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Full text of "Elementary Calculus"

ELEMENTARY CALCULUS 



BY 
FREDERICK S. WOODS 

AND 

FREDERICK H. BAILEY 

PIlOlfMSSOUS 0V MATHEMATICS IN THK MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY 




GINN AND COMPANY 

ItORVOK NI3W YORK OHtCAGO LONDON 
ATLANTA AM.A8 COLUMBUS SAN PBANOISOO 



COPYRIGHT, 1922, BY FREDERICK 8 WOODS 

AND FREDERICK H BAILEY 

ALL BIGHTS RESERVED 

PRINTED IN THB UNITED STATES OF AMERICA 

I 
82611 



153 



GINN AND COMPANY . PRO- 
PRIETORS BOSTON > V.S.A. 



PKEFACB 

This book is adapted to the use of students in the first year 
in technical school or college, and is based upon the experience 
of the authors m teaching calculus to students in the Massa- 
chusetts Institute of Technology immediately upon entrance. 
It is accordingly assumed that the student has had college- 
entrance algebra, including graphs, and an elementary course 
in trigonometry, but that he has not studied analytic geometry. 

The first three chapters form an introductory course in 
which the fundamental ideas of the calculus are introduced, 
including derivative, differential, and the definite integral, but 
the formal work is restricted to that involving only the poly- 
nomial. These chapters alone are well fitted for a short course 
of about a term. 

The definition of the derivative is obtained through the 
concept of speed, using familiar illustrations, and the idea 
of a derivative as measuring the rate of change of related 
quantities is emphasized. The slope of a curve is introduced 
later. This is designed to prevent the student from acquiring 
the notion that the derivative is fundamentally a geometric 
concept. For the same reason, problems from mechanics are 
prominent throughout the book. 

With Chapter IV a more formal development of the subject 
begins, and certain portions of analytic geometry are introduced 
as needed. These include, among other things, the straight line, 
the conic sections, the cycloid, and polar coordinates. 

The book contains a large number of well-graded exercises for 
the student. Drill exercises are placed at the end of most sec- 
tions, and a miscellaneous 'Set of exercises, for review or further 
work, is found at the end of each chapter except the first. 



ui 



iv PREFACE 

Throughout the book, the authors believe, the matter is pre- 
sented in a manner which is well within the capacity of a first- 
year student to understand. They have endeavored to teach 
the calculus from a common-sense standpoint as a very useful 
tool. They have used as much mathematical rigor as the 
student is able to understand, but have refrained from raising 
the more difficult questions which the student in his first 
course is able neither to appreciate nor to master. 

Students who have completed this text and wish to continue 
their study of mathematics may next take a brief course in 
differential equations and then a course in advanced calculus, 
or they may take a course m advanced calculus which includes 
differential equations. It would also be desirable for such stu- 
dents to have a brief course in analytic geometry, which may 
either follow this text directly or come later. 

This arrangement of work the authors consider preferable to 
the one for a long time common in American colleges by 
which courses in higher algebra and analytic geometry precede 
the calculus. However, the teacher who prefers to follow the 
older arrangement will find this text adapted to such a program. 

F. S. WOODS 
F H BAILEY 



CONTENTS 



CHAPTER I. EATES 

SECTION PAGE 

1 Limits . 1 

2 Average speed . . . .... 3 

3. True speed ... . 5 

4. Algebiaic method . .... . 8 

5 Acceleiation . ....... 9 

G Rate of change ........ .11 

CHAPTER II. DIFFERENTIATION 

7 The derivative . . . . . .... .15 

8. Differentiation of a polynomial ... 18 

9 Sign of the derivative ... . . . .... 20 

10 Velocity and acceleration (continued) . ... 21 

11 Rate of change (continued) . ... ... 24 

12 Graphs . .... .... 27 

13 Real roots of an equation 30 

14 Slope of a straight line . .... .31 

15 Slope of a curve ... . 36 

16 The second derivative . . , .39 

17. Maxima and minima . . ... 41 

18 Integration ... ... 44 

19 Area . 47 

20 Differentials ...*... 50 

21. Appioximations ... . ... 53 

General exercises .... . . . . - . .55 

CHAPTER III. SUMMATION 

22. Area by summation . . . ... 00 

23. The definite integral .... . 62 

24. The general summation problem .66 

25. Pressure 68 

26 Yolume . 71 

General exercises ... 76 

T 



vi CONTENTS 

CHAPTER IV. ALGEBRAIC FUNCTIONS 

SECTION PAGE 

27 Distance between two points ... . . . . 79 

28. Circle ... . .... ... 79 

29. Parabola ... 81 

30 Parabolic segment . 83 

81. Ellipse ... 85 

32 Hyperbola . . 87 

33. Other curves 91 

34 Theorems on limits . . .93 

35. Theorems on derivatives .... 94 

36. Formulas 101 

37 Differentiation of implicit functions ... 102 

38 Tangent line ... . 104 

39. The differentials dx, dy, ds 106 

40. Motion in a curve ... ... 107 

41. Related velocities and rates Ill 

General exercises . 113 



CHAPTER V. TRIGONOMETRIC FUNCTIONS 

42. Circular measure . .... ... 119 

43 Graphs of trigonometric functions . . ... 121 

44. Differentiation of trigonometric functions 124 

45. Simple harmonic motion . .... 127 

46 Graphs of inverse trigonometric functions 130 

47 Differentiation of inverse trigonometric functions . . 131 

48. Angular velocity . . ... . ... . 185 

49. The cycloid . . .... ... . 137 

50. Curvature 189 

51 Polar coSrdinates , 142 

53. The differentials dr, dO, ds, in polar coordinates . . . 146 

General exercises . . . . 149 



CHAPTER VI. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

53. The exponential function .... 154 

54. The logarithm . . 154 

55. Certain empirical equations 159 

56. Differentiation . . 163 

57. The compound-interest law 166 

General exercises 168 



CONTENTS vii 
CHAPTER VII. SERIES 

SECTION PAGE! 

58. Power series . . . . . 172 

59. Maclaurm's series . 173 

60. Taylor's seuos . . . 177 

General exercises . . ... 179 

CHAPTER VIII. PARTIAL DIFFERENTIATION 

61 Partial differentiation . . . 181 

62. Higher partial denvatives . 184 

63. Total differential of a function of two variables . . . 185 
64 Rate of change ... .189 

General exercises . . . 191 

CHAPTER IX. INTEGRATION 

65. Introduction . . . 194 

66. Integral of u n . . 195 

67-68 Other algebraic integrands 199 

69 Integrals of trigonometric functions ... 205 

70. Integrals of exponential functions . . ... 207 

71-72. Substitutions . . . 208 

73-74 Integration by parts . . . 212 

75 Integration of rational fractions ... . 216 

76. Table of integrals . . 217 

General exercises ... .... . 220 

CHAPTER X. APPLICATIONS 

77. Review problems . ... 225 

78. Infinite limits, or integrand 229 

79. Area in polar cobrdmates . 230 

80. Mean value of a function . .... 233 

81. Length of a plane curve 235 

82. Work 237 

General exercises ... . . ... 239 

CHAPTER XI. REPEATED INTEGRATION 

83. Double integrals ... . . . . 244 

84. Area as a double integral . . . . . . 246 

85. Center of gravity . 249 



yiii CONTENTS 

SECTION PAGE 

86 Center of gravity of a composite area . . . 255 

87 Theoiems . . . 257 
88. Moment of inertia 260 

89 Moments of inertia about parallel axes . 266 

90 Space cooidinates 269 

91. Certain sui faces . .... 271 

92. Volume . . . .... .277 

93 Center of gravity of a solid . 282 

94 Moment of inertia of a solid . . 283 

General exeicises . . . . . 286 

ANSWERS 291 

INDEX 315 



ELEMENTARY CALCULUS 



* CHAPTER I 
RATES 

1. Limits. Since the calculus is based upon the idea of a 
limit it is necessary to have a clear understanding of the word. 
Two examples already familiar to the student will be sufficient. 

In finding the aiea of a circle m plane geometry it is usual 
to begin by inscribing a regular polygon in the circle. The area 
of the polygon differs from that of the circle by a certain 
amount. As the number of sides of the polygon is increased, 
this difference becomes less and less. Moreover, if we take any 
small number 0, we can find an inscribed polygon whose area 
differs from that of the circle by less than e\ and if one such 
polygon has been found, any polygon with a larger number of 
sides will still differ m area from the circle by less than e. The 
area of the circle is said to be the limit of the area of the 
inscribed polygon. 

As another example of a limit consider the geometric progres- 
sion with an unlimited number -of terms 



The sum of the first two terms of this series is 1, the sum 
of the first three terms is If, the sum of the first four terms 
is 1-J, and so on. It may be found by trial and is proved in 
the algebras that the sum of the terms becomes more nearly 
equal to 2 as the number of terms which are taken becomes 
greater. Moreover, it may be shown that if any small number 
e is assumed, it is possible to take a number of terms n so that 
the sum of these terms differs from 2 by less than e. If a value 
of n has thus been found, then the sum of a number of terms 

1 



2 RATES 

greater than n will still differ from 2 by less than e. The 
number 2 is said to be the limit of the sum of the first n terms 
of the series. 

In each of these two examples there is a certain variable 
namely, the area of the inscribed polygon of n sides in one case 
and the sum of the first n terms of the series in the other case 
and a certain constant, the area of the circle and the number 2 
respectively. In each case the difference between the constant 
and the variable may be made less than any small number e by 
taking n sufficiently large, and this difference then continues 
to be less than e for any larger value of n. 

This is the essential property of a limit, which may be defined 
as follows: 

A constant A is said to be the limit of a variable Xif, as the vari- 
able changes its value according to some law, the difference between 
the variable and the constant becomes and remains less than any 
small quantity which may be assigned. 

The definition does not say that the variable never reaches its 
limit. In most cases in this book, however, the variable fails to 
do so, as in the two examples already given. For the polygon is 
never exactly a circle, nor is the sum. of the terms of the series 
exactly 2. Examples may be given, however, of a variable's 
becoming equal to its limit, as in the case of a swinging pendulum 
finally coming to rest. But the fact that a variable may never 
reach its limit does not make the limit inexact. There is nothing 
inexact about the area of a circle or about the number 2. 

The student should notice the significance of the word 
" remains" in the definition. If a railroad train approaches a 
station, the difference between the position of the train and 
a point on the track opposite the station becomes less than any 
number which may be named ; but if the train keeps on by the 
station, that difference does not remain small. Hence there is 
no limit approached in this case. 

If X is a variable and A a constant which X approaches as a 
limit, it follows from the definition that we may write 

X**A + e, (1) 



SPEED 3 

where e is a quantity (not necessarily positive) which may be 
made, and then will remain, as small as we please. 

Conversely, if as the result of any reasoning we arrive at a 
formula of the form (1) where X is a variable and A a constant, 
and if we see that we can make e as small as we please and 
that it will then remain just as small or smaller as X varies, 
we can say that A is the limit of X. It is in this way that we 
shall determine limits in the following pages. 

2. Average speed. Let us suppose a body (for example, an 
automobile) moving from a point A to a point B (Fig. 1), a 
distance of 100 mi If the automobile takes 5 hr. for the trip, 
we are accustomed to say that it has traveled at the rate of 
20 mi. an hour. Everybody knows A p Q B 

that this does not mean that the ' ' ' 

automobile went exactly 20 mi. in 

each hour of the trip, exactly 10 mi. in each half hour, exactly 
5 mi in each quarter hour, and so on. Probably no automobile 
ever ran in such a way as that. The expression " 20 mi. an 
hour " may be understood as meaning that a fictitious automobile 
traveling in the steady manner* just described would actually 
cover the 100 mi. in just 5 hr. ; but for the actual automobile 
which made the trip, " 20 mi. an hour " gives only a certain 
average speed. 

So if a man walks 9 mi. in 3 hr., he has an average speed of 
3 mi. an hour. If a stone falls 144 ft. in 3 sec., it has an average 
speed of 48 ft. per second. In neither of these cases, however, 
does the average speed give us any information as to the actual 
speed of the moving object at a given instant of its motion. 

The point we are making is so important, and it is so often 
overlooked, that we repeat it in the following statement: 

If a "body traverses a distance in a certain time, the average speed 
of the body in that time is given ly the formula, 

, distance 

average speed = 

time 

but this formula does not in general give the true speed at any 
given time. 



4 KATES 

EXERCISES 

1. A man runs a half mile m 2 mm and 3 sec. What is his 
average speed m feet per second ? 

2. A man walks a mile m 25 min. What is his average speed in 
yards per second ? 

3. A train 600 ft. long takes 10 sec. to pass a given milepost. 
What is its average speed in miles per hour ? 

4. A stone is thrown directly downward from the edge of a 
vertical cliff Two seconds afterwards it passes a point 84 ft down 
the side of the cliff, and 4 sec after it is thrown it passes a point 
296 ft. down the side of the cliff What is the average speed of the 
stone in falling between the two mentioned points ? 

5. A railroad train runs on the following schedule : 



Boston 




10 00 A M. 


Worcester 


(45 mi.) 


11.10 


Springfield 


(90 mi ) 


12.35 P M. 


Pittsfield 


(151 mi.) 


225 


Albany 


(201 mi.) 


355 



Find the average speed between each two consecutive stations and 
for the entire trip. 

6. A body moves four times around a circle of diameter 6 ft in 
1 min. What is its average speed in feet per second ? 

7. A block slides from the top to the bottom of an inclined 
plane which makes an angle of 30 with the horizontal. If the top 
is 50 ft. higher than the bottom and it requires f mm. for the block 
to slide down, what is its average speed in feet per second ? 

8. Two roads intersect at a point C B starts along one road 
toward C from a point 5 mi. distant from C and walks at an average 
speed of 3 mi an hour. Twenty minutes later A starts along the 
other road toward C from a point 2 mi. away from G At what 
average speed must A walk if he is to reach C at the same instant 
that B arrives ? 

9. A man rows across a river $ mi. wide and lands at a point 
mi. farther down the river. If the banks of the river are parallel 
straight lines and he takes ^ hr. to cross, what is his average speed 
in feet per minute if his course is a straight line ? 



SPEED 5 

10. A trolley car is running along a straight street at an average 
speed of 12 mi. per hour. A house is 50 yd. back from the car track 
and 100 yd. up the street from a car station. A man comes out of 
the house when a car is 200 yd away from the station What must 
be the average speed of the man m yards per minute if he goes in 
a straight line to the station and arrives at the same instant as 
the car ? 

3. True speed. How then shall we determine the speed at 
which a moving body passes any given fixed point P in its 
motion (Fig. 1) ? In answering this question the mathema- 
tician begins exactly as does the policeman in setting a trap for 
speeding. He takes a point Q near to P and determines the 
distance PQ and the time it takes to pass over that distance. 
Suppose, for example, that the distance PQ is ^ mi. and the 
time is 1 min. Then, by 2, the average speed with which 
the distance is traversed is 

mi. 
= 30 mi. per hour. 

hr - 

This is merely the average speed, however, and can no more 
be taken for the true speed at the point P than could the 20 mi. 
an hour which we obtained by considering the entire distance 
A3. It is true that the 30 mi. an hour obtained from the 
interval PQ is likely to be nearer the true speed at B than 
was the 20 ini. an hour obtained from AS, because the interval 
PQ is shorter. 

The last statement suggests a method for obtaining a still 
better measure of the speed at P ; namely, by taking the interval 
PQ still smaller. Suppose, for example, that PQ is taken as 
fa mi. and that the time is 6J sec. A calculation shows that the 
average speed at which this distance was traversed was 36 mi. 
an hour. This is a better value for the speed at P. 

Now, having seen that we get a better value for the speed at 
P each time that we decrease the size of the interval PQ, we 
can find no end to the process except by means of the idea of a 
limit denned in 1. We say, in fact, that the speed of a moving 
body at any point of its path u the limit approached ly the average 




6 RATES 

speed computed for a small distance beginning at that point, the 
limit to be determined by taking this distance smaller and smaller. 

This definition may seem to the student a little intricate, and 
we shall proceed to explain it further. 

In the case of the automobile, which we have been using for 
an illustration, there are practical difficulties in taking a very 
small distance, because neither the measurement of the distance 
nor that of the time can be exact. This does not alter ^ 
the fact, however, that theoretically to determine the speed 
of the car we ought to find the time it takes to go an 
extremely minute distance, and the more minute the dis- 
tance the better the result. For example, if it were possi- 
ble to discover that an automobile ran ^ in. in -^^-g- sec., 
we should be pretty safe in saying that it was moving at p 
a speed of 30 mi. an hour. _p 

Such fineness of measurement is, of course, impossible ; 
but if an algebraic formula connecting the distance and 
the time is known, the calculation can be made as fine as 
this and finer. We will therefore take a familiar case in 
which such a formula is known ; namely, that of a falling body. 

Let us take the formula from physics that if s is the distance 
through which a body falls from rest, and t is the time it takes 
to fall the distance s t then 

s = 16* 3 , (1) 

and let us ask what is the speed of the body at the instant 
when t SB 2. In Fig. 2 let be the point from which the body 
falls, % its position when t 2, and 7 its position a short time 
later. The average speed with which the body falls through the 
distance PP Z is, by 2, that distance divided by the time it 
takes to traverse it. We shall proceed to make several succes- 
sive calculations of this average speed, assuming 7^ and the 
corresponding time smaller and smaller. 

In so doing it will be convenient to introduce a notation as 
follows : Let ^ represent the time at which the body reaches 2?, 
and t a the time at which it reaches J%. Also let ^ equal the 
distance OP^ and a the distance OP a . Then s^-s^P^ and 



SPEED 7 



tf 2 ^ is the time it takes to traverse the distance P^. Then the 
average speed at which the body traverses Ufa is 

So Si 



(2) 



Now, by the statement of our particular problem, 



Therefore, from (1), s 1= 16 (2) 2 = 64. 

We shall assume a value of t a a little larger than 2, compute 
s a from (1), and the average speed from (2). That having been 
done, we shall take t z a little nearer to 2 than it was at first, and 
again compute the average speed This we shall do repeatedly, 
each time taking t a nearer to 2. 

Our results can best be exhibited in the form of a table, as 
follows : 



t a -L 



2.1 70.56 .1 656 656 

201 646416 01 .6416 64.16 

2 001 64.064016 .001 064016 64 016 

2.0001 64 00640016 .0001 .00640016 64 0016 

It is fairly evident from the above arithmetical work that as 
the time i? 3 ^ and the corresponding distance s s s x become 
smaller, the more nearly is the average speed equal to 64. 
Therefore we are led to infer, in accordance with 1, that the 
speed at which the body passes the point J$ is 64 ft. per second. 

In the same manner the speed of the body may be computed 
at any point of its path by a purely arithmetical calculation. In 
the next section we shall go farther with the same problem and 
employ algebra. 

EXERCISES 

1. Estimate the speed of a falling body at the end of the third 
second, given that s = 16 t z , exhibiting the work in a table. 

2. Estimate the speed of the body in Ex. 1 at the end of the 
fourth, second, exhibiting the work in a table. 



8 RATES 

3. The distance of a falling body from a fixed point, at, any th 
is given by the equation s = 100 + 16 t*. Estimate the speed of t 
body at the end of the fourth second, exhibiting the work in a tab 

4. A body is falling so that the distance traversed in the tiuu 
is given by the equation s = 16 1 2 + 10 1 Estimate tho speed of 1,1 
body when t = 2 sec , exhibiting the work in a table. 

6. A body is thrown upward with such a speed that at. any tin 
its distance from the surface of the earth is given by tho equsitu 
s = 100 1 16 t a . Estimate its speed at the end o a sexumd, exhibi 
ing the work in a table. 

6. The distance of a falling body from a fixed point a,t any tiin 
is given by the equation s = 50 -f- 20 1 + 16 1*. Estimate its spee 
at the end of the first second, exhibiting tho work in a table. 

4. Algebraic method. In this section we shall show how it i 
possible to derive an algebraic formula for the speed, still con 
fining ourselves to the special example of the falling body whos< 

equation of motion is ., , 2 ^ 

s = i.o t . (1 

Instead of taking a definite numerical value for f t , wo dial 
keep the algebraic symbol t r Then 



Also, instead of adding successive small qimntit,ie,s to t to 
get 2 , we shall represent the amount added by the al 
symbol^. That is, , . , . 



and, from (1), s z== 16 ( 

Hence * 2 - ^ = 16^+ A) 2 - 10 tf. 82 

This is a general expression for the distance / t V* in Ffj. 
Now t z -t : = Ti, and therefore the average speed with 
body traverses %P S is represented by tho expression 



It is obvious that if Ji is taken smaller and smidH Iho nver- 
age speed approaches 32 ^ as a limit, In i'aufc, tho quantity 83^ 



ACCELEKATION 9 

satisfies exactly the definition of limit given in 1. For if e 
is any number, no matter how small, we have simply to take 
10 Ti < e in order that the average speed should differ from 32 ^ 
by less than e ; and after that, for still smaller values of h, this 
d iff 01 once remains less than e. 

We have, then, the result that if the space traversed by a 
1 ailing body is given by the formula 



the speed of the body at any time is given by the formula 



It may be well to emphasize that this is not the result which 
would be obtained by dividing s by t. 

EXERCISE 

Find the speed in each of the problems in 3 by the method 
explained in this section. 

5. Acceleration. Let us consider the case of a body which is 
supposed to move so that if s is the distance in feet and t is the 
time in seconds, s = t s (V) 

Then, by the method of 4, we find that if v is the speed in 
feet per second, = 3 i 2 



We see that when t = 1, v = 3 ; when t = 2, v 12 ; when t = 3, 
v = 27 ; and so on. That is, the body is gaming speed with each 
second. We wish to find how fast it is gaining speed. To find 
this out, let us take a specific time 



(,-4. 



The speed at this time we call v lt so that, by (2), 

i= = 3 (4) 2 = 48 ft. per second. 
Take * 2 =5; 

then v = 3 (5) 2 ~ 75 ft. per second. 



10 RATES 

Therefore the body has gained 75 48 = 27 units of speed in 
1 sec. This number, then, represents the average rate at which 
the body is gaining speed during the particular second con- 
sidered. It does not give exactly the rate at which the speed 
is increasing at the beginning of the second, because the rate 
is constantly changing. 

To find how fast the body is gaining speed when ^ = 4, we 
must proceed exactly as we did in finding the speed itself. 
That is, we must compute the gain of speed in a very small 
interval of time and compare that with the time. 

Let us take t.= 4.1. 

a 

Then 2 = 50.43 

and ^^=2.43. 

Then the body has gained 2.43 units of speed hi .1 sec., which 

2 43 
is at the rate of -~- 24.3 units per second. 

Again, take 2 =4.01. 

Then v 2 = 48.2403 

and ^^=.2403, 

A gain of ,2403 units of speed in .01 sec. is at the rate of 
' . =24.03 units per second. We exhibit these results, and 
one other obtained in the same way, in a table: 



4,1 60.43 .1 2.48 24.3 

4.01 48.2403 .01 .2403 24,08 

4.001 48.024003 001 .024003 24.003 

The rate at which a body is gaining speed is called its 
acceleration. Our discussion suggests that in. the example before 
us the acceleration is 24 units of speed per second. But the 
unit of speed is expressed in feet per second, and so we say 
that the acceleration is 24 ft. per second per second. 



KATE OF OHANG-E 11 

By the method used in determining speed, we may get a 
general formula to determine the acceleration from equation (2). 
We take , , r 



Then v a = 

and v z ~Vi= 

The average rate at which the speed is gamed is then 



h, 

and the limit of this, as h becomes smaller and smaller, is 
obviously 6t t . 

This is, of course, a result which is valid only for the special 
example that we are considering. A general statement of the 
meaning of acceleration is as follows: 

. , ,. ,. ., , change in speed 
Acceleration = limit of . 5 

change in time 

EXERCISES 

1. If s = 4 1*, find the speed and the acceleration when t = # r 

2. If s = rf 8 + 1*, find the speed and the acceleration when t = 2. 

3. If s = 3 t z + 2 1 + 5, how far has the body moved at the end 
of the fifth second ? With what speed does it reach that point, and 
how fast is the speed increasing ? 

4. If s = 4 i 8 + 2 1 3 + 1 + 4, find the distance traveled and the 
speed when t = 2. 

5. If s = $ t 6 + 1 -t- 10, find the speed and the acceleration when 
t=z>2 and when t = 3. Compare the average speed and the average 
acceleration during this second with the speed and the acceleration 
at the beginning and the end of the second. 

6. If s = at + &, show that the speed is constant. 

7. If at 2 + to + G, show that' the acceleration is constant. 

8. If s = at* + bt* + ct +/, find the formulas for the speed and 
the acceleration. 

6. Rate of change. Let us consider another example which 
may be solved by processes similar to those used for determining 
speed and acceleration. 



L2 



KATES 



m the 



A stone is thrown into still water, forming ripples which 
,ravel from the center of disturbance in the form of circles 
Tig. 3). Let r be the radius of a circle and A its area. Then 

J "J /-I N 

A=irr. (1) 

We wish to compare changes in the area with changes 
adius. If we take r t = 3, then A^ 9 TT; and if we take 
,hen ^t a = 167T. That is, a change 
)f 1 unit in ?*, when r 3, causes 
i change of 7 TT units in A. We are 
^empted to say that A is increas- 
ng TTT times as fast as r. But 
jefore making such a statement it 
s well to see whether this law holds 
'or all changes made in r, starting 
'rom r 1 = 8, and especially for small 
jhanges in r. 

We will again exhibit the calcu- ^ IQ 3 

ation in the form of a table. Here 

^=3, A^Qnr, and r z is a variously assumed value of r not 
nuch different from 3. 




1 

01 

.001 



9.61 TT 
9.0601 TT 
9,006001 f 



.1 
.01 
001 



A,,- A! 

.01 TT 
.0001 7T 

000001 IT 



r a - r x 

ITT 
6.01 TT 
6 001 TT 



The number in the last column changes with the number 
a r^ Therefore, if we wish to measure the rate at which A is 
3hanging as compared with r at the instant when r = 3, we must 
bake the limit of the numbers in the last column. That limit is 
abviously 6 TT. 

We say that at the instant when r 8, the area of the circle is 
changing 6 TT times as fast as the radius. Hence, if the radius 
Is changing at the rate of 2 ft. per second, for example, the area 
is changing at the rate of 12 TT sq. ft. per second. Another way 
of expressing the saine idea is to say that when, r ts 3, the rate 



RATE OF CHANGE 13 

of change of A with respect to r is 6 TT. Whichever form of expres- 
sion is used, we mean that the change in the area divided by 
the change in the radius approaches a limit 6 TT. 

The number 6 TT was, of course, dependent upon the value 
r = 3, with which we started. Another value of ^ assumed at 
the start would produce another result. For example, we may 
compute that when r x = 4, the rate of change of A with respect 
to r is 8-rr; and when r^= 5, the rate is 10 ?r. Better still, we 
may derive a general formula which will give us the required 
rate for any value of r^. 

To do this take 



Then A 2 = TT (r* + 2 rji + A 2 ) 

and A z - A 1 = TT (2 r x h + 7i 3 ) ; 

A A 

SO that 2 l = 2 TH* + 7i7T. 

T V 
'z 'l 

The limit of this quantity, as Ti is taken smaller and smaller, is 



Hence we see that from formula (1) we may derive the fact 
that the rate of change of A with respect to r is 2 TIT. 

EXERCISES 

1. In the example of the text, if the circumference of the circle 
winch bounds the disturbed area is 10 ft and the circumference is 
increasing at the rate of 3 ft. per second, how fast is the area 
increasing ? 

2. In the example of the text find a general expression for the 
rate of change of the area with respect to the circumference. 

3. A soap bubble is expanding, always remaining spherical. If 
the radius of the bubble is increasing at the rate of 2 in. per second, 
how fast is the volume increasing ? 

4. In Ex 3 find the general expression for the rate of change 
of the volume with respect to the radius. 

5. If a soap bubble is expanding as in Ex. 3, how fast is the 
area of its surface increasing 9 



14 BATES 

6. In Ex. 5 find the general expression for the rate of change 
of the surface with respect to the radius 

7. A cube of metal is expanding under the influence of heat. 
Assuming that the metal retains the form of a cube, find the rate of 
change at which the volume is increasing with respect to an edge. 

8. The altitude of a right circular cylinder is always equal to 
the diameter of the base. If the cylinder is assumed to expand, 
always retaining its form and proportions, what is the rate of change 
of the volume with respect to the radius of the base ? 

9. Find the rate of change of the area of a sector of a circle of 
radms 6 ft with respect to the angle at the center of the circle. 

10. Find the rate of change of the area of a sector of a circle 
with respect to the radius of the circle if the angle at the center 

7T 

of the circle is always j. What is the value of the rate when the 
radius is 8 in. ? 



CHAPTER II 
DIFFERENTIATION 

7. The derivative. The examples we have been considering 
in the foregoing sections of the book are alike in the methods 
used to solve them. We shall proceed now to examine this 
method so as to bring out its general character. 

In the first place, we notice that we have to do with two 
quantities so related that the value of one depends upon the 
value of the other. Thus the distance traveled by a moving 
body depends upon the time, and the area of a circle depends 
upon the radius. In such a case one quantity is said to be 
a function of the other. That is, a quantity y is said to be a 
function of another quantity, x, if the value of y is determined ly 
the value of x. 

The fact that y is a function of x is expressed by the equation 

y=/<v>, 

and the particular value of the function when x has a definite 
value a is then expressed as /(a). Thus, if 

f(x) = x*- 3 aM-42j + l, 
/(2)=2-3(2) 3 +4(2) + l = 5, 
/(0)=0-3(0) + 4(0) + 1 = 1. 

It is in general true that a change in x causes a change in 
the function y, and that if the change in x is sufficiently small, 
the change in y is small also. Some exceptions to this may be 
noticed later, but this is the general rule. A change in a; is 
called an increment of x and is denoted by the symbol Ao? (read 
" delta x "). Similarly, a change in y is called an increment of 
y and is denoted by Ay. For example, consider 



15 



16 DIFFERENTIATION 

When = 2, # = 12. When a? =2.1, y = 12.71. The change 
in x is .1, and the change m y is .71, and we write 

Aa = .l, Ay = .71. 

So, in general, if a^ is one value of a;, and x z a second value 

of #, then 

Ax = xs, i x l1 or #2=0^4- Are; (1) 

and if y x and y 2 are the corresponding values of y, then 

ty=y z -Vv <* y 2 =y a 4Ay. (2) 

The word increment really means "increase," but as we are 
dealing with algebraic quantities, the increment may be nega- 
tive when it means a decrease. For example, if a man invests 
$1000 and at the end of a year has $1200, the increment of his 
wealth is $200. If he has $800 at the end of the year, the 
increment is $200. So, if a thermometer registers 65 in the 
morning and 57 at night, the increment is ,8. The incre- 
ment is always the second value of the quantity considered minus 
the first value. 

Now, having determined increments of x and of y, the next 
step is to compare them by dividing the increment of y by 
the increment of x. This is what we did in each of the three 
problems we have worked in 3-6. In finding speed we began 
by dividing an increment of distance by an increment of time, 
in finding acceleration we began by dividing an increment of 
speed by an increment of time, and in discussing the ripples in 
the water we began by dividing an increment of area by an 
increment of radius. . 

The quotient thus obtained is - That is, 

Ax 

Ay _ increment of y _ change in y 
AJC increment of x change m x 

An examination of the tables of numerical values in 3, 5, 6 
shows that the quotient ^ depends upon the magnitude of Asc, 
and that in each problem it was necessary to determine its limit 



DEEIVATIVE 17 

as A# approached zero. This limit is called the derivative of y 
with respect to x, and is denoted by the symbol -~- We have then 

fy v -j. Ay v , change in v 
~ = limit oi ~ = limit of . . a 
ax Ax change in x 

At present the student is to take the symbol -^ not as a 

dx 

fraction, but as one undivided symbol to represent the deriva- 
tive. Later we shall consider what meaning may be given to 

dx and dy separately. At this stage the form -^ suggests simply 

A (%>*Xs 

the fraction ^, which has approached a definite limiting value, 

The process of finding the derivative is called differentiation, 
and we are said to differentiate y with respect to x. From the 
definition and from the examples with which we began the book, 
the process is seen to involve the following four steps : 

1. The assumption at pleasure of Ax. 

2. The determination of the corresponding Ay. 

Aw 

8. The division of Ay by Aa: to form -*- 

Ax 

4. The determination of the limit approached by the quotient 
in step 3 as the increment assumed in step 1 approaches zero. 

Let us apply this method to finding -J*- when y - - Let x t 

be a definite value of x, and y,- the corresponding value of y. 

x i 

1. Take Ax = 7i. 
Then, by (1), x^x^h. 

2. Then ^l^-J.; 

11 h 

whence, by (2), Ay- 



3. By division, T^ = . . 7 

J Ax asf+AiS 



18 DIEFEBENTIATION 

4. By inspection it is evident that the limit, as h approaches 

zero, is 5 which is the value of the derivative when x-=x*. 

xl 

But x^ may be any value of x ; so we may drop the subscript 1 and 
write as a general formula 

dx 3? 

EXERCISES 

Pind from the definition the derivatives of the following ex- 
pressions : 

6. y-rf + i. 
2 



2. y = *+2a*+i. e -y=2+* 

3. y = a: 4 ~a 8 . 7. y = "+ \x*+ x - 5. 

i =i 8 = 3a;2 + 1 . 

y ~~ x s y ~~ x 

8. Differentiation of a polynomial. We shall now obtain for- 
mulas by means of which the derivative of a polynomial may be 
written down quickly. In the first place we have the theorem : 

The derivative of a polynomial is the sum of the derivatives of 
its separate terms. 

This follows from the definition of a derivative if we reflect 
that the change in a polynomial is the sum of the changes in its 
terms. A more formal proof will be given later. 

We have then to consider the terms of a polynomial, which 
have in general the form #af. Since we wish to have general 
formulas, we shall omit the subscript 1 in denoting the first 
values of x and y. We have then the theorem: 

If y =s ax*) where n is a positive integer and a is a constant, then 



(1) 

dx ^ ' 



To prove this, apply the method of 7 : 
1. Take Aa?=A; 

whence jc a = x + h. 



POLYNOMIAL 19 

2. Then y 3 =ax2 = a(z + A)"; 

whence Ay == a (x + A)" are" 



.4 

3. By division, ^ = a(nx^- l + n ^ n ~ 1 ^ tf-'h + . . . + A"" 1 )- 

ZA3/ j > 

4. By inspection, the limit approached by > as Ji approaches 
zero, is seen to be anx n ~ l . 

Therefore ~ = anx n ~\ as was to be proved. 
dx 

The polynomial may also have a term of the form ax. This 
is only a special case of (1) with w = l, but for clearness we 
say explicitly: 

If y = ax, where a is a constant, then 

- (2) 

dx 

Finally, a polynomial may have a constant term. c. For this 
we have the theorem: 

If y = G, where c is a constant, then 

f^O. (3) 

ax 

The proof of this is that as c is constant, A<? is always zero, 
no matter what the value of A is. Hence 



and therefore = 0. 

ax 

As an example of the use of the theorems, consider 

y 
We write at once 

^ 
ax 



20 DIFFERENTIATION 

EXERCISES 

Find the derivative of each of the following polynomials : 

1. 3.2 + a- 3. 6. x 1 + 7 x s + 21 8 - 14 as 

2. o; 8 +2a; + l. 7. x*- x* + 4a - 1 

3. o: 4 + 4a5 8 +6a; 2 +4a; + l. 8. 3 

4. x 6 + |a: 4 +2a; 2 +3. 9. aa 

5. a; 6 -4a: 4 + a: 2 -4a; 10. a + bx z + ca; 4 + ex 6 . 



i 

9. Sign of the derivative. If ~ is positive, an increase in the 

dx , dy . 

value of x causes an increase in the value of y. If ~ is negative, 

dx 

an increase in the value of x causes a decrease in the value of y. 
To prove this theorem, let us consider that -jj- is positive. 

Then, since ~- is the limit of -> it follows that is positive 
dx A A 

for sufficiently small values of Aa;; that is, if A# is assumed 
positive, ky is also positive, and therefore an increase of x 

causes an increase of y. Similarly, if -j- is negative, it follows 

Aw 
that is negative for sufficiently small values of Aa: ; that is, 

if Arc is positive, A?/ must be negative, so that an increase of x 
causes a decrease of y. 

In applying this theorem it is necessary to determine the 
sign of a derivative. In case the derivative is a polynomial, this 
may be conveniently done by breaking it up into factors and 
considering the sign of each factor. It is obvious that a factor 
of the form x a is positive when x is greater than a, and 
negative when x is less than a. 

Suppose, then, we wish to determine the sign of 



There are three factors to consider, and three numbers are im- 
portant ; namely, those which make one of the factors equal to 
zero. These numbers arranged in order of size are 3, 1, and 6. 
We have the four cases : 

1. x< 3. All factors are negative and the product is 

Dative. 



VELOCITY AND ACCELERATION 21 

2. 3 < x < 1, The first factor is positive and the others 
are negative. Therefore the product is positive. 

3. 1 < x < 6. The first two factors are positive and the last 
is negative. Therefore the product is negative. 

4. x > 6. All factors are positive and the product is positive. 
As an example of the use of the theorem, suppose we have 

# = X s - 3; y*-Q x + 27, 

and ask for what values of x an increase in y will cause an 
increase in y. We form the derivative and factor it. Thus, 



^==3a a -6a3-9==3(a: + r)(a;-3). 

//* 

Proceeding as above, we have the three cases: 

1. x< 1. -^ is positive, and an increase in x therefore 

dx 
increases y. _ 

2. 1 < x < 3. -^ is negative, and therefore an increase in x 
, dx 

decreases y. _ 

3. a; > 3. -^ is positive, and therefore an increase in x in- 
creases y. 

These results may be checked by substituting values of x in 
the derivative. 

EXERCISES 

Pind for what values of x each of the following expressions will 
increase if x is increased, and for what values of x they will decrease 
if x is increased : 



1. a! > -4aj + 6. 6. 

2. 3a a +10a; + 7. 7. x a - x*- 5x + 5 

3. l + Ssc-a: 2 . 8. 1 + 6x +12a: 2 + 8oj 8 . 

4. 7_3a._3a; 2 9. 6 + 60; + 6a a - 2 3 - 3 

5. 2 je 8 +3a; a - 12 + 17. 10. 12 - 12a - 6a; 2 + 4a; 8 + 



10. Velocity and acceleration (continued). The method by 
which the speed of a body was determined in 4 was in reality 
a method of differentiation, and the speed was the derivative of 
the distance with respect to the time. In that discussion, how- 
ever, we SQ arranged each problem that the result was positive 



22 DIFFERENTIATION 

and gave a numerical measure (feet per second, miles per hour, 
etc.) for the rate at which the body was moving. Since we may 
now expect, on occasion, negative signs, we will replace the word 
speed by the word velocity, which we denote by the letter v. 
In accordance with the previous work, we have 

da -, N 

* a) 

The distinction between speed and velocity, as we use the 
words, is simply one of algebraic sign. The speed is the numer- 
ical measure of the velocity and is always positive, but the 
velocity may be either positive or negative. 

From 9 the velocity is positive when the body so moves that 
s increases with the time. This happens when the body moves in 
the direction in which s is measured. On the other hand, the 
velocity is negative when the body so moves that s decreases 
with the time. This happens when the body moves in the direc- 
tion opposite to that in which s is measured. 

For example, suppose a body moves from A to B (Fig. 1), a 
distance of 100 mi., and let P be the position of the body at a 
time t, and let us assume that we know that AP = 4 1. If we 
measure s from A, we have 



ds 

whence v = - = 4. 

dt 

On the other hand, if we measure s from J5, we have 



whence v - = 4. 

dt 

"We will now define acceleration by the formula 

dv 

*=w 

in full accord with 5 ; or, since v is found by differentiating s, 
we may write 



VELOCITY AND ACCELERATION 23 

where the symbol on the right indicates that * is to be differ- 
entiated twice in succession. The result is called a second 
derivative, 

A positive acceleration means that the velocity is increasing, 
but it must be remembered that the word increase is used in 
the algebraic sense. Thus, if a number changes from 8 to 
5, it algebraically increases, although numerically it decreases. 
Hence, if a negative velocity is increased, the speed is less. Simi- 
larly, if the acceleration is negative, the velocity is decreasing, 
but if the velocity is negative, that means an increasing speed. 

There are four cases of combinations of signs which may 
occur : 

1. v positive, a positive. The body is moving in the direction 
in which s is measured and with increasing speed. 

2. v positive, a negative. The body is moving in the direction 
in which s is measured and with decreasing speed. 

8. v negative, a positive. The body is moving in the direction 
opposite to that in which s is measured and with decreasing 
speed. 

4. v negative, a negative. The body is moving in the direc- 
tion opposite to that in which 8 is measured and with increasing 



As an example, suppose a body thrown vertically into the air 
with a velocity of 96 ft. per second. From physics, if s is meas- 
ured up from the earth, we have 



From this equation we compute 

v = 96 - 32 t, 



When t< 3, v is positive and a is negative. The body is going 
up with decreasing speed. When t > 3, v is negative and a is 
negative. The body is coming down with increasing speed. 

On the other hand, suppose a body is thrown down from a 
height with a velocity of 96 ft. per second. Then, if 8 is measured 



24 DIFFERENTIATION 

down from the point from which the body is thrown, we have, 
from physics, 8 = Wt+lQt*, 

from which we compute 

v = 96 + 32 1, 

= 32. 

Here v is always positive and a is always positive. There- 
fore the hody is always going down (until it strikes) with an 
increasing speed. 

EXERCISES 

In the following examples find the expression for the velocity 
and determine when the body is moving in the direction in which 
s is measured and when in the opposite direction . 

1. s = t* 3t + 6. 3. s = t* Qt* + 24* + 3. 

2. s = 10* - tf 4. 's = 8 + 12* - 6i 2 + < 8 . 

5. s = 25 4 - 



In the following examples find the expressions for the velocity 
and the acceleration, and determine the periods of time during which 
the velocity is increasing and those during which it is decreasing . 



6. s = 3z5-4i + 4 8. s = %t* 2 2 

7. s = 1 + 5t - #. 9. s = t* - 5t* + St + 1. 

10. s = 1 + 4 + 2t* t 8 . 



11. Rate of change (continued). In 6 .we have solved a 
problem in which we are finally led to find the rate of increase 
of the area of a circle with respect to its radius. This problem 
is typical of a good many others. 

Let x be an independent variable and y a function of x. 
A change Aa; made in x causes a change Ay in y. The fraction 

compares the change in y with the change in x. For exam- 
ple, if Aa; = .001, and Ay = .009061, then we may say that the 
change in y is at the average rate of ' = 9.061 per unit 

change in x. This does not mean that a unit change in a? would 
actually make a change of 9.061 units in y, any more than the 



RATE OF CHANGE 



25 



statement that an automobile is moving at the rate of 40 mi. an 
hour means that it actually goes 40 mi. in an hour's time. 

The fraction then gives a measure for the average rate at 
which y is changing compared with the change in x. But this 
measure depends upon the value of A#, as has been shown in 
the numerical calculations of 6. To obtain a measure of the 
instantaneous rate of change of y with respect to x which shall 
not depend upon the magnitude of A#, we must take the limit 

of ~ t as we did in G. 
Ax 

We have, therefore, the following definition : 

Tlie derivative -j- measures the rate of change ofy with respect to a\ 
ax 

Another way of putting the same thing is to say that if -^ 
has the value m, then y is changing m times as fast as x. 

Still another way of expressing the same idea is to say that 
the rate of change of y with respect to x is defined as meaning 
the limit of the ratio of a small change 
in y to a small change in x. 

We will illustrate the above general 
discussion, and at the same time show 
how it may be practically applied, by the 
following example, which we will first 
solve arithmetically and then by calculus. 

Suppose we have a vessel in the shape of 
a cone (Fig 4) of radius 3 in. and altitude 
9 in. into which water is being poured at 
the rate of 100 cu. in. per second. Re- 
quired the rate at which the depth of the 
water is increasing when the depth is 6 in. 

From similar triangles in the figure, if h is the depth of the 

water and r the radius of its surface, r -. If Fis the volume 
of water, w , . i , ,^ 




FlG. 4 



We are asked to find the rate at which the depth is increasing 
when h is C in. Let us call that depth 



so that 7^= 6. Then 



26 DIFFERENTIATION 

y = 8 TT. Now we will increase \ by successive small amounts 
and see how great an increase in V^ is necessary to cause that 
change in \ ; that is, how much water must be poured in to raise 
the depth by that amount. The calculation may be tabulated as 
follows . 

A h AF AT 

A /i 

.1 .407 IT 4 07 ir 

01 04007 TT 4.007 v 

.001 0040007 TT 4.0007 TT 

The limit of the numbers in the last column is evidently 4 TT. 
Therefore the volume is increasing 4 IT times as fast as the 
depth. But, by hypothesis, the volume is increasing at the rate 
of 100 cu. in. per second, so that the depth is increasing at the 

rate of - = 7.96 in. per second. 

47T 

We have solved the problem by arithmetic to exhibit again 
the meaning of the derivative. The solution by calculus is much 
quicker. We begin by finding 

dV 1 , 3 

aTe**' 

This is the general expression for the rate of change of V 
with respect to A, or, in other words, it tells us that V is instan- 
taneously increasing \ rf times as fast as li for any given A. 
Therefore, when A = 6, V is increasing 4 TT times as fast as A, 
and as V is increasing at the rate of 100 cu. m. per second, A is 

increasing at the rate of - = 7.96 in. per second. 

4 7T 

EXERCISES 

1. An icicle, which is melting, is always in the form of a right 
circular cone of which the vertical angle is 60 Find the rate of 
change of the volume of the icicle with respect to its length. 

2. A series of right sections is made in a right circular cone of 
which the vertical angle is 90 How fast will the areas of the sec- 
tions be increasing if the cutting plane recedes from the vertex at 
the rate of 3 ft. per second ? 



GRAPHS 27 

3. A solution is being poured into a conical filter at the rate of 
5 cc per second and is running out at the rate of 1 ce. per second 
The radius of the top of the filter is 10 cm. and the depth of the 
filter is 30 cm Find the rate at which the level of the solution is 
rising in the filter when it is one fourth of the way* to the top. 

4. A peg in the form of a right circular cone of which the ver- 
tical angle is 60 is being driven into the sand at the rate of 1 m. 
per second, the axis of the cone being perpendicular to the surface 
of the sand, which is a plane. How fast is the lateral surface of the 
peg disappearing in the sand when the vertex of the peg is 5 in. 
below the surface of the sand? 

5. A trough is in the form of a right prism with its ends equi- 
lateral triangles placed vertically The length of the trough is 10 ft 
It contains water which leaks out at the rate of cu ft per minute. 
Find the rate, in inches per minute, at which the level of the water 
is sinking in the trough when the depth is 2 ft. 

6. A trough is 10 ft. long, and its cross section, which is vertical, 
is a regular trapezoid with its top side 4ft. in length, its bottom 
side 2 ft , and its altitude 5 ft. It contains water to the depth of 
3 ft , and water is running in so that the depth is increasing at the 
rate of 2 ft. per second. How fast is the water running in ? 

7. A balloon is in the form of a right circular cone with a hemi- 
spherical top. The radius of the largest cross section is equal to 
the altitude of the cone. The shape and proportions of the balloon 
are assumed to be unaltered as the balloon is inflated. Find the 
rate of increase of the volume with respect to the total height of 
the balloon. 

8. A spherical shell of ice surrounds a spherical iron ball concen- 
tric with it. The radius of the iron ball is 6 in. As the ice melts, 
how fast is the mass of the ice decreasing with respect to its thickness ? 

12. Graphs. The relation between a variable x and a function 
y may be pictured to the eye by a graph. It is expected that 
students will have acquired some knowledge of the graph in 
the study of algebra, and the following brief discussion is given 
for a review. 

Take two lines OX and 07 (Fig. 5), intersecting at right 
angles at 0, which is called the origin of codrdinates. The line 
called the axis of as, and the line 0rthe axis of y ; together 



28 DIFFERENTIATION 

they are called the coordinate axes, or axes of reference. On OX 
we lay off a distance OM equal to any given value of x, measur- 
ing to the right if x is positive and to the left if x is negative. 
From M we erect a perpendicular MP, equal in length to the 
value of y, measured up if y is positive and down if y is negative. 

The point P thus determmed is said to have the coordinates 
x and y and is denoted by (x, y). It follows that the numerical 
value of x measures the distance Y 

of the point P from OY, and the 
numerical value of y measures the 
distance of P from OX. The coor- 
dinate x is called the abscissa, and 
the coordinate y the ordinate. It is p 

evident that any pair of coordinates 
(x, y) fix a single point P, and that 

any point P has a single pair of M -^ 

coordinates. The point P is said to _, g 

be plotted when its position is fixed 

111 this way, and the plotting is conveniently carried out on 
paper ruled for that purpose into squares. 

If y is a function of x, values of x may be assumed at pleasure 
and the corresponding values of y computed. Then each pair of 
values (x, y) may be plotted and a series of points found. The 
locus of these points is a curve called the graph of the function. 

It may happen that the locus consists of distinct portions not 
connected in the graph. In this case it is still customary to say 
that these portions together form a single curve. 

For example, let - ... _ 

r y~bz ar. (1) 

We assume values of x and compute values of y. The results 
are exhibited in the following table : 

-i o l 2 8 4 6 ' o 



2/-60 4 6 6 4 0-0 

These points are plotted and connected by -a Smooth curve* 
giving the result shown in F-ig. 6. This -curve should have the 



GEAPHS 29 

property that the coordinates of any point on it satisfy equa- 
tion (1) and that any point whose coordinates satisfy (1) lies 
on the curve. It is called the graph both of the function y and 
of the equation (1), and equation (1) is called the equation of 
the curve. 

Of course we are absolutely sure of only those points whose 
coordinates we have actually computed. If greater accuracy is 
desired, more points must be found y 

by assuming fractional values of x. 
For instance, there is doubt as to the 7 . 

shape of the curve between the points 6 / \ 

(2, 6) and (3, C). We take, therefore, 4 ' > 

#=2^- and find / = 6|. This gives 3 ( 

us another point to aid us in draw- 2 

ing the graph. Later, by use of the 1 

calculus, we can show that this last -1 0^1 2 8 4 5 
point is really the highest point of " -2 

the curve. -s 

The curve (Fig. 6) gives us a ~ 4 

graphical representation of the way "^ 

in which y varies with x. We see, for 
example, that when x varies from 1 
to 2, y is increasing; that when x 

varies from 3 to 6, y is decreasing; and that at some point 
between (2, 6) and (3, 6), not yet exactly determined, y has its 
largest value. 

It is also evident that the steepness of the curve indicates in 
some way the rate at which y is increasing with respect to x. 
For example, when # = 1, an increase of 1 unit in x causes an 
increase of 6 units in y ; while when a =1, an increase of 1 unit 
in x causes an increase of only 2 units m y. The curve is 
therefore steeper when x = 1 than it is when x = 1. 

Now we have seen that the derivative - measures the rate 

ax 

of change of y with respect to x. Hence we expect the derivative 
to be connected in some way with the steepness of the curve. 

We shaft tihmwfnro /iianr.. *Hr. ^ QTL m gs \%. and" 15. 

USc Lib B'lore Q 

515N22 ! 13838 

in IIIIIIIIIIIIIIIIIMIII ~ 



30 DIFFERENTIATION 

1 EXERCISES 

Plot the graphs of the following equations : 

1. y = 2x -|-3. 4. ?/ = a; 2 -5a; + 6. 7. y = zc 8 . 

2. y = 2x + 4. 5. y = a; 2 + 4 a; + 8 8. y = x* 4aj a 

3. y 5. 6. y 9 3 a; a; 2 . 9. j/ = a: 8 1. 

10. What is the effect on the graph of y mx + 3 if different 
values are assigned to m ? How are the graphs related ? What 
does this indicate as to the meaning of m ? 

11. What is the effect on the graph of y = 2x + J if different 
values are assigned to b ? What is the meaning of & ? 

12. Show by similar triangles that y = mx is always a straight 
line passing through 0. 

13. By the use of Exs. 11 and 12 show that y = mx + b is always 
a straight line 

13. Real roots of an equation. It is evident that the real roots 
of the equation f(x) determine points on the axis of x at 
which the curve y=f(x) crosses or touches that axis. More- 
over, if x 1 and 2 ( t <* 2 ) are two values of x such that f(x^) 
and/( 2 ) are of opposite algebraic sign, the graph is on one side 
of the axis when x = x i , and on the other side when ic = a; 2 . 
Therefore it must have crossed the axis an odd number of times 
between the points x = z l and x = x z . Of course it may have 
touched the axis at any number of intermediate points. Now, if 
f(x) has a factor of the form (x a) 1 , the curve y =/(#) crosses 
the axis of x at the point x = a when k is odd, and touches the 
axis of x when k is even. In each case the equation /(#) = is 
said to have Jc equal roots, x a. Since, then, a point of crossing 
corresponds to an odd number of equal roots of an equation, and a 
point of touching corresponds to an even number of equal roots, 
it follows that the equation f(x) = has an odd number of real 
roots between x^ and x z if /(#,) and / (:c 2 ) have opposite signs. 

The above gives a ready means of locating the real roots of 
an equation in the form /(z) = 0, for we have only to find two 
values of x, as x l and a? 2 , for which f(x) has different signs. We 
then know that the equation has an odd number of real roots 



STEAIGHT LINE 31 

between these values, and the nearer together x l and # 2 , the 
more nearly do we know the values of the intermediate roots. 
In locating the roots in this manner it is not necessary to con- 
struct the corresponding graph, though it may be helpful. 

Ex. Find a real root of the equation X s + 2 x 17 = 0, accurate to two 
decimal places. 

Denoting x 3 + 2 x 17 by f(x) and assigning successive integial values 
to x, we find/(2) = 5 and/(3) = 16 Hence there is a leal root of the 
equation between 2 and 3. 

We now assign values to x between 2 and 3, at intervals of one tenth, 
as 2 1, 2 2, 2.3, etc., and we begin with the values nearei 2; since /(2) is 
nearer zeio than is/(3). Proceeding in this way we find jf (2.3) = 233 
and/(2 4) = 1.624 , hence the root is between 2.3 and 2.4. 

Now, assigning values to a? between 2 3 and 2.4 at intervals of one hun- 
diedth, we find /(2.31) = - .054 and /(2.32) = 127, hence the root as 
between 2 31 and 2.32. 

To determine the last decimal place accurately, we let x = 2.315 and 
find /(2.315) = .037. Hence the root is between 2 31 and 2 315 and is 
2.31, accurate to two decimal places 

If /(2 315) had been negative, we should have known the root to be 
between 2 315 and 2.32 and to be 2 32, accurate to two decimal places. 

EXERCISES 

Find the real roots, accurate to two decimal places, of the follow- 
ing equations : 

1. a 8 +2a;-6 = 0. 4. j*- 4a 8 + 4 = 0. 

2. a 8 + o;+ll = 5. a: 8 - 3ic 2 + 60: - 11 = 0. 

3. *-lla: + 6 = 0. 6. 



14. Slope of a straight line. Let LK (Figs. 7 and 8) be any 
straight line not parallel to OX GS OY, and let J? (a^, y^ and 
P 2 (# 2 , y 3 ) be any two points on it. If we imagine a point to 
move on the line from P 1 to P^ the increment of x is # 2 x and 
the increment of y is y^y^ We shall define the slope as the 
ratio of the increment ofy to the increment ofx and denote it by m. 

We have then, by definition, 



82 



DIFFERENTIATION 



A geometric interpretation of the slope is easily given. For 
if we draw through JFJ a line parallel to OJf, and through J^ a 
line parallel to OF, and call R the intersection of these lines, 
then x z x l = P 1 E and y^y^RP^ Also, if $ is the angle 
which the line makes with OX measured as in the figure, then 

: tan 0. C ) 



It is clear from the figures as well as from formula (2) that 
the value of m is independent of the two points chosen to define 
it, provided only that these are on the given line. We may there- 
fore always choose the two points so that y^y^ is positive. 









EIG. 7 



FIG. 8 



Then if the line runs up to the right, as in Fig. 7, # 2 a^ is 
positive and the slope is positive. If the line runs down to the 
right, as in Fig. 8, x z o^ is negative and m is negative. There- 
fore the algebraic sign of m determines the general direction in 
which the line runs, while the magnitude of m determines the 
steepness of the line. 

Formula (1) may be used to obtain the equation of the line. 
Let m be given a fixed value and the point I^(x^ y^) be held 
fixed, but let Ji^ be allowed to wander over the line, taking on, 
therefore, variable coordinates (a?, y). Equation (1) may then 
be written y-y^m(x- ^). (3) 

This is the equation of a line through a fixed point (a^, y^ 
with a fixed slope m, since it is satisfied by the coordinates of 
any point on the line and by those of no other point. 



STRAIGHT LINE 33 

In particular, P^ (x^ y^) may be taken as the point with coor- 
dinates (0, i) in which the line cuts OY. Then equation (3) 
becomes y = mx + l. (4) 

Since any straight line not parallel to OX or to OY intersects 
OY somewhere and has a definite slope, the equation of any 
such line may be written in the form (4). 

It remains to examine lines parallel either to OX or to OF. If 
the line is parallel to OX, we have no triangle as in Figs. 7 and 8, 
but the numerator of the fraction m (1) is zero, and we there- 
fore say such a line has the slope 0. Its equation is of the form 

y = , (5) 

since it consists of all points for which this equation is true. 

If the line is parallel to OY, again we have no triangle as in 
Figs. 7 and 8, but the denominator of the fraction in (1) is zero, 
and m accordance with established usage we say that the slope of 
the line is infinite, or that m=oo. This means that as the position 
of the line approaches parallelism 
with OY the value of the fraction 
(1) increases without limit. The 
equation of such a line is 

x = a. (6) 

Finally we notice that any equa- 
tion of the form 

^ + 5y + (7=0 (7) 

always represents a straight line. This follows from the fact that 
the equation may be written either as (4), (5), or (6). 

Tbe line (7) may be plotted by locating two points and 
drawing a straight line through them. Its slope may be found 
by writing the equation in the form (4) when possible. The 
coefficient of x is then the slope. 

If two lines are parallel they make equal angles with OX. 
Therefore, if m l and m z are the slopes of the lines, we have, 
from (2), m z =m,. (8) 

If two lines are perpendicular and make angles <^ and ^> s 
respectively with OX, it is evident from Fig. 9 that < a = 90+ ^ ; 




34 DIFFERENTIATION 

whence tan $ == ~ cot $ = Hence, if m^ and m a are th 

slopes of the lines, we have 




It is easy to show, conversely, that if equation (8) is satis 
fied by two lines, they are parallel, and that if equation (9) i 
satisfied, they are perpendicular. Therefore equations (8) am 
(9) are the conditions for parallelism and perpendiculant 
respectively. 

Ex. 1. Find the equation of a straight line passing through the pom 
(1, 2) and parallel to the straight line determined by the two points (4, 2 
and (2, - 3) 

By (1) the slope of the line determined by the two points (4, 2) am 

_ g _ 2 5 

(2, 3) is = - Therefore, by (3), the equation of the requiiei 

line is n R / i\ 

y - 2 = i (a; - 1), 

which i educes to x 2y 1 = 0. 

Ex. 2. Find the equation of a straight line through the point (2, 3 
and perpendicular to the line 2 a: 3 ?/ + 7 = 

The equation of the given straight line may be written in the forn 
y = 3 x + , which is form (4). Therefore m = . Accordingly, by (0) 
the slope of the required line is By (3) the equation of the requim 
line is 

y + 3 = -|(a;-2), 
which reduces to 3 x + 2 y - 0. 

Ex. 3. Find the equation of the straight line passing through the poinl 
( 3, 3) and the point of intersection of the two lines 2 a? y 3 = anr 



The coordinates of the point of intersection, of the two given lines musl 
satisfy the equation of each line. Therefore the codrdinates of the poinl 
of intersection are found by solving the two equations simultaneously, 
The result is (1, - 1) 

We now have the problem to pass a straight line through the points 

(- 3, 3) and (1, - 1). By (1) the slope of the required line is J* + 1 = - 1. 
Therefore, by (3), the equation of the line is 



which reduces to x + y = 0. 



STRAIGHT LINE 35 

EXERCISES 

1. Find the equation of the straight line which passes through 
(2, - 3) with the slope 3 

2. Find the equation of the straight line which passes through 
( 3, 1) with the slope $ 

3. Find the equation of the straight line passing through the 
points (1, 4) and (f , ). 

4. Find the equation of the straight line passing through the 
points (2, 3) and (- 3, - 3). 

5. Find the equation of the straight line passing through the 
point (2, 2) and making an angle of 60 with OX. 

6. Find the equation of the straight line passing through the 
point Q, |) and making an angle of 135 with OX. 

^7. Find the equation of the straight hue passing through the 
point ( 2, 3) and parallel to the line x + 2 y + 1 = 0. 

8. Find the equation of the straight line passing through the 
point ( 2, 3) and perpendicular to the line 3 a; + 4 ?/ 12 =0. 

9. Find the equation of the straight line passing through the 
point (, ) and parallel to the straight line determined by the two 
points (f, |) and (J, - ) 

10. Find the equation of the straight line passing through 
(i ~ i) an( ^ perpendicular to the straight line determined by the 
points (2, 1) and (- 3, 5). 

11. If /? is the angle between two straight lines which make angles 
#j and < 2 (<j!> 2 > <j) respectively with OX, prove from a diagram similar 
to Fig 9 that /3 = <f> 2 < r If tan t = m^ and tan tj> z = w a , prove by 
trigonometry that 

tan ft = . f l 
1 + m 2 Wj 

12. Find the angle between the lines # 2?/-fl=;0 and 



13. Find the angle between the lines 2x 4^ + 5 = and 

y 6= 0. 

14. Find the angle between the lines y = Sas + 4 and a?H-3y-f-7=0. 

15. The vertex of a right angle is at (2, 4) and one of its sides 
passes through the point (- 2, 2). Find the equation of the other side. 

16. Find the foot of the perpendicular from the origin to the line 



36 DIFFERENTIATION 

15. Slope of a curve. Let An (Fig. 10) be any curve serving 
as the graphical representation of a function of a\ Let 1\ be any 
point on the curve with coordinates y; l = OM^ t //,= yl/j/J. TU\o 
l^xM^ and draw the perpendicular J/ a /j, fixing the point / a ' 
on the curve with the coor- 

M in Q^OQ *T* (iT(/T Q/ " 1\/T T* * s O 

LiiixajUco ^jj"""" '-'' 2* ,yjj """ -*'*j*j" jr ^r> 

Draw ^J? parallel to OX. /R 

Then 

PR W.M".. = ArK. p 

V" -K 

7jj J5 = Jlf.,Pa M, 1 1 = A?/, ^ 

\ a , r>t> / 

and 




Ax 

* / 

Draw the straight line / ^ ^ 

j, prolonging it to form a JT IU . to 

secant J? Then, by 14, -^ is the slope of the secant J>8, and 

may be called the average slope of tho curve between the points 
JJ and P v 

To obtain a number which may be used for tho actual slope 
of the curve at the point / it is necessary to uso tho limit 
process (with which the student should now bo familiar), by 
which we allow kx to become smaller and smaller and the point 
P z to approach P l along the curve. Tho result is tho derivative 
of y with respect to #, and we have tho following result : 

The slope of a curve at any point u yiven, ly the value of tfo 

derivative -^ at that point. 

dx f 

As this limit process takes place, tho point / approaching t!u 
point Py it appears from the ligure thai, the secant 7A' approaches 
a limiting position 1{T. The line 1\T is called a tanymt to 1,lu> 
curve, a tangent Iriny then by deflnitwn the Una approached an tt 
limit by a secant throuc/h two point* of the curve a tht* two point* 
approach coincidence. It follows that tho slope* of tho tangent is 
the limit of the slope of the secant, Therefore, 



slope of a curve at any point in the mme an the dope of tfta 
*angent at that point, 



SLOPE OF A CUEVE 37 

From this and 9 AVO may at once deduce the theorem : 

If the derivative is positive, the curve runs up to the right. 
If the derivative is negative, the curve runs down to the right. Jf 
the derivative is zero, the tangent to the curve is parallel to OX. 
If the derivative is infinite, the tangent to the curve is perpendic- 
ular to OX. 

dif 

The values of x which make ~~ zero or infinite are of par- 

dx 

ticular interest m the plotting of a curve. If the derivative 
changes its sign at such a point, the curve will change its cliiec- 
tion from down to up or from up to down. Such a point will 
be called a turning-point. If y is an algebraic polynomial, its 
derivative cannot be infinite; so we shall be concerned in this 
chapter only with turning-points for which 

^ = 0. 
Ax 

They are illustrated m the two following examples: 

Ex. 1. Consider equation (1) of 12, 
y = 5 x a; 8 . 

Plere ^ = 5 - 2 a: = 2^- 

dx \2 

Equating ~ to zero and solving, we have x = - as a possible turning- 

point It is evident that when x<- , is positive, and when #>- ~ is 

2 dx 2 dx 

negative Therefoie x = \ corresponds to a turning-point of the curve at 
which the lattei changes its direction from up to down It may be called 
a high point of the cmve 

Ex. 2. Consider 



Here ^ = |(* a - 2* - 3) = |(a: - 3) (a + 1). 

(IX o o 

Equating f -~- to zero and solving, we have x = 1 and x = 3 as possible 

dv 
turning-points From the factored form of -^, and reasoning as in 9, 

we see that when y < 1, ~ is positive ; when 1< c < 8, ^r is negative ; 



38 



DIFFERENTIATION 



when x > 3, is positive. Therefore both x 1 and x = 3 give turmng- 

dx 

points, the former giving a high point, and the latter a low point- 
Substituting these values of x in the equa- 
tion of the curve, we find the high point to 
be ( lj 4f ) and the low point to be (3, |). 
The graph is shown in Fig 11 

It is to be noticed that the solu- 
tions of the equation do not 
^ dx 

always give turning-points as illus- 
trated in the next example. 

Ex. 3. Consider 
Here 




dx 



= x z - 6 x + 9 = (x 3) 2 







Solving = 0, we have x = 3 ; but since the derivative is a perfect 
dx 

square, it is never negative Therefore x = 3 does not give a turning-point, 
although when x = 3 the tangent to the curve is 
parallel to OX. The curve is shown in Fig. 12. 

The equation of the tangent to a curve 
at a point (a^, y x ) is easily written down 

We let ( -^ ) represent the value of -^ at 
VaaJ/! ,, . dx 

the point (a^, y x ). Then m = f-^j, and, n I, I X 



from (3), 14, the equation of the tangent is 




Ex. 4. Find the equation of the tangent at (1, 1) to the curve 



We have 
and 



dx 



Therefore the equation of the tangent is 



wljich reduces to 



1 = 0. 



SECOND DERIVATIVE 

From (2), 14, it also follows that if $ is the angle which 
the tangent at any point of a curve makes with OX, then 



EXERCISES 
Locate the turning-points, and then plot the following curves : 

1. 2/ = 3a a -f-4a: + 4. 4. 2/ = a- 8 - 6 # a + 9 a; + 3. 

2. ?/=3 + 3a!-2a; a . 5. 2/ = (2o; 8 + 3a: 2 -12a;- 20) 

3. 2/ = ce 8 -3a: 2 -j-4. 6. ?/ = 2 + 9ce -f 3a; a x* 

' 7. Find the equation of the tangent to the curve y 3 2x -f- a 2 
at the point for which x = 2 

8. Find the equation of the tangent to the curve 2/=l+3a: as 2 3a; 8 
at the point for which x = 1. 

9. Find the points on the curve y = a 8 + 3a: 2 3a; + 1 at which 
the tangents to the curve have the slope 6 

10. Find the equations of the tangents to the curve 

y = x s + 2x* - x + 2 
which make an angle 135 with OX 

11. Find the equations of the tangents to the curve y =a5 8 -f-a; a 2x 
which are perpendicular to the line 3: + 2y + 4 = 

12. Find the angle of intersection of the tangents to the curve 
y = oj 8 -}- as 2 2 at the points for which x = 1 and x = 1 respectively. 

16. The second derivative. The derivative of the derivative 
is called the second derivative and is indicated by the symbol 

-=-(] or -, We have met an illustration of this in the 
ax \dx/ ax* 

case of the acceleration. We wish to see now what the second 
derivative means for the graph. 

Since ~ is equal to the slope of the graph, -we have 
dx 

d*y d , , ^ 
j& ^ -=- (slope). 
dor dx 




40 DIFFERENTIATION 

From this and 9 we have the following theorem : 

If the second derivative is positive, the slope is increasing as x 
increases ; and if the second derivative is negative, the slope is de- 
creasing as x increases. 

We may accordingly use the second derivative to distinguish 
between the high turning-points and the low turning-points of 
a curve, as follows: 

If, when x a, - = and -r is positive, it is evident that 

/7 du 

-^ is increasing through zero ; hence, when x < a, -~ is nega- 

dx d <lr 

tive, and when x > a, -j- is positive. The point for which x = a 

Cu3s 

is therefore a low turning-point, by 9. 

d'ii d^y 

Similarly, if, when x = a, -^- and -~ is negative, it is 

du 
evident that -& is decreasing through zero ; hence, when x < a, 

j CbX J 

-Z- is positive, and when x > a, -?- is negative. The point for 
dx dx 

which x = a is therefore a high turning-point of the curve, by 9. 
These conclusions may be stated as follows: 

If -j- and -=-* is positive at a point of a curve, that point 

^ "^ " rlai fj^itl 

is a low point of the curve. If -J- = and -~ is negative at a 

CliK ClX 

point of a curve, that point is a high point of the curve. 

In addition to the second derivative, we may also have third, 

d?u d^ti 
fourth, and higher derivatives indicated by the symbols -,'{> '^ 

etc. These have no simple geometric meaning. 

EXERCISES 

Plot the following curves after determining tlieir high and low- 
points by the use of -^ and -r : 
dx dor 

= 3a; 8 x 2 3. y = 7 18x 3a- a H- 4,r 8 , 



MAXIMA AND MINIMA 41 

17. Maxima and minima. If f(a) is a value of /() which 
is greater than the values obtained either by increasing or by 
decreasing x by a small amount, /(ff) is called a maximum value 
of f(p). If /() is a value of /(#) which is smaller than the 
values of f(j) found either by increasing or by decreasing x by 
a small amount, /(a) is called a minimum value of /(a?). 

It is evident that if we place 



and make the graph of this equation, a maximum value of /(of) 
occurs at a high point of the curve and a minimum value at a 
low point. From the previous sections we have, accordingly, 
the following rule for finding maxima and minima: 

To find the values of x which give maximum or minimum lvalues 
of y, solve the equation 

^ = 0. 
dx 

If x = a is a root of this equation, it must bo tested to see 
whether it gives a maximum or minimum, and which. We have 
two tests : 

TEST I. If the sign of -^ changes from + to as x increases 

Cv*& 

through a, then & = a (jives a maximum value of y. If the sign of 

-~- changes from to + as x increases through a, tJien x = a gives 
(K& 

a minimum value of y. 

H'U d^ii 

TEST II. If x = a makes -^ = and -~~ netHttive. then # = # 

dx do* , , 

/y y / ff *aj 

gives a maximum value of y, Jf x a mal(e -~ = and -~~. 

ty^fj cfj 

positive, then x a gives a minimum value of y. 

Either of these tests may be applied according to convenience, 
It may be noticed that Test I always works, while Test II fails 

d\i 
to give information if ---^ = when x a. It is also frequently 

CtJs 

possible by the application of common sense to a problem to 
determine whether the result is a maximum or minimum, and 
neither of the formal tests need then be applied. 



42 DIFFERENTIATION 

Ex. 1. A rectangular box is to be formed by cutting a square from 
each, corner of a rectangular piece of cardboard and bending the resulting 
figure. The dimensions of the piece of cardboard being 20 in by 30 m., 
required the largest box which can be made. 

Let x be the side of the square cut out. Then, if the cardboard is bent 
along the dotted lines of Fig 13, the dimensions of the box are 30 - 2 x, 
20 2 x, x. Let V be the volume of the box. 
Then p = (20 - 2 a:) (30 - 2 *) 

8Q&3D 



7V 

= 600 - 200 x + 12 a; 2 . "j 

dx 



fiy I - . ------- 1 - 

Equating ___ to zero, we have 



13 



whence x = 25 5 = 3.9 or 12 7. 



The result 12.7 is impossible, since that amount cannot be cut twice 
fiom the side of 20 in The result 3.9 corresponds to a possible maximum, 
and the tests are to be applied. 

dV 

To apply Test I we write in the factored form 
dx 



dV 

when it appears that - changes from + to , as x increases through 3.9 
ax 

Hence x = 3.9 gives a maximum value of V. 

d s V 

To apply Test II we find - = - 200 + 24 x and substitute x = 3.9. 

dyr 

The result is negative. Therefore x = 3.9 gives a maximum value of V. 

The maximum value of V is 1056 + cu. in., found by substituting 
x =s 3.9 in the equation for V. 

Ex. 2. A piece of wood is in the form of a right circular cone, the 
altitude and the radius of the base of which are each equal to 12 in. What 
is the volume of the largest right circular cylinder that can be cut from 
this piece of wood, the axis of the cylinder to coincide with the axis of 
the cone ? 

Let x be the radius of the base of the required cylinder, y its altitude, 

and V its volume. Then 

V= vx*y. (1) 

We cannot, however, apply our method directly to this value of V, since 
it involves two variables x and y. It is necessary to find a connection 



MAXIMA AND MINIMA 43 

between x and y and eliminate one of them. To do so, consider Fig. 14, 
which is a cross section of cone and cylinder. From smiilai triangles we have 

FE _AD t 
EC DC' 



that18 ' 
whence 



y 12 x. 
Substituting in (1), we have 



whence 



dV 
dx 



24 vx 8 irx*. 



dV 



Equating to zero and solving, we find 

dx B 

x = or 8. The value x = is evidently not a 
solution of the problem, but x = 8 is a possible 
solution. 

Applying Test I, we find that as x increases through the value 8, 




dx 



changes its sign from + to . Applying Test II, we find that 



24 ir 6 irx is negative when x = 8. Either test shows that x = 8 

(IX 

corresponds to a maximum value of V. To find V substitute x = 8 in the 
expression for V. We have V = 256 ir cu. in. 



EXERCISES 

1. A piece of wire of length 20 in. is bent into a rectangle one 
side of which is a:. Find the maximum, area. 

2. A gardener has a certain length of wire fencing with which to 
fence three sides of a rectangular plot of land, the fourth side being 
made by a wall already constructed. Required the dimensions of 
the plot which contains the maximum area. 

3. A gardener is to lay out a flower bed in the form of a sector 
of a circle. If he has 20 ft. of wire with which to inclose it, what 
radius will he take for the circle to have his garden as large as 
possible ? 

4. In a given isosceles triangle of base 20 and altitude 10 a rec- 
tangle of base x is inscribed. Find the rectangle of maximum area. 

6. A right circular cylinder with altitude 2 as is inscribed in a 
sphere of radius a. Find the cylinder of maximum, volume, 



44 DIFFEEENTIATION 

6. A rectangular box with a square base and open at the top is 
to be made out of a given amount of mateiial. If no allowance is 
made for the thickness of the material or for waste in construction, 
what are the dimensions of the largest box that can be made ? 

7. A piece of wire 12 ft. in length is cut into six portions, two 
of one length and four of another Each of the two former portions 
is bent into the form of a square, and the corners of the two squares 
are fastened together by the remaining portions of wire, so that the. 
completed figure is a rectangular parallelepiped Find the lengths 
into which the wire must be divided so as to produce a figure of 
maximum volume 

8. The strength of a rectangular beam, varies as the product oE 
its breadth and the squaie of its depth Find the dimensions of the 
strongest rectangular beam that can be cut from a circular cylindri- 
cal log of radius a inches 

9. An isosceles triangle of constant perimeter is revolved about 
its base to form a solid of revolution. What are the altitude and 
the base of the triangle when the volume of the solid generated is 
a maximum ? 

10 The combined length and girth of a postal parcel is 60m. 
Find the maximum volume (1) when the parcel is rectangular with 
square cross section , (2) when it is cylindrical 

11. A piece of galvanized iron 5 feet long and a feet wide is to be 
bent into a U-shaped water drain I feet long. If we assume that the 
cross section of the drain is exactly represented by a rectangle on 
top of a semicircle, what must be the dimensions of the rectangle 
and the semicircle in order that the drain may have the greatest 
capacity (1) when the drain is closed on top 9 (2) when it is open 
on top ? 

12. A circular filter paper 10m. in diameter is folded into a 
right circular cone. Find the height of the cone when it has the 
greatest volume 

18. Integration. It is often desirable to reverse the process of 
differentiation. For example, if the velocity or the acceleration 
of a moving body is given, we jftay wish to find the distance 
traversed ; or if the slope of a curve is given, we may wish to 
find the curve. 



INTEGKATION 45 

The inverse operation to differentiation is called integration, 
and the result of the operation is called an integral. In the case 
of a polynomial it may be performed by simply working the 
formulas of differentiation backwards. Thus, if n is a positive 
integer and , 



then 



The first term of this formula is justified by the fact that if it 
is differentiated, the result is exactly aa?. The second term is 
justified by the fact that the derivative of a constant is zero. 
The constant may have any value whatever and cannot be 
determined by the process of integration. It is called the constant 
of integration and can only be determined in a given problem by 
special information given in the problem. The examples will 
show how this is to be done. 

Again, if 

^ = , 
dx 

then ;/ = ax + C. (2) 

This is only a special case of (1) with n = 0. 
Finally, if 



= a Q r n + a^" l + + a n ^x + a n , 
ax 



C8) 



Ex. 1. The velocity v with which a body is moving along a straight 
line AB (Fig. 15) is given by the equation 



How far will the body move in the J'IG. 15 

time from * = 2 to * = 4? 

If when t = 2 the body is at P v and if when t = 4 it is at P B , we are 
to find P,P S . 

(IV 

By hypothesis, -j = 10 1 + 5. 

Therefore s - B i* + 5 1 + C. (1) 



46 



DIFFERENTIATION 



have first to determine C. If s la measured from P v it follows that 
when t _ 2> s - o. 

Therefore, substituting in (1), we have 

= 8(2) 2 + 5 (2) + C; 
whence C=-42, 

and (1) becomes s = 8 i 3 + 5 1 - 42. (2) 

This is the distance of the body from P l at any time /. Accordingly, il> 
remains for us to substitute t = 4 m (2) to find the loquiied distance 1\I\, 
Thei e results _ 2 _ 42 _ 106> 



If the velocity is in feet per second, the required distance is in foot. 

Ex. 2. Required the curve the slope of which 
at any point is twice the abscissa of the point. 



By hypothesis, -j- = 2 x 

CIX 

Therefore y = JB* + C 



(1) 



Any curve whose equation can be derived 
fiom (1) by giving C a definite value satisfies 
the condition of the problem (Fig. 16) If it 
is required that the curve should pass through 
the point (2, 3), we have, from (1), 

3 = 4 + <?; 

whence C = 1, 

and therefore the equation of the curve is 




PICK 10 



But if it is required that the curve should pass through ( 8, 10), 
we have, from (1), 10 = 9 + C; 

whence C = 1, 

and the equation is y = 2 + 1 

EXERCISES 

In the following problems v is the velocity, in feet per second, of 
a moving body at any time t 

l. If v = 32 1 + 30, how far will the body move in the time from 



AEEA 47 

2. If v = 3 1* + 4 1 + 2, how far will the body move in the time 
from t = 1 to t = 3 f 

3. If v = 20 + 25, how far will the body move in the fourth 
second ? 

4. If v = t 2 2 1 + 4, how far will the body move in the fifth 
and sixth seconds ? 

5. If v 192 32 1, how far will the body move before v 9 

6. A curve passes through the point (1, 1), and its slope at 
any point (x,y) is 3 more than twice the abscissa of the point. 
What is its equation ? 

7. The slope of a curve at any point (aj, y) is Go; 2 + 2x 4, and 
the curve passes through the point (0, 6) What is its equation ? 

8. The slope of a curve at any point (x, y) is 4 3# jc 2 , and 
the curve passes through the point ( 6, 1). What is its equation ? 

9. A curve passes through the point (5, 2), and its slope at any 
point (05, y) is one half the abscissa of the point What is its equation ? 

10. A curve passes through the point ( 2, 4), and its slope at 
any point (x, y) is a; 2 x + 1. What is its equation ? 

19. Area. An important application of integration occurs in 
the problem of finding an area bounded as follows : 

Let US (Fig. 17) be any curve with the equation y =/(&), and 
leitkED and BC be any two ordinates. It is required to find the 
area bounded by the curve RS, y 
the two ordinates J3D and J3C, 
and the axis of x. 

Take -MP, any variable or- 
dinate between MD and J3<7, 
and let us denote by A the 
area EMPD bounded by the 
curve, the axis of #, the fixed 
ordinate J8I>, and the variable Q jj jf jy" B -^ 

ordinate MP. 

jUlG J7 

It is evident that as values 

are assigned to a; =OJf, different positions of MP and correspond- 
ing values of A are determined. Hence A is a function of x for 

dA 

which we will find -r 
ax 




48 DIFFERENTIATION 

Take MN=Ax and draw the corresponding ordinate NQ. 
Then the area MNQP=&A. If L is the length of the longest 
ordinate of the curve between MP and NQ, and s is the length 
of the shortest orduiate in the same region, it is evident that 

s A# < Au4 < L Aa;, 

for L&x is the area of a rectangle entirely surrounding AJ, and 

is the area of a rectangle entirely included m A^t. 
Dividing by A&, we have 

. A.1 r 
s < < i. 
Aa; 



As Aa; approaches zero, JVQ approaches coincidence with MP, 
and hence * and L, which are always between NQ and MP, 
approach coincidence with MP. Hence at the limit we have 

~ = MP^ y =/(*). (1) 

Therefore, by integrating, 

A=F(x) + C, (2) 

where F(x) is used simply as a symbol for any function whose 
derivative is /(#) 

We must now find C. Let OE = a. When MP coincides with 
ED, the area is zero. That is, when 

x = a, A = 0. 
Substituting in (2), we have 



whence C = F(a), 

and therefore (2) becomes 

^=^(00-^(0). (8) 

Finally, let us obtain the required area JSBCD. If OB ft, thin 
will be obtained by placing x = b in (3). Therefore we have, 

finally> JFta). (4) 



AEEA 49 

In solving problems the student is advised to begin with 
formula (1) and follow the method of the text, as shown in 
the following example: 

Ex. Find the area bounded by tho axis of x, the curve y = ^x z , and the 
ordmates x = 1 and x = 3. 

In Fig 18, BE is the line x = l, CD is the line x = 3, and the required 
area is the a.re&J3CDE Then, by (1), 

~dx = S X *' 
whence A ~ \ x 8 + C 

' When a; = 1, A = 0, and therefore 

whence C = -J, 
and A = J a: 3 % 

Finally, when ^ 

x s=3, 




EXERCISES 

1. Find the area bounded by the curve y = 4 a 1 a; 2 , the axis of 
x, and the lines x = 1 and a: = 3. 

2. Find the area bounded by the curve y = s* + 8 a; + 18, the 
axis of x, and the lines x = 6 and x = 2. 

3. Find the area bounded by the curve y = 1C + 12 x x s , the 
axis of x, and the lines x = 1 and a; = 2 

4. Find the area bounded by the curve y + a; 2 9 = and the 
axis of x. . 

5. Find the area bounded by the axis of x and the curve 
y = 2 x a; 2 . 

6. Find the smaller of the two areas bounded by the curve 
y = 5 x* x a , the axis of x, and the line x = 1. 

7. Find the area bounded by the axis of x, the axis of y, and the 
curve Ayx 3 605 + 9. 

8. Find the area bounded by the curve y = x*~ 2 3 <taj + 8 
and the axis of x. 



50 DIFFERENTIATION 

9. If A denotes the area bounded by the axis of y, the curve 
a; =/(y), a fixed line y ~ b, and any variable line parallel to OX, 
prove that 7 



10. If A denotes an area bounded above by the curve //=/(*")> 
below by the curve y = F(x), at the left by the fixed line x = (t, 
and at the right by a variable ordinate, prove that 



20. Differentials. The derivative has been denned as the limit 

of and has been denoted by the symbol ~. This symbol 
Aa? clx 

is in the fractional form to suggest that it is the limit of a 
fraction, but thus far we have made no attempt to treat it as 
a fraction. 

It is, however, desirable in many cases to treat the derivative 
as a fraction and to consider dx and dy as separate quantities. 
To do this it is necessary to define dx and dy in such a manner 
that their quotient shall be the derivative. We shall begin by 
defining dx, when x is the independent variable; that is, the 
variable whose values can be assumed independently of any oilier 
quantity. 

We shall call dxth& differential of x and define it as a change 
in x which may have any magnitude, but which is generally 
regarded as small and may be made to approach aero as a 
limit. In other words, the differential of the independent variable 
x is identical with the increment of x ; that is, 

dx sss A*. (1) 

After dx has been defined, it is necessary to define dy so 
that its quotient by dx is the derivative. Therefore, if y =/(V) 

and -j- =/'(;), we have 

dx dy 

t 

That is, the differential of the function y is equal to the derivative 
times the differential of the independent variable x. 



DIFFERENTIALS 51 

In equation (2) the derivative appears as the coefficient of dx. 
For this reason it is sometimes called the diff&renticd coefficient. 

It is important to notice the distinction between dy and Ay. 
The diff erential dy is not the limit of the increment A#, since both 
dy and ky have the same limit, zero. Neither is Ay equal to a very 
small increment A#, since it generally differs in value from Ay. 
It is true, however, that when dy and Ay both become small, they 
differ by a quantity which is small compared with each of them. 
These statements may best be under- 
stood from the following examples : 

Ex. 1. Let A be the area of a square with 
the side x so that 



If a; is increased by Ao; = dx, A is increased 
by A.4, where 

A4 = (a? + dxf - x* = 2 x dx + (dx) a . & ^ 

Now, by (2), dA = 2zdx, FNJ. ig 

so that A.I and dA differ by (dx)' 2 

Referring to Fig 19,. we see that dA is represented by the rectangles (1) 
and (2), while A.4 is represented by the rectangles (1) and (2) together 
with the square (3) , and it is obvious from the figuie that the square (8) 
is very small compared with the rectangles (1) and (2), provided djc is 
taken small. For example, if x = 5 and dx = .001, the rectangles (1) 
and (2) have together the area 2 z die = .01 and the square (3) has the 
area .000001. 

Ex. 2. Let s - 16 t s , 

where $ is the distance traversed by a moving body in the time t 
If t is increased by Al = dt, we have 

As = 10 (t + dt)* - 1C & - 32 tdi + 16 (dt}\ 
and, from (2), ds = 32 1 dt ; 

so that As and ds differ by 16 (dt) z The terra 16 (rf/) a is very small com- 
pared with the term BZtdt, if dt is small. For example, if f = 4 and 
dt - .001, then 32 tdt = ,128, while 16 (eft) 2 .000016 

In this problem As is the actual distance traversed in the time dt, and 
da is the distance which would have been traversed if the body had moved 
throughout the time dt with the same velocity which it had at the begin- 
ning of the time dt. 



52 



DIFFERENTIATION 



In general, if y=f(%) and we make a graphical representa- 
tion, we may have two cases as shown in Figs. 20 and 21. 

In each figure, WN = PR &v = dx and RQ=ky, since HQ 
is the total change in y caused by a change of dx = M.N in x. 
If PT is the tangent to the curve at P, then, by 15, 



so that, by (2), dy = (tan HPT) (P7?) = R T. 





R 



M 



N 



C M 



N 



FIG. 20 



PIG 21 



In Fig. 20, dy < A#, and in Fig. 21 dy >&y; but in each case 
the difference between dy and Ay is represented m magnitude 
by the length of QT. 

This shows that RQ = Ay is the change in y as the point JP is 
supposed to move along the curve /=/(^') while RT dy is 
the change in the value of y as the point P is supposed to move 
along the tangent to that curve. Now, as a very small arc docs 
not deviate much from its tangent, it is not hard to see graphi- 
cally that if the point Q is taken close to P, the difference between 
RQ and RT, namely, QT, is very small compared with RT. 

A more rigorous examination of the difference between the 
increment and the differential lies outside the range of this book. 

EXERCISES 

1. If y = x 6 - 3ic 2 + ix + 1, find dy. 

2. If y = x 4 + 4 8 - x* + 6.-B, find dy. 

3. If V is the volume of a cube of edge x, find both AF and dV 
and interpret geometrically. 



APPROXIMATIONS 53 

4. If A is the area of a circle of radius r, find both Avl and dA. 
Show that A.4 is the exact area of a ring 1 of width dr, and that dA 
is the product of the inner circumference of the ring by its width. 

5. If V is the volume of a sphere of radius r, find AF and dV. 
Show that A V is the exact volume of a spherical shell of thickness 
dr, and that dV is the product of the area of the inner surface of 
the shell by its thickness. 

6. If A is the area described in 19, show that dA = ydx Show 
geometrically how this differs from A/i 

7. If s is the distance traversed by a moving body, t the time, 
and v the velocity, show that ds = vdt. How does ds differ from A& 9 

8. If y = x z and x = 5, find the numerical difference between 
dy and A?/, with successive assumptions of dx = 01, dx, = .001, and 
dx = .0001. 

9. If y = x 8 and x = 3, find the numerical difference between dy 
and Ay for dx = .001 and for dx = 0001 

10. For a circle of radius 4 in. compute the numerical difference 
between dA and AJ. corresponding to an increase of r by .001 in. 

11. Eor a sphere of radius 3 ft. find the numerical difference 
between dV and A V when r is increased by 1 in. 

21. Approximations. The previous section brings out the fact 
that the differential of y differs from the increment of y by a 
very small amount, which becomes less the smaller the incre- 
ment of x is taken. The differential may be used, therefore, to 
make certain approximate calculations, especially when the ques- 
tion is to determine the effect upon a function caused by small 
changes in the independent variable. This is illustrated in the 
following examples: 

Ex. 1. Find approximately the change in the area of a square of side 
2 in. caused by an increase of .002 in. in. the side. 

Let x be the side of the square, A its area. Then 
A = x 2 and dA = 2 xdx. 

Placing x 2 and dx =* .002, we find dA = .008, which is approximately 
the required change in the area. 

If we wish to know how nearly correct the approximation is, we may com- 
pute A4 = (2.002) 8 - (2) 2 = .008004, which is the exact change in A. Our 
approximate change is therefore in error by .OOOQOAj a very email amount. 



54 DIFFERENTIATION 

Ex. 2. Find approximately the volume of a sphere of radius 1.9 in. 

The volume of a sphere of radius 2 in. is -^ TT, and the volume of the 
required sphere may be found by computing the change in the volume of 
a sphere of radius 2 caused by decreasing its radius by .1. 

If r is the radius of the sphere and V its volume, we have 

F= Tjr 8 and d V = 4 vr*dr. 



Placing r = 2 and dr .1, we find dV= 1.6 TT. Hence the volume 
of the required sphere is approximately 



To find how much this is in error we may compute exactly the volume 
of the required sphere by the formula 

V= ^7r(1.9) 8 = 9.1453 ir. 

The approximate volume is therefore in error by .0786 TT, which is less 
than 1 per cent of the true volume. 

EXERCISES 

1. The side of a square is measured as 3 ft. long. If this length 
is in error by 1 in , find approximately the resulting error in the area 
of the square. 

2. The diameter of a spherical ball is measured as 2 in., and the 
volume and the surface are computed. If an error of ^ in. has boon 
made in measuring the diameter, what is the approximate error in 
the volume and the surface ? 

3. The radius and the altitude of a right circular cone are meas- 
ured as 3 in. and 5 in. respectively. What is the approximate error 
in the volume if an error of ^ in. is made in the radius ? What is 
the error in the volume if an error of ^ in. is made in the altitude ? 

4. Find approximately the volume of a cube with 3.0002 in. on 
each edge. 

5. The altitude of a certain right circular cone is the same as 
the radius of the base. Find approximately the volume of the cono 
if the altitude is 3.00002 in. 

6. The distance * of a moving body from a fixed point of its 
path, at any time t, is given by the equation $ = 16 t a -f- 100 1 50. 
Find approximately the distance when t s= 4.0004. 

7. Find the approximate value of v? + x 2 when x =s 1,0003. 

8. Find approximately the value of as* + * a + 4 when = .99989. 



GENERAL EXERCISES 55 

9. Show that the volume of a thin cylindrical shell is approxi- 
mately equal to the area of its inner surface times its thickness. 

10. If V is the volume and S the curved area of a right circular 
cone with radius of its base r and its vertical angle 2 a, show that 
V s= I 7n <s ctn a and S = irr 1 esc a Thence show that the volume of 
a thin conical shell is approximately equal to the area of its inner 
surface multiplied by its thickness. 

GENERAL EXERCISES 

Find the derivatives of the following functions from the definition 
3 + 2x 1 6. VJB> 

' ' * 




2. ! 4. 

a x 

8. Prove from the definition that the derivative of - is , 1 

C C 

9. By expanding and differentiating, prove that the derivative 
of (2 x + 5) 8 is 6(2a; + 5) 2 . 

10. By expanding and differentiating, prove that the derivative 
of (a 2 + 1) 8 is 6 a; (a 2 + 1) 2 . 

11. By expanding and differentiating, prove that the derivative 
of (a; + a) n is n(x + a)"" 1 } where % is a positive integer. 

12. By expanding and differentiating, prove that the derivative 
of (a; 2 + a s ) n is 2 nx(x i + a 2 )"- 1 , where n is a positive integer. 

13. Find when x*+ 8 a; 8 4- 24 a 8 + 32 a- + 16 is increasing and 
when decreasing, as x increases. 

14. Find when 9 x* 24 of 8 x* -f- 32 x + 11 is increasing and 
when decreasing, as x increases. 

15. Find a general rule for the values of x for which ace 2 + bx + e 
is increasing or decreasing, as x increases. 

16. Find a general rule for the values of x for which x* a?x + 1 
is increasing or decreasing, as x increases. 

17. A right circular cone of altitude x is inscribed in a sphere of 
radius a. Find when an increase in the altitude of the cone will cause 
an increase in its volume and when it will cause a decrease, 

A-B 



* HINT, In these examples make use of the relation vA'vIt = 



56 DIFFERENTIATION 

18. A particle is moving in a straight line in such a manner that 
its distance x from a fixed point A of the straight line, at any time t, 
is given by the equation x t s 9 & + 15 1 + 100. When will the 
particle be approaching A ? 

19. The velocity of a certain moving body is given by the equation 
v = t z 7 1 -f- 10. During what time will it be moving in a direction 
opposite to that in which s is measured, and how far will it move ? 

20. If a stone is thrown up from the surface of the earth with a 
velocity of 200 ft. per second, the distance traversed in t seconds is 
given by the equation s = 200 1 16 z5 a . Find when the stone moves 
up and when down 

21. The velocity of a certain moving body, at any time t, is given 
by the equation v = t 2 8 1 + 12 Find when thu velocity of the 
body is increasing and when decreasing. 

22. At any time t, the distance s of a certain moving body from a 
fixed point in its path is given by the equation s = 16 24 1 -f 9 tf 2 f. 
When is its velocity increasing and when decreasing ? When is its 
speed increasing and when decreasing ? 

23. At any time t, the distance of a certain moving body from a 
fixed point in its path is given by the equation s t 6 6 1* -j- 9 1 + 1 
When is its speed increasing and when decreasing ? 

24. A sphere of ice is melting at such a rate that its volume is 
decreasing at the rate of 10 cu in. per minute. At what rate is the 
radius of the sphere decreasing when the sphere is 2 ft. in diameter ? 

25. Water is running at the rate of 1 cu. ft. per second into a 
basin in the form of a frustum of a right circular cone, the radii of 
the upper and the lower base being 10 in and 6 in. respectively, and 
the depth being 6 in. How fast is the water rising in the basin wluw 
it is at the depth of 3 in. ? 

<. 26. A vessel is in the form of an inverted right circular cone the 
vertical angle of which is 60. The vessel is originally filled with 
liquid which flows out at the bottom at the rate of 3 cu. in. per minutw. 
At what rate is the inner surface of the vessel being exposed when 
the liquid is at a depth of 1 ft. in the vessel ? 

27. Find the equation of the straight line which passes through 
the point (4, 6) with the slope . 

28. Find the equation of the straight line through the points 
(2, - 3) and (- 3, 4). 



GENERAL EXERCISES 57 

29. Find the equation of the straight lino determined by the points 
(2, -4) and (2, 4). 

30. Find the equation of the straight line through the point, 
(1, 3) and parallel to the line a? 2 y -f- 7 = 

31. Find the equation of the straight lino tli rough the point 
(2, 7) and perpendicular to the lino 2 . -f 4 // + 9 = 

32. Find the angle between the straight lines 2,r -|- 3// + H = 
and y + 3 x -f- 1 = 0. 

Find the turning-points of the following curves and draw the 
graphs 

33. y=3 as-ai" 

34. 2/ = 16c a -40a- + 25 

35. y } (x 2 - 4 x - 2). 

36. y = a* - 6 a* - 15 x + 5. 

37. y = 3- 3 + 3 as* 9 a: 14. 

38. Find the point of intersection of the tangents to the curve 
y = x + 2 a; 2 x s at the points for which y = 1 and a? = 2 
respectively 

39. Show that the equation of the tangent to the curve y = o" 2 
+ 2bx + G at the point (x 1} y^) is y = 2 (ca^-f /;), a,r, a + 0. 

40. Show that the equation of the tangent to the curve y a* a 
+ ax + fi at the point (a^, y a ) is y = (3 a a + )*" 2 , -|_ />. 

41. Find the area of the triangle included between the coordintitc 
axes and the tangent to the curve y = a; 8 at the point (2, 8). 

42. Find the angle between the tangents to the curve //=-2./'" 
-f- 4 a; 2 x at the points the abscissas of which arc 1 and 1 
respectively. 

43. Find the equation of the tangent to the curve y =s x n 3 a; 9 
-+ 4.x 12 which has the slope 1. 

44. Find the points on the curve ;// = 3 .r, B 4 a*, 2 at which the 
tangents are parallel to the lino x y = 0. 

V 1 46. A length I of wire is to be cut into two portions which aw to 
be bent into the forms of a circle and a square respectively. Show 
that the sum of the areas of those figures will bo least when the wire 
is cut in the ratio IT : 4. 



58 DIFFERENTIATION 

46. A log in the form of a -frustum of a cone is 10 ft. long, the 
diameters of the bases being 4 ft. and 2 ft. A beam with a square 
cross section is cut from it so that the axis of the beam eoiuuidos 
with the axis of the log. Find the beam of greatest volume that 
can be so cut 

47. Required the right circular cone of greatest volume which can 
be inscribed in a given sphere. 

48. The total surface of a regular triangular prism is to be Jt. Find 
its altitude and the side of its base when its volume is a maximum. 

49. A piece of wire 9 in. long is cut into five pieces, two of ouo 
length and three of another Each of the two equal pieces is bent 
into an equilateral triangle, and the vertices of the two triangles are 
connected by the remaining three pieces so as to form a regular 
triangular prism. How is the wire cut when the prism has the largest 
volume ? 

50. If t is time in seconds, v the velocity of a moving body in 
feet per second, and v = 200 32 1, how far will the body move in 
the first 5 sec. ? 

51. If v = 200 32 1, where v is the velocity of a moving body 
in feet per second and t is time in seconds, how far will the body 
move in the fifth second ? 

62. A curve passes through the point (2, 8), and its slope at 
any point is equal to 3 more than twice the abscissa of the point. 
Find the equation of the curve. 

53. A curve passes through the point (0, 0), and its slope at any 
point is oj a 2 x + 7. Find its equation. 

54. Find the area bounded by the curve y + s a 16 = and llio 
axis of ao, 

5 5 . Find the area bounded by the curve y = 2 as 8 15 a? a 4. 3G + 1 , 
the ordinates through the turning-points of the curve, and OX. 

56. Find the area between the curve y x a and the straight lint? 
y x -f- 6. 

57. Find the area between the curves y =* jc a and y = 18 a 8 . 

58. The curve y = ax* is known to pass through the point (h, Jc), 
Prove that the area bounded by the curve, the axis of a-, and the 
line x = h is hk. 

59. Compute the difference between &A and dA for the area 4 of 
a circle of radius 5, corresponding to an increase of .01 in the radius. 



GENERAL EXERCISES 59 

60. Compute the difference between AF and dV for the volume V 
of a sphere of radius 5, corresponding to an increase of 01 in the 
radius 

61. If a cubical shell is formed by increasing each edge of a cube 
by dx, show that the volume of the shell is approximately equal to 
its inside surface multiplied by its thickness. 

62. If the diameter of a sphere is measured and found to be 2 ft , 
and the volume is calculated, what is the approximate error in the 
calculated volume if an error of in has been made in obtaining 
the radius ? 

63. A box in the form of a right circular cylinder is 6 in deep 
and 6 in. across the bottom. Find the approximate capacity of the 
box when it is lined so as to be 5 9 in deep and 5 9 in across the 
bottom. 

64. A rough wooden model is in the form of a regular quadran- 
gular pyramid 3 in. tall and 3 in. on each side of the base. After it 
is smoothed down, its dimensions are all decreased by .01. What is 
the approximate volume of the material removed ? 

65. By use of the differential find approximately the area of a 
circle of radius 1.99. What is the error made in this approximation ? 

66. Find approximately the value of a; 6 + 4 C 8 + a when x = 3.0002 
and when x = 2.9998. 

67. The edge of a cube is 2.0001 in. Find approximately its surface. 

68. The motion of a certain body is defined by the equation 
s = i 8 + 3 i* + 9 1 27. Find approximately the distance traversed 
in the interval of time from t = 3 to t = 3.0087. 



CHAPTER III 
SUMMATION 

22. Area by summation. Let us consider the problem to find 
the area bounded by the curve y ^ # 2 , the axis of r, and the 
ordmates x = 2 and x 3 (Fig. 22). This may be solved by the 
method of 19 ; but we wish to show that it may also be considered 
as a problem in summation, since the area is approximately equal 
to the sum of a number of rectangles constructed as follows : 

We divide the axis of x between x = 2 and x 3 into 10 parts, 

3 2 
each of which we call A#, so that Ax = r =.1. If x l is the 

first point of division, x a the second point, and so on, and 
rectangles are constructed as shown in the figure, then the 
altitude of the first rectangle is (2) 2 , that of the second rec- 
tangle is z 2 =|(2. 1) 2 =.882, and so on. The area of tho 
first rectangle is ^(2) 2 Ao;=.08, that of the second rectangle is 
J 1 2 Aa; = |(2.1) 2 A=.0882, and so on. 

Accordingly, we make the following calculation: 

s = 2, -|(2) 3 Aa;= .08 

0^=2.1, %(xy&x= .0882 

# 2 =2.2, (z 2 ) 2 Az = .0968 

z 8 =2.3, l(xy&x= .1058 

z 4 =2.4, %(xy&x= .1152 

aj a =2.5, O 6 ) 2 Aa= .1250 

a; 6 =2.6, %(x^&x= .1352 

z 7 =2.7, |(r 7 ) 2 A^= .1458 

z 8 =2.8, (z 8 ) 2 Aa;= .1508 

tf=2.9, f Ar= -1G82 



. ^ 

i his is a first approximation to the area. 

For a better approximation the axis of x between #= 2 and 05= 3 
may be divided into 20 parts with Aa;= .05. The result is 1.241 8. 

60 



ABJEA 



61 




FIG. 22 



If the base of the required figure is divided into 100 parts with 
Az = .01, the sum of the areas of the 100 rectangles constructed 
as above is 1.26167. 

The larger the num- 
ber of parts into , which 
the base of the figure is 
divided, the more nearly 
is the required area ob- 
tained. In fact, the re- 
quired area is the limit 
approached as the number 
of parts is indefinitely "~ Q 
increased and the size of 
A approaches zero. 

We shall now proceed to generalize the problem just handled. 
Let LK (Fig. 23) be a curve with equation y =/(), and let 
OEa and OB b. It is re- 
quired to find the area bounded 
by the curve LK, the axis of #, 
and the ordmates at E and B. 

For convenience we assume 
in the first place that a < b 
and that f(x) is positive for 
all values of x between a and b. 
We will divide the line EB 
into n equal parts by placing 

and laying off the 




E MI M e 



Jf J/ Mj M t B 



FIG. 28 



n 



lengths EM^- 
Let 



_ 1 J5=Aa: (in Fig. 23,n=9). 



. ., Jf^A* parallel to OX. 
/(a)Aa;=:the area of the rectangle 
/(#].) AOJ = the area of the rectangle 
/(* 2 ) As? = the area of the rectangle 



_ 1 ), and 
Then 



alsc 



the area of the rectangle 



62 SUMMATION 

The sum 

/(a) A* +/(aOA* +/(aA* + . . . +/(^_ 1 )Aa; (1) 



is then the sum of the areas of these rectangles and equal to 
the area of the polygon EDR^R^ . . . R^^^li^B. It is evident 
that the limit of this sum as n is indefinitely increased is the 
area bounded by ED, EB, BC, and the arc DC. 

The sum (1) is expressed concisely by the notation 



where S (sigma), the Greek form of the letter S, stands for the 
word " sum," and the whole expression indicates that the sum 
is to be taken of all terms obtained from/(,)Aa; by giving to i 
in succession the values 0, 1, 2, 3, ., n 1, where x ~a. 
The limit of this sum is expressed by the symbol 

C b 

Ja 

where /is a modified form of S. 

Hence / /(#)<fa; = Lim^/(a; f )A:c=: 

i/a isO 



area 



It is evident that the result is not vitiated if ED or J3C is of 
length zero. 

23. The definite integral. We have seen in 19 that if A is 
the area EBCD of 22, 



where F(x) is any function whose derivative is/(). Comparing 
this with the result of 22, we have the important formula 



r h 

J a 

The limit of the sum (1), 22, which is denoted by C /(*) dx, 

Jn 

is called a definite integral, and the numbers a and b are called 



DEFINITE INTEG-KAL 68 

the lower limit and the upper limit*, respectively, of the definite 
integral. 

On the other hand, the symbol I f(x)dx is called an indefinite 

integral and indicates the process of integration as already de- 
fined in 18. 

Thus, from that section, we have 

+ 0, 
G, 

and, in general, \ f(x)dx=F(x)+ C, 

where F(x) is any function whose derivative is /(#) 

We may therefore express formula (1) in the following rule : 

To find the value of \ f(x)dx, evaluate \f(x)dx, substitute 

x = b and x: = a successively, and subtract the latter result from 
the former. 

It is to be noticed that in evaluating I f(x)dx the constant 

of integration is to be omitted, because if it is added, it dis- 
appears in the subtraction, since 



In practice it is convenient to express F^ F^a) by the 
symbol [^ ()]> so that 



Ex. 1. The example of 22 may now be completely solved. The required 

area is 

27 8 10 



* The student should notice that the word "limit" is here used In a 
quite different from that in which it is used when a variable is said to approach 
a limit (1). 



64 



SUMMATION 



The expression /(a;) dx which appears in formula (1) is called 
the element of integration. It is obviously equal to dF(x). In 
fact, it follows at once from 19 that 

dA=ydx = f(x) dx. 

In the discussion of 22 we have assumed that y and dr are 
positive, so that dA is positive. If y is negative that is, if the 
curve m Fig. 23 is below the axis of x and if dx is positive, 
the product ydx is negative and the y 

area found by formula (1) has a nega- 
tive sign. Finally, if the area required 
is partly above the axis of x and 
partly below, it is necessary to find 
each part separately, as in the follow- 
ing example: 

Ex. 2. Fmd the aiea bounded by the 
curve y = x 3 a; 2 6 x and the axis of a- 

Plotting the curve (Fig 24), we see that 
it crosses the axis of x at the points 
B (-2, 0), 0(0, 0), and C(3, 0). Hence 
part of the area is above the axis of x and 
part below. Accoidingly, we shall find it 
necessary to solve the pioblem in two parts, 
first finding the aiea above the axis of x 
and then finding that below To find the 
first area we proceed as in 22, dividing 
the area up into elementary rectangles fox- 

each of which 

dA = ydx = (a. 8 a; 2 - 6 *) dx , 

whence A- C\x*- x*-Qx)dx = [I* 4 - jj 3? 




o 



FIG. 24 



Similarly, for the area below the axis of x we find, as before, 
dA = tlx = x s x s 



But in this case y = x s jc z G x is negative and hence dA is negative, 
for we are making x vary fiom to 8, and therefore dx is positive, Tlioio- 
foie we expect to find the result of the summation negative. In faot, 

we have -s 

A - I (X s - x* - 6 x^dx = [| a: 4 ~ J as* ~ 
J o 

-^(3)-3(3)]-0 = -15i 



DEFINITE INTEGRAL 



65 



.8 



As we aie asked to compute the total aiea bounded by the ciuve aud 
the axis of x, we discard the negative sign in the last summation aud add 
5J and 15 f, thus obtaining 21 ^ s as the required result. 

If we had computed the definite integral 

i 8 x z 6 a.) dx, 

we should have obtained the icsult 10 fy, which is the algebraic sum of 
the two portions of area computed separately. 

Ex. 3. Find the area bounded by the two curves y = .r 2 and y = 8 a; 2 . 
We diaw the curves (Fig 25) 

?/ = r a f\ "\ 

y x (.*-} 

and y = 8 - a; 8 , (2) 

and by solving their equations we 
find that they mtcisect at the points 
P x (2, 4)aud> 3 (-2,4) 

The reqiuiod area OP^JIP^O is evi- 
dently twice the area OJ\liO, since 
both ciuves aie symmetiical with le- 
spect to OY. Accoidmgly, wo shall 
find the area OP^BO and multiply it 
by 2. This lattei area may be found 
by subtracting the aic>a ON^P^O from 
the area ON^P-^BO, each of these areas 
being found as in the pievious example ; 
or we may proceed as follows : 

Divide ON^ into n parts dx, and 
through the points of division draw 
stiaight lines paiallol to OY, intersect- 
ing both curves. Let one of those lines 
be M^ Q l Q 2 Through the ] 101 nts Q : and 
(2 2 draw straight linos parallel to OX 

until they meet the next vortical line to the right, thcnoby forming 
the rectangle Q^SQy, The aroa of such a rectangle may be takon as 
dA and may be computed as follows: its base is ih % and ids altitude IH 
Q^ = A/^ M^Q,! = (8 j 8 ) j" u =8 2 .c a ; for M^Q^ is tho ordinato 
of a point on the curve (2) and J/j^ the ordinate of a point ou (1). 

Thcicfore , , , 



A 



Finally, tho required area is 2(10) = 21 } 




= C \R - 2 r a ) rl,r =[83-- ?, 3*] 



66 SUMMATION 

EXERCISES 

1. Find the area bounded by the curve 4 y as 2 2 = 0, the axis 
of x, and the lines x 2 and x = 2. 

2. Find the area bounded by the curve y = a? 7 X* + 8 a + 16 
the axis of x, and the lines x = 1 and as = 3 

3. Find the area bounded by the curve y = 25 x 10 a; 2 -f- oi 8 and 
the axis of a. 

4. Find the area bounded by the axis of x and the curve 
y = 25 - a 2 . 

5. Find the area bounded by the curve y=4aj 2 4aj -3 and 
the axis of x. 

6. Find the area bounded below by the axis of x and above by 
the curve y = # 8 4 a 2 4 x + 16. 

7. Find the area bounded by the curve y = 4 as 8 Sec 2 9cc+18 
and the axis of #. 

8. Find the area bounded by the curve x* + 2y 8=0 and tho 
straight line x-)-2y 6 = 0. 

9. Find the area bounded by the curve 3y aj 2 = and the 
straight line 2x + y 9 = 

10. Find the area of the crescent-shaped figure bounded by the 
two curves y = x* + 7 and y = 2 x z -f- 3 

11. Find the area bounded by the curves 4y=sa3 a 4 a; and 
a 2 4o: + 42/ 24 = 0. 

12. Find the area bounded by the curve o; + 3 = 2/ a 2y and 
the axis of y. 

24. The general summation problem. The formula 

F(a) (1) 



rf 

J a 

has been obtained by the study of an area, but it may be given 
a much more general application. For if f(x) is any function 
of x whatever, it may be graphically represented by the curve 
y =/(). The rectangles of Fig. 23 are then the graphical rep- 

resentations of the products f(z)dx, and the symbol f f(x)dx 

J a 



GENERAL PROBLEM 6T 

represents the limit of the sum of these products. We may 
accordingly say: 

Any problem which requires the determination of the limit of the 
sum of products of the type f(x) dx may be solved by the use of 
formula (1). 

Let us illustrate this by considering again the problem, already 
solved in 18, of determining the distance traveled in the time 
from t = t t to t = 3 by a body whose velocity v is known. Since 

ds 
"*#' 

we have ds = vdt, 

which is approximately the distance traveled in a small interval 
of time dt. Let the whole time from t = t^ to t = t z be divided 
into a number of intervals each equal to dt. Then the total dis- 
tance traveled is equal to the sum of the distances traveled in 
the several intervals dt, and hence is equal approximately to the 
sum of the several terms vdt This approximation becomes better 
as the size of the intervals dt becomes smaller and their number 
larger, and we conclude that the limit of the sum of the terms vdt 
is the actual distance traveled by the body. Hence we have, if s 
is the total distance traveled, 

a = I vdt. 

A 
If, now, we know v in terms of t, we may apply formula (1). 

Ex. If v = 16 1 + 5, find the distance traveled in the time from t = 2 to 
* = 4. 

We have directly 

s = f *(16 1 + 5) dt = [8 1* + 5 *] = 106. 

EXERCISES 

1. At any time t the velocity of a moving body is 3 & 4- 2 1 f t. per 
second. How far will it move in the first 6 sec.? 

2. How far will the body in Ex. 1 move during the seventh second ? 

3. At any time t the velocity of a moving body is 6 + 5 1 & ft. per 
second, Show that this velocity is positive during the interval from 
t = 1 to t f= 6, and find how far the body moves during that interval. 



68 SUMMATION 

4. At any time t the velocity of a moving body is 4)5 3 24 f +11 ft. 
per second. During what interval of time is the velocity negative, and 
how far will the body move during that interval ? 

6. The number of foot pounds of work done in lifting a weight in 
the product of the weight in pounds and the distance in feet through 
which the weight is lifted. A cubic foot of water weighs 021 lb. 
Compute the work done in emptying a cylindrical tank of depth 8 ft. 
and radius 2 ft , considering it as the limit of the sum of the jnort'rt 
of work done in lifting each thin layer of water to the top of the tank. 

25. Pressure. It is shown m physics that the pressure on 
one side of a plane surface of area A, immersed in a liquid tit a, 
uniform depth of h units below the Kin-face of the liquid, is equal 
to wJul, where w is the weight of a unit volume of the liquid. 
This may be remembered by noticing 

that wliA is the weight of the column " 
of the liquid which would be supported 
by the area A. 

Since the pressure is the same in 
all directions, we can also determine ^ IQ 2Q 

the pressure on one side of a plane 

surface which is perpendicular to the surface of the liquid and 
hence is not at a uniform depth. 

Let ABC (Fig. 26) represent such a surface and RS the line of 
intersection of the plane of ABC with the surface of the liquid. 
Divide ABC into strips by drawing straight lines parallel to Jttf. 
Call the area of one of these strips dA, as in 28, and the depth of 
one edge h. Then, since the strip is narrow and horizontal, tho 
depth of every point differs only alightly from 7i, and tho pres- 
sure on the strip is then approximately wMA, Talcing P m tho 
total pressure, we write ^p_ j 3* 

The total pressure P is the sum of tho pressures on the several 
strips and is therefore the limit of the sum of terms of tho 
form whdA, the limit being approached as the number of tho 
strips is indefinitely increased and the width of each indefinitely 
decreased. Therefore / 

P= I wlidA, 




PEESSUKE 69 

where the limits of integration are to be taken so as to include 
the whole area the pressure on which is to be determined. To 
evaluate the integral it is necessary to express both h and dA 
111 terms of the same variable. 

Ex. 1. Find the pleasure on one side of a rectangle BCDE (Fig 27), 
wheie the sides BC and ED aiu each 4 ft. long, the sides BE and CD are 
each 8 ft. long, immersed in watei so that the plane of the rectangle is 
peipendiculai to the surface of the 

water, and the side BC is paiallel & 

to the surface of the watei and 2 ft. 
below at 

In Fig. 27, LK is the line of inter- 
section of the surface of the water 

and the plane of the rectangle. Let M * N 

be the point of intersection of 

LK and BE produced Then, if x is 

measured downward from along E D 

BE, x has the value 2 at the point B ^ 

and x has the value 5 at the point E. 

We now divide BE into parts dx, and through the points of division 
draw straight lines parallel to BC, thus dividing the given rectangle 
into elemental y rectangles such as MNP.S 

Therefore dA = area of MNR S = MN - MS - 4 dx. 

Since MN is at a distance x below LK, the pressure on the elementary 
rectangle MNRS is approximately wx(4: dx'). Accoidmgly, we have 



and P = f G 4 nxdx = [2 vxffe* 2 w(6) - 2 w(2) 2 = 42 20 

t/2 ^ 

For water, w - 02 J Ib = ^ T. 

Hence we have finally 

P = 2625 Ib. = 1 T % T 

Ex. 2. The base CD (Fig. 28) of a triangle BCD is 7 ft., and its altitude 
from B to CD is 5 ft This triangle is immersed in water with its plane 
perpendicular to the suiface of the water and with CD parallel to the sur- 
face, and 1 ft below it, B being below CD Find the total pressure on one 
side of this triangle. 

Let LK represent the line of intersection of the plane of the triangle 
and the surface of the water. Then B is 6 ft below LK Let BX be per* 
pendicular to LK and intersect CD at T We will measure distances 
from B in the direction BX and denote them by x. Then, at the point B, 
x has the value , and at T, x has the value 5. 



TO SUMMATION 

Divide the distance BT into parts dx, and through the points of divi- 
sion draw straight lines parallel to CD, and on ouch of these linen n 
lower base construct a rectangle such as MNJR8, where J'l and F sire 
two consecutive points of division % 

onBX. L K 

Then BE = x, G \ 

EF = dx, 
and, by similar triangles, 



whence 



CD BT 

MN x 




and 

Then dA = the area of MNRS = J xdx, 

Since B is C ft. below LK, and BE = x, it follows that E is (fl - a;) ft. 
below LK. 

Hence the pressure on the rectangle is approximately 

dP = (& a" ?*) (6 or) w = (-V 1 war $ ?/JJK S ) rfj?, 

/*" 
/o 
= (105 to - iJA w) - = JL^ - to - 2010^ Ib. = Ii4 T. 



EXERCISES 

s/ 1. A gate in the side of a dam is in the form of a square, 4 ft. 

on a side, the upper side being parallel to and 1C It. bolow the surfaoo 

of the water in the reservoir. What is the pressure on the gate '( 

v* 2. Find the total pressure on one side of a triangle of base 6 ft. 

and altitude 6 ft., submerged in water so that the altitude is vertical 

and the vertex is m the surface of the water. 

</ 3. Find the total pressure on one side of a triangle of base 4 ft. 

and altitude 6 ft., submerged in water so that the base is horizontal, 

the altitude vertical, and the vertex above the base and 4 ft, from tlio 

surface of the water. 

e base of an isosceles triangle is 8 ft. and the equal sidas 
6 ft. The triangle is completely immersed in water, its baae 
allel to and 6 ft. below the surface of the water, its alti- 
g perpendicular to the surface of the water, and its vertex 

uc.ijmg ,uove the base. Find the total pressure on one side of the 

triangle. 



VOLUME 71 

v 6. Find the pressure on one side of an equilateral triangle, 6 It. 
on a side, if it is partly submerged in water so that one vortex is 
one foot above the surface of the water, the corresponding altitude 
being perpendicular to the surface of the water. 

\s. The gate in Ex. 1 is strengthened by a brace which runs 
diagonally from one corner to another. Find the pressure on each 
of the two portions of the gate one above, the other below, the 
brace. 

7. A dam is in the form of: a trapezoid, with its two horizontal 
sides 300 and 100 ft. respectively, the longer side being at the top ; 
and the height is 15 ft. What is the pressure on the dam when the 
water is level with the top of the dam ? 

8. What is the pressure on tho dam of Ex. 7 when the water 
reaches halfway to the top of the dam ? 

9. If it had been necessary to construct the dam of Ex. 7 with 
the shorter side at tho top instead of the longer side, how much 
greater pressure would the dam have had to sustain when the 
reservoir is full of water ? 

10. The center board of a yacht is in the form, of a trapezoid in 
which the two parallel sides are 3 ft and 6 ft., respectively, in length, 
and the side perpendicular to these two is 4 ft. in length. Assuming 
that the last-named side is parallel to the surface of the water at 
a depth of 2 ft., and that the parallel sides are vertical, find the 
pressure on one side of the board. 

11. Where shall a horizontal lino be drawn across tho gate of 
Ex. 1 so that tho pressure on the portion above tho lino shall equal 
the pressure on the portion below ? 

26. Volume. The volume of a solid may be computed by di- 
viding it into n elements of volume, dV^ and taking the limit* 
of the sum of these elements as n is increased indefinitely, tho 
magnitude of each element at the same time approaching aero. 
The question in each case is the determination of the form of 
the element dr. We shall discuss a comparatively simple case 
of a solid such as is shown in Fig, 29. 

In this figure let Olf be a straight line, and let tho distance 
of any point of it from be denoted by h. At one end tho solid 
is bounded Tby a plane perpendicular to OH at <7, where 00 ~ a> 



72 



SUMMATION 




and at the other end it is bounded by a piano perpendicular to 
OH at B, where OS = 5, so that it has parallel bases. 

The solid is assumed to be such that the area A of any plane 
section made by a plane perpendicular to Oil at a point distant 
h from can be expressed as a func- 
tion of h. 

To find the volume of such a solid 
we divide the distance CB into n parts 
dh, and through the points of division 
pass planes perpendicular to OH. We 
have thus divided the solid into slices 
of which the thickness is dh 

Since A is the area of the base of a 
slice, and since the volume of the slice 
is approximately equal to the volume 
of a right cylinder of the same base 
and thickness, we write 

dV=Adli. 

The volume of the solid is then the limit of the sum of terms of 
the above type, and therefore 

/"' 
F= I Adh. 

Jo. 

It is clear that the above discussion is valid even when one 
or both of the bases corresponding toh=a and 7t = fi, respectively, 
reduces to a point. 

Ex.1. Let OY (Fig. 30) 
be an edge of a solid such 
that all its sections made 
by planes perpendicular to 
OFaie rectangles, the sides 
of a rectangle in a plane 
distant y fiorn being re- 
spectively 2 y and y z We 
shall find the volume in- 
cluded between the planes j, IO go 
# = and?/ = 2 J. 

Dividing the distance from # = 0toy = 2& into n parts <li/, and passing 
planes perpendicular to OY, we form rectangles such as MffttS, wlioro, if 





VOLUME 73 

OM = y, MN - y* and MS = 2 y Hence the area MNRS = 2 y s , and the 
volume of the elementary cylinder standing on MNRS as a base is 

thatls ' 

Therefore V = 

Ex. 2. The axes of two equal light circular cylinders of radius o inter- 
sect at right angles. Required the volume common to the two cylinders. 

Let OA and OB (Fig. 31) be the axes of the ^ 

cylmdeis and OY the common perpendicular to 
OA and 013 at then point of intersection 0. Then 
OA D and OBD are quadrants of two equal chcles 
cut from the two cylinders by the planes through 
OY perpendicular to the axes OB and OA, and 
OD = a Then the figure represents one eighth 
of the requned volume. 

We divide the distance OD into n parts dy, 
and through the points of division pass planes per- 
pendicular to OY Any section, such as LMNP, 
is a square, of which one side NP is equal to 

V OP Z ON* OP = a, being a radius of one of the cylinders, and hence, 
as ON= y, _ 

NP = Va 2 - f 

Accordingly, the area of LMNP = a z y z , and the volume of the ele- 
mentary cylinder standing on LMNP as a base is 




whence V = f "(a a - y^ dy - [a 2 // - J y] = a 8 . 

Hence the total volume is - 1 / a 8 . 

t 

This method of finding volumes is particularly useful when 
the sections of the solid made by parallel planes are bounded 
by circles or by concentric circles. Such a solid may be gen- 
erated by the revolution of a plane area around an axis in its 
plane, and is called a solid of revolution. We take the following 
examples of solids of revolution : 

Ex. 3. Find the volume of the solid generated by revolving about OX 
the area bounded by the curve ?y 2 = 4 *, the axis of x, and the line x = 8. 

The generating area is shown in Fig. 32, where AB is the line a- = 8. 
Hence OA s= 3, 



74 



SUMMATION 



Divide OA into n parts dx, and through the points of division pass 
straight lines parallel to OY, meeting the curve. When the area is revolved 
about OX, each of these lines, as MP, JVQ, etc., generates a circle, the plane 
of which is perpendicular to OX The area ^ 
of the circle generated by MP, for example, 
is irMP z , which is equal to iry z = 7r(4 x), if 



Hence the area of any plane section of 
the solid made by a plane perpendicular to 
OX can be expressed in terms of its dis- 
tance from 0, and we may apply the pievious 
method for finding the volume. 

Since the base of any elementary cylinder 
is 4 TTX and its altitude is dx, we have 



Hence V=f*4 vxdx = [2 ** = 18 . 




Ex. 4. Find the volume of the ring surface generated by revolving about 
the axis of x the area bounded by the line y = 5 and the curve y = 9 x~ 

The line and the curve (Fig 33) are 
seen to intersect at the points P l (2, 5) ^ 

and P z (2, 5), and the ring is generated 
by the area P^BP^P^ Since this area is 
symmetrical with respect to OY, it is evi- 
dent that the volume of the ring is twice 
the volume generated by the area AP Z BA. 
Accordingly, we shall find the latter volume 
and multiply it by 2. 

We divide the line OM Z = 2 (M z being 
the projection of P 2 on OX) into n parts 
dx, and through the points of division draw 
straight lines parallel to OY and intersect- 
ing the straight line and the curve One 
of these lines, as M QP, will, when revolved 
about OX, generate a circular ring, the 
outer radius of which is MP = y = Q~x* 
and the inner radius of which is MQ *=.y = 5 
Hence the area of the ring is jjf M 'M 




Accordingly, 



= TT (56 - 18 x 9 + a?<). FIG- 

dV = TT (56 - 18 w 2 + a;*) dx 



S3 



Accordingly, the volume of the ring is 2 (70$ w) = 140$ TT. 



VOLUME 75 

EXERCISES 

1. The section of a certain solid made by any plane perpendicular 
to a given line Oil is a circle with one point in OH and its center 
on a straight line OB intersecting Oil at an angle of 46 If the 
height of this solid measured from along Oil is 4ft., find its 
volume by integration. 

2. A solid is such that any cross section perpendicular to an 
axis is an equilateral triangle of which each side is equal to the 
square of the distance of the plane of the triangle from a fixed point 
on the axis. The total length of the axis from the fixed point is 5. 
Find the volume. 

3. Find the volume of the solid generated by revolving about OX 
the area bounded by OX and the curve y = 4 as a 2 . 

4. Find the volume of the solid generated by revolving about OX 
the area included between the axis of x and the curve y 2 3 x* 

5. Find the volume of the solid generated by revolving about the 
line y = 2 the area bounded by the axis of y, the lines x = 3 
and y = 2, and the curve y = 3 a: 2 . 

6. On a spherical ball of radius 5 in. two great circles are drawn 
intersecting at right angles at the points A and B* The material of 
the ball is then cut away so that the sections perpendicular to AB 
are squares with their vertices on the two great circles. Find the 
volume left. 

7. Find the volume generated by revolving about the line x = 2 
the area bounded by the curve if 8 a?, the axis of a?, and the 
line x = 2. 

8. Any plane section of a certain solid made by a plane perpen- 
dicular to OF is a square of which the center lies on OY and two 
opposite vertices lie on the curve y = 4 as 2 . Find the volume of the 
solid if the extreme distance along OYis 3. 

9. Find the volume generated by revolving about OY the area 
bounded by the curve y 2 8 x and the line x = 2. 

10. Find the volume of the solid generated by revolving about OX 
the area bounded by the curves y = 6 x as 2 and y s= flj 8 6 a? + 10. 

11. The cross section of a certain solid made by any plane perpen- 
dicular to OX is a square, the ends of one of whose sides are on the 
curves 16 y = y? and 4 y = y? 12, Find the volume of this solid 
between the points of intersection of the curves. 



T6 SUMMATION 

GENERAL EXERCISES 

1. The velocity in feet per second of a moving body at any 
time t is t* 4 1 + 4 Show that the body is always moving in 
the direction in which s is measured, and find how far it will move 
during the fifth second 

2. The velocity in feet per second of a moving body at any time 
t is t* 4 1. Show that after t = 4 the body will always move in 
the direction in which s is measured, and find how far it will move 
in the time from t = 6 to t = 9. 

3. At any time t the velocity in feet per second of a moving 
body is t z 6 1 + 5 How many feet will the body move in the 
direction opposite to that in which s is measured? 

4. At any time t the velocity in miles per hour of a moving body 
is t* 2 1 3. If the initial moment of time is 12 o'clock noon, how 
far will the body move in the time from 11.30 A M. to 2 p M. ? 

y 5 . Find the area bounded by the curves 9 y = 4 x z and 45 9 y = a: 3 . 
6. Find the total area bounded by the curves ?/ 2 =4a; and 
T/ 2 = 4 a 8 4 ax. 

'' 7. Find the total area bounded by the curve y = x B and the 
straight line y = 4 x. 

8. Find the total area bounded by the curve y = x (x 1) (03 3) 
and the straight line y = 4 (a; 1). 

9. ABCD is a quadrilateral with A 90, J3 = 90, AB 5 ft., 
BC = 2 ft., A D 4 ft. It is completely immersed in water with AB 
in the surface and AD and BC perpendicular to the surface. Find 
the pressure on one side 

10. Prove that the pressure on one side of a rectangle completely 
submerged with its plane vertical is equal to the area of the rectangle 
multiplied by the depth of its center and by w (consider only the 
case in which one side of the rectangle is parallel to the surface). 

11. Prove that the pressure on one side of a triangle completely 
submerged with its plane vertical is equal to its area multiplied by 
the depth of its median point and by w (consider only the case in 
which one side of the triangle is parallel to the 1 surface). 

12. The end of a trough, full of water, is assumed to be in the 
form of an equilateral triangle, with its vertex down and its plane 
vertical. What is the effect upon the pressure on the end if the 
level of the water sinks halfway to the bottom? 



GENERAL EXERCISES 77 

13. A square 2 ft on a side is immersed in water, with one vertex 
in the surface of the water and with the diagonal through that 
vertex perpendicular to the surface of the water. How much greater 
is the pressure on the lower half of the square than that on the 
upper half? 

14. A board is symmetrical with respect to the line AJB, and is of 
such a shape that the length of any line across the board perpendic- 
ular to AB is twice the cube of the distance of the line from A. 
AD is 2 ft. long The board is totally submerged in water, AB being 
perpendicular to the suiface of the water and A one foot below the 
surface. Find the pressure on one side of the board. 

15. Find the pressure on one side of an area the equations of whose 
boundary hues are x = 0, y = 4, and v/ 2 = 4 x respectively, where the 
axis of x is taken in the surface of the water and where the positive 
direction of the y axis is downward and vertical. 

16. Find the volume generated by revolving about OX the area 
bounded by OX and the curve 4 y = 16 a; 2 . 

I,- 17. Find the volume generated by revolving about OX the area 
bounded by the curve y = a 2 + 2 and the line y = 3. 

18. Find the volume generated by revolving about OX the area 
bounded by OX and the curve y = 3 x x 8 . 

19. Find the volume generated by revolving about the line y = 1 
the area bounded by the curves 9 y = 2 x* and 9 y = 36 2 a; 2 . 

20. An axman makes a wedge-shaped cut in the trunk of a tree. 
Assuming that the trunk is a right circular cylinder of radius 8 in., 
that the lower surface of the cut is a horizontal plane, and that the 
upper surface is a plane inclined at an angle of 45 to the horizontal 
and intersecting the lower surface of the cut in a diameter, find the 
amount of wood cut out. 

21. On a system of parallel chords of a circle of radius 2 there 
are constructed equilateral triangles with their planes perpendicular 
to the plane of the circle and on the same side of that plane, thus 
forming a solid. Find the volume of the solid. 

22. Show that the volume of the solid generated by revolving 
about OY the area bounded by OX and the curve y = a Ix* is 
equal to the area of the base of the solid multiplied by half its altitude. 

23. In a sphere of radius a find the volume of a segment of one 
base and altitude /*. 



78 SUMMATION 

24. A solid is sucli that any cross section perpendicular to an axis 
is a circle, with its radius equal to the square root of the distance of 
the section from a fixed point of the axis. The total length of the 
axis from the fixed point is 4. Find the volume of the solid. 

25. A variable square moves with its plane perpendicular to the 
axis of y and with the ends of one of its diagonals respectively in 
the parts of the curves y* = 16 x and ^ = 4 x, which are above 
the axis of x Find the volume generated by the square as its plane 
moves a distance 8 from the origin. 

26. The plane of a variable circle moves so as to be perpendicular 
to OX, and the ends of a diameter are on the curves y = a; 2 and 
y = 3 cc 2 8 Find the volume of the solid generated as the plane 
moves from one point of intersection of the curves to the other. 

27. All sections of a certain solid made by planes perpendicular 
to OF are isosceles triangles. The base of each triangle is a line 
drawn perpendicular to OY, with its ends in the curve y = 4 a: 2 . The 
altitude of each triangle is equal to its base Find the volume of 
the solid included between the planes for which y = and y 6. 

28. All sections of a certain solid made by planes perpendicular 
to Y are right isosceles triangles One leg of each triangle coincides 
with the line perpendicular to OY with its ends in OY and the curve 
y* 4 x. Find the volume of the solid between the sections for which 
y = and y = 8. 

29. Find the work done in pumping all the water from a full 
cylindrical tank, of height 15 ft. and radius 3 ft., to a height of 
20 ft. above the top of the tank. 

30. Find the work done in emptying of water a full conical 
receiver of altitude 6ft and radius 3ft., the vertex of the cone 
being down. 



CHAPTER IV 



ALGEBRAIC FUNCTIONS 

27. Distance between two points. Let P 1 (a^, / x ) and P z (r z , / a ) 
(Fig. 34) be any two points in the plane XOY, such that the 
straight line P^P Z is not parallel either to OX or to OY. Through 
P! draw a straight line parallel 
to OJT, and through P 2 draw a 
straight line parallel to OY, and 
denote their point of intersection 
by R. 

iPlTATl T^ 7? Al 1 * "-" 1" - ty 

JL UOll JL^J.V * A-l*t/ t'g " " *t*j 



and 

In the right triangle P^RP Z 



whence 
If 




C 1 ) 



is parallel to 0JT, and the formula reduces to 

J?J = JBa- ! (2) 

In like manner, if ^-x^ P^ is parallel to OF, and the 
formula reduces to 



28. Circle. Since a ciVc?e is the locus of a point which is 
always at a constant distance from a fixed point, formula (1) 
27, enables us to write down immediately the equation of P 
circle. 

Let C(h, 7c) (Fig. 35) be the center of a circle of radius r. 
Then, if P(x, y) is any point of the circle, by (1), 27, a? and 
y must satisfy the equation 

-^ a =^ (1) 

79 



80 ALGEBKAIC FUNCTIONS 

Moreover, any point the coordinates of which satisfy (1), 
must be at the distance r from C and hence be a point of the 
circle. Accordingly, (1) is the equation of a circle. 

If (1) is expanded, it becomes 




! -2Aa;-2^ + A 2 +F r 2 =0, (2) 

an equation of the second degree with no term in xy and 

with the coefficients of a; 2 and / a 

equal. 

Conversely, any equation of the 
second degree with no xy term and 
with the coefficients of a; 2 and y* 
equal (as 

O 
where A, G, F, and are any con- ^ 1Q g5 

stants) may be transformed into 

the form (1) and represents a circle, unless the number cor- 
responding to r z is negative (see Ex. 3, page 81), in which 
case the equation is satisfied by no real values of x and y and 
accordingly has no corresponding locus. 

The circle is most readily drawn by making such transfor- 
mation, locating the center, and constructing the circle with 
compasses. 



Ex.1. 

This equation may be written in the form 

(x*-2x ) + (y2-4y ) = 0, 

and the terms in the parentheses may be made perfect squares by adding 
1 in the first parenthesis and 4 in the second parenthesis As we have 
added a total of 5 to the left-hand side of the equation, we must add an 
equal amount to the right-hand side of the equation The result is 

Cr a - 2a: + 1) + (f - 4?/ + 4) = 5, 
which may be placed in the form 

(-l) + <y-a)-5, 
the equation of a circle of radius V5 with its center at the point (1, 2). 



CIRCLE 81 

Ex. 2. 9 a: 2 + 9 f - 9 x + 6 y - 8 = 0. 

Placing 8 on the right-hand side of the equation and then dividing by 
9, we have ** + ,*-*+ 3, = |, 

which may be treated by the method used in Ex. 1. The result is 



the equation of a circle of ladius JV5, with its center at (\, i). 
Ex.3. 9a; 2 + 9y 2 -6a;-|-12y + ll = 

Proceeding as in Ex. 2, we have, as the transformed equation, 



an equation which cannot be satisfied by any real values of js and y, since 
the sum of two positive quantities cannot be negative Hence this equation 
corresponds to no real curve. 

EXERCISES 

1. Find the equation of the circle with the center (4, 2) and 
the radius 3 

2. Find the equation of the circle with the center (0, 1) and 
the radius 5. 

3. Find the center and the radius of the circle 

9 = 0. 



4. Find the center and the radius of the circle 

- 6v - 15 = 0. 



5. Find the equation of the straight line passing through the 
center of the circle 



and perpendicular to the line 

2aj + 3v/ 4 = 

6. Prove that two circles are concentric if their equations differ 
only in the absolute term 

29. Parabola. The locus of a point equally distant from a fixed 
point and a fixed straight line is catted a parabola. The fixed 
point is called the focus and the fixed straight line is called the 
directrix, 





82 ALGEBRAIC FUNCTIONS 

Let F (Fig. 36) be tho focus and A\V the directrix of a 
parabola. Through F draw a straight line perpendicular to 
ftS, intersecting it at JP, and let this lino be the axis of jr. 
Let the middle point of T>F be taken as 0, tho origin of coordi- 
nates, and draw the axis QY. Thon, if tho distance, PA* is 2 <, Urn 
coordinates of F are (0, 0) and tho equation of H$ is sr -~ - r. 

Let /^(.r, y) be any point of tho parabola, and draw Uu 
straight line FP and tho straight lino NP gy 
perpendicular to US. 

Then NP = + , 

and, by 27, FP- V(' <0 2 + //* /J 

whence, from tho definition of tho parabola, 

(a-*)** if- Or +<0" 
which reduces to ,?/ 2 = 4 6'a;. (1 ) ** p x(li ,w 

Conversely, if the coordinates of any point /* satisfy (1), it 
can be shown that the distances FP and NP aru equal, and 
hence P is a point of the parabola. 

Solving (1) for y in terms of a-, we havo 

y = 2VS, (2) 

We assume that e is positive. Thon it is evident that if a 
negative value is assigned to #, y is imaginary, and no correspond- 
ing points of the parabola can bo located. All poKitivo vahu'M 
may be assigned to JR, however, and hence tho parabola IICH 
entirely on the positive sidtj of tho axis OF. 

Accordingly, wo assign positive valuos to #, compute tho nor- 
responding values of y, and draw a smooth curve through tho 
points thus located. 

It is to be noticed that to every value tWHigncd to w lhro aro 
two corresponding valuos of ?/, equal in magnitude and opposite 
in algebraic sign, to which there correspond two points of tho 
parabola on opposite sides of OX and equally dintant from it. 
Hence the parabola is ttymmMoal with rospect to CUT, and ac- 
cordingly OX is called the gunk of the parabola. 

The point at which its axis intersects a parabola is called the w 
tew of the parabola. Accordingly, is the vertex of the parabola. 



PARABOLA 83 

Returning to Fig. 36, if F is taken at the left of with the 
coordinates ( c, 0), and RS is taken at the right of with the 
equation x = c, equation (1) becomes 

f = -kcx (3) 

and represents a parabola lying on the negative side of OY. 
Hence we conclude that any equation in the form 

f=fa, (4) 

where k is a positive or a negative constant, is a parabola, with 

(k 
Z' 

and its directrix the straight line x'-^* 

Similarly, the equation x*= Icy (5) 

represents a parabola, with its vertex at and with its axis coin- 
ciding with the positive or the negative part of OY, according 
as & is positive or negative. The focus is always the point 

0, -^ j and the directrix is the line y = - whether k be positive 

^"Y "i 

or negative. 

30. Parabolic segment. An important property of the parabola 
is contained in the following theorem : 

The square of any two chords of a parabola which are perpen- 
dicular to its axis are to each other as their distances from the 
vertex of the parabola. 

This theorem may be proved as follows : 
Let %(%!, #j.) and P z (x^ 7/ 3 ) be any two points of any parab- 
ola /= Tex (Fig. 37). 

Then y*= lex^ 

and yl 7c a ; 

y\ X* 
whence , ~ 

yl ** 

whence 



84 



ALGEBRAIC FUNCTIONS 



From the symmetry of the parabola, 2# 1 = 
But a? = OM^ and x,== OM Z , and hence 
(1) becomes 



and 2# 2 = 




FIG. 37 



and the theorem is proved. 

The figure bounded by the parabola 
and a chord perpendicular to the axis 
of the parabola, as Q^OP^ (Fig. 37), 
is called a parabolic segment. The 
chord is called the lase of the segment, the vertex of the 
parabola is called the vertex of the segment, and the distance 
from, the vertex to the base is called the altitude of the segment. 

EXERCISES 
Plot the following parabolas, determining the focus of each : 

1. !/* = 8x 3. 2/ 2 =:6a5. 

2. cc 2 =42/. 4. aj 2 = -7y. 

5. The altitude of a parabolic segment is 10 ft., and the length of 
its base is 16 ft. A straight line drawn across the segment perpen- 
dicular to its axis is 10 ft. long. How far is it from the vertex of 
the segment ? 

6. An arch in the form of a parabolic curve, the axis being 
vertical, is 50 ft. across the bottom, and the highest point is 15 ft. 
above the horizontal. What is the length of a beam placed horizon- 
tally across the arch 6 ft. from the top ? 

7. The cable of a suspension bridge hangs in the form of a 
parabola. The roadway, which is horizontal and 400ft. long, is 
supported by vertical wires attached to the cable, the longest wire 
being 80 ft. and the shortest being 20 ft. Find the length o a 
supporting wire attached to the roadway 75 ft. from tho middle. 

8. Any section of a given parabolic mirror made by a plane 
passing through the axis of the mirror is a parabolic segment of 
which the altitude is 6 in. and the length of the base 10 in. Find 
the circumference of the section of the mirror made by a piano 
perpendicular to its axis and 4 in. from, its vertex 



ELLIPSE 85 

9. "Find the equation of the parabola having the line x = 3 as its 
directrix and having its focus at the origin of cooidmates. 

10. Find the equation of the parabola having the line y 2 as 
its directrix and having its locus at the point (2, 4). 

31. Ellipse. The locus of a point the sum of whose distances 
from two fixed points is constant is called an ellipse. The two 
fixed points are called the foci. 

Let F and F' (Fig. 38) be the two loci, and let the distance 
F'F be 2 c. Let the straight line determined by F' and F be 
taken as the axis of x, and the 
middle point of F'F be taken 
as 0, the origin of coordinates, 
and draw the axis OY. Then 
the coordinates of F' and F 
are respectively ( c, 0) and 
(*, 0). 

Let P(x, y) be anj' point 
of the ellipse, and 2 a repre- n' 

sent the constant sum of its 00 

Ju ICr uO 

distances from the foci. Then, 

from the definition of the ellipse, the sum of the distances F'P 
and FP is 2 a, and from the triangle F'PF it is evident that 
2 a > 2 G ; whence a > c. 




By 27, 

and FP= 

whence, from the definition of the ellipse, 

V(s-M) 2 +# 2 W(a-c) a +/= 2 - (1) 

Clearing (1) of radicals, we have 

O 2 - c a > 2 + ay = a 4 - aV. (2) 

Dividing (2) by a* aV, we have 



86 ALGEBRAIC FUNCTIONS 

But since ><?, a* c* is a positive quantity which may be 
denoted by J 3 , and (3) becomes 



Conversely, if the coordinates of any point P satisfy (4), it 
can be shown that the sum of the distances Jf'P and FP ia 2 a, 
and hence P is a point of the ellipse. 

Solving (4) for y in terms of #, we have 

y=s >/-** (5) 

It is evident that the only values which can be assigned to x 
must be numerically less than a; for if any numerically larger 
values are assigned to r, the corresponding values of y are 
imaginary, and no corresponding points can be plotted. Hence 
the curve lies entirely between the lines x = a and x = a. 

We may, then, assign the possible values to #, compute the 
corresponding values of y, and, locating the corresponding points, 
draw a smooth curve through them. As in the case of the pa- 
rabola, we observe that OX is an axis of symmetry of the ellipse. 

We may also solve (4) for x in terms of y, with the result 

as = 5^-^. (0) 

From this form of the equation we find that the ellipse lies 
entirely between the lines ,y = --iandy = 5 and is symmetrical 
with respect to OY. 

Hence the ellipse has two axes, A' A and B'JB (Fig. 88), which 
are at right angles to each other. But A 1 A = 2 a and JR'13 = 2 1) ; 
and since a > 5, it follows that A' A > B'B, Hence A' A is called 
the major axis of the ellipse, and fl'JJ is called the minor axis 
of the ellipse. 

The ends of the major axis, A' and A, are called the wrtwes of 
the ellipse, and the point midway between the vertices is called 
the center of the ellipse ; that is, is the center of the ellipse, 
and it can be readily shown that any chord of the ellipse which 
passes through is bisected by that point. 



ELLIPSE 87 

From the definition of />, c=* Va a #*, and the coordinates of 
the foci are ( V<r A 9 , 0). 

OF 
The ratio ----- (thai is, the ratio of the distance of the focus 

from the center to the distance of either vertex from the center) 
is called the eccentricity of the ellipse and is denoted by t>. lint 

M'^VtfCjS, (7) 

Vrt a i* 

and hence c - - - > (8) 

\ / 

whence it followa that the eccentricity of an ellipse is always 
less than unity. 

Similarly, any equation in form (4), in which I' 2 > a 2 , represents 
an ellipse with its center at 0, its major axis on 6>J", and its 
minor axis on CLY. Then the vertices are the points (0, &), 



~ 

the foci are the points (0, vV *), and e = 

In either case the nearer the foci approach coincidence, the 
smaller e becomes and the more nearly b = a. Hence a drde 
inat/ le considered an an ellipse with dointxdfnt fovi and equal axes. 
Its eceentrieity is, of course, zero. 

EXERCISES 

Plot tho following ellipses, finding the vertices, the Toci, and the 
eceentrieity of eaoh : 

1. 9 ^ a + 1 if 144. 3. 3 tc 8 + 4 ?/ 2. 

2. 9 w! 8 + 4 ?/ 30. 4. 2 a; 3 + 3 y a s= 1. 



6. Find tho equation of the ellipse winch lias its foci at tlie points 
( 2, 0) and ((>, 0) and which has the sum of the dis lances of any 
point on it from the foci equal to 10. 

6. Find the equation of the ellipse having its food at the points 
(0, 0) and (0, ) and having the length of its major axis equal to 7. 

32. Hyperbola. The locus of a point the difference of whose 
, distances from two faced points is constant is called a hyperbola. 
The two fixed points are called the foci* 



88 



ALGEBRAIC FUNCTIONS 



Let F and F' (Fig. 30) be the two foci, and lot tho distance 
F'F bo 2 0. Let the straight line determined by F' and F bo 
taken as the axis of .P, and the middle point of F'F be taken 
as 0, the origin of coordinates, and draw the axis O Y. Then 
the coordinates of F 1 and 
F are respectively ( <?, 0) 
and (c, 0). 

Let jP (a;, ?/) be any point 
of the hyperbola and 2 a 
represent the constant dif- 
ference of its distances from 
the foci. Then, from the 
definition of the hyperbola, . , 
the difference of the dis- ' 
tances F'P and FP is 2 a, 
and from the triangle F'PFii is evident that 2 < 2 , for the 
difference of any two sides of a triangle IH loss than the third 
side ; whence a < a. 




1?ia * 80 



By 27, 

and 

whence either 



;*+; 



2 (1) 

or V(* + 0'+^-V(? ::: ^+7-2a, (2) 

according as JRP or ^"/* in the groatcsr diHtance. 

Clearing either (1) or (2) of radicals, wo obtain tho aamo 
result : 



Dividing (3) by a* V, we have 

i it _ "1 fA.\ 

' "T a * * \*J 

cr r 

But since <c, 2 * is a negative quantity whiuh may bo 
denoted by ft a , and (4) becomes 



HYPJEEBOLA 89 

Conversely, if the coordinates of any point P satisfy (5), it 
can be shown that the difference of the distances F'P and FP 
is 2 a, and hence P is a point of the hyperbola. 

Solving (5) for y in terms of #, we have 

y = -^i?-d\ (6) 

In this equation we may assume for x only values that are 
numerically greater than a, as any other values give imaginary 
values for y. Hence there are no points of the hyperbola be- 
tween the lines t = a and x = a. The hyperbola is symmetrical 
with respect to OX. 

As the values assigned to x increase numerically, the corre- 
sponding points of the hyperbola recede from the axis OX. We 
may, however, write (6) m the form 

5? 

(7) 

Now if y x and i/ a are the ordinates of points of (7) and of 
the straight lines y = -x respectively, then 

w 7 

oa 




whence Lim (# a 



Hence, by prolonging the straight lines and the curve indefi- 
nitely, we can make them come as near together as we please. 

Now, when a straight line has such a position with respect 
to a curve that as the two are indefinitely prolonged the dis- 
tance between them approaches zero as a limit, the straight line 
is called an asymptote of the curve. It follows that the lines 

y = -x and y => as are asymptotes of the hyperbola (Fig. 39). 

(t CL 

If we had solved (5) for a in terms of y, the result would 
have been 

(8) 



90 ALGEBRAIC FUNCTIONS 

from which it appears that all values may be assigned to #, and. 
that OT is also an axis of symmetry of the hyperbola. 

The points A' and A in which one axis of the hyperbola inter- 
sects the hyperbola are called the vertices, and the portion of the 
axis extending from A' to A is called the transverse axis. The 
point midway between the vertices is called the center ; that is, 
is the center of the hyperbola, and it can readily be shown 
that any chord of the hyperbola which passes through O is 
bisected by that point. The other axis of the hyperbola, which 
is perpendicular to the transverse axis, is called the conju- 
gate axis. This axis does not intersect the curve, as is evident 
from the figure, but it is useful in fixing the asymptotes and 
thus determining the shape of the curve for large values of x. 

From the definition of 5, c = Va 2 + 6 2 , and the coordinates of 
the foci are (V 2 +5 2 , 0). Therefore 



If we define the eccentricity of the hyperbola as the ratio 

OF , , _ 

> we have a = ^+* ( 1 0) 



a quantity which is evidently always greater than unity. 
Similarly, the equation 2 



is the equation of a hyperbola, with its center at 0, its trans- 
verse axis on OF, and its conjugate axis on OX. Then the ver- 
tices are the points (0, 5), the foci are the points (0, V& 2 +a 2 ), 

the asymptotes are the straight lines y = -#, and e = 7" 

Cv 

If 5 = a, in either (5) or (11), the equation of the hyperbola 
assumes the form 

a 2 -2/ 2 =a 2 or y-a^=:a a , (12) 

and the hyperbola is called an equilateral "hyperbola. The equa- 
tions of the asymptotes become y =* x ; and as these lines 
are perpendicular to each other, the hyperbola is also called a 
rectangular hyperbola. 



CURVES 
EXERCISES 



91 



Plot the following hyperbolas, finding the vertices, the foci, the 
asymptotes, and the eccentricity of each . 



2. 9o; a 4?/ a =36. 5. 

3. 32/ a -2jc a =6 6. 

7. Find the equation of the hyperbola having its foci at the points 
(0, 0) and (4, 0), and the difference of the distances of any point on 
it from, the foci equal to 2. 

8. The foci of a hyperbola are at the points ( 4, 2) and (4, 2), 
and the difference of the distances of any point on it from the foci 
is 4. Eind the equation of the hyperbola, and plot. 

33. Other curves. In the discussion of the parabola, the ellipse, 
and the hyperbola, the axes of symmetry and the asymptotes 
were of considerable assistance in constructing the curves ; more- 
over, the knowledge that there could be no points of the curve 
in certain parts of the plane decreased the labor of drawing 
the curves. We shall now plot the loci of a few equations, 
noting in advance whether the curve is bounded in any direc- 
tion or has any axes of symmetry or asymptotes. In this way 
we shall be able to anticipate to a con- ^ 

siderable extent the form of the curve. 



Ex. 1. (y + 3) a = (x - 2) a (a; + 1). 
Solving for y, we have 



In the first place, we see that the only 
values that may be assigned to x are greater 
than 1, and hence the curve lies entirely on 
the positive side of the line x = 1. Further- 
more, corresponding to every value of a?, there 
are two values of y which determine two points at equal distances from 
the line y = 8. Hence we conclude that the line y 8 is an axis of 
symmetry of the curve. 

Assigning values to x and locating the points determined, we draw the 
curve (JFig. 40). 




< 49 



92 ALGEBKAIO FUNCTIONS 

Ex. 2. xy = 4. 

4. 

Solving foi y, we have y = - 

x 

It is evident, then, that we may assign to x any real value except xero, 
in which case we should be asked to divide 4 by 0, a process that cannot 
be carried out. Consequently, there can be no point of the curve on the 
line # = 0; that is, on OY. We may, 
however, assume values for x as near 
to zero as we wish, and the nearer they 
are to zero, the nearer the corresponding 
points are to OF; but as the points 
come nearer to OY they recede along 
the curve. Hence OF is an asymptote x^ 

of the curve. 

If we solve for x, we have 



Fm. 41 
and, reasoning as above, we conclude 

that the line y = (that is, the axis OX) is also an asymptote of the curve. 

The curve is drawn in Fig. 41. It is a special case of the curve 
xy = k, where k is a ical constant which may be either positive or negative, 
and is, in fact, a i octangular hyperbola leferiod to its asymptotes as axes. 

It is customaiy to say that when the denominator of a fraction is SMTO, 
the value of the fraction becomes infinite The curve just constructed 
shows graphically what is meant by such 
an expiession. 

Ex. 3. sry + 2 or + y 1 = 0. 
Solving for y, we have 



y 



from which we conclude that the line 
a? = 1 is an asymptote of the curve. 
Solving for x, we have 




Fio. 42 



2+y 

from which we conclude that the line ?/= 2 is also an asymptote of the curve. 

We accordingly draw these two asymptotes (Fig. 42) and the curve 
through the points determined by assigning values to either x or y and 
computing the corresponding values of the other vai table. 

The curve is, in fact, a rectangular hyperbola, with the lines x 1 
and y SB 2 as its asymptotes. 



CUEVES 



Ex.4. 



a- 8 
X 



Solving for y, we have 

Vx* 
5 ' 
2a~a; 



whence it is evident that the curve is sym- 
metrical with respect to OX The lines x 
and x = 2 a, corresponding to tlie values of a? 
which make the numuiatoi and the denomi- 
nator of the fraction under the ladical sign 
lespectively zero, divide the plane into three 
stups; and only values between and 2 a 
can be substituted for y, since all other values 
make y imaginaiy. It follows that the curve 
lies entirely in the strip bounded by the two 
lines x and x 2 a. 

By the same reasoning that was used in 
Exs 2 and 3, it can be shown that the line 
x = 2 a is an asymptote of the curve. 

The curve, which is called a cuboid, is drawn 
in Fig. 43. 



EXERCISES 



Plot the following curves : 

1. 2/ a =o; 8 

2. 2/ a =aj a (te + 4). 

3. 2/ a =4(o;-8). 

4. y a =a: a 5cc-f 6. 
6. y z = a) (a? 2 4). 

6. / a =aj 8 




FIG. 43 



7. y a = 

8. 7/ 2 =4 a?. 

9. xy 5. 

10. 3y oj// = 

11. xij 2aj + 



34. Theorems on limits. In. obtaining more general formulas 
for differentiation, the following theorems on limits will be 
assumed without formal proof : 

1. The limit of the sum of a finite number of variables is equal 
to the. sum of the limits of the variables. 

2. The limit of the product of a finite number of variables is 
egual to the product of the limits of the variables. 



94 ALGEBRAIC FUNCTIONS 

3. The limit of a constant multiplied by a variable is equal to 
the constant multiplied by the limit of the variable. 

4. The limit of the quotient of two variables is equal to the 
quotient of the limits of the variables, provided the limit of the 
divisor is not zero. 

35. Theorems on derivatives. In order to extend the process 
of differentiation to functions other than polynomials, we shall 
need the following theorems : 

1. The derivative of a constant is zero. 
This theorem was proved m 8. 

2. The derivative of a constant times a function is equal to the 
constant times the derivative of the function. 

Let u be a function of x which can be differentiated, let c be 
a constant, and place ,. _ . 

* U -^ G&6i 

Give x an increment Ax, and let AM and Ay be the corre- 
sponding increments of u and y. Then 

Ay = c (u + AM) cu =s c AM. 

TT Ay AM 

Hence - = <, 

Ax Ax 

and, by theorem 3, 34, 

T . Ay T . AM 

Lim -T- C Lim -- 

Ax Ax 

Therefore ~i~ c T~ 

ax dx 

by the definition of a derivative. 
Ex. 1. y = 5 (a; 3 + 3 x z + 1). 



3. The derivative of the sum of a finite number of functions is 
egual to the sum of the derivatives of the functions. 

Let M, v, and w be three functions of x which can be differen- 
tiated, and let 



DERIVATIVES 95 

Give x an increment Aa;, and let the corresponding increments 
of u, t>, w, and y be AM, Av, Aw, and A#. Then 

Ay = (M H- Aw + v + At) + w + Aw) (w + v + w) 
= Aw + Aw + Aw ; 

1 Ay AM , Aw , Aw 

whence ^ = -7 + T~~ + 7 

A# Aa; Aa; Aa; 

Now let Aa; approach zero. By theorem 1, 34, 

, . Ay T . AM , T A0 . 7 Aw 
Lini ^ = Lim - - + Lim - + Lim - ; 
Aa; Ax Ax Ax 

that is, by the definition, of a derivative, 

dy __ du dv dw 
duo dx dx dx 

The proof is evidently applicable to any finite number of 
functions. 

Ex. 2. y = x* - 3 x s + 2 x z - 7x. 



4. The derivative of the product of a finite number of functions 
is equal to the sum of the products obtained "by multiplying the 
derivative of each factor by all the other factors. 

Let u and v be two functions of x which can be differentiated, 
andlet y = uv. 

Give x an increment Aa;, and let the corresponding increments 
of u, v, and y be Aw, Av, and A#. 

Then Ay (it + Aw) (v + Av) uv 

= u Av + v AM + AM Av 

, Ay Av Au . AM A 

and = w * + T~ + T-" ^ 

Aa? Ax Ax Ax 



96 ALGEBRAIC FUNCTIONS 

If, now, Ax approaches zero, we have, by 34, 

r Aw T . Aw , T Aw , T . AM T . A 
Lim -~ = u Lim - + v Lim - + Lim - Lim Aw. 
Ao; Aa; Aa; Are 

But Lim Aw = 0, 

-. ,, P dy dv , du 

and theretore -^- = w 4- w 

ax ax ax 

Again, let y = uvw. 

Regarding uv as one function and applying the result already 
obtained, we have 



dy dw , d(uv) 
-JL = m - + w -+ 
dx ax at 

dw \ du . < 
= uv-=--)-w\u + v 
dx \_ dx t 



tu? 
dw dv du 

= UV + UW -f VW -r- 

ax dx dx 

The proof is clearly applicable to any finite numbers of factors. 
Ex. 3. y = 3a:- 



dx 

= (3 x - 5) (z" + 1) (3 a; 2 ) + (3 x - 5)a; 8 (2 a;) + (a: 2 + I)* 8 (3) 
= (18 a; 8 - 25 a; 2 + 12 a: - 15)a: 2 

5. The derivative of a fraction is equal to t7ie denominator times 
the derivative of the numerator minus the numerator times the deriva- 
tive of the denominator, all divided ty the square of the denominator. 

nt 

Let y -> where u and v are two functions of x which can be 
v 

differentiated. Give x an increment Aa;, and let Aw, Av, and A?/ 
be the corresponding increments of w, v, and y. Then 

. _ u + Aw _ u _ v Aw w Aw 
v-f-A'y w w a +'uAw 

An Az> 
w- -- u- 

and % = A.; A^^ 

Ao: v a + w Aw 



DERIVATIVES 97 

Now let A* approach zero. By 34, 

T . A-M T . A?J 
v Liin - -- u Lim - 
A/ A Aa; 

Lim - = ,, T . , > 

Aa; tr+vLmiAv 

du dv 

01 _________ rilj _ l rty , 

efa/ <fcc difo 

whence - = 

dx v* 

T 8 1 

Ex.4. y = ~ ~ 

J x* + l 

(^_(a, 2 + l)(27:)-(a; 2 -l)2r_ 4 .1? 
dx (x* + I) 3 (a,' J + I) 2 

6. TJie derivative of the nth power of a function is obtained ly 
multiplying n times the (nT)th power of the function ly the 
derivative of the function. 

Let y = u n , where u is any function of x which can be differ- 
entiated and n is a constant. We need to distinguish four cases : 

CASE I. When n is a positive integer. 

Give x an increment Aa, and let AM and A?/ be the corre- 
sponding increments of u and y. Then 

Ay = (u 4- Aw)" u n ; 
whence, by the binomial theorem, 



A?/ n _iAw , n(n 1) B _ 2A AM , 

^L = nu n l ~ H ft y w w 2 Au- h . 

Aa; Aa? 2 Aa; ' Aa; 

Now let Aa;, AM, A?/ approach zero, and apply theorems 1 and 

2, 34. The limit of ^ is ^ the limit of ~ i s ^*, and the 
Aa; a* Aa; ajc 

limit of all terms except the first on the right-hand side of 
the last equation is zero, since each contains the factor AM. 
Therefore d du 

^. = nw n ~ 1 - T - < 
ax dx 



08 ALGEBKAIC FUNCTIONS 

CASE II. When n is a positive rational fraction. 

ny 

Let n = where p and q are positive integers, and place 



By raising both sides of this equation to the gth power, we have 

y=u. 

Here we have two functions of x which are equal for all 
values o x. 

Taking the derivative of both sides of the last equation, we 
have, by Case I, since p and q are positive integers, 

^-1^==*-^. 
yy dx P dx 

Substituting the value of y and dividing, we have 

f^itf- 1 **. 

* dx q dx 

Hence, in this case also, 

dy _ ,du 
-^- nu n ~ l 
dx ax 

CASE III. When n is a negative rational number. 
Let n m, where m is a positive number, and place 

ys=U~ m s= -- 
9 u m 



Then ^= ** (by 5) 

dx u zm 

mu m ~ l 

= (by Cases I and II) 

u 

mu~ m ~^~r 
dx 

Hence, ui this case also, 

d_ _i^w 
dx dx 



DERIVATIVES 99 

CASE IV. When n is an irrational number. 

The formula is true in this case also, but the proof will not 
be given. 

It appears that the theorem is true for all real values of n. 
It may be restated as a working-rule in the following words: 

To differentiate a power of any quantity, bring down the exponent 
as a coefficient, write the quantity with an exponent one less, and 
multiply ly the derivative of the quantity. 



Ex. 5. y = (x s + 4 x* - 5 x + 7) s . 



^ = 3 (a: 8 + 4 a 2 - 5 x + 7) 2 -f (a; 3 + 4 x z - 5 x + 7) 

tlJG U3S 

= 3 (So; 2 + Bx - 5) (.e 3 + 4a: 2 - 5 x + 7) 3 . 



Ex. 6. y = Vtf + -. = ^ + x- 

* 



dx 3 

2 3 

*' 



Ex. 7. y = (a: + l)Va; 8 



= (z + 1) [ J (a; 2 + I)" * - 2 ar] + (a 2 + 



1 
(a 2 + !) 

2 a 8 + x 4- 1 




Y /Li 



100 ALGEBRAIC FUNCTIONS 

7. If y is a function of x, then x is a function of y, and the 
derivative of x with respect to y is the reciprocal of the derivative 
of y with respect to x. 

Let A# and A?/ be corresponding increments of x and y. It is 
immaterial whether b*.x is assumed and A?/ determined, or A# 
is assumed and Aa; determined. In either case 

A.r 1 
A^~"Ay' 
Aa; 

. _. A,r 1 

whence Lun = A ; 

Ay T A?/ 

' Lim - 

Aas 

dx 1 
that is, -- = - 



8. If y is a function of u and u v' function of a*, then y is 
a function of x, and the derivative of y with respect to x is equal 
to the product of the derivative of y with respect to u and the 
derivative of u with respect to x. 

An increment Aa; determines an increment Aw, and this in turn 
determines an increment Ay, Then, evidently, 

A?/ __ A?/ Aw 
Ax Aw A:# 

, T . Aw , . A?/ T . AM 

whence Lim ~ = Lim - Lim . ; 

A:B A?t Arc 

, , , . dy dy du 

that is, ^ .X. . 

dx du dx 



Ex. 9. w a u 3 + 8 u + 1, whoro w = - 



1+ 



The same result is obtained by fmbHtituting in the expression for y the 
value of u in terms of a; and then differentiating. 



DERIVATIVES 101 

36. Formulas. We may now collect our formulas of differen- 
tiation in the following table : 

^ = 0, (1) 

dx 

rf co = tf *, C2 ) 

dx dv 

d(u4-v) _du dv n 

dx dx dx 

d Cuv) dv , du , , , 

\ - U T + V 1T' ( 4 ) 

dx dx dx 



, /?/\ du dv 

d(-) v- -- M 

\v/ dx dx 






dy^y. d* t /g-) 

dx du dx ^ ^ 

dy 

^ = (9) 

dx dx 

du 

Formula (9) is a combination of (7) and (8). 
The first six formulas may be changed to corresponding for- 
mulas for differentials by multiplying both sides of each equation 

by dx. They are 7 n _< - v 

J J da = 0, (10) 

(11) 

Zw, (12) 

d (uv) udv + v du, (1 3) 

w\ vdu udv 



102 ALGEBRAIC FUNCTIONS 

EXERCISES 
Find -~ in each of the following oases : 




2. y s (a; 2 - 2 1 + 3) (a: 3 +60: + 9). 
14. 



H-l 
"** 



7 . 

8 X?/ = (4 jr fl + 3 x + I) 3 - 

4 X* + 1. 




16. > re V<.) 

17 ' '/ = (' + 
18. 




37. BijEferentiatlon of implicit functions. Consider any equa- 
tion containing two variables a? and ,?/, If one of them, EH jr, in 
chosen as the independent variable and a value is assigned in 
it, the values of y are determined, Hence tho given equation 
defines y as a function of rr. If the equation is solved for y in 
terms of a, y is called an earpKoit function of x, If the equation 
is not solved for y, y is (jailed an implidt function of #* For 
example, ^+8aM-4y + 4^ + 2^ + 4^0, 
which may be written 

# 9 + (4 aj + 2)y + (8 .-K 9 -|- 4 4- 4) 0, 
defines y as an implicit function of a?. 

If the equation is solved for y, the renult 

# 2j lVa^T8 
expresses y as an explicit function of #. 



IMPLICIT FUNCTIONS 103 

If it is required to find the derivative of an implicit function, 
the equation may be differentiated as given, the result being an 
equation which may be solved algebraically for the derivative. 
This method is illustrated in the following examples : 

Ex. 1. z 2 + 2 = 5. 



If x is the independent variable, 



that is, 2 x + 2 y- = 0, 

dx 

whence Jl = -. 

dx y 

Or the derivative may be found by taking the differential of both sides, 
as follows: + /) = d(5) = 0; 



that is, 2 xdx + 2 ydy = 0, 

whence ^=_. 

dx y 

It is also possible first to solve the given equation f or y, thus . 

y = V5-ff 2 ; 
whence 



rfa; V5 a: 8 
a result evidently equivalent to the result previously found. 

The method of finding the second derivative of an implicit 
function is illustrated in the following example : 

Ex. 2. Find ^f if a a + f = 5 

(IJS 

We know from Ex. 1 that ^L = _ 2 . 

dx y 

Therefore fSUfLft 

dor ax \y/ 




.*-*(!) 



_ ?y a + a- 8 _ 5 
f 2/ 8 ' 

since ^ 2 + a; 2 = 5, from the given equation 



104 ALGEBRAIC FUNCTIONS 

EXERCISES 

"Find ~ from each of tho following equations : 

CtJC 

1. a* + / - 3 *y = 0. 3. v/ = ~~- 

j -|- // 



2. aj 2 // -f- 4 V = 8 a 9 . 4. V // + ' + V ^ ,r -- . 

Pind ^ and l -~. from each oJ' tho following (wuiationa : 



6. 2x a 

6. 4 a- 3 9v/ a =3C. 

7 . ai+jf-a. io - 

8. a* + * = <A 11. 



38. Tangent line. Let J^C^'j, ^O bo a clioHou point, of any 
curve, and lot (^) bo the value of '/- when aj=sw ( . and ;/-://,. 

V ^V v / // >* ^ ^ 

/x..\ VWi <M 

Then f-^J is the slopo oC tho curve at, iho point J^ and also 
\dx/\ 

the slopo of tho tan gout line ( 1H) to tho ourvo at that point, 
Accordingly, tho equation of the tang-out line at J{ m ( 15) 



Ex. 1. Find tho equation (if tho lungojit lino to tho parabola / 8 ss JJ .r at 
tho point (3, ). 

By differentiation wo have 



whence //SB.'-., 

. ti // 

Hence, at tho point (3, JJ), tho Hlopn of tho tangont lino is J, and itH 
equation is y-3-1 (*-) 

or ar 2 ?/ + 3 = 0. 

The angle of mteraeotion of two curves is tho angle betwoon 
their respective tangontu at the point of iutenwotiun. The 
method of finding tho anglo of intersection i illustrated in the 
example on. tho following pago* 



TANGENT LINE 105 

Ex. 2. Find the angle of intersection of the cncle # 2 + # 2 = 8 and of 
the parabola ar 2 = 2 y. v 

The points of intersection are P l (2, 2) 
and />(- 2, 2) (Fig 44), and fioiu the sym- ^ 

nietry of the diagram it is evident, that the 
angles of intersection at P l and P z aie the 
same. 

Diffeientiating the equation of the circle, 

we have 2 JT + 2 y = 0, whence = ; 

' (Ijc ux y 

and differentiating the equation of tlTb pa- 

labola, we find = i. Fid 44 




Hence at P i the slope of the tangent to the circle 1 is 1, and the slope 
of the tangent to the parabola is 2 

Accordingly, if /? denotes the angle of intersection, by Ex 11, p 35, 



or y = tan~ 1 3. 

EXERCISES 

1. Find the equation of the tangent lino to the ourve x 8 
+ 16 y - 8 = at the point (2, 2). 

2. Find the equation of the tangent lino to the p,urve 5 a; 2 4 a- 2 // 
= 4 y 8 at the point (2,1). 

3. Find the point at which the tangent to the curve 8 y = a; 8 at 
(1, &) intersects the curve again. 

4. Find the angle of intersection of the tangents to the curve 
y z = x 9 at the points for which x = 1 

9T If 

5. Show that the equation of the tangent to the ellipse ~j + ^ = 1 
at the point (a^, y,) is + -M = L 

6. Show that the equation of the tangent to the hyperbola 
| - |j = 1 at the point fo, ^) is ^ - %& = I. 

Y. Show that the equation of the tangent to the parabola ?/ = kx 

1: 
at the point (x lt y^ is y$ = ^ (as + a 1 ,). 

Draw eacli pair of the following curves in one diagram and deter- 
mine the angles at which, they intersect : 



106 ALCKEBRAK! FUNCTIONS 

8. a: 9 + y j = S, a- 9 + if - U ^ H- 4 >/ - f> 0. 

9. a; a = 3 #, U // a 8 a-. 

10. y fl ~ 4 '; <* 'I' //" -- fi 

11. y = 2.r, .<//- 18. 

12. x 4 // J 0, .r - - -1 j' 4 if 0. 

13. aj a +y 9 ~2, l >, .<' a -//- ". 

39. The differentials <?#, dy, ds. On any given <uw li-t tin- 
distance from somo Jixod initiul point moasmvd alnnjjf the cur\i 
to any point P bo donoiwd by , whcro is posit.ivt* if /' lifs in 
one direction from the initial point, and negative if /' lies in I In- 
apposite direction. Tho (jlit)ico of tho ^ 

positive direction is purely arbitrary. ./ 

We shall take as the positive direc- 
tion of the tangent that which shown 
the positive direction of the ourvo, 
and shall denote the angle between // 

the positive direction of (L\ and the ' 

positive direction of the tangent by 0, 

Now for a fixed curve and a fixed 
initial point the position of a point /' p, (Ji 

is determined if a is given. Hence & 
and y, the coordinates of J\ are functions of which in 
are continuous and may bo differentiated. \V Hhall now Mhuw tlmt 
dx , dy . , 

-7- COS G>, - aa BUI O, 

ds </ 

Let arcP0=Afl (Fig. 45), whenj 7* and Q are so <hnm thai 
As is positive. Then 1'R AJJ and A*<y A^/, and 



'/f 




As arc /'( tiro /%; chord /</ 

" 00 



Aa 

chord PQ 



MOTION IN A CURVE 107 

We shall assume without proof that the ratio of a small chord 
to its arc is very nearly equal to unity, and that the limit of 

"=1 as the point Q approaches the point P along the 



curve. At the same time the limit of RPQ = <f>. Hence, taking 
limits, we have dx d 

^ = cos & ^ = sm ^ (1) 

ttrO l*O 

If the notation of differentials is used, equations (1) become 

dx = ds cos 0, dyds' sin $ ; 

whence, by squaring and adding, we obtain the important equation 

ds = dx + dy . (2) 

This relation between the differentials of #, y, and s is often rep- 
resented by the triangle of Fig. 46. This figure is convenient as a 
device for memorizing formulas (1) and (2), but it should be borne 
m mind that RQ is not rigorously y 
equal to dy ( 20), nor is PQ rigor- 
ously equal to ds. In fact, RQ = Ay, 
and PQ = As ; but if this triangle is 
regarded as a plane right triangle, 
we recall immediately the values of 
sin^>, cos<, and tan< which have 
been previously proved. 

40. Motion ia a curve. When a o JL 

, , . .IT FIG. 46 

body moves m a curve, the discus- 

sion of velocity and acceleration becomes somewhat complicated, 
as the directions as well as the magnitudes of these quantities 
need to be considered. We shall not discuss acceleration, but 
shall notice that the definition for the magnitude of the velocity, 
or the speed, is the same as before (namely, 

ds 

V=S 3I' 
dt 

where s is distance measured on the curved path) and that the 
direction of the velocity is that of the tangent to the curve. 





108 ALGEBRAIC I UNCTIONS 

Moreover, as the body moves along a curved path through a 
distance PQ=&s (Fig. 47), x changes by an amount PII&K, 
and y changes by an amount 
RQ=&y. We have then 

Lim = -^ = <y = velocity of 

A at 

the body in its path, 

Ax dx , 

Lim = -7- = v x = component 
At dt 

of velocity parallel to OX, 

Lim ^ = ^ = v v = component 

At dt J r FIG 47 

of velocity parallel to Y. 

Otherwise expressed, v represents the velocity of P, v r the 
velocity of the projection of P upon 0-3T, and v v the velocity 
of the projection of P on OY. 

Now, by (8), 36, and by 39, 

_ dx _ dx ds 
x dt ds dt 

= V COS <, (1) 

, dy dy ds 

and v. . ~ = -* 

y dt ds dt 

= v sm <p. (2) 

Squaring and adding, we have 

v*=v*+ V ;. (3) 

Formulas (1), (2), and (3) are of especial value when a par- 
ticle moves in the plane XO F, and the coordinates x and y of 
its position at any time t are each given as a function of t. The 
path of the moving particle may then be determined as follows : 

Assign any value to t and locate the point corresponding to 
the values of x and y thus determined. This will evidently be 
the position of the moving particle at that instant of time. In 
this way, by assigning successive values to t we can locate 
other points through which the particle is moving at the corre- 
sponding instants of time. The locus of the points thus deter- 
mined is a curve which, is evidently the path of the particle. 



MOTION IN A CURVE 109 

The two equations accordingly represent the curve and are 
called its parametric representation, the variable t being 1 called a 
parameter.* By (9), 36, the slope of the curve is given by 
the formula j al 



dx d v x 
dt 

In. case t can be eliminated from the two given equations, the 
result is the (#, /) equation of the curve, sometimes called the 
Cartesian equation; but such ehmmation is not essential, and 
often is not desirable, particularly if the velocity of the particle 
in its path is to be determined. 

Ex. 1. A particle moves m the plane XOY so that at any time t, 
a; = a + bt, y = c + dt, 

where a, &,' c, and d are any real constants Determine its path and its 
velocity in its path 

To determine the path we eliminate t from the given equations, with 

the result 

d f .. 
y c = -(x d), 

the equation of a straight line passing through the point (a, c) with the 

i d 
slope - 

In this case the path may also be determined as follows Fiom the given 

equations we find dx l dt, and dy ddt\ whence = '- As the slope of 

/ rA *' 

the path is always the same (that is, -I, the path must be a straight line 

which passes through the point (a, c) the point determined when t = 0. 
To determine the velocity of the particle in its path we find, by differ- 
entiating the given equations, 

dx , di/ . 

v *=jr l > v = dt = d > 

whence, by (3), v = V& 2 H- d\ 

Hence the particle moves along the straight line with a constant velocity. 

* It may be noted m passing that the parameter in the parametiic represen- 
tation of a curve is not necessarily time, but may be any third variable in terms 
of which a and y can be expressed. 



110 ALGEBRAIC FUNCTIONS 

Ex. 2. If a projectile starts with an initial velocity v in an initial direc- 
tion which makes an angle a with the axis of x taken as horizontal, its 
position at any time t is given by the parametnc equations 
x v t cos a, y = v t sin a \ gt z . 

JTind its velocity in its path. 

dx 

We have v x = = v cos a, 

at 

dy 
v v = ~f t ~' 

Hence = Vi> 2 gv t sin a + g*P. 

EXERCISES 

1. The coordinates of the position of a moving particle at any 
time t are given by the equations x = 2 1, y t s Determine the path. 
of the particle and its speed in its path 

2. The coordinates of the position of a moving particle at any 
time t are given by the equations x t 2 , y = t + 1 Determine the 
path of the particle and its speed in its path. 

3. The coordinates of the position of a moving particle at any 
time t are given by the equations x = 2 1, y == f t $ t*. Determine 
the path of the particle and its speed in its path. 

4. At what point of its path will the particle of Ex. 3 be moving 
most slowly ? 

5. The coordinates of the position of a moving particle at any 
time t are given by the equations x = ? 3, y t 9 + 2. Determine 
the path of the particle and its speed in its path 

6. The coordinates of the position of a moving particle at any 
time t are given by the equations x = 4 &, y 4 (1 t) 2 . Determine 
the path of the particle and its speed in its path. 

7. Find the highest point in the path of a projectile. 

8. Find the point in its path at Trhich the speed of a projectile 
is a minimum. 

9. Find the range (that is, the distance to the point at which 
the projectile will fall on OX), the velocity at that point, and the 
angle at which the projectile will meet OX. 

10. Show that in general the same range may be produced by 
two different values of a, and find the value of which produces 
the greatest range. 

11. Find the (a, y) equation of the path of a projectile, and plot. 



VELOCITIES AND RATES 111 

41. Related velocities and rates. Another problem of some- 
what different type arises when we know the velocity of one 
point iii its path, which may be straight or curved, and wish to 
find the velocity of another point which is in some way con- 
nected vviUi tho first but, in general, describes a different path. 
The method, in general, is to form an equation connecting the 
distances traveled by the two points and then to differentiate 
tho equation thus formed with respect to the time t. The result 
is an equation connecting the velocities of the two points. 

Ex. 1 A lump is 00 ft. above tho ground. A stone is let drop from a 
point on tho same lovol us the lamp and 20 ft, away from. it. Find the speed 
of the stone's shadow on the ground 
at the end of 1 HOC., assuming that the 
distance traversed by a falling body in 
the time t is 1(5 < a . 

Lot A C (Fig, 48) be tho surface of 
tho ground which is assumed to be a 
homontal plane, L the position of the 
lamp, the point from which the stone 
was dropped, and S the position of the _ 
stone at any time t. Then Q is the posi- ^ I0 4g 

tion of the shadow of S on the ground, 

LSQ being a straight line. Let OS = x and BQ, = ?/. Then L = 20, BO = 60, 
and BS = 60 a?. In the similar triangles LOS and SBQ, 





whence y SB ~ 20. (2) 

7 ^ 3 

We know x =16 < a , whence ~ = 82 1 ; and wish to find ?, the velocity of Q. 
at at 

Difterontiating (2) with respect to t, we have 



dt x* dt 

When / s= 1 sec,, x =s 16, and ~ =s 82 ; whence, by substitution, we find 

at 

~& ss 150 ft. per second. 
dt 

The result is negative because y is decreasing as time goes on. 

In 6 and 11, if the rate of one of two related quantities 
was known, we were able to find the rate of the other quantity. 



112 ALGEBRAIC FUNCTIONS 

This type of problem may also be solved by the same method 
by which the problem of related velocities has been solved. 
We shall illustrate by taking the same problem that was used 
in 11. 

Ex. 2. Water is being poured at the rate of 100 cu. in. per second into 
a vessel in the shape of a right circulai cone of radius 3 in. and altitude 
9 iu. Required the rate at which the depth of the water is increasing when 
the depth is 6 in. 

As in 11, we have V = ^ irh a ; 

dV 1 , z d1i 
whence -^ = * ^ 

We have given = 100, h = 6 ; from which we compute 

^ = 25 =796 . 
dt v 

EXERCISES 

1. A point is moving on the curve y 2 = sc s The velocity along 
OX is 2 ft. per second What is the velocity along OY when x = 2 ? 

2. A ball is swung in a circle at the end of a cord 3 ft long so 
as to make 40 revolutions per minute. If the cord breaks, allowing 
the ball to fly off at a tangent, at what rate will it be receding from 
the center of its previous path 2 sec after the cord breaks, if no 
allowance is made for the action of any new force ? 

3. The mside of a vessel is in the form of an inverted regular 
quadrangular pyramid, 4 ft square at the top and 2 ft deep. The 
vessel is originally filled with water which leaks out at the bottom 
at the rate of 10 cu. in. per minute. How fast is the level of the 
water falling when the water is 10 in. deep ? 

4. The top of a ladder 20 ft long slides down the side of a ver- 
tical wall at a speed of 3 ft per second. The foot of the ladder slides 
on horizontal land. Find the path described by the middle point of 
the ladder, and its speed in its path. 

5. A boat with the anchor fast on the bottom at a depth of 40 ft. 
is drifting at the rate of 3 mi per hour, the cable attached to the 
anchor slipping over the end of the boat At what rate is the cable 
leaving the boat when 50 ft of cable are out, assuming it forms a 
straight line from the boat to the anchor ? 



GENERAL EXERCISES 113 

6. A solution is being pouied into a conical filter at the rate of 
5 cc per second and is running out at the rate of 2 cc. per second. 
The radius of the top of the filter is 8 cm and the depth of the filter 
is 20 cm. Find the rate at which the level of the solution is rising 
in the filter when it is one third of the way to the top. 

7 . A trough is in the form of a right prism with its ends isosceles 
triangles placed vertically. It is 5 ft long, 1 ft across the top, and 
8 in. deep It contains water which leaks out at the rate of 1 qt. 
(57| cu. in ) per minute Find the rate at which, the level of the 
water is sinking in the trough when the depth is 3 in. 

8. The angle between the straight lines AB and BC is 60, and 
AB is 40 ft. long. A particle at A begins to move along AB toward 
B at the rate of 5 ft per second, and at the same time a particle at 
B begins to move along BC toward C at the rate of 4 ft per second. At 
what rate are the two particles approaching each, other at the end 
of 1 sec * 

9. The foot of a ladder 50ft. long rests on horizontal ground, and 
the top of the ladder rests against the side of a pyramid which makes 
an angle of 120 with the ground. If the foot of the ladder is drawn 
directly away from the base of the pyramid at the uniform rate 
of 2 ft. per second, how fast will the top of the ladder slide down 
the side of the pyramid ? 

GENERAL EXERCISES 

Plot the curves : 

1. 3 a 2 + 7 f = 21 9. ? / 2 (4 + or 2 ) = cc 2 (4 - a 2 ). 

2. 4?/ a == 9.r. z a, x 

* 10. ?/- = a; a - . 

3. 9a; 2 y*=lG. a + x 

4. 2/ a - 2 y = a; 8 + 2 a? - 1. n 

_ 8 a 8 12. 

y -rf+4a"' IS. 

6. 



7, (7/-a J ) 2 =9-a ! 2 > == ^T4* 

8. (* + yf f(y + 2). 15. xY + 36 = 16 f 

Find the turning-points of the following curves and plot the curves 

16. y = (2 + oO(4 - a-,) 2 . = (x - I) 3 

17. y = (x + 3) 2 (a: - 2). ' J JB + 1 ' 



a 03 , 
- at 
x 



114 ALGKEBBAIC FUNCTIONS 

20. Find the equation of the tangent to the curve f = x z ~- 

/ 3 a 6a\ 
the point ( -g-> ~6/ 

21. Find the equation of the tangent to the curve x* + y* = * 
at the point (a;.,^, 3^). 

22. Prove that if a tangent to a parabola ?/ 2 = 7o has the slope ?;?, 

its point of contact is (j- 5, ^-) and therefore its equation is 

A 4 VTE 7/t' J& tfvj 



a 2 

JE ?/ 

23. Prove that if a tangent to an ellipse + ~ = 1 has the slope m, 

ft U 

its point of contact is ( * m ,^ , & 2 ) and therefore 
^ \ Vrt 2 ?^ 2 + 6 2 V arm 2 + 67 

its equation is y = wcc Va 2 m 2 + i 2 

24. Show that a tangent to a parabola makes equal angles with 
the axis and a line from the focus to the point of contact. 

25. Show that a tangent to an ellipse makes equal angles with the 
two lines drawn to the foci from the point of contact 

Find the angles of intersection of the following pairs of curves . 



26. , = 



~ 



28. a; 2 = 

29. a: 2 -4y-4=0, a? + 12y - 36 = 0. 

30. 2/ ! =o; 8 } 2/ ! =(2-a;) 8 . 
31. 

32. 

33. The coordinates of a moving particle are given by the equa- 
tions x tf 3 , y = (1 tf 2 )^. Find its path and its velocity in its 
path. 

34. A particle moves so that its coordinates at the time t are 

2 

x = 2t, y = . . Find its path and its velocity in its path. 

35. A projectile so moves that x = at, y = bt ^g^. Find its 
path and its velocity in its path. 



GENERAL EXERCISES 115 

36. A body so moves that x = 2 + tf$, y = 1 -f t. Find its path 
and its velocity in its path. 

37. A particle is moving along the curve if 4. x; and when as = 4, 
its ordmate is increasing at the rate of 10 ft. per second. At what 
rate is the abscissa then changing, and how fast is the particle moving 
in the curve ? Where will the abscissa be changing ten times as fast 
as the ordmate ? 

38. A particle describes the circle o; a +y a =a 2 with a constant 
speed v . Find the components of its velocity. 

39. A particle describes the parabola y 2 = 4 ax in such a way that 
its a-component of velocity is equal to ct Find its y-component of 
velocity and its velocity in its path. 

40. A particle moves so that x = 2 1, y = 2 "Vt & Show that it 
moves around a semicircle in the time from t = to t = 1, and find 
its velocity in its path during that time. 

41. At 12 o'clock a vessel is sailing due north at the uniform rate 
of 20 mi. an hour. Another vessel, 40 mi. north of the first, is sailing 
at the uniform rate of 15 mi an hour on a course 30 north of east. 
At what rate is the distance between the two vessels diminishing at 
the end of one hour ? What is the shortest distance between the 
two vessels ? 

42. The top of a ladder 32 ft. long rests against a vertical wall, 
and the foot is drawn along a horizontal plane at the rate of 4 ft. 
per second in a straight line from the wall. Find the path of a 
point on the ladder one third of the distance "from the foot of the 
ladder, and its velocity in its path. 

43. A man standing on a wharf 20 ft. above the water pulls in a 
rope, attached to a boat, at the uniform rate of 3 ft. per second Find 
the velocity with which the boat approaches the wharf 

44. The volume and the radius of a cylindrical boiler are expand- 
ing at the rate of .8 cu. ft. and .002 ft. per minute respectively. How 
fast is the length of the boiler changing when the boiler contains 
40 cu. ft. and has a radius of 2 ft. ? 

45. The inside of a cistern is in the form of a frustum of a regular 
quadrangular pyramid. The bottom is 40 ft. square, the top is 60 ft. 
square, and the depth is 10 ft, If the water leaks out at the bottom 
at the rate of 5 cu. ft. per minute, how fast is the level of the water 
falling when the water is 5 ft, deep in the cistern ? 



116 ALGEBRAIC FUNCTIONS 

46. The inside of a cistern is in the form of a frustum of a right 
circular cone of vertical angle 90. The cistern is smallest, at the 
base, which is 4 ft in diameter. Water is being poured in at the rate 
of 5 cu ft. per minute. How fast is the water rising in tho cistern 
when it is 2 ft deep ? 

47. The inside of a bowl is in the form of a hemispherical sur- 
face of radius 10 in If watei is running out of it at the iat,o of 
2 eu in. per minute, how fast is the depth of the water decreasing 
when the water is 3 in deep ? 

48. How fast is the surface of the bowl in Ex 47 being exposed ? 

49. The inside of a bowl 4 in deep and 8 in. across the top is in 
the form of a surface of revolution formed by revolving a parabolic 
segment about its axis Water is running into the bowl at the rate 
of 1 cu in per second How fast is the water rising in the bowl 
when it is 2 in deep ? 

50. It is required to fence off a rectangular piece of ground to con- 
tain 200 sq ft , one side to be bounded by a wall already constructed. 
Find the dimensions which will require the least amount of fencing. 

51. The hypotenuse of a right triangle is given. Find the other 
sides if the area is a maximum 

52. The stiffness of a rectangular beam varies as the product of the 
breadth and the cube of the depth. Find the dimensions of the stiffest 
beam which can be cut from a circular cylindrical log of diameter 18 in. 

53. A rectangular _plot of land to contain 384 sq. ft. is to be in- 
closed by a fence, and is to be divided into two equal lots by a fence 
parallel to one of the sides What must be the dimensions of tho 
rectangle that the least amount of fencing may be required ? 

54. An open tank with a square base and vertical sides is to havo 
a capacity of 500 cu ft Fiud the dimensions so that the cost of 
lining it may be a minimum 

55. A rectangular box with a square base and open at the top is 
to be made out of a given amount of material. If no allowance is 
made for thickness of material or for waste in construction, what are 
the dimensions of the largest box which can be made ? 

^ 56. A metal vessel, open at the top, is to be cast in the form of a 
right circular cylinder. If rt is to hold 27 TT cu in , and the thickness 
of the side and that of the bottom are each to be 1 in, what will be the 
inside dimensions when the least amount of material is used ? 



GENERAL EXERCISES 117 

57 . A gallon oil can (231 cu in ) is io be made in the form, of a 
right chcular cylinder. The material used for the top and the bottom 
costs twice as much per square inch as the material used for the 
side What is the radius of the most economical can that can be 
made if no allowance is made for thickness of material or waste in 
construction ? 

58. A tent is to be constructed in the form of a regular quadran- 
gular pyiauucl Find the ratio of its height to a side of its babe when 
the air space inside the tent is as great as possible for a given wall 
surface 

59. It is required to construct from two equal circular plates of 
radius a a buoy composed of two equal cones having a common base. 
Find the radius of the base when the volume is the greatest 

60. Two towns, A and I>, are situated respectively 12 mi. and 
18 mi. back from a straight river from which they are to get thoir 
water supply by means of the same pumping-station. At what point 
on the bank of the river should the station be placed so that the least 
amount of piping may be required, if the nearest points 011 the river 
from A and B respectively are 20 mi. apart and if the piping goes 
directly from the pumping-station to each of the towns 9 

61. A man 011 one side of a river, the banks of which are assumed 
to be parallel straight lines mi apart, wishes to reach a point on 
the opposite side of the river and 5 mi. further along the bank If 
he can row 3 mi. an hour and travel on land 5 mi. an hour, find the 
route he should take to make the trip in the least time. 

62. A power house stands upon one side of a river of width I miles, 
and a manufacturing plant stands upon the opposite side, a miles 
downstream. Find the most economical way to construct the con- 
necting cable if it costs m, dollars per mile on land and n dollars a 
mile through water, assuming the banks of the river to be parallel 
straight lines. 

63. A vessel A is sailing due east at the uniform rate of 8 mi. 
per hour when she sights another vessel B directly ahead and 20 mi. 
away. B is sailing in a straight course S. 30 W at the uniform rate 
of 6 mi per hour. When will the two vessels be nearest to each other? 

64. The number of tons of coal consumed per hour by a certain 
ship is 0.2 + 0.001 v* } where v is the speed in miles per hour. Find 
an expression for the amount of coal consumed on a voyage of 
1000 mi. and the most economical speed at which to make the voyage. 



118 ALGEBRAIC FUNCTIONS 

65. The fuel consumed by a certain steamship in an hour is pro- 
portional to the cube of the velocity which would be given to the 
steamship in still water. If it is required to steam a certain distance 
against a current flowing a miles an hour, find the most economical 
speed. a 2 

66. An isosceles triangle is inscribed in the ellipse ^ 4- ~~ = 1, 

Ct () 

(a > 6), with its vertex in the upper end of the minor axis o the 
ellipse and its base parallel to the major axis Determine the length 
of the base and the altitude of the triangle of greatest area whiuli 
can be so inscribed. 



CHAPTER V 
TRIGONOMETRIC FUNCTIONS 

42. Circular measure. The circular measure of an angle is the 
quotient of the length of an arc of a circle, with its center at 
the vertex of the angle and included between its sides, divided 
by the radius of the arc. Thus, if 6 is the angle, a the length 
of the arc, and r the radius, we have 

-~ CD 

The unit of angle in this measurement is the radian, which 
is the angle for which a r in (1), and any angle may be said 

to contain a certain number of radians. But the quotient - in 

r 

formula (1) is an abstract number, and it is also customary to 
speak of the angle 6 as having the magnitude - without using 
the word radian. Thus, we speak of the angle 1, the angle -|, 
the angle > etc. 

In all work involving calculus, and in most theoretical work 
of any kind, all angles which occur are understood to be ex- 
pressed in radians. In fact, many of the calculus formulas would 
be false unless the angles involved were so expressed. The 
student should carefully note this fact, although the reason for 
it is not yet apparent. 

From this point of view such a trigonometric equation as 

y = sin x (2) 

may be considered as defining a functional relation between two 
quantities exactly as does the simpler equation y a; 3 . For 
we may, in (2), assign any arbitrary value to x and determine 
the corresponding value of y. This may be done by a direct 

119 



120 TRIGONOMETRIC FUNCTIONS 

computation (as will be shown in Chapter VII), or it may be 
done by means of a table of trigonometric functions, in which 
case we must interpret the value of x as denoting so many radians. 
One of the reasons for expressing an angle in circular measure 
is that it makes true the formula 

*T bill fv *4 xrtx 

Lim - = 1, (3) 

A-+0 fl 

where the left-hand member of the equation is to be read 

" the limit of ^-y as h approaches zero as JB 

a limit." 

To prove this theorem we proceed as 

follows: ->. Ti ' 

Let h be the angle A OB (Fig. 49), r the ^ % \ x j // 

radius of the arc AB described from as "" x ^j/ 

a center, a the length of AB, p the length B ' 

of the perpendicular BC from B to OA, 

and t the length of the tangent drawn from B to meet OA 

produced in D. 

Revolve the figure on OA as an axis until B takes the position 

B'. Then the chord BGB'=2p, the arc BAB' =2 a, and the 

tangent B f D=tliG tangent BD. Evidently 

BD + DB' > BAB' >BCB'; 
whence t>a>p. 

Dividing through by r, we have 




r r r 
that is, tan h > h > sin h. 

Dividing by sin A, we have 



cos h sin A ' 

or, by inverting, cos h < ^ ^ < 1. 

h 



GRAPHS 121 

Now as h approaches zero, cos h approaches 1. Hence r 

ft 

which lies between cos h and 1, must also approach 1 ; that is, 



A - o n 

This result may be used to find the limit of k as It 

approaches zero as a limit For we have l 

c\ a fl n /I 'I/ 

2 sin 3 - sin 2 - sm- 
1 cos h _ 2 2 _ h 2 

A ~ h I 2 7t 

2 2 , 

. h 

sm- 

Now as h approaches zero as a limit, approaches unity, 

by (3). Therefore h 

r 1 cos h A . , . 

Lira =0. (4) 

A-O 7i V ' 

43. Graphs of trigonometric functions. We may plot a trigo- 
nometric function by assigning values to x and computing, or 
taking from a table, the corresponding values of y. In so doing, 
any angle which may occur should be expressed in circular 
measure, as explained m the previous section. In this connec- 
tion it is to be remembered that TT is simply the number 3.1416, 
and that the angle w means an angle with that number of radians 
and is therefore the angle whose degree measure is 180. 

The manner of plotting can be best explained by examples. 

Ex. 1. y = a sin fa- 
it is convenient first to fix the values of x which make y equal to aero 
Now the sine is zero when the angle is 0, IT, 2 w, 3 IT, TT, 2 TT, on , iu 
general, kv, where k is any positive or negative integer. To make y = 0, 
therefore, we have to place bx = kv, whence 

2w IT n TT 2ir 8w 

* > r- > U, ~) -- , ~j > '. 

b b b b b 

The sine takes its maximum value + 1 when the angle has the values 
if 5 TT OTT TT 5 TT TT 

e>'~2"' ~n~' eifcc< a * * s > * n *k is case > "when x = > , ~, etc. For these 

values of or, w = a 



122 



TRIGONOMETRIC FUNCTIONS 



O rr . 

The sine takes its minimum value 1 when the angle is '-, - . etc. , 

q 17 _ ii 2 

that is, in this case, when x = ' - etc Foi these values of r, y = a. 

20 2o 

These values of x for which the sine is 1 lie halfway between the 
values of x for which the sine is 

7 




FIG 50 

These points on the graph are enough to determine its general shape 
Other values of x may be used to fix the shape more exactly The graph 
is shown in Fig 50, with a = 3 and I = 2 The curve may be said to repre- 
sent a wave. The distance from peak to peak, -, is the wave length, and 
the height a above OX is the amplitude 

Ex 2. y = a cos Ix. 

As in Ex 1, we fix first the points for which y = Now the cosine of 
an angle is zero when the angle is -> , , etc ; that is, any odd 
multiple of \> We have, therefore, y = when 

At 



26' 



STT 



26' 
Y 



2b 




FIG. 61 

Halfway between these points the cosine has its maximum value + 1 
or its minimum value - 1 alternately, and y = a. The graph is shown 
in Fig 51, with a = 3and& = 2 



GKRAPHS 



128 



Ex. 3. y = a sin (bx + c). 

We have y = when bx + c = 0, TT, 2 TT, 3 rr, c>tc.; that is, when 



7 T 



2ir 




]?i&. 52 

Halfway between these values of x tho HHIQ has its maximum value + 1 
and its minimum value 1 alternately, and y db a. Tho tmvvo is tho 

same as in Ex. 1, but is shifted ~ units to the loft (Fig. 52). 
Ex, 4. y sin a; + 4 sin 2 ,x. 

The graph is found by adding the ordinatcs of tho two curves y = sin x 
and y ^ sin 2 x, as shown in Fig. 58. 

Y 




sin x+i sin 8*0 



3?io. 68 

EXERCISES 
Plot the graphs of the following equations : 

1. y us 2 sin 3 a:. 6. y 

2. 7/ = 3oos 7 - y 

8, y 

3. Ssinfaj ~V 9. 2/ 

^x 10. y 

-j. 11. y 



tan 2 SB, 



SOOCB. 



4. ^2008 



vers jr. 



6. y *a 3 sin (as 2), 



12, 



gin 



4. s i n 



124 TKIGONOMETKIC FUNCTIONS 

44. Differentiation of trigonometric functions. The formulas 
for the differentiation of trigonometric functions are as follows, 
where u represents any function of x which can be differentiated : 

d . du 

sinM = cosM - 
dx dx 

d du 

cosM = sinw > 
dx dx 

d . 2 du 

tan u = sec u -> 
dx dx 

A ct n W = -csc 2 w^, (4) 

dx dx 

d , du XCN 

sec u = sec u tan u ( o) 
dx dx 

d , du X /J N 

esc u = esc u ctn u (6) 
dx dx 

These formulas are proved as follows: 

1. Let y sin w, where u is any function of a; which may be 
differentiated. Give x an increment A and let AM and Ay be 
the corresponding increments of u and y. Then 

Ly = sin (u -f- Aw) sin w 

= sin w cos AM + cos u sin A% sin u 
= cos u sin AM (1 cos Aw) sin u ; 

. AV sin AM 1 cos AM . 

whence - = cos M sm M. 

Au AM AM 

Now let Ar and therefore AM approach zero. By (3), 42, 

T sinAit ^ , , , A ^ , T . 1 cosAw A m . . 
Lim - = 1, and, by (4), 42, Lim A = 0. Therefore 
AM Aw 

<fo/ 

~ = cos u. 

du 

But by (8), 36, ^ = K 
dx dudx 

and therefore -^ = cosw-^ 

aa: dx 



3 

2. To find cos w, wo write 

= sin u 



rrn d d / 

Then -7- cos u = -~ sm 

?23 tfo 



/7T \ 

?/ ) 

\2 / 



'7T 

= coal 



3. To find --- tan 7/, wo write 
djc 

, sin ?/ 

tan?< = 

cos ?/, 

,, . a 

Then - tan u = ~ 



dx cos u 

d , d 

cos u - sin ? sin u -~ cos ?* 



eos 4 tt 

VVH . 

cos 2 w ^ y 

f?. 



;i 

4. To find ~^~ cinu. wo write 
j? 

, COS 7^1 

ctn u = -i 
sin u 

rn , d . d cosw 

I hen ctn ?/ = .- -: 

d.r, ax Bin u 



, d d . 

sin u - cos w cos u -=- fiin M 




126 TRIGONOMETBIC FUNCTIONS 

5. To find sec u, we write 
ax 

1 , ^1 

secw = - = (COSM) \ 
cos u 

Then sec w = - (cos )~* cos u (by (6), 36) 

sum <2 



j 

6. To find esc w, we write 
aa; 

esc u = - = (sin. u)~\ 
sin M 

3 7 

Then esc u = (sin w)~ 2 sin u (by (6), 36) 



- 

Ex. 1. y = tan 2 a; tan 3 a; = tan 2 x (tan a:) 8 . 

^ = sec 2 2 x-j- (2 a) - 2 (tana) tana; 
dx dx^ ' v J dx 

= 2 seo 2 2 a; 2 tan a; sec 2 a:. 
Ex. 2. y = (2 sec*ar + 3 sec a a;) sin x. 
= sin a: ^8 sec 8 *-! (see a) + 6 sec x ~- (sec a) J + (2sec*a: + 3 sec 2 a:)-^(sina;) 

= sin x (8 sec*a; tan x + Q sec 2 a; tan a:) + (2 sec*a; + 3 sec s a:) cos x 
= (1 cos s a:) (8 sec 6 a; + 6 sec 8 a;) + (2 sec s a: + 3 sec a;) 
= 8 sec 6 a: 3 sec ar. 

EXERCISES 
nd in each of the following cases : 



y = 2 tan|- 6. y = & sin 8 5a! - sin5 . 

sw.*2x. 7. ^rs 

s a 5aj. 8 . . 



SIMPLE HARMONIC MOTION 127 



9. 2/ = cos 8 -2cos. 11. y 

2 . , o * x ,o sec a* + ton as 

10. y - ctn- + 2 ctn -. 12. y = 



13. y = sin (2 a? + 1) cos (2 a; 1). 

14. y = tan 8 3 a; 3 tan 3 x + 9 ie. 

15. y = see 2 a: tan 2 x. 

16. y = ^ (3 cos 6 2 8 5 cos B 2 os). 

17. sin 2 a + tan 3 ?/ = 0. 

18. asy + ctn xij = 0. 

45. Simple harmonic motion. Let a particle of mass m move 
in a straight line so that its distance s measured from a fixed 
point in the line is given at any time t by the equation 

s s=a G sin Si, (1) 

where c and b are constants. We have for the velocity v and 
the acceleration a v ^ cbcos ^ (2 ) 

a sss c6 2 sin Ztf. (3) 

When 0, 8 a and the particle is at (Fig. 54), When 

t = -j^r* s =s c and the particle is at A t where OA** o, 
2 o 

When t is between and v is positive and a is negative, 

>a y 

so that the particle is moving from to A with decreasing speed. 

When t is between and T v is i A 

25 o 

, , ,. ,-, , ]?i<*. 64 

negative and a is negative, so that 

the particle moves toward with increasing speed. When 

71" 

==--, the particle is at 0. 

8 

As * varies from ? to -^?? the particle moves with decreasing 
b 26 

speed from to J?, where OJ? =* o. 

Finally, as varies from |^ to -, the particle moves back 

2o o 

from J5 to (9 with increasing speed. 



128 TRIGONOMETRIC FUNCTIONS 

The motion is then repeated, and the particle oscillates between 
B and A, the time required for a complete oscillation being, as 

27T 

we have seen, =- The motion of the particle is called simple 
harmonic motion. The quantity c is called the amplitude, and tho 

2 7T 

interval -= after which the motion repeats itself, is called 

the period. 

Since force is proportional to the mass times the acceleration, 
the force F acting on the particle is given by the formula 

F kma = kmcb s sm bt= Jcmb*s. 



This shows that the force is proportional to the distance s 
from the point 0. The negative sign shows that the force pro- 
duces acceleration with a sign opposite to that of s, and there- 
fore slows up the particle when it is moving away from arid 
increases its speed when it moves toward 0. The force is there- 
fore always directed toward and is an attracting force. 

If, instead of equation (1), we write the equation 

s = c8mb(t-t Q ), - (4) 

the change amounts simply to altering the instant from which the 
time is measured. For the value of s which corresponds to t t 
in (1) corresponds to t = t l +t in (4). Hence (4) represents 

simple harmonic motion of amplitude c and period 
But (4) may be written * 

s = G cos fa sin bt o sin bt cos bt, 
which is the same as 

s = A sin bt + B cos bt, 
where A = c cos bt , B = c sin bt . 

A and B may have any values in (5), for if A and B are given, 
we have, from the last two equations, 

c=V^ a +5 2 , tanfa = --, 
which determines G and t in (4). 



SIMPLE HARMONIC MOTION 129 

Therefore equation (5) also represents simple harmonic motion 

2 7T 

with amplitude VA z +jB a and period -7 

In particular, if in (5) A = and 33 = c, we have 

8 = G COS It. (6) 

If in (4) we place t = - ' , it becomes 

*j & 

* =<! cos fi (-*), (7) 

which differs from (6) only in the instant from which the time 
is measured. 

EXERCISES 

1. A particle moves with constant speed v around a circle. 
Prove that its projection on any diameter of the circle describes 
simple harmonic motion 

2. A point moves with simple harmonic motion of period 4 sec. 
and amplitude 3 ft Find the equation of its motion. 

3. Given the equation s = 5 sin 2 1 Find the tune of a complete 
oscillation and the amplitude of the swing. 

4. Find at what time and place the speed is the greatest for the 
motion defined by the equation s = G sin It Do the same for the 
acceleration. 

5. At what point in a simple harmonic motion is the velocity zero, 
and at what point is the acceleration zero ? 

6. The motion of a particle in a straight line is expressed by the 
equation s = 5 2 cos 3 *. Express the velocity and the acceleration 
in terms of s and show that the motion is simple harmonic. 

7. A particle moving with a simple harmonic motion of amplitude 
5 ft has a velocity of 8 ft. per second when at a distance oj! 3 ft. 
from its mean position. Find its period. 

8. A particle moving with simple harmonic motion has a velocity 
of 6 ft. per second when at a distance of 8 ft. from its mean position, 
and a velocity of 8 ft. per second when at a distance oC 6 ft. from its 
mean position. Find its amplitude and its period. 

9. A point moves with simple harmonic motion given by the 
equation *= # sin et. Describe its motion, 



130 TRIGONOMETRIC FUNCTIONS 

46. Graphs of inverse trigonometric functions. The equation 

x = sin y (1) 

defines a relation between the quantities x and y which may be 
stated by saying either that x 'is the sine of the angle y or that 
the angle y has the sine x. When we wish to use the latter form 
of expressing the relation, we write in place of equation (1) 
the equation y = sm -^, (2) 

where 1 is not to be understood as a negative exponent but as 
part of a new symbol sin" 1 . To avoid the possible ambiguity 
formula (2) is sometimes written 

y = arc sin x. 

Equations (1) and (2) have exactly the same meaning, and 
the student should accustom himself to pass from one to the 
other without difficulty. In equation (1) y is considered the 
independent variable, while in (2) x is the independent variable. 
Equation (2) then defines a function of x which is called the 
anti-sine of x or the inverse sine of x. It will add to the clearness 
of the student's thinking, however, if he will read equation (2) 
as " y is the angle whose sine is x." 

Similarly, if a = cos#, then y = GOS~' L x; if x ioxiy, then 
y tan" 1 a; ; and so on for the other trigonometric functions. We 
get in this way the whole class of inverse trigonometric functions. 

It is to be noticed that, from equation (2), y is not completely 
determined when x is given, since there is an infinite number 

of angles with the same sine. For example, if # = - #=> > 

STT 18 IT , _.. . t 2 . & 

-~> ft > etc. 1ms causes a certain amount of ambiguity in 

using inverse trigonometric functions, but the ambiguity is re- 
moved if the quadrant is known in which the angle y lies. We 
have the same sort of ambiguity when we pass from the equa- 
tion x = y* to the equation y = Va;, for if x is given, there 
are two values of y. 

To obtain the graph of the function expressed in (2) wo 
may change (2) into the equivalent form (1) and proceed as 



GRAPHS 131 

in 43. In this way it is evident that the graphs of the inverse 
trigonometric functions are the same as those of the direct func- 
tions but differently placed with reference to the coordinate 

axes. It is to be noticed particularly 
that to any value of x corresponds an 
infinite number of values of y. 



X 



FIG. 55 FIG 56 

Ex. 1. y = sin- 1 * 

From this, x = sin y, and we may plot the graph by assuming values of 
y and computing those of a (Fig. 55). 

Ex. 2. y = tan- 1 *. 

Then x == tan y, and the graph is as in Fig. 56. 

EXERCISES 

Plot the graphs of the following equations : 

1. y tan- 1 2 x. 3. y = sin" 1 (a: 1). 5. y =1+ cos" 1 :*;. 

2. y =3 ctn-^cc. 4. y = tan- 1 (* + 1). 6. y = ^tan" 1 ^ 
7. ?/ = cos-^jc 2). 8. y sin-^S x 4. 1) - - 

47. Differentiation of inverse trigonometric functions. The 

formulas for the differentiation of the inverse trigonometric 
functions are as follows: 

1. -7- sin"" 1 ^ = . -r- when sin"" 1 ^ is in the first or the 

ax vl w a <*x ,. , , , , 

x u lourth quadrant ; 

i diL 

i when sin-^f is in the second or 

A/ "I M_ QI^ QiX ,t i t i 

x w the third quadrant. 



132 TKIGONOMETKIC FUNCTIONS 

2. - cos" 1 ^ = , -7- when cos" 1 ** is m the first or the 

^ x vlw second quadrant; 

i (Jtjf 

= when cos" 1 ^ is m the third or the 

VI - u 2 dx f ourth quadrant 

d L. 1 du 

O ~z 





dx 1 + w z ax 

A d . _j 1 du 

4. ctn u= 3 
dx 1 + 

5. sec" 1 ?* = when sec" 1 ^ is in the first or the 

dx u^/u 2 -l dx 



= when sec" 1 u is in the second or 

uvu I ^ ie f our tli quadrant. 

fj J. el/Mi 

6 CSC" I M= . -7- when csc" 1 ^ is in the first or 

x uvu 1 |.j ie third quadrant; 

= , -r- when csc"" 1 ^ is in the second or 



The proofs of these formulas are as follows: 
1. If y = sin~X 

then sin y = u. 

TT i P A A dy du 

Hence, by 44, co S2 ,^ = -; 

cfc 1 du 



, 
whence 



-- -- 
dx cos y dx 



But cos y = Vl u 2 when # is m the first or the fourth quad- 
rant, and cosy = Vl u 2 when y is m the second or the third 
quadrant. 

2. If y = cos" 1 ^, 

then cos y u- 

Hence _ sin ^ = ^ ; 

9 dx dx 

whence * 



eta sin # 



DIFFERENTIATION 183 

But sin y Vl M a when y is in the first or the second quad- 
rant, and am y = Vl u 2 when y is in the third or the fourth 
quadrant. 

3. It' y tan' 1 it, 
then tan y = u. 

it 2 d?t d u 

1 1 ence sec u ~^~ ; 

/yif>- /yVp 

dv*t Lt*v 

, dy 1 du 

whence -~ = 

dx 1 + u dx 

4. If yssctir 1 !/, 
then ctn ?/ = w. 

V T 41 (iV (/ ( V UV 

, (??/ 1 du 

whence ~ = = ; -7- 

a 1 + w" are 

5. If yssseo'X 
then sec ?/ = M. 

1 1 ence sec y tan ?y -~ = -r- ; 

ti*y* ft'V 

. d'i/ 1 <??* 

whence - = , -=-* 

djc sec ;y tan y dx 

But Rcoyss?/, and tan,y = V?^ a --l when ?/ is in the first or 
the third quadrant, and tan ?/:=-~-vV 1 when y is in the 
second or the fourth quadrant. 

M T$ /j/ rtiart '""'J'^f 

then esc y = ?/,. 

_ r . dy du 

Hence esc y otn y j~ =s y ; 

rvM/ \4iJu 

whence -f- = , -r- 

?a; esc y ofo&y dx 



134 



TRIGONOMETRIC FUNCTIONS 



But cscy = w, and ctn^=Vw 2 1 when y is in the first or 
the third quadrant, and ctn# = Vw 2 1 when y is in the 
second or the fourth quadrant. 

If the quadrant in which an angle lies is not material in a 
problem, it will be assumed to be in the first quadrant. This 
applies particularly to formal exercises m differentiation. 

Ex. 1. y = sin" 1 Vl x z , where y is an acute angle 

dv 1 . .fLn. a>\| ^ 

L-(l-a 2 ) dx 



dx 



Vl - 



This result may also be obtained by placing sin- 1 Vl x z = cos" 1 a:. 
Ex. 2. y = sec" 1 V4a: a + 4ar + 2. 



dy dx 

dx V4 x* + 4 x + 2 V(4 a,- 2 + 4 x + 2) - 1 

~ ~2a; 2 + 2a,-- 



EXERCISES 
Find -7^ in each of the following cases : 



2. y = sin" 1 -. 

_ -ift 3 
. y _ sin g 

i 3a; 

4. y = cos -1 



11. y = COS~ 



2 



5. y = tan~ 1 

6. y = tan" 1 Vcc 2 2 x. 

7. y = ctn- 1 ^- 

8. y = sec" 1 5 a. 

9. y = csc~ 1 2aj. 

, . a; + 6 
10, y-tan-> 



13. y = tan" 1 Va 3 - 1 + 




16. y = 

2\a a;/ 
18. y ss Vl aj 9 +a5003~ 1 Vl w a 



ANGULAR VELOCITY 135 

48. Angular velocity. If a line OP (Fig. 57) is revolving in 
a plane about one of its ends 0, and in a time t the line 
OP has moved from an initial position OM to the position OP, 
the angle MOP = 6 denotes the amount of rotation. The rate 
of change of with respect to t is 
called the angular velocity of OP. The 
angular velocity is commonly denoted 
by the Greek letter < ; so we have 
the formula JQ 




In accordance with 42 the angle 8 
is taken in radians ; so that if t is in 
seconds, the angular velocity is in 

radians per second. By dividing by 2-Tr, the angular velocity 
may be reduced to revolutions per second, since one revolution 
is equivalent to TT radians. 

A point Q on the line OP at a distance r from describes 
a circle of radius r which intersects OM at A. If s is the length 
of the arc of the circle A Q measured from A, then, by 42, 

s = rO. (2) 

ds 

Now -r- is called the linear velocity of the point $, since it 
dt 

measures the rate at which s is described ; and from (2) and (1), 

ds d0 



showing that the farther the point Q is from. the greater is 
its linear velocity. 

Similarly, the angular acceleration, which is denoted by oc, is 

denned by the relation , , 2 ,, 

J aft> a o 



This is connected with the linear acceleration -=-5 by the 



formula 



136 



TRIGONOMETRIC FUNCTIONS 



Ex 1. If a wheel revolves so that the angular velocity is given by the 
formula u> = 8 1, how many revolutions will it make in the time from t = 2 
to t = 5 ? 

We take a spoke of the wheel as the line OP Then we have 

dQ = 8 tdt 

Hence the angle through which the wheel revolves in the given time is 
B = C 6 B tdt = [4 fif =100 - 16 = 84. 

1/2 

The result is in radians. It may be reduced to revolutions by dividing by 
2 ir. The answer is 13.4 revolutions 

Ex. 2. A particle traverses a circle at a uniform rate of n revolutions 
a second Determine the motion of the projection of the particle on a 
diameter of the circle 

Let P (Fig 58) be the particle, 
OX the diameter of the circle, and 
M the projection of P on OX Then 

x = a cos 6, 

where a is the radius of the circle. 
By hypothesis the angular velocity 
of OP is 2 nir radians per second 
Therefore , 





o> = = 2 mr ; 
dt 




whence 



FIG. 68 



If we consider that when t = 0, the particle is on OX, then (7 = 0. 

Therefore /, 

x = a cos 8 = a cos 2 rart = a cos <at. 

The point M therefore describes a simple harmonic motion In fact, 
simple harmonic motion is often defined in this way 

EXERCISES 

1. A flywheel 4ft in diameter makes 3 revolutions a second. 
Find the components of velocity in feet per second of a point on the 
rim when it is 6 m above the level of the center of the wheel. 

2. A point on the rim of a flywheel of radius 5 ft which is 3 ft. 
above the level of the center of the wheel has a horizontal component 
of velocity of 100 ft per second Find the number of revolutions 
per second of the wheel. 



CYCLOID 1 

3. If the horizontal and vertical projections of a point descr 
simple harmonic motions given by the equations 

x = 5 cos 3t, y = 5 sin 3 1, 
show that the point moves in a circle and find its angular velocity 

49. The cycloid. If a wheel rolls upon a straight line, ea 
point of the rim describes a curve called a cycloid. 

Let a wheel of radius a roll upon the axis of #, and let 
(Fig. 59) be its center at any time of its motion, JV its point 





N 

FIG. 59 

contact with 6L3T, and P the point which describes the cycloi 
Take as the origin of coordinates, 0, the point found by rollu 
the wheel to the left until P meets OX. 

Then ON= arc PN. 

Draw MP and CN, each perpendicular to OX, PR parallel 
OX, and connect C and P. Let 

angle NGP <j>. 
Then x = OM = 0-ZV- JfJV 



= a<j> a sin <. 



a a cos $ 

Hence the parametric representation ( 40) of the cycloid 
a&=a(<f> sin<), 
# = (! -cos<). 



138 TRIGONOMETRIC FUNCTIONS 

If the wheel revolves with a constant angular velocity to = -^ 
we have, by 40, 

i) = a, C\. cos <f>") -*- = Q>* (1 cos <p), 

?; = a sm <j> -If- = am sw. <f> ', 
at 

whence v a = a 2 co 2 (2-2 cos <) = 4 a 2 6) 2 sin 2 1 

v = 2 aco sm ^> 



as an expression for the velocity in its path of a point on the 
run of the wheel. 

EXERCISES 

A 

1. Prove that the slope of the cycloid at any point is ctn ^- 

2. Show that the straight line drawn from any point on the rim 
of a rolling wheel perpendicular to the cycloid which that point is 
describing goes through the lowest point of the rolling wheel. 

3. Show that any point on the run of the wheel has a horizontal 
component of velocity which is proportional to the vertical height of 
the point 

4. Show that the highest point of the rolling wheel moves twice 
as fast as either of the two points whose distance from the ground as 
half the radius of the wheel 

5. Show that the vertical component of velocity is a maximum 
when the point which describes the cycloid is on the level of the 
center of the rolling wheel. 

6. Show that a point on the spoke of a rolling wheel at a distance 
b from the center describes a curve given by the equations 

x = a<f> b sin <, y = a b cos <j>, 

and find the velocity of the point in its path. The curve is called a 
trochoid. 

7. Find the slope of the trochoid and find the point at which the 
curve is steepest. 

8. Show that when a point on a spoke of a wheel describes a 
trochoid, the average of the velocities of the point when in its highest 
and lowest positions is equal to the linear velocity of the wheel 



CUJftVATUKE 



139 



50. Curvature. If a point describes a curve, the change of 
direction of its motion may be measured by the change of the 
angle < ( 15). 

For example, in the curve of Fig. 60, if AJ%= s and ./J^ = As, 
and if fa and fa are the values of <j& for the points P l and P z 
respectively, then </> 2 ^ is the total change of direction of the 
curve between J? and P y If Y 

fafa = A<, expressed in 
circular measure, the ratio 

is the average change 
As 

of direction per linear unit 
of the arc PJ\. Regarding 
as a function of s, and 

taking the limit of as 
As 

As approaches zero as a 

limit, we have -f-> which is called the curvature of the curve at 
as 

the point P. Hence the curvature of a curve is the rate of change 
of the direction of the curve with respect to the length of the arc. 

If is constant, the curvature is constant or uniform ; other- 
ds 

wise the curvature is variable. Applying this definition to the 
circle of Fig. 61, of which the Y 

center is C and the radius is a, 




FIG CO 



we have A< 

As = a Arf>. 

di 
-- 



P^CP^, and hence 



Therefore - = - 
-j As a 

Hence - and the circle is 
as a 

a curve of constant curvature equal 
to the reciprocal of its radius. 

The reciprocal of the curva- 
ture is called the radius of cur- 
vature and will be denoted by p. Through every point of a curve 
we may pass a circle with its radius equal to p, which shall have 
the same tangent as the curve at the point and shall lie on the 




FIG. 61 



140 TRIGONOMETRIC FUNCTIONS 

same side of the tangent. Since the curvature of a circle is 
uniform and equal to the reciprocal of its radius, the curvatures 
of the curve and of the circle are the same, and the circle shows 
the curvature of the curve in a manner similar to that in which 
the tangent shows the direction of the curve. The circle is 
called the circle of curvature. 

From the definition of curvature it follows that 

_ ds 

If the equation of the curve is in rectangular coordinates, 

ds 

by (9), 36, p = ~ 

dx 
To transform this expression further, we note that 

,2 , 2 . ,2 



whence, dividing by dx* and taking the square root, we have 



Since <j> = tan- 1 , (by 15) 



dx 3 

< 



dx ^ . fdy\ 



\dx, 

Substituting, we have p = ^ 2 x 

d y 

dx* 

In the above expression "for p there is an apparent ambiguity of 
sign, on account of the radical sign. If only the numerical value 
of p is required, a negative sign may be disregarded. 



CUBVATUEE 141 

3,2 y2 

Ex. 1. Find the radius of curvature of the ellipse + *j = 1. 

dy & a a: 

Here -f = - -=- 

da; a a y 



Therefore P 

Ex. 2. Find the radius of curvature of the cycloid ( 49). 

We have -^ = a(l cos<) = 2asin 2 2, 

a<p <s 

d?/ . n A i 

^ = a sin /> =s 2 a sm J cos ^ 
d<p it & 

Therefore, by (9), 36, 

f^ = ctn^. 
dx 2 



and 



EXERCISES 

1. Find the radius of ourvature of the curve y* -fa a*. 

222 

2. Find the radius of curvature of the curve x* + y * = 

3. Find the radius of curvature of the curve y taa-^a; 1) 
at the point for which x 2 

/ 7T\ 3 

4. Show that the circle ( x -r- ) + y 2 = 1 is tangent to the curve 

y = sin a; at the point for which x = -jr, and has the same radius of 
curvature at that point. 

B. Find the radius of curvature of the curve x = cos t, y = cos 2 1, 
at the point for which t = 0, 

6. Find the radius of curvature of the curve x = a cos < -{- 
a< sin A, y =s a sin <]!> a<j! cos <. 

7. Prove that the radius of curvature of the curve as = o.cos 8 <, 



lias its greatest value when <j> = 



TT 




142 TRIGONOMETRIC FUNCTIONS 

51. Polar coordinates. So far we have determined the posi- 
tion of a point in the plane by two distances, x and y. We may, 
however, use a distance and a direction, as follows : 

Let (Fig. 62), called the origin, or pole, be a fixed point, and 
let OM, called the initial line, be a fixed line. Take P any point 
in the plane, and draw OP. Denote OP by r, and the angle M OP 
by 6 Then r and 6 are called the polar coor- 
dinates of the point P(r, 0), and when given 
will completely determine P. 

For example, the point (2, 15) is plotted 
by laying off the angle MOP =15 and meas- 
uring OP= 2. 

OP, or r, is called the radius vector, and 
6 the veotorial angle, of P. These quantities may be either 
positive or negative. A negative value of 6 is laid off in the 
direction of the motion of the hands of a clock, a positive angle 
m the opposite direction. After the angle 6 has been constructed, 
positive values of r are measured from along the terminal 
line of 0, and negative values of r from along the backward 
extension of the terminal line. It follows that the same point 
may have more than one pair of coor- 
dinates. Thus (2, 195), (2, -165), 
(-2, 15), and (- 2, - 345) refer to 
the same point. In practice it is usu- 
ally convenient to restrict to positive 
values. 

Plotting in polar coordinates is facili- 
tated by using paper ruled as m Figs. 64 
and 65. The angle 6 is determined from " ElG 63 
the numbers at the ends of the straight 

hnes, and the value of r is counted off on the concentric circles, 
either toward or away from the number which indicates 6, 
according as r is positive or negative. 

The relation between (r, 0) and (x, y) is found as follows : 

Let the pole and the initial line OM of a system of polar 
coordinates be at the same time the origin and the axis of # of a 
system of rectangular coordinates. Let P (Fig, 63) be any point 




POLAR COORDINATES 



143 



of the plane, (a;, #) its rectangular coordinates, and (r, 0) its 
polar coordinates. Then, by the definition of the trigonometric 
functions, 



XJ X 

cos 6 = - 



sm = 2-. 

Whence follows, on the one hand, 

x = r cos 

y = r sin i 
and, on the other hand, 

sin0 = 



cos = 



(1) 
(2) 



By means of (1) a transformation can be made from rectangular 
to polar coordinates, and by means of (2) from polai to rectangular 
coordinates 

When an equation is given in polar coordinates, the corre- 
sponding curve may be plotted by giving to 6 convenient values, 
computing the corre- 
sponding values of r, 
plotting the resulting 
points, and drawing a 
curve through them. 

Ex. 1. r a cos $ 

a is a constant which 
may be given any con- 
venient value We may 
then find from a table of 
natural cosines the value 
of r which corresponds 
to any value of 9. By 

plotting the points cor- J * 

xesponding to values of 6 FIG 04 

from to 90, we obtain 

the arc ABCO (Fig. 64). Values of 6 from 90 to 180 give the arc ODEA. 
Values of 6 from 180 to 270 give again the arc ABCO, and those fiom 270 
to 860 give again the arc ODEA. Values of 6 greater than 360 can clearly 
give no points not already found. The curve is a circle. 




aw 



MS 



144 



TRIGONOMETRIC FUNCTIONS 



Ex. 2. r= a sin 35. 

As 6 increases from to 30, r increases from to a ; as 9 increases 
from 30 to 60, r decreases from a to , the point (r, 0) traces out the loop 
040 (Fig. 65), which is evidently symmetrical with respect to the radius 
OA. As 6 increases from 
60 to 90, r is negative 
and decreases from to 

a ; as increases from 

90 to 120, r increases from isoj 

a to ; the point (r, 0) 
traces out the loop OBO 
As & increases from 120 
to 180, the point (r, 6} 
traces out the loop OCO. 
Larger values of & give 
points already found, since 
sin 3 (180 + 5) = - sm 3 6. 
The three loops are congiu- 
ent, because sm,3 (60 + 0) = 
sin30 This curve is called 
a rose of three leaves. 



r = i a V2 cos 2 0. 




Ex. 3. r 2 = 

Solving for r, we have 



Hence, corresponding to any values of 9 which make cos 2 9 positive, there 
will be two values of r numerically equal and opposite in sign, and two 
corresponding points of the curve symmetrically situated with respect 
to the pole If values are assigned to 9 which make cos 2 9 negative, the 
corresponding values of r will be 
imaginary and there will be no 
points on the curve. 

Accordingly, as 9 increases | ^ | $f 

from to 45, r decreases numer- 
ically from aV2 to 0, and the 
portions of the curve in the first 
and the third quadrant are con- 
structed (Fig 66) ; as 9 increases from 45 to 135, cos 2 9 is negative, and 
there is no portion of the curve between the lines 9 = 45 and 9 = 135 , 
finally, as 9 increases from 135 to 180, r increases numerically from to 
aV2, and the portions of the curve in the second and the fourth quadiant 
are constructed The curve is now complete, as we should only repeat the 
curve already found if we assigned further values to 0, it is called the 
lemniscate. 




FIG 06 




GRAPHS 145 

Ex. 4. The spiral of Archimedes, 
r = a8. 

In plotting, 6 is usually considered 
in cncular measuie When &= 0, r = , 
and as 6 mci eases, ? inci eases, so that 
the cuive winds an infinite uiuubei of 
times around the ongin while leced- 
ing from it (Fig 67) In the flgme the 
heavy line represents the poitiou of 

the spiral coriespondmg to positive values of 0, and the dotted line the 
portion coriespondmg to negative values of 6 

EXERCISES 
Plot the graphs of the following curves 

a 

1. r = a sm 6 9. r = a sin 8 - 

2. r = asm 20 

10. ?' 2 = a 2 sin 

3. r = a cos 30 

11. ?* = 



4. r = a sin-- 12. ,, = a (l _ cos 2 0). 

13. r = a(l+2cos20) 

5 - r=acos 2 ' 14. r =atan0 

6. 7- = 3 cos + 5 15. r = a tan 2 

7. ?' = 3 cos + 3.* 1 

16 r = t 

8. r = 3 cos + 2 ' 1 + cos 

Find the points of intersection of the following pairs of curves : 

17. r = 2 sm 0, ? = 2-x/S cos 0. 

18. r 2 = a 2 cos 0, ;- 2 = a 2 sin 2 

19. r = 1 + sin 0, r = 2 sin 

20. r 2 =a 2 sin0, ' 2 =a 2 sm30 

Transform the following equations to polar coordinates . 

21 . .r?/ = 4. 23. x z + y*-2ay = 0. 

22 . a; 2 + f - 4 </a? - 4 ay = 24. (a* + 2/ 2 ) 2 = a a ( a - Z/ 2 ) 
Transform the following equations to rectangular coordinates : 

25. r= ft sec 6 27. r = atan0. 

26. r = 2 n cos 28. r = a cos 2 

* The curve is called a cardioid 

t The curve is a parabola with the ongui at the focus. 



146 TRIGONOMETRIC FUNCTIONS 

52. The differentials <?r, d9, ds, in polar coordinates. We have 
seen, in 39, that the differential of arc in rectangular coordinates 
is given by the equation 

(1) 



If we wish to change this to polar coordinates, we have to 

place 

x = r cos 0, y r sin 6 ; 

whence dx = cos 6dr r sin 6 d6, 

dy = sin Qdr + r cos 6d6. 
Substituting in (1), we have 

ds* = dr z + r*d6 z . (2) 

This formula may be remembered by means of an " elemen- 
tary triangle " (Fig. 68), constructed as follows : 

Let P be a point on a curve r =/(0), the coordinates of P 
being (r, 0), where OP = r and MOP = 6. Let Q be increased 
by an amount d0 t thus determining another 
point Q on the curve. From as a center 
and with a radius equal to r, describe an 
arc of a circle intersecting OQ in R so that 
O.E = OP = r. Then, by 42, PR, = rdd. Now 
EQ is equal to Ar, and PQ is equal to As. -^ 6g 

We shall mark them, however, as dr and da 
respectively, and the formula (2) is then correctly obtained by 
treating the triangle PQJR as a right triangle with straight-line 
sides. The fact is that the smaller the triangle becomes as Q 
approaches P, the more nearly does it behave as a straight-line 
triangle ; and in the limit, formula (2) is exactly true. 

Other formulas may be read out of the triangle PQR. Let us 
denote by i/r the angle PQJR, which is the angle made by the 
curve with any radius vector. Then, if we treat the triangle PQR 
as a straight-line right-angle triangle, we have the formulas : 




. (3) 

dr k J 



DIFFERENTIALS 147 

The above is not a proof of the formulas. To supply the 
proof we need to go through a limit process, as follows: 

We connect the points P and Q by a straight line (Fig. 69) 
and draw a straight line from P per- 
pendicular to OQ meeting OQ at S. 
Then the triangle PQS is a straight- 
line right-angle triangle, and therefore 




, 
chord PQ 

SP arcPg 

arc PQ ' chord PQ' 

FIG. 69 
Now angle POQ = A0, arc PQ = As, 

and, from the right triangle OSP, SP = OP sin POQ = r sm A0. 
Therefore 

rsuiA * arcP r 



Ag chord P<2 A0 As chord PQ 

Now let A0 approach zero as a limit, so that Q approaches P 
along the curve. The angle SQP approaches the angle OPT, 

where PT is the tangent at P. At the same time ap- 

A/9 rlfi 

proaches 1, by 42 ; - approaches by definition ; and 

T>/~) ^ 

aic '' approaches 1, by 39. In this figure we denote the 

CllOlXl JL ty 

angle OPT by >/r and have, from (4), 

. . dB ,CN 

smf = r , (5) 

which is the first of formulas (3). It is true that in Fig. 69 we 
have denoted OPT by -^ and that in Fig. 68 ^ denotes OQP. 
But if we remember that the angle OQP approaches OPT as a 
limit when Q approaches P, and that in using Fig. 68 to read off 
the formulas (3) we are really anticipating this limit process, the 
difference appears unessential. 

The other formulas (3) may be obtained by a limit process 
similar to the one just used, or they may be obtained more 



148 TEIGONOMETEIC FUNCTIONS 

quickly by combining (5) and (2). For, from (2) and (5), 
we have 



whence cos -\lr = . (6) 

ds 

By dividing (5) by (6) we have 

rdd /7N 

CO 

dr 

In using (7) it may be convenient to write it in the form 

tan i/r = , (8) 

dr 

d0 
since the equation of tho curve is usually given in the form 

t?rt 

r =/(#), and - is found by direct differentiation. 
cLu 

Ex. Find the angle which the cmve 1 = a sin 4 makes with the radius 
vector 6 = 80 

Here ^ = 4 a cos 4 0. Therefore, fiom (8), tan ^ = " sin 4 ^ = i tan 'I 6 
do ^ T 4 cos 4 6 4 

Substituting 6 = 30, we have tan ^ = * tan 120 = - | V3 = - 4 MO 
Therefore j/r = 156 35'. 

EXERCISES 

1. Find the angle -which the curve r = a cos 3 makes with the 
radius vector 6 = 45 

2. Find the angle which the curve r = 2 + 3 cos makes with the 
radius vector 6 = 90 

A 

3. Find the angle which the curve r = a 2 sin a makes with the 
initial line. 

9 6 

4. Show that for the curve r = a, sin 8 5 j i/r = - 

o 3 

5. Show that the angle between the cardioid r = a(l cos 0) and 
any radius vector is always half the angle between the radius vector 
and the initial line. 



GENEHAL EXERCISES 149 

6. Show that the angle between the lemniscate r 2 = 2 . 2 cos 2 9 

TT 

and any ladms vector is always plus twice the angle between the 

radius vector and the initial line 

7. Show that the curves r a = a a sin 2 6 and ?- 2 = a 2 cos 2 6 inter- 
sect at right angles 

GENERAL EXERCISES 

Find the graphs of the following equations : 

a 1 4- 1 * o / L ^ 

1. ?/ = 4sm - 5. y = Ssmlas + TT ) 

2. y = cos (2 x 3) 6. vy a = tan x. 

3 . y = tan- 7- y = 2cos2(o! - 2) 

4. ys= Jsin2aj + ^8in8a5. 8. ?/ = 3cos3h + ^ 

Find -f^ in the following cases : 

CtttXj 



9. y = 2x 
10. y = ,J tan (3 K + 2) + J tan 8 (3 x + 2). 

12. tan (x + ?/) + tan (x y) = 0. 

13. y=3ctn fi | + 5ctn 8 |. 21. y = 

14. ?/ = csc a 4 x + 2 ctn 4 x 22. ?/ = 

16. ?/ = sm a 4 x cos*2 a; 23> V cos ~ a .a_|_ 4 > 

- cos 8 - 2 cos ~ 24. v = ctn" 1 Vsc 2 2x. 



17. ?/ = 

2 



18. ?y=atan 8 2a! ^tan2C+03. 25. 2/ = csc" 1 1 

t SC "y" X 

19. y = sin 







20. ?/ = cos" 1 3 

28. A particle moves in a straight line so that s = 6 

Show that the motion is simple harmonits and find the center about 
which, the particle oscillates and the- amplitude of the motion. 



150 TRIGONOMETRIC FUNCTIONS 

20 
?/ 

29. A particle moves on the ellipse i + ^ = 1 so that its projeq- 

tion upon OX describes simple harmonic motion given by x = a cos kt. 
Show that its projection upon OF also describes simple harmonic 
motion and find the velocity of the particle in its path. 

7T 

30. A particle moving with simple harmonic motion of period -r 

has a velocity of 9 ft per second when at a distance of 2 ft. from its 
mean position. Find the amplitude of the motion. 

31. A particle moves according to the equation s = 4 sin \t + 
5 cos t. Show that the motion is simple harmonic and find the 
amplitude of the swing and the time at which the particle passes 
through its mean position. 

32. Find the radius of curvature of the curve y = x sin- at the 

2 x 

point for which x = 

7T 

33. Find the radius of curvature of the curve ?/ = at the 

7!! 

point for which x = TT 

34. Find the radius of curvature of the curve y a sin' 1 ' V, 2 .r* 

CL 

at the point for which x = r 

35. Find the radius of curvature of the curve x => a cos <, y = I sin <{>, 

IT 

at the point for which <j> = -r- 

Plot the graphs of the following curves . 

a 

36. r ! =a 2 sm|- 41. r =1-20. 

42. r a = 

37. r a = a 2 sin40 .. , 

43. r*= 

38. r =a(l-sin0). . 

v ' 44. r =l + sm s . 

39. r =a(l+cos20). ^ 

30 

40. r = a(l + 2 sin 0). 45. r =* 1 + sin -7p 

Find the points of intersection of the following pairs of curves ; 

46. r*=3cos20, ?- s = 2cos a 0. 

47. r BS a cos 0, r 2 BB a 2 sin 2 0. 

48. r 2 sin 0, 7-' B* 4 sin 2 0. 

49. r ~a(l + sin 20), r 2 =? 4 a'siii 2 0. 



GENERAL EXERCISES 151 

Transform the following curves to polar coordinates : 

50. = " 



Transform the following curves to ^-coordinates 

52 ?' 2 =2a a sm20. 53. r = a(l cos0) 

54. Find the angle at which the curve r = 3 + sin 2 6 meets the 
circle r 3 

55. Find the angle of intersection of the two curves r = 2 sin 6 
and r 2 = 4 sin 2 

56. Find the angle of intersection of the curves r = a cos and 
r = a sin 2 B. 

57. If a ball is fired from a gun with the initial velocity v , it 

as? 

describes a path the equation of which is i/ = x tan a ' 

r i J 



where a is the angle of elevation of the gun and OX is horizontal 
What is the value of a when the horizontal range is greatest ? 

58. In measuring an electric current by means of a tangent galva- 
nometer, the percentage of error due to a small error in reading is 
proportional to tan x + ctn x. For what value of x will this percent- 
age of error be least ? 

59. A tablet 8 ft high is placed on a wall so that the bottom of 
the tablet is 29 ft. from the ground. How far from the wall should 
a person stand in order that he may see the tablet to best advantage 
(that is, that the angle between the lines from his eye to the top and 
to the bottom of the tablet should be the greatest), assuming that 
his eye is 5 ft. from the ground ? 

60. One side of a triangle is 12 ft. and the opposite angle is 36 
Find the other angles of the triangle when its area is a maximum 

61. Above the center of a round table of radius 2 ft is a hanging 
lamp. Plow far should the lamp be above the table in order that 
the edge of the table may be most brilliantly lighted, given that 
the illumination varies inversely as the square of the distance and 
directly as the cosine of the angle of incidence ? 

62. A weight P is dragged along the ground by a force F. If 
the coefficient of friction is k, in what direction should the force be 
applied to produce the best result ? 



L52 TRIGONOMETRIC FUNCTIONS 

63. An open gutter is to be constructed of boards in such a way 
,hat the bottom and sides, measuied on the inside, are to be each 
3 in wide and both sides are to have the same slope How wide 
should the gutter be across the top in oider that its capacity may 
je as great as possible ? 

64. A steel girdei 27 ft long is to be moved on rollers along a 
Dassageway and into a corridor 8 ft. in width at right angles to the 
Dassageway. If the horizontal width of the girder is neglected, how 
vide must the passageway be in order that the girder may go around 
-he corner 9 

65. Two particles are moving in the same straight line so that 
iheir distances from a fixed point are respectively x = a cos kt and 

e' = acosud -f qOj & and a being constants Find the greatest 

hstance between them 

66. Show that for any curve in polar coordinates the maximum 
Liid the minimum values of r occui in general when the radius vector 
s perpendicular to the curve. 

67. Two men aie at one end of the diameter of a circle of 40 yd 
adius. One goes directly toward the center of the circle at the 
miform rate of 6 ft. pei second, and the other goes around the 
.ircumference at the rate of 2 TT ft per second How fast are they 
eparatmg at the end of 10 sec. 'f 

68. Given that two sides and the included angle of a triangle are 
> ft , 10 ft , and 30 respectively, and are changing at the rates of 
- ft , 3 ft , and 12 per second respectively, what is the area of the 
riangle and how fast is it changing ? 

69. A revolving light in a lighthouse mi offshore makes one 
evolution a minute If the line of the shore is a straight line, how 
ast is the ray of light moving along the shore when it passes a 
ioint one mile from the point nearest to- the lighthouse ? 

70. BC is a rod a feet long, connected with a piston rod at C, and 
t B with a crank AB, b feet long, revolving about A. Find C's 
elocity in terms of All's angular velocity. 

71. At any time t the coordinates of a point moving in the ajy-plane 
re x = 2 3 cos t, y = 3 + 2 sin t Find its path and its velocity in 
,s path. At what points will it have a maximum speed? 

72 . At any time t the coordinates of a moving point are x = 2 sec 3 1, 
= 4 tan 3 1. Find the equation of its path and its velocity in its path. 



GENERAL EXERCISES 153 

73. The parametric equations of the. path of a moving particle are 
O3 = 2cos 8 <, 2/=2sin 8 < If the angle < increases at the rate of 
2 radians per second, find the velocity of the particle in its path 

74. A particle moves along the curve y = smo3 so that the 
a-component of its velocity has always the constant value a Find 
the velocity of the particle along the curve and determine the points 
of the curve at which the particle is moving fastest arid those at 
which it is moving most slowly 

75. Find the angle of intersection of the curves y = smx and 
y = cos x 

76. Find the angle of intersection of the curves y = sma; and 

/ , ^ 
2/ = sm(a: + ^ ) 

77. Find the angle of intersection of the curves y since and 
y = cos 2 x between the lines x = and ce = 2 TT. 

78. Find the points of intersection of the curves T/ = since and 
y = sin 3 x between the lines x = and x = TT Determine the angles 
at the points of intersection. 

79. Find all the points of intersection of the curves y = cos x and 
y = sin 2 x which, lie between the lines x = and x = 2 TT, and 
determine the angles of intersection at each of the points found. 



CHAPTER VI 
EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

53. The exponential function. The equation 

/ = *, 

where a is any constant, defines y as a function of x called the 
exponential function. 

If x = n, an integer, y is determined by raising a, to the nth 
power by multiplication. 

If #=*- a positive fraction, y is the jth root of the plli 

power of a. 

If x is a positive irrational number, the approximate value of 
y may be obtained by expressing oc approximately as a fraction. 

If x = 0, y = a=l. If x =- m, y = a~ m = 
The graph of the function is readily found. 

Ex. Find the gi aph of y = (1 5y. By giving convenient values to x 
we obtain the curve shown in Fig 70 To determine the shape of the 
curve at the extreme left, we place a equal to a large negative number, 

say x = - 100 Then y = (1.5)- = ^ J , 

which is very small It is obvious that the 

larger numencally the negative value of x 

becomes, the smaller y becomes, so that the 

curve appi caches asymptotically the negative _. 

portion of the a:-axis. O 

On the other hand, if # is a large positive 
number, y is also large. FIG. 70 

54. The logarithm. If a number N may be obtained by placing 
an exponent L on another number a and computing the result, 
then L is said to be the logarithm of .2V to the base a. That is, if 

N= d>, (1) 

then L = log a JV. (2) 

154 




LOGARITHMS 155 

Formulas (1) and (2) are simply two different ways of ex- 
pressing the same fact as to the relation of ,JV" and L, and the 
student should accustom himself to pass from one to the other 
as convenience may demand. 

From these formulas follow easily the fundamental properties 
of logarithms; namely, 

M = log 



=log a N, (3) 

log a l=0, 

log a - = -log JV. 

Theoretically any number, except or 1, may be used as 
the base of a system of logarithms. Practically there are only 
two numbers so used. The first is the number 10, the use of 
which as a base gives the common system of logarithms, which 
are the most convenient for calculations and are used almost 
exclusively in trigonometry. 

Another number, however, is more convenient in theoretical 
discussions, since it gives simpler formulas. This number is 
denoted by the letter e and is expressed by the infinite series 



where 21=1x2, 31=1x2x8, 41=1x2x3x4, etc. 
Computing the above series to seven decimal places, we have 

e = 2,7182818.... 

An important property of this number, which is necessary in 
finding the derivative of a logarithm, is that 



156 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

To check this arithmetically we may take successive small 
values of h and make the following computation : 

When &=.!, (1+ A)*= (1.1) 10 = 2.59374. 

When h =.01, (1+ A)*=(1.01) 100 = 2.70481. 

When h = .001, (1 + hy= (1.001) 1000 = 2.71692. 

When h =.0001, (1+ hy= (1.0001) 10000 = 2.71815. 



i 



Working algebraically, we expand (1+A) A by the binomial 
theorem, obtaining ... . _._. ... 

- 11 if i~1 V - 2 

A* 



2! 8! 



where JS represents the sum of all terms involving A, 7i a , 7i 8 , etc. 
Now it may be shown by advanced methods that as h approaches 
zero, B, also approaches zero ; so that 



When the number e is used as the base of a system of loga- 
rithms, the logarithms are called natural logarithms, or Napierian 
logarithms. We shall denote a natural logarithm by the symbol 
In*; thus, 

N ~*> (4) 

then L = In N. 

Tables of natural logarithms exist, and should be used if 
possible. In case such a table is not available, the student 

* This notation is generally used by engineers. The student should fenow 
that the abbreviation "log" is used by many authors to denote the natural 
logarithm. In this book " log " is used for the logarithm to the base 10. 



LOGAEITHMS 



157 



may find the natural logarithm by use of a table of common 
logarithms, as follows- 

Let it be required to find In 213. 

If x = In 213, 

then, by (4), 213 = e v ; 

whence, by (3), log 213 = a? log e, 



or 



log 213 2.3284 
Iog2.7183 0.4343 



Certain graphs involving the number y 
e are important and are shown in the 
examples. 

o 

Ex. 1. y = In or. 

Giving x positive values and finding y, we 
obtain Fig 71. 

Ex. 2. y = e-*? I FIG. 71 

The curve (Fig 72) is symmetrical with respect to OY and is always 
above OX. When a; = 0, y = 1 As a: increases numerically, y decreases, 
approaching zero. Hence OX is an asymptote ^ 





FIG. 72 



FIG. 73 



Ex.3. y = 



This is the curve (Fig. 73) made by a cord or a chain held at the ends 
and allowed to hang freely. It is called the catenary. 

Ex. 4. y e ~ *" sin Ix. 

The values of y may be computed by multiplying the ordmates of the 
curve y = <s~ oa! by the values of sin bx for the corresponding abscissas. Since 
the value of sinZw oscillates between 1 and -1, the values of e- 



158 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 



cannot exceed those of e-** Hence the graph lies in the portion of the 
plane between the curves y = e~ ax y 
and y = e ~ ax . When a: is a mul- 

tiple of -, y is zero. The graph 

therefore crosses the axis of x an 
infinite number of times Fig. 74 
shows the graph when a = 1, b = 2 TT 



Ex. 5. y e x 

When x approaches zeio, being 
positive, y increases without limit -1 '' 
When x approaches zero, being neg- 
ative, y approaches zeio , for example, when 

x = Ttnjff' y = e1000 ' and when x ~ ~~ "W 
11 = e - 1000 = - The function is therefore 

J glOOO 

discontinuous for x 

The line y - 1 is an asymptote (Fig. 75), 
for as x increases without limit, being posi- 

tive or negative, approaches 0, and y 
appioaches 1. 

Ex. 6. r = e a . 

The use of r and 6 indicates that we are 
using polar coordinates. 

When 6 = 0, r 1. As increases, r in- 
creases, and the curve winds around the origin 
at increasing distances from it (Fig 76). 
When 6 is negative and increasing numer- 
ically without limit, r approaches zero. 
Hence the curve winds an infinite number 
of times aiound the origin, continually ap- 
proaching it The dotted line in the figure 
corresponds to negative values of 6 

The curve is called the logarithmic spiral 

EXERCISES 

Plot the graphs of the following equations 

! y = ()* 5 - y = x&x - 




?a* 



74 



X 



75 




M 



3. y= B 

4. y 



7. y = log2o:. 
1 
'a: 



8. = 



FIG. 76 



9. y = log sin x. 

10. y = log tan ar. 

11. y = e~ 2ar sin4ai. 

12. y = e~ x cos 3 a;. 

13. r=e-* e . 



EMPIKICAL EQUATIONS 159 

55. Certain empirical equations. If x and y are two related 
quantities which are connected by a given equation, we may 
plot the corresponding curve on a system of ^-coordinates, and 
every point of this curve determines corresponding values of 
x and y. 

Conversely, let x and y be two related quantities of which 
some corresponding pairs of values have been determined, and 
let it be desired to find by means of these data an equation con- 
necting x and y in general. On this basis alone the problem 
cannot be solved exactly. The best we can do is to assume that 
the desired equation is of a certain form and then endeavor to 
adjust the constants in the equation in such a way that it fits 
the data as nearly as possible. We may proceed as follows : 

Plot the points corresponding to the known values of x and y. 
The simplest case is that in which the plotted points appear to 
lie on a straight line or nearly so. In that case it is assumed 
that the required relation may be put in the form 

y = mx + *, (1) 

where m and I are constants to be determined to fit the data. 
The next step is to draw a straight line so that the plotted 
points either lie on it or are close to it and about evenly dis- 
tributed on both sides of it. The equation of this line may be 
found by means of two points on it, which may be either two 
points determined by the original data or any other two points 
on the line. 

The resulting equation is called an empirical equation and 
expresses approximately the general relation between x and y. 
In fact, more than one such equation may be derived from the 
same data, and the choice of the best equation depends on the 
judgment and experience of the worker. 

Ex. 1. Corresponding values of two related quantities x and y are given 
by the following table : 

ai 1 2 4 6 10 



y 1.8 2.2 2.9 3.9 6.1 
Find the empiucal equation connecting them. 



160 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

We plot the points (a;, y) and draw the straight line, as shown, in 
Fig. 77. The straight line is seen to pass through the points (0, 1) and 
(2, 2). Its equation is therefore 

T7" 

?/ = 5 x + 1, 
which is the required equation 

111 many cases, however, 
the plotted points will not 
appear to lie on or near a 
straight line. We shall con- 
sider here only two of these ~o i 
cases, which are closely con- - FIG ?7 

nected with the case just 

considered. They are the cases in which it may be anticipated 
from previous experience that the required relation is either 
of the form y=db*, (2) 

where a and b are constants, or of the form 

y=atf, (3) 

where a and n are constants. 

Both of these cases may be brought directly under the first 
case by taking the logarithm of the equation as written. Equa- 
tion (2) then becomes 

log y = log a, + x log 6. (4) 

As log a and log I are constants, if we denote log y by y', 
(4) assumes the form (1) in x and y\ and we have only to 
plot the points (#, y'~) on an o^'-system of axes and determine 
a straight line by means of them. The transformation from. (4) 
back to (2) is easy, as shown in Ex. 2. 

Taking the logarithm of (3), we have 

log y = log a + n log x. (5) 

If we denote logy by y' and logo; by x\ (5) assumes the form 
(1) in. a/ and /, since log a and n are constants. Accordingly 
we plot the points (V, y r ~) on an a/?/-system of axes, determine 
the correspond ing straight line, and then transform back to (8), 
as shown in Ex. 3. 



EMPIKICAL EQUATIONS 161 

Ex. 2. Corresponding values of two related quantities x and y are given 
by the following table . 

a; 8 10 12 14 16 18 20 



y 3.2 40 73 98 152 240 364 

Find an empirical equation of the form y = dbf* 

Taking the logarithm of the equation y = ob*, and denoting log y by y', 

we have ' 

if = log a + x log i. 

Determining the loganthm of each of the given values of y, we form a 
table of corresponding values of x and y', as follows : 

x 8 10 12 14 16 18 20 

if = logy 5051 6028 8633 9912 1.1818 1.3909 15611 

We choose a large-scale plotting-paper, assume on the y'-axis a scale 
four times as laige as that on the 
or-axis, plot the points (x, /), and 

20 

J, 

point. Its equation is 10 > 



tour times as laige as tnat on tne * 

or-axis, plot the points (x, /), and i j 

draw the straight line (Fig 78) 20 I 

H-iivM-in-li +.Vo -fifaf. o.nrl tliA HTvKh i e 



through the first and the sixth 15 



' = .08858 x - .20354. 5 



m 



2 4 6 8 10 12 14 16 18 20 * 
Therefore log a = 20354 = FlG ?g 

9 7965 - 10, whence a = .626 ; and 

log & = .08858, whence & = 1 22. Substituting these values in the assumed 
equation, we have 



as the required empirical equation The result may be tested by substitxit- 
ing the given values of x in the equation. The computed values of y will 
Le found to agree fairly well with the given, values. 

Ex. 3. Corresponding values of pressure and volume taken from an 
indicator card of an air -compressor are as follows: 

p 18 21 205 33.5 44 62 

v .035 .656 .476 .897 .321 .243 

Find the relation between them in the tormpv" ~ c, 



162 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 



Writing the assumed relation, in. the form p = cv~ n and taking the 
logarithms of both sides of the equation, we have 



log p =n log v + log c, 



or 

where 



y = log jtjj x = log v, and 6 = log c. 
The corresponding values of x and y are 

- 4012 -.4935 - 6144 
1 5250 1 6435 1 7994 



x = logu 1972 
y = logp 1 2553 



- 2549 - .3233 
1 3222 1 4232 



For convenience we assume on the ar-axis a scale twice as large as that on 

the y-axis, plot the points (x, ?y), and diaw the straight line as shown in 

Fig 79 The construction should 

be made on large-scale plotting- 

paper. The line is seen to pass 

through the points (- 05, 1 075) 

and ( .46, 1.6) Its equation is 

therefore 

y=- 128 x +101 

Hence n 1 28, log c = 1 01, 
c = 10 2, and the required rela- 
tion between p and v is 




pv 



i as = 10 2 



) -.55-50- lt5-W-SS-SO~25-0-15 10- OS 

FIG. 79 



EXERCISES 



1. Show that the following points lie approximately on a straight 
line, and find its equation : 



4 

21 



46 



13 

7 



20 
12 



22 

12 9 



25 
14,5 



30 
18.2 



2. For a galvanometer the deflection D, measured in millimeters 
on a proper scale, and the current /, measured in microamperes, are 
determined in a series of readings as follows : 

D 291 48.2 72.7 92.0 118.0 140.0 165.0 109.0 
I 0.0493 0.0821 0123 0.154 0197 0234 0.274 0328 



Find an empirical law connecting D and /. 



EMPIRICAL EQUATIONS ' - , 163 

3. Corresponding values of two related quantities x and y are 
given in the following table 

x 01 03 0.5 0.7 0.9 1.1 1 3 1.5 

y 3316 4050 4046 0.6041 0.7379 9013 1 1008 1.3445 

Find an empirical equation connecting x and y in the form y = ab*. 

4. In a certain chemical reaction the concentration c of sodmm 
acetate produced at the end of the stated number of minutes t is 
as follows : 

* 1 2 8 4 5 

c 00837 00700 00586 00492 00410 

Find an empirical equation connecting c and t in the form c = atf 

5. The deflection a of a loaded beam with a constant load is 
found for various lengths I as follows 

Z 1000 900 800 700 600 

a 7.14 5.22 3.64 242 1.50 

Find an empirical equation connecting a and I in the form a = nlf 1 

6. The relation between the pressure j and the volume v of a gas 
is found experimentally as follows : 

p 20 23.5 31 42 59 78 

v 0019 0.540 0.442 0358 0277 0219 

Find an empirical equation connecting p and v in the form pv n = c. 

56. Differentiation. The formulas for the differentiation of 
the exponential and the logarithmic functions are as follows, 
where, as usual, u represents any function which can be differen- 
tiated with respect to #, In means the Napierian logarithm, and 

a is any constant: 

d , log a e du ,1 . 

-r-log a tt = fi a-- r (1) 

das u da 

d , 1 du ^ o ^ 

--- In u = - 3-, (2) 

aa? u ax 



164 EXPONENTIAL AND LOaAEITHMIC FUNCTIONS 



dx" ~- ~dx (3) 

d u B du 

dx dx' ^ -' 

The proofs of these formulas are as follows: 



.j 
To find - log a M place y = log M. 

Then, if u is given an increment AM, y receives an increment 
Ay, where 

Ay = log a (M + AM) - log M 



Aw, A , AM\ A 

) ' 
u / 



the transformations being made by (3), 54. 



Then 

AM 



Now, as AM approaches zero the fraction may be taken 
A of 54. u 



as A of 54. 

u 

Hence Lim fl 4- Y "= e. 

A-0\ M / 



and 



2. If y = hi u, the base a of the previous formula is e ; and 
since log e=l, we have 

dy _ 1 du 
dx u dx 



DIFFERENTIATION 165 



3. If # = ", 

we have In y = In a u = u In a. 

Hence, by formula (2), 

1 dy , du 
-- = lna ; 
y ax ax 

. dy ul du 

whence ~ = a In a - 

ax ax 

4. If y = e w the previous formula becomes 

^ = e ^. 
dx dx 

Ex. 1. y = In (a; 2 -4 a + 5). 

dy _ 2 x 4 
da; a; 2 4 a; + 5 



Ex.2. y = '- 
Ex. 3. y = e- 



^ = cos bx (e~ 
dx dx 



EXERCISES 

Find in each of the following cases : 
dx 

-- I~sin2aj 



2. y*= 

3. JJ^-. . 

4. y = a sln ""X 12. y = e" a>e sin 3 cc. 

5. / 



. y^ln 

6. y = lnV2a; a +6a; + 9. * 

, a; - 3 14. y - 6 8a (9 a 2 - 6ai + 2). 
7 . w == -J In - 5 

x + 3 _ 16. y e* x (2 sin a: - cos ). 
8. = ln(flj + Vaj 8 + 4). a: 9 



9. 3/ = 

17. y = sec x tan 03 + In (sec a? -f- tan *). 

. VaTTl-1 

18. v In 



166 EXPONENTIAL AND 'LOGARITHMIC FUNCTIONS 

57. The compound-interest law. An important use of the ex- 
ponential function occurs in. the problem to determine a, function 
whose rate of change is proportional to the value of the function. 
If y is such a function of #, it must satisfy the equation 



where ~k is a constant called the proportionality factor. 
We may write equation (1) in the form 

ld V-7 f . 

r /fc , 

ydx 

whence, by a very obvious reversal of formula (2), 56, we have 

In y = kx 4- (7, 

where C is the constant of integration (18). 
From this, by (1) and (2), 54, 



Finally we place e c =A, where A may be any constant, since 
C is any constant, and have as a final result 

(2) 



The constants A and Jc must be determined by other condi- 
tions of a particular problem, as was done in 18. 

The law of change here discussed is often called the compound- 
interest law, because of its occurrence in the following problem : 

Ex. Let a sum of money P be put at interest at the rate of r% per annum. 

A 

The interest gained in a time Ai is -Pr^ Ai, where A* is expressed in 
years. But the interest is an increment of the principal P, so that we have 



In ordinary compound interest the interest is computed for a certain 
interval (usually one-half year), the principal lemaming constant during 
that interval The interest at the end of the half year is then added to the 
principal to make a new principal on which interest is computed for th.9 



COMPOUND-INTEREST LAW 167 

next half year. The principal P therefore changes abruptly at the end of 
each half year. 

Let us now suppose that the principal changes continuously ; that is, 
that any amount of interest theoretically eained, in no matter how small 
a time, is immediately added to the puncipal. The average rate of change 
of the principal in the peuod Ai is, fiorn 11, 

* = lL m 

Ai 100 ( } 

To obtain the rate of change we must let A< approach zero in equation 
(1), and have dp ? , 



From this, as in the text, we have 

r 

P=Ae lo . (2) 

To make the problem conci ete, suppose the original principal were $100 
and the rate 4%, and we ask what would be the principal at the end of 14 yr. 
We know that when t = 0, P = 100. Substituting these values in (2), we 
have A = 100, so that (2) becomes 

JL, 1 

P = 100e 100 =100e 20 

Placing now t = 14, we have to compute P = 100 ei e The value of P 
may be taken from a table if the student has access to tables of powers of e 
In case a table of common logarithms is alone available, P may be found 
by first taking the logarithm of both sides of the last equation. Thus 

logP = loglQO + it loge = 24053; 
whence P = $254, approximately 

EXERCISES 

1. The rate, of change of y with respect to x is always equal to 
\ y, and when x = 0, y = 5. Find the law connecting y and x, 

2. The rate of change of y with lespect to x is always 0.01 times ?/, 
and when x = 10, y = 50. Find the law connecting y and x. 

3. The rate of change of y with respect to x is proportional to y, 
When x = 0, y = 7, and when x = 2, y = 14 Find the law connect- 
ing y and x. 

4. The sum of $100 is put at interest at the rate of 5% per annum 
under the condition that the interest shall be compounded at each 
instant of time. How much will it amount to in 40 yr.? 



168 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

6. At a certain date the population of a town is 10,000. Forty 
years later it is 25,000. If the population increases at a rate which 
is always proportional to the population at the time, find a general 
expression for the population at any time t. 

6. In a chemical reaction the rate of change of concentration of 
a substance is proportional to the concentration at any time. If the 
concentration is y^ when t = 0, and is T T when t = 6, find the law 
connecting the concentration and the time. 

7. A rotating wheel is slowing down in such a manner that the 
angular acceleration is proportional to the angular velocity. If the 
angular velocity at the beginning of the slowing down is 100 revolu- 
tions per second, and in 1 min. it is cut down to 50 revolutions per 
second, how long will it take to reduce the velocity to 25 revolutions 
per second ? 

GENERAL EXERCISES 

Plot the graphs of the following equations : 

i i 

1 - y = ($)"* 4 - y el ~ x - ? y e?. 

2. y = e 1 -*. 5. y = (e* + -*). 8. y = aser*. 

3. y = e S GOSX. 6. y= ~ 9. y=*y?&-*. 

J e'+e"* J 

10. For a copper-nickel thermocouple the relation between the 
temperature t in degrees and the thermoelectric power p in micro- 
volts is given by the following table : 

t 50 100 160 200 

P 24 25 26 26.9 27.5 

Find an empirical law connecting t and p. 

11. The safe loads in thousands of pounds for beams of the same 
cross section but of various lengths in feet are found as follows : 

Length 10 11 12 18 14 16 

Load 123.6 121.5 111.8 107.2 1018 90.4 

Find an empirical equation connecting the data. 



GENERAL EXERCISES 169 

12. In the following table s denotes the distance of a moving 
body from a fixed point in its path at time t 

t I 2 4678 

s 10 4 6400 0.1024 0410 0164 

Find an empirical equation connecting s and t in the form s = ab*. 

13. In the following table c denotes the chemical concentration of 
a substance at the time t . 

t 2 4 6 8 10 

c 00060 00048 0.0033 0.0023 00016 

Find an empirical equation connecting c and t in the form c = ah*. 

14. The relation between the length I (in millimeters) and the 
time t (in seconds) of a swinging pendulum is found as follows : 

I 634 805 90.4 1013 107.3 1406 



t 0.806 0892 0.960 1010 1038 1.198 

Find an empirical equation connecting I and t in the form t Td n . 

15. For a dynamometer the relation "between the deflection 8, 

2ir 

when the unit 6 = -r-rri and the current I, measured in amperes, is 

as follows : 
6 40 86 120 160 201 240 280 320 362 



I 0.147 0215 0.252 0293 0.329 0360 0.390 0.417 0442 

Find an empirical equation connecting J and 6 in the form I k$" 

16. In a chemical experiment the relation between the concen- 
tration y of undissociated hydrochloric acid and the concentration x 
of hydrogen ions is shown in the table 

x 1.68 1.22 784 426 0.092 047 0096 0.0049 



y 1 82 676 216 074 0085 0.00815 0.00036 00014 

Find an empirical equation connecting the two quantities m the 
form y = kx n . 



170 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

17. Assuming Boyle's law, pv = c, determine o graphically from 
the following pairs of observed values : 

39.92 42.17 45.80 48 62 51 80 (J0.47 CMS 1)7 



40.37 38.32 85.32 33.29 31.22 2080 
Tind -J- in each of the following cases : 

CCSs 

. Bx-2 
18. = 



19. y = In sin a-. 

e _ e - 

20. ?/ = tan" 1 

& 

21. y = In (2as + V4 aj a l) + 2a osc"^ a;. 

_a 

22. y = x z e x . 

23. = l 



24. y = I tan 2 oas + In cos ax. 
25 . y = x tan- 1 ^ $ In (1 + a; 2 ). 



26. A substance of amount x is being decomposed at a rate which 
is proportional to a;. If x = 3 12 when t = 0, and cc == 1.30 whun 
t = 40 min., find the value of x when t = 1 hr. 

27. A substance is being transformed into another at a rate whiith 
is proportional to the amount of the substance still imtranafovnuid, 
If the amount is 50 when t = 0, arid 15.6 when t = 4 hr., lind how 
long it will be before y^ of the original substance will rumain. 

28. According to Newton's law the rate at which tho temperature 
of a body cools in air is proportional to the difference botwoou tho 
temperature of the body and that of the air. If the tempwraturo of 
the air is kept at 60, and the body cools from 130 to 120" in 300 sue., 
when will its temperature be 100 ? 

29. Assuming that the rate of change of atmospheric prnssuro p 
at a distance h above the surface of the earth is proportional to tho 
pressure, and that the pressure at sea level is 14.7 Ib. per square inch 
and at a distance of 1COO ft. above soa level is 18.8 Ib. per square 
inch, find the law connecting^ and h. 



GENERAL EXERCISES 171 

30. Prove that the curve y = e~ Zx sm Sec is tangent to the curve 
y = e- 2 * at any point common to the two curves. 

31. At any time t the coordinates of a point moving in a plane 
are x = a~' 2t cos 2 1, y o~' 2t sin 2 1. Find the velocity of the point at 
any time t Find the rate at which the distance of the point from 
the origin is decreasing. Prove that the path of the point is a loga- 
rithmic spiral. 

32. Show that tho logarithmic spiral r = e" 9 cuts all radius vectors 
at a constant angle 

33. Find the radius of curvature of the curve y = e~ Za sin 2x at 

7T 

the point for which x = 

ft ( 5 

34. Show that tho catenary y n^+e n and the parabola 

1 ^ 

y = a -f- - a; 3 have tho same slope and the same curvature at their 

i Q> 

common point. 

35 . Find the radius of curvature of the curve x =s e t sin t, y = e* cos t. 

36. Show that the product of the radii of curvature of the curve 

X 

y =s ae " at the two points for which x = a is a?(e + e" 1 ) 8 . 

37. Find the radius of curvature of the curve ?/ = In a; and its 
least value. 

38. Find the radius of curvature of the curve y = e r cosaj at the 

7T 

point for which x = -= 



\ 

n ) 



CHAPTER VII 
SERIES 

58. Power series. The expression 

a fl + a : x + ajK? + a$+ ajf-\ ---- , (1) 

where a Ql a^ a z , are constants, is called a power series in x. 
The terms of the series may be unlimited in number, in which 
case we have an infinite series, or the series may terminate after 
a finite number of terms, in which case it reduces to a polynomial. 

If the series (1) is an infinite series, it is said to converge 
for a definite value of x when the sum of the first n terms 
approaches a limit as n increases indefinitely. 

Infinite series may arise through the use of elementary opera- 
tions. Thus, if we divide 1 by 1 a; in the ordinary manner, 
we obtain the quotient 



and we may write 



LX 



(2) 

^ ' 



Similarly, if we extract the square root of 1 + x by the rule 
taught in elementary algebra, arranging the work as follows; 



l+x 



2 + - 



V 


1 4 




a+ I 


a 2 
4 






2? 


a; 8 x* 




4" 


"8 + 64' 



172 



POWER SERIES 173 

the operation may be continued indefinitely. We may write 



The results (2) and (3) are useful only for values of x for 
which the series in each case converges. When that happens 
the more terms we take of the series, the more nearly is their 
sum equal to the function on the left of the equation, and in 
that sense the function is equal to the series. For example, the 
series (2) is a geometric progression which is known to con- 
verge when x is a positive or negative number numerically less 
than 1. If we place x = % in (2), we have 

"- 



which is true in the sense that the limit of the sum of the terms 
on the right is f . If, however, we place x 3 in (2), we have 



which is false. A reason for this difference may be seen by 
considering the remainder in the division which produced (2) 
but which is neglected in writing the series. This remainder is 

after n terms of the quotient have been obtained ; and if 



l-x 

x is numerically less than 1, the remainder becomes smaller and 
smaller as n increases, while if x is numerically greater than 1, 
the remainder becomes larger. Hence in the former case it may 
be neglected, but not in the latter case. 

The calculus offers a general method for finding such series 
as those obtained by the special methods which led to (2) and 
(3). This method will be given in the following section. 

59. Maclatirin's series. We shall assume that a function can 
usually be expressed by a power series which is valid for ap- 
propriate values of x, and that the derivative of the function may 
be found by differentiating the series term by term. The proof 
of these assumptions lies outside the scope of this book. Let us 
proceed to find the expansion of sin x into a series. We begin by 
writing $oLv s *A+Bx + Cx*+Dx*+Ex*+Fx*+ - - -, (1) 

where -4, J?, (7, etc. are coefficients to be determined. 



174 SERIES 

By differentiating (1) successively, we have 



- 5 . 4 

cosa;=3.2. J> + 4.3 . 2.Je + 5. 
sin z = 4- 3- 2- J+5. 4- 3 . 2 . Jfc 
-3 - 



By substituting x = in equation (1) and each of the fol- 
lowing equations, we get 

.4 = 0, 5=1, C=0, 3.2.D = -1, J?=0, 5.4.3.2.^ = 1; 

whence JL = 0, 5 = 1, tf= 0, -O = -^ T #=<>, ^=^- 

o I o 

Suhstitutuig these values in (1), we have 

8 , X 6 SQ^ 

sma; = a; __ + __..., (2) 

and the law of the following terms is evident. 

The above method may obviously be used for any function 
which may be expanded into a series. We may also obtain a 
general formula by repeating the above operations for a general 
I auction /(a;). 



We place f(x) = A + Bz + C3?+J)x*+lSz l + - (3) 

and, by differentiation, obtain in succession 



f"(x) = 3 ! Z> + 4 3 2 . Ex+ . ., 



where f f (x), /"()/'"(); and/ v (a;) represent the first, second, 
third, and fourth derivatives of /(#). 

We now place a; = in these equations, indicating the results 
of that substitution on the left of the equations by the symbols 



MACLAUKIN'S SERIES 175 



> f" ()> etc - We thus determine A, B, C, D, E, etc., 
and, substituting in (8), have 

A*) =AOHAO>+^AoX+^/"(OX+ ^<oy+- - - (4) 

This is called Madaurin's scries. 

Ex. 1. Find the value of sin 10 to foui decimal places. 

Wo tuny use sones (8), but have? to remembei ( 42) that x must be in 

eiivuhii measure. Hence wo i>lace x = -~~ 17458, where wo take five 

loO 

significant figmes in ordei to insure accuracy in the fomth significant 
iigiU'O of the result.* 

Subntibutmg in (2), we have 

. TT ,_,_ f.17458) 8 , 
sin = .17451}- v ' + - 

ia O 

= .17458 - 00089 = .17804. 

Ilonco to four decimal places am 10= .1736. 

We have u.scd only two terms of the series, since a rough calculation, 
which may bo made with a = 2, shows that the third term of the series 
will not affect the fourth decimal place. 

Ex. 2. Find the value of sin 61 to four decimal places. 

In radians the angle 01 is ~-.TT= 1.0647. If this nuxnbei were sub- 

luO 

Htituted in the sonos (2), a great many terms would have to be taken to 
include all which affect the first four decimal places. We shall therefore 

find a series for 8in(~+oA and afterwards place o: = j~(=l). We 
chow the angle ~^(~ (10) because it is an angle near 61 for which we " 

know the nine and cosine. The series may b(5 obtained by the method by 
which (2) was obtained, For variety we shall use the general formula (4). 

Wo have then . \ i _. 

/(O) . tnn| = |V; 



This JH not a general rule. In other cases the student may need to cany 
two or men three more significant figures in the calculation than are needed 
in the result. 



176 SERIES 

Therefore, substituting in (4), we have 



In this we place x = -^- = .01745 and perform the arithmetical calcula- 
loO . . 

tion. We have sin 61= sin(^ + ~}= 8746. 

\o 180/ 

Ex. 3. Expand In (1 + a) 

The function In x is an example of a function which cannot be expanded 
into a Maclaurm's series, since if we place /(a;) = In a;, we find /(0),/'(0), 
etc to be infinite, and the series (4) cannot be written We can, however, 
expand ln(l + a;) by series (4) or by using the method employed in obtain- 
ing (2). The latter method is more instructive because of an interesting 
abbreviation of the work We place 

ln(l + a;) = A + Bx + Cx z + Dx 3 + Eat + . 

Then, by differentiating, 

= B + 2 CJB + 3 Dx* + 4 Ex s + - . 



1 + x 
But we know, by elementary algebra, that 

- - =1 x + x z x s + 
1 + x 

Hence, by comparing the last two series, we have 

JB=1, C = -k Z>=, E=-%, etc. 

By placing x = in the first series, we find In 1 = A, whence .4 = 0. We 
have, therefore, ^ 8 ^ 

In (1 + a;) = x -_ + _-_+.... 

EXERCISES 

Expand each, of the following functions into a Maclaurin's series : 
1. d*. 2. cossc. 3. tan a;. 4. sin" 1 ^. 

5. tan" 1 ^ 6. sinf-j + a?). 7. ln(2 + aj). 

8. Prove the binomial theorem 

t . ... i , n(n 1) . n(n 'IVw 2) , . 

(a + o;) n =a' l +wa n - 1 aj4- v / a n ~ 2 a5 2 + v O y . v / a n - 8 a5*4--- . 

i\ O\ 

9. Compute sin 5 to four decimal places 
10. Compute cos 62* to four decimal places. 



TAYLOR'S SERIES 177 

60. Taylor's series. In the use of Maclaurin's series, as given 
in the previous section, it is usually necessary to restrict our- 
selves to small values of x. This is for two reasons. In the 
first place, the series may not converge for large values of x\ 
and in the second place, even if it converges, the number of 
terms of the series which it is necessary to take to obtain a 
required degree of accuracy may be inconveniently large. This 
difficulty may be overcome by an ingenious use of Maclaurin's 
series as illustrated in Ex. 2 of the previous section. We may, 
however, obtain another form of series which may be used when 
Maclaurin's series is inconvenient. 

Let f(x) be a given function, and let a be a fixed value of 
x for which the values of /(#) and its derivatives are known. 
Let * be a variable, or general, value of x which does not differ 
much from a ; that is, let x a be a small number, positive or 
negative. We shall then assume that/(a;) can be expanded in 
powers of the binomial x a ; that is, we write 

f(&=A+(x-a) + C(x-a)*+I)(x-a) 8 -\ , (1) 

and the problem is to determine the coefficients -4, B, C, . 
We differentiate equation (1) successively, obtaining 



In each of these equations place x = a. We have 
f(a)~A, /< = , /"(a) =2,1 (7, etc.; 

f" (ch f" (a\ 

whence 4=/(a), B =/'(), C^^f 2 ' #= 8 , ' et <>- 

Substituting in equation (1), we have as the final result 

a)^ - - .. (2) 



This is known as Taylor's series. Since, as has been said, it 
is valid for values of x which make x a a small quantity, the 



SERIES 

function /(a?) is said to be expanded in the neighborhood of 
x a. It is to be noticed that Taylor's series reduces to 
Maclaurin's series when a = 0. Maclaunn's series is therefore 
an expansion in the neighborhood of x 0. 

Ex. Expand In a; in the neighborhood of x = 3. 

Here we have to place a = 3 in the general formula. The calculation of 
the coefficients is as follows : 

/():= In a, /(3) = ln3, 



and therefore 

1 f~ QNfl 

BT ( - 3 ) 



This enables us to calculate the natural logarithm of a number near 3, 
provided we know the logarithm of 3. For example, let us have given 
In 3 = 1 0986 and desire In 3. Then x - 3 = , and the series gives 



= 1.0986 + 1667 - .0139 + 0015 - .0002 + . . . 
= 1 2527. 

The last figure cannot be depended upon, since we have used only 
four decimal places in the calculation. 

EXERCISES 

Expand each of the following functions into a Taylor's series, 
using the value of a given in each case : 

1. In*, a = 5. M TT 

4. cos a;, a - 

2. j a = 2. 



. TT 

3. sin x, a = 




8. Compute sin 46 to four decimal places by Taylor's series. 

9. Compute cos 32 to four decimal places by Taylor's series. 
10. Compute e 1 - 1 to four decimal places by Taylor's series. 



GENERAL EXEECISES 179 

GENERAL EXERCISES 
Expand each of the following functions into series in powers of a; 

1. ln(l a?). ,1+as . fir 

' 4. ln= 6. sm(v + 

2. seco;. 1 -a? \b 

3. ,~ 6. cos(-7r + ). 7. /z 3' 



8. Verify the expansion of tan a? (Ex 3, 59) by dividing the 
series for sin x by that for cos x. 

9. Verify the expansion of sec x (Ex. 2) by dividing 1 by the 

series for cos x. 

1 x 

10. Expand 1 by Maclaurin's series and verify by dividing 

JL ~T~ tC 

the numerator by the denominator 

11. Expand e* cos x into a Maclaurin's series, and verify by 
multiplying the series for & by that for cosaj. 

12. Expand e^sime into a Maclaurin's series, and verify by 
multiplying the series for tf* by that for sin a;. 

13. Expand e T ln(l+aj) into a Maclaurin's series, and verify by 
multiplying the series for e 00 by that for ln(l-|- K). 

14. Compute cos 15 to four decimal places 

15. Compute sin 31 to four decimal places 

16. Compute e* to four decimal places by the series found in 
Ex. 1, 59. 

17. Using the series for ln(l+ x~), compute Inf to five decimal 
places. 

18. Using the series found in Ex. 4, compute In 2 to five decimal 
places, and thence, by aid of the result of Ex. 17, find In 3 to four 
decimal places. 

19. Using the series found in Ex. 4, compute In | to five decimal 
places, and thonce, by aid of the first result of Ex. 18, find In 6 to 
four decimal places. 

20. Using the series found in Ex. 4, compute ln$ to four decimal 
places, and thence, by aid of the result of Ex. 18, find In 7 to three 
decimal places. 

21. Compute the value of TT to four decimal places, from the ex- 

1 IT 

pansion of sin- 1 a5 (Ex. 4, 59) and the relation sin- 1 ^ = -g 



180 SEEIES 

22. Compute the value of IT to four decimal places, from the ex- 
pansion of tan- 1 ;*; (Ex. 5, 59) and the relation tan- 1 - + 2 tan" 1 - = 

23. Compute -v/17 to four decimal places by the binomial theorem 
(Ex. 8, 59), placing a = 16, x = 1. 

24. Compute "\/26 to four decimal places by the binomial theorem 
(Ex. 8, 59), placing a = 27, x =- 1. 



/"* X o-\r\ /* 

25. Obtain the integral / dx in the form of a series 

pansion. J x 

26. Obtain the integral I e'^dx in the form of a series 
pansion. ^ 

f*x C&C 

27. Obtain the integral / in the form of a series expansion. 

Uo * ~T" * 

28. Obtain the integral / 5 in the form of a series expansion. 

i/o x 



expansion 



CHAPTER VIII 
PARTIAL DIFFERENTIATION 

61. Partial differentiation. A quantity is a function of two 
variables x and y when the values of x and y determine the 
quantity. Such a function is represented by the symbol /(#, #). 
For example, the volume V of a right circular cylinder is a 
function of its radius r and its altitude h, and in this case 



Similarly, we may have a function of three or more variables 
represented by the symbols /(a;, #, z),f(x, y, 2, w), etc. 

Consider now /(*, jr), where x and y are independent varia- 
bles so that the value of x depends in no way upon the value 
of y nor does the value of y depend upon that of x. Then we 
may change x without changing y, and the change in x causes 
a change in /. The limit of the ratio of these changes is the 
derivative of / with respect to x when y is constant, and may 

/ JTJ?\ 

be represented by the symbol ( ) v^ 

\ dx J v 
Similarly, the derivative of / with respect to y when x is 

constant, is represented by the symbol i-j- Y These derivatives 

\dy/x 

are called partial derivatives of / with respect to x and y re- 
spectively, The symbol used indicates by the letter outside 
the parenthesis the variable held constant in the differentiation. 

When no ambiguity can arise as to this variable, the partial de- 

ftf flf 
rivatives are represented by the symbols and ^-, thus: 



dx \dx A*-*!) Aa; 

+ A ^> 
Ay 

181 



182 PARTIAL DIFFERENTIATION 

So, in general, if we have a function of any number of variables 
f(x, y, ., 2), we may have a partial derivative with respect to 
each of the variables. These derivatives are expressed by the sym- 

QJ? rvj* A / 

' ' ' ' *' orsometimesb y/*C*> y> - - * 



To compute these derivatives we have to apply the formulas 
for the derivative of a function of one variable, regarding as 
constant all the variables except the one with respect to which 
we differentiate. 

Ex. 1. Consider a perfect gas obeying the law v = We may change 

P 
the temperature while keeping the pressure unchanged. If A* and AJ> are 

corresponding increments of t and v, then 



P P P 

, 8v c 

and = - . 

dt p 

Or we may change the pressure while keeping the temperature un- 
changed If Ap and Ai? are corresponding increments of p and v, then 

A 

and *L = -. 

dp p* 

Ex. 2. /= a; 8 - 3 x*y + y, Ex. 3. /= sin (re 2 + y 2 ), 

?L = 3 a; 2 - 6 xy, %- = 2 a; cos (x* + w 2 ), 

-- v ^ y 



Ex. 4. In differentiating in this way care must be taken to have the 
functions expressed in terms of the independent variables. Let 

x r cos 9 t y = r sin 6 



Then , 

or or 

fa A fy A 

- = r sin 6, ~ = r cos 6, 
c9 dd 

where r and 6 are the independent variables. 



PARTIAL DERIVATIVES 



183 




Moreover, since r 



8r 
Sx 



where x and y are the independent variables. 

fix 



= sin0, 



(2) 



It is to be emphasized that in (1) is not the reciprocal of in (2). 






ftr 



j /ON r i 
and,in(2), = ( 



T . j /-IN 
In fact, in (1), = 
^ ' dr 

and because the variable held constant is differ- 
ent in the two cases, there is no reason that 
one should be the leciprocal of the other. It 
happens in this case that the two are eqiial, but 
this is not a general rule. Graphically (Fig. 80), 
if OP = r is increased by PQ = Ar, while 6 is 
constant, then PR = Aa: is determined Then 

9j, fdx\ PR 

= ( ) 

\drja 




Sr 



T n 

Lim. - - = cos 0. 
PQ, 



Moreover (Fig. 81), if OM = x is increased by 
MN-PQ, = Aa;, while y is constant, then R Q= Ar 

, . -, m, 

is determined Then 



dr /dr\ T . 
= ( ) = Lim 
8x \dx/v 



ST f^y fix 

cos0. It happens here that = But -5? 
- or ox co 

in (1), and , in (2), are neither equal nor 
reciprocal. 




EXERCISES 



n 



Jx 

3-1 , 
t J35 ZS.M 



and ~ in each of the following cases 



6 . , = 

C y 

2ajv 
6. sm - Ji 



9, If * = ln<X - 2xy + 2/ a + 3aj - 

K 2 

10. If = Va; 2 + 2/ 2 e- r , prove a; + y 



prove ^ 

*. 



= 



184 PAETIAL DIFFERENTIATION 

62. Higher partial derivatives. The partial derivatives of 
/(#, y) are themselves functions of x and y which may have 
partial derivatives, called the second partial derivatives of /(a-, y~). 

rrn, d /3A /2A 2 /9A 2 /9A T> 4. -4. vi 

They are r ( )' ( TT- ) TT-( )' ^ ut ^ mav b shown 
17 dx\dxj 8y\dx/ 8x\dy/ dy\dy/ J 

that the order of differentiation with respect to x and y is imma- 
terial when the functions and their derivative fulfill the ordinary 
conditions as to continuity, so that the second partial derivatives 
are three in number, expressed by the symbols 



a/8/\ 3/3A a 2 / 

dx\Zy) dy\dx] dxcy 



Similarly, the third partial derivatives of /(#, /) are four in 
number; namely, 



\ j 3 /gA 

"" " 



So, in general, T; signifies the result of difEerentiating 



y) p times with respect to a?, and g times with respect to 
, the order of differentiating being immaterial. 



In like manner, p 9r signifies the result of differentiating 



/(#, y, g) jp times with respect to #, q times with respect to y, 
and r times with respect to 2, in any order. 



TOTAL DIFFERENTIAL 185 

EXERCISES 

1. If = (x z + 2/ 2 ) tan- 1 ^ find 

x & 

2. If e"sm(x y), find ^-3 

3V 

3. If = 



Verify ^-^] = -^-[^- m each of the following cases : 
17 &e\<ty/ By\faJ 

i x 

4. * = a:?/ 2 + 2 ye- 1 . 6. s = sm- 1 -. 

y 

K g - y + y 7 - * 

O. w I is> , 

te y Vflj 2 4- j/ 2 

, 82 &* 

8. If^ = tan- 



9. If In (a; 2 a 2 /), prove a 2 ji g-g = 0. 

O2 / -r/\ 

10. If V = i cos w<, prove n*r ^^ ^ + w (w + ^)ii = ^- 



63. Total differential of a function of two variables. In 20 
the differential of a function of a single variable, y =/(), is 
defined by the equation ay=f'(x)dx, (1) 

where /* (#) is the derivative of y. 

But /(aO=Lim; (2) 



and hence, according to the definition of a limit ( 1), 

|2=/0) + e, (8) 

where e denotes the difference between the variable ^ and its 

limit f (x) and approaches zero as a limit as A > 0. 
Multiplying (3) by As, we have 

Ay=/(a;)Aa? + Ac. (4) 

But Az = dx and Ay =/( + Aas) -/(), so that (4) may be 
written in the form 

-/(SB) =/ () * + e <n (5) 



186 PARTIAL DIFFERENTIATION 

In the case of a function of two variables, /(, y), if x alone 
is changed, we have, by (5), 

f(x + Az, y) -/(a, y) = ^dfc + e^ (6) 

the theory being the same as in the case of a function of one 

OJJ 

variable, since y is held constant. The term dx may be denoted 
by the symbol d x f. 

Similarly, if x is held constant and y alone is changed, we have 

/(a, y + Ay) -/(*, y) = ^ dy 4- e 2 cZy, (7) 

df 

and dy may be denoted by the symbol d f. 
dy 

Finally, let x and y both change. Then. 
A/=/CB+AZ, y+Ay)-/(a;, y) 

==/(+ Aas, y+ Ay)-/(z+ As, y)4-/(z+ Az,y) -/(,y> (8) 
Then, by (6), 

/(a + As, y) -/O, y) = | <fc + ^db; (9) 

and similarly, by (7), 

/(as + As, y + Ay) -/<> + As, y) = ^dy + e/rfy, (10) 

OJ? * 

where is to be computed for the value (a? + Aa;, y). But if 

of y 

is a continuous function, as we shall assume it is, its value 
oy 

for (x + A#, y) differs from its value for (x, y) by an amount 
which approaches zero as dx approaches zero. Hence we may 
write, from (8), (9), and (10), 



A/= dx + y + ^rfg + e 3 rfy, (11) 

cj? ay? 

where both ^ and ^- are computed for (x, y\ 
dx dy ^ aj 

We now write df = j dx + % dy, (1 2) 

ox dy 



so that A/= 4f+ e^ + e t dy, (13) 

and ?f is called the total differential of the function, the expres- 
sions dfc/ and ^ being called the partial differentials. 



TOTAL DIFFERENTIAL 187 

It is evident, by analogy with the case of a function of a 
single variable, that a partial differential expresses approximately 
the change in the function caused by a change in one of the 
independent variables, and that the total differential expresses 
approximately the change in the function caused by changes in 
both the independent variables. It is evident from the defini- 

tion that , ,. , ,, , ,. ._, A ,. 

v f. (14) 



Ex. The period of a simple pendulum with small oscillations is 



whence g 

Let I = 100 cm with a possible error of ^ mm in measuring, and 
T = 2 sec. with a possible error of ^ $ y sec. in measuring. Then <ll = -^ 



Moreover, dg = - dl =j- dT, 

and we obtain the largest possible error in g by taking dl and dT of oppo- 
site signs, say dl = ^, dT = -j^. 

Then dg = ^ + *r a = 1.05 v* as 10.36. 

The ratio of error is 

<!l = ^ _ 2 ~ = .0005 + .01 - .0105 = 1.05%. 
g I T 

EXERCISES 

1. Calculate the numerical difference between A and dz wlien 
s=s 4 icy a 8 y a , a = 2, ?/ = 3, Aas = cZaj s= 01, aud Ay = dy =.001 

2. An angle < is determined from the formula < = tan" 1 ^ Toy 

x 

measuring the sides x and y of a right triangle. If x and, y are 
found to be 6 ft. and 8 ft. respectively, with a possible error of one 
tenth of an inch, in measuring each, iind approximately the greatest 
possible error in <. 

3. If C is the strength of an electric current due to an electro- 
motive force E along a circuit of resistance R, by Ohm's law 



188 PARTIAL DIFFERENTIATION 

If errors of 1 per cent are made in measuring E and R, find 
approximately the greatest possible percentage of error in com- 
puting C 

4. If F denotes the focal length of a combination of two lenses 
in contact, their thickness being neglected, and / x and / 2 denote the 
respective focal lengths of the lenses, then 

I = l , i 

* A /.' 

If / x and/ 2 are said to be 6 in and 10 in respectively, find approx- 
imately the greatest possible error in the computation of F from the 
above formula if errors of .01 in. in / t and 0.1 in. in f z are made 

5. The eccentricity e of an ellipse of axes 2 a and 2b (a > I) is 
given by the formula . 



The axes of an ellipse are said to be 10 ft. and 6 ft. respectively. 
Find approximately the greatest possible error in the determination 
of e if there are possible eriors of .1 ft. m a and .01 ft. in 5. 

6. The hypotenuse and one side of a right triangle are respectively 
13 in and 5 in. If the hypotenuse is increased by .01 in., and the 
given side is decreased by 01 in , find approximately the change in 
the other side, the triangle being kept a right triangle 

7. The horizontal range R of a bullet having an initial velocity of 
v , fired at an elevation a, is given by the formula 



Ji = 

9 



Find approximately the greatest possible error in the computation 
of R if v = 10,000 ft per second with a possible error of 10 ft per 
second, and a 60 with a possible error of 1' (take g = 32). 

8. The density D of a body is determined by the formula 



* , 

w w' 



where w is the weight of the body in air and w 1 the weight in water. 
If w = 244,000 gr. and w'= 220,400 gr., find approximately the 
largest possible error in D caused by an error of 5 gr. in w and an 
error of 10 gr. in w', 



HATE OF CHANGE 189 

QJ7 

64. Rate of change. The partial derivative - gives the rate 

ox 

of change of / with respect to x when x alone varies, and the 
partial derivative -j- gives the rate of change of / with respect 

to y when y alone varies. It is sometimes desirable to find the 
rate of change of/ with respect to some other variable, t. Ob- 
viously, if this rate is to have any meaning, x and y must be 
functions of t, thus making/ also a function of t. Now, by 11, 

the rate of change of / with respect to t is the derivative . 

at 

To obtain this derivative we have simply to divide df, as given 
by (12), 63, by dt, obtaining m this way 



dt 8x dt By dt 

The same result may be obtained by dividing A/, as given by 
(11), 63, by At and taking the limit as At approaches zero as 
a limit. 

Ex. 1 If the radius of a right circular cylinder is increasing at the rate 
of 2 in per second, and the altitude is increasing at the late of 3 in per 
second, how fast is the volume increasing when the altitude is 15 in. and the 
radius 5 in ? 

Let V be the volume, r the radius, and h the altitude. Then 



_ , 

y W ' dt ~ dr dt dh dt 

dr . nd7i 
+ 7ir a -~. 
dt dt 

By hypothesis, ~ = 2, ~ = 3, = 5, 7t = 15. Therefore ~ = 375 IT cu. in. 

T dt dt dt 

per second. 

The same result may be obtained without partial differentiation by ex 
pressing V directly in terms of t. Foi, by hypothesis, r = 5 + 2 1, 7i =15 + 3 1 
if we choose t = when r = 5 and 7i = 15. Therefore 

7= (8 75 + 375 t + 120 i a +12 * 8 )7r; 
whence ~ = (375 + 240 1 + 36 2 ) *-. 

When t 0, *- = 375 TT cu. in. per second, as before. 

dt 



190 PARTIAL DIFFERENTIATION 

Ex. 2. The temperature of a point in a plane is given by the formula 

1 



The rate of change of the temperature in a direction parallel to OX is, 
accordingly, &*__ 2* 

to ~"~ (a: 2 + jr 2 ) 2 ' 

which gives the limit of the change in the temperature compared with a 
change in x when x alone varies 

Similarly, the rate of change of u in a direction parallel to OF is 
Jw _ 2 ?/ 



Suppose now we wish to find the rate of change of the tempei ature in a 
direction which makes an angle a with OX From Pig. 82, if P^(x v y^) is 
a fixed point, and P (x, y) a moving point _ 

on the line through P x making an angle a 
with OX, and s is the distance P^P, we have p 

P! R = x x 1 




jo 
whence x = x t + s cos a, " 

y = y^ + s sin a, ^. 

, dx dy 

and - = cos a, ^ = sma. PIGI 82 

Replacing t by s in formula (1), and substituting the values of -- and 

dii d s 

~ which we lust found, we have 

as 

du du du 

= cos + sin a 

as dx dy 

_ _ 2 x cos ex + 2 y sin ex 



Formula (1) has been written on the hypothesis that x and y 
are functions of t only. If x and y are functions of two vari- 
ables, t and s, and (1) is derived on the assumption that t 
alone varies, we have simply to use the notation of 61 to write 
at once 



a 

which may also be written as 



dt dx dt %y dt 



GENERAL EXERCISES 191 

EXERCISES 

1. If 2 = e tan x } x = sin tf, y = cos t, find the rate of change of 
with respect to t 

1 -f- x 

2. If = tan" 1 -: ~ ) x = sin t, y cos t, find the rate of change 

of with respect to t when t 

jL 

3. If V= (a * 6""*) cos ay, prove that V and its deiivatives m 
any direction are all equal to zero at the point ( 0, ) 

4. If F= ) find the rate of change of V at the point 

Va; 2 + y z 
(1, 1) in a direction making an angle of 45 with OX. 

5. If the electric potential F-at a,ny point of a plane is given by 
the formula V= In Vcc a + y 2 , find the rate of change of potential at 
any point (1) in a direction toward the origin ; (2) in a direction at 
right angles to the direction toward the origin 

B. If the electric potential F at any point of the plane is given 

-\/(g> _ ff*) 2 + W 2 

by the formula F= In , * ? find the rate of change of 

V(o;H-a) 2 -|-y 2 b 

potential at the point (0, a) in the direction of the axis of y, and at 
the point (a, a) in the direction toward the point ( a, 0). 

GENERAL EXERCISES 

, TJ! xy 1 80 fa A 

1. If = sin , . > prove x -z -- y 5- = 0. 
xy + 1 i ox ^ oy 

"i /?/ /i 

2 ' If = S 



3. If at = f + ye* prove oj 2 -f y 

r\2 O2 

4. If # = e-y cos a (7c cc), prove that jr-j 4- ^ = 0. 



B. If 3 = e-c-wx s i n / CCj p rov e that f t = <*.- 1 

6. If * = e-**sin (TO?/ + x Vft 2 , 2 - 7c 2 ). prove that 
v ' x * 

2 P 

a W' 

rtrr fc /IN *T. * ^^ , ! Sfr , 1 A 

7. If F= <?**cos (a In r), prove that -5-5- 4- -~ H- 1 -5-75 == 0. 

N ' or r 0>' ir 0* 



192 PAETIAL DIFFEKENTIATION 

8. A right circular cylinder has an altitude 8 ft. and a radius 
6 ft. Find approximately the change in the volume caused by de- 
creasing the altitude by .1 ft. and the radius by .01 ft. 

9. The velocity v, with which vibrations travel along a flexible 
string, is given by the formula 



where t is the tension of the string and m the mass of a unit length. 
of it Find approximately the greatest possible error in the compu- 
tation of v if t is found to be 6,000,000 dynes and m to bo .005 gr. 
per centimeter, the measurement of t being subject to a possible 
error of 1000 dynes and that of m to a possible error of .0005 gr. 

10. The base AB of a triangle is 12 in. long, the side AC is 10 in., 
and the angle A is 60. Calculate the change in the area caused 
by increasing A C by 01 in and the angle A by 1. Calculate also 
the differential of area corresponding to the same increments. 

11. The distance between two points A and B on opposite sides 
of a pond is determined by taking a third point C and measuring 
AC = 90 ft , BC = 110 ft , and BCA = 60. Find approximately the 
greatest possible error in the computed length of AB caused by 
possible errors of 4 in. in the measurement of both AC and BC. 

12. The distance of an inaccessible object A from a point B is 
found by measuring a base line B C 100 ft , the angle CBA =s<x= 45, 
and the angle BCA = 0= 60. Find the greatest possible error in 
the computed length of AB caused by errors of 1' in measuring both 
a and 

13. The equal sides of an isosceles triangle are increasing at the 
uniform rate of .01 in. per second, and the vertical angle is increas- 
ing at the uniform rate of .01 radians per second. How fast is the 
area of the triangle increasing when the equal sides are each 2 ft. 
long and the angle at the vertex is 45 ? 

14. Prove that the rate of change of * = In (a; -J-Vjc a + f) in the 
direction of the line drawn from the origin of coordinates to any 
point P(x, y) is equal to the reciprocal of the length of OP. 

15. The altitude of a right circular cone increases at the uniform 
rate of .1 in per second, and its radius increases at the uniform rate 
of .01 in. per second How fast is the lateral surface of the cone 
increasing when its altitude is 2 ft. and its radius 1 ft.? 



GENERAL EXERCISES 193 

*t - ffi ~\ JL* > 

16. Given = tan" 1 ; 1- tan" 1 Find the general expres- 
sion for the derivative of along the line drawn from the origin of 
coordinates to any point. Find also the value of this derivative at 
the point (1, 1). 

17. In what direction from the point (3, 4) is the rate of change 
of the function tt kzy a maximum, and what is the value of that 
maximum rate ? 

18. Find a general expression for the rate of change of the func- 
tion u s= &-v sin x + - e-*v sin 3 cc at the point ( -77 ) ). Find also the 

3 \ 3 / 

maximum value of the rate of change. 



CHAPTER IX 
INTEGRATION 

65. Introduction. In 18 and 23 the process of integration 
was defined as the determination of a function when its deriva- 
tive or its differential is known. We denoted the process of 

integration by the symbol /; that is, if 

then Cf(x) dx = F(x) + C, 

where C is the constant of integration (18). 

The expression f(x) dx is said to be under the sign of inte- 
gration, and/(a;) is called the integrand. The expression J<\J'} + (' 
is called the indefinite integral to distinguish it from the definite 
integral defined in 23. 

Since integration appears as the converse of differentiation, 
it is evident that some formulas of integration may bo found 
by direct reversal of the corresponding formulas of differentia- 
tion, possibly with some modifications, and that the correctness 
of any formula may be verified by differentiation. 

In all the formulas which will be derived, the constant C will 
be omitted, since it is independent of the form of the integrand; 
but it must be added in all the indefinite integrals found by 
means of the formulas. However, if the indefinite integral is 
found in the course of the evaluation of a definite integral, the 
constant may be omitted, as it will simply cancel out if it has 
previously been written in ( 23). 

The two formulas / * 

I cdu = c I du (1) 

ft 

and 

i >, - - .^ 

/ 

194 



](du + dv + dw-\ ) = Cdu + Cdv + Cdw + . (2) 



INTEGEAL OF w* 195 

are of fundamental importance. Stated in words they are as 
follows : 

(1) A constant factor may be changed from one side of the sign 
of integration to the other. 

(2) The integral of tJie sum of a finite number of functions is 
the sum of the integrals of the separate functions. 

To prove (1), we note that since c du d(cu), it follows that 



I cdu = I d (cu) cu = c I du. 



In like manner, to prove (2), since 

du -H dv + dw + . . . = d(u + v + w +...), 
we have 

/ (du + dv + dw H- ) = I d (u + v + + ) 

u + v + w+ 

I du -H I dv + I dw + . . . 

The application of these formulas is illustrated in the follow- 
ing articles. 
66. Integral of u". Since for all values of m except m = 



/w m \ 

or d )=w m ~ 

\m/ 

it follows that Cu m ~ l du = . 
J m 

Placing m == n +1, we have 



/or aZZ values of n except n s= 1. 

In the case %= 1, the expression under the sign of inte- 

gration in (1) becomes , which is recognized as 



Therefore 



T = lnw. (2) 



196 INTEGRATION 

In applying these formulas the problem is to choose for u 
some function of x which will bring the given integral, if pos- 
sible, under one of the formulas. The form of the integrand 
suggests the function of x which should be chosen for u. 

Ex. 1. Find the value of JY aa; 2 + Ix + - + J dx 
Applying (2), 65, and then (1), 05, we have 



fft 

J \ 



= afx z (lx + bj'xflx + cf ( ~- + k JV 



The fiist, the second, and the fourth of these integrals may be evaluak'tl 
by formula (1) and the thud by foirnula (2), wheie u = x, the icsults beiny 

11 & 

respectively - ax 9 , - bx*, -- , and c In a.. 



- , 
o & 



Therefore C lax* + lx + - + -^ } dx = ~ ax* + J lx z + c In i - - + C 
J \ x a,*/ 3 2 x 

Ex. 2. Find the value of J*(z 2 + 2)ar<fo. 

If the factors of the integrand are multiplied together, we have 
f (a; 2 + 2) xdx = f(x s + 2 a;) dx, 

which may be evaluated by the same method as that used in Ex 1, the 
result being z 4 + x z + C. 

Or we may let x 2 + 2 = , whence 2xdx = du, so that xdx = % du. Hence 



22 



Comparing the two values of the integral found by the two methods of 
integration, we see that they differ only by the constant unity, which may 
be made a part of the constant of integration. 

Ex. 3. Find the value of C(ox z + 2 bz) a (ox + &) dx. 

Let GKC 2 + 2 bx = u Then (2 ax + 2 fydx = du, so that (ox + &) dx = J rfw. 

Hence J" (oa; 2 + 2 fcc) 8 (aa: + l)dx = C %u*du 

1 / 7 1 u 4 , 

=s / M B du = . + C 
2J 24 



INTEGRAL OF M" 197 

Ex. 4. Find the value of C 4 < + *] dx 
J ax 2 + 2 fa; 

As in Ex. 3, let ax z + 2bx = u. Then (2 ax + 2 b) dx =* du, so that 



Ilence 



= 2 In w + C 

s= 2 In (az 2 + 2 to) + 



Ex. 5. Find the value of C (e + 

Let e * + & = u. Then e ax adx = tfu. 

Hence J(e + 6)cdiB J^ jf 



If the integrand is a trigonometric expression it is often pos- 
sible to carry out the integration by either formula (1) or (2). 
This may happen when the integrand can be expressed m terms 
of one of the elementary trigonometric functions, the whole 
expression being multiplied by the diffeiential of that function. 
For instance, the expression to be integrated may consist of a 
function of smo; multiplied by cosEefe, or a function of cos a; 
multiplied by ( sma^a;), etc. 

Ex. 6. Find the value of [ Vsm x coa s xdx. 

Since d(smx) = oosxrfx, we will separate out the factor cosrcrfar and 
express the rest of the integrand in terms of sin a:. 

Thus -Vsinxcos s xdx = Vsirue (1 sm s ;c) (cosxdx). 

Now place sin x = M, and we have 




198 INTEGRATION 

Ex. 7. Find the value of Jsec 6 2xdz. 

Since d(tan 2 a:) = 2 sec 2 2 xdx, we separate out the factor sec 8 2 rdc and 
expiess the rest of the integrand in terra a of tan 2 a: 

Thus 



sec 2 x dx = sec* 2 x (sec 3 2 

= (1 + tan 8 2 a:) 2 (sec 3 2 a:r/ 

= (1 + 2 tan 2 2 ar +*tan*2 a;) (se 



Now place tan 2z=su, and we have 
fsec 6 2 xdx = 



= Jtan2a: 

EXERCISES 
Find the values of the following integrals 

*6 M 4- sec 3 ace 
tan ax 



+ C. 



. C 



-j=\dx. 

C( r 1 \^ 
J [x-vx j=]dx. 

J \ x~vx/ 

r?**- 

C tffa 
"J^T 

i. C(x* + l)*xdx. 
I Vo 4 + 4x 8 t7a;. 

/. 

c 

J 2 a; -f sin 2 a;' 

in f 1-cosa; 
!. I . . .,dx. 
J (aj sinaj) 4 



-/: 



12. 
13 



dx. 



&r. 



sin ax 



rfa;. 



e 8 *^ 
e* x + & 



_ , 1 + COS 2 G , 

9. I - - - ote. 



fif 

J 1 + cos aa 
14. I cos 8 2 as sin 2 OJ$B. 

16. I sin 3 3 a; cos 3 xdx. 

16. / sin (x + 2) cos (a -|- 2) Ja;. 

17. / cos^3 x sin 3 *c?a;. 

18. I sec 4 3ic^a5. 
J 

19. rctn 2 (2a;-|-l)cso !1 (2aj 

"*/' 



ALG-EBKAIC INTEGRANDS 199 

67. Other algebraic integrands. From the formulas for the 
differentiation of sm~ 1 M, tan" 1 ^, and sec-'w, we derive, by re- 
versal, the corresponding formulas of integration : 

du 



== sin" 1 ?/, 



du 
and 



These formulas are much more serviceable, however, if u is 

/ 

replaced by - (a > 0). Making this substitution and evident 

fit 

reductions, we have as oar required formulas 

du . ,u .^ N 

== sin" 1 - (1) 

<J <J /y NX 

a u a 



L 



_, - ON 

= -tan- 1 -> (2) 

2 s y 



and 



+ a 2 a a 

du 



/du 1 *u s Q ~ 

^-sec" 1 -- (3) 

wV w a - a* a- 

Referring to 1, 47, we see that sin" 1 - must be taken in the 

Uf 

first or the fourth quadrant; if, however, it is necessary to 



u 



have sin" 1 - in the second or the third quadrant, the minus sign 
must be prefixed. In like manner, in (3), sec" 1 - must be taken in 



the first or the third quadrant or else its sign must be changed. 

/fix 
__ Letting 2 
Vj) 4 a: 2 

r fa _. f \& u 

J -y/9 _ 358 J V9 ~ M 3 



/fix 
__ Letting 2 x = u, we have du = 2 dx ; 

whence dx = ^ du, and Vj) 4 a: 2 




200 INTEGRATION 

dx 



Ex.2 Find the value of f , * Jf we let V:JW~ M, UH-J 
J jt V;} j, a - 4 



t 

du = V3 rfj; ; whence <ir = -= f/, and 

Va 

/^j _ r __ <lu 
x V3 x 2 -4 ^ w vV~-~4 



/ rfr 

Ex. 3. Find the value of I /. 
17 



Since V4 a; js z = Vd (r 2) 2 , we may lot w =s a 2 ; whttiusH ^/.r - */, and 

C ^ C _ ~ t 

J - 2 "^ ^* 



S1J1, 



-i + C' 

6 

x -~ 2 



Ex. 4. Find the value of C <lx 

J 2 .r + 3 u, + 5 

We may first write the integrand in the form 



1 1 = 1 1 

2 aJ+x+ 2'* 



and let u = x + I Then du = dj., 

J 2 2 + 3 x + 5 = <>J (.r + ^) B +"Y 



= ~f 

2J i 



1 X s ton-'JL 



4 4 

_ 2 t 4tt 

-~r~ au v" f 5f +r 



- 

Vai 



/ 



ALGEBBAIC INTEGRANDS 201 

5z-2 , 



Separating the integrand into two fractions 

5z _ 2 
2o; a 4-3 2a; 2 +3' 
and using (2), 65, we have 

2 / 5xdx r 2dx 



__ / 
J 



If we let rz = 2 a; 2 + 3, then du = 4 artfo 

5a;c?a; 5 /vM 5, 5 



and if we let u V2 x, then rfu = V2 dx 

, r 2dx AT / du /= 1 . . it Vo. ,a;Vo 

and I i4 = V2 I - = V2 - tan- 1 ^ = tan- 1 - 
J2a; 2 +3 Ju*+$ V3 3 3 



Ex. 6. Find the value of 



There is here a certain ambiguity, since tan- 1 V and tan~ 1 ( 1) have 
each an infinite number of values. If, however, we remember that the graph 
of tan~ 1 a; is composed of an infinite number of distinct parts, or Iranches 
(Fig 56, 46), the ambiguity is removed by taking the values of tan- 1 Vs 
and tan- 1 ( 1) from the same branch of the graph For if we consider 

- 1 ^ tan- x o and select any value of tan~ 1 a, then if & = a, 



/ 

| 

Ja 



a X + 1 

tan~ l 6 must be taken equal to tan- 1 ^, since the value of the integral is 
then zero. As & varies from equality with a to its final value, tan- x & will 
vaiy from tan -1 a to the nearest value of tau- a &. 

The simplest way to choose the proper values of tan- 1 ?* and tan~ 1 is 

to take them both between and Then we have 



/'Va dx _ v __ / TT\ _ 7 
./-i a 2 + 1 ~ 3 \ i/~ 1 



TT 
12* 

The same ambiguity occurs in the determination of a definite integi al 
by (1), but the simplest way to obviate it is to take both values of sin- 1 

7T 7T 

between -^ and The proof is left to the student. 

B 



202 



EXERCISES 
Find the values of the following integrals : 

r dx r dx 

l ' J Vl6 - 9 a? ' J 



V5 x - 3 a/ 
dx dx 



/dx C B 
a - 12. /, - 

/dx C (?SK 

13 I 
a:V4a; 2 -9 ' J 3-e 3 - 4 a; + 2* 

ra.r + 11 

x 1^ 



16 ' 

7 

' 
a 



^ 
J 



10 
' 



r / * 19 r 

J VGas-a* 19 ' J_ 

T ^ o 
J V6*-4a* 



68. Closely resembling formulas (1) and (2) of the last section 
m the form of the integrand are the following formulas ; 



+o. (1) 

/du> -. / / 
^r^^ (u+ ^ u --^ (2) 

and 



These formulas can be easily verified by differentiation and 
this verification should be made by the student ' 



ALGEBRAIC INTEGRANDS 203 

/f/1" 
. 

Letting V2 a; = u, we have du = "v2dx ; whence dx = du, and 

,;= du 



= _JL T rfu 

= -i- In [M + Vu*-8] + C 

Va 

= 4= ln C' ^2 + Va x a - 3] + a 



rfe 



/(/ 
~ ~ 
Vy a B 

As in Ex. 4, 07, we may write tho integrand in the form 

Va Vx a -f- 3 a? Va V(j; + ^) a ~ $ 
and lot M = x + S ! whence rfw = rfx, 

/__.4!L _JL T - ^- ; 
V3 a: a + 1 x VJJ ^ V(,^ + *j) a - J 

V8^ Vu a ^ 

" vl n w 
"vl 



~ -Lin (3 + 2 + VO a; 2 + 12 *) +K> 
Vi) 



where C = -- In 8 + A". 

Va 



/ 



o i IK 

z a;" 8 + * JLO 



Writing the integrand in the form 

I ^j 1 ,..^ "S 

we let * x Hh 1 ; whence rf ss rfa?. 



204 INTEGRATION 

r dx _ 1 f* <lf 

J 2s* + a: -15 ~ 2 J (* + i) a W" 



2 2(-V) + V 

llnljli + C 
11 a + 3 



It I >1 ' 

11 X + i) 

where C = -j^ In 2 + K. 

EXERCISES 

Find the values of the following integrals : 



1 r *" 11 r~ /* ~ 
j.i - . * J.J.. i , - 

J V^T2 J 3^ + 6 

2 c dx 12 r ^ 

J V9rf-l' Jo*-3(B + 

3 r *" 13 r rfa! 

' J V3 2 -4 ' J *+- 

* C dx C <fa 

' J Va^a" ' J 4if-2* - 

5. f ^ IB f* rf f- 

' J V3jc a +2a3 + 3 ' J V5-""4' 

6 C dx r* dx 

J***-M ' J VOSTT' 

r dx 

' I oa -i' 1 

^ 2a ^- 1 



'-a- 8' 



TRIGONOMETRIC FUNCTIONS 205 

69. Integrals of trigonometric functions. Of the following for- 
mulas for the integration of the trigonometric functions, each 
of the first six is the direct converse of the corresponding for- 
mula of differentiation ( 44), and the last four can readily be 
verified by differentiation, which is left to the student. 

/ sin udu cos M, (1) 

I cos udu = sin w, (2) 

I seu"ud (, = tan u, (8) 

/Qscfudu = ctn w, (4) 

v J 

sec u tan u du = sec it, (5) 

esc u ctn udu = esc u, (6) 

I tan udu = hi sec u, (7) 

I ctn udu = In sin it, (8} 

| sec udu = In (sec u -h tan it), (9^ 

I esc udu = In (esc it ctn u) . 

Ex. 1. Find the value of Tsui 7a.v7a;. 

If we let 11=7,1:, 

then du7dr; 

whence dx = \ du, 

and Cs\n7x dx = T sin (| f/u) 

=3 ^ I sin udu 

= \ cos + C* 
= J-cos7a? + C, 



206 INTEGRATION 

Ex. 2. Find the value of f sec (2 x + 1) tan (2 x + 1) dx 
If we let w = 2 a; -f 1, then du = 2 dx, 

and Tsec (2 x + 1) tan (2 x + 1) cfo = f sec M tan du 

= ^secu + C 

= sec (2 a; + 1) + (7. 

Often a trigonometric transformation of the integrand facili- 
tates the carrying out of the integration, as shown in the 
following examples: 

Ex. 3. Find the value of f coa z axdx 

Since cos s ax - (1 + cos 2 ax), 

Ccos*ax dx = f ( J + J cos 2 aar) dx 

= Tdk + ^ Tcos 2 aacfo 
J i / 

= o x + T~ sin 2 aa: + C 1 , 

A T CZ 

the second integral being evaluated by formula (2) with M = 2 ax. 
Ex. 4. Find the value of f Vl + cos arcfo. 

Since cos x = 2 cos 2 - 1, 

ii 

Vl + cosa;=v / 2cos|, 

- ib 

and f Vl + cos xdx ~ C V^ cos - </j; 



Ex. 5. Find the value of Aan 2 3 xdx, 
Since tan 2 3a; sec a 3 - 1, 

Jtan 2 3 arcfa: = J*(sec 2 3 a: - 1) da: 

= Csec*3xdx Cdx 

= J-tan3a; a;, 
bhe first integral being evaluated by formula (3) with v*= 



TRIGONOMETRIC FUNCTIONS 207 

EXERCISES 

Find the values of the following integrals : 



,, 1. /sin (3 a; 2) die. 


13. I cos^dte. 




t 2. | cos (4 2x)dx. 


/y 3. jpxa 
14. I (sin^ + cos-j dx. 


3. 1 sec(3a: l)tan(3a; l)da; 


-/(-- 


* 2aA s 
tan T j 


i 4. / sec a Tdo;. 
J 4 


16. fain 2 ! cos a | 


dx. 


* f+ 3a; j 
^6. 1 tan -%- dx. 


17./^ 1+ eos = 


5x 


6. I ctno'ajcfa;. 


18. 1 Vl COS ' 




7 .Jcse ( 2* + s)<fc. 


c* . 

19. I sin oxdx. 
Jo 




S-Jcscfctnl^ 


C\ ai 
20. I tan 75 aa;. 

Jo 2 




9. I sec(4a; + 2)<c. 


Tw / 
21. J tan ( cc "|~ 

v/O ^ 


.=)*. 



22. 

a;. 



v'10. Jose 9 (3 2 a:) da 

rr 

,, /*cos2o; , 23. I ( 

.yll. I : dx. ] ir 

* J smaj J~i 

r . * r*i 

12. I sin a 7rdaj. 24. J 

J 2 Jis 

70. Integrals of exponential functions. The formulas 

/ It J M XI \ 

and | a v du*=-. a* (2) 

J In a 

are derived immediately from the corresponding formulas of 
differentiation. 



208 INTEGRATION 

Ex. 1. Find the value of C$ x djc 

If we let 3 x = u, we have 

Ce 8x dx 



/*/B 
JLrrfa;. 
x z 

x i i 

If we place V 5 = 5"" and let = u, we have 

. _/*. 



EXERCISES 

!Fmd the values of the f ollowing integrals : 

1. ie^^dx. 6. f(a + -)<&. 9. f 10* dx. 



//g2x _ a-2as / 

e^ajtfo;. 6. / _ a , rf. 10. / 2 00 

11. C\**-*dx. 

. f V ~ *"'rfaj. 

Jo ^+0"* 



8. 

4. e^f+txfa Qt +e"dx. 12 



71. Substitutions. In all the integrations that have been 
made m the previous sections -we have substituted a new vari- 
able M for some function of x, thereby making the given integral 
identical with one of the formulas. There are other cases in 
which the choice of the new variable u is not so evident, but 
in which, nevertheless, it is possible to reduce the given integral 
to one of the known integrals by an appropriate choice and wh> 
stitution of a new variable. We shall suggest in this section a few 
of the more common substitutions which it is desirable to try, 

I. Integrand involving powers of a + U. The substitution of 
some power of 2 for a -f- bx is usually desirable. 



SUBSTITUTIONS 209 

Ex. 1. Find the Value of f ^^ , . 
J (1+2*)* 

Here we let 1 + 2 x = s? ; then x = ^ ( 8 - 1) and dx = % z*dz. 

Therefore C x * dx = - f(z< - 2 * + z) tls 

17 (1+2 a?)* 8J 



Eeplacing a by its value (1 + 2 #)* and simplifying, we have 

= (1 + 2 B) * ( " 12 * + 2 ^ + c ' 



II. Integrand involving powers of a + ?;#". The substitution 
of some power of for a + frr 71 is desirable if the expression 
under the integral sign contains x n ~ l d& as a factor, since 
= bnx n ~ 1 cfo. 



/x/j.2 J. x 2 
rfa;. 
a; 

We may write the integral in the form 



and place a a + o 2 = a . Then xdx = sdz, and the integral becomes 



Replacing s by its value in terms of x, we have 

x 2 Va a + a 2 + a 

Ex. 3. Find the value of fjs 6 (l + 2 spfidx. 
We may write the integral in the form 



and place 1 + 2 a; 8 = 2 . Thou xhlx = ifisdz, and tho now integral in z ia 
fr J(*< - * a ) ds = n^ s (3 * 2 ~ 5) + C. 

Replacing z by its value, we have 

Cx 6 (1 + 2 a)icfa ~ ^ (1 + 2 3 8 )8 (3 a 8 - 1) + C. 



X 



210 INTEGRATION 

III. Integrand involving Va 2 ar 2 . If a right triangle is 
constructed with one leg equal to x and 
with the hypotenuse equal to a (Fig. 83), 
the substitution x = a sin is suggested. 

Ex. 4. Find the value of fVa 8 a: 2 da;. _, gg 

Let a; = a sin <. Then dx = a cos <f> d<f> and, from the triangle, V 2 a; a 
= a cos <. 

Therefore C^/a?x*dx = a 2 f cos 2 





But 
and 



for, from the triangle, sin d> = - and cos <& = 

a a 

Finally, by substitution, we have 

J 2 \ a/ 

IV. Integrand involving V^+a 3 . If a right triangle is 
constructed with the two legs equal to x 
and a respectively (Fig. 84), the substitu- 
tion x = a tan <f> is suggested. 

Ex. 5. Find the value of f dx 

i/ /* o * ^o\ $ 



Let i = a tan < Then dx = a sec 8 <tf< and, from the triangle, V# a + a a 
= a sec <f>. 




Therefore f 

But, from the triangle, sin <f> = * ; so that, by substitution, 

Va; 8 + flr a 



* 



SUBSTITUTIONS 211 

V. Integrand involving V# a a 2 . If a right triangle is 
constructed with the hypotenuse equal to 
x and with one leg equal to a (Fig. 85), 
the substitution x = a sec <#> is suggested. 

-*f r 

r a 

Ex. 6. Find the value of / a 8 Vo: 2 a?dx " ^^ ge 




Let as = a sec <. Then dx = a sec < tan < d< and, from the triangle, 
Va; 2 - a 2 = a tan 



Therefore fa 8 Vj; 2 - a 2 dx = a 

= a G J"(tan 2 </> + tan 4 <) sec 2 



.J. _ j 

Exit, from the triangle, tan <j> = , so that, by substitution, we have 






J> V* 2 - a a rf.c = T V (2 a a + 3 a; 2 ) V(a; 2 - a 2 ) 8 + C 

We might have written this integral in the form fa; 2 Va; fl cPfxdx) and 
solved by letting z a = a; 2 a 2 . 

72. If the value of the indefinite integral is found by substitu- 

X6 
f(x)dx may be 

performed in two ways, differing in the manner in which the 
limits are substituted. These two ways are shown in the solutions 
of the following example: 

Ex. Find f Va"--.ida;. 

Jo 

By Ex. 4, 71, 

Va x*dx = i/a Va - x* + 

Therefore C " 




212 INTEGRATION 

Or we may proceed as follows Let x = a sin < When x = 0, = ; 
and when x = a, < = , so that < varies from to - as x vanes fiom to a. 
Accordingly, " v 




4 

The second method is evidently the better method, as it obviates the 
necessity of replacing z in the indefinite integral by its value in tenna of 
x before the limits of integration can be substituted. 

EXERCISES 

Find the values of the following integrals : 

1. \ X . * 6. I ~~ . 11 




'/(*-* 

7. r **< 

J (3- 

/, 

\J & 

9. f * 

J (V 5 - 



5 - I . . , 10. I aV2a; $dx. 15 




73. Integration "by parts. Another method of importance in 
the reduction of a given integral to a known type is that of 
integration by parts, the formula for which is derived from the 
formula for the differential of a product, 

d (uv) = udv + v du. 
From this formula we derive 

uv = / udv + f vdUy 
which is usually written in the form 

I udv = uv I vdu. 



INTEGEATION BY PAETS 213 

In the use of this formula the aim is evidently to make the 
original integration depend upon the evaluation of a simpler 
integral. 

Ex. 1. Find the value of Cxe*dx, 

If we let x = u and cPdx = dv, we have du = dx and u = e*. 
Substituting in our formula, we have 

j x&dx = are 1 j e x dx 



It is evident that in selecting the expression for dv it is desirable, if 
possible, to choose an expression that is easily integrated. 



Ex. 2. Find the value of Csm 



Here we may let sin- 1 x = u and dx = do, -whence du = rx and v = x. 
Substituting in our formula, we have "^ 1 ~~ x * 

xdx 



sin-i-xdx = x sin -1 a: { 

J Vl-o; 2 



= x sin- 1 ^ + l a; 2 + C, 
the last integral being evaluated by (1), 66 

Sometimes an integral may be evaluated by successive inte- 
gration by parts. 

Ex. 3. Find the value of Cx*e*dx. 

Here we let x z = u and eFdx = do. Then du ^xdx and = e 31 . 

Therefore C 



The integral f xeF dx may be evaluated by integration by parts (see Ex. 1), 
so that finally 

Ja;V= dx = x*<*> - 2 (ar - 1) e* + C = eP (a; 2 - 2 x + 2) + (7. 

Ex. 4. Find the value of je" sin&zda:. 

" sin bx = u and' c** rfz = dv, we liave 

= - *<"* sin &K fc 8 * cos Ba; ?. 
a v 



214 INTEGRATION 

In the integral Ce 01 cos bxdx we let cos bx = u and e ax dx dv, and havi 

/I b /* 

e"* cos bxdx - e"* cos fa + - / e** sin Sards. 
a a w 

Substituting this value above, we have 

/e"* sin bxdx = - e 01 sin &e -- ( - eP* cos &r + - fc * sin bx dx\ 
a a\a aJ ) 

Now bringing to the left-hand member of the equation all the terms 
containing the integral, we have 

(1 + ) fe * sin bxdx = - e"* sin bx -- -e 031 cos bx, 
\ a?/ J a a? 

t. r __ , , fi** (a sin bx b cos bx) 

whence I e"* sin bxdx = v ' 

J a z +b* 

Ex. 5. Find the value of fVa; 2 + a?dx. 



g 



Placing Va; 3 + c 2 = w and <?a; = fo, whence rf = 
we have Va; B + o a 



and w = a, 



Since a; 2 = (a^ + a 2 ) a 2 , the second integral of (1) may be written as 



/a; 2 + a 2 



which equals 



z 2 + 



a s /* 



Evaluating this last integral and substituting in (1), we hare 
whence JVa; 2 + a?dx =s $ [a Var 2 + a a + a a In (x + 

74. If the value of the indefinite integral f/O) ** is found 
by integration by parts, the value of the definite integral 

r* 

I f(z)dv may be found by substituting the limits a, and I in 

va 

the usual manner, in the indefinite integral. 



INTEGRATION BY PARTS 215 

IT 

Ex. Find the value of C*x*sinxdx. 
Jo 

To find the value of the indefinite integral, let a; 2 = and sin a; da; = dv. 
Then J a; a sm xdx = a? 2 cos x + 2 J x coaxdx. 

In / a; cos a, dx, lot a; = u and cos xdx = do. 

Then \x cos x dx x sin x i sin x dx 

= x sin a; + cos x. 
Finally, \ve have 

I x*ainxdx = ar a cosa, + 2 a? sin a: + 2 cos a; + C. 

IT 

"" P Ha 

Hence C*j? sinarda; a; 2 cos x + 2 x sin x + 2 cos a; 

= -2. 

The better method, however, is as follows: . 

l$.f(x)dx is denoted by udv, the definite integral \ f(x)dx 

stb Jo. 

may be denoted by I udv, where it is understood that a and b 

t/a 

are the values of the independent variable. Then 

/& . s*b 

I udv=z\uv~\ n I vdu. 

Jo, v a 

To prove this, note that ib follows at once from the equation 

, /^6 />fc /*! /& 

** / / / / 

i/a yo /o vo 

Applying this method to the problem just solved, we have 

IT 

v r -13 ir 

r'5'a; a sina?afa;= a; 2 cosa; + 2 C^xciOBxdx 

/* * j 

Jo 

["|J W 

2a;sina: 2 C*&inxdx 

w 

STT-f 2 cos a: I 
L Jo 



216 INTEGRATION 

EXERCISES 

Find the values of tho following intftfra 



. CynP'tlr. 5. I .rscc" 1 !}^//^. 9. I ./ *-V.r. 

. I stP<i* L (?x- 6. I (In Hindoos ;'(/.r. 10, / ,i"'lu.i'f/>. 

. COOB- I X<?JK. 7. I/' 9 "Von .r /.i'. il. / Mr "J, !/,!. 

| tair~ 1 3a:</J3 8. / ,r uos 3 '^ <Ar. 18, / VPOH JJuv/.i". 



4. 



75. Integration of rational fractions, A rtttwixtt fwrfiHit i.s it 
fraction whose numerator and denominator ur polynomials. Il 
can often be integrated by expressing it UH tho twin of 
fractions whoso denominators are faot-oi'H of th 
of the original fraction. Wo .shall illustrate only tin* inw in 
which the degree of the numerator is lews thun tho th'gm* of tin* 
denoiuinator and in which the ftu'torH of tho tlmitmunutur m* 
all of the first degree and all different, 

Ex. Find tho value of f-~^ f ^}'\" </'' 

The factors of tho denominator arc x + tt, x 2, a<l x + 2, Wo ftHwtmo 



(x + 8)(j- 4) ""a + a-B^ + a 

whojo J, /?, and ^7aro constants to bo (lotonniucd. 

Clearing (1) of fmclions by uiuUiplyuig by (ja + n)(x 9 4), w 



or 



l, /J, and (7 firo to bo datidrmmnd so thai Iho righWrnml memhr 
of (3) .shall bo identical with th k'ft-hand nuwnbcr, lhn <'<*iHj5iHtu( wf 
powers of a: on the two sides of tho equation muni Iw tujual. 

Therefore, equating the ooeffieimitH of liku powors of / in (Jl), 
theequatzons 



- 1 A + <t 7i ~ G ~ li, 
whence we find ^t = ~ 2, Ji 2, r, Y = 1, 



RATIONAL FRACTIONS 217 

Substituting these values in (1), we have 

a? 8 + 11 x +14 _ 2 2 1 



= - 2 In (x + 3) + 2 In (a; - 2) + In (a, + 2) + C 



EXERCISES 

Find the values of the following integrals 

i r a;+i 7 A r 

*' J ?-"o7+8 '^ 4 ' J ( 



2) (*- 



Bas+l 



T __ 2 -5a! + 5 T 

' J ( iC -.l)(a J ~2)( a! -3) rfa; - 6> J 



76. Table of integrals. The formulas of integration used in 
this chapter are sufficient for the solution of most of the prob- 
lems which occur in practice. To these formulas we have added 
a few others. In some cases they represent an integral which 
has already been evaluated, and in other cases they are the 
result of an integration by parts. In all cases they can be 
verified by differentiating both sides of the equation. 

These collected formulas form a brief table of integrals which 
will aid in the solution of the problems in this book. It will be 
noticed that some of the formulas express the given integral only 
in terms of a simpler integral. 

I. FUNDAMENTAL 

1. | cdu=sc I du. 

2. I (du + Av + dw ) = I du + | dv + I dw . 
8. / udv sauv I vdu. 



213 INTEGRATION 

II. ALGEBRAIC 

4. / u n du = . (n = 1) 

J n + l 

C du 

5. I = mu. 



du 1 



p^ = ln 
r 

A I 

' J 

_ r du _ 1 , u a 

' \ ~o 5 o ^ ! ' 

J u a za u-\-a 

8. / vfls u du = ( u ~\a? 
J V 2V 

9. JWa 2 -w a dM = -i.O a _t 
ju^a*-u*du=>- w + 2 



sm~- 



19. 

w 



10. 

12. 
13. 

I __ / t>. 

n n 

/ . _ 

14. 

15. 
16. 



/du 
Va 2 - 

fl 2 /* 

L 

s* 

I u^/u z a z du = % (w 2 



TABLE 219 



_. du 1 ,M 

20. I . =-seo- 1 - 



/, 

. I V2 au u*du = - (w a) V2 au w 2 + a 2 sin" 1 

/. 



22. / t ""* = sm- 1 - 

V2 aw - w 2 a 

III. TRIGONOMETRIC 

23. I smudu=x cosw. 



ft . /* . , , u 1 
24. I sm waw == - -r sm 2 w. 
J 24 

/I w, 1 /* 
sm n udu = sin""^ cos w H I sin"" 2 wcZM. (n 3= 0) 
w w J ^ J 

26. | cos udu SB sin w. 

/w 1 
cos 2 wdw == - + T sill 2 w. 
2 4 

/I 7i _ i / 
coa*udu BBS - cos n - 1 w sin w H I cos" ~ a w^. (w ^ 0) 
n n J 

29. I tan udu In sec w. 

/tan'*"''^ C H /\\ 

tan w w?w s= rf I tan w "" udu. (n l^* 0) 
n-1 J 

81. f ctnw rfw sss In sin w. 

/r*i"n " *** i/ /** 
ctn n w^ - r - I ctn 11 - 3 wc?w. (n - 1 + 0) 

n-1 J 

88. I sec wrfti a= In (sec t* + tan w). 

34. Jseo'wtfw =s tan M, 

85, Jcso WC^M In (cso w - ctn ). 

86, I cso a wc?'M! ctaa w. 



220 INTM! RATION 

37. I sec u tau u du HUC w. 



38. I oso ? ctn w tft* = <IBO ?f. 



/" 

/ 

/Hin mfl ?M'0,s" 1 U H - I r . 

Hill m WCt)S n ?/,(Zjts=S" '-'-.-,} / ,sh'"M riLS* "'/0/, 

m + w / 1 ><</ 

( i i it a ) 

I || I U 1 |'ll"*1/ I"*/ VU*' O/ //')/ " - - j., U4_ i I Wtll // 4*1 i 1 !* J/ */Jjf 

* v I niix c* ^v'Ci (v vvCv "*-"*^ T i Ji i * F**' I ~ *i** 

J si , 



n I , j M111 
41 . I Hill"' W C50S" U dlt,*x 



42. Tsii 



m + 1 
IV. EXPONENTIAL 



r<'"?MM". 

4. Ca*du**-~a H . 
J In a 



43. 
44. 
45. 
46. 



A* 



, 



I WCOH" H M 
< H- 1 

{/ f 1 / > 



GENERAL EXERCISES 
Find the vaJuos of tho following mtgralH t 

I. J (3 aj 8 4- 4 - ^ 3 ~ ^ flfaj. 4, f(j f 

. f2aj a 
J 



GENERAL EXERCISES 221 

7. f(2- 



9. f i+ * 2 * w- r 

J (3x + xrf J 

r^-. 27. r 

J as - 1 J ( 

// 
n-(8.-l)orf(8.-l)*, " J ( 



IS. 



13. 



dx 
dx 



/C dx 
GSG^4; C l/C 30 I * 

/ "\/ 1\ 'V* r - - 11 '>* 

i/ v ti a/ " t*/ 

11. fsec 8 (-2Han 8 (:-2)daJ. 31. C , dx 

J V ' J V34-2aj-o; a 

15. / ctn (a; 1) sec 4 (a; 1) dx. 32 . | 

J J W . 

1C. f oso 8 2ajctu s 2a;^sB. 33. J 

17. J tan'Saj-yseoSa-rZa-. 34 . T- & 

/H -U-Jlu-.iniLJ.UI 
(itn 2 IB V(!HO '2xdx. 35. 

/-.. -.-.IIU.J. rt 

CMC* fiaJ"V(!t,u 5 "/?. 3g / _ 

* I . .<!() 
J ' 

/* **" *? *"" $$, o*y I 

siu*4aj " J 4. s 

/CHC 4 5 C /* 
^ 38t / 
v tan 8 Sir J >!' 

M.J- *. v 39 '/9> + 1 



19 

20. 



222 INTEGRATION 

58. 



* 



54. 

55, 



/ 



42. I 


69 ' / 9 a -7' 


/dx 
x Vcc* 6 


60. / f . 

,y oar 4 


c dx 

J x Vce 6 4 


61 -/4*6^ 


45. I 
J Vl~12a; 2 -4aj 4 


"/Wl"* 


46. fl^T^dx 


r dx 


., r 4-3oj , 
47. 1 . dx 

J V9-2a; 2 


-fl7C 


/dx 


&5 'J9x*-(>x- 


Va: 2 -7 


49. f dx 

J V4a: 2 +3 


661 J 4a; 2 -4aj- 


O.J V3 * . 


67. r 2 da; 


' J V4a; 4 -5' 


i dx 


52 f f*^ . 

J Vo; 8 +7 


69. | 

/ 5 2 + 2x '. 



53. / --'*,&, 70t / ^ 



/ 

/^fal "71 I 

V3a: 2 + l ' /J "J 

/<a; ^ 

V2a; a - 3aj' 72 ' J seo(a!- |)<fa. 

56. /^ dx _ ^8^43;^ 

/ v4.'r a _L /Lo _l_7 ' d< I ?T~ <W5> 

* yv * a; T*a:-f-< ^y cos 2 x 

/OnK SI t 

. . w . i / sin 2 a/ cos 

/w a / 4 I I -- _^ j_ 

"V n <T^ _I_ /Lrw'i * ^ * I I , ***"' ' " 
voa+4a;-l J \ Sinas COSJB 



75. 



Cam*~dx 



GENERAL EXEECISES 

91. Cx5*+*dx. 



223 



. ri 008405, 

76. / da;. 
J l+cos4aj 


92. 1 xVda;. 


/" cos4a; 7 
77. j da;. 
J cos 2 x sin 2 a; 


/cc 
a? tan" 1 ^*- 


78. da;. 
x x 


94. 1 aj 3 sm3fl3daj. 


i cso ~ H~ ctn jr 




79. 1(860*2 a? tan*2 a;) <fcc. 
80. I Vl + sm 2 a; da;. 
81. 1 Ve s daJ. 


95. I (In 2a:) a da5. 
96. | In (3 ; + V9o; a 4) 

97. r a xdx 

J ' 


82. I xtyefidx. 


M -/"i-". * 


88 -/i- 


,/ "i 


s*. r "^ '. 


100. 1 


J (ox -j~l)* 
86. 1 a; 6 *^a5 8 + 2 das. 


/"a; 8 -6 j a -9 x + 24 

i/ "~" """ ow 


86. 1 ^ - 


102. r a y -?~-- 

Ja V aj a - 9 


87. r 2 ^ 1 - 


103 ' Ji V4 - a; 8 * 


8s - r r 

J (l-aj 3 ) 8 


^jTe^.-.- 


/ 3 daj 


105. 1 no' H i ' ft' 



89. 



106. 



224 



107. 



INTEGEATION 



ptan(as- 
J \ Q 



na 
108. 



109. 



110. 



111 



112 



Bx , Bx 



114. | 

i/S 



* . 

a; V a; 8 4- 9 



115 - 



116. 



118. 



ofo: 



JV 

t/Q 



I ~~zz" 
J* e 

J^J* g tan-la 


119. 1 

*sQ 

120. 1 

Jo 


l+x* dX 

/a" JP^tC 
4 






123. I acHarr^cfaj. 



. 1 

Jo 



CHAPTER X 



APPLICATIONS 

77. Review problems. The methods in Chapter III for de- 
termining areas, volumes, and pressures are entirely general, 
and with our new for- y 

mulas of integration we can 
now apply these methods 
to a still wider range of 
cases. 

Ex. 1. Find the area of the 

x* i/" 
ellipse -^ + ~ = 1. 

It is evident from the sym- 
metry of the curve (Fig. 86) 
that one fourth of the required 

aiea is bounded by the axis of p IG< gg 

y, the axis of x, and the curve. 

Constructing the rectangle MNQ,P as the element of area dA, we have 



dA**ydx = --VtP-aPdx. 
a 

Hence 







5ST 

Ex.2. Find the area bounded by the ~ 

axis of x, the parabola y 9 =a kx, and the 

straight line y + Sff-frssO (Fig. 87), 

c 

The straight line and the parabola intersect at the point C ( j s ) > and 

/, \ \4 -i/ 

the straight line intersects OX at B ( - , j . Draw CD perpendicular to 0Z. 
If we construct the elements of area as in Ex, 1, they will be of different 



226 



APPLICATIONS 



form according as they aie to the left or to the right of the line CD] fc 
on the left of CD we shall have 

dAydx k*x*dx, 
and on the right of CD we shall have 

dA = ydx = (k 2 a:) dx 

It will, accordingly, be necessary to compute sepaiately the areas ODi 
and DBC and take their sum. 

Area 

Area DBC = J* 3 (X: - 2 x) dx = \kx - a; 2 ! 2 = & k*. 

1 4 

Hence the required area is 7 S k s It is to be noted that the area DBC 
since it is that of a right triangle, 
could have been found by the formulas ^ 

of plane geometry ; for the altitude 

D C = | and the base DB = l~^-~, 
2 z 2 4 4 



and hence the area = 



16 



Or we may construct the element 
of area as shown in Fig 88 

Then, if x 1 and z 2 are the abscissas 
respectively of P a and P t , 







. 

2 4 3k 48 

Ex. 3, Let the ellipse of Ex 1 be represented by the equations 
x = a cos <f>, y = l sin <. 

Using the same element of area, and expressing y and dx in terms of 

we have , , ,, . .... 

dA = (b sin </>) ( a sin 



As x vanes from to a, </ varies from to ; 

A 



hence 



=4 f 

t/O 



=- 4 



EEVIEW PROBLEMS 



227 



It is evident from foimula (1), 23, that the sign of a definite integral 
is changed by interchanging the limits. Hence 



A = 




irab 



Ex. 4. Find the volume of the ring solid generated by revolving a 
ciicle of radius a about an axis in its plane b units fiorn its center (b > a) 

Take the axis of i evolution Y 
as OY (Fig. 89) and the line 
through the center as OX. Then 
the equation of the ciicle is K 

(r-&) 2 + y 2 =a 2 . 

' M"-- 

A straight line parallel to OX 
meets the circle in two points . 
P v where a, = x i b Va 2 y*, and 
P 2 , where x = x z = b + Va 2 ij\ 
A section of the requned solid 
made by a plane through P 1 P 2 
perpendicular to OY is bounded 
by two concentric circles with 
radii MP l = x v and M P z = x z respectively. Hence, if dV denotes the 








PIG. 89 



element of volume, 



dy= 



The summation extends from the point L, wheie y = a, to the point If, 
where y = a. On account of symmetry, however, we may take twice the 
integral from y = to y = a. Hence 



V- 2 C 
Jo 




PIG. 90 



Ex. 5. Find the pressure on a 
parabolic segment, with base 2 b and 
altitude a, submerged so that its 
base is in the surface of the liquid 
and its axis is vertical 

Let RQ.C (Fig. 90) be the parabolic segment, and let CB be drawn 
through the vertex C of the segment perpendicular to RQ, an the surface of 
the liquid. According to the data, RQ = 2 ft, CB-a. Draw LN parallel 
to Tti, and on LN as a base construct an element of area, dA. Let 

CM=x. 



228 APPLICATIONS 

Then dA *=(LN)ilje. 

r> , * 
But, from 30, 

whence 

a 

2/; I 
and therefore dA = - <to. 

* 

The depth of Zi^V below the surface of the liquid is CH ( W - ,r ; 
hence, if 10 is the weight of a unit volume of tho liquid, 

tip sz 1 $* (a x) wdxt 
a* 

, .. r a 2biii i, x , 

and P - { r j? 1 ( - .r) of.c 

Jo 



EXERCISES 

1. Find the area of an arch of tho curve y = sin *. 

w / * _ > \ 

2. Find the area bounded by tho catenary y = A v'"Hh " % fcliw 
axis of a?, and the lines r = /*. 

3. Find the area included between tho curve ?/ = - .. * . ., and 
.... * 7 r + -1 rt^ 

its asymptote. 

4. Find the area of one of the closed figurus boundtul by Uit* 
curves if = 16 a? and ?/ 2 = sc 8 . 

6. Find the area bounded by the curve 2/ a sss2(ce 1) and tilio 
Iine2a; 3?/ = 

6. Find the area between the axis of SB and ono arch of tht 
cycloid a; = a(j!> sin ^), y = a(l COH 0). 

7. Find the volume of tho solid generated by revolving about OY 
the plane surface bounded by OY and the curve* a$ -j- ^ s 3, 

8. Any section of a certain solid made by a piano pM'pwulinulnr to 
OX is an isosceles triangle with tho ends of its busa r8tint? on Ihw 



ellipse 1 4- '4 = 1 an d its altitude equal to tho distance of tho plane 
from the center of the ellipse. Find the total volume o tho Bulid, 

9, Find the volume of the solid formed by revolving about tho 
line 2 y + a the area bounded by 0110 arch of tho curve y m sin as 
and the axis of x. 



REVIEW PROBLEMS 229 

10. Find the volume of the solid formed by revolving about the 
line y + a = the area bounded by the circle a 8 + t/ 2 = a 2 . 

11. Find the volume of the solid formed by revolving about the 
line x a the area bounded by that line and the curve ay 2 = x 3 . 

12. A right circular cone with vertical angle 60 has its vertex at 
the center of a sphere of radius a Find the volume of the portion 
of the sphere included in the cone. 

13. A trough 2 ft. deep and 2 ft. broad at the top has semielliptical 
ends If it is full of water, find the pressure on one end. 

14. A parabolic segment with base 18 and altitude 6 is submerged 
so that its base is horizontal, its axis vertical, and its vertex m the 
surface of the liquid Find the total pressure. 

15. A pond of 15 ft. depth is crossed by a roadway with vertical 
sides. A culvert, whose cross section is in the form of a parabolic 
segment with horizontal base on a level with the bottom of the 
pond, runs under the road. Assuming that the base of the parabolic 
segment is 4 ft. and its altitude is 3 ft., find the total pressure on 
the bulkhead which temporarily closes the culvert. 

1. Find the pressure on a board whose boundary consists of a 
straight line and one arch of a sine curve, submerged so that the 
board is vertical and the straight line is in the surface of the water. 

78. Infinite limits or integrand. There are cases m which it 
may seem to be necessary to use infinity for one or both, of the 
limits of a definite integral, or in which the integrand becomes 
infinite. We shall restrict the discussion of these cases to the 
solution of the following illustrative examples : 

Ex. 1. Find the area bounded by the curve y = -^ (Fig. 91), the axis of 
x, and the ordinate x = 1. x 

It is seen that the curve has the axis of x as an asymptote ; and hence, 
strictly speaking, the required area is not completely ^ 
bounded, since the curve and its asymptote do not 
intersect. Accordingly, in Fig. 91, let OM=l and 
0JV = ft (& > 1) and draw the ordinates MP and 

' 




If the value of b is increased, the boundary line NQ, moves to the right; 
and the greater b becomes, the nearer the area approaches unity. 



230 APPLICATIONS 

We may, accordingly, define, the area boundt'd by the eune, Ihe u\i* of r, 
and the oidinate .r = 1 as tlw hunt of the area Mj\Ql*\w It tm'tcnst's indeti 
nitely, and denoto it by the symbol 

-'f = Lim f ft '('' .- 1. 



Ex. & Find the, area bounded by the eurvt' // - ^ (KU?. tiiJ), tlw 
axis of X) and the oidinatoH JK and u' -- . v "" " '" 



Sinco the line u = <t is an jwyinptntc' of Ihc eum*, //~~- * when i 
fmthtiimoH 1 , ilic area in not, ntneUy speukiuj-, hounded. We IIH ( \, lun 
find the area bounded on thci ri^hl by the tirdinuto 
a; = a Ji, where ft is a Htuall quantity, with the lesult 



If A M), Hin~ l 



a -A 

' (I 

7T 




Hence wo may regard J as tlui vahu of th> area 
required, and express it by tlw 

if **(/" "*~~* M ""jj" 1 

ii * Ftu. UiJ 

Ex. 3. Find the valuo of J ' "f, 
Proceeding as in Ex, 1, w( plaoo 

/ f(JG * . /* * w J* 

/ - -7- -a Lim / , . 
^i V / h **<' i Vj 

But 



an expression which increases Indeflnifcely m * Voo } hrn^o th* givt*n 
integral has no finite value, 

We accoidmgly conclude that in ranli aawt w rntiut dftermiuo n limit, 
and that the problem has no solution if wt* cannot find a 



79. Area in polar coordinates, Let (Fig, 98) he thn po! mid 
OH the initial line of a system of polar noSniinatai (r, &}< nf> 
and 0% two fixed radius vectors for which 0*0 and Bmi 
respectively, and P& any curve for which the aquation Ift 
Requirod the area JJOE 



AREA IN POLAR COORDINATES 



231 



To construct the differential of area, dA, we divide the angle 
into parts, dd. Let OP and OQ be any two consecutive 
radius vectors ; then the angle POQ = d0. With as a center 
and OP as a radius, we draw the arc of a circle, intersecting OQ 
at R. The area of the sector 



It is obvious that the re- 
quired area is the limit of the 
sum of the sectors as their 
number is indefinitely in- 
creased. Therefore we have 



and 




FIG. 93 



This result is unchanged 
if P l coincides with 0, but 
in that case OP l must be tangent to the curve. So also P z may 
coincide with 0. 

Ex. 1. Find the area of one loop of the curve r = a sin 3 (Fig 65, 51) 

As the loop is contained between the two tangents 0=0 and 6 = , the 
required area is given by the equation 8 



= f 
Jo 



12 



Ex. 2. Find the area bounded by the lines 0= ~ and 0=~t the curve 



~ 
44 

r = 2 a cos 0, and the loop of the curve r = a cos 2 6 which is bisected by 
the initial line. 

Since the loop of the curve r = a cos 2 Q is tangent to the line OL 



(Fig. 94), for which = - ~, and the line ON, for which 6 



f it is evi- 



dent that the required area can be found by obtaining the area OLMNO, 
bounded by the lines OL and ON and the curve r 2 a cos 6, and subtract- 
ing from it the area of the loop. The area may also be found as follows : 
Let OPjPg be any radius vector cutting the loop r = a cos 2 Q at P t and 
the curve r = 2 a cos 6 at P a . Let OP l = r t and 0P 2 r v Draw the radius 



232 



APPLICATIONS 



vector OQ^jj, making an angle dd with OP^P^ With OP l and OP S as- 
radii and as a center, constiuct arcs of circles niti'iaoetouft O(^l^ at A\ 
and J2 2 respectively. Then the area ot the sector J\01fi IH ^ rfilQ and tho 
area of the sector P Z OR S is ^ r| dd. We ^y 

may now take the area P 1 / > 3 JB a /' 
and have 



Then ^ = 

/ __ 7T ,_ 

4 

01, since the required area is symmetrical 
with respect to the line OM, wo may place 




(>2-*i)0. FIG, 04 

From the curve r = 2 a cos0, we have rf = 4 a 8 cos s 0, and from the curve 
r = a cos 2 0, we have r a = a 2 cos 2 2 0; so that finally 



- a a cos 2 2 ^) <W 

e-?-'22 

2 8 



EXERCISES 

1. Find the total area of the lemniscato ?' a = 2 2 cos 2 0, 

2. Find the area of one loop of the curve r =s a sin n 0, 

3. Find the total area of the cardioid r = a(l-f- COB 0). 

4. Find the total area bounded by the curve ? a 6 + 3 <uw 0. 

5. Find the area of the loop of the curve r* t a" cos 2 non 3 
which is bisected by the initial line. 

6. Find the area bounded by the curves r = a cos 8 and r == ., 

7. Find the total area bounded by the curve r = 3 + 2 COB 4 0. 

8. Find the area bounded by the curve ?cos a ^=sl and the 

7T * 

lines = and = 75 

. 

9. Find the area bounded by the curves r ^ 6 + 4 cos and 
r s= 4 cos 

10. Findtheareaboundedbythecurvesr=acos0andr a =:a 8 cos2^ 



MEAN VALUE 



233 



B 



80. Mean value of a function. Let f(x) be any function of x 
and let y =/() be represented by the curve AB (Fig. 95), where 
OM=a and ON=b. Take the points M^ M z , , M n _ r so as 
to divide distance MN into 
n equal parts, each equal to 
dx, and at the points M, M^ M z , 
' -, M n _ l erect the ordinates 

2/0' y t y> ' &.-! Tlien the 

average, or mean, value of these 
n ordinates is 



M 



This fraction is equal to 



ndx 



, dx-\ 
b a 



If n is indefinitely increased, this expression approaches as a 
limit the value 



This is evidently the mean value of an " infinite number " of 
values of the function / (x) taken at equal distances between 
the values x = a and x = 6. It is called the mean vafote of the 
function for that interval. 

Graphically this value is the altitude of a rectangle with the 
base MN which has the same area as MNBA which equals 



/ 

/a 



We see from the above discussion that the average of the 
function y depends upon the variable x of which the equal 
intervals dx were taken, and we say that the function was 
averaged with respect to x. If the function can also be averaged 
with respect to some other variable which is divided into equal 
parts the result may be different. This is illustrated in the 
examples which follow. 



234 APPLICATIONS 

Ex. 1. Find the mean velocity of a body falling from rest duimg the 
tune t i if the velocity is averaged with respect to the tune. 

Here we imagine the time from to ^ divided into equal intervals dt 
and the velocities at the beginning of each interval averaged. Proceeding 
as in the text, we find, since v gt, that the mean velocity equals 

1 



Since the velocity is gt^ when t = t v it appears that in this case the 
mean velocity is half the final velocity. 

Ex. 2 Find the mean velocity of a body falling from rest through a 
distance s 1 if the velocity is averaged with respect to the distance. 

Here we imagine the distance from to s 1 divided into equal intervals 
ds and the velocities at the beginning of each interval aveiaged Pro- 
ceeding as in the text, we find, since v = V2 y$, that the mean velocity is 




Since the velocity is v^2 gs v when s = s v we see that in this case the 
mean velocity is two thirds the final velocity. 

EXERCISES 

1. Find the mean value of the lengths of the perpendiculars 
from a diameter of a semicircle to the circumference, assuming the 
perpendiculars to be drawn at equal distances on the diameter. 

2. Find the mean length of the perpendiculars drawn from the 
circumference of a semicircle to its diameter, assuming the perpen- 
diculars to be drawn at equal distances on the circumference 

3. Find the mean value of the ordmates of the curve y = sm x 

7T 

between x = and x = -5- , assuming that the points at which the 
ordmates are drawn are at equal distances on the axis of x. 

4. The range of a projectile fired with an initial velocity V Q and 

v^ 
an elevation a is sin 2 a. Find the mean range as a varies from 

TT ff 

to 7p averaging with respect to a. 

6. Find the mean area of the plane sections of a right circular 
' cone of altitude h and radius a made by planes perpendicular to the 
axis at equal distances apart. 



LENGTH OF PLANE CURVE 235 

6. In a sphere of radius a a series of right circular cones is 
inscribed, the bases of which are perpendicular to a given diameter 
at equidistant points. Find the mean volume of the cones 

7. The angular velocity of a certain revolving wheel varies with 
the time until at the end of 5 mm. it becomes constant and equal to 
200 revolutions per minute If the wheel starts from rest, what is its 
mean angular velocity with respect to the time during the interval 
in which the angular velocity is variable ? 

8. The formula connecting the pressure p in pounds per square 
inch and the volume v in cubic inches of a certain gas is pv = 20. 
Find the average pressure as the gas expands from 2^- cu in. to 5 cu. in. 

9. Show that if y is a linear function of aj, the mean value of y 
with respect to x is equal to one half the sum of the first and the 
last value of y in the interval over which the average is taken. 

81. Length, of a plane curve. To find the length of any 
curve AB (Fig. 96), assume n 1 points, J?, J?, , -_ r be- 
tween A and B and connect each pair of consecutive points by 
a straight line. The length of AB is 
then defined as the limit of the sum 
of the lengths of the n chords AQ, 
%%, ]%PV -, X^.iB as n is increased 
without limit and the length of each 
chord approaches zero as a limit. By 
means of this definition we have already 
shown ( 39 and 52) that " FIG. 96 

d8**^dy?+df (1) 

in Cartesian coordinates, and 

<2 = V,fr a 4-rW (2) 

in polar coordinates. 

Hence we have s = I ^dx*+ dy* (3) 

and s => C-Vdr* + r*d6\ (4) 

To evaluate either (3) or, (4) we must express one of the 
variables involved in terms of. the cither, or both in terms of a 
third* The limits of integration may then be determined. 




236 APPLICATIONS 

Ex. I. Find the length of the parabola y 2 = kx from the vertex to the 
point (a, 6). 

From the equation of the parabola we find 2 ydy = kdx Hence formula 
(3) becomes either 



Either integral leads to the result 



Ex. 2. Find the length of one arch of the cycloid 

x = a(<jf> sm<), y = a (1 cos <). 

We have dx a (1 cos <) dtp, dy a sin <j> d$ ; 

whence, from (1), ds = a V2 2 cos <j>d<j> = 2a sin 2 rf<ji>. 

/* 27r <i 

Therefore s 2 a | sin d<& = 8 a. 

/o 2 

EXERCISES 

v 1. Find the length of the curve 3y*(x I) 8 from its point of 

intersection with OX to the point (4, 3"). 

af - --^ 
w 2. Find the length of the catenary y -~\e a + e ") from x = 

to ai = h. 



v 3. Find the total length of the curve 

4. Find the total length of the curve x = a cos 8 ^>, y = a sin 8 <. 

5. Find the length of the curve 

x = a cos <f> -}- a<l> sin <, y = a sin < a< cos <, 
from < = to j! = 4 TT. 

6. Find the length of the curve x e~*cos t, y= e-*sin#, between 

7T 

the points for which t = and t = -^ . 



7. Find the length of the curve r = a cos 4 j from the point on 

the curve for which 6 ~ to the pole 

8. Find the total length of the curve r = a (1-|- cos 5). 

9. Show that the length of the logarithmic spiral re a between 
any two points is proportional to the difference of the radius vectors. 
of the- points, 



WORK 237 

82. Work. By definition the work done in moving a body 
against a constant force is equal to the force multiplied by the 
distance through which the body is moved. If the foot is taken 
as the unit of distance and the pound is taken as the unit of 
force, the unit of measure of work is called & foot-pound. Thus 
the work done in lifting a weight of 25 Ib. through a distance 
of 50 ft. is 1250 ft-lb. 

Suppose now that a body is moved along 0JT(Fig. 97) from 
A (x = a) to B (x = b") against a force which is not constant but 
is a function of a;, expressed by /(). Let the line AB be 
divided into intervals each equal A i j/W J> x 

to dx, and let one of these inter- 
vals be MN, where OM x. Then 

the force at the point M is /(a?), and if the force were con- 
stantly equal to f(x) throughout the interval MN, the work 
done in moving the body through MN would be/ (x) dx. This 
expression therefore represents approximately the work actually 
done, and the approximation becomes more and more nearly 
exact as MN is taken smaller and smaller. The work done in 
moving from A to B is the limit of the sum of the terms f(x) dx 
computed for all the intervals between A and B. Hence we have 



and 



/> 
= / 

Jit 



Ex. The force which resists the stretching of a spring is propor- 
tional to the amount the spring has been already sti etched. Foi a cer- 
tain spring this force is known to be 10 Ib. when the spring has been 
stretched & in. Find the work done in stretching the spring 1 in. from 
its natural (unstretched) length. 

If F is the force reqmied to stretch the spring through a distance x, 
we have, from the statement of the problem, 



and since F ~ 10 when x = %, we have k = 20. Therefore F = 20 x. 
Reasoning as in the text, we have 



w~C\ 

Jo 



238 APPLICATIONS 

EXERCISES 

1. A positive charge m of electricity is fixed at 0, The repulsion 
on a unit charge at a distance x from is -3 Find tho work done 

33 

m bringing a unit charge from infinity to a distance a from 0. 

2. Assuming that the force required to strotdi a wiro from the 

*t* 

length a to the length a + x is proportional to -'j and that a foixio 

if* 

of 1 Ib stretches a certain wire 36 in in lungth to a length .0,'i in. 
greater, find the work done in stretching that wiro from M in, to 40 in. 

3. A block slides along a straight line from O against a resistance 

ka? 
equal to . a ? where 7c and a are constants and or, is the distance 

(K j" Cb 

of the block from at any time. Find the work done in moving the 
block from a distance a to a distance a V from 0. 

4. Find the foot-pounds of work done in lifting to a height of 
20 ft above the top of a tank all the water contained in a full cylin- 
drical tank of radius 2 ft. and altitude 10 ft. 

5. A bag containing originally 80 Ib. of sand is liftod through a 
vertical distance of 8 ft. If the sand leaks out, at such a rato that while 
the bag is being lifted, the number of pounds) of sand lost is wjual to a 
constant times the square of the number of foot through which the bag 
has been lifted, and a total of 20 Ib. of sand i& lost during the lifting, 
find the number of foot-pounds of work done in lifting tho bag, 

6. A body moves in a straight line according to the formula c ef, 
where x is the distance traversed in a time t. If tho resistance of the 
air is proportional to the square of tho velocity, find tho work done 
against the resistance of the air as the body moves from != to ft* a. 

7. Assuming that above the surface of the earth the form* of the 
earth's attraction varies inversely as tho square of the diBtanca from 
the earth's center, find the work done in moving a weight of w pounds 
from the surface of the earth to a distance a miles above the surfaca. 

8. A wire carrying an electric current of magnitude C is bent 
into a circle of radius a. The force exerted by the current upon a 
unit magnetic pole at a distance a from the center of the circle in a 
straight line perpendicular to the plane of the circle is known to be 

27rCa s 

(a?+ ^ ' done in brin g in 8 a ttnit magnetic pole from 

infinity to the center of the circle along the line just mentioned, 



GENERAL EXEKClSES 239 

9. A piston is free to slide in a cylinder of cross section S The 
force acting on the piston is pS, where p is the pressure of the gas 
in the cylinder, and is 7.7 Ib. per square inch when the volume v is 
2 5 cu in. Find the work done as the volume changes from 2 cu. in. to 
6 cu. in., according as the law connecting p and v is (1) pv k or 

(2)X* = * 

GENERAL EXERCISES 

1. Find the area of the sector of the ellipse 4oj 2 +9y 2 =36 
cut out of the first quadrant by the axis of x and the line 2 y = x. 

2. Find the area of each of the two parts into which the area of 
the circle a; 3 + if = 36 is divided by the curve y 2 = a; 8 . 

3. Find the area bounded by the hyperbola xy = 12 and the 
straight line x + y 8 = 0. 

4. Find the area bounded by the parabola a; 2 =4 ay and the 

8 a 8 



5. Find the area of the loop of the curve ay 2 = (x a) (x 2 a) z . 

6. Find the area of the two parts into which the loop of the 
curve y* = cc 2 (4 a;) is divided by the line x y = 0. 

7. Find the area bounded by the curve ary* + a 2 6 2 = a 2 2/ a and its 
asymptotes. 

8. Find the area bounded by the curve 2/ 2 (aj a + & 2 ) = <&& and its 
asymptotes. 

9. Find the area bounded by the curve a? = a cos 6, y = b sin 8 

10. Find the area inclosed by the curve x = a cos 8 0, y = a sin 8 0. 

11. Two parabolas have a common vertex and a common axis, but 
lie in perpendicular planes. An ellipse moves with its plane perpen- 
dicular to the axis and with the ends of its axes on the parabolas. 
Find the volume generated when the ellipse has moved a distance 
h from the common vertex of the parabolas. 

12. Find the volume of the solid formed by revolving about the line 
a; = 4 the figure bounded by the parabola y*= 4 x and the line x = 1. 

13. A right circular cylinder of radius a, is intersected by two 
planes, the first of which is perpendicular to the axis of the cylinder 
and the second of which makes an angle with the first. Find the 
volume of the portion of the cylinder included between these two 
planes if their line of intersection is tangent to the circle cut from 
the cylinder by the first plane. 



240 APPLICATIONS 

fn O 

., w. u^ U.UU.UJLU w*.^c* U w ~~ ~ + y* a? as a base, 

an isosceles triangle is constructed with its altitude equal to the 
ordinate and its plane perpendicular to the plane of the curve. Find 
the volume generated as the triangle moves from x = a to x = a. 

15. Find the volume of the solid generated by revolving about 

8 a 8 
the line OF the figure bounded by the curve y = 2 . and the 

line y o,. 

16. Find the volume of the solid formed by revolving about the 
line x = 2 the plane area bounded by that line, the parabola y z = 3x, 
and the lines y = 3. 

17. Find the volume formed by revolving about the line x = 2 
the plane figure bounded by the curve y 3 = 4 (2 a;) and the axis of y. 

18. The sections of a solid made by planes perpendicular to OF 
aie circles with one diameter extending from the curve ^ 4 a; to 
the curve y 2 = 4 4 x. Find the volume of the solid between the 
points of intersection of the curves. 

19. The area bounded by the circle x* + $ 2 ax = is revolved 
about OX, forming a solid sphere Find the volume of the two parts 

into which the sphere is divided by the surface formed by revolving 
3.8 

the curve y 2 = about OA' 

y 2a x 

20. Find the volume of the two solids formed by revolving about 
Y the areas bounded by the curves ar 8 -f # 2 = 5 and g/ 2 = 4 ai. 

21. Find the volume of the solid formed by revolving about OX 
the area bounded by OX, the lines x = and x == a, and the curve 

z 

y = x + ae a . 

22. The three straight lines OA, OB, and OC determine two planes 
which intersect at right angles in OA. The angle A OB is 45 and the 
angle AOC is 60. The section of a certain solid made by any plane 
perpendicular to OA is a quadrant of an ellipse, the center of the 
ellipse being in OA, an end of an axis of the ellipse being in OB, and 
an end of the other axis of the ellipse being m OC. Find the volume 
of this solid between the point and a plane perpendicular to OA at 
a distance of two units from 0. 

23. The section of a solid made by any plane perpendicular to OX 
is a rectangle of dimensions a? and sm x, x being the distance of the 
plane from O. Find the volume of this solid included between the 
planes for which x = and x = IT. 



GENERAL EXEECISES 241 

24. An oil tank is in the form of a horizontal cylinder the ends 
of which are circles 4 ft. in diameter. The tank is full of oil, which 
weighs 50 Ib. per cubic foot Calculate the pressure on one end of 
the tank. 

25. The gasoline tank of an automobile is in the form of a hon- 
zontal cylinder the ends of which are plane ellipses 20 in high and 
10 in broad. Assuming w as the weight of a cubic inch of gasoline, 
find the pressure on one end of the tank when the gasoline is 15 in 
deep. 

26. A horizontal gutter is U-shaped, a semicircle of radius 3 in , 
surmounted by a rectangle 6 in wide by 4 in. deep If the gutter is 
full of water and a board is placed across the end, how much pressure 
is exerted on the board ? 

27. The end of a horizontal gutter is in the form of a semicircle 
of 3 in. radius, the diameter of the semicircle being at the top and 
horizontal. The gutter receives water from a roof 50 ft above the 
top of the gutter. If the pipe leading from the roof to the gutter is 
full, what is the pressure on a board closing the end of the gutter ? 

28. A circular water main has a diameter of 5 ft. One end is closed 
by a bulkhead, and the other is connected with a reservoir in which 
the surface of the water is 20 ft above the center of the bulkhead 
Find the total pressure on the bulkhead. 

29. Find the area of a loop of the curve ?* a = a? sin n6. 

30. Find the area swept over by a radius vector of the curve 

IT 

r a tan as $ changes from to 

4 4 

31. Find the area inclosed by the curve r = H .. and the 

. J 1 cos $ 

curve r = H , 
1 + cos 6 

32. Find the area bounded by the circles r=a cos 6 and r= a, sin 6. 

33. Find the area cut off from one loop of the curve r 2 = 2 a 2 sin 2 6 
by the circle r a 

34. Find the area of the segment of the cardioid r = a (1+ cos 0) 
cut off by a straight line perpendicular to the initial line at a dis- 
tance | a from the origin 0. 

35. Find the area cut off from a loop of the curve r = a sm 3 6 by 

. , . , 

the circle r s= 



242 APPLICATIONS 

36. Find the area cut off from the lemniscate r 2 = 2 a 2 cos 2 by 

_/*> 
the straight line r cos = 75 ' 

37. Find each of the three areas bounded by the curves r = a 
and r = a (1 + sin 0). 8 

38. Find the mean height of the curve y 2 . 2 between the 
lines x = 2a and aj = 2 a. 

39. A particle describes a simple harmonic motion defined by the 

(792i?/ \ 
~9~) 

during a complete vibration is half the maximum kinetic energy iL 
the average is taken with respect to the time. 

40. In the motion defined in Ex 39 what will be the ratio of 
the mean kinetic energy during a complete vibration to the maxi- 
mum kinetic energy, if the average is taken with respect to the 
space traversed? 

41. A quantity of steam expands according to the Iawj0v 08 =; 2000, 
p being the pressure in pounds absolute per square foot Find the 
average pressure as the volume v increases from 1 cu. ft. to 5 cu. ft. 

a 2 

42. Find the length of the curve y = a In 2 . from the origin 

a ^ 

to the point for which x = ~ 



43. Find the length of the curve y = Li' x - between the points 
for which x =sl and x = 2 respectively. ~~ 

44. Find the total length of the curve x = a cos 8 <, y = b sin*<. 

a 

45. Find the total length of the curve r = a sin 8 ^ 

46. Find the length of the spiral r = aO from the pole to the end 
of the first revolution. 

JA 

47. If a center of force attracts with a magnitude equal to r> 

SC# 

where x is the distance of the body from the center, how much work 
will be done in moving the body in a straight line away from the 
center, from a distance a to a distance 8 a from the center ? 

48. A body is moved along a straight line toward a center of 
force which repels with a magnitude equal to 7ccc when the body 
is at a distance x from the center. How much work will be done 
in moving the body from a distance 2 a to a distance a from the 

ter? 



GENERAL EXERCISES 243 

49. A central force attracts a body at a distance x from the center 

7c 
by an amount -j Find the work done in moving the body directly 

away from the center from a distance a to the distance 2 a. 

50. How much work is done against hydrostatic pressure in rais- 
ing a plate 2 ft. square from a depth of 20 ft. to the surface of the 
water, if it is kept at all times parallel to the surface of the water ? 

51. A spherical bag of radius 5 in. contains gas at a pressure 
equal to 15 Ib. per square inch. Assuming that the pressure is in- 
versely proportional to the volume occupied by the gas, find the 
work required to compress the bag into a sphere of radius 4 in. 



CHAPTER XI 
REPEATED INTEGRATION 
83. Double integrals. The symbol 



nv t 
f(x, y) dxdy, (1) 

_ 4 

in which a and b are constants and y^ and y^ are either con- 
stants or functions of x, indicates that two integrations are to 
be carried out in succession. The first integral to be evaluated is 



where x and dx are to be held constant. The result is a func- 
tion of x only, multiplied by dx; let us say, for convenience, 

;) dx. 
The second integral to be evaluated is, then, 



F(x) dx, 

which is of the familiar type. 
Similarly, the symbol 

6 /*x s 

f(x, y) dydx, (2) 

where a and b are constants and ^ and # g are either constants 
or functions of #, indicates first the integration 



in which y and dy are handled as constants, and afterwards 
integration with respect to y between the limits a and b. 

244 



ft n 2 

Ex. 1. Evaluate / I xythdy. 

The first integral is 

j scydxily 

The second integration is 



DOUBLE INTEGRALS 245 

s /.a 



Ex. 2. Evaluate f f "' O 2 + y*)<lr.dii 

Jo Ji-x ^ ' 



The ni*st integration is 



The second integration is 






Ex.3. Evaluate f^" f**jpdyt?jc. 

Jo i/o 

The first integration is 

v 3 r n j/ a 

( u y*<lytlx = ?/ 2 ^/y * = /-.//. 

1/0 L Jo <l ' 

The second integration is 

r 2a ^ y= rir a = ! a 4 

Jo 4a '' LaOoJo 6 



A definite integral in one variable has been shown lo be the 
limit of a sum, from which we infer that formula (1) involves 
first the determination of the limit of a sum with respect to ;/, 
followed by the determination of the limit of a sum with respect 
to x. The application of the double integral comes from it 
interpretation as the limit of a double summation, 

How, such forms arise in practice will be illustrated in the 
following sections, 

EXERCISES 

Find the values of the following integrals ; 

a"%/ / 8 r* 

^dydx, 3. / / 

Ji J,, 

n\ 
xydoady. 4. 





' 



246 

5. 



REPEATED INTEGRATION 



dydx 



8. 



/ /-as 

| I 

c/O ^/o 

rf r 

Jo /o 







r 8 



a(l + coi S) 




? sin Odddr. 



r*d$dr. 



r cos 6dOdr 



84. Area as a double integral. Let it be required to find an 
area (such as is shown in Fig. 98) hounded by two curves, with 
the equations y^~f^(x) and y 2 =/ 2 () intersecting in points for 








which x a and z=b respectively. Let the plane be divided into 
rectangles by straight lines parallel to OX and OT respectively. 
Then the area of one such rectangle is 



where dx is the distance between two consecutive lines parallel 
to O Y 3 and. where dy is the distance between, tw,o consecutive 
lines parallel to OX. The sum of the rectangles which are either 



AKEA AS DOUBLE INTEGRAL 247 

wholly or partially within the required area will be an approx- 
imation to the required area, but only an approximation, because 
the rectangles will extend partially outside the area. We assume 
as evident, however, that the sum thus found becomes more 
nearly equal to the required area as the number of rectangles 
becomes larger and dx and dy smaller. Hence we say that the 
required area is the limit of the sum of the terms dxdy. 

The summation must be so carried out as to include every 
rectangle once and only once. To do this systematically we 
begin with any rectangle in the interior, such as PQJRS, and add 
first those rectangles which lie in the vertical column with it. 
That is, we take the limit of the sum of dxdy, with x and dx 
constant and y varying from y^f^(x) to 2/ 2 =/ fl (X). This is 
indicated by the symbol 



/" 

vt/. 



x = [/ 2 (x) -/,()] dx. (2) 



This is the area of the strip TUVW. We are now to take 
the limit of the sum of all such strips as dx approaches zero 
and x varies from a to 5. 

We have then 

^ <= /"<>,- ^ ^ = f V, (x) -/ x <V>] dx. (3) 

Jo. J a 

If we put together what we have done, we see that we have 

n s 
dxdy. (4) 

_. 

This discussion enables us to express the area as a double 
integral. It does not, however, give us any more convenient way 
to compute the area than that found in Chapter III, for the result 
(2) is simply what we may write down at once for the area of a 
vertical strip (see Ex. 3, 28). 

If it should be more convenient first to find the area of a 
horizontal strip, we may write 



n* 
dy dx. (5) 



248 



BEPEATED INTEGRATION 



Consider a similar problem in polar coordinates. Let an 
area, as in Fig. 99, be bounded by two curves r^/X^O 
and r 2 =/ s (0), and let the values of corresponding to the 
points B and be l and a respectively. The plane may be 
divided into four-sided figures by circles with centers at and 
straight lines radiating 
from 0. Let the angle 
between two consecu- 
tive radii be d6 and 
the distance between 
two consecutive circles 
be dr. We want first 
the area of one of the 
quadrilaterals such as 
PQRS. Here OP = r, 



= dr, and the angle 
POS=d0. By geom- 
etry the area of the 
sector POS=^r 2 d0 and 

A 

the area of the sector 




FIG. 99 



therefore PQRS = J (r + dr)*d6 - J r*d0 = rdrd0 + % (dr)*d0. 
Now as dr and d0 approach zero as a limit the ratio of the second 
term in this expression to the first term also approaches zero, 
since this ratio involves the factor dr. It may be shown that the 
second term does not affect the limit of the sum of the expression, 
and we are therefore justified in writing as the differential of area 

dA = rd0dr. (6) 

The required area is the limit of the sum of these differ- 
entials. To find it we first take the limit of the sura of the 
quadrilaterals, such as PQRS, which lie in the same sector UO V. 
That is, we integrate rd0dr, holding and dd constant and al- 
lowing r to vary from r^ to r. We have 



/ 

^r. 



(7) 



which is the area of the strip TUVW. 



CENTER OF GRAVITY 249 

Finally we take the limit of the sum of the areas of all such 
strips in the required area and have 



(8) 
If we put together what we have done, we may write 

dOdr. 9 



/*s /"a 

= I I r 
t/0 t J>i 



It is clear that this formula leads to nothing which has not 
been obtained in 79, but it is convenient sometimes to have 
the expression (9). 

85. Center of gravity. It is shown in mechanics that the cen- 
ter of gravity of n particles of masses m^ w 2 , -, m n lying m a 
plane at points whose coordinates are^, y x ), (x z , / 2 ), , (#, y^) 
respectively is given by the formulas 



= 



ss 

\-m n 



This is the point through which the resultant of the weights 
of the particles always passes, no matter how the particles are 
placed with respect to the direction of the earth's attraction, 

We now wish to extend formulas (1) so that they may be 
applied to physical bodies in which the number of particles may 
be said to be infinite. For that purpose we divide the body 
into n elementary portions such that the mass of each may be 
considered as concentrated at a point (or, y). Then, if m is the 
total mass of the body, the mass of each element is dm. We have 
then to replace the m's of formula (1) by dm and to take the 
limit of the sums involved in (1) as the number n is indefinitely 
increased and the elements of mass become indefinitely small. 
There result the general formulas 

/ xdm I ydm 

- X = J , 9 ,J . (2) 

I dm I dm 



250 



REPEATED INTEGRATION 



To apply these formulas we consider first a slender wire so 
fine and so placed that it may be represented by a plane curve. 
More strictly speaking, the curve may be taken as the mathe- 
matical line which runs through the center of the physical wire. 
Let the curve be divided into elements of length ds. Then, 
if c is the area of the cross section of the wire and D is its 
density, the mass of an element of the wire is Dads. For con- 
venience we place DC = p and write 



where p is a constant. If this is substituted in (2), the constant 
p may be taken out of the integrals and canceled, and the 
result may be written in the form 



sx 



= ixds, sy = I yds, 



(3) 



where a on the left of the equations is the total length of the 
curve. These formulas give the center of gravity of a plane curve. 

Ex. 1. Find the center of gravity of one fourth of the circumference 
of a circle of radius a. 



Here we know that the total length is 
\ rrax = Cxds, vay = Cyds. 

To integrate, it is convenient to in- 
troduce the central angle <j> (Fig 100), 
whence x = a cos <j>,y = a sin <f>, ds = ad<j>. 

7T 

Then |irox= l*a*cos<f>dd>, 
Jo 



, 
whence 



we , so that, from (3), we have 



2a 




FIG. 100 



Consider next a thin plate, which may be represented by a 
plane area. Strictly speaking, the area is that of a section 
through the middle of the plate. If t is the thickness of the 
plate and _D its density, the mass of an element of the plate 
with the area A A is DtdA. For convenience we place Dt = p 
and write 

dm pdA, 



CENTER OF GEAVITY 



251 



where p is a constant. If this is substituted in (2) and the p's 
are canceled, we have 



i = I xdA, AyI ydA, 



(4) 



where A is the total area. These formulas give the center of 
gravity of a plane area, 

Ex. 2. Find the center of gravity of the area bounded by the parabola 
y* = kx, the axis of x, and the ordmate through the point (a, b) of the 
parabola (Fig 101) 

We place dA= dxdy in (4) and have 
A js = C Cx dx dy, Ay = Cfy dx dy. 

To evaluate, we choose the element dxdy 
inside the area in a general position, and first 
sum with respect to y along a vertical strip. 

We shall denote by y l the value of y on the _ i . 

parabola, to distinguish it from the general ^ 
values of y inside the area The first integra- 
tion gives us, therefore, respectively 




C Vi xdxdy = xy 1 dx and C l ydxdy \y* dx, 

so that we have Ax Cxy^dx, Ay = C^y* dx- 

J / 

On examination of these results we see that each contains the factor 
y^dx (which is the area ( 22) of an elementary vertical strip), multiplied 
respectively by a? and ^ y v which are the coordinates of the middle point of 
the ordmate y r These results are the same as if we had taken dA = y t dx in 
the general foimula (4), and had taken the point (x, y) at which the mass 
of dA is concentrated as (z, y x ), which is in the limit the middle point of 
dA. In fact this is often done in computing centers of gravity of plane 
areas, and the first integration is thus avoided. 

Kow, from the equation of the parabola y% = kx, and to complete the 
integration, we have to substitute this value for y t and integrate with 
respect to x from x = to x = a We have 



AS = 

* from the 



/** yfco" 

Ay i i kxdx <- 



of the curve, k = and, by 23, A ~ - < 

fl O 



252 REPEATED INTEGRATION 

Substituting these values and i educing, we have finally 
x = | a, y=&b 

In solving this problem we have carried out the successive mtegratioii.s 
separately, in oider to show clearly just what has been done. If now we 
collect all this into a double integral, we have 

Ax=f a f ''xdxdy. /ly=f"f 'ytlxda. 
Jo Jo Jo Jo 

Ex. 3. Find the center of gravity of a sextant of a circle of radum a 
To solve this problem it is convenient to place the sextant so that the 

axis of v bisects it (Fig. 102) and to use ._ 

polar coordinates. 

From the symmetry of the figure the 

centei of gravity lies on OX, so that we 

may write at once y = 0. To find a take r 

an element of area ? dddr in polar cooi- 

dinates and place x = r cos We have 

then, from (4), 
a 

/" r ^ 
/ r 2 cos 6 dddr, 
rr Jo 
~6 

where A = lira?, one sixth the area of a 
ciicle In the first integration 6 and dO 
are constant, and the summation takes 
place along a line radiating from with 

r varying from to a The angle then varies from - - to ~ , and thus the 
entire area is covered The solution is as follows . 6 




liraPx =J 8 a 3 cos Odd 
whence 



6 

_ 2a 



Consider now a solid of revolution formed by revolving the 
plane area (Fig. 103) ABCD about Y as an axis. It is assumed 
that the equation of the curve CD is given. It is evident from 
symmetry that the center of gravity of the solid lies on OY, so 
that we have to find only y. 

Let dFbe any element of volume. Then dm = pdV, where p is 
the density and is assumed constant. Substituting in (2), we have 

(5) 



CENTER OF GRAVITY 



253 



Let the solid be divided into thin slices perpendicular to 
as was done in 26, and let the summation first take place over 
one of these slices. In this summa- 
tion y is constant, and the result 
oL the summation is therefore y 
times the volume of the slice. It is 
therefore u(^r'^^y)- We have now 
to extend the summation over all 
the slices. This gives the result 



r b 
Vy = / ir^ydy, (G) 

/ 






TIG. 103 



where OA = a and 07? = &. 

It is to be noticed that this result 
is what we obtain if we interpret 
dm in (2) as the mass of the slice and consider it concentrated 
at the middle point of one base of the slice. 

Ex. 4. Find the centei of gravity of a light cucular cone of altitude 6 
and radius (Fig. 104) 

This is a solid of revolution formed by revolv- 
ing a light triangle about OF. However, the 
equation of a straight line noed not be used, as 

X CL 

similar triangles are simpler. We have - = - > 

y b 

whence x -y. The volume V is known to be 
b 

J 7r 8 6. Therefore, from (G), we have 



f*mP aj 

= J - fdy = 




whence 



FIG. 104 



EXERCISES 



1. Show that the center of gravity of a semicircunifereuce of 

o a 
radius a lies at a distance of from the center of the circle on 

7T 

the radius which bisects the semicircumference. 

2, Show that the center of gravity of a circular arc which subtends 

an. angle <x at the center of a circle of radius a lies at a distance sin ~ 
from the center of the circle on the radius which, bisects the arc, 



a 



254 EEPEATED INTEGRATION 

- 

3. A wire hangs so as to form the catenary y ~ (/* -1- "). 

Find the center of gravity of the piece of the curve between Iho 
points for which x = and x = a. 

4. Find the center of gravity of the arc of the cycloid 
x = a(<[> sm$), y = a(l coB<f>), between the first two sharp 
points. 

5. Find the center of gravity of a parabolic segment of base, 2 b 
and altitude a. 

6. Find the center of gravity of a quadrant of the area of a circle. 

7. Find the center of gravity of a triangle. 

8. Find the center of gravity of the area bounded by the ourvo 
y = sin x and the axis of x between x = and x = IT. 

9. Find the center of gravity of the plane area bounded by tho 
two parabolas y* = 20 x and a? = 20 y. 

10. Find the center of gravity of a figure which is composed of 
a rectangle of base 2 a and altitude I surmounted by a semicircle 
of radius a 

11. Find the center of gravity of the area bounded by the Ihvst 
arch of the cycloid (Ex. 4) and the axis of aj. 

12. Show that the center of gravity of a sector of a circle lies at 

4 a a 
a distance ~ sin- from the vertex of the sector on a line bisecting 

the angle of the sector, where a is the angle and a the radius. 

13. Find the center of gravity of the area bounded by the cardioid 



14. Find the center of gravity of the area bounded by the curve 
r = 2 cos & + 3. 

15. Find the center of gravity of a solid hemisphere. 

16. Find the center of gravity of a solid formed by revolving 
about its altitude a parabolic segment of base 26 and altitude a. 

17. Find the center of gravity of the solid formed by revolving 
about OF the plane figure bounded by the parabola y*** fee, the axis 
of y, and the line y = k. 

18. Find the center of gravity of the solid bounded by the sur- 
faces of a right circular cone and a hemisphere of radius & with the 
base of the cone coinciding with the base of the hemisphere and the 
vertex of the cone in the surface of the hemisphere 



CENTER OF GRAVITY 255 

86. Center of gravity of a composite area. In finding the cen- 
ter of gravity of a body which is composed of several parts the 
following theorem is useful: 

If a body of mass M is composed of several parts of masses 
M^ Mf - , M n , and if the centers of gravity of these parts are 
respectively (x f ^), (z a , / ? ), .,(# w , #), then the center of grav- 
ity of the composite body is yiven by the formulas 

+ J 

C ; 



We shall prove the theorem for the % coordinate. The proof 
for y is the same. 

By 85 we have, for the original body, 



MX 



- fxdm, ' (2) 



where the integration is to be taken over all the partial masses 
M v Jf a , , M n into which the body is divided. But we have also 



where the subscripts indicate that the integration in each case 
is restricted to one of the several bodies. But formula (2) can 
be written r C C 

MX = / x^m^ I x^dm^-}- + / ^dm n ; 

and, by substitution from (3), the theorem is proved, 

x. Find the center of gravity of an area bounded by two circles one 
of which is completely inside the other. 

Let the two circles be placed aa in Fig. 105, where the center of the 
larger circle of radius a is at the origin, and the center of the smaller 
circle of radius o is on the axis of $ at a distance o from the origin* 



256 



REPEATED 1NTE( 1 RAT U )K 



The area which can be considered as composed of two ]>jut.n i that of 
the larger circle, the two paits being, fust, the smaller eirde and, wroucl, UN- 
irregular ring whose centei of gravity 

we wish to find Now the center of 
gravity of a circle is known to be at 
its center, Theiefore, in the formula 
of the theoiem, we know (T, J/), which 
is on the left of the equation, to bo 
(0, 0), and (~c v yj to be (c, 0), and wish 
to find (7 2 , T/ z ). 

Since we are dealing with aieas, 
we take the masses to be equal to the 
ai eas, and have, accordingly, M = ira z 
(the mass of the larger ciiclc), M = irb z 
(the mass of tho smallei cncle), and 
My, = if ( a ft a ) (the mass of tho ring). 
Substituting in the formula, we havo 




jor> 



= IT/A- + TT (<r - A- 1 ) r a ; 



whence, by solving for ,r a 



It is unnecessary to find y i} since, by Hyiniuctiy, Mi coutiM of 
lies on OX 

EXERCISES 

1. Show that if there are only two component masses fl/^ mul. JHf^ 
in formulas (1) of the theorem, the center of gravity of tho oomjxmto 
mass lies on the line connecting the centers of gravity of tho com- 
ponent masses at such a point as to divide that hue into sogmtmlH 
inversely proportional to the masses. 

2. Prove that if a mass M^ -with center of gravity (.r^ //j) haw <m(, 
out of it a mass M z with center of gravity (i? 9 , y a ), tho c-oiiUu 1 of gravity 
of the remaining mass is 



Mi-Mt 



J/ t - M t 



and r s aro tangtnit oxtonuJly. Find 



3. Two circles of radii 
their center of gravity 

4. Find the center of gravity of a hemispherical hol3 boundod 
by two concentnc hemispheres of radii r^ and r^ 

5. Place r 9 = ^ + A^ m Ex. 4, let A?' approach xero, and thus iind 
the center of gravity of a hemispherical surface. 



CENTER OF GRAVITY 25Y 

6. Find the center of gravity of a hollow right circular cone 
bounded by two parallel conical surfaces of altitudes 7^ and A 2 
respectively and with their bases in the same plane. 

7. Place \= /^-H A/i in Ex 6, let A/t approach zero, and thus 
find the center of gravity of a conical surface 

8. Find the center of gravity of a carpenter's square each aim 
o which is 15 in on its outer edge and 2 in wide 

9. From a square of edge 8 in a quadrant of a circle is cut out, 
the center of the quadrant being at a corner of the squaie and the 
radius of the quadrant being 4 in. Find the center of gravity of the 
figure remaining 

10. Two iron balls of radius 4m and 6 in respectively are con- 
no uted by an iron rod of length 1 in. Assuming that the rod is a 
cylinder o radius 1 in., find the center of gravity of the system. 

11. A cubical pedestal of side 4 ft. is surmounted by a sphere of 
radius 2 ft Find the center of giavity of the system, assuming 
that the sphere rests on the middle point of the top of the pedestal. 

87. Theorems. The folio wing theorems involving the center 
oi' gravity may often ho used to ad- Q 

vantage in finding pressures, volumes x 

c>I solids of revolution, or areas of 
surfaces of revolution. 

I. The total pressure on a plane sur- 
faae immersed in liquid in a vertical 
position is equal to the area of the sur- 
face multiplied ly tlw pressure at its 
center of gravity. 

Let the area he placed a in Fig 106, ETO< 106 

where the axis of K is in the surface 

of the liquid and where the axis of y is measured downward. 
Then, by 25, - 

/* I yO p a- fl! i)^> '*' 

which may be written IIH a double integral in the form 

p = CCwydydx => w nydxdy, (2) 



258 BEPEATED INTEGRATION 

In fact, this may be written down directly, since the pressure on 
a small rectangle dxdy is its area, dxdy, times its depth, T/, times w. 
Moreover, from 85, we have 



Ay 



--CCydxdy. (3) 



By comparison of (2) and (3) we have 

P = wyA. 

But wyis the pressure at the center of gravity, and the theorem 
is proved for areas of the above general shape. If the area is 
not of this shape, it may be divided into such areas, and the 
theorem may be proved with the aid 
of the theorem of 86. 

Ex.1. A circular bulkhead which closes y="b 
the outlet of a reservoir has a radius 8 ft , 
and its center is 12 ft below the surface of 
the water. Find the total pressure on it 

Here A = 9 it and the depth of the center 
of gravity is 12. Therefore 

P = 108 irw = -V-Trtons = 10.6 tons. 

II. The volume generated by revolving 
a plane area about an axis in its plane 
not intersecting the area is equal to the area of the figure multiplied 
by the circumference of the circle described by its center of gravity. 

To prove this take an area as in Fig. 107. Then, by 26, if 
V is the volume generated by the revolution about OY, 




(4) 
which can be written as a double integral in the form 

F=27rf C\dydx. (5) 

t/ i/, 

By 85, J&= C C\dydx; 

Ja u/jTj 

and, by comparison of (4) and (5), we have 

V~ 
which was to be proved. 



THEOREMS OF PAPPUS 259 

Ex. 2. Find the volume of the nag surface formed by revolving about 
an axis in its plane a circle of radius a whose center is at a distance c from 
the axis, where c > a. 

We know that A = -sra 2 and that the center of gravity of the circle is 
at the center of the circle and therefore describes a circumference of length 
2 ire. Therefoie F = 2ir 2 ffl 2c. 

III. The area generated by revolving a plane curve about an 
axis in its plane not intersecting the curve is equal to the length 
of the curve multiplied by the circumference of the circle described 
by its center of gravity. 

To prove this we need a formula for the area of a surface of 
revolution which has not been given. It may be shown that if 

S is this area, then r 

S^Zirlxds. (6) 

A rigorous proof of this will not be given here. However, the 
- student may make the formula seem plausible by noticing that 
an element ds of the curve will generate on the surface a belt 
of width ds and length 2 irx. The product of length by breadth 
may be taken as the area of the belt. 
Moreover, by 85, we have 

sx I xds; (7) 

and comparing the two equations (6) and (7), we have 



which was to be proved. 

Ex. 3. Find the area of the ring surface described in Ex. 2. 

We know that s - 2 TTO and that the center of gravity of a circumfer- 
ence is at its center and therefore describes a circumference of length 2 ire 
Therefore 5 = 4 ^ac. 

Theorems II and III are known as the theorems of Pappus. 

EXERCISES 

1. Find by the theorems of Pappus the volume and the surface 
of a sphere. 

2, Find by the theorems of Pappus the volume and the laterax 
surface of a right circular cone. 



260 .REPEATED INTEGRATION 

3. Find by the theorems of Pappus the volume generated by 
revolving a parabolic segment about its altitude. 

4. Find by the theorems of Pappus Iho volume generated 
by revolving a parabolic segment about its base. 

5. Find by the theorems of Pappus the volume geru>ratod by 
revolving a parabolic segment about the tangent at its vortex 

6. Find the volume and the surface generated by revolving a 
square of side a about an axis in its piano perpendicular to one. of 

its diagonals and at a distance b\b> ;= } from it,s center. 

V V2/ 

7. Find the volume and the area generated by revolving' a right 
triangle with legs a and b about an axis in its plane ]>urullel to the 
leg of length a on the opposite side from the hypotenuse and at a 
distance c from, the vertex of the right angle. 

8. A circular water main has a diameter of 4 ft. One end IH 
closed by a bulkhead, and the other is connected with a reservoir 
in which the surface of the water is 18 ft. above the center of the 
bulkhead Find the pressure on the bulkhead. 

9. Find the pressure on an ellipse of semiaxos a. and I completely 
submerged, if the center of the ellipse is o units below the surface of 
the liquid. 

10. Find the pressure on a semiellipse of semiaxes a and I (a > />) 
submerged with the major axis in the surface of the liquid and the 
minor axis vertical. 

11. Find the pressure on a parabolic segment submerged with the 
base horizontal, the axis vertical, the vertex above the base, and the 
vertex c units below the surface of the liquid. 

^ 12. What is the effect on the pressure of a submerged vertical area 
in a reservoir if the level of the water in the reservoir is raised by c feet ? 

88. Moment of inertia. The moment of inertia of a particle 
about an axis is the product of its mass and the square of its 
distance from the axis. The moment of inertia of a number 
of particles about the same axis is the sum of the moments of 
inertia of the separate particles abont that axis. From these 
definitions we may derive the moment of inertia of a thin plate. 

Let the surface of the plate be divided into elements o 
area cU Then the mass of each element is pdA, where p is the 
product of the thickness of the plate and ita density. Lot M bo 



MOMENT OF INEETIA 261 

the distance of any point in the element from the axis about 
which we wish the moment of inertia. Then the moment of 
inertia of clement is approximately 

I?p dA. 

We say " approximately " because not all points of the element 
are exactly a distance R from the axis, as R is the distance ot 
some one point in the element. However, the smaller the ele- 
ment the more nearly can it be regarded as concentrated at one 
point and the limit of the sum of all the elements, as their 
size approaches zero and their number increases without limit, 
is the moment of inertia of the plate. Hence, if I represents the 
moment of inertia of the plate, we have 



= f 



(1) 

If in (1) we let /o = l, the resulting equation is 

1= C&dA, (2) 



where I is called the moment of inertia of the plane area. When 
dA in (1) or (2) is replaced by dxdy or rdrdd, the double 
sign of integration must be used. 

Ex. 1. Find the moment of inertia of a rectangle of dimensions a and 6 
about the side of length & 

Let the rectangle be placed as in Fig. 108. Let it be divided up into 
elements dA ~ dfdg. Then a- is the distance of some point in an element 
from OY. Hence, in (2), we have y 
JR SB x and dA = fatly. Therefore 

y=b 



We first sum the rectangles in 
a vertical strip, w y ranges from 
to &. We have 



This is the moment of ineitia of j. IG< IQ% 

the strip MN, and might -have been 

written down at once, since all points on the left-hand boundary of the 
strip are at a distance x from OY and since the area of the strip is Idx 



262 KEPEATED INTEGRATION 

The second integration gives now 

Jo 

If, instead of asking for the moment of inertia of the area, we had asked 
for that of a plate of metal of thickness t and density D, the above result 
would be multiplied by p = Dt. But m that case the total mass M of the 

plate is aab, so that we have 

I \ Ma? 

Ex. 2. Find the moment of inertia of the quadrant of an ellipse 

r 2 7/ 2 

}. >L 1 (a > 6) about its major axis 

If we take any element of area as dxtly, we find the distance of its 
lower edge from the axis about which we wish the moment ot ineitia 
to be y (Fig 109). Hence R = y and ^ 

I 

It will now be convenient to sum first 
with respect to x t since each point of a 
horizontal strip is at the same distance from 
OX "We therefore write 

J-ffffy**. FIG. 109 

Now, indicating by a^ the abscissa of a point on the ellipse to distinguish 
it from the general x which is that of a point inside the ellipse, we have 
for the first integration 





For the second integration 

/ 

To integrate, place y = I sin <j> Then 



TT 

= afc 8 f * s 
/o 



16 



If, instead of the area, we consider a thin plate of mass M, the above 
result must be multiplied by p, where M = irabp , whence 



The polar moment of inertia of a plane area is defined as the 
moment of inertia of the area about an axis perpendicular to 
its plane. This may also be called conveniently the moment 



MOMENT OF INERTIA 263 

of inertia with respect to the point in which the axis cuts the 
plane of the area, for the distance of an element from the axis 
is simply its distance from that point. Thus we may speak, 
for example, of the polar moment of inertia with respect to 
an axis through the origin perpendicular to the plane of au 
area, or, more concisely, of the polar moment with respect to 
the origin. 

If the area is divided into elements dxdy, and one point in 
the element has the coordinates (#, #), the distance of that 
point from the origin is Vi?+]A That is, in (2), if we place 
(lA dxvly and J2 a =o; 2 +y 2 , we shall have the formula for the 
polar moment of inertia with respect to the origin. Denoting 
this by J , we have 

(3) 



This integral may be split up into two integrals, giving 

(4) 



where the change in the order of the differentials in the two 
integrals indicates the order in which the integration may be 
most conveniently carried out. 

The first integral in (4) is the moment of inertia about OY 
and may be denoted by / ; the second integral is the moment 
of inertia about OX and may be denoted by I K . Therefore 
formula (4) may be written as 



so that the problem of finding the moment of inertia may be 
reduced to the solving of two problems of the type of the 
first part of this section. 

Ex, 3. Find the polar moment of inertia of an ellipse with respect to 
the origin, 

In Ex. 2 we found I K for a quadrant of the ellipse. For the entire 
ellipse it is four times as great, since moments of inertia are added by 
definition. Hence 



264 REPEATED INTEGEATION 



By a similar calculation I y - 
Theiefoie ^0=4 ffa * ( 2 + &*) 

If the area is leplaced by a plate of mass Af, this result gives 
/ = \. M(a* + I") 

If polar coordinates aie used, the element of area is rd6d 
and the distance of a point in an element from the origin is 
Hence, in (2), dA = rdBdr and R = r. Therefore 

r- (6 

In practice it is usually convenient to integrate first wit 
respect to r, holding 8 constant. This is, in fact, to find tr 
polar moment of inertia of a sector with vertex at 0. 

Ex.4. Find the polar moment of inertia of a cncle with lespect to 
point on its circtimference 

Let the circle be placed as in Fig 110 Its equation is then (Ex 1, 5] 
r = 2 a cos 0, where a is the radius If we take any element rdQdr an 
find I for all elements which lie in the 
same sector with it, we have to add the 
elements i*dQdr, with r ranging from to 
t v where r t is the value of ? on the cncle , 
and therefore > l = 2 a cos We have 



f Vrffltft = i 7 * (IB = 4 a 4 

/o 1 

We have finally to .rm these quantities, 
with 1 anging from to + - We have 

** * 




I =J*la*cQ&6d6= I 

~I TIG. 110 

If M is the mass of a ciicular plate, this result, multiplied by p, 



Ex. 5. Find the polar moment of inertia of a cucle with respect to i 
centei. 

Heie it will be convenient first to find the polar moment of inertia of 
ring (Fig 111). We integrate first with respect to 6, keeping r constan 
We have 



MOMENT OF INERTIA 265 

which is the approximate area of the ring Swrdr multiplied by the 
square of the distance of its inner cucumference from the center We 
then have, by the second integration, ^ 



= f a 
Jo 



If M is the mass of a circular plate, this 
result, multiplied by p, gives 

I / / f\ S \ U \\ \ 

X 




The moment oE inertia of a solid 
of revolution about the axis of revo- 
lution is the sum of the moments of 
inertia of the circular sections about 
the same axis ; that is, of the polar ]FIG< m 

moments of inertia of the circular sections about their centers. 
If the axis of revolution is OY, the radius of any circular 
section perpendicular to OF is x and its thickness is dy. Its 
mass is therefore pirx*dy ; and therefore, by Ex. 5, its moment 
of inertia about Y is \ pirtfdy. The total moment of inertia of 
the solid is therefore 

X\pTr \ x*dy. 

Ex. 6. Find the moment of inertia of a circular cone about its axis. 
Take the cone as in Ex. 4, 85. Then we have 



But, if M is the mass of the cone, we have M = 
Therefore / = fa Ma*. 

EXERCISES 

1. Find the moment of inertia of a rectangle of base & and alti- 
tude a about a line through its center and parallel to its base. 

2. Find the moment of inertia of a triangle of base b and altitude 
a about a line through its vertex and parallel to its base. 

3. Find the moment of inertia of a triangle of base & and alti- 
tude a about its base. 

4. Find the moment of inertia of an ellipse about its minor axip 
and also about its major axis, 



266 REPEATED INTEGRATION 

5. Find the moment of inertia of a trapezoid about its lower base, 
taking the lower base as b, the upper base as a, and the altitude as 7*. 

6. Find the moment of inertia about its base of a parabolic seg- 
ment of base b and altitude a. 

7. Find the polar moment of inertia of a rectangle of base I and 
altitude a about its center. 

8. Find the polar moment of inertia about its center of a circular 
ring, the outer radius being r z and the inner radius r r 

9. Find the polar moment of inertia of a right triangle of sides 
a and b about the vertex of the right angle. 

10. Find the polar moment of inertia about the origin of the area 
bounded by the hyperbola xy=6 and the straight line aj+y 7=0. 

11. Find the polar moment of inertia about the origin o the area 
bounded by the curves y=x z and y = 2 as 2 . 

12. Find the polar moment of inertia about the origin of the area 
of one loop of the lemniscate r 2 = 2a? cos 2 0. 

13. Find the moment of inertia of a right circular cylinder of 
height Ji, radius r, and mass M, about its axis. 

14. Find the moment of inertia about its axis of a hollow right cir- 
cular cylinder of mass M, its inner radius being r v its outer radius r a , 
and its height h. 

15. Find the moment of inertia of a solid sphere about a diameter. 

16. A ring is cut from a spherical shell whose inner and outer radii 
are respectively 5 ft. and 6 f t , by two parallel planes on the same 
side of the center and distant 1 ft and 3 ft respectively from the 
center Find the moment of inertia of this ring about its axis 

17. The radius of the upper base and the radius of the lower base 
of the frustum of a right circular cone are respectively r^ and r# and 
its mass is M. Find its moment of inertia about its axis. 

89. Moments of inertia about parallel axes. The finding of a 
moment of inertia is often simplified by use of the following 
theorem ; 

The moment of inertia of a body about an oasis is equal to its 
moment of inertia about a parallel axis through its center of gravity 
plus the product of the mass of the body ly the square of the 
distance between the axes. 




MOMENT OF INEKTIA 267 

We shall prove this theorem only for a plane area, in the 
two cases in winch the axes lie in the plane of the figure or 
are perpendicular to that plane. We shall also consider the 
mass of the area as equal to the v 

KT * 

area, as in 88. 

Case I. Wlien the axes lie in tJie 
plane of the Jiyure. 

Let the area be placed as in 
Fig. 112, where the center of grav- 
ity (^ #0 lf} taken as the origin 
(0, 0) and where the axis of y is 
taken parallel to the axis LK, L 
about which we wish to find the 

moment of inertia. Let x be the distance of an element dxdy 
from OF, and ^ its distance from LK. Then, if I g is the 
moment of inertia about OF, and J z the moment of inertia about 
LK, we have ^^ 

1=11 x*dxdy, It i I xldxdy. (1) 

JJ JJ 

Moreover, if a is the distance between OF and LK, we have 

so that, by substituting from (2) in the second equation of (1), 
we have rf * **. np 

Ji- x*dxdy + 2 a I I xdxdy + a* I dxdy. (3) 

Now, by 84, ndxdy^Ai by 85, \( xdxdy = Ax = 0, since 

by hypothesis x = ; and, by (1), the first integral on the right 
hand of (8) is I g . Therefore (8) can be written 

7,-7,+ oU, (4) 

which proves the theorem for this case. 

Case II. When the axes are perpendicular to the plane of 
the figure. 

We have to do now with polar moments of inertia. Let the 
area be placed as in Fig. 118, where the center of gravity is 



268 REPEATED INTEGRATION 

taken as the origin, and P is any point about which we wish 

the polar moment of inertia. Let I g be the polar moment of 

inertia about 0, and I p the polar moment of inertia about P. 

Draw through P axes PX' and , 

PY 1 parallel to the axes of coor- 

dinates OX and OY. Let I x and 

I tl be the moments of inertia 

about OX and OY respectively, 

and let 1^ and J tf / be the moments 

of inertia about PX' and PY'. 

Then, by (5), 88, / / 




J 
J.-- 

Moreover, if (a, 6) are the coordinates of P, we have, by 
Case I, 1^ = 1^ a *A, /,=/ + VA. (6) 

Therefore, from (5), we have 

/,==/,+ <V+&V> (7) 

which proves the theorem for this case also. 

The student may easily prove that the theorem is true also 
for the moment of inertia of any solid of revolution about an 
axis parallel to the axis of revolution of the solid. 

Ex. Find the polar moment of inertia of a circle with respect to a point 
on the circumfeience. 

The center of gravity of a circle is at its center, and the distance of any 
point on its circumference from its center is a. By Ex 5, 88, the polar 
moment of a circle about its center is |ira*. Therefore, by the above 
theorem, I, = i* + o (F) t *. 

This result agrees with Ex 4, 88, where the required moment of inertia 
was found directly 

EXERCISES 

1. Find the moment of inertia of a circle about a tangent. 

2. Find the polar moment of inertia about an outer corner of 
a picture frame bounded by two rectangles, the outer one being of 
dimensions 8 ft. by 12 ft, and the inner one of dimensions 5 ft. by 9 ft. 



SPACE COORDINATES 269 

3. Find the moment of inertia about one of its outer edges of a 
carpenter's square of which the outer edges are 15 in. and the inner 
edges 13 in 

4. Find the polar moment of inertia about the outer corner of 
the carpenter's square in Ex. 3. 

5. From a square of side 20 a circular hole of radius 5 is cut, 
the center of the circle being at the center of the square. Find the 
moment of inertia of the resulting figure about a side of the square. 

6. Find the polar moment of inertia about a corner of the square 
of the figure in Ex. 5. 

7. Find the moment of inertia of a hollow cylindrical column 
of outer radius r a and inner radius 1\ about an element of the inner 
cylinder. 

8. Find the moment of inertia of the hollow column of Ex. 7 
about an element of the outer cylinder. 

9. Find the moment of inertia of a circular ring of inner radius ^ 
and outer radius ?- a about a tangent to the outer circle. 

10. A circle of radius a has cut from it a circle of radius 75 tangent 

i 

to the larger circle Find the moment of inertia of the remaining 
figure about the line through the centers of the two circles 

11. Find the moment of inertia of the figure in Ex 10 about a 
line through the center of the larger circle perpendicular to the line 
of centers of the two circles and in the plane of the circles. 

90. Space coordinates. In the preceding pages we have be- 
come familiar with two methods of fixing the position of a 
point in a plane ; namely, by Cartesian coordinates (x, y), and 
by polar coordinates (r, 0). If, now, any plane has been thus 
supplied with a coordinate system, and, starting from a point 
in that plane, we measure another distance, called 0, at right 
angles to the plane, we can reach any point in space. The quan- 
tity will be considered positive if measured in one direction, 
and negative if measured in the other. We have, accordingly, 
two systems of space coordinates. 

1. Cartesian coordinates. We take any plane, as -3T0F, in which 
are already drawn a pair of coordinate axes, OX and OF, at 
right angles with each other, Perpendicular to this plane at 



2TO 



BEPEATED INTEOKATION 



z 




the origin we draw a third axis OZ (Fig. 114). If P is any 
point of space, we draw PM parallel to OZ, meeting the plane 
XO Y at Jf, and from M draw a line par- 
allel to OF, meeting OX at L. Then for 
the point P (#, y, z), OL x, LM y, 
and MP = z. It is to be noticed that 
the three axes determine three planes, 
XOY, YOZ, and ZOX, called the coor- 
dinate planes, and that we may just as 
readily draw the line from P perpendic- 
ular to either the plane YOZ or ZOX and 
then complete the construction as above. 

These possibilities are shown in Fig. 115, where it is seen that 
x = OL = NM= SE = TP, with similar sets of values for y and s. 

2. Cylindrical coordinates. Let XOY 
be any plane in which a fixed point 
is the origin of a system of polar coor- 
dinates, and OX is the initial line of that 
system (Fig. 116). Let OZ be an axis 
perpendicular to the plane XOY at 0. 
If P is any point in space, we draw 
from P a straight line parallel to OZ 
until it meets the plane XOY at M. y 
Then, if the polar coordinates of M in 
the plane XOY are r = OM, 6 = XOM, and we denote the dis- 
tance MP by s, the cylindrical coordinates of P are (r, 0, g). 

It is evident that the axes OX and 
OZ determine a fixed plane, and that 
the angle 6 is the plane angle of the 
dihedral angle between that fixed plane 
and the plane through OZ and the 
point P. If SP is drawn in the latter 
plane perpendicular to OZ, it is evident 
that OM= SP = r and OS = MP = z, 
The coordinate r, therefore, measures 
the distance of the point P from the axis OZ, and the coordinate 
2 measures the distance of P from the plane 




Z 




116 



SUEFACES 



271 



If the line OX of the cylindrical coordinates is the same as 
the axis OX of the Cartesian coordinates, and the axis OZ is the 
same in both systems, it is evident, from 51, that 

a; = r cos 0, y == r sin 0, = z. (1) 

These are formulas by winch we may pass from one system 
to the other. It is convenient to notice especially that 

(2) 



91. Certain surfaces. A single equation between the coordi- 
nates of a point in space represents a surface. We shall give 
examples of the equations of certain surfaces which are impor- 
tant in applications. In this connection it should be noticed 
that when we speak of the equation of a sphere we mean the 
equation of a spherical surface, and when we speak of the 
volume of a sphere wo mean the volume of the solid bounded 
by a spherical surface. The word sphere, then, indicates a sur- 
face or a solid, according to the context. Similarly, the word 
cone is used to denote either a conical surface indefinite in 
extent or a solid bounded by a conical surface and a plane 
base. It is in the former sense that we speak of the equation 
of a cone, and in the latter sense 
that we speak of the volume of 
a cone. In the same way the 
word cylinder may denote either 
a cylindrical surface or a solid 
bounded by a cylindrical surface 
and two piano bases. This double 
use of these words makes no con- 
fusion in practice, as the context 
always indicates the proper mean- 
ing in any particular case. 

1. Sphere will center at origin. Con- 
sider any sphere (Fig. 117) with its center 
at the origin of coordinates and its radius 
equal to a. Let P be any point on the surface of the sphere. Pass a plane 
through P and OZ, draw PS perpendicular to OZ, and connect and P. 
Then, using cylindrical co6rdinates, in the right triangle OPS, OS=z, SP^r, 
and OP as a, Therefore . 8s= a 9 . (1) 




FIG. 117 



272 



REPEATED INTEGRATION 



This equation is satisfied by the cyhnducal coordinates of any. point on 
the surface of the sphere and by those of no other point Jt is therefore 
the equation of the sphere in cylindrical 
coordinates. 

By means of (2), 90, equation (1) 

*"** *+? + *,* (>) 

which is the equation of the sphere in 
Cartesian coordinates. 

2. Sphere tangent at origin to a cooi 1 - 
dinate plane. Consider a sphere tangent 
to the plane XOY at (Fig 118). Let 
P be any point on the surface of the 
sphere Let A be the point in which the 
axis OZ again meets the sphere Pass a 
plane through P and OZ , connect A. and 
P, and P , and diawPS perpendicular 
to OZ Then, using cylindrical coordi- 
nates, OS = z, SP = i, and OA = 2 a, 
where a is the radius of the sphere 

Now OA P is a right triangle, since it is inscribed in a semicircle, 
and PS is the perpendicular from the vertex of the right angle to the 
hypotenuse. Therefore, by elementary plane geometry, 

~SP*=OS' SA = OS(OA - 08) 
Substituting the proper values, we have 




X 



FIG. 118 



Z 



(3) 

which is the equation, of the sphere in cylindrical 
coordinates. 

B J ( 2 ) 90 > equation (3) becomes 

# a + f y a +2 2 2 a? = 0, (4) 

which is the equation of the sphere in Cartesian 
coordinates. 

3 Right circular cone. Consider any right circular 
cone with its vertex at the origin and its axis along 
OZ (Fig 119). Let a be the angle which each element 
of the cone makes with OZ. Take P any point on the surface of the cone, 
pass a plane through P and OZ, and draw PS perpendicular to OZ. Then 

SP 
SP = r and OS = z. But = tan SOP - tan a Therefore we have 

C/o 

r = gtano (5\ 

\ V J 

as the equation of the cone in cylindrical coordinates, 




FIG. 119 



SURFACES 



273 



By 2, 90, equation (.">) becomes 




0, (6) 

as the equation of the cone in Cartesian coordinates, 

As explained above, we have heie used the word cone in the sense of a 
conical surface If the cone is a solid with its altitude h and the radius of 

its base a, then tan a - In this case equation % 

li 

(5) or (6) is that of the curved surface of the 
cone only. 

4. Surface of revolution. Consider any sur- 
face of revolution with OZ the axis of revolution 
(Fig 120). Take P any point on tho smface 
and pass a plane through P and OZ In the 
piano POZ draw OR peipendicular to OZ and, 
fiom P, a straight line perpendicular to OZ 
meeting OZ in S If we regard OR and OZ as 
a pair of i octangular axes foi the plane POZ, 
tho equation of the cuive CD in which the 

plane POZ cuts the surface is FIG 120 

=/(r) f (7) 

exactly as y ~f(x) is the equation of a curve in 12 

But CD is the same curve in all sections of the surface through OZ 
Therefore equation (7) is true for all points P and is tho equation of the sur- 
face in cylindrical coordinates. Whan the plane POZ coincides with the 
plane X OZ t r is equal to x, and equation (7) becomes, z 

for that section, z= .ff y .\ > /-g\ 

Hence we have the following theorem : 

The equation of a surface of revolution formed by 
revolving about OZ any curve in the plane XOZ may 
be found in cylindrical coordinates by writing rfor x 
in the equation of the curve. 

Tho equation of the surface in Cartesian coor- 
dinates may thtn be found by placing r = Vr* + y*. 
For example, the equation of the surface formed 
by revolving the parabola 2 =4o: about OZ as 
an axis is z a as 4 ? in cylindrical co&rdinates, or 
s*sBlO(.e fl + y 8 ) in Cartesian coordinates. 

5. Cylinder. Consider first a right circular cylinder with its axis along 
OZ (Fig. 121). From any point P of the surface of the cylinder draw 7 J S 
perpendicular to OZ. Then 8P is always equal to , the radius of the 
cylinder. Therefore, for all points on the surface, 

reo, (9) 




121 



274 



REPEATED INTEGRATION 



which is the equation of the cylinder in cylindrical cobrdinates. Reduced 
to Cartesian coordinates equation (9) becomes 

x*+y*=a?, (10) 

the equation of the cylinder in Cartesian coordinates. 

More generally, any equation in x and y only, or in r and 6 only, repre- 
sents a cylinder. In fact, either of these equations, if interpreted in tho 
plane XOY, represents a curve, but if a line is drawn from any point in 
this curve perpendicular to the plane XOY, and P is any point on tins lino, 
the coordinates of P also satisfy the equation, since z is not involved in 
the equation. As examples, the equation y a =4:x xepresputs a parabolic 
cylinder, and the equation r = a sin 3 & represents a cylinder whoso base is 
a rose of three leaves (Fig. 65, p. 144). 

6 Ellipsoid Consider the surface defined by the equation 



c 2 



(11) 



If we place z = 0, we get the points on the surface which Ho in tho XO I' 
plane These points satisfy the 
equation . * 

- + *= = ! (12) _.^ 



and therefore form an ellipse. 

Similarly, the points in the 
ZOX plane lie on the ellipse 

S + 7 3 = 1 > (13) 

and those in the YOZ plane lie 
on the ellipse 



6 s c 2 



(14) 




FIG. 122 



The construction of these 
ellipses gives a general idea of 
the shape of the surface (Fig 122). To make this more precise, let UH 
place * = j in (11), where z x is a fixed value. We have 



which can be written 



c 2 



==1, 



(16) 



SURFACES 275 

As long as z < c 3 , equation (16) represents an ellipse with semiaxes 



a v'l " and b A/1 ^ By taking a sufficient number of these sections 

we may construct the ellipsoid with as much exactness as desired. 

If z = c 8 in (16), the axes of the ellipse reduce to zero, and we have a 
point If Sj a > r a , the axes are imaginaiy, and there is no section. 

1 Elliptic paraboloid. Consider the surface 



where we shall assume, foi defimteness, that c is positive. 
If we place z = 0, we get 2 

- 2 + f 8 = > 

^ ft A /ift 



(17) 



(18) 



which is satisfied in real quantities only by x and y = 0. Therefore the 
XOY plane simply touches the surface at 

the origin 

If we place z = c, we get the ellipse 



1 

' 



(19) 



which lies in the plane c units distant from 
the XOY plane 

If we place y = 0, we get the parabola 



and if we place a? = 0, we get the parabola 




FIG. 123 



The sections (10), (20), and (21) determine the general outline of 
surface. For more detail we place z = s x and find the ellipse 



c c 

so that all sections parallel to the XOY plane and above it are ellipses 
(Fig. 123). 

8. Elliptic cone. Consider the surface , 



I 3 -!! 



0. 



(23) 



Proceeding as in 7, we find that the section z = is simply the origin 
and that the section, z = cis the ellipse 



276 



REPEATED INTEGRATION 



If we place x 0, we get the two straight lines 

i =-t*- 

and if we place y = 0, we get the two stiaight lines 

a = -z 
c 



(25) 



(20) 



The sections we have found suggest a cone with an elliptic baso. To 
prove that the surface really is a cone, we change equation (SJiJ) to cylin- 
cotirdmates, obtaining 

^ + E^\ r a = ~- nm 

2 /i2 / /(2 V <* 

B I/ / C 

Now if 6 is held constant in (27), the coefficient of r z is constant, and thi 

equation may be written T 

1 J r = kz, (28) 

which is the equation of two straight hues in the plane through OK 
Z Z 




FIG 124 




determined by & = const Hence any plane trough OZ cuts the surface 
two straight lines, and the surface is a cone (Fig. 124). 
9. Plane Consider the surface 



n 



Ax + ED + Cz -f D = 0. /29\ 

The section z = is the straight hne ffff (P lg 1 25 ) with the equation 
Ax + Vy + DssQ, (80) 

the secfcon y = is the straight line LH with the equation 

At + a+DszO, (31) 

and the section , = ia the straight hne L/C with the equation 
By + Cz + T) = 0. 



VOLUME 277 

The two lines (31) and (32) intersect OZ in the point L (O, 0, J , 

unless C = 0. Assuming for the present that C is not zero, we change 
equation (29) to cyhndncal cooidmates, obtaining 

(.4 cos 6 + B sin d)r + Cz + D = 0. (33) 

This is the equation of a stiaight line LN in the plane B = const It 
passes through the point L, which has the cyhndncal coordinates r = 0, 

2 _ ~i ; and it meets the line KH, since when z = 0, equation (33) is the 



same as equation (30). Hence the sui face is covei ed by straight lines which 

pass through L and meet KH. The locus of such lines is clearly a plane 

We have assumed that C in (29) is not zero. If C = 0, equation (29) is 

Q. (34) 



The point L does not exist, since the lines corresponding to HL and KL 
are now parallel. But, by 5, equation (34) lepresents a plane parallel to 
OZ intersecting XOYm the line whose equation is (34) Theiefoie we 
have the following theorem 

Any equation of the first degree represents a plane. 

92. Volume. Starting from any point (#, y, z) in space, we 
may draw linos of length dx, dy, and dz in directions parallel 
to OX, OF, and OZ respectively, and on these lines as edges 
construct a rectangular parallelepiped. The volume of this fig- 
ure we call the element of volume dV and have 

(1) 



For cylindrical coordinates we construct an element of volume 
whose base is rd6dr( 84), the element of plane area in polar 
coordinates, and whose altitude is dz. This figure has for its 
volume dV the product of its base by its altitude, and we have 

d7=rd0drdz. ( 2 ) 

The two elements of volume dV given in (1) and (2) are 
not equal to each other, since they refer to differently shaped 
figures. Each is to be used in its appropriate place. To find 
the volume of any solid we divide it into elements of one of 

these types. , 

To do this in Cartesian coordinates, note that the a-coordinate 
of any point will determine a plane parallel to the plane YOZ 



278 



REPEATED INTEGRATION 




and x units from it, and that similar planes correspond to the 

values of y and & We may, accordingly, divide any ruquirotl 

volume into elements of volume as follows: 
Pass planes through the volume parallel to Y0% ami <ta* 

units apart. The result is to divide the required volume into 

slices of thickness dx, one of which # 

is shown in Fig. 126. Secondly, 

pass planes through the volume 

parallel to JTO^and dy units apart, 

with the result that each slice is 

divided into columns of cross sec- 
tion dxdy. One such column is 

shown in Fig. 126. 
Finally, pass planes through the 

required volume parallel to XQY 

and dz units apart, with the result 

that each column is divided into 

rectangular parallelepipeds of dimensions das, dy, and fe, One 

of these is shown in Fig. 126. 
It is to be noted that the order followed in the above 

explanation is not fixed and that, in fact, the choice of be- 
ginning with either a or y or 4 and the subsequent order 

depend upon the particular volume z 

considered. 

A similar construction may be 
made for cylindrical coordinates. 
In this case the coordinate & 
determines a plane through OZ. 
We accordingly divide the volume 
by means of planes through OZ 
making the angle d0 with each 
other. The result is a set of slices 
one of which is shown in Fig. 127 Fro. 127 




VOLUME 279 

Finally, these columns are divided into elements of volume 
by planes parallel to XOY at a distance dz apart. One such 
element is shown in Fig. 127. 

When the volume has been divided in either of these ways, 
it is evident that some of the elements will extend outside the 
boundary surfaces of the solid. The sum of all the elements 
that are either completely or partially in the volume will be 
approximately the volume of the solid, and this approximation 
becomes better as the size of each element becomes smaller. 
In fact, the volume is the limit of the sum of the elements. 
The determination of this limit involves in principle three in- 
tegrations, and we write 



= I | | dxdyds 



(3) 
or V= CCCrd6drdz. (4) 

In carrying out the integrations we may, in some cases, find 
it convenient first to hold z and dz constant. We shall then 
be taking the limit of the sum of the elements which lie in a 
plane parallel to the XOY plane. We may indicate this by 
writing (3) or (4) in the form 

V=* Cdz CCdxdy or V= Cdz CCrdOdr. (5) 

But, by 84, 1 1 dxdy = A and \\rd6dr~A, where A is 

the area of the plane section at a distance from XO Y. Hence 
(5) is r 

F= / Adss, (6) 

in agreement with 26. ** 

Hence, whenever it is possible to find A by elementary means 
without integration, the use of (6) is preferable. This is illus- 
trated in Ex. 1. 

In some cases, however, this method of evaluation, is not 
convenient, and it is necessary to carry out three integrations. 
This is illustrated in Ex. 2. 



280 



REPEATED INTEGRATION 



Ex. 1. Fmd the volume of the ellipsoid + 



=1. 



By 6, 91, the section made by a plane parallel to XOY is an ellipse 

with semiaxes a %/! ^ and Z> %/! - ^ Therefore, by Ex. 1, 77, its area 

/ z a \ c 

is waft (1 1 Hence we use formula (6) and have 

\ c 2 / 

r e l s a \ 4 

V = irab I 1 1 -]ds=:- irdbc. 

J_ c \ c 2 / 3 

Ex. 2. Find the volume bounded above by the sphere a: 2 -f ?/ a + s 2 = 5 and 
below by the paraboloid a; 2 + y z = 4 z (Fig 128). 

As these are stu faces of i evolution, this example may be solved by 
the method of Ex. 1, but in so doing we need two integrations one 
foi the sphere and the other foi the paraboloid We shall solve the 
example, however, by the other method in order to illustrate that method 

We fii st reduce our equations to cyhn- ^ 

dncal coordinates, obtaining lespectively 

r* + z* = 5 (1) 

and r 1 = 4 z (2) 

The surfaces intersect when r has the 
same value in both equations; that is, 

when z* + 4 z = 5, (3) 

which gives z=loiz = 5 The latter 

value is impossible , but when z = 1, we 

have r = 2 in both equations Theiefore 

the surfaces intersect in a circle o radius 2 in the plane z = 1 

lies duecHly above the circle r = 2 in the XOY plane. 

We now imagine the element tdQdrdz inside the surface and, holding 
r, 0, dd, dt constant, we take the sum of all the elements obtained by varying 
z inside the volume These elements obviously extend from 2 = s l in the 

lower boundary to z = z 2 in the upper boundary. From (2), z 1 = and, 
from (1), 2 2 V5 r 8 . The first integration is therefore 




This circle 



rdddr 




MOdr. 



We must now allow 6 and r so to vary as to cover the entire circle ? == 2 
above which the required volume stands. 

' If we hold Q constant, r varies from to 2. The second integration is 
therefore 



VOLUME 281 

Finally, 9 must vary from to 2 IT, and the third integration is 
/5V5 



If we put together what we have done, we have 

/>2ir /.a ft 

F=f / / 

Jo Jo t/, 



EXERCISES 

1. Find the volume bounded by the paraboloid = y? + ^ and 
the planes x = 0, y = 0, and g = 4. 

& C^ 9*2 

2. Prove that the volume bounded by the surface - = -5 4- fa 

and the plane * = c is one half the product of the area of the base 
by the altitude. 

3. Find the volume bounded by the plane = and the cylinders 
SB B -f f = 2 and y* = a* az. 

4. Find the volume cut from the sphere r 2 4- 2 = a 2 by the cylinder 
r = a cos 6. 

5. Find the volume bounded below by the paraboloid r* = a and 
above by the sphere r 2 4- 2 2 a# = 

6. Find the volume bounded by the plane XOY, the cylinder 
a-a {> 2/a _ 2 ax = 0, and the right circular cone having its vertex at 0, 
its axis coincident with OZ, and its vertical angle equal to 90 

7. Find the volume bounded by the surfaces r a = &, = 0, and 

y sa a COS 0, 

8. Find the volume bounded by a sphere of radius a and a right 
circular cone, the axis of the cone coinciding with a diameter of the 
sphere, the vertex being at an end of the diameter, and the vertical 
angle of the cone being 90 . 

9. Find the volume of the sphere of radius a and with its center 
at the origin of coordinates, included in the cylinder having for its 
base one loop of the curve i* ;= a 2 cos 2 6. 

10, Find the volume of the paraboloid a? 4- f = 2 * cut off by the 
plane #~x 4-1. 

11. Find the volume of the solid bounded by the paraboloid 

1 and the plane * = *. 



dm 



282 EEPEATED INTEGRATION 

93. Center of gravity of a solid. The center of gravity of a solid 
has three coordinates, x, % z, which are defined by the equations 

I xdm I ydm I 

*= J ' ' v = J r ' - 

I dm I dm I 

where dm is the mass of one of the elements into which the 
solid may be divided, and #, y, and a are the coordinates of the 
point at which the element dm may be regarded as concentrated. 
The derivation of these formulas is the same as that in 85 
and is left to the student. 

When dm is expressed in terms oi space coordinates, the 
integrals become triple integrals, and the limits 'of integration 
are to be substituted so as to include the whole solid. 

We place dm = pdF, where p is the density. If p is constant, 
it may be placed outside the integral signs and canceled from 
numerators and denominators. Formulas (1) then become 

7x= CxdV, Vy=CydV, Vz*= CzdV. (2) 

Ex. Find the center of gravity of a body bounded by one nappe of a 
right circular cone of vertical angle 2 a and a sphere of radius a, the center 
of the sphere being at the vertex of the cone. 

If the center of the sphere is taken as the origin of coordinates and the 
axis of the cone as the axis of 2, it is evident from the symmetry of the 
solid that x = y = Q. To find z we shall use cylindrical coordinates, 
the equations of the sphere and the cone being respectively 

r 2 + z 2 =s o a and r z tan a. 

As in Ex. 2, 92, the surfaces intersect in the circle r a sin a in 
the plane z = a cos a. Therefore 

/>2r a a tin a />Va 8 r 8 

V=\ I / rdOdrdz = S Tra 8 (1 - cos a) 

JO /o t/rotnar o \ ./ 

n nZv n a sin a /-\/a 2 r* 

and \zdV-\ \ \ rzdOdrdz = Iwa 4 sin a cr. 

t/ i/O /0 t/rctnrt w 

Therefore, from (2), 5 s? f a (1 + cos a). 



CENTER OF GRAVITY 283 

EXERCISES 

1. Find the center of gravity of a solid bounded by the paraboloid 

y 7 / 

- = ' -f 7; and the plane = c 
c a 2 1r L 

2. A ring is cut from a spherical shell, the inner radius and the 
outer radius of which are respectively 4 ft. and 5 ft., by two parallel 
planes on the same side of the center of the shell and distant 1 ft 
and 3 ft. respectively from the center. Find the center of gravity of 
this ring. 

3. Find the center of gravity of a solid in the form of the frustum 
of a right circular cone the height of which is h, and the radius 
of tho upper base and the radius of the lower base of which are 
respectively r^ and ?' 2 . 

4. Find the center of gravity of that portion of the solid of 
Ex. 2, p. 73, which is above the plane determined by OA and 
OB (Fig. 31). 

5. Find the center of gravity of a body in the form of an octant 

Qn A 

a i * ej* 

of the ellipsoid ~ + 73 + -= = 1. 
e a? o* <? 

6. Find the center of gravity of a solid, bounded below by the 
paraboloid az r* and above by the right circular cone * + f = 2 a. 

7. Find the center of gravity of a solid bounded below by the 
cone * SB r and above by the sphere r* + # 2 = 1. 

8. Find the center of gravity of a solid bounded by the surfaces 
JB = 0, i* + * a ** i 3 , and r = 



94. Moment of inertia of a solid. If a solid body is divided 
into elements of volume c?F, then, as in 88, the moment of 
inertia of the solid about any axis is 

1= CtfpdV** p C&dT, (1) 

where JB is the distance of any point of the element from the 
axis, and p is the density of the solid, which we have assumed 
to be constant and therefore have been able to take out of the 
integral sign. If M is the total mass of the solid, p may be 
determine^ from the formula JtfwpK 



284 REPEATED INTEGRATION 

If the moment of inertia about OZ, which we shall call 1^ is 
required, then hi cylindrical coordinates JR = r and dV= r$6drdz, 
so "that (1) becomes 

I z = p CCCr*d0 drds. (2) 

If we use Cartesian coordinates to determine /,,, we have 
and dV= dxdydz, so that 

'xdydz. (3j 

Similarly, if I y and I x are the moments of inertia about OY 
and OX respectively, we have 

f\ f\ SI 

(4) 

In evaluating (2) it is sometimes convenient to integrate 
with respect to z last. We indicate this by the formula 

I a = p Cdz CCr 9 dO dr. (5) 



But I I i^dddr is, by 88, the polar moment of inertia of a 

plane section perpendicular to OZ about the point in which OZ 
intersects the plane section. Consequently, if this polar moment 
is known, the evaluation of (5) reduces to a single integration. 
This has already been illustrated in the case of solids of revolution. 

A similar result is obtained by considering (3). In fact, the 
ease with which a moment of inertia is found depends upon a 
proper choice of Cartesian or cylindrical coordinates and, after 
that choice has been made, upon the order in which the integra- 
tions are carried out. 

Equation (3) may be written in the form 

>WCy*dxdydt, (6) 

and the order of integration in the two integrals need not be 
the same. Similar forms are derived from (4). 

The theorem of . 89 holds for solids. This is easily proved 
by the same methods used in that section. 



MOMENT OF INERTIA 285 

Ex. Find the moment of inertia about OZ of a cylindrical solid of 
altitude h whose base is one loop of the curve r a sin 3 6, 

The base of this cylinder is shown in Fig 65, p. 144. We have, from 
formula (2), - In8fl 



wheie the limits are obtained as follows: 

First, holding r, 6, d6, dr constant, we allow z to vary from the lower 
base s = to the nppei base z = li, and integrate. The result phr a dOdi is 
the moment of inertia of a column such as is shown in Fig 127. We 
next hold and d& constant and allow r to vary from its value at the 
origin to its value on the curve r = a sin 3 0, and integrate. The result 
\ /7i 4 ain 4 3 Qd& is the moment of inertia of a slice as shown in Fig. 127. 
Finally, we sum all those slices while allowing to vary from its smallest 

value to its largest value ^ The result is zz 



The volume of the cylinder may be computed from the formula 

a" 1 *- /idBinSO nh 

V = I f i rdBdrdz = A 

Jo Jo Jo 

Therefore Af= -faphePir and J s =JI/ 2 . 



EXERCISES 

1. Find the moment of inertia of a rectangular parallelepiped 
about an axis through its center parallel to one of its edges. 

2. Find the moment of inertia about OZ of a solid bounded by 
the surface t= 2 and $ = r. 

3. Find the moment of inertia of a right circular cone of radius a 
and height h about any diameter of its base as an axis. 

4. Find the moment of inertia aboxit OZ of a solid bounded by 

2 Q 

the paraboloid * =* ~ + ^ and the plane = e. 

5. Find the moment of inertia of a right circular cone of height li 
and radius a about an axis perpendicular to the axis of the cone at 
its vertex. 

6. Find the moment of inertia of a right circular cylinder of 
height h and radius a, about a diameter of its base. 

7. Find the moment of inertia about OZ of the portion of the 
sphere ?* 4- *" ~ a a out out by the plane and the cylinder 
r ** a cos 6. 



286 REPEATED INTEGRATION 

8. Find the moment of inertia about OX. of a solid 'bounded by 
the paraboloid r* and the plane = 2. 

9. Find the moment of inertia about its axis of a right elliptic 
cylinder of height h } the major and the minor axis of its base being 
respectively 2 a and 2 b. 

10. Find the moment of inertia about OZ of the ellipsoid 
t+t + t =i 

I ' 7,2 ' 2 

^ GENERAL EXERCISES 

n n n 

1. Find the center of gravity of the arc of the curve x* -I- y* = *, 
which is above the axis of x. 

2. A wire is bent into a curve of the form 9y 2 = a? 8 . Find the 
center of gravity of the portion of the wire between the points for 
which x = and x = 5 respectively. 

3. Find the center of gravity of the area bounded by the curve 
ay* = a? and any double ordmate. 

4. Find the center of gravity of the area bounded by the axis 
of x, the axis of y, and the curve j/ 2 = 8 2 x> 

5. Find the center of gravity of the area bounded by the curves 
y = x 9 and y = -5 > the axis of x, and the line x = 2 

6. Find the center of gravity of the area bounded by the axes 
of x and y and the curve x = a cos 8 ^, y = a sin s < 

7. Find the center of gravity of the area bounded by the ellipse 

jg2 & 

~z + a =1 ( a > *) tlie <Hfle as 2 + y 2 = a , and the axis of y. 



8. Find the center of gravity of the area bounded by the parabola 
v? = 8 y and the circle cc a + f = 128 

9. Find the center of gravity of the area bounded by the curves 
** ~ a (y - &) = 0, a 2 ay = 0, the axis of y, and the line = o, 

^ 10. Find the center of gravity of an area in the form of a semi- 
circle of radius a surmounted by an equilateral triangle having one 
of its sides coinciding with the diameter of the semicircle. 

11. Find the center of gravity of an area in the form of a rec- 
tangle of dimensions a and I surmounted by an equilateral triangle 
one side of which coincides with one side of the rectangle which is 
b units long. 



GENERAL EXERCISES 287 

12. Find the center of gravity of the segment of a circle of 
radius a cut off by a straight line b units from the center. 

13. From a rectangle b units long and a units broad a semicircle 
of diameter a units long is cut, the diameter of the semicircle 
coinciding with a side of the rectangle. Find the center of gravity 
of the portion of the rectangle left. 

14. Find the center of gravity of a plate in the form of one half of 
a circular ring the inner and the outer radii of which are respectively 
' x and > 

15. In the result of Ex. 14, place r z = r^ + AT- and find the limit as 
Ar vO, thus obtaining the center of gravity of a semicircumference. 

16. Find the center of gravity of a plate in the form of a T-square 
10 in across the top and 12 in. tall, the width of the upright and 
that of the top being each 2 in. 

17. From a plate in the form of a regular hexagon 5 in. on a side, 
one of the six equilateral triangles into which it may be divided is 
removed. Find the center of gravity of the portion left. 

18. Find the center of gravity of a plate, in the form of the ellipse 

r o/ 

~ + 75 = 1 (a > i), in which there is a circular hole of radius c, 

the center of the hole being on the major axis of the ellipse at a 
distance d from its center. 

19. Find the center of gravity of the solid formed by revolving 

!K 2 ?/ 2 

about OY the surface bounded by the hyperbola -j j$ =1 and the 
lines ?/ = and y = &. a 

20. Find the center of gravity of the solid generated by revolving 
about the line % = a the area bounded by that line, the axis of a;, and 
the parabola y a = hx. 

21. Find the center of gravity of the segment cut from a sphere of 
radius a by two parallel planes distant respectively \ and h z (h z > 7^) 
from the center of the sphere. 

22. Find the moment of inertia of a plane triangle of altitude a and 
base b about an axis passing through its center of gravity parallel to 
the base. 

23. Find the moment of inertia of a parallelogram of altitude a 
and base b about its base as an axis, 



288 BEPEATED INTEGRATION 

24. Find the moment of inertia of a plane circular ring, tho inner 
radius and the outer radius of which are respectively 3 m and 5 in., 
about a diameter of the ring as an axis. 

25. A square plate 10 in. on a side has a square hole 5 in. on a 
side cut in it, the center of the hole being at the center of tho jilafco 
and its sides parallel to the sides of the plate. Find the moimmt of 
inertia of the plate about a line through its center parallel to ouo 
side as an axis. 

26. Find the moment of inertia of the plate of Ex. 25 nbout one of 
the outer sides as an axis. 

27. Find the moment of inertia of the plate of Ex. 20 about one 
side of the hole as an axis. 

28. Find the moment of inertia of the plate of Ex 2tf about ono 
of its diagonals as an axis 

29. A square plate 8 in. on a side has a circular hole 4 in. in 
diameter cut in it, the center of the hole coinciding with tho cuntor 
of the square Find the moment of inertia of the plate about a lino 
passing through its center parallel to one side as an axis. 

30. Find the moment of inertia of the plate of Ex. 29 about a 
diagonal of the square as an axis 

31. Find the moment of inertia of a semicircle about a tangent 
parallel to its diameter as an axis. 

32. Find the polar moment of inertia of the plate of Ex. 25 about 
its center. 

33. Find the polar moment of inertia of the entire area bounded 
by the curve T- 2 = a? sm 3 6 about the pole. 

34. Find the polar moment of inertia of the area bounded by tho 
cardioid r = a ( 1 + cos ff) about the pole. 

35. Find the polar moment of inertia of that area of tho circle 
r = a which is not included in the curve r a, sin 2 6 about tho 
pole. 

36. Find the moment of inertia about OF of a solid bounded by 
the surface generated by revolving about OY the area bounded by the 
curve </ x, the axis of y, and the line y 2. 

37. A solid is in the form of a hemispherical shell the inner 
radius and the outer radius of which are respectively and r , Find 
its moment of inertia about any diameter of the base of the shell as 
an axis. 



GENERAL EXERCISES 289 

38. A solid is in the form of a spherical cone cut from a sphere 
of radius a, the vertical angle of the cone being 90. Find its 
moment of inertia about its axis. 

39. A solid is cut from a hemisphere of radius 5 in by a right 
circular cylinder of radius 3 in, the axis of the cylinder being 
perpendicular to the base of the hemisphere at its center Find its 
moment of inertia about the axis of the cylinder as an axis. 

40. An anchor ring of mass M is bounded by the surface generated 
by revolving a circle of radius a about an axis in its plane distant 
b(b > <i) from its center. Find the moment of inertia of this anchor 
ring about its axis. 

oj2 ?/" 

41. Find the moment of inertia of the elliptic cylinder + 77 =1 

a* (r 

(a > ft), its height being h, about the major axis of its base. 

42. Find the center of gravity of the solid bounded by the cylinder 
= 2 a cos 0, the cone ss = r, and the plane x = 

43. Find the moment of inertia about OZ of the solid of Ex. 42. 

44. Find the volume of the cylinder having for its base one loop 
of the curve r = 2 a cos 2 0, between the cone = 2 r and the plane 
= 

46. Find the center of gravity of the solid of Ex. 44. 

46. Find the moment of inertia about OZ of the solid of Ex. 44. 

47. Find the volume of the cylinder having for its base one loop 
of the curve r ss a cos 2 & and bounded by the planes = and 
ft =s x + 2 a. 

48. Find the moment of inertia about OZ of the solid of Ex. 47 

49. Find the volume of the cylinder r = 2 a cos included between 
the planes * = and * =s 2 x + a. 

60. Find the moment of inertia about OZ of the solid of Ex. 49. 

61. Through a spherical shell of which, the inner radius and the 
outer radius are respectively r l and r a , a circular hole of radius 
a (a < fj) is bored, the axis of the hole coinciding with a diameter 
of the shell. Find the moment of inertia of the ring thus formed 
about the axis of the hole. 



ANSWERS 

[The answers to some problems are intentionally omitted.] 

CHAPTER I 
Page 4 ( 2) 

1. 21^. 4. 100ft per second 7. 2J 

2. l^jj. 5. 88 07 mi. per hour for entire trip. 8. l^mi per hour. 

3. 40J\>. 6. 1,26. 9. 08.4. 

Page 5 ( 2) 
10. 106 8. 

Page 7 ( 3) 

1. 06 ft. per second. 2. 128 ft. per second. 

Page 8 ( 3) 

8. 128 ft. per second. 5. 68 ft per second. 

4. 74 ft pci bccoud. 6. 52 ft. per second. 

Page 11 ( 5) 

1, 12 ii ; 24 j. 8. 85 ; 82 ; 6. 5. 5, 4, when t = 2 ; 10, 6, when t = 3 

2. 16 j 14. 4, 42 5 57. 8. 8ai a + 2M + c ; 6at + 26. 

Page 13 ( 6) 

1. i- sci. ft. per second. 2, 0. 3. 87rr a cu. in. per second. 

TT 2?r 

4, 4wr a . 5, 167irsq. in. per second. 

Page 14 ( 6) 

6. 8irr. 7. 8 (edge) 3 . 8, birr*. 9. 18. 10. 2ir. 

CHAPTER II 
Page 18 ( 7) 

3. 4fc B -2. 4. :- 5. 2 ; 



8, -,_ S . 7. tf + B+1. 8, 8-~ 

201 



292 ANSWERS 

Page 21 ( 9) 

1. Increasing if x > 2 ; decreasing if x < 2. 

2. Increasing if x > g , decreasing if x < |. 

3. Increasing if x < ; decreasing if x > . 

4. Increasing if a < , decreasing if sc > . 

5. Increasing if aj < 2 or x > 1 , decreasing if 2 < K < 1. 

6. Increasing if x < 5 or x > S , decreasing if 5 < x < 8. 

7. Increasing if < 1 or a: > ; decreasing if 1 < x < fj. 

8. Always increasing. 

9. Increasing iftc< lor ^<a<l; decreasing if 1 < x < J or a; > 1. 
10. Increasing if x > 1 , decreasing if x < 1. 

Page 24 ( 10) 

1. When t < - 1 or t > I , when !<<!. 

2. When 5 ; when t > 5. 

3. When t < 2 or t > 4 ; when 2 < i < 4. 

4. Always moves in dnection in winch a is measxirod. 

5. When t > f , when < $. 

6. Always increasing. 

7. Always decreasing 

8. Increasing when t > 2 , decreasing when t < 2. 

9. Increasing when t > , decreasing when i < $ 
10. Increasing when < < f , decreasing when < > |. 

Page 26 ( 11) 

1. TrA 8 . 2. 6 irh sq. ft. per second. 



Page 27 ( 11) 

8. 2 cm. per second. 5. 0.20 TT 

4. 20 9 sq in. per second. 6. 64 cu. ft per second. Tl 3 8 (t tal hol ff ht >*' 

8. 4 TT (2 + 12 ( + 36) , t is thickness. 
Page 31 ( 13) 

1. 1.46. 3 0.46; 2.05. 5. 2.41. 

2. - 2 07. 4. 1.12 ; S 93. G. 2.52. 

Page 35 ( 14) 

1. 8a;-y_9 = 0. 6, + y + l = *12. tan-ij. 

2. 2x + Sy + 3 = 7. aj + 2y + 8 = 0. 18, tan-*12. 

3. 21aj-2y-13 = 0. 8. 4x- Sy-1 = 0. 14.*. 

4 - J/ + S = ,_ -M.-4y-.6-0. W. i.-Sy-M-o. 

6. V3a;-y-2V3-2= ! 0. 10. 5a: - Qy - 4 = 0. W, (-Itfr, 2ft)* 

*The symbol tan- 1 jj- represents the angle whose tangent is \ (cl. 46). 



ANSWEES 



293 



Page 39 ( 15) 

1. (- I >|). 

2, (2, 4J). 

3. (0, 4), (2, 0). 

4, (1, 7), (3, 3). 

6. (- 2, 0), (1, - 0). 
6. (-1, - 3), (3,29). 

Page 43 ( 17) 

1. aSsq.in. 

2. Length is twice breadth. 

Page 44 (5 17) 

6, Depth is one-half side of base. 

7, 2 portions 4 ft. long ; 4 portions 1 ft. long 

-o jjt. 

8, Breadth = 



9. (- 3, 10), (1, 2) 

10. 278 + 27^-86 = 

11. 18a5 
18. tan-ijj 



3. 5ft. 

4. 50. 



- - 28 = 0. 



5. 



, .. 
; depth = 






9, Altitude = ^-f- 5 base - ? (P - perimeter) 

4 4 

10, 2000 cu. in. ; 2547 cu. in 
1L Height of rectangle = radius of semicircle , semicircle of radius, 

5 

12. -7= in. 
VS 

Page 46 ( 18) 

1, 426ft. 

Page 47 ( 18) 

8. 40ft. 5. 676ft. 

7. y = 2x 8 + x a -4x+6. 



3, 05ft. 

4, 88$ ft. 

Page 49 ( 19) 
1. 7$. 2. 1 

Page 53 ( 20) 

8. 0.0001 ; 0.000001 ; 0.00000001 

9. 0.000009001 ; 0.000000090001. 




3, 62J. 4. 36. 5. 



Page 54 

1. 72 sq. in. 






8, 



16 
Sir 



cu. in. 



6. 



10. 000003 sq. in. 

11, 456 58 ou. in. 



4. 27.0054 ou. in. 

5. 28.2749 cu m. 

6. 606 0912. 

7. 0012. 

8. 5.99934 



294 ANSWERS 

Page 55 (General Exercises) 



_ 

' 



o 2a A 4 x g _ j t 13. Increasing if x > 2 , 

' (a - x) 2 " ' (x 2 + I) 2 ' ' 2 Vajs ' decreasing if < - 2. 

14. increasing if f<o;<ora;>2, decreasing if a; < f or <j < x < 2. 

15. Increasing if x > - , decreasing if y, < - . 

2 o, 2(t 

16. Increasing if x < ^= or x > ~ , decreasing if ^= < x < = 

V3 V3 V3 V3 

4/7 4 ft 

17. Increase if x < - , decrease if x > - 

3 o 

Page 56 (General Exercises) 

18. 1< t < 5 19. 2 < * < 5 , 4. 

20. Up when < i < 6 J , down when 6 < < 12. 

21. Increasing when t > 4 , decreasing when t < 4. 

22. D increasing when < 3, v deci easing when t > 3 , speed increasing when 

2<i<3ori>4, speed decreasing when !S<2or3<i<4 

23. Increasing when !<<2ori>3, decreasing when *<lor2<i<3 

24. 0055 in. per minute 

25. 8 6 in. per second 27. x + 2 y + 6 = 0. 

26. 1 sq in per minute. 28. 7x+ 



Page 57 (General Exercises) 

29. x- 2 = 31. 2x-y + 3 = 0. 83. (-, 8|). 

80. x - 2y - 7 = 0. 32. tan-l 34. (lj, 0) 

35. (2, - 2). 41. 10. 

36. (- 1, 13), (5, - 95). 42. tan-i T V 

37. (- 3, 13), (1, - 19). 43. x - y - 11 = 

38. (- 4, 20). 44. (1, - 1), (- J, - 



Page 58 (General Exercises) 

46. 6f ft. long 52. y - x 2 + Bx - IB 

47. Altitude of cone is | radius of sphere. 53. y = x 8 x* + 7 x 

54.85^. 



i** a ^ *u 

48. Altitude= ; sideof base =^_ M 2g 

49. 2 pieces 3 in long, 3 pieces 1 in. long 56. 20 1 
50 600ft. 57. 72. 

51. 56ft 59. 0.0003. 

Page 59 (General Exercises) 

60. 0.00629. 64. 0.09 cu in 67. 24.0024 sq. in. 

62. 288 TT cu in. 65. 0.0003. 69, 0.4698. 

63, 161. 16 cu. in, 66. 854.1028j 353.8972, 



Page 66 ( 23) 

1. 8J 3. 52 T S . 

2. 23,} 4. 166f r 

Page 67 ( 24) 
1. 160ft 

Page 68 ( 24) 

4. When m 

Page 70 ( 25) 
1. 8$ T. 

Page 71 ( 25) 

6. Approx. 2418 Ib. 
6. 



ANSWEES 


29i 


CHAPTER IH 




6. 5 7. 36i 
6. 42 1 8. 2J. 


|. 9. 96. 11. 42 J 
10. 10|. 12. 10. 



2. 140ft. 



; 83ft. 



3. 57Jft. 



5, 8000irft.-lb. 



2. 2JT. 



3. 3T. 



Page 75 ( 26) 

1. 21 TT. 
625V8 

~~ 



7. 585}T. 
11. 2 1 ft. from upper side 



8. 84 A *. 

4. l^w. 

5. 6577r. 



cu in. 



7. 8 

8. 2| 



Page 76 (General Exercises) 

1. 6Jft. 5. 20. 

2. 81ft. 

3. 10 ft "' ~3~ 

4. 8JJ mi- ' 8. 

Page 77 (General Exercises) 

13. Twice as groat. 17. 

14, Jft T. 

16. 16 to. 35 

16. 68 ^TT 19. 96 IT. 

Page 78 (General Exercises) 

24. STT. 26. 115J. 26. 

29. 728,049 ft.-lb. 



Page 81 ( 28) 

1. & a + i/ 8 - 8as -f 



CHAPTER IV 

11 cs 0. 
0. 
6, 8a~ 



9. 234| T 
10. 2 T. 



9. 25f IT 

10. 213 JTT. 

11. 38| 



8. 

9. JfT 

12. Reduced to | original 
pressure 



20. 34lcu.m. 

21.^-1 

28. (aft 2 - Jft")ir 

27. 9 28. 204| 

80. 5301 ft Jb. 



3. (-3, 5); 5. 

*.(- *.!); 



296 ANSWERS 

Page 84 ( 30) 

M-2,o). MO, -if) 7. as/, ft 

a- (o.i). 5. 8H ft. . iom/5 

MH,0). 6. lOVlOft. 8. i m 

Page 85 ( 30) 

9. y 2 + 6a;-9 = 0. 10. x*-4x-12y + 16= 0. 

Page 87 ( 31) 



2- (0, 3), (0, V6), 



5. 9x 2 + 25^a - 36x - 189 = 

6. 49a; 2 + 24 y2 _ 12 0y _ 144 _ . 

Page 91 ( 32) 

I- (3,0), ( 



4 . o, f 



2 - (2,0), (Vl3, 0), 3a;2y=0; 1 

3. (0, V2), (0, V5), 

4. (2V,0); (4,0), 



Page 102 ( 36) 

1. 18s 2 + 22a;_3 

8). 9- 



_ 
' ~(a; 2 -9) 2 ' 11. -. 



21 



12. 
13. 



7 2a! _i + ^ 

* 2 ^' 14 . 

8. ' 



ANSWERS 297 



15 -" . -. 18. 

' 3i tr - i>) \' (x ~"a) (JR - 6) V(a a + x 2 ) 8 

ii . M 1 jr8 

11 *"' . 19. , 22. 

v'W-.tfi (x + l)Vx a -l V(x 2 +9) 8 

' r ~ ] . 20. 




v*--* vV + 8)* V(l + x)* 

Page 104 ( 37) 

<tt/ je a Vi/ + x Vy x __x 2 2a 8 x 

* ""' U , 1/& ai 

a/// g !_. 8> " i' ' 

'* *" --J i ^ ,,'S* ""ftw* 8u 8 ' xa 2X2 



e ' -3 ! 



Page 105 (38) 

Sa O. 2, x~7y + 5 = 0. 3. (-2.-1). 4. 



PAgel06(88) 

-. 10, tan-18. 12. 

8.*". . 5 5 ton" 1 A- 11. tan-i|. 13. 

"i * 

Page 110 (40) 

1. x-r 8 5 V44.9tK ,. fo a Bin ) 

a, x^d/-!)' 4 , V4tf+i. \ 20 20 

3. x*-0x + = 0j SVll7-48t + l6i a . B a sill 2a 

4. (8,1). 9 ' 



'4* 
11. yaa 

\ "aj 



8V i Moond. 2. 12.6 ft, per second. 3, ^u, perminute. 

4, Circle; ! 5 -fl, per secona (x- distance of point from *dl). 

x 
, 2,64 ft. per second. 

Pftge 113 (41) permi nute. 8. 6.6 ft. per second. 

6. 0.18 cm. por second. ?, 0.21 in. per mumi*. 
, ^ti^t.per 8 ccond, where, is the distance of top of ladder, and IB 

^dtonoe of foot of ladder, fromDase of pyramid. 



298 



ANSWERS 



Page 114 (General Exercises) 

20. 3 

21. x 



8 



27. tan- 

' ^ " 



28. -; tan- 17. 
2i 



30. tan-i . 
5 

35. 



81. -, tan- 1 -- 
2' 2 

~ 7T 

On i n 11 

32. -,lan 

33. x$ + ifi 
= 2 ato - <W8? 



' OJ 2 -1- 4 ' (i 2 + I) 2 
Page 115 (General Exercises) 

37. 20 ft per second , 

lOVEft per second, (100,20). 

1 a ' v a 
39. Velocity in path = VCKC + a 5 

40. 



Page 116 (General Exercises) 

46. 08 ft. per minute 

47. 01 in. per minute. 

48. |f sq in per minute 

49. 04 in. per second. 

50. Length is twice breadth 
51 Other sides equal. 

Page 117 (General Exercises) 
57. 2 64 in 58. 



41. 6 8 mi per houi , 28 8 mi 



42. 
43. 



t 4 |250-3i a _ 
' 3 \ 04- 1 3 
ft pei second, whore 



3s 

Vs 2 - 400 

s is length of lope from man 
to boat. 

44. 06 ft per minute 

45. sJu ft per minute. 



52. Breadth, 9 m , depth, 9V3 in. 

53. Length = | bieadth 

54. Side of base, 10 ft , 
depth, 5 ft 

55. Depth = ^ side of base. 

56. Radius, 3 in. , height, 3 in. 



1 

Vi' 



60. 8 mi from point on bank nearest to A. 

bm , , bn , 

a mi. on land , mi in water 



59. 



61. 4 1|} mi tiavel ou land 



Vn 2 m 2 

63. l^f-hr. 

Page 118 (General Exercises) 



Vn 2 - 



mr 



64. v/lOO mi per hour. 



Page 126 ( 44) 

1. 15 cos 5 x 

2. sec 2 -. 

3. 2sm 2 2zcos2x. 

4. 5 sin 10 x. 



65. Velocity m still water = mi per hour 

66. Base = aVI , altitude = | b 

CHAPTER V 



6. 5 sin 2 5 x cos 8 5 aj. 

_ o60!, &X 

7. 5 sec 2 tan . 

2 2 

8. -8csc 8 3a;ctn3x. 



ANSWERS 



299 



Page 127 ( 44) 

9. hin- 1 . 



10. 



11. 



12. 2 sec x (sou x + tana-) 2 . 
Page 129 ( 45) 

. S_. Mil--- 

3. 7T, 5. 

4. At moan point of motion ; at 

extreme points of motion. 



2 



13. 2 cos 4 x. 

14. 9tan 4 3aj. 

15 . 2 sec 2 x (sec 2 2 x + tan 2 2 x) 
16 sm 8 2 a; cos 2 2 a; 

17. - ? } 

18. -?. 



5. At extieme points of motion, at 

moan point of motion. 

6. 2V(fc - 3) (5 - s) ; 4(4-8) 

7. TT. 

8. 10, 2?r. 



Page 134 ( 47) 




3 


7. JfL. 


1 


8. ( * 


1 


0. - V !L 


8 " 




' Vx-x a 


x v 4 x a 


8 
4 ' Vl2x- ( .)x a 


10 ' n 


i 


5, - 1 


11. - 


2 + 2 x + a 


(x + 1) "V 


1 
n * 


12. - , l 




18. 



Page 136 ( 48) 

1. 1)3. = 9.42 ft. per second ; % = =F 86.40 ft. per second. 

Page 137 ( 48) 

3. 8 radians per unit of time. 

Page 138 ( 49) 

a b cos <t) ' 

Page 141 ( 50) 

17V17 



2. 63 




6. 



300 



ANSWERS 



Page 145 ( 51) 

17. Origin , (V3, j). 18. Origin , ( a^j|, |) . 19. Origin , (2, | 



21. ^8^20 = 

22. r 

23. r 

24. r 



25. x - a = 

26. x 2 + V* - 2 ax = 
27. 

28. 



Page 

Page 

9. 
10. 
11. 



18. 
19. 



148 ( 52) 

149 (General Exercises) 

2(l + sec 2 2x). 

sec 4 (3x + 2). 

cos 2 (2- 3 x). 

sec 2 (x y) + sec 2 (x + y) 

sec 2 (x y) sec 2 (x + y) 

1 .x 



1. IT tan-* 1 



3. 0. 



21. 



20. - 



tan 4 2 a;. 

l _ 
(x + 1) Vx 

- 1 

V2 + x - x 2 

27. 



2(x + l)Vx 

1 

xV49x 2 -l 
4 



ig t 

5 5 
14 8csc 2 4x(ctn4x + l) 

15. a tan ox sec 2 ox. 

16. 8 cos 8 2 x sin 4 a; cos 6 x. 

1 

. x)Vx 3 2x 

2 

V3-4x-4x 2 
2~3x 

Vo x 2 



24. 



25. 



26. 



Page 

29. 
30. 
31. 



(x 2 -l)Vx 2 -2 
150 (General Exercises) 

fc Va 2 sm 2 ^ + 6 2 cos 2 ^. 

2^_ 

V41, 



5 = 3; 2 



35. 



j 47. Origin ; / ^ 

""*" O * / A. 

34. 2aVi. 48. Origin; (^,tan-i 2 ). 

49. Origin; 
Page 151 (General Exercises) 



51. - = 



COS(? 



52. 

53. (x 8 + y 2 ) 2 + 2 aa;(x a + y 2 ) - a 

54, tan-i. 



s= 0. 



ANSWERS 301 

65. 0) tan-12 57. -. 59 IflVift 

56. 0, :,ton-i8V5. Kg* 6 ' 72 ' 

2 68 '4 61. V2ft 

62. At an angle tan- 1 /*; with ground. 

Page 152 (General Exercises) 

63.12m. 65. a. 68. 15 sq ft , 10.04 sq. ft. per second. 

64. sVBft. 67. 0.1 ft per second. 69. 26 7 mi. per minute. 

nn /i /i , & 2 8in0COS(? \ . . ,.., , 

70. / b sin & + , ) times angular velocity of AS. where = angle 
\ Va a -6 a sin 2 0/ 

CL4.B. 

71, <* ~ 2 ) 2 + fa - 3 ) 2 - 1 ; V9 sin 2 * + 4 cosat , where I = (2 k + 1) | 
78. 5^-^ = 1 ; OsecSiVtan 2 3< + 4sec 2 3<. 

Page 153 (General Exercises) 

73. sin 20. 75. tan-^Vi 

74. aVl+cos 2 o5, fastest when 75. tan- 1 ^ 

x = far; most slowly when 77. 0, tan- 1 3 Vs. 
a5 = (2A! + l)-. 78. tan-i^, tan-i4V2 

2 79. tan-i3; tari-ij. 

CHAPTER VI 
Page 162 ( 55) 

(The student is not expected to obtain exactly these answers , they are given 
merely to indicate approximately the solution.) 

1. y =0.62 x- 0.70. 2. 1 = 0.00172) 

Page 163 ( 55) 

3. ^=s0.80(2.7)^. 4. c=0.010(0,84). 5. a=0.0000000048Z3'o. 6 j" = 10. 



Page 165 ( 56) 



1 1 12. e- 2ac (3cos3a; 2sin8x). 

" ~* < n j. 1 

13, Ctll"" 1 !!/. 



14. 

3. 2 a; a** -Una. "' Vi^"+~4* 16, 

4. a n ^ _., 9 16. 

e" + e- 



5| rf j. A -. i ' 10. - 4 sec 2*. W- 2 see's. 



20:4.8 n 2(e 8a! e- 2a? ) ig. J; . 

- - - "' ' eaa+e-is* ' asVx + l 



302 

Page 167 ( 57) 



ANSWERS 



2. ?/ = 45.22e oola: . 



4. 



Page 168 ( 57) 

5. P = 10000 e 022 9* 6. c = 0.01e- n44B <. 7, 2 mm, 

Page 168 (General Exercises) 
*10. p = 0.018 1 + 24. *11. Load = 102 - 6 6 length. 

Page 169 (General Exercises) 



*12. s = 25 (0 40)' *14 t = 1 Vl. 
*13. c = 010 (0.83)* *15. 1= 0.023 V0 

Page 170 (General Exercises) 


18. ?/ = 


"17. pv = 1620 


21. 2 csc- 1 2 a;. 


25. tan~^x. 


18 l . 


2 

22. 2 (x + 1) e 


26 0.898 


' 9 a: 2 - 4 


19. ctn x 


23. * . 


27. 16 8 hr 


o 


g2 i . fijX 


28. 1000 sec 


20. 








24. a tan 3 ox. 


29. p = 147i 



Page 171 (General Exercises) 
31. 2"\/2e~ 2i , 2e~ 2 * 33. 

87. (l + 1) * 



35. Vic*. 



x ' 2 



38. 



Page 176 ( 59) 



CHAPTER VII 



3.8 

- 



2. 



, , a; 8 . 2s 5 , 17x7 
3. a; + H --- h 

8 15 ^ 315 



2i 41 61 



4- ^ 

"4" "" -L. - ____ - ____ __ JL 

2 32-4 6 + 2 . 4 6 7 



e. 

V2 



- 
2 2 2 s 3 



3 T 5 7 T ' 

X ~2l~3! + " j 
* Statement in regard to answeis to exeicises in 55 is true of tins answer, 



9. 0.0872 
10, 4695, 



ANSWERS 303 

Page 178 ( 60) 



5 2 5 2 8 5 8 

___ + .___ _____ + . 



8 



/ ir\ 8 
1 V V 



, 

oy 2 21 2 si 



6 E + i to 

+ 



2 2! 2 31 



8. 0.7193 9, 0.8480 10. 8.0042 

Page 179 (General Exercises) 






_ + 

' 5 ~ 2 * 5 ' 2 1 2_ ' 8 I 

a j.^ - 1 ^!_I^ ?!. 
1 z %~ x a '21 2*31 



10. 1 - 2cc + 2 2 - 2fc 8 + . . 14. 9659. 

11 i j. 2aj 8 4a* 15. 0.61CO. 

u . a H . x -- + ... 



T 21 8! 5!^ ' 18. 0.69815; 10986. 

13 3.4.?! + ! + ^!+..., 19> - 22 314; 1.6094. 

Page 180 (General Exercises) 

88, 2.0805. 2e a;8 a 6 _ a 7 

M, 2.9625. ' "*" 3 5(2 I) "" 7(8 !) 



304 ANSWERS 

CHAPTER VIII 



Page 183 ( 61) 

8j/ 2 , . 

xy x 2 ' xy y 3 



5. y X 



6. 



---, 
(K + y)* x + y ' 



s._L_, *_. 7 i * 

x 2 +2/ 2 ' x 2 +i/ 2 y ' tf 

y x 1 
4. , > , s. , ; 

VI x a y 2 Vl x 2 ^ 2 v x 2 + y a 

Page 185 ( 62) 

x 2 y* 
1. -5 5- 2. e" sm (x y) 3, 

Page 187 ( 63) 

1. 000061 2. 0012. 3, 2^ 

Page 188 ( 63) 

4.0.018m 5.00105 6.0015m. 7. 6320ft. 8.0.0064. 

Page 191 ( 64) 

1. -2. 2. -| 4. - 5. - J . o. 6. 0;0 

Vx 2 + y 2 
Page 192 (General Exercises) 

8. - 14.33 cu ft. 10. 0.5655 sq. in.; 11. 3.0 in 13. 2.206 sq. in. per second 
9l 1735 05756sq.m 12. 035m, 15. 4.4 uq. in. per second. 

Page 193 (General Exercises) 

16. _iyJ5. 17 /y-tan-ia fit ia J V5 . 

5 a tan A , &*. is. -cos a . 'Sina, 1. 

CHAPTER IX 
Page 198 ( 66) 



. -p if* -p uiiiij,. 7. i(x* + 4)a. 14 1 , . 

2. g \(3x+7)x4 8. llnW-im. ' ~a ln ( 1 + c ^). 



9. ^In(2x + sin2x) 14. ~Jcos*2x. 

10. - 1 W- 

4. i nz __! L. 3(x-smx) 16 

a: 2. 758 .1 

1)]. a 17. _ 

12. ln(x 8 + Sx 2 + 1). ig, 

i. 20. 



ANSWERS 



805 



Page 202 ( 67) 

, 1 ,3x 
1. -sin- 1 - 
3 4 



i 



.sin- 1 



Vs 

4. sec -1 xV8. 
i a 

5. 



' V7 V? 



V21 



4x-3 
3 

6x-5 



11. -^-sm-i 

Vs 5 



V3 
*5 



-=-. 
VlO 



12. 



13. 



V5 



V2 



V2 



16.? 



17. ~. 
4 



Page 204 ( 68) 



3. 



5. 



. Jln(8a+Vo"-l). 
-p 
n 
-r- 



15 5sm- 1 --2V4-x 2 . 

ft 

18. v. 19. ~- 2C 

36 



11. 



In (3 



1 . 2x-6 
20 U 2x + 5' 



7. 



8. 



9. 



- 
2 VlC 8X+V15 



-- 7 : 
2 Vi5 



~ 



w, 4 to -.. 
Page 207 ( 69) 

1, - ^ COB (S 05-2). 

2, ~Jsin(4~2x). 

3, Jsec(8x 1). 

4, 4tan?. 

4 
2 8x 



6. $ln sin 6 as, 
7. 



3a; + 5 



12. ^Lln-*"- 7- 
V5 2 x - 8 + V5 



' V33 2 x + 5 -t- V33_ 

1 . 4x-l-Vl3 

14. =rln _ 

2 VlS 4x-l + Vl3 



8 + V5 



15. 
16, 

18. -4=1] 
V2 

19. ln; 

20. iln 



8. 2 esc - 

9. ^ln [sec (4 x + 2) + tan (4 a + 2)]. 

10. ctn(8-2a). 

11. In (esc x etnas) + 2 cosx. 



13. 



14, a cosx. 



306 



ANSWERS 



15. 3 tan - sec - 

\ O d 1 


X. io> -waaui, aa, -in 

V2 2 1 + V2 




19 s - ^. 


16. |(x sin x) 


\J* ii 7J* 

20. In j S3< 4* 


, w 4V2 3x 
17. - Sill 
8 4 


21. V3-l-^- 24. 3 
iii 


Page 208 ( 70) 




1, l e 2a: + 5 


5. ^(eaae-a^+aa;. 3 


Sai -5 * 


6. eF + e- 35 . ' 2 In 2 


S^iic i 


7. ln(#-l)-x. n. i (e-1) 


e+l 


g. 2(e^ fi~^). c a j, i 


go + &z c + to 


99 12- I" --- ' 


' ft(H-lnc) 


' In 10 


Page 212 ( 72) 





_ 
4. Vx 2 1 tan- 1 V.e 2 1 



8. - 



5. 



x ,x 

6 sin- 1 -- 

V4-X 2 2 

9. - 



4Vx 2 -4 
10. |. 11.. 12.- 18. ^(9\/8-10V5). 14.. 15. 2 

Page 216 ( 74) 

1. (8 x 1) e 3 * 7. J (2 cos + sin a) e 2 *. 

8. J(x 2 + 2xsincc + 2cosx). 

9. 4-2Ve 

10. J (81 In 8 -96). 

11. i(-2). 

12. l(7T-2). 



2. 



3. zcos-ix Vl x 2 

4. xtan- 1 3a-j^ln(l+ 9a 2 

5. la^Beo-^x 

6. sin z (In sin a; 1) 

Page 217 ( 75) 



4. In 



X 2 -l 

2) 2 



2. 



Page 220 (General Exercises) 



--. 
a; 2 as 2 



5. In' 



ANSWERS 



307 



Page 221 (General Exercises) 

T. /, (2 + C ST ) S . 9. -\/8 a; + x a 

8. ij (1 + 2s + a;*)*. 10. z a + a: + In (a - 1). 

11. n \, [B fdu (2 z - 1) - 3 sm (2 1 1) ] . 

12. - un - /3 COB* - + 4 cos 2 - + B\. 
8 f>\ 5 5 / 

13. a 1 ,) ctn 4 x (] 5 + 10 ctn 2 4 x + 3 ctn 4 4 a) . 

14. ,V, t 8 HucB (- c - 2) - 5 suc (a- - 2)]. 

15. i tau (jt - 1) + In tan (.r - 1) 

16. Ju (7 c\sc fi 2 x 5 c'sc 7 2 tr) . 
11 i(MoSa--7)v / 8ee8j. 

18. Vcac2x 

19. ~?% (8 ctn 5 x + 8 ctn 3 5 a:) Vctn 6 x 

20. T jt,j(lB3 34sin a 4o; + Own 4 4 a) \/wn4a5. 

21. - 7 ' fl (9 ctn 5 x + 4 ctn 6 a 1 ) \/ctn 8 5 x. 



23, sin- 3 



V5 



1 . 2 sr + 8 

- sec- 1 -! 
5 



24. ^sin-i-^-. 

s aVa 

,. JL O 35 

25 - - i ppf**" J- i , * 

' V6 V6 



4 2 

21 i B ec-i^ti. 
2 2 

28. J-sec-i^li. 

'sVs Vs 



5 
2;r-5 



,2J!-1 



OB 
35. 



Vl6 



30. sin- 1 



31. 



00 i ,23; + 8 

33. = sin- 1 = 

V2 V8 



36. 



_ 

Vo 



38 



39. 



Vn 



Vc 



Vn 



2V6 V5 



Page 222 (General Exercises) 
41. 



-ft 1 . a -2 

42 - Rl Tl **" 1 -i"T rrm-i 

2 2 

1 1-2 

43. -i-soc-i^-. 

aVo V6 

1 9-S 

. - JL nv 

** "^T SGC "*"* * 




48. ln(oj-|-Vx a -7). 
49. 




45, 4 1 



46.1 



VlO 



51. 

62. i Inj^-h Va lfi + 7). _ 

53, 2 V(c a + 4 + In (a; 4- V a + 4). 

54. V8ic a +l -- 



4r1 

V21 



55, -i= In 



8 + 



-Ox). 



308 ANSWERS 

1 

57. =ln(5x + 2 + V25x 2 + 20 x 5) 
VH 

58. ln(3x-4 + V9x 2 - 24x + 14) 

4 V6 x V& - 2 

i ^ ^ 

62. 



59. 



- 

6V? 8x+V7 

^in^?- 2 . 

4 Vo x V5 + 2 



2V21 



63. ln-. 
6 x 

.. 1, 2x-5 

64. -In 

5 x 



.^ 
2V6 

70. l 



__ 1 , x-1 
bo. in 



12 3x + 
4V6 2x- 



67. 



15 



5" x + 2 
Page 223 (General Exercises) 



75 



3 3 .a; x 9 . 2a; 

-x -sm 8 -cos- sin . 

8 4 3 3 16 3 



68. ta 

71. (tan3x ctnSx). 

72. In [sec(x-g) + tan ( aj - )] 

73. cos2x. 

74. ln(secx + tanx). 



Gx-3) 



76 Jtan2x x 


as. ^ (x a + 4) taiv-i- a 


i. 


77. |(sm2x cos2x). 


94. ^ (2-9 x 2 ) cos 3 a; + 


x sin 3 x 


78. x -f- 2 ( ctn esc 1 

\ 9 O / 


95. x[(ln2x) 2 -21n2x- 


f2] 


\ / 
79. tan2x x 


96. x In (8 x +V9 x 2 - 4)- J V x a - 4. 


80 smx cosx 


1 4/ 0\ 


/ * 


81. 2Ve= 


98 In '* ' 




82. f v^ 1 . 


'2 (x + 8) 8 ' 




83. x Jln(l + e 2z ). 


99. In fa" 2 ) 2 . 




V4x 2 -9 


85. ^ (2 x 8 + x s - 6) -v^x 8 + 2. 


100. ^ i n I 2 x + J ) ( 2 x + s ) : 

8 2r 1 




86. f Vx + 3 


u At ifi ~~" JL 




87. ^x- 1 l s ln(2x + l). 


101. lin^-^fa + S)^ 




88. Z ~ 


2 (x - 3) 2 




3(l-x 2 )! 


102.^ 

12 




89. a8 
27(4x 2 +9)f 


103. *. 
12 




90 (2x 2 + 25)Vx 2 -25 


104 -^Infl. 




1875 x 8 


105. 4 In 4 





91. 



(In 5)8 ' 



106. 



ANSWERS 809 

Page 224 (General Exercises) 

107. V5-2 112. 6? 118. 

V5 ' 113. 7. 119 . 

108 '0 114.^11X2. 

109>c2 ~ e - 1 9 + 4V2 WO- 

110. ~^j- . ' ^^/g n 14 ' 121. 2 - In 3. 

, I / ! \ 116. STT 122. 

Til ... ........ V A" i t ft*/ 

' lu 5 ' 117. UvS. 123. 



CHAI?TEB X 



Page 228 ( 77) 




1. 2. 


4. 


(h A\ 


5. 


e e y 


6. 


3, 47ra a 


7. 



Page 229 ( 77) 

10. 2w 2 a 8 . 2 ?H^V2 A/8^ 13> T ^ T 16< 

11. Jjfira 8 ' ~8" 14. 259jw. 16. JTTW. 

Page 232 ( 79) 

1. 2a 2 . T 2 5 a a V2 7, HTT. ^ 

2 7m_ 2 'JJ," 3 8. J. 10. 4-(7T-2). 

4n' 4. ~~. 6. |7ra 2 9. 407T. * 

2 

Page 234 ( 80) 

1. ^L. ; ft is radius of semicircle. 2. ; a is xadius of semicircle. 

2 2u 3 

IT 7T0 

Page 235 ( 80) 

6. 7ra 8 . 7. 100 revolutions per minute. 8. 5.64 Ib. per square inch, 

Page 236 ( 81) 

8), 4. Ca. 7. 

I 5. STr'a. 



310 ANSWERS 

Page 238 ( 82) 

l m vka 5 686 1 ft.-lb 

'a '12 6. 2kca 2 , k is the con- 

2, 22f ft.-lb 4. 196,350 ft.-lb stant ratio 

7. - mi.-lb ; B is radius of earth in miles 8. 2 irC 

R + a 

Page 239 ( 82) 

9. 1.76ft.-lb., 1 56ft.-lb. 

Page 239 (General Exercises) 

1 Ssm-i| 3. 16 -12 In 3 6. 1^ , 15 T \ 

~ 



, 
5_ 

_. 9V3 
247T + - . 

5 15 

10 ^TTCE 2 _ 

11. ^ Trfc 2 Vi^jj , fe t and k z are the values for k in the equation j/ 2 = 



12. -.. 13. 

15 

Page 240 (General Exercises) 
14. dfrrf. 15.4FGn4-l) 16 17. 



19. 



5 



., 

15 15 ' 23. _ 4 

Page 241 (General Exercises) 
24. 400irlb. 28. 123 T 31. 21 J. 



. 
26. 441b. n 8 



34. -(8ir+9V3) 35. ' 

J.D ' 

Page 242 (General Exercises) 

861 -4 37. ^(8+7T), (57T-8), (8-7T) 

88 ' T* ^ * 41.950 42. -(In 9-1). 



ANSWERS 311 

48, | 

45. -"" 

2 8a3 

Page 243 (General Exercises) 

q T. 

49. ~. 50. 50,000 ft.-lb. 51, 438,1 ft.-lb. 

a a* 

CHAPTER XI 
Page 245 ( 83) 

1.80 -In 8. 2. In 3. 8. 2Jf. 4. IT 1 

Page 246 ( 83) 

6. 14. ' T" ' 4 "~ ' ' ~6~" 

10, ~ (22 - TT). 11, Va (IT V2 - 4). 12. . 

y 30 

Page 254 ( 85) 

(2 cz (i (f^ 4- 4 c" 1^\ 4 a 'N/S 

~~r> . /0 . ) 6. On axis of qxxadrant, from 
e+l 4e(e 8 --l) / ? . . ' Sv 
' . v 7 center of circle. 

4. lira, ~ r -\ 7. Intersection ot medians. 

8. /-, -\. 

5, On axis, ~ from vertex. \ 2 8 / 

10. On axis, distant 4 a * + 3 f + 0fca fr0 m L!!' 



16. On axis, distant (radius) from base. ir. On axis, distant - from base. 

a 

16. On axis, distant - from base. 18. Middle point of axis. 

Page 256 ( 86) 

3, On line of centers, distant (ri *"*,? from center of circle of radius r,. 

r x 2 + r| 

4, On axis of shell, distant ^ a ~~ *' from common base of spherical surfaces. 

5, Middle point of axis. 8 ( r *" ^ ) 

Page 257 ( 86) 

6, On axis, a 1 - distant from base, 

7, On axis, $ of distance from vertex to base. 

8, (4$, 4|), the outer edges of the square being; taken as OJTand OF. 



312 ANSWERS 

9. On axis, distant 4 8 from corner of square 

10. On axis, distant 3.98 m from center of cylinder in direction of larger ball. 

11. On axis, distant 8.4 ft. from base of pedestal. 

Page 260 ( 87) 

3. base x altitude 

4. ^ , a = altitude and 26 = base of segment. 

15 

5- 7rg ; a = altitude and 26 = base of segment. 
5 



6. 27ra 2 6, 8ira6. 

7. ^(b + Sc), 7r[2ca + 26c + 6 2 + (6 + 2c) Va 2 + 6 2 ] 



ii 4u>a6 /Kr4-q\ 


3 

a = altitude and 26 = base of segment 12. cw (area) . 


11. (0 C -t- O CM , 

15 
Page 265 ( 88) 


1. j\Ma*. 


2. %Ma z . 


3. Ma 2 4iJfaMJft 


Page 266 ( 88) 






^.A 2 (6+3a) 
6(6 + a) 


9. 
10. 


JJf(a 2 +6 2 ). 14. ^Jlf(r 2 2 +r 2 ). 
145 f- 15. Jfa 2 ,a=radms. 


6. ^.Ma 2 


11 


8 1J' 16. 6757r. 


7. 1 J 3 3f(a2 + 6 2 ) 


12. 
13. 


T~ 17 S 3f ^ ~ TI 


iJfr 2 . ' 10""r- ri 8 



Page 268 ( 89) 

1 McP, a = radius. 

Page 269 ( 89) 



3. 2284$. 
4. 4569. 
5. 44,990 

Page 281 ( 92) 
1. 2?r. 

S . 

4 


6. 89,980 
7 ^Jf(r 3 2 - 
8. llf^H 

B 77ra 8 
o. 
6 



9 

10. 
11. 



4) 6 32a8 

9 

7 Sffffl4 
326 

8. Tra 8 9. (37T + 20 16V2). 10. u, 



ANSWERS 313 

Page 283 ( 93) 



2t On axis of ring, distant 2ft. from / Q a * 

center of shell. 6- f^i -^ V 

3. On ax 1S , distant 7i ( r ' + 2 Vt + '.") / 6 + 8 

4W + V. + T/) ?. (0,0, + 
from upper base 

, , 8/00 8 2 ( 2b2 - 2 ) \ 

4. On axis, from base. Ot ( u u . ( 

' 8 \ 8[b 8 -(b 2 -a a )&]/ 



Page 285 ( 94) 

1. ^5 Jlf (a 2 + 6 2 ), where Jf is mass, and a and & are the lengths of the sides 

perpendicular to the axis. 

2. Atf M. 

8, jf M(3 a 2 -1- 2 W) 5. ^ M(a + 4 /i) 2 a 2 (15 TT - 26) 

4. J Jfcf (a 8 + 6 2 ). 6. ^ M (8 a 2 + 47i). ' 25 (3 TT - 4) 

Page 286 ( 94) 

14.9T 

8. i^.. 
3 

Page 286 ( General Exercises) 



. 

8. .. 9. }flf(a 2 + & 8 ). 10. 

3 



. f , oV a; = 
\ 7 / 



6 is the ordhiale. 852 



10. On axis, distant _ from base of triangle and away from 



. 
semicircle. 

11, On axis, distant 4a *+ 2ab -^ + & * from base. 



Page 287 (General Exercises) 

18, On axis of segment, distant ^ T.IL1'" ^ from center 

of circle, 8 ( 7m8 



314 ANSWERS 

18. On axis, distant , , ~~ from center of semicircle. 
8(86 -Tra) 



. ! J- 7* f J- f^\ 

14. On axis of plate, distant * v ' a 1J from center of circles. 

gp V 2""" I/ 

15. On axis, distant 1 from center of circle 

7T 

16. On axis of square, 8 in. f lorn bottom. 

17. On axis, distant - in. from center of hexagon 

o 

c 2 d 

18. On axis, distant from center of ellipse 

ab ~ C 96 

19. On axis of solid, distant from smaller base 

16 

20. On axis, distant from base ^ of distance to top. 

21. On axis of segment, distant l ~ * 2 ~" ll '-" fmrn ofmtnr 

4-T^n^fh Ji ~\ f] 8 'ii **"* \jouuv3i 

of sphere 21 2 

22. T^lfa 2 23, 



Page 288 (General Exercises) 

24. *j-2l. a* 

25. JJL5 u 31> S7 ( 15 ^ - 32 )J 34. 



16 

26. *fM. a 1S radius. 

27. &&M 35. 



32. %JiM. 30 

29 XPJli_4 4 86 ' 

30, uyi_4Tr] 33. ~. 37t 



Page 289 (General Exercises) 



38- 3Q-18-5V2;. 42. (^,0, 

39. J/i&M 43. 

40. 



1 1 ly/OLO . y in, 44 




48 ' 32 + 3465 ' 49 ' 3ira8 80. 



51. ^ __ 

^13838 
103 




INDEX 

(The numbeis refer to the pages) 



Abscissa, 28 
Acceleration, 9, 21, 186 
Algebraic functions, 70 
Amplitude, 128 

Angle, between curve and radius 
vector, 140 

between curves, 104 

between straight lines, 35 

vectonal, 142 

Angular velocity and acceleration, 185 
Anti-sine, 180 
Approximations, 53, 187 
Arc, differential of, 106, 146 
Archimedes, spual of, 145 
Area, as double integral, 246 

of ellipse, 225 

of plane curve, 47, 225 

in polar coordinates, 280 

by stimulation, 60 

of suiface of revolution, 259 
Asymptote, of any curve, 80, 02 

of hyperbola, 90 
Average value See Mean value 
Axis, of cooidmates, 28 

of ellipse, 86 

of hyperbola, 00 

of parabola, 82 

Cardioicl, 145 

Cartesian equation, 100 

Cartesian space coordinates, 260 

Catenary, 157 

Center of _ gravity, of any solid, 282 

of circular arc, 258 
' of composite area, 255 

of half a parabolic segment, 25f 

of plane area, 251, 



Center of gravity, of plane curve, 
250 

of quarter circumference, 250 

of right circular cone, 253 

of sextant of circle, 252 

of solid of revolution, 252 
Circle, 79, 148 

of cuivature, 140 
Circular measure, 119 
Cissoid, 98 

Compound-interest law, 166 
Cone, circular, 272 

elliptic, 275 

Constant of integration, 45, 194 
Coordinates, 27 

cylindrical, 270 

polar, 142 

space, 269 
Curvatuie, 189 
Curves, 91 
Cycloid, 137 
Cylinder, 278 
Cylindrical codrdinates, 270 

Definite integral, 62, 194 
Derivative, 15 

higher, 40 

partial, 181 

second, 39 

sign of, 20, 40 
Differential, 50 

of arc, 106, -146 

of area, 64 

total, 185 

Differential coefficient, 51 
Differentiation, 15 

of algebraic functions, 94 



815 



316 



INDEX 



Differentiation, of exponential and 
logarithmic functions, 163 

of implicit functions, 102 

of inverse trigonometric functions, 
131 

partial, 181 

of polynomial, 18 

of trigonometric functions, 124 
Directrix of parabola, 81 
Distance between two points, 79 
Double integration, 244 

e, the number, 155 
Eccentricity, of ellipse, 87 

of hyperbola, 90 
Element of integration, 64 
Ellipse, 85 

area of, 225 
Ellipsoid, 274 

volume of, 280 
Elliptic cone, 275 
Elliptic paraboloid, 275 
Equation of a curve, 29 
Equations, empirical, 159 

parametric, 109 

roots of, 30 

Equilateral hyperbola, 90, 92 
Exponential functions, 154 

Palling body, 6, 8 
Focus, of ellipse, 85 

of hyperbola, 87 

of parabola, 81 
Force, 128 

Formulas of differentiation, 101, 124, 
131, 163 

of integration, 195, 199, 202, 205, 

207, 217 

Fractions, partial, 216 
Functions, 15 

algebraic, 79 

exponential, 154 

implicit 102 

inverse trigonometric, 130 

logarithmic, 154 

trigonometric, 119 



Graphs, 27 

of exponential functions, 157 
of inverse tiigonometnc functions, 

130 

of logarithmic functions, 157 
in polar coordinates, 142 
of trigonometric functions, 121 

Hyperbola, 87 

Implicit functions, 102 
Increment, 16 
Indefinite integral, 63, 194 
Infinite integrand, 229 
Infinite limits, 229 
Integral, 45, 194 

definite, 62, 194 

double, 244 

indefinite, 68, 194 
Integrals, table oJ, 217 
Integrand, 194 
Integration, 45, 194 

collected formulas, 217 

constant of, 45, 194 

by partial fractions, 216 

by parts, 212 

of a polynomial, 45 

repeated, 244 

by substitution, 208 
Inverse sine, 130 
Inverse trigonometric functions, 130 

Lemniscate, 144 
Length of curve, 235 
Limit, 1 

of 



of (1 + h)\ 156 

theorems on, 93 
Limits of definite integral, 63 
Line, straight, 81 
Linear velocity, 135 
Logarithm, 154 

Napierian, 156 



INDEX 



317 



Logarithm, natural, 156 
Logarithmic spual, 168 

Maclaurm's series, 173 
Maxima and minima, 41 
Mean value, 233 
Measure, circular, 119 
Moment o inertia, 200 

of circle, 204 

polar, 202 

ol quadrant of ellipse, 262 

of rectangle, 201 

of solid, 283 

of solid of revolution, 265 
Moments of inertia about parallel axes, 

260 
Motion, in a curve, 107 

simple harmonic, 127 

Napierian logarithm, 166 

Ordinate, 28 
Origin, 27, 142 

Pappus, theorems of, 259 
Parabola, 81, 146 
Parabolic segment, 83 
Paraboloid, 275 
Parallel lines, 38 
Parameter, 109 

Parametric representation, 109 
Partial differentiation, 181 
Partial fractions, 216 
Parts, integration by, 212 
Period, 128 

Perpendicular linos, 84 
Plane, 276 

Polar coordinates, 142 
Polar moment of inertia, 202 
Pole, 142 
Polynomial, derivative of, 18 

integral of, 46 
Power series, 172 
Pressure, 68 

theorem on, 267 
Projectile, 110 



Radian, 119 

Eadms of curvature, 139 
Eadius vector, 142 
Bate of change, 11, 189 
Revolution, solid of, 73 
surface of, 259, 273 
Hoots of an equation, 30 
Rose of three leaves, 144 

Second derivative, 89 

sign of, 40 

Segment, parabolic, 83 
Senes, 172 

Maclaurm's, 173 

power, 172 

Taylor's, 177 
Sign of derivative, 20, 40 
Simple harmonic motion, 127 
Slope, of curve, 86 

of straight line, 31 
Solid of revolution, 73 
Space coordinates, 269 
Speed, average, 3 

true, 6 

Sphere, 271, 272 
Spiral, logarithmic, 158 

of Archimedes, 145 
Straight line, 81 

Substitution, integration by, 208 
Summation, 66 
Surfaces, 271 

of revolution, 273 

Table of integrals, 217 
Tangent line, 88, 104 
Taylor's series, 177 
Total differential, 185 
Trigonometric functions, 119 
Trochoid, 138 
Turning-point, 37 

Value, mean, 238 
Vector, radius, 142 
Vectonal angle, 142 
Velocity, 21, 107 
angular, 135 



318 

Velocities, related, 111 
Vertex, of ellipse, 86 

o hyperbola, 90 

ot paiabola, 82 

of parabolic segment, 84 



INDEX 



Volume, of any solid, 277 

of solid with parallel bases, 71 
of solid of revolution, 73 

Work, 237