of
the xy plane is given by T = k/(x 2 + ?/ 2 ),
find the rate of change of the tempera-
ture at the point (3, 4) in the direction
such that B = 120.
SOLUTION. Since T = k/(x 2 + ?/ 2 ), we
have
FIG. 145
dT
dr
2kx
r+]^ji--
2ky
sin 6.
At the point (3, 4) for = 120, or cos 6 = - 1/2, sin = V3/2, this becomes
dT 3/c 4kV3
dr
625
625
= - 0.006 k.
PROBLEMS
Find the directional derivative for each function defined below. (Nos. 1-8.)
l.s = l/(z 2 + ?/), at (2, 3) toward (- 1, - 1). Ans. 36/[5(13) 2 ].
2. T = 2/ 2 tan 2 x, at (gr/3, 2), = 5 T/6.
3. Q = logVz 2 - 2 ?/ 2 , at (- 2, 1) toward (- 6, - 2). Ans. 7/5.
4. z = tan-Ky/s), at (1, 1) toward (- 2, - 3).
5. z = tan -1 [(l - z)/2/] -f tan- ! [(l + z)/?/], at (1, 1) toward the origin.
Ans. 6/(5V2).
6. F = log (x 2 + 2/ 2 ) 1/3 , at (a, b) toward the origin.
7. z ~ e~ ax cos ay, at [0, 7r/(4 a)] toward [ ir/(4 a), 0]. Ans. a.
8. 2 = e~r 2 cos x, at (TT, 1) toward the origin.
224 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
Find extreme values for the directional derivatives desired in each of the
following cases. (Nos. 9-18.)
9. z = log [(z 2 + */ 2 )/(z 2 - 2/ 2 )] at (2, - 1). Ans. Max. for S = tan^2.
10. z = e* 2 ' 2 * at (2, 1).
11. z = tan- 1 (y/x) at (5, 1). Ans. Max. for 2 = tan -1 (- 5).
12. T = (a: 2 + t/ 2 ) t&n~ l (y/x) at (1, 1).
13. z = log (z?/ 2 - x 2 y) at (1, - 1). Ans. Max. for 04 = tan'K- 1).
14. z = e x tan" 1 ?/ at (0, 1).
15. z = e- a * sin ay at [0, 7r/(4 a)]. Ans. Min. for = 7 7r/4.
16. z = ?/ 2 tan 2 a; at (w/3, 2 )
17. z = log (z 2 -j- 2/ 2 ) sin (a; + 2/) at (7r/4, Tr/4). Ans. Max. for = ir/4.
18. u = e- sin x -f (l/3)e- 3 " sin 3 z at (ir/3, 0).
19. The electrical potential at a point is given by P = logVx 2 -f y 2 -
What is its rate of change at (0, 4) toward (3, 0)? In what direction from
(0, 4) is its rate of change a maximum? Ans. 1/5, 7r/2.
20. In what direction from (3, 1) does the function e 2x tan" 1 [x/(3 y)~\ have
zero rate of change?
21. In what direction from ( l /2, 1, 2%) is the surface z 4 z 2 y 2
rising most rapidly? Ans, tan -1 2, in third quadrant.
22. Use Problem 19 to find
(a) dP /dr at (2, 3) in the direction of the x axis.
(b) dP/dr at (2, 3) in the direction which makes vr/4 with the y axis.
(c) dP/dr at (2, 3) toward (5, - 1).
(d) dP /dr at (2, 3) to be maximum.
23. Show that the directional derivative of log (x -f- Vz 2 -}- y 2 ) at any
point P toward the origin O is numerically the same as the reciprocal of PO.
24. A portion of a hill may be represented by 4 x 2 + y 2 -f 2 9 0.
(a) Sketch the surface with the z axis directed upward. (6) What grade has
a road through the point (1, 2, 1) with = ?r/4? (c) What direction has a
contour line at (1, 2, 1)? (d) What direction at (1, 2, 1) is the steepest?
(cO What direction has a 2% grade at (1, 2, 1)?
25. On a hill represented by z G 2 x~ y 2 find the direction of the
contour line at (1, 1, 3). Also the direction of steepest grade.
Am. tan-H 2); tan"^!^), in third quadrant.
26. At any point of the surface z 4- r = what direction has no rise or fall;
in what direction is the fall most rapid; what direction has a 3% rise?
137. Maxima and Minima of a Function f(x,y). A function of
two independent variables has a maximum value for the values
x = x\j y = y\ provided that, for all values of Ax and Ai/ that are
sufficiently small numerically, but not both zero, the quantity
137] PARTIAL DIFFERENTIATION 225
f(xi t yi) is greater than f(xi + Az, y\ + Ay). For a minimum,
f(xi, yi) must be less than/(#i + Arc, y\ + Ay).
Geometrically, the function z = /(x, y) represents some surface,
and if z\ = f(xi, y\) is a maximum or minimum value of the
function, the tangent plane to the surface at the corresponding
point PI will be parallel to the xy plane. That is, its equation
will be z Zi = 0. The equation of the tangent plane at the
point Pi is ( 133):
Now, if Pi is a high point or a low point, these two equations must
be identical; hence
< - I---
These conditions are necessary but are not sufficient. Each
derivative should be tested on both sides of the critical point PI.
Complete tests which are sufficient are discussed in more advanced
courses.
PROBLEMS
Find the maximum or minimum values of each of the following functions
of two variables. (Nos. 1-6.)
1. /(, y) * 2 - 3 x + 4 2/ 2 + 4 y + 1. Ans. - 9/4 (min.).
2. 0(r, s) = r 2 + 4 r + 2 ,s> 2 - * + 5.
3. f(z, = 5 + 6 z - 4 z 2 - 3 J 2 . Ans. 11/4 (max.).
4. /(x, 2) = l+6z-9z-2a; 2 -2z 2 .
5. 0(r, s) = r 2 - 3 rs + 3 s 2 -f 4 r - 10 s + 6. Ans. - 10/3 (min.).
6. /(x, y) = 3 z 2 + 2 z?/ + 3 y* - 6 x + 4 y.
Find the high point or the low point of each of the following quadric sur-
faces. (Nos. 7-11.)
7. z = 4 - (x - 2) 2 - (y + 3) 2 . Ans. (2, - 3, 4) (high).
8. z = 2z 2 -3z + 3i/ 2 -62/-fl.
9. 2 = 2z 2 -6z2/ + 5?/ 2 -:r-h32/ + 2.
Arcs. (-2, -3/2, 3/4) (low).
10. z - 3 x 2 + 5 xy 4- 4 2/ 2 - 7 x - 2 y + 6.
11. 2 = z 2 22/ 2 3s-f52/ 1- Ans. Neither max. nor min.*
* The point (3/2, 5/4, - 1/8) satisfies the conditions z x = 0, z y = but the
section made by the plane x = 3/2 is concave upward at this point, while the
section made by the plane y = 5/4 is concave downward.
226 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
12. What should be the dimensions of a rectangular open tank with a fixed
surface area so as to have a maximum capacity?
13. Divide a number N into three parts such that their product shall be the
greatest possible. Ans. N/3, N/3, N/3.
14. What dimensions should an open rectangular tank of given volume
have in order that its inner surface shall be a minimum?
138. Envelopes. The curves given by f(x, y, c) = for all
values of c constitute a one-parameter family of curves. An
illustration of such a family of curves is the different paths of
projectiles fired with the same muzzle velocity in the same vertical
plane but at all angles of elevation from to 180.
Two curves of the family, obtained by assigning to c the values
c and c + Ac, may intersect in one or more points. As Ac
each intersection approaches a corresponding limiting position P
for each value of c. The locus of the points P, if they exist, as c
varies is called the envelope of the one-parameter family of curves.
Since P is the limiting position of the intersection of the curves
f(x t y, c) = and f(x, y, c + Ac) = as Ac 0, its coordinates
satisfy both
(1) f(x,V,c) =0,
and
/ 2 ) j; m /(X ?/, c + Ac) - f(x, y, c) ^ df(x, y, c) = Q
Ac dc
Hence equations (1) and (2) may be considered as parametric
equations of the locus of P.
To obtain the rectangular equation of the envelope, eliminate
the parameter c between these two equations.
If equations (1) and (2) are solved for x and y, the envelope
may be written in the form
(3)
Hence the slope of the envelope is
(4) 2H
( ) dx
138]
PARTIAL DIFFERENTIATION
227
The slope of a curve of the family (1) for a definite value of c is
given by the equation
(5)
df(x, y, c) ^dy ^ Q
dx
dx ' dy
But for any point (x, y) of the family (1)
(6) df(x, y, c) = - dx + dy + - dc = 0.
If we also impose the condition that (x, y) lie on the envelope,
we have, from (2) and (3),
dx =
dy = d\l/>
Therefore (6) becomes
(7)
.
dx ~~ dy d<f>
whence, from (4) and (5), the envelope and any member of the
family of curves have the same slope at their intersection. This
proves that each member of the family of curves is tangent to the
envelope of the family.
EXAMPLE
Find the envelope of the family of lines y = mx + p/m where p is a con-
stant. Show that the envelope is tangent to each line of the family.
SOLUTION. Here the function is
f(x, y, m) = y -mx -
therefore
0,
dm
1 2 = -
7W 2
This last equation gives m =b
Substituting in the first equation and
simplifying, we have the parabola
y = %Vpx, or y z = 4 px,
as the envelope of the family of lines.
Figure 146 shows the envelope and some of the lines of the family.
To show that the envelope is tangent to the lines, we eliminate y between
FIG. 146
228 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. IX
the equation of the envelope and that of the lines. This gives
m z x* + 2 px + = 4 px, or mV - 2 px + -^ = 0.
m 7TI
The left-hand side of this last equation is a perfect square; hence the two
points where any one of the lines meets the parabola have the same abscissa,
or the lines are tangent to the envelope. This method of showing tangency
is suggested rather than finding the slopes of the curves and comparing them
at intersections.
PROBLEMS
Find the envelope of each of the following families of curves, name the
envelope, and show that the tangency condition is satisfied. (Nos. 1-14.)
1. y = m*x + 2m. Arts, xy + 1 = 0.
2. x cos -f- y sin = p, 6 the parameter.
3. 6 x cos m + 4 y sin m = 24. Ana. z 2 /16 + */ 2 /36 = 1.
4. y = mx Va*m* + 6 2 .
5. x 2 + (y - k) 2 = 4. Ana. x = d= 2.
6. y = px -f- 1 + 4/p.
7. y = mx am 2 . Arw. 4 a^/ = z 2 .
8. 2 raz + m 2 ?/ = 5.
9. x 2 + (y - A;) 2 = fc 2 . Ans. y = 0.
10. y = fcz 2 + 1A.
U. y =* w 2 s/2 + 1/cwi. ^/w. 8 c 2 ^ 3 = 27 a?.
.2. ?/ 2 = /; 3 3 px.
,3. (x - fc) 2 -f (y - fc) 2 = 4 fc. Ans. (x - yY = 4(x + T/ + 1)
14. (a; a) 2 -f ?/ 2 = 2 ap p 2 , a the parameter.
15. Find the envelope of the hypotenuse of a right triangle of constant
area c. ' Ans. 2 xy =* c.
16. If the sum of the major and minor axes of a family of ellipses is constant,
what is the envelope of the family?
17. Find the envelope of a family of lines the sum of whose intercepts on
the coordinate axes is a constant k. Ans. (x + y fc) 2 = 4 xy.
18. Find the envelope of the normals of the parabola y 2 = 2 px.
19. Find the envelope of the circles through the origin with centers on
x 2 = 2 y. Ans. x*(y + 1) + y 9 = 0.
20. A line-segment of fixed length moves with its extremities on the coordi-
nate axes. Find the envelope.
CHAPTER X
INTEGRATION
139. Integration. Notation. Up to this point we have been
interested in the rate of change of a given function with respect
to its variables. However, it is desirable to be able to reverse the
process of differentiation in order to find the function whose rate
of change is givfcn. For instance, we may want to know the
distance a moving body has traveled in a given time when its
velocity is known; or to find the equation of a curve when its slope
is given. This inverse of the operation of differentiation is called
integration.
The inverse problem then is : Given the derivative of an unknown
function of one independent variable with respect to that variable,
to find the function. Familiarity with the formulas of differential
calculus should enable us to write down at once many functions
which have given rates of change.
THEOREM. // two functions have the same derivative, they can
differ only by an additive constant.
PROOF. Let/(x) and g(x) be two functions which have the same
derivative and let
*(*) = /(*) -
Then, by hypothesis,
*'(*) = f'(x)-g'(x) =0.
That is, the rate of change of <j>(x) with respect to x is zero for all
values of x\ hence <(#) is a constant.
If the rate of change of a function y with respect to its variable x
is given by /(x), that is, if dy = f(x)dx then the function
y = F(x) + C,
whose derivative is f(x) is called an indefinite integral of f(x).
This relation between f(x) and F(x) + C is denoted by the symbol
(1) ffWdx = F(x) + C,
229
230 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. X
and is read: The integral of f(x)dx is F(x) + C. Hence
d[F(x) + C] =/(x)dz,
and
^ (x}dx = F(x) + C
indicate inverse operations.
The function f(x) is called the integrand. The constant C is
called the constant of integration, and may have any value what-
ever. It is determined in each problem by means of information
in addition to the rate of change of the unknown function.
The expressions t fau n du j fa du t fa du/u may also be written
afu n du, afdu, a f du/u because
1 i, ^ A
d(au} = a ' du >
and d(a log u) = a - rf(log u). Thus a constant factor may be
removed from the integrand and placed before the integral sign. This
operation is usually very desirable, as the simplified integrand is
more easily handled during the process of integration.
From the forms above, we have the formulas:
/C nu n+l
au n du = a I u n du = r - + C, n ^ - 1.
J n + 1
(II) la du = a I du = an + C.
When n = 1, the expression under the integral sign in
formula (I) becomes a du/u and so we have a third integral
formula,
(HI) fa = alogu + C, u > 0.
U u
Since the differential of the sum of a finite number of terms
like au n is the sum of their differentials, we have
(IV)
ff n (u)du
F, (II) +/,(!!) + ... + F B (u) + C,
if each f t (u)du is dFi(u) and if i = 1, 2, n, where n is finite.
139] INTEGRATION 231
EXAMPLES
1. Find /5 dx.
SOLUTION. Formula (II) gives fb dx = 5f dx = 5 x -f C.
2. Find /(x 3 + 2 z 2 - 7) dz.
SOLUTION. Formulas (I), (II), and (IV) give
^ + ?-~ -7x + C.
3. Find /2(z 2 + a 2 ) 2 zcfo.
SOLUTION. Expanding the parenthesis, we have
f 2(x 2 + a 2 ) 2 xdx = 2f(x 5 + 2 a 2 * 3
However, a result is more readily obtained if the student recognizes that
2 x dx d(x 2 -f- a 2 ). In that case the integral is seen to be in the form of
fau n dn where a = 1, u = x' 2 + a 2 , du = 2 x dx and n = 2. This, makes it
possible to write
/( r 2 I n 2\3
(x 2 + a 2 ) 2 2 x ,/* = (X ^ a; - -f
In these two seemingly different results, terms involving x occurring in
the one are identical with those in the other; the results differ at most by
an additive constant. This merely means that the constant of integration C
does not represent the same quantity in both cases.
4. Find fx*<lx/(x - 3).
SOLUTION. When the integrand is an improper algebraic fraction, it should
be reduced to a mixed number before integration. Thus,
= f + ^f + 9* 4- 27 log ( - 3) + C.
Here we use formula (HI) to integrate the fractional part of the integrand.
Evidently, a = 27, u = x 3, and du = dx.
PROBLEMS
Integrate each of the following expressions.
1. f (I - x - x*)dx. Ans. x - x z /2 -
2. y*(10 -It + t z )dt.
3. f (1/2) (< 3 /3 - 3 * 2 + 5 t)dt.
Ans.
232 DIFFERENTIAL AND INTEGRAL CALCULUS [On. X
4. f(l + 2/x* + 3/x*)dx.
5. f(Vgt - VW*)dt. Ans. (2/3)^ 3 / 2 - (3/5)JV'V/ + C.
6.
. 7. f (2Vt + '3\
4n. (4/3)* 3 / 2 -f (9/4)* 4 / 3 -f (16/5)* 5 / 4 + C.
8.
9. /" "LlL^ ffc. Ans. - (4/3)or 3 - log z + C.
J x*
10.
11. (a 2 / 3 - x***)*dx. Ans. a*i*x - (6/5)a 2 / 3 x 6 / 3 + (3/7)a; 7 / 3 + C.
, 13. ___ ^ ^ ns . - X 3 _ 3x 2/ 2 _ 3 3; _ 3 log ( x - 1) + C.
/ 1 x
yxM^ + l
^ x -f- 1
- 15. /*^~~ dx. Am. x 2 -x + (2/3) log (3 x + 2) -f C.
~~
X
16.
/() T z - 14 X ,
I ~.r dx.
J 3x 1
140. The Integral of a Power of a Function. Due to the great
importance of the integration formulas given in the preceding
article, we shall add a group of problems which are somewhat
different from those of that article. There we used the formulas
involving u n , but in each problem u was represented by a single
letter. We now stress the fact that u may represent any function
of a variable and consequently that du is then the differential of that
function and not the differential of the original independent variable/
Failure to notice and understand this causes the beginner in
integration a great amount of trouble. Hence, in applying
formulas (I), (II), (III), and (IV), the student must examine
the form f(x)dx in ff(x)dx for the purpose of separating it into
two parts, one of which shall represent w n , and the other du, of
the formulas. In choosing the function which may represent u,
such natural groupings as quantities under radical signs, or in
140] INTEGRATION 233
parentheses, or denominators of fractions, should be considered.
It is of utmost importance, however, that, having chosen the
group of symbols to represent u n , the remaining part of the expres-
sion under the integral sign shall differ at most by a constant factor
from du.
EXAMPLES
1. Integrate fx 2 (x* -f 3) 1 ' 8 dx.
SOLUTION. Since x* -f- 3 is a radicand, we examine the remaining factors
of the expression to see if they form the differential of x* -f- 3 to within a
constant factor. That is, setting u = x 3 -f 3, we get du = 3 x 2 dx. Hence
the remaining factors x 2 dx are equal to du/3. Now substituting these values
for (a; 3 + 3) 1/3 and x 2 dx,we may write the original integral fx 2 (x* -f- 3) 1/3 dx
in the form yV /3 -du/3, or (1/3) yV /3 -du. Integrating, we have
fx 2 (x*
? /
4/3
= (o+3)</ 3 + C.
If the student can see what constant factor is needed to change the given
integral into the form of a formula, he can multiply and divide by this constant
and a change of variable is not necessary. Thus, we may write
Cx 2 (x 3 + 3) 3 dx = f (x* + 3) 1 / 3 (3 x 2 dx)
j 3 j
(z 3 -f 3) 4 / 3 , n
directly without explicitly using the variable u. However, even then u is used
implicitly.
sin0
~ -n- i /" sin ,
2. Find I r -de.
J 1 -f- cos 6
SOLUTION. Any fractional integrand should make the student recall
the expression du'/u. Hence we shall try setting u = 1 + cos 0, whence
du = sin d0, and substitution gives
/
= - log (1 -f cos 0) + C.
Without the explicit use of w, this is readily written
sin ^ Cd(l -h cos 0)
~ -
/
i
- log
3. Find y % sec 4 d0.
SOLUTION. An integrand which is trigonometric should be examined for a
factor which is the derivative of some one of the trigonometric functions.
234 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. X
If such a factor is present and the remaining factors of the integrand can be
expressed in terms of the trigonometric function whose derivative is present,
the integral can readily be put into the form of one of the formulas.
In this example, sec 2 6 do is the differential of tan 0; hence we shall express
the remainder of the integrand in terms of u = tan 6. This is readily possible
as the remaining factor is sec 2 0, which is equal to 1 -f- tan 2 6. Substitution
gives
3
f sec 4 dO = f (1 + w 2 ) du = u -f ^ + C.
tan 3
A
tan
C.
PROBLEMS
Integrate each of the following expressions by means of suitable substitutions
for u and du.
2.
3. fx*Vx* - 3 dx.
4. fxVx* 1 dx.
'* f
x+2
x 2 + 4 x
cos0
f r
J 1 + sin
^7 C
J
dx.
de.
2 tan - 3
8. J sin cos d0.
^9. /sec 2 tan d0.
10. y*cos0 (1 - sin 2 0)d0.
/ x (/x
IL ^ (x - I) 3 / 2
r , sec 2 30
^ Q ta
13. J x ctn x 2 esc x 2 dx.
14. *
15.
. (2/3) (1
C.
L b (a -f bu*-
(1/4) (ic 8 -3) 4 / 3 + C.
Ans. (1/2) log (x 2 + 4 a;) + C.
. (1/2) log (2 tan - 3) + C.
An.
. - (1/2) esc Z 1 + 6.
- (1/10) (3-2 x>y"> + C.
INTEGRATION
235
19. j cos 0V I sin e d$.
sin 2 x
20-
(3 -4 cos 2s) 1 / 2
dr.
Arcs. - (2/3) V(l - sin I?) 3 + C.
1 Jo-} (^~u cert an)
141. Integrals of Exponential and Trigonometric Functions.
Additional integration formulas are obtained by the inverse of
certain known formulas of differentiation. Thus
(V)
(VI)
a u du =
J u log a
fe u du = e u + C.
C.'
In these formulas we must realize that the differential factor du
is the differential of u which represents any function of one variable
and occurs as the exponent only of a or e.
EXAMPLES
1. Evaluate f% xe**~ l dx.
SOLUTION. The u of formulas (V) and (VI) appears as the exponent of
a or e; hence we set u ~ x 2 1. Then du = 2 x dx or x dx = du/2. Substi-
tution gives
/3 xe**-> dx
C.
2. Evaluate /8 1 -* dx.
SOLUTION. Here we must set u = 1 x f whence du = dx and using
formula (V), we have
logS
3. Evaluate /^Tll^
,rf 1
SOLUTION. Since the derivative of an exponential function contains the
same function, we are able to use the formula for fdu/u only if the numerator
has the same exponential forms as the denominator. For that reason exponen-
tial fractions as integrands are not treated in the same manner as algebraic
236 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. X
fractions. We proceed as follows. Since the numerator of the integrand is
not the derivative of the denominator, we start to reduce the fraction to a
mixed expression. To do this, write the integrand in the form
14-3 e znx
- 1 4- e 2 "* '
and divide the denominator into the numerator once so that we have
1 4- 3 e** = 4 e**
- 1 4- e 2ox ~ 1 + e 20 * '
Then
f*&-\ dx " /"("" * + e e -*l) f k
= - x 4- 1 log (c 2 - - 1) 4-C,
because the numerator is 2/a times the derivative of the denominator of the
fractional part of the integrand. This part may be changed into terms of u
where u is set equal to the denominator. It is well actually to make such
substitutions until the work becomes very familiar.
The two things to notice carefully in this solution are: first, we arrange
the members of the fraction in ascending powers of e; and second, we do not
continue the division after the remainder becomes the same, except for a constant
factor, as the derivative of the denominator.
From the derivatives of the trigonometric functions we have
the inverse formulas
(VII) / sin u du = cos u + C.
(VIII) I cos u du = sin u + C.
(IK) /sec 2 u du = tan u + C.
r
(X) : esc 2 u du = - ctn u + C.
(XI) / sec u tan u du = sec u + C.
(XII) / esc u ctn u du = esc u + C.
In these formulas the student must observe that u represents a
function of some variable, say x or 6, and du is the differential of
that function.
141] INTEGRATION 237
EXAMPLES
1. Find /cos (1-3 x)dx.
SOLUTION. Since 1 3 x replaces u in formula (VIII), we set u 1 3 x.
Then du 3 dx or dx = dw/3. Substituting, we get
cos (13 x)dx = y (cos u) (
cos
Therefore
y cos (1-3 x)c/x = - j- sin (1 - 3 x) -f C.
2. Find fx esc (2 x 2 - 3) ctn (2 x 2 - 3) dx.
SOLUTION. Using formula (XII), we set u = 2 x 2 3. Then
du = 4 x dx, or x dx
Substituting, we have
f x esc (2 x 2 - 3) ctn (2 x 2 - 3) dx = j y esc u ctn
= - esc (2 x 2 - 3) + C.
PROBLEMS
Find the function represented by each of the following integrals.
1. f cos (2-3 x)dx. Arts, - (1/3) sin (2 - 3 x) + C.
2. y sin (3-2 x)dc.
3. y (2 -f sec 2 0)rf0. Arts. 2 + tan + C.
4. y esc 20 ctn 2 0^0.
5. y 4 sin cos d0. Arcs. cos 2 -f C.
6. y x sec 2 z 2 tan 2 a; 2 dx. ,
7. y Vl 4- cos Oc/3) (te. ^ns. 6\/2 sin (/6) + C.
8. 3 sin 3 x cos 3 x dx.
. (1/2) log ( 4- 3) - (1/2) cos 2 a; -f C.
10. f[e* + x^cos (2 x 2 - 5)]dx.
/x*^ 1
(x e* e~ x -f e* sin e*)rfx. ^Ins. - e x + e~* cos e* + C.
12. [2e~ 2 * + (2x)-]rfx.
238 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. X
Arts. - (1/4) ctn2z -f C.
e
^ 4 *
. (1/6) (e** -f
afev? ^
13. fdx/(l - cos 4 x).
14. f (c~* x 4- ze* 2 - z 2 e l -*Vz.
15. y (c 2 * 4- l)*e**dx.
16. j (e az 4- x ae 4- 2 a ex }dx.
17. y (x 2 4- 2^ 4- sec 2 2 x)c/x.
Ans. x 3 /3 4- 2* /log 2 4- (1/2) tan 2 x 4- C.
18. /6 )X (sin x 4- cos x) dx/Vc x sin a; 3. 2
on / .
4rU. f , - Q \ o
j r
. (1/2) log (1 + e 2 *) + C.
21. fdx/(e* x + 1). Ans. x - (1/4) log (e 4 * 4- 1) 4- C.
22. fdx/(2 - e x ). ^ ^(/ e ^"^
23. y ( 3z - l)^/(e 3 * 4- 1). ^/w. - x 4- (2/3) log (e 3 * 4- 1) 4- C.
24. f dy/(?> a").
25. /"V2(l - cos3o;)dx. Ans. - (4/3) cos (3 x/2) 4- C.
C .
/csc 2 csc ctn
26. I (w.
J ctn csc -
27. yYlO 8 * 4- 3 z l 4- ~
28. fxe* z *- z cos 3 e 3 * 2 - 2 dz.
CHAPTER XI
METHODS OF INTEGRATION DEFINITE
INTEGRALS
Although we have many formulas of integration from the
inverse of known differentiation formulas, there frequently occur
integrals which cannot be identified as the differentials of known
functions. For that reason we need methods whereby such inte-
grands can be changed into forms more readily identified. A few
such methods will now be given.
142. Additional Trigonometric Integrals. The following four
formulas are readily derived as shown below.
/VTTTN d j Csmudu
(XIII) I tan u du = I
J J cosu
= log (cos u) + C, or log (sec u) + C.
(XIV) Cctnudu = f C *fau U = lQ g ( sin u) + c - ~ -
/vxr\ C j r sec u(sec u + tan u) du
(XV) I sec u du = I p-
J J sec u + tan u
= log (sec u + tan u) + C.
esc u(csc u ctn u) du
(XVI) f csc u du = C
esc u ctn u
log (esc u ctn u) + C.
If an integrand is of the form
sin m u cos n u,
where m or n is a positive odd integer, we can reduce it to powers
of either sin u multiplied by d(sin u) or cos u multiplied by d(cos u).
EXAMPLE
1. Find /cos 2 sin 3 d0.
SOLUTION. Since the exponent of sin is a positive odd integer, we factor
out sin dB. This leaves cos 2 sin 2 0, which can be changed into terms involv-
ing only powers of the cos as follows.
cos 2 sin 2 = cos 2 0(1 cos 2 0) = cos 2 cos 4 0.
239
240 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
Then let u = cos 0, whence du - sin de. Therefore, substituting, we have
/cos 2 e sin 3 e de = - f (K? - u*) du = - \ -f ~ + C
J o o
- 3 cos 5 0-5 cos 8 , r
15 "*"
If m and n are both positive even integers, the double angle formu-
las can be used to reduce the degree, and this operation must be
repeated until an odd power appears. Then treat as explained
above when the exponent is a positive odd integer.
EXAMPLE
2. Find /cos 2 2 de.
SOLUTION. Replace cos 2 2 e by its equal (1 -f cos 4 0)/2. This makes
/"cos 2 20d0 = ~Al-r-cos40)d0
/ <&'
sin 40
2" 1 " 8 H "*
In the second term of the new integrand we let u = 40 and then d0 = <w/4.
Hence the final result. ?,
For integrands of the forms
tan n u or ctn n u,
where n is an odd positive integer, a reduction to sines and cosines
is desirable. This makes the integrand come under the type
explained in Example 1 above.
However, if n is an even positive integer, factor out tan 2 ^ or
ctn 2 u and replace by sec 2 u 1 or esc 2 u 1 and repeat the
operation until all terms are of the form tan m u sec 2 u du
or ctn w u esc 2 u du.
EXAMPLE
3. Find /ctn 4 d0.
SOLUTION. The exponent of ctn is an even integer and hence we factor
out ctn 2 and replace it by esc 2 1. This gives
f ctn 4 de = f (ctn 2 esc 2 - ctn 0) de.
Replacing the second term of this new integrand by (esc 1 1), we have
f ctn 4 de = f (ctn 2 esc 2 - esc 2 B -f 1) de.
142] FORMAL INTEGRATION 241
Now let u = ctn 0, whence du esc 2 d6. This makes
fctn*6dO = f (- u* + 1) du + fd8
. . , _ ctn 8 .
= -f- ctn h C.
o
Many of the more usual trigonometric integrands can be trans-
formed into one of the types discussed in this article.
PROBLEMS
Evaluate each of the following integrals.
1. f tan 2 OdB. Ana. - (1/2) log cos 2 + C.
2. /ctn (0/2)d0. ^-^ lf*L&-t-4rK*-0} +
3. f sec 3 do. Ans. (1/3) log (sec 3 + tan 3 0) + C.
4. /csc (1 - 20)d0.
5. /sin 2 (2/3) x dx. Ans. (l/2)[x - (3/4) sin (4 x/3)] + C.
6. /cos 2 (1-3 x)dx.
7. /tan 2 (2 x - l)cfo. Ans. (1/2) tan (2 z - 1) - x + C.
8. /ctn 2 (3 - 20)d0.
* 9. /sin 3 (2 x + 3)dx.
Ans. (1/6) cos 8 (2 x -f 3) - (1/2) cos (2 x + 3) + C.
10. /cos 3 (2s - 3)<&.
11. fx tan 8 3 z 2 dx. Ans. (l/6)[(l/2) tan 2 3 x 2 - log sec 3 x 2 ] + C.
12. /x 2 ctn 3 (1 - x 3 )dx.
13. /sin 2 2 cos 2 20d0. Ans. (1/64) (80- sin 8 0) -f C.
14. f8m*20coa*20d0.
15. /tan 4 <#. , Ans. (1/3) tan 3 - tan -f- -f C.
16. /sec 4 (0/2)d0.
17. /cos x cos 2 3 efo. Ans. sin x - (2/3) sin 8 x + (7.
18.
242 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
19. f tan 3 2 sec 4 28dB. Ans. (1/24) (3 tan 4 2 + 2 tan 6 2 (?) + C.
20. /sin 4 (3-2 z)cfo.
21. /sin 5 (z/3Mz.
Ans. - 3 cos (z/3) + 2 cos 3 (z/3) - (3/5) cos 5 (s/3) + C.
22. fdx/Vl + cos 2 s.
23. fd0/Vl - cos 40. Ans. (1/2 \/2) log (esc 2 - ctn 2 0) -f C.
24.
25. / (sec 2 - cos 2 0) 2 ^0.
A/w. (1/2) tan 2 - 3 0/2 + (1/8) sin 4 -f C.
26. C V(l - cos 3l) 3 d0.
y ^
27. /sin 3 cos 3 / 2 0(/0. Ans. - (2/5) cos fi / 2 -f- (2/9) cos 9 / 2 + C.
28. A;os 3 2 sin~ 2 / 3 2 </0.
29. /sin 4 (2 x/3) cos 3 (2
s. (3/2) [(1/5) sin 5 (2 z/3) - (1/7) sin 7 (2 s/3)] -f- C.
30. /sin 2 2 x cos 4 2 x dx.
31. / Vsin 2 x cos 5 2 z dx.
Ans. (1/3) sin 3 ^ 2 2 x - (2/7) sin 7 / 2 2 x + (1/11) sin 11 / 2 2 x + C.
32. /cos 3 (30/2)^0/^8111 (30/2).
143. Substitutions for Radicands. If the integrand is rational
except for a radical of the form
or
the substitution of u for this radical will make the new integrand
rational. Integral powers of the radicals mentioned above permit
the same substitution.
i 143] FORMAL INTEGRATION 243
If the radical is of the form
this same substitution is satisfactory, if only powers of x m are
present, together with x m ~ l dx.
EXAMPLES
1. Evaluate fx\/Z x - 4 dx.
SOLUTION. Let u = v/2 x 4, then u 2 = 2 x 4 and dx = u du. Also
x = (u 2 + 4)/2. Substituting, we have
y*x\/2x -4 dx = i /" (i/ 2 -f
In terms of the original variable this result is
(3 x 2 - 2 x - 8) (2 x - 4) 1 / 2 + C.
lo
2. Evaluate
SOLUTION. Let u = (3 + x) 1/3 , then x = ?^ 3 3 and dx = 3 w 2 c?w. Also
+ a;2 = i 4. ( M s _ 3)2 = u 6 - 6 w 3 -f 10. Therefore
= ~ (5 x 2 - 18 x + 101) (3 -f x) 2 ' 3 + C.
3. Evaluate f . X ^ /2
SOLUTION. Let tt = (8 + x 2 ) 1 ' 2 , then w 2 = 8 -f x 2 and udu = xdx.
Also x 2 = u 2 8. Separating x 3 dx into x 2 and x dx and substituting, we get
r &&* r(u* -$}udu r
J (8+^)3/2 = 7 ^ -y (1 -8u 2 )rfw
, 8 , 16 -h x 2 , ^
244 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
PROBLEMS
Find the values of each of the following integrals. Sf
1. fxVx* + 4 dx. Am. (1/3) V(x* + 4) 3 + C.
2. fxdx/(\ + 4x 2 ) 3 / 2 .
3. /"x 3 (l + 2 x 2 )'/ 2 dx. Ans. (1/30) (3 x 2 - 1)(1 + 2 x 2 ) 3 / 2 -f C.
4. f xV2 x 2 - 1 dx.
5. VVl -e*dx. Aws. - (2/3) (1 - e*) 3 / 2 + C.
6. (4 -f
7. yV (1 -f 2 x 6 ) 1 / 3 rfx. Ans. (1/75) (3 z 6 - 1)(1 + 2 x 6 ) 3 / 2 + C.
8. J xVx -f 4 dx.
9. fx*V2 -3xdx.
Ans. - (2/27) (2 - 3 x) 3 / 2 [4/3 - (4/5) (2 - 3 x) + (1/7) (2-3 x) 2 ] + C.
10. x rfx/^4 - 6 x 2 .
11. / xV3 x + ldx. Ans. (1/28) (4 x - 1)(3 x + I) 4 / 3 + C.
12. / x 6 V 5 - 2 x 3 dx.
ax
5 H- ax*
a sin ax ] dx. Ans. v 5 4- as 2 + cos ax -f- C'.
15.
Ans. - (3/7) (2 - xj^^x 2 + 3 + 18) + C.
16. fx*dx/Vl ~x 2 .
17. /x^ ax/(2 - 3 x 2 ) 8 /*. An*. - (2/45) (2 - 3 ) l /*(8 + 3 x 8 ) + C.
19. y (1 + 2 x 3 ) cte/^1 - 2 x - z 4 .
~ 2 * - s<)/ 4 + C.
144] FORMAL INTEGRATION 245
20. f x* XVQ x 2 - 4.
Arts. (1/3) (z 2 - 25) 1 > 2 (z* + 50) -f- C.
/x 2 - 25
22. f x* dx/(4 x -f 1)</ 2 .
144. Integration by Trigonometric Substitutions. If the inte-
grand contains a radical of the form
(1) V=t a 2 dz u*
the substitution of a trigonometric function for u will rationalize
this radical. If the radical is
(a) \/a 2 u 2 , set u = a sin 9 ;
(b) vV + u 2 , set u = a tan 6;
(c) Vw 2 a 2 , set u = a &ec 6.
These substitutions make the radicands become some constant
multiplied by 1 sin 2 0, 1 + tan 2 0, and sec 2 1, respectively.
As each of these functions represents the square of a monomial,
the expressions are rationalized.
If the radical is of the form
(2) Va + bx dz ex 2 ,
we can factor out the constant c and complete the square so as to
get
\/C V=fc fc db (x =t rf) 2
The substitution of u for (x d) changes this radical to form (1).
Or directly, if the radical is
(a) Vk - (x zb d) 2 , set (x d) = Vk sin 6;
(b) Vft+ (xitd) 2 , set (xrtd)
(c) V~ *+ (xdbd) 2 , set (x =fc d) = Vft sec 6.
246 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
EXAMPLES
1. Find r dX
x 2 + a 2
SOLUTION. The combination of two positive signs suggests that we set
x a tan 0. Then Vx 2 -f- a 2 becomes a sec and dx = a sec 2 dd. There-
fore
/, = C ( L^. = r sec 0d0 = i c) g ( S cc -f tan 0) -f C.
Vx 2 -f 2 / a sec /
But tan = x/a and hence sec = Vx 2 -f a' 2 /a. These values in the result
above give
r~-J^=- = log X + Vx ' 2 + a2 + C = log (x + V^T^" 2 ) -f A',
J Vx 2 4- a 2 a
where K = C log a.
2. Evaluate / ?=======.
/ V9 r _ r2
SOLUTION. Completing the square under the radical, we have
V2x - x 2 = Vl - (x - I)*.
This suggests that we set x I = sin 0. Then V2 .c x 1 = cos and
dx = cos d0. Substituting, we have
= sin- (*-!)+ C.
3. Evaluate / v/2 -f 4 x - 3 x 2 dz.
SOLUTION. Factor out 3 and complete the square, then V2 + 4 x 3 x 2
becomes V5V2/3 + (4/3)x x 2 , which may be written in the form
V3V(10/9) - (x - 2/3) 2 .
Now set x - 2/3 = (VlO/3) sin 0, whence dx = (VlO/3) cos Odd and
the new radical becomes ( VlO/3) cos 6. Therefore
y V2 -f 4 x 3 x 2 f/x = V3 J ~ o~ cos ^ cos do = - J* cos 2 6 dO
10\/3 r\ /t . o^jn 10V3/0 sin20\ , ^
+C
C.
144] FORMAL INTEGRATION 247
These trigonometric substitutions can be used to change quad-
ratic forms into more suitable expressions even if there is no
radical involved, as is illustrated in the following examples.
4. Evaluate
SOLUTION. Write 3 x 2 -f 5 as 3(.r 2 + 5/3) and set x = Vo73 tan 0.
Then 3(x 2 -f 5/3) = 5 sec 2 6 and dx = v/5/3 sec 2 dO. Substituting, we get
dx r V573 sec 2 dO _ 1 /5 r tan" 1
~
r dx = r 3 sec 2 dO _ 1 5 r tan" 1 xt$/)
J*& + l~J 5sec 2 ~5\3^ ~ V
5sec
X
5. Evaluate f- J X _
SOLUTION. Write 2 x 2 - 3 as 2(x 2 - 3/2,) andjhen let x = x/3/2 sec 0.
This makes 2 .e 2 3 = 3 tan 2 0, and cfo = V3/2 sec tan </0. Whence
/dx r \/3/2 sec tan 0d0 1 /7T / . ;/1
4 r r - .;: = / --- ' . .. n -- = 77 V b / CSC dO
2 x 2 3 ./ 3 tan 2 J
= \/6 log (esc - ctn 0) -f (7
1 , f . xV-2 - V3 , .,
= o V( ""-V 2 -^rl + c -
This result can be rationalized in so far as x is concerned by writing it in the
form
-og
12
1 K,
= T - Vb log
-f v3
The first six problems below may be used as formulas of inte-
gration if desired.
PROBLEMS
Show that each of the following formulas (Nos. 1-6) is correct.
1. f - = sin- 1 h C, when a > \u\.
J Va 2 u' 2 a
du
!. / v 2 " ^ = log (u + Vu* - a 2 ) + C, when ti > |a|;
= log (- u - Vu~ - a 2 ) -h C, when u < - \a\.
248 DIFFERENTIAL AND INTEGRAL CALCULUS [Ca. XI
3 ' = log (u + v/=rT72) + c "
4. f du = l
J u 2 + a 2 a
5.
+ a 2 a a
a -
log + C ' When
Evaluate each of the following integrals.
7. fdx/V2 - x 2 . Ans. sin' 1 (x/\/2) + C.
8. <lx/(x* + 4).
9, fdx/Vlx* -f 7. Arcs. (1/2) log (2s + V4 x a -f 7) -f- C.
10. <L
11. fzdx/VG -4x 2 . Anslsin- 1 (2 x/V6) -f C.
12. fdy/(y*- 1).
13. fsde/oVze* ~ 3. Ans. V3 scc~ l ((?V273) + C.
14. fz<to/(0* -9).
15. fdit/Vi +28 -2s 2 . Arw. (l/\/2) sin' 1 [(2 s - 1)/V3] + C.
16. ftlx/V2x* 4- 6 a: + 7.
17. fdx/Vlx -f 3x 2 .
*. (1/V3) log (3 a? + 2 + V9 x 2 -f 12 x) -f- C.
18. fdx/(Zx* + 5x + 4).
19. fdx/(3 x* - 2 x). Arw. (1/2) log [(3 x - 2)/3 x] + C.
20. fdx/V^x - 6 - x 2 .
21. fdx/Vx~ -h 6x -f 13. Ans. log (x -f- 3 + Vx 2 + 6 x -f 13) + C.
144] FORMAL INTEGRATION 249
22. fdx/^S - 4x 2 + 4z.
23. fdx/V4x + z 2 . Ans. log (x + 2 + Vx* + 4 x) -f C.
24.
25. fde/0*Vd* - 4. /Ins. Vfl* _ 4/4 + c.
26. fdx/xVa* + x*.
27. ^Vl - x 2 rfs. Arcs. (l/8)[sin- 1 a? - zVl - a: 2 (1-2 z 2 )] -f C.
28. fx*dx/(3 -xtyi*.
If the numerator of a fraction with a quadratic denominator
involves the form ax + 6, we suggest the following operations.
EXAMPLES
* i-. i x /" 2 z 3 .
1. Evaluate / 3 ^ _ 2x+2 fc.
SOLUTION. Factor out the 3 from the denominator and complete the
square, so that
f 2x-Z _ 1 C
J 3z 2 -2x+2 X ~3J (x 2 -
(2 z - 3) r
= 1 f (2a? -
3 J (x - 1/c
2 x/3 + 1/9) -f 2/3 - 1/9
-3)dx
1/3) 2 + 5/9
Then, as previously, set x 1/3 = (\/5/3) tan 0. This substitution gives
x = 1/3 + (V5/3) tan 6 and dx = (V5/3) sec 2 0d0. Rewriting the integral
above, we now have
-^ sec 2 Ode
o
/(2s- 3)dx = 1 I
3x 2 -2x-f2 31
1 . 9 x 2 - 6 z + 6 7V5 . , 3 x - 1
lo - an " 1 -
1 3
where K = C + o 18 c '
5 O
250 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
2. Evaluate f _**+$**=.
J V5 + 3 x - 4 x 2
SOLUTION. Factoring out 4 and completing the square in the denominator,
we have
/_J?J!L : v dx ~ i C-^ ( A X IL?ML
V5~+3z - 4^2 2J V89/64 - (x -
V 89/64 - (x - 3/8) 2
Now sot x - 3/8 = (A/89/8) sin 0, then x = 3/8 + (V89/8) sin and
dx = (V89/8) cos dO. By substitution,
/O J
/4 + 4 89
v/89
~8~ (
COS (10
8 5 0--~cos* + C.
Then since = sin^CCSx - 3)/V89] and cos = Vl - (8 x - 3) 2 /89,
this becomes in terms of x
/
(2x + 3}ftx 15 . . 8 x - 3 1 /r-ro - J~l . n
~-=^=^==== = - sin' 1 -^^ -- - V5 + 3 x - 4 x 2 + C.
V5 + 3 x - 4 x 2 8 V89 ^
PROBLEMS
Evaluate each of the following integrals.
1. f (x - l)(tx/V4 - x' 2 . Ans. - V4 - x 2 - sin" 1 (x/2) + C.
2. (3x - l)dx/Vx 2 + 4.
3. /G r )X - 2)r/x/(3x 2 - 4).
^4ns. (5/6) log (3 x 2 - 4) - (1/2 V3) log [(xV3 - 2)/(z\/3 4- 2)] + C.
4. f(2x - 3)ete/(5a: 2 + 7).
5. "(3
ylns. (3/16) (7 sin- 1 [(8 x - 3)/3] + (4/3) V3x - 4z 2 ) + C.
6 f(3 - 2 t)dt/(4 t* - 2 t + 3).
145] FORMAL INTEGRATION 251
7. f (x + l)cfa/V8 + 4 x + x 2 .
s. Vx2 -f 4 3 + 8 - log (x + 2 + Vx 2 + 4x4-8) + C.
8. * (2 5 + 3)<fe/\/15 + 10s - 5*.
9. (3 x - 2)dx/ V5 x 2 - 10 a; - 15.
Arcs. (1/VS) log (x - 1 + Vx 2 - 2 x - 3) + (3/V5)Vz 2 - 2 x - 3 + C.
10. (4 - 3)</0/ Vl + 3~0~- 3 2 .
11. (5 x - 2)dx/Vs -
4ns. (1/4) sin- 1 [(2x - l)/3] - (5/2) V^ - x 2 + x + C.
12. y (4 + 3)cto/V-2 -30 + 40 2 .
145. Integration by Parts. The formula for the differential of
the product u*v gives a very useful integration formula. Since
d(u'v) = U'dv + Z;C/M,
we get, by integrating each term,
I d(u'v) = u-v I U'dv + / y*dw.
This is used most frequently when written in the form
(XVII) Cu dv = uv - jv du.
This formula allows us to replace a difficult integral, represented
in the formula by f u dv, by uv fv du, where f v du may be
more readily evaluated. Obviously, the usefulness of the formula
depends upon the proper choice of u and dv. Since we need both
du and v , the function chosen to represent dv should be one which is
readily integrated. Also, it is advisable to choose as the function u
that part of the integrand which is simplified by differentiation.
Formula (XVII) is specially useful when the integrand is com-
posed of the product of two types of functions; that is, the product
of an algebraic and a trigonometric, a logarithmic and an exponen-
tial, an inverse trigonometric and an algebraic function, or other
such combinations. Such integrals lend themselves to integration
by parts f as the application of this formula is called.
252 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
Some general suggestions can be made which will usually help
the student to satisfactory choices for u and dv.
(a) An integrand composed of the product of a positive integral
power of some variable x and either a trigonometric or an exponential
function of the variable will generally be simplified by letting u repre-
sent the power of x and dv the remainder of the expression, provided
this remainder can be readily integrated. This choice is made
because the differential of a power of a variable is of lower degree
than the original, while the differentials of trigonometric or
exponential functions are no simpler than the original functions.
It may be necessary to apply the formula several times before the
form fv du becomes a simple integral.
(6) An integrand which involves a logarithmic or inverse trigono-
metric function is simplified by letting u represent such a function.
This choice removes such functions since the differential of log u
or of any inverse trigonometric function is an algebraic function.
(c) If the integrand is the product of an exponential and a trigono-
metric function, such as e and either sin bx or cos bx, the formula
must be applied twice and the final relation solved for the desired
integral. In such repeated applications, set u equal to either of
the functions for the first application, and in the second set u equal
to the function appearing for the derivative of u in the first applica-
tion.
EXAMPLES
1. Integrate fxe^dx.
SOLUTION. The integrand is the product of a power of x and the
exponential e zx . Therefore we set u = x and dv e* x dx. Whence du = dx
and v = (l/2)e 2 *. Substituting in formula (XVH), we get
/I 1 /*
xe**dx * s xe ** ~" o J e
2. Evaluate fx sin" 1 x dx.
SOLUTION. Since sin~ l x is present, we let u = sin" 1 x and dv = x dx.
Then du = cfo/Vl - s 2 and v = z 2 /2. These give
fxsm-ixdx = i^sin-^ - \f-^=>
J 2 2*/ Vl x 1
145] FORMAL INTEGRATION 253
The second integral is similar to those we have solved by means of a trigo-
nometric substitution. Hence we let x sin 0, so that dx ~ cos do and
l x 2 cos 0. Therefore the second integral / . becomes
J v 1 x 2
/s
.
v 1
sin 2 cos d0
_ sin 29 _
~2 4~ +C>
But = sin" 1 a: and sin 2 = 2 sin cos 2 xVi x 2 . Therefore
/X*dx 1 . , I n. - r . ~
--=== = s sin" 1 x - - x VI - x* -f C,
v 1 x 2 ^ -^
which makes
/\ 11 _
x sin" 1 x dx = - x 2 sin" 1 x -. sin" 1 x -f 7 x v 1 x 2 -f- C.
J 44
3. Find /e 31 cos 2 x dx.
SOLUTION. This comes under (c) above and we let u e 3z or cos 2 x
as differentiation does not simplify either function. If u = c 3x , we may
write dv = cos 2 x dx, whence du, = 3 e 3 *cte and t; = (1/2) sin 2 x. These give
/I 3 /
e 3 * cos 2 x dx = - e 3 * sin 2 3 - - / e 3 * sin 2 cb.
4 6 J
This second integral is no simpler than the original and so we repeat the
operation by lotting u = c 3 * and dv = sin 2 x dx. Then du = 3 e 3 * c?x and
v (1/2) cos 2 x. These in the second integral of (1) give
/ e* x cos 2 x dx = - c s * sin 2 x - - c 3z cos 2 x + ^ J e3x cos 2 x dx
= s ^ 3x sin 2 x + j e 31 cos 2 x ^ y* e 3x cos 2 x dx.
Transposing the last integral to the left and dividing by its resulting coefficient,
13/4, we have
/V* cos 2 x dx = ~ (2 sin 2 x -f 3 cos 2 x) -f C.
/ lo
4. Evaluate y* Va 2 -f x 2 dx.
SOLUTION. This is not a mixture of two types of functions; however, it is
readily integrated by parts.
254 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XI
Thus, let u = Va 2 ~f a: 2 and cto = cfo. Whence dw = xdx/Va 2 -f x 2
and y = #. Therefore
a 2 -f x 2 dx = xa' 2 + z 2 -
Transposing and dividing by 2, we get
5. Evaluate y*seo 3 do.
SOLUTION. Tins is anotlior common integral which may be integrated by
parts.
Set u = sec 6 and dv = sec 2 do. Then du = sec tan do and t; = tan 9.
Therefore
/* sec 3 r/0 = sec tan J s<;c tan 2 d0
= sec tan - ^ sec (sec 2 - 1) dO
= sec tan y sec 3 J0 -f ^ soc e/0
= sec tan sec 3 </0 4- log (sec -f tan 0).
Whence
/"sec* (10 = - [sce. tan -f log (sec + tan 0)] -f C.
J z
In the last three examples the constant of integration has not been added
until all other operations have been completed. This is done merely to
prevent the necessity of changing the form of the constant several times.
PROBLEMS
Evaluate each of the following integrals.
1. y tan~ l z dx. Ans. x tan" 1 x - (1/2) log (1 + x*) + C.
2. y sin" 1 2 x dx.
3* Jx sin x dx. Ans. x cos x H- sin x -f- C.
146] FORMAL INTEGRATION 255
4. J x cos"" 1 x dx.
5. f(x + 2)e*+*dx. Ans. e*+* (x + 1) -f C.
6. J log s ds.
7. f log t dt/t*. Ans. - (1/2 * 2 ) log < - l/(4 * 2 ) -f C.
8. fu^du.
9. y*x 2 log 2 x dx. 4na. (1/9) z 8 (3 log 2 x - 1) + C.
10. /V sin 2 rf0.
11. f x* sin" 1 2xdx.
4ns. (l/3)x sin- 1 2 x + (1/36) (1 -f 2x 2 )Vl -4x 2 -f o\
12. fe 2t sintdt.
13. y e 3j: cos 3 x rfx. Ans. (l/6)c 3a! (sin 3 x -f cos 3 x) + C.
14. fV^o*- 12 e?0.
15. fV4x 2 + l dx.
Ans. (l/2)[>V4.c 2 + 1 -f (1/2) log (2x + V4 a; 2 + 1)] -f C.
16. f 0* co^ 3 dO.
17. y x sec- 1 3 x dx. Ans. (l/2)x 2 sec' 1 3 x - (1/18) V9x 2 - 1 + C.
18. sec 3 3 dO.
19. d0/ (sin 8 20).
An5. (l/4)[- esc 2 ctn 2 -f- log (esc 2 - ctn 2 0)] + C.
20. yV (logx)*dx.
21. y*x 3 Va 2 -x a dx. Ans. -[5x 8 (a* - Z 2 ) 3/2 + 2 (a 2 - z 2 ) 6 > 2 ]/15 -f C.
22. fxe~*dx/(\ -x) 2 .
146. Integration of Rational Fractions. An improper rational
fraction should be reduced to a mixed number before integration.
The proper fraction resulting from division may be separated into
partial fractions whose denominators are the factors of the original
denominator.
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
We shall explain the method of partial fractions for three differ-
ent types by means of examples.
(a) When the factors of the denominator are linear and not repeated.
EXAMPLE
1. Evaluate f \ ** 7 \ X \ * dx.
J x 3 -f- 3 x 1 4 x
SOLUTION 1. This is a proper fraction and the factors of the denominator
are x, x + 4, and x 1. Therefore, we assume that
5x* - 7 x + 8 _ A B C
^ ' ,v,3 I J ~.2 A -^ v- ' n* J_ A ' .
+ 3 z 2 - 4 x x + 4 ^ x - I 1
where A, B, and C are unknown constants. Clearing (1) of fractions and
collecting the coefficients of the powers of x on the right, we have
(2) 5 x 2 - 7 x + 8 s (A + B + Ox 2 -f (3 A - B + 4 C)z - 4 A.
/"or tfwo polynomials to be identically equal, it is necessary that the coefficients
of corresponding powers of the variable be equal. Equating coefficients, we get
[ A+B+ C - 5,
(3) I 3A- + 4C=-7,
[ - 4 A = 8.
Solving equations (3) simultaneously, we find that A = 2, B = 29/5,
C = 6/5. Hence
/5s -7s +8 // 2
J *'+3*'-4* ^ = 7 V - i + 6(S
29 6
/rfx 29 r_dx_ 6 f dx
- ~ 2 J ^^"sJ 7+1 + sJ i~=l
9Q fi
= - 2 log a: -f =F log (x + 4) -f ~ log (*-!)+ C.
O D
SOLUTION 2. In clearing (1) of the preceding solution of fractions we get
(4) 5 x 2 - 7 x + 8 a 4(s + 4)(? - 1) + #3(3 - 1) + CX* + 4).
relation must be true for all values of x, we proceed to use special
values so as to get equations involving A, B, and C which are more quickly
solved than those of (3) above. Thus we notice that x = removes the
terms that involve B and C, and so the resulting equation, 8 = 4 A, is
solved at once, and A 2; similarly x = 1 removes A and B terms and
146] FORMAL INTEGRATION 257
we get 6 = 5 C, or C 6/5; also x = - 4 gives 116 = 20 5, or B = 29/5.
The remainder of the solution is the same as that given above.
(b) When the factors of the denominator are linear and some are
repeated.
EXAMPLE
SOLUTIONS. Assume
4 x 2 -- 7 x -f 10 A Bx + C
(x + 2) (3 x - 2) 2 ~ x -f 2 l (3 x - 2) 2
But B.
c + C Y - #'(3 3 - 2) + C',
where B' = /3, and C v/ = (2 B
+ 3 C)/3.
Hence
we have
4 x 2 - 7 x + 10
A B' C'
U>;
(x + 2)(33 - 2) 2 a
. _i_ 2 3a; 2 (3# 2) 2
Here we have assumed that the original fraction is separated into two
proper fractions having the factors of the original denominator as their denomi-
nators. Then we show that the last fraction can be separated into two
fractions and so all integral powers of each factor of the given denominator up to
and including the power to which it appears in the denominator may be used as
denominators of partial fractions. We also point ou ', that the numerators remain
constants. Clearing (5) of fractions and using either method outlined in
Example 1, we find that A = 5/8, B' = - 13/24, C' = 8/3. Therefore
r 4rr 2 - 7 x + 10 7 5, , . . 13. 8 .
J (x+Wx-^W dx = 8 bg (X + 2) - 72 Iog (3 * - 2> - 9(8* -2) + C '
(c) When some factors of the denominator are of the second degree.
y v i *
3. Evaluate
EXAMPLE
5^ - 12 ,
SOLUTIONS. Since x 1 is a repeated linear factor, we shall assume
partial fractions of the types A/(x 1) and B/(x I) 2 . Then, since the
other factor of the denominator is a quadratic, we assume a numerator Cx -\- D
which is the next lower degree. Thus
s 3 -3:r 2 -f5a;-12_ A B Cx -f D
(x - l) 2 (z 2 4-3x-2)~x-l""(a:-l) 2 3 2 -r-33
and, as shown in the other examples, the solution is completed.
258 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
A rational trigonometric function of an angle x may be trans-
formed into a rational algebraic fraction in u by the substitution
(1) tan = u.
i< or tnen
Sin ~ = . , COS ~ = ~.
2 Vl + u 2 % Vl + u 2
whence
(2)
2u 1 - u 2
" m ^ 1 + w 2 * C0 ^ 1 + u 2
Also
(3)
2du
x - 2 tan X da; - 1 . 2
EXAMPLE
4. Evaluate y*sec x dx.
n cv 1 1 -H H 2 , , . , . . ,
SOLUTION. Since sec x = - = - - ; and dx = - : - > the integral
cos x 1 w 2 1 -}- w 2
becomes 2 fdu/(\ w 2 ), which is readily integrated.
PROBLEMS
1. Prove the following formula by the use of partial fractions.
du 1 f u a i i ^ i I 1 - a u ,
/du 1 f u a i
S^T' = 2l log lT+^' l
Evaluate each of the following integrals.
2. fdx/(x* -6x + 5).
3. (2 x - l)dx/(x 2 - 1). Ans. log (x - l)i/(x -f 1) 3/J + C.
4. /(3x-f 2)dx/(x -f- x).
5. y*ete/(6x - 9x 2 -h 15). ATW. (1/3) log (x -f l)/(5 - 3 x)" 1 / 8 + C.
6. y*(2x + l)dx/(x -x).
7. fdx/(l - sin x). Ans. 2/[l - tan (x/2)] -f C.
8. fdx/(\ + cos x).
147] FORMAL INTEGRATION 259
9. y2x<fx/[(x+2)(x 2 -l)].
Ans. log (a -f l)(z - l) 1/3 (x + 2)~ 4 / 8 + C.
10. y dx/[x(x 2 + 2 x + 1)3.
11. f(2x + l)dx/[x(x 2 + 1)]. Arw. 2 tan- 1 x + log x(x 2 + I)- 1 / 2 -f C.
12. y (* + 3)dx/[(x + l)(x 2 + 1)].
13. /"cfe/(4 + 2 cos a;). Ans. (1/VS) tan-Y-^^ -f C.
47 V V3 /
14. fdB/(Z - 2 sin (9).
15. y (x + 3)^/[a:(x 2 - Qx + 5)]. Ans. Iogz 3 / 6 (z - 5) 2 / 6 (x - I)- 1 -f C.
16. f(3x z -x + 8)dx/t(x -f l)(o; 2 -f 5)].
17. fx*dx/(x* - 1). Arj^. a;/3 +x + (1/2) log [(x - l)/(x + 1)] -f- C.
18. y (1 + 2 3 - 3 x*)dx/(2 x 3 + 2 x).
19. y* (x 4 -f- l)<fo/(z 3 -f x). Aris. (1/2) x 2 + log [x/(x 2 + 1)] + C.
20. f csc 6d0.
21. y* (ctn c)/(l - cos 0).
Ans. - (1/4) ctn 2 (0/2) - (1/2) log tan (0/2) + C.
22. y (tan dO)/(l - cos 0).
23. (x 4 + 11 x - 6)rfx/(x 3 + 8).
Ans. (l/2)x 2 + log [ V x * - 2 x + 4/(a? + 2)] + C.
24. y (x 4 -f- 2 x 3 + 2 x 2 + 2 x -f l)dx/(x 3 -f x 2 + x + 1).
25. y (sin 2 Ode) /(I- sin 0). Ans. cos - -f 2/[l - tan (0/2)3 -f C.
147. Use of Integral Tables. The student should realize by
now that integration is largely an individual process and readily
accomplished or not according to one's choice of a suitable method.
We have tried to offer enough suggestions as to methods to
enable one to evaluate most of the more usual integrals. However,
a Table of Integrals is often necessary or at least very convenient.
Such tables do not have all possible integrals in them, hence the
transformations previously proposed may be needed to bring the
given integral into the form which may be found in the tables.
260 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
Often this may be done by merely replacing some function of the
variable present by a new variable. Then again it may be neces-
sary to use the formula for integration by parts or the process out-
lined for rational fractions before the given integral is changed into
one which may be found in the tables at the end of this book.
148. The Definite Integral. Usually, as in the case of a mov-
ing body, we are not interested in the whole distance the body has
traveled but in the distance it has traveled during a definite interval
of time. This is also generally true in all problems of integration.
Thus, consider the function y derived from the relation
(1) dy = dF(x) =/(*)dx.
We have
(2) y = ff(x)dx = F(x) + C.
For x = a this function has the value F(a) + C, and f or x = b
its value is F(b) + C. Hence, if x has changed continuously
from the value a to the value 6, the total change in the function is
(3) [F(b) + C]- [F(a) + C] = F(b) - F(a).
This difference of the values of the indefinite integral for x a
and x = b is called the definite integral of f(x)dx between the
limits a and 6. It is written
(4) ( b f(x)dx = *(*)? s F(b) - F(a).
Notice that since the constant of integration disappears in taking
the difference of the two values of the indefinite integral, it is not
written when the integral has limits.
EXAMPLES
1. If dy = 2 x*dx, find the change in y from x = 2 to x = 4.
SOLUTION. Using integral notation, we have
y = f2x 2 dx.
Then the change in y is represented by
r 4 o *j 2 *n 4 i28 i6 0*1 *
J 2 2 x*dx -y- J - y = 37J units.
148]
FORMAL INTEGRATION
261
2. The velocity of a falling body is gt ft. /sec. How far will the body fall
during the third and fourth seconds? (g = 32.2 ft./sec. 2 )
SOLUTION. Since v = ds/dt = gt t we have ds = gt dt. Therefore
8 = fgtdt.
Then the change in s, or the distance fallen, is
s 2 - Si = f 2 4 gt dt = I gt*\ = 6 a =
193.2 ft.
If a change of variable, due to a substitution to facilitate inte-
gration, is used, the limits must be changed to correspond to the
new variable. However, if such new limits are not of convenient
form for evaluating the integrated function, this function may be
changed back to the original variable arid the original limits used.
In such cases, the limits should appear on the integral sign during
the process of integration but written so as to refer to the original
variable, as illustrated in Example 4 below.
3. Evaluate
'-a/2 -
SOLUTION. In this example we set u a sin 0, then du = a cos 6 dd
and N/a 2 w 2 = a cos 6. When u = a/2, tho givon substitution makes
sin = 1/2, or = 7r/6, and when u = a/2, sin = 1/2, so = ?r/6.
Now the example can be written
a oos dO
[
e
1
\!
if
IB
i
/\
k
-/
JTT
\[
I
]
In general an integral has one and only
one value at each limit, but here the
value 6 = sm" 1 (u/a) has an infinite num-
ber of values at each limit. We must be
careful in selecting the values to be used.
A part of the graph of s\ir l (u/a) is
shown in Fig. 147. In the integral u is
supposed to vary continuously from
a/2 to a/2. This condition is satisfied if we choose values for
6 or sin" 1 (u/a) from the part AB of the curve. That is, for neg-
ative values of u, we take sin" 1 (u/a) between 7r/2 and 0, and
for positive values of u, we use sin" 1 (u/a) as evaluated between
and 7T/2.
FIG. 147
262 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
4. Evaluate f xVx 2 dx.
J2
SOLUTION. To integrate this example set u = Vx 2. Then u* = x 2,
2udu = cr, and x = u 2 -f 2. Also for x = 2, u = 0, and for z = 5, w = V3.
Hence
The substitution used in such integrals, here u \/x 2,
must be one which changes in one direction throughout the interval
of integration. Thus, as x increases continuously from 2 to 5, the
new function u increases throughout the interval from to
Care must be taken in substitution so that this is true.
The original limits may be used if the following notation is carried out:
f xVx 2 dx = 2 f (u* + 2 u z )du
/2 /r = 2
= 2 R (x - 2)*/ 2 + \ (x - 2)'/T
L^ J2
= 38\/3.
5
5. Find the value of / -? ;
/-aX 2 + a 2
SOLUTION. Let x = a tan 0, then f/ = a sec 2 d0, and the denominator,
x 2 + ^ 2 = 2 sec 2 0. Also for x = a, tan 0= 1, or = ir/4, and for
x = a, tan = 1, so = *-/4. Therefore
r (Jx r^ 74 scc2 g <fr ^ l r^ 74 de = i^T /4
%/~o x 2 -}- a 2 /-T/4 a 2 sec 2 a ^-/4 a J -/4
~ ?a "~ \ Ta/ " 2^ *
Again from d0 = dx/(x* -f- a 2 ) we get
1 ""1*^ = ^ /*" ^/JF 1 x""l^*
a Jx= a J a x* -f- a 2 a ctj a
which has an infinite number of values at each limit. Consider the graph of
= tan' 1 -
148]
FORMAL INTEGRATION
263
of Fig. 148. We must be careful to select proper values for the integral. In
the integral x varies continuously from a through to -f o. This requires
that we move along a single branch from
x = a to x a. It is most convenient
to use the branch which passes through the
origin; then for a negative value of x we
must choose or tan" 1 (x/a) between ir/2
and 0, and for a positive value of x the
value of tan" 1 (x/a) must be between
and 7T/2. -"^7"" ^X
6. Evaluate /
Jb
dx
'*> xVx* - a*
SOLUTION. This form of integral suggests
the substitution x a sec 0. This makes
Vx 2 a 2 = a tan 6, dx = a sec tan dO\
therefore
FIG. 148
r
=*ca gee tan
= 6 a sec a tan a >
1 r~I c 1 F i c -i ^~]
_ 1 sec" 1 ~ ~ ~ sec scc I *
a 1 aj6 a L a a J
Here again
! i - T T
- sec~ l -
a aj6
Fia. 19
say b
has an infinite number of values at
each limit. Fig. 149 shows [the graph
of y = = S ec" 1 (x/a). Since x varies
continuously from 6 to c, must vary
continuously and to do this sec" 1 (x/a)
can be taken between and ir/2 for posi-
tive values b and c. Then for negative
c, the values of sec" 1 (x/a) should be taken between
and
Evaluate
PROBLEMS
of the following definite integrals.
. 3.
3.
'(9 -
.3 - (3/2) \/3.
2M
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
IS.
16.
17.
18.
19.
20.
21.
22.
DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
f a x*dx/(2a -x).
'O
f V2 d/(4 6* + 3). Ans. 7T/4A/3.
J -1/2
I X (IX / X ~\~ o.
Jo
f 5 dx/(2 x* - 5 x + 3). ATW. log (7/4).
(4x / 3/5)(13\ / 2 - 2).
. T/6.
Vl -h arfs.
/o . _
jQi\/\. -. P2
-1 '
/* 1/3 rfx/V4 9 x 2 .
/-1/3
/7T/2
I ^ sin d8.
JQ
2 dx/V* -2x*.
7).
^2 , -
I v2x x 2 da;.
/i
f 1/2 dx/V7 +4
J-l/2
f 15 (9 -
//4
I cos 3 2 x dx.
'O
/* x 2 Vl - x* dx.
J-l/2
/* sin" 1 f tt.
/-1/2
. 1.
Ans. ir/(6V3)
, 7T/8/
12 - V3/2.
V3 - 1).
24. f* Gain* 9 dO.
JQ
150] FORMAL INTEGRATION 265
149. Interchange of the Limits of the Definite Integral. Since
the symbol
rf(x) dx represents F(b) - F(a),
_
where
J/(*) dx = F(x) + C,
and
rf(x) dx represents F(a) F(b) 9
_
it is evident that
ff(x) dx = - j^V(x) dx.
150. Subintervals of Integration. Since
dx = F(b) - F(a) and f V() ^ = ^(0 - F(6),
it is evident that
rr b r c
*s a t/6
A SUPPLEMENTARY GROUP OF INTEGRALS
1. /fo/4 - 3)dy. Ans. (4/3) (y/4 - 3) 8 + C.
r x*dx
Am. (l/48)[8 x 3 - 6 x 2 + 6 x - 3 log (2 x -f 1)] + C.
4. fZx*dx/(l -x).
s. (l/2)[5 log (x 2 - 3) -I- cos 2 x] + C.
6. fy(a + by z )- 1 dy.
7. /" x dx/Va 2 x 2 . Ana. Va 2 x 2 + C.
8. /" (*/V6 - ox 2 + sin ox) dx.
266 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
9. f 5 bx dx/(S a - 6 6x 2 ). Ans. - (5/12) log (8 a - 6 6s 2 ) -f C.
10. fx*dx/Va* -fx 3 .
11. y ctn 2 (1 -20) esc 2 (1 -20) d0. Ans. (1/6) ctn 3 (1 - 2 0) -f C.
12. f to.n* sec* de.
13. y sec 8 (2 (?) tan 2 (2 0) d0.
An*. (1/210) [35 tan 3 (2 e) + 42 tan 6 (2 0) -f 15 tan 7 (2 0)] -f- C.
14. y (1 + 2 x 3 ) dx/v 7 ! - 2 x - x 4 .
15. fdx/(\ -f e 3 *). Ans. x - (1/3) log (1 -f e 3 *) + C.
16. fe* z dx/(l - e*).
17. y cos a0 d0/(l -f- sin o0). Ans. (I/a) log (1 -f- sin a0) -f C.
18. / sin ax cos ax dx/(l cos 2 ax).
19. y* (x + 2) 2 dx/x 3 . Ans. log x - 4/x - 2/x 2 + C.
20. V&' 2 cos 2 -h a 2 sin 2 sin e cos d0.
21. fxdx/\ / x~^l.
22. fVx + 1 dx/x.
23. fxV2x + Ids.
24.
25.
26.
Ans. (2/3) (x -f 2)Vx - 1 -f
Ans. (1/15) (6 x 2 + x - l)V2x -f 1 C7
Ans. - (l/3)(x 2 -f-
27. /x 2 r/x/(x 2 -f I) 2 . Ans. (1/2) tan' 1 x - 2(a;2 X + i}
28. f (3 x - 2) dx/V9 - x 2 .
29. y*sin 5 0cos 2 0d0.
Ans. (1/105) cos 3 (35 - 42 cos 2 + 15 cos 4 0) + C.
30. y sin 1 / 3 (x/3) cos 3 (x/3) dx.
31. /"log x dx/x. Ins. (1/2) 0og x) 2 + C.
150]
FORMAL INTEGRATION
267
32. flogxdx/x*.
33. fcos*(x/5)dx.
Ans. (1/16) [20 sin (x/5) cos 3 (x/5) -f 15 sin (2 x/5) -f 6 a] + C.
34. cos 2 (1 - 2 a?) sin 2 (1 - 2 a) dx.
35. ye* dx/(l - e*) 2 .
36. e 2x cos 3 x da:.
Arcs. 1/(1 - e*) -f C.
37. fdx/V'Sx* + 4x - 10.
Arw. (1/V3) log (3 a; + 2 + \/9 x 2 + 12 x - 30) + C.
38. y*x log (x 2 + 1) dx.
39. y*(a 2 -x*)*t*dx.
Ans. (l/8)[(a 2 - x 2 ) 1 / 2 (5 a 2 a: - 2 x 3 ) + 3 a 4 sin- 1 (x/o)] + C.
40. f x sec~ l (x/2) dx.
41. fdx/V2 x 2 - 3. Ans. (1/V2) log (x -f \/x 2 - 3/2) + C.
42. fdx/Vl +4 a; - z 2 .
43. fdx/(x 2 + 6 x - 13).
17
44. fGdx/(x* - 4).
3 ~
-f 3 -f V22
-f- C.
45.
(x 2 + 1)].
Ans. log (x + l)-'/*(x 8 + I) 1 / 4 -f (1/2) tan~ l x -f C.
46. fdx/(7 -f 4x -
47. fx*dx/V2x* -f 1. Arcs. (1/9) (2 x 3 -f l) J / 2 (x 3 - 1) -f C.
48. fdx/V2x* +3x.
49. y x dx/^2 x -f 6. Arcs. (l/7)(2 x + 6) 3 / 4 (2 x - 8) -f C.
50. fdx/Vx* - 3x -f 1.
51. fdx/V7 +5x -2x 2 . Arcs. (l/>/2) sin" 1 [(4x - 5)/9] -f C.
52. f cos* 0d8/(l - cose).
53. y x cos~ l (2 x) rfx.
Arcs. (x 2 /2) cos- 1 (2x) -f (1/16) sin' 1 (2 x) - (x/8)Vl -4x 2 -f C.
268 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
54. yVz 2 - 12 dx.
55. fyVaV* - j/ 2 / 3 dy.
Ans. - (1/35) (8 a 4 / 3 -f- 12 a 2 /V/ 3 + 15 2/ 4 / 3 )(a 2 / 3 - y 2 / 3 ) 3 / 2 -f (7.
56. /V - 5 i/ 8 -f 6 y - 4) ^/(y 3 - 4 y 2 - 5 y).
57. fy*dy/l(y-l)(y* + lft.
Ans. (1/2) log (y - 1) -f (1/4) log (y* -f 1) -f- (1/2) tan-* y + C.
58. fVy*~^idy/y*.
59. fdx/V(x* - 9) 3 . Ans. - x/(9\/z 2 - 9) -f C.
60. y (x 4 4- 2 x 3 4- 2 x 2 -f 2 x + 2) dx/[(x -f l)(z 2 -f 1)].
61. f 5 x dx/Vl - x 4 . Ans. (5/2) sin' 1 x 2 + C.
62. y*cos 3 (2 x - !)<&.
63. \/2/ 2 - Idy/y. Ans. (y - I) 1 / 2 - tan' 1 (y 2 - I)'/ 2 + C.
64.
+2x -3x*.
65. y tan 8 (2 - 3 x) rfo?.
Ans. - (1/6) sec 2 (2 - 3 x) - (1/3) log [cos (2-3 x)] -f C.
66. y tan 3 (2 s/3) sec 4 (2 x/3) dx.
67. y* tan dff/(l 4- cos ^). Ans. - log [1 - tan 2 (0/2)] -f C.
68. /* ctn dO/(l -f sin 0).
69. j sin mx cos nz dx t mj^n.
Use sin (A + B) -f sin (A - B) = 2 sin A cos B.
. 1 Fcos (m -f n)x cos (m n)x~] n
Ans. I - ; - -p - I -f- C.
2L m + n m n J
70. J cos mx cos nx dx, m ^ n.
71. J sin mx sin nx dx, m^n.
A ! r s ^ n ( m + n)x sin (m n)x"| ^
^nLnS. p: I - ; - - I -f- C.
2[_ m + n m n J
72. y a 2 6 2 rf0/(a 2 sin 2 6 + b 2 cos 2 0).
73. \/sin~0 de/cos*i* e. Ans. (2/3) tan 8 / 2 -f C.
75.
-f Vx)/xdx.
An*. (1/2) x a tan- 1 x - x/4 -f (1/8) tan- 1 2 x -f C.
150]
FORMAL INTEGRATION
269
76. j x sin~ J 3 x dx.
77. /"sin 4 (3 - 2 x) dx.
Ans. [4 sin 3 (3 - 2 x) cos (3 - 2 x) - 6(3 - 2 3) +3 sin (6-4 x)]/32 + C.
78. cos 3 3 x dx/(l - cos 3 x).
/7T/2
79. I sin 6 cos 0d0.
JQ
** [4 x/(x* + 4) - s/2] <te.
80.
'O
81. cos 5 (2 a;) dx.
a)
82. f 5 dx/(2x* -x -3).
/2
83. y a \/a 1 / 3 +
* 4
84.
sec 2
+
85. sin 0(1-0 + sin (?)
^o
90. f e Vl + log x dx/x.
92. /" 2 z 2 \/4 -x*dx.
72
_J O ~ ^
93. J dx/Ve~ 2x + 2 e~ x .
94. /* a <fo/[>Va 2 + x 2 ].
'a
95. If dy/dx = sec 2 x, what is / %?
/x ""O
^r /
96. y 2/ dx for a: = v 2 ry y 1 .
97. A
Arcs. 1/2.
'
Ans. 4/15.
Ans. 4(11 A/2 - 4)a 7 / 6 /35.
Aws. 2 - /2.
Ans.S/3.
Ans. 5 Tr/16.
\/3/2 + (1/2) log \^ + l -
V3 - 1
Ans. 3 - \/3.
Ans. 1.
+ (dx/dy)* dy for a; = V2 ry - y*.
Ans. vr/2.
270 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XI
98. yVl + (dx/dy)*dy for x = V2 ry - y*.
99. f a Vl + (dy/dx)* dx for z 2 / 3 + y 2 '* = a 2 / 3 . Ans. 3 a/2.
100. yVl + (dy/dx)* dx for z/ = 6 + Va 2 - x 2 .
^0
101. fxyVl + (dy/dx) 2 dx for 9 x* + 25 i/ 2 = 225. Ans. 245/8.
102. f a yVl + (dy/dx)* dx for 8 a 2 ?/ 2 = x 2 (a 2 - x 2 ).
^o
103. fJj'W dy itf(y) = (d/dy)f(y). Ans. f(x) - /(a).
104. f X J"(y)(x - y) dy. (Use parts.)
/a
105. jf* C/ /// (T/)(X - 2/) 2 /2!] dy.
Ans. f(x) - /(a) - f'(a)(x - a) - f"(a)(x - a) 72!.
/x
fn+i (y}(x y) n dy/nl. Get several terms.
_
CHAPTER XII
APPLICATIONS AND INTERPRETATIONS OF
INTEGRALS
151. Curves with Given Properties. If the equation of a
curve is given as F(x t y) = or 0(r, 0) = 0, we can find the
differentials of these functions. The derived equation will usually
involve x, y, dx, and dy or r, 0, dr, and do, and is called a differential
equation.
On the other hand, if a differential equation is given which we
can reduce to the form
y = f 2 (x)dx,
or
0i (r) dr = 2 (0) d,
in which the variables are said to be separated, we may be able
to integrate each member of such a relation arid thereby obtain
the equation of the curve except for the constant of integration. Some
additional information is necessary to find the constant so as to
know definitely the equation of the curve. If the constant of
integration cannot be found definitely, we can consider it as a
parameter and then the equation represents a family or system of
curves, one curve for each definite value assigned the constant.
Such a system of curves is called a one-parameter system as the
constant of integration plays the role of the parameter.
EXAMPLES
1. If the slope of a curve at any point is the square of the reciprocal of
the abscissa of the point and if the curve passes through (2, 4), find its equation.
SOLUTION. We are given that
m dy - l
(1) Tx~^
which written in differential notation becomes
(2) <*/=
271
272 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XII
Integrating each member of this equation, we have
(3) y = - \ + C,
where C represents both constants of integration, one of which might have
been written on each side of the equation.
The equation (3) represents a family of curves such that the slope of each
satisfies the condition (I). To find the member of this family that passes
through (2, 4), we impose the condition that (2, 4) satisfy (3) and thereby find
that C = 9/2. Therefore the desired equation is
x 2 '
which is a rectangular hyperbola.
2. Find the equation of the curve which is perpendicular to the line joining
any point of itself to the point (3, 4), if it passes through the origin.
SOLUTION. Let P\ represent the point (3, 4) and P(x, y} any point of the
curve. The slope of the curve at P is the
negative reciprocal of that of the line P\P.
Therefore
dy _ x 3
dx = ~ y - 4'
whence, separating the variables, we find
(y - 4)dy = (3 - x)dx.
Integration gives the system of curves
FIG. 150
Since the curve we want passes through
the origin, we substitute x = 0, y = in this equation and find that C = 0.
Clearing and transposing, we obtain
x 2 + 7/ 2 - 6 x - 8 y = 0,
a circle with (3, 4) as its center.
3. Find the equation of the curve through the point (a, 7r/4), from which
is derived the relation dr/dQ = (a 2 /r) cos 2 6.
SOLUTION. Separating the variables r and 0, the given relation can be
written as follows,
rdr = a 2 cos 2 Odd.
Integrating, we get
- sin 2 + C.
151] APPLICATIONS OF INTEGRALS 273
Then substitution of (a, 7r/4) in this gives
a 2 a 2 . TT .
- = - sin 72 + C,
whence C = 0. Therefore the desired equation is
r 2 = a 2 sin 2 0.
PROBLEMS
1. The slope of a curve at any point is x 2 and it passes through (1, 5).
Find its equation. Ans. 3 y = x 3 -f- 14.
Find the equation of the curve which satisfies the following conditions.
(Nos. 2-6.)
2. Slope 5 and passes through the point ( 2, 8).
3. Slope 2 z, through the point (1, 3). Ans. y = x 2 -f 2.
4. Slope 2x/(l + x 2 ) t through the point (0, 0).
5. Slope 2 2/ r through the point (0, 5). Ans. 2 x = log Q//5).
6. Slope ?/ 2 , through the point (1, 5).
7. The rate of change of the slope of a curve with respect to a: is 2 6 x.
Find its equation if it passes through (1, 1) and ( 1, 5).
Ans. y = I 2x + x 2 x*.
8. The slope of a curve is proportional to the ordinate at any point. If it
passes through (1, 3) with the slope 3/2, what is its equation?
9. The slope of a curve at any point is three units less than the square of
twice the abscissa of the point. If it passes through (2, 7), what is its equation?
Ans. 3i/ = 4z 3 -9z-r-7.
10. The rate of change of the slope of a curve with respect to x is 3 x.
Find its equation if it passes through the points (2, 3) and ( 2, 11).
11. Find the equation of the curve for which d 2 y/dx 2 = 5, and m = 3 at
the point (2, 4). Am. 2 y = 5 x' 2 - 14 x.
12. The same as Problem 11 for d z y/dx* = 4 2 re, if y has a minimum
value at the point (1, 2).
13. The same as Problem 11 for dm/dx cos x, and m 3 at the point
(7T/2, 2). Ans. y = 2 x - cos x -f 2 - TT.
14. The same as Problem 1 1 for dm/dx 2 sin x, and m = 4 at the point
(T/2,2).
15. The same as Problem 11 for y" = 4 # 2 , and m 1 at the point
(2, - 10/3). Ans. 12 y = 24 x* - x* - 52 x - 16.
16. Find the polar equation of the curves which make an angle with each
radius vector equal to that which locates the radius vector.
274 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XII
17. Find the equation in polar coordinates of the curves which make the
same angle with the radius vector to each point. Ans. r ce^.
18. Find the equation of the curve through (0, c) which has its tangent of
constant length c. The length of the tangent to a curve is ?/Vl -f (dx/dy) 2 .
19. Find the equation of the curve through (1, 5) which intersects the
curves y = (1/2) log (1 x 2 ) -f- c at right angles.
Ans. 2 y = 2 log x - x* + 11.
152. Straight Line Motion. If the distance a body moves
along a straight line is given as a function of the time, we have
found the velocity and acceleration to be ds/dt and d 2 s/dt*, respec-
tively. We are now able to reverse the problem. That is, given
the velocity or acceleration of a body as a function of the time, we
may find the distance the body has moved and its direction of
motion along its path at any time t. The constants of integration
which appear during the inverse operations are determined by
initial conditions or additional information.
EXAMPLES
1. If the acceleration of a body is 6 3 t, find the distance the body will
move after t while the velocity is increasing; given the initial condition
v when t = 0.
SOLUTION. The velocity increases during the interval in which the accelera-
tion is positive. For this example this interval begins at t = and continues
until t = 2. Hence the distance moved during the interval from t = to
t = 2 is desired. The acceleration
Then
dv = (6 - 3 t)dt,
Since v when t = 0, the constant of integration C is zero.
Before evaluating s over the interval from t to t 2, we must see if
the motion is in the same direction during the interval. Since the direction
of motion may change when v is zero, we see from
that the direction of motion can change only at t = or t 4. Hence it
152] APPLICATIONS OF INTEGRALS 275
does not change between t = and t = 2. Now, writing s as a definite integral,
we have
=12 -4=8 units.
2Jo
2. If v = 2 3 t -f- 2 , find the distance traveled by a particle during the
interval from t = 1 to t = 5.
SOLUTION. The velocity is zero at t = 1 and t = 2. Since it is negative'
from t = I to t = 2, the motion of the particle during that time is opposite to
the direction in which positive s is measured. To find the total distance
traveled, we must compute 5 for the intervals from t = 1 to t = 2 and from
t = 2 to t 5, and add the two results numerically. That is, let
si = f (2 - 3 t + t*)dt and s 2 = f* (2 - 3 t + t*)dt.
%/l /2
Whence
and
Therefore the total distance traveled is
= 13 units.
U O
PROBLEMS
1. The velocity of a particle along a straight line is 2 t 4. How far
does it move from t = 1 to t = 4? Ans. 5 units.
2. The same as Problem 1, with v = t 2 - 6 t -f- 5, t = 0, t = 3.
3. The same as Problem 1, with v = t z - 6 t + 5, t = 3, < = 7.
s. 16 units.
4. The same as Problem 1, with v = * 2 + 2 * - 3, J = 0, t = 3.
5. The same as Problem 1, with v = 10 - 7 * -f * 2 , t = 0, * = 6.
Ans. 15 units.
6. The same as Problem 1, with v = -2t 2 + llt-l2,t = 3,t = 7.
7. The same as Problem 1, with v = 12(J 3 4 2 -f 3 0, t = 0, J = 4.
Ans. 96 units.
276 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XII
8. The velocity of a body moving along a straight line is 3 t t 2 . How
far does it move while the velocity is positive?
9. The velocity of a body along a straight line is 10 7 t -f t z . How far
does it move in the direction opposite to that in which the distance is measured?
Ans. 4% units.
10. The same as Problem 9 for v = t 2 - 3 t + 2.
11. The acceleration of a body along a straight line is t* 4 t. How much
does its velocity change during the interval in which the acceleration is nega-
tive? Ans. 10% units.
12. The velocity of a body along a straight line is t 2. How far apart are
its positions at t = 1 and t = 4? How far has it moved?
13. The same as Problem 12, with v = t 2 - 6 t -f 8, t = 1, t = 3.
Ans. 2/3 units; 2 units.
14. The same as Problem 12, with v = 2 t 2 - 7 t -f 5, t = 0, t = 2.
15. The velocity of a body along a straight line is 18 3 t. How far will it
move from t = 2 until it stops? Ans. 24 units.
16. If v = e 3 e 2t , how far does the body move after t = while v > 0?
153. Projectiles. Let the point of projection of a projectile
be taken as the origin of coordinates, the x axis horizontal, and
the y axis directed upward. Sup-
pose the initial velocity is VQ and
the angle of elevation is a.
If we assume that the force of
gravity alone acts on the projec-
tile, there is no horizontal force
and hence the horizontal velocity
dx/dt is constant and the horizon-
tal acceleration d 2 x/dt 2 is zero.
The vertical force acting down-
ward being that of gravity, we
have
FIG. 151
(1)
= =
dt 2 dt
dy' = - gdt.
Integrating (1), the vertical component of the velocity is found
to be
(2)
tf.&.-fi + C.
153] APPLICATIONS OF INTEGRALS 277
The initial conditions are
(3)
t = o, x = 0, y = 0,
dx dy
= VQ COS , -77 = V Sin Of,
at at
whence C of (2) is V Q sin a.
Then, since the horizontal component is constant, these com-
ponents of the velocity are
,A\ dx dy . ,
(4) -3- = i> cos a, - = 0$ + t^o sin a.
Integrating relations (4), we have
x = v Q t cos a + Ci,
= ^ sin a - </ 2 + C 2 .
A
The initial conditions # = 0, ?/ = 0, t = make Ci = C 2 = 0.
Hence the path of the projectile is given parametrically by
(5) x = Vot cos , y = Vo* sin a - gt 2 .
As the velocity of the projectile in its path is
V = \/V X 2 + Vy 2 ,
we have, from (4),
v \/v Q 2 cos 2 a + v Q 2 sin 2 a 2 v gt sin a +
or
(6) v = Vvo 2 - 2 z; ^^ sin +
In the following problems the student should derive the paths
needed by integrations and not by substitution in formulas (5).
PROBLEMS
1. Find the rectangular equation of the path of a projectile of 200 ft. /sec.
initial velocity at an elevation of v/Q. Ans, y = x/V3 <ja; 2 /60000.
2. Find the time of flight on a horizontal plane of a projectile of initial
velocity V Q and elevation a.
278 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
3. Show that the range of a projectile on a horizontal plane is (i> 2 sin 2 a) /g.
What elevation makes the range a maximum? Ans. 7r/4.
4. To what maximum height will a projectile rise?
5. When is the velocity of a projectile least? Ans. I = (^ sin ot)/g.
6. Find the ranges on an inclined plane of inclination 6, the projectile being
fired directly up and down the plane.
7. Find the elevation for a maximum range for Problem 6.
Ans. ex = 0/2 + 7T/4.
8. A plane flying 100 mi./hr. at a height of 1/2 mile will drop a bomb on a
target. How long before passing over the target should the bomb be released?
9. If it is 60 ft. from the position of a pitcher's hand when he releases a
ball to the batter, and if the ball leaves his hand horizontally and 5 ft. above
the ground, how slowly may it leave his hand to pass by the batter 2 ft. above
the ground? Ans. 80V3 ft./sec.
10. A projectile of initial velocity v ft./sec. is to hit a target k ft. away on
the same plane as the point of projection. What angle or angles of elevation
are necessary?
11. Assuming a = 60, v 75 ft./sec., g - 32 ft./sec. 2 , find v t , v v , and v
of a projectile if x = 20 ft. Ans. 37.5 ft./sec.; 47.89 ft./sec.; 60.82 ft./sec.
12. Solve Problem 11 if y = 30 ft. instead of x = 20 ft.
13. Find the envelope of the paths of projectiles fired from a given point
with the same muzzle velocity but with different angles of elevation.
Ans. 2 y = v<?/g - gx*/vj.
154. The Law of Natural Growth. If bacteria are allowed
to grow naturally with sufficient food supply, the increase per
second in the number in a cubic unit of culture is proportional
to the number in that cubic unit. The law of natural growth may
be derived from this type of rate of change. Thus, if the rate of
change of a quantity y with respect to an independent variable t
is proportional to y itself, we have
dt
Therefore
154] APPLICATIONS OF INTEGRALS 279
Integrating, we find
log y = kt + log C,
where the constant of integration is taken as log C so as to simplify
the next operations. Then
log y - log C = log ~ = kt,
and hence
C = **'
or
(1) y = Ce*'.
This is the law of natural growth.
Some instances in which this law applies are: The rate of
decrease of air pressure as we leave the surface of the earth is
proportional to the pressure at each height. The rate of increase
of the number of bacteria in a culture at any time is proportional
to the number present at that time. The rate of change of the
difference of the temperatures of a body and a cooling flow of air
or liquid is proportional to that difference. This law applies even
to compound interest if the interest is assumed to be continuously
compounded. Such is the case for an investment which is con-
tinually increasing, or decreasing at a given rate.
The student should derive the law by integration to fit each
problem given below.
EXAMPLES
1. When light enters a medium, its rate of absorption with respect to
the depth of penetration t is proportional to the amount of light incident on a
unit area at the depth. Find the law connecting L and t if the incident light
is I/o and the emerging light after passing through a thickness ti is LI.
SOLUTION. We are given that
= - kL
Whence
= - k dt
and integrating, we have
log L = - kt + log C.
280 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
Therefore
L = Ce~ kt .
But when t = 0, L - L , then C = Z/ . Also when t = t\ t L - LI] hence
Li = Loe-*\ or
whence
L = 1
2. Suppose an amount of money A at a rate of interest r is changing
continuously by instantaneous compounding. Find the law connecting A
and the time t it is at interest.
SOLUTION. We have, to begin with, the compound interest law
( r \nt
l+
where A Q is the original amount invested, r the rate of interest, t the number
of years, and n the number of compoundings per year. If the student is
not familiar with this law he can readily derive the special cases where n = 1,
2, and thereby see how the law is derived for any n.
We can rewrite the law in the form
r\n/r-|r
) ].
To let the compounding occur instantaneously is to let n > oo , and we have
A = Aoe rt ,
since lim (1 -f- r/n) n/r = e. This is the natural growth law because
n > oo
dA , . A
- = r(A^) = rA.
The example as stated is equivalent to an investment of value A which is
continually increasing at a rate r per year.
PROBLEMS
1. Sugar decomposes at a rate proportional to the amount present. If
50 Ibs. becomes 16.4 Ibs. in 4 hrs., when will 0.05 Ib. remain?
Ans. In 24.79 hrs.
2. If the rate of growth of a quantity is proportional to itself and if it
doubles in 10 yrs., when will it treble?
3. The rate of change of v with respect to t is proportional to v. Ifv = 10.5
units at t = 0, and 26.25 units at t = 5, find its value at t = 6.
Ana. 31.53 units.
155] APPLICATIONS OF INTEGRALS 281
4. A substance s decomposes at a rate proportional to itself. If s = 3.24
units when t = 0, and 1.62 units when t 3, find s for t = 5.
5. A sheet hung in the wind loses its moisture at a rate proportional to
the moisture remaining. If one half is lost in 1 hr. what part remains after
5 hrs.? Check by a progression. Ans. 1/32 of the original amount.
6. A rotating wheel is slowing down in such a way that the angular
acceleration is proportional to the angular velocity. If o> = 75 r. p. s. at the
beginning of the slowing down and in 2 min. it is 3 r. p. s., at what time is
co - 30 r. p. s.?
7. A cube of dry ice evaporates at a rate proportional to the surface of the
block. How long before a 6 in. cube reduces to a 1 in. cube, if an 8 in. cube
reduces to 6 inches in 1 hr.? Ans. 2.5 hrs.
8. If the population changes at a rate proportional to itself, what popu-
lation will a city of 40,000 have in 20 years, if 20 years ago it was 25,000?
9. The rate of decrease of a quantity is proportional to the quantity.
If 1/4 disappears in 2 hrs., when will 1/10 remain? Ans. At t = 16 hrs.
10. The rate of decrease of a quantity is proportional to the quantity.
If 1/5 disappears in 2 hrs., when will 1/9 remain?
11. If 100 Ibs. of one substance is transformed into another at a rate
proportional to the amount untransformed, and after 3 hrs., 51.2 Ibs. is not
transformed, derive a formula for the amount transformed at any time.
Ans. s = 100(4/5)'.
12. The funds of an institution over $300,000 are appreciating in value at
the rate of 4% per annum. If the original amount is $800,000, when will it be
doubled?
13. The same as Problem 8; if the population increased from 1600 to 2000
in the past 4 years, what will it be 10 years hence? Ans. 3494.
14. The same as Problem 11 for 12 Ibs. becoming 9 Ibs. in 2 hrs. Also find
how much is transformed at the end of 4 hrs.
15. For a constant temperature, the rate of change of the pressure of the
atmosphere with respect to the height above sea level is proportional to itself.
Find the pressure at a height of 5000 feet if it is 11.8 Ibs. at 6000 feet and 15 Ibs.
at sea level. Ans. 12.28 Ibs.
155. The Area under a Curve. Let RS, Fig. 152, be a portion
of the curve whose equation may be represented by y = f(x).
Let DB be an ordinate at a fixed distance a from the y axis
and let MP be the ordinate of a variable point P on the curve.
The area bounded by DMPB is represented by A. It is evident
282 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XII
that the value of A depends upon the position of the variable ordinate
MP and hence is a function of x.
Now let P move to Q so that x increases an amount Ax and A
increases the corresponding amount AA or MNQP. The value of
A A is such that
Y
(1) MP -Ax < &A < NQ-Ax
if the curve rises from P to Q.
The inequality signs will merely be
reversed if the curve falls from
PtoQ.
Dividing relation (1) by Ax, we
have
C(b,0)
FIG. 152
/
MP < = < NQ.
Arc
Now let Ax > so that the ordinate 7VQ approaches MP in
position and value, then, since AA/Ax remains between the two,
it follows that
(2)
(8 52, iv).
That is, the rate of change of A with respect to x is the ordinate
of the bounding curve at the extreme right-hand boundary. It is
convenient to think of A as generated by the line-segment MP
as it moves from the initial position DB determined by x = a
to some final position.
Relation (2) gives us the differential equation
dA = y dx,
and by integration we find
(3) A = fy dx = J/Ot) dx = F(x) + C.
The expression fy dx of course has a meaning when either y
is expressed as a function of x or when dx is replaced by its value
in y and dy. The variable whose differential occurs under the
integral sign is called the variable of integration; if the integral
has limits, they refer to the variable of integration. Any other vari-
ables appearing in the integrand must be expressed in terms of the
variable of integration before the integral can be evaluated.
156]
APPLICATIONS OF INTEGRALS
283
The area between the two fixed ordinates DB and CE may now
be evaluated. To determine C in (3), we have x = a and A =
when P is at B. Hence C = - F(a), and A = F(x) - F(a).
When P is at #, 3 = b and A = F(6) - F(a). Therefore ( 148)
(4)
Similarly, the area between the curve x g(y), the y axis,
and the lines y = c and y = d is
(5)
/
_
The formulas (4) and (5) are subject to the definite restriction
that the curve y = f(x) does not cross the x axis, and the curve
= o(y) does not cross the y axis between the limits of integration.
The reason for this is explained at the close of the following article.
156. The Area between Two Intersecting Curves. Consider
the area A between the two curves MN, y = fi(x), and RS,
y = f*(x), from their intersection at x = a to the variable vertical
line PiP 2 . If the corresponding areas between each curve and the
x axis are AI and A 2 , then
(1) A = AI AI,
and
dA
dx
N
dx
dx
or
(2)^-1,. -*-/,(*) -Mx).
Therefore the rate of change of A
with respect to x is the vertical
boundary PiP*. From (2) we have
(a,0)
(b,0)
FIG. 153
(3) dA = [/ 2 (z) -fi(x)]dx
and hence the area between the intersections at x = a and x = b is
(4)
A=
-/i 00] dx.
284 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
The student should set up and use formula (4) in such problems
instead of finding the area between each curve and the x axis.
If two curves intersect in more than two points, the limits
of (4) refer to any two consecutive intersections. The reason for
this is evident if we consider the two curves of Fig. 154. Suppose
the curves y = J\(x) and y = Ji(x) intersect in the three points
B, C, and D. Then the area AI
bounded by the two curves between
B and C is generated by the line
PiP 2 , which is / 2 (o;) f\(x) in
length. But the area A 2 between
C and D bounded by the two
curves is generated by P 2 Pi, which
has the length /i(x) - f 2 (x). Then
dA l =
while
FIG. 154
These differentials differ in sign. If we integrate [/ 2 (x) fi (x)] dx
from point B to point D we get A\ A 2 . That is, we get the
algebraic sum of Aiand ( A 2 )
since fz(x) fi(x) is positive
from B to C and negative from
CtoD.
EXAMPLES
1. Find the area between the
curve 3y = x 3 and the x axis from
x = - 2 to x = 3.
SOLUTION. The desired area is
composed of the two parts BOC and
ODE. (See Fig. 155.) The area
BOC is generated by the line
P\P'i = y = x 3 /3 as P2 moves
from C to 0. Hence the area of FIG. 155
BOC = I jT (- x 3 ) dx = - x 4 /12]_ = 4/3 sq. units.
Similarly ODE is generated by PzPi = y = s 3 /3 and its value is
= 81/12 = 6Ji sq. units.
Therefore the desired area is 8^ square units.
O1P S
156]
APPLICATIONS OF INTEGRALS
285
2. Find the area bounded by the two curves y
SOLUTION. The desired area is gener-
ated by PiPz as it moves from to B. t
Hence
dA = (2/2 - yi) dx
Therefore
A = f l ( X v* - x 2 ) dx
dx.
A figure is an essential part of the solution
of any area problem.
PROBLEMS
FIG. 156
1. Find the area between the lines y = 3 x, y = 0, and x = 6 and check
by the use of some formula. Ans. 54 sq. units.
2. Find the area under y = 2 x x 1 and above y = 0.
3. Given the lines x -f y = and 2x y 2=0. Express any verti-
cal line-segment from the first line to the second as a function of x.
Ans. 3 x - 2.
4. In Problem 3, express any horizontal line-segment from the second line
to the first as a function of y.
5. What is the length of a horizontal section of the area enclosed by
y 2 = 4 x + 4 and x 2 = 4 y + 8. Ana. 2V y -f 2 + 1 - 2/ 2 /4.
6. Find in terms of x the length of a vertical section of the area enclosed
by t/2 _, 4 x anc i x 2 -j- y.
Find the area enclosed by the following boundaries. (Nos. 7-20.)
7. y x*, y = 0, x 1, x = 4. Ans. 21 sq. units.
8. y = x* + 1, x + y = 1.
9. x = y 2 4, x = 0. .Ans. 32/3 sq. units.
10. x = y 2 4- 2 1/, x + 2 y = 0.
11. y = 5 x - z 2 , 2 ?/ = 5 x - x*. Ans. 125/12 sq. units
12. i/ 2 = x + 6, 3 = 4 - 2/ 2 .
13. # = 2 cos 0, y = 2 sin 0. Ans. 4 TT sq. units.
14. x = 4 cos 6, y = 3 sin 6.
286 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
Ans. 2(e 1/e 2 ) sq. units.
Ans. c 2 1 sq. units.
Ans. ir/2 1 sq. units.
15. y = e~*/ 2 , x = - 2, z = 4.
16. y = e~ x , y = e x , x = 2.
17. y = log *, z = 0, ?/ = 0, ?/ = 2.
18. y = x log x, y = x x 2 .
19. x = sin y, y = 0, x = 1.
20. y = sin x, y = 1, x = 0, x = TT.
Find the area enclosed by the loops of the following curves. (Nos. 21-23.)
21. y z = x 2 (x + 2). Ans. (32/15) \/2 sq. units.
22. y* = (x - l}(x - 3) 2 .
23. x 2 = y% + 4). Ans. 4096/105 sq. units.
Find the area bounded by the following curves. (Nos. 24-27.)
24. The ellipse x 2 /a 2 + 2/V& 2 = 1.
25. An arch of y sin 2 (2 x) and the x axis. Ans. ir/4 sq. units.
26. The coordinate axes and the parabola Z 1/2 + y 112 = a 1 / 2 .
27. The equilateral hyperbola x 2 y 2 a 2 and the double ordinate
through the point P\ on the curve. Ans. Xiiji a 2 log [(^i -f- y\)/a\.
28. Prove that the area of a parabolic segment formed by a line-segment
perpendicular to its axis is two-thirds of the rectangle which circumscribes the
segment.
157. Some Improper Integrals. A definite integral whose
integrand becomes infinite for some value or values of the variable
of integration in the interval of integration or which has an infinite
<a,0)
<c,0)
Fia. 157
(6,0) X
FIG. 158
limit is called an improper integral. If we consider the geometric
interpretation of such an integral, the reason for this is at once
evident. The definite integral, i.e., J^f(x)dx f represents geo-
157] APPLICATIONS OF INTEGRALS 287
metrically the number of square units in that part of the xy plane
which is bounded by y = f(x), x = a, x = b, and the x axis, and
this exists for all continuous functions f(x) in the interval
a i x ^ b. If, however, the integrand becomes infinite for
some x between x = a and x b, the curve y = /(x) has a vertical
asymptote and the area represented by the integral has its bound-
ary recede to infinity and it is no longer bounded. Figure 157
shows such an unbounded area due to the asymptote at x = c.
Again, if a limit of the integral is infinity, the bounding curve
has a horizontal asymptote, thereby making the area unbounded
also. Figure 158 shows such an area. Such improper integrals
are defined as follows:
(a) // f(x) becomes infinite at one of the limits of integration,
say at x = b, then
(1) f V(*) dx - lim f b h f(x) dx, h > 0.
Ja h *-Qj a
(b) If f(x) becomes infinite at x = c where a < c < 6, then
(2) f b f(x)dx^lim r~ hl f(x)dx+lim F f(x)dx,
J a A! *-Ot/ a h t >0t/c-h/i,
where hi and h z are positive.
(c) // one limit is infinite, say the upper one, then
(3) f "/(*) dx ES lim f b f(x) dx.
Ja &-* 00 Ja
These three definitions and the methods of evaluation of such
integrals are illustrated by the following examples.
EXAMPLES
1. Find the area under the curve y or 1 / 2 from x = to x 2.
SOLUTION. The curve y = x~ 112 has x as a vertical asymptote. The
method is then to find the value of the shaded area in Fig. 159 and evaluate
the limit of that area as h > 0, if possible.
A = f 2 x-v*dx = lim f 2 x-v*dx - lim ( 2 x
/O h-+QJh h-+Q
= lim (2V2 2V/0 = 2\/2 sq. units.
h >o
At first the student might think that the area under such a curve is always
288 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
infinite since its boundary recedes to infinity along the y axis. But for this
curve such is evidently not true and it is essential to evaluate the integral in
each case before a conclusion is reached.
1 (40)
FIG. 159
FIG. 160
/* uX
- -
-1 (x - I)
SOLUTION. The graph of the curve represented by ?/ = l/(x I) 2 is
shown in Fig. 160. It has a vertical asymptote, x 1. Hence
dx . .. /2 dx , , _
/2 dx ,. /l~~^i dz
= lim ( - -^j] l ~ hl
/I 1
since l//ii and 1/^2 >
/ios no meaning.
(b,0)X
FIG. 161
and hztO. In this case the area under the
3. Find the area under the curve
y = 3 x~ 2 from x = 1 to x = .
SOLUTION. As this example involves a
horizontal asymptote, we evaluate the
shaded area in Fig. 161 and then find the
limit of such an area as 6 > o , if possible.
Thus
/3 x~~ 2 dx = lim f 3 x" 2 dx
fc *oo J\
-3 lim (-IT)
6 > oo \ X Jl/
- 3 Hi
158] APPLICATIONS OF INTEGRALS 289
since 1/6 as b > . Hence this area exists and is 3 square units. A
figure is very important in all such problems as it shows the places where the area
is not bounded.
PROBLEMS
Evaluate, if possible, each of the following improper integrals. (Nos. 1-14.)
Ans. 2. 2.
. Cdx/V3 - x. Ans.
'
. . . 4. (e* + 1) dx/(e* - 1).
'O 'O
/2/3 , _ /3
5. / t dx/V - 9 x\ 6. / c/x/(2 a; - I) 2 .
^ 1/3 'O
/Ins. 2 7T/9.
7. f*dx/(2x - 3) 2 / 3 . 8. f*dx/(3 - x) 2 .
/i ./o
Ans. (3/2)(^5 + 1).
9. y a; 8 dx/(x 2 - 1). Ans. - oo. 10. f*2xdx/(x* - I) 4 / 3 .
11. f V ^dx/(4 - x*). Ans. oo. 12. C*dx/(Z - 5 x) 5 / 3 .
/ 2 A)
13. f l dx/x(l + x 2 ). Ans. oo. 14. /"Vl + x 2 dc/a;.
'O ^0
See whether the curves below define definite areas, and if so, find the area.
15. 2/ 2 (4 - x) = x 2 , x = 4, y = 0. Ans. 32/3 sq. units.
16. T/(Z - 2) 2 / 3 = 1, x = 0, x = 4, y = 0.
17. 2/(z - 4) 3 = 8, y = 0, x = 1, x = 4. Ans. No.
18. y(x - 2) 2 = 3, y = 0, re - 2, x = 4.
19. 2/(l - x) = 1, x - 0, x = 1, y = 0. Ans. No.
20. 2/(l - x 2 ) 3 / 2 = x, x = 0, x = 1, 2/ = 0.
21. 2/ 3 (z - I) 2 = 8 x\ y = 0, x = 0, x = 3.
Ans. 9[>^2 + 1/2] sq. units.
22. xy*(l + x*) = 2, y = 0, x = 0, x = 4.
23. 2/ = log x, x = 0, y = 0. Ans. 1 sq. unit.
24. The cissoid y 2 = z 3 /(2 a x) and its asymptote x = 2 a.
158. Volume Generated by a Portion of a Plane. We have
seen that the definite integral I f(x)dx may be interpreted as
the area of a portion of the xy plane which is generated by a
290 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
vertical line-segment, one end of the line-segment remains on the
curve y = f(x) and the other on the x axis, as it moves from
x = a to x = b.
Now suppose the form of a system of parallel cross-sections
of a solid is known and that the area of any one of the sections
can be expressed in terms of the distance of the section from
some fixed point of the solid. We may then consider such a solid
as generated by a plane cross-section as it moves throughout the
length of the solid.
EXAMPLE
The cross-sections of a certain solid are squares whose sides are equal to
the squares of the distances of their planes from one end of the solid. Find
the volume of the solid if it is 8 inches long.
SOLUTION. Take the origin as the pointed end of the solid and assume the
cross-sections perpendicular to the x axis. Consider the portion of the solid
between two cross-sections of distances
x and x -f Az from 0. It is evident,
since the solid increases in cross-section as
x increases, that
x*-Ax < AF < (x + A.r) 4 -Az,
where AF is the part of the solid between
the two cross-sections.
Dividing through by Ax we have
* 4 < < (X + Ax)4 '
FIG. 162
,. AF dV
hm - 55
Ax *o Az ax
Now let Ax > 0; we conclude that
4 . ( 52, IV.)
Therefore dV = x*dx\ and, integrating between the limits x - and x - 8,
we have
V = jf x*dx = y T = 6553| cubic inches.
PROBLEMS
1. Planes perpendicular to the x axis cut a solid in circular sections with
diameters joining y - x and y = 2 x. Find the volume of the solid between
the planes x = and x = 5. Ans. 125 Tr/12 cu. units.
2. Planes perpendicular to the x axis cut circles with diameters from
y = ltoz 2 = 5 y from a solid. What is the volume between the inter-
sections of the line and parabola?
158] APPLICATIONS OF INTEGRALS 291
3. A variable square moves with its plane perpendicular to the y axis
and with the ends of one diagonal on the ellipse 4 x 2 + ?/ 2 = 16. Find the
volume generated. Ans. 128/3 cu. units.
4. Plane sections of a solid perpendicular to the x axis are squares with
the ends of a diagonal on 16 y x 2 and 4 y = x 2 12, respectively. Find
the volume between the intersections of the curves.
5. A variable square moves with its plane perpendicular to the y axis
and with the ends of one side on y = x and y 2 = 4 x, respectively. What
volume is generated between the intersections of the line and the parabola?
Ans. 32/15 cu. units.
6. A variable equilateral triangle moves with its plane perpendicular to
the x axis. One side extends in the xy plane between the parabolas y 2 4 x
and 2/ 21 ~ 16 x. What volume is generated from the origin to x = 2?
7. An equilateral triangle moves so that its plane is perpendicular to the
y axis. One end of its base is on the y axis and the other end on the circle
x* + y z = a 2 . What volume is generated? Ans. (2/3) a 3 \/3 cu. units.
8. Sections of a solid perpendicular to the x axis are squares with a side
the double ordinate of x 2 + 4 y 2 = 4. What volume lies between x = 2
and x = 2?
9. In Problem 8, use the y axis, double abscissas, and y = to y = 1 .
Ans. 32/3 cu. units.
10. The base of a solid may be represented by the area bounded by y 2 = 6 x
and y 2 = 6 3 x. Cross-sections by planes perpendicular to the y axis are
equilateral triangles. Find the volume of the solid.
1 1. The same as Problem 10, except that the sections are squares.
Ans. 128/15 cu. units.
12. The same as Problem 10, except the sections are isosceles triangles
with altitudes the same as their bases.
13. Sections of a solid perpendicular to some line are squares whose sides
are the square roots of the distances to the sections from one end of the line.
Find the volume if the line is 6 units long. Ans. 18 cu. units.
14. A variable circle moves with its plane perpendicular to the x axis, its
center on y = sin 2 x and an end of a diameter on the x axis. What volume
is determined by one arch of the sine curve?
15. A chip of constant angle 6 is cut from a tree of radius r feet by a hori-
zontal saw-cut and an oblique axe-cut. What is the volume of the chip if it
extends halfway through the tree? Ans. (2/3)r 3 tan cu. ft
292 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
ADDITIONAL PROBLEMS
1. A curve passes through the point (0, 1) and its slope at any point is
3 5 x 3 x 2 . Find the equation of the curve.
Ans. 2?/=2 + 6a;-5x 2 ~2x 3 .
2. Find the equation of the curve through the origin whose slope at any
point is x/(l x 2 ).
3. A curve of slope cos 3 x passes through the point (u-/3, 2). What is its
equation? Ans. 3y = sin 3 x + 6.
4. Find the equation of the curve whose slope is xy 2 if it passes through
the point (4, 3).
5. Find the equation of the curve whose rate of change of slope with
respect to x is 2 -f 4 x, if the curve has a low point at (1, 3).
Ans. 3 y = 3 z 2 + 2 x 3 - 12 x + 16.
6. Find the equation of the curve through (1, 6) which intersects the
curves 2 y = log c(l + # 2 ) at right angles.
7. An object moving in a straight line has an acceleration proportional to
the square root of its velocity. It has an initial velocity of 16 feet per second
and comes to rest after 6 seconds. How far does it move? Ans. 32 ft.
8. If the acceleration of a particle along a straight line is 2/v and its
velocity is 3 units/sec, when t = 0, how far does it move during the next 4
seconds?
9. If the acceleration of a body is proportional to its velocity, find an
expression for the distance traveled if s = and v 1 unit /sec. when t = 0.
Ans. s = (l/k)(e kt - 1) units.
10. A man on a bank of a river 60 ft. above the water can throw a stone
with initial velocity of 100 ft. /sec. What angles of elevation should be used to
hit a spot on the water 125 ft. from the bank?
11. An investment of $4000 is depreciating at the constant rate of 4% per
year. When will the investment be worth $3000? Ans. In 7.19 years.
12. For a constant temperature, the rate of change of the pressure of the
atmosphere with respect to the height above sea level is proportional to itself.
Find the pressure at a height of 4500 ft. if it is 11.9 Ibs. at 6000 ft. and 15 Ibs.
at sea level.
13. Find the equation of the paths of a point whose rate of change of ordi-
nate with respect to its abscissa is proportional to the ordinate. Which one
of the curves passes through (0, 2) with the slope 4? Ans. y = 2 e 2 *.
14. If the rate of change of the slope of a curve with respect to the ordinate
at any point is proportional to the slope, find the path of the point.
Find each of the areas bounded as follows. (Nos. 15-20. )
15. y = 6 x x*, y - 0. Ans. 36 sq. units.
158] APPLICATIONS OF INTEGRALS 293
16. y = x\ y = 6 - z 2 .
17. x = 4 y 2 , x = 7/ 2 -f 1. Ans. 2VB sq. units.
18. 4 y = x 2 , 4 i/ = a; 3 , a; = 2.
19. y 2 = 2 px and its latus rectum. Ans. (2/3) p 2 sq. units.
20. The loop of y 2 = x 3 - z 4 .
Evaluate, if possible, each of the following improper integrals. (Nos. 21-26. )
21. f dx/(2x-l) 2 . Ans. >.
'O
22. f 4 x* dx/V5 - x 2 .
J ^l
23. y* 4 <&/(3 - 2z) 2 / 3 . Ans. 4.738.
24. 2
25. x dx/(l + a:) 3 / 2 . Ans. -
IT/2
26. / sec 4 (9 (fd.
/IT
6. /
/o
27. The cross-section of a solid perpendicular to a line A B is a right triangle
with its hypotenuse equal to the distance of the section from A. If AH is 4
units and one acute angle of the section is always ?r/6 , find the volume of the
solid. Ans. 8V3/3 cu. units.
28. Sections of a solid perpendicular to the x axis are squares with sides as
double ordinates of if =4 x. What is the volume between x = and x = 2?
29. A variable circle moves with its plane perpendicular to the x axis and
a diameter extends from 16 y = x 2 to 4 y = x 2 12. What volume is there
between the intersections of the curves? Anx. 9.6 TT cu. units.
30. A variable circle moves with its plane perpendicular to the y axis. It
passes through the y axis and its center is on the ellipse x 2 /a? + y*/b 2 = 1.
What volume is generated?
31. A trough 10 ft. long has cross-sections which are parabolic segments of
altitude and base each 4 ft. What is its volume? Ans. 320/3 cu. ft.
32. The retarding effect of fluid friction on a rotating disc is proportional
to the angular velocity of the disc. Find an expression for the angular velocity
at any time.
33. A pair of shears is wanted which has a constant cutting angle. What
equation may represent the edge of one blade? Ans. r = ce^.
294 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XII
34. Water flows out of a hemispherical basin through a hole at the bottom
in such a manner that the volume of the water at any time decreases at a rate
proportional to the square root of the depth of the water remaining in the
basin. What volume remains at any time and at what depth does the surface
of the water fall most slowly? Ans. (7r/i 2 /3)(3r h) cu. units; h = 2r/3.
CHAPTER XIII
THE DEFINITE INTEGRAL AS THE LIMIT OF A SUM
159. An Approximation of the Value of a Definite Integral.
A function, whose rate of change is known, has been found by
reversing the operation of differentiation. If this reversal is
inconvenient or impossible, an approximation of any desired
accuracy may be found for the value of the function or integral if
it is a definite integral. This can be done because we have seen
that any proper integral may be represented geometrically by a
definitely bounded area.
C b
Suppose the value of the proper integral I f(x) dx is desired.
/a
If we sketch the curve y /(x), as shown in Fig. 163, the numeri-
cal value of the integral is the num-
ber of square units in the area
MNPQ, regardless of what geomet-
ric or physical meaning the integral
may have. Obviously this area
may be approximated by finding
the sum of the areas of a system
of rectangles either inscribed under
the curve or circumscribed over it.
In constructing the rectangles shown
in Fig. 163, we take their width Ax
as any convenient divisor of the in-
terval MN. It appears that the resulting approximation becomes
better as we decrease the width of the rectangles and consequently
increase their number.
If the interval b a is divided into n equal parts, b = a + n Ax.
The lengths of the n rectangles will then be either /(a), /(a + Ax),
...,/[a + (n - 1) Ax], or /(a + Ax),/(a + 2 Ax), ,/(&).
EXAMPLE
/4
(26 x 2 )dx. Note that they become
better as the number of rectangles is increased.
295
M (a, 0)
FlG 163
296 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XIII
SOLUTION. The curve over the area to be evaluated is the parabola
y 26 x z . The three rectangles shown in the figure have a combined
area of (22 -f 17 -f 10), which is 49 units. Now suppose six rectangles under
the curve from x = 1 to x = 4. Their width is 1/2, their heights are 23.75,
22, 19.75, 17, 13.75, 10 respectively, and
their combined area is 53.125 units. Such
treatment may be continued until any
desired accuracy is acquired. Thus it
appears that the value of the given inte-
gral is the limit approached by such
sums as the width Ax of each rectangle
approaches zero as a limit. The exact
value is
(26 - x*)dx = (26 x - x*/3)
= 57 units.
We shall show in the next ar-
ticle that, if the curve y ~ f(x) is
continuous in the interval a r^.x < b, the value of the definite
is
so
15
10
s
Y
[N
\
\
\.
1 2 3 4 5^ X
FIG. 164
Xb
- f(x)dx is such a limit.
PROBLEMS
Draw a figure representing each of the following integrals and use Ax =0.5
unit to approximate the value of each. Either inscribed or circumscribed
rectangles may be used. The figure should help you decide which to use.
Ans. 5.77 units.
Ans. 0.465 unit.
r . _
. / dx/Vl -f- x 2 .
4. f 5 dx/ V4 x 2 - 5.
5. dx/(2 x 2 + 3) 1 / 1 .
J 2
6. (4 x 2 + I) 1 / 8 dx.
7. /" 2 (1 -f x 2 ) 2 / 3 <&.
/i
8. f l (2 - x*)*'* dx.
Ans. 1.104 units.
An*. 4.34 units.
160] DEFINITE INTEGRAL AS LIMIT OF A SUM 297
160. The Definite Integral as the Limit of a Sum. Since in
most applications the meaning of the definite integral is something
very different from an area, it is desirable to prove the following
analytic theorem without any reference to its possible geometric
interpretation. The importance of the theorem in its wide range
of applications is such that it is generally called the fundamental
theorem of the integral calculus.
THEOREM. Let F(x) and its derivative f(x) be continuous func-
tions of x in the interval a =j x ^ b. Divide this interval into n
sub-intervals A#i, Ax 2 , , Ax n , where Ax- > 0.
Let Xi be any value of the variable in the corresponding sub-interval
Ax. Then the limit of the sum of differential products
as n increases indefinitely and each Ax t decreases and approaches
zero, is the value of the definite integral
C b
Ja
dx s F(V) - F(a).
In symbols this conclusion is written
lim /(Xi)Ax t = f(x)dx = F(b) -
n > oo i=l
PROOF. Let the values of the variable for the points of sub-
division be aij a^ a^j , a n _i. That is, a + A^i = a\, ai + Aa?2 = a^
- - , a n -i + A;r n = b. Then by the Theorem of Mean Value,
90, we have
F(x + &x) -F(x) =/(2)-Ax,
for some x in the interval Ax. Or, since f(x) is a continuous
function,
F(x + Ax) - F(x) = [/(*') + e]A*
for any x' in the interval Ax, where lim 6 = 0.
As
Applying this theorem successively to each of the sub-intervals,
we have the relations
298 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
- F(a) =
F(a 2 ) F(OI) =
F(a 3 ) - F
The sums of corresponding members of these give, for any num-
ber w,
*=i
Now let n increase indefinitely by making each A# t approach
zero. Each change; in the value of n produces a new set of e's such
that* lim e l = 0. Then we have
Ax t >-0
F(b) - F(a) = lim I" f) /(,) Az v + f) e.-'A
Now consider the second sum of the right-hand member. Among
the e'& there is an e k} such that |^ for a given n is greater than or
at least equal to any other Ci . Then since
n
6i Ax = 61 Ax + <? 2 Ax 2 + + e n Ax n ,
t=i
n n n
E ^'AXt ^ Z kt * Ax t < e^ X! Ao:.
i=l t=l tl
n
But by hypothesis X! Ao;* = 6 a for all values of n and e t - ->
t = i
as Ao;,- 0. Hence
n
lim I XI e'Ai I ^lim [e k (b a)] = 0,
if the interval b a is finite. Therefore
b f(x)dx
n > oo t = 1
s C b f(x)dx = lim
*/ a n > oo
* This is due to a theorem in a more advanced course in which such a function
as f(x) is known as uniformly continuous.
161] DEFINITE INTEGRAL AS LIMIT OF A SUM
299
When the interval b a is not finite, the result is an improper
integral which has been treated in the preceding chapter.
The quantities Ax t - and e< are infinitesimals, being variables
each of whose limits is zero. If an infinitesimal Ao^ is multiplied
by some finite number not zero, say /(x,-), the product is still
an infinitesimal and the two infinitesimals /(x t ) 'Ax and Ax t - are
of the same order. But the products i Ax t of two infinitesimals
are infinitesimals of higher order than either e or AJC (85).
n
Since the lim c l Ax t - was proved to be zero, these infinitesimals
n + i=l
A# t
6i Ax t of higher order than the first could have been omitted with-
out affecting the value of the limit of the sum which gives the
integral. This is of very great importance because often the
element of integration cannot be expressed exactly as f(x l ) Ax,-,
but includes some additional infinitesimal of higher order. How-
ever, such infinitesimals may be discarded, since the limit of their
sum as the number increases indefinitely is zero in all cases with
which we shall deal.
In applications of the fundamental theorem, the expression for
the element of integration is simplified if the Ax\ are taken equal
and if the x t are chosen as divisional values of the variable. As
this choice is possible under the general proof given above, we shall
use it in illustrations.
161. Geometric Illustration of the Fundamental Theorem.
Suppose AB is the graph of the continuous curve y = /(x) from
x = a to x = b. The rectangles of width Ax under the arc have
as the sum of their areas
(1) /(a)- Ax + /(a + Ax) -Ax
+ 2Ax)-Ax +
+ /(& - Ax)- Ax,
(a, o) o
a quantity which is less than the
area under the arc.
The rectangles over the arc of the
curve have as the sum of their areas FIG. 165
(2) /(a + Ax) Ax + /(a + 2Ax)-AxH ----- h/(&)-
which quantity is greater than the area under the arc.
(b t o)
300 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
But these two sums differ in only the first term of (1) and the
last term of (2). Their difference is then
(3) [/(&)- /(a)]Az,
which is the area of the shaded rectangle of Fig. 165. Obviously,
this difference approaches zero as a limit as A# > and hence each
sum approaches the area under the arc as a limit and again we have
/
t/a
f(x)dx = lim
n oo
Ax -0
Iii this discussion we have assumed that along the arc AB, as
x increases from a to b, y increases continuously.
If y decreases as x increases, the same situation would occur
except the rectangles under the curve would be represented by (2)
above and those over the curve by (1). Finally, if the arc AB
has a finite number of maxima and minima, the reasoning given
above would be applied to each segment of arc along which y
steadily increases or decreases as x increases.
162. Geometric Illustration Using Polar Coordinates. The
same geometric reasoning may be applied if the integral is expressed
in polar coordinates. For instance, the area OPQ may be approxi-
mated by the sum of a set of
c i rcu i ar sectors of angle A0 either
inscribed in the arc PQ or cir-
cumscribed about it. One such
sector has the area r\ 2 'A0/2 and
the difference of the two sums is
evidently the shaded area, which
is represented by
This difference approaches zero as a limit when A0 > and hence
the limit of either sum is the area of OPQ. We may write then
A =lim ErS-M
n_-co t=i A
A0 *-0
Consequently, any integral of the form / n f(0)dO may represent
t/00
an area and is the limit of the sum of a set of circular sectors, if /(0)
163] DEFINITE INTEGRAL AS LIMIT OF A SUM 301
is computed from the expression r 2 /2, where r = 0(0) is the equa-
tion of the bounding curve, such as PQ.
Since any proper integral in polar coordinates may accordingly
be interpreted as an area or has the numerical value of the square
units of a definite area, the student should see that the theorem
of the preceding article may have different types of illustration.
Also, assumptions analogous to those made in the discussion
involving x and y must be made here concerning continuity and
variation of r and 6.
163. Areas. The idea of the definite integral as the limit of
a sum permits us to set up the expression for the area bounded by
given curves as the limit of the sum of elements of the area. Each
element is a rectangle if the bounding curves are given in rectangular
coordinates, a sector of a circle if the equations of the curves are
expressed in polar coordinates. In either system of coordinates
the student must be careful to choose, as elements of the desired
area, rectangles or sectors which have the same characteristics through-
out the area, that isj every such element has its area represented by
the same type of differential product. In the case of rectangular
coordinates this means that all rectangles needed for a good approx-
imation of the desired area must have their corresponding ends
touching the same curves. It may be necessary to use horizontal
rectangles to bring this about. If all elements cannot be made
of the same type, it is necessary to divide the area into parts over
each of which typical elements may
be used. In polar coordinates all
sectors must reach from the pole to
the same curve, and, if the element
is the difference of two sectors, the
corresponding ends of such elements
must touch the same curves.
EXAMPLES
1. Find the area between the curve
y ~ 4 x z 2 and the x axis.
SOLUTION. A figure is an essential part
of a solution of this type of problem.
Solving for the intersections of y 4 x x* and y
the x axis, x = and x = 4.
FIG. 167
0, we have as limits along
A typical element shown by the shaded area is
302 DIFFERENTIAL AND INTEGRAL CALCULUS [Cii. XIII
where < k t < 1 and /r t A?/ t Ax of higher order than y t Ax may be discarded
in taking the lim S AA t . Therefore
n > oo i=l
A = lim
n * co i
Ax M)
/*4
= /
/0
Then
A = J (4 z - 3 2 )<te,
since y refers to the ordinate of the given parabola. Whence
(X A \ ~1 4 64 32
2 x 2 - --- I = 32 - = V square units.
1/Jo o o
2. Find the area bounded by the two curves y = x 5 x 2 and y x 2 .
SOLUTION. Solving simultaneously, we find the intersections of the curves
at x = 0, and x = 2. Let 2/2 and yi represent any corresponding ordinates of
the parabola and the cubic respectively. Then the shaded area is
where both ki and k 2 lie between and 1, and where (kz A?/ 2 fri Ay\) l Aar,
of higher order than (t/ 2 Z/i) Ax, may be discarded in the limit giving A.
Whence
A =li
o
= / "
/()
Therefore
- r 2
~ ^0
16
3
(2 x 2 - x*)dx
16
4
- square units.
o
FIG. 168
The expression for the element
of area AA r should always be set
up before integration and care
must be taken to let y\ represent
an ordinate of one curve and 7/2 the corresponding ordinate of the
other curve. Then if read positively, as y 2 y\ in Fig. 168, and
if the sum is taken in the positive direction along the x axis, the
result of integration is always positive. Such an element must be
set up for each distinct part of a composite area.
163] DEFINITE INTEGRAL AS LIMIT OF A SUM
303
3. Find the area of the cardioid r = a(l 4- cos 0).
SOLUTION. The shaded element of area is
= r> 2 A0 - /c t [r t 2 - (r, + Ar t ) 2 ]A0, < fo < 1,
where (l/2)fc,(2r t Ar -f Ar t ) A0 is of higher
order than (l/2)r t 2 A0. Whence
A = lim X r^r t 2 -A01 = 2 f'^
n-^oo i=l |_2 J /<) 2
A0 >"0
Therefore
A = a-
(1 -f 2 cos0 + cos 2 0)d0
'U
= a 2 ( 9 + 2 sin + -? + ^J
V 24
= 3 7ra 2 /2 square units.
FIG. 169
4. Find the area outside the circle r = 2 a cos and inside the cardioid
r = a(l + cos 0).
SOLUTION. Let r^ and n represent any corresponding radius vectors of the
cardioid and circle respectively. Evidently
the element of area has different values
in the first and second quadrants. The
shaded area in the first quadrant is
A where r t A0 is an infinitesimal of higher order
than (r 2 2 ri 2 ) t A0.
In the second quadrant the shaded area is
and t A0 is of higher order than (r 2 2 ) t A0.
FIG. 170
Therefore
A =Hr
2
l
*
(1 +
= a 2 /"* C(l + cos 0) 2 - 4 cos 2 0]dO H- a 2
^0 7T
= a 2 [2 sin - | - j sin 2 0T /2 + a 2 [~2 sin + ^ + ~ sin 2 0T
L J 4 J L 24 J T /2
= a2 ( 2 - l) + 2 (if ~ 2 ) = ^ S( l uare units -
304 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XIII
We call attention to the fact that an element like A<A above can
only be used if r 2 and n have the same sign for any 6 in the interval
of integration; otherwise r 2 and r\ would lie in different quadrants
and AAi would not represent an element of area between the
curves.
PROBLEMS
Find the area bounded by the following curves. Draw each figure.
1. x 1 y -\- 1, x y + 1 = 0. Ans. 4J^ sq. units.
2. y 2 -x = 10, x -2y + 2 = 0.
3. xy = 8, 2 x -f y = 10. Ans. (15 - 16 log 2) sq. units.
4. The parabola y 2 = 2 px and its latus rectum.
5. One arch of y = cos 2 2 x and the x axis. ylns. T/4 sq. units.
6. y sin x and i/ = cos x between two consecutive intersections.
7. r = a cos 3 6. Ans. 7ra 2 /4 sq. units.
8. r = a sin 2 0.
9. Smaller loop of r = 1/2 - cos 6. Ans. (1/8) (2 TT - 3V3) sq. units.
10. Larger loop of r = 1/2 -f sin 0.
11. Find the area enclosed by r 2 = a 1 sin 2 0. Ans. a* sq. units.
12. Prove that the entire area enclosed by the loops of the curve
r a sin kO equals one-fourth, or one-half, the area of the circle of radius a,
according as k is an odd or an even integer.
13. Find the area inside r = 2 sin and outside r = 2(1 sin 0).
Ans. 4(V3 7T/3) sq. units.
14. Find the area inside r = 2 and outside r = 4(1 cos 6).
15. Find the area of the limagon r = 3 -f 2 cos 9. Ans. 11 TT sq. units.
16. Find the area of the loop of the curve y 2 = x 2 -f x*.
17. Find the area within the hypocycloid x 2 ' 3 -f- 2/ 2/3 = a 2 ' 3 .
Ans. 3 ?ra 2 /8 sq. units.
18. Find the area of the curve z 2 /a 2 -f r/ 2 / 3 /& 2/3 = 1-
19. Find the area which is enclosed by the x axis and one arch of the cycloid
x a (0 sin 0), y = a(l cos 0). Ans. 3 Tra 2 sq. units.
20. Find the area under one arch of the curve, x = a0, T/ = a(l cos 0).
21. Find the area between the witch y = 8 o 3 /(# 2 + 4 a 2 ) and its asymptote,
y - 0. 4ns. 4 xa 2 sq. units.
22. A goat is tied to a staple on the outer side of a circular fence of radius
a units by a rope of length *a units. Find the area over which the goat can
graze.
164. Solids of Revolution. The volume of a solid generated
by revolving the area bounded by plane curves about some line in
the plane may be evaluated very easily by use of the fundamental
164] DEFINITE INTEGRAL AS LIMIT OF A SUM
305
theorem. The following examples give very satisfactory elements
of volume; one or more such elements can be used in all evalua-
tions of volumes of revolution.
EXAMPLES
1. Find the volume generated by the area under 2 y = x 3 from the origin
to the point (2, 4) when revolved about the x axis.
SOLUTION. The shaded area when revolved
about y = generates
A F t = TT^-AZ
< A; < 1,
where irk(2yi AT/* H-Ai/t )Az, the second term,
may be discarded, since it is an infinitesimal
of higher order than iry^'&x. Therefore we
have
r 2
V = lim L 7rt/ t 2 'Ax = 7T I i/ 2 cfo
n+>ao - 1 /0
32 i
cubic units.
FIG. 171
Since the elements of volume may be considered as vy^-Ax without affecting
the value of the limit of their sum, we may consider this type of element as
a circular disc generated by ?/ t Ax as it
2 4) turns about the x axis.
2. Find the volume generated by revolv-
ing the area of Example 1 about the y
axis.
SOLUTION. Let the shaded area gen-
erate the element of volume; since this
element may be considered as the difference
of two concentric circular discs, or a washer,
we have
= 7r(2) 2 At/ - 7r^ 2 A2/,
discarding infinitesimals of higher order than
the first. Therefore
V = lim 2 O(4 - x
i = 1
(4 -
= * f* [4 _ (2 y)V*yy = ^ cubic units.
'o o
In each of these examples the element of volume has been
306 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
generated by a rectangle of variable length perpendicular to the
axis of revolution. However, the elements of volume assumed
different forms. If the rectangle is taken parallel to the axis of
revolution, the volume AF which it generates is a thin cylindrical
shell
Consider the second example when AF t is generated by the shaded area in
Fig. 173. Then AF, = 2 TTZ,?/, A:r except for infinitesimals of order higher
than the first. Therefore, applying the fundamental theorem, we get
V lim X AF t = 2 TT / xy dx ~ TT / x 4 dx '~~- cubic, units.
n *co=l /<) /U f)
This solution shows that the cylindrical shell element can be considered as the
product of the length of the rectangle, the width of the rectangle*, and the
inner circumference of the shell.
2 X
FIG. 173
Fio. 174
Figure 174 shows a similar element for Example 1. In this case the funda-
mental theorem gives for F, in cubic units,
V = lim
> = 2 T
(2 - x)dy =2
PROBLEMS
[2 - (2 j/)'/'] dy =
1. Find the volume of a sphere as a solid of revolution. Set up the
integral using (a) disc elements, (b) cylindrical shells.
Ans. 4 7ra 3 /3 cubic units.
2. Find the volume of an ellipsoid of revolution by using (a) and (b) of
Problem 1.
3. Find in several ways the volume of a paraboloid of revolution of
altitude h units and diameter of base 2 6 units. Ans. irhb*/2 cubic units.
165] DEFINITE INTEGRAL AS LIMIT OF A SUM 307
4. Revolve the area under y = log x about the x axis and find the volume
between x = 1 and x = e.
Find the volumes generated by the areas bounded by the curves below
when revolved about the given line. Use different types of elements. (Nos.
5-17.)
5. y = x, x 4, y about y = 2. Ana. 1607T/3 cu. units.
6. xy = 12, x = 2, x = 4, y about x 1.
7. y = x 2 , x = 2, y = about x = 0. Ans. 8 IT cu. units.
8. ?/ 2 = 4 x, x = 4 about = - 2.
9. x 2 = 4(1 - ?/), y = about ?/ = 3. Arcs. 2087T/15 cu. units.
10. .n/ = 12, y = 0, x = 2, z = 3 about y = - 7.
11. ?/ = x 3 , x = 1, y = about x = 2. yl^s. 7^/5 cu. units.
12. The loop of ?/ 2 = .r 4 (x + 4) about the ?y axis.
13. A circle of radius 3 units about a line in its plane that is 7 units from
its center. Ans. 126 ?r 2 cu. units.
14. y = 8/0 2 + 4), y = 1, x = about y = 0; x = 0.
15. ?/ = 2 e 2 *, ?/ = e x t x = 0, x = 1 about y = 1.
Ans. (7T/2) (2 e 4 - 5 e 2 -f- 4 e - 1) cu. units.
16. One arch of y = sin x, ?/ = 0, about y = 2.
17. One arch of x = sin 2 y t x = 0, about x = 2.
Ans. ?r(16 -f- 7r)/4 cu. units.
18. A vessel has the form generated by revolving x* = 4 y about x = 0.
If 4 cu. units of liquid leak out per minute, at what rate is its depth changing?
19. What volume is generated if r = a(l sin 6} is revolved about its axis?
Ans. 8 7ra 3 /3 cu. units.
20. Prove that the volume formed by rotating the area under y = sin x
from x = to x = IT about the y axis is four times the volume generated when
the same area is rotated about the x axis.
165. Length of a Plane Curve.
Let s represent the length of a curve
y = /fa) from the point P(a, 6)
to Q(c, d). Divide the interval
a i x i c of the x axis into n equal
parts Ax t and erect ordinates to the
curve from each point of division.
The broken line joining the upper
ends of these ordinates has as its
length an approximate value of the
length of the arc PQ; when the number of segments increases in-
definitely and each segment approaches zero as a limit, the length
of the broken line approaches the length of the arc PQ as a limit.
FIG. 175
308 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
If each segment of the broken line is represented by As the
length of the broken line is given by
= E X/AoT t 2 + Ai/, 2
, I
Now, by the Theorem of Mean Value, 90, A^/Aa;* is equal to
the value of the derivative (tyi/dxi for some point (#, J/) of the
curve in the interval Arc t . Therefore
i- ^ * 1- v- /i , /dy\ z A
s = hm As< = Inn L \ M + ( 7/;~ ^;
or
(1) 5 =
if no ordinate of the curve meets the arc more than once.*
If y is used as the variable of integration, the same process,
when AT/J is factored from the radical, gives
if no abscissa meets tho arc of the curve more than once.*
The integral giving the length of a curve may be written
(3) s = J*
+
Then if x and y are each expressed in terms of a third variable,
dx and dy can be replaced by their values in terms of that variable.
In polar coordinates, by the substitution of their values for dx
and d\j in (3), we have
(4) s = /\/dr 2 + r 2 d6 2 ,
and we choose as the variable of integration the one which makes
the integration simpler. In using (4), the dependent variable
must be a single-valued function of the variable of integration.*
* If this condition is not satisfied, consider smaller pieces of the arc for each of
which tho condition holds.
165] DEFINITE INTEGRAL AS LIMIT OF A SUM
309
EXAMPLES
1. Find the length of the curve 9 x 2 = 4(1 + j/ 2 ) 3 from (2/3, 0) to the
point (10\/5/3, 2).
SOLUTION. It appears easier to
solve for x than for y. We have
therefore
whence
^ = 2 y
if x > 0. Then
Vl
FIG. 176
(1+2 y*)dy
2. Find the length of one-half of one arch of the cycloid x 2 (0 sin 6),
y = 2(1 cos 6}.
SOLUTION. Since the curve is given by parametric equations, we shall
use the differential form (3) of the formula for the length of a curve 1 and simplify
after substituting for dx and dy. That is,
s = f ~" Vdx* + dy* = f" x/4(l - cos 0)* + 4 sin 2 B dO
= 2 f* V'2 - 2 cos 6 d6 = 4 C" sin ^dd.
Jo Jo 2
= 42 cos - =8 units.
L 2jo
3. Find the length of the curve r = a(l sin 0}.
SOLUTION. Using the form Vdr* + r' 2 d6 2 , we have, from symmetry,
s = 2 f ' * Va 2 cos 2 B + a 2 (l - sin 6)* dO
3r/2
/r x
V 2 - 2 s
-/2
sin 6 dO
z sin ^ cos - t
37T/2
FIG. 177
8 a units.
310 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
PROBLEMS
1, A curve has ds = (1 -f x*)dx. What is its length from x = 1 to x = 3?
Ans. W% units.
Find the length of each of the following curves. (Nos. 2-21.)
2. y 2 = 4(1 + z 2 ) 2 ' 3 from x = to x = 3.
3. The semi-cubical parabola ?/ 2 - (x - 2) 3 from (2, 0) to (G, 8).
/ins. 9.07 units.
4. The semi-cubical parabola ?/ 3 = s 2 in the first quadrant from y to
V = 4.
5. // = log sec x from (0, 0) to (ir/3, log 2).
yi/i.s. log tan (5 ir/12) units = log (2 + \/3) = 1.317 units.
6. y = a log x from (1, ()) to (a, a log a).
7. ?/ = log (I - x" 1 ) from a; = to = 1/2. j4ws. log 3-1/2 units.
8. y log esc x from x 7r/3 to x = ir/2.
9. a; 2 / 3 4- 2/ 2/3 = a 2/3 . Aws. Ga units.
10. The hypocycloid x a cos 3 0, y a sin 3 0.
11. One arch of the cycloid x - a (0 sin 0), y = a(l cos 0).
Ans. 8 a units.
12. y = 2 log [4/(4 - x 2 )] from x = to x = 1.
13. y = 4 log [167(10 - z 2 )] from ^ = to x = 2.
yl^s. 4 log 3 2 units.
14. The cardioid r = 2(1 - sin 0).
15. r = a cos 4 (0/4) from = to = 2 TT. ^Ins. 8 a/3 units.
16. The cardioid r = a(l -f- cos^?).
17. r = a cos 3 (0/3) from = to = 3 TT. ^4n5. 3?ra/2 units.
18. x c~ l cos 2 t, y c~ l sin 2 from = 0, to t = 7r/4.
19. a; = c - (J / 2) ' cos t, y = e - (3/2)< sin ^, from t = 2 a/3 to t = a.
Ans. (Vl3/3)(e- a - e~^^ a ) units.
20 The involute x = a (cos / + / sin y ~ a (sin i t cos /) from t =
to * = ti.
21. The catenary y = (a/2)(e x / -f e~ I/a ) from # == to z = ari.
A?is. (a/2)(e ll/0 e-*i/ a ) units.
166. Mean Value of a Function. The limit of the average
value of a finite number of values of a function of x, taken at
equal intervals of x, as their number increases without limit is
called the mean value of the function.
166] DEFINITE INTEGRAL AS LIMIT OF A SUM 311
Let q(Xi), i = 0, 1, 2, , n 1, represent n values of the func-
tion q(x) such that n-Ax = k, the interval of x. Then the average
value of the q(xC) is
n
Using the summation symbols and replacing n by its value
k/Ax, we may write this in the form
I n-l
Average of g(z) = 7 Z gO) Ax-.
AC 1 =
Hence, from the definition above,
i w ~ i i r b
Mean value of q (x) = T lim q(x l )-Ax = r I q(x)dx,
/C n _^ oo i = It is a
where a and ?) are derived from the interval over which x is allowed
to vary in computing the average value of q(x t ).
EXAMPLES
1. Find the mean value of the ordinate to the curve ?/ = cos x if the ordi-
nates are erected at equal intervals from x = to x w/2,
SOLUTION. Assume n such ordmatos in tlie given interval. Then calling
the average ordinate A, we have by defi-
nition
A =
2/0 + yi + 3/2 -f- - + yn-i
, n-l
Hence
= - 2 T/ A, since n Ax = -
" i = ^
n-l
Mean ordinate = - lirn 2 2A'A:e
2 r*'/ 2 , 2
- I cos a: aa: = ^. units.
FIG. 178
2. Find the mean value of the area of squares constructed on a set of
parallel chords of a circle if the chords are at equal intervals along the circum-
ference of the circle.
SOLUTION. Suppose the square A BCD constructed on AB, any one of the
312 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
chords. Such a chord is 2 x\ in length and hence each square has the area
4 Xi\ The equal intervals are As t and n As = wr. Therefore
-f 4 xS -f + 4 *JLi
Average area =
Hence
n-l
n-l
Mean area = - - Iim 2 x t 2 - As
4
--
irr
x*ds
FIG. 179
since ds for the circle is equal to (r/x}dy. Accordingly, we have
Mean area = ^ I -- (yVr 2 y 2 -f r 2 sin" 1 J
= 2 r 2 square units.
PROBLEMS
1. Find the mean vortical width of tho area bounded by y = x* and
x y 2 with respect to, or at, equal intervals of x. Ans. 1/3 unit.
2. Find tho moan vertical width of the loop of y 2 ~ z 2 (4 x) if measured
at equal intervals along the x axis.
3. Linos are drawn parallel to the x axis in tho area bounded by xy = 6,
y = x, ij = 2. If these lines arc equally spacod along tho y axis, what is their
mean length? Ans. [3 log (3/2) - 1]/(V6 - 2) units.
4. What is the mean value of x log x with respect to x from x = 2 to
5. The same as Problem 4 for 2 (log x) 2 from x = 1 to x - 4.
Ans. (1/12)[32 (log 4) (log 4 - 1) + 15] units.
6. What is the mean value of the square of the ordinate of y = -f- V4 x 2
with respect to x?
7. What is the mean length of the radius vector of r = 4 sin 2 9 for one
loop with respect to 0? Ans. 8/ir units.
8. The same as Problem 7 for r = a(l -f cos 0) in the first quadrant.
9. Tho same as Problem 7 for r 2 = a 2 (l -f cos 0) in the first quadrant.
166] DEFINITE INTEGRAL AS LIMIT OF A SUM 313
10. Find the mean area of isosceles triangles inscribed in a circle of radius
10 units if they have a common vertex and successive bases cut off equal
segments on the diameter through that vertex.
11. The same as Problem 10 for equal arcs cut off by successive bases.
Ana. 200/7T sq. units.
12. What is the mean volume of cylinders inscribed in a sphere if their
altitudes have a common difference?
13. The same as Problem 12 for diameters with a common difference.
Ans. 7r 2 a 3 /8 cu. units.
14. Find the mean area of circles on the double ordinates of 4 x 2 + y z 36
as diameters, if they are spaced equally along the x axis.
15. What is the mean area of right triangles in r = a cos 0, with the hypote-
nuses along the initial line and radius vectors at equal intervals of from
to 7T/2 as one side? Ans. a 2 /2 TT sq. units.
16. Find the mean value of the widths of the parabolic segment y z 2 x 1
x = 3, (a) if the widths are taken at equal intervals along the y axis; (b) the
x axis. Why are the results different?
17. Rectangles arc inscribed in the ellipse x z /a 2 -f y*/b z =1. If the
altitudes are at oqual intervals along the x axis, prove that the moan value of
the area is the same as when the bases are at equal intervals along the y axis.
18. Find tho mean ordinate of a quadrant of a circle if the ordinates are at
equal intervals along (a) the arc; (b) the x axis; (c) the y axis.
ADDITIONAL PROBLEMS
Find the areas bounded as follows. (Nos. 1-9.)
1. r = a sin 4 0. Ans. 7ro 2 /2 sq. units.
2. y = x\ y = 8 - x\
3. r = a(l + cos0). Ans. 3 7ra 2 /2 sq. units.
4. 4 x 2 - 9 y 2 -f 36 = 0, above y = 1.
5. z 2 /9 -t- 7/V16 = 1. Ans. 12 ir sq. units.
6. r 2 = 9 sin 2 0.
7. Outside r = a, and inside r =* 2 a sin 2 0.
Ans. a*(V3 -f 27T/3) sq. units.
8. Outside r = a sin 0, and inside r = a (3 4- 2 sin 0).
9. Inside r = 5 sin and outside r = 2 -f- sin 0.
Ans. \/3 -f 8 7T/3 sq. units.
Find the volume of the solid of revolution generated by rotating each of the
areas denned below about the line specified. (Nos. 10-15.)
10. y 2 = 8 x, x * 2 about x = - 2.
314 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIII
11. y 2 = x, x -f y = 2 about the y axis. Ans. (14%) TT cu. units.
12. 16 J = // 2 , 4 x - y/ 2 - 12 about x = 1.
13. y = z 2 -f 1, ?/ = 4- 3 about a: = 2. Ans. (13J^) TT cu. units.
14. y c~ x from x to = *> about ?/ = 0.
15. A parabolic segment of altitude a units and base 2 6 units about its
base. Ans. 16 ira 2 6/15 cu. units.
Find the length of each of the following curves. (Nos. 16-17.)
16. x = /, // = log set; from = to t = ir/4.
17. /y - *VO -f l/(2x) from z = 1 to x = 3. 4ws. 4% units.
18. Find the perimeter of the area bounded by ?/ = e x + e~ x and ?/ = 2.
19. A suspended wire is in the form of a parabola. Its ends are 2 I units
apart on a horizontal line and its dip is h units. Find its length.
Ans. \// 2 -f- 4 h* -f- (/Y2 h) log [(2 /i -f- V/ 2 + 4 A 2 )/q units.
20. Find the mean vertical width of the loop of 4 y~ = x 2 (4 2 x) with
ro.spoet to .r.
21. Find tho_mean value of the square of the ordinates of the semicircle
y = -f V'9 x- if they are equally spaced along the .r axis. Am. 6 sq. units.
22. Problem 21 except the ordmatcs are equally spaced along the arc.
23. Find the mean velocity of a falling body with respect to the distance
fallen, if it falls from rest for 5 seconds. Ans. (3J)gr ft. /sec.
24. Find tho mean volume of cylinders inscribed in a right circular cone of
altitude <S in. and radius 4 in., if the upper bases are equally spaced along the
axis of the cone.
25. Find the mean volume of cylinders inscribed in a sphere if the angles
at the center subtended by the altitudes have a common difference.
Ans. 4a 3 /3 cu. units.
26. Find the volume common to two equal circular cylinders if their axes
meet at right angles.
27. When a liquid flows through a straight pipe of radius a, the velocity of
flow at a distance r from the center of the pipe is &(a 2 r 2 ). Find the mean
value of the velocity along a diameter. Ans. 2 a 2 /3 units per unit of time.
28. Oil is poured into a tank containing water so that the density of the
mixture is decreasing at the rate of 0.2 Ib. per minute. Find the mean value
of the density of the mixture with respect to the time during 5 minutes.
Ans. 62 Ibs.
29. Assuming that oil flowing in a straight pipe of inside radius a moves in
straight lines parallel to the axis of the pipe, and that the velocity at a distance
r from the axis is t\>(l r 2 /a 2 ), where r is the velocity at the center, find what
volume of oil will flow past a given point per second.
166] DEFINITE INTEGRAL AS LIMIT OF A SUM 315
30. A carpenter chisels a square hole of side r inches through a round post
of radius r inches; the axis of the hole meets that of the post at right angles
and two sides of the hole are parallel to the axis of the post. How much wood
is cut away?
31. Two equal circular cylinders of radius 3 inches intersect so that their
axes meet at an angle of 60. Find the volume of the part common to the
two cylinders. Ana. 96 v 3 cu. in.
32. Given the curve y (log.T)/V\r. Find the volume of the solid of
revolution formed by revolving about the x axis the urea bounded by the
curve, the x axis, the ordinate through the maximum point, and that through
the inflection of the curve.
33. Find the mean horizontal width of the area bounded by y = x t/2 and
y = x 3 with respect to y. Ans. 5/12 unit.
34. Problem 33 with respect to x along the first curve.
35. Problem 33 with respect to x along the second curve.
Ans. 5/14 unit.
CHAPTER XIV
MULTIPLE INTEGRATION
167. Areas by Double Integration. If the bounding curves
of a plane area are given in rectangular coordinates, we may divide
the area into small elements of dimensions Ax by Ay by means of
two sets of parallel lines at intervals of Ax and Ay along the x axis
and the y axis, respectively. The
sum of the areas of those elements
which lie entirely within the bound-
ary may be used as an approxima-
tion of the enclosed area. The
limit of such a sum as both Ax and
Ay approach zero as a limit is the
area enclosed by the curves. That
is,
A = lim X X Ay*Az.
FIG. 180 jg
This double sum symbol means H(XAy)A:r. Therefore the
m n
expression lim X! X Ay -Ax means lim X (lim X Ayi)Ax ? , which
may be written in the form
lim X
tn---a> j
Azv K)
2/2 \
dyJAx,',
where the fundamental theorem has been applied to the quantity
in the parenthesis. During this process x } and Ax 7 - are kept con-
stant. Now, considering the parenthesis of the last expression as
the/(x ; ) of the fundamental theorem, since 7/1 and 2/2 are functions
of Xj alone, we have, by a second application of the theorem,
(1)
A = lim X] Ay- As
Av-M)
316
f ( f^
J a \/t/,
dx.
167] MULTIPLE INTEGRATION 317
This is called a double integral; it gives the value of ^1 and is usually
written in the form
nVi
dydx.
-L
It should be observed that the second integral sign belongs with
the first differential in such double integrals. Also it must be noted
that during the first integration, which gives the area of the strip
MN, x does not vary. The limits on the second integral sign are
usually variables defined by the y fi(x) equations representing
the curves bounding the area, but the limits on the first integral
sign are always constants, the extreme values of the last variable of
integration for the area being evaluated.
If the fundamental theorem is applied to the double sum in
reverse order, we get
(2) A = lim Ax- Ay = f Pd* dy.
Az *0 t/c t/j
Reversing the order of integration is sometimes very desirable,
as it may simplify a difficult integral.
Performing the first integration of (1), we have
c b
A = I (2/2 - yi)dx,
Ja
which is the form of the single integral representing the desired
area. This fact should help the student to write the proper limits
for the double integral.
If the plane area is bounded by curves given in polar coordinates,
it may also be evaluated by double integration.
Let the area OPQ be divided into small four-sided elements by
means of concentric circles and radial lines at intervals of Ar and
A0 respectively. The element shaded in Fig. 181 is
,
= rvAr-A0 + (1/2) Ar 2 -A0.
2
The quantity (1/2) Ar -A0 is an infinitesimal of higher order than
318 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
7VAr-A0, as it can be written (l/2)(Ar)(ArA0); hence it can
be omitted from the element. Then we have
(3) A = lim r t .Ar.A0
A0+0
rt rf (0 >
= / / rdrd6,
Joi Jo
Q(r,,e s )
where r = /(0) is the equation of
the curve PQ.
We shall illustrate the evaluation
of areas by double integration by solving examples previously
solved by single integration in 103. The student should compare
the two methods and observe that in the new solution of each ex-
ample the final integral obtained is the integral used in the pre-
vious solution.
EXAMPLES
1. Find the area bounded by y and y = 4 x 2 .
SOLUTION. Suppose the desired {iron divided into elements Ay by Ax as
shown in Fig. 182. Then A A = Ay>Ax, and
NA
\
A lim SS
= r 4 r 4x
*A) Jo
dy dx,
since yi = and r/ 2 = 4 x x 2 and the
extreme limits of .r are and 4.
Integrating with respect to r/, we have
A =
dx
FIG. 182 where the integrand represents the strip
which was used as the element in single
integration. The final integration gives
/i = f 2 .c 2 Y ) TT square units.
167]
MULTIPLE INTEGRATION
319
25 Find the area bounded by the two curves y = x 3 x* and y = x*.
SOLUTION. The limits for x are found to be and 2 by solving the equations
of the curves simultaneously.
Hence,
since AA
= AJ/-AX,
Y 1
A = lim
S 2 AT/ -Ax
j
1
= fo
y* s ^ cty dx.
i
Whence
v
H
J lx \
I yry
A- r
A)
y\dx \
7 >
&y /-~f
^f^i-^ y
x2
= /
A)
(2 x 2 - x 3
\ 4 .. XT)
)ax = - square units. f ^
^3? 2 x
3. Find the area enclosed by the car-
dioid r = a(l + cos 0). Fio. 183
SOLUTION. Here we may take as the
element of area AA ; the infinitesimal r t Ar A^ as explained above. Hence
lim
n, n *>
Ar->0
TO n
V V
rir /a(H-cosO)
= 2 / /
/o */o
The first integration with respect to r gives
/7r/r2\"~la(l -j-COfl 0)
A = 2 / ( T> ) dO
A) V2/JO
Fio. 184 - a*f " (1 -f 2 cos + cos 2 0)^0.
Integrating this, we get, as previously,
A = ' square units.
%j
4. Find the area outside the circle r = 2 a cos and inside the cardioid
r a(l + cos0).
SOLUTION. With A At/ represented by n Ar A0, the area desired in the first
and fourth quadrants is given by
m n /ir/2 /^(H-coa 6)
320 DIFFERENTIAL AND INTEGRAL CALCULLS [Cn. XIV
and that in the second and third quadrants is
r r r a(l+cosd)
Therefore the total area is
/jr/2 /.a(l-fcos0)
A = 2 / / rdrde
JQ J 2 a cos 9
+ 2 y * y a 8 r c/r c#,
A
and integration with respect to r gives
A = o 2 /"* (1 -f 2 cos - 3 cos 2 6)dB
*/o
Fia. 186 4- a 2 C" (1 + 2 cos -f cos 2 0>#.
'ir/iJ
These integrals with the combined value 7ra 2 /2 square units appeared in the
solution of this example by single integration.
PROBLEMS
Find by the method of double integration each of the areas bounded as
Mows. (Nos. 1-5.)
1. The semi-cubical parabola 2/ 2 * x 3 and the line y = 2 x.
Ans. 16/5 sq. units.
2. Inside the circle x 2 4 ay -h y 2 = and outside the parabola x 2 = 2 ay.
3. The segment of r = 8 and r cos = 4 which does not include the pole.
Ans. 16(4 7T/3 V3) sq. units.
4. Inside r = 2 a(l cos 0) and outside r = a.
5. Between the parabola x* = 2 ay and the witch y a 3 / (z 2 4- a 2 ).
Ans. o 2 (T/2 1/3) sq. units.
6. In Problems 1 and 2, reverse the order of integration, change the limits
and check the results.
Find each of the following areas by double integration. (Nos. 7-19.)
7. Between y = xe~ x and the x axis. What part of this area lies between
the high point and the point of inflection?
Ans. 1 sq. unit; (2/e 3/e 2 ) sq. units.
8. The loop of 2 y f = z*(2 - x).
168]
MULTIPLE INTEGRATION
321
9. Between the circles r a sin 6 and r = a cos 0.
Ans. a?(ir 2)/8 sq. units.
10. Outside the circle r = 3a/2 and inside the cardioid r =* a(l -f cos0).
11. Enclosed by x*y = log x, y = and x = c 1 / 2 .
Ans. (1 1/Ve) sq. units.
12. Under one arch of the cycloid x a(0 sin 0), y a(\ cos 0).
13. Enclosed by the hypocycloid x = a cos 3 0, y = a sin 3 0.
s. 3 7ra 2 /8 sq. units.
14. Between x lf2 -f y 1/2 = a 1 / 2 and x + y = a.
15. Enclosed by the lemniscate r 2 = a 2 cos 2 0.
16. Enclosed by the three-leafed rose r = a sin 3 0.
Ans. a 2 sq. units.
17. Enclosed by the hyperbolic spiral rO = 2 c and any two vectors n and
r 2 drawn to it. Ans. c(r\ r 2 ) sq. units.
18. Between the axes, the catenary y = (a/2)(e xla -f e-* /a ), and a; = 6.
19. The area formed by the x axis and any arch of y = sin # is divided into
two parts by the curve y = cos x. Find the area of each part.
Ans. 2 \/2, and 2 -f V2 sq. units.
20. PI and P 2 are any two points on the hyperbola xy = k. Prove that
the area between the arc PiP2 and the x axis is equal to the area between the
same arc and the y axis.
168. Solids of Revolution by Double Integration. If the area
which generates the volume of a solid of revolution is divided into
elements Ay by Ax and the element
in the jth row and the iih column
is revolved about the line y = c, the
volume generated is
-Ax,
= 2
+ 7T'A2/ 2 -Az,
2
where TT A?/ Ax is of higher order than
2 xr ; Ai/ Ax. Hence by the funda- ~|~
mental theorem we have
1
Y
^^rr^
^
J
-*<
-: 3!
fl
V*
/
1
^t
/
/
L
\X
7
/
y
^
1
X
r *
3
4
)
i _
-
a
r-
fi
6 X
c)
Fia. 186
lim
2
= 1 ,; 1
Ay-Ax
/6 /y,
= 2;r / I
/a t/y.
dx.
322 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
Similarly for polar coordinates, if an area is revolved about the
initial line the circular ring generated is 2 ?rr t 2 sin ; 'Ar*A0 except
for infinitesimals of higher order and therefore
V = lim 27rr l 2 sin0 / -Ar-A0 = 2 TT f f'* r 2 sin dr dti.
m, n * j 1 = 1 t/ v r !
In this element sin becomes cos if the volume has the line
6 = 7T/2 as its axis.
EXAMPLE
Solve the first example given in 104 on volumes of revolution.
SOLUTION. In this example r t j y v , so
VI
Therefore
m n
V= lim D H 2 7r?/t Ay Ax
m, n -<-n ; = 1 i= 1
FIG. 187
7T 7-2 32 7T , .
= -r I x 6 ax = ^ cubic units.
4 /o 7
PROBLEMS
Find by double integration the volumes of each of the solids defined below.
(Nos. 1-20.)
1. A sphere by rectangular coordinates; by polar coordinates.
Ans. 47ra 3 /3 cu. units.
2. Area between x 2 = 4 y and y 4, revolved about y -f 4 = 0.
3. Area between 4x = 16?/ y 1 and x = 0, revolved about y 0.
Aws. (2730%) TT cu. units.
4. Area between y # 2 , x 2, and ?/ = 1, revolved about x = 0.
5. Area between 16 y 2 = (x -h 4) 3 and its tangent at (12, 16), revolved
about y 0. A MS. (113%) TT cu. units.
6. Area between x 1 8 ?/ + 23 = and x z + y - 13 = 0, revolved about
x =0.
169] MULTIPLE INTEGRATION 323
7. Area bounded by ?/ 2 = x(x l)(x 2) from x = 2 to x 3, revolved
about y 0. Ans. 9 7r/4 cu. units.
8. Area between y 2 4 x 4- 4 and .r = 3, revolved about J -f- 4 = 0.
9. Area between x = (y -{- l) z , z = 1, and ?/ = 1, revolved about y 0.
/iw.s*. 11 TT/G cu. units.
10. Area between 5 y 2 80 16 x and x = 0, revolved about x 4- 1 =0.
11. Area in a parabolic segment 8 units high, with a base of 10 units, re-
volved about the base. Ans. (lillVa) * cu. units.
12. Area of a circle of radius 2 units, revolved about a line 3 units from the
center.
13. Area between one arch of y sin 3 x amis = 0, revolved about y 4-1=0.
Ana. 7r(8 -f- TT)/() cu. units.
14. Area in the first quadrant bounded by y(x z -f- 4) =8, x = 0, and
y = 1, revolved about the y axis.
15. Area bounded by y(2 x I) 2 = 4, x = 0, z = 1/2 and ?/ = 0, revolved
about ?/ = 0. Ans. Not finite.
16. A football, if a section containing a seam is an ellipse 12 in. by 8 in.
17. Area bounded by y 2 c- f , y <* x , x 0, and x = 1, revolved about
y = 1. Ans. (ir/2) (2 ^ - 5 e 2 -f 4 e - 1) cu. units.
18. Area t>ounded by a cardioid, revolved about its axis.
19. Area bounded by r = a cos 0, revolved about = ir/2.
Ans. ?T 2 a 3 /4 cu. units.
20. Area bounded by the first-quadrant loop of r = 2 sin 2 6, revolved about
6 = 7T/2.
21. The arc of the curve xy = x y joining the origin to any point
PI(XI, T/I) of the curve bounds with the x axis and the lino x = Xi an area A.
The same arc bounds with the y axis and the line y = y\ an area B. Show
that the volumes generated when A is revolved about y and B about
x = are equal for any point PI.
22. Derive the general double integration element for polar coordinates if
an area bounded by n = 4>i(6) and r z = $2(0) is revolved about a line parallel
to the polar axis; perpendicular to the polar axis.
169. Non-symmetric Volumes by Double Integration. Just as
we have the problem of finding the area bounded by given curves,
we also have that of finding the volume bounded by given sur-
faces.
324 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XIV
Suppose the surface 8 of Fig. 188 is projected upon the xy plane
and the volume under this surface bounded below by the plane and
laterally by the cylinder of projection
is desired. Divide the base of the re-
quired solid into elements Ay by Ax.
The volume of the column over the
element of area in the ith row and the jth
strip may be considered as an element
of volume AF,,. Then
^ y AVij = Zij Ay Ax + k Az i3 Ay Ax,
where \k\ < 1, and where z = f(x, y)
is the equation of the surface S and
kAz^ Ay Ax is of higher order than
z vj Ay Ax. Therefore
F- lim
(*> rv*
=11 f(x, y)dy dx.
*^a t//
The rectangular column used may be
taken perpendicular to any other plane
of reference and the order of integra-
tion may be reversed if the operations
are thereby simplified for certain solids.
When the integration is performed
with respect to y, the x 3 - and Ax are
held constant so that the part of the
required volume between the planes
x = Xj and x = x,- + Ax is evaluated.
The result of this integration is an ex-
pression under the remaining integral
sign for the slice shown in Fig. 189.
FIG. 189
The second integration
is the limit of the sum of such slices as Ax
the required volume.
0, and this limit is
EXAMPLES
^ Find the volume under the surface 2 z x 2 + 2/ 2 which is above the
xy plane and inside the cylinder x 2 + y 2 = 4.
169]
MULTIPLE INTEGRATION
325
SOLUTION. Any element A?/ Ax in the xy plane supports a column repre-
sented by
AF,y - ,/ Ay Ax -f fc-A2.yAyAx, |fc| < 1.
Therefore
m n
V = lim i] i] z t? A?/'Ax.
= 1 t = l
Considering y v variable, the limit of the sum of columns for x/ and Ax constant
is taken from the xz plane to the cylindrical surface y \/4 x 2 for that
part of the volume in the first octant. This first integration gives a slice as
shown in Fig. 190. To find the total volume in the first octant we treat such
slices as elements and find the limit of their
sum from the yz plane to the extreme point
on the x axis and the circular cylinder as
Ax > 0. That is, if we substitute for Zij its
value in terms of x and y, we have
7 = lim v V ! /-* ..ON A.. .-
m, n
Therefore
Fia. 190
= 4 TT cubic units.
^ 2. Find the volume bounded by the plane z = x and the paraboloid of
revolution 2 = x 2 -f- y 2 .
SOLUTION. Each column of cross-section A?/ Ax is considered as beginning
at the paraboloid and extending up to the plane. The length of such a
column is
(z 2 + /c 2 Az 2 ), 7 - (zi + fcAsi)*/, |/c 2 |, | fa | ^ 1,
where zi and 21 represent the z coordinates of the points of the two bounding
surfaces whose projections coincide at any point (x,- f y t ) below the desired
volume.
Figure 191 shows the first octant and for this part of the volume the inte-
gration with respect to y is from y = to the intersection of the two surfaces,
326 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
which is located where x = x 2 -f y 2 or y = Vx x 2 . The limits for x are
determined by the points common to z x, z = x 2 + 2/ 2 , and y 0, which are
(0, 0, 0) and (1, 0, 1). Therefore, the symmetry of the solid gives
yr m n
V = lim 2 S [*, - (xS + ;
m, n *-oo 1=1 ; = 1
(x - x 2 - y*)dy dx
= 2 f f
'O *^0
FIG. 191
j -.
o/0
Completing the square inside the parenthesis
and setting x 1/2 = (1/2) sin 0, we have
^ cos 4 (9 d0
> 16
= oo cubic units.
oZ
170. Volumes Defined by Cylindrical Coordinates. The prob-
lem of evaluating the volume bounded by {surfaces is in general
greatly simplified if the surfaces are symmetric with respect to
some coordinate axis so that cylindrical coordinates may be used.
Then we divide the volume into elements by means of coaxial
cylinders around and planes through the axis of symmetry. The
cross-section of such a column is the same as the element of area
in polar coordinates so that we may use 7vArA0 to represent it.
Then if the z axis is the axis of symmetry we have
AV t ] = (22 + A; 2 'Az 2 Zi /CiA2i) t - ; 7vAr'A0,
N> M ^i-
Therefore
m n
V = lim E (22 - *i)i,rvAr-A0
Ar *0
A8 >0
n
__
(z 2 - Zi)r dr dd,
where V is the volume between the two surfaces z =/i(r) and
171]
MULTIPLE INTEGRATION
327
EXAMPLES
M 1. Find the volume required in Example 1 of the preceding article.
SOLUTION. The paraboloid is 2 z = r 2 and the cylinder is r = 2. There-
fore, since 22 = r 2 /2 and z\ 0, z
TO n
m, n *co j = 1 i= 1
Ar ^0
A0 >0
= 4 C f^-rdrdO
Jo Jo 2
"JQ \4/Jo /o
= 4 TT cubic units.
^ 2. The second example of the preceding A >
article represents a solid bounded by the FIG. 192
paraboloid z = r 2 and by the plane z r cos 0.
Setting the two values of z equal, we find from r 2 = r cos 0, that and
cos are the limits for r. Therefore
m n
V = lim 2
m, n > > 3 = 1 i = 1
7T/2 / cos
H (r t cos ; r 2 )r\ Ar A0
f> IT/ 4 f> COS U
= 21 I (r 2 cos 6 - r*)dr d0
t/Q /Q
~~ "Jo L 3 4 Jo
I /-ir/2
= 7, / cos 4 6 dO
b/o
FIG. 193
= ^ cubic units.
The first of these examples is very greatly simplified by the change of
coordinates and the second is simplified in the integration. In many problems
the student must decide whether to use rectangular coordinates, or cylindrical
coordinates. Hence it is important to know the advantages of each system.
171. Special Volumes by Double Integration. It is advisable
to use double integration when considering a solid whose cross-
sectional area is not readily expressed in terms of the variable of
integration. This method is illustrated in the solution of the
following example.
328 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
EXAMPLE
What is the volume of a solid with circular base whose sections perpendicular
to a given diameter of the base are parabolic segments with the height of each
equal to its base?
SOLUTION. Take the base in the xy plane
with the given diameter as the y axis. The
variable of integration for single integration
would then be y and the student will probably
have some trouble expressing the slice perpen-
dicular to the y axis in terms of y. The sec-
tion PQR is in the plane y = y 3 . Then
2 KP = RQ = 2 x lt by hypothesis. With x and
z variable in the plane y = y } we have NM = x
and NQ = 2 x % z. Therefore the parabolic
segment makes
(2 x t ) 2
2 x,
FIG. 194
2 x = V4 x, 2 - 2 3*2.
The volume of the horizontal element shown in Fig. 194 is therefore
(\/4~x 2 2 jc,.z \
l -f- k'Ax) Az-Ay, < k < 1,
2i '
whence, since x 2 + 2/> 2 ri2 >
V - Jim 2 S AVij
m Al^to 3 ~ *
: 2 f f 2 2 y * \/4 a 2 4 if 2 zVa? ?/ 2 ^ J?/
rt /* f (4 a 2 - 4 ?/ 2 - 2 zx/a 2 "- w 2 ) 3 /n 2V a 2 -2/ 2 _.
/o L ~ -T/TZ"! "^Jo 2
32 a 3
, .
cubic units.
PROBLEMS
Use an element such that the volumes bounded as below may be found by
double integration. Write the equations where not given.
1. A sphere.
2. An ellipsoid of revolution.
Ans. 4 7ra 3 /3 cu. units.
172] MULTIPLE INTEGRATION 329
3. That part of the cylinder r = 2 sin 6 between the planes z =
and 2 z - 4 + r sin 0. Ans. 5 ir/2 cu. units.
4. Below by z = 0, above by the cone r 2, laterally by the cylinder
r = 2 sin 6.
5. The plane 2 = 8 and the cone z = 2 r. Ans. 128 ir/3 cu. units.
6. That part of the cylinder r = cos between the cone z r and the
plane 2 = 0.
7yuThe paraboloid 4 x 2 -f y* = 42 and the plane 2 = 4.
Ans. 16 TT cu. units.
8. The paraboloid y* + & = 2 x and plane x -f y = 1.
9. The plane 2 = 0, and the cylinders x 2 -f ?/ 2 = 4 and x 2 = 4 2 2.
A/is. 6 TT cu. units.
10. The paraboloid 4 z 2 -f 2 2 = 4 ?/ and plane y = 2.
11. In the first quadrant by the paraboloid 5 2 = x 2 + 4 ?/ 2 and the
planes x y, z = 0, and x = 0. Ans. (25/8) sin-^l/VS) cu. units.
12. The cylinder r = a(l + cos 0) and the planes 2=0 and 2 = r cos 0.
13. The cylinder r - a sin 6 and the planes r cos -f 2 = a and = Tr/2,
2=0. Aws. a 3 (3 TT - 2)/24 cu. units.
14. The plane 2 = and*the cylinders x 2 + 2/ 2 = 1 and x 2 = 4 2.
15. The cylinders r = \/cos~0, r 2 cos 2 = 12, and the plane 2 = 0.
Ans. 13 7T/32 cu. units.
16. The paraboloid x 2 + 2/ 2 =42, the cylinder x 2 -f y 2 = 4 z, and the plane
0=0.
17. The cylinder y = x 2 and the planes x4-y-r-2=2,a;=0, and 2=0
in the first octant. Ans. 17/20 cu. units.
18. The planes 2 = rb 1 and the cone 2 2 = 8 x 2 -f t/ 2 .
19. The paraboloid 2/-f4o; 2 -f92 2 = and the plane ?/ + 1 =0.
Ans. 7T/12 cu. units.
20. The dome 2 = a r 3 /a 2 and the plane 2 = 0.
21. (z/a) 1 / 2 + (y/6) 1/2 -f (2/c) 1 / 2 = 1 and the coordinate planes.
Ans. afrc/90 cu. units.
22. A circular paraboloid of altitude a units and radius of base b units.
172. Volumes by Triple Integration. The evaluation of the
volume of a solid bounded by surfaces whose equations are known
may be done by triple integration. If given in rectangular coordi-
330 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
nates, assume the desired volume divided by three sets of parallel
planes, one set parallel to the yz plane and Ax apart, another
parallel to the xz plane and Ay apart, and the third set parallel to
the xy plane, Az apart. These planes divide the solid into small
cubical blocks Az*Ay*Ax in volume together with the Jrregular
blocks that do not lie wholly within the solid. The sum of the
volumes of those blocks which lie wholly within the solid is an
approximation for the desired volume. This approximation is
better the nearer the planes of each set are to each other. Also
the limit of such a sum is the volume of the solid if the dimensions
of the blocks approach zero as a limit. That is,
V = Hm E E L Az-Ay> Ax = f P C** dz dy dx,
Az-M) J a Jy. J z.
A.y-K) l l
where z\ and 2 2 may be functions of x and y or of either or constants ;
7/1 and 2/2 may be functions of x or constants; the limits of x are
necessarily constants.
If cylindrical coordinates are used, similar reasoning gives the
volume as expressed by
m n I
V = lim Z r r A2-Ar-A0
Ar *0
m*',
-
r dz dr dO,
where the limits of z may be functions of r and 6} those for r may
be functions of 0] those for 6 are necessarily constants.
EXAMPLES
. 1. Find the volume requested in Example 1 of 169.
' SOLUTIONS, (a) RECTANGULAR COORDINATES. Here AF = Az'&y-Ax.
Also the limits of z are the paraboloid and the xy plane, or z\ 0, and
02 = (x* 2 + 2/; 2 )/2. The limits of y for one-fourth of the volume are from
the xz plane to the cylinder, or y\ 0, and 2/2 = \/4 x t 2 . The extreme
values of x for the first octant are and 2. Therefore
V - lim Z22AZ.A2/.A* = 4 f* f ^* C^^dz dy dx.
Ae-*o ^o JQ JQ y
172]
MULTIPLE INTEGRATION
331
Integrating this with respect to 2, we get the expression for the volume as a
double integral, as given in the article mentioned above.
FIG. 195
FIG. 196
(6) CYLINDRICAL COORDINATES. See Fig. 196. In this representation
Zi 0, z 2 = r t 2 /2 and n = 0, r 2 = 2, while 0i = and #2 = ?r/2 gives one-
fourth of the volume. Therefore
m n I
lim
H 2
, m, n *-oo
Az H)
Ar H)
A0 >0
= 4 I i i r dzdr d6,
and integration with respect to z gives the double integral used in 170.
These examples should make the student realize that if we use
fewer integrations than the number of dimensions of the quantity
in question, we assume one or more integrations in so doing.
That is, we choose as an element of the quantity one that is com-
posed of more elemental units and thereby assume integrations.
Therefore, areas and any quantity that depends upon two dimen-
sions may be evaluated by either single or double integration;
volumes and three dimensional quantities may be evaluated by
single, double, or triple integration. After one integration of a triple
integral the resulting expression is the same one which would have
been formed to solve the problem by a double integral. After another
integration the result is just the form necessary to solve the problem
by a single integral. The proper choice of limits will always pro-
duce this result. *
332 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
2. Find the volume bounded by the paraboloid z x 2 + 4 y 2 and the
ylinder z = 6 3 y 2 .
SOLUTION. The element of volume is
Hence we have
V = lim S2A2-A?/.A:r
Az *0
cfc fify cte.
Ax *0
The limits of y are found by setting the
values of z for the two surfaces equal. Thus,
x 2 + 4 y 2 6 3 ?/ 2 , so 7 ?/ 2 = 6 x 2 , or
y = V((T- z 2 )/7. Then the limits for
x are found from [this relation between a;
and ?/ when y = 0. The first integration
gives
= 4
Therefore
Let x = V6 sin 0, so that dx = V6 cos </0 and (6 - x 2 ) 1 / 2 = \/6 cos 0.
Then
96
V7
r"/ 2 24 T3 1 l 71 ^
/ cos 4 cte = - - + sin 2 4- - sin 4
^0 V7 L.2 8 Jo
V7
18 7T
V7
cubic units.
173. Mass. If p represents the density of a solid at a point,
the element of mass in rectangular coordinates is pA2'Ai/Aa;.
When the solid is homogeneous p is a constant and should be placed
in front of the integral; however, for a variable density p must
be replaced by some function of the coordinates before any inte-
gration is performed.
In cylindrical coordinates the element of mass AM is
173] MULTIPLE INTEGRATION 333
PROBLEMS
Use triple integration to find the volumes bounded by the following surfaces.
Show a figure for each. Write any necessary equations. (Nos. 1-7.)
1. The plane 2 = 4 and the paraboloid z r 2 . Ans. 8 TT cu. units.
2. The cylinder r a cos and the sphere r 2 -{- 2 2 = a 2 .
3. The paraboloid 2 z = 4 r 2 and above the cone 6 z = 5 r.
Ans. 352 Tr/3 5 cu. units.
4. That part of the cylinder r = 2 cos between the paraboloid
r 2 = 9 3 z above and the plane 2 = below.
5. That part of the sphere r 2 + z 2 = 2 inside the cylinder r 2 = a 2 cos 2 0.
Ans. (2/9) (20 -f 3 TT - 16V2)a 3 cu. units.
6. That part of the cylinder r = 4 sin bounded below by z and
above by the cone r = 2 z.
7. Inside the sphere r 2 -f z 2 = 4 and outside the cylinder r = 2 sin 0.
Ans. 16(3 TT + 4)/9 cu. units.
Name each of the following surfaces, sketch it, and find the desired volume.
(Nos. 8-22.)
8. Below ?/ 2 = 4 2 2, above 2 = 0, and within x 2 -f- y 2 4.
9. Bounded above by r 2 -f- 2 2 = 4, below by z = r.
Ans. 8 ?r(2 - \/2)/3 cu. units.
10. Bounded above by r 2 -f 2 2 = 5, below by r 2 = 4 z.
11. Enclosed by x 2 -f y 2 = 2 2 and y = z 1. Ans. 9 T/4 cu. units.
12. Bounded above by r 2 + 2 2 = 2 z, below by r 2 = 2.
13. Inside the cylinder r 2 cos 2 0, outside the cone z 2 r, and above
2=0. Ans. 128/9 cu. units.
14. The part of r 2 + 2 2 = a 2 inside r = a cos 2 0.
15. Bounded below by r = 32, above by a sphere of radius 2 units with its
center at the pole. Ans. (16 7r/3)(l - 1/VTo) cu. units.
16. The larger volume bounded by a cone with a vertical angle of 90 at
the center of a sphere, and the sphere.
17. Bounded by r = cos 0, z 0, and z = 2 r cos 6. Ans. ?r/4 cu. units.
18. Bounded by z 0, r cos 0, and z - r, with z > 0.
19. Bounded by z 0, r 2 = 4 2, and r = 4 cos 0. Ans. 6 TT cu. units.
20. Bounded by & + 4 z 2 = 8 ?/, and y = 1.
334 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
21. Ellipsoid 2 /a 2 -f y^/b* -f z 2 /c 2 = 1. Ans. 4 7ra&c/3 cu. units.
22. Bounded by z = 0, x* -f- y 2 = az, and z 2 H- y 2 = ax.
23. Three equal cylinders of radius a units have their axes along three
mutually perpendicular lines which meet in a point. Find the volume common
to the three cylinders. Ans. 8(2 V2)a 3 cu. units.
24. Find the volume enclosed by the surface of z 2 / 3 + y 2 ' 3 + 2 2/3 = a 2 / 3 .
25. The density of a cone at a point varies as the square of the distance
of the point from the axis. If the cone has a radius of a and altitude /?, find its
mass. Ans. irka 4 h/lO units.
Use triple integration to find the masses of the solids bounded "as follows.
(Nos. 26-32.)
26. A hemisphere, if p is proportional to the distance of any point from the
base.
27. A cone of radius 1 unit and altitude 2 units if the density at every point
is equal to the square of its distance from the vertex. Change the order of
integration and the limits, and check. Ans. 9 rr/5 units.
*
28. An ellipsoid of revolution, if p equals the square of the distance of any
point from the axis of revolution.
29. A right circular cylinder, if the density at any point is proportional to
the square of its distance from a diameter of the base.
Ans. k7ra z h(4 h* + 3 a 2 )/12 units.
30. The same as Problem 29 for a right circular cone.
31. Bounded by z = r 2 and z = 4, if p is proportional to the square of the
distance of any point from the line z = 0, 6 = 7r/2. Ans. 208 irk/3 units.
32. Bounded by z = x 2 -f- 4 y 2 and z = 4, if p is proportional to the dis-
tance of any point from the y axis.
174. Moment of Inertia. The moment of inertia of a particle
of mass m about a point is defined as the product of the mass m
by the square of the distance R from the point to the particle.
Thus, 7 = mR*. Also the moment of inertia of an aggregate of
particles about a point is the sum of the moments of inertia of all
of the particles about the point.
Consider now a thin plate of metal of uniform density and thick-
ness. If we have its boundary given by y = /(#) and divide the
plate into small elements AT/ by Ax and take P(x % , ?/,) a point of
each element, the moment of inertia of any element about the
origin is approximately p(x t 2 + 7/ ; 2 ) -Ay-Ao;, where p is the mass
174]
MULTIPLE INTEGRATION
335
per unit of area of the plate. Hence the moment of inertia of the
whole plate about the origin is
/=
nv*
"i
where the limits of integration are taken so as to cover the area of
the plate.
The moment of inertia about an axis is similarly defined with R
representing the distance of a par-
ticle from the given axis. Thus
the relation (1) for the x and y
axes gives
(2)
f*b
= f
*J a
y*dydx,
x'dydx.
FIG. 198
The moment of inertia of a thin
plate about a point of its plane
is the sum of its moments of inertia about two perpendicular lines
of the plane through the point. This follows immediately from
(1) and (2) since I = I x + I y .
If the density is not uniform we have p = F(x, y) which must
remain a part of the integrand.
These definitions permit us to find the moment of inertia
about a point (polar moment), or about an axis, of a thin plate
(area), a small wire (arc), or a solid (volume). In each case, we
divide the mass M of the whole into elements AM,- and define I as
follows :
(3)
7= lim ;/?, 2 -AM, =
where limits are taken so as to cover the whole quantity under
consideration. Of course for areas and volumes double and triple
336 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
integration respectively are implied in the definition. The student
must be careful to represent by AM{ an element of any quantity
being considered multiplied by its mass per unit of the quantity.
Also be sure that Ri is the distance
of this element from the point or
line with respect to which the mo-
ment of inertia is desired.
EXAMPLES
1. Find the moment of inertia about
the pole of a thin uniform plate in the
shape of one loop of the curve whose equa-
tion is r a cos 3 0.
SOLUTION. HereAM t = pAA t , and A/li
we have seen is r l Ar A0 except for infini-
= r t and p is assumed constant. Hence
FIG. 199
tesimals of higher order. Also R t
lim
c
p a
-
2 2 p.r 2 (rvAr-A0)
J 1 t = 1
r. a cos 3
r 3 dr dO
'-7T/6
cos 4 3 6 d8
p a
"
4 /.ir/6 i r 11 -i
j ^ I 1 + 2 cos 6 -f i -f ;* cos 12 01 dB
16 L 2
PTT a 4
24 -,/
units.
2. Set up the moment of inertia about the polar
axis of the solid bounded by z = r 2 and z a.
SOLUTION. In this case, R 13 2 r t 2 sin 2 ; + %*?
Since z varies from the paraboloid to the plane,
r from the z axis to the intersection of the paraboloid
and the plane where it is \/a, and from to 2 TT, we
have
/
/ = 4y y J^ P (r* sin 2 + z z )r dz dr dB.
The integrations are readily performed.
FIG. 200
174] MULTIPLE INTEGRATION 337
PROBLEMS
Find the moments of inertia of each of the areas bounded as follows. (Nos.
1-10.)
1. xy = 12, y = 0, x = 2, x = 4, about x = 0. Ans. 72 p units.
2. 2/ 2 = 2(x - 4), 3 = 3 y, about x = 0.
3. A loop of the four-leafed rose r = a cos 2 0, about the pole.
Ans. 3 ?ra 4 p/64 units.
4. The loop of t/ 2 = x 2 (2 - x) t about a; = 0.
5. A semicircle, about (a) the tangent parallel to its diameter; (6) a tan-
gent perpendicular to the diameter. Ans. (a) a 4 p(15 ^ 32) /24 units.
6. 3 ?/ 2 = 4 x, 3 y = 2 x, about g/ = 0.
7. x 2 = 2/, y = z + 6, about ?/ = 0. Ans. 415 2 5 g p units.
8. The smaller area bounded by y = cos x, y = Q and # = vr/4, about
x = 0.
9. 7/2 = i _ x> x o, about x = 0. Ans. 32p/105 units.
10. r = 4 sin 0, (a) about the pole; (b) about = 0.
Find the moments of inertia of the areas and solids bounded as follows.
(Nos. 11-20.)
11. A semicircle of radius 2 units, about a line parallel to its diameter and
3 units from the diameter (two cases). Ans. 4(5 w _8) p units.
12. Area in the first quadrant inside r 2 a sin 2 and outside r a, about
6 = 7T/2.
13. Area in x 2 + 2/ 2 = 2 above x 2 = y, about the ?/ axis.
Ans. p(57r - 8) /20 units.
14. Area enclosed by y = 2 z x 2 and ?/ = 0, about y = 4 if the density
at a point is proportional to the distance of the point from y = 4.
15. The semicircle y ~ + ^<& x 2 , about y = 2 a.
Ans. a 4 p(51 TT - 64) /24 units.
16. The area y = 0, y = e* x , x = 0, s = 1, about 3 = 0.
17. The solid bounded by the cone z = r and the plane 2 = 4, about the
polar axis. Ans. 256 wp units.
18. The solid z = x 2 -f 2/ 2 , 2 = x, about the a; axis.
19. A right circular cone of height h units and radius of base a units, about
its axis. Ans. -ira*hp/lQ units.
20. The right circular cone of Problem 19, about its vertex.
338 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIV
21. The slant height being given, find the right circular cone having the
maximum moment of inertia about its axis. Ans. r = 2 h.
22. Find the moment of inertia of the solid bounded by z r 2 and 2=4,
about the line -jr/2 in the polar plane.
23. Find 7* for the solid bounded by r = 4 cos 0, z = 0, and z = 3.
Ans. 36 TTP units.
24. The density at any point of the hemisphere z -f Va 2 r 2 is &
times the distance of the point from the circular base. Find h for the hemi-
sphere. Find Iz for the part of the hemisphere within the cylinder r = a cos 0.
ADDITIONAL PROBLEMS
Find the areas bounded as follows. (Nos. 1-5.)
1. One loop of the four-leafed rose r 2 sin 2 8. Ans. ir/2 sq. units.
2. Outside the circle r = a and inside the cardioid r = a(l -f- cos 0).
3. Inside r = sin and outside r == sin 2 0. Ans. Tr/16 sq. units.
4. Bounded by y = e 2 *, y = 0, x 1, and x = 2.
5. Bounded by the ellipse 4 x 2 + 6 ?/ 2 = 9. Ans. 9 7r/2 V6 sq. units.
Find the volumes of revolution generated as follows. (Nos. 6-11.)
6. y = z 2 , x = 1, T/ = 0, (a) about x = 0; (6) about y = 0.
7. ?/ 2 = foe, x = 0, 2/ = a, about z = 0. Ans. 7ra 5 /5 k 2 cu. units.
8. x 2 -f- 4 = 4 y, x = 0, y x, about the x axis.
9. y = x 3 , x = 1, i/ = 0, about x -f- 2 = 0. Ans. 7 7r/5 cu. units.
10. One arch of x cos 4 1/ and x = 0, about x = 2.
11. One arch of x sin 2 ?/ and a; = 0, about the ^/ axis.
Ans. 7r 2 /4 cu. units.
Find the following volumes. (Nos. 12-16.)
12. The volume bounded above by r 2 -f 2 2 = 4 and below by r 2 =2 z.
13. The volume of y 2 -f x 2 = 2 z cut off by 2 z -f 2/ =6.
Ans. (625/64) TT cu. units.
14. The volume bounded by r 2 = 4 z, r = 4 cos 0, and 2=0.
15. The volume of x 2 -f 2/ 2 = oz> z 2 -f- 2/ 2 = 2 ax, and 2 = 0.
Ans. 3 7ra s /2 cu. units.
16. The volume under y 2 = 4 2 z, above 2 = 0, and within the cyl-
inder x* + 2/ 2 = 4.
174] MULTIPLE INTEGRATION 339
17. The volume between two surfaces is given by
A f*/ 2 r a/ v2 rV a z-r* j , ,_
4/1 / r dz dr do.
/O JQ Jr
Write the equations of the surfaces and sketch them.
Ans. The solid lies above z = r, below r 2 -\- z 2 a 2 .
18. What is left of a sphere of radius 5 in. if a circular hole is cut through
its center, if the diameter of the hole is 8 in.?
19. A rectangular sheet of metal is of variable density. The density along
any line parallel to an end is proportional to the square of the distance of the
line from the end. When the distance is 4 units, the density is 2. Find the
mass of a sheet measuring 8 by 10 units. Ans. 333 J^ units.
20. The same as Problem 19 except that the density is proportional to the
cube of the distance and is 36 at a distance of 6 units.
21. Find the mass of a thin plate bounded by y 2 = 4 x and x = 1, the density
at any point P being proportional to xy 2 . Ans. 32 fc/21 units.
Find the moment of inertia of the following areas and solids. (Nos. 22-31.)
22. The solid bounded by z = r 2 and z = c, about = in the polar
plane.
23. The solid bounded above by z = 6 r and below by z r 2 , about
= in the polar plane. Ans. 376 irp/3 units.
24. A right circular cone, about a diameter of its base.
25. The area within one loop of r = 2 a cos 2 and outside r = a, about the
pole. Ans. a 4 p(20 w + 21 V3)/48 units.
26. 7z of a cylinder of altitude h units with one loop of r = a sin 3 as its
base, if p is proportional to the square of the distance of a point from = in
the polar plane.
27. A paraboloid of revolution of altitude a units and radius of base
b units, about its axis. Ans. b 4 airp/Q units.
28. The paraboloid of Problem 27 about a diameter of its base.
29. I x of the mass common to x 2 -f y* -f z 2 =4 a 2 and the cylinder of radius
a units with its axis along the z axis. Ans. 2 a 5 7rp(128 51 \/3)/15 units.
30. The area of a loop of r 2 = 4 sin 3 0, about the pole.
31. The area of y = e*, x = 0, x = 2, y = 0, about the origin.
4ns. p(e 8 + 18 e 2 - 19) /9 units.
32. What does k C^ 3 P f^ ~ * (y 2 + * 2 ) dx dy dz represent?
/0 ^2 '1
340 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XIV
33. Prove that the moment of inertia of an elliptical plate about either axis
is one-fifth its mass times the square of the length of that semi-axis.
34. A hollow sphere has an inside radius of 6 in. and outside radius of 10 in.
At any point, the density varies inversely as the distance of the point from the
center; and on the outer surface, the density is 2J^. Find the mass of the
sphere.
CHAPTER XV
ADDITIONAL APPLICATIONS OF THE
FUNDAMENTAL THEOREM
175. Introduction. In the two preceding chapters we have
shown how the integral as the limit of a sum may be used either
as a single sum or as a multiple sum. The following applications
involve both ideas and although the illustrative examples exemplify
the methods we think advisable to use, the student will often find
that several methods are open to him.
176. Area of a Surface of Revolution. A surface of revolution
may be considered as generated by the arc of a plane curve revolved
around a line in its plane. Suppose this line to be the y axis and let
z = /(?/) from y = a to y = b be
the arc of the curve. Divide the
interval a < y < b along the y axis
into n segments of length A?/.
Planes perpendicular to the axis
at the points of division will
divide the surface into narrow
bands. Each band is generated
by a As t of the curve z = f(y). If
Zi represents the ordinate of any
point of the element As, the prod-
uct 2 7r2 t ''As; is approximately the
area of the corresponding band,
and the limit of the sum of such products as As
of the surface. That is,
FIG. 201
is the area
(1)
S = lim
i = 1
--
zds
There are similar formulas for surfaces of revolution about the
341
342 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
other coordinate axes. Thus, if a curve in the xy plane whose
equation is y f(x) is revolved about the x axis, the formula for
the area of the surface is
(la)
s =
If the axis of rotation is any line parallel to the x axis, as y fc,
the radius in (la) is not y, but y k. Similar formulas hold
if the axis of rotation is parallel to the y axis or to the z axis.
In polar coordinates formula (la) becomes
(2)
= 2 IT Cr sin
+
for surfaces with the initial line as an axis. If the axis of rotation
is the line 8 = ir/2, the radius in (2) becomes r cos 9.
EXAMPLES
1. Find the surface of a hemisphere by considering it generated by a
quadrant of a circle.
SOLUTION. Suppose the circular arc has
the equation x 2 -f- y 2 = a 2 , x ~ 0, y ~ 0.
Then dx/dy y/x and the limits for y are
and a. Therefore
lim 2
n> co { = 1
Ast K)
2 TT
-y = a
/ x ds
* / y =
/a
dy
v
= 2 Tra 2 square units.
2. Find the area of the surface formed by revolving one arch of the cycloid
about the x axis.
SOLUTION. As a: = a(0 sin 0), y = a(l cos 0) we have
dy/dx = (dy/dO)/(dx/d6) = (osin0)/[a(l - cos 0)] = ctn (6/2).
Hence Vl + (dy/dx) 2 = Vl -f ctn 2 ((9/2) = esc (0/2).
176] ADDITIONAL APPLICATIONS 343
Therefore
n 2* a
S = lim .2 27T'2/<-As l = 2 ira 2 C (1 - cos0) 2 csc^d0
/27r 9 e r 2 * 6
4 sin 4 -.csc ^. dO = 8 ira 2 I sin 3 - c
44 JQ 2
= 8 Tra 2 f" -
o . 2 ,
2 cos - + 5 cos 3 = = 5 square units.
PROBLEMS
Find the area of the surface of each of the solids of revolution generated as
follows. (Nos. 1-15.)
1. By revolving the parabola y" 1 = 2 ax between the extremities of the latus
rectum about y = 0. Ans. 2 va?(2V2 l)/3 sq. units.
2. By revolving a: 2 = 6 y from the origin to ij = 9/2 about x 0.
3. By revolving about the y axis, (a) the circle x 2 -f- y 2 = 4 #, (6) the
semicircle of this circle between x and x = 2.
A?is. (a) 16 -7T 2 sq. units; (6) 8 TT(TT 2) sq. units.
4. By revolving the hypocycloid x 2/3 -f 7/ 2/3 = a 2/3 about an axis.
5. By revolving the cardioid r = a(l -}- cos (?) about the polar axis.
ylns. 32 7ra 2 /5 sq. units.
6. By revolving one loop of the lemniscate r 2 = a 2 cos 2 about the polar
axis.
7. By revolving one arch of y = sin # about ?/ = 0.
Ans. 2 7r[V2 + log (1 + V2)] sq. units.
8. By revolving the part of y log x which lies in the fourth quadrant
about x 0.
9. By revolving the catenary y = (a/2)(e xla -f e~ xla ) from x = to x = a
about x = 0. Ans. 2 7ra 2 (l l/e) sq. units.
10. By revolving the curve 4 y -f x 2 log z 2 = from x 1 to x 4
about the y axis.
11. By revolving the tractrix, for which dy/dx - y/Va z y 2 , from PI to
P 2f about the x axis. Ans. 2 7ra(2/i 2/2) sq. units.
12. The ton/s formed by revolving the circle (x a) 2 + y z r 2 about the
y axis.
13. By revolving the loop of the curve 3 ay z x(a x) 2 about the x axis.
Ans. 7ra 2 /3 sq. units.
344 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
14. By revolving the semi-cubical parabola ay 2 = rr 3 from the origin to
x = 4 a/3 about x = 0.
15. By revolving the cubical parabola 9 ?/_== z 3 from (0, 0)_tp (3, 3) about
the y axis. Ans. 3 TT [3 VlO -f log (3 -f VlO)]/3 sq. units.
16. Prove that the surface produced by rotating r 2 = a 2 sin 2 about either
axis, 0, or ?r/2 is equal to the surface of the circumscribed sphere.
17. Find the areas of the ellipsoids of revolution formed by revolving the
ellipse x 2 /a z -f- y*/b 2 1, (a > 6) about the x axis; about the y axis.
Ans. 2 Trb[b -f- (a 2 /c) sin" 1 ^/^)] sq. units;
2 7ta[a -j- (6 2 /c) log {(a + c)/6J] sq. units.
177. Force Exerted on a Submerged Surface by a Fluid. To
find the force exerted on a submerged surf ace of variable depth we
use the physical principle that the pressure at a point within a fluid
is the same in all directions and is the same on units of surface of a
common depth below the surface of the fluid. The force on a
surface of area A submerged to the depth of h units is the weight
of the column of fluid which could be supported by the area, that is,
whA, where w is the weight of a cubic unit of the fluid. We now
illustrate a method of applying these principles.
EXAMPLES
1. Find the force on the vertical end of a parabolic gutter if it is full of water,
the gutter being (> inches across the top and 4 inches deep.
SOLUTION. Let Fig. 203 represent an end of the gutter. Suppose it to be
divided into narrow strips by means of
straight lines parallel to the surface AC.
Let A/t be the width of any one of these
strips with parallel sides l % and / -f A/ t . The
area of such a strip is l t >Ah -\~ fcA/ t -A/i
(see Fig. 203), where < k < 1. Now if h v
represents the distance of the strip from
the bottom of the gutter, the force on the
strip is w(li -f- fc'A/ t -)(4 hi) Ah, approxi-
mately. The approximation is due to the
fact that (4 h t ) is not the depth of all
parts of the strip considered. However, the
force on the strip differs from the expression approximating it by infinitesimals
of higher order than the first. Therefore
lim 2 w(li + fc-AZ)(4 - hjAh ** w f 1(4 - h)dh.
n~*. t = l ^0
177] ADDITIONAL APPLICATIONS 345
To express I in terms of h, we have from the parabolic segment Z 2 /36 = h/4
or I = 3VVi. Whence
F = 3 w T 4 (4 /I*/ 2 - /i*/ 2 )d/i = 3 w (-^ - ^- 2> )T
JQ \ o 5 /Jo
128 w
= = Ibs.
o
We may also solve this example by using coordinate axes. If we choose B
as origin, the coordinates of C are (3, 4), and the equation of the parabola is
4 x 2 = 9 y. Using any point (x, y) to set up the element of integration, we
may write
/* /*
F = w I 2 z (4 ?/)cfr/ = 3 w / (4 y)y llz dy,
A) A)
which gives the same result as that found above.
2. A hemispherical bowl is full of water; find the total force exerted on the
bowl.
SOLUTION. Consider the surface of the bowl as formed by revolving a quad-
rant of the circle x' 2 -f z 2 a? about the z
axis. Thon the element As t of arc generates
a narrow band whose area we may represent
approximately by 2 wx % - As t . This area is at
a depth of z l and hence the approximate force
on the band is 2 TTWX^ZI As t . Therefore
n
F = lim 2
n > co i = 1
~X = l
= 2 irW I
** x =*
= a 2 we get ds = ^ 1 + (~~ Y
From x 2
Whence
F = 2 ?rwa J xdx irwa* Ibs.
PROBLEMS
Find the force on one face of the submerged surfaces bounded or described
as follows. (Nos. 1-20.)
1. A parabolic segment of altitude 12 units and base 8 units, if its axis is
vertical and the vertex, above the base, is 5 units below the surface.
Am. 3904 w/5 Ibs.
2. A parabolic segment of altitude 6 units and base 12 units, vertex down
and axis vertical, with the base 5 units below the surface.
346 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XV
3. A parabolic segment of altitude 3 units and base 4 units, plane vertical,
axis horizontal and 1 unit below the surface. Ans. 135 w/ 16 Ibs.
4. The parabolic segment of Problem 3 except that the axis is 3 units be-
low the surface.
5. A semicircular plate, diameter horizontal, plane vertical, and' tangent
parallel to the diameter in the surface. Ans. ahv(3 T 4)/6 Ibs.
6. The same as Problem 5 with radius 2 units and diameter 8 units below
the surface.
7. A circular plate, plane vertical, radius 5 units, and center 10 units
below the surface. Ans. 250 TTW Ibs.
8. The same as Problem 7 with radius 4 units and center 2 units below
the surface.
9. A quarter of a circle of radius 2 units, plane vertical, one radius hori-
zontal along the upper edge of the plate and 5 units below the surface.
Ans. u>(15 TT -f 8)/3 Ibs.
10. A semicircular plate in a vertical plane, radius 4 units, diameter along
the upper edge and 3 units below the surface.
11. The end of a full trough which is a right isosceles triangle with a base of
4 units. Ans. 8 w/3 Ibs.
12. An isosceles triangle, plane vertical, base below the vertex. Base 6 in.,
sides 5 in., base 3 in. below the surface.
13. A right triangle in a vertical plane, one leg horizontal, 5 units long and
4 units below the surface. The vertical leg projects out of the liquid 2 units.
Ans. 31 J wlbs.
Water Surface
14. An isosceles triangle in a verti-
cal plane. The base is 6 units and the
sides are 5 units. The base is above
the vertex and 1 unit below the sur-
face.
15. The same as Problem 14 with
the base vertical and the upper end is
2 units below the surface.
Ans. 60 w Ibs.
FIG. 205
16. A semi-ellipse, a = 3, b = 2
units, plane vertical, the major axis
horizontal and along the lower edge. The highest point 2 units ^below the
surface,
17. Find the force on one face of a semi-ellipse as shown in Fig. 205.
Ans. 16 w& TT - 2)/3 Ibs.
178] ADDITIONAL APPLICATIONS 347
18. An elliptic plate, major axis vertical, a = 5, b = 2 units, and the center
is 3 units below the surface.
19. A square with sides a units has one diagonal vertical and the upper end
is in the surface. Compare the force on the upper and the lower halves.
Ans. I to 2.
20. The same as Problem 19 but the upper end of the diagonal is 6 units
below the surface, also find the force.
21. A horizontal cylindrical tank has a radius of 2 ft. What is the force
on an end if 3 ft. of liquid is in the tank? Ans. (8 7r/3 -f- 3V3) w Ibs.
22. A horizontal tank of 5 ft. radius is sealed and water is forced in until
it is half full; the upper half of the tank is then under the pressure of one
atmosphere, equivalent to 34 ft. of water. Find the force on the inside of one
end due to air pressure and weight of water.
23. A trapezoidal dam has its upper base of 300 ft. in the surface, its lower
base of 100 ft. is 15 ft. below the surface. Find the force. Ans. 585 }J tons.
24. The area y sin x, y = 0, ^ x = IT has y = horizontal and 3 units
below the surface. Find the force.
25. An oil cup is a paraboloid of revolution 6 in. in height and 2 in. radius
of base at the top. If the cup is full of oil, what is the force on its curved
surface? Ans. 0.0739 w Ibs.
26. An ellipsoid of revolution with semi-axes a 6, b = 4 is half full of
water. What is the force on its curved surface?
178. Work. Suppose that a body is moved a distance s along a
line against a variable force, and that the force is some function of s,
say /(s). Now divide the distance s into small intervals As,-, and
let/(s t ) represent the force exerted at the beginning of each inter-
val As. Then the work over each interval of the line is approxi-
mately f(st) As;, and the sum of such products is an approxima-
tion of the work done. Hence we have
W= lim ./(,) As, =
n >o> =!
As t K)
where b a represents the total distance through which the body
is moved along the line.
When work is done against the force of gravity, the force exerted
is the weight of the body and the distance is the vertical distance
through which it is lifted.
348 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
EXAMPLES
1. If the force necessary to stretch a spring is proportional to the amount
the spring is stretched and is 15 Ibs. when the spring is stretched 1 inch, find
the work done in stretching the spring 3 inches.
SOLUTION. We have given that F % ks ly where ,s' t represents the distance
the spring is stretched. Also, since F = 15 Ibs. when s I inch, the formula
gives k 15. Therefore F t = 15 s,. Then the element of work is given by
AJF t = 15 s t As and hence the total work is
W
lim
n
S
> co i =3 l
r*
= ^J*
s ds = 67.5 in. Ibs.
2. A conical vessel is 16 ft. across the top and 12 ft. deep. If it has water
in it to a depth of 10 ft., find the work necessary to pump the water to a height of
4 ft. above the top of the vessel.
SOLUTION. The work done to raise a hori-
zontal layer of water to the point of delivery
is AWi w(l6 z l ) 7rr t 2 -A2 t approximately,
where w is the weight of a cubic foot of water.
But, from similar triangles, r t = 2 2 t /3,
and hence
W = lim 24 wirz* (16
' = 1
,10
FIG. 206
4 Wit f 1U
= ^Tpy (16 - z)z 2 ^2
TT/16^ 3 ^\1 10
A3 ~4yJo ~
34000 TTW
27
ft. Ibs.
PROBLEMS
1. A boat moves in a straight line according to the law s = 3 2 + t. If
air and water resistance is equal to the square of its velocity, find the work done
against resistance from t to t = 5 units. Ans. 38,480 units of work.
2. In hoisting ore from a mine, the load consists of (a) the weight M of
the car and contents, (b) the weight of the cable at m Ibs. /ft. What work is
done in hoisting a distance of h ft. from the bottom of a mine k ft. deep?
3. Find the work done in moving a body from = 0toz = 10 units if the
force necessary at any point is x/(100 + x 2 ) 6 / 2 units.
Ans. (4 - \/2)/12,000 unit of work.
4. A body is moved in a straight line according to the laws = 2 t 2 -f t
against a resistance proportional to its velocity, find the work done against this
resistance from t to t = 3 units.
178] ADDITIONAL APPLICATIONS 349
5. The position of a body is given by s t cos t. If its motion is resisted
by a force equal to its velocity, what is the work done in overcoming this resist-
ance from t = to t = 7T/2 units. Ans. ir(6 + 7r 2 )/48 units of work.
6. The same as Problem 5 if s = t cos t sin t.
7. A tank in the form of a paraboloid of revolution, altitude 3 ft. and
radius of top 2 ft., is full of water. What work is necessary to pump the water
to the top? Ans. 6 TTW = 1178 ft. Ibs.
8. The same as Problem 7, but the water is to be pumped to a delivery
point 6 ft. above the top.
9. A cistern in the form of a paraboloid of revolution, altitude 4 ft., radius
of top 3 ft., contains 3 feet of water. What work is done in lowering the water
level 1 ft., by pumping it to a point 2 ft. above the top of the cistern?
Ans. 19.5 TTW = 4875 7r/4 ft. Ibs.
10. A conical vessel is 4 ft. deep and the diameter of its top is 4 ft. Find
the work necessary to empty the vessel, if it was full of liquid, by pumping
it 2 ft. above the top of the vessel.
11. A conical vessel is full of a liquid. It is 4 ft. high and the radius of the
bottom is 6 ft. What work is necessary to pump the contents to a point 2 ft.
above the vertex? Ans. 240 irw ft. Ibs.
12. A full hemispherical bowl is 6 ft. in diameter. Find the work neces-
sary to pump the liquid to a point 5 ft. above the top.
13. A full cylindrical tank 10 ft. long and 4 ft. across an end lies horizon-
tally. What work is necessary to pump the contents to a point 20 ft. above
the top of the tank? Ans. 880 irw ft. Ibs.
14. A hemispherical bowl of radius 4 ft. has its contents discharged at a
point 1 ft. above tho top. Find the work necessary to lower the depth of the
water from 3 ft. to 2 ft.
15. A cistern filled with liquid is in the shape of a paraboloid of revolution
10 ft. deep and 6 ft. across the top. Find the work required to pump the con-
tents to a point 4 ft. above the top. Ans. 330 vw ft. Ibs.
16. A trough has semi-elliptical ends. It is 2 ft. deep, 2 ft. across the top,
and 8 ft. long. What work is required to pump the contents to the top if it
was full at the start?
17. A cylindrical tank 10 ft. high and 10 ft. in diameter stands on a platform
50 ft. high. Find the depth of the water when one-half of the necessary work
has been done to fill the tank from the ground level through a pipe in the
bottom. Ans. 5.23 ft.
18. The weight of a body varies inversely as the square of its distance from
the center of the earth, if the body is above the surface of the earth. Find the
work done in lifting P Ibs. of material to a height h miles above the surface.
350 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
19. A cubical tank 4 ft. across, 6 ft. deep, and 12 ft. long, is full of oil weigh-
ing *50 Ibs. per cubic foot. Find how much the surface of the oil is lowered
when one-third of the necessary work has been done to pump the oil to a deliv-
ery point at the top of the tank. Ans. 2\/3 ft.
20. An oil tank in the form of an inverted paraboloid of revolution is
12 ft. deep and 4 ft. across the bottom. If it is full of oil weighing 55 Ibs. per
cubic foot, how much is the surface lowered when one-half of the work neces-
sary to pump the oil to the top has been done?
179. Work as Change of Kinetic Energy. If a particle of mass
m moves in a straight line so that its acceleration is a, the
force it exerts is defined as ma. Now since d*s/dt 2 = a, and
d 2 s _ d(ds/dt) __ dv ds __ dv
dT 2 ~ dt ~~ ds eft ~~ V ds*
we have the relation
In differential form this is
Fds = mv dv y
and integrating from s = si to s = s 2 , calling v\ and v 2 the corre-
sponding limits for v, we find
f
Fds =
Therefore, since (1/2) mv 2 is the kinetic energy of the particle of
mass m and velocity v, the work done in the interval Si < s < s 2
is equal to the change in the kinetic energy in the same interval.
PROBLEMS
The following solids of given density are rotating about their axes with an
angular velocity o>. Find the work each can do in coming to rest. (Nos. 1-5.)
1. A cylinder of radius a and altitude h units.
Ans. Mv 2 /2 units. (M = mass, v = oo>.)
2. A right circular cone of radius a and altitude h units.
3. A sphere of radius a units. Ans. Mv 2 /5 units.
180] ADDITIONAL APPLICATIONS 351
4. A hollow cylinder with inner radius ai and outer radius a^ if the density
at every point varies inversely as the distance from the axis.
5. A paraboloid of revolution of height h and radius of base a units.
Ans. Mv 2 /Q units.
6. A hollow cylinder closed at both ends and a solid cylinder have the
same radius, altitude, mass, and color. How can these be distinguished
without taking a section?
7. A cylinder of given altitude, radius, and density is rotating with a given
angular velocity. If a frictional force of 4 Ibs. is applied on its surface, how
many revolutions will it make before coming to rest? Ans. pr 3 /ta> 2 /32.
8. A sphere of radius 12 inches and density lYi makes 15 revolutions per
second about a diameter. What frictional force applied at the equator will
bring the sphere to rest after 10 revolutions?
9. Assume that near the muzzle of a gun the resultant force on the base of
a shell is R(l -f kx~ 5/ *) tons, where x is the distance in inches that the shell has
traveled down the bore, and R and k are constants. What expression repre-
sents the change in the kinetic energy at the muzzle if the bore is shortened
from 121 inches to 100 inches? Ans. R(2l + 0.059 k) in. tons.
10. The velocity of a 200 Ib. sled is 60-4 t, measured in feet per second.
Find the mean value of its kinetic energy as to time from t ~ to t = 10
seconds.
180. Centroids of Plane Areas. The point at which a thin
sheet of metal can be balanced is called the center of mass of the
sheet; that is, it is the centroid of the area represented by the
metal sheet. Since an area may be balanced at its centroid, the
sum of the moments of the elements of the area about any line
through its centroid must be zero. By the moment (static
moment) of an element of area about a line we mean the product
of the number of square units in the element by its distance from
the line. Also the moment of a finite area about a line is the same
as the sum of the moments of all the elements of the area about that
line.
The centroid is that point at which the total area may be assumed
concentrated without affecting the moment of the area about a line.
Therefore, if (x, y) is the centroid, A the area, M x the moment
about the x axis, and M v that about the y axis, we have
(1) M x = A-jj, M v = A-x.
352 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XV
These equations permit us to find x and y if the area and its
moments can be found.
To find MX for a given area, we shall consider the shaded element
of Fig. 207, which is parallel to the
y axis. Its moment with respect
to the x axis is defined as the
product of its area AA, which is
(3/2 y\) Ax, approximately, and
the ordinate of its centroid, which
is (t/ 2 + &2 Ay 2 + t/i + ki At/Oi/2,
^ where |/c 2 | and Ifeil < 1. Hence
(a,0) ^ lri PZ1 J/ " - "' "~
where e< as Arc if Aj/i, A?/2 * also. Therefore
(2)
= lim
As the distance from the y axis to the centroid of the shaded
element is o\- + k*Ax, k < 1, we have
(3) M y = lim (|/ 2 - yOiX^Ax = T (y 2 - ^i)xdx.
n *-> i=l ^
Aa:-*0
r 6
We have seen that A = I (j/ 2 j/i)etc; therefore
t/o
(4)
- Vi)x dx
nit
I fa-y
t/a
/ G/2 - yjdx
*/a
180]
ADDITIONAL APPLICATIONS
353
These results may also be derived by double integration if we
consider the elements Ay -Ax and the points (z<, ?/,), the centroids
of the elements. We have at once
x dy dx
(5)
n V *
- -i
/& /I/2
/ / y
- a Vi
dx
pb s*v*
/ / dy
*J a ^J y l
dx
(a,0)
FIG. 208
Evidently the first integration of formulas (5) gives the single
integration formulas (4).
If the area is more easily found by means of a horizontal
element, formulas corresponding to (4) can be derived. It is then
usually advisable to interchange the order of integration in
using the formulas (5).
EXAMPLE
Find the centroid of the area bounded by the parabola x = y 2 and
y = 2 - x.
SOLUTION. Using a horizontal element of length l t and width AT/, we have
and
y =
FIG. 209
/I
-a (2-y-i^
("-S-OL
354 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
tes of the centroid may be set up as d
f f y xdxdy /*/*,
*/-2*V _ . _ J-2Jy*
The coordinates of the centroid may be set up as double integrals as follows:
y **
181. Centroids Using Polar Coordinates. The clement
Ti Ar A0 with centroid (r t , 6,) has this centroid r t sin 0, and r t cos 0,-
distant from the polar axis and the radius vector = ?r/2, respec-
tively. Hence if 7 cos and 7 sin represent these distances for
the centroid of an area, we have
f f 'r 2 cose</rrf9 f f ' r? sin 6 e/r </6
_ - /a /r, . .. t/a /rj
r cos 6 = - - - , r sin 6 =
-j , - -^
I I rdrdt I I rdrdO
t/a t/r t / /rj
EXAMPLE
Find the centroid of the part of the cardioid between = and = w/2.
SOLUTION. Using the formulas given
above, we have
ir/2 /<*(! + COS 0)
/ir2 /<* COS
I I r 2 cos dr dO
. ^o /o
r cos =
rdrde
FIG. 210
A) *M)
a 3 /T/2
^- / (1 + cos 0)* cos dO
o /0
tr<
T|~^ + 4 sin - sin 3 -f sin 2 + ~ ^9 |
3 L 8 o2 Jo
i . e
16 + o T
16 -f 67
rsm i
/7T/2 /a(l+cosO) /3 /7r/2
/ I r*sin0drd0 -^ I (1 -f cos 0) 3 sin0 <#
JQ JQ __ o /0
T 7 " /" a ^ rdrdO Q T a 2
^o A) 8
10 a
+ 3*-'
182]
ADDITIONAL APPLICATIONS
355
182. Centroid of an Arc of a Plane Curve. The centroid of an
arc may be found by considering the moments of elements As
with centroids (x it 7/ t ), about the axes. As in previous articles, we
have
lim
i = 1
X =
lim ]
n * co i = 1
f
ds
and
n-*> i=i * / yds
.
lim As t I ds
y =
EXAMPLE
Find the centroid of a quadrant of the circular arc of radius a.
SOLUTIONS, (a) Assume the circle given by x 2 -f y 2 = a 2 . Then
T^
Therefore
/x - dx at dy
y _ Ja
X a a , r a dx
- dx a I
- y J( * v a 2 x z
,JL
sin" 1 -
ajo
2a
(0,0)
FIG. 211
From symmetry, y = x = 2 a/ir.
(6) Suppose we use the fact that the circle can be represented by the para-
metric equations x a cos 0, y a sin 0, and that ds a d0; then we may
write
y ds
*
tfsinedd a?( - cos 6)
2a
fell
^1 T/2
00
Jo
and similarly for 2.
If possible, axes of symmetry should be used in locating centroids.
356 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
183. Centroid of a Volume. Considering the moments of ele-
ments of a volume about the coordinate planes, we have in a simi-
lar manner (AM y ),-,j = &V lJ i x t , etc. The same type of reasoning
as in previous articles gives
and similarly
I I I y dzdydx
y =
III dzdydx>
/ / I z dzdydx
fff dzd dx ' Z fff dzdydx
EXAMPLE
rt Find the centroid of the solid bounded above by the plane z = 4 and below
by the paraboloid z = x 2 + 2/ 2 -
SOLUTIONS, (a) From symmetry we have x = 0, y = 0. Then
_ o /: : _ A
z dz dy dx
FIG. 212
Jo Jo
f* p / * = **lf dydx
/o JQ 2 Jx 2 +y 2
r r 4-^ ( 4 _ z2 _ ,j rf rfx
/o ^o
~
(4 -
These integrals can be evaluated by means of the substitution x = 2 sin 0, but
the length of this solution leads us to suggest another.
184] ADDITIONAL APPLICATIONS 357
(6) In cylindrical coordinates, the equation of the paraboloid is z = r 2 .
Then
/ir/2 /*2 /4 1 /-ir/2 /2
4 / / / zr dz dr de f- / / (16 - r 4 )r rfr dB
- _ JQ *^o ^r 2 _ 2/o ^o
2 /*7r/2 /2 /4 ~ /-7T/2 /-2 '
4 / I I rdzdrdO i / (4 - r 2 )r rfr d0
A) A) 'r 2 A) ^0
r"\12
-M
6/ Jo
/ir
/
A)
(c) Still another method is to consider the solid as generated by revolving
z m x 2 about the z axis. Then, taking disc-shaped elements
we have
/-4 /-4 2 3-j4
7T / J 2 2 rf2 / Z*dz ~ Q
. = /o _ = /o _ = _^Jp = 8 ^
_ _ _
I a; 2 c/3 / 2 dz \
Jo JQ 2 Jo
The student should consider the several possible ways of solving such prob-
lems before attempting to apply any one method.
184. Centroids of Non-homogeneous Bodies. The results of
the preceding articles enable us to find the centroids of thin plates,
wires, and solids of uniform density. However, if the material
under consideration has its density variable according to some
law, the elements A^L,, As t , AF t become p&A t , pAs t , pAVi respec-
tively, where p is expressed as a function of one or more of the
variables involved. Then
I pxds I py ds
J pds J*p ds
and similar relations for areas and volumes. These formulas
evidently reduce to the relations previously given if p is constant.
PROBLEMS
Find the centroid of each of the areas bounded as follows. (Nos. 1-3.)
1. By y 2 = 4 x, and 2 x - y = 4. Ans. (8/5, 1),
2. A quadrant of a circle.
358 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
^ 3. A quadrant of an ellipse. Ans. (4 a/3 TT, 4 6/3 TT).
4. Prove that the centroid of the area under any parabolic arch is two-fifths
the distance from the base to the vertex.
Find the centroid of each of the areas bounded as follows. (Nos. 5-15.)
5. By y = cos x and the coordinate axes, from x to x = ir/2.
Am. [(7T/2) - 1, 7T/8].
6. The cardioid r = a(l cos 0).
7. The limacon r = 3 -f 2 sin 0. Ans. (20/11, Tr/2).
8. The coordinate axes and the parabola x llz -f- y 112 = 1/2
9. A quadrant of the hypocycloid x = a cos 3 0, y = a sin 3 0.
Ans. x = y = 256 a/315 TT.
10. The cissoid ?/ 2 = a; 3 / (2 a x) and its asymptote x = 2 a.
11. One loop of r = a cos 2 0. Ans. [(128 a V2)/105 TT, 0].
12. One arch of the cycloid x = a(0 sin 0), ?/ = a(l cos 0) and the
x axis.
13. Between y = z log x and the x axis. Ans. (4/9, 8/27).
14. Between y = xe~ x and its asymptote.
15. The area in the loop of the curve y* = 4 x 2 x 3 above the x axis.
Ans. (16/7, 5/4).
Find the centroid of each of the following arcs of curves. (Nos. 16-18.)
16. The semicircular arc of y Va? x 2 .
17. The arc of the cycloid in Problem 12, from = to = ir.
Ans. (4 a/3, 4 a/3).
18. The arc of the cardioid r = a(l + cos 0) above the polar axis.
Find the distance of the centroid of the following surfaces of revolution
from the plane of the base. (Nos. 19-21.)
19. The lateral surface of a cone of altitude h, with a radius of base a units.
Ans. h/3 units.
20. A hemisphere of radius a units.
21. The paraboloid 2 ft. high, with a radius of base 4 ft.
A 50 - 12V2 ,. ,.
Ans. ^ = 0.944 ft.
oo
Find the centroid of each of the following solids. (Nos. 22-23.)
22. A cone of altitude h, and radius of base a units.
23. A hemisphere of radius a units. Ans. 3 a/8 units from center.
185] ADDITIONAL APPLICATIONS 359
24. Prove that the centroid of a paraboloid of revolution of altitude h is
h/3 units from the base.
Find the centroid of the following solids. (Nos. 25-29.)
25. Formed by rotating the area included by x 2 = 4 ?/, x 4, and y =
about y = 0. Ans. x 10/3 units.
26. Formed by rotating y = sin x about y from x to x = ir/2.
27. Formed by rotating one arch of y = cos x about x = 0.
4ns. y = (TT + 2)/16 units.
28. Formed by rotating about the y axis the area bounded by the ellipse
x 2 /a 2 -f y-/b z = 1 in the first quadrant.
29. Included between the sphere 2 2 + r 2 = 12 and the paraboloid z = r 2 .
s. 2 = 2.10 units.
30. Prove that the wedge above the xy plane cut from the cylinder
x 2 -f y 2 = a 2 by |/ = 2 and 2 = has for its centroid y = 2 2 = 3 ira/16.
185. Centroids of Composite Areas. If an area is made up
of several parts which are separated, we need nothing new to
find the centroid of the whole. The moment of the whole is the
sum of the moments of the several parts, and the formulas
M x A*y, My = A*x
still hold. The only difference from the earlier use is that now
A = A, + A, + A 3 + -, M X = M X ' + M x " + M x f " + . -,
where M x f = A\y\, M x " = A-^y^ M "' = A^j^ . Hence we
have
(Ai + A 2 + A z + -)i? = AMI + A 2 y2 + A 3 y 3 H ---- ,
or
(i)
1=1
where Ai and A l y t for each i may be found by either method for
single areas. Similarly,
360 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
EXAMPLE
Find the centroid of the area of a square of side s after a circle of radius r
has been cut out of it.
SOLUTION. Take the coordinate axes so that the center of the square is at
the origin and the center of the circle on the x axis at (a, 0). The area of the
square may be considered as composed of the circle and that part of the square
outside the circle. The centroid of the square is (0, 0) and that of the circle is
(a, 0); their areas are s 2 and ?rr 2 , respec-
tively. The other area is s 2 Trr 2 ; assum-
ing its centroid to be at (x 2 , $2) and using
formula^ (1) and (2) above, we have
(.s 2 -
=
-r- (s 2 -
whence
To get 7/2, we have
Trr 2 -0-
FIG. 213
Trr 2 -f- (s 2 - Trr 2 )
52 = 0.
We observe that any pair of variables (x t , y % ) of the formulas of this article may
be found if the others are known.
PROBLEMS
1. Find the centroid of that part of an equilateral triangle remaining after
a circle has been cut out of it.
Ans. [(4 Trr 2 a)/(4 Trr 2 - sV3), 0], if (0, 0) and (a, 0) are centroids of
the triangle and circle, respectively.
2. A square is cut from an ellipse. What is the centroid of the part of the
ellipse remaining?
3. An ellipse is cut from a circle. Find the centroid of the remaining area.
Ans. [abc/(ab - r 2 ), 0].
4. A right circular cylinder has a right circular cone removed. What is the
centroid of that part of the cylinder remaining?
5. A sphere is removed from a right circular cone. What is the centroid
of the remaining part of the cone?
Ans. [4^7(4?^ - R*H), 0, 0], where r refers to sphere, R and H
refer to cone, (a, 0, 0) is centroid of sphere and (0, 0, 0) is the
centroid of cone.
186] ADDITIONAL APPLICATIONS 361
6. Cut a sphere from a regular tetrahedron and find the centroid of the
remainder.
7. A right circular cone is removed from a sphere. What is the centroid
of the remainder? Ans. [cr*h/(4 a 3 r 2 /0, 0, 0].
186. Attraction. The law of gravitation is as follows: Any two
particles attract each other with a force directly proportional to the
product of their masses and inversely proportional to the square of the
distance between them. Two particles of masses m\ and m%, respec-
tively, separated by a distance x exert upon each other a force F
given by the formula
(1) F =
where K is the gravitational constant of proportionality.
We are now in a position to calculate the attraction between
bodies of known masses.
EXAMPLE
Calculate the attraction between a thin straight wire of mass m per unit of
length and a particle of mass M at a point P not on the wire.
SOLUTION. Suppose that the particle of mass M is at the point P on the
y axis and that the wire lies along the x axis from A to B. Divide the wire into
elements &x of mass m&x. Then the
approximate force between the particle and x t
any element is JCraAf Az/n 2 . The com- A(-a,oi C(a,0) '\&x B(b,0)
ponents of this force which are perpen-
dicular and parallel to the wire are
KmM'Ax n . KmM'Ax .
- - - cos 0, and - - - sin t ,
respectively. Therefore the sums of such
elements of force are approximate values
for the two forces perpendicular and parallel
to the wire. That is, an approximation for
the force perpendicular to the wire is
* Km M*Ax cos t -
and hence the component of the attraction in that direction is given by
/m , ,. KmM*&X'COs6i v ,- /& cos dx
(2) A = lira L - 1 - ~ KmM I - -
n-*co - n 2 ./-a f 2
362 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
Similarly, the component parallel to the line is
(3) B = KmM f b * m * dx -
J-a r 2
The values of A and B are readily found since, from Fig. 214, r 2 = d 2 -f z 2 ,
cos = d/Vd 2 -f z 2 , and sin e = x/Vd 2 -f x*.
Of course, there is no parallel component if P is on the perpendicular bisector
of the wire, and if P is on the wire at the parallel component is due to the
segment C(a, 0) to B(b, 0). Hence relation (3) reduces to
rs nf & ~~ a
= KmM r
ab
(4) B = KmMf a b ~ = KmM ( - ^
PROBLEMS
1. Find the attraction of a quadrant of a circular wire on a particle of
mass m at its center. Ans. 2 KmM^/2/Tra 2 .
2. Find the attraction on a particle of mass m in the line of a" wire of length
L whose density varies as the square of the distance from the end nearer the
particle.
3. Find the attraction of a circular plate on a particle of mass m on the
line through its center perpendicular to the plane of the plate.
Ans. 2 KmM(l - d/Vd* + 2 )/a 2 .
4. Find the attraction of a quadrant of a circular wire on a particle of
mass m which lies on the circumference of which the wire is a part, and which is
opposite the mid-point of the wire.
5. Find the attraction of a square sheet of density p on a particle on the
line perpendicular to the square at its center.
6. Find the attraction of a hemispherical shell on a particle at the center.
7. Find the attraction of a right circular cylinder on a particle at a dis-
tance d along its axis from one base.
Ans.2KmM[h
8. Find the attraction of a right circular cylindrical shell on a particle at
the center of one base.
ADDITIONAL PROBLEMS
1. Find the force on a circular valve of radius 2 ft. placed vertically with
its center 20 ft. below the surface of the water. Ans. 80 ww Ibs.
186]
ADDITIONAL APPLICATIONS
363
2. Find the force on the shaded area shown in Fig. 215. The vertical
parabolic segment is submerged to the top of the shaded area.
T 3. Find the force on a square flood gate
8] ft. on a side if one side is in the surface,
by finding the force on each of the parts
into which it is separated by a diagonal.
Ans. 2% tons, 5J^ tons.
4. Find the force on the face of the part
of an ellipse in the first quadrant if a 4 ft.,
6=2 ft., and the upper end of the vertical
minor axis is 1 ft. below the surface.
5. Find the force on the lower half of
the ellipse in Problem 4.
Ans. 4 u;(9 TT 4- 8)/3 Ibs.
6. A plate is in the shape of a parabolic
segment, altitude 5 ft., base 6 ft., plane
vertical, axis horizontal, and 4 ft. below the surface. Find the force on
one face of (a) the complete segment, (6) the part below the axis, (c) the strip
from 4 to 5 ft. below the surface.
7. A cross-section of a channel may be represented by a segment of
y = az 4 . If the water is 6 ft. deep and 10 ft. across the top, find the force on
the dam across the channel. Ans. 128 to Ibs.
8. A flood gate is a trapezoid with top base 16 ft., bottom base 10 ft.,
and the altitude 4 ft. The top is 4 ft. below
r t Le 1 ^ e sur ^ acc - What is the force on the gate?
9. Find the force on the plane area shown
in Fig. 216 if it is vertical.
8 , Ans. 1744 w/3 Ibs.
FIG. 215
'90*
90^
10. A triangular plate has a side horizon-
tal, 6 ft. long and 2 ft. below the surface.
Its altitude is 3 ft. and the vertex is down.
Show that the force on one face is independ-
ent of the shape of the triangle.
11. A vessel is made by revolving a
parabolic segment of altitude 6 ft., and base
4 ft. about its axis. Find the work required
to lower the level of the water from 5 ft. to
3 ft. by pumping to 8 ft. above the top.
Ans. 476 ww/$ ft. Ibs.
12. A horizontal cylindrical tank of length 10 ft. and radius 6 ft. is half full
of a liquid weighing 50 Ibs./cu. ft. Find the work necessary to deliver the
liquid to a point 2 ft. above the tank.
Fia. 216
364 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XV
13. Three cisterns are each filled with a liquid. One has the shape of a
hemisphere, another the shape of a paraboloid of revolution, and the third
the shape of a right circular cone. Each has a radius of a, and the same
capacity. Compare the units of work required to pump each cistern empty.
Ans. 27 : 32 : 36.
14. A full semi-elliptical reservoir of revolution is 5 ft. .deep and 6 ft. across
the top. How much work is required to empty it at the top?
15. A hemispherical cistern is full of water. Two men are to pump it out,
each doing half of the work. How deep is the water when the first man has
finished his work? Ans. r(l - V(2 - \/2)/2) units.
16. A circular disc of uniform thickness weighs 100 Ibs., and its radius is
10 inches. Find the work it can do in coming to rest if it spins on an axis
through its center and perpendicular to its plane with a rim velocity of 20 feet
per second.
17. A thin circular disc of radius a spins in oil about its center with a rim
velocity of v. If the oil exerts a resistance on each small element of the disc
proportional to the product of the area of the element and its velocity, find
the total retarding force on the disc. Ans. 4 kirvcP/Z units.
18. A slender bar of length L lies along the positive side of the x axis with
the nearer end at a distance d from the origin. If the origin is a center of
attraction such that each particle of the bar is attracted by a force inversely
proportional to the square of its distance from the origin, find the total attrac-
tion.
19. The moment of inertia of a plane area about an axis in its plane is equal
to its moment of inertia about a parallel axis through its centroid added to the
product of the area by the square of the distance between the axes. To show
this, take (2, y) of the area at the origin.
20. Prove Problem 19 for a solid of revolution and an axis parallel to its
axis of revolution.
21. Show that the force on a plane surface submerged in a vertical position
in a fluid is equal to the area of the surface multiplied by the pressure at its
centroid.
22. Find the centroid of one-half of the curve x 2f * + 2/ 2/3 = 2/3 .
23. Find the centroid of the area in the first quadrant bounded by y = sin x,
y = cos x, and x = 0.
Ans. [(TT - 2V2)/(4 - 2V2), 1/(4V2 - 4)].
24. Find the centroid of each of the areas given below.
/ \ I *^ ~ * y> /i\ \y == Ao X) , ^ 14 x y ,
jz 2 = 4 4 y. [y = x 2 2 x. j y = x x 2 + 4.
25. Find the centroid of a loop of the lemniscate r 2 = a 2 cos 2 0.
Ans, (7ra\/2/8, 0).
186] ADDITIONAL APPLICATIONS 3C5
26. The circles (x rt h) 2 -f- y* = a 2 for h < a form two crescents. Prove
that the centroids of these crescents are at [rt Trh/(ir 2 -f sin 2 0), 0] where
= cos" 1 (h/a).
27. Revolve the area under y log x from x 1 to x = 3 about the
x axis and find the volume and centroid of the volume generated.
Ans. 1,03 TT cubic units; x = 2.41 units,
28. Revolve the area under y = sin 2 2 x from x = to x = ir/2 about the
axis and find x for the solid by integration.
29. Show that the area generated by revolving an arc of a plane curve about
a line in its plane not cutting the arc is the length of the arc multiplied by the
circumference of the circle described by the centroid of the arc. (A THEOREM
OF PAPPUS.)
30. Show that the volume formed by revolving a plane area around a line
in its plane not cutting the area is equal to the area multiplied by the circum-
ference of the circle described by its centroid. (PAPPUS.)
31. (a) Use Problem 29 to find the surface of a torus.
Ans. 4 7r 2 a6 sq. units.
(6) Use Problem 30 to find the volume of a torus.
Ans. 2 7r 2 a 2 6 cu. units.
32. A square has a side 2 a. Find the centroid of the figure obtained by
adding to the square a semicircle having a side of the square as a diameter.
Find the centroid if such a semicircle is cut out.
33. Given the parabola z 2 = 2 py and any line y mx + & meeting the
parabola in the points A and B. Through C, the midpoint of AB, draw a line
parallel to the axis of the parabola meeting the curve at D. Show that the
centroid of the area AGED lies on the line CD.
CHAPTER XVI
INFINITE SERIES WITH CONSTANT TERMS
187. Infinite Series. An indicated sum of n terms formed
according to some law is called a series of n terms. If the number
of terms increases indefinitely, the series is called an infinite series.
The geometric series
(1) a + ar + ar 2 + ar 3 + + ar n ~ l
is an illustration of a series of n terms. We know that the sum of
the first n terms of this series is
. . _ a(l - r n ) _ a _ ar n
i _ r i _ r i _ r
The sum of other series of a finite number of terms may often be
expressed conveniently, but the sum of an infinite series will be
defined as follows :
(3) S = lim S n ,
n >-
where S n is the sum of the first n terms of the series. If this limit
does not exist, we say that the series has no sum.
If we consider the geometric series as an infinite series, we write
it
(4) a + ar + ar 2 + + ar"- 1 + ,
and then its sum is
(5) S = lim S n = lim
This limit is a/(l r) if \r\ < 1, as lim ar n = if |r| < 1. But
n *<
if |r| > 1, the quantity ar n increases indefinitely with n and the
series has no sum unless a is zero. Also, if |r| = 1, the series has no
sum unless a is zero.
366
189] INFINITE SERIES WITH CONSTANT TERMS 367
188. Convergence and Divergence. If the limit of the sum
of the first n terms of a series as n increases indefinitely exists, the
series is said to be convergent; if the limit docs not exist the series
is said to be divergent. Hence the geometric series given above is
convergent if | r | < 1 , and divergent if | r \ > 1 and a ^ 0. ^
Convergent series are of greater importance than divergent ones
in elementary applications. For this reason, although we cannot
always find lim S n) it is necessary to know whether such a limit
?}.-_>. CO
exists. Some methods of determining the existence of this limit
will now be given.
189. Comparison Test for Convergence or Divergence. To
prove the validity of this test we need the following theorem on
limits, which we shall not prove.
THEOREM. // S n is a variable quantity which increases (decreases)
steadily as n increases but never becomes greater (less) than some
fixed finite number S as n * <x> , then S n approaches a limit L which
is not greater (less) than 8. Hence if the sum of the first n terms of
an infinite series of positive terms is always less than a definite
number, whatever n may be, this sum has a limit and the series is
convergent.
Since each term of any series is a finite number and since the
sum of any finite number of such terms is a finite number, it is
evident that the convergence or divergence of a series is not affected
by discarding or adding a finite number of terms. Now we may state
the comparison test as follows:
A series S^ of positive terms is convergent if after some term each
of its terms is less than or equal to the corresponding term of a series
Si of positive terms which is known to be convergent. Likewise,
$2 is divergent if after some term each term is equal to or greater than
the corresponding term of a series Si known to be divergent.
PROOF. Let
(1) a\ + 2 + 3 +
be a positive term series known to be convergent. Suppose
(2) ui + u 2 + u 3 +
is a positive term series such that u k < a k .
368 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVI
If A n and U n are the sums of the first n terms of (1) and (2),
respectively, and if lim A n = A,Jt is evident that
n +>oo
(3) [/ n <4
Since the series (2) is made up of positive terms, U n increases
as n increases; but, by (3), it is always less than- AJ This proves
that U n has a limit, by the theorem stated above, and hence series
(2) is convergent.
The proof for divergence is obvious.
To apply the test, we must use a convergent series whose terms
are greater than or equal to the corresponding terms of the series
being examined, or a divergent series whose terms are less than
or equal to those of the unknown series. Evidently, to show that
the terms of the unknown series are greater than the corresponding
terms of a convergent series, or less than those of a divergent series,
is to show nothing about the unknown series.
EXAMPLE
Test by comparison the series
5,5.55. 5
4 ^ 6 10 ^ 18 ^ 2 + 2
TEST. Since 5 is a factor of each term, we consider the series
1111 *
4 6 10 18 2 + 2
Comparing the n-th term of this series with the n-th term of the convergent
geometric series
S + i + S 4 """ 1 "^" 1 ""'
we assert that
2 + 2 = 2* '
for, if we clear of fractions, we have
2 n < 2 n + 2 or < 2,
which is true for all values of n. Therefore the series is convergent.
The student should notice that the n-th terms of the two series are compared
and those only.
190] INFINITE SERIES WITH CONSTANT TERMS 369
190. Cauchy's Ratio Test. // in a series of positive terms
the ratio of the (n + 1 )-$t term to the n-th term has a limit L as n > oo
the series is convergent for L < 1, divergent for L > 1. The test
fails for L = 1.
PROOF. Let the series to be tested be
(1) Ui + U 2 + U 3 + + U n + U n+ i + .
By hypothesis, lim -^ = L. This means that corresponding to
n >- ^n
any preassigned positive number , no matter how small, there
exists a value of the index, say A;, such that the inequality
(2) L - 6 < -i- 1 < L + e
H n
is satisfied for all values of n i k.
CASE I. L < 1.
Since L is less than 1 , we can find a number r which lies between
L and 1, so that L < r < 1. If we choose (r L)/2 as the e
to be used in (2), we have, from the right side of the inequality,
(3) !f=i < J + ^ < r for all n> k.
u n & &
From (3), by using n fc, k + 1, k + 2, , k + p, , we find
u k+ i < ru k
ru k +i < r z u k
rUk+2 < r*u k
u k+P < ru k +j^,i < r p u k
Hence each term, after the first, of the series
(4) u k + u k +i + ^+2 + + u k + p +
is less than the corresponding term of the series
(5) u k + ru k + r*u k H 1- r p u k -\ .
But the series (5) is a geometric series with ratio r < 1 ; hence
it converges (as shown in 187 and 188).
370 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XVI
Therefore, by applying the theorem of 189, we see that the
series (4) also converges. Consequently, the series (1) converges.
CASE II. L > 1.
Since L is greater than 1, we can find a number R which lies
between 1 and L, so that 1 < R < L. If we choose (L R)/2
as the to be used in (2), we have, from the left side of the inequality,
(6) ^i > ~ + ~ > R for all n > k.
11 n *> **
From (6), by using n = k, k + 1, k + 2, , k + p, , we find
Uk+i > Ruk
Uk+2 > Ruk+i > R*u k
R p u k
Hence each term, after the first, of the series
(7) U k + U k +i + Uk+2 + + Uk+p + ' ' '
is greater than the corresponding term of the series
(8) u k + Ru k + R 2 u k + + R*u h + -.
But the series (8) is a geometric series with ratio R > 1 ; hence
it diverges (as shown in 187 and 188).
Therefore, by applying the Theorem of 189, we see that the
series (7) also diverges. Consequently, the series (1) diverges.
CASE III. 1=1.
Neither of the sets of inequalities given above hold in this case
and the test fails to determine whether the series is convergent or
divergent. The examples below show this to be true.
(a) Try the ratio test on the harmonic series
190] INFINITE SERIES WITH CONSTANT TERMS 371
TEST. The n-th term is 1/n, the (n + l)-st term is l/(n + 1);
therefore
n *a> \ 1/n n >.oo n >o ~t I *
n
This series will be shown to be divergent in the next article.
(6) Use the ratio test on
1 I 1 I 1 I ... , 1 . ...
1-2 ^2-3 ^3-4^ ^n(n + l)
TEST. The ratio u n+ i/u n = n(n + l)/(n + 1) (n + 2) =n/(n + 2).
Whence
r n-fl r r .
lim ( = lim -r- = lim - x = 1.
n *oo \ n / n ^.oo ~\ n ^o
-. ,
n
This series is convergent, as will be shown later.
EXAMPLES
1. Use the ratio test on
!> L_L -L. H + 2 J
' 2 ^ "^* "-
TEST. The (n + l)-st term is (n + 3)/[(n + 2) -3 n ]; therefore
n + 3 n-f 2
(n + 2)3V (n +
n 2 -f4n+3 1 ..
I 3^(n -f l)(n + 3)
* J = JllSL - 3(n + 2) -
(n2 -f 4 n + 4) 3 n - 1 , 4 , 4 "" 3 '
""
and hence the series is convergent.
2. Use the ratio test on
+ + ,.,,
2 ' 2 a 2* 2 n
TEST. The ratio
372 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XVI
Therefore
lim H*i =i lim (n
and the series is divergent.
191. The Integral Test. If we have a series of positive terms
where u n is expressed as somef(n), and if the function f(x) is defined
for all x i a > and if /(x) decreases steadily as x increases from
a to oo , the series converges if f<T f(x)dx is finite and diverges if the
integral is infinite.
We illustrate the test by means of some examples.
EXAMPLES
1. Use the integral test on the harmonic series
~ + 23"*"In
TEST. P>om the form of the n-th term we have the function f(x) l/x,
where x replaces n in that term.
The graph of the function from x 1
to x b is shown in Fig. 217. The area
under this curve as b > > is repre-
sented by
U,0) (2,0) (3,0) (4,0) (5,0) (6,0)
/ dx ~1
= lim (log x)
v -^~ . X b-^co & Jl
= lim (log 6) =
FIG. 217 Since this area is infinite, construct
rectangles with unit bases and with
altitudes equal to the ordinates of the curve at x = 1, 2, 3, , n. These
rectangles circumscribe the curve as shown in the figure. Also the area of each
rectangle is equal in numerical value to the corresponding term of the series
being tested. As these combined areas are greater than that under the curve
from x = 1 to x = n -\- 1, we have
Therefore
S- lim S n > lim r +1 ^= =0,
n *< n *< 'I X
and the series is divergent.
2. Apply the integral test to the series whose n-th term is 1/n* where
k* 1.
iy 1 J UN JP liN 1 1 fclt.JLUL.& W 1 1 n V-/U1N d I A1N 1 1 J^itiVlfc
TEST. CASE I. k < I.
The curve to be used is the graph of the function f(x) = 1/z*. The area
under the curve from x = 1 to x = >
,.
lim
r b dx r I xl ~ k \~\ b
I -r- = hm ( - -- r I
i a:* &-*. \1 k/Ji
1 -
lim (b l ~ k - 1)
is is infinite, we circumscribe "o
rectangles as in the previous example.
These show that
FIG. 218
Hence
lim
lim (n
and the series is divergent.
CASE II. h > 1.
The area under the graph of
from x = 1 to x
lim
x k
lim
lim
1 -
..
= hm
1 A; 6^00 A; 1
A: 1 '
a finite number. Since the area is
finite, construct rectangles with unit
bases and altitudes equal to the
ordinates of the curve at x = 1, 2, 3,
, n + 1> which are inscribed under
the curve, as shown in Fig. 219.
The area of each rectangle r is nu-
merically equal to the corresponding
term of the series being tested. There-
fore the area under the curve' from
x 1 to x n is greater than the
combined areas of the first n rectangles except for the first rectangle, which is
not under that part of the curve. Hence we have
(1,0) (2,0) (3,0) (4.0) (5,0) (6,0)
FIG. 219
_
n k ~ l
/ n dx __ n k
. ~x~ k ~ k - 1
374 DIFFERENTIAL AND INTEGRAL CALCULUS [Cii. XVI
and therefore
S 1 = lim (S n 1) < lim / ^ - -r r ,
n >oo n >< *1X K> 1
S < 1 +
and the series is convergent.
These examples give a group of series of the form
1+1 + 1 + .. . + 1+...
J- -t- 2 * + 3 * -I- + n * -t- ,
which, together with any finite multiples of themselves, are con-
vergent if k > 1 and divergent if k < 1. These may be used for
comparison with unknown series.
The integral test may be readily applied when the series is a
positive term series with an ft-th term which is integrable.
If the f(x) derived from the n-th term becomes infinite after
x = 1, we merely use that part of the graph of the function which
is to the right of any vertical asymptotes. This will make the
first rectangle constructed correspond to some term after the first
in the series and hence the test will compare the sum of a series
with an integral where some of the first terms of the original series
have been discarded.
t
192. Alternating Series. This type of series has its alternate
terms negative.
THEOREM. // after some term each term of an alternating series
is less than or equal to the preceding, that is, Uk+i ^ u k) and if
lim u n = 0, the series is convergent.
n >o>
PROOF. Let
S n = U\ U 2 + Us U* + + ( l) n ~ l U n ,
where each u k ^ UK+\. Now consider n even and write S n in the
two formsjbelow:
(1) S n = (Ui - U 2 ) + (U 3 - U 4 ) + + (U n -i - U n ),
(2) S n = Ui - (tta Ms) ' (u n -2 - U n -\) U n .
193] INFINITE SERIES WITH CONSTANT TERMS 375
Since Uk+i ^ u k , each parenthesis is positive or zero, and so,
from (1), S n is positive and increases with n or remains unchanged.
The relation (2) shows that S n is less than u\. Hence S n has a
limit.
Considering an odd number of terms
Sn+l = S n + (
we find that this sum has the same limit as S n since
lim Sn+i = lim S n + lim (
but by hypothesis lim ( l) n w n +i = 0. Therefore the series is
n *>oo
convergent.
It is evident that the error in using n terms of a convergent alter-
nating series to represent its sum is less than the (n + l)-st term.
This is because
(3) S - S n = U n +l ~ U n+2 + Uv+3 ',
and this when written as (2) above shows that the difference
S S n is less than u n +i.
193. Absolute Convergence. A series is said to be absolutely
convergent if the series derived from it by making all terms
positive is cpnvergent. If a series is convergent but becomes
divergent when all terms are made positive, it is said to be con-
ditionally convergent. Thus
I _!
3 3 4 3
is absolutely convergent but
is conditionally convergent.
THEOREM. // a series is absolutely convergent^ it is a convergent
series.
PROOF. Let
S n = ai + a 2 + a 3 + + a n .
d/O DLX " Ji.JUi.rS 1 1A.L. AISJJ liN IJlAjrttAl-i ^A1A_/U.L(UO [^
By hypothesis
S = lim S n = lim (|ai| + |a 2 | + |a 8 | + + K|) = C,
n >< n >
a finite number. But
S n = (| a, | + |a y | + + K|) - (| a* | + |a| + + |a,|),
where the first parenthesis contains only positive terms and the
second only negative terms of the series. But
*
lim (|a,| + |a,| + |a,| + + |o|) = C,
n >
and hence
lim (| a,- 1 + K-| + ) < lim (|ai| + |a s | H ---- + |a n |) = C,
n *>m n >oo
and also
lim (|a*| + |ai| + ---)^lmi (|ai| + |a 8 | + + |a n |) = C.
tt__ >.co n *>oo
Since each of these limits exists, then
lim S n = lim (|a t -| + |a/| + )- lim (I a* + \a>i\ + )
71 >> oo 71 > ot> ft > oo
exists and the series is convergent, which was to be proved.
This theorem allows us to use the Cauchy ratio test for series
some of whose terms are negative. In so doing, we consider the
absolute value of the ratio of the (n + l)-st term to the n-th term,
i.e., \Un+i/u n \.
A very important necessary condition for convergence is that
lim u n = 0. This condition is however not sufficient, as is exempli-
n *<
fied by the harmonic series.
EXAMPLE
Test the series
333'
3 ~~2! + 5!~4! + '"
for convergence.
TEST. This is an alternating series and hence, if we can show that the
(n + l)-st term is less than the n-th and that the limit of the n-th term as
n increases indefinitely is zero, it is convergent.
193] INFINITE SERIES WITH CONSTANT TERMS 377
The numerical value of the n-th term isM = 3/n!, also | u n+l \ = 3/(n -f 1) !.
Therefore set
(Ti + l)! n!
whence
3n! <3(n + 1)!,
or
1< n + 1,
which is true for all n.
Also lim (3/n !) = 0. Therefore the series is convergent.
n >>
PROBLEMS
Test the following series for convergence or divergence. Several tests may
be applied by the student in most of the problems. Test the alternating series
for absolute convergence. (Nos. 1-11.)
! o ~1~ 7 ~1~ 7- + * ' ' ~^~ o~~ + Ans. Divergent.
jj 4 u & 71
7 1 , 2* , 3* ,
jr c
_ __ __ __ -
2 ^ 2 5 + 1 ^ 3 6 + 1 ^
3. 2 + ^2 + 2Ts "^ 374 + ' ' * ^ ns - Divergent.
4 1 + 1 , JL . 1 +...
4 ' 2 + 5 + 10 + 17 + '
5. 1 + ^ + ^ + ;p -r- Ans. Convergent.
, , 8 , 12 , 16 ,
7. ^ + 3 + 2^32 + 33 + * ' * + 23n-i + ' ' Ans. Convergent.
1 2 2 3 2
9t 2 ^" 2 3 -f 1 ~^~ 334-1 + * * * -Aw. Divergent.
10 1 . 8 27
.
1 -f 2 16 + 2 81 + 2 ""
' 3 "*" 5 ~^" 7 "^" 9 "^" * * "" ^ n8 * Divergent.
378 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVI
Use the integral test on each of the following series. (Nos. 12-22.)
12. 1+1+1 + ....
13. 1 H -- 7= H 7= -}-. ,4ns. Divergent.
V2 \/3
1S * J72 + 374 + 5Te + ' "' ^ ns ' Convergent.
4 6
16 -
1 3^=1:
17 ' 5 + 15 + A + k + ' - ' ' Ans ' Conver ent -
18. i+-L ++ l4.....
19 - x + 2 3 /* - 1 + 33/2 - i + 43/2 _ i + ' ' ' ^ ns - Convergent.
20. sin 5 + ^sin 5 + Lin 5 H ---- .
^ o
21. e" 1 -f e~ 2 -h ^~ 3 + Ans. Convergent.
22. 2 e- 2 + 3 e~ 3 -f 4 e~ 4 + .
Test each of the following series for convergence and for absolute conver-
gence.
23. Q ~ 5 + 7 ~~ n + * * * Ans. Conditionally conv.
24 ' X - 2 + 3 " I + ' * '
1 2 2 3 2
' ' * Ans. Conditionally conv.
26
13 4. i ~~ 23 a- i "" 33 4.
*
___ - - ___ -
2-3 2 2 -6" r 2 3 ll 2 4 -18
1357
27. o~5"T'y~Q~r'"''- Arw. Divergent.
193] INFINITE SERIES WITH CONSTANT TERMS 379
12 1
29 ' 2 ~~ 2^32 + 2^52 - * ' * Ans. Absolutely conv.
1 2,3
3 * 2-5 + 10-""
31. 1 + T Ans. Absolutely conv.
149
32. r: jr 5 jr -f- ;
1-2 8 - 2 ' 27 - 2
33. j-^ 27^ + ^74 ~~ 475 + Ans. Conditionally conv.
CHAPTER XVII
POWER SERIES AND SOME APPLICATIONS
194. Power Series. Series of the form
-\ ,
where ao, fli, #2, a, are constants, are called power series.
A power series may be convergent for only some values of the
variable or for all values. We state without proof a theorem
which we shall need.
ABEL'S THEOREM. // a power series converges for x = c, it
converges for any x such that \x\ < c.
All of the values of the variable for which a power series is
convergent constitute the interval of convergence of the series.
The ratio test may be used to find the interval of convergence
in most cases. The values of the variable for which the limit
of \u n +i/u n \ = 1 are called the end points of the interval; to de-
termine whether the series is convergent for them or not, we must
use some other constant-term series test.
EXAMPLES
1. Find the interval of convergence of
_ *! 4. *! _ ?L 4. . . .
SOLUTION. For this series \,u n \ = | x n /n \ and therefore
lim
u n
Whence the series converges for values of x such that \x\ < 1, that is, for
1 < x < 1. The series diverges for \x\ > 1 or if x > 1 and x < 1.
For \x\ = 1 the test fails and we must use the values x = 1 in the series
in order to get constant term series to test for convergence by some other test.
If we let x = 1 in the series we have
380
194] POWER SERIES AND SOME APPLICATIONS 381
which is divergent, as it is the harmonic series except for sign. Then for
x = 1 the series becomes
which is convergent since
lim - = and j T < ~ for all n.
n *> n n -j- 1 n
The values 1 for x are the end points of the interval of convergence of
this series and we have shown that x = I belongs to the interval but x = 1
does not. Therefore the interval of convergence of the series is
- 1 < x < 1.
2. Find the interval of convergence of
SOLUTION. As \u n \ \x 2n ~ 1 /(2n 1)!| we have
lim
Un+l
U n
= lim
j2n+l / X 2n ~ l
(2~n + I)!/ (2n - 1)!
1
^ 77: r rr
2 n(2 n + 1)
=
for all finite values of x. Therefore the series converges for all finite values
of x. Such an interval is written as follows:
oo < X < oo.
3. Find the interval of convergence of
x - S 1 (r - 5) 2 1 (x - 5)*
5 " "^ 2 % 5 2 "^ 3 " 5 s + ' ' '
SOLUTION. Here |w n | == | (x 5) n / n *5 n | and therefore
lim
n *>
Un+l
r
lim
x
(x 5) n+1 n5 n
Un
(
5
n + l)^ 71 " 1 " 1 (x
lim .- ..-
-5)
x -5
b
n __^oo n + 1
5
The series is convergent then if | (x 5)/5| < 1. That is, for
or
or
-5<z-5<5,
< x < 10,
382 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
and divergent for x < and x > 10. Now set x = in the series. We have
~~ 1+ 2~3+4 '
which we have shown in Example 1 to be convergent by the alternating series
test. For x = 10, the series becomes
1 + 2 + 3 H ,
which is divergent. Therefore the interval of convergence is
< x < 10.
The interval of convergence is of great importance in that such a series
does not represent some f(x) outside the interval of convergence. Even in
the interval of convergence the series should not be used for f(x) unless the
remainder after n terms approaches zero rapidly.
PROBLEMS
Find the interval of convergence of each of the following series. Be sure to
test the end points of each interval.
2> 2 + 2^3 + 3^4 + " '
3. 1 +-*- + - + .... Am. - 1 < x < 1.
1-2 2*3
x x 2 x* j 4
' ~ + "~" + "**
x x 2 x' x^
2 "*" 3^22 ~ 5^3 + 77^ - Ans. - 2 < x < 2.
7. | + -^= + -^= + . Ana. - 5 < x < 5.
5 5 2 V2 5 3 V3 ~~
195] POWER SERIES AND SOME APPLICATIONS 383
10 i I * - 2 i (' " 2 >* . <* ~ 2)> i . . .
10. i + 3 + 5 + 7 + .
11 * + l I 2( * + 1)2 I 3( * + 1)3 I A, *,<?***
11. | -- h - J5 - -H -- f, - + ' Ans. -5<z<3.
i? i | * + 3 . (s+3)' , (s + 3)' ,
+ + ~~ r + + -
14 i _i_ (2>! . (x + 2) (x + 2)'
1 ~" < ' * h " 2 ' '
16 g-H _ (a: + I) 2 (s + I) 3
3 3 2 V2 3V3
i . .
1 -- __^_ _^ -- __ -- 1
. __ -- ___ _ __
19. Is 1 + \ (x - 1) + \ (x - I) 2 + | (x - I) 3 + convergent for
o O J
x = 0, - 1, 1 , ! f 5? Ans. Only for x = 3/4.
195. Taylor's and Maclaurin's Theorems. Since a power
series becomes a convergent constant-term series for any value
of the variable within its interval of convergence, its sum is a
function of the variable within the interval.
To find a series which represents a given function for some
interval is the important problem which is now presented.
Suppose we have a continuous function f(x) with continuous
first, second, , and (n + l)-th derivatives, denoted by /'(or),
/"(#)> *)/ n+1 (^)> in the interval from x = a to x = 6, inclusive.
(A) Using the Extended Theorem of Mean Value.
The proof of the law of the mean given in 90 can be restated,
for a function f(x) which has a continuous derivative, by setting
-/(a) _ K
b-a ~ Kl >
384 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
or
(la) f(b ) - j (a ) - Kl(b - a) = 0,
so that the auxiliary function there used becomes
(2a) *(*) = /(&) - f(x) - K,(b - x),
which reduces to the left side of (la) when x = a. This leads, as
in 90, to the final formula proved there, which may be written
in the form
(3a) /(&)=/() +/'(*i)(& -a), a < x, < b.
Proceeding in a precisely analogous manner, let us write
-/(<*) -/'(a) (6 -a)
(b - a}* - *>
or
(Ib) /(&) - /(a) - f'(a)(b - a) - K 2 (b - a) 2 = 0.
If we now examine the auxiliary function
(2b) <j>(x} = /(6) - /(x) - f(x)(b - x) - K,(b - x
we observe that 0(a) = by (Ib), and 0(6) = identically.
Hence, by Rolle's theorem (89), we must have 0'(:r 2 ) = for
some value x 2 between a and b. But, from (2b),
*'(*) = -f(x) +/'(x) -T(x)(6 - x)
whence, since ^(0:2) = 0, we have
so" that 2 K 2 = f"(x<i), since b x 2 can not be zero. Therefore
(Ib) may be written in the form
f(b) = /(a) + /'(a) (6 - a) + /"(*,) Jl , a < x, < b.
In the same manner, if we commence with
(1) /<&) - /(a) - /'(a) (6 - a) - /"(a) (6 ~ f a)2 ----
- /-(a) ^ - #+,(& - a)-H- = 0,
195] POWER SERIES AND SOME APPLICATIONS 385
and use the auxiliary function
(6 - xY
(2) *(*) = /(&) - /(*) - f'(x)(b -x)- /"(*)
2!
we see again that $(a) = and <(&) = 0, so that <t>'(x n+ i) = 0,
where x n +\ lies between a and 6. Proceeding exactly as above,
we thus obtain the formula
(3) /(6) = /() + /'(a) (6 - a) + /"(a) ^^ +
where a < # n +i < 6.
It is evident that the same formula would be obtained by using
in the place of 6 any value of x greater than a. Replacing bbyx
in (3), we may then write
(4) /(*) = f(a) + /'(a) (x- a) + /"(a) ( * ~ g)2 +
a n X ^ ,
^ -h/ ^n + lj (n+1) J '
where a < x n+l < x < 6. The same formula holds also if 6 < a,
if all the inequalities throughout are reversed, and if finally
a > x n +i > x > 6.
This is known as Taylor's theorem and the function /(x) is said
to be expanded about or m f/ie neighborhood of x = a.
If we set a = in Taylor's theorem, we have
(5) /(x) =/(0)
r +l
, .x. , where
which is known as Maclaurin's theorem for/(z), and f(x) is said
to be expanded in the neighborhood of x = 0.
386 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
(B) Using Successive Integrations*
We form the expression ( 150, Problems 103-106)
(6)
and apply integration by parts, setting u = (x y) n /n\ and
dv = f n + l (y)dy. Then du = [- (x - y) n - l /(n - l)\]dy and
v = f n (y). Therefore
Another integration by parts gives
Successive integrations give
- /" <) > ;, a - /' (a) (x - a) + /<z) - /(a) ;
* This method of proof may be omitted if desired.
195] POWER SERIES AND SOME APPLICATIONS 387
whence, by transposition,
(7) /(*) =/(a) +/'(a)(x - a) +f"(a)^~^+ -
Then a = gives
(8) /(*) = /(O)
These formulas are Taylor's and Maclaurin's theorems with an
integral for the last term.
(C) Using Undetermined Coefficients.
We shall here adopt another point of view in the derivation of
Maclaurin's series for a function /(x) which possesses derivatives of
all orders but complete proofs will not be attempted.
If a power series
(9) Oo + aix + a 2 x* + a 3 x 3 + + a n x n + a n+l x n+1 +
converges for any value of x different from zero, as x = p, it will
converge in an interval p < x < p, where p > ( 194).
In more advanced works in mathematics, it is shown that if the
power series (9) converges within a certain interval p < x < p,
with p > 0, to a function /(#), then the power series which is
obtained by termwise differentiation will converge within the same
interval to the derivative of /(#). By repeated applications of this
theorem, we can then see that if a convergent power series is
repeatedly differentiated termwise the successive derived series
will converge to the successive derivatives of f(x) within the
interval p < x < p.
We shall assume that there is a power series which converges
to the function f(x) for all values of x within the interval
p < x < p, with p > 0. Let
(10) f(x) = oo + aix + a 2 z 2 H ---- + a n x n + a n+l x
n + l
388 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
Our problem is to determine the values of the coefficients OQ, i,
#2, *, tt n , By the aforementioned property of power series,
the successive derived series will converge to the successive deriva-
tives of f(x) within the interval p < x < p. Forming these
successive derived series, we have
f'(x) = ai + 2 a& + 3 cw* H + na n x n ~ l
+ (n + l)a n +\x n H ,
f"(x) = 2 1 a 2 + 3 2 a 3 x + h n(n - 1) a n x n ~ 2
+ (n + 1) n*a n +ix n - 1 H ,
/"'(x) = 3 2 1 a s + + n(n - l)(n - 2) a n z- 3
(11) + (n + l)n(n - 1
/()(ir) = n(n - l}(n - 2) 3 2 - 1 a n
+ (n+ l)n(n - l)(n -2)
4 3 2 a n+ iz + -,
Since the series in (10) and (11) converge for x = to the
values of f(x) and its successive derivatives, respectively, at x 0,
we have, by letting x = 0:
(12)
/(O) = oo
r (o) = fll
/"(O) =2!a 2
/ r// (0) =3!a 3
/(n)(Q) = n \a n
By solving the equations of (12) for the coefficients Oo, ai, a 2 ,
a n , , we have
"
oo = /(O), ai = /'(O), a 2 -
t
Therefore we obtain
(13) f(x) = /(O) + /' (0)*
196] POWER SERIES AND SOME APPLICATIONS 389
This is Maclaurin's series for /(#). For a given function, the
coefficients of the series can be obtained and the interval of con-
vergence of the resulting power series can be obtained by the
methods of 194. It can be shown that the series obtained
actually converges within this interval to the function f(x) pro-
vided that the remainder E n \ of 196 approaches zero as n
becomes infinite; but the proof is beyond the scope of this book.
196. Applications of Taylor and Maclaurin Theorems. We say
that we compute the values of some functions by using series
into which the functions have been expanded. Such series should
have their terms decrease rapidly, since we never use the infinite
series in computation problems. At best only a few terms are used,
and unless the terms decrease rapidly more terms will have to be
computed. The most important thing is the value of the remainder
of the series when (n + 1) terms have been used as an approxi-
mation. In (4) and (5) of 195, this remainder is the last term
whose value depends on f n+l (x n +i) where a < x n +\ < x ^b (or,
if b < a, 6 i x < x n+ i < a). An upper limit for the value of the
remainder after n + 1 terms may be found by using the maximum
value of |/ n+1 (#) I i n the interval from a to x. Suppose that the
largest numerical value of f n+l (x) within the interval of integration
is given by x = c. Then |/ n+1 ( c ) I represents that greatest nu-
merical value. Then we see that the remainder or error, which
we shall denote by E n +i, for the Taylor expansion, satisfies the
inequality
(1) |n +1 |=i
(*-
+ 1) !
where c is the value for which | f n ^ 1 (x) \ is a maximum for all values
of x under consideration; and, for the Maclaurin expansion,
(2) | E n+l | =i
f n+l (c) 7
(n + 1) !
Similar formulas can be derived also from (7) and (8) of 195.
Three important questions arise in using series for numerical
calculations, namely:
(a) What error is made in using a given number of terms?
390 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
(6) How many terms must be used to get a value of the function
correct to a given number of decimal places?
(c) Within what interval of the variable will a given number of
terms give required accuracy?
We illustrate methods of answering these questions by means
of special examples.
Since a Taylor series is composed of powers of (x a), the
rapidity of its convergence depends upon the numerical value of
(x a). Hence, in expanding a function, we choose a so that
| x a | is as small as possible and at the same time so that the
values of /(),/' (a), /" (a), etc., are easily computed.
EXAMPLES
1. Find a series suitable for evaluating cos 31.
SOLUTION. A suitable series means one for which \x a| is small and
/(), /'()> etc., are readily computed. Since 30 is near the angle to be used
and sin 30 and cos 30 are readily written down, we take a = Tr/6. Then
* = cosx,
f(x) = -sinz,
/"Or) = -cosx,
Hence, by substitution in the formula (4) of 195, we have
,
cos.- - --- --- - +
-5 --- 2 --- y^T - 2 . 3 |
For x = 31 7T/180, the first three terms give
, 18 V3
cos31 =--
--360 -- 4
= 0.866025 - 0.008727 - 0.000132 H
= 0.85717 .
196] POWER SERIES AND SOME APPLICATIONS 391
This is correct to five decimals, for the error after three terms, Es, satisfies the
inequality
I ft | < - sin c < j = 0.00000088.
2. How many terms of the expansion of log x in the neighborhood of
x = 3 are needed to get log 3.5 correct to four decimal places?
SOLUTION. In relation (1) of this article we have \E n +i\ < 0.00005, a = 3,
x 3.5, and n is to be determined. Since
the maximum value of | f n+1 (x) \ in the interval 3 ^ x ^ 3.5 is evidently
i
Then we must have
The smallest possible value of n satisfying this inequality is found by trial.
In \Enn\ ^ (0.5)+V[3 w+l ( + 1)] the value n = 2 gives \E 3 \ ^0.00154;
n=3 gives |# 4 | ^ 0.00019; n = 4 gives | # 6 1 < 0.000026. Thus n=4
makes the error as small as desired. Since (n -f 1) terms precede the term
representing the value of the remainder E n+i of the series and we have found
n 4, it is necessary to use five terms of the series.
3. What accuracy is given if six terms of the series of Example 2 are used?
SOLUTION. As n 5 if six terms are used, we have
= 0.00000357.
1 ' =3 6 6
Therefore six terms will give log 3.5 correct to at least five decimals.
4. Find the interval within which two terms of the series of Example 2
will give an error not exceeding 0.005.
SOLUTION. Since n -j- 1 = 2 and f(x) = log x, we need f"(x) I/a: 2 .
Therefore
IE |-=L. (* - 3)"
c 2 2
where c lies between 3 and x. To make | E 2 1 ^ 0.005 we must determine x so
that
I 1 (x - 3) 2
rt < 0.005.
392 DIFFERENTIAL AND INTEGRAL CALCULUS [Cm. XVII
There are two cases to consider.
CASE I. WHEN x > 3. In this case c = 3, and we have
.
whence x = 3.3.
CASE II. WHEN x < 3. The maximum value of | f"(x) \ is now 1/z 2 and
| x 3 1 is 3 x. Therefore
-Si 0.005,
whence x S 2.727.
Therefore the complete interval is
2.727 < x < 3.3.
However, we may consider the two cases simultaneously so as to give an
interval symmetric about x = 3. To do this we let | x 3 1 = h and
c 3 -|- c, where | c | ^ /&. Then the inequality used above becomes
1 ^ ^ 0.005.
(3 + e) 2 2
The maximum value of I/ (3 H~ e) 2 is for e = h as this makes 3 + e least, so
that our inequality becomes
_
(3 - A) 2 ='
whence A i 0.2727. Hence | x 3 1 = 0.2727 and the symmetric interval is
- 0.2727 < x - 3 < 0.2727,
or
2.7273 < a; ^ 3.2727.
PROBLEMS
1. Assume e x = a -f &z + ex 2 + dx* -f- , and find a, 6, c, d, by the
method of undetermined coefficients.
2. Expand 3t/ 2 7?/-|-9ina series of powers of y 3.
3. Expand 2z 3 3z 2 8z-flOina series of powers of x + 2.
*. - 2 -f 28(z -f 2) - 15(x -f 2) 2 + 2(a: + 2) 3 .
Expand each of the following functions to four terms, first in a power series
of (x a) (Taylor), and second in a power series of x (Maclaurin). (Nos. 4-8.)
4. (1 db *).
196] POWER SERIES AND SOME APPLICATIONS 393
5. sin x.
/ \ sin a, , ^ n cos a , \ ,
sin a + cos a (x - a) jr-p (x - a) 2 r-r- (x - a) 3 -f >
x - x 3 /3 ! + x 5 /5 ! - x 7 /7
6. cos x.
7. e-.
L - x + x 2 /2 ! - x 3 /3 !+.
8. log (1 x).
Expand each of the following functions to three terms using both Taylor's
and Maclaurin's series. (Nos. 9-10.)
9. e s'm x f
e sin a[i _|_ cos a . ( x a ) -|. ( C os s a sin a) - ^-= f- ),
. 1 + x + x 2 /2 ! -.
10. e* cos x.
11. Write the Maclaurin expansion of e~ x2 to four terms and find the inter-
val of convergence. Ans. oo <x < oo.
12. Expand the following functions in a power series of x.
(a) sin" 1 x. (d) log cos x.
(6) a*. (e) log (\/x 2 + 1 - x).
(c) tan- 1 x. (/) tan (x + T/4).
13. Write a Taylor's series suitable for evaluating sin 29 and approximate
the value of sin 29 by using three terms. Ans. 0.4849.
14. Same as Problem 13 for cos 62.
15. Same as Problem 13 for sin 59. Ans. 0.8572.
16. What is an upper limit (\E n+l |) of the error in Problem 13?
17. What is an upper limit (\E n+l |) of the error in Problem 14?
Ans. 0.00001.
18. What is an upper limit (|# n +i|) of the error in Problem 15?
19. Write several terms of the Maclaurin series for e x and find an upper
limit for error in e' 02 if four terms are used. Ans. (10)~ 8 (0.68),
394 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XVII
20. How many terms would make the result in Problem 19 correct to four
places?
21. Find | E n +\ \ if log x is expanded about x = 3 and three terms are used
to approximate log 3.11. Ans. |# 3 | < 0.00002.
22. How many terms must be used to get the result in Problem 21 correct
to four decimals?
23. Derive a series for tan" 1 0.1 and evaluate to four decimals.
Ans. 0.0997.
24. Derive a series for tan" 1 0.2 and evaluate to four decimals.
25. Derive a series and evaluate cos 29 to three decimals. Ans. 0.875.
26. Derive a series and evaluate sin 29 40' to four decimals.
27. Derive a series and evaluate cos 46 to four decimals. Ans. 0.6947.
28. Write the Taylor's series for e~ x about x 1 and find the interval of
convergence.
29. Use Problem 28 to find an upper limit to the error in e~ 1 - 1 if three terms
are used. Ans. \E*\ < 0.00006.
30. Write series for log (1 x) and evaluate log 0.9 to three decimal places.
31. How many terms of the Maclaurin series for log (1 x) are necessary
to approximate log 1.05 to four decimal places? Ans. 3 terms.
32. Find an upper limit of the error involved in Problem 31.
33. Evaluate log 1.1 to three decimal places. Ans. 0.095.
34. Evaluate cos 61 to four decimal places.
35. What is an upper limit to the error in cos 12 if four terms of the
Maclaurin series is used? Ans. \E*\ < (10)~ 7 .
36. If log 2 = 0.6931, compute log 2.1 to three decimal places.
37. Within what limits will three terms of the Maclaurin expansion of
log (1 + x) give an error not greater than 0.0001?
Ans. - 0.0717 < x < 0.0669.
38. Within what limits will three terms of the Maclaurin series for cos x
give an error not greater than 2 units in the fourth decimal place?
39. Within what interval do four terms of the series for sin x about ir/6 give
results correct to five decimals? Ans. 0.399 g x < 0.628.
197] POWER SERIES AND SOME APPLICATIONS 395
40. Using (1) of 196, find the unknown in each of the following cases if
f(x) sin x and a = 7r/3.
(a) n = 3, x = 61 30' expressed in radians, \E*\ = ?
(b) x = 58 45', |# n +i| ^0.0001, n = ?
(c) n = 3, |# 4 | ^0.0005, a: = ?
41. The same as Problem 40 for/(z) = log x, a = 3.
(a) n = 3, z = 3.2. 4ns. |#4| < 0.000005.
(6) x = 3.2, !#+!! < 0.0001. Ans. n = 2.
(c) n = 3, |# 4 | ^ 0.0005. Ans. 2.48 ^ x < 3.63.
197. Newton's Method of Approximating the Roots of an
Equation. Taylor's series may be used to solve equations approxi-
mately. A method due to Newton
is illustrated by what follows.
Suppose f(x) = is a given
equation. Let Fig. 220 represent
the graph of f(x) near a root. By
trial we locate a first approxi-
mation to the root, such as x = a.
The slope of the tangent at Q is
/'(a), which from the figure is equal
to /(a)/ (a x). Therefore x, a
second] approximation to the root, J 220
is given by
x = a
But this is the value of x obtained by replacing J(x) by the first two
terms of Taylor's expansion. It may be used as a new value of a
to obtain the third approximation, and so on.
EXAMPLE
Find to three decimals the positive root of x 2 cos x = 0.
SOLUTION. By graphical methods or by use of tables we find that there is
a root between 0.8 and 0.9. We now write the equation with all terms in the
left-hand member and set
f(x) x 2 cos x = 0.
If we replace f(x) by the first two terms of Taylor's expansion about 0.8,
we have
/(*) = /(0.8) +/'(0.8)(* - 0.8) = 0,
396 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
if we discard higher powers of x 0.8, which itself is less than 0.1. This
becomes
0.64 - cos 0.8 + (1.6 + sin 0.8) (z - 0.8) = 0,
0.64 - 0.6967 + (1.6 + 0.7 174) (x - 0.8) = 0,
x - 0.8 = 0.024,
x = 0.82,
if we carry the division to two decimal places.
To get a third decimal, we repeat the operation, using the expansion about
0.82. This gives
f(x) = 0.6724 - cos 0.82 + (1.64 + sin 0.82)(x - 0.82) = 0,
x - 0.82 = 0.0041,
x = 0.8241,
so we conclude that x = 0.824 is the solution desired. If the fourth decimal is
more than 5, either increase the third decimal by unity or better still repeat
the operation using three decimals in the expansion.
PROBLEMS
Solve each of the following equations to three decimals. (Nos. 1-13.)
1. cos x - 10 x = 0. Ans. 0.0995.
2. e* /2 -f x = 2 for its smallest positive root.
3. e x + x = 2 for the root near 0.4. Ans. 0.443.
4. 2 cos x = x 2.
5. e 2* _ x = 2 . Ans. - 1.981, 0.448.
6. x 3 + 4 x - 7 = 0.
7. 3 log x - e* + 4 = for the root near 1.7. Ans. 1.731.
8. x 3 = sin (x + 2).
9. sin 2 x + x = 1.4. Ans. 0.529.
10. 2 x cos x = x 2 - 1.2.
11. x 2 - 2 x 4- log a; = 0. Ans. 1.690.
12. tan 3 x = 1 3 x for its smallest positive root.
13. sin (x/2) - 2 x + 8 = 0. Ans. 4.403.
14. Find a high and a low point on the curve y x sin x.
15. Solve the equation e*'* + cos 2 x = x 2 2. Ans. 1.293, etc.
16. Solve the equation e~* -f sin x = 1.235 for the root between 1.2 and 1.3.
199] POWER SERIES AND SOME APPLICATIONS 397
198. Approximations to Definite Integrals. The quantity
represented by jf 6 f(x)dx is well known when the indefinite
integral F(x), which has f(x) as its derivative with respect^to
Xj can be found. We recall that the numerical value of J^ b f(x)dx
is F(b) F (a). However, F(b) F(a) cannot be found if we
cannot find F(x) and also if we do not know the exact form of /(x),
but merely know its values for special values of x. We shall
develop two methods of evaluating J^ b f(x)dx approximately which
depend upon the ability to find numerical values of f(x) for various
values of x in the interval a i x ^ 6. These methods are there-
fore applicable whether or not the form of F(x) is known.
199. The Prismoidal Formula. Let f(x) be expanded in a
Taylor's series in the neighborhood of x = a. If we represent
f(x) by the first four terms of this expansion, we have
(1) /Or) = A + B(x - a) + C(x - a) 2 + D(x - a) 3 ,
where A, B, C, and D represent the fractional coefficients of the
Taylor's series.
Under this assumption, for f a b f(x)dx we write
r b
(2) / [A + B(x - a) + C(x - a) 2 + D(x - a)*]dx
_ A < - a, + S + Q* +
- a) + 2 C(t - .)
From (1) we find the values of f(x) for x = a, (a + 6)/2, and
b in terms of A f B, C, D to be
(3)
/(a) = A,
a) C(b a) 2 D(b a) 3
= A |
2 ' 4 ' 8
f(b) = A + B(b - a) + C(6 - a) 2 + D (b -a)'.
If we form the combination
398 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
from relations (3) we get the quantity in the last bracket in
equation (2). Whence
This is the prismoidal formula for approximating to the value of the
definite integral J^ b f(x)dx. The accuracy of this value depends
upon the value of (x a), as we have discarded powers greater
than the third in assuming the value (1) for/(#).
The student will notice that this value of the definite integral
depends upon the values of f(x) only for x a, (a + 6)/2,
arid 6; that is, at the ends and in the center of the interval of
integration. For this reason the formula is applicable if f(x)
is known or if its value can be found for these special values of the
variable of integration.
Evidently the prismoidal formula gives an exact result if
f(x) is a quadratic function, in fact even if it is a cubic function,
as the four terms assumed for f(x) make a cubic function. How-
ever, since generally /(x) is not a cubic, it is necessary to consider
means of making the approximation better. Due to the fact that
the formula will in general give more accurate results when b is
near to a, there is an immediate means of improving its accuracy.
We may divide the interval from a to b into several smaller inter-
vals and apply the formula to each.
As an aid to judicial division, a
graph of f(x) should be used.
Smaller intervals in which the
variation of f(x) is fairly uniform
should be chosen for the several
applications of the formula. If the
figure represents /(x) over the inter-
val a to 6, it would be advisable to
apply the formula over at least the
three intervals a to c, c to d, and
FIG. 221 d to b for improved accuracy.
(6,0) X
200. Simpson's Rule. This is merely a special case of the
suggestion given above for better accuracy. The interval a to &
is divided into smaller intervals of equal length, the prismoidal
formula is applied to each and the various results are combined
200] POWER SERIES AND SOME APPLICATIONS 399
into one formula. Since the smaller intervals are equal, suppose
we represent their length by 2 Ax. Then we have
/& /a +2 Ax /a +4 Ax
(1) / f(x)dx = / /(x)rfx + I f(x)dx
*J a /a */a-f2Ax
+6 Ax s*b
f(x)dx + ---- h I f(x)dx.
.-4 Ax /6-2 Ax
Applying the prismoidal formula to each integral on the right-
hand side of (1), we obtain
/i6 O A/v.
(2) / f(x)dx = ^=- [/(a) + 4/(a + Ax) + f(a + 2 Ax)]
/a
O A~
+ ^ 5 [/(a + 2 Ax) + 4/(o + 3 Ace) + /(a + 4 Ax)]
Ar
[/(o + 4 Ax) + 4/(o + 5 Ax) +/(a + 6 Ax)]
+ + ~ lf(b - 2
or
r& Ar
(3) J f(x)dx = ^ [/(a) + 4/(a + Ax) + 2/(a + 2 Ax)
+ 4/(a + 3 Ax) + 2/(a + 4 Ax) +
+ 2/(6 - 2 Ax) + 4/(6 - Ax) +/(&)].
This formula is called Simpson's /?u/e /or approximating a
definite integral.
Again we suggest that the length of the sub-intervals (2 Ax)
taken be small for the sake of accuracy. The student is also
reminded that the quantity 2 Ax must be chosen so as to divide
the interval (6 a) exactly.
In applying any of these methods when the form of f(x) is not
known and the function is given only by a table of values, it zs
necessary to plot the known values of the function and then draw a
smooth curve through these points. The values of f(x) at the special
points needed are then read from the graph.
400 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
201. Integration of a Series. A third method of approxi-
mating the definite integral when integration is impossible or
inconvenient is to expand f(x) in a power series and integrate
termwise. This method may be used to any desired accuracy if
the interval of integration is within the interval of convergence,
the end points of the interval of convergence being excluded. Rapidly
decreasing terms with the remainder approaching zero are necessary
to make this method useful.
EXAMPLES
1. Find the length of the curve y = x* from (1, 1) to (2, 4).
SOLUTIONS. The element of arc As t is Vl + 4 z t 2 Aa: and the desired quan-
tity is represented by / Vl 4- 4 x 2 dx.
'l
(a) SOLUTION BY PRISMOIDAL FORMULA.
Since f(x) = Vl 4- 4 z* f and a = 1, (aj- 6)/2 = 1.5, 6=2, we have
/(a) = V5,/[(a 4- 6)/2] = V10, /(&) = Vl7. Therefore
r 2
/ 1
+ 4vio 4-
(6) SOLUTION BY SIMPSON'S RULE.
Suppose 2 Ax = 0.5. Then /(a) = x/5,jf(a + Ax) =
/(a + 2 Ax) = A/10, /(& - Ax) = (1/2) V53, and /(&) = VI 7. Therefore
__ __ _ _ _
( V5 + 2V29 -h 2V10 4- 2 V53 4- V17) = 3.16786.
The value of this integral correct to six significant figures is 3.16784.
A(x)
90
80
_^--~" -
i
*-*^^
/
r- 1
kx.
/
"N
60
50
40
SO
10
123456789 1C
FIG. 222
2. At intervals of 6 ft. the areas in
square yards of the cross-sections of a
railway cut are 70, 88, 94, 93, 87, 76.
How many cubic feet of earth must be
removed in making the cut between the
two end sections given?
SOLUTIONS, The element of volume
is A(xi)Ax where A(xt) is a cross-
section of the cut at a distance z<
yards from one end. Since A(x*) is
only given by a table of values, we
must use a graph to approximate the
integral representing the desired volume,
/* 10
that is, / A(x)dx.
'O
201] POWER SERIES AND SOME APPLICATIONS 401
(a) SOLUTION BY PBISMOIDAL FORMULA.
We have given A (a) = A(0) = 70, A(b) = A(10) = 76; from the graph
we get A[(a -f &)/2] = A (5) = 93.5. Therefore
V = f W A(x)dx = ^ [70 -h 4(93.5) + 76] = 866.67 cu. yds.
/o o
(b) SOLUTION BY SIMPSON'S RULE.
If we take 2 Ax = 2, the table gives A(0) = 70, A (2) = 88, A (4) = 94,
A (6) = 93, A (8) = 87, A (10) = 76. From the graph we estimate that
A(l) = 79, A(3) = 91.5, A(5) = 93.5, A(7) - 89, A(9) = 80. Therefore
V - g [70 + 4(79) + 2(88) + 4(91.5) + 2(94) + 4(93.5) + 2(93)
+ 4(89) -f 2(87) + 4(80) + 76] = 867.33 cu. yds.
/*!
3. Evaluate I [(cos x)/x~\dx.
/0.2
SOLUTION. Since cos x - 1 x*/2 1 -f- x 4 /4 ! z fl /6 ! + s and is con-
vergent for all x, and the remainder has the limit zero, we have
x . x* x 5
21 + 4i - 6i
,
dx
= 1.37961.
PROBLEMS
The answers given are approximations.
1. The width of a sheet of tin is w ft. at a distance x feet from one end.
Approximate the area of the sheet defined by:
wii.
0.4
0.8
3.3
3.7
4.1
4
x ft.
0.7
1.1
1 8
2.1
3.2
4
Ans. 10.5 sq. ft.
2. A circular pin has the length 20 in. Diameters x in. from one end are
given by:
d in.
2
4
5
6
8
x in.
5
10
15
20
Write the integral for its approximate volume and evaluate.
402 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
3. The diameters of a buoy at x ft. from one end are:
xft.
2
5
6
7
9
Dft.
2
4
6
8
10
9
Approximate its volume. Ans. 96 TT cu. ft.
4. At 10-ft. intervals, right sections of a vessel are 2.67, 3.89, 5.18, 8.76,
9.13 sq. ft. Approximate its volume.
5. The area bounded by y = 0, x = 2, x 6 and the curve defined by
X
2.0
2.5
3
4.5
5.5
6
y
10.2
8.3
3.8
5.7
2.1
3.6
is desired. Ans. 19.6 sq. units.
6. What volume is formed if the area of Problem 5 is revolved about the
ixis?
x axis?
7. Approximate the length of the parabola y 2 x 2 from (1, 2) to (2, 8).
Use Ax = 1/4. Evaluate by integration. Ans. 6.09 units; 6.086 units.
8. At 6 ft. intervals the area in square yards of cross-sections of a railway
cut are 75, 88, 91, 87, 74. Approximate the volume of earth removed.
9. Approximate the length of x 2 - t/ 2 -f 9 = from (0, 3) to (4, 5). Use
Ax = 0.5. Ans. 4.553 units.
Approximate the following integrals as suggested or by several methods.
(Nos. 10-22.)
10. J -r x 2 , using x = 1/4, 1/2, 3/4 as division points.
11. f 3 ^ 2 -, using x = 1/4, 1/2, 3/4 as division points. Ans. 0.380.
/ 1/2 dx
120 J rrn 2 ky Simpson's rule, using five values of x.
13. / , using four sub-intervals.
/O.I xe*' *
/ T /3
v2 -f tan x dx by prismoidal formula; also'usir
ils. " /6
/T/2
vlOsinx dx, using four sub-intervals.
16. C ~p- (to, usuig five values for ^.
/o v
Am. 1.276.
Tusing four sub-inter-
vals.
Ans. 0.89 TT.
201] POWER SERIES AND SOME APPLICATIONS 403
/1/2
17. / e~ xl dx, using three terms of a series. Ans. 869/1792.
A)
X.7T/2 _
18. I v cos x dx, using three terms of a series.
A)
dx. Ans. 1.63.
21. f* Vl + x 3 dx. Ans. 2.13.
Ji
22. f Vl + log x dx.
/2
23. Given y e~ xf2 . (a) Approximate its length from x = to x = 2.
(6) Revolve the curve about the x axis and calculate approximately the
volume of the solid generated. Ans. (a) 2.10 units, (6) 2.72 cu. units.
24. Show that the prismoidal formula gives the exact area between the
parabola y = ax z + bx -f c, the x axis, and the ordinates x = A and
x = /* + 2 Ax.
25. Approximate the area of the surface generated by revolving the arc of
x* y 2 = 9 from x = 3 to x 5 around the y axis. Ans. 108.28 sq. units.
26. If P is the resultant force of the pressure on the piston of an engine when
a weight has been raised to a height of h feet, find the work done in raising the
weight (a) two feet; (6) eight feet, if
h (feet)
5
1
1.5
2
3
4
5
6
7
8
P (Ibs.)
100
110
110
110
100
73
54
44
38
34
30
27. A is the area of the water plane of a vessel at a distance x feet above
the keel. If
x (feet)
2
4
6
8
10
A (sq. ft.)
2690
3635
4320
4900
5400
find the displacement of the vessel for a draught of 10 feet.
Ans. 36,537 cu. ft.
28. An oil barrel has the following diameters D inches at the given distances
x inches from one end. Approximate its volume.
x inches
4
7
11
15.5
20
24
28
31
D inches
21
23
23 5
24
24 25
23.75
23.5
22.5
21.25
29. A body weighing 1610 Ibs. was lifted vertically by a rope, there being a
damped spring balance to indicate the pulling force of F Ibs. When the body
404 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XVII
had been lifted x feet from its position of rest, the force was recorded auto-
matically as follows:
X
11
20
34
45
55
66
76
F
4010
3915
3763
3532
3366
3208
3100
3007
Find the probable velocity in feet per second at x 30 feet and at z = 72 feet.
[?m> 2 /2 = f* (F - 1610)<te]. Ans. 51.62 ft./sec., 73.92 ft./sec.
A)
202. Calculation of IT. If we expand tan* 1 x by Maclauriri's
series we have
(1)
, .
tan" 1 x = x $ + -= -=
3 5 7
2n - 1
This is readily shown to be convergent for every value of x in
the interval 1 ^ x < 1, and the remainder, which is less than
| z 2n+1 /(2 n + 1)|, approaches zero for all values of x in the inter-
val of convergence. Whence, letting x = 1,1/2, and 1/3 and using
the relation
= tan- 1 (1) = tan- 1 + tan" 1
we have, from (1), that
4L2
(' "
13
= 0.78540
to 5 significant figures. Therefore TT = 3.14160, approximately.
The two series for tan" 1 (1/2) and tan" 1 (1/3) are used here
because they converge more rapidly than does the series for
tan- 1 (1).
203] POWER SERIES AND SOME APPLICATIONS 405
203. Computation of Logarithms. The Maclaurin expan-
sions of log (1 + x) and log (1 x) allow us to write
X 3 X*
or
Now set x = l/(2 ?/ + 1) and we get
3(2
1 - x
whence from (1)
log' 1+ ^
1 (2 n -
or
-L. T * -L 1 _L
(2 ) log ( y + 1 ) = log y -p 2 ,. _i_ i ~r o /^ ,, _i_ i \3 '
LZy + 1 3(2 y + I) 6
1
(2n~
This series is called the logarithmic series, since, given log y,
we have a means of computing log (y + 1) at once to any desired
accuracy. The region of convergence of (2) is y > and the
convergence is rapid after y = 3. Of course the remainder has
the limit zero over the region of convergence.
Logarithms computed by means of series (2) are to the base e;
we readily convert them to the base 10 by means of the relation
l oglo ff = log AVlog 10 = (log AO/2.30259 = 0.43429 log N.
406 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
PROBLEMS
1. Show that the series for tan- 1 ^ about x = is not desirable for comput-
ing TT if x 1.
2. Compute TT to four decimals using
tan-'(l) = 2tan-'(l/2) - tan-^l/T).
3. Form other trigonometric formulas like that of Problem 2 for tan -1 (l)
and compute TT to four places.
4. Compute by formula (2) the natural logarithms of the following num-
bers. Find to as many decimals as requested. (NOTE. Use the laws of
logarithms.)
(a) 1, (6) 2, (c) 7, (rf) 17, (e) 24, (/) 33, (g) 37.
5. Compute by formula (2) the common logarithms of the following num-
bers. Find to as many decimals as requested.
(a) 1/2, (6) 1, (c) 3, (d) 4, (e) 25, (/) 34, (g) 41.
6. Compute the value of TT by taking 4 terms of the series for sin^x and
evaluate for x = 1/2.
204. Sin x and Cos x Expressed as Exponential Functions.
The Maclaurin expansions for e v , sin i/, and cos y are convergent
for all real values of y. These expansions, which should be
memorized by the student, are
CD " = l + y + i + f, + fi + '-'>
(2) sin2/ = 2/ _! + J>!_J/? + ...,
(3) cos 2 / = l--j + ! - + ....
Now if we assume that these three series represent the functions
for y = ix, where i = \/~ 1 and x is real, we have
(A\ f ix - (i _ ! 4. j?L _ . . A _L { ( x _ j?l 4. ^! _ . . .V
W e ~ V 2 ! + 4 ! )* l \ X 3 ! + 5 ! y
But the two parentheses are just (3) and (2), and therefore
(5) e ix = cos x + i sin x.
204] POWER SERIES AND SOME APPLICATIONS 407
Similarly we get
(6) e~ ix cos x i sin x.
These relations give
(7) sin x = ~* ,
(8) cos x = ^
Then since e x+iy = e x *e iy , we may use complex numbers for y
in (1) and, under the same assumption, find that
(9) e x+lv = e x (cos y + i sin y).
These relations between trigonometric and exponential expres-
sions are of special importance in certain applications.
ADDITIONAL PROBLEMS
Find the interval of convergence of each of the following scries. (Nos. 1-7.)
Ans. - 1 < x < 1.
_ x - 4 (ar - 4) 2 (x - 4) 3
2. - + +
4. (2/3) (x - 3) + (3/4) (a: - 3) 2 + (4/5) (x - 3) 3 +
r 1 (x I) 3 ('r I) 5
5. Z-^-i - (x l> + -=-H ----- . Ana. < x < 2.
o o 7
j+2 _ (x + 2) 2 (a; + 2) 3 ____
3 32. 2 1 / 2 S^S 1 / 2 """
Solve each of the following equations by Newton's method. (Nos. 8-16.)
8. x 3 + 2 x 2 - 7 = 0.
9. x 3 - 2 z 2 + cos a? = 0. 4ns. 0.764, etc.
10. 2 tan 3 x -= 1 - x.
408 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVII
11. log x -f x 2 = 2 sin x. Ans. 1.291.
12. xe~* cos 2 x = 0.
13. e'l* = cos (x/3) -f 0.6. Arcs. 0.885, etc.
14. 1 - z 2 = sin (x -f 2).
15. e*' 2 -f cos 2 z = x 2 - 2. Arcs. - 1.294, etc.
16. cos TTX + V x 0. How many roots?
17. Find an upper limit to the error if four terms are used for cos x about
x = for cos 12. Ans. 0.00008.
18. The same as Problem 17 for sin x about x = w/Q for sin 32.
19. The same as Problem 17 for log (2 + x) for three terms to get log 2.1.
Ans. 0.00004.
20. Derive the scries for e x for x near 2. How many terms are needed to
get e 1 - 98 to four decimals?
21. How many terms of the Maclaurin series for log (2 -j- x) must be taken
to get log 2.3 to three decimals? Ans. Four terms.
22. Find tan~ 1 (l/4) to three places by using a suitable series.
23. Evaluate log 1.992 from the series for log x about x 2. Ans. 0.689.
24. Find the interval of convergence of the series in Problem 23 and eval-
uate log 2.01.
25. Find an upper limit for x which will permit the approximation of
(1 + x) 4 to within 1 in 1000 by two terms of the Maclaurin expansion.
Ans. 0.0129.
26. Within what positive interval do three terms of the Maclaurin expan-
sion of e x give a maximum error of not more than 1 in 1000?
27. Evaluate to three decimals by using sufficient terms of a series each of
the following:
(6) J sin 1/2 x dx,
(d) f* cos x 2 dx.
JG
Ans. (a) 1.09, (b) 0.161, (c) 0.503, (d) 0.756.
28. Find the length of the chord of an arc of radius 200 feet subtending 3
at the center (a) by trigonometric methods, (b) by the approximation formula
that the length of the chord equals rd r0 3 /24, where 6 is the central angle
and r is the radius. Compare the two results. Also derive the approximation
formula.
204] POWER SERIES AND SOME APPLICATIONS 409
29. Show that the difference between the length of a circular arc and Us
chord is approximately r0 3 /24. Also show that the error in the approximation
cannot exceed r0 5 /1920.
30. A horizontal cylindrical tank, 6 feet long and 3 feet in diameter, has
6 cubic feet of water in it. Find the depth of the water correct to two decimals,
by solving an equation by Newton's method.
31. Draw the graphs of y = x, y = x - x 3 /3 !, y = x - x 3 /3 ! + arV5 !,
with the same coordinate axes and compare them with the graph of y = sin x
from x - 1 to x = 1.
32. Obtain the binomial theorem by expanding (a + x) n . What is its
interval of convergence?
33. The length of the cable of a suspension bridge is given by
~l
where I is the length of span, w the weight per unit length horizontally, P the
tension in the cable at its lowest point. In general w/P is small. Derive
the following formula, which is more convenient for calculation:
L ~" ' \PJ 24 ~\PJ 768'
34. Can the following functions be expanded in Maclaurin series? Why?
(a) log x\ (b) esc x\ (c) ctn 0; (d) esc 2 0.
35. By means of the Maclaurin expansion of e x compute
(a) e 1 / 2 ; (b) e 1 / 3 ; (c) e 1 / 5 ; (d) e~ l .
Ans. 1.6487; 1.3956; 1.2214; 0.36788.
36. Use the binomial theorem to find
(a) (1.002) 9 ; (6) (0.8) 10 ; (c) (2.002) 8 ; (d) (1.997) 12 .
CHAPTER XVIII
HYPERBOLIC FUNCTIONS
205. Definitions. Certain combinations of e x and e~ x represent
functions analogous to the trigonometric functions. Their geo-
metric representation is related to the equilateral hyperbola in a
manner similar to that in which the trigonometric functions are
related to the circle. Hence the name hyperbolic functions.
They are defined as follows.
(1)
sinh x =
x - - x
cosh x =
sinh x _ e*__ e^
cosh x ~ e x 4- e~ x '
^, cosh x e
ctnh x =
sech x
~
sinh x e x
1 2
csch x =
cosh x e x + e~ x '
1 2
sinh x e x e~ x '
and are read hyperbolic sine of x, etc.
206. Functional Relations. From the definitions we have at
once the relations
sinh ( AC) = sinh x,
cosh ( x) = cosh x,
tanh ( x) = tanh x,
ctnh ( x) = ctnh x,
sech ( x) = sech x,
csch ( x) = csch x.
(2)
410
206]
HYPERBOLIC FUNCTIONS
411
Squaring and combining so as to eliminate e x and e~~ x we obtain
the relations
(3)
cosh 2 x sinh 2 x = 1,
tanh 2 x + sech 2 x = 1,
ctnh 2 x - csch 2 x = 1.
Then from the definitions of sinh x and cosh x we have by
addition and subtraction
(4)
cosh x + sinh x = e x ,
cosh x sinh x = e~ x .
From relations (4) we derive the functions of the sum of two
variables as follows :
By definition,
sinh (x + y) =
px+y __
e x e y
= |[(cosh x + sinh z)(eosh y + sinh y)
(cosh x sinh #)(eosh y sinh y)].
Whence, expanding and collecting, we have
sinh (x + y) = sinh x cosh y + cosh x sinh y.
Replacing y by y and using relations (2), we get
sinh (x y) = sinh x cosh y cosh x sinh y.
This and similar operations give
(5)
sinh (x =fc y) = sinh x cosh y zt cosh x sinh y,
cosh (x db y) = cosh x cosh y it sinh x sinh y,
tanh x zb tanh y
ctnh x ctnh y
412 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XVIII
Letting y = x and using the positive signs in (5), the double
variable formulas appear as
sinh 2 x = 2 sinh x cosh x,
cosh 2 x = cosh 2 x + sinh 2 x
= 2 cosh 2 x 1
= 2 sinh 2 x + 1,
2 tanh x
(a)
tanh 2 x =
ctnh2x
1 + tanh 2 *'
1 + ctnh 2 x
2 ctnh x '
Then, from the third formula for cosh 2 x, we find
cosh 2 x 1
sinh x = d=
or, replacing x by x/2 in this and analogous forms, we obtain
(7)
sinh;
tanh
=w
->/
-^
coshx - 1
> 0.
'cosh x 1
cosh x + 1 '
tosh x + 1
If we set P = x + y and Q = x y and use sinh (a;
and cosh (x y) as in trigonometry, we have
(8)
sinh P + sinh Q = 2 sinh
cosh P + cosh Q = 2 cosh
sinh P sinh Q = 2 cosh
cosh P cosh O = 2 sinh
cosh
cosh
sinh
( P ~
2
P ~
.208]
HYPERBOLIC FUNCTIONS
413
207. Derivatives of the Hyperbolic Functions. Since
sinh x (e x e~ x )/2 we have
d
^v ""-&V 2
and if u = f(x) we have
d /e* - e~ x \ e* + <r
,
= C08h X >
d , - i N i du
j- (smh u) = cosh u -j-
ax ax
Thus, and similarly, we get
(9)
d . . , , - du
-j- (sinh u) = cosh u -7- ,
d , u N . . du
-=- (cosh u) = sinh u -^- ,
v y '
^- (tanh u) = sech 2 u -,- ,
-j- (ctnh u) = csch 2 u-r- ,
, < ^ * ,. ^
(sech u) = sech u tanh u
v
^-
dx
(csch ti) = csch u ctnh u - .
v y
208. The Inverse Hyperbolic Functions. The inverse of the
hyperbolic function is called an antihyperbolic function. Thus, if
x = sinh y
then
y sinh" 1 x, etc.
Since x = sinh y = (e v e~ v )/2 y we have
2x = e v e-",
or
- 2 xe" - 1 = 0.
414 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XVIII
Therefore
e v = x \/l + x 2 .
But e v > for all finite values of ?/, and so the lower sign is impos-
sible; hence
y sinh" 1 x = log (x + \/l + a; 2 ).
Thus, and similarly, we have
sinh" 1 x = log (x + Vl + x 2 ),
(10)
cosh" 1 x = log (x db \/* 2 - 1), * > 1,
ctnh-' x = log
I * I > 1,
sechr 1 x
csch" 1 * =
(1 _l_ \/l x'A
- 1,
, x > 0,
By differentiation of the values for the inverse functions we get,
since y = sinh" 1 u,
and then
u = sinh y
du , dy
-j- = cosh y -j- .
But cosh 2 y sinh 2 y = 1, or cosh y =
relations we have
+ u 2 . From these
1
du
Tx
cosh y dx
209] HYPERBOLIC FUNCTIONS
Similarly we obtain the following formulas:
du
(11)
du
S
s; <*->>-"<> -
du
E
< ctoh "'> "
du
II) = -
415
209. Integration by Hyperbolic Substitutions. The derivative
formulas (11) suggest some integration formulas which may be
very easily derived by hyperbolic substitutions. For example,
consider
sinh 2 u + a 2 = a cosh
Set x = a sinh u, then dx = a cosh u du and
So the integral becomes
"a cosh
/
.
ainlr* - + (7.
416 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVIII
This and similar substitutions enable us to obtain readily
= sinh- 1 - + C,
(12)
/dx - . x .
, = zfc cosh- 1 - +
V* 2 - a 2 <*
/dx I , - . x , -
-^ 5 = - tanh- 1 - + C,
a 2 x 2 a a
/dx
xVa 2 -
= - sech- 1 - + C
a a
= - cosh- 1 - + C,
a x
dx
.
x\/a 2
= - - sinh- 1 - + C.
a x
The relations (10) allow us to write formulas (12) with logarith-
mic results.
210. Series for Hyperbolic Functions. Since the Maclaurin
expansions of e x and er* are
T 2 T 3
.-l + * + gl + - + ...,
we may write
sinh x = =
cc 2 a; 3
or
(13)
211] HYPERBOLIC FUNCTIONS
In like manner
We have previously assumed ( 204) that
(ix)* (ix) 6
417
. ,. . .
sm (ix) = ix-
3!
5!
sin (ix) = i sinh x.
cos (ix) = cosh x.
cosh (ix) = cos x,
sinh (ix) = i sin x
Therefore
(is)
Similarly
(16)
Also the relations
(17)
and
(18)
are readily derived.
211. The Relation to the Equilateral Hyperbola. The circle
with its center at the origin and radius a may be represented
parametrically by jy
x = a cos 6,
The sector OAB has its area K
represented by
a 2 0i/2, where 0\ = cos^Oci/a).
So cos- 1 (xi/a) = 2 K/a\ For the
unit circle the parameter 6 becomes
2 K or twice the area of the sector. FlG - 223
The equilateral hyperbola with its center at the origin and semi-
transverse axis a along the x axis, may be represented parametri-
cally by the equations
x = a cosh u y
y = a sinh u,
418 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVIII
since, eliminating u by means of cosh 2 u sinh 2 u = 1, we have
Y x 2 - y 2 = a 2 .
Now consider the hyperbolic sec-
tor A OB. Its area is
K =
~ I Vx 2 - a 2
Ja
FIG. 224
- a 2 log (x + Vx 2 - a 2 )]*'
i2/_i _ Xiyi
2 2
[log (xi + VxS - a 2 ) - log a]
Vxi 2 a 2
whence, by the second of relations (10), we have
TS a U 1 X l
K = -x- cosh" 1
2 a
This may be written
and for a = 1 the parameter u becomes 2 ^T, jus^ as the parameter
does in the circle.
We note then two things :
(a) The cos" 1 (x/a), or 6, the parameter in the equations of the
circle, is represented by twice the area of the circular sector divided
by the square of the radius a.
(6) The cosh~ l (x/a), or u, the parameter in the equations of the
equilateral hyperbola, is represented by twice the area of the hyperbolic
sector divided by the square of the semi-transverse axis a.
212. Relations between Hyperbolic Functions of u and Trigono-
metric Functions of 8. Such relations are derived at once from
Fie. 225. Here BC is tangent to the circle at B t and CP is per-
212]
HYPERBOLIC FUNCTIONS
419
pendicular to OA. So to each there is a point B on the circle
which fixes a point P on the hyperbola. The parametric equations
of the hyperbola given in the preceding article and Fig. 225 give
a cosh u = OC = OB sec = a sec 0.
The definitions of the hyperbolic functions and the identities
(3) of the first article of this chapter
together with
cosh u = sec 8
give
sinh u = tan 6,
tanh u = sin 9,
ctnh u = esc 9,
csch u = ctn 9,
sech u = cos 9.
FIG. 225
In these relations the variable 6 is called the gudermannian of u
and is denoted by
9 = gd u.
PROBLEMS
1. All formulas not expressly derived in this chapter may be used as
exercises.
2. Derive
(a) sinh (x iy), (b) cosh (x iy), (c) tanh (x db iy),
in terms involving sinh x, cosh z, sinh y, cosh y, sin x, cos x, sin y, cos y.
3. Show that
(a) (cosh x + sinh z) 5 = cosh 5 z + sinh 5 x.
(b) (cosh x -f- sinh x) k = cosh kx -f- sinh fcz.
4. Derive sinh (x it/) and evaluate for x = 3, y = 3.
5. Derive cosh (x iy) and evaluate for x = 4.2, y = 3.
s. - 32.9 - 4.70 i.
6. Derive cosh (x iy) and evaluate for x - 4, y = 3.
420 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XVIII
7. Derive cosh (x iy) and evaluate for x = 3, y 2.
Am. - 4.19 + 0.11 1.
8. Derive cosh (x - iy) and evaluate for x - 2, y = 3.
9. Problem 4 for * = 1, y = 3. Aws. - 1.16 - 0.22 t.
10. Problem 4 for x = 1,6, # = 0.8.
11. Problem 4 for x = 4, y =* 0.5. Ans. 23.96 - 13.08 L
12. Problem 4 for x = 3, y = - 2.
Integrate each of the following integrals by means of hyperbolic substitu-
tions, (Nos. 13-23.)
13. f . dx Ans. sinh- 1 (x/2) + C.
J V4 + X 2
14 f dx .
J Vx 2 -3
15. /"-i^s- Ana. - ^tmk-i (-?-)+ C.
J & - 3 V3 VV3/
4 x 2
18.
(x + 4)\/9 -8x-
dx
17. r / g . Ans. - J sinh-i f 2 ) + C.
J x\/4 + x 2 2 W
.
Vx* - 4 x + 1
19. f - /* An5. I cosh-i -4-r + C.
7 / -- 2 5 x 4
21. f . Ans. - \ sintr* -^ + 0.
J (x - l)Vx 2 -2x + 26 5 *-l
22. /- ,
7 x V2 - x 2 - 4 x
/dx 1
====== Ans. TTZT sinh~ ;
xv 14 + x 2 - 6 x V 14
CHAPTER XIX
EXACT DIFFERENTIALS AND LINE INTEGRALS
213. Exact Differentials. A function of two variables, say
f( x > 2/)> nas as ^s differential the known expression
where /(#, y) and its partial derivatives are supposed to be con-
tinuous in the interval under consideration.
However, every expression
(2) fi(x, y)dx+f*(x, y)dy
of the same type as (1) need not be the differential of some function
F(XJ y). Writing (2) in the form
(3) M dx + N dy,
it is the differential of some F(x, y) if M = dF/dx and N = dF/dy,
since
But, under our assumptions of continuity,
d_/dF\ ^ d_/
dy\dxj dx
\dy
hence we find that M and N must satisfy the relation
if (3) represents the differential of F(x, y).
It can be shown that not only is (4) necessary for (3) to repre-
sent the differential of some function, but it is also sufficient.
That is, if M and N satisfy (4) there is always an F(x, y) for which
(3) is the differential.*
* Goursat-Hedrick, Mathematical Analysis, Vol. I, 151-152.
421
422 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIX
If we have an exact differential, it is often desirable to find the
original function. An example will illustrate a method of finding
the function.
EXAMPLE
Is (2 x 4- 5 y 7)dx + (5 x 8 y + 3)dy an exact differential? If so,
find the original function.
SOLUTION. Here M = 2x + 5y 7 and N = 5 x 8 y -f- 3. Since
dM/dy 5 = dN/dx, the expression is exact. Therefore
which we obtain by differentiation with respect to x, assuming y constant.
Then, integrating with respect to x, regarding y as constant, we get
where the constant of integration is represented by f(y), since y is assumed
constant. Now differentiate this expression for F(x, y) with respect to y.
Then
dF d_ f(
Since N = dF/dy t we have
5x4- ~f(y) =5x-8t/4-3,
or
~/(2/) = ~ 8 y 4- 3.
Therefore
and we have
F(x, y) = x 2 + 5 xy - 7 x - 4 y* + 3 y 4- C.
PROBLEMS
Integrate the differentials below if they are exact.
1. x dy -f- (y - 7)dx. Ans. xy 7 x + C.
2. (3 y - S)dx 4- (3 x 4- 7)dy.
3. (x 4- y)dx 4- (x - y)dy. Ans. (x 2 i/ 2 )/2 4- xv 4- C.
214] EXACT DIFFERENTIALS AND LINE INTEGRALS 423
4. (3 x - 2 y)dx + (2x4-3 y)dy.
5. (xy cos xy 4 sin xy)dx 4 x 2 cos xy dy.
6. (e z cos y - l)dx e* sin y dy.
7. (e*/x)dx + e* \Qgxdy.
8. (x - l/y)dy 4 (y - l/x)dx.
9. (xy 4 x 2 )<fo 4 (x 2 /2)dy.
10. (sin 2 x y cos z)cte 4 sin x rfy.
11. (e x sin y y)dx 4 (e x cos y x 2)dy.
Ans. x sin xy 4 C.
Ans. e v log x 4- C.
Ans. x 3 /3 + z 2 y/2 4 C.
12. (z/\/z 2 4 2/ 2 )<Zz 4 (t//Vx 2 4 y*)dy.
13. e* cos x d y 4 ye* sin z rfx.
14. xe~*tv dy - (ye~ z/v 4 x*)dx.
+ y 2 -4 Q//(zVz 2 4 2/ 2 4- x 2 4
. e x sin y xy 2 y -}- C.
Ans. Not exact.
Ans. log (x 4- Vx 2 4 y 2 ) 4 C.
16. (xy -4 x sin xy)dy^+ (y*/2 4 y sin xy 4-
17. (l/x 4 y/x z ev /x )dx - dy/xev f *.
18. dx/Vx - y 2 4- [(x -
19. xe*^ sin y dx -{- (e xv cos y 4
20. [(2 xy 4 D/y]dx 4 [(y - x)/y 2 ]dy.
214. Line Integrals. The expression
Ans. log x 4- e-" /aj 4 C.
Ans. Not exact.
f
*/ (a,
is called a /me integral and has a meaning when some relation
between x and y is known. This relation, say y = F(#), must
represent a curve through the points (a, 6) and (c, d) used as the
limits of the integral. In general, the value of a line integral
changes when the line or curve y = F(x) of integration is changed,
but if fi(Xj y)dx +/2(#, y)dy is exact, the integral has the same
424 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIX
value for all curves joining (a, b) and (c, d). This is evident
because in that case
C (Ctd} - C z *j -
I [ J i \X) y)ctx ~\~ J%\X) y)dy\ == I v^z == z?, z\y
J(a, 6) J Zl
where z is the integral of the exact differential expression
fi(x, y)dx + / 2 (#, y)dy and z<i arid z\ are determined by the
points (a, 6) and (c, d).
EXAMPLES
1. Assume that at each point of the xy plane there acts a force which varies
as we go from point to point. Let us find the work done in moving a body
along the curve C from A to B.
SOLUTION. Let PT be the tangent to the curve C at any point P, and let
PR be the direction of the force F at P. Then the work done in moving the
body a distance As t is given approximately by P\ cos t As t ,* so that the work
in going from A to B along C is
2 F,
= J F cos rfs,
where / means the integral along the
c
curve C, that is, the variables in the
*X integral are related by means of the equa-
tion of C.
Therefore we have
FIG. 226
W J F(c,Q8 <f> cos a -j- sin <f> sin a)ds,
since B = < a. But the x-component of the force F is F cos a, which we
shall represent by F x ; and the ^/-component is /"\ = Fsma. Also
dx and sin </> ds = rf?/. Therefore
cos
or the work is a line integral along the curve C.
* Work is defined as the product of the force exerted and the distance the force acts
\n a given direction.
214] EXACT DIFFERENTIALS AND LINE INTEGRALS 425
The point A may coincide with B, thus making C a closed curve; in this
case, if F x dx + F y dy is an exact differential, it is easy to see that
f c (F x dx + F v dy) = 0.
2. Derive a formula employing a line integral for the area bounded by a
closed curve.
SOLUTION. Consider the curve below which has not more than two values
of y for each value of x and not more than
two values of x for each value of y. Now
let NNi ?/i, NNz yz t 881 = xi, and t G
SS 2 = 2. Then the area A of the portion
of the plane enclosed by the curve is
/
(yz - y\}dx
V.
/& /&
y 2 dx I yidx
- "a
/*Q, /&
= / y 2 dx I yidx
Jb Ja
(0,c)
(0,0)
N
FIG. 227
(b,o) X
this last integral being taken around the curve so that the area lies to the left.
Similarly, we have
r d
A = I (x 2 - xi)dy
JG
/d fc
Xzdy -f I
Jd
= x dy,
where again the integral is taken so that the area lies to the left.
Adding the two values for A, we have
a line integral giving the area of the closed curve.
3. Apply the formula of Example 2 to find the area of the circle x = a cos 6,
y = a sin 6.
SOLUTION. Here dx = a sin 6 do and dy = a cos dO. Therefore the
formula above becomes
426 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XIX
PROBLEMS
Find the values of each of the following line integrals. (Nos. 1-3.)
1. J ' [2xy dx + (x* y*)dy] along (a) a straight line; (6) a parabola
with its axis on the x axis; (c) a parabola with its axis on the y axis. Show that
the integral is independent of the path. Ans. 128/3; M v = N x .
2. J ' (y dx x dy) along the paths used in Problem 1.
/(4, 2)
[z dx/Vx 2 + i/ 2 + y dy/Vx* + t/ 2 ] along (a) a straight line;
(6) x = t 2 , y = t; (c) a broken line consisting of a part of the x axis and a
perpendicular to it. Ans. 2\/B.
4. Find the area of an ellipse by means of a line integral.
5. By means of a line integral, find the area enclosed by the hypocycloid
x = a cos 8 0, y = a sin 3 6. Ans. 3 wa 2 /S sq. units.
6. Evaluate J*(l/2)(x dy - y dx) along (a) the parabola y 2 = 4 x from
(1, 2) to (0, 0); (b) the straight line from (0, 0) to (1, 2). The difference of
these two values gives the area enclosed by the parabola and the line.
7. Find the area of the ellipse x 2 cos 0, y 4 sin 0.
Ans. 8 TT sq. units.
8. Find the area enclosed by the curve x = a(l cos 6), y = a sin 6.
9. Evaluate J (xy dx y*dy) from (1, 1) to (4, 8) along the curve
x = t\ y = t\ Ans. 134f.
ADDITIONAL PROBLEMS
Integrate each of the following functions if it is exact. (Nos. 1-9.)
1. (x + y + a)dx + (x - y + b)dy.
Ana. x*/2 - y*/2 + xy + ox + by + C.
2. (cos y + 2 x)dx x sin y dy.
3. (x 2 + y*)dy + 2xy dx. Ans. x*y + 2/ 3 /3 + C.
4. 2 a; dy - y 3 (ty + 2 y dx.
5. x(* + 2 y)cZx + (x 2 - y*)dy. Ana. x/3 - y'/3 + x*y + C.
6. (e* a)(fy
214] EXACT DIFFERENTIALS AND LINE INTEGRALS 427
7. (2 y 2 3 x)dx 4 xy dy. Ans. Not exact.
8. (2 x + y/x*)dx - (1/x + 2 cos 2 y)dy.
x dx , y dy A XT , ,
9. - H - Ans. Not exact.
Vz 2 -
'
10. Find the value of / ' (x*dx + y <fy) along the curve x = tf, y = t 2 .
'o,
, 2)
11. Find the value of y (y dx - x dy)/Vx* - j/ 2 along the curve
x 2 - y 2 = 4. 4ns. - 2 log (1 + V2).
12. Determine the value of
/(i, i)
y C(2 x H- y 6*)dx + (cos t/ + x e
along a straight line joining the two limits. Is this value independent of the
path?
13. Evaluate f ' [(zy -f x*)dx + x*dy~], (a) along a line through the
origin, (b) along a parabola with its vertex at the origin and its axis along the
y axis. Is the integral independent of the path? Ans. 5/3; 11/6. No.
14. Find the value of / ' [(y x)dx -f- y dy], (a) along a straight line,
'(o, 0)
(6) along a broken line consisting of a part of the y axis and a line perpendicular
to it, (c) along a broken line consisting of a part of the x axis and a line perpen-
dicular to it.
CHAPTER XX
SOME DIFFERENTIAL EQUATIONS OF THE
FIRST ORDER
215. Definitions. In many cases it is easier to find relations
between the rates of change of functions than between the actual
functions. Such relations are equations which involve differen-
tials or derivatives, and are called differential equations. A
differential equation which has a single independent variable is
called an ordinary differential equation. If several independent
variables occur, so that partial derivatives are present, the equa-
tion is known as a partial differential equation.
The order of a differential equation is the same as that of the
derivative of highest order in it.
The degree is the degree of the derivative of highest order after
the equation has been made rational and integral in the derivatives.
Thus
dx*
_ ( v .
~~ \ J + dx)
is an ordinary differential equation of the second order and of the
second degree. The equation
dx* r
is of the third order and first degree.
An illustration of a partial differential equation of the second
order is
dx* <ty 2 " '
216. Derivation of a Differential Equation from Its Primitive.
The equation y 2 = 2 kx represents a one-parameter family of
parabolas, since k may be considered to have any value whatever.
If we differentiate this equation with respect to
428
216] FIRST-ORDER EQUATIONS 429
2 y dy = 2 k dx. Eliminating k between the two relations, we find
the first-order differential equation
(1) 2xdy = ydx.
This equation is called the differential equation which arises
from eliminating the arbitrary constant k between y 2 = 2 kx
and 2 y dy = 2 k dx. It must represent some characteristic
common to all of the parabolas of the family, since it is inde-
pendent of k.
Similarly, suppose we have
(2) y = Ci sin x + c% cos x
and wish to eliminate the constants c\ and 02. To do this we need
two additional equations which may be derived by repeated differ-
entiation. Thus
dij
(3) -~- = Ci cos x c 2 sin x,
and
/^\ d 2 y
(4) y~ = Ci sm x c% cos x.
Eliminating Ci and c 2 between equations (2), (3), and (4), we get
The equation (2) with two constants gives rise to a differential
equation of the second order.
If we have an equation in x and y involving three arbitrary con-
stants we need three additional equations to eliminate the con-
stants. These are obtained by repeated differentiation and will
introduce derivatives of the third order.
The original equation connecting x and y is called the primitive
of the derived differential equation. Thus
y = Ci sin x + Cz cos x
is the primitive of the equation
3+-*
430 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
By extending this reasoning, it appears that a primitive involving
n arbitrary constants gives rise to a differential equation of the
n-th order. Also, that a differential equation of the n-th order cannot
have more than n arbitrary constants in its solution. For if there
were more, say n + 1, elimination of them would seem to lead to a
differential equation of order n + 1 instead of n. Similarly, it
cannot have less than n constants. Hence an equation of order n
has a solution or primitive with n arbitrary constants.
PROBLEMS
Form the differential equation of lowest order which arises from elimination
of constants in each of the following primitives.
2. log x -f log (y -f 1) = c.
7. y =ci sin x -f c 2 cos x + x*.
Am. x dy 2 y dx = 0.
Ans. x dx + (x dy y dx)e vfx = 0.
Ans. -j-^ + y = x 2 + 2.
217. Solutions of a Differential Equation. A solution or
integral of a differential equation is any relation between the
variables which satisfies the equation. Thus y = cos x + c is a
solution of dy + sin xdx = 0; and, as c may have any value
whatever, the differential equation has an infinite number of solu-
tions. Each solution may be represented by a curve in the xy
plane called an integral curve.
218] FIRST-ORDER EQUATIONS 431
The second order equation
has the solution y = c\ cos x + c 2 sin x, where c\ and c* are
arbitrary constants. This solution may evidently represent a
double infinity of integral curves.
A solution of a differential equation which involves the maxi-
mum number of arbitrary constants is called the general or
complete solution. Solutions obtained by giving particular values
to the arbitrary constants of the general solution are called particu-
lar solutions. Thus y = c\ sin x + cz cos x is the general solution
of d 2 y/dx 2 + y = 0, while each of the equations y = 3 sin x and
y = 2 sin x cos x are particular solutions.
Methods of solving the more general types of the first order and
first degree follow.
218. Variables Separable. The first-order equation of first
degree can always be written in the form
(1) Mdx + Ndy = 0.
If it is of the type where variables are separable, each of the
quantities M and TV can be resolved into two factors such that
x does not occur in one factor and y does not occur in the other.
That is, we may write
(2) M = /,(*) -My), N = /,(*) .
where any of the functions /, may be constants. Then (1 ) becomes
fi(x) -h(y)dx+f 3 (x) -f t (y)dy = 0.
Dividing through by fa(y) -fs(x), we get an equation of the form
(3) f(x)dx + g(y)dy = 0.
The solution of (3) is given at once by
(4) ff(x)dx + fg(y)dy = c.
432 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
EXAMPLE
Solve the equation xVl -f y* dx + yVl -f x 2 dy = 0.
SOLUTION. Here M = xVl J/ 2 , JV = 2/Vl + z 2 where each has two
factors of the types necessary for separation of the variables. Whence
dividing by Vl -H 2/ 2 Vl H- 2 we have
d
~
Integration of each term gives the solution as
Vl -f x 2 + VT+Y 2 = c.
219. Exact Differential Equations. An exact differential
equation is one which has been formed from its primitive by
differentiation without any additional operations of reduction.
The test for exactness and the method of integration are given in
213 on exact differentials. The solution of an exact differential
equation is the integral of the exact differential set equal to a constant.
EXAMPLE
The equation x dy + y dx 2 x dx = has the solution xy x 2 = c.
220. Homogeneous Equations. If f(x, y) becomes x n *F(v)
when we set y = vXj then f(x y y) is said to be homogeneous in
x and y of the n-th degree. Thus e xly is homogeneous of degree
zero, since e xly = e xlvx = e llv = x e ll \ The function (xy + y 2 ) 112
becomes x(v + v 2 ) 1 ' 2 and is accordingly homogeneous of the first
degree.
If M and N are homogeneous functions of the same degree
the equation M dx + Ndy = is homogeneous. In that case
either y = vx or x = vy will make the variables separable. This is
because the equation becomes, for y = vx,
x n -fi(v)dx + x n -fa(v)(vdx + xdv) = 0,
or
[/i(0 + vf*(v)]dx + xf 2 (v)dv = 0;
and if the function fi(v) + vf 2 (v) ^ the result may be written
dx f*(v)dv
220] FIRST-ORDER EQUATIONS 433
where the variables have been separated. If the bracket is zero
the equation reduces to dv = 0; hence v = c or y = ex is a solution
of the equation.
Sometimes the substitution y = vx gives a function of v which is
not readily integrated, when x vy will give a more desirable
function.
EXAMPLE
Solve the equation (x 2 -\- xy)dy y 2 dx 0.
Solution. Let x vy; then dx v dy + y dv. Therefore
( v iy* 4. V y*)dy - y*(v dy + y dv) = 0,
or
dy _ dv
y ~ v z '
Integrating, we have
log y = - - + log c,
or
Whence y ce~ vlx t since v x/y.
If the integration introduces a logarithm it is desirable to call the constant
of integration log c so that the two logarithms may be combined as shown in
the solution.
PROBLEMS
Integrate each of the following equations. (Nos. 1-19.)
1. (x xy)dy -f- (y + xy)dx = 0. Ans. log xy + x y = c.
2. xVl y 2 dx yVl x 2 dy = 0.
i (6 x - 2 y + l)dx - (2 x - 2 y + 3)dy - 0.
Ana. 3 x 9 + y 1 2 xy + x 3 y c.
,X . /
4. x 2 V 1 -f y dx + y*vl - x dy = 0.
"-^S. (y - z 2 l)dx + dy = 0. Ans. 3 xy x 3 3 x = c.
6. 6 ajy da; + 3 x 2 dy 4 x*y dx = x 4 dy 15 y 2 dy.
. 7. (2 y cos 2 x -f 2 sin 2 x)dx + sin 2 a; dy = 0.
. y sin 2 x cos 2 x c.
434 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
8. (y 2 - xy)dx + (x 2 -f xy)dy = 0.
. 9. (y 2 e xv -f cos x)dx -f- (2 y -f e** -f- x^ e af ')dy = 0.
Ans. ye zv -f- sin a; -f- y 2 = c.
10. x 2 dy xy dx = (?/ 2 xy)dy.
11. 2 xy dy = (I y 2 )dx. Am. logx(l y 2 ) = c.
12. (6 - 2 y + 5)dx + (2 y - 2 a; + 3)<fy = 0.
13. 2y 2 dx + (4xy - 3 y 2 )dy = 0. Ans. 2 xy 2 - ?/ 3 = c.
14. (3 x - 4 2/)dx -(4x-f7 y)dy = 0.
15. e v dx sin x dx -\- xe v dy = 0. Ans. xe v -f- cos x = c.
16. [#*(! -f x) -f y cos xy~$x + (4y -\- x cos X2/)cfa/ = 0.
17. x 2 dy + y*dx + 2 xy(dx -f Ji/) - 3 x 2 dx = 0.
Ans. x 2 y }- xy 2 x 3 = c.
18. (t/e* + 2 x)dx - (2 - e*)dy = 0.
19. (2 x - 3 y - y sin x)dx + (cos x - 3 x)dy = 0.
Ans. x 2 3 xy + ?/ cos x = c.
20. The rate of change of the population of a city is proportional to the
population at any time. Find the equation expressing its growth. If the
population doubles in 60 years, how long will it take to treble?
21. The tangent to a member of a system of curves has its x intercept equal
to three times the abscissa of the contact. What is the system?
Ans. xy 2 = c.
Orthogonal Trajectories. Let F(x, y, dy/dx) = be the differential equation
obtained from the one-parameter family of curves /(x, y, c) 0. Then the
differential equation F(x, y, dx/dy) has for its solution a one-parameter
family known as the orthogonal trajectories of the given system, since each
curve of one system meets every curve of the other system at right angles.
Find the orthogonal trajectories of each of the following families. (Nos.
22-23.)
22. The parabolas i/ 2 = kx.
23. The system of circles x 2 + 2/ 2 2 kx =* 0. Ans. x 2 + y 2 cy = 0.
x 2 v 2
24. Prove that the conies 2 , , -f , 2 , = 1, where a and b are fixed
from a self -orthogonal system.
25. By solving a first-order differential equation, show that the diameters
of the system of circles x 2 -f y 2 = a 2 are the orthogonal trajectories.
26. Solve dy/dx -f xy = x completely for the curve through (1, 1).
221] FIRST-ORDER EQUATIONS 435
27. The length of the tangent drawn from any point P(x, y) of a tractrix is
a units. Show that the differential equation of the curve is Va 2 y 2 dy = y dx.
If (0, a) is a point of the curve, prove that its equation is
221. Linear Equations of the First Order. A linear differential
equation is one that is of the first degree in the dependent variable y
and its derivatives. Thus
is a linear equation in y of the second order if PI, P 2 , and P 3
are functions of x only or are constants. We are at present
interested only in linear equations of the first order, the general
type for which is
where P and Q are functions of x or are constants.
The substitution y = uv permits the solution of this equation;
the method being indicated in the examples given below.
In the equation Mdx + Ndy = 0, of course, either variable may
be considered the dependent variable. Therefore, an equation
may sometimes be put in the form
where P' and Q f are functions of y or are constants. Then we
use the substitution x = uv and the method is the same in both
cases.
EXAMPLES
1. Solve (x + 1) dy - 2 y dx = (x + I) 4 dx.
SOLUTION. Since this equation is of the first degree in y and dy, we write it
in the form (1) of this article. Whence we get
436 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XX
Now set
(2) y = w>\
then
/ov dy dv du
(6) j- = U -, -- r V -r-
v ' dx dx dx
Substituting (2) and (3) in (1) and collecting the coefficients of v and dv/dx,
we have
... dv . fdu 2 u
Now the relation (2) introduces two unknowns and hence two conditions are
needed to determine them. One condition is necessarily that y = uv shall
satisfy equation (1); the other we may impose at will. We shall assume, as
this second condition, that the coefficient of v in (4) be zero. Whence
du 2dx
u x -f 1
Integrating, we find
log u = 2 log (x + 1),
or
u = (x + I) 2 -
Here we choose the constant of integration as zero. This is permissible,
as the general solution of (5) is not needed. It is only necessary to add the
constant after the final integration in the solution.
This value of u makes the transformed equation (4) reduce to
dv = (x + l)dx.
Whence
Then finally
= 2^
V-w- 5 (
222] FIRST-ORDER EQUATIONS 437
2. Solve ds + (s - 1) tan t dt = 0.
SOLUTION. The equation may be written
Setting s uv gives
-r -f s tan t - tan t.
dv , fdu . , \
37 + u tan t] v = tan t.
From
-T: + u tan = 0, = tan t dt.
dt u
whence
log u = log cos t, u = cos .
This value of u reduces the transformed equation to
cos t -r = tan t, or dv = tan sec t dt.
Therefore
v = sec t + c.
The general solution is then
s = 1 + c cos t.
222. Bernoulli's Equation. This equation may be written in
the form
^ + /i (x) ?=/,(*) -IT,
and may be solved by the method given in the preceding article
for the solution of the linear equation of the first order.
EXAMPLE
Solve (y xy*)dx dy = 0.
SOLUTION. This may be written
CD |---**
Set y uv, then dy/dx = u dv/dx -f v du/dx and we have
fn \ dv . fdu
(2) W _ + ^__
By making the coefficient of v equal zero, we get
u = e*.
438 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
This value of u in the equation (2) gives
dv 9 , dv ,
e* -7- = xe* x v 2 , or -r = xe'dx.
dx ' v 2
Integrating, we have
= (x l)e x -f c, or t; = 7
v (x - \}e x - c
Substituting in y = wv, we find the general solution to be
& 1 1
?/ = or - = x 1 ce~*.
u (x \)e x - c* y
223. Integrating Factors. Some equations which are not
exact can be made so by multiplying the whole equation by some
function. Such a multiplier is called an integrating factor. There
is no convenient general method for finding integrating factors.
A close inspection of the terms of the equation will sometimes lead
to the proper factor. In so doing, certain differentials should be
kept in mind. Thus
/T\ ?/ (\T T /7?/
fx\ = yax xay
\yj y 2
x\ _ ydx-x dy
j ( i x \ y dx xdy
d [ tan" 1 - ) = - =-: ~ ,
\ y/ x* + ij 2 '
j ( . . x\ y dx x dy
d [ sin" 1 - ) = - - " ,
\ I/ / II / 9 9
\ (/ / '/ \/ ?y ~~ i?*
and many others. The numerators of each of these given are the
same and if it occurs in an equation, either I/?/ 2 , l/#2/, l/# 2 >
V(z 2 + 2/ 2 ), or 1/2/vV x 2 may be the integrating factor
needed.
Suppose the equation M dx + N dy = is given and we wish
to find whether or not there is an integrating f actor 0(x), that is,
a function of x alone. If there is such a 0(x),
<t>(x) Mdx + <t>(x) Ndy =
223] FIRST-ORDER EQUATIONS 439
is exact; hence
dy dx
or
^ V '
dy dx ^ v } dx
This may be written
This shows that there is such a function only if
1 P M _ ^1
N\_dy dx]
is a function of x alone. If this is the case and if we call it f(x),
we have
whence <j> (x) e is the integrating factor.
Similarly, there is an integrating factor which is a function of
y alone if [l/M][(dN)/(dx) - (3M)/(dy)] is & function of y alone.
s*
Calling this /(?/), the integrating factor is e y y .
There are many other known integrating factors, but it is not
necessary to include them here.
EXAMPLES
1. Solve by finding an integrating factor
x dy y dx + 2 2/ dy + 2/ 2 # dx 0.
SOLUTION. The terms x dy y dx suggest one of the factors mentioned at
the beginning of this article. If we use \/xy we have
440 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
which is exact since dM/dy = dN/dx. Integrating, we have
log y - log x + xy = log c,
or
y cxe~ xv .
2. Integrate x dy y dx = (x 2 + y 2 ) ll2 dx.
SOLUTION. If we divide through by the radical (x 2 -f y 2 ) 112 we have
xdy-ydx = ^
Vx 2 + y 2
The first term of the numerator on the left suggests the need of x 2 as its
denominator, so we have
x dy y dx
Vx 2 + y 2
x 2
This may be written
Integrating, we get
logjj+^/l+dyj^ogs + logc,
therefore
PROBLEMS
Integrate each of the following equations. (Nos. 1-28.)
1. x dy y dx x dx. Ans. y x log ex.
2. dy/dx + y = e*.
3. ch/ - (x# + 2 x)dx = 0. Arw. y = ce* 2 / 2 2.
4. (y - x)dy 4-2/^ = 0.
5. 3 Ids -f- 6 s dt = te ds. -Ans. * 2 s = ce/ 8 .
6. ^ 4 x dx + xy dx = 0.
7. dx + (x tan y sec y)dy = 0. Ans. x = sin y + c cos y.
8. x 4 dy/dx + x a (2 x - l)y - 1.
9. dy - 2 XT/ dx 4- 4 xy 2 (fx = 0. Ans. y(c + 2 e* 2 ) = e* 2 .
223] FIRST-ORDER EQUATIONS 441
10. (v 2 4 \Ydu 4- [4(v* -f l) 2 vu - l]dv = 0.
11. xy*dy 4- i/ 3 dx - y dx -h 2 7/ 2 dx 4 x dy = 0.
Ans. xy x/y 4- 2 x = c.
12. (x + l)dy/dx - 2 y = e*(x + I) 3 .
13. x(x dy -\- y dx) -f y(x dy y dx) + 2 x 3 y sin x 2 dx = 3 x 2 y 3 dy.
Ans. log XT/ -f y/z cos x 2 = y 8 -f c.
14. (x + 6 xy Vx 2 4 y 2 )dx + (y 4- 3 xVx 2 + y 2 )dy = 0.
15. x 2 dy 4- (1 - 2 x)y dx = x 2 dx. Ans. y = x 2 (l 4- ce 1 ^).
16. (x 3 ?/ 2 - y)ete - (x 2 ?/ 3 4 )dj/ = 0.
17. xdy/dx + y + x*y* == 0. Ans. xy^(c 4 x 2 /2) = 1.
18. dy (xy x)dx = 0.
19. dy/dx 4- xy = x?/ 2 . ^ins. 2/(ce*'/ 2 4 1) = 1.
20. dy + y cos x dx = e* dx.
21. (x 4 x*y}dy = i/ dx. Ans. x(c y*) =2 y.
22. x dx = 7/[sec (x 2 + 2/ 2 ) l]dt/.
23. (2 - xi/)/y dx - (2 4- xy)x dy = 0. Ans. 2 log (X/T/) - xy = c.
24. x (dy/dx) 4 y 4- x*y*e* = 0.
25. 2 y dx + x d// xT/e* 1 ' sin x dx 4- x 2 ye*v cos xdy -{- xy*e xv cos x dx =0.
Ans. 2 log x 4- log ?/ 4- e x " cos x = c.
26. (Vx 2 - y 2 + x)dx 4 (^x 2 y 2 - y)dy = 0, and find the curve
through the point (1, 0).
27. (1 4- z 2 ) dy/dx 4- ^ = tan~ l x, and find the curve through the origin.
Ans. y - tan- 1 x - 1 4- e- tan ~ la? .
28. dy/dx 4- 2/ cos x = (1/2) sin 2 x, and find the curve through (ir/2, 6).
29. What family of curves cut the circles x 2 4- y 2 = o 2 at 60? Which
of these curves passes through (1, 1)? Ans. log xy V3 log (x/y) 0.
30. Find the orthogonal trajectories of x 2 + y 2 2 kx == a 2 , where a is fixed.
31. Solve Vl x 2 dy/dx 4- 2/ = sin" 1 x for the curve through (0, 1).
Ans. y = sin~ l x - 1 + 2 e-*~ 1 *.
32. Solve a linear equation of your own choice or any one of this list and
add a constant after the first integration. What happens to this constant in
the final solution?
442 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
224. Clairaut Equations. If a differential equation can be
arranged in the form
(i) y = P*+f(p),
where p = dy/dx, it is called a Clairaut equation. To solve such
an equation, differentiate it with respect to x, so that
/o\ . .
(2) p .p + s + ._._ |
then
Now, discarding the bracket, which involves no differentials,
we have
dp
-~ = 0, or p = c.
Therefore
is the solution of (1).
225. Singular Solutions. It is interesting to note that the
envelope of the particular solutions derived from equation (4)
of the preceding article is also a solution of equation (1) of that
article. This is true in general for differential equations; such
solutions are called singular solutions, because they cannot be
derived from the complete solutions by giving the constants of
integration special values.
These singular solutions have little importance in elementary
applications and so will not be discussed here.
PROBLEMS
Solve each of the following equations. In each case show that the envelope
6f the general solution is also a solution.
1. y = px -f 2 p 2 . Ans. y = ex -f 2 c 2 .
2. y = px + e*>*.
225] FIRST-ORDER EQUATIONS 443
3 '
4. y = x + toa
5. 2 x 3 ~ - 4 x 2 2/ = a -, (Let a; 2 = u). Ans. y = ex* - oc 2 .
**-.>.
7. 1 - 2 y j- = 4 e 2 * (^ Y, (Let e 2 * = ti). Ans. e 2 * = cy + c 2 .
ADDITIONAL PROBLEMS
Integrate each of the following equations. (Nos. 1-26.)
1. (1 4- x)y* dx - x z dy = 0. Ans. 2x*-2xy-y = 2 cx*y.
2. (x - \)dy -2ydx = (x - I) 6 dx.
3. dy/dx -\- y/x x 2 . Ans. 4 xy = x 4 + c.
5. (iy/dx ay/x = (x + l)/x. Ans. y = x/(l a) I/a + cx a .
6. Ids - s dt = te* dt.
7. x rf?/ y dx == x 2 e 2:r dx. Ans. 2 2/ = xe 2 * + ex.
8. (1 + x*)dy/dx + 2xy = lx*.
9. (x 2 + y 2 - a 2 )x dx + (x 2 - ?/ 2 - 6% dy = 0.
Ans. x 4 - ?/ 4 - 2 a 2 x 2 - 2 6V +
10. e v dx y sin y dy -{- xe v dy 0.
11. (x sin x^/ \)(x dy -\- y dx) = y dy + cos x?/ dx.
Ans. x cos x^ -f- xy -j- y 2 /2 = c.
12. xdy y dx = x 2 ?/ 2 sec (y/x)dy.
13. x cos 2 y dx -\- esc xdy 0. Ans. sin x + tan y x cos x = c.
14. (tw - v 3 )cto + du = 0.
15. dx + (x ?/)di/ = 0. Ans. x = y I 4- ce~v.
16. [(x - t/)e" /af + xjlx + xe*l x dy = 0.
17. xy-f- = y 2 + 1. Ans. i/ = Vex 2 - 1
444 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XX
18. y|? + p-2lf-0.
19. dy + 4 xy dx = x 2 dy. Ans. (1 z 2 ) 2 = cy.
20. (?/ 2 -I- x + \)y dy - (t/ 2 + l)ete = 0.
21. dy/dz = e*+*. Ans. ce* + e*** + 1=0.
22. (?/ 2 - 4 rc 2 )^x + xy dy = 0.
23. (y + a - l)x dx - (x* - l)dy = 0. Aws. |/ = x* - 1 + cVx* - 1.
24. (4 x + y)d + (* - 4 y)dy = 0.
25. t rfs - t*s 3 dt = s eft. Ans. sVc - i 4 = t V2.
26. v 2 du - u^dv uv du.
27. Solve the linear equation dy/dx +fi(x) -y =ft(x). This solution is
often used as a formula.
Ans. y = e~S f
28. Apply the formula derived in Problem 27 to solve Problems 3 and 15 of
this group.
CHAPTER XXI
DIFFERENTIAL EQUATIONS OF HIGHER ORDER
226. Some Special Types of Higher Order. Certain equations
of t\ e second order which occur frequently in elementary applica-
tions will now be discussed.
CD = K X) > or
These equations are usually solved by repeated integration with
respect to x. However, if we set d n ~ l y/dx n ~ l = p, the equation
becomes of the first order.
EXAMPLE
Solve = xe*.
dx 3
SOLUTION. One integration gives
Integrating again, we have
j- = (x l)e x e x -\- c\x -f C2
= (x - 2)e x + cix + c 2 .
The final integration gives the solution as
,
y = (x - 3)e* + - + c z x -f c 3 .
We reduce this equation to one of the first order by means of the
substitution dy/dx = p. This substitution is made because the
dependent variable is missing, and p is used as a dependent vari-
445
446 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XXI
able, since we use for d 2 y/dx z the value dp/dx. The resulting
equation
dp f/ ,
j- = f(x, p)
is of the first order in p and x.
EXAMPLE
SOLUTION. Set p = dy/dx, then rfp/rfx = d*y/dx*, and we have
in which the variables are separable. Hence, we write it in the form
Integrating, we have
log P = - 2 log (z 2 - 1) + log ci,
or
p = d(:r 2 - I)- 1 / 2 .
Therefore
= i -- ,
V x 2 - 1
and integrating we have the solution as
y = ci log (x -f vz 2 - 1) -f c 2 .
A special equation of this type occurs in 237 of Chapter XXII, where
the suspension bridge is treated. This same substitution leads there to a
solution involving the hyperbolic cosine.
(ni) ^ =
Again we reduce the order of the equation by a substitution.
Setting p = dy/dx, we have
d*y _ dp _ dp dy _ dp
dx 2 dx dy dx dy
226] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 447
and we are able to remove x, which occurs only implicitly. The
equation becomes
which is of the first order in p and y.
EXAMPLE
Solve y -r~
SOLUTION. Let p dy/dx and p dp/dy = d 2 y/dx*. Then we have
dp
WTy'V = y >
or
^v _? = ^!,
dy y p'
which may be solved by the Bernoulli equation method presented in 222.
Thus let p uv\ then dp/dy = u dv/dy + v du/dy, and we have
(dv v\ . du v
_ i^_ u== ^_.
dy y) dy uv
Obtaining v so that the coefficient of u is zero, we find that
v = y.
This makes the equation reduce to
u du = dy,
whence
u = A/2 y -f- 2 CL
But p dy/dx uv and so
Therefore
dy
and integration gives the solution- as
V y + Ci +
(IV)
448 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
This equation is merely a special form of III but it is of sufficient
importance to be solved separately. Since dy/dx is missing, we
merely set p dp/dy for d 2 y/dx 2 arid have
p dp = f(y)dy.
This gives for one integration
P 2 = zf^dy + c,
a result in which the variables are separable.
This type of equation occurs in problems on motion in which
the acceleration along the path is proportional to the distance
traversed. Such an equation is solved as the example of this
type.
EXAMPLE
Solve d*s/dt* = db k*s.
SOLUTION. Setting d*s/dt z equal to p dp/ds, where p is ds/dt, this equation
becomes
p dp = k*s ds.
Integration gives
(,/,.\ 2
^J = rt k*(s* + Cl ).
This is called the energy integral since ds/dt = v and therefore, if we multiply
the equation by m/2, it becomes
1 mfc 2 , . ,
- my 2 = d= -y (s 2 + Ci).
That is, the kinetic energy of a body of mass m and acceleration k*s is given by
This equation must be solved as two types, depending upon the sign before fc 2
(a) If the sign before k 2 is positive, the energy integral reduces to
whence
226] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 449
Integrating, we have
log (8 + Vfi 2 + Cl ) = kt 4 log C 2 ,
or, solving for s,
s = oe**' 4
where a = c 2 /2 and 6 = ci/2 c 2 .
This solution may also be written in the form
s = A sinh (=t A-Q 4- # cosh (=t Art),
where B + A = 2 a and B - A =26.
(6) If the sign before k 2 is negative, the constant Ci must be negative and
such that |ci| > s 2 ; otherwise ds/dt would be imaginary. Writing Ci as
( a 2 ) the energy integral becomes
dt '
Hence
~-JL==kdt.
Va 2 s 2
Integration gives
sin- 1 ^ = kt + c,
or
s = a sin ( Art 4* c)
= A sin Art 4- # cos Art,
where A = a cos c and 5 = a sin c.
We have seen this last equation as that which defines simple harmonic
motion. Hence, simple harmonic motion is the only possible motion in which
the acceleration along the path is proportional to the distance from a fixed point,
if the constant of proportionality is negative. (See 104.)
PROBLEMS
Solve each of the following equations. (Nos. 1-18.)
1. d?y/dx z = \/x. Ans. y = x log x x 4- c& + c 2 .
2. d*y/dx* = 1/Vl - x\
3. d*y/dx* sec 2 ax. Ans. y (I/a 2 ) log sec ax 4- c& 4- c 2 .
4. d*y/dx* - x* = 0.
5. (1 x*)d*y/dx* z cfy/da; = 0. Ans. y = Ci sin" 1 x 4- c 2 .
450 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
6. d*y/dx* + dy/dx = (dy/dx)\
7. sd*s/dt* 4- (ds/<#) 2 = 1.
Ans. Vs 2 Ci 2 = 4- C2, according as ds/d ^ 1.
8. d*u/dv* = [14- (dw/dv) 2 ] 1 / 2 .
9. (1 4- x)d*x/dy* + (dx/dy)* = 0. Ana. x*/2 + x = c& + c 2 .
10. (1 -f v)d*u/dv* - du/dv = 0.
11. y d*y/dx* - (dy/dx)* + (dy/dx) 3 - 0. Ans. y - d log y = x + c 2 .
12. d 2 x/d?/ 2 - m 2 z = 0.
13. d 2 7//dz 2 + dy/dz = sin 2 z.
?/ = c\ 4- 02^"* (cos 2 x -\~ 2 sin 2 z)/10.
14. yd*y/dx* (dy/dx)* = 7/ 2 log y.
15. d 2 s/cft 2 - ds/dt = 3t, if s = 0, and ds/cfe = 1 at t = 0.
= 4(e< - 1) -- (3/2)J 2 - 3 t.
16. dy/dx + Vx 2 4- (dy/dx) 2 - x d^y/dx* =0, if y = 3, and dy/dx = 2 at
= 1.
17. 2 y d*y/dx* -f 2(dy/dx)* = y dy/dx, if x = 0, y = 2, and dy/da; = 3.
Ans. y 2 4- 20 = 24 e*/ 2 .
18. (1 - y)d*y/dx* + (dy/dx) 2 = 0, if y = 3, and dy/dx = 2 at x = 1.
19. Due to f Fictional resistance, the angular acceleration of a certain wheel
is negative and is proportional to its angular velocity. If the wheel slows
30% in 20 sec., when will it slow 50%? Ans. 38.87 sec.
20. The angle made by the position of a pendulum with the vertical
satisfies approximately the equation d 2 6/dt z = ( g/l)0, where g and I are
constants and the independent variable t is the time. Find 6 in terms of t
if the pendulum is released with zero velocity at t from the position == 0.1
radian.
21. Show that the circle is the only curve for which the curvature at every
point is a constant. (HINT: Take the curve in such a position that y' =
at x = 0.)
22. Solve d*y/dx* m 2 y = by the substitution dy/dx p] now solve
it again by using y e kx .
23. The acceleration of gravity varies inversely as the square of the distance
from the center of the earth. With what velocity must a body be projected
from the surface of the earth to escape? (Use as radius of the earth 21-10 6 ft.)
Ans. 7 miles per second.
227] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 451
24. Prove that a particle sliding without friction from the highest point of a
circle in the vertical plane along any chord will reach the circumference in the
same time it would fall along the vertical diameter. (HINT: Acceleration
along a chord with inclination of 6 is d*s/dt 2 g sin 6.)
227. Linear Equations with Constant Coefficients. Linear
differential equations with constant coefficients are of the form
n ~ 2
dy
where a , ai, a 2 , , a n are constants.
lff(x) = 0, the equation is called homogeneous; and if f(x) ? 0,
it is said to be complete. In this treatment, we shall limit the
discussion almost entirely to equations not higher than the second
order, as such equations occur most frequently in the elementary
applications.
THEOREM. // y = u and y v are solutions of the homogeneous
equation
then y c\u + Czv, where Ci and c 2 are independent of x, is also a
solution.
PROOF. If y = u is a solution of (1), on substituting u for y,
we have
/n\ . du, 2
(2) a ^ + a 1 ^,+ a U = >
which can be written
,n d*(du) d(ciu) , A
(3) flo dx2 + i ^ -- h a 2 (ciii) = 0,
since c\ is independent of x. In the same manner we get
fA\ d*(c 2 v) ,
(4) a ~ + ai
The addition of corresponding members of (3) and (4) gives
a, d(ClM + C2t>) + a 2 (c lM + c 2 .) = 0,
452 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
which shows that y = c\u + c z v is a solution of equation (1).
This is moreover the general solution since it contains two arbitrary
constants.
This method of proof evidently may be used for the correspond-
ing theorem for an equation of the n-th order.
228. Solution of the Homogeneous Equation of the Second
Order. A very simple method of solution is to set y uv and
carry through operations similar to those for the solution of linear
equations of the first order. The one difference is that here we
determine u so that the coefficient of dv/dx is zero instead of the
coefficient of v.
EXAMPLES
1. Solve d*y/dx* + 3 dy/dx - 4 y = 0.
SOLUTION. Set y = uv, then dy/dx u dv/dx -\- v du/dx, and therefore
d*y/dx* = u d*v/dx* + 2(du/dx) (dv/dx) + v d*u/dx 2 . Substituting these in
the equation, we get
(d 2 u , du . \ , /_ du . \ dv . d*v _
_|_3 4 M )j, + f2~-+3w)- 7 --hM- r -5 = 0.
dx 2 dx ) \ dx / dx dx 2
Since the coefficient of v is in form the same as the left side of the original
equation, we would not gain anything by imposing the condition^that this
coefficient vanish in order to determine u. Hence we turn to the next coeffi-
cient and determine u so that the coefficient of dv/dx is zero. Therefore we let
or
Whence
du 3 ,
= - dx.
u 2
or
u = e~ 3 */ 2 .
As previously, we take the constant of integration as zero since there
are still two integrations to perform. This value of u changes the uv equation
into the form
P
L4
""" o e ^ ^ I v ~r j._o U.
228] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 453
Collecting and dividing out the factor e~ 3 */ 2 , which is not zero, we get
dx* = ~4~'
which has been solved in 226, IV. As formerly, we let d^v/dx* = p dp/dv
and get
, 25 v.
p dp = j- dv.
Integration gives
4
Therefore, since p - dv/dx,
and, integrating, we have
dv 5 ,
r, dx,
v 2 -f- Ci 2
log (v + Vv 2 -f cO = - + log c 2 ,
whence
t; -j- Vf 2 -|- Ci = Cye 6x l 2 ,
and solving for t;,
where fci = c 2 /2, & 2 = Ci/2 c 2 . Hence
y = uv = e~ z
= &iC z -f- /C2C~ 4x
is the general solution if the positive sign is taken for 5 x/2. The negative
sign merely interchanges ki and kz, which are the arbitrary constants.
2. Solve d*y/dx* - 4 dy/dx + 4 y = 0.
SOLUTION. Set y = uv. Substituting in the given equation, we have
(d*u . du . A \ . / du A \ dv . d 2 v n
35 4 3 h4tt)y-f(2-i 4^)3-4-^ 3-r = 0.
rfa; 2 dx / \ dx J dx dx*
As in the previous example, obtain u so that the coefficient of dv/dx is zero.
That is, set '
dx ~~ U ""
Then
454 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
and substitution gives
^ =
dx* U<
Therefore
v = Cix + Cj.
The general solution is then
y = (cix -f c 2 )e 2 *.
3. Solve d*y/dx* + dy/dx + 2 y = 0.
SOLUTION. Set y = uv and the equation becomes
(^ 2 w , rfw . \ . ( n du . \dv . d*v _
T-^+j-+2w); + l2j~ + M)3-+i*j-; = 0.
dx 2 dx ) \ dx Jdx dx*
Obtaining u so that
we have
u =* e-^/ 2 ,
This reduces the equation above to
_ TH
dx* * ~ 4 J
which we have solved previously.
Setting d 2 v/dx 2 = p dp/dv and integrating, we have
Therefore, since p dv/dx,
dv V7
and
. . v
sin" 1
Whence
g\/7 4-
.
v = Ci sin
2
i sin T rt ^y- J + fc a cos ( ^y ),
229] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 455
where k\ = Ci cos (c 2 /2) and & 2 = ci sin (c 2 /2). Then the solution is
y = c-"{ fcl sin ( **2) + fc, cos
These three examples illustrate the only three possible forms which may
arise from the integration of the second order linear equation with constant
coefficients. That is, the value of d*v/dx z is either or k*v.
If, in the equation ay" + by' + cy 0, we substitute e mx for y
we obtain e mx (am 2 + bm + c) =0. Hence y = e mx will be a
solution only if m is a root of the equation
(1) am 2 + bm + c = 0.
If these roots are real and distinct, say Wi and w 2 , then y = e m i x
and y = e m * x are particular solutions and the general solution is
y = cie m i x + c 2 e m :c . Compare Example 1.
If mi is a double root of (1), we would obtain thus only one par-
ticular solution. However, as in Example 2, the u in the sub-
stitution y = uv becomes the particular solution e m i* and as a
result d 2 v/dx 2 becomes zero. Hence the substitution y = e^ x ^v
where v = c\x + c 2 will give the general solution y = e m * x (c\x + c 2 ).
If (1) has complex roots, the substitution y e mx is possible,
and the general solution obtained may be changed by means of (9)
in 204 into a form similar to that in Example 3.
229. The Complete Equation of the Second Order. We shall
give several methods for solving the complete equation with
constant coefficients. The first given, while not general, is one
that is readily applicable to the types most frequently met with
in elementary applications. This method requires that we prove
the theorem given below :
THEOREM. The general solution of a complete linear equation
with constant coefficients of the form
is given by y = w + z where y w is the general solution of the
homogeneous equation
d*y dy ,
a 5J + ai ^ + a ^ =
and y = z is any particular solution of the complete equation.
456 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
PROOF. Since y w is a solution of the homogeneous equation,
we have
dw . ~
Also, by hypothesis, y = z is a solution of the complete equation;
therefore
d*z . dz .
Adding the corresponding members of these two equations, we
have
d 2 (w + z) d(w + z) / i \ tt \
a dx* + ai dx + a ^ W + * } = fM '
which shows that y = w + z is a solution of the complete equation.
It is the general solution because w contains two arbitrary con-
stants.
The solution w of the homogeneous equation is called the com-
plementary function of the complete equation.
This method of proof applies readily to equations of higher order.
230. A Method for Finding a Particular Integral. As our
solutions of the homogeneous equation of 228 are general and give
us the complementary function for any complete equation, we
need only a method of finding a particular integral of the complete
equation. We give here the method of undetermined coefficients.
Although the method is not general, it is sufficient in all cases where
the right-hand member f(x) contains only such terms as have a
finite number of distinct derivatives; that is, terms like x n (n a
positive integer), e*, sin ax, cos ax, constants, and products of
any of these. The form of f(x) is most frequently of this type in
the equations which arise from elementary applications to physics
and engineering. The following examples illustrate the method.
EXAMPLES
1. Solve d*y/dx* + dy/dx 6 y = sin x.
SOLUTION. The complementary function is found as in 228 to be
2/o = he-** + k&**.
230] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 457
To find a particular solution, we examine the right-hand member. This
suggests that there may be a solution of the form
7/1 a sin x.
On substituting this value for y on the left-hand side, we get
7 a sin x -j- a cos x sin x.
Since this equation must be true for all values of x, we may equate the corre-
sponding coefficients to determine the constant a. But this gives
- 7 a = 1, a = 0,
which are not consistent equations. Hence y a sin x is not a particular
solution.
Since derivatives of y appear on the left-hand side, suppose we try as a
value for y\ the, sum of such terms as appear in the right-hand member together
with all such terms as can be derived from them by differentiation^ each term
having an undetermined coefficient. In this example sin x and cos x are the
only possible terms; therefore let
2/i = a sin x + b cos x.
Substituting this in the complete equation, we get
(a 7 b) cos x (b -f- 7 a) sin x = sin x.
Equating coefficients, we have
a 76 = 0, 6 7 a = 1,
which determine a and 6. Solving them, we obtain
_7_ b== __L.
Therefore a particular solution is
__ 7 sins cos a;
2/1 50 50"
The general solution is
4- If -3* 4- Ic 2* __ ^ sm x _ cos x
y 2/0 T y\. i -r 2 Q 50 *
2. Solve d*y/dx* - 2 dy/dx + 2 y = x 2 + sin 2 x.
SOLUTION. The complementary function may be found, as previously
explained, to be
#o = e*(k\ sin x + fo cos x).
458 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
The terms in the right-hand member are x* and sin 2 x. Their derivatives
are then x, a constant, and cos 2 x, except for coefficients. Therefore we set
2/1 = a sin 2 x }- b cos 2 x -\- ex 2 -{- dx + e.
Substituting this in the equation, we get
(4 b - 2 a) sin 2 x - (4 a + 2 b) cos 2 x -f 2 cz 2 + (2 d - 4 c) x
+ (2 e - 2 d + 2 c) = x 2 + sin 2z.
Equating corresponding coefficients, we have
4&-2a = l f 4 a -f 2 6 = 0, 2 c = 1, 2d-4c = 0, 2e-2d + 2c = 0,
which are sufficient to determine the unknown constants if the equations are
consistent. Solving, we have a = 1/10, b 1/5, c = 1/2, d 1, e = 1/2,
and therefore ?/i = - (sin 2 z)/10 + (cos 2 z)/5 -f x 2 /2 + x -f 1/2 is a
particular solution of the given equation. The general solution then is
y = 2/o + 2/i or
/f . T N sin 2 x cos 2 rr z 2 .1
T/ = e x (A:i sin x + /c 2 cos x) ^ 1 ^ (- + ^ -f- ^
3. Find the general solution of d 2 y/dx 2 -\- dy/dx 6 y = x -f- e 2;c .
SOLUTION. The complementary function was found in the first example
to be 2/0 = fcie~ 3 * + A^e 2 *. We notftce <to a term like one of those in this comple-
mentary function is also found in the right-hand member. The method of choos-
ing 7/1 as given above fails to give a particular integral when a term in the
right-hand member is also a term in the complementary function. The
method merely leads to inconsistent equations in the undetermined coefficients.
For that reason we need the following suggestion. // a term in the right-hand
member is a term in the complementary function, replace that term and its deriva-
tives by x n times each, choosing n 1, 2, until the method of undetermined
coefficients leads to consistent equations. Thus, to solve the equation above for
a particular integral, we select the following terms:
2/i = axe 2 * -f- ex + d.
Substituting this in the equation, we get
5 ae* x - 6 ex + (c - 6 d) = e 2 * + x.
Therefore we have
5o=l, -6c = l, c~6d = 0,
which give
231] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 459
The general solution is therefore
- K Ce2x x I
231. Variation of Parameters. Another method of obtaining
the general solution of the complete equation is to consider the
constants in the complementary function as undetermined func-
tions of the independent variable. Since there are n of the" con-
stants and the only condition on them is that the complementary
function must satisfy the homogeneous equation, there are n I
further conditions which may be imposed. The conditions are
imposed in such a manner as to simplify the work as the solution
proceeds. The method as applied to a second-order equation will
now be given. The same is readily seen to apply for any order.
It may also be used if n particular solutions are known, even if the
equation does not have constant coefficients.
EXAMPLE
Solve by variation of parameters the equation
5j( + y = sec2 *
SOLUTION. This is evidently an example where the method of undetermined
coefficients will not work, as sec 2 x has an infinite number of distinct deriva-
tives. However, we find the complementary function to be
2/o = c\ sin x -\- Cz cos x.
We now think of c t and c 2 as parameters which make y a solution of the
complete equation. This leaves one additional condition which may be
imposed on the parameters. Differentiating ?/o, we have
j-^ = Ci cos x Cz sin x + sin x -f cos x -
dx dx dx
Now, as the additional condition on Ci and c 2 , let us assume that the sum of
the last two terms of dyo/dx be equal to zero. That is, let
/i\ dc\ . dc z ~
(1) sm x -j h cos x -7- = 0.
ax ax
Then the second derivative of y Q is
-3-7 = Ci sin x 02 cos x +- cos x -~ sin x -r-^ .
dx 2 dx dx
460 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
Substituting for yo and d^y^/dx* in the original equation and simplifying,
we have
dc\ . dct
(2) cos x -3 -- sin x -r- = sec 2 x.
dx dx
Solving (1) and (2) simultaneously for dci/dx and dc 2 /dx, we find
dci dci
-r- = sec x, -r- = sec x tan x.
dx ' dx
Therefore
ci = log (sec x + tan x) + ki, c% = sec x + fo.
Substituting these values in y Q , we find the general solution in the form
y k\ sin x + &2 cos x + sin # log (sec x -f- tan x) 1.
232. A General Method for Solving the Complete Equation of
the Second Order. Consider the equation
V * 3* + ' + * -/<*>
If we set y = uv, this becomes
d 2 u . du .
Now suppose we impose the condition on u that shall make the
coefficient of v in (2) zero. That is, let
/o\ cPu . du .
(3) aQ ~dtf +ai ~fa + a2U = *
Thie equation is readily solved for u since it is the homogeneous
equation solved in 228. From (3) we determine a particular
value for u, say u = <t>(x).
Using this value of u, equation (2) reduces to
(4) 2a [*(*)] + arf(*) + a^f*) = /(*),
where 0(x) and /(x) are both known functions. Now (4) is of
form II of 226, and writing p = dv/dx and dp/dx = d*v/dx*,
this reduces to a linear equation of the first order and the solution
is readily completed.
232] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 461
EXAMPLES
1. Solve d?y/dx* -f 2 dy/dx -f y = x.
SOLUTION. Setting y = uv and determining u so that the coefficient of v
in the transformed equation is zero, we have
u = er*.
This value of u makes the transformed equation reduce to
dx 2 ~~ " *
Therefore, integrating this twice, we have
v = xe* 2 e x -i- CiX -{- c^
whence the complete solution is given immediately by
y uv = x 2 -\- (ciX -\- Cz)e- x .
2. Solve d*y/dx 2 + dy/dx - 6 y = x -f &*-
SOLUTION. If y is set equal to uv, this equation is transformed into the
following:
(d 2 u du \ / du \ dv d z v _
^ + ^~ bW > / !; + Vrfi + W > /^ + W ^~ :C + ea: '
If w makes the coefficient of v zero, we have as previously u = e* x . This
value of u makes the equation above reduce to
d 2 v dv 211
Now let dv/dx = p and d*v/dx 2 = dp/dx and
g + 5 p - - + 1.
This is a linear equation of the first order in p and so is readily solved as in
221. The solution is
e -2x e -2x I
2 --- 9- + 5
Integrating, we have
__
g- "" "36" 5 "" ~5~
whence the general solution y = uv is
462 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
233. Systems of Linear Equations with Constant Coefficients.
If the equations have only one independent variable and as many
dependent variables as there are equations, we proceed as follows.
Differentiate the given equations until the original and derived
equations are sufficient for eliminating all but one of the dependent
variables and their derivatives. We shall limit our discussion to
two first-order equations so as to make the derived equation not
higher than the second order. Systems of higher order are treated
in the same manner.
EXAMPLE
Solve the system
dy
~TJ~T % == sin ty
dx ,
SOLUTION. This system has only dy/dt and y to eliminate, and as we
need n -f 1 equations to eliminate n quantities, it is necessary to differentiate
one of the equations to get a third equation. We use the second because
differentiation does not introduce a higher derivative of y than already exists
in the equations. This gives
d*x , dy . .
_ + j = _ sm< .
Eliminating dy/dt and y between the three equations, we get
g-*= -2sin<,
which is a linear equation with constant coefficients. Solving this equation,
we have z as a function of / with two constants of integration. Substituting
this value of x in the second equation, we get y at once.
The method may be applied to equations of higher order if repeated differen-
tiation is used. Enough equations can always be found because in general
only one new derivative is introduced when two equations are differentiated.
PROBLEMS
Solve the following equations.
1. d*y/dx* - 7 dy/dx + 12 y = 0. Ans. y = de* x + c z e Sx .
2. d*y/dx* + 4 y = 0.
3. d*y/dx* + 6 dy/dx -f 9 y = 0. Ans. y = e- 3x (dx + c 2 ).
233] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 463
4. d 2 y/dx 2 - dy/dx + y = 0.
5. d*s/dt* - 3 ds/dt = 2 - 6 t. Ans. s = d -f c 2 e + * f -
6. d*u/dv* - 7 dw/dv -f 10 u = 3 t;e 2 >.
7. d 2 0/d* 2 -f 4 d0/d* + 4 = cos t.
Ans. = e- 2 '(ci* + c 2 ) + (3 cos t -f 4 sin 0/25.
8. d 2 y/dx 2 + dy/dx - 6 y = sin 2 x.
9. d*x/dy 2 - 7 dx/dy + 12 x = e 3 *. Ans. x = Cie 4 * -f c 2 e 3 " ye 3 ".
10. d*y/dx* + 6 (fy/da; -f 9 ?/ = e* sin j.
11. d 2 i//(ia; 2 + 4 dy/dx -f 13 y = sin x.
Ans. y = (ci cos 3 x + c 2 sin 3 x)e~ 2 * -f (3 sin x cos x)/40.
12. d 2 y/dx 2 2 dy/dx -f- y = x + 2z
13. d 2 x/dy 2 - dx/dy + x = y - 1.
Ans. x = e"/ 2 [d cos (l/2)y V5 + c 2 sin (l/2)y V3] -f y.
14. d 2 y/dx 2 -f 9 y = 2 sin 3 x.
16. d 2 y/dx 2 + 4 y = tan 2 x.
17. d 2 x/dy 2 5 dx/dy -f 6 x = 3 e 3 ". Ans. x = Cie 2 " + c 2 e 3 * -f 3 ye 3 ".
18. d 2 y/dx 2 + 4 y = esc x.
19. d 2 y/d* 2 + y = 3 sec 2 1.
Ans. y = ci sin ^ -f- c 2 cos t where Ci = 3 log (sec t -f tan t) -f fa,
c 2 = fc 2 3 sec .
20. d 2 y/dx 2 - 2 dy/dx -f y = e*/x.
J dy/cft - x = 2,
< dy/di + dx/di = 2. Ans. x = Cie-', y = c a + 2 *
dy/dZ -j- x = cos t,
22
{
23. J dy/dt + * = 0>
dx/dt 4-3x-|-4y = 0. Ans. x
-f 3 x = e 2 ',
dx/dt + 3 y = e~ 2< .
25. d 2 i//dx 2 - 5 dy/dx -f 6 y = x + e x , if # = 0, dy/dx *= 1 at x = 0.
Ans. y = (58e 8aj - 81e 2a! + ISe* + 6 x + 5)/36.
26. d 2 y/dx 2 - dy/dx - 2 y e* -f e-*, if y = 1, dy/dx - 1/6 at x - 0.
27. d 2 y/dx 2 + 6 dy/dx + 5 y = e 2 *, if y = 1/6, dy/dx = - 1/6 at x = 0.
Ans. y = (Be- 6 ' + 7e~*
28. d 2 s/dJ 2 + o 2 s = cos at, if s = 0, ds/dt ~ 1 at t = 0,
464 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XXI
29. d*s/dP - 6 ds/dt - 16 s = sin t, if s = 0, ds/dt = at t = 0.
4ns. s = (e 8 ' - 13e- 2 ' + 12 cos t -34 sin 0/650.
30. d*y/dx* + 3 d*y/dx* + 3 cty/(fo + */ = 5 e* sin z.
234. An Approximate Graphical Solution of the First-Order
Equation. The first-order equation may be written in the form
which has as its general solution a one-parameter family of curves.
To determine approximately the graph of one curve of the family
we take PI(XI, yi) as a point of the curve. The coordinates of PI
substituted in (1) give dy\/dx\ or the slope of the curve at the
point PI. Through PI draw a line-segment with this slope;
then take a convenient Ax from P\ and construct the resulting Ay
to the line through PI. This locates a point P% near PI. Find
the slope at P% by substituting x% and y% in (1) and continue the
operations. The smooth curve through the points so determined is an
approximate solution. The solution is more nearly correct, the
smaller Ao: is taken. This is evident because each point after the
first is located on the tangent to the curve through the previous
point. Also it is evident that large values of dy/dx are not con-
ducive to a good approximation. If /(z, y) is n-valued there are
n curves.
EXAMPLE
Find graphically the approximate solution of dy/dx xy = which passes
through the point (1, 1).
SOLUTION. Suppose we use Ax = 0.1
and thereby find the following table
and resulting figure.
Pi: (1, 1),
P 2 : (0.9,0.9),
(0.8, 0.82),
P 3 :
P 4 : (0.7,0.75),
P 6 : (0.6,0.7),
P: (0.5, 0.66),
Pi: (0.4,0.63),
P 8 : (0.3,0.6),
P: (0.2,0.58),
P, : (0.1, 0.57),
Pn: (0, 0.56).
m\ = 1.
w 2 = 0.81.
m z = 0.66.
mi = 0.53.
w 8 = 0.42.
m = 0.33.
mi = 0.25.
w 8 = 0.18.
m 9 = 0.12.
mio = 0.06.
The accuracy of this may be observed from the fact that the ordinary solution
would locate Pn at (0, 0.61).
236] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 465
235. An Approximate Graphical Solution of the First
Order Equation by Means of Trajectories. In the equation
/(#, y, dy/dx) = 0, substitute in succession dy/dx = Ci, c 2 , c 3 ,
and plot the resulting curves. Across each curve draw a system
of parallel lineal elements (short line-segments) with their slope
equal to the value of dy/dx which determined the curve. These
lineal elements give the directions of the integral curves as they
cross each of the curves determined by the values Ci, c 2 , c 3 , %
If we start at a point on any curve and go in the direction of the
lineal elements that cross it and 'approach each successive^curve in
the direction of the elements across it, we can draw approximate
solutions of the equation. Of course
the values c t - should be at most
consecutive integers in order to have
enough trajectories for an accurate
system of solutions.
EXAMPLE
Solve by trajectories x 2 y dy/dx = 0.
SOLUTION. Let dy/dx = 0, 1, 2, 4, 6, 8,
10. The resulting equations give the pa-
rabolas shown in Fig. 229. The lineal ele-
ments show the value of dy/dx for each
parabola. The heavy curves are approxi-
mate solutions of the equation.
FIG. 229
236. Approximate Graphical Solution of Equations of the Second
Order. This equation may be written
(1)
> dx
If we substitute xi, y if and dyi/dxi in equation (1) and solve
the resulting equation for d*yi/dxi*, its sign tells us whether the
integral curve is concave upward or downward at the point
PI(ZI, 2/i). In doing this we select a point and assign a slope
mi at the point and thus determine the value of d z y/dx 2 for these
assumed values. Then dyi/dxi and d*t/i/dzi 2 allow us to find
the radius of curvature #1 at PI. Whence we may lay off Ri
along the line through PI with slope 1/Wi, above P! if
d*yi/dxi* > 0, below PI if d*yi/dx < 0. Now use fa as a
466 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
radius to construct a small arc Pfz of the circle of curvature
of the curve at PI. Then find from the graph the coordinates
of P 2 and the slope of the arc PiP2 at P%. Using these coordi-
nates and this slope, repeat the operations. The smooth curve
through the points P t - is the desired approximate solution. The
points Pi should be located with the help of the graph paper or
even more accurately by use of trigonometry. Moreover, the arcs
should be kept very short. If /(#, y, dy/dx) is n-valued there
are n curves.
EXAMPLE
Obtain a graphical solution of the differential equation
SOLUTION. Take as PI the point (0, 5) and let mi = 0. These give the
following table and figure (Fig. 230) :
Pr. (0,5), mi = 0, ft = -5.
P 2 : (0.95, 4.9), m 2 = - ~ t R 2 = - 4.5.
P 3 : (1.35, 4.8), m 3 = - ~ , fl, = - 4.07.
P 4 : (1.6, 4.7), nn = - ~ , R, = - 3.78.
P.: (2.05, 4.5), m 6 = - ^ , R, = - 3.28.
P 6 : (2.4, 4.3), me = - |, fa = - 3.0.
P 7 : (2.8,4.0).
I 2 3 X
FIG. 230
PROBLEMS
Solve graphically each of the following equations.
1. xy - x* + dy/dx = 0, (-1, 0).
2. x dy/dx -3/2 = 0, (1, 1).
3. ydy/dx+x = 0, (0,2).
4. x dy/dx y sin x = 0, (1, 1.5).
236] DIFFERENTIAL EQUATIONS OF HIGHER ORDER 467
5. x 2 dy/dx + y = 0, by trajectories.
6. x 2 y* dy/dx = 0, by trajectories.
7. x d*y/dx 2 - x 2 dy/dx - y = 0, (1, 1), mi = 2.
8. (1 + y)d 2 y/dx 2 + (dy/dx)* = 3 x - 10, (1/2, 1/2), m, = 5.
9. y 2 d*y/dx 2 - 4 x dy/dx + y = 0, (0, 3), mi = 0.
10. a: d 2 y/dx 2 + dy/dx + x = 0, (1, - 1/4), m x = - 1/2.
11. 2 3 + 2 y dy/dx = 3, (1/2, 1), Ax = 0.4.
12. 2 y dy/dx - 3 x 2 + 1 = 0, (1, 0), Ax = 0.5.
13. y dy + (1 x 2 )dx = 0, by trajectories.
14. y 2 dx 2 x dy = 0, by trajectories.
15. dy/dx = 1/x -I- 1/y, (1, 1).
16. dy/dx = Vx 2 + y 2 , (2, 1).
ADDITIONAL PROBLEMS
Solve each of the following equations. (Nos. 1-31.)
1. d*y/dx 3 sin x. Ans. y = cos x + c\x* + C2X
2. d 3 y/dx 3 = e 2 * + x.
3. d*y/dx z = x cos x.
Ans. y = x 4 /4 ! + sin x -{- dx 2 + c&
4. d 2 y/dx 2 - y dy/dx = 0.
5. d 2 y/dx 2 - dy/dx + 1=0.
6. x d 2 y/dx 2 + dy/dx + x = 0.
7. d 2 y/dx 2 = (dy/dx) 2 + 1.
8. y d 2 y/dx 2 - (dy/dx) 2 = y 2 .
9. (dx/dy) 2 1 = x d 2 x/dy 2 . Ans. CiX + Vc^x 2 + 1 = e c i<+ c >.
+ c& + c 3 .
Ans. y = de x -f x + c 2 .
Ans. y = log sec (x + Ci) + c 2 .
10. d 2 y/dx 2 = VT+ (dy/dx) 2 .
11. (1+ 2/)d 2 y/dx 2 - (dy/dx) 2 = 0.
Ans. log (!+</)= dx + c a .
12. 2 y d 2 y/dx 2 -f 2 (dy/dx) 2 = y dy/dx for dy/dx = 3 at (0, 2).
13. x d 2 y/dx 2 =Vl + (dy/dx) 2 for dy/dx = at (1, 1/4).
Ans. y = (l/2)(x 2 /2 -logx).
14.~ x d 2 y/dx 2 + dy/dx = x for dy/dx = 3/2 at (1, 0).
468 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXI
15. 2 x dy/dx d 2 y/dx 2 = (dy/dx) 2 - 1 for dy/dx = 2 at (1, 1).
Ans. 9 y = 2(3 x + I) 3 / 2 - 7.
16. d 2 s/d* 2 + 4 - cos 2 t.
17. d 2 3/d/ 2 - ds/d* = t. Ans. a - ci + ctf* - * - * 2 /2.
18. d 2 x/dy 2 -f dx/dy - 2 * * ye*.
19. d 2 y/dt 2 + y = sec .
Ans. y GI cos + c 2 sin -f- cos / log cos t -f sin t.
20. d 2 y/dx 2 -f 8 dy/dx + 15 y = 2 e" 3 -.
21. d 2 s/d* 2 ~ 3 <fcM + 2 5 = e*< - 3 .
Ans. s = cie*' + c 2 e< + te 2< - 3^/2 - 9/4.
22. dV<ft 2 - 6 cfc/<tt -h 9 x == e 3 '/^ 2 .
23. 4 d 2 2//rfx 2 + y = 0, if dy/dx *= 3 at x = TT, y = 2.
Ans. y = 2 sin (l/2)z - 6 cos (l/2)x.
24. d 2 y/dx 2 4 dy/dx -f 5 y - for the curve through (0, 0) with slope 1.
25. d 2 y/dx 2 - 6 dy/dx -f 9 y = 2 e 3 ' + sin 2 x.
Ans. y = cie 3 * + C 2 xe 3x + x 2 e 3 * + (5 sin 2 x -f 12 cos 2 x)/169.
J dx/d* - dy/d/ = e",
\ d*y/dt* = 2 dx/d* - x + .
27. d 3 y/dx 3 + 2 d 2 y/dx 2 + 5 dy/dx = x.
Ans. y = ci + e~ x (c 2 sin 2 x + c 3 cos 2 x) + x 2 /10 2x/25.
28. d 2 y/dx 2 -j- tan x dy/dx = sin 2 x.
29. d 2 y/dx 2 - 5 dy/dx + 6 y = e*, (n ^ 2, 3).
Ans. y = cie 2 * -f- c 2 e 3 * -f e nx /(n 2 5 n + 6).
,30. d*y/dx* - 2d 3 y/dx 3 -f d 2 y/dx 2 = x.
31. (1 - x)d 2 y/dx 2 -f x dy/dx y = (1 x) 2 . (Find particular inte-
grals by inspection for the left-hand member set equal to zero and then use
variation of parameters.) Ans. y = Cix -f c 2 e* + x 2 + 1.
32, Curve of Pursuit. A point Q starts at the origin and moves along the
x axis with constant speed u. At the same time a second point P starts at
A(0, a) and moves at a constant speed u/k in a direction always toward Q.
Find the equation of the path of P and, assuming k < 1, the point at which P
catches Q.
CHAPTER XXII
APPLICATIONS OF DIFFERENTIAL EQUATIONS
SOME APPLICATIONS IN STATICS
237. The Form of Suspended Cables. Suppose the cable of
Fig. 231 to be in equilibrium under the action of its weight and
tensions T at two points A and B on a horizontal line. Our
problem is to determine its form.
Take the y axis vertical and through the lowest point C of the
cable. Any segment CP is in equilibrium under the tangential
tensions H and T at C and P, respectively, and the weight ws,
where w is the weight of the cable per unit length and s is the length
of the arc CP. Since CP is in equilibrium, the horizontal com-
ponents of force are equal; that is,
(1)
T cos B = H,
and the equality of vertical com- T
ponents gives
(2)
T sin B = ws.
On dividing (2) by (1) we have
(3) tan 9 = -^ = ~ - s,
from which, by differentiation,
... d*y _ w ds
dx 2 H dx
Fia. 231
But ( 101)
and therefore
dx
dx 2
469
470 DIFFERENTIAL AND INTEGRAL CALCULUS [Cn. XXII
the positive sign being chosen because the curve is concave upward.
Substituting dy/dx = p and integrating, we get
w
log (p + Vp 2 + 1) = jj x + k,
where k = 0, since p = when x = 0. Solving for p, we find
71
P
dy \( %
= = I pH
e
dx
( %. x - A
I pH /> H \
\ e e )'
from which
" s x , ~~ 77
= e +e
or
.
z +
H l,
= - cosh
To fix the position of the x axis, we take & 2 = 0; then T/ = H/w
when a: = 0, or 0(7 is equal to the length of the cable which has as
its weight the horizontal tension at the lowest point C. The curve
assumed by the cable is a catenary.
Now consider the situation if the cable has vertical rods attached
which suspend a road-bed AB
carrying a uniformly distributed
load. Let the weight of the road-
bed and load be large compared with
the weight of the cable and vertical
rods. Disregarding the weight of
the cable and rods, show that the
cable of Fig. 232 takes the form of
FIG. 232 ^ ne parabola y = wx 2 /2 H. Here w
is the weight of the road-bed and
load per foot of road-bed, and H is the horizontal tension at its
lowest point.
238] APPLICATIONS OF DIFFERENTIAL EQUATIONS 471
PROBLEMS
1. If along the roadway of a suspension bridge, w 2000 Ibs. per ft. per
cable and if // = 2,500,000 Ibs., derive the equation of the cable.
Ans. y = z 2 /2500.
2. If the cable of Problem 1 sags 100 ft. at its mid-point, what is the span
length?
3. What is the direction of the cable of Problem 1 at the ends? At the
quarter span points? Ans. 2/5; 1/5.
238. Bending of Columns and Beams. When a cantilever
beam (one supported at only one end) is bent as in Fig. 233, the
fibers on its upper part are stretched
and those on the lower are compressed.
There is a curve along which the
fibers are neither compressed nor ex-
tended. This curve is called the
neutral axis. In mechanics it is
shown that M = EI/R where M is the
bending moment at any point of the
axis P(XI, ?/i ), / is the moment of
inertia of a cross-section of the beam
about an axis through P and perpen-
dicular to the xy plane, E is Young's
modulus for the material, and R is the radius of curvature of the
neutral axis at P.
But R = [1 + (dy/dx) 2 ]* f2 /(d 2 y/dx 2 ) may be replaced approxi-
mately by l/(d 2 y/dx 2 ) if dy/dx is small, which gives the approx-
imate relation
JL
FIG. 233
(1)
- = M
1 o ~"~^ ***
dx 2
PROBLEMS
1. For a concentrated load W Ibs. at the end of the cantilever beam shown
above, the bending moment at P is W(l x\). Derive the equation of the
neutral axis (elastic curve).
2. Determine the deflection at the free end of the neutral axis in Problem 1.
3. Find the slope of the elastic curve of Problem 1 at its free end-point.
4. A timber cantilever beam 4 in. by 4 in. in cross-section projects 120 in.
from the face of a brick wall. A load of 200 Ibs. is placed at the end. If
E = 1,500,000 Ibs. per sq. in. and 7 = 64/3, determine the equation of the
472 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XXII
clastic curve. What is the deflection at the free end? Also what is the slope
of the curve at the free end?
5. Let a beam of homogeneous material and constant cross-section be
fastened horizontally at one end. Also suppose a load distributed over it
uniformly such that the weight of load and beam is w Ibs. per foot. Then the
bending moment at P due to the shaded element at Q is w(x Xi)dx, and the
total bending moment at P due to the part from P to the end of the beam is
(Fig. 233)
M
f\x -
Jx*
(I -
Hence, from (1),
2 El
(I - z,).
The negative sign is used because the curve is concave downward. Find y\
and show that the deflection is wl*/8 El at the end.
6. Let the weight of a beam be w Ibs. per foot and let a weight of Q Ibs.
be applied at the end. Show that
Find y and the deflection at the free end.
7. The beam OAC of homogeneous material and uniform cross-section has
forces W applied at the center of its ends. Find an expression for the force
which will produce bending.
Y Disregarding the weight, we have
from which
FIG. 234
y = csin (\jj x y
\ IF "* /
since y when x = 0. Evidently c represents the maximum deflection,
and is not zero if bending occurs. Then, since dy/dx at x = 1/2,
W l
which means that
I
lEl 2
?
2'
or
239] APPLICATIONS OF DIFFERENTIAL EQUATIONS 473
if bending occurs. This last formula is known as Euler's formula for the
strength of a column and may also be obtained as follows. The value of x
which locates C is / and it locates the end of an arch of the sine curve. Hence
VW/EI I = T or W = 7r 2 #/// 2 .
8. If a beam is fastened at both ends,
there must act at each end a bending
moment k to keep the slope at those
points zero. Then, in addition to the
moment Wy which tends to produce bend-
ing, there exists a moment k ^opposing
bending. Therefore
FIG. 235
Show that y = (k/W)[l - cos (VW/EIx)] and that W = 4 7r 2 E//Z 2 will
produce bending.
SOME APPLICATIONS TO DYNAMICS
239. Simple Pendulum. An important case of motion in a
plane is that of a simple pendulum. Assume the bob to be
suspended by a very light wire or rod, the weight of which we shall
neglect. Resolving forces along the tangent at B and disregarding
the resistance of the air, we get
d 2 s
m -^ = - mg sin 0,
where s = aO.
If 6 is kept so small that sin 6 is ap-
proximately equal to 0, this gives
B
FIG. 236
the solution of which is
= Ci sin
Z tj
dt*
+ c 2 cos
If = when t = and if do is the maximum displacement,
then
= sin
^ t\ .
474 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XXII
At the maximum displacement
g . TT . TT a
' = =
from which the period is T = 2 ir\/a/g.
For the seconds pendulum, this makes a = 0/4 ?r 2 .
240. Non-Periodic Motion of Bodies in Resisting Media.
When a body moves in a medium, that medium offers a resistance
depending on the velocity, but the exact law of dependence is not
known.
EXAMPLE
Suppose a body falling from rest in air meets with resistance proportional
to the velocity. Let the resistance equal c when the velocity is unity. Then,
since effective accelerating force equals the impressed force minus the resist-
ance, we have
dv
ffl = ffio CD .
dt
or
=
dt ~ g V '
where k = c/m. This gives
_
~dt~
since v = when t = 0. Then
(kt + e-*< - 1),
since s = when t = 0.
Observe that as t increases v + g/k and s - (gt/k g/k 2 ), which are
expressions for motion with a constant velocity.
PROBLEMS
1. Suppose that in the example above the resistance of the air had varied
as the square of the velocity. Then dv/dt = g kv 2 . Show that
e<V0* ,
and
-)
Observe that as t increases v - Vg/k and s-* (l/k}(t^gk log 2), which
are expressions for motion with a constant velocity.
241] APPLICATIONS OF DIFFERENTIAL EQUATIONS 475
2. A constant force E begins to act at time t = on a boat of mass M .
Assuming that the water offers a resistance which is proportional to the velocity
v and which equals R when the velocity is unity, find v in terms of t. The
effective accelerating force is Mdv/dt t the impressed force is E, and the resist-
ing force is Rv. Therefore
,,dv , . _
Note that v * E/ R as t increases and that this is a particular solution of the
differential equation.
3. Suppose in the problem above that the boat is moving with a velocity
ro when t = and the propelling force E is suddenly removed. Find expres-
sions for the velocity and the distance the boat will move.
241. Periodic Motion of Bodies in Media. A mass M is
supported by a spiral spring fixed at the end A. Let be the
position of equilibrium of the mass and let the stiffness of the spring
be such that a force of 1 Ib. elongates it c ft. Disregarding the
resistance of the air, the force on M is equal to q/c, by Hooke's
law, where q is the displacement from the position of equilibrium.
Therefore
ft)
U;
which has the solution
t
q ki sin + k% cos
. -
vMc
If the motion arose from releasing the mass at
time t = after it had been pulled a ft. below its
position of equilibrium, then ki = 0, k 2 = a, and
t
q = a cos
The motion is simple harmonic of period 2 w\/Mc. We see
from the expression for the period that the frequency of the vibra-
tions is increased by decreasing the mass or increasing the stiffness
of the spring.
Let the mass M be immersed in a fluid which offers a resistance
476 DIFFERENTIAL AND INTEGRAL CALCULUS [On. XXII
proportional to the velocity and which equals R when the velocity
is unity. Then the resisting force is
i -jrr
c dt
and the equation becomes
a homogeneous linear equation with constant coefficients.
In the two cases just considered the system when once started
in motion was left to itself; in such cases, vibrations are called
free oscillations.
Let us now suppose that, in addition to the frictional force and
the stiffness of the spring, an outside force E acts on M. Then
the equation becomes
(3) M*3 = E -Z-R%,
^ / dt 2 c dt
a complete linear equation with constant coefficients. In this
case, the vibrations are called forced oscillations^ provided E is a
periodic function of the time.
242. Free Oscillations. If R 2 c > 4 M, the solution of equa-
tion (2) in 241 is
(a) q = e- m iw>(nw?'
where k = V (R 2 c - 4 M)/4 M 2 c. If R 2 c = 4 M, the solution is
(6) q = e-ww^at + b).
If R 2 c < 4 M y the solution is
(c) q = e-*"< 2 ">[mi sin (B) + m 2 cos (ft)],
where I = V(4M - J? 2 c)/4M 2 c.
In case (c), the motion is that of free oscillations of period
4 TrAf \/c/V4 M R 2 c, which have a decreasing amplitude, or
are damped, due to the factor e~ Rt/( * M) . The period may be
increased by making R 2 c -* 4 M.
In cases (a) and (6), the motion is non-oscillatory.
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