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MATHEMATICAL TEXTS
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EDITED BY
PERCEY F. SMITH, PH.D.
PROFESSOR OF MATHEMATICS IN THE SHEFFIELD
SCIENTIFIC SCHOOL OF YALE UNIVERSITY
SIR
ELEMENTS OF THE
DIFFERENTIAL AND INTEGRAL
CALCULUS
Revised Edition
By WILLIAM ANTHONY GRANVILLE, PH. D., LL. D.
Formerly President of Gettysburg College
PERCEY F. SMITH, PH.D.
and WILLIAM RAYMOND LONGLEY, PH.D.
Professors of Mathematics, Yale University
GINN AND COMPANY
BOSTON NEW YORK CHICAGO LONDON ATLANTA DALLAS COLUMBUS SAN FRANCISCO
COPYRIGHT, 1941, BY WILLIAM ANTHONY GRANVILLE, PERCEY F. SMITH
AND WILLIAM RAYMOND LONGLEY
ALL EIGHTS RESERVED
641.10
COPYBIGHT, 1904, 1911, BY WILLIAM ANTHONY GRANVILLE AND PERCEY F. SMITH
COPYRIGHT, 1929, 1934, BY WILLIAM ANTHONY GRANVILLE, PERCEY P. SMITH
AND WILLIAM RAYMOND LONGLEY
&tfienarum ffiregg
GINN AND COMPANY PRO-
PBIETOKS BOSTON U.S.A.
PREFACE
The alterations in the text in this revision consist of verbal
changes in the details of some of the proofs and discussions, and in
the addition of a chapter on Hyperbolic Functions. This chapter
has been written with the same attention to clearness and com-
pleteness that marks all other sections of the book. Also, cylindrical
coordinates have been employed to broaden the applications of
double integration.
The problems have in general been completely revised, and in
some respects their appeal and interest have been increased. Some
applications to the mathematics of economics will be found in the
problems.
Additional problems for the use of superior students have been
added at the end of most chapters.
Answers to a good many of the problems are given in the text.
Some of the answers are purposely omitted in order to give the
student greater self-reliance in checking his work. Teachers who
desire answers to the other problems should communicate with the
publishers.
The labor of the authors will be amply repaid if this revised edition
meets with the generous and well-nigh universal favor accorded Gran-
ville's Calculus since its first appearance.
PERCEY F. SMITH
WILLIAM R. LONGLEY
CONTENTS
DIFFERENTIAL CALCULUS
PAGE
CHAPTER I. COLLECTION OF FORMULAS 1
Formulas from elementary algebra and geometry, 1 Formulas from plane
trigonometry, 2 Formulas from plane analytic geometry, 3 Formulas
from solid analytic geometry, 4 Greek alphabet, 6
CHAPTER II. VARIABLES, FUNCTIONS, AND LIMITS 7
Variables and constants, 7 Interval of a variable, 7 Continuous vari-
ation, 7 Functions, 8 Independent and dependent variables. 8 Notation
of functions, 8 Division by zero excluded, 9 Graph of a function ; con-
tinuity, 10 Limit of a variable, 10 Limiting value of a function, 11 Theo-
rems on limits, 1 1 Continuous and discontinuous functions, 12 Infinity (oo),
13 Infinitesimals, 17 Theorems concerning infinitesimals and limits, 17
CHAPTER III. DIFFERENTIATION 19
Introduction, 19 Increments, 19 Comparison of increments, 20 Deriva-
tive of a function of one variable, 21 Symbols for derivatives, 22 Dif-
ferentiable functions, 23 General rule for differentiation, 23 Interpretation
of the derivative by geometry, 25
CHAPTER IV. RULES FOR DIFFERENTIATING ALGEBRAIC
FORMS 28
Importance of the General Rule, 28 Differentiation of a constant, 29
Differentiation of a variable with respect to itself, 29 Differentiation of
a sum, 30 Differentiation of the product of a constant and a function, 30
Differentiation of the product of two functions, 30 Differentiation of the
product of n functions, n being a fixed number, 31 Differentiation of a
function with a constant exponent. The Power Rule, 32 Differentiation
of a quotient, 32 Differentiation of a function of a function, 37 Differen-
tiation of inverse functions, 38 Implicit functions, 39 Differentiation of
implicit functions, 40
CHAPTER V. VARIOUS APPLICATIONS OF THE DERIVATIVE . 42
Direction of a curve, 42 Equations of tangent and normal ; lengths of sub-
tangent and subnormal, 43 Maximum and minimum values of a func-
tion ; introduction, 47 Increasing and decreasing functions. Tests, 50
Maximum and minimum values of a function ; definitions, 52 First
method for examining a function for maximum and minimum values.
Working rule, 53 - Maximum or minimum values when f'(x) becomes
infinite and f(x) is continuous, 55 Maximum and minimum values.
Applied problems, 57 Derivative as the rate of change, 64 - Velocity in
rectilinear motion, 65 Related rates, 67
viii CONTENTS
PAGE
CHAPTER VI. SUCCESSIVE DIFFERENTIATION AND APPLICA-
TIONS . ... 73
Definition of successive derivatives, 73 Successive differentiation of im-
plicit functions, 73 Direction of bending of a curve, 75 Second method for
testing for maximum and minimum values, 76 Points of inflection, 79
Curve-tracing, 81 Acceleration in rectilinear motion, 83
CHAPTER VII. DIFFERENTIATION OF TRANSCENDENTAL
FUNCTIONS. APPLICATIONS 86
Formulas for derivatives; second list, 86 The number c. Natural loga-
rithms, 87 Exponential and logarithmic functions, 89 Differentiation of
a logarithm, 90 Differentiation of the exponential function, 91 Differen-
tiation of the general exponential function. Proof of the Power Rule, 92
Logarithmic differentiation, 93 The function sin j*, 97 Theorem, 98
Differentiation of sin v, 99 The other trigonometric functions, 99 Dif-
ferentiation of cos ?;, 100 Proofs of formulas XV XIX, 100 Comments,
101 Inverse trigonometric functions, 105 Differentiation of arc sin r,
106 Differentiation of arc cos r, 106 Differentiation of arc tan v, 107
Differentiation of arc ctn v, 108 Differentiation of arc sec r and arc esc v,
108 Differentiation of arc vers r, 109
CHAPTER VIII. APPLICATIONS TO PARAMETRIC EQUATIONS,
POLAR EQUATIONS, AND ROOTS 115
Parametric equations of a curve. Slope, 115 Parametric equations.
Second derivative, 119 Curvilinear motion. Velocity, 120 Curvilinear
motion. Component accelerations, 121 Polar coordinates. Angle between
the radius vector and the tangent line, 123 Lengths of polar sub-
Uingent and polar subnormal, 126 Real roots of equations. Graphical
methods, 128 Second method for locating real roots, 130 Newton's
method, 131
CHAPTER IX. DIFFERENTIALS 136
Introduction, 136 Definitions, 136 Approximation of increments by
means of differentials, 137 Small errors, 138 Formulas for finding the
differentials of functions, 140 Differential of the arc in rectangular coordi-
nates, 142 Differential of the arc in polar coordinates, 144 Velocity as
the time-rate of change of arc, 146 Differentials as infinitesimals, 146
Order of infinitesimals. Differentials of higher order, 148
CHAPTER X. CURVATURE. RADIUS AND CIRCLE OF CURVA-
TURE . . . .149
Curvature, 149 Curvature of a circle, 149 Formulas for curvature; rec-
tangular coordinates, 150 Special formula for parametric equations, 151
Formula for curvature ; polar coordinates, 1 51 Radius of curvature, 1 52
Railroad or transition curves, 152 Circle of curvature, 153 Center of cur-
vature, 157 E volutes, 158 Properties of the evolute, 162 Involutes and
their mechanical construction, 163 Transformation of derivatives, 166
CONTENTS ix
PAGE
CHAPTER XL THEOREM OF MEAN VALUE AND ITS APPLICA-
TIONS . . . . 169
Rolle's Theorem, 169 Osculating circle, 170 Limiting point of intersec-
tion of consecutive normals, 171 Theorems of Mean Value (Laws of
the Mean), 172 Indeterminate forms, 174 Evaluation of a function
taking on an indeterminate form, 174 Evaluation of the indeterminate
form -> 175 Evaluation of the indeterminate form , 17$ . Evaluation
of the indeterminate form QO, 178 Evaluation of the indeterminate
form oc - oo, 179 Evaluation of the indeterminate forms 0, 1, 00,
180 The Extended Theorem of Mean Value, 182 Maxima and minima
treated analytically, 182
INTEGRAL CALCULUS
CHAPTER XII. INTEGRATION; RULES FOR INTEGRATING
STANDARD ELEMENTARY FORMS . . . 187
Integration, 187 Constant of integration. Indefinite integral, 189
Rules for integrating standard elementary forms, 190 Formulas (3), (4),
(5), 193 Proofs of (0) and (7), 198 Proofs of (8) (17), 200 Proofs of
(18) (21 ), 203 Proofs of (22) and (23), 211 Trigonometric differentials,
213 Integration of expressions containing Vu- ?/ 2 or V ' u' z a 2 by
trigonometric substitution, 221 Integration by parts, 223 Comments, 227
CHAPTER XIII. CONSTANT OF INTEGRATION 229
Determination of the constant of integration by means of initial condi-
tions, 229 Geometrical signification of the constant of integration, 229
Physical signification of the constant of integration, 233
CHAPTER XIV. THE DEFINITE INTEGRAL 237
Differential of the area under a curve, 237 The definite integral, 237
Calculation of a definite integral, 239 Change in limits corresponding to
change in variable, 240 Calculation of areas, 241 Area when the equa-
tions of the curve are given in parametric form, 242 Geometrical repre-
sentation of an integral, 244 Approximate integration ; trapezoidal rule,
245 Simpson's rule (parabolic rule), 247 Interchange of limits, 249
Decomposition of the interval of integration of the definite integral, 250
The definite integral a function of its limits, 250 Improper integrals. In-
finite limits, 250 Improper integrals. When // <(x) is discontinuous, 251
CHAPTER XV. INTEGRATION A PROCESS OF SUMMATION . . 254
Introduction, 254 The Fundamental Theorem of Integral Calculus, 254
Analytical proof of the Fundamental Theorem, 257 Areas of plane curves ;
rectangular coordinates, 258 Areas of plane curves ; polar coordinates,
262 Volumes of solids of revolution, 265 Length of a curve, 271
Lengths of plane curves ; rectangular coordinates, 272 Lengths of plane
curves ; polar coordinates, 274 Areas of surfaces of revolution, 277 Solids
with known parallel cross sections, 283
x CONTENTS
PAGE
CHAPTER XVI. FORMAL INTEGRATION BY VARIOUS DEVICES 289
Introduction, 289 Integration of rational fractions, 289 Integration by
substitution of a new variable ; rationalization, 296 Binomial differentials,
299 Conditionsof rationalization of the binomial differential, 302- Transfor-
mation of trigonometric differentials, 303 Miscellaneous substitutions, 305
CHAPTER XVII. REDUCTION FORMULAS. USE OF TABLE OF
INTEGRALS 307
Introduction, 307 Reduction formulas for binomial differentials, 307
Reduction formulas for trigonometric differentials, 312 Use of a table of
integrals, 315
CHAPTER XVIII. CENTROIDS, FLUID PRESSURE, AND OTHER
APPLICATIONS 320
Moment of area ; centroids, 320 * Centroid of a solid of revolution, 323
Fluid pressure, 325 Work, 328 Mean value of a function, 333
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER XIX. SERIES 338
Definitions, 338 The geometric series, 339 - Convergent and divergent
series, 340 General theorems, 341 Comparison tests, 342 Cauchy's test-
ratio test, 345 Alternating series, 347 Absolute convergence, 348
Summary, 348 Power series, 350 - The binomial series, 353 Another
type of power series, 354
CHAPTER XX. EXPANSION OF FUNCTIONS 357
Maclaurin's series, 357 Operations with infinite series, 362 Differentia-
tion and integration of power series, 365 Approximate formulas derived
from Maclaurin's series, 367 Taylor's series, 369 - Another form of
Taylor's series, 371 Approximate formulas derived from Taylor's series, 372
CHAPTER XXL ORDINARY DIFFERENTIAL EQUATIONS ... 375
Differential equations order and degree, 375 Solutions of differential
equations. Constants of integration, 376 Verification of the solutions of
differential equations, 377 Differential equations of the first order and of
the first degree, 378 Two special types of differential equations of higher
order, 387 Linear differential equations of the second order with constant
coefficients, 390 Applications. Compound-interest law, 399 - Applications
to problems in mechanics, 402 Linear differential equations of the nth
order with constant coefficients, 407
CHAPTER XXII. HYPERBOLIC FUNCTIONS 414
Hyperbolic sine and cosine, 414 Other hyperbolic functions, 415 - Table
of values of the hyperbolic sine, cosine, and tangent. Graphs, 415 Hyper-
bolic functions of v + w, 417 Derivatives, 420 Relations to the equi-
lateral hyperbola, 420 Inverse hyperbolic functions, 423 Derivatives
(continued), 425 Telegraph line, 428 Integrals, 430 Integrals (con-
tinued), 432 The gudermannian, 435 Mercator's Chart, 438 Relations
between trigonometric and hyperbolic functions, 440
CONTENTS xi
PAGE
CHAPTER XXIII. PARTIAL DIFFERENTIATION 444
Functions of several variables. Continuity, 444 * Partial derivatives, 445
Partial derivatives interpreted geometrically, 446 The total differential,
449 Approximation of the total increment. Small errors, 451 Total de-
rivatives. Rates, 455 - Change of variables, 457 - Differentiation of implicit
functions, 458 Derivatives of higher order, 462
CHAPTER XXIV. APPLICATIONS OF PARTIAL DERIVATIVES 466
Envelope of a family of curves, 466 The evolute of a given curve con-
sidered as the envelope of its normals, 469 Tangent line and normal plane
to a skew curve, 471 Length of arc of a skew curve, 473 Normal line and
tangent plane to a surface, 475 Geometric interpretation of the total dif-
ferential, 477 - Another form of the equations of the tangent line and
normal plane to a skew curve, 480 Law of the Mean, 482 Maxima and
minima of functions of several variables, 483 - Taylor's theorem for func-
tions of two or more variables, 488
CHAPTER XXV. MULTIPLE INTEGRALS 491
Partial and successive integration, 491 - Definite double integral. Geomet-
ric interpretation, 492 Value of a definite double integral taken over a
region S, 497 - Plane area as a definite double integral. Rectangular coordi-
nates, 498 Volume under a surface, 501 Directions for setting up a double
integral, 503 Moment of area and centroids, 503 - Theorem of Pappus,
504 - Center of fluid pressure, 506 - Moment of inertia of an area, 508
Polar moment of inertia, 510 Polar coordinates. Plane area, 512
Problems using polar coordinates, 514 General method for finding the
areas of curved surfaces, 517 - Volumes found by triple integration, 521
Volumes using cylindrical coordinates, 524
CHAPTER XXVI. CURVES FOR REFERENCE 531
CHAPTER XXVII. TABLE OF INTEGRALS 538
INDEX 553
DIFFERENTIAL CALCULUS
CHAPTER I
COLLECTION OF FORMULAS
1. Formulas from elementary algebra and geometry. For the con-
venience of the student we give in Arts. 1-4 the following lists of
formulas. We begin with algebra.
(1) Quadratic AJT* + B.r + C = 0.
Solution. 1. By factoring : Factor Ax' 2 4- Bx 4- r, set each factor equal
to zero, and solve for x.
2. By completing the square : Transpose r, divide by the coefficient
of J 2 , add to both members the square of half the coefficient of x, and
extract the square root. _ R
3. By the formula x =
Nature of the roots. The expression Ii 2 4 AC beneath the radical in
the formula is called the di^crlmhuint. The two roots are real and unequal,
real and equal, or imaginary, according as the discriminant is positive,
zero, or negative.
(2) Logarithms
log ab = Jog a -f log b. log a 7 ' = n log a. log 1=0.
'a = -
(3) Binomial theorem (n being a positive integer)
Jog ^ = log a log b. log v'a = - log a. log a a = 1.
|
(4) Factorial numbers. n! = [n = l-2-3-4---(w- l)w.
In the following formulas from elementary geometry, r or R
denotes radius, a altitude, B area of base, and s slant height.
(5) Circle. Circumference = 2 Trr. Area = Trr 2 .
(6) Circular sector. Area = \ r 2 a, where a = central angle of the sector
measured in radians.
1
DIFFERENTIAL CALCULUS
(7) Prism. Volume = Ba.
(8) Pyramid. Volume \ Ba.
(9) Right circular cylinder. Volume ?rr 2 a. Lateral surface = 2 Trra.
Total surface 2 irr(r + a).
(10) Right circular cone. Volume = J irr 2 a. Lateral surface = Trrs.
Total surface irr(r -f s).
(11) Sphere. Volume = :! Trr 3 . Surface = 4 Trr 2 .
(12) Frustum of a right circular cone. Volume = J ?ra(fi 2 -f r 2 -f fir).
Lateral surface irs(R -f r).
2. Formulas from plane trigonometry. Many of the following for-
mulas will be found useful.
(1) Measurement of angles. There are two common methods of measur-
ing angular magnitude ; that is, there are two unit angles.
Degree measure. The unit angle is ;i ( l of a complete revolution and is
called a degree.
Circular measure. The unit angle is an angle whose subtending arc is
equal to the radius of that arc, and is called a radian.
The relation between the unit angles is given by the equation
180 degrees TT radians (TT 3.14159 ),
the solution of which gives
1 degree = T^ = 0.0174
1 80
radian ; 1 radian = 57.29 degrees.
From the above definition we have
Number of radia.ns In an angle =
subtend hiq ore
radius
These equations enable us to change from one measurement to another.
(2) Relations
ctn .r
1
tan
sec x . csc r ~
, sin x ,
tan j* ; ctn .r
cos JT
cos x
sin x
cos x sin x
sin 2 x -f cos 2 x = 1 ; 1-1- tan 2 x = sec 2 .r ; 1 -f ctn 2 x = esc 2 x.
(3) Formulas for reducing angles
Angle
Sine
Cosine
Tangent
Cotangent
Secant
Cosecant
x
sin .r
cos .r
tan x
ctn x
soc .r
CSC X
90 - x
cosx
sin jc
ctn x
tanx
CSC X
secx
90 -f a-
COS JT
sin .r
ctn .r
tan x
CSC J"
sec x
180 - x
sin .r
cos x
tan x
ctn .r
sec .r
CSC X
180 -f jc
sin j-
COS X
tan x
ctn x
sec x
CSC X
270 - r
cos x
sin .r
ctn j-
tan .r
esc .r
sec x
270 4- f
cos .r
sin x
ctn jc
t an x
CSC X
sec x
360 - y
sin jc
cos x
tan .r
ctn .r
sec x
esc a;
COLLECTION OF FORMULAS {
(4) Functions of (x -f y) and (x y)
sin (x -f- y) sin x cos y + cos x sin y.
sin (.r #) = sin .r cos # - cos .r sin y.
cos (x -f /y) = cos .r cos // - sin x sin //.
cos (.r - #) = cos .r cos // -f sin .r sin ?/.
tan (jr + y} - ^L^^UL . tau ^ __ } = _lan .r - tan y .
1 - tan jr tan // * 1 -f tan .r tan y
(5) Functions of 2 x and | x
sin 2s = 2 sin j- cos jc ; cos 2 .r cos 2 jc - sin 2 x ; tan 2 .r - __r tan ^ ..
1 tan 2 x
fl^L-' :
\ -
; cos =, : tun =
- - \ - 2
sin-' j' - i - 1 cos 2 .r ; COS L> .r -- J + .1 cos 2 j-.
(6) Addition theorems
sin jr -f sin // = '2 sin i (.r -f- //) cos \ (.r ?/).
.sin x sin // -. 2 cos .1 (.> -f //)sin ! (.r //).
cos .r -f cos // L* co^ \ ( ./ -f // ) cos \ (j- ?/).
cos.r- cos// ^ i? yiu ', (./ ' //)sin ^(.r - jy).
(7) Relations for any triangle
Law of sine*. -. = - - -^~
sin .4 sin /> sin C
Law of cos hi ex. a 2 h- -f r 2 2 be cos A.
Formulas for area.
sin (/> 4- T)
- a) (a />)(.-; r), where ,s ?, (a -f b -f- r).
3. Formulas from plane analytic geometry. The more important for-
mulas are given in the following list.
(1) Distance between two points /*i(xi, yi) and P2(x2, yz)
Slope of P^Pz.
Midpoint. x-\ (x, + x 2 ), ?/ = 2 (2/i -f 2/2).
(2) Angle between two lines
1 -f mim2
(For parallel lines, wi = W2 ; for perpendicular lines, mint = 1.)
DIFFERENTIAL CALCULUS
(3) Equations of straight lines
Point-slope form. y y\ = m(x xi).
Slope-intercept form. y mx + b.
Two-point form. y "" yi = y ' z ~ ^ -
x x\ x% Xi
Intercept form. - -f ~ = 1.
(4) Perpendicular distance from the line Ax + By -f C = to PI(JCI, yi)
(5) Relations between rectangular and polar coordinates
(6) Equation of the circle
Center (h, k). (JT - h)' 2 + (y - k) 2 = r 2 .
(7) Equations of the parabola
Vertex the origin. y' 2 = 2 px, focus (\ p, 0).
x' 2 2 py, focus (0, i /;).
Vertex (h, k). (y k) 2 = 2 p(x h), axis // = /c.
(x h}' 2 2 p(y k}, axis x /i.
Axis ///r y-axiii. y ylj^ 2 -f C.
(8) Equations of other curves
Ellipse with center at the origin and with foci on the x-axis. (a > b)
Hyperbola with center at the origin and with foci on the x-axis.
^1 _ 111 i
a 2 b 2 ~
Equilateral hyperbola with center at the origin and with the coordinate
axes for asymptotes.
xy = C.
See also Chapter XXVI.
4. Formulas from solid analytic geometry. Some of the more impor-
tant formulas are given.
(1) Distance between PI(XI, yi, z\) and P2(*2, 1/2, 22)
d = V(xi - x 2 ) 2 -h (yi - l/ 2 ) 2 + (21 - z 2 ) 2 .
COLLECTION OF FORMULAS
(2) Straight line
Direction cosines : cos a, cos (3, cos 7.
Direction numbers: a, b, c.
Then
a 6 r
cos 2 a- -f cos 2 ft -f cos 2 7 = 1.
a
cos a =
cos ft -
cos 7 =
i 2 + b' 2 + r 2
b
Va 2 -f h 2
r
Vu 2 ~+ 6- + r 2
For the line joining (jc\, ?/i, cO and (j- 2 , ?/ L ., -.,)
cos or _ cos ft __ cos 7
(3) Two lines
Direction cosines : cos <rv, cos ft, cos 7 ; cos <v', cos ft', cos 7'.
Direction numbers: a, b, r; a', //, r'.
If = angle between the lines,
cos = cos a cos <V -f cos ft cos ft' -f cos 7 cos 7',
c/a' 4- W/ -f re'
cos t/ =
2 + b- + r 2 Va /2 "-f // 2 -f r' 2
Parallel lines. - = ~ = -^ .
a' 6' r x
Perpendicular lincx. aa r -f 6/>' -f rr' ~ ().
(4) Equations of the straight line with direction numbers a, b t c pass-
ing through (xi, z/i, 21)
x - n _ y - 7/1 _ z - :\ t
a b c
(5) Plane. For the plane Ax -f Ky + Cz + J>-() the coefficients A, B, C
are the direction numbers of a line perpendicular to the plane.
Equation of a plane passing throuyJi (>i, //j, ^ij and perpendicular to the
line with direction numbers A, B, C.
A(x - XL) + B(y - ?/,) -f C(z - 2,) = 0.
(6) Two planes
Equations : Ax + By -f Cz 4- D = 0,
A'x+ B'y+ C'z+ D' = ().
Direction numbers of the line of intersection :
EC' - CB' t CA' - AC', AB' - BA'.
DIFFERENTIAL CALCULUS
If 6 = angle between the planes, then
AA' + BB' + CC'
cos =
B' 2
C /2
(7) Cylindrical coordinates.* The distance z of a point P(x, y, z) from
the A' 7-plane and the polar coordinates (p, 6) of
its projection (x, y, 0) on the X T-plane are called
the cylindrical coordinates of P. The cylindrical
coordinates of P are written (p, 6, z).
If the rectangular coordinates of P are x, y, z,
then, from the definitions and the figure, we have
x = p cos 0,
p 2 -
+
y P sin 0, 2
9 arc tan -
x
(8) Spherical coordinates.* The radius vec-
tor r of a point P, the angle between OP and
the z-axis, and the angle between the projec-
tion of OP on the AT-plane and the .r-axis are
called the spherical coordinates of /'. </> is called
the colatitude and 6 the longitude. The spherical
coordinates of P are written (r, (/>, 0).
If the rectangular coordinates of P are x, y, z,
then, from the definitions and the figure, we have
= r sin c/> cos
f * =:
?/ = r sin sin $, 2 = r cos ;
6 = arc tan -> </> = arc tan
^
5. Greek alphabet
LETTERS
NAMES
LETTERS
NAMES
A a
Alpha
I
t
Iota
B ft
Beta
K
K
Kappa
r 7
Gamma
A
X
Lambda
A 6
Delta
M
M
Mu
E
Epsilon
N
V
Nu
z r
Zeta
2
Xi
H 77
Eta
Omicron
e e
Theta
n
IT
Pi
LETTERS NAMES
P p Rho
Z d s Sigma
T r Tau
T i> Upsilon
<f> Phi
X x Chi
^ i/' Psi
12 a Omega
* For a discussion of cylindrical and spherical coordinates see Smith, Gale, and
Neellry's "New Analytic Geometry, Revised Edition" (Ginn and Company), pp. 320-322.
CHAPTER II
VARIABLES, FUNCTIONS, AND LIMITS
6. Variables and constants. A variable is a quantity to which an
unlimited number of values can be assigned in an investigation.
Variables are denoted usually by the later letters of the alphabet.
A quantity whose value is fixed in any investigation is called a
constant.
Numerical or absolute constants retain the same values in all prob-
lems; as 2, 5, V?, TT, etc.
Arbitrary constants are constants to which numerical values may
be assigned, and they retain these assigned values throughout the
investigation. They are usually denoted by the earlier letters of
the alphabet.
Thus, in the equation of a straight line,
x and y are the variable coordinates of a point moving along the
line, while the arbitrary constants a and b are the intercepts, for
which definite values are assumed.
The numerical (or absolute) value of a constant a, as distinguished
from its algebraic value, is represented by |a|. Thus, | 2| = 2 = |2|.
The symbol |a| is read "the numerical value of a."
7. Interval of a variable. Very often we confine ourselves to a por-
tion only of the number system. For example, we may restrict our
variable so that it shall take on only values lying between a and b.
Also, a and b may be included, or either or both excluded. We shall
employ the symbol [a, 6], a being less than b, to represent the num-
bers a, b, and all the numbers between them, unless otherwise stated.
This symbol [a, 6] is read u the interval from a to b."
8. Continuous variation. A variable x is said to vary continuously
through an interval [a, 6] when x increases from the value a to the
value b in such a manner as to assume all values between a and 6
in the order of their magnitudes, or when x decreases from x = 6 to
x = a, assuming in succession all intermediate values. This may be
illustrated geometrically by the diagram on page 8.
8 DIFFERENTIAL CALCULUS
The origin being at O, lay off on the straight line the points A
and B corresponding to the numbers a and b. Also, let the point
P correspond to a particular
value of the variable x. Evi- g ^ - ^
dently the interval [a, b\ is rep-
resented by the segment AH. As x varies continuously through
the interval [a, H the point P generates the segment AB when x
increases, or the segment BA when x decreases.
9. Functions. When two variables are so related that the value of
the first variable is determined when the value of the second variable
is given, then the first variable is said to be a function of the second.
Nearly all scientific problems deal with quantities and relations of
this sort, and in the experiences of everyday life we are continually
meeting conditions illustrating the dependence of one quantity on
another. For instance, the weight a man is able to lift depends on
his strength, other things being equal. Similarly, the distance a boy
can run may be considered as depending on the time. Or we may
say that the area of a square is a function of the length of a side, and
the *o!ume of a sphere is a function of its diameter.
10. Independent and dependent variables. The second variable, to
which values may be assigned at pleasure within limits depending
on the particular problem, is called the independent variable, or
argument; and the first, variable, whose value is determined when
the value of the independent variable is given, is called the dependent
variable, or junction.
Frequently, when we are considering two related variables, it is
in our power to fix upon either as the independent variable; but
having once made the choice, IM> change of independent variable is
allowed without certain precautions and transformations. For ex-
ample, the area of a square is a function of the length of its side.
Conversely, the length of a side is a funrlJO'i of the area.
11. Notation of functions. The symbol /(:r) is used to denote a func-
tion of J, and is read / of x. In order to distinguish between different
functions, the prefixed letter is changed, as F(.r), (/>(;r), /'(*), etc.
During any investigation a functional symbol indicates the same
law of dependence of the function upon the variable. In the simpler
cases this law takes the form of a series of analytical operations upon
the variable. Hence, in such a case, the functional symbol will indicate
the same operations or series of operations applied to different values
of the variable. Thus, if
J(x) =r j2 - 9 .r + 14,
then f(y) = y 2 9 y + 14.
VARIABLES, FUNCTIONS, AND LIMITS 9
Also /(&+l) = (6 + l) a -9ifc + l) + 14 = & 2 -7& + 6,
/(0)=0 2 -9. 0+14-14,
/(- 1) = (- 1)2 - 0(- 1) + 14 - 24,
/(3) = 3-- 9-3 + 14 --4.
12. Division by zero excluded. The quotient of two numbers a and
6 is a number x such that a = />.r. Obviously, division by zero is
ruled out by this definition. For if l> o, and we recall that any
number times zero equals zero, we see that .r (ioes not exist unless
a = 0. But, in this case, jc may be any number whatever. The forms
a
()' ()'
are, therefore, meaningless.
Care should he taken not to divide hy zero inadvertently. The fol-
low ing fallacy is an illustration.
Assume that a b.
Then, evidently, ah = <i~.
Subtracting />-, ab b- a' 2 b' 2 .
Factoring, b(<i b) = (a -f />)(</ />).
Dividing by a b, l> a + b.
But a - b ;
therefore b 2 />,
or 1 --- 2.
The ahsurd result is due to the fuel that we divided by a b 0.
PROBLEMS
1. Given f(s) = jr* - 5 s- - 1 .r -} 20, show that
/(I) = 12, /(5) = 0, /(()) - - 2/C}), f^} ^ - r )/(- 1).
2. Itfis) = 4 - 2 s- -f r 1 , find /(O ),/(!),/(- l),/(2 ),/(-- 2).
3. If F(0) = sin 2 + cos 6, find /-'(O), /M TT), F(TT).
4. Given /(.r) ,r :( 5 jr' 2 4 jr -f 20, show that
/(/-f 1) = / ;! - 2 ^ - 11 /+ 12.
5. Given /(//) ?/ 2 2 y -f 6, show r that
f(y -f //) = ?/ L> - 2 ?/ -f f> -f 12 O/ ~ 1 J/' + h 2 -
6. Given /(j*) = ?'*> -f 3 j-, show that
fU -f /O - f(s) = Ms 2 + 1 ;// -f 3 ^ 2 -f R
7. Given /(r) = -> show that /(V -f /O - /O) - -- -
^
.
Jf J^ ~T" X/
8. Given 0(z) = 4* f show that <f>(z + 1) - <t>(z) = 3 0().
10 DIFFERENTIAL CALCULUS
9. If <f>(x) = a', show that 0(y) - </>(*) = 0(y + 2).
1 x
10. Given 0(x) = log show that
11. Given /(x) = sin x, show that
f(x -f 2 h) f(f) = 2 cos (x -f /?) sin /?,.
HINT. Use (6), p. .'}.
13. Graph of a function; continuity. Consider the function x 2 ,
and let
(1) y = x' 2 .
This relation gives a value of y for any value of x ;
that is, y is defined by (1) for all values of the inde-
pendent variable. The locus of (1), a parabola (see
figure), is called the yraph of the function x 2 . If x
varies continuously (Art. 8) from x a to x = />, then
y will vary continuously from y a- to y/ = &-', and
the point /'(x, ?/) will move continuously along the graph from the
point (a, a 2 ) to (h, />-). Also, a and b may have any values. We then
say, "the function x- is continuous for all
values of x."
Consider the function - and let
x
(2) y = --
This equation gives a value of y for any
value of x except x = (Art. 12). For x =
the function is not defined. The graph, the locus
of (2), is an equilateral hyperbola (see figure). If x increases con-
tinuously through any interval [a, />] which does not include x 0,
then '// will decrease continuously from - to p and the point P(x, y)
will trace the graph between the corresponding points (a, -)> (?>, ^-
Then we say, "the function - is continuous for all values of x except
x 0." There is no point on the graph for x = 0.
These examples illustrate the concept of continuity of a function.
A definition is given in Art. 17.
14. Limit of a variable. The idea of a variable approaching a limit
occurs in elementary geometry in establishing a formula for the area
of a circle. The area of a regular inscribed polygon with any number
of sides n is considered, and n is then assumed to increase indefinitely.
VARIABLES, FUNCTIONS, AND LIMITS 11
The variable area then approaches a limit, and this limit is defined
as the area of the circle. In this case the variable v (the area) in-
creases constantly, and the difference a r, where a is the area of
the circle, diminishes and ultimately becomes less than any preas-
signed number, however small.
The relation illustrated is made precise by the
DEFINITION. The variable r is said to approach the constant / as
a limit when the successive values of r are such that the numerical
value of the difference r / ultimately becomes and remains less
than any preassigned positive number, however small.
The relation defined is written lim r -~ /. For convenience, we shall
use the notation r /, read, "r approaches / as a limit," or, more
briefly, "r approaches Z." (Some authors use the notation v == /.)
ILLUSTRATIVE EXAMPLE. Lot the valuer of r be
2 + 1, li -t i. 12 -f |, .... 2 + 1, - f
without end. Then, obviously, lim r - 2, or r U.
If we mark on a straight line, as in Art. S, the point /, correspond-
ing to the limit /, and from L lay off on each side a length 6, however
small, then the points determined by /' \\ill ultimately all lie within
the segment corresponding to the interval [/ e, / -1 e|.
15. Limiting value of a function. In applications, the situation that
usually arises is this. We have a variable r, and a given function z
of r. The independent variable r assumes values such that v /.
We then have to examine the values of the dependent variable z,
and, in particular, determine if z approaches a limit. If there is a
constant a such that lim z = a, then the relation described is written
lim z = a,
r- I
read, "the limit of z, as v approaches I, is a."
1G. Theorems on limits. In calculating the limiting value of a func-
tion, the following theorems may be applied. Proofs are given in Art. 20.
Suppose u, ?', and w are functions of a variable j, and suppose that
lim u = A, lim v B, lim w = C\
x - a x - a -r - a
Then the following relations hold.
(1) lim (u + v - w) = A -f B - C.
jc- a
(2) lim (uvw) = ABC.
x-* a
u A
(3) lim " = - 9 if B is not zero.
jt-a V B
12 DIFFERENTIAL CALCULUS
Briefly, in words, the limit of an algebraic sum, of a product, or
of a quotient is equal, respectively, to the same algebraic sum, product,
or quotient of the respective limits, provided, in the Last named, that the
limit of the denominator is not zero.
If c is a constant (independent of rj and B is not zero, then, from
the above,
c c
(4) lim (u + c) = A + c, lim cu cA, lim - =
x * a JT-* a a- -> a V Jj
Consider some examples.
1. Prove lim <JT'* -f- 4 a-) - 12.
JT *
Solution. The given function is the sum of .c- and 4 jc. We first find the limit-
ing values of these two functions. By rji,
lim jr' 2 = 4, since j : ' = x x.
X li
By (4), lim 4 a- = 4 lim x 8.
J" ~ X - li
Hence, by (1), the answer is 4 -f 8 -- 12.
~ : ' 9 r t
2. Prove lim ~ ~ -
Solution. (Considering the numerator, lim i^ :> 9) 5, by (2) and (4). For
the denominator, lim '.: 1 U) - 4. Hence, by '.'}), we have the re<juired result.
17. Continuous and discontinuous functions. In Ex. 1 of the pre-
ceding article, where it was shown that
lim (.r- + 4 .r) = 12,
.r 1!
we observe that the answer is the ralue of the function for x 2.
Thai is, the limiting value of the function when ;r approaches 2 as a
limit is equal to the value of the function for jc = 2. The function is
said to be continuous for .r = 2. The general definition is as follows.
DEFINITION. A function /(.r) is said to be continuous for r = a if
the limiting value of the function when .r approaches a as a limit is
the value assigned to the function for .r = a. In symbols, if
lim /(*)=/ (a),
x * a
then /(.r) is continuous for x a.
The function is said to be discontinuous for x = a if this condition
is not satisfied.
Attention is called to the following two cases of common oc-
currence.
VARIABLES, FUNCTIONS, AND LIMITS 13
CASE I. As an example illustrating a simple case of a function con-
tinuous for a particular value of the variable, consider the function
J x ' X *>
For z = 1, f(x) = /(I) = 3. Moreover, if x approaches 1 as a limit,
the function f(x) approaches 3 as a limit (Art. 16). Hence the
function is continuous for x = 1.
CASE II. The definition of a continuous function assumes that
the function is already defined for x-o. If this is not the case, how-
ever, it is sometimes possible to assign such a value to the function
for x = a that the condition of continuity shall be satisfied. The
following theorem covers these cases.
Theorem. If f(x) is not dcjiucd for x = a, and if
lim f(x) = B,
x a
then f(x) mil be continuous for x = a if tt is assumed as the value of
f(x) for x = a.
1> A
Thus, the function ~
is not, defined for x 2 (since then there would be division by zero).
T.ut for every other value of :r,
and lim (x + 2) = 4 ;
therefore lim
Although the function is not defined for x = 2, if we arbitrarily as-
sign to it the value 4 for x 2, it becomes continuous for this value.
A function f(x) is said to be coHtitiuoux in an niter ml whew it is
continuous for all valves of x in this in terra I.*
In the calculus we have to calculate frequently the limiting value
of a function of a variable ?' when r approaches as a limit a value a
lying in an interval in which the function is continuous. This limit-
ing value is the value of the function for r = a.
18. Infinity (oo). If the numerical value of a variable v ultimately
becomes and remains greater than any preassigned positive number,
* In this book we shall deal only with functions which are in general continuous, that is,
continuous for all values of x, with the possible exception of certain isolated values, our
results in general being understood as valid only for those values of x for which the function
in question is actually continuous.
14 DIFFERENTIAL CALCULUS
however large, we say v becomes infinite. If v takes on only positive
values, it becomes positively infinite; if negative values only, it
becomes negatively infinite. The notation used for the three cases is
lim v = QO , lim v = + oo , lim r = 00 .
In these cases v does not approach a limit as defined in Art. 14. The
notation lim v = QO, or v > oo, must be read " v becomes infinite/'
and not "v approaches infinity."*
We may now write, for example,
lim - = oo,
x *<)
meaning that - becomes infinite when x approaches zero.
Referring to Art. 17, it appears that if
lim/(.r) = 00,
x - a
that is, if f(x) becomes infinite as x approaches a as a limit, then
f(x) is discontinuous for x a.
A function may have a limiting value when the independent
variable becomes infinite. For example,
lim i = 0.
X oc X
And, in general, if /O) approaches the constant value A as a limit
when x > oo, we use the notation of Art. 17 and write
lim /(*) = A.
x -- oc
Certain special limits occur frequently. These are given below. The
constant r is not zero.
Written in the form, of limits Abbreviated form often used
(1) lim - = oo. =oo.
r - V
(2) limczj = oo. c oo = oo.
r -* oo
/o\ r V OO
(o) Jim - = oo. = oo.
r -- oo C C
(4) lim - = 0. - = 0.
" 00
* On account of the notation used and for the sake of uniformity, the expression
p * -f oo is sometimes read " P approaches the limit plus infinity." Similarly, r * oo is
read "r approaches the limit minus infinity," and v is read " r, in numerical value,
approaches the limit infinity."
This phraseology is convenient, but the student must not forget that infinity is not a
limit, for infinity is not a number at all.
VARIABLES, FUNCTIONS, AND LIMITS 15
These special limits are useful in finding the limiting value of the
quotient of two polynomials when the variable becomes infinite.
The following example will illustrate the method.
ILLUSTRATIVE EXAMPLE. Prove lim "L
"*" ~ tS
. - _.
5 x - x- - 7 x* 7
Solution. Divide numerator and denominator by x*\ the highest power of x
present in either. Then we have
J* - J* - < X-*
The limit of each term in numerator or denominator containing x is zero, by (4).
Hence, by (1) and (3) of Art. H> we obtain the answer. Jri any similar case the
first step is therefore as follows.
I>iride both numerator and denominator by the highest power of the variable
occurring in cither.
If u and v are functions of jr, and if
lim u = A, lim r -= 0,
x * a a- - (i
and if A is not zero, then
r u
lim - =: oo.
r . tf
This notation provides for the exceptional case of (3), Art. 16,
when B = and A is not zero. See also Art. 20.
PROBLEMS
Prove each of the following statements,
Proof. lim ^ = lim
x
[Dividing both numerator and denominator hy x 2 .]
The limit of each term in numerator and denominator containing x is zero, by
(4). Hence, by (1) and (3), Art. 16, we obtain the answer.
2. lim X = 2.
r 4 t 2 + 3 / + 2 1
3 - J ja + 2 / - 6 = "" 3*
. ,. r 2 /? 4- 3 x/z 2 -f h* x
4. lim r . _ , 2 = -
16 DIFFERENTIAL CALCULUS
. ,. 3 h + 2 xh* + s'W 1 _ ,. 4j/ 2 -3 _ _
M 4-3*A-2^ = - 27 7 ',! 2i^ + 3* ~ "
6- lim ft'****'- 4 **'' = 1. 8. Km */-*/* + ! = 3.
A-.O 2 z(2z - /r; J x-,* 2^* + 4z- 7
9 jj m ft"- 7 "' -f Qi^" ] + + a, < = ao
', . ' &./" -I- h.r" '+ + & 60
-f
. r
, . , (L'" 4 cs* -f
, r ar 4 -f ^^ -f r A
11. 11 in - --- : - r- = 0.
, r ^^ 4- kr 1 ' -f ^ l
12. lim - ----- ; - 7 - = oo.
.1 . fj (if -f f'J" -f JS -f {/
-
13. lim- ------ ~
.s . a * a-
, ]j rn _^Z_ Hj_ nj ,n i^ ( n positive integer)
//
S- (>_ 5
.7 ..... " ~*
* - 4 4
ir. i: m S* 4 S-
lc. 11 III ........ .
i r
16. lim
Proof. The limiting value cannot" bo found by substituting h = Q, for we then
obtain (Art. lli) the indeterminate form - We then transform the expression in
a suitable manner as indicated below, namely, rationalize the numerator.
\ __x -f h - \_JT \ jr -f // -f \ J J -f ^ - .r ]_
\ JT
TT ,. \ .r + // \ r ,. 1 1
J Ten ce lim -- - ---- = lim : --- - = = -
/* *> /z /*-o\ j- -f h + \ x 2v
17. Given /(j) = .r 2 , show that
/j -* W
18. Given /(j) = a.r 2 -f 6.r -f r, show that
19. Given /(or) - show that
*~o h
20. If f(x) = x*, find lim /(* + *)
VARIABLES, FUNCTIONS, AND LIMITS 17
19. Infinitesimals. A variable r which approaches zero as a limit
is called an infinitesimal. This is written (Art. 14)
lim r = or r 0,
and means that the numerical value of r ultimately becomes and
remains less than any preassigned positive number, however small.
If lim r = /, then lim (r /) = (); that is, the difference between a
variable and its Inn it is an infinitesimal.
Conversely, if the different bit ween a variable and a constant is an
infinitesimal, then the variable approaches the constant as a limit.
20. Theorems concerning infinitesimals and limits. In the following
considerations all variables are assumed to be functions of the same
independent variable and to approach their respective limits as this
variable approaches a fixed value a. The constant t is a preassigned
positive number, as small as we please, but not zero.
We first prove four theorems on infinitesimals.
I. An algebraic sum of u infinitesimals is an infinitesimal, n being
a fixed number.
For the numerical value of the sum will become and remain less
than e when the numerical value of each infinitesimal becomes and
remains less than -
n
II. The product of n constant c by an infinitesimal is an infinitesimal.
For the numerical value of the product will be less than e when
the numerical value of the infinitesimal is less t han ,---
kl
III. The product of n infinitesimals is an, infinitesimal, n being a
fixed number.
For the numerical value of the product will become and remain
less than e when the numerical value of each infinitesimal becomes
and remains less than the r/th root of e.
IV. If lim v = /, and I is not zero, ihen the quotient of an infinitesimal
i by v is also an infinitesimal.
For we can choose a positive number r, numerically less than /,
such that the numerical value of v ultimately becomes and remains
greater than r, and also such that the numerical value of i becomes
and remains less than ce. Then the numerical value of the quotient
will become and remain less than .
Proofs of the theorems of Art. 16. Let
(1) u A i, v B = j, w C = k.
18 DIFFERENTIAL CALCULUS
Then i, j, k are functions of x, and each approaches zero as x > a\
that is, they are infinitesimals (Art. 19). From equations (1), we
obtain
(2) u -f v w ( A + B C) = i + j k.
The right-hand member is an infinitesimal by theorem I above.
Hence, by Art. 19,
(3) lim (u -h v w) = A -f- B C.
From (1), we have u = A -{- i, v=B+j. By multiplication and
transposing AH we get
(4) ur - A B = Aj -f Bi -f ij.
By the above theorems I III the right-hand member is an infinitesi-
mal, and hence
(5) lim uv = AB.
? a
The proof is readily extended to the product urw.
Finally, we may write
f("\ 1* _ 1 A -\- i _ A _ Bi Aj m
v B~~ B+j B ~ B(B +j)'
The numerator in (G) is an infinitesimal by theorems I and IT. By
(3) and (4), lim /*(/*+./) = B' 2 . Hence, by theorem IV, the right-
hand member in (6) is an infinitesimal, and
/r^ V U A
(7) hm - =
j- - v B
Hence the statements in Art. 16 are proved.
CHAPTER III
DIFFERENTIATION
21. Introduction. We shall now proceed to investigate the man-
ner in which a function changes in value as the independent variable
changes. The fundamental problem of the differential calculus is to
establish a measure of this change in the function with mathematical
precision. It was while investigating problems of this sort, dealing
with continuously varying quantities, that Newton* was led to the
discovery of the fundamental principles of the calculus, the most
scientific and powerful tool of the modern mathematician.
22. Increments. The increment of a variable in changing from one
numerical value to another is the difference found by subtracting the
first value from the second. An increment of x is denoted by the
symbol Ax, read "delta x." The student is warned against reading
this symbol "delta times :r."
Evidently this increment may be either positive or negative t
according as the variable in changing increases or decreases. Similarly,
Ay denotes an increment of //,
A(f> denotes an increment of 0,
A/(J) denotes an increment of /(j), etc.
If in y f(x) the independent variable x takes on an increment Ax,
then A?/ will denote the corresponding increment of the function f(x)
(or dependent variable y).
The increment Ay is always to be reckoned from the definite
initial value of y corresponding to the arbitrarily fixed initial value
of x from which the increment Ax is reckoned. For instance, consider
the function
* Sir Isaac Newton Cl 642-1 727), an Englishman, was a man of the most extraordinary
genius. He developed the science of the caloulas uml**r the name of Fluxions. Although
Newton had discovered and made use of the new science 5ts early as HJ70, his first published
work in which it occurs is dated 1687. having the title " Philosonhiao Naturalis Principia
Mathematical' This was Newton's principal work. Laplace said of it, " It will always remain
preeminent above all other productions of the human mind." See frontispiece.
t Some writers call a negative increment a decrement.
19
20
DIFFERENTIAL CALCULUS
Assuming x = 10 for the initial value of x fixes y = 100 as the
initial value of y.
Suppose x increases to x = 12, that is, Ax = 2 ;
then y increases to y 144, and Ay = 44.
Suppose x decreases to x = 9, that is, Ax = 1 ;
then y decreases to y = 81, and Ay = 19.
In the above example, ?/ increases when x increases and y decreases
when x decreases. The corresponding values of Ax and Ay have
like signs. It may happen that // decreases as x increases, or the
reverse ; in either case Ax and A// will then have opposite signs.
23. Comparison of increments. Consider the function
(1) = 3-2.
Assuming a fixed initial value for x, let x take on an increment Ax.
Then y will take on a corresponding increment A?/, and we have
or
Subtracting (1 )
(2)
?/ + Ay/ (x + Ax) 2 ,
y + Ay/ = x 2 + 2 x Ax + (Ax) 2 .
// = x 2
Ay/= 2x Ax+ (Ax) 2
we get the increment A// in terms of x and Ax.
To find the ratio of the increments, divide both members of (2)
by Ax, giving
-V = 2 x + Ax.
Ax
If the initial value of x is 4, it is evident (Art. 16) that
Let us carefully note the behavior of the ratio of the increments of
x and // as the increment of x diminishes.
Initial
New
Increment i Initial
New
, ,
Increment 1 Av
Value of x
Value of x
A*
Value of y
Value of y
Ay Ax
\
4
r>.o
1.0
16
25. I 9.
9.
4
4.8
0.8
! 1G
23.04
7.04
8.8
4
4.6
0.6
16
21.16
5.16
8.6
4
4.4
0.4
16
19.36
3.36
8.4
4
4.2
0.2
16
17.64
1.64
8.2
4
4.1
0.1
16
16.81
0.81
8.1
4
4.01
0.01
16
16.0801
0.0801
8.01
DIFFERENTIATION 21
It is apparent that as Ax decreases, A?/ also diminishes, but their
ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01,
illustrating the fact that the value of -'( can be brought as near to
8 as we please by making A.r sufficiently small. Therefore
lim ^-' = 8.
Az-0 AJ
24. Derivative of a function of one variable. The fundamental
definition of the ditlerential calculus is as follows.
The derivative* of ct function /,> //,v limit of the ratio of the increment
of the function to the increment of the independent variable, when the
latter increment varies and approudtcx zero ax a limit.
When the limit of this ratio exists, the function is said to be differ-
cntiable, or to possess a deriralire.
The above definition may be given in a more compact form sym-
bolically as follows. Given the function
0) ?/ = /U),
and consider x to have a fixed value.
Let take on an increment A.r ; then the function y takes on an
increment A?/, the new value of the function being
(2) ?/ + A// - /(.r -f A/).
To find the increment of the function, subtract (1) from (2), giving
(3) Av/-/(j + A.r)-/M.
Dividing both members by the increment of the variable, AJ,
(A\ A# = f(.r + A:r) f(jr)
( } Ax " A*
The limit of the right-hand member when AJ > (} is, from the
definition, the derivative of /(:r), or by (1 ), of ?/, and is denoted by the
symbol Therefore
,,, ^ dy r /(x + A*)-/(*)
(A) = hm
dx AX-O Ax
defines the derivative of y [or f(n)] with respect to x.
From (4) we get also
dy r m Ay
- = lim -
dx AX-.O AJC
* Also called the differential coefficient or the derived function.
22 DIFFERENTIAL CALCULUS
Similarly, if u is a function of t, then
= lim = derivative of u with respect to t.
at A/ -* o A
The process of finding the derivative of a function is called differ-
entiation.
25. Symbols for derivatives. Since Ay and Ax are always finite and
have definite values, the expression
Ax
is really a fraction. The symbol
dv
ax
however, is to be regarded not as a fraction but as the limiting value of
a fraction. In many cases it will be seen that this symbol does possess
fractional properties, and later on we shall show how meanings may
be attached to dy and dor, but for the present the symbol ^ is to be
considered as a whole. x
Since the derivative of a function of x is in general also a function
of y, the symbol f'(x) is also used to denote the derivative of f(x).
Hence, if f , ,
!/=/(*),
we may write -^ = /'(j),
which is read "the derivative of y with respect to x equals / prime
of x" The symbol ^
Tx
when considered by itself is called the differentiating operator, and
indicates that any function written after it is to be differentiated with
respect to x. Thus,
-JL or y indicates the derivative of y with respect to x ;
-T- f(x) indicates the derivative of f(x) with respect to x ;
(2 x 2 + 5) indicates the derivative of 2 x 2 + 5 with respect to x.
dy
y' is an abbreviated form of -?-
ax
The symbol D c is used by some writers instead of - If, then,
ax
DIFFERENTIATION 23
we may write the identities
It must be emphasized that the variable, in the essential step of
letting Ax > 0, is Ax, and not x. The value of the latter is assumed
fixed from the start. To emphasize that x = xo throughout, we may
write
26. Differentiable functions. From the Theory of Limits it is clear
that if the derivative of a function exists for a certain value of the
independent variable, the function itself must be continuous for that
value of the variable.
The converse, however, is not always true, functions having been
discovered that are continuous and yet possess no derivative. But
such functions do not occur often in applied mathematics, and in this
book only different! able functions arc considered, that is, functions that
possess a derivative for all values of the independent variable save at
most for isolated values.
27. General Rule for Differentiation. From the definition of a deriv-
ative it is seen that the process of differentiating a function y = f(x)
consists in taking the following distinct steps.
GENERAL RULE FOR DIFFERENTIATION*
FIRST STEP. In the Junction replace x by x + Ax, and cakulate the
new value of the function, y + Ay.
SECOND STEP. Subtract the given value of the function from the new
value and thus find Ay (the increment of the function).
THIRD STEP. Divide the remainder Ay (the increment of the function)
by Ax (the increment of the independent variable).
FOURTH STEP. Find the limit of this quotient when Ax (the incre-
ment of the independent variable) varies and approaches zero as a limit.
This is the derivative required.
The student should become thoroughly familiar with this rule by
applying the process to a large number of examples. Three such
examples will now be worked out in detail. Note that the theorems
of Art. 16 are used in the Fourth Step, x being held constant.
called the Four-Step Rule.
24 DIFFERENTIAL CALCULUS
ILLUSTRATIVE EXAMPLE 1. Differentiate 3 x' 2 4- 5.
Solution. Applying the successive steps in the General Rule, we get, after placing
y = 3 x'-' 4- 5.
Firxf Step. y 4- A^/ = 3(x 4- Ax) 2 4- 5
= 3 x*' 4- ft x Ax 4- 3( Ax)- 4- 5.
Second Step, y 4- A?/ - 3 x- -f 6 x Ax -f 3 (Ax)- 4- 5
]/ = 3 x 1 ' 4_5
A?/ = fix- Ax 4- 3(Ax)*
'/Vnrd S/f'/). -~ - 6x4-3- Ax.
Fourth Step. In the right-hand member let Ax * 0. Then by (A)
Or y' - (3 x- 4- 5) = 6 x.
ILLUSTRATIVE FA AMPLE 2. Differentiate x' 1 - 2 x 4- 7.
Solution. Place y = x ' - 2 x 4 7.
Fir*/ tf/cyj. // + A// = (x 4- A.r) ' - 2(x -f Ax) 4- 7
--- x 1 4 3 x- Ax -f 3 x ( Ax)- 4- (Ax) f - 2 x - 2 Ax 4- 7.
Second Step. // 4- A// = x ! 4- 3 jr-' Ax 4- 3 x ( Ax) J 4- ( Ax) ( - 2 x - 2 - Ax 4- 7
// Z,ZJ -2x 4- 7
A// -- 3 x- : Ax 4- 3 x ( Ax) 2 4- ( Ax)-' - 2 Ax
Third Step. -- - 3 x " 4- 3 x Ax 4- (Ax) ' - 2.
Fourth Step. In the right-hand member let Ax > 0. Then by (A)
Or y' - -j- (x' ! - 2 x 4 7) = 3 x- - 2.
ILLUSTRATIVE KXAMPLE 3. Differentiate--
x
Solution. Place // ~
x-
A'JS/ Step. y f A//
(x 4 A.r) a
Second Stej>. y 4~ A//
(x 4- Ax)-
, c _ -c Ax (2 x 4-
(x 4- Ax)- x- x j (x 4- Ax) 2
Third Step. ^-=^- c . J__AL_.
Ax .r-'Cr + Ax)- 5
Fourth Step. In the right-hand member let Ax * 0. Then by
dr x-\x)- x <!
DIFFERENTIATION
25
PROBLEMS
Use the General Rule in differentiating each of the following functions.
1. i/ = 2 -3 -r. An*. //' = - 3.
2. # = mx -f 6.
3. // = ax 2 .
4. s = 2t- t' 2 .
5. y = ex 3 .
6. y = 3 J - j- 3 .
7. M = 4 r 2 -f 2 r*
8. // = x 4 .
12. ?/ = -
9.P^ ;
10. y =
11. s =
-1
3
//'
= m.
-L ^ .r
</'
= 2 ax.
f?
*'
= 2 - 2 /.
' P "~ -f 2
?/ '
= 3 ex' 2 .
H ,__^/-f/>
?/'
= 3 - 3 j' 2 .
(7-f /)
M'
= 8 r -f 6 r'-.
15 w--'"'" 1 " 1
?/'
= 4 x\
j*
ip
2
16 w - -1 -
/0
(0 + !)-
///
r> j-
17. -/ -.
-4;
<ty-
li_
(.r j -f a :> ) 2
/
I' 2
10,,
lo. // -
8.T
19. ?/ = 3 ar a - 4 j- - 5.
20. 8 = a/ 2 + ft/ + r.
21. w = 2r* - 3 r 2 .
22. ?/ - ar s + bx' 2 -f r.r
23. p=(a-bOY 2 .
24. ?/= (2 - jr)(l ~ 2 j-
25. i/=
4 - s*
26. x - (r;
d.
27. // -
29. // =
a + t>x' 2
a -f
28. Interpretation of the derivative by geometry. We shall now
consider a theorem which is fundamental in all applications of the
differential calculus to geometry. It is />><
necessary to recall the definition of the } ^
tangent line to a curve at a point P on
the curve. Let a secant be drawn through
P and a neighboring point Q on the curve
(see figure). Let Q move along the curve
and approach P indefinitely. Then the
secant will revolve about P, and its limiting position is the tangent
line at P. Let
(1) y=f(?)
be the equation of a curve AD. This curve is the graph of /(r)
(see figure).
26 DIFFERENTIAL CALCULUS
Now differentiate (1) by the General Rule and interpret each step
geometrically from the figure (p. 25). We choose a point P(x, y)
on the curve, and a second point Q(x + Ax, y + Af/) near P, also
on the curve.
FIRST STEP. y + Ay = f(x + Ax) = NQ
SECOND STEP. y + Ay = f(x + Ax) = NQ
y =/(x) _
THIRDSTEP. A^ /(x + Ax) - /(x) = Jgg. |g
Ax Ax MN PR
= tan Z #/ J Q = tan </>
= slope of secant line PQ.
At this point, therefore, we see that the ratio of the increments Ay
and Ax equals the slope of the secant line drawn through the points
P(x, y) and Q(x + Ax, y + Ay) on the graph of /(or).
Let us examine the geometric meaning of the Fourth Step. The
value of x is now regarded as fixed. Hence P is a fixed point on the
graph. Also, Ax is to vary and approach zero as a limit. Obviously,
therefore, the point Q is to move along the curve and approach P as a
limiting position. The secant line drawn through P and Q will then
turn about P and approach the tangent line at P as its limiting
position. In the figure,
(/> = inclination of the secant line PQ,
r = inclination of the tangent line PT.
Then lim </> = r. Assuming that tan is a continuous function (see
Ar~
Art. 70), we have, therefore,
FOURTH STEP. ^ ~ /'(x) = lim tan <t> = tan r,
ttX Ar-^O
= slope of the tangent line at P.
Thus we have derived the important
Theorem. The value of the derivative at any point of a curve is equal
to the slope of the tangent line to the curve at that point.
It was this tangent problem that led Leibnitz* to the discovery of
the differential calculus.
* Gottfried Wilhelm Leibnitz (1646-1716) was a native of Leipzig. His remarkable
abilities were shown by original investigations in several branches of learning. He was
first to publish his discoveries in calculus in a short essay appearing in the periodical Acta
Eruditorum at Leipzig in 1684. It is known, however, that manuscripts on Fluxions
DIFFERENTIATION 27
ILLUSTRATIVE EXAMPLE. Find the slopes of the tangents to the parabola y = x*
at the vertex and at the point where x = $ .
Solution. Differentiating by the General Rule (Art. 27), we get
(2) 3* = 2 x = slope of tangent line at any point (x, y) on curve.
To find slope of tangent at the vertex, substitute x = in (2) ,
giving dy
dx~'
Therefore the tangent at the vertex has the slope zero ; that
is, it is parallel to the x-axis and in this case coincides with it.
To find the slope of the tangent at the point P, where x = i,
substitute in (2), giving ,
that is, the tangent at the point P makes an angle of 45 with the x-axis.
PROBLEMS
Find by differentiation the slope and inclination of the tangent line to
each of the following curves at the point indicated. Verify the result by
drawing the curve arid the tangent line.
l.y = x' 2 - 2, where x = 1. Am. 2 ; 63 26'.
2. y = 2 x \ x 2 , where x 3. 4. y 3 + 3 x x : \ where x 1.
4
3. y = ~, where x = 2. 5. y = .r ;< 3 x 2 , where x 1.
X -L
6. Find the point on the curve y = 5 x x 2 where tbe inclination of
the tangent line is 45. Anx. (2, 6).
7. Find the points on the curve y x* -f x where the tangent line is
parallel to tbe line y 4 x. Ans. (1, 2), ( 1, 2).
In each of the three following problems find (a) the points of intersection
of the given pair of curves ; (b) the slope and inclination of the tangent
line to each curve, and the angle between tbe tangent lines, at each point
of intersection (see (2), p. 3).
8. y = 1 x 2 , Ans. Angle of intersection = arc tan $ = 53 8'.
y = x 2 1.
9. y = x 2 , 10. y = x*-3x,
z-2/ + 2 = 0. 2z + ?/ = 0.
11. Find the angle of intersection between the curves 9 y = x 3 and
y = 6 + 8 x - x 3 at the point (3, 3). Ans. 21 21'.
written by Newton were already in existence, and from these some claim Leibnitz got the
new ideas. The decision of modern times seems to be that both Newton and Leibnitz
invented the calculus independently of each other. The notation used today was intro-
duced by Leibnitz.
CHAPTER IV
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS
29. Importance of the General Rule. The General Rule for differen-
tiation, given in the last chapter (Art. 27) is fundamental, being found
directly from the definition of a derivative, and it is very important
that the student should be thoroughly familiar with it. However, the
process of applying the rule to examples in general has been found
too tedious or difficult : consequently special rules have been derived
from the General Rule for differentiating certain standard forms of
frequent occurrence in order to facilitate the work.
It has been found convenient to express these special rules by
means of formulas, a list of which follows. The student should not
only memorize each formula when deduced, but should be able to
stale the corresponding rule in words.
In these formulas v/, i\ and w denote functions of x which are
differentiate.
FORMULAS FOR DIFFERENTIATION
dx
dx
II -7- = 1.
dx
... d , , . du dv dw
III (U + V W) = + 3 -- -T--
dx dx dx dx
_ d . . dv
IV -j- (ci>) ^ c
dx dx
_. d dv du
v
VI
Via
du dv
v __ u _
dx dx
-(-) =
dx\v/
28
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 29
du
VHa E(i?)=T'
VTTT ^ ^ dv u - * .
VIU 5x = 5U " 5J ' J' k em S a function of v.
IX = -T-, y being a function of x.
dy
30. Differentiation of a constant. A function that is known to have
the same value for every value of the independent variable is con-
stant, and we may denote it by
// = c.
As x takes on an increment AJ, the function does not change in
value, that is, A?/ = 0, and
But lim ~ -~ 0.
AX . oA.r dx
dx ~~ *
The derivative of a constant is zero.
This result is readily foreseen. For the locus of y c is a straight
line parallel to OX, and its slope is therefore zero. But the slope is
the value of the derivative (Art. 28).
31. Differentiation of a variable with respect to itself
Let y = x.
Following the General Rule (Art. 27), we have
FIRST STEP. y + A?/ = x + Ax.
SECOND STEP. A?/ = A.r.
THIRD STEP. -~ = 1.
FOURTH STEP. ( -~ = 1.
dx
!=
The derivative of a variable with respect to itself is unity.
This result is readily foreseen. For the slope of the line y = x is
unity.
30 DIFFERENTIAL CALCULUS
32. Differentiation of a sum
Let y = u + v w?.
By the General Rule,
FIRST STEP. y + ky = u + Aw + v + A?; w Aw.
SECOND STEP. A?/ = A?/ + Av Aw?.
THIRD STEP. ^^ + ^___^.
Ax Ax Ax Ax
Now (Art. 24),
: <to, lim ^! == ^, ] im ^^.
e/x A/ - o Ax dx AX - o Ax dx
Hence, by (1), Art. 16,
FOURTH STEP. ^ ^ + ^ _ ^_ ; .
c/x dx rfx dx
m d , . . du dv dw
HI .'. (u + v - w) = + - -
dx dx dx dx
A similar proof holds for the algebraic sum of any number of
functions.
The derivative of the algebraic sum of n functions is equal to the same
algebraic sum of their derivatives, n being a fixed number.
33. Differentiation of the product of a constant and a function
Let y = cv.
By the General Rule,
FIRST STEP. # + A?/ = c(v + Ar) = cv + cAt?.
SEC:OND STEP. A// = cA?\
THIRD STEP. ^-c^*
Ax Ax
Whence, by (4), Art, 16,
FOURTH STEP. ?=^:r-
ax ax
,.i(, = c|.
The derivative of the product of a constant and a function is equal to
the product of the constant and the derivative of the Junction.
34. Differentiation of the product of two functions
Let y = uv.
By the General Rule,
FIRST STEP. y + Ay = (u + Aw) (v + Av).
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 81
Multiplied out, this becomes
y + A// ur + wAr + ?>Aw + AwAv.
SECOND STEP. Ay = wAr + rAw + A//Ar.
THIRD STEP. ^ - w ^ + r ^~ + Aw ^
Ax Ax Ax Ax
Applying (2) and (4), Art. 16, noting that lim A// = 0, and hence
that the limit of the product A. - is zero, we have
FOURTH STEP. ^ = ?/ + r
ax rfx dx
V . d dv du
' * dx ~ dx dx
The derivative of the product of tiro functions is equal to the first
function titnes the derivative of the second, plus the second function
times the derivative of the first.
35. Differentiation of the product of n functions, n being a fixed
number. When both sides of V are divided by UP, this formula
assumes the form
A. (, <\ ( llL ^!1
djc __ djr dx
UV U V
If, then, we have the product of 71 functions
we may write
fl-r " ^ _ df dx
V n
ZJ dp f(r;^..., n )
ax , ax
Vi V'2 /'.'jr/i ' ' * v n
dv dux dv:\ (
_ dx , dx dx_ ,
Multiplying both sides by viv<2 - - v n , we get
+ (V\V2 '
32 DIFFERENTIAL CALCULUS
The derivative of the product of n functions, n being a fixed number,
is equal to the sum of the n products that can be formed by multiplying
the derivative of each function by all the other functions.
36. Differentiation of a function with a constant exponent. The
Power Rule. If the n factors in the above result are each equal to v,
VI "dx*" dx
When v x, this becomes
Via
dx
We have so far proved VI only for the case when n is a positive
integer. In Art. 65, however, it will be shown that this formula
holds true for any value of n, and we shall make use of this general
result now.
The derivative of a function with a constant exponent is equal to the
product of the exponent, the function with the exponent diminished by
unity, and the derivative of the function.
This rule is called the Power Rule.
37. Differentiation of a quotient
Let y = -> (t; ^ 0)
v
By the General Rule,
FIRST STEP. y + Ay/ = u + ^ u -
y v + A?;
SECOND STEP. Av = ^M^ - - = **/*-?;**.
+ Ac v 0(0 + A)
THIRD STEP. A?/ - Ar A *-
Ax '(( + At')
Applying il)-(4), Art. 16,
FOURTH STEP.
dy
du
v- i
dx
dv
l d~x
dx
d (u\
v' 2
v du_ i
dv
dx\vf
v*
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 33
The derivative of a fraction is equal to the denominator times the
derivative of the numerator, minus the numerator times the derivative
of the denominator, all divided by the square of the denominator.
When the denominator is constant, set r = c in VII, giving
du
Vila !(H
dx\c
We may also get VII a from IV as follows :
du
d / u\ 1 du dx
dx V c ' c dx c"
The derivative of the quotient of a function by a constant is equal to
the derivative of the function divided by the constant.
PROBLEMS*
Differentiate the following functions.
1. y = x*.
Solution. -/~ = ~r (x <{ ) = 3 x' 2 . Ans. By VI a
In = 3.]
2. y = ax 4 - bx 2 .
Solution. ^ = -f (ax 4 - bx 2 ) = ~ (ax 4 ) - -f (bx 2 ) by III
dx dx dx dx
= a (x 1 ) 6 (x*') by IV
= 4 ax :i - 2 6 x. A ns. By VI a
3. y = x* + 5.
Solution. & = -7- (rr*) -f (5) by III
= j :r>. Ans. By VI a and I
4. y = ig-^ + 8>^.
Solution. ^ = ~ (3 x 1 ^) - ~ (7 x~^) -f (8 x*) by III
ax dx ax dx
= : V y x^ + I x-J + V ~^- An *- By IV and VI a
* When learning to differentiate, the student should have oral drill in differentiating
simple functions.
34 DIFFERENTIAL CALCULUS
5. y - (x 2 - 3) 5 .
Solution. & = 5(x 2 - 3)< -f (x 2 - 3) by VI
ax ax
[p = x 2 - 3, and n = 5.]
= 5(x 2 - 3) 4 2 x = 10 x(x 2 - 3) 4 . An*.
We might have expanded this function by the Binomial Theorem ((3),
p. 1), and then applied III, etc., hut the above process is to be preferred.
by VI
Solution. = (a 2 - * - (<* -
7. y = (3 r 2 + 2) l + 5 jc 2 .
Solution. ^ = (3 & + 2) y- (1 + 5 a*)* 4- (1 + 5 x^i -f (3 x* +2) by V
dx c/x dx
[u - 3 a-'- 1 + 2, and r - (1 + 5 j' J )l]
= (3x 2 -f 2)i(l -f 5x 2 )~^y- (1 -f5x 2 )4 (1 -f 5x 2 )^Gx by VI etc.
= (3 x 2 4-2H1 -f 5x 2 )' 2 5x + 6x(l -f 5x 2 )2-
g
V 1 -f 5 x 2 v 1 -f 5 x 2
_
. A . r/// _ _ H _ hv VIT
Solution. -~ = - ;; - " - DV v X1
dx - x-
_ 2 r((7 2 - :r 2 ) -f xfa 2 -f T 2 )
[Multiplying both numerator and denominator by (a 2
(a 2 - x 2 )^
Prove each of the following differentiations.
9. ~ (3 .r 4 - 2 x 2 -f 8) = 12 a- 3 - 4 x.
dx
10. 4 (4 + 3 x - 2 jfi) = 3-6 x 2 .
dr
11. 4 (a/ 6 ~ 5 W 3 ) = 5 a/ 4 - 15 fa 2 .
ar
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 35
16. - (2j- J -f 4
17.^U 8 -a s ;=s.
i Q d i a -f bx -f <~.r~
18 ' d? i 7
v;
' = f. JT ' - X
20. ; =
21. V =
a + bt + r/ 2
-p=-
Vaj-
23. /(/) = (2 - 3 /-) :l .
_
4V.r
2V?
a
(4 - 9 ,
<f/?/ __ .r
^r~ /., 2 . 2 J
(2 - 5
^ = ( a - ~
dj: - 2 I ^*
30. s = <Va 2 + < z .
dy _ 2 a -h 3 bx
ds< 2 Va 4- ^
ds _ a 2 -f 2 f 2
31. =
o x
2 a
36
DIFFERENTIAL CALCULUS
33. y =
35. r =
36. // =
37.
38.
Va 2 +
X
X
,__ /a 2 + x' 2
~ Va' 2 -x' 2 '
,_ 3/2+in.
'- \^TJn
39.
40.
41.
77 -
dy __ 4 a 2 x
dx~~ (a 2 -x 2 ) 2 *
d?/ _ a 2
^ x 2 Va 2 -f x 2
rfjy _ a 2
d:r (a 2 - x 2 )*'
dr _6 6 - 10 2
de~
d^ r
^~ (l+rx)Vr
<// 2 a 2 r
^ (a 2 -o- 2 )Va 4
d,s- 4
Ul (2-h30 a (2~
do: ?/
dx a-y
Differentiate each of the following functions.
42. /(.r) = V2.r
43. // =
44. // =
47. y = a- 2 V5 _
45. ,s-
Va - bx
V -f ft/ .
48. // =
50. 2/=
46. r = -
V^H-3 JT
In each of the following problems find the value of ^ for the given
value of .r.
52. y = (x 2 - o-) ; .r = 3.
do-
Am. 540.
53. // =
VJ ; x = 64.
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 37
54. y = (2 j-V* + (2 .rV' ; .r = 4. Arcs. g.
55. y = Vif+77 1 ' ; j- = L\
,_ Vl6 -f Ii .r
57. 7/ =
58. ?/ = jr\/8 - .r 2 ; .r = 2. 0.
59. // = .r-Vl + .*; .r = 2. 20.
60. // = (4 - .r-V ; .r = U. 63. // = j-N/.S -f 2 .
61- // = rri~7~> ; * r ^ - 64. // =-
j-= J. 65. // =
38. Differentiation of a function of a function. It sometimes hap-
pens that y, instead of being defined directly as a function of J, is
given as a function of another variable r, which is defined as a func-
tion of x. In that case y is a function of jc through v and is called a
function of a function.
2 v
For example, if y = ^ --- r, >
~
and
then ?/ is a function of a function. P>y eliminating v we may express y
directly as a function of JT, but in general this is not the best plan
when we wish to find -
dx
If y = f(v) and v = <(JT), then '// is a function of # through 0. Hence,
when we let x take on an increment AJ, v will take on an increment Av
and y will also take on a corresponding increment A?y. Keeping this
in mind, let us apply the General Rule simultaneously to the two
functions
amj ^
FIRST STEP. y+&y=f(v
SECOND STEP. y+&y=f(v
y
y =f(v + A?;) -/O) At; = <f>(x + Ax)
THIRD C^TFP Ay /(^ + A.) -/QQ Ag_(a; + Ag)
IHIRDbTEP. AJH - A?; AJ"~ Aar
38 DIFFERENTIAL CALCULUS
The left-hand members show one form of the ratio of the increment
of each function to the increment of the corresponding variable, and
the right-hand members exhibit the same ratios in another form. Be-
fore passing to the limit let us form a product of these two ratios,
choosing the left-hand forms for this purpose.
This gives -J/ . _J?, which equals -r^-
Av Ax Ax
Write this T* = *-f'-
Ax Av Ax
FOURTH STEP. When Ax - 0, then also Ai> * 0. Passing to the
limit,
(A) *y = d JL.Q. By (2), Art. 16
v ' dx dv dx
This may also be written
(B) -^=/'00-f.
// y =/(?;) and v </>(x), tlie derivative of y with respect to x equals
the product of the derivative of y with respect to v by the derivative of v
with respect to x.
39. Differentiation of inverse functions. Let y be given as a func-
tion of x by means of the relation
y =/(*).
It is often possible in the case of functions considered in this book
to solve this equation for x, giving
that is, we may also consider y as the independent and x as the
dependent variable. In that case
/(x) and <t>(y)
are said to be inverse functions. When we wish to distinguish between
the two it is customary to call the first one given the direct function
and the second one the inverse function. Thus, in the examples which
follow, if the second members in the first column are taken as the
direct functions, then the corresponding members in the second
column will be respectively the inverse functions.
y = x 2 + 1, x = \ly - 1.
y = a x , ar = log a y.
y = sin x, x = arc sin y.
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 39
Let us now differentiate the inverse functions
y=J(x) and x = 0(^/)
simultaneously by the General Rule.
FIRST STEP. y+Ay=f(x + Ax) x + Ax = 4)(y + Ay).
SECOND STEP, y + Ay = f(x + Ax) x + Ax = (y + Ay)
JL
_
=f(x + Ax)-f(x) Ax
THIRD STEP Ay^f(x+Ax}-.f(x) r AT ^ <(?/ + A?/)
Ax Ax Ay Ay
Taking the product of the left-hand forms of these ratios, we get
AJ/ AT
Ax Ay '
A//
FOURTH STEP. When Ax - 0, then also, in general, Ay - 0. Pass-
ing to the limit,
(C) r = ' by (3), Art. 16
dx dx
dy
or
The derivative of the inverse function is equal to tlie reciprocal of the
derivative of the direct function.
40. Implicit functions. When a relation between x and y is given
by means of an equation not solved for y, then y is called an implicit
function of x. For example, the equation
(1) x* - 4 y =
defines y as an implicit function of x. Evidently x is also defined by
means of this equation as an implicit function of y.
It is sometimes possible to solve the equation defining an implicit
function for one of the variables and thus obtain an explicit function.
For instance, equation (1) may be solved for y, giving
. i r 2
y 4 x ,
showing y as an explicit function of x. In a given case, however,
such a solution may be either impossible or too complicated for
convenient use.
DIFFERENTIAL CALCULUS
41. Differentiation of implicit functions. When y is defined as an
implicit function of s, it was explained in the last article that it
might be inconvenient to solve for y in terms of y or x in terms of y
(that is, to find y as an explicit function of x, or x as an explicit
function of y).
We then follow the rule :
Differentiate the terms of the equation as given, regarding y as a Junc-
tion of x, and solve for
ax
This process will be justified in Art. 231. Only corresponding
values of x and y which satisfy the given equation may be substi-
tuted in the derivative.
Let us apply this rule in finding from
us
cu f ' + 2 .r'ty - ix = 10.
Then -^+^)-^
I ( ) /V.M >' I C . '2. 7 ^ . T> .' __ /I
/ O -*'} T , () \ // "* ^* ' /? >
rf//__ ?/ 7 6aj- e> G^ 1 '//
The student should observe thai in general the result will contain
both j 1 and v/.
PROBLEMS
Find -'- for each of the following functions.
2. y = 2 - ?/ 2 , ?( = j-a - .r.
An*. <^ = ^.
dr V
- 1).
5. 15 .r = 15 y + 5 ?/ : < + 3 y fi .
6. ,= V + \/ A
RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 41
7. y 2 = 2 px. 13. .r 3 4- 3 .r 2 ?/ -f ?/ :< = r 1 .
8. x 2 + ?/ 2 - r 2 . 14. JT -f 2V^7/_+ ?/ = a.
9. 6 2 x 2 -f a 2 ?/ 2 = a 2 6 2 . 15 ' J ' 2 + a ^- r .' / + ?/ 2 = ft3 -
10. Vx -f V# = Va. ' * ' * '
o 3 17. r;.r :l 3 /> 2 .r// -f r?/ 3 = 1.
12. ^ - 3 arj/ + j/ 3 = 0. 18 '
Find the slope of each of the following curves at the given point.
19. jc' 2 -f jry + 2 y' 2 = 28 ; v-. '*). Awx. - }.
20. .r< ~ 3 xy' 2 + ?/' = 1 ; (2, - 1 ). - J.
21. V2~7 -f VlTjy = 5 ; (2, 3). 23. .r' 1 - a.n/ -f 3 a//- = 3 a :t ; (a, a).
22. jc 2 - 2V7// - ir = 52 ; (8, 2). 24. .r'-' - .r \fxlj - 2 /r' = 6 ; (4, 1).
25. Show that the parabolas //-' ~ 2 pj- f />- and // : ' ~ p' 2 2 ;>.r inter-
sect at right angles.
26. Show that the circle ./-' -f //-' 12 x (> // -f 25 ~ is tangent to the
circle s- -f if 2 + 2 jr -f it = 10 at the point (2, 1).
27. At what angle does the line ?/ = 2 .r cut- the curve ^ 2 xy -f 2 ?/ 2 = 28 ?
28. If f(x) and </>(//) are inverse functions, show that the graph of </>(.r)
may be found as follows: construct the graph of fix) and rotate it
around the origin 9(T counterclockwise.
ADDITIONAL PROBLEMS
1. The vertex of the parabola y 2 = 2 px is the center of an ellipse.
The focus of the parabola is an end of one of the principal axes of the
ellipse, and the parabola and ellipse intersect at right angles. Find the
equation of the ellipse. Ans. 4 x 2 + 2 y 2 p 2 .
2. A circle is drawn with its center at (2 a, 0) and with a radius such
that the circle cuts the ellinse b' 2 x 2 + a 2 ?/ 2 = a 2 b 2 at right angles. Find the
radius. AUK. r 2 - 2(3 a 2 -f b 2 ).
3. From any point P on an ellipse lines are drawn to the foci., Prove
that these lines make equal acute angles with the normal at P.
4. Prove that the line Bx -f Ay = AB is tangent to the ellipse
b 2 x 2 4- a 2 ?/ 2 = a 2 b 2 if, and only if, # 2 a 2 -f ,4 2 /> 2 - A 2 B 2 .
5. Find the equation of the tangent to the curve x m y n = a m + n at any
point. Prove that the portion of it intercepted between the axes is divided
in the ratio at the point of contact. Ans. myiix x\) + nx\(y #,) = 0.
n
6. If A* is the slope of a tangent to the hyperbola b 2 x 2 - a 2 ?/ 2 = a 2 b 2 ,
prove that its equation is y - kx Va^k 2 - b 2 , and show that the locus of
the points of intersection of the perpendicular tangents is x 2 -f y 2 = a 2 b 2 .
CHAPTER V
VARIOUS APPLICATIONS OF THE DERIVATIVE
42. Direction of a curve. It was shown in Art. 28 that if
=/(*)
is the equation of a curve (see
figure), then
-~ = slope of the line tangent
to the curve at P(x, y).
The direction of a curve at
any point is defined as the direction of the tangent line to the curve
at that point. Let r inclination of the tangent line. Then the
slope = tan r, and
-~ = tan r = slope of the curve at any point P(x, y).
At points such as D, F, H, where the direction of the curve is
parallel to the x-axis and the tangent line, is horizontal,
r = ; therefore -^ = 0.
ax
At points such as A, B, G, where the direction
of the curve is perpendicular to the x-axis and
the tangent line is vertical, I o
T = 90; therefore -& becomes infinite.
ILLUSTRATIVE EXAMPLE 1. Given the curve y = * x 2 -f 2 (see figure).
3
(a) Find the inclination r when x = 1.
(b) Find r when x = 3.
(c) Find the points where the direction of the curve is parallel to OX.
(d) Find the points where T = 45.
(e) Find the points where the direction of the curve is parallel to the line
2x -3 y = 6 (line AB).
Solution. Differentiating, -f- = x 2 2 x = tan T.
ax
42
VARIOUS APPLICATIONS OF THE DERIVATIVE
43
(a) When x=l,tanr=l-2 = -l; therefore r = 135. Ans.
(b) When x = 3, tan 7 = 9-6 = 3; therefore r = 71 34'. Ans.
(c) When r = 0, tan r = ; therefore x' 2 - 2 x = 0. Solving this equation, we
get x = or 2. Substituting in the equation of the curve, we find y = 2 when
x = 0, y = f when x - 2. Hence the tangent lines at 0(0, 2) and D(2, j) are
horizontal. -Ans.
(d) When r_= 45, tan r = 1 ; therefore r 2 - 2 x = 1. Solving this equation, we
get x= 1 v 2 = 2.41 and 0.41, giving two points where the slope of the curve
(or tangent) is unity.
(e) Slope of the given line = I ; therefore x' 2 - 2 x = J. Solving, we get
x = 1 VS = 2.29 and - 0.29, giving the abscissas of the points F and E where
the direction of the given curve (or tangent^ is parallel to the line AB.
Since a curve at any point has the same direction as its tangent
line at that point, the angle between two curves at a common point
will be the angle between their tangent lines at that point.
ILLUSTRATIVE EXAMPLE 2. Find the an^le of intersection of the circles
(A) x 2 + y 2 -4;r = l f
(B) x 2 + y* - 2 y = 9.
Solution. Solving simultaneously, we find the points of intersection to be
(3, 2) and (1, -2).
Let mi = slope of the tangent to the circle A at (or, j/),
and W2 = slope of the tangent to the circle R at (jr. y).
Then from (A), mi = jg =
and from (B), m^ ~r
ax
by Art. 41
By Art. 41
Substituting x = 3, y = 2, we have
mi = J slope of tangent to (A) at (3, 2).
7H 2 = 3 = slope of tangent to (/?) at (3, 2).
The formula for finding the angle between two lines whose slopes are m\
and m2 is
tan = 7^ - ^- (2), Art. 3
f 1 -f 7W1///2
3
Substituting,
tan 6 =
^ 1 ; /. = 45". Arw.
This is also the angle of intersection at the point (1, 2).
43. Equations of tangent and normal; lengths of subtangent and sub-
normal. The equation of a straight line passing Y
through the point (xi, yi) and having the slope
m is y -_y 1 = m ( x -. Xl ). (3), Art. 3
If this line is tangent to the curve AB at the
point PI(XI, y\), then m is equal to the slope of
the curve at (xi, y\). Denote this value of m by mi. Hence at the
point of contact PI(ZI, y\) the equation of the tangent line TPi is
M N
44
DIFFERENTIAL CALCULUS
The normal being perpendicular to the tangent, its slope is the
negative reciprocal of rwi ((2), Art. 3j. And since it also passes
through the point of contact Pi(r\, y\), we have for the equation
of the normal PiN,
(2) y |/i = (x jfi).
m\
That portion of the tangent which is intercepted between the
point of contact and OX is called the Ivtnjth of ihe, tauywt (= TP } ),
and its projection on the r-axis is called the length of the subtangent
(= TM). Similarly, we have the length of rite, y\
normal (= P\K) and the length of the sub-
normal (= yi/A r ).
/i f i)
In the triangle TP\M 9 tan r~m\
therefore
7V17 ' _f
(3)
= length of sub tangent.
In the triangle 1\1P } N, tan r = in\
therefore
(4)
71 / A"* =
= length of subnormal.
The length of the tangent (TPi) and the length of the normal
(PjAO may then be found directly from the figure, each being the
hypotenuse of a right triangle having two legs known.
When the length of subtangent or subnormal at a point on a curve
is determined, the tangent and normal may easily be constructed.
PROBLEMS
1. Find the equations of tangent and normal and the lengths of sub-
tangent, subnormal, tangent, and normal, at the point (a, a) on the cissoid
^a- x
Solution.
dx y(2 a - x)~
Substituting x a, ?/ = a, we have
rr 2 = slope of tangent.
-
a, equation of tangent.
Substituting in (1) gives
U = 2x-
Substituting in (2) gives
2 y -f x = 3 a, equation of normal.
* If the subtangent extends to the ritrht of 7\ we consider it positive; if to the left,
negative, If the subnormal extends to the right of M, we consider it positive, if to the
Jeft, negative.
VARIOUS APPLICATIONS OF THE DERIVATIVE 45
Substituting in (3) gives TM = ( ~ = length of subtangent.
Substituting in (4) gives MX = 2 a = length of subnormal.
Also, PT = V(TM)* + (MP^ = x- + <|2=-\5 = length of tangent,
\
4
and PN = V(MN) 2 + (MP)* - \ 4 ' -fa- = a \ 5 = length of normal.
Find the equations of the tangent and normal at the given point.
2. y = ar 3 - 3 x; (2, 2). .4//,s. 9 .r - // - 16 = 0, .r -f 9 ?/ - 20 = 0.
9 T 4- 1
3. T/ = ^L1. (2, 5). 7 j- - y -9 = 0, .r + 7 <// - 37 = 0.
o X
4. 2x 2 -xy + y- = 16; (3, 2).
5. 7/ 2 + 2 ?/-4 jr-f 4 = 0; (1, -2).
6. Find the equations of the tangent and normal at (x\, y\) to the
ellipse b' 2 x 2 -f a 2 ?/ 2 a~b' 2 .
Ann. b-fix -h a-y\y = a L> /> 2 , a~ii\x b~jr\y jr]y\((i~ h 2 ).
7. Find the equations of the tangent and normal, and the lengths of
the subtangent and subnormal, at the point u*i, ?/i ) on the circle s- -f ir T~.
Ans. JC\JT -f 7/1 y = r' 2 , .r\y ijix ~ 0, * JT\.
' ' xi
8. Show that the subtangent to the parabola /r ~ 2 ;>.r is bisect (d at
the vertex, and that the subnormal is constant and equal to ;>.
Find the equations of the tangent and normal, and the lengths of the
subtangent and subnormal, to each of the following curves at the points
indicated.
9. ay x 2 ; (a, a). A us. 2 jr y a, x 4- 2 // ~ 3 a, ^ 2 a.
10. x 2 - 4 ?y 2 = 9 ; (5, 2). f> .r - 8 ?/ = 9, 8 x + 5 ?/ = 50, V, J.
11. 9x 2 + 4 2/ 2 = 72; (2, 3).
12. xy + ? y 2 + 2 = 0; (3, - 2).
13. Find the area of the triangle formed by the j-axis and the tangent
and the normal to the curve y = 6 jr - .r' 2 at the point (5, 5). A us. A p.
14. Find the area of the triangle formed by the //-axis and the tangent
and the normal to the curve //- = 9 j- at the point (5, 2).
Find the angles of intersection of each of the following pairs of curves.
15. y 2 = x 4- 1, x 2 + y 2 = 13. Am. 109 39 r .
16. y = 6 - a* 2 , 7 x 2 + y 2 = 32.
Ans. At ( 2, 2), 5 54'; at ( 1, 5), 8 58'.
17. y = x 2 ,y*-3 y = 2 jr.
18. z 2 + 4 ?/ 2 = 61, 2x 2 -y* = 41.
46 DIFFERENTIAL CALCULUS
Find the points of contact of the horizontal and vertical tangents to
each of the following curves.
19. y = 5 x - 2 x 2 . Ans. Horizontal, (|, )
20. 3 y' 2 - 6 y - x - 0. Vertical, (- 3, 1).
21. x 2 + 6 xy + 25 y 2 - 16. Horizontal, (3, - 1), (- 3, 1).
Vertical, (5, - ), (- 5, f ).
22. x 2 - 8 xy + 25 y 2 = 81.
23. x 2 - 24 x?/ + 169 t/ 2 = 25.
24. 169 x 2 + 10 x?/ -I- t/ 2 = 144.
25. Show that the hyperbola x 2 - y 2 = 5 and the ellipse 4 x 2 + 9 ?/ 2 = 72
intersect at right angles.
26. Show that the circle jr' 2 + ?/ 2 = 8 ax and the cissoid (2 a - x)?/ 2 = x 3
(a) are perpendicular at the origin;
(b) intersect at an angle of 45 at two other points. (See figure in
Chapter XXVI.)
27. Show that the tangents to the folium of Descartes x : ^ -f if = 3 axy
at the points where it meets the parabola y 2 = ax are parallel to the //-axis.
(See figure in Chapter XXVI.)
28. Find the equation of the normal to the parabola y 5 x -f x 2 which
makes an angle of 45 with the x-axis.
29. Find the equations of the tangents to the circle x 2 + y 2 = 58 which
are parallel to the line 3x 1 y 19.
30. Find the equations of the normals to the hyperbola 4 x 2 - y' 2 = 36
which are parallel to the line 2 x 4- 5 y = 4.
31. Find the equations of the two tangents to the ellipse 4 x 2 + y' 2 = 72
which pass through the point (4, 4). Ans. 2 x + y = 12, 14 x + y = 60.
32. Show that the sum of the intercepts on the coordinate axes of the
tangent line at any point to the parabola x^ + ifi = a* is constant and
equal to a. (See figure in Chapter XXVI.)
33. Show that for the hypocycloid x* + y$ = cfi the portion of the
tangent line at any point included between the coordinate axes is constant
and equal to a. (See figure in Chapter XXVI.)
34. The equation of the path of a ball is y = x - -^- ; the unit of
distance is 1 yd., the x-axis being horizontal, and the origin being the
point from which the ball is thrown, (a) At what angle is the ball thrown?
(b) At what angle will the ball strike a vertical wall 75 yd. from the
starting point? (c) If the ball falls on a horizontal roof 16yd. high, at
VARIOUS APPLICATIONS OF THE DERIVATIVE
47
what angle will it strike the roof? (d) If thrown from the top of a build-
ing 24 yd. high, at what angle will the ball strike the ground ? (e) If
thrown from the top of a hill which slopes downward at an angle of 45,
at what angle will the ball strike the ground ?
b
35. The cable of a suspension bridge hangs in
the form of a parabola and is attached to support-
ing pillars 200 ft apart. The lowest point of the cable is 40 ft. below
the points of suspension. Find the angle between the cable and the sup-
porting pillars.
44. Maximum and minimum values of a function ; introduction. In
a great many practical problems we have to deal with functions
which have a greatest (maximum) value or a least (minimum)
value,* and it is important to know what particular value of the
variable gives such a value of the function.
For instance, suppose that it is required to find the dimensions of
the rectangle of greatest area that can be inscribed in a circle of
radius 5 inches. Consider the circle in the following figure. Inscribe
any rectangle, as ED.
Let CD = x ; then DE = VlOO - x-, and the area of the rectangle
is evidently
(1) A = j VlOO -:r L> .
That a rectangle of maximum area must exist may be seen as follows.
Let the base CD (= x) increase to 10 inches (the diameter) ; then
the altitude DE = VlOO x- will decrease to zero and the area will
become zero. Now let the base decrease to
zero ; then the altitude will increase to 10
inches and the area will again become zero.
It is therefore evident by intuition that there
exists a greatest rectangle. By a careful study
of the figure we might suspect that when the
rectangle becomes a square its area would be
greatest, but this would be guesswork. A
better way would evidently be to plot the
graph of the function (1) and note its behavior. To aid us in draw-
ing the graph of (1), we observe that
(a) from the nature of the problem it is evident that x and A must
both be positive ; and
(b) the values of x range from zero to 10 inclusive.
Now construct a table of values and draw the graph, as in the
figure on page 48.
* There may be more than one of each, aa illustrated on page 55,
48
DIFFERENTIAL CALCULUS
What do we learn from the graph?
x
*
1
9.9
2
19 G
:j
2S.G
4
3G.G
r
4,'J.O
G
1.0
7
19.7
H
cj
1S.O
10
'> 10 or'
(a) If carefully drawn, we may find quite accurately the area of
the rectangle corresponding to any value of x by measuring the length
of the corresponding ordinate. Thus,
when x =- OA1 = 3 in.,
then A 1\1 f> = 28. 6 sq. in. ;
and when x = ON ~~ 4\ in.,
then A = A< Q ~ about 39.8 sq. in. (found by measurement).
(b) There is one horizontal tangent (A\S Y ). The ordinate TH from
its point, of contact T is greater than any other ordinate. Hence this
observation : One of the inscribed rectangles has evidently a greater area
than any of the others. In other words, we may infer from this that
the function defined by (1) has a maximum value. We cannot find
this value (= 7/7") exactly by measurement, but it is very easy to
find by the calculus. We observed that at T the tangent was
horizontal ; hence the slope will be zero at that point (Art. 42). To
find the abscissa of T we then find the derivative of A with respect
to x from (1), place it equal to zero, and solve for x. Thus we have
(1)
.4 = arVlOO - jc' 2 ,
100 - 2 x 2 100 - 2 x 2
dA_
dx VlOO - x 2
= 0.
VlOO - x 2
Solving, x = 5 V2.
Substituting, we get DE = VlOO x 2 = 5 V&
VARIOUS APPLICATIONS OF THE DERIVATIVE
49
Hence the rectangle of maximum area inscribed in the circle is a
square of area
A = CD X DE = 5V2 X 5 V2 = 50 sq. in.
The length of HT is therefore 50.
Take another example. A wooden box is to be built to contain
108 cu. ft. It is to have an open top and a square base. What must
be its dimensions in order that the amount of material required shall
be a minimum ; that is, what dimensions
will make the cost the least ?
Let x = length of side of square base in feet,
and y = height of box.
Since the volume of the box is given, how-
ever, y may be found in terms of jr. Thus,
Volume - x-'u - 108 ; .-.?/ = ^
.r~
We may now express the number ( M) of square feet of lumber
required as a function of jr as follows.
Area of base = JT~ sq. ft.,
432
and
(2)
Area of four sides = 4 .r// - sq. ft. Hence
M = x' 2 +
X
M
1
433
2
220
3
153
4
124
r>
111
6
108
7
111
8
118
9
129
10
143
is a formula giving the number of square feet required in any such box
having a capacity of 108 cu. ft. Draw a graph of (2), as in the figure.
50 DIFFERENTIAL CALCULUS
What do we learn from the graph ?
(a) If carefully drawn, we may measure the ordinate correspond-
ing to any length (= x) of the side of the square base and so deter-
mine the number of square feet of lumber required.
(b) There is one horizontal tangent (RS). The ordinate of its
point of contact T is less than any other ordinate. Hence this obser-
vation : One of the boxes evidently takes less lumber than any of the
others. In other words, we may infer that the function defined by
(2) has a minimum value. Let us find this point on the graph ex-
actly, using the calculus. Differentiating (2) to get the slope at any
point, we have
dM -9r 432
j <<-" +' -- rr *
ax x 2
At the lowest point T the slope will be zero. Hence
that is, when x = 6 the least amount of lumber will be needed.
Substituting in (2), we see that this is
M = 108 sq. ft.
The fact that a least value of M exists is also shown by the follow-
ing reasoning. Let the base increase from a very small square to a
very large one. In the former case the height must be very great and
therefore the amount of lumber required will be large. In the latter
case, while the height is small, the base will take a great deal of
lumber. Hence M varies from a large value, grows less, then in-
creases again to another large value. It follows, then, that the graph
must have a "lowest" point corresponding to the dimensions which
require the least amount of lumber and therefore would involve the
least cost.
We will now proceed to the treatment in detail of the subject of
maxima and minima.
45. Increasing and decreasing functions.* Tests. A function y =f(x)
is said to be an increasing function if y increases (algebraically) when
x increases. A function y f(x) is said to be a decreasing function
if y decreases (algebraically) as x increases.
The graph of a function indicates plainly whether it is increasing
or decreasing. For instance, consider the graph in Fig. a, p. 51.
* The proofs given here depend chiefly on geometric intuition. The subject of maxima
and minima will be treated analytically in. Art. 125.
VARIOUS APPLICATIONS OF THE DERIVATIVE
51
\
As we move along the curve from left to right the curve is rising ;
that is, as x increases the function (= y) increases. Obviously, Ay
and Ax agree in sign.
On the other hand, in the graph of
Fig. 6, as we move along the curve from
left to right the curve is falling ; that is,
as x increases, the function (== //) always
decreases. Clearly, Ay and Ax have op-
posite signs.
That a function maj r be sometimes increasing and sometimes
decreasing is shown by the graph (Fig. c) of
As we move along the curve from left to
right the curve rises until we reach the point
A, falls from A to B, and rises to the right of
B. Hence
(a) from x = oo to x = I the function is
increasing ;
(b) from x = 1 to x = 2 the function is de-
creasing ;
(c) from x = 2 to x + oo the function is
increasing.
At any point, such as C, where the function is increasing, the
tangent makes an acute angle with the j-axis. The slope is positive.
On the other hand, at a point, such as D, where
the function is decreasing, the tangent makes
an obtuse angle with the j-axis, and the slope is
negative. Hence the following criterion :
A function is increasing when its derivative is
positive, and decreasing when its derivative is
negative.
For example, differentiating (1) above, we
have
(2) g = f( x ) = 6 z 2 - 18 * + 12 = 6(z - l)(z - 2).
When x < 1, /'(*) is positive, and f(x) is increasing.
When 1< x < 2, f'(x) is negative, and /(or) is decreasing.
When x > 2, f'(x) is positive, and f(x) is increasing.
These results agree with the conclusions arrived at above from
the graph.
FIG. b
52 DIFFERENTIAL CALCULUS
46. Maximum and minimum values of a function; definitions. A
maximum value of a function is one that is greater than any value
immediately preceding or following.
A minimum value of a function is one that is less than any value
immediately preceding or following.
For example, in Fig. c, Art. 45, it is clear that the function has a
maximum value MA (= y = 2) when x = 1, and a minimum value
NB (=-y = 1) when x = 2.
The student should observe that a maximum value is not neces-
sarily the greatest possible value of a function nor a minimum value
the least. For in Fig. r it is seen that the function (= y) will have
values to the right of B that are greater than the maximum MA, and
values to the left of A that are less than the minimum NB.
If /OO is an increasing function of jr when .r is slightly less than a,
and a decreasing function of x when jr is slightly greater than a, that
is, if /'Or) changes sign from + to as x increases through a, then
f(x) has a maximum value when x a. Therefore, if continuous,
}'(x) must vanish when j = a.
Thus, in the above example (Fig. r), at (\ J'(x) is positive; at
A, /'(.r) = ; at />, /'(:r) is negative. /
On the other hand, if /Or) is a decreasing
function when j is slightly less than a, and an
increasing function when x is slightly greater
than a, that is, if /'Or) changes sign from to +
as x increases through a, then /Or) has a mini-
mum value when r = a. Therefore, if contin- jj \ x
uous, /'Or) must vanish when r = a. ' FlG - c
Thus, in Fig. c, at D, J'(x) is negative; at B, f(x) = 0; at E,
f f (x] is positive.
We may then state the conditions in general for maximum and
minimum values of /Or).
/(jc) is a maximum if /'(*) = and /'(*) changes sign from + to .
f(x) is a minimum if /'(*) = and/'(x) changes sign from to -f.
The values of the variable satisfying the equation f'(x) = are
called critical values; thus, from (2), Art. 45, x = 1 and x 2 are the
critical values of the variable for the function whose graph is shown
in Fig. c. The critical values determine turning points where the
tangent is parallel to OA".
To determine the sign of the first derivative at points near a
particular turning point, substitute in it, first, a value of the variable
sliglitly less than the corresponding critical value, and then one
VARIOUS APPLICATIONS OF THE DERIVATIVE 53
slightly greater. If the first sign is + and the second , then the
function has a maximum value for the critical value considered.
If the first sign is and the second +, then the function has a
minimum value.
If the sign is the same in both cases, then the function has neither
a maximum nor a minimum value for the critical value considered.
For example, take the above function in (1), Art. 45,
(1) y = f(jr) = 2 JT* - 9 JT- + 12 jc - 3.
Then, as we have seen,
(2) f'(jr) = 6(j- l)(j-2).
Setting /'(y) = 0, we find the critical values y = 1, :r = 2. Let us
first test x = 1. We consider values of jc near this critical value to be
substituted in the right-hand member of (2), and observe 1 the signs
of the factors. (Compare Art. 45.)
When x < 1, /'(*) = (-)(-) - +.
When x > 1, f'(jr) - (+)(-) - -.
Hence /(r) has a maximum value when x ~ 1. By the
table, this value is ?/=/(!) = 2.
Next, test y = 2. Proceed as before, taking values of x now near
the critical value 2.
W f hen .r < 2, /'(/) - (+)(-) - -.
When x > 2, fix) = (+)(+.) = +.
Hence /(y) has a minimum value when x 2. By the table above,
this value is y =/(2) = 1.
We shall now summarize our results in a workruy ride.
47. First method for examining a function for maximum and mini-
mum values. Working rule.
FIRST STEP. Find the first derivative of the function.
' SECOND STEP. Set the first derivative equal to zero and solve the resulting
equation for real roots. These roots are the critical values of the variable.
THIRD STEP. Considering one critical value at a time, test the first
derivative, first for a value a trifle less * and then for a value a trifle greater
than the critical value. If the sign of the derivative is first + and then ,
the function has a maximum value for that particular critical value of the
variable; but if the reverse is true, then it has a minimum value. If the
sign does not change, the function has neither.
* In this connection the term "little less/' or " trifle less," means any value between the
next smaller root (critical value) and the one under consideration ; and the term " little
greater," or "trifle greater," means any value between the root under consideration and
the next larger one.
54 DIFFERENTIAL CALCULUS
In the Third Step, it is often convenient to resolve /'(a*) into
factors, as in Art. 46.
ILLUSTRATIVE Ex AMPLE 1. In the first problem worked out in Art. 44 we showed
by means of the graph of the function
A = xVlOO -x 2
that the rectangle of maximum area inscribed in a circle of radius 5 in. con-
tained 50 8(4. in. This may now be proved analytically as follows by applying the
above rule.
Solution. /(x) = x\ 100 - x 2 .
r,- , a, /// N 100 - 2 X 2
First Step. f'(x) =
VI 00 -x 2
Second Step. Setting f'(x) 0, we have
x^r 5 \ '2 = 7.07,
which is the critical value. Only the positive sign of the radical is taken, since,
from the nature of the problem, the negative sign has no meaning.
Third Slcp. When x < 5\"2, then 2 x-' < 100, and /'(x) is -K
When x > 5\ , then 2 x 2 > 100, and f (xj is -.
Since the sign of the first derivative changes from + to , the function has a
maximum value /(5V2) 5\ 2 5\ 2 = 50. An*.
ILLUSTRATIVE EXAMPLE 2. Examine the function (x l) 2 (x -f !)' for maxi-
mum and minimum values.
Solution, /(x) = (x-l) 2 (x-f IP.
First Step. f'(x) = 2(x - l)(x + 1 )'< + 3(x - l) 2 (x + 1) 2 = (x -l)(z + l) 2 (5x- 1).
Second Step, (x - l)(x + 1) 2 (5 x - 1) = 0.
Hence x 1, 1, \, are the critical values.
Third Step. /'(x) = 5(x - D(x -f 1 ) 2 (x - -J).
Examine first the critical value x = 1 (C in figure).
When x < 1, /'(x) = 5(-)(+) 2 (-f ) = -. "
When x > 1, /'(x) = 5(+ )(+) L '(-f-) = +.
Therefore, when x 1 the function has a minimum value /(I) = ( ordinate
of D.
Examine now the critical value x I (B in figure).
W r henx< U'(x) = 5(-)(+) 2 (-) = -K
When x > I /'(x) - 5(-)(+) 2 (+) - -.
Therefore, when x \ the function has a maximum value /(J) =1.11 (= ordi-
nate of B).
Examine lastly the critical value x = 1 (A in figure).
When x < - 1, /'(jO = 5(-X~) 2 (-) = +.
When x > - 1, /'(x) = 5(-)(+ ) 2 (-) = +.
Therefore, when x 1 the function has neither a maximum nor a minimum
value.
VARIOUS APPLICATIONS OF THE DERIVATIVE
55
48. Maximum or minimum values when /'(*) becomes infinite and
f(x) is continuous. Consider the graph in the figure. At B, or G,
FIG. d
f(x) is continuous and has a maximum value, but /'(a*) becomes in-
finite, since the tangent line at B is parallel to the -//-axis. At E, f(x)
has a minimum value, and /'(/) again becomes infinite. In our dis-
cussion of all possible maximum and minimum values of /(j), we
must therefore include as critical values also those values of x for
which /'(x) becomes infinite, or, what is the same thing, values of x
satisfying
The Second Step of the rule of the preceding section must then
be modified as indicated by (1 ). The other steps are unchanged.
In Fig. d above, note that /'(/) also becomes infinite at A, but the
function is neither a maximum nor a minimum at A.
ILLUSTRATIVE EXAMPLE. Examine the function a - b(x - r) for maxima and
minima.
Solution.
2b
Since x = c is a ^^^1 va i u e for which ^ - (and f'(x] = oo), but for which
f(x) is not infinite, let us test the function for maximum and minimum values when
X = C * When x < c, f'(x) = -f .
When x > c, f'(x) = -.
Hence, when x = c = OM, the function has a maximum value /(c) = a = MP.
56 DIFFERENTIAL CALCULUS
PROBLEMS
Examine each of the following functions for maximum and minimum
values.
1. r { 6 x 2 + 9 x. Ans. x = 1, gives max. = 4.
x = 3, gives min. = 0.
2. 10 + 12 .r - 3 jr 2 - 2 jr 1 . a- = 1, gives max. = 17.
x 2, gives min. = 10.
3. 2 :r { + 3 j* 2 -f 12 .r 4. No max. or min.
4. /< + 2.r 2 - 15 j*- 20.
5. 2 .r 2 .r 4 . a* = 0, gives min. = 0.
x 1, gives max. = 1.
6. x 4 4 .r. x = 1, gives min. 3.
7. r 4 - j' 2 + 1.
8. 3 .r 4 - 4 ^ - 12 .r 2 . a- = - 1, gives min. = - 5.
x = 0, gives max. = 0.
x = 2, gives min. 32.
9. x* 5 r 4 . x = 0, gives max. = 0.
x = 4, gives min. = 256.
10. 3 JT* - 20 jr 3 .
2 rr*
11. j-2 + ^__. x a, gives min. = 3 a 2 .
12. 2, -g-
a 4
13. r 2 + a: = a, gives min. = 2 a 2 .
14 ____, x = a, gives min. = J.
' a* 2 4- a 2 x = a, gives max. = .
15.
16 - ^ + a*
T 2 -4- rr "
17. x , + ^
1Q /O I v^2/1 o*^2
ID. ^~T~Uy ^JL U^ .
19. (2 + x) 2 (l - x)*.
2
20. ?> + r(j* a) . x = a, gives min. = 6.
21. a b(x cr. No max. or min.
VARIOUS APPLICATIONS OF THE DERIVATIVE 57
22. (2 + *)*(! - x) 5 . Ans. x = l, gives min. = 0.
x 1, gives max. = V4 = 1.6.
23. a- (a + -r) 2 (a - x)*. .r = - a, gives max. = 0.
.r = \ a, gives min. = ~ jj a ft .
-r ~ /, a, gives max. = ^JJ a r> .
j* o, gives neither.
24. (2 a: - a)*(j - a)*. x - J a, gives max. = J a.
r = a, gives min. = 0.
* i <*> gives neither.
25 3* + 2 x ~ 0, gives max. - J.
' :r 2 -f 2 j- + 4 ' J- = - 4, gives min. = - J.
26 ^ 2 + ^ + 4 T .'i, gives max. ~ - 5.
^ ~f~ 1 u* 1, gives min. ,'{.
27 ^ 2 -f r + 4 a- = - *J, gives max. = .3.
* x 2 + 2 .r + 4 j- 2, gives min. =- [;.
(.r-
. _
- _ Drives max. -- ------ -
a -f /> 4 />
----- > . .
a -f b a
(a -
.
gives max. =
a
on (a x) 3 a . . ., 7 o
30. ^ _^_. x = - givers mm. = *? a 2 .
a 2 x
.
x 2 x -f 1
49. Maximum and minimum values. Applied problems. In many
problems we must first construct, from the jjiven conditions, the
function whose maximum and minimum values are required, as was
done in the two examples worked out in Art. 44. This sometimes
offers considerable difficulty. No rule applicable in all cases can be
given, but in many problems we may be guided by the following
General directions
(a) Set up the function whose maximum or minimum value is in-
volved in the problem.
(b) // tho resulting expression contains more than one variable, the
conditions of the problem will furnish enough relations between the vari-
ables so that all may be expressed in terms of a single one.
58 DIFFERENTIAL CALCULUS
(c) To the resulting function of a single variable apply the above rule
(p. 53) for finding maximum and minimum values.
(d) In practical problems it is usually easy to tell which critical value
will give a maximum and which a minimum value, so it is not always
necessary to apply the third step.
(e) Draw the graph of the function in order to check the work.
The work of finding maximum and minimum values may fre-
quently be simplified by the aid of the following principles, which
follow at once from our discussion of the subject.
(a) The maximum and minimum values of a continuous Junction
must occur alternately.
(b) When c is a positive constant, cf(x) is a maximum, or a mini-
mum for such values of z, and such only, as makef(x) a maximum or a
minimum.
Hence, in determining the critical values of x and testing for max-
ima and minima, any constant factor may be omitted.
When c is negative, cf(x) is a maximum when f(x) is a minimum,
and conversely.
(c) // c is a constant, f(x) and c + f(x) have maximum and mini-
mum values for the same values of x.
Hence a constant term may be omitted when finding critical values
of x and testing.
PROBLEMS
1. It is desired to make an open top box of greatest possible volume
from a square piece of tin whose side is a, by cutting equal squares out
of the corners and then folding up the tin to form the sides. What should
be the length of a side of the squares cut out ?
Solution. Let x = side of small square depth of box ;
then a 2 x side of square forming bottom of box,
and volume is V = (a 2 x) 2 j,
which is the function to be made a maximum by varying x.
Applying the rule, p. 53,
p:
rtV
First Step. -~- = (a - 2 x) 2 - 4 x(a - 2 x} = a 2 - 8 ax + 12 x 2 .
Second Step. Solving a 2 8 ax + 12 or 2 = gives critical values x - and -
It is evident from the figure that x = - must give a minimum, for then all the
tin would be cut away, leaving no material out of which to make a box. By the usual
test, x jj is found to give a maximum volume -- Hence the side of the square to
be cut out is one sixth of the side of the given square.
VARIOUS APPLICATIONS OF THE DERIVATIVE
59
The drawing of the graph of the function in this and the following problems is
left to the student.
2. Assuming that the strength of a beam with rectangular cross sec-
tion varies directly as the breadth and as the square of the depth, what
are the dimensions of the strongest beam that can be sawed
out of a round log whose diameter is d!
Solution. If x = breadth and y = depth, then the beam will
have maximum strength when the function xy- is a maximum.
From the figure, y' 2 = d~ x' 2 ; hence we should test the function
First Step. /'Or) = ~ 2 x- + d' 2 - x 2 - d' 2 - 3 x' 2 .
Second Step, d 2 3 x 2 = 0. .-. x = -~ = critical value which gives a maximum.
Therefore, if the beam is cut so that
Depth = \ j{ of diameter of log,
and Breadth \ \ of diameter of log,
the beam will have maximum strength.
3. What is the width of the rectangle of maximum area that can be
inscribed in a given segment OAA f of a parabola?
HINT. If Or = h, BC = h-x and PP' = 2 y ; there-
fore the area of rectangle PDD'P' is
2(h-x)y.
But since P lies on the parabola y 2 = 2px, the function to
be tested is
f(x) = 2(h-x)\ f 2px.
AUK. Width = | h.
4. Find the altitude of the cone of maximum volume that can be
inscribed in a sphere of radius r.
HINT. Volume of cone = J irx-y. But
X 2 = BC x CD = y(2 r - y} ;
therefore the function to bo tested is
TT A\
f(y) ^ y*(2 r y).
Ans. Altitude of cone = $ T.
5. Find the altitude of the cylinder of maximum volume that can be
inscribed in a given right cone.
HINT. Let AC r and BC = h. Volume of cylinder = irx 2 y.
But from similar triangles ABC and DBG,
Hence the function to be tested is
f(y)=y 2 y(h-y} 2 .
Ans. Altitude = | h.
60 DIFFERENTIAL CALCULUS
6. If three sides of a trapezoid are each 10 in. long, how long must the
fourth side be if the area is a maximum? Ans. 20 in.
7. It is required to inclose a rectangular field by a fence, and then to
divide it into two lots by a fence parallel to one of the sides. If the area of
the field is given, find the ratio of the sides so that the total length of fence
shall be a minimum. Ans. 2 3.
8. A rectangular garden is to be laid out along a neighbor's lot and is
to contain 432 sq. rd. If the neighbor pays for half the dividing fence,
what should be the dimensions of the garden so that the cost to the owner
of inclosing it may be a minimum? Ans. 18 rd. x 24 rd.
9. A radio manufacturer finds that he can sell .r instruments per week
at p dollars each, where 5 x 375 3 p. The cost of production is
(500 4- 15 x -f ?, x 2 } dollars. Show that the maximum profit is obtained
when the production is about 30 instruments per week.
10. In Problem 9 suppose the relation bet-ween x and /; is
JT= 100-1:
Show that the manufacturer should produce only about 25 instruments per
week for maximum profit.
11. In Problem 9 suppose the relation between x and p is
T 2 = 2500 - 20 p.
How many instruments should be produced each week for maximum profit?
12. The total cost of producing x articles per week is (ax 2 -f bx -f c)
dollars and the price (p dollars) at which each can be sold is /) = /3 a x 2 .
Show that the output for maximum profit is
x _ Va 2 -f 3a-qi- b) - a
3 a
13. In Problem 9 suppose a tax of t dollars per instrument is imposed
by the government. The manufacturer adds the tax to his cost and de-
termines the output and price under the new conditions.
(a) Show that the price increases by a little less than half the tax.
(b) Express the receipts from the tax in terms of / and determine the
tax for maximum return.
(c) When the tax determined in (b) is imposed, show that the price is
increased by about 33 per cent.
14. The total cost of producing x articles per week is (ax 2 -f bx -f c)
dollars, to which is added a tax of t dollars per article imposed by the
government, and the price (p dollars) at which each can be sold is p = p ax.
Show that the tax brings in the maximum return when t = J(/3 b) and
that the increase in price is always less than the tax.
NOTE. In applications in economics a, b, c, a, are positive numbers.
VARIOUS APPLICATIONS OF THE DERIVATIVE 61
15. A steel plant is capable of producing x tons per day of a low-grade
steel and y tons per day of a high-grade steel, where // = - If the
10 x
fixed market price of low-grade steel is half that of high-grade steel, show
that about 5| tons of low-grade steel are produced per day for maximum
profit.
16. A telephone company finds there is a net profit of if 15 per instru-
ment if an exchange has 1000 subscribers or less. If there are over 1000
subscribers, the profits per instrument decrease lc for each subscriber
above that number. How many subscribers would give the maximum net
profit? AUK. 11250.
17. The cost of manufacturing a given article is p dollars and the num-
ber which can be sold varies inversely as the >/th po\\er of the selling price.
Find the selling price which will yield the greatest total net profit.
* tlV
A //,s\ *
n 1
18. What should be the diameter of a tin can holding 1 qt. (58 cu. in.)
and requiring the least amount of tin (a) if the can is open at the top?
(b) if the can has a cover?
-vV = 5.29
\ 7T
Ans. (a) -v = 5.29 in. ; (b) ; = 4.20 in.
\ 7T \ 7T
19. The lateral surface of a right circular cylinder is 4 TT sq. ft. From
the cylinder is cut a hemisphere whose diameter equals the diameter of
the cylinder. Find the dimensions of the cylinder if the remaining volume
is a maximum or minimum. Determine whether it is a maximum or a
minimum. .4//s. Radius 1 ft., altitude = 2 ft. ; maximum.
20. Find the area of the largest rectangle with sides parallel to the
coordinate axes which can be inscribed in the figure bounded by the two
parabolas 3 // 12 - s- and 6 // - s 2 - 12. Ans. 16.
21. Two vertices of a rectangle are on the .r-axis. The other two vertices
are on the lines whose equations are y 2 jr arid 3 jc -f y 30. For what
value of y will the area of the rectangle be a maximum? Am. y 6.
22. One base of an isosceles trapezoid is a diameter of a circle of radius
a, and the ends of the other base lie on the circumference of the circle.
Find the length of the other base if the area is a maximum. Am. a.
23. A rectangle is inscribed in a parabolic segment with one side of the
rectangle along the base of the segment. Show that the ratio of the area
1
of the largest rectangle to the area of the segment is /-'
24. The strength of a rectangular beam varies as the product of the
breadth and the square of the depth. Find the dimensions of the strong-
est beam that can be cut from a log whose cross section is an ellipse of
semiaxes a and b. Ans. Breadth = 2 />\/A ; depth = 2 en/I-
62 DIFFERENTIAL CALCULUS
25. The stiffness of a rectangular beam varies as the product of the
breadth and the cube of the depth. Find the dimensions of the stiff est
beam that can be cut from a cylindrical log whose radius is a.
-4ns. a X aVi.
26. The equation of the path of a ball is y - mx - m " x ", where
800
the origin is taken at the point from which the ball is thrown and m is the
slope of the curve at the origin. For what value of m will the ball strike
(a) at the greatest distance along the same horizontal level? (b) at the
greatest height on a vertical wall 300 ft. away? Ans. (a) 1 ; (b) $.
27. A window of perimeter p ft. is in the form of a rectangle surmounted
by an isosceles right triangle. Show that the window will admit the most
light when the sides of the rectangle are equal to the sides of the right
triangle.
28. The sum of the surfaces of a sphere and a cube being given, show
that the sum of their volumes will be least when the diameter of the sphere
is equal to the edge of the cube. When will the sum of the volumes be
greatest ?
29. Find the dimensions of the largest rectangle which can be inscribed
in the ellipse ~-}~ ^~ = 1. Ans. a\/2 X &V5.
a- 6*
30. Find the area of the largest rectangle which can be drawn with
its base on the .r-axis and with two vertices on the witch w r hose equation
is ?/ = ., 8ff> * (See figure in Chapter XXVI.) Ans. 4 a 2 .
ir* -f 4 a 2
31. Find the ratio of the area of the smallest ellipse that can be cir-
cumscribed about a rectangle to the area of the rectangle. The area of an
ellipse is irab, where a and b are the semiaxes. Ans. | TT-
32. The two lower vertices of an isosceles trapezoid are the points
( 6, 0) and (6, 0). The two upper vertices lie on the curve jc 2 + 4 y 36.
Find the area of the largest trapezoid which can be drawn in this way.
Ans. 64.
33. The distance between the centers of two spheres of radii a and b,
respectively, is c. Find from what point P on the line of centers AB the
greatest amount of spherical surface is visible. (The area of the curved
surface of a zone of height h is 2 irrh, where r is the radius of the sphere.)
|
Ans. -jp^ 5 units from A.
a* + b*
34. Find the dimensions of the largest rectangular parallelepiped with
a square base which can be cut from a solid sphere of radius r.
Ans. h = $ rV3.
VARIOUS APPLICATIONS OF THE DERIVATIVE 63
35. Given a sphere of radius 6 in. Calculate the altitude of each of
the following solids :
(a) inscribed right circular cylinder of maximum volume ;
(b) inscribed right circular cylinder of maximum total surface ;
(c) circumscribed right cone of minimum volume,
Ans. (a) 4 V3 in. ; (b) 6.31 in. ; (c) 24 in.
36. Prove that a conical tent of a given capacity will require the least
amount of canvas when the height is V2 times the radius of the base.
Show that when the canvas is laid out flat it will be a circle with a sector
of 152 9' cut out. How much canvas would be required for a tent 10 ft.
high? Ans. 272 sq. ft.
37. Given a point on the axis of the parabola ?/ 2 2 px at a distance a
from the vertex ; find the abscissa of the point on the curve nearest to it.
A HK. JT = a p.
38. Find the point on the curve 2 y = x 2 which is nearest to the point
(4, 1). Am. (2, 2).
39. If PQ is the longest or shortest line segment which can be drawn
from P(a, b) to the curve y f(x), prove that PQ is perpendicular to the
tangent to the curve at Q.
40. A formula for the efficiency of a screw is E = ~ a " > where
h -f tan
is the angle of friction and h is the pitch of the screw. Find // for maximum
efficiency. -4ns. h = sec tan 6.
41. The distance between two sources of heat A and /?, with intensi-
ties a and b respectively, is I. The total intensity of heat at a point P
between A and B is given by the formula
~ \" /
where x is the distance of P from A. For what position of P will the tem-
perature be lowest ? a ij
Ans. x = T -
42. The lower base of an isosceles trapezoid is the major axis of an
ellipse ; the ends of the upper base are points on the ellipse. Show that
the maximum trapezoid of this type has the length of its upper base
half that of the lower.
43. An isosceles triangle with vertex at (0, b) is to be inscribed in the
ellipse b' 2 x' 2 + a' 2 y 2 a 2 b 2 . Find the equation of the base if the area of
the triangle is a maximum. Ans. 27/4-6 = 0.
44. Find the base and altitude of the isosceles triangle of minimum area
which circumscribes the ellipse b 2 x 2 4- a 2 ?/ 2 = a 2 6 2 , and whose base is
parallel to the z-axis. Ans. Altitude = 36, base = 2 aV.
64 DIFFERENTIAL CALCULUS
45. Let P(a, b) be a point in the first quadrant of a set of rectangular
axes. Draw a line through P cutting the positive ends of the axes at A
and B. Calculate the intercepts of this line on OX and OY in each of the
following cases.
(a) when the area OAK is a minimum;
(b) when the length AH is a minimum;
(cj when the sum of the intercepts is a minimum;
(d) when the perpendicular distance from O to AB is a maximum.
An*, (a) 2 a, 2 b ; (b) a + <6\ b + cPb* ;
(c) a 4- V^, fe + VS; (d) 2l^, 2!^!.
a 6
50. Derivative as the rate of change. In Art. 23 the functional
relation
(1) y = r 2
gave as the ratio of corresponding increments
When x = 4 and Ar = 0.5, equation (2) becomes
(3. -.
Then, we say, the average rate of change of y with respect to x
equals 8.5 when jr increases from x 4 to x = 4.5.
In general, the ratio
(A) -^- average rate of change of y with respect to x when x changes
X from x to x + A.r.
Constant rate of change. When
(4) y = ax + b,
we have -r& = a.
Ax
That is, the average rate of change of y with respect to x equals a,
the slope of the straight line (4), and is constant. In this case, and
in this case only, the change in y (A?/), when x increases from any
value x to .r + A.r, equals the rate of change a times Ao*.
Instantaneous rate of change. If the interval from x to x + Ax
decreases and A.r -> 0, then the average rate of change of y with
respect to x in this interval becomes at the limit the instantaneous
rate of change of y with respect to x. Hence, by Art. 24,
(B) -f- = instantaneous rate of change of y with respect to x for a
definite value of x.
VARIOUS APPLICATIONS OF THE DERIVATIVE 65
For example, from (1) above,
(o) , ^ Z x.
When x = 4, the instantaneous rate of change of y is 8 units per
unit change in x. The word "instantaneous" is often dropped in (B).
Geometric interpretation. Let the graph of
(6) #=/(*)
be drawn, as in the figure. When x in-
creases from OM to OA 7 , then y increases
from MP to NQ. The average rate of
change of y with respect to x equals the
slope of the secant line PQ. The instanta-
neous rate when x = OM equals the slope of the tangent line PT.
Hence the instantaneous rate of change of y at P(x, y) is equal to
the constant rate of change of y along the tangent line at P.
When x XD, the instantaneous rate of change of //, or /(.r) r in (6)
is f'(xo). If x now increases from .TO to xo + A:r, the exact change in
y is not equal to/'(jo)A.r, unless f'(x) is constant, as in (4). We shall
see later, however, that this product is equal to A//, nearly, when Ax
is sufficiently small.
51. Velocity in rectilinear motion. Important, applications arise
when the independent variable in a rate is the time. The rate is
then called a time-rate. Velocity
in rectilinear motion affords a - 4 < 2 ^ ^
. , A o r i> f B
simple example.
Consider the motion of a point P on the straight line AB. Let
s be the distance measured from some fixed point, as O, to any posi-
tion of P, and let t be the corresponding elapsed time. To each value
of t corresponds a position of P and therefore a distance (or space) s.
Hence s will be a function of /, and we may write
s=f(t).
Now let t take on an increment A ; then s takes on an increment
As, and A
n ) = the average velocity
v ' Az
of P when the point moves from P to P', during the time interval A.
If P moves with uniform motion (constant velocity), the above ratio
will have the same value for every interval of time and is the velocity
at any instant.
66 DIFFERENTIAL CALCULUS
For the general case of any kind of motion, uniform or not, we
define the velocity (time-rate of change of s) at any instant as the limit
of the average velocity as At approaches zero as a limit ; that is,
<o .-
The velocity at any instant is the derivative of the distance (= space)
with respect to the time, or the time-rate of change of the distance.
When v is positive, the distance s is an increasing function of t,
and the point P is moving in the direction AB. When v is negative,
s is a decreasing function of /, and P is moving in the direction BA.
(Art. 45.)
To show that this definition agrees with the conception we al-
ready have of velocity, let us find the velocity of a falling body at
the end of two seconds.
By experiment it has been found that a body falling freely from
rest in a vacuum near the earth's surface follows approximately the
law
(2) s - 16.1 /-,
where s = distance of fall in feet, t time in seconds. Apply the
General Rule (Art. 27) to (2).
FIRST STEP. s + As=lfi.l(J + A/) 2 = 1 6.1 Z 2 + 32.2 AZ+16.1(A/) 2 .
SECOND STEP. A* = 32.2 t-At + 16.
As
THIRD STEP. -r 1 = 32.2 1+ 16.1 At = average velocity throughout
the time interval At.
Placing t = 2,
As
(3) = 64.4 + 16.1 At = average velocity throughout the
time interval At after two
seconds of falling.
Our notion of velocity tells at once that (3) does not give us the
actual velocity at the end of two seconds ; for even if we take At very
small, say , o ( y or nuio of a second, (3) still gives only the average
velocity during the corresponding small interval of time. But what
we do mean by the velocity at the end of two seconds is the limit of
the average velocity when At diminishes toward zero ; that is, the ve-
locity at the end of two seconds is, from (3), 64.4 feet per second.
Thus even the everyday notion of velocity which we get from experi-
ence involves the idea of a limit, or, in our notation,
/ As\
v = lim )= 64.4 ft. per second.
Ai-oVAt/
VARIOUS APPLICATIONS OF THE DERIVATIVE 67
52. Related rates. In many problems several variables are in-
volved each of which is a function of the time. Relations between
the variables are established by the conditions of the problem. The
relations between their time-rates of change are then found by
differentiation.
As a guide in solving rate problems use the following rule.
FIRST STEP. Draw a figure illustrating the problem. Denote by x,
p, z, etr. the quantities which vary with the time.
SECOND STEP. Obtain a relation bet-ween the variables involved which
will hold true at any instant.
THIRD STEP. Differentiate with respect to the time.
FOURTH STEP. Make a list of the given and required quantities.
FIFTH STEP. Substitute the known quantities in, the result found by
differentiating (third step), and solve for the unknown.
PROBLEMS
1. A man is walking at the rate of 5 mi. per hour toward the foot of
a tower 60 ft. high. At what rate is he approaching the top when he is
80 ft. from the foot of the tower?
Solution. Apply the above rule.
First Step. Draw the figure. Let x distance of the man from the foot, and
y = his distance from the top, of the tower at any instant.
Second Step. Since we have a right triangle,
y y = x 2 + 3600.
Third Step. Differentiating, we get
**%=**%'<*
m 4y = xdx
U ' dt y dt'
- This means that at any instant whatever
(Rale of change ofy] (-) times (rate of change ofx).
Fourth Step. x 80, -~ = 5 mi. an hour
at
= - 5 x 5280 ft. an hour.
_ ?
dt
Fifth Step. Substituting in (1),
f = 80
100
= 4 mi. per hour. Ans.
& = _ JjL x 5 x 5280 ft. per hour
dt 100
68
DIFFERENTIAL CALCULUS
2. A point moves on the parabola 6 y = x 2 in such a way that when
x 6 the abscissa is increasing at the rate of 2 ft. per second. At what
rate is the ordinate increasing at that instant?
Solution. First Step. Plot the parabola.
Second Step. 6 y = x' 2 .
Third Step. 6^ = 2x^.or
) -..
dt~3 dt
This moans that at any point on the parabola
(Rate of ckanac of ordinate) [r] liws ( rafe f change of abscissa).
Fourth Step. x = 6.
~ - 2 ft. per second.
dt
Fifth Step. Substituting in (2),
= 4 ft- P^r second,
^ = ~ x 2 = 4 ft. per second. Ans.
at o
From the first result we note tliat at the point / J (G, 6) the ordinate changes twice
as rapidly as the abscissa.
If we consider the point /"( t>, 6) instead, the result is
the minus sign indicating that the ordinate is decreasing
as the abscissa increases.
3. A circular plate of metal expands by heat so
that its radius increases at the rate of 0.01 in. per
second. At what rate is the surface increasing
when the radius is 2 in. ?
Solution. Let x = radius and y area of plate. Then
y = 7r.r~.
du rt dx
That is, at any instant the area of the plate is increasing in square inches 2 ?nr
times as fast as the radius is increasing in linear inches.
Substituting in (U),
= 2 TT x 2 x 0.01 = 0.04 TT sq. in. per second. Arts.
dt
4. An arc light is hung 12 ft. directly above a straight horizontal
walk on which a boy 5 ft. tall is walking. How fast is the boy's shadow
VARIOUS APPLICATIONS OF THE DERIVATIVE 69
lengthening when he is walking away from the light at the rate of 168 ft.
per minute?
Solution. Let x = distance of the boy from a point directly under the light L,
and y length of boy's shadow. From the figure,
y : y + x = 5 : 12,
or y = 8 x.
T^.- .. ,. t/v 5 dx
Differentiating, d/ 7 d/ '
that is, the shadow is lengthening v> as fast as the boy is
walking, or 120ft. per minute.
5. A point moves along the parabola //-' = 12 jc in such a way that its
abscissa increases uniformly at the rate of 2 in. per second. At what point
do the abscissa and ordinate increase at the same rate? Ans. (3, 6).
6. Find the values of .r for which the rate of change of
^ - 12 jc 2 + 45 .r - 13
is zero. /b/.s\ 3 and 5.
7. A barge whose deck is 12 ft. below the level of a dock is drawn up
to it by means of a cable attached to a ring in the floor of the dock, the
cable being hauled in by a windlass on deck at the rate of 8 ft. per minute.
How fast is the barge moving towards the dock when 10 ft. away?
Ann. 10 ft. per minute.
8. A boat is fastened to a rope which is wound about a windlass 20 ft.
above the level at which the rope is attached to the boat. The boat is
drifting away at the rate of 8 ft. per second. How fast is it unwinding the
rope when 30 ft. from the point directly under the windlass?
Ann. ().Gf) ft. per second.
9. One end of a ladder 50 ft. long is leaning against a perpendicular
wall standing on a horizontal plane. Suppose the foot of the ladder to be
pulled away from the wall at the rate of 3 ft. per minute, fa) How fast is
the top of the ladder descending when its foot is 14 ft. from the wall?
(b) When will the top and bottom of the ladder move at the same rate?
(c) When is the top of the ladder descending at the rate_of 4 ft. per minute?
Ans. fa) I ft. per minute; fb) when 25 V2 ft. from the wall;
(c) when 40 ft. from the wall.
10. One ship was sailing south at the rate of 6 mi. per hour; another
east at the rate of 8 mi. per hour. At 4 P.M. the second crossed the track
of the first where the first was 2 hr. before, fa) How was the distance
between the ships changing at 3 P.M. ? fb) How at 5 P.M. ? (c) When was
the distance between them not changing ?
Ans. (a) Decreasing 2.8 mi. per hour; (b) increasing
8.73 mi. per hour; (c) 3.17 P.M.
70 DIFFERENTIAL CALCULUS
11. The side of an equilateral triangle is o in. long, and is increasing at
the rate of k in. per hour. How fast is the area increasing?
An.s. \ afcV3 sq. in. per hour.
12. The edges of a regular tetrahedron are 10 in. long and are increasing
at the rate of 0.1 in. per minute. Find the rate of increase of the volume.
13. If at a certain instant the two dimensions of a rectangle are a and 6
and these dimensions are changing at the rates ra, n respectively, show r that
the area is changing at the rate an -f bm.
14. At a certain instant the three dimensions of a rectangular parallele-
piped are 6 in., 8 in., 10 in., and these are increasing at the respective rates
of 0.2 in. per second, 0.3 in. per second, 0.1 in. per second. How fast is the
volume increasing?
15. The period (Psec.) of a complete oscillation of a pendulum of
length / in. is given by the formula P 0.324 VL Find the rate of change
of the period with respect to the length when / = 9 in. By means of this
result approximate the change in /' due to a change in / from 9 to 9.2 in.
AJIK. 0.054 sec. per inch ; 0.0108 sec.
16. The diameter and altitude of a right circular cylinder are found at a
certain instant to he 10 in. arid 20 in. respectively. If the diameter is in-
creasing at the rate of 1 in. per minute, what change in the altitude will
keep the volume constant? A?IK. Decreasing 4 in. per minute.
17. The radius of the base of a certain cone is increasing at the rate of
3 in. per minute and the altitude is decreasing at the rate of 4 in. per
minute. Find the rate of change of the total surface of the cone when the
radius is 7 in. and the altitude is 24 in.
An*. Increasing 96 TT sq. in. per minute.
18. A cylinder of radius r and altitude // has a hemisphere of radius r
attached to each end. If r is increasing at the rate of .] in. per minute, find
the rate at which // must change to keep the volume of the solid fixed at
the instant when r is 10 in. and Jt is 20 in.
19. A stone is dropped down a' deep shaft and after t sec. another stone
is dropped. Show that, the distance between the stones increases at the
rate of ty ft. per second.
20. A gas holder contains 1000 cu. ft. of gas at a pressure of 5 Ib. per
square inch. If the pressure is decreasing at the rate of 0.05 Ib. per square
inch per hour, find the rate of increase of the volume. (Assume Boyle's
Law : pv = c.) Ans. 10 cu. ft. per hour.
21. The adiabatic law for the expansion of air is PV IA = C. If at a
given time the volume is observed to be 10 cu. ft. and the pressure is
50 Ib. per square inch, at what rate is the pressure changing if the volume
is decreasing 1 cu. ft. per second?
Ans. Increasing 7 Ib. per square inch per second.
VARIOUS APPLICATIONS OF THE DERIVATIVE 71
22. If y = 4 x r 3 and .r is increasing steadily at the rate of A unit per
second, find how fast the slope of the graph is changing at the instant when
x = 2. An*. Decreasing 4 units per second,
23. Water flows from a faucet into a hemispherical basin of diameter
14 in. at the rate of 2 cu. in. per second. How fast is the water rising
(a) when the water is halfway to the top? (b) just as it runs over? (The
volume of a spherical segment is ], irr >2 h -f ^ ?r/r ! , where // is the altitude of
the segment.)
24. Gas is escaping from a spherical balloon at the rate of 1000 cu. in.
per minute. At the instant when the radius is 10 in., (a) at uhat rate is the
radius decreasing? (b) at what rate is the surface decreasing?
AUK. (b) 200 sq. in. per minute.
25. If r denotes the radius of a sphere, tf the surface, and V the volume,
x , , ,. dV rdS
prove the relation = -
dt 2 dt
26. A railroad track crosses a highway at an angle of (10". A locomotive
is 500 ft. from the intersection and moving away from it at the rate of
60 mi. per hour. An automobile is 500 ft. from the intersection and moving
toward it at the rate of 30 mi. per hour. What is the 1 rate of change of the
distance between them?
Ans. Increasing 15 mi. per hour or IftV.'J mi. per hour.
27. A horizontal trough 10 ft. long has a vertical section in the shape of
an isosceles right triangle. If water is poured into it at the rate of H cu. ft.
per minute, at what rate is the surface of the water rising when the water
is 2 ft. deep? Aws. I ft. per minute.
28. In Problem 27, at what rate must water be poured into the trough
to make the level rise I ft. per minute when the water is 3 ft. deep ?
29. A horizontal trough 12 ft. long has a vertical cross section in the
shape of a trapezoid, the bottom being 3 ft. wide and the sides inclined to
the vertical at an angle whose sine is ;?. Water is being poured into it at
the rate of 10 cu. ft. per minute. How fast is the water level rising when
the water is 2 ft. deep?
30. In Problem 29, at what rate is water being drawn from the trough
if the level is falling 0.1 ft. per minute when the water is 3 ft. deep ?
31. The ^-intercept of the tangent line to the positive branch of the
hyperbola xy = 4 is increasing 3 units per second. Let the -//-intercept be
OB. Find the velocity of B at the end of 5 sec., the x-intercept starting at
the origin. Am. f| unit per second.
32. A point P moves along the parabola y 2 x so that its abscissa in-
creases at the constant rate of k units per second. The projection of P on
the .r-axis is M. At what rate is the area of triangle OMP changing when
P is at the point where x = a? Ann. 2 fcVa units per second.
72 DIFFERENTIAL CALCULUS
ADDITIONAL PROBLEMS
1. Rectangles inscribed in the area hounded by the parabola y 2 = 16 x
and its latus rectum and such that one side always lies along the latus
rectum serve as the bases of rectangular parallelepipeds whose altitudes
are always the same as the side parallel to the .r-axis. Find the volume of
the largest parallelepiped. Ans. A j*j&Vb = 73.27.
2. An ellipse symmetrical with respect to the coordinate axes passes
through the fixed point (//, k). Find the equation of the ellipse if its area
is a minimum. Am. k 2 jr 2 + h' 2 y 2 2 h 2 k 2 .
3. The curve j :< 3 /// + //'* = has a loop in the first quadrant sym-
metric with respect to the line // = jr. Isosceles triangles having a common
vertex at the origin and bases along the line j- -f y ~ a are inscribed in this
loop. Find the value of a if the area of the triangle is a maximum.
Am. J(l + Vl3) =2.303.
4. At a point /' in the first quadrant on the curve // = 7 jr 2 a tangent
is drawn, meeting the coordinate axes at A and B. Find the position of P
which makes A B a minimum. Am. Ordinate */-.
5. The cost of erecting an office building is $50,000 for the first story,
$52,500 for the second, $55,000 for the third, and so on. Other expenses
(lot, plans, basement, etc.) are $350,000. The net annual income is $5000
for each story. How many stories will give the greatest rate of interest on
the investment? Am. 17.
6. For a certain article the increase in the number of pounds consumed
is proportional to the decrease in the tax on each pound. If the consump-
tion is in Ib. when there is no tax and n Ib. when the tax is / dollars per
pound, find the tax which should be imposed on each pound to bring in the
most revenue.
7. A triangle ABC is formed by a chord BC of the parabola y ~ kx 2
and the tangents AB and AC at each extremity of the chord. If BC re-
mains perpendicular to the axis of the parabola and approaches the vertex
at the rate of 2 units per second, find the rate at which the area of the
triangle is changing when the chord BC is 4 units above the vertex.
8. A vertical cylindrical tank of radius 10 in. has a hole of radius 1 in.
in its base. The velocity with which the water contained runs out of the
tank is given by the formula v 2 2 gh, where h is the depth of the water
and g is the acceleration of gravity. How rapidly is the velocity changing ?
.4r?8. Decreasing y^ g ft. per second per second.
9. A light is 20 ft. from a wall and 10 ft. above the center of a path
which is perpendicular to the wall. A man 6 ft. tall is walking on the path
toward the wall at the rate of 2 ft. per second. When he is 4 ft. from the
wall, how fast is the shadow of his head moving up the wall ?
Am. f ft. per second.
CHAPTER VI
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS
53. Definition of successive derivatives. We have seen that the
derivative of a function of x is in general also a function of x. This
new function may also be diil'erentiable, in which case the derivative
of the first derivative is called the second derivatire of the original
function. Similarly, the derivative of the second derivative is called
the third derivative ; and so on to the ni\\ deriratire. Thus, if
V - 3 r>,
<LM ._ 19 ~3
, i ^ x ,
ax
dx\ drv/.r
Notation. The symbols for the successive derivatives are usually
abbreviated as follows.
rf 2 ?/
If y =/(j-), the successive derivatives are also denoted by
, _ . . . .
= - , , dxn - .
In the example given above, the notation y = 3 x 4 , y' = 12 x 3 ,
I/'' = 36 x 2 , T/'" = 72 r, ?y w = 72 is most convenient.
54. Successive differentiation of implicit functions. To illustrate the
process we shall find ^ f rom the equation of the hyperbola
(1) b 2 x 2 - a 2 ?/ 2 = a 2 6 2 .
73
74 DIFFERENTIAL CALCULUS
Differentiating with respect to x (Art. 41),
2 b*x - 2 a' 2 y ^ = 0,
(9) ^L ^ -
dx a-y
Differentiating again, remembering that y is a function of x,
d~y _ dx
dx* ^
Substituting for ^ its value from (2),
., ,i, /fcVN
a 4 '//- r/ 4 v/ ;{
Hut, from the given equation, b-x 2 a-y~ = a-6 2 .
PROBLEMS
Prove each of the following differentiations.
2. = >/" + W. ^ ~ iL^!___.
"'* 8(a-f6/)^
d 2 // __ 4 a?>-
3_ y \LJ-.y^. . *j< __
(i bx dx' 2 (a b.
_ /~~ ; r; d~u a 2
4. // Va- -f r 2 . - =
5. // =
6. =
fi\~
dl (a-' + r
d-y _ 2 a' 2
(i + r dx' 2 (a + x) 3
7. f(s) =
8. y =
'2 t -h 1
r-< *?
.r-fl
<ir 2 y3
d' 2 y __ _ 4 a 2
dx' 2 //*
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 75
9. j* 2 -f- */ 2 = r 2 .
10. y 2 = 4 ru*.
11. fc-.r 2 -f a 2 // 2 = a' 2 b 2 .
12. or-' -f 2 Jixy -f fy/ 2 1.
13. JT* + ?/ :{ ^ 1.
14. .r + 2 T 2 ^ 2 - - 4
a 4 ?/'
(//.r -f by)*'
= a 4 .
djr-
V
</.r 2
In Problems 15 25 find the values of //' and //'' for the given values of
the variables.
16. ;// = V25 - 3 x ; j = 3.
17. ?/ = x Vx 2 + 9 ; j* = 4.
18. j- 2 - 4 ?/ 2 = 9 ; j- = 5, ?/ = 2.
19. x 2 -f 4 xy 4- z/ 2 -f 3 = ; x = 2, ?/ = -
20. ?/= (3 -a: 2 ) 4 ; j- = 1.
21. ?/ = Vl 4-
2 J-; j- =
22. ?/ =
23. ?/ =
4 ; x = 2.
2 ; x =
In each of the following problems find
24. ?/ 2 + 2 .r/y = 1 G ; .r = 3, y ^ 2.
25. .r' { - .r/y 2 -f //" = 8 ; x- 2, y = 2.
27. ?/-
a: 2 H- a 2
29. y j-Vr/"' - x 2 .
30. // 2 - 4 xy - If).
31. x :i - 3 ajry -f ?y :{ = ?> 3 .
28. T/ = V2 - 3 a:.
55. Direction of bending y
of a curve. When the point \
P(x, y) traces a curve, the
slope of the tangent line at
P varies. When the tan-
gent line is below the curve
(Fig. a), the arc is concave upward; when above the curve (Fig. 6),
the arc is concave downward. In Fig. a the slope increases when
/
FIG. a
X
x
FIG. 6
76 DIFFERENTIAL CALCULUS
P describes the arc AP'. Hence f'(x) is an increasing function of x.
On the other hand, in Fig. b, when P describes the arc QB, the slope
decreases, and J'(-jr) is a decreasing function. In the first case, there-
fore, f"(x) is positive; in the second case, negative (Art. 45). Hence
we have the following criterion for determining the direction of
bending at a point.
The graph of y = f(x) is concave upward if the second derivative of
y with respect to x is positive, and concave downward if this derivative
is negative.
56. Second method for testing for maximum and minimum values.
At A in Fig. a of the preceding section, the arc is concave upward,
and the ordinate has a minimum value. That is, /'(x) = and /"(/)
is positive. At B in Fig. b, f'(x) and f"(x) is negative.
We may then state the sufficient conditions for maximum and
minimum values of f(x) for critical values of the variable as follows
j(x) is a maximum if f'(x) and/"(*) a negative number,
/(jt) is a minimum if f'(x) = and/"(*) a positive number.
Following is the corresponding working rule for applying this test
for maximum and minimum values.
FIRST STEP. Find the first derivative of the function.
SECOND STEP. Set the first derivative equal to zero and solve the result-
ing equation for real roots in order to find the critical values of the variable.
THIRD STEP. Find the second derivative.
FOURTH STEP. Substitute each critical value for the variable in, the
second derivative'. If the result is negative, then the function is a maxi-
mum for that critical value; if the result is positive, the function is a
minimum.
When /"(/) = or does not exist, the above process fails, al-
though there may even then be a maximum or a minimum ; in that
case the first method given in Art. 47 holds, being fundamental.
Usually the second method does apply, and when the process of
finding the second derivative is not too long or tedious, it is generally
the shortest method.
ILLUSTRATIVE EXAMPLE 1 . Let us now apply the above rule to test analytically
the function ,
M = x~ -f
x
found in the example worked out on page 49.
Solution. /(;r)=.r 2 +
439
First Step. f(x)=2x-~>
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 77
Second Step.
Third Step.
Fourth Step.
Hence
; -^=.
x = 6, critical value.
/"(6) = -f .
/(6) = 108, minimum value.
ILLUSTRATIVE EXAMPLE 2. P'xamine r* - 3 x- - 9 x 4 5 for maxima and
minima. Use the second method.
Solution. /(*) = ** - 3 x- -9x45.
First Step. /'(*) = 3 x 2 - 6 x - 9.
Second Step. 3 x 2 - 6 x - 9 = ;
hence the critical values are x = 1 and 3.
Fourth Step. /"(- 1) = - 12.
.-./( 1) = 10 (ordinate of /I) = maximum value.
/"(3) = 4 12. /. /(3) = 22 (ordinate of #) minimum value.
PROBLEMS
Examine each of the following functions for maxima and minima.
An*. JT - 2, gives max. 2.
x ~ 0, gives min. 2.
x 1 , gives max. 6.
x = 1, gives min. = 2.
0) x 0, gives max. a 3 ,
x = a, gives min. = 0.
x - 2, gives max. = 22.
x = 1, gives min. = 5.
x \, gives max. = \.
jr | , gives min. = $.
x = 0, gives max. = 2.
x = 1, gives min. = 3.
x = 2, gives min. 30.
1. x 3 4- 3 x 2 - 2.
2. x 3 - 3 x + 4.
3. 2 x 3 - 3 ox 2 4- a 3 . (a ;
4. 2 4- 12 x 4- 3 x 2 - 2 x 3 .
5. 3 x - 2 x 2 - -
6. 3 x 4 - 4 x 3 - 12 x 2 4- 2.
7. x 4 - 4 x 2 -f 4.
ax
x = 0, gives max. 4.
x = V2, gives min. = 0.
x = a, gives max. = |.
x a, gives min. = \.
9. x 3 4- 9 x 2 4- 27 x 4- 9.
10. 12 x 4- 9 x 2 - 4 x 3 .
11. x 2 (x-4) 2 .
12. * +
13. x-
78 DIFFERENTIAL CALCULUS
14. A rectangular box with a square base and an open top is to be made.
Find the volume of the largest box that can be made from 1200 sq. ft. of
material. Ans. 4000 cu. ft.
15. A water tank is to be constructed with a square base and an open
top, and is to hold 125 cu. yd. If the cost of the sides is $2 a square yard,
and of the bottom $4 a square yard, what are the dimensions when the
cost is a minimum? Ans. A cube of edge 5 yd,
16. A rectangular flower bed is to contain 800 sq. ft. It is to be sur-
rounded by a walk which is 3 ft. wide along the sides and 6 ft. wide across
the endtt. If the total area of the bed and walk is a minimum, what are the
dimensions of the flower bed? Ans. 20 ft. x 40 ft.
17. A rectangular field to contain a given area is to be fenced off along
the bank of a straight river. If no fence is needed along the river, show that
the least amount of fencing will be required if the field is twice as long as
it is wide.
18. A trough is to be made of a long rectangular piece of tin by bending
up U\o edges so as to give a rectangular cross section. If the width of the
piece is 14 in., how deep should the trough be made in order that its carry-
ing capacity may be a maximum? Ans. 3.5 in.
19. A window composed of a rectangle surmounted by an equilateral
triangle is 15 ft. in perimeter. Find its dimensions if it admits the maxi-
mum amount of light. An*. Rectangle is 3.51 ft. wide and 2.23 ft. high.
20. A solid wooden sphere weighs ?/> Ib. What is the weight of the
heaviest right circular cylinder which can be cut from the sphere?
Ans. r^lb.
V3
21. The slant height of a right circular cone is a given constant a.
Find the altitude if the volume is a maximum. A a
-ri/io. 7^
V3
22. An oil can is to be made in the shape of a cylinder surmounted by a
cone. Show that for a given capacity the least material is required if the
altitude of the cylinder is equal to the altitude of the cone.
23. Ciiven the parabola // 2 = 8 jc and the point P(6, 0) on the axis, find
the coordinates of the points on the parabola nearest to P. Ans. (2, 4).
24. A given isosceles triangle has a base of 20 ft. and an altitude of 8 ft.
What are the dimensions of the maximum inscribed parallelogram, one side
coinciding with the base of the triangle, if the acute angle of the parallelo-
gram is arc tan ^? Ans. 5 ft. X 10 ft.
25. A miner wishes to dig a tunnel from a point A to a point B 200 ft.
below and 600 ft. to the east of A. Below the level of A it is bed rock and
above A is soft earth. If the cost of tunneling through earth is $5 and
through rock is $13 per linear foot, find the minimum cost of a tunnel.
Ans. $5400.
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 79
26. A sheet of paper for a poster is to contain 16sq. ft. The margins
at the top and the bottom are to be (\ in., and those on the sides 4 in.
What are the dimensions if the printed area is to be a maximum?
.4 MS. 1.90ft. x 3.27 ft,
27. An electric current flows through a coil of radius r and exerts a
force F on a small magnet the axis of which is on a line drawn through
the center of the coil and perpendicular to its plane. This force is given
by F _, w here .r is the distance to the magnet from the center
(r 2 + .r 2 )*
of the coil. Show that F is a maximum for .r \ r .
57. Points of inflection. A point of inflection (or inflectional point)
on a curve separates arcs having opposite directions of bending (see
Art. 55).
In the figure, B is a point of inflection. When the tracing point,
on a curve passes through such a point, the second derivative will
change sign, and if continuous must vanish at the point. Hence we
must have
(1) At points of inflection, /"(*) = 0.
Solving the equation resulting from (1) gives the abscissas of the
points of inflection. To determine the direction of bonding in the
vicinity of a point of inflection, test f"(jr) for val-
ues of x first a trifle less and then a trifle greater
than the abscissa at that point.
If /"(#) changes sign, we have a point of in-
flection, and the signs obtained determine if the
curve is concave upward or concave downward in the neighborhood.
The student should observe that near a point where the curve is
concave upward (as at A) the curve lies above the tangent, and at
a point where the curve is concave downward (as at C) the curve
lies below the tangent. At a point of inflection (as at B) the tangent
evidently crosses the curve.
Following is a rule for finding points of inflection of the curve whose
equation is y = f(x). This rule includes also directions for examining
the direction of bending.
FIRST STEP. Findf"(x).
SECOND STEP. Setf"(x) = 0, and solve the resulting equation for real
roots.
THIRD STEP. Testf"(x) for values of x first a trifle less and then a
trifle greater than each root found in the second step. If f"(x) changes
sign, we have a point of inflection.
80 DIFFERENTIAL CALCULUS
When f"(x) = +, the curve is concave upward + .*
Whenf"(x) = , the curve is concave downward ^^^.
Before the Third Step it is sometimes convenient to factor f"(x).
It is assumed that f'(x) and J"(x) are continuous. The solution
of Problem 2, below, shows how to discuss a case where f'(x) and
f"(x) are both infinite.
PROBLEMS
Examine the following curves for points of inflection and direction of
bending.
1. 77 = 3 j- 4 - 4 x 3 + 1.
Solution.
First Step.
Second Step.
/(x) = 3 x 4 - 4 x'< 4- 1.
/"(x) = 36 x 2 - 24 x.
36 J' 2 - 24 x ~ 0.
. x 3 and x are the roots.
Third Step.
tl
/"(x) =36T(x- I).
When x < 0,/"(x) = +.
When 3 >x> 0,/"(x) = -.
Therefore the curve is concave upward to the left and concave downward to
the right of x (A in figure).
When ()<x < 3,/"(x) = -.
When x > l,f"(x) = -f.
Then^fore the curve is concave downward to the left and concave upward to
the right of x = rj (/ in figure).
Hence the points .4(0, 1) and B($, \\] are points of inflection.
Th*> curve is evidently concave upward everywhere to the left of 4, concave
downward between ,4(0, 1) and (, A}), and concave upward everywhere to
the right of B.
2. (y-2}*= (jr-4).
Solution. y = 2 -f (x - 4)*.
Second Step. When x = 4, both first and second derivatives become infinite.
Third Step.
When x < 4, = +.
dx 2
W T hen x > 4, ^ = -.
dr 2
* This may easily be remembered if we say that a vessel shaped like the curve where
it is concave upward will hold (+) water, and where it is concave downward will spill
( ) water.
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 81
We may therefore conclude that the tangent at (4, 1^ is perpendicular to the
x-axis, that to the left of (4, 2) the curve is concave upward, and to the right of
(4, 2) it is concave downward. Therefore (4,2) is a point of inflection.
3. y = x 2 . Ans. Concave upward everywhere.
4. y = 5 2 x x 2 . Concave downward everywhere.
5. y = x 3 . Concave downward to the left and
concave upward to the right of
(0, 0).
6. y = x 4 . Concave upward everywhere.
7. y = 2 x 3 - 3 x 2 - 36 x + 25. Concave downward to the left and
concave upward to the right of
jc ~ \
8. 2/=24x 2 -x 4 . 9. ?/^x + 1 -.
* ' x
58. Curve-tracing. The elementary method of tracing (or plotting)
a curve whose equation is given in rectangular coordinates, and one
with which the student is already familiar, is to solve its equation for
y (or x), assume arbitrary values of x (or //), calculate the correspond-
ing values of y (or r), plot the respective points, and draw a smooth
curve through them, the result being an approximation to the re-
quired curve. This process is laborious at, best, and in case the equa-
tion of the curve is of a degree higher than the second, it may not be
possible to solve the equation for y or x. The general form of a curve
is usually all that is desired, and the calculus furnishes us with
powerful methods for determining the shape of a curve with very
little computation.
The first derivative gives us the slope of the curve at any point ;
the second derivative determines the intervals within which the curve
is concave upward or concave downward, and the points of inflection
separate these intervals; the maximum points are the high points,
and the minimum points are the low points on the curve. As a guide
in his work the student may follow the
Rule for tracing curves, using rectangular coordinates
FIRST STEP. Find the first derivative; place it equal to zero; solve
to find the abscissas of the maximum and minimum points. Test these
values.
SECOND STEP. Find the second derivative; place it equal to zero;
solve to find the abscissas of the points of inflection. Test these values.
THIRD STEP. Calculate the corrcspomling ordinates of the points
whose abscissas were found in the first two steps. Calculate as many
82
DIFFERENTIAL CALCULUS
more points as may be necessary to (jive a good idea of the shape of the
curve. Make a table such as is shown in tlie problem worked out below.
FOURTH STEP. Plot the points determined and sketch in the curve to
correspond with the results shown in the table.
If the calculated values of the ordinatcs are large, it is best to
reduce the scale on the ?/-axis so that the general shape of the curve
will be shown within the limits of the paper used. Coordinate plot-
ting paper should be employed. Results should be tabulated as in
the problems solved. In this table the values of x should follow one
unother, increasing algebraically.
PROBLEMS
Truce the following curves, making use of the above rule. Also find
the e([iiations of the tangent and normal at each
point of inflection.
1. y rr .r :{ - 9 jr 2 4- 1M x - 7.
Solution. Use the above rule.
Ft rat Step. y' - 3 x'~ - 18 x -f 24,
(lx- - lS.r -f- 24 - 0,
x = 2, 4.
Second Step. y" = G x - 18,
G.r - 18 = 0,
Third Step. x - 3.
X
V
l/'
if'
Remarks
Direction of Curve
(1
7
4
1 ,
- concave down
'_!
III
max.
J
\\
11
_
I>t . of irifl.
1
J)
-f
inin.
1
(i
""
+
+
r" concave up
Fourth Step. Plotting the points and sketching in the curve, we get the figure
shown.
To find the equations of the tangent and normal to the curve at the point of
inflection /V3, 11), use formulas (1), (2\ Art. 43. This gives 3 x + y 20 for the
tangent and 3 ?/ x 30 for the normal.
2. ;] // = .r< - 3 x' 2 - 9 x -f 11.
An*. Max. ( 1, J /); min. (3, -\f-) ; point of inflection, (1, 0) ;
tangent, 4 jc + y -4 ; normal, .r 4 // 1 = 0.
3. 6 y = VI - 24 x - 15 :r 2 - 2 j-\
A//S. Max. ( 1, ^) : min. (- 4, - 5) ; point of inflection,
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 83
4. y = x 4 - 8 x 2 .
Am. Max. (0, 0) ; min. ( 2, - 16) ; points of inflection,
(?,Vi*, -Y>.
5. y 5 x J*.
Am. Max. (1, 4); min. (- 1, - 4) ; point of inflection, (0, 0).
*-?Ti-
Am. Max. (V3, V3) ; min. ( ~ V3, - V3 ) ; points of inflection,
(-3, -ij), (0, 0), (3, H).
7. ?/ = /< + 6 .T 2 . 18. (I If ^ .r- + -
8. ?/ = 4 + 3 .r - .r<.
9. 3 ?/ = 4 .r< - 18 r 2 + 15 x. 19- '-'// ^ .'"' -f ~
10. ?/ = O - a) : * -f b.
11. 12 // = (.r - I) 1 - 24 (.r -1)2. 20. a-'// ^ .r< -f ^,.
12. // = r 2 (9 - -r L> ).
13. ?/ = 2 .r> - 5 .r 2 . 21. // = - - -"-' - .,-
.r- -1 1 a-
14. // = 3 .r> - 5 .r\
15. ?/ = .r r> - 5 .r 1 . 22. ?/ - - ''-- -
(.r f <i)-
16. y = jr(s- - 4) 2 .
4 23. .r 1 '// - (.r-' { 1 )'.
59. Acceleration in rectilinear motion. In Art. 51 velocity in rec-
tilinear motion was defined as the time-rate of change of the distance.
We now define acceleration as the time-rute of climiye of the velocity.
That is,
dv
(A) Acceleration = a
(h
From (C), Art. 51, we obtain also, since v --*
Referring to Arts. 45, 47, and 56, we have the following criteria
which apply to a definite instant / = / :
If a > 0, v is increasing (algebraically).
If a < 0, v is decreasing (algebraically).
If a > and v = 0, s has a minimum value.
If a < and v = 0, s has a maximum value.
If a = and changes sign from + to (from to +) when t
passes through to, then v has a maximum (a minimum) value when
84 DIFFERENTIAL CALCULUS
In uniformly accelerated rectilinear motion, a is constant. Thus
in the case of a body falling freely under the action of gravity only,
a = 32.2 ft. per second per second. Namely, from (2), Art. 51,
8 = 16.1 1 2 , v = = 32.2 t, <* = = 32.2.
PROBLEMS
1. By experiment it has been found that a hody falling freely from rest
in a vacuum near the earth's surface follows approximately the law
s ~ 16.1 t 2 , where = space (height) in feet, t = time in seconds. Find
the velocity and acceleration (a) at any instant; (b) at end of the first
second ; (c) at end of the fifth second.
Solution. (1) s = 16.1/ 2 .
ds
(a) Differentiating, ~ = 32.2 t,
or, from (C), Art. 51, (2) v - 32.2 / ft. per second.
Differentiating again, -r. = 32.2,
or, from (A) above, (3) a 32.2 ft. per (sec.) 2 ,
which tells us that the acceleration of a falling body is constant ; in other words, the
velocity increases 32.2 ft. per second every second it keeps on falling.
(b) To find and </ at the end of the first second, substitute / 1 in (2) and (3).
Then T 32.2 ft. per second, a = 32.2 ft. per (sec. ) 2 .
(c) To find rand a at the end of the fifth second, substitute t = 5 in (2) and (3).
Then v = 161 ft. per second, a = 32.2 ft. per CsecJ 2 .
Given the following equations of rectilinear motion; find the position,
velocity, and acceleration at the instant indicated.
2. * 4 1- ~ (> t ; f 2. AUK. * 4, r 10, a = 8.
3. .s- -= 120 { - lf> t'- ; / = 4. * = 224, r - - 8, a = - 32.
4. .r - 32 / - 8 /' ; / = 2. s = 32, r = 0, a = - 16.
5. // = (J /-' - 2 t' { ; t = 1. y = 4, r = 6, a = 0.
.
7. / = 16 t' 2 - 20 / -f 4; / = 2.
8. // = 100 - 4 / - 8 / 2 ; / = 3.
9. * = V5/ -f ; ^ = 5.
V5f
10. * = \ 7 3 / 4- 2 ; / = 2.
SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 85
In the following problems find the acceleration at the instant indicated.
ll.r = 80-32/:f = 0. An.. -32. 18r = ^_., = 1 .
12. r = 4 / 2 -!()/; / = 2. 6. / + 1
Given the following equations of rectilinear motion ; find the position
and acceleration when the particle first comes to rest.
14. * = 16 / 2 - 64 / + 04. AHS. s = 0, a = 32.
15. *= 120 /- 16 / J . Pt 20
17. ,s- 5/4- ~.
16. * = 3 c*t - /'. * + 1
18. A ball thrown directly upward moves according to the law
,s- = 80 /- If) /-.
Find (a) its position and velocity after 2 sec. and after 3 sec. ; (b) how
high it will rise; (c) how far it will move in the fourth second.
19. If the equation of a rectilinear motion is ,s = V / -f 1, show that the
acceleration is negative and proportional to the cube of the velocity.
20. The height (*ft.) reached in /sec. by a body projected vertically
upward with a velocity of ?M ft. per second is given by the formula
8 = r\t j gt~. Find a formula for the greatest height reached by the body.
21. In the preceding problem suppose r\ ~ 1(>0, g 32. Find (a) the
velocity at the end of 4 sec. and at the end of (5 sec.; tb) the distance
moved during the fourth second and during the sixth second.
22. A car makes a trip in 10 min. and moves according to the law
,s- = 250 f 2 :\ t\ where / is measured in minutes and in feet. Ca) How far
does the car go? (b) What is its maximum speed? (c) How far has the
car moved when its maximum speed is reached?
Anx. (a) 12,500 ft.; (b) 1924 ft. per minute; (c) 0944 ft.
ADDITIONAL PROBLEMS
1. Trace the curve (4 - 2 x + s-)tj 2 x - s 2 , arid find the equations
of the tangent and normal at each point of inflection.
Ans. Max. (1, \). Point of inflection CO, 0): tangent, x 2 y = ;
normal, 2 x + y 0. Point of inflection (2, 0) : tangent, x + 2 y 2 = ;
normal, 2 x y 4 = 0.
2. A certain curve (the tractrix) is such that the length of every tangent
from its point of contact Pur, y} to its intersection A with the x-axis is
the constant r (A P c). Show that
( \ 4lL V /| x d 2 ij _ c 2 y
3. Determine k so that the normals at the points of inflection of the
curve y = k(x 2 3) 2 will pass through the origin. ... 7. _ 1
_ri/lc>. K> .
4V2
CHAPTER VII
DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS.
APPLICATIONS
We consider now functions such as
sin 2 -jr, 3% log (1 + s~),
called transcendental Junctions, as distinguished from the algebraic
functions hitherto discussed.
60. Formulas for derivatives; second list. The following formulas,
grouped here for convenience of reference, will be derived in this
chapter, and, with the formulas of Art. 29, comprise all formulas for
derivatives used in this book.
dv
d dx 1 dv
X ^ (lnl = T~ ITx On^log^)
Xa
v dx
XI -^-(a") = a"lna^-
dx dx
XI a f <"> = ?
dx dx
XII 4. <!") = 1*1" 1^ + inu.u*^.
dx dx dx
XIII (sin v) cos v
d , , dv
XIV ^ (cos v) = - sin v -
XV (tan v) sec 2 v
c?x dx
XVI (ctn y) = - esc 2 v ^~-
^ ^ dv
XVII (sec r) = sec i> tan i; .
dx clx
86
TRANSCENDENTAL FUNCTIONS 87
XVin (esc v) = esc v ctn v
dx dx
& / \ . dv
(vers v) = sin v -
dv
XX (arc sin i;) = ._
rf* Vl - i;
XXI (arc cos v) ~
dx -^/i _ v z
XXII 4- (a 1 " ^n v) = 7-^-
dx 1 -f ^
XXIII (arc ctn i;) = - 7-^..
XXIV ^- (arc sec v) =
dv
dx
dv
XXV ~ (arc csc ") = -
XXVI -- (arc vers v) =
61. The number e. Natural logarithms. One of the most important
limits in the calculus is
(1) Km (1 + ar = 2.71828 -.
x ->
This limit is denoted by c. To prove rigorously that such a limit e
exists is beyond the scope of this book. For the present we shall
content ourselves with plotting the locus of the equation
(2) y=(l + xf
and showing graphically that as x -+ the function (1 + x) x (= y)
takes on values in the near neighborhood of 2.718 and therefore
e = 2.718 approximately.
88 DIFFERENTIAL CALCULUS
As x from the left, y decreases and approaches e as a limit.
As x * from the right, y increases and also approaches e as a limit.
X
V
X
y
10
1.2710
f>
1.4810
2
1.7320
1
2.0000
.5
2.2500
-.5
4.0000
.1
2.5W7
j
2.8680
.01
2 7018
- .01
2.7320
.001
2.71<;!>
- .001
2.7195
The fact expressed in (1) is used in Art. 63.
As jr > -f oo, // approaches 1 as a limit ; and as .r > 1 from the right,
?/ increases without, limit. The lines ?/ 1 and f = 1 are asymptotes.
In Chapter XX we shall show how to calculate the value of c to any
number of decimal places.
Natural, or Napierian, logarithms are those which have the number
e for base These logarithms play a very important role in mathe-
matics. To distinguish between natural logarithms and common
logarithms when the base is not explicitly stated, the following no-
tation will be used.
Natural logarithm of r (base c) = In r.
Common logarithm of r (base 10) = log r>.
By definition, the natural logarithm of a number A T is the exponent
x in the equation
(3) e j - A; that is, .r - In A 7 .
If .r == 0, A r = 1, and In 1 - 0. If x = 1, AT = e, and In e = 1.
If j, _._ _ oo^ (h en ,\ T - 0, and we write In = oo.
The student is familiar with the use of tables of common loga-
rithms, where the base is 10. The common logarithm of a number N
is the exponent // in the equation
(4) 10" = A T , or // = log N.
Let us find the relation between In A 7 and log A 7 .
In (3), take logarithms of both members to the base 10. Then we
have, from (2), p. 1,
(5) x log c = log A' .
Solving for x, which equals In \, by (3), we get the desired relation
_ log W
~~ log e
TRANSCENDENTAL FUNCTIONS 89
That is, we obtain the natural logarithm of any number by dividing
its common logarithm by log e.
Equation (A) may be written
(6) log .V = log c - In .Y.
Hence the common logarithm of a number is obtained by multiplying
its natural logarithm by log e. This multiplier is called the modulus
(= ji/) of common logarithms.
By tables, log e - 0.4343, and -i- = 2.303.
log e
Equation (4) may now be written
(7) In ;V - 2.303 log A T .
Tables of natural logarithms should be at hand.
62. Exponential and logarithmic functions. The function of jr de-
fined by
(1) ij = e jr (c = 2.718 )
is called an exponential function. Its graph is shown in the figure.
The function is an increasing function for all values of j, as we shall
see later, and it is everywhere continuous.
From (1), we have, by definition, y
(2) y = In y.
The functions e f and In y are inverse functions
(Art. 39). Interchanging x and // in (2), we have ,
(3) y - In y,
in which y is now a logarithmic function of x. The graph is shown in
the figure. The function is not defined for negative values of x, nor
for x = 0. It is an increasing function for all
values of x > 0, and is everywhere continuous.
That is (Art. 17), for any value a of x greater
than zero
(4) lim In x = In a.
x-*a
When x > 0, then y * oo, as remarked above. The ?y-axis is an
asymptote in the graph.
The functions a x and log,, x (a > 0) have the same properties as
e r and In x and graphs similar to the above.
90 DIFFERENTIAL CALCULUS
63. Differentiation of a logarithm.
Let y = In v. (v > 0)
Differentiating by the General Rule (Art. 27 j, considering r as the
independent variable, we have
FIRST STEP. y + A// = In (v + Arj.
SECOND STEP. Ay/ = In (r -f Ar) Jn ?'
By(2),p.l
THIRD STEP. ^ = -i-in A +- )
A?' A/? \ r /
We cannot evaluate the limit of the right-hand member as it
stands by Art. 16, since the denominator Ar approaches zero as a
limit. But we can rewrite the equation as follows:
I M ull iplymtr l>y -
L r J
P,v(2),p.l
The expression following In is in the form of the right-hand mem-
ber of (2), Art. 61, with .r =
r
FOURTH STEP. ~ ^- - in r = -
a." r /;
r
When Ar-*0, - 0. Honco lim fl + V }? = r, by]
r Ar-(A ' /
(1), Art. 61. FsiiiK (1), Art. (>2, wo havc the result. J
Since r is a function of / and it is required to differentiate In v
with respect to .r, we must use formula (A), Art.. 38, for differentiating
a function of a function, namely,
dy_ _ dy_ (ir
<7,r dv di
Substituting the value of ^ from the result of the Fourth Step,
we get
dv
v d ,. . dx 1 dv
X (In v) = = -
dx v v dx
TRANSCENDENTAL FUNCTIONS 91
The derivative of the natural logarithm of a function is equal to the
derivative of the function divided by the function (or the reciprocal of the
function times its derivative}.
Since log v log e In r, we have at once (IV, Art. 29)
v
Xa
dx v dx
64. Differentiation of the exponential function
Let // = a 1 . (a > 0)
Taking the logarithm of both sides to the base e, we get
In y v In a,
r = lM = -L.ln V .
In a In a
Differentiating with respect to y by formula X,
^^_L .1.
(/-// In a y
From (C), Art. 39, relating to inverse functions, we get
*="*
or
(1) ^ = lna-a".
tfV
Since ?? is a function of .r and it is required to differentiate a r with
respect to j% we use formula (A), Art. 38. Thus we get
dii i ., dv
-~ = In a a' -y-
aa* ax
XI /. (a^)-lna-^- -
ax ax
When a e, In a = In e = 1, and XI becomes
The derivative of a constant with a variable exponent is equal to the
product of the natural logarithm of the constant, the constant with the
variable exponent, a,nd the derivative of the exponent.
92 DIFFERENTIAL CALCULUS
65. Differentiation of the general exponential function. Proof of the
Power Rule
Let y = u''. (n > 0)
Taking the logarithm of both sides to the base e,
In '// = v In u t
or y = e'^". By (3), Art. 61
Differentiating by formula XI a,
^ = <'"" 4- ("ton)
ax ax
to + l nw .rfr\ byVandX
u (Ijr (Is/
<-'- 1 s+-""s-
7Vzt' derivative of a function 'with a variable exponent is equal to the
sum of the two results obtained bij first differentiating by VI, regarding
the exponent as eonstant, and ayam differentiating by XI, regarding the
function (is constant.
Let v = n, any constant ; then XII reduces to
f (,,") = WM " i^.
rf.r rf^
Thus we have shown that the Power Rule VI holds true for any
value of the constant n.
ILLUSTRATIVE EXAMPLE 1. Differentiate y In (:r 2 4- a).
Arts.
x 2 + a
2 x
ILLUSTRATIVE EXAMPLE 2. Differentiate y log '
Solution. By (2), p. 1, we may write this
2, = log 2 r - log (1 + J 2 ).
Then ^ = o j. _ EJL (i + a-2) bylllandXfl
lhen dx 2 j dx l+x 2 <te
TRANSCENDENTAL FUNCTIONS 93
ILLUSTRATIVE EXAMPLE 3. Differentiate y = a- { *\
Solution. & = In a a*** ~ (3 or 2 ) by XI
= 6 x In a a**'. Ans.
ILLUSTRATIVE EXAMPLE 4. Differentiate y be^ 4 z2 .
Solution. & = b 4- (c r * * * 2 ) by IV
dx dx J
= be fy +**(c*+x 2 ) by XL a
ILLUSTRATIVE EXAMPLE 5. Differentiate y = .r' T .
Solution. 3^ = fz'' ' -- (.r) + r' T In z (r 1 ) by XII
dx dx dx
= (^x^ l -f x' J In a* <'
Ans.
66. Logarithmic differentiation. Instead of applying X and X a at
once in differentiating logarithmic functions, we may sometimes
simplify the work by first making use of one of the formulas of (2)
on page 1. It is important that these formulas should be used when-
ever this is possible.
ILLUSTRATIVE EXAMPLE 1. Differentiate y = In Vl x 2 .
Solution. By using (2), p. 1, we may write this in a form free from radicals, as
follows :
--
ILLUSTRATIVE EXAMPLE 2. Differentiate y In '
Solution. Simplifying by means of (2), p. 1,
y = i [In (1 + * 2 ) - In (1 - ;
A
Then i* = - - I by III and X
dx J
In differentiating an exponential function, especially a variable
with a variable exponent, the best plan is first to take the natural
94 DIFFERENTIAL CALCULUS
logarithm of the function and then to differentiate. Thus Illustrative
Example 5, Art. 65, is solved more elegantly as follows :
ILLUSTRATIVE EXAMPLE 3. Differentiate y x fT .
Solution. Taking the natural logarithm of both sides,
In y = e r In x. By (2), p. 1
Now differentiate both sides wit h respect to .r.
dx
T
( l
c/ (In jr] -\
= <- - 1 + in x
- In .
, ,*T
( >
r d (^)
by X and V
by X and XI a
or
dx
ILLUSTRATIVE EXAMPLE 4. Differentiate ?/ - (4 x'-' 7) V *
Solution. Taking the natural logarithm of both sides,
In y - (2 4- vV~' - 5 ) In (4 x j - 7 ).
Differentiating both sides with respect to .r,
= ( 2 f V^^5) -L- f l n (4 r2 - 7)
rfj: 4 j- - i
4 j--' - 7
In the case of a function consisting of a number of factors it is
sometimes convenient to take the natural logarithm and simplify
by (2), p. 1, before differentiating. Thus,
ILLUSTRATIVE EXAMPLE 5. Differentiate y = -\/
V (j -
-nor -2)
Solution. Taking the natural logarithm of both sides,
In y = i[ln (J - 1) + In (x - 2) - In (x - 3) - In (x - 4)].
Differentiating both sides with respect to j,
i^/ = ir_i_ , _! i L
ydx 2[ar-l j-2 x-3 x~4
TRANSCENDENTAL FUNCTIONS 95
PROBLEMS
Differentiate each of the following functions.
1. ?/- In (ax + b}. " dli
6. y=\ri*x\= (Inx) 3 ].
7. ij = In (2 ^ - :>> x 2 +
9. ?/ = In
dx ax -h ft
2. ?, = In (a.r-> + 6). 'A' = _JL_.
dx ox 2 + ft
3. ? / = ln(ax + ft)'. ^- 2fl -
dx ax +
4. ?/ = In ax n .
dx
.] ..
5. y = In x :? .
dx x
( lli ^ ^ n2 x
dx ~ x
dy __ 6 x(x 1)
</.r~2.r :{ - 3^ 2 -f 4"
8. //-log-. ^ :
J" dc
>r 2 d?/.
s 2 dx x(l -h a- 2 )
10 ln V9 - ^> .r 2 ^ = ~ 2 J
dx 9-2 x 2
11. ?/ - In (a.rVc7T7). ^ = 2 a + 3 y .
^ 2 r(a + x)
12. /O) - r In .r. /'(. r ) = 1 + In #.
13. /(.r) = In (j- -f- VfTT 2 ). /'(^) = ]
Vl
i __ i^ /u -r (// a, L r aft
. > in
- ^ (// a 2 - ft 2 / 2
15. f(x) = .r 2 In x 2 . f'(jc) = 2 y(l -f 2 In x).
16. y = <>"*. ^ = w -.
dx
17. j/ = 10- ^ = w 10- In 10.
dx
. y e _ xe r .
19. ?/ = -. ^ = _1.
^ J dx e z
20 <? e^ ^ s -. f
dt 2^/1'
21. r = ft 2 ''. ^ = 2 ft 2 " In ft.
d//
96
DIFFERENTIAL CALCULUS
22. u = se".
23. v = -
u
24. /, - ^
25. /y In (.rV)
26. ?/ = ~~ry
27. 7/ = .r-Y '.
28. // = ~(c' 1 - t
dv e"(u -1)
du
= .
ds JT
dy _ 2 c j
g = c "(2 ,-**).
a = |(,' + <- :) -
fir U"
30.*^.
2 - 4 InJ
3Lf(j . )== in^il^:.
V J-- -f 1 -f JT
HINT. First rationalize the denominator.
32. y == s r . ^^ s '- // = r ' r (
^
/'(,)= - 7 ==-
Vx 2 -f 1
35. ?/ =
.r \/3 .r + a
36. y =
37. // = .r"
V4 + >r 2
3 s + a 2 x
1 , J
^-^a
mb
In Problems 38 47 find the value of -^ for the given value of x.
Ans. y' = %.
38. y - In (jc-
39. /y = log (4 .r - 3); J
40. y - JT In V.r -f 3 ; -r
41. y = xe- 2r \ .r = J.
. rt In .r 2 4
42. 7/ = -y- ; .r = 4.
' = 0.3474.
y' = 1.4319.
7/' = 0.
t/' = - 0.0483.
TRANSCENDENTAL FUNCTIONS
97
43. y =
44. y = log V25 - 4 x ; x = 5.
45. n = 10 x7 ; x-4.
Find -~ for each of the following functions.
djc' 2
.r'V.r a + 9
,
= 4.
\20
52. // = In
48. \i In cjc.
49. ?/ = f".
50. // x In a*.
51. 2/ = <' r -'.
Differentiate each of the following functions.
.r -f a
53. y = ~
-
54. In
55.
In Va 2 jc 2
58. (^ : In V
59. 10' log/.
60. (ttf)'".
61. 2 s ,s 2 .
57. In ' 6$
Vl> / -f 3
67. The function sin x. The graph of
(1) y = sin x
is shown in the figure. Any value of x is assumed to be the measure
of an angle in radians (Art. 2).
n
i T 2
y = - l
Thus for x = 1, y = sin (1 radian) = sin 57 18' = 0.841. The
function sin x is defined and is continuous for all values of x. It is
important to note that sin x is a periodic function with the period
For
2 TT) = sin x.
That is, when the value of x is increased by a period, the value
of y is repeated.
98
DIFFERENTIAL CALCULUS
The property of periodicity has the following interpretation in the
graph on page 97 : The portion of the curve for values of x from Oto2w
(arc OQBRC in the figure) may he displaced parallel to OX either to
the right or left a distance equal to any multiple of the period 2 TT, and
it will be part of the locus in its new position.
68. Theorem. Before differentiating sin x (Art. G9) it is necessary
to prove that
,. sin x -
(B) hm = 1.
x -> %
This limit cannot be found by Art. 16. We
proceed by geometry and trigonometry.
Let O be the center of a circle whose radius
is unity. Let x angle AOM measured in radian?. Since the radius
is unity, arc AM = x, also.
Lay off arc AM f = arc AM and draw 71/7' and M'T tangent to
the circle at M and M' respectively. From geometry,
MM' < arc MAM 1 < MT ~\- M'T.
Or, by trigonometry,
2 sin x < 2 x < 2 tan x.
Dividing through by 2 sin x, we get
1 <
1
sin x cos x
Replacing each term by its reciprocal and reversing the inequality
signs, we have
1 ^ sin x ^
1 > > cos x.
Therefore when x is small, the value of
sin x
lies between 1 and cos x.
But when x * 0, lim cos x = cos = 1, since cos x is continuous for
x = (see Art. 17). Thus we have proved (B).
It is interesting to note the behavior of this function from its
graph, the locus of equation
oiri x
x
The function is not defined for x = 0. Let us, however, assign
the value 1 to it for x = 0. Then the function is defined and is
continuous for all values of x (see Art. 17).
TRANSCENDENTAL FUNCTIONS 99
69. Differentiation of sin v
Let y = sin v.
By the General Rule, Art. 27, considering v as the independent
variable, we have
FIRST STEP. y + A// = sin (v + Ar).
SECOND STEP. Ay = sin (v + A?0 sin r.
The right-hand member must be transformed in order to evaluate
the limit in the Fourth Step. To this end, use the formula from
(6), p. 3,
sin A - sin B = 2 cos \ (A + B) sin ] (A - B),
setting A = v + Ar, B = 0.
Then ?, ( A + #) = r + J A/?, " ( A - B) = ?> A?;.
Substituting,
sin + A/0 sin v = 2 cos (v + *> Ar) sin ?, A?;.
/ \ \ \ n
Hence A// ~ 2 cos ( ?' + -7^- j sin
sin
Av
,2
THIRD STEP.
FOURTH STEP. -p = cos v.
dv
/sinf\
Since lim I - ] 1, by Art. 68, and lim cos (M f ) - coi
Ar .Ol Ar I A,. .0 V -'
Substituting this value of ( -~ in (4), Art. 38, we get
a t/
r/v/ dv
-~ = cos z;
ax ax
XIII /. (sin v) cos v
The statement of the corresponding rules will now be left to the
student.
70. The other trigonometric functions. The function cos x is de-
fined and is continuous for any value of x. It is periodic, with the
period 2 TT. The graph of y _ cos x
is obtained from the graph of Art. 67 of sin x by taking the line
x = i TT as the #-axis.
100
DIFFERENTIAL CALCULUS
The graph of y = tan x
(see figure) shows that the function tan x is discontinuous for an
infinite number of values of the independent variable x; namely,
when x = (n + ^)TT, where n denotes any positive or negative integer.
In fact, when x > \ TT, tan x be-
comes infinite. Hut from the rela-
tion tan (TT + x) = tan x, we see that
the function has the period TT, and
the values x = (n + ])ir differ from
2 TT by a multiple of the period.
The function ctn x has the period
TT. It is defined and is continuous
for all values of x except x nir, n being any integer as before.
P'or these values ctn x becomes infinite. Finally, secx and esc x
are periodic, each with the period 2 TT. The former is discontinuous
only when x (n + 2)^, the latter only when x mr. The values
of x for which these functions become infinite determine vertical
asymptotes in the graphs.
71. Differentiation of cos v
Let y = cos v.
P>y (3), p. 2, this may be written
Differentiating by formula XIII,
du
Jl = c
= sin
I Since cos (^ r\= sin r, by (3), p. 2. 1
XIV
d f . . du
.'. (cos u) = sin v
dx dx
72. Proofs of formulas XV-XIX. These formulas are readily de-
rived by expressing the function concerned in terms of other functions
whose derivatives have been found, and differentiating.
Proof of XV. Let y = tan v.
By (2), p. 2, this may be written
cost?
TRANSCENDENTAL FUNCTIONS 101
Differentiating by formula VII,
cos r (sin r) sin r -r 1 (cos r)
rf^_ dx _ do* y
dx cos- v
cos 2 r -7- + sin 2 v -r~
_ dx dx
dr
- -^- - sec 2 f- ; - Using (2), p. 2
cos- r dx
XV
. . dx VM ^ , - ^ . dx
To prove XVI-XIX, differentiate the form as given for each of
the functions below.
XVI. ctn v = ^ XVII. sec /? - -^- XVIII. esc v = -^
tan ?> cos /> sin v
XIX. versine r = vers r = 1 cos v.
The details are left as exercises.
73. Comments. In the derivation of formulas I-XIX it was neces-
sary to apply the General Rule, Art. 27, only for the following.
TTT d f ^ </'// , <]P (I'M" 1 Al 1
III -- (u + v w) -T- + -T Algebraic sum.
dx as d.r dx
TT d , ^ dv , dvx -,-> , ,
V -r- (uv) = w -7- + ' T" * Product,
dr ax ax
dn dr>
VII -(-)= Quotient.
dx \ v ' v' 2
VIII <!& = <!&. <to Function of a function.
ax dv dx
IX -~ = -r - - Inverse functions,
ax ax
a 7 ?/
cfe
X 4- HOK 1 = Logarithm.
ax ?;
XIII jLfein^^eos,,^. Sine,
dx dx
Not only do all the other formulas deduced depend on these, but
all we shall deduce hereafter depend on them as well. Hence we see
102 DIFFERENTIAL CALCULUS
that the derivation of the fundamental formulas for differentiation
involves the calculation of only two limits of any difficulty, namely,
lim 5!5-? = i by Art. 68
t - V
I
and lim (1 + v) v = e. By Art. 61
PROBLEMS
Differentiate the following functions.
1. y = sin u.r 2 .
Solution. & = cos oar 2 -f (ax*) by XIII
dx dx
[v-a^.]
= 2 ax cos ax' 2 . Ans.
2. ?/ tan Vl x.
Solution. ^ - sec-' \ 1 - x ~ (1 - x)* by XV
= sec- \ 1 - x \ ( 1 x)~ *( 1)
2 \ 1 - x
3. y = cos 3 x.
Solution. This may also be written
y- (cos.r) :{ .
^ = !i ( cos J^-T-C cosy) by VI
dx dx
[r = cos.v and ?? 3.J
= 3 cos- T ( - sin J-) by XIV
= 3 sin x cos 2 x. Ans.
4. ?/ = sin ?u: sin" jc.
Solution. -^ = sin nx ~ (sin x) n 4- sin" r ~ (sin wx) by V
Cw* fix ax
[?< = sin ??T and r sin" a-.]
= sin nx n (sin a:)"" 1 -r- ( s i n x )
ax
4- sin" x cos nx -r- (nx) by VI and XIII
dx
= w sin nx sin"" 1 x cos x + n sin" x cos nx
= w sin" -1 x(sin nx cos x 4 cos nx sin x)
= n sin" ~ ! x sin (H 4- l)x
TRANSCENDENTAL FUNCTIONS
103
5. y sin ax.
6. y = 3 cos 2 .r.
7. s = tan 3 t.
= 2 ctn --
17. P -
18. y = sin 2 x cos x.
19. ?/ = In sin ax.
20. ?/ = In Vcos 2 x.
21. y e a<f sin frx.
22. s f ' cos 2 /.
23. y In tan -
24 . ?/ =
y r = a cos or.
?/' = 6 sin 2 x.
s' = 3 sec- 3 /.
dn o ?'
- = esc- -
dv 2
y f = 4 sec 4 .r tan 4 x.
p' (76 esc bO ctn bO.
?/' sin .r cos .r.
c/.s- - sin 2 /
dp _ SOC L> .H ^
d ^ ? (tan30) S
dx V sec .r
7/' cos x x sin x.
/'(0) = tan 2 0.
(jj) _ cos sin
To "" 0-
?/' 2 cos 2 x cos x sin 2 x sin x.
y' a ctn ax.
T/ = - tan 2 x.
y' = c" r (a sin bx 4- fo cos ?>x).
' = - c '(2nm 2 / -f cos 2 0-
/ i A X 9 X
7/' . ctn - sec 2 -
. / = .
\ 1 sin x
25. f(0) = sin r0 + a) cos (6 - a). f'(0) = cos 2 (9.
26. /(x) = sin 2 (IT - x). !'(*) = - 2 sin (TT - x) cos (TT - x).
27. p = i tan 3 - tan -f 6. p' = tan 4 0.
28. y = x**. ^r = * Hin *(?^x 4. C os x In x)-
CLX \ X /
29. i/ = (cos x) x .
CLX \ X
V' = ?/On cos x - x tan x).
104 DIFFERENTIAL CALCULUS
Find the second derivative of each of the following functions.
d 2 y
30. ?/ = sin kx. Arts. -~ k 2 sin kx.
dx 2
31. p = \ cos 20. ^ = - cos 2 (9.
d$~
32. u = tan v. = 2 sec 1 ' r tan r.
(/?"
33. ?/ = x cos x. ~-^( 2 sin j* .r cos j.
<ir-
*>/i _ sin r c/ L '// _ 2 sin x 2 x cos r x 2 sin .r
J^ ?/ * i- ------- - - .
x dx 2 x- {
d -' s*
35. .s = r' cos /. ---- ~ 2 r' sin /.
a/ fc
36. = r ' sin 2 /. ^ = - r '(3 sin 2/4-4 cos 2 /).
<//-
37. y/ c nr sin ftj-. ^ - c" r \ (a- - />'-') sin bx 4 2 a?> cos
</j*"
Find - from each of the following equations.
38. 7, - cos (.r -
sin (x //) 1
<Lr ("' cos (.r 4 //)
40. cos // In (x 4 //).
dn - 1
dx I 4 (x 4 //) sin //
In Problems 41-50 find the value of ~ for the given value of x (in
radians).
41. -// = x cos .r ; x = I. Ans. y' 1.841.
42. y = x sin ; x = 2. //' = 1.381.
43. y - In cos j- ; x - 0.5. ?/' = 0.546.
44. y - -; x = - 0.5. ?/ = - 3.639.
45. ?/ = sin j- cos 2 .r ; .r 1. ?/' 1.754.
46. ?/ = In Vtan :r ; .r \ TT. y' = 1.
47. # = r* sin x ; x - 2. ?/ = 3.643.
48. t/ = 10c-'cos TTX; x = 1. ?/' = 3.679.
49. y = 5 f 2 s in -^ ; .r = 2. ?/' = - 21.35.
50. y = 10 c" 10 sin 3 x ; j- = 1. y' = - 27.00.
TRANSCENDENTAL FUNCTIONS 105
74. Inverse trigonometric functions. From the equation
(1) ?/ = sin x.
we may read "x is the measure of an angle in radians whose sine
equals //." For a central angle in a circle with radius unity, x equals
also the intercepted arc (see Art. 2). The statement in quotation
marks is then abbreviated thus
(2) x arc sin //,
read " x equals an arc whose sine is //." Interchanging x and y in (2),
we obtain
(3) y arc sin .r,
and arc sin x is called the inverse xine function of :r. It is defined
for any value of x numerically less than or equal to 1. From (1)
and (2), it appears that sin x and arc sin // are inverse functions
(Art. 39).
Equation (,'J) is often written // sin ' .r, read "the inverse sine of x."
This notation is inconvenient, for the reason that sin- ' .r, as thus written,
might be read as sin s with the exponent 1.
Consider the value of y determined by x = \ in (3), that is, by
(4) y - arc sin \.
One value of// satisfying (4) is // I TT, since sin J TT ~ sin 30 = \.
A second value is y \\ TT, since sin ; r ; TT sin 150 ?,. To each of
these solutions any multi])le of 2 /r may be added or subtracted.
Hence the number of vain ex of // satisfying (4) ix without limit. The
function arc sin x is then said to be " multiple-valued/'
The graph of arc sin x (see figure) shows this property
well. When x = OM, y=MI> lt ,17/V, Ml**, -, MQi,
For most purposes in the calculus it is allowable
and advisable to select o'tie of the many values of y. We
select, then, the value between ?, TT and \ TT, that is,
the smallest numerical value. Thus, for example,
(5) arc sin J = TT, arc sin = 0, arc sin ( 1) = \ TT.
The function arc sin x is now single-valued, and if
(6) y = arc sin x, then \ir~y~\ir. /
In the graph we confine ourselves to the arc QOP.
106 DIFFERENTIAL CALCULUS
In the same manner each of the other inverse trigonometric func-
tions may be made single-valued. Thus, for arc cos x, if
(7) y = arc cos x, then ^ y ~ TT.
As examples,
arc cos \ = ?, TT, arc cos (- \ ) = \ TT, arc cos ( 1) = TT. rj
From (6) and (1) we now have the identical relation
(8) arc sin x + arc cos x = TT.
In the graph of arc cos x (see figure), we confine our-
selves to the arc QP\P.
Definitions establishing a single value for each of
the other inverse trigonometric functions are given
below.
75. Differentiation of arc sin u
Let y = arc sin r ; ( \ TT i y ^ \ TT)
then v = sin ?y.
Differentiating with respect to // by XIII,
dv =
therefore ~ - ^~- By (C), Art. 39
Since r is a function of .r, this may be substituted in (A), Art. 38,
_
dx cos '// dx
-os y \ 1 sin-'// \ 1 r H , the positive si^n of tho radical bpinpf)
taken, since cos // is positive for all values of y between and inclusive.
XX .', (arc sin v) =
If y = arc sin :r, \f = - =
.
dx Vl - v 2
The graph is the
arc QP of the figure. The slope becomes infinite at Q and
P, and equals 1 at 0. The function increases (y' > 0)
throughout the interval x = ~ 1 to or = 1.
76. Differentiation of arc cos u
Let y arc cos v; (0 g y ^ TT)
then = cos ?/.
TRANSCENDENTAL FUNCTIONS
Differentiating with respect to y by XIV,
107
therefore
By (O, Art. 39
But since r is a function of .r, this may be substituted in 04),
Art. 38, giving rf?/ l rfr j
'dx" "~
sin
.
dx
iin // = \ 1 -- cos"// = \ 1 r-', the plus sitfn of the radical being taken, |
ince sin y is positive for all values of // between and TT inclusive. |
XXI
, .
.-. (arc cos in
dx
dy
dx
If y = arc cos J, then y'
1
When x increases from
1 to + 1 (arc PQ of the first figure on page
106). y decreases from TT to (//' < 0).
77. Differentiation of arc tan v. Let
(1) y = arc tan r; then
(2) v = tan ?/.
The function (1) becomes single-valued if
we choose the hast mmcrical value of //, that
is, a value between - 1 TT and i TT, corresponding to arc
the figure. Also, when v -> - oo, yy -' - 2 *" I when v ""' + ' ?y ""' 2
Or, symbolically,
(3) arc tan (+ oo) - \ TT, arc tan (-00) = - \ IT.
Differentiating (2) with respect to y by XV,
of
sec~ y ,
dy
and
By (C), Art. 39
dv sec 2 y
Since v is a function of x, this may be substituted in (A), Art. 38,
giving dy ^ 1 <fo = _j_jg..
c/x "~ sec 2 ?y dx 1 + v' 2 dx
[sec 2 ?/ = 1 + tan'-' y = 1 + *' L '.]
dv
XXII
108
DIFFERENTIAL CALCULUS
1
If y = arc tan x, then ?/' = , and the function is an increasing
function for all values of x.
The function arc tan - furnishes a good
jr.
example of a discontinuous function. Con-
fining ourselves to one branch of the graph of
y arc tan -
we see that as jr. approaches zero from the left, ?/ approaches \ TT as a
limit, and as :r approaches zero from the right, ;// approaches -f i ?r as
a limit. Hence the function is discontinuous when x (Art. 17). Jts
value for x can be assigned at pleasure.
78. Differentiation of arc ctn v. Following
the method of the last article, we get
dv
XXIII ~ (arc ctn i;) =
<**
1 -f i^
The function is single-valued if,
when // arc ctn v, < // < TT,
corresponding to the arc AB of the figure.
Also, if v -> + oo, // - > () ; if r - . - oo, // , jr. That is, symbolically,
arc ctn (-f oc ) = ; arc ctn i oo; = TT.
79. Differentiation of arc sec v and arc esc v. Let
(1) y arc sec r.
This function is defined for all values of r except those lying be-
tween - 1 and + 1. To make the function single- valued (see figure),
when v is positive, choose // between and J TT (arc AB) ;
when r is negative choose // between - TT and - 1 TT (arc CD).
Also, if v + oo, // > 1 TT ;
^^> // ^> 7T.
Solving (1), ? = sec ?/.
Differentiating with respect to y by XVII,
if
dr
T- = sec vy tan y ;
therefore
.
sec y tan y
By (C), Art 39
y v '' u Ot7
TRANSCENDENTAL FUNCTIONS
109
Since v is a function of z, this may be substituted in (A), Art. 38,
giving
^ 1 dr _ 1 (h
dx sec y tan y dx
v~ 1
I sec y = r,
I since tan
= r, and tan y = V sec 2 y 1 = \ r-' 1, the plus sign of the radical being taken,"!
y is positive for all values of y between and ^ and between - 7T and ~-
XXIV /. - (arc sec v) =
dv
Differentiation of arc esc v. Let
2/ = arc esc v ;
then v esc //.
Differentiating with respect to y by XVIII
and following the method of the last section,
we get
dv
XXV - ( arc
dx
v) =
- 1
The function y = arc esc r is defined for all values of v except those
lying between 1 and + 1, and is many-valued. To make the func-
tion single-valued (see figure above),
when v is positive, choose y between and \ TT (arc AB) :
when v is negative, choose y between TT and \TT (arc CD).
80. Differentiation of arc vers v }
27T
Let y = arc vers v ; *
then v = vers y.
Differentiating with respect to y by XIX,
dv
\
a v
therefore
dv sin y
By (C), Art. 39
* Defined only for values of r between and 2 inclusive, and many-valued. To make
the function single- valued, y is taken as the smallest positive arc whose versed sine is 0;
that is, y lies between and TT inclusive. Hence we confine ourselves to arc OP of the graph.
110 DIFFERENTIAL CALCULUS
Since v is a function of x, this may be substituted in (4), Art. 38.
K ivin dji = 1 dv = 1 dv t
Ax ~ sin y dx V2 c r' 2 dx
ffjin y ~ V'l - COR- // ~ V] - < 1 - versv/^ = V'J r r a , the plus sign of the radical!
[being taken, since sin y is positive for all values of y between arid TT inclusive.J
XXVI
/. (arc vers v) =
dv
PROBLEMS
Differentiate the following functions.
1. y arc tan ax' 2 .
.
Solution.
<( -\
dx (ax " }
dx 1 -f (
[r -
. Ans.
2. ?/ = arc sin (3 jr 4 X H ).
-
Solution.
byXXn
by XX
3. y = arc sec ~
./* *" JL
\ 1 - 9 .r-' -f 24 x 4 - 16 x r ' vl- x' 2
Solution.
x- 4- 1 ix- + 1\-
by XXIV
i .r^-'-l I
x* - 1)2 j- (r= 4- O2 a;
__ __
= _. __!!_. Ans.
x 2 + 1
4. i/ = arc cos--
a
5. y = arc sec 2! .
a
dx
a 2
c lu a
dx xVx 2 a 2
TRANSCENDENTAL FUNCTIONS
111
6. y = arc ctn -
a
7. y = arc sec -
8. ?/ = arc esc 2 x.
9. y arc sin Vx.
10. 6 = arc vers p 2 .
11. y x arc sin 2 .r.
As. 22 = ^4-
ax a* -f .r 2
dy __ - 1
dx ~ Vl - .r 2
</?/ - 1
<k* .rV4 x' 2 - 1
^^ 1
d.r ~~ 2Vx .r 2
N? = 2
</p V2 p 2
dy
<iy - o ,
--- arc sin 1 x -I .
</x Vl - 4 x 2
JLZ. y x~ arc cos x. ~
dx
j x arc cos x ~- .
Vl - x 2
13. /(M) = u^/d 2 >/ -' -h a' 2 arc sin -
/'( M ) = 2Va 2 - u' 2 .
f'(r\ /^"- r
14. f(x) Va~ j h (i arc sin
f (j) -\la + x
dr 2 // 2
i r ' " ~\ / ' ''
(Z
du Va 2 - ?^ 2
</?' // 2
Va 2 - w 2 a
</?/ , ., .,,.2
(a- v/-)^
dr Va 2 ?-/ 2
w , Vfl-' - ?^ 2
I/. ?' arc sin H
a ?/
c/?^ ?^ 2
c/?' V2 au u 2
10 / 1 '^ \ I A. /O ^ <J
IJj. t? :_ (i arc COS I J -r V ^ a// M .
V a/
1Q , _ , a + r
du u
19. </> arc tan
dr 1 -h r 2
dx _ y
20. x = r arc vers - V2 ry y~.
r
21. y ^ x 3 arc tan x 4- ?, log (r 2 -hi)-/, x 2 .
In Problems 22 27 find the value of ~ for
22. y x arc sin x ; x L
23. i/ = x arc cos x ; x %.
d(l V2 ry - y' 2
~i = x 2 arc tan x.
f/X
the given value of x.
Ans. y' 1.101.
^'=2.671.
y' 0.285.
x
25. y Vx arc ctn - ; x = 4.
arc sec 2 x. j
T/' = - 0.054.
y' - 0.053.
<6U. ?/ , > * - 1
Vx
27. # = x 2 arc esc Vx ; x = 2.
y' = 2.142.
112 DIFFERENTIAL CALCULUS
Differentiate each of the following functions.
28. arc sin Vx. 33. arc sec Vx.
2
29. arc tan - 34. e* arc cos x.
x
30. x arc cos - 35. In arc tan x.
01 arc ctn 2 x 0/ , / : -
31. 36. V arc sin 2 x.
x
32. arc versa-*). 37 arc cos V^
Vx
PROBLEMS
Sketch the following curves, and find the slope at each point where
the curve crosses the axes of coordinates.
1. u In .r. AUK. At (1, 0), m = 1.
2. y =r log jr. At (1, 0), m = 0.434.
3. ?/ = In (4 - JT). At (3, 0), m = - 1 ;
at x = 0, m = - J.
_
4. // = In V4 - x' 2 .
5. Show that if ?/ = J, a(r"' -f c a ), then ?/" = $-
2
a*
Find the angles of intersection of each of the following pairs of curves.
6. // = In (.r + 1 ), y = In (7 - 2 x). Ans. 127 53'.
7. ?/ = In (JT + 3), y = In (5 - x' 2 ).
8. // - sin xjy - cos j. 109 28'.
9. y tan j% ?/ = ctn jr. 53 8'.
10 ?/ = cos j*, ?/ = sin 2 x.
Find the maximum, minimum, and inflectional points on the following
curves and draw their graphs.
ll.y = x\njr. Ans. Min. (-> - -V
\e e)
10 _ J Min. (e, e) ;
In x inflectional point, (e 2 , J e 2 ).
13. ?/ = In (8 x - x 2 ). Max. (4, In 16).
14. y = xe x . Min. (- 1, --);
inflectional point, ( 2, -V
15. y =
TRANSCENDENTAL FUNCTIONS 113
16. A submarine telegraph cable consists of a core of copper wires
with a covering made of nonconducting material. If u* denotes the ratio
of the radius of the core to the thickness of the covering, it is known that
the speed of signaling varies as .r-ln-- Show that the greatest speed
1 J
is attained when JT = -p-
Vf
17. What is the minimum value of // ac kj + be Ax ? Anx. 2Vob.
18. Find the maximum point and the points of inflection of the graph
of y = f~ J ~, and draw the curve. f
Am. Max. (0, 1) ; points of inflection, ( -~. p
\ V2 Vf
19. Show that the maximum rectangle with one side on the .r-axis
which can be inscribed under the curve in Problem 18 has two of its
vertices at the points of inflection.
Find the maximum, minimum, and inflectional points for the range
indicated, and sketch the following curves.
20. y I x sin x ; (0 to 2 TT).
Am. Min. (\ TT, - 0.3424); max. (jf TT, 3.4840);
inflectional points, (0, 0), (TT, \ TT), (2 TT, TT).
21. y = 2 x tan x ; (0 to ?r).
An*. Alax. ( 1 , TT, 0.571); min. (] TT, 5.712);
inflectional points, (0, 0), (TT, 2 TT).
22. y = tan x 4 x ; (0 to TT).
AW.S-. Min. f; 1 , TT, - 2.457); max. (J TT, - 10.11);
inflectional points, (0, 0), (TI, 4 TT).
23. i/ = 3 sin x 4 cos x ; (0 to 2 TT).
AUK. Max. (2.498, 5); min. (5.G40, - 5);
inflectional points, (0.927, 0), (4.069, 0).
24. y = x -f cos 2 x ; (0 to TT).
25. ?/ = sin TTJC cos TTJ ; (0 to 2).
26. 0= J.r + sin2.r; (0 to TT).
27. y = x - 2 cos 2 :r ; (0 to TT).
28. y = TTx + sin irx ; (0 to 2).
29. Show that the maximum value of y = a sin x 4- b cos x is Va 2 4- 6 2 .
30. The turning effect of a ship's rudder is shown theoretically to be
k cos 6 sin 2 6, where 6 is the angle the rudder makes with the keel, and
k is a constant. For what value of is the rudder most effective ?
An*. About 55.
114 DIFFERENTIAL CALCULUS
31. Into a full conical wineglass of depth a and generating angle a
there is carefully dropped a sphere of such size as to cause the greatest
overflow. Show that the radius of the sphere is
a sin a
sin a. + cos 2 u
32. Find the dimensions 01 the cylinder of maximum volume which
can be inscribed in a sphere of radius 6 in. (Use the angle 6 subtended
by the radius of the base* of the inscribed cylinder as a parameter. Then
r = 6 sin 6, h = 12 cos 0.)
33. Solve* Problem 32 if the convex surface of the cylinder is to be a
maximum, using the same parameter.
34. A body of weight W is dragged along a horizontal plane by meant
of a force P whose line of action makes an angle JT with the plane. The
magnitude* of the force is given by the equation
m sin jc + cos jr
where m denotes the coefficient of friction. Show that the pull is least
when tan x = m.
35. If a projectile is fired from (> so as to strike an inclined plane which
makes a constant angle cv with the* horizontal at (>, the range is given
by the formula 2 ,,, ( . O[ , ^ (Q _ (v)
A ---- - --
(/ cos- cv
where ?> and y are constants and is the angle of elevation. Calculate the
value of giving the maximum range* up the plane. AHX. 9 } TT + -I CY.
36. For a square-headed screw with pitch and angle of friction </> the
efficiency is given by the formula
p __ tan
tan (0 + 0) +/
where / is a constant. Find the value of for maximum efficiency when
is a known constant angle.
ADDITIONAL PROBLEMS
1. The curves ?/ = jr In .r and // jr In (1 jc} intersect at the origin and
at another point A. Find the angle of intersection at .4. Ans. 103 30'.
2. Sketch the following curves on the same axes and find their angle
of intersection. , - { x ( 2 x
?/ = In ('- - 1 ), ?/ = In (:* .r - ' - 1 ). Am. 32 28'.
3. The line A B is tangent to the curve whose equation is y = e x + 1
at A and crosses the .r-axis at B. Find the coordinates of A if the length
of AB is a minimum. Ans. (0, 2).
CHAPTER VIII
APPLICATIONS TO PARAMETRIC EQUATIONS, POLAR
EQUATIONS, AND ROOTS
81. Parametric equations of a curve. Slope. The coordinates jr and
y of a point on a curve are often expressed as functions of a third
variable, or parameter, /, in the form
(1)
Each value of t gives a value of .r and a value of // and determines a
point on the curve. Equations (1) are called parametric equations
of the curve. If we eliminate / from equations (1), we obtain the
rectangular equation of the curve. For example,
[_// = rsin/
are parametric equations of the circle in the
figure, t being the parameter. For if we elim-
inate t by squaring and adding the results, we
have . , .>.,,.> JN
3- -f ?/- = r-(cos- / + sin- /) r~,
the rectangular equation of the circle. It is evident that if t varies
from to 2 TT, the point 7'O, //) will describe a complete circumference.
Since, from (1), y is a function of /, and t is a function (inverse)
of x, we have
by (A), Art. 38
dx
dydt
'' dt rfx
dj J_.
' dt' dx y
dt
by (C), Art. 39
that is,
04)
dt
By this formula we may find the slope of a curve whose parametric
equations are given.
115
116
DIFFERENTIAL CALCULUS
ILLUSTRATIVE EXAMPLE 1. Find the equations of the tangent and normal,
and the lengths of the subtangent and subnormal to the ellipse*
x = a cos <
at the point where $ = 45.
Solution. The parameter being (/>, -77 = a sin </>, -77 & cos
acp a<p
Substituting in (A), # = - /J COS ff = - - ctn = slope at any point = m.
dx a sm a
Substituting (/> 45" in the given equations (13), we get xi = \ a\ 2, y\ ^ b\'2
as the point of contact, and the slope m becomes
a a
Substituting in f 1) and (2), Art. 43, and reducing, we get
bx -f ay V2 ab -.equation of the tangent,
V2(ax - by) a 2 - b' 2 = equation of the normal.
Substituting in (3) and (4), Art. 43,
\ b^ / -\ = - 2 x - = length of subtangent,
i &V2I - i = J ~ V " length of subnormal.
\ a/ 2 a
ILLUSTRATIVE EXAMPLE 2. Given the equations of the cycloid f in parametric
form,
f 4 ) j x <i( ,
v ' l# = cos v),
being the variable parameter; find the lengths of the subtangent, subnormal,
and normal at the point (.n, //i) whore = 6 } .
* As in the figure, draw the major and minor auxiliary circles of the ellipse. Through
two points /* and C on the same radius draw 7>M parallel toOY'and DP parallel to OA'.
These lines will intersect in a point / > (j, //) on the ellipse, y
because _ . .,
y = OH = OB cos 4> = a cos />
and l/ = AP = OD = Or sin <jf> = 6 sin <,
or - = cos and j- = sin </>.
Now squaring and adding, we get
-f %- cos 2 </) + sin 2 (/> = 1,
o 2 ft-
the rectangular equation of the ellipse. <j> is some-
times called the eccentric angle of the point P of the
ellipse.
fThe path described by a point on the circumference of a circle which rolls without
sliding on a fixed straight line is called a cycloid. Let the radius of the rolling circle be a, P
the tracing point, and M the point of contact with the fixed line X, which is called the
PARAMETRIC AND POLAR EQUATIONS
117
Solution. Differentiating, = a (I - cos t?), ~ = a sin0.
du du
Substituting in (A), Art. 81,
dy sin . ,
dx = 1 - cos = m = slope at any polnt
When = 0i, y = y\ = a (I cos 0i), m = m\ =
sin 0t
1 cos 0i
Following Art. 43, we find (see figure at foot of tnis page)
cos 0i ) J
7W = subtangent =
sn
NM = subnormal = a sin ft.
MP = length of normal = a\ 2(1 - cos 0,) = 2 a sin \ 0,. By (5), p. 3.
In the figure, PA a sin 6\ (if = 0,) = the subnormal NM as above. Hence
the construction for the normal PM and tangent Ptt is as indicated.
Horizontal and vertical tangents. From (4), and referring to Art. 42,
we see that the values of the parameter / for the points of contact of
these tangent lines are determined thus :
Horizontal tangents : solve -& = f or t.
Vertical tangents : solve -~ = for t.
ILLUSTRATIVE EXAMPLE 3. Find the points of
contact of the horizontal arid vertical tangents to
the cardioid (see figure)
'% a cos - J a cos 2 - J a
^ y = a sin - \ a sin 2 0.
\
dx
Solution. -73 = a( sin i
du
+ sin 2 0) ; ^| = a(cos - cos 2 0).
Horizontal tangent*. Then cos - cos 20 = 0. Substituting (using (5), p. 3)
cos 20 = 2 cos 1 ' 0-1, and solving, we get -- 0, 120", 240".
Vertical tangents. Then - sin 6 -f sin 20 = 0. Substituting (using (5), p. 3)
sin 2 = 2 sin cos 0, and solving, = 0, 60, 180, 300.
base. If arc PM equals OM in longth, then P will touch at if the circle is rolled to the left.
We have, denoting the angle 7TM by 0,
x = ON - OM - NM = aO - a ain = a(0 - sin 0),
y - NP = fl/C _ AC - a - a cos - a(l - cos 6),
the parametric equations of the cy-
cloid, the angle 6 through which
the radius of the rolling circle turns
being the parameter. OD = 2 ira
is called the base of one arch of the
cycloid, and the point V is called
the vertex. Eliminating 6, we get
the rectangular equation
118 DIFFERENTIAL CALCULUS
The common root = should he rejected. For both numerator and de-
nominator in (A) become zero, and the slope is indeterminate (see Art. 12). From
(5), x = y = when = 0. The point is called a cusp.
Substituting the other values in (5;, the results are as follows:
Horizontal tangents: points of contact (- \ a, _-J a\ 7 3).
Vertical tangents: points of contact (\ a, -i \ </\/3 ), ( 2 a, 0).
Two vertical tangents coincide, forming a "double tangent" line.
These results agree with the figure.
PROBLEMS
Find the equations of the tangent and normal, and the lengths of the
subtangent and subnormal, to each of the following curves at the point
indicated.
7\inynit \ormal Subt. Subn.
1. j- = / 2 ,7/ = 2/-f 1 ; t-1. Am. .r-?/ + 2 = (), :r + ?/-4 = 0, 3, X.
2. x = f\ y = 3 / ; / = - 1 . .r - ?/ - 2 = 0, .r -f y -f 4 - 0, - 3, - 3.
y-r2 = Q, G.r- y- 35 = 0, -6, - -J.
= 0, - 1, - 9.
5. x = cos 2 0, // = sin ; = J TT.
6. x rr /a, // = 2 - / ; / = 1 . 11. j* = tan 0, // = ctn ; = 4 TT.
7. 3 j- = t'\ '1 y = /'-' ; / = 4 J!. 12. jr = - ,'{ r ', // = 2 r' ; / = 0.
8. j- = 6 / - /'*, ?/ = 2 / + 3 ; t - 0. 13. .r -m 3 cos n', ?/ - 5 sin a- ; a = J TT.
9. , r = r, ?/ = / :{ -f 3 t; 1 = 1. 14. j- = sin 2 0, ?y = cos ; - J TT.
10. r = p?/ = 2/; / = -!. 15. r=-ln(/-2),3?/ = /; / = 3.
In each of the following problems plot the curves and find the points
of contact of the horizontal and vertical tangents.
16. JT = 3 t /-\ ?/ = / + !. Ann. Horizontal tangents, none ; vertical
tangents, (2, 2), (- 2, 0).
17. x = 3 - 4 sin 0, // = 4 + 3 cos 6.
Ann. Horizontal tangents, (3, 1), (3, 7);
vertical tangents, (7, 4), ( 1, 4).
18. jc = t~ - 2 /, // = r - 12 /. 20. x = sin 2 f, y = sin t.
19. s = 1i + r cos 0, .y A- + r siri 0. 21. .r = cos 4 0, y = sin 4 0.
In the following curves (figures in Chapter XXVI) find lengths of
(a) subtangent, (b) subnormal, (c) tangent, (d) normal, at any point.
rt _ m , l a = a (cos f -f / sin /),
22. The curve ; . . '
t ?/ a (sin / / cos n.
4r?s. (a) 11 ctn /, (b) ?y tan /, (c) -^- , (d) --
sin / cos f
PARAMETRIC AND POLAR EQUATIONS 119
23. The hypocvcloid (astroid) / x = f a c . OR ''f ' (Figure, p. 533)
I// = 4 sin-*/.
.4?js. (a) - // ctn t, ( ?/ tan t, (c) -~ (d) -^
sin / cos t
_, r . , fx~ rcos/,
24. The circle ^ . /
[^/ = r sin /.
nc rm. j- -j f x = </(- cos/ cos2 /), (Figure,
25. The cardioid { \ . . . ./' h ..
^ // = a (2 sin / sm J () P- 11
i "1+P
26. The folium J 7., (Figure, p. 533)
I 2/= J^jT^-
r - ^ COS /,
27. The hyperbolic spiral ! (Figure, p. 534)
i ?/ = y sin <.
L f
82. Parametric equations. Second derivative. Usin^ y/' as symbol
for the first derivative of y with respect to :r, then (A), Art. 81, will
give y' as a function of /,
(1) V' =
To find the second derivative y", use this formula (A) again, re-
placing y by y'. Then we have
dt
ifx=/(0,asin (1), Art. 81.
ILLUSTRATIVE EXAMPLE. Find y" for the cycloid (see Illustrative Example 2,
Art. 81)
fx = a(d -sin 6),
]^y a(l cos 0).
Solution. We found y' = l ^"^' and ^| = a(1 " cos ^
Also, differentiating,
c^_' __ (1 - cos 6) cos sin 2 __ cos ,j_. __ L__.
^~ l-cos^) 15 (l-coc0J- (l-cos(9)
Substituting in (),
- -- Ans.
Note that y" is negative, and the curve therefore is concave downward, as in the
figure for the cycloid, p. 117.
120 DIFFERENTIAL CALCULUS
PROBLEMS
1. In each of the following examples find -~ and -7-^ in terms of t.
ax ax 2
(a)x = t-l,y = t* + l. Ans. = 2 /, ^ 2.
-. _ _._-
/^ T _ o , 7 . __ <_!. (e) x = a cos /, |/ = b sin f.
^c; x - L i, y - 3 (f) x = 2(1- sin 0, = 4 cos *.
/<ix T _ P >)J = (_ 2 . (K) x = sin '> ?/ = sin 2 <
v ; 6 y 2 (h) x = cos 2 /, y = sin ?.
2. Show that the curve r = sec ^, ?/ = tan ^ has no point of inflection.
3. In each of the following examples plot the curve and find the maxi-
mum, minimum, and inflectional points :
(a) x = 2 a ctn 0, y = 2 a sin 2 6.
Arts. Max. (0, 2 a) ; points of inflection, ( => "V"
(b) x = tan /, y siri t cos /.
Ans. Max. (1, -j); min. ( 1, \}\
p>oints of inflection, ( V3, - V (0, 0), /V3, V
, , ,
4 / \ 4 /
83. Curvilineai motion. Velocity. When the parameter / in the
parametric equations (1), Art. 81, is the time, and the functions /(/)
and </>(/) are continuous, if /, varies continuously the point P(jr, ?/) will
trace the curve or path. We then have a curvilinear motion, and
(1) *=/(0, 0=0(0
are called the equations of motion.
The velocity v of the moving point P(x, ?/) at any instant is de-
termined by its horizontal and vertical components.
The horizontal component v x is equal to the velocity along OX
of the projection M of P, and is therefore the time rate of change of x.
Hence, from (C), Art. 51, when s is replaced by .r, we get
In the same way the vertical component v v ,
or time rate of change of ?/, is
m *-J.'
Lay off the vectors v x and v v from P as in the figure, complete the
rectangle, and draw the diagonal from P. This is the required vector
PARAMETRIC AND POLAR EQUATIONS 121
velocity v. From the figure, its magnitude and direction are given
by the formulas .
v x dx
~dt
Comparing with (4), Art. 81, we see that tan r equals the slope
of the path at P. Therefore the direction of v lies along the tangent
line at P. The magnitude of the vector velocity is called the speed.
84. Curvilinear motion. Component accelerations. In treatises on
mechanics it is shown that in curvilinear motion the vector accelera-
tion a is not, like the vector velocity, directed along the tangent, but
toward the concave side of the path of motion. It may be resolved
into a tangential component, a t , and a normal component, a n , where
_dv t _?/2
at ~dt' a "~jT
(R is the radius of curvature. See Art. 105.)
The acceleration may also be resolved into components parallel to
the axes of coordinates. Following the same plan used in Art. 83
for component velocities, we define the component accelerations paral-
lel to OX and OY,
/m dv Xm dv,,
(F) , = , a v - -
Also, if a rectangle is constructed with vertex P and sides a x and
a vy then a. is the diagonal from P. Hence
(G) a
which gives the magnitude (always positive) of the vector accelera-
tion at any instant.
In Problem 1 below we make use of the equations of motion of a
projectile, which illustrate very well this and the preceding article.
PROBLEMS
1. Neglecting the resistance of the air, the equations of motion for a
projectile are
x = vi cos <t> t, y = 0i sin <t> t 16.1 1 2 ;
where v\ = initial velocity, <f> = angle of projection with
horizon, and t = time of flight in seconds, x and y being "of
measured in feet. Find the component velocities, com-
ponent accelerations, velocity, and acceleration (a) at any instant ; (b) at
the end of the first second, having given Vi = 100 ft. per second, </> = 30.
122 DIFFERENTIAL CALCULUS
Find (c) the direction of motion at the end of the first second; (d) the
rectangular equation of the path.
Solution. From (C) and (Z>),
(a) TJ = PI cos </> ; v v v\ sin <p 32.2 /.
Also, from (), = v rr - 64.4 /*>, sin </> -I- 1036.8 t 2 .
From (F) and (G), a., = ; a v 32.2 ; a 32.2, direction downward.
(b) Substituting t 1, t'i 100, < = 30' in these results, we get
r, = 86.6 ft. per sec. a x = 0.
v y = 17.8 ft. per sec. a, = - 32.2 ft. per (sec.) 2 .
v (speed; = 88.4 ft. per sec. a 32.2 ft. per (sec.; 2 .
(c) r = arc tan - arc tan ^ 11" 37' = angle of direction of motion with
the horizontal. v * 8(j ' 6
(dj When v\ = 100, 4> 30, the equations of motion become
x - 50 /V3, y - 50 / - 16.1 /-.
Eliminating /, the result is y ^ - x~, a parabola.
\ 3 <5
2. Show that the rectangular equation of the path of the projectile
in Problem 1 is ir ^
y = .r tan - ^ (1 + tan 2 0).r 2 .
3. If a projectile be given an initial velocity of 160 ft. per second in a
direction inclined 45 with the horizontal, find (a) the component velocities
at the end of the second and fourth seconds ; (b) the velocity and direc-
tion of motion at the same instants.
Ans. (a) When t 2, ?v 113.1 ft. per sec., v y = 48.7 ft. per sec.,
when / 4, r x = 113.1 ft. per sec., r v = 15.7 ft. per sec. ;
(b) when / = 2, r = 123.1 ft. per sec., r = 23 18',
when / = 4, r = 114.2 ft. per sec., r = 172 6'.
4. With the data as in Problem 3 find the greatest height reached by
the projectile. If the projectile strikes the ground at the same horizontal
level from which it started, find the time of flight and the angle of impact.
5. A projectile with an initial velocity of 160 ft. per second is hurled
at a vertical wall 480 ft. away. Show that the highest point on this wall
that can be hit is at a height above the o*-axis of 253 ft. What is </> for
this height? Ans. = 59.
6. If a point referred to rectangular coordinates moves so that
x a cos t + b and y = a sin t -f c,
show that its velocity has a constant magnitude.
PARAMETRIC AND POLAR EQUATIONS 123
7. If the path of a moving point is the sine curve
I .r = at,
\ u b sin a/,
show (a) that the .r-component of the velocity is constant; (b) that the
acceleration of the point at any instant is proportional to its distance from
the /-axis.
8. Given the equations of motion .r = t\ // - (t - 1)-'. ( a ) Find the
equation of the path in rectangular coordinates, (b) Draw the path with
the velocity and acceleration vectors for / -- A, / ~- 1, / 2. (c) For what
values of the time is the speed a minimum? (d) Where is the point when
its speed is 10 ft. per second ?
Ans. (a) Parabola, .r- 1 + i/~ = 1 ; (c) / = ; (d) (16, 9).
9. In uniform motion (speed constant) in a circle, show that the ac-
celeration at any point P is constant in magnitude and directed along
the radius from P toward the center of the circle.
10. The equations of a curvilinear motion are j- ~ '2 cos 2 /, // 3 cos /.
(a) Show that the moving point oscillates on an arc of the parabola
4 ?/- - 9s - 18 0. Draw the path, (b) Draw the acceleration vectors
at the points where r 0. (c) Draw the velocity vector at the point
where the speed is a maximum.
Given the following equations of curvilinear motion, find at the
given instant r r , v in r ; a x , <v, ; , c\ ; position of point (coordinates) ; direction
of motion. Also find the equation of the path in rectangular coordinates.
11. x = t 2 , y = 2t; / = 2.
12. x = 2 /, ?/ = /<; t= 1.
13. x = P, y = /-'; ? = 2.
14. x = 3 /, y = t' 2 - 3 ; / = 3.
15. x- 2 -t,y= 1 -f /-; / = 0.
16. x = a cos /, y = a sin / ; f = j TT.
17. x = 4 sin /, y 2 cos / ; / J TT.
18. x sin 2 t, y 2 cos / ; t = 1, w.
19. x = 2 sin t ; y = cos 2 f ; / = 1 T.
20. x = tan t; y = ctn / ; t = \ TV.
85. Polar coordinates. Angle between the radius vector and the tan-
gent line. Let the equation of a curve in polar coordinates p, 6, be
CD p
124
DIFFERENTIAL CALCULUS
We proceed to prove the
Theorem. // \f/ is the angle between the radius vector OP and the
tangent line at P, then
(H) tan $ =
where
.
dd
Proof. Through P and a point
Q(p + Ap, + A0) on the curve near
P draw the secant line AB. Draw PR
perpendicular to OQ.
Then (see figure) OQ = p+ Ap, angle POQ = A0, PR = p sin A0,
and OR = p cos A0. Also,
PR PR __ p sin A0
" RQ ~ OQ OR ~~ p + Ap p cos A0
(2)
Denote by ^ the angle between the radius vector OP and the
tangent line PT. If we now let A0 approach zero as a limit, then
(a) the point Q will approach P ;
(b) the secant AB will turn about P and approach the tangent
line PT as a limiting position ; and
(c) the angle PQR will approach ^ as a limit.
Hence
, rtx , . v p sin A0
(3)
v }
, . v
tan ^ = hm
Y
o p + Ap p cos A0
To get this fraction in a form so that the theorems of Art. 16 will
apply, we transform it as shown in the following equations:
p sin A0
p sin A0
~
Pd
fSince from (5), p. 3, p - p cos A0 = p(l cos A0 ) = 2 p sin 2 -1
sin A0
p.
A0
A0
A0
[Dividing both numerator and denominator by A0 and factoring.]
PARAMETRIC AND POLAR EQUATIONS
When A0 - 0, then, by Art. 68,
125
sin
lim
= 1, and lim
=L
Also,
\f\
lim sin = 0,
Hence the limits of numerator and denominator are, respectively,
p and p'. Thus (H) is proved.
To find the slope (tan r in the figure), proceed as follows. Take
rectangular axes OX, OY, as usual. Then for P(x, y) we have
(4) 3 p cos 0, ]1 = p sin 0.
Using (1), these equations become parametric equations of the curve,
being the parameter. The slope is found by (4). Thus, from (4),
-? = p ' cos - p sin 0, -& = p' sin + p cos 0,
du du
Slope of tangent = tan r
p' sin 4 P cos
p' cos p sin 6
Formula (7) is easily verified for the figure* on page 124. For, from 1he triangle
Substituting
OPT, T = 4-
Then tan r = tan (0 f ^) =
Uin
1 tan tan ^
tan ""' * 7 > tan \L = and reducing, we have (7).
COS0 P
ILLUSTRATIVE EXAMPLE 1. Find tan ^ and the
slope for the cardioid p a(\ cos 0).
Solution. --- = p' a sin 6. Substituting in (77)
and (7),
dd
i P (i(l
tan \I/ t = -^
P a si
tan r =
2 a sin \ cos J
= tan J 0. ((5), p. 3)
a sin 2 + "(1 - cos 0) cos
a sin cos 6 a(l - cos i
cos cos 2
) sin
; = tanl!0. ((5), (6), p. 3)
sin 2 - sin 6
At P in the figure, ^ = angle OPT = = J angle XO7\ If the tangent
line PT is produced to cross the axis OX, forming with it the angle T, we have
angle XOP = 180 - angle OPT + r.
Therefore r = i - 180, and tan r = tan 0, as above ((3), p. 3).
NOTE. Formula (H) has been derived for the figure on page 124. In each prob-
lem, the relations between the angles i/s T, and 6 should be determined by examin-
ing the signs of their trigonometric functions and drawing a figure.
126
DIFFERENTIAL CALCULUS
To find the angle of intersection (/> of two curves C and C" whose
equations are given in polar coordinates, we may proceed as follows :
Angle TPT' = angle OPT 1 - angle OPT,
or <j) = \//' \[/. Hence
/ n * JL tan ^r' - tan t/r
(7) tan^ = z-p
' 1 + tan t/r' tan r/r
where tan \[/' and tan \{/ are calculated by
(//) from the equations of the curves and
evaluated for the point of intersection.
ILLUSTRATIVE EXAMPLE 2. Find the angle of
intersection of the curves p = a sin 20, p a cos 2 0.
Solution. Solving the two equations simultaneously, we get, at the point of
intersection,
tan 2 = 1, 2(9 = 45, = i>L > J.
From the first curve, using (H),
tan \[/ f = i tan 2 = J, for = 22.^,
From the second curve,
tan ^ = - J cot 20 = -?,, for = 22\.
Substituting in (/),
tan </> = *
-f
= $. /. </> = arc tan
The curves are shown in Chapter XXVI.
86. Lengths of polar subtangent and polar subnormal. Draw a line NT
through the origin perpendicular to the radius vector of the point P
on the curve. If PT is the tangent and PN the normal
to the curve at P, then
OT length of polar subtangent,
and ON length of polar subnormal,
of the curve at P.
In the triangle OPT, tani// = Therefore
P
(1) OT = p tan^ - p 2 ~ = length of polar subtangent.*
In the triangle OPN, tani/' = -- Therefore
ON
(2)
ON =
tan \[/
dp
= - length of polar subnormal.
do
* When increases with p, - is positive and i/' is an acute angle, as in the above figure.
tip
Then the subtangent OT is positive and is measured to the right of an observer placed at O
and looking along OP. When - is negative, th^ subtangent is negative and is measured
dp
to the left of the observer.
PARAMETRIC AND POLAR EQUATIONS 127
The length of the polar tangent (= PT) and the length of the polar
normal ( PN) may be found from the figure, each being the hypotenuse
of a right triangle.
ILLUSTRATIVE EXAMPLE. Find the lengths of the polar subtangont and polai
subnormal to the lemniscate p~ = a 2 cos 2 (figure in Chapter XX YJ).
Solution. Differentiating the equation of the curve, regarding p as an implicit
function of 6,
-,
dO dO
Substituting in (1) and (2), we get
Length of polar sub tangent =
- ,
a -sin 2
Length of polar subnormal = - " L> ff.
P
If we wish to express the results in terms of 0, find p in terms of from the given
equation and substitute. Thus, in the above, p = i </ \ cos 20; therefore the
length of the polar subtangent = a ctn 2 Veos~2~0.
PROBLEMS
1. In the circle p a sin 0, find \p and r in terms of #.
A//*. t=0,T = 20.
2. In the parabola p = a sec- - show that r -f ^ -- TT.
3. Show that t/' is constant in the logarithmic spiral p e< l . Since the
tangent makes a constant angle with the radius vector, this curve is also
called the equiangular spiral. (Figure, p. 534)
4. Show that tan \p = in the spiral of Archimedes, p = aO. Find
values of ty when = 2 TT and 4 TT. (Figure, p. 5.*J4)
-4 w.s'. ^ ^ 80 57' and 85 27'.
Find the slopes of the following curves at the points designated.
5. p = a(l - cos 0)', = ~- Am. -1.
6. p = a sec 2 ; p 2 a. 3.
7. p = a sin 4 ; origin. 0, 1, oo , 1.
8. p 2 = a 2 sin 4 ; origin. 0, 1 , oo , 1.
9. p a sin 3 ; origin. 0, V3, V3.
10. p = a cos 3 ; origin.
11. p = acos20; origin. 14. p ad; 6 = ~-
13. p =
128 DIFFERENTIAL CALCULUS
Find the angle of intersection between the following pairs of curves.
17. p cos 6 = 2 a, p 5 a sin 0. Am. arc tan if.
18. p = a sin 6, p = a sin 2 0. _
j4w. At origin, 0; at two other points, arc tan 3V3.
n
19. p sin = 2 a, p = a sec- -. y4ws. 45 .
20. p = 4 cos 0, p = 4(1 - cos 0). 60.
21. p = (5 cos 0, p = 2(1 + cos 0). 30.
22. p = sin 0, p = cos 2 0. and arc tan ^
23. p' 2 sin 2 = 4, p-' = 16 sin 2 0. 60.
24. p = a(\ + cos 0), p = 6(1 - cos 0).
25. p = sin 2 0, p = cos 2 + 1.
26. p 2 sin 2 = 8, p = 2 sec 0.
Show that the following pairs of curves intersect at right angles.
27. p = 2 sin 0, p 2 cos 0.
28. p - a0, p0 a.
29. p - a(l + cos 0), p = a(l - cos 0).
30. p 2 sin 20 = a 2 , p- cos 20 = 6 2 .
31. p = a sec- y p = /> esc' 2 -
32. Find the lengths of the polar subtarigent, subnormal, tangent, and
normal of the spiral of Archimedes, p = aO.
Ann. Subtangent = > tangent = - Va 2 + p 2 ,
subnormal = a, normal = Va 2 -f p 2 .
The student should note the fact that the subnormal is constant.
33. Find the lengths of the polar subtangent, subnormal, tangent, and
normal in the logarithmic spiral p = a ft .
A NS. Subtangent = ~-> tangent = p^/l-f tj
subnormal = p In a, normal = p Vl -f In 2 a.
34. Show that the reciprocal spiral p0 = a has a constant polar sub-
tangent.
87. Real roots of equations. Graphical methods. A value of x which
satisfies the equation
(1) /(J)=0
is called a root of the equation (or a root o//0)). A root of (1) may
be a real number or an imaginary (complex) number. Methods of
determining real roots approximately will now be developed.
PARAMETRIC AND POLAR EQUATIONS
129
Location and number of the roots.
FIRST METHOD. If the graph of /fr), that is, the locus of
(2) 0=/fr)
is constructed, following Art. 58, the intercepts or? the x-axis are the
roots. From the figure, therefore, we know at once the number of
roots and their approximate values.
ILLUSTRATIVE EXAMPLE. Locate all real roots of
(3) !'< - 9 j-2 f 24 JT - 7 = 0.
Solution. The graph has been con-
structed in Art. 58, Problem 1. It
crosses the axis of x between and 1.
Hence there is one real root between
these values, and then 4 are no other real roots.
The table gives the values of /(()) and /(I ), show-
ing a change of sign.
The table of values of x and // used in
plotting the graph may locate a root exactly,
namely, if y = for some value
of x. If not, the values of '// for
two successive values x = a,
x b may have opposite signs.
The corresponding points
X
a
XQ
b
y
Kb)
=
P(a,f(a)), Q(b,f(b)) are, therefore, on opposite sides of the r-axis, and
the graph of (2) joining these points will cross this axis. That is, a root
XQ will lie between a and b.
An exact statement of the principle involved here is as follows.
// a continuous function /fr) changes sign in an interval a < x < b
and if its derivative does not change sign, then the equation f(x) = has
one root, and only one, between a and b.
Location of a root by trial depends upon this principle. If a and 6
are not far apart, a further approximation can be found by interpo-
lation. This amounts to determining the intercept on the x-axis of the
chord PQ. That is, the portion of the graph joining P and Q is re-
placed, as an approximation, by the chord.
ILLUSTRATIVE EXAMPLE (CONTINUED). The root
between and 1 may be located by calculation more
closely between 0.3 and 0.4. See table. Let 0.3 + *
be this root. Then, by interpolation (proportion),
__ = -583 = Q32 DiflF. 0.1 1.807
0.1 1.807'
Hence x 332 is a second approximation. This is the intercept on the x-axis
of the line that joins the points Q(0.4, 1.224) and P(0.3, - 0.583), which He
_z f(x) = y
0.4 E224
0.3 + z(root)
0.3 - 0.583
130 DIFFERENTIAL CALCULUS
on the graph of (3;. In the figure, MP = - 0.583,
NQ = 1.224, drawn to scale. The abscissas of M and
N are 0.3 and 0.4 respectively. Also, MC = z, and the
homologous sides of the similar triangles MPC and
PQIt give the above proportion.
For an algebraic equation, of which (3) is *"
an example, Homer's method is best adapted
to calculating a numerical root to any desired
degree of accuracy, as explained in textbooks on algebra.
88. Second method for locating real roots. The method of Art. 58 is
well adapted to constructing quickly the graph of f(jr). By this graph
the roots are located and their number determined. In many ex-
amples, however, the same result is attained more quickly by drawing
certain intersecting curves. The following example shows how this
is done.
10 ' 20" '10 '" Id 50 ItiO' 7U ' M)"( )
1 -H c
Rad ion
I I'JO 200 'X.
ILLUSTRATIVE EXAMPLE. Determine the num-
ber of real roots U in radians) of the equation
(1) ctnj--j- = 0,
and locate the smallest root.
Solution. Transpose and write (1) thus :
(2) ctri x x.
If we draw the curves
(3) y ctn x and y x
on the same axes, the abscissas of Ihc points of in-
tersection will be roots of (1\ For, obviously,
eliminating y from (3) gives equation (1), from
which the values of x of the points of intersec-
tion are to be obtained.
y ctn X
x (degrees)
x (radians;
y
OO
10
.175
5.67
20
.349
2.75
30
.524
1.73
40
.698
1 19
45
.785
1.000
50
.873
.839
GO
1.017
.577
70
1 222
.364
80
1.396
.176
90
1.571 | -0
PARAMETRIC AND POLAR EQUATIONS
131
In plotting it is well to lay off carefully both scales (degrees and radians ^ on OX.
Number of solutions. The curve y - ctn .r consists of an infinite number of branches
congruent to AQB of the figure i,see
Art. 70;. The line y = x will obviously
cross each branch. Hence the equation
(1) has an infinite number of solutions.
Using tables of natural cotangents
and radian equivalents of degrees, we
may locate the smallest root more closely
as shown in the table. By interpolation
we find x = 0.860. Ann.
A
(degiee*
x
(radians)
ctn *
CtU X - A
r>r
O.STIJ
O.SIU)
o '/;n
It)
root
1 Sfi. r >
0.8(19
4 01 1
DiiT
0.018
- 0.0.18
The Second Method may be described as follows.
Transpose certain selected terms of f(.r] so that it becomes
(4) /iW-MO.
Plot the curves
(5) y=fiW, 2/=/u(ar)
on the same axes, choosing suitable scales (wot necessarily the same on
both axes).
The number of points of intersection of these curres equate the num-
ber of real roots of f(x) = 0, and the abscissas of these points are the roots.
The terms selected in (4) can often be chosen so that one or both
of the curves in (5) are standard curves.
For example, to locate the real roots of
.r : < + 4 x - 5 = 0,
write the equation r { = r> - 4 x.
The curves in (5) are now the standard curves
y = JC' A , y ~ f> 4 x,
a cubical parabola and a straight line.
As a second example, consider
2 sin 2x4-1- x z = 0.
Write this in the form sin 2 x = I O 2 - 1 )
Then the curves in (5) are the standard curve
y = sin 2 x
and the parabola y = \ (** ~ l )
89. Newton's method. Having located a root, Newton's method
affords a procedure to calculate its approximate value.
The figure shows two points
132 DIFFERENTIAL CALCULUS
on the graph of /(x) on opposite sides of the x-axis. Let PT be
the tangent line at P (Fig. a). The intercept a' of this line on the
x-o, x^a'\ j X [ / = &' = & X
(,
FIG. a FIG. 6
x-axis is, obviously, an approximate value of the intercept of the
graph and hence of the corresponding root of f(x) == 0. Newton's
method determines the x-intercept of PT.
We find this intercept a' as follows. The coordinates of P are
xi = a, '//i /(a). The slope of PT is mi =f'(a). Hence the equation
of PT'is ((1), Art. 43)
0) y-f(a)^f'(a)(x-a).
Putting ?/ = and solving for x(= a') gives Newton's formula for
approximation
,-a-M.
Having found a' by (K), we may substitute a' for a in the right-
hand member, obtaining f , ,,
<-<-$&
as a second approximation. The process might be continued, giving
a sequence of values , ,
U, U- , u/ , Li ) * ' '
approaching the exact root.
Or the tangent may be drawn at Q (Fig. b). Then replacing a
in (K) by &, we obtain b', and from b' we obtain 6" etc., giving values
6', 6", 6'", -
approaching the exact root.
ILLUSTRATIVE EXAMPLE. Find the smallest root of
ctn x x =
by Newton's method.
Solution. Here /(*) = ctn x j,
/'(x) = - csc' J x - 1 = - 2 - ctn 2 x.
PARAMETRIC AND POLAR EQUATIONS 133
By the illustrative example of Art. 88, we take a = 0.855. Then, by the table in
Art ' 88 ' Afl>= 0.014.
Also, /'(a) = - 2 - ^0.869)* = - 2.76.
Hence, by (K), a' = 0.855 -f ~~ = 0.860. Ans.
If we used 6 = 0.873 in (K), then
b' - 0.873 - ~^~ - 0.861.
By interpolation we found x = 0.860. The above results are valid to three
places of decimals.
From the figures on page 132 we observe that the graph crosses
the o>axis between the tangent PT and the chord PQ. Hence tlie exact
root lies between the value found &// Newton's method and that found by
interpolation. This statement is, however, subject to the reservation
that -f"(x) = has no root between a and b, that is, that there is no
point of inflection on the arc PQ.
PROBLEMS
Determine graphically the number and approximate location of the real
roots of each of the following equations. Calculate each root to two decimals.
1. JT* + 2 jr - 8 = 0. A //*. 1 .07.
2. .r' - 4 .r + 2 = 0. - 2.21, 0.54, 1.07.
3. .r 3 - 8 x - 5 = 0. - 2.44, - 0.00, 3.10.
4. .r a - 3 .f - 1 = 0. - 1.53, - 0.35, 1.88.
5. ^ - 3 jr* + 3 = 0. - 0.88, 1.35, 2.53.
6. ^ + 3 x 2 - 10 = 0. 1.49.
7. .r 3 - 3 x' 2 - 4 .r + 7 = 0. - 1.71, 1.14, 3.57.
8< j.3 + 9 j-2 _ 5 .r - 8 = 0. - 2.70, - 1 .30, 2.12.
9. 2 x* - 14 .r 2 -f 2 x -f 5 = 0. - 0.51, 0.71, 0.80.
10. x* -f 8 JT - 12 = 0. - 2.30, 1.22.
11. x 4 - 4 ^ - 6 x 2 + 20 .r -f 9 = 0. - 2.10, - 0.41, 2.41, 4.16.
12. r 4 -I- 4 :r< - 6 x' 2 - 20 x - 23 = 0. - 4.00, 2.00.
Determine graphically the number of real roots of each of the following
equations. Calculate the smallest root (different from zero), using both
interpolation and Newton's formula.
13. cos x + x 0. Ans. One root ; x = - 0.739.
14. tan x x 0. Infinite number of roots.
15. cos 2 x - x = 0. One root ; x = 0.515.
134 DIFFERENTIAL CALCULUS
16. 3 sin x - x = 0. Ans. Three roots ; x = 2.279.
17. 2 sin x - x 2 - 0. Two roots ; x = 1.404.
18. cos x - 2 x 2 = 0. Two roots ; x = 0.635.
19. ctn x -f x 2 = 0. Infinite number of roots ; x = 3.032.
20. 2 sin 2 x - x 0. Three roots ; x = 1.237.
21. sin x + x - 1 = 0. One root; x = 0.511.
22. cos x -f x 1 = 0. One root ; f 0.
23. c' x cos j* = 0. Infinite number of roots ; x 1.29.
24. tan x log x 0. Infinite number of roots ; x 3.65.
25. ( ' -f .r - 3 = 0. One root ; x = 0.792.
26. sin 3 x cos 2 .r = 0. Infinite number of roots ; x = 0.314.
27. 2 sin J .r cos 2x^0. Infinite number of roots ; .r = 0.517.
28. tan x 2 c* 0. Infinite number of roots ; x = 1.44.
29. The inner radius (r) and outer radius (R) in inches of the hollow
steel driving shaft of a steamer transmitting 77 horse power at a speed
of N revolutions per minute satisfy the relation 7i 4 r 4 = -^- If
77 = 2500, N = 160, r = 6, find R. Nir "
30. A cylindrical shell with a hemispherical end has a diameter d in.,
and contains V cu. in. The length of the cylindrical part is // in. Show
that (P + 3 hd' 2 = ^-^ - Given // = 20, V = 800, find d. Ans. d = 6.77
7T
31. The quantity of water Q cu. ft. per second flowing over a weir of
width 7> ft. is given by Francis's formula
where 77 is the height of the water (the head) above the crest of the weir.
Given Q 12.5, H = 3, find 77. (Solve the formula for the factor 77^ and
then plot.) Ans. II = 1.23.
32. If V cu. ft. is the volume of 1 Ib. of superheated steam at a tem-
perature 7" F. and pressure P Ib. per square inch,
V = 0.6490 ^-~~4?.
P P S *
Given V = 2,8, T = 420, find P. A
33. The chord r of an arc .s in a circle of radius r is
given approximately by the formula
- 24 2
If r = 4 ft., c = 5.60 ft., find s. Am. s = 6.23 ft.
PARAMETRIC AND POLAR EQUATIONS 135
34. The area u of a circular segment whose arc s subtends the central
angle x (in radians) is u = % r' 2 (x sin x) Find the value of x if r = 8 in.
and u = 64 sq. in. Ans. x = 2.554 radians.
35. The volume V of a spherical segment
of one base of height CD = h is
V= ir(rh' 2 - i/? 3 ).
Find h if r = 4 ft., V = 150 cu. ft.
Ans. h =4.32 ft.
36. The volume V of a spherical shell of
radius R and thickness / is
Derive this result. If R = 4 ft. and V is one
half the volume of a solid sphere of equal
radius, find t. Ans. t = 0.827 ft,
37. A solid wooden sphere of specific gravity S and diameter d sinks
in water to a depth h. Let x - and show that 2 .r :{ \\ x- -f tf 0.
(See Problem 35.) Find x for a maple ball for which S 0.786.
Ans. 0.702.
38. Find the smallest positive value of for which the curves p cos
and p = e~ e intersect. Find the angle of intersection at this point.
Ans. = 1. 29 radians; 29 (> .
ADDITIONAL PROBLEMS
1. Find the angle of intersection of the curves p = 2 cos and p ~ e
at the point of intersection farthest from the origin.
Ans. Point of intersection is = 0.54 radian ; 75 56'.
2. Show that the curve p = a sin 4 \ cuts itself at right angles.
3. Any radius vector of the cardioid p = a(l + cos 0) is OP. From the
center C of the circle p = a cos a radius of the circle CQ is drawn parallel
to OP and in the same direction. Prove that PQ is normal to the cardioid.
4. A square, one of whose diagonals lies along the polar axis, is cir-
cumscribed about the cardioid p = a(l - cos 0). Show that its area is
fi(2 + V3)o 2 .
5. The path of a particle is the ellipse p = gp The particle
i. *~ C COS u
moves so that the radius vector p describes area at a constant time rate.
Find the ratio of the velocities of the particle at the ends of the major axis.
A 1 -
Ans. -
1 + e
CHAPTER IX
DIFFERENTIALS
90. Introduction. Thus far we have represented the derivative of
y=f(x) by the notation
^-f'fcr)
dx~ J(X) '
We have taken special pains to impress on the student that the
symbol
d/n
dx
was to be considered not as an ordinary fraction with dy as numerator
and dx as denominator, but as a single symbol denoting the limit of
the quotient
All
Ax
as Ax approaches zero as a limit.
Problems occur where it is important to give meanings to dx
and dy separately, and this is especially useful in applications of the
integral calculus. How this may be done is explained in what follows.
91. Definitions. If f'(x) is the derivative of f(x} for a particular
value of JT, and Ax is an arbitrarily chosen increment of x, then the
differential of J(x), denoted by the symbol df(s), is defined by the
equation
(A) d/(x)=
If now /(;r) = :r, then f'(x) = 1, and (4) reduces to
dx = Ax.
Thus, when x is the independent variable, the differential of x(= dx)
is identical with Ax. Hence, if ?/ f(x), (A) may in general be written
in the form
(B) dy=f'(x)dx*=^-dx.
dx
* On account of the position which the derivative f'(x) here occupies, it is sometimes
called the differential coefficient
136
DIFFERENTIALS 137
The differential of a function equals its derivative multiplied by the
differential of the independent variable.
Let us illustrate what this means geomet-
rically.
Draw the curve y =/(x).
Let/'(.r) be the value of the derivative at P. ~-4
Take dx = PQ, then
dy = f'(x)dx = tan r - PQ - ^ PQ = QT.
t y
Therefore dy, or df(x), is the increment (= QT) of the ordinal e of the
tangent corresponding to dr.
This gives the following interpretation of the derivative as a
fraction.
// an arbitrarily chosen increment of the independent variable j/or
a point P (x, y) on the curve y = f('jc) be denoted by dx, then in the
derivative
dy denotes the corresponding increment of the ordinate of the tangent
line at P.
The student should note especially that the differential (= dy)
and the increment (= A?/) of the function corresponding to the same
value of d'jr(=A'jr) are not in general equal. For, in the figure,
dy = QT, but A'# = QP'.
92. Approximation of increments by means of differentials. From
Art. 91 it is clear that Ay(= QP' in the figure; and r///(- QT) are
approximately equal when dx(= PQ) is small. When only an ap-
proximate value of the increment of a function is desired, it is usually
easier to calculate the value of the corresponding differential and use
this value.
ILLUSTRATIVE EXAMPLE 1 . Find the volume approximately of a spherical shell
of outside diameter 10 in. and thickness ^ in.
Solution. The volume y of a sphere of diameter x is
(1, V=l*x\
Obviously, the exact volume of the shell is the difference AV between the
volumes of two solid spheres with diameters 10 in. and 9J in. respectively. Since
only an approximate value of AV is required, we find dV. From (1) and (),
d V = \ 7TX 2 dx, since = 2 ^ 2 -
138 DIFFERENTIAL CALCULUS
Substituting x = 10, dx = - J, we obtain dV - 19.63 cu. in., approximately,
neglecting the sign, which merely means that V decreases as x decreases. The
exact value is AV r ~ 19.4 cu. in. Note that the approximation is close, for dx is
relatively Kmall t that is, small a* compared with x (= 10*. The method would be
worthless otherwise.
ILLUSTRATIVE EXAMPLE 2. Calculate tan 4G', approximately, using differen-
tials, given tan 45 = 1, sec 45 = \ 7/ 2. 1 = 0.01745 radians.
Solution. Let y tan x. Then, by (J3),
When x changes to x + dx, y will change to y -f dy, approximately. In (1),
substitute x - \ IT (45"), dx = 0.0175. Then dy = 0.0350. Since y - tan 45 = 1,
y -f dy 1.0350 ~ tan 46 ", approximately. Anx.
(Four-place tables give tan 46 r = 1.0355.)
93. Small errors. A second application of differentials is afforded
when small errors in calculation are to be determined.
ILLUSTRATIVE EXAMPLE 1. The diameter of a circle is found by measurement
to be 5.2 in., with a maximum error of 0.05 in. Find the approximate maximum
error in the area when calculated by the formula
(1) A \ TTX". (x diameter)
Solution. Obviously, the exact maximum error in A will be the change (AA)
in its value found by (1) when x changes from 5.2 in. to 5.25 in. The approximate
error is the corresponding value of dA. Hence
dA - i irxdx - I TT x 5.2 x 0.05 - 0.41 sq. in. Ans.
Relative and percentage errors. If di( is the error in u, then the ratio
(2) =- the relative error ;
v ' u
(3) 100 = the percentage error.
The relative error may be found directly by logarithmic differen-
tiation (Art. 66).
ILLUSTRATIVE EXAMPLE 2. Find the relative and percentage errors in the
preceding example.
Solution. Taking natural logarithms in (1),
log A log \ TT -f- 2 log x.
~.~ ,. .. 1 dA 2 , dA 2dx
Differentiating, "TT"^"' an T~ = --
A dx, x A x
Substituting x - 5.2, dx = 0.05, we find
Relative error in .4 = 0.0192 ; percentage error = l-fifa %. Ans.
The errors in calculation considered here are due to small errors
in the data upon which the calculation is based. The latter may
arise from lack of precision in the measurements or from other causes.
DIFFERENTIALS 139
PROBLEMS
1. If A is the area of a square of side x, find dA. Draw a figure show-
ing the square, dA, and A A. Aus. dA = 2 x dx.
2. Find an approximate formula for the area of a circular ring of
radius r and width dr. What is the exact formula ?
Ans. dA = 2 TIT dr ; A A = ir(2 r + Ar)Ar.
3. What is the approximate error in the volume and surface of a cube
of edge 6 in. if an error of 0.02 in. is made in measuring the edge?
Ans. Volume, 2.16 cu. in. ; surface, 1.44 sq. in.
4. The formulas for the surface and volume of a sphere are S = 4 -rrr 2
and V J Trr 3 . If the radius is found to be 3 in. by measuring, (a) what
is the approximate maximum error in S and V if measurements are accu-
rate to 0.01 in.? (b) what is the maximum percentage error in each case?
Ans. (a) S, 0.24 TT sq. in. ; V, 0.36 TT cu. in.;
(b) s, I <;'< ; v, 1 ;,.
5. Show by means of differentials that
L (approximately).
x -f djc x x 2
6. Find an approximate formula for the volume of a thin cylindrical
shell with open ends if the radius is r, the length h, and the thickness L
Ans. 2 irrht.
7. A box is to be constructed in the form of a cube to hold 1000 cu. ft.
How accurately must the inner edge be made so that the volume will
be correct to within 3 cu. ft.? Ans. Error ^ 0.01 ft.
8. If ?/ = a: 5 and the possible error in measuring x is 0.9 when x = 27,
what is the possible error in the value of ?/? Use this result to obtain ap-
proximate values of (27.9)^ and (26.1)". Am. 0.2; 9.2; 8.8.
Use differentials to find an approximate value of each of the following
expressions.
9. V66. 11. ^120. 13. T^. 15. \/35.
1 A __
10. V98. 12. \^1010. 14. /
7=' - -
V51
17. If In 10 2.303, approximate In 10.2 by means of differentials.
Ans. 2.323.
18. If e 2 = 7.39, approximate e 2 - 1 by means of differentials. Ans. 8.13.
19. Given sin 60 = 0.86603, cos 60 = 0.5, and 1 = 0.01745 radians,
use differentials to compute the values of each of the following functions
to four decimals : (a) sin 62 ; (b) cos 61 ; (c) sin 59 ; (d) cos 58.
Ans. (a) 0.8835; fb) 0.4849; fc) 0.8573; fd) 0.5302.
140 DIFFERENTIAL CALCULUS
20. The time of one vibration of a pendulum is given by the formula
t* , where t is measured in seconds, g = 32.2, and /, the length of the
V
pendulum, is measured in feet. Find (a) the length of a pendulum vibrat-
ing once a second ; (b) the change in / if the pendulum in (a) is lengthened
0.01 ft. ; (c) how much a clock with this error would lose or gain in a day.
An*, (a) 3.26 ft. ; (b) 0.00153 sec. ; (c) - 2 min. 12 sec.
21. How exactly must the diameter of a circle be measured in order
that the area shall be correct to within 1 per cent? Arts. Error ^ ' ; .
22. Show that the relative error in the volume of a sphere, due to an error
in measuring the diameter, is three times the relative error in the radius.
23. Show that the relative error in the ?rth power of a number is
n times the relative error in the number. -
24. Show that the relative error in the nth root of a number is - times
the relative error in the number.
25. When a cubical block of metal is heated, each edge increases
r ^ per cent per degree increase in temperature. Show that the surface
increases fa per cent per degree, and that the volume increases f\ per cent
per degree.
94. Formulas for finding" the differentials of functions. Since the
differential of a function is its derivative multiplied by the differen-
tial of the independent variable, it follows at once that the formulas
for finding differentials are the same as those for finding derivatives
given in Arts. 29 and 60, if we multiply each one by dx.
This gives
I d(c ) - 0.
II d(j) - dy.
III d(u -f- v ?r) du + dv dw.
IV d(r/) = fd/>.
V d('ur) u dv + v du.
VI d(v n ) = w n ~ } dv.
Via d(jr") = nj? n - } dx.
vn
Vila rf(") = y-
dp
X
v
XI d(a ! ') = a r In a dv.
XI a d(e*) = e*dv.
XII d(w.") = vu r ~ } du + In u u r dv.
XIII d(sin r) = cos vdv.
DIFFERENTIALS 141
XIV d(cos r) = sin vdv.
XV d(tan r) = sec 2 rdv. Etc.
XX d(arcsinr) = --^L==. Etc.
The term "differentiation" also includes the operation of finding
differentials.
In finding differentials the easiest way is to find the derivative as
usual, and then multiply the result by dx.
ILLUSTRATIVE EXAMPLE 1. Find the differential of
Solution.
dy = d( X + * } - ( x * + :
\ar a + 3/
( x' 2 + 3 )dx (x + 3 )2 .r (te ( 3 - G j: - J- 1 )
. .
(x 2 + 3)- (x- -f 3)-
ILLUSTRATIVE EXAMPLE 2. Find </// from
h' 2 x 2 ~ a~y~ ~ a' 2 b' 2 .
Solution. 2 6 2 x dx - 2 a' 2 y dy ~ 0.
, b-jc j 4
/. aij a/. Ans.
<*' 2 y
ILLUSTRATIVE EXAMPLE 3. P'ind dp from
p~ = a' 2 cos 2 #.
Solution. 2 p dp = a' 2 sin 20-2 <i#.
, a' 2 sin 2 , n
.". op = --- (/. >inK.
P
ILLUSTRATIVE EXAMPLE 4. Find d[arc sin (3 / - 4 /')_).
Solution. d[arc sin (3 / - 4 / :< )
\ 1 - (3 / - 4 /')' \l-r- 4
PROBLEMS
Find each of the following differentials.
1. y x* 3 x.
ax
3. T/ Vax -f- 6.
-f 6
, - C/-I2 __ O r2
4/00 7 _ ' '* t-t A
. y = xv a 2 x 2 . ay =
a 2
5. s = ae bt . ds abc ht dt.
i j dv
6. u = In rr. aw =
v
7. p sin ad. dp = a cos aO d6.
142
DIFFERENTIAL CALCULUS
8. y = In sin x.
9. p cos 0.
10. = r f cos TT/.
,4ns. </?/ = ctn
f/p- (cos - 0sin 8)(W.
(In r'(cos TT/ Trsin
Find the differential of each of the following functions.
12. v = VV -f 1.
10 ,,, r==r
lo. // /
Va~ ~ x-
19. If jr 2 -f ?r = a-, show that f/?/ =
Find
20. 2 JT'~ -f 3 .r?/ -f 4 ?/ 2 = 20.
in terms of .r, //, and dx from each of the following equations.
_ _ (4 jr 4 3 ?/)dx
3x48//
21. x :{ + 6 x/r 4 2 //" = 10. 24. x : ' -f /' = a\
22. x -f 4 Vx// 4 12 // = a. 25. x - ?/ = r j4 ^.
23. Vx 4- V?/ = Va. 26. sin (x //) - cos (x 4 ?/).
27. The legs of a right triangle are found by measurement to be
14.5 ft. and 21.4 ft. respectively. The maximum error in euch measure-
ment is 0.1 ft. Find the maximum error in degrees in calculating the
angle opposite the smaller side by using the formula for the tangent of
that angle.
95. Differential of the arc in rectangular coordinates. Let s be
the length of the arc AP measured from a fixed point A on the
curve. Denote the increment of s (= arc PQ) by As. The following
proof depends on the assumption that, as Q ap-
Ay
arc
From the figure, ^
(1) (Chord PQ) 2 = (A*) 2 + (A?/) 2 .
Multiplying and dhdding by (As) 2 in the left-hand member and
dividing both members by (Ax) 2 , we get
(2)
/Chord PQV 2 / As V_
V As
\ 2 /A8\ 2 _
/ VAx/ ~
DIFFERENTIALS
143
Now let Q approach P as a limiting position ; then Ax > and
we have
(3)
\dx
Multiplying both members by dr% we get the result
(C) ds 2 = dx 2 + dy 2 .
Or, if we extract the square root in (3) and multiply both members
by dr, i
From (C), we may readily show also that
i
All these forms are useful.
From (>), since
I-T sec*"7%
we obtain r/,s = sec T djc, assuming the angle r to be acute. Hence
we may easily prove
(F)
dx du
= cos r, -~ = sin r.
ds ds
^ tan T COB r = sin r.
f/8 I
For later reference, we add the formulas, setting y f = -J-
i y 1
(G) COST j> sinr = 2 p
(1 -f y' 2 )" 2 (1 + y r2 )*
If the angle r is obtuse (?/' < 0), a negative sign must be placed before
the denominators in (G) and before cos r in (F). y j
In the accompanying figure, I'Q = A:r = rfy,
FT is tangent at P, and r is acute. Angle
PQT is a right angle.
Therefore Q T = tan r rfj = r/y/. By Art. 91
Then
/> 7 1 = W/r 2 + rf?/- - rfs. By (C)
The figure will help in memorizing the relations above.
The assumption made at the beginning of this article is proved
in Art. 99.
144
DIFFERENTIAL CALCULUS
96. Differential of the arc in polar coordinates. From the relations
(1) j = p cos 0, y = p sin 6
between the rectangular and polar coordinates of a point, we obtain,
by V, XIII, XIV, of Art. 94,
(2) dx = cos dp p sin d6, dy = sin dp + p cos dO.
Substituting in (C), Art. 95, reducing, and extracting the square
root, we obtain the result
(H) ds = Vd/> 2 + p 2 d0 2 .
This may be written
r /^*\2 o
dO.
The figure is drawn so that the angle \f/
between the radius vector OP and the tan-
gent line PT is acute (Art. 85). Also p, SO,
and Sp (= OP' - op) are positive. Take
p for the independent variable. Then
Sp = dp. In the right triangle PQT, take
PQ = dp. Then QT = tan ^ r/p.
But
Therefore
hence
/
tan ^ = p
-7-
P
= p-p
dp
PT = Vrfp- +
L> = rfs.
ILLUSTRATIVE EXAMPLE 1. Find the differential of the arc of the circle
x 2 + y* = r a .
Solution. Differentiating, ^ = _.
aj: y
To find rfs in terms of a* we substitute in (/)), giving
To find ds in terms of y we substitute in (), giving
ILLUSTRATIVE EXAMPLE 2. Find the differential of the arc of the cycloid
x = a (6 - sin 0), y = a(l - cos 0), in terms of and dd. (See Illustrative Ex-
ample 2, Art. 81.)
Solution. Differentiating,
dx = a(l - cos 6)d6, dy = a sin dd,
DIFFERENTIALS 145
Substituting in (C)
ds' 2 - a" (I - cos 6}~d6" -f a 2 sin- dO* - 2 a-(l - cos 0}dQ' 2 .
From (5), Art. 2, 1 - cos 6 = 2 sin- .\ 0. Hence ds = 2 a sin J <J0. A?ks.
ILLUSTRATIVE EXAMPLE 3. Find the differential of the arc of the cardioid
p = a(l cos 0) in terms of 6.
Solution. Differentiating, ( - = a sin 6.
du
Substituting in (/) gives
cte = LaHl-cos0j- + a 2 sin- 0]adt? = a^
PROBLEMS
For each of the following curves find ds- in terms of jr and dr.
1. 2 y = .r 2 . Ans. ds- = Vl + j* dr.
2. 7/ 2 = 2 px. (/.s - ^ dr.
3. & 2 x 2 -f a 2 // 2 - a 2 6 2 . ds - \
\
>, c 4 , o ; (^ -f-
4. 6 x?y = a* 4 -f 3. d^ = -- -
5. ?y ~ In sec jr. ds- = sec jr djc.
6. a 2 // = JT\ 9. 2 ?/ = c j -f J .
7. a?/ 2 a* :? . 10. // = sin jr.
8. V.r -f V/y = Va. 11. // = cos 2 x
For each of the following curves find ds- in terms of ?/ and dy.
,
:j?y
14. x^ J 4- ?y^ J = a*. <is = ^J~ d?/.
15. a 2 ?y = x 3 .
16. ?y 2 - 2 .r - 3 ?/ = 0. 17. 2 x// 2 - y 2 -4 = 0.
For each of the following curves find ds, sin r, and cos r in terms of
t and dt.
18. x = 2 f -f 3, y t 2 2. 20. x a sin /, y a cos J.
19. x = 3 / 2 , ?y = 2 P. 21. x = 4 cos f, ?/ 3 sin /.
146 DIFFERENTIAL CALCULUS
For each of the following curves find dx in terms of 6 and dO.
22. p a cos 0. A tin. dn a dO.
23. p ~ 5 cos 4- 12 sin 0. <& = 13 d8.
24. p = 1 - sin 0. ds = V2 - 2 sin e/0.
25. p 3 sin 4 cos 0. 30. p a cos 0.
26. p = 1 + cos 0. 31. p = 4 sin- { f-
4
27. p^sec 2 ^- 32. p =
Z 1 '
28. p = 2 - cos 0. 33. p - :
29. p = 2 4- 3 sin 0. 34. p =
1 + cos i
4
3 cos i
4
1 - 3 cos 6
97. Velocity as the time-rate of change of arc. In the discussion of
curvilinear motion in Art. 83, the velocity, or, more correctly, the
speed v was given by (),
(1) fl 2 = r x *+7y>.
By (C) and (D) in Art. 83, r, - > r, = d -%-
a/ a/
Substituting in (1), using differentials and (C), Art. 95, the result is
Extracting the square root, taking the positive sign, we have
ds
V = 3T
Hence, in curvilinear motion the speed of the moving point is the
time-rate of change of the length of arc of the path.
This statement should be compared with the definition of velocity
in rectilinear motion as the time-rate of change of distance (Art. 51).
98. Differentials as infinitesimals. In applied mathematics differ-
entials are often treated as infinitesimals (Art. 20), that is, as vari-
ables approaching zero as a limit. Conversely, relations between
infinitesimals are frequently established in which these are replaced
by differentials. The "principle of replacement " involved here is
very useful.
DIFFERENTIALS 147
If x is the independent variable, we have seen that A.r = dr, and
thus AJ may be replaced by djc in any equation. If A.r > 0, so will
also dy 0. On the other hand, A// and rf// are not in general equal.
But, when x has a fixed value and A.r (= dx] is an infinitesimal, so
also is A?/, and, from (B), Art. 91, dy as well. Furthermore it is easy
to prove the relation
(1) lim^-l.
JLX-O W?/
Proof. Since \im=& =/'(?),
AX ^ o A.r
we may write ^ = /'(.r) + /, if lim i = 0.
-AT Aj- .0
Clearing of fractions, and using (B),
A// = dy + i A.r.
Dividing both members by A//, and transposing, the result is
dy _ 1 - AJ
i ^ _^
A?/ A//
Hence lim ^ = 1 , or also lim ^ = 1 Q. E. D.
Ar -, A'// Ax-0"?/
We now state, without proof, the
Replacement Theorem. //? problems involving only the ratios of in-
finitesimals which simultaneously approach zero, an 'infinitesimal may
be replaced by a second infinitesimal so related lo it that the limit of their
ratio is unity.
From the above theorem, Ay/ may be replaced by dy, and, in
general, any increment by the corresponding differential.
In an equation which is hom.ogeneom in infinitesimals the above
theorem is simple in application.
ILLUSTRATIVE EXAMPLE 1. By (5), p. 3, if x - \ i, 1 - cos i = 2 sin 2 J i. Let
{ be an infinitesimal. Then, by (B), Art. 68, sin i may be replaced by i, sin 2 J i
by ^ i 2 , and therefore 1 cos i by J i 2 . Also tan i ( cos i sin i) may be replaced
by i.
ILLUSTRATIVE Kx AMPLE 2. In (1), Art. 95, all quantities are infinitesimals
ultimately, since A.r -0. The equation is homogeneous, each term being of the
second degree. By the theorem, we may replace the infinitesimals as follows :
Chord PQ by arc PQ - As, and As by ds ; A// by ety ; and Ax by dx.
Then (1) becomes ds 2 - dx- + d?y 2 , that is, (C).
148
DIFFERENTIAL CALCULUS
99. Order of infinitesimals. Differentials of higher order. Let i and
; be infinitesimals which simultaneously approach zero, and let
lim 4 = L.
i
If L is not zero, ? and j are said to be of the same order.
If L = 0, j is said to be of higher order than i.
If L becomes infinite, j is said to be of lower order than i.
Let L 1. Then j i is of higher order than i.
[hm(^) = Iim^-l) = hmj- 1 = O.j
The converse also is true. In this case (L = 1), j is said to differ
from i by an infinitesimal of higher order.
For example, dy and AJ are of the same order if f'(jr] neither vanishes nor
becomes infinite. Then A// and AJ are of the same order, but A/y - dy is of higher
order than A.r. For this reason dij is called the "principal part of A?/." Obviously
powers of an infinitesimal i are of higher order than i.
ILLUSTRATIVE EXAMPLE. Prove the assumption of Art. 95,
Solution. In the figure we have, by geometry,
chord PQ <' arc PQ < PT + TQ.
Therefore, by division,
1 <
chord PQ " chord PQ chord PQ
Now
and hence
chord PQ = sec </> A.r, PT = sec r AJ, TQ = Ay - dy,
PT
A// - dy
Ax
chord PQ sec <' chord PQ
lim /_Z^_\ = lf ^limj TQ ^ \ = . Jim / ^rc PQ \ = L
Differentiate of higher order. Let // =/(j). The equation
defines the second differential of //. If //" neither vanishes nor becomes
infinite, d-y is of the same order as Ax 2 and therefore of higher order
than dy. In a similar manner d*y, - - , d n y may be defined.
PROBLEM
In triangle ABC the sides a, b, c are infinitesimals which simultaneously
approach zero, and r is of higher order than b. Prove lim r = !
CHAPTER X
Let us find the
CURVATURE. RADIUS AND CIRCLE OF CURVATURE
100. Curvature. In Art. 55 the direction of bending of a curve
was discussed. The shape of a curve at a point (its ilatness or sharp-
ness) depends upon the rate of change of direction. This rate is called
the curvature at the point and is denoted by K.
mathematical expression for K.
In the figure, P f is a second point on
a curve near P. When the point of con-
tact of the tangent line describes the
arc PP'(= As), the tangent line turns
through the angle AT. That is, AT is
the change in the inclination of the tan-
gent line. We now set down the follow-
ing definitions. ^ r
= average curvature of the arc PP'.
The curvature at P ( K) is the limiting value of the average curvature
when P' approaches P ax a limiting position, that is
(A)
K = lim = = curvature at P.
AS-. oAs ds
In formal terms the curvature is the rate of change of the inclination
with respect to the arc (compare Art. 50).
Since the angle AT is measured in radians and the length of arc As
in units of length, it follows that the unit of curvature at a point is one
radian per unit of length.
101. Curvature of a circle
Theorem. The curvature of a circle a,t any point equals the reciprocal
of the radius, and is therefore the same at
all points.
Proof. In the figure the angle AT be-
tween the tangent lines at P and P' equals
the central angle PCP' between the radii
As
_. = 1,
As ~ As ~As~ R
149
CP and CP'. Hence
AT angle
150 DIFFERENTIAL CALCULUS
since the angle PCP' is measured in radians. That is, the average
curvature of the arc PP' is equal to a constant. Letting As > 0, we
have the result stated in the theorem.
From the standpoint of curvature, the circle is the simplest curve,
since a circle bends at a uniform rate. Obviously, the curvature of a
straight line is everywhere zero.
102. Formulas for curvature ; rectangular coordinates
Theorem. When the equation of a curve is given in rectangular co-
ordinates, then
(B) /; = __,,
where y' and //" are, w>/w/?/W//, the Jirst and second derivative* of y
with respect to jr.
Proof. Since r arc tun //, (// ( -^)
\ dx'
differentiating, we have
(1 ) ' y l _/>!_ . P>y XXII, Art. 60
(Lc 1 -| //'"
Hut
(-) Y- (1 -h //''-' A l>.v (o>, Art. 95
Dividing (1) by (2) gives (B). Q.E.D.
EXKIUUSK. If ?/ is the indo])endent variable, show that
(O A' - "*" ,
(1 4- *' 2 )*
where j j and jc" are, respectively, the first and second derivatives
of .r with respect to //.
Formula (C) can be used as an alternative formula in cases where
differentiation with respect to y is simpler. Also, (B) fails when y' be-
comes infinite, that is, when the tangent at P is vertical. Then in (C)
.r' - and K - - .r".
Sign of K. Choosing the positive sign in the denominator of (B),
we see that A* and //" have like signs. That is, A' is positive or nega-
tive according as the curve is concave upward or concave downward.
ILLUSTKATIVIO EXAMPLE 1. Find the curvature of the parabola y 2 = 4 x
(a) at the point (1, '2); (Iri at the vertex.
> f 1 /'>\ '7 -,/'
Solution. ii' = -< v " = ^-(:) = -.rJK..
// ' dx\ij/ y 2
(a) When x = 1 and // ^ 2, then //' = !, y" = - L Substituting in
8
" = 8 \2~ 0.177. Hence at (1, 2) the curve is concave downward and the
RADIUS AND CIRCLE OF CURVATURE 151
inclination of the tangent is changing at the rute 0.177 radian per unit arc. Since
0.177 radian = 10 7', the angle between the tangent lines at P( 1. 2) and at a point
Q such that arc PQ ~ \ unit is approximately 10 .
(b) At the vertex (0, 0), //' becomes infinite. Hence use (C).
' I " I ( J-ll 1 r 1
2 ' 2 dy 2 2
ILLUSTRATIVE EXAMPLE 2. Find K for the cycloid (see Art. 81)
x a(0 sin 0], // (!-- cos 0).
Solution. In Illustrative Example 2, Art. 81, we found
, _. sin
.1 cos 6
>)
Hence 1 +?/'-- T -
1 cos
Also, in the Illustrative Example, Art. 82, it was shown that
_ 1
Substituting in (B), ' c(} ~ (>(>s 0) ~
1 1
2 a\ 2 2 cos 4 a sin \
103. Special formula for parametric equations. From equation
Art. 81, we have, by differentiation,
djr d-y
,- dy f dtdf 2 dtdl' 2
(lj ~j "^^ '
(ft)
Whence, using (B), Art. 82, and substituting in (B), Art. 102, and
reducing, we obtain
where the accents indicate derivatives with respect to /, ; that is,
, dx ,, d 2 x , di/ ,, d-y
'?' . x t II 11 ~ ~
w* Ti ** lj' '* ]i ' 7i'>
dt dt 2 di dt~
Formula (D) is convenient, but it is often better to proceed as in
Illustrative Example 2, Art. 102, finding //' as in Art. 81, '//" as in
Art. 82, and substituting directly in (B).
104. Formula for curvature ; polar coordinates
Theorem. When the equation of a curve is given in polar coordinates,
p2 . 2 P' 2 PP"
(E) K = P ^^ *f-,
(P 2 + P' 2 )" 2
152 DIFFERENTIAL CALCULUS
where p' and p" are, respectively, the first and second derivatives of p
with respect to 0.
Proof. By (/), Art. 85, r = + ^.
Hence
<> s-'+S-
Also, by (#), Art. 85, ^ = arc tan
Hence
Then, by (1),
Dividing (2) by (3) gives (). Q.E.D.
ILLUSTRATIVE EXAMPLE. Find the curvature of the logarithmic spiral p = e" 9
at any point.
Solution. ( -t~ - p' = ac at) = ap ; ^ = p" </->"" = a- p.
Substituting in (),
p\ 1 -f a-
105. Radius of curvature. The radius of curvature R at a point
on a curve equals ( he reciprocal of the curvature at that point. Hence,
from (),
(F) Jf = I == (l2l.
y
ILLUSTRATIVE EXAMPLE. Find the radius of curvature at any point of the
/ _ i\
catenary y =- -\r a -f r / (figure in Chapter XXVI).
/ - -\ / - r \
a . ?/ ' = I\ ( , ._ ,, uj . y " = J_ \ ffl ^_ tf a) = JL.
2 2 a a-
1 -fl/' 2 ^ 1 +7\f"-f / =Mfa+f
2
Solution.
= - Ans.
a
106. Railroad or transition curves. In laying out the curves on a
railroad it will not do, on account of the high speed of trains, to pass
abruptly from a straight stretch of track to a circular curve. In
order to make the change of curvature gradual, engineers make use
of transition curves to connect the straight part of a track with a
RADIUS AND CIRCLE OF CURVATURE 153
circular track. This curve should have zero curvature at its point
of junction with the straight track and the curvature of the circular
track where it joins the latter. Arcs of cubical parabolas are generally
employed as transition curves.
ILLUSTRATIVE EXAMPLE. The transition curve on a railway track has the shape
of an arc of the cubical parabola // = J ,r l . At what rate is a car on this track
changing its direction (I mi. = unit of length) when it is passing through (a) the
point (3, 9)? (b) the point (2, J)? (c) the point (1, J)?
Solution. ^ = ^. ^ = 2x.
dx dx 2
Substituting in (B), A" =
(1 -f
(a) At (3, 9), K = radian per mile = 28' per mile. ATM.
(812)2
(b) At (2, 5), K = -~ radian per mile = 3 16' per mile. Ans.
(17)3
(c) At (1 , I \ K = -^ = U radian per mile - 40" 30' per mile. Am.
(2)2 V2
107. Circle of curvature. Consider any point /' on the curve (\
The tangent line drawn to the curve at P has the same slope as the
curve itself at P (Art. 42). In an analogous /0
manner we may construct for each point of the
curve a tangent circle whose curvature is the /
same as the curvature of the curve itself at that ',
point. To do this, proceed as follows. Draw \
the normal to the curve at P on the concave
side of the curve. Lay off on this normal the dis-
tance PC = radius of curvature (= K) at P. With c as a center draw
the circle passing through P. The curvature of this circle is then
*-*
which also equals the curvature of the curve itself at P. The circle
so constructed is called the circle of curvature for the point P on
the curve.
In general, the circle of curvature of a curve at a point will cross
the curve at that point. This is illustrated in the above figure.
(Compare with the tangent line at a point of inflection (Art. 57).)
Just as the tangent at P shows the direction of the curve at P, so
the circle of curvature at P aids us very materially in forming a geo-
metric concept of the curvature of the curve at P, the rate of change
of direction of the curve and of the circle being the same at P.
154
DIFFERENTIAL CALCULUS
In a subsequent section (Art. 114) the circle of curvature will
be defined as the limiting position of a secant circle, a definition
analogous to that of the tangent
given in Art. 28.
ILLUSTRATIVE EXAMPLE. Find the ra-
dius of curvature at the point (U, 4 ) on the
equilateral hyperbola xy 12, and draw the
corresponding circle of curvature.
~ i * dy
Solution. - -
ax
d^y 2 v
~~l - *.
dx- jr*
For W. 4).
ix .'J dx' 2 9
. K Jl^lLi*j=5..v
M -4
The circle of curvature crosses the curve at two points.
iLLi'STHATivE K.VAMi'LE 2. Find /i at (2, 1 ) for the hyperbola
j j -f 4 /// - 2 //-' - 10.
Solution. I )iiferent iiitui^, regarding // us an implicit function of x, we get
.r } 2 // f 2 JTIJ' - 2 ////' - ().
Differentiating again, regarding ij and //' as implicit functions of x, we get
1 4- 4 //' - 2 .//"-' -f 2(.r - //)//' - 0.
Substituting the given values .r 2, // 1, we find ?/' -- 2, y" - l --.
Hence, by (F), U - ?| \/5. .\.s.
The method of this example (namely, regarding // and //' as implicit functions
of .r) can often be used to advantage when only the numerical values of ?/' and y"
are required, and not general expressions for them in terms of x and y.
PROBLEMS
Find the radius of curvature for each of the following curves at the
point indicated. I )raw the curve and the corresponding circle of curvature.
1. 2 // -- .r j ; i(), 0). An*. 7i = 1.
2. 6 // = .r ; ; (2, .',). ft = g Vs.
3. ?/ L> ~ .r ; ; (1, 1).
4. // sin .r ; ^ TT, 1).
5. // = r j ; (0, 1).
6. .r 2 - 4 // 2 = 9 ; (5, 2). 8. y = 2 sin 2 x ; ( J TT, 2).
7. // 2 = ^-f8; (1,8). 9. y = tanx; (| TT, 1).
RADIUS AND CIRCLE OF CURVATURE 155
Calculate the radius of curvature at any point (.r\, y\] on each of
the following curves.
10.,, = ^. A,:,. K =
(>
11. jr' = 2 JU-.
12. Wx* - aW - a*b*. *
13. b-jr- -f a-y- a' 2 b' 2 .
14. ,* + ** = *. /, = ^' +/>'
a*
15. j-3 + 2/ S = a I A* = 3 (an //,)*.
16. x = r arc vers - V2 rz/ z/~. /i 2\/2 n/i.
17. ?y In sec x. A* ~ sec jv
18. If the point of contact of the tangent line at (2, 4) to the parabola
?/* 8 x moves along the curve a distance As ().l, through what angle,
approximately, will the tangent line turn? (Use differentials. )
19. The inclination of the curve 27 // .r { at the point .\(IJ, 1) is 45.
Use differentials to find approximately the inclination of the curve at the
point H on the curve such that the distance along the curve from A to B
is A.s- = 0.2 units.
Calculate the radius of curvature at any point (pi, (M on each of
the following curves.
20. The circle p = a sin 0. An*. R \ a.
3
21. The spiral of Archimedes p - ad. (Fig., p. 5:M) K = -^ _^ 2 ^
22. The cardioid p = a (I - cos 0). (Fig., p. 5153; # = 3^ a Pi-
a 2
23. The lemniscate p 2 = a 2 cos 2 0. (Fig., p. 5.'52) A' = ^-^
24. The parabola p = a sec 2 J 0. (Fig., p. f>:J7) R = 2a sec 3 J 0j.
25. The curve p = a sin 3 J 0. ^ = 4 sin 2 J 0^
m r> _flf5-4cos0,) j
26. The tnsectnx p = 2 a cos a. ^ ( fi - 7
27. The equilateral hyperbola p 2 cos 2 = a 2 .
a(l - e 2 ) p aft
'
.3
oo TU -
28. The conic p =
-
1 e cos fl - -' cos L ) {
156 DIFFERENTIAL CALCULUS
Find the radius of curvature for each of the following curves at the
point indicated. Draw the curve and the corresponding circle of curvature.
29. jr = 2 /, ?/ = t 2 - 1 ; / = 1. A MS. ft = 4>/2.
30. jr = 3 t 2 , // = 3 / - / : < ; t = 1. tf = .
31. .r = 2 r ', // = 2 f ' ; / = 0. 72 = 2 V2.
32. -r = a cos /,, ?/ = a sin / ; / t\. ft a.
33. jr ~ 2 /, ?/ = - ; / - 1 . 36. jr = 2 sin /, ;// = cos 2 / ; t = TT.
34. .r - /- -f 1, // = t ' - 1 ; / = 1 . 37. x - tan /, // = ctn / ; / - \ IT.
35. x 4 cos /, ?/ = 2 sin /;// = !. 38. r / sin /, // 1 cos / ; / TT.
39. Find the radius of curvature at any point (f = /i ) on the hypocy-
cloid j = a cos- 1 /, // a sin :! /. ,4//,s. ft ;i a sin /i cos /i.
40. Find the radius of curvature at any point (t = t\) on the involute
of the circle .r = aloe* / + / sin /;,
// ~ a (sin / t cos /). An*, ft at\.
41. Find the point on the curve ?/ = r' where the curvature is a
maximum. An$. s 0.347.
42. Find the points on the curve 3 y ~ .r ! 2 x where the curvature
is a maximum. ,4^/.^. jr 0.931.
43. Show that the radius of curvature becomes infinito at a point of
inflection.
44. Given the curve // 3 jr .r\
(a) Find the radius of curvature at the maximum point of the curve
and draw the corresponding circle of curvature.
(b) Prove that the maximum point of the curve is not the point of
maximum curvature.
(c) Find to the nearest hundredth of a unit the abscissa of the point
of maximum curvature. An*, .r 1.01.
45. Find the radius of curvature at each maximum and minimum
point on the curve // ~ x 4 2 .r L> . Draw the curve and the circles of cur-
vature. Find the points on the curve where the radius of curvature is a
minimum.
46. Show that at a point of minimum radius of curvature on the
curve y = f(x] we have
'^Vl
,dx) \
47. Show that the curvature of the cubical parabola 3 a 2 ?/ = x 3 in-
creases from zero to a maximum value when x increases from zero to
4 _____
I aV125. Find the minimum value of the radius of curvature.
An*. 0.983 a.
RADIUS AND CIRCLE OF CURVATURE
157
108. Center of curvature. The tan pent line at P(r, y) has the
property that x, y, and y f have the same values at P for the tangent
line and the curve. The circle of curvature at P has a similar prop-
erty; namely, x, y, y', and //" have the same values at P for the
circle of curvature and the curve.
DEFINITION. The center of curvature (a, /j) for a point P(x, y) on
a curve is the center of the circle of curvature.
Theorem. The coordinates (a, /j) of the center of curvature for
P(x, y) are
y y
Proof. The equation of the circle of curvature is
(1) Cr-a^+fo-jCO-^/i'-,
\\here K is given by (F). Differentiating (1),
/ x a ,, R~
(2)
?/" = -
y-P " (y
From the second of these equations, after substituting the value
of R from (F), we obtain
- , ' y* I
From the first of equations (2), we get, using (3),
Solving in (3) for /3, in (4) for a, we
have (G). Q.E.D.
EXERCISE 1. Work out (G) di-
rectly from the accompanying figure,
using (G), Art. 95. (a = x R sin r,
/3 = y -f- R cos r, etc.)
EXERCISE 2. If x' and x" are, re-
spectively, the first and second deriv-
atives of x with respect to y, derive
(G) in the form
a ~ x" x"
Formulas (H) may be used when y' becomes infinite, or if dif-
ferentiation with respect to y is simpler.
158
DIFFERENTIAL CALCULUS
ILLUSTRATIVE EXAMPLE. Find the coordinates of the center of curvature of
the parabola y' 2 4 px corresponding (a) to any point on the curve; (b) to the
vertex.
1 "'
Solution. Use (//). Then x' = -j- . x" =
y' 2 + 4
Hence a = x
= 3 x + 2 p,
Therefore (a) ( 3 x -f 2 p, ~~~^ t } Ls the center of
V 4 p-/
curvature corresponding to any point on the curve.
(b) (2 p, 0) is the renter of curvature corresponding
to the vertex (0, Oj.
From Art. 57 we know that at a point of inflection (as Q in the
next figure)
Therefore, by (B), Art. 102, the curvature K = 0; and from (F),
Art. 105, and (G), Art. 108, we see that in general a, 0, and R in-
crease without limit as the second derivative approaches zero, unless
the tangent line is vertical. That is, if we sup-
pose P with its tangent to move along the curve
to P', at the point of inflection Q the curvature
is zero, the rotation of the tangent is momen-
tarily arrested, and as the direction of rotation
changes, the center of curvature moves out in-
definitely and the radius of curvature becomes
infinite.
109. Evolutes. The locus of the centers of
curvature of a given curve is called the evolnte
of that curve. Consider the circle of curva-
ture at a point P on a curve. If P moves along the curve, we may
suppose the corresponding circle of curvature to roll along the curve
with it, its radius varying so as to be always equal to the radius of
curvature of the curve at the point P. The curve CCi described by
the center of the circle is the evolute of PP 7 .
Formulas (G) and (77), Art. 108, give the coordinates of any point
(a, /3) on the evolute expressed in terms of the coordinates of the
corresponding point (x, y) of the given curve. But y is a function
of jc ; therefore these formulas give us at once the parametric equation
of the evolute in terms of the parameter x.
RADIUS AND CIRCLE OF CURVATURE
159
To find the ordinary rectangular equation of the evolute we elimi-
nate x and y between the two expressions and the equation of the
given curve. No general process of elimination can be given that
will apply in all cases, the method to be adopted depending on the
form of the given equation. In a large number of cases, however,
the student can find the rectangular equation of the evolute by taking
the following steps.
General directions for finding the equation of the evolute in rec-
tangular coordinates.
FIRST STEP. Find a and /3from (G) or (H), Art. 108.
SECOND STEP. Solve the two resulting equations for y and y in terms
of a and /3.
THIRD STEP. Substitute these values of x and y in the given equation
and reduce. This gives a relation between the variables a and (3 which
is the equation of the evolute.
ILLUSTRATIVE EXAMPLE 1. Find the equation of the evolute of the parabola
V=4p*.
- , , C,,l
dx y dx-
a = 3 ar + 2 p, = -
Solution.
First Step.
Second Step.
Third Step. (4
Remembering that a denotes the abscissa and p the
ordinate of a rectangular system of coordinates, we see
that the evolute of the parabola AOB is the semicubical parabola DC'E, the
centers of curvature for O, P, PI, P> 2 being at C', C, d, C 2 respectively.
160
DIFFERENTIAL CALCULUS
ILLUSTRATIVE EXAMPLE 2. Find the equation of the evolute of the ellipse
^2 X 2 4. n -iyi = aw.
c . ..
Solution.
-- = -- .
dc a 2 !/
T-T
a 2 ?/ 3
p- / ei
/" tm Mtfp. a =
Second Step, x
./ 1 <* \*
"\a 2 - b' z )
(au; + (/^j = (a 2 - 6 2 )% the
c?quation of tho evolute KHK'H' of the ellipse
A^/l'/^'. /, A 1 ', ^/', // are the centers of curvature corresponding to the points
A, A', K, It', on the curve, and (\ ( v , ("" correspond to the points P, P', P".
ILLUSTRATIVE EXAMPLE a. The paranietric equations of a curve are
(1)
/ 2 4- 1
: 4 '
Find the equation of the evolute in parametric form, plot the curve and the
evolute, find the radius of curvature at the point where / = 1, and draw the cor-
responding circle of curvature.
X
Solution.
Substituting in (G) and reducing gives
(2) ^^^1
By (A), Art. 81
By (B), Art, 82
-f 3 t
calculate x, ^ from (1) and a, j8 from
(2), and tabulate the results.
Now plot the curve and its evolute.
The point (J, 0) is common to the
given curve and its evolute. The given
curve (semicubical parabola) lies en-
tirely to the right and the evolute
entirely to the left of x J.
The circle of curvature at A ( Ji , Jl ) ,
where f = 1, will have its center at
A'( k, 1,} on the evolute and ra-
dius = AA'. To verify our work find
the radius of curvature at A. From
(F), Art. 105, we get
LE OF CURVATURE 161
Assuming values of the parameter t, we
t ! x | y i
ft
.> '
o 1.
'* i
2 I
L } ,
i , _: n
- Y
f !
~ 3 , 1
i -. 3 a
i
',
i i
r,
!'
() >
1 ,
r, i a
r.
i 1
A I -Si
;*
4 : 3 n
3 l| ~4~
.. !i
''< 5 i
This should equal the distance
AA' = "
By (1), Art. 3
ILLUSTRATIVE EXAMPLE 4. Find the parametric equations of the evolute of the
cycloid
V ;
r.r = </</
COS/).
Solution. As in the Illustrative Example of Art. cSl?, we get
d// _ sin / d' 2 y 1
dx 1 cos/ dx~ a(\ cos/) 2
Substituting these results in formulas (G), Art. 108, we get
(4)
f a ait 4 sin /),
1 jg - t/d - cos /). Ann.
NOTE. If we eliminate / between equations (4), there results the rectangular
equation of the evolute OO'Q* referred to the axes O'a and O'fi. The cocirdinates
of with respect to these axes are (~ mi, - 2 a). Let us transform equations (4) to
the new set of axes OX and OY. Then
a x ?ra, ft = y 2 a.
Also, let t t' TT.
Substituting in (4) and reducing, the ^
equations of the evolute become
(5)
x = a(t' - sin /'),
y = a(l cos /')
Since (5) and (3) are identical in o
form, we have:
The evolute of a cycloid is itself a cycloid whose generating circle equals that of the
given cycloid.
162
DIFFERENTIAL CALCULUS
110. Properties of the evolute. The evolute has two interesting
properties.
Theorem 1. The normal at P(x, y) to the given curve is tangent to
the evolute at the center of curvature C(a, p) for P. (See figures in the
preceding article.)
Proof. From the figure,
(Ij a = x Rsin T,
ft = y + R COS T.
The line PC lies along the normal at
P, and the
(2) Slope of PC -
x a tan r
slope of normal at P.
We show now that the slope of the evolute equals the slope of PC.
Note that j/j
Slope of evolute = -
da
since a arid fi are the rectangular coordinates of any point on the
evolute.
l^et us choose as independent variable the length of arc on the
given curve ; then x, //, R, T, a, are functions of 8. Differentiating
(1) with respect to s gives
(8)
(4)
da d'jr
dr
dR
)
-- # cos r __ sin T J
as ds as ds
rf/i dy . dr , rf/2
j = j ~~ ^ sm T T" H- cos r ~T~'
ds ds ds ds
But ( ~r = cos T, -^ = sin r, from Art. 95 : and - = .
rf.s ds ds R
Substituting in (3) and (4), and reducing, we obtain
(5)
Dividing the second equation in (5) by the first gives
da dR
= - sin r
rfs r/.s*
rf/J dR
~ = COST-T--
(6)
= - ctn T - --
da. tan r
= slope of PC.
Q.E.D.
Theorem 2. The length of an arc of the evolute is equal to the dif-
fer erne between the radii of curvature of the given curve which are tangent
to this arc at its extremities, provided tfiat along the arc of the given curvt
R increases or decreases.
RADIUS AND CIRCLE OF CURVATURE
163
Proof. Squaring equations (5) and adding, we get
<" (DXiNff-
But if s' = length of arc of the evolute,
ds'~ da- + dfi-,
by (C), Art. 95, if s = s', x = a, y = /i Hence (7) asserts that
dff
(8)
= ^T ' or i =
ds
Confining ourselves to an arc on the given curve for which the
right-hand member does not change sign, we may write
(9)
= 4-1 or -,~ = -
dR~ ' ^ w dtf
That is, Jfee rate of change of the arc of the evolute with respect to R
i s _|_ i or _ i. Hence, by Art. 50, corresponding increments of s' and
R are numerically equal. That is,
(10) s' - s' = (R- flu),
or (first figure, p. 159) Arc CCi = (PiCi - PC).
Thus the theorem is proved.
In Illustrative Example 4, Art.109, we observe that at O', R = ;
at P v , R = 4 a. Hence arc O'QQ V = 4 a.
Tfce length of one arch of the cycloid (as OO'Q V ) / eight times the
length of the radius of the general ing circle.
111. Involutes and their mechanical construction. Let a flexible
ruler be bent in the form of the curve r,r.,, the evolute of the curve
PI Po, and suppose a string of length
Ru, with one end fastened at Co, to
be stretched along the ruler (or
curve). It is clear from the results
of the last article that when the
string is unwound and kept taut, the
free end will describe the curve PjP<>.
Hence the name evolute.
The curve PiP 9 is said to be
an involute of CiCo. Obviously any
point on the string will describe an
involute, so that a given curve has
an infinite number of involutes but
only one evolute.
164 DIFFERENTIAL CALCULUS
The involutes PiP*, /Y/V, /Y'/V are called parallel curves since
the distance between any two of them measured along their common
normals is constant.
The student should observe how the parabola and the ellipse on
pages 159, 160 may be constructed in this way from their evolutes.
PROBLEMS
Find the radius and center of curvature for each of the following
curves at the given point. Check your results by proving (a) that the
center of curvature lies on the normal to the curve at the given point,
and (b) that the distance from the given point to the center of curvature
is equal to the radius of curvature.
1. 2py = x' 2 ; (0, 0). Am. (0, p).
2. x' 2 + 4 y' 2 = 25 ; (3, 2). (iVb ~ H)-
3. r j -i/ ; '= 19 ; (3, 2). (W, 5?)-
4.x?/ = 6; (2, 3). (-?i )
5. // = <"; (0, 1). (- ~ 3 )-
6. y = cosr ; (0, 1). (0, 0).
7. ?/ = lnr; (1, 0). ( 3 ~ 2) -
8. = 2 sin 2. r; (J TT, 2). (i TT, )-
9. (/ -f 6) :i + ^?/ 2 = ; (-3, 3). (- 13, 8).
10. 2?/ = x 2 -4; (0, -2).
11. xy = x' 2 + 2; (2, 3).
12. y = sin TT.T ; (J, 1).
13. ?/ = -?, tan2o-; (-J TT, J).
Find the coordinates of the center of curvature at any point (x, y)
of each of the following curves.
A 3 <r -f 2
14. ?/ 2 = 2 '
a 4 -f 15 ?/ 4 ,. a 4 ?/ - 9
15. ^ = a^. a= 2 - . P= 4
16. 6 2 x 2 - a 2 ?/ 2 = a 2 5 2 . a =
2 +
17. j.8 + y = a. a = x 4- 3 x?/,
RADIUS AND CIRCLE OF CURVATURE 165
18. Find the radii and centers of curvature for the curve xy 4 at the
points (1, 4) and (2, 2). Draw the arc of the evolute between these cen-
ters. What is its length ? __
Ans. At (1, 4), A>, = V Vl7, a = V, = V ;
at (2, 2), A> 2 = 2 \/2, a = 4, = 4;
A>, - ft, = 5.938.
Find the parametric equations of the evolute of each of the following
curves in terms of the parameter t. Draw the curve and its evolute, and
draw at least one circle of curvature.
19. x = 3/ 2 , = 3f-? :i . Am. a = %(l + 2 t' 2 - /*), = - 4 <.
20. x = 3 t, # = f 2 - 6. a = - J f :l , = 3 /-' - jj.
21. x = 6 - * 2 , y = 2 f. t* = 4 - 3 / 2 , = - 2 P.
22. a- = 2 /, i/ = J 2 - 2. a - - 2 /*, = 3 f 2 .
23. x = 4 f, y = 3 4- J 2 . = - /\ 0=11 + 3 / 2 .
24. a- = 9 - f 2 , ?/ = 2 f. = 7 - 3 / 2 f = - 2 P.
3 12/ 4 + 9 . 27 + 4 / 4
25.* = 2*, ?/ = r - 4 t .> ' ^= el
/^2 fo'2\
26. j* = a cos f, ?/ = ?> sin L a - cos :i /,
z
27^ x _ a COS 3 ^ a a cos-* / + 3 a cos / sin 2 ,
?y = a sin :i f. = '* " cos 2 / sin / + a sin :{ iL
28. x = a(cos t + / sin /), ot = a cos /, = a sin t.
y = a(sin / t cos /).
29. x = 4 - / 2 , ?/ = 2 /.
30. JT = 2 /, ?/ = 16 - / 2 .
31. x = / 2 , ?y = i ^.
32. x = 1 cos f, y = t - sin /.
33. x = cos 4 /, ?/ = sin 4 /.
34. x = a sec , ?/ = b tan f.
35. x = cos t, y = .
36. x = 6 sin t, y = 3 cos J.
37. x = 3 esc /, i/ = 4 ctn J.
38. x = a(t + sin 0-
7/ = a(l cos/)-
39. x = 2 cos t + cos 2 i.
y = 2 sin ? + sin 2 ^.
166 DIFFERENTIAL CALCULUS
40. Show that in the parabola x* H- y 2 = a 2 we have the relation
+ (3 = X(x + y).
41. Given the equation of the equilateral hyperbola 2 xy = a 2 , show
that - *
2
From this derive the equation of the evolute
(+ 0)1 - (a- 0)3 = 2 a*.
112. Transformation of derivatives. Some of the formulas derived
above independently can be deduced from others by formulas which
establish relations between derivatives. Two cases will be presented
here.
Interchange of dependent and independent variables.
NOTATION. Let ?/' = 2/" = ! = f, etc.,
, dx ,, dx' d 2 x .
x' = , x" = -, etc.
dy dy dy z
By IX, Art. 29,
XT // 'f
Now y" -~~ = -7
dx x'
Using (/), we Ret ^ - - ~ >
(/// x' 2
dy"
Again, y'" = *f = *3L.
dx x/
A,tt ,y.r~rrf Q ,
Using (/), ?- = - ~
dy x' 4
v ' v '" O v "|
.-.y'" = -i X -*x '
Y '5
And so on for higher derivatives. By these formulas equations in
y'> y", '//'"> etc. can be transformed into equations in x 1 , x", x'", eta
RADIUS AND CIRCLE OF CURVATURE 167
ILLUSTRATIVE EXAMPLE. Transform (B), Art. 102, into (C) in that article.
Solution. Using (/) and (/) above,
1L
x"
m '
Transformation from rectangular to polar coordinates. Given the
relations
(1) x = p cos 6, y = p sin 6
between the rectangular and polar coordinates of a point. If the
polar equation of a curve is p=f(0), then equations (1) are para-
metric equations for that curve, being the parameter.
NOTATION. The independent variable is 0, and x', .r", y', y", p', p"
denote successive derivatives of these variables with respect to 0.
Differentiating (1),
(2) x' = p sin + p' cos 0, y' = p cos + p' sin ;
(3) z" = -2p'sin0 + (p"-p)cos0, <//"-2p'cos0 + (p"-p)sin0.
By formulas (1), (2), (3), equations in j, ?/, x', ?/, r", ?/" may be
transformed into equations in p, 0, p', p".
ILLUSTRATIVE EXAMPLE. Derive (), Art, 104, directly from (/)), Art. 10,'J.
Solution. Taking numerator and denominator in (D) separately, substituting
from (2), (3), and reducing, we obtain the results
X 'y _ y> x " = p + 2 p' 2 - PP" ; ^ /1! + //" 2 = P" + P /2 -
Putting these values in (D) gives ().
PROBLEMS
In Problems 1-5 interchange the dependent and independent variables,
4 d 2 * /dx\ 2 A
Ans. z -?/(-7-) =0.
rf V(]f
4
168
DIFFERENTIAL CALCULUS
dy
~
6. Transform
by assuming x = p cos 6, y = p sin 6.
A~<. P 2
~ - - - - =
7. Transform the equation y~ - - ( -~ -f - - - = by assuming
8. Transform the equation s 2 ~? + 2 .c ~^ -f ^7 y = by assuming .r = -
cLr 2 dr .r- ' /
A d' 2 ij . v A
Ans. -7*7 4- a v = 0.
~
ADDITIONAL PROBLEMS
1. Given the curve or 3 cos f 4- cos 3 t, y = 3 sin t sin 3 t. Find the
parametric equations of the evolute. Find the center of curvature for
/ = arid show that it coincides with the corresponding point on the given
curve. Ans. a = 6 cos i - 2 cos 3 t, = 6 sin t 4- 2 sin 3 t.
2. If It is the radius of curvature at any point of the ellipse
b 2 jc 2 -f a*// 2 = a' 2 b' 2 and D the perpendicular distance from the origin to
the tangent drawn at this point, prove that RD* = a 2 b 2 .
3. Find the equations of the evolute of the parabola ?/ 2 = 4 x t using x
as a parameter. Find the points of the parabola for which the correspond-
ing centers of curvature are also points of the parabola. Hence find the
length of the part of the evolute inside the parabola.
Ans. (2, 2V2); 4(V7 - l).
4. (a) At every point (jc, y} of a certain curve, its slope is equal to
?/ and the curve passes through the point (2, 0). Verify that the
equation of the curve is log (1 -f- y) = 1 V5 jc 2 .
(b) Find the curvature of the curve at this point and draw a small
portion of the curve near it. Ans. K = oVVH.
(c) Draw the circle of curvature at this point. Ans. cv = f , ft = |.
5. The slope of the tangent to a certain curve C at any point P is
given by -jr- = - where .s is the length of arc (measured from some fixed
point) and a is a constant. The center of curvature of C at P is P 1 '. De-
note the radius of curvature of C at P by R and the radius of curvature of
the evolute of C at P' by 7i'. Prove
R =
a 2
CHAPTER XI
THEOREM OF MEAN VALUE AND ITS APPLICATIONS
113. Rolle's Theorem. A theorem which lies at the foundation of
the theoretical development of the calculus will now be explained.
Let y=J(x) be a single-
valued function of x, con-
tinuous throughout the in-
terval [a, 6] (Art. 7) and
vanishing at its extremities
(/(a) = 0, f(b) = 0). Sup-
pose also that /(x) has a
derivative/' (2) at each in-
terior point (a < x < 6) of
the interval. The function will then be represented graphically by
a continuous curve as in the figure. Geometric intuition shows us at
once that for at least one value ofx between a and b the tangent is par-
allel to the x-axis (as at P) ; that is, the slope is zero. This illustrates
Rolle's Theorem, ///(x) is continuous throughout the interval [a, b]
and vanishes at its extremities, and if it has a derivative /'(x) at every
interior point of the interval, then f'(x) must vanish for at least one
value of x between a and b.
The proof is simple. For/(x) must be positive or negative in some
parts of the interval if it does not vanish at all points. But in this
special case the theorem is obviously true. Suppose, then, that /(x)
is positive in a part of the interval. Then /(x) will have a maxi-
mum value at some point within the interval. Similarly, if f(x)
is negative, it will have a minimum value. But if f(X) is a maxi-
mum or minimum (a < X < b), then f'(X) = 0. Otherwise, /(x)
would increase or decrease as x passes
through X (Art. 51).
The figure illustrates a case in which Rolle's
Theorem does not hold ; f(x) is continuous through-
out the interval [a, 6]. f'(x), however, does not
exist for x = c, but becomes infinite. At no point
of the graph is the tangent parallel to the x-axis.
169
170
DIFFERENTIAL CALCULUS
We give first two applications of Rolle's Theorem to geometry.
114. Osculating circle. If a circle be
drawn through three neighboring points
Po, Pi, P'2 on a curve, and if Pj and
P 2 be made to approach P f , along the
curve as a limiting position, then this
circle will in general approach in mag- t x
nitude and position a limiting circle
caUed the osculating circle of the curve
at the point PO.
Theorem. The osculating circle is identical with the circle of curvature.
Proof. Let the equation of the curve be
U) ?/=/(*);
and let x ( , xi, x>> be the abscissas of the points Po, PI, Po respectively,
(a', fi f ) the coordinates of the center, and R f the radius of the circle
passing through the three points. Then the equation of the circle is
and since the coordinates of the points P , PI, P-2 must satisfy this
equation, we have
(2)
Now consider the function of x defined by
F(*) = (x~ a') 2 + (y - i3 f ) 2 - R' 2 ,
in which y is defined by (1).
Then from equations (2) we get
) = 0, F(X!) = 0, F(x 2 ) = 0.
Hence, by Rolle's Theorem (Art. 113), F'(x) must vanish for at
least two values of x, one lying between xo and x\, say x f , and the
other lying between xi and x^ say x" ; that is,
F'(x') = 0, F'(x") = 0.
Again, for the same reason, F"(x] must vanish for some value of
x between x' and x" , say x-^ ; hence
F"(z 3 ) = 0.
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 171
Therefore the elements a', 0', R f of the circle passing through the
points Po, PI, P2 must satisfy the three equations
F(zo) = 0, F'(x') = 0, F"(.r.i) = 0.
Now let the points PI and P L > approach P as a limiting position ;
then xi, ? 2 , x', x", x 3 will all approach :r () as a limit, and the elements
a, /3, R of the osculating circle are therefore determined by the three
equations F(* ) = 0, F'(a- ) = f F"(x )=0;
or, dropping the subscripts, by
(3) (x - a) 2 + (y - 0) 2 = ff-',
(4) (x - a) + (y - ]8)'//' - 0, differentiating (3).
(5) 1 + if 2 + (y - J3)y" = 0, differentiating (4).
Solving (4) and (5) f or x a and ?/ 0, we get (?/" ^ 0),
(6) x a = > 7/ rf ~ ^-
?/ , '//
Solving (6) for a and 0, the result, is identical with (G), Art. 108.
Substituting from (6) in (3), and solving for R, the result is (F),
Art. 105. Hence the osculating circle is identical with the circle of
curvature.
In Art. 28 the tangent line at P was defined as the limiting posi-
tion of a secant line drawn through P and a neighboring point Q
on the curve. We now see that the circle of curvature at P may be
defined as the limiting position of a circle drawn through P and two
other points Q, R on the curve.
115. Limiting point of intersection of consecutive normals
Theorem. The center of curvature (^ for a point P on a curve is the
limiting position of the intersection of the normal to the curve at P with
a neighboring normal. C(a f /*j
Proof. Let the equation of a curve be c[ <tt '^
(1) =/(*)
The equations of the normals to the
curve at two neighboring points Po and
D ^ A'.^o'l/o)
^ are Oo - y) + ('//o - ?/)/'Oo) = 0,
If the normals intersect at C'(a', /3'), the coordinates of this point
must satisfy both equations, giving
172 DIFFERENTIAL CALCULUS
Now consider the function of x defined by
<t>(x) = (x - a') + (y- p')y' f
in which y is defined by (1).
Then equations (2) show that
</>(z ) = 0, 0(xi) = 0.
But then, by Rolle's Theorem (Art. 113), <'(j) must vanish for
some value of x between XQ and x\, say x'. Therefore a' and /3' are
determined by the two equations
If now PI approaches P as a limiting position, then x' approaches
x , giving
and f"(a', 0') will approach as a limiting position a point C(a, ]8)
on the normal at /V Dropping the subscripts and accents, the last
two equations are
(x-a) + (y- &)y' = 0,
1 + y' 2 + (y - fry" = 0.
Solving for a and /3, the results are identical with (G), Art. 108. Q.E.D.
116. Theorems of Mean Value (Laws of the Mean). For later appli-
cations we need the
Theorem. ///Or) and F(x) and their first derivatives are continuous
throughout the interval [a, 6"|, and if, moreover, F'(x) does not vanish
within the interval, then for some value x = xi between a and 6,
--
Proof. Form the function
Evidently </>() = </>(?>) = 0, and Rolle's Theorem, Art. 113, may
be applied. Differentiating,
This must vanish for a value x xi between a and 6.
~ /(a) r*/
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 173
Dividing through by F'(XI) (remembering that F'(XI) does not
vanish), and transposing, the result is (4). Q.E.D.
If F(x) = x, (A) becomes
b - a J v A>
In this form the theorem has a simple *
geometric interpretation. In the figure
the curve is the graph of /(x). Also,
OC = a, CA=/(a),
OD = b, DB=f(b).
Hence
r/_\
of chord AB.
b a
Now f'(xi) in (B) is the slope of the curve at a point on the arc
AB, and (B) states that the slope at this point equals the slope of
AB. Hence there is at least one point on the arc AB at which the tan-
gent line is parallel to the chord A B.
The student should draw curves (as the first curve in Art. 113)
to show that there may be more than one such point in the interval,
and curves to illustrate, on the other hand, that the theorem may
not be true if f(x) becomes discontinuous for any value of x between
a and 6, or if f'(x) becomes discontinuous (as in the second figure of
Art. 113).
Clearing (B) of fractions, we may also write the theorem in the form
(C) /(&)=/(<*) + (b - <0/' (*0-
Let 6 = a -r Aa ; then b a = Aa, and since x\ is a number lying
between a and 6, we may write
Xl =a + 6- Aa,
where 6 is a positive proper fraction. Substituting in (C), we get
another form of the Theorem of Mean Value,
(D) f(a + Aa) -/(a) = Aa/' (a + Aa). (0 < < 1)
PROBLEMS
1. Verify Rolle's Theorem by finding the values of x for which f(x)
and/'fa) vanish in each of the following cases.
(a) f(x) = x 3 - 3 x. (e^ f(x) = sin irx - cos TTX.
(b) f(x) = 6 x 2 - x 3 . (f) f(x) = tan x - x.
(c) f(x) = a + bx -f ex 2 . (g) f(x) = x In x.
(d) f(x) = sin x. (h) /(x) = xe*.
174 DIFFERENTIAL CALCULUS
2. Given f(x) = tan x. Then /(O) = and /(ir) = 0. Does Rolle's
Theorem justify the conclusion that f'(jc) vanishes for some value of x
between and TT? Explain your answer.
3. Given (y -f l) :i = x 2 . Then y = when x = 1 and // = when
x = + 1. Does Rolle's Theorem justify the conclusion that y' vanishes
for some value of x between - 1 and + 1 ? Explain your answer.
4. In each of the following cases find x\ such that
f ( b) = /() + (b-a)f'(jci).
(a) f(x) = x-, a 1, b 2. Ans. Xi = 1.5.
(b) /(x) = Vx, a = 1, b = 4. x, = 2.25.
(c) /(x) = 6", a = 0, b = 1. Xi = In (<? - 1) = 0.54.
(dj /(x) - -, r; = 1, 6 = 2.
(e) /(x) = In x, a 0.5, b 1.5.
5. Given f(x) = - a - 1,6 = 1. For what value of Xi, if any, will
x
f(b) = /() + (b- )/'(x!)?
6. Given /(x) = x^, a = 1, /> 1. For what value of x,, if any, will
/(ft) =/(*;) -f (fc-aWi)?
117. Indeterminate forms. When, for a particular value of the
independent variable, a function takes on one of the forms
L\ 2?, X oo, oo - oo, 0, oc", I- t
it is said to be jndetcrvmuite, and the function is 'not defined for that
value of the independent variable by the given analytical expression.
For example, suppose we have
where for some value of the variable, as x = a,
/(a) - 0, F(a) = 0.
For this value of x our function is not defined and we may there-
fore assign to it any value we please. It is evident from what has
gone before (Case II, Art. 17) that it is desirable to assign to the
function a value that will make it continuous when x = a whenever
it is possible to do so.
118. Evaluation of a function taking on an indeterminate form. If
the function f(x) assumes an indeterminate form when x = a, then ii
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 175
exists and is finite, we assign this value to the function for x = a,
which now becomes continuous for jc a (Art. 17).
The limiting value can sometimes be found after simple transfor-
mations, as the following examples show.
ILLUSTRATIVE EXAMPLE 1. Given /(j) ^ ~ ^. Prove lim f(x] =4.
Solution. /(2) is indeterminate. But, dividing numerator by denominator
fix} = x + 2. Then lim (x + 2) = 4.
z - 2
ILLUSTRATIVE EXAMPLE 2. Given /(x) = sec x tan j\ Prove lim f(x) = 0.
.r - ^ TT
Solution. /(*) ir> indeterminate (*> - oc). Transform as follows:
sec x - tan x = L^J^LZ = L^L . Lin* __ cor x ^
fos j- cos .r 1 -f sin .r 1 4 sin x
Hence the limit is 0.
See also Art, 18. General methods for evaluating the indeter-
minate forms of Art. 117 depend upon the calculus.
119. Evaluation of the indeterminate form - (iiven a function of
ff.r)
the form j~~- such that /(a) and F(a) ~ 0. The function is
indeterminate when x a. It is then required to find
We shall prove the equation
Proof. Referring to (4), Art. 116, and setting 6 = -jr, remembering
that /(a) = F(a) = 0, we have
m=m- <<-<'>
If x > a, so also .TI -> a. Hence, if the right-hand member of (1)
approaches a limit when r\ -> a, then the left-hand member will ap-
proach the same limit. Thus (E) is proved.
From (), if /'(a) and F'(a) are not both zero, we shall have
(2) i i
W
Rule for evaluating the indeterminate form - Differentiate the nu-
merator for a new numerator and the denominator for a new denominator.
The value of this new fraction for the assigned value of the variable will
be the limiting value of the original fraction.
176 DIFFERENTIAL CALCULUS
In case it happens that /'(a) = and F'(a) = 0, that is, the first
derivatives also vanish for x = a, then () can be applied to the ratio
F'(x)
and the rule will give us ij m
x-aF(x) F"(a)
It may be necessary to repeat the process several times.
The student is warned against the very careless but common mistake
of differentiating the whole expression as a fraction by VII.
If a = oo, the substitution jc = - reduces the problem to the evaluation
of the limit for z 0. z
Thus lim = lim = li m
Therefore the rule holds in this case also.
ILLUSTKATIVK EXAMPLE 1. F^ove lim ^H!^ = n ,
x .0 JT
Solution. l^tf(x)=^Kinnx l F(x) = jr. Then/(0) =0, F(0) = 0. Therefore, by (),
Hm /(l = lim Hl = Hm ncoanr = w
1 - b
ILLUSTRATIVE EXAMPLE 2. Prove lim
*"* 3 ? + 2 _ 3
-' -jr^-x + 1 2
Solution. Let -/(.r) - ^ - 3 x + 2, F(.r) =** -x* - x + 1. Then /(I) =
=0, Therefore, by (),
ILLUSTRATIVE EXAMPLE 3. Prove lim - t _ = 2.
Solution. Let /(x) = e* - e~ r - 2 x, F(x) = x - sin x. Then /(O) = 0, F(0) =
Therefore, by (),
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 177
PROBLEMS
Evaluate each of the following indeterminate forms by differentiation.*
1. lim ~ 2 ~ 1C
z-4
2. lim
_ o x sin
e v + sin y 1
x :4X 2 -f x-20
jr-a 1
c M a" na"
,. In x n
3. lim - - 1.
. ,. e e~ o
4. lim : -- 2.
x -. o sin .r
2.
. v In sin JT i
6. hm - - 2 - ~ 4-
x- -
7. lim ^- I= ln r
,. ^ arc sin i
8. hm - -- - *
0-*o sin- 1 6
15 lim tan 6 + sec - 1
15. hm ^ e _ ^ ^ + l
,/> v r sin ;
16. hm
. o
x-,0 sin :i
18. Given a circle with center at O, radius r,
and a tangent line A T. In the figure, A M equals
arc AP, and B is the intersection of the line
through M and P and the line through A and O.
Find the limiting position of B as P approaches
A as a limiting position. An*. OB = 2 r.
* After differentiating, the student should in every case reduc the resulting expression
to its simplest possible form before substituting the value of the variable.
178 DIFFERENTIAL CALCULUS
120. Evaluation of the indeterminate form ~. In order to find
lim
x-^a
when both f(x) and F(x) become infinite when x > a, we follow the
same rule as that given in Art. 119 for evaluating the indeterminate
form -. Hence
Rule for evaluating the indeterminate form ^. Differentiate the
numerator for a new 'numerator mid the denominator for a new denomi-
nator. The value of this new fraction for the assigned value of the mi /-
able will be the limiting value of the original fraction.
A rigorous proof of this rule is beyond the scope of this book.
ILLUSTRATIVE KXAMPLE. Prove lim -- n/r - 0.
.r C-.C JT
Solution. I*t /(j) = In r, F(JT) = osc .r. Then /(O) - - o , F(Q) - oo . Hence,
by the rule, I
,. f(jr) ,. f'(jf) .. .r _ .. siri L 'j
lim $-L - ) im - J -f~- = hm - - lim -
j , o f (jc) j ,<)/<'(>) j. , (, -- c'.-.c J ctn -i*
rni I /I^ 1- " " I' " ' /I <-l T7 T^
Then, by (A), hm - - lim - : - - 0. Q.E.D.
j- o JT cos jr , o cos j; J - sin x
121. Evaluation of the indeterminate form oo. If a function
f(y) - </>(**') takes on the indeterminate form oo for x a, we write
the Bi en function
so as to cause it to take on one of the forms -r or ^, thus bringing it
under Art. 119 or Art. 120.
As shown, the product f(jr) - c/>(x) may be rewritten in either of
the two forms given. As a rule, one of these forms is better than
the other, and the choice will depend upon the example.
ILLUSTHATHK KXAMPLE. Prove lim (sec 3 J* cos 5 .r) = .
X -+ /, 7T
Solution. Since sec -J TT = oo , cos J IT = 0, we write
o c 1 c C()S 5 X
sec 3 .r cos 5 J ~ - 7- cos 5 x = -
cos 3 .r cos 3 j"
Let/(j) = cos 5 or, F(x) - cos 3 jr. Then/(| TT) = 0, F(\ TT) = 0. Herice, by (),
-3sm3.r
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 179
122. Evaluation of the indetei ruinate form oo oo. It is possible in
general to transform the expression into a fraction which will assume
either the form - or ~-
ILLUSTRATIVE EXAMPLE. Prove lim (sec .r tan x] 0.
X - t l , JT
Solution. We have sec \ TT tan J T = x ~ x . .-. indeterminate.
^ y o * 1 sni - v 1 sin *"
Bv (2), p. 2, sec jr tan .r = ~~
cos j* cos x cos r
= I - sin jc, F(rt = cos j-. Then/(\ TT) = 0, F(\ TT) - 0. Hence, by (E),
p (.r) __ ^ 0;
X- '.TT ^H-
,.
hm
- ~ s ' in
Q.E.D.
PRO3LEMS
Evaluate each of the following indeterminate forms.
2. lim
ctn x
j , o ctn 2 r
,. tan 3 ^
3. hm --
^ _ tan 6
4. lim
c s
5. lim : --
x - .r. In x
, .
6. hm
7. lim
ctn x
In sin 2 x
A
Ans. 0.
,o In sin .r
8. lim jc In sin jr.
JT-
rt T 7T . 7T0
9. lim --- tan - -
10. lim r sin -
x -. oo ^
11. lim (TT 2 .r)tan x.
12. lim (1 - tan 0) sec 2 0. Ann. 1.
-4.
13. lim
J^
14. lim
JT-. 1
"J ._ ^
In j* In ^
0.
15. lim
1 2 :
_sin j <^ 1 cos<5
00 .
16. lim
__!/ L'l.
?/ - 1 In y\
oo .
17. lim
J--+
_sin- r .r 2 ]
1.
18. lim
j" 4- In j-
s In .A*
0.
19. lim
ft .(j
19 esc 2 0.
A 7T".
20. lim
a (J
ctn 2 jr
ctn 15 j*
a.
21. lim
^-^jtanr*-
22. lim (sec 5 <? tan 6.)
i-
i-
180 DIFFERENTIAL CALCULUS
TT 7T 1 r* _
" ' t\f* i *( '
23. lim .
4 ^*
24. limf-4-
28 .
123. Evaluation of the indeterminate forms 0, 1, 00. Given a func-
tion of the form f(x)+ (T)
In order that the function shall take on one of the above three
forms, we must have, for a certain value of x,
/(x)=0, 0Cr)=0, giving 0;
or /(*) = !, 0tr) = oo, giving I 00 ;
or /(j) = oo, (f)(x) = 0, giving 00.
Let y=f(x)+ (f) .
Taking the natural logarithm of both sides,
Invy = 000 In /(a-).
In any of the above cases the natural logarithm of y (the function)
will take on the indeterminate form
-oo.
Evaluating this by the process illustrated in Art. 121 gives the
limit of the logarithm of the function. This being equal to the loga-
rithm of the limit of the function, the limit of the function is known.
For if limit In // a, then lim // - e".
ILLUSTRATIVE EXAMPLE 1. Prove lim x z 1.
j .
Solution. The function assumes the indeterminate form when x = 0.
Let y = x 1 ;
then In y a* In j = oo, when x = 0.
r> * u. 1 ni i In x oo
By Art, 121, In ?/=-- - , whenz = 0.
By Art, 120, lim = lim
r - 1 J- -
Therefore lim In y - 0, and lim y- lim x* = e = 1. O.E.D.
' -* X () X -
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 181
i 2
ILLUSTRATIVE EXAMPLE 2. Prove lim (2 - j-) tan 5 WJr = f *
jr-l
Solution. The function assumes the indeterminate form l when jr = 1.
Let y = (2 - .r) tttn * ^ ;
In i/ = tan i 7r.r In (2 - j) - oo . 0, when j = 1 .
By Art. 121, ,n^!
By Art. 119, lim ln (2 ~ ^ = hm L> ~ ' r
.T-.I ctn i TTJ .r-i - \ TT csc- .^ Trr TT
o i 2
Therefore lim In y - -, and lim y ._ lim (2 - .r) tan ^ ^ = r 77 " O E D
3--1 7T x .1 ' j. .1 " v
ILLUSTRATIVE EXAMPLE 3. Prove lim (c-tn j-) sinj " = i.
j- *o
Solution. The function assumes the indeterminate form 00 when x ~ 0.
Let y=- (ctnx) sinj ;
then In y sin x In ctn j- = oc , when j = 0.
By Art. 121, In = ^^ r = , whon x = 0.
csc j: oo
- esc- j*
By Art. 120, lim l!L*!L = ] im _ ilH _ ^ n m LL. = .
x -*o csc J j-^o csc x ctn .r T .<> cos'*' .r
Therefore lim In y 0, and lirn y = lim (ctn .r) Hinj " = c = 1. Q.E.D.
- -
PROBLEMS
Evaluate each of the following indeterminate forms.
1. lim (sinx) tanx . Ans. 1. g i[ m / COS -V
x - I ' z . r \ y/
2. lim /| + lY. e*. 9- Hm (cos ^ .
S-limx^. J- 10. Hm (COB ?
4. lim fl -h -V. c a . 11. lim (r^ + 2 x)^
i/--* x V y/ T-. o
5. lim (1 -f sin x) ctnT . r. 12. lim (x -4- l) ctnx .
x-O x-0
- /I \sina;
6. lim (c x -f x) x . e 2 . 13. lim ( j '.
1
7. lim (1 + ntY. c\ 14. lim (I -f r) lnr .
/ - J- -
182 DIFFERENTIAL CALCULUS
124. The Extended Theorem of Mean Value. Let the constant R
be defined by the equation
(1) J(b) -f(o.) -(b- a)}' (a) - (& - a ^ R = -
Let F(y) be a function formed by replacing b by x in the left-hand
member of (1) ; that is,
(2) F(x) = f(x) -J(a) - (x - a)/' (a) - \(x - a)*R.
From (1), F(b) = ; and from (2), F(a) = ; therefore, by Rolle's
Theorem (Art. 113), at least one value of x between a and b, say x h
will cause F'(x) to vanish. Hence, since
F'(x)=J'(x)-f'(<i)-(x-a)R,
we get F'(Xi) =f'(xi) ~J'(o) - (x l - a)R = 0.
Since F f (x\ ) and F'(a) = 0, it is evident that F'(r) also satisfies
the conditions of Eolle's Theorem, so that its derivative, namely F"(x),
must vanish for at least one value of x between a and x\, say x*, and
therefore x<2 also lies between a and b. But
F"(x) =f"(jr) - R\ tlierefore F"(jrJ =f /f (^) - R = 0,
and R=--f"(xJ.
Substituting this result in (1 ), we get
' "
a) 2 /"(^). (a
By continuing this process we get the general result,
(G)
Equation (G) is called the Extended Theorem of Mean Value, or the
Extended Law of the Mean.
125. Maxima and minima treated analytically. By making use of
Art. 116 and the results of the last section we can now give a general
discussion of maxima and minima of functions of a single independent
variable.
Given the function /(.r). Let h be a positive number as small as
we please ; then the definitions given in Art. 46 may be stated as
follows.
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 183
If, for all values of :r different from a in the interval |a h, a + H
(1) /(x) /(fl) = a negative number,
then f(jc) is said to be a majriwinn -whew x = a.
If, on the other hand,
(2) /(.r) /(a) = a positive number,
then /(.r) is said to be a minimum when x == a.
We begin with an analytical proof of the criterion on page 51.
A function is increasing when the derivative is positive, and decreas-
ing when the derivative is negative.
For, let y=f(x). When A;/ is small, numerically, and the
A.r
derivative /'(.r) will agree in sign (Art. 24). I jet /'(/) > 0. Then,
when Ax is positive, so is A?/, and when A.r is negative, so is A?/.
Therefore f(jr) is increasing. A similar proof holds when the deriv-
ative is negative.
The truth of the following statement is now easily deduced.
///(a) is a maximum or minimum value of f(x), then f (a) = 0.
For, if /'(a) 5^ 0, f(x) would increase or decrease as x increases
through a. But then f(a) is neither a maximum nor a minimum.
We seek general sufficient conditions for maxima and minima.
Consider the following cases.
I. Let /'(a) = 0, and/"(a)^ 0.
From (f), Art. 124, replacing 6 by n and transposing /(a),
(3) /(a-) -/(a) = ( *~ a}2 f"(.r<2). (a < x>> < x)
Since /"(a) ^ 0, and /"Or) is assumed as continuous, we may
choose our interval \a h, a + h\ so small that /"(/) will have the
same sign as /"(a). Also, (/ a)' 2 does not change sign. Therefore
the second member of (3) will not change sign, and the difference
/Or) -/(a)
will have the same sign for all values of x in the interval [a h, a + h],
and, moreover, this sign mil be the same as the sign o//"(a). It there-
fore follows from our definitions (1) and (2) that
(4) /(a) is a maximum iff'(a) and f"(a) = a negative number;
(5) /(a) is a minimum if f'((i) = and j"(a) = a positive number.
These conditions are the same as those in Art. 56.
184 DIFFERENTIAL CALCULUS
II. Let /'(a) =/"(a; - 0, and /'"(a) ^ 0.
From (G), Art. 124, putting n = 3, replacing b by r, and trans-
posing /(a),
(6) f(x) - /(a) = ,4 te - a) 3 /"W (a < a* < a-)
12
As before, /'"(or) will have the same sign as f'"(a). But (x a) :i
changes its sign from to -f as x increases through a. Therefore
the difference ., . fi
- f(a)
must change sign, and /(a) is neither a maximum nor a minimum.
III. Let /'(a) =/"(a) - =/<"-'>(a) =0, and/ (n) (a)^ 0.
By continuing the process as illustrated in I and II, it is seen that
if the first of the derivatives of f(x) which does not vanish f or x = a
is of even order (= ri), then
(H) /(a) is a maximum if / (n) (a) = a negative number;
(/) /(a) is a minimum if f (n) (d) = a positive number.*
If the first derivative of f(x) which does not vanish for x = a is of
odd order, then /(a) will be neither a maximum nor a minimum.
ILLUSTRATIVE EXAMPLE 1. Examine x* - 9 x' 2 -f 24 x - 7 for maximum and
minimum values.
Solution. /(x) = z< { - 9 x' 2 + 24 x - 7.
/'Or) = 3 a; 2 - 18 z + 24.
Solving, 3 x 2 -18x4- 24 =
gives the critical values x = 2 and z = 4. /./'(2) = 0, and /'(4) = 0.
Differentiating again, /"(*) ~ 6 x 18.
Since /"(2) = - 6, we know, from (//), that/(2) = 13 is a maximum.
Since /"(4) = -f 6, we know, from (/), that /(4) = 9 is a minimum.
ILLUSTRATIVE EXAMPLE 2. Examine e* + 2 cos x 4- e~ x for maximum and mini-
mum values.
Solution. /(r) = e x -f 2 cos .r -f f-*,
/'(j) = <?* - 2 sin x - e- 1 = 0, for x = 0,t
f"(x} = e x 2 cos x -f <?"* = 0, for x 0,
/"'(z) = g* -f 2 sin x - e~ x = 0, for x = 0,
/"( = e' -f 2 cos x -f- <r T = 4, for x = 0.
Hence, from (/), /(O) = 4 is a minimum.
* As in Art. 46, a critical value x = a is found by placing the first derivative equal to
zero and solving the resulting equation for real roots.
t x = is the only root of the equation e* 2 sin r e~ x 0.
THEOREM OF MEAN VALUE AND ITS APPLICATIONS 186
PROBLEMS
Examine each of the following functions for maximum and minimum
values, using the method of the last section.
1. x 4 4 x 3 + 5. Arts, jc = 0, gives neither,
x = 3, gives min. = 22.
2. x 3 + 3 x 2 + 3 x, x = - 1, gives neither.
3. x 3 (x - 2) 2 . x = 0, gives neither,
x = L gives max. = 1.11,
x = 2, gives min. = 0.
4. x(x-l) 2 (x+l) 3 .
5. Investigate 4 x 6 - 15 x 4 + 20 x 3 - 10 x 2 at x = 1.
6. Show that if the first derivative of /(x) which does not vanish for
x = a is of odd order (= n), then/(x) is an increasing or decreasing function
when x = a, according as /W (a) is positive or negative.
ADDITIONAL PROBLEMS
1. If y = e x -f e~ x , find dx in terms of ?/ and dy. Ans. dx = ^ .
V;// 2 - 4
2. Prove that
T- In (3 x + 2 + V9 x 2 -f 12 x) =
.
Vi) jf- -f 12
3. Prove that
Vx- -f 1 i
= x 2 Vx 2 -f 1.
4. Show that the curve x = t* + 2 * 2 , ?/ = 3 / 2 + 4 f has no point of
inflection.
5. Prove that the points of intersection of the curves 2 ?/ x sin x and
y = cos x are points of inflection of the first curve. Sketch both curves on
the same axes.
6. Given the damped harmonic motion
s =r ae~ ht sin ct,
where a, 6, and c are positive constants ; prove that the successive values
of t for which v = form an arithmetic progression and that the correspond-
ing values of s form a decreasing geometric progression.
7. The abscissa of a point P moving on the parabola // - ax 2 is increas-
ing at the rate of one unit per second. Let O denote the origin and let T
be the intersection of the x-axis with the tangent to the parabola at P.
Show that the rate of increase of the arc length OP is numerically equal
TP
to the ratio
186 DIFFERENTIAL CALCULUS
8. Let MP be the ordinate at any point P on the catenary (see page
532). The line MA is drawn perpendicular to the tangent at P. Prove
that the length of MA is constant and equal to a.
9. The curve x 2 y -f 12 y = 144 has one maximum point and two points
of inflection. F"ind the area of the triangle formed by the tangents to the
curve at these three points. Arts. 1.
10. Given In 6 = 1.792 and In 7 = 1.946. Calculate In 6.15 first by
interpolation and second by differentials. Show graphically that the true
value lies between the two approximations.
11. Given the ellipse b 2 jr 2 + a 2 ?/ 2 = a' 2 b' 2 , find the length of the shortest
tangent intercepted between the coordinate axes. Ans. a 4- b.
12. Given the area in the first quadrant bounded by the curves y 2 = x
and y' 2 = x 3 . A rectangle with sides parallel respectively to the axes is
drawn within the limits of this area. The width of the rectangle (measured
horizontally) is J and one of the diagonals has an extremity on each curve.
Find the area of the rectangle of maximum area which can be constructed
in this way. Ans. 0.019.
13. Rectangles are drawn with one side along the ./--axis, a second side
along the line x = * and one vertex on the curve ?/ = c~ r \ Find the area
of the largest of these rectangles. A/?s. e~ - 2r > = 0.7788.
14. Find the maximum and minimum values of y, if
X X
y ae a 3 x 2 ac .
Ans. Max. = a ; min. a(l 3 log 2).
15. Given x 2 -f 3 xy -h 2 y 2 5 x 6 y -h 5 0, find the maximum and
minimum values of y. Ans. Max. 1 ; min. = 5.
INTEGRAL CALCULUS
CHAPTER XII
INTEGRATION ; RULES FOR INTEGRATING STANDARD
ELEMENTARY FORMS
126. Integration. The student is already familiar with the mutu-
ally inverse operations of addition and subtraction, multiplication
and division, raising to a power and extracting roots. In the ex-
amples which follow, the second members of one column are, respec-
tively, the inverse of the second members of the other column.
y = a*, x = log,, // ;
y = sin x, x = arc sin ;//.
From the differential calculus we have learned how to calculate
the derivative f'(x) of a given function f(jr), an operation indicated by
-f A') =/'(*),
dx '
or, if we are using differentials, by
df(jr)=f'U)dx.
The problems of the integral calculus depend on the inverse
operation, namely :
To find a function f(x) whose derivative
(1) /'(*) - 4>W
is given.
Or, since it is customary to use differentials in the integral cal-
culus, we may write
(2) dj(x) = f'(x)dx = cf>(x)dx
and state the problem as follows :
Having given the differential of a function, to find the function itself.
187
188 INTEGRAL CALCULUS
The function /(x) thus found is called an integral of the given
differential expression, the process of finding it is called integration,
and the operation is indicated by writing the integral sign * /in front
of the given differential expression ; thus ^
(3) ff'(x)dx=f(x } ,
read the integral of f'(x)dx equals f (or). The differential dx indicates
that x is the variable of integration. For example,
(a) If /(x) = x a , then/'(x)dx = 3 x' 2 dx, and
(b) If /(x) = sin x, then/'(x)dx = cos xdx, and
cosxcfx = sin x.
/r
(c) If /(x) = arc tan x, then/'(x)rfx = .,, and
1 "T~ X"
- arc tan x.
+ x~
Let us now emphasize what is apparent from the preceding ex-
planations, namely, that
Differentiation and integration are inverse operations.
Differentiating (3) gives
(4) ' dff'(x)dx=J'(x)dx.
Substituting the value of f'(x)dx [= tf/(x)] from (2) in (3), we get
(5) fdf(x)=f(x).
i /*
Therefore, considered as symbols of operation, 3- and / - dx are
dx J
inverse to each other ; or, if we are using differentials, d and / are
inverse to each other. r J
When d is followed by I they annul each other, as in (4), but
when I is followed by d, as in (5), that will not in general be the
case. The reason for this will appear at once from the definition of
the constant of integration given in the next section.
* Historically this sign is a distorted S, the initial letter of the word sum. See Art. 155.
INTEGRATION 189
127. Constant of integration. Indefinite integral. Prom the preced-
ing section it follows that
snce
rf(:r 3 ) = 3 x 2 dx, we have f 3 .r~dr = x 3 ;
since d(x* + 2) = 3 x~dx, we have f 3 x-dx = x 3 + 2 ;
since d(x 3 - 7) = 3 x-djr, we have / 3 x 2 dx = x 3 7.
In fact, since d(x* + c) = 3 .r-rfjr,
where C is any arbitrary constant, we have
3 a- 2 rfjp = a- 3 + C.
f
A constant C arising in this way is called a constant of integration, a
number independent of the variable of integration. Since we can give
C as many values as we please, it follows that if a given differential
expression has one integral, it has infinitely many differing only by
constants. Hence r
Jf'(x)dx=f(x) + C;
and since C is unknown and indefinite, the expression
/Or) + C
is called the indefinite integral of f f (x)dx.
It is evident that if </>(#) is a function the derivative of which is
f(x), then <f>(x) + C, where C is any constant whatever, is likewise
a function the derivative of which is /(a;). Hence the
Theorem. // two functions differ by a constant, they have the same
derivative.
It is, however, not obvious that if 0(#) is a function the derivative
of which is /(#), then all functions having the same derivative f(x)
are of the form
_j_ ^
where C is any constant. In other words, there remains to be
proved the
Converse theorem. // two functions have the same derivative, their
difference is a constant.
Proof. Let <t>(x) and \l/(x) be two functions having the same deriva-
tive /(x). Place
F(x) = </>(x) \l/(x) ; then, by hypothesis,
(1) F *(x) =
190 INTEGRAL CALCULUS
But from the Theorem of Mean Value (Z>), Art. 116, we have
F(x + Ax) - F(x) = Ax F'(x + Ax). (0 < 6 < 1)
.-. F(x + Ax) - F(x) = 0,
[Since by (1) the derivative of F(x) is zero for all values of x.J
and F(x + Ax) = F(x).
This means that the function
F(x) =
does not change in value at all when x takes on the increment Ax,
that is, <t>(x) and \f/(x) differ only by a constant.
In any given case the value of C can be found when we know the
value of the integral for some value of the variable, and this will be
illustrated by numerous examples in the next chapter. For the pres-
ent we shall content ourselves with first learning how to find the
indefinite integrals of given differential expressions. In what fol-
lows we shall assume that every continuous function has an indefinite
integral, a statement the rigorous proof of which is beyond the scope
of this book. For all elementary functions, however, the truth of
the statement will appear in the chapters which follow.
In all cases of indefinite integration the test to be applied in veri-
fying the results is that the differential of the integral must be equal
to the given differential expression.
128. Rules for integrating standard elementary forms. The differen-
tial calculus furnished us with a General Rule for differentiation
(Art. 27). The integral calculus gives us no corresponding general
rule that can be readily applied in practice for performing the inverse
operation of integration.* Each case requires special treatment, and
we arrive at the integral of a given differential expression through
our previous knowledge of the known results of differentiation. That
is, we must be able to answer the question, What function, when dif-
ferentiated, will yield the given differential expression ?
Integration, then, is essentially a tentative process, and to expedite
the work, tables of known integrals are formed called standard forms.
To effect any integration we compare the given differential expression
with these forms, and if it is found to be identical with one of them,
the integral is known. If it is not identical with one of them, we
strive to reduce it to one of the standard forms by various methods,
many of which employ artifices which can be suggested by practice
* Even though the integral of a given differential expression may be known to exist, yet
it may not be possible for us actually to find it in terms of known functions.
INTEGRATION 191
only. Accordingly a large portion of our text will be devoted to the
explanation of methods for integrating those functions which fre-
quently appear in the process of solving practical problems.
From any result of differentiation may always be derived a formula
for integration.
The following two rules are useful in reducing differential expres-
sions to standard forms.
(a) The integral of any algebraic sum of differential expressions
equals the same algebraic sum of the integrals of these expressions
taken separately.
Proof. Differentiating the expression
/ du + I dv I dw,
u, v, w being functions of a single variable, we get
du + dv - dw. By III, Art. 94
(1)
/. / (du + dv- dw) = (du + Cdv - Cdw.
(b) A constant factor may be written either before or after the inte-
gral sign.
Proof. Differentiating the expression
gives adv. By IV, Art. 94
(2) .'. I adv= a I dv.
On account of their importance we shall write the above two rules
as formulas at the head of the following list of M Standard Elementary
Forms."
STANDARD ELEMENTARY FORMS
(1) i(du + dv- dw) = idu + I dv - Cdw.
(2) I ad v = a I dv.
(3) Cdx = x + C.
/ v n + l
V " dV = ^+l + C ' (*-!)
192 INTEGRAL CALCULUS
Cdv
(5) J T = lnv+C
= In v + In c = In cv.
(6)
(7) C
(8) / sin v </v = cos v + C.
(9) I cos v dv = sin v + C.
_/
(10) I sec 2 i; c/v = tan v + C.
(11) f esc 2 i; rfi; = - ctn v + C.
(12) I sec i; tan v dv = sec v+C.
(13) I esc i; ctn v dv = esc v+C.
(14) / tan v dv = In cos v In sec v + C.
(15) I ctn v dv In sin v + C.
(16) / sec v dv = ln (sec v + tan v) + C.
(17) I esc v dv = In (esc v ctn v) + C.
(18) f *" 2 = iarc tan-+ C.
J v 2 + a 2 a G
fffv v
x = arc sin - + C.
Va 2 - v 2
INTEGRATION 193
129. Formulas (3), (4), (5). These are easily proved.
Proof of (3). Since d(x + C) = dr, II, Art. 94
we get I dx = x + C.
Proof of (4). Since
( 7 ,n + 1 \
j-j-j- + Cj = tcdr, VI, Art. 94
/ ? ,n-H
V n dv = - T + (7.
W+ 1
This holds true for all values of n except n = 1. For when
n 1, (4) involves division by zero.
The case when n = 1 comes under (5).
Proof of (5). Since ,/.,
d(lnr+D =-f X, Art. 94
v
we get f = In w + r.
J v
The results we get from (5) may be put in more compact form if
we denote the constant of integration by log c c. Thus
C^l = ] n v + i n c = i n CVf
J v
Formula (5) states that if the expression under the integral sign is a
fraction whose numerator is the differential of the denominator, then the
integral is the natural logarithm of the denominator.
ILLUSTRATIVE EXAMPLES*
Work out the following integrations :
7-6
-6 + 1
/ 7- r
1. Cx^dx = -j- r + C = *- + C, by (4), where v = x and n = 6.
J o + 1 /
2.J Vxdz =fx*dx = 2L+C = z*+C, by (4), where = a: and n = J.
* When learning to integrate, the student should have oral drill in integrating simple
functions.
194 INTEGRAL CALCULUS
3 Cdx = r x . 3dx = ^ + c = _J_ + Cib y (4), where r = x and
' J y* J X
4./ax*dx = a/Vdx = <^+C. By (2) and (4)
5./(2x 3 - 5x 2 -3x
x -/5 x 2 d(x -/3 xrfx +4 dx by (1)
= 2/x 3 dx - 5/x 2 dr - 3/xdx + 4/dx by (2)
NOTE. Although each separate integration requires an arbitrary constant, we
write down only a single constant denoting their algebraic sum.
=f2 ax'^dx - fbx~' 2 dx -f/3 rx^dx by (1)
= 2 a fx~*dx - bfx~ 2 dx + 3 cfx$dx by (2)
= 2a .2-6. + 3r.^ + C by (4)
= 4 aVx -f - + 1 rx 5 ' 4- C.
./ 5
HINT. First expand.
Solution. This may be brought into the form (4). For Comparison with (4).
insert the factor 2 b' 2 after the integral sign before x dx, and v a' 2 + b' 2 x'\ n = - 2 >
its reciprocal fo/orr the integral sign. Those operations dv = 2b*xdx.
balance each other by (2).
: = 5 L / % (a + 6x)l(26*zd I )[=^ 5 /*i* = ^5 + C, by (4)]
( fl 2 + & 2 X 2)1
"^
3 ^
NOTE. The student is warned against transferring any function of the variable
from one side of the integral sign to the other, since that would change the value
of the integral.
By < 2)
INTEGRATION 195
This resembles (5). If we insert the factor 2 c 2
after the integral sign and its reciprocal before it,
Comparison with (5).
v = 6 2 + c 2 * 2 , dv = 2 c 2 x dx.
the value of the expression will not be changed.
C x dx 3 a C2c 2 xdx [ 3 a Cdv 3 a , , , , /rv l
Hence 3 a I - 7 = ^7 I r^ ~ = ^r~, I "~ = ;r^ ln p + ^ b y ^ 5 >
J 6 2 + c 2 x 2 2 o-j 6^ + c-jc 2 L 2 r - j v 2 c 2 j
Solution. First divide the numerator by the denominator. Then
x -f 1 x + 1
Substituting in the integral, using (1), and integrating gives the answer.
11. f 2 / ~ I dx = x - In (2 JT -f 3) 2 -f T.
J 2 x + 3
9 3. _ J ^
Solution. Dividing, ^ - r = 1 - - - Substitute and use (1) etc.
-- --
The function to be integrated is called the integrand. Thus in
Illustrative Example 1, p. 193, the integrand is x (] .
PROBLEMS
Verify the following integrations.
1. fx* dx = ^ + C. 6. J*3 a?/ 2 d?/ = a?/ 3 4- c:.
l* = _? +c .
t" t
*
5
3. / .r 5 dx = =- + C. / - " ~ 8
= V2.r + r.
. r
INTEGRAL CALCULUS
= ~- 6x 4- 5 In x 4- C.
2Va by
x -h I) 2 cfcr = ^r 4 4- ~ "^ 7 + c -
/* 4 x' 2 dx _
V Vx :i 4- 8 ~
r 6 z dz 1
V(5-32r 2 ) 2 S-Sz^"*"
r / 2 rf/ _ _
"J (a 4- 6/ 3 ) 2
4- C\
4- C.
INTEGRATION 197
36 '/ x* = a" 3 " +C '
/f* ^QlTl 1*^3 Rin*^ T"
sin 2 x cos x dx = I (sin x) 2 cos x dx = l " ; -f C = ^-^ -f C,
/ o o
HINT. Use (4), making v = sin x, dt> = cos x dx, n = 2.
oo r - j sin 2 aj ^ , /,
38. / sin ax cos ax ax = h C .
J 2 a
39. Jsin 2 x cos 2 2 x dx = - CQS ^ 2 ' r -f C.
40. ftan | see- 1 dx = tan 2 ~ -f r.
4i / cosaxdr = 2 V6 4- sin a.r ^ ^
" J V6 + sin ax a
rf
42.
tan
44 Cr'<i* = ln(2 + r)
V 2 + a- :! 3
_ In (a + M z ) .
~
. f f d< _
5> J ^+1^ ~
26
= In (^ + 3
In (;/" + 4 ?/)
> L-.
' lie . - In (a + be 9 ) .
+ be e ~ b +
Ar . r sin x dx i /- N , /-t
49> i - In (1 cos x) + T .
j 1 cos x
50. f s f^/ ( ^ =li n (a + b tan ^ 4- C.
J a -h o tan ?/ o
51.
. f (2 ^ + ^ )rf:r = 2 x - In (x + 2) -f C.
J x + 2
62 . /Lt|* = f - + 3 In (X + 1) + C.
55. d = 2 In
6
198
INTEGRAL CALCULUS
Work out each of the following integrals and verify your results by
differentiation.
r 2xdx
56. 177===;.
> V 6 - 5 x 2
Solution.
/6 - 5 x*
Verification, d [- ft (6 - 5 x*)l + ('[ = - -ft 3S(6 - 5 x 2 )4(- 10 x)dx
2 x dx
57. / O'* -f 3 x 2 )dx.
j
58.
66.
67,
68
/ X(/J
v Vi^^ 5 '
C ' <lt
-J ITFT1
sin (iC (10
71. f_EL___
J cos a^? + b
esc 2 (&fr
72
130. Proofs of (6) and (7). These follow at once from the corre
spending formulas for differentiation, XI and XI a, Art. 94.
/^ x *
ba' 2T dx =
Solution. Cba* r dx = b fa 2 ' dx.
h
By (2)
This resembles (6). Let r = 2 x ; then d?? = 2 dx. If we then insert the factor
2 before dx and the factor J before the integral sign, we have
= . + C. By (6)
INTEGRATION 199
PROBLEMS
Work out the following integrals.
1. f 6 e** dx = 2 c* r + C. 4. f 10' dx = r-^- + C.
J J In 10
2. JV dr = nr " + C. 5. JV* </// = ^^ + C.
7. fU" + f ~)(ir = a(r - r" 'V -f C
r( - -"\ 2 n/ ^
8.J Vf a ~ r n ) dx - %\c (i - c a )
*-ffC xZ dx = I c r2 + C.
10. f<' Bin x cos x dx = r sin x + T.
11. /Vanfl sec 2 ^ f/^ C i&nf) + C.
12. V? (// = 2 VP -f T.
13. dx = + C.
1 -f In a
Work out each of the following integrals and verify your results by
differentiation.
ID
. i e> e~* (Ljr,.
' J \ e x J '
J b' A <
17
f-f-
oo Cc T dx
m.f*xe--**r.
23 J r' - 2
18
'/tl'
24. ix(c r ' 2 -f 2)dx.
29. C(e 2T ) 2 dx.
19
. Cc ax dx.
?^ r^" 3 ^
30. /V os2x sin 2 xdx.
25.J v - dr.
21
. f z V 3 dx.
26. ft 2* dt.
p 2 ^.
J e lA
200 INTEGRAL CALCULUS
131. Proofs of (8)-(17). Formulas (8) (13) follow at once from the
corresponding formulas for differentiation, XIII, etc., Art. 94.
Proof of (14). f tan v dv = f
'sin r dv
cos v
C sin r dv
_ r
J
COS 7'
r)
COS V
In cos v + C by (5)
= In sec r + r.
[Since - In COM r In - In 1 -f In see r In sec r.
wc r J
Proof of (15). fctn v d, = f**^ = f ^^
J J sin v J sin v
- In sin v + C. By (5)
Proof of (16). Since sec v = sec v f
sec v + tan v
sec v tan v + sec 2 ?>
sec v + tan ?;
/, Tsec r tan ?' + sec 2 v ,
sec r dr ~ I ! dv
J sec* v + tan r
(sec r + tan r)
Proof of (17). Since esc v = esc v
sec ?> + tan r
=- In (sec v + tan r) + f. By (5)
esc r ctn r
esc r ctn r
esc v ctn 7 1 + esc 2 v
esc r ctn v
" 2M dv
/, f esc r ctn ?' + esc 2 v
esc v ay = I !
J esc r ctn v
/(/(esc ?> ctn v)
esc r ctn v
= In (esc v ctn r) + C. By (5)
An alternative form of (17) is
esc v dv = In tan %v+C (see Problem 4, p. 201 )
INTEGRATION
201
ILLUSTRATIVE EXAMPLE 1. Prove the following integration.
Jsin2a*dj = - co *- fl flx +r.
Solution. This resembles (8). For let r = 2 ax ; then dr 2 a dr. If we now
insert the factor 2 o before ds and the" factor before the integral sign, we get
f sin 2 ax c?x - -i- f sin 2 aj- 2 a dr [ = f si
J 2 aj L - <*J
sin r eft' -- - ~- cos r 4- T, by (8)
- - .
2 a 2 (i
iLLfSTRATH'E EXAMPLE 2. Prove the following integration.
f (tan 2 x - 1 )'-' dx = J tan 2 x + In cos 2 -f- ( \
Solution. (tan 2 x - 1 )* = tan-' 2 x - 2 tan 2 s + 1 .
tan 2 2 x = sec- 2 x - 1 .
Hence, substituting,
| (tan 2 .S F 1 )- ds | (sec'-' 2 *? 2 tan 2 ^-V/x / sec" 2 x dx 2 | tan 2 d.
Now let r = 2 x. Then d?' 2 ds. Using (10) and (14), the steps are as follows.
By (2), Art. 2
I sec* 2 x ds = \ I sec- 2 x d(2 x)
ftan 2 x dx = J ft an 2 x d(2 x)
= I I sec 1 '
r dr -- \ tan ?
= ^ tan 2 x.
v dv .', 1 n cos r
I In cos 2 x.
PROBLEMS
Verify the following integrations.
r ds sin rns -h C.
m
2. / tan bs ds = - In sec bs -f ( \
8./U
/^
sec as djc - In (sec as -f tan <ur) + C.
a
4. f esc v dv In tan J r -f ^ Y .
5. fsec 3 / tan / f// = ?, sec 3 / -f C.
/I
esc ay ctn a?/ c/// -- esc ay + C.
7. fcsc 2 3 j: ds = - J ctn 3 j- + C.
8. fctn I dr = 2 In sin ^ + C.
9. fs' 2 sec 2 J :i dx - J tan x :i H- C.
10.
sin- x
= - ctn x -f C.
202 INTEGRAL CALCULUS
11. f-^-- = tan x + r.
J COS- X
12. f (tan -f ctn Or dO = tan - ctn 6 + C 1 .
13. I (sec tan <f)) 2 d$ ~ 2 (tan sec 0) + C.
14 J LL = ctn jr + esc J -f C.
'J 1 -f OOB.T
HINT. Multiply both numerator and denominator by 1 cos x and reduce
before integrating.
15. f = tan r - sec jr + C.
J 1 -f sin j'
16. ( . Sin ** = - In (1 -f cos ) + C.
./ 1 -f cos ,s
. .
./ 1 -f tan jr
18. I j - cos ^' J djc I sin r L> -f (\
19. f(.r 4~ sin 2 .r)(/.r i(.r 2 cos 2 .r) -f C.
20. /" ^l^^L, = 2V4-COS.,- + C .
J >/4 _ cos j>
21. rn + cos^dj = }n(x + sin r) + r _
J jr -f sin j^
22. / -J^L^L^ = Vl -f 2 tan fl -f r.
; V 1 -f 2 tan 6
Work out each of the following integrals and verify your results by
differentiation.
23. /"sin ~ dr. 29. |sec 2 2 or (Lr.
24. Jcos ( b -f flj-V/x. 30. Jtan | dr.
r ^i f f//
25.J esc* (a - brtdjr. 01 -J tan 5 ^
ftw r </> . a</> 7 , 33.
27. I esc y- ctn -y 1 (/0. J ctn 7 ?/
/> Q . /"sin V.T c/j*
28. J r x ctn f rfj. M - J y^
35.
sin 2 3 t
INTEGRATION 203
43. f (1 - esc //) 2 dy.
J
O.T f a rfjr
37. I -
J cos" bj"
45. (V-^
J 1 sin .r
"./ 3 -f cos !2 .r
47. f 0? / ( f .
' Vu -j- fo sin f
csc ctn (10
rcsc c
'J 5 4
49 r cscj\r
V -
csc
csc\r(/.r
ctn .r
38. | (sec 2 0- csc^
39. I (tan </> -f sec </>)-
40. f(tan 4 s ctn^
41. f(ctn.r - I) 2 dr.
42. I (sec / I) 2 (//.
132. Proofs of (18)-(21). Formulas (18) and (20) follow easily
from the corresponding formulas for differentiation.
Proof of (18). Since , r \
(II - 1
d(- arc tan - + rW ~ ^^ - -r^-:, by XXII, Art. 60
\a a / a
we get
r rf' 1 i 4 ?? i ^
-n - "= " arc ian " + r
./ ?'~ + " ' a
Proofs of (19) and (19 a). We prove (19) first. By algebra, we have
1
1
Hence
Then
v a r -f a
1 1
a 2
_J LI.
/) a r -f a \
= 7T- ln ( ?? - a )-7r- ln
^1 d Ll (I
= _L i n
2a
To prove (19 a), by algebra,
1
i + C '
2 a
by(1)
By (2), Art. 1
a + v a v a
204
INTEGRAL CALCULUS
The rest of the proof proceeds as above.
NOTE. The integrals in (19) and (19 a) satisfy the relation
r dv _ _ C dv
J r 2 - '-' ~ J a' 2 - v' 2 '
Hence either formula can be applied in any given case. Later we shall see that
one form must be chosen in many numerical examples.
Proof of (20). Since
arc sin- + (j =
a,
*(-
\a
we get
>K )'
/dr . v . ,,
-^ arc sin - + ( .
Va 2 - r' 2 a
=> by XX, Art. 94
Proof of (21). Assume r = a tan z, where z is a new variable ;
differentiating, dv = a sec" z dz. Hence, by substitution,
dr C <J sec 2 z dz C sec 2 z dz
r dr __ C a see- z dz __ C sec-sera
./ V?' 2 + a 2 J Va 2 tan 2 z + a~ J Vtan 2 z + 1
I sec z dz In (sec z + tan z) + r
by (16)
- In (tan z + Vtan- a +1) + r. By (2), Art. 2
But tan z = -; hence,
a
P* = = ln(*+ /ir
J V?' 2 + a- \ ^ a ~
In (r + Vr 2 + a~) In a + c.
Placing r = In a + r, we get
I = In (v + Vtr' + a 2 ) + C.
In the same manner, by assuming v a sec z, dv = a sec z tan z dz,
we get
/dv ra sec a tan z dz r
i \ , \ sec z ar
V r 1 a 1 J va'-sec^z a L J
= In (sec a + tan z) + c by (16)
= In (sec z + Vsec 2 z - 1) + c by (2), Art. 2
INTEGRATION 205
ILLUSTRATIVE EXAMPLE. Work out the following integration.
f_*_ = l arc taii^ + r.
J 4 x' 2 + 9 6 3
Solution. This resembles (18). For, let f ' = 4 r* and a* = 9 ; then r = 2 x
dv = 2dt, and a = 3. Hence if we multiply the numerator by 2 and divide in front
of the integral sign by 2, we get
r_j_-ir 2d * r=!f_*l_:=-Larctan2+r. By (18)1
J 4 z~ + 9 2J (2 x) 2 + (3) 2 L 2j r 2 + a 2 a J
= - arc tan -- -f C.
o <>
PROBLEMS
Work out the following integrals.
1 r-^_ = larctanf + C.
J x 2 -f 9 3 3
f , dg = In ( + V.s^ - 16)
2 - 6
6 f
J
dx
Vl6 - 9 x 2 ^
= - arc sin '- + C.
9 r_lL = arc tan f + C.
_ ln EL + C .
or
f ax dx = a arc tan |! 4. C .
' J x 4 -f 6 4 2 & 2 ?> 2
i f. ^ = - arc tan ( T-^ ) + C.
J ( - 2) 2 -f 9 3 \ o
206 INTEGRAL CALCULUS
15. f dy = i In (ay + Vl + ay) 4- C.
J Vl + aV 2 a
18. f <** = arc sin ("*\ + C .
J V4 - (u + 3j 2 \ 2 y
Work out each of the following integrals and verify your results by
differentiation.
17 f (L
r*
on f 3d?/
r 6 < dt
J V9 -
IP f <ly
16 a* 2
~"'J 9 ?/ 2 - 16
23 f d *
r sin d0 B
15. 1 L
J V ( J y' 2
in r ^
T4
J V4 * 2 + 5
*&O. 1 . '
J V4 -f cos 2 6
r dx
/" (/.f
25
J V^ 4 4
" J m 2 + (x + n) 2
n r d
~ ' J 25 r- -
B1 f 7d>
-4
J \/5 X 2 + 3
30 'J 4 - (2 u - I) 2
r 2
~ 'J Vl -V 2r
31 '/5 - x"
The standard formulas (18)-(21) involve quadratic expressions
(7^'ia-, a- ?; L> ) with two terms only. If an integral involves a
quadratic expression containing three terms, the latter may be re-
duced to one with two terms by completing the square, as shown in
the following examples.
ILLUSTRATIVE EXAMPLE 1. Verify the following:
= 1 arc tan ^4_1 + (\
+ 2 x -f 5
Solution. x'-' + 2 x + 5 = x- -f 2 j- -f 1 + 4 - (x -f I) 2 -f 4.
. r ^ = r ^
" J jr 2 -f 2 j + 5 J (r + I) 2 + 2 2
This is in the form (18). For let r = x + 1, and a = 2. Then dt; = dx.
Hence the above becomes
/-; - ; = - arc tan - -f C ~ - arc tan ~t y C.
r 2 + a* a a 2 2
/9 /fv. 9 y. 1
^- 2 arc sin :1 - :i + C.
V2 +x - x' 2 *
Solution. This is in the form (20), since the coefficient of x' 2 is negative.
Now 2 -f a* x 2 = 2 (x' 2 x -f J) 4- 1 = S (x J) 2 .
Let v = x I, a ?. Then dt> = dr.
f_2dr = 2 r dr = 2 f d; = 2 arcsin 2 + c> by (20)
^v2+x-x 2 ^ v| - (j - J)-' JVa-'-r 2 a
INTEGRATION
ILLUSTRATIVE EXAMPLE 3. f- - ~ - = ~ In 3 x ~~ 3 + C.
J 3 x 2 + 4 x 7 10 3x4-7
Solution. 3z 2 +4x-7 = 3(x 2 + 1 x - j) = 3 (or 3 + J * + I - V)
= 3[(:r+{jV-Vl.
. f <* T - C dx = \ C dr
' 'J 3 x 2 + 4 x - 7 J 3[(x -f J) 3 - J 3j r 3 - a-'
by form (19), if r = x -f , a = *j, since also dr = (ir. Then wo have
PROBLEMS
Work out the following integrals.
4. f . dj< = = arc sin (2 s - 3) -f C.
J Va x - jc* - 2
' o 2
J 2 x 2 2 x
dx
/2 x - x 2
arc sin (x 1) -f C.
/2 as + s' 2
= = ln
+ 3 y + 1 ~ V5 A " \2 I/ -f 3 + V5
^ = 2 arc tan /2 + l\
l-fx-fx 2 V3 \ V3 /
207
208 INTEGRAL CALCULUS
16 . f Jfa = 1 arc tan (2^tl) + C.
J4x J 4-4x4-5 4 \ 2 /
= arc tan =
Vll \ Vll
17. f . * = | arc sin (*Z*\ + C.
J V2 - 3 x - 4 x 2 2 \ V41 /
Work out each of the following integrals and verify your results by
differentiation.
10 jc 09
'
f 4( ^
V Vx 2 - 4 x -f 13
.,
31
r _ a
. I .
J Vi> 2 -
Vl - t - 2 P
r* 4- 4x4-1*
2 w 2 4-2 w 4- l"
r x 2 dfx
36 -J 9 j.o _ 3 ^ _ i"
37.
15 4- 4 t - t 2
S8./ :
V9 x 2 4- 12 x -f 8
28 f 39 - r y dx =
*' J r 2 _ 2 r - 3 ^ V4 x 2 - 12 x + 7
When the integrand is a fraction of which the numerator is an
expression of the first degree while the denominator is an expression
of the second degree or the square root of such an expression, the
integral can be reduced to standard forms, as shown in the following
examples*
INTEGRATION
209
ILLUSTRATIVE EXAMPLE 1. Prove the following :
3x " 1 r = V4^-4-9-lln (2 x
9) 4- C.
Solution. Multiply through by dr and apply (1).
C 3x-l ^ = r 3jdr __ r dJ
J V4 x 2 -f 9 > V4 x a -h 9 J ^4 jr 2
Then by (4) and (21) we obtain the answer.
ILLUSTRATIVE EXAMPLE 2.
-f 9
Solution. 3x 2 +4z-7=r 3[U -f S V - V'l, by Illustrative Example 3, p. 207.
Let v = x + 1. Then x = v - ?, and dr = dr. Substituting,
r 2 x - 3 ^ _ T2 ( r - "i ) - 3 1 C<r^ ,,
j3x^4x~7 dr ~J air^-V) J ^-^
- 1 r 2 r (/r _ 12 r _ ^'~
~aj f 5 - v 9 J r a - V'
Using (5) and (19), and substituting back t> = jr + 3, we have the above result.
PROBLEMS
Work out the following integrals.
-f 2
~ *)<** = _ VF^T^ - arc sin j- -f C.
10 f(2x
1U 'Jx 2 +
210
INTEGRAL CALCULUS
12.
13.
14.
!^J I ln(1 _ 6 ,_ 9:r2)
- 6x 9 r 2 6
4 \ 3 x -h 1 + V2
+ C.
= Vx 2 + 2 a- -f 2 In (x + 1 + Vx 2 + 2 x) + C.
V4 x - x 2
arc sin
-f r.
15.
/27 -f 6 x - x 2
. = - V27 -f 6 x - x 2 + 3 arc si
sn
B . f fy-+ ~')</x _ ,.
'J Vl9 - 5 x + x 2
17 rjyizjjd
' J V4 x 2 - 4 x
-I- V In (r - -f Vl9 - 5 x -h .r 2 ) + C.
-f 5
1*
- :| In (2 x - 1 + V4 x 2 - 4 x + 5) -f C.
.. ^ ^ J ~ ^ ^^' r - = 2 Vl2 x ~ 4 x 2 5
'J Vl2x~4 x 2 - 5
Q / 9 X 3\
+ ~arcsinr- r ) -f C.
Work out each of the following integrals and verify your results by
differentiation.
19 -/^T^'
20 jCi^-4V/x.
21 'jT-:ij- 2 '
2g -/yo-^'
23 r( 4 - r - lw - r .
^ V3 + 5r-
27.
(3 - 4
V3 x - a- 2 - 2
2 8 . f (5^ + 2)^ .
J Vx- + 2 a- 4- 5
30 f(8 - 3 x)<ir
' J x a + x + 1
QI r (J" + 4)dr
24. r^4-
J X- -
32.
(2r +
25 r(i4J^.
^ V3j'-x 2
2 x 2 + 2 x -h 1
(3
o, f (3a- + )riJ-
33 ' J 9 x* - 3 x - 1
f. (6 - r)(fa
34
V4x 2 - 12 x + 7
INTEGRATION
133. Proofs of (22) and (23). To prove (22), substitute
t' = a sin z.
dv = a cos z dz,
211
Then
and
Hence
Va 2 v 2 = Va 2 a 2 sin 2 z = a cos z.
fVa 2 -z; 2 <fo = a 2 f cos 2 z dz = f (cos 2z+ l)dz by (5), Art. 2
To obtain the result in terms of r, we have, from above,
z = arc sin - and sin 2 z 2 sin z cos z = 2 - a "~ ??
a a a
Substituting, we obtain (22).
Proof of (23). By substituting r = a tan z, we show (see Art. 132)
readily that
(1) I V?' 2 + a 2 dr = I a sec z a sec- z dz = a 2 I sec* r dz.
In a later section it is proved that
(2) f sec 3 zdz = l sec z tan z + I In (sec z -f tan z) + r.
we derive, from (1) and (2),
Since tan z = - and sec- z =
a
(3) fx/V + a 2 d - I V?- + a 2 + ~ In (r + v/r-' + a-) + C' 9
where C'=C-^- In a. Hence (23) is proved when the positive sign
holds.
By substituting v = a sec z, we obtain (see Art. 132)
(4) I vV 2 a 2 dv = | a tan z a sec z tan z dz = a 2 j tan 2 z sec z dz
= a 2 ( sec 3 2 dz a 2 J sec z dz.
Comparing (4) with (2), we have
(5) f Vfl 2 a 2 dtf = y sec z tan z - ^ In (sec z + tan z) + C.
But sec z = - and hence tan z =
Substituting in (5),
we obtain (23) for the negative sign before a 2 .
212 INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 1. Prove the following:
f \/4 - 9 x 2 dx = 7 \/4 - 9 x 2 4- 1 arc sin -^ + C.
Solution. Compare with (22), and let a 2 = 4, r = 3 x. Then dp = 3 dx. Hence
f \/4 - 9 x' 2 dx = 7 f V4 - 9x 2 3 dx = - f v a* - t> 2 d0.
,/ 3 J <> /
Using (22), and setting P = 3 x, a 2 = 4, we have the answer.
ILLUSTRATIVE EXAMPLE 2.
f \/3 x 2 -f 4 x - 7 dx
= 1(3 x -f 2)\ / 3x* + 4z-7- 25-^2 In (3 x 4- 2 -f V9 x 12 + 12 x - 2l) -f C.
1 18
Solution. By Illustrative Example 3, p. 207,
3 x 2 -f 4 x - 7 = 3[(x -f I) 2 - V] = 3 ( r2 - fl2 )
if v = x 4 <j , a = /} . Then d?? dr.
/. C\'\lx* + 4 x - 7dx = V^ f\ t' 2 - a 2 dr.
Using (23), and setting r = x 4- , = . we obtain the answer.
PROBLEMS
Verify the following integrations :
1. JVl -4x*dx = ^ Vl - 4 .r 2 4- 1 arc sin 2 x 4- C.
2. fVl 4- 9.r 2 dx = ? Vl -f- 9 J 2 4- 1 In (3 a- 4- Vl 4- 9 .r 2 ) 4- C.
/ li D
3. /\ /~ - 1 cij: = 7 Vx 2 -4 - In (a- 4- V.r L> - 4) 4- C.
J \ 4 4
4. fV25 - 9 x 2 dx - 7 V25 - 9 x' 2 + ~ arc sin ^ 4- C.
/ 2t t> o
5. fV4 :r 2 4- 9 dx = - V4 x 2 4- 9 4- 7 In (2 x 4- V4 x 2 -f 9) 4- C.
./ 24
7. TV3 -2x- x 2 dx = 21--l V3 - 2 x - x 2 4- 2 arc sin ^- -f C.
8. rV5- 2x4-x 2 rf.r = ^-^ V5 - 2 x 4- x 2
x 2 dx =
4- 2 In (x - 1 -f V5-2x4-x 2 ) 4- C.
2 x - x 2 + arc sin (x - 1) 4- C.
10.JVlO-4x4-4x 2 dx = 2J 4 ~ 1 VlO - 4 x -f 4 x 2
+ 5 In (2 x - 1 4- VlO-4x-4-4x 2 ) 4- C.
INTEGRATION 213
Work out each of the following integrals and verify your results by
differentiation.
11.JV16-9 x 2 dx. 16. fV5 - 4 x - J a dr.
12. JV4 + 25 x 2 dx. 17. JVs 4 2 JT +7* dr.
13. JV9 r 2 - 1 rfr. 18. (Vr- - 8 .r + 7 dr.
14. JV8 - 3 .r 2 dr. 19. f V4 - 2 .r - .,- dr.
is. f V5 -f 2 .r- dr. 20. rVJ^^TTTn dr.
134. Trigonometric differentials. We shall now consider some trigo-
nometric differentials of frequent occurrence which may be readily
integrated by transformation into standard forms by means of simple
trigonometric reductions.
Example I. To find \ sin" 1 u cos n u (hi.
When either m or n is a positive odd integer, no matter what the
other may be, this integration may be performed by means of simple
transformations and formula (4),
it- -T
For example, if m is odd, we write
sin m ?/ = sn m ] u sin u.
Then, since m 1 is even, the first term of the right-hand mem-
ber will be a power of sin 2 w, and can be expressed in powers of cos 2 u
by substituting sin -> u = 1 _ Cos2 M>
Then the integral takes the form
(1) I (sum of terms involving cos ?/) sin u du.
Since sin u du r/(cos M), each term to be integrated is of the
form v n dv if v = cos u.
Similarly, if n is odd, write cos" u = cos 7 " 1 u cos u, and use the
substitution cos 2 ?/ = 1 sin 2 u. Then the integral becomes
(2) I (sum of terms involving sin u) cos u du.
214 INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 1 . Find J sin 2 x cos 5 x dx.
Solution. I sin 2 x COB* x dx I sin 2 x cos 4 x cos x dx
= fsin 2 x(l - sin 2 x) 2 cos x dx by (2), Art. 2
= I (sin 2 x 2 sin 4 x -f sin" x) cos x dx
= f (sin x)' 2 cos x dx 2 I (sin x) 4 cos x dx + | (sin x) fi cos x dx
sin' 1 x 2 sin fl x
By(4)
5 -
Here r = sin x, dr - cos x dx, and n ~ 2, 4, and 6 respectively.
ILLUSTRATIVE EXAMPLE 2. Prove I sin' J x dx = I cos :i J x - 2 cos J x -f C.
Solution, l^t J x = ?/. Then x = 2u, dx = 2du. Substituting,
(3) /sin-' \ x dx - 2 j sin :! w du.
Now f sin-' w du j sin 2 ?/ sin ?i dw - I (1 - cos 2 w) sin ?* d?^
~ / sin u du I cos 2 ?/ sin w d?/ = cos u -f \ cos r< ?^ 4- C.
Using this result in the right-hand member of (3) and substituting back u x,
we have the answer.
PROBLEMS
Work out the following integrals.
1. fsin : < j- dx = ?, cos : ' JT cos JT -f C.
2. fsin 2 B cos 8 dO = J sin-^ ^ 4- r\
3. | cos 2 sin (/> d</> i cos :< $ + C.
4. Tsin-'^ 6 j* cos B .r f/.r = ^ T sin' 1 6 jr -f r\
5. fcos^ 2 8 sin 2 <f fl = - J cos 4 2 fl + C.
6. r^~^ d.r - esc x - \ csc :} j- + C.
J sin 4 JT
7. f 51 ^ rf0 = sec + cos + C.
J COS 2
8. fcos 4 J sin 3 jc d.r = - \ cos 5 JT + \ cos 7 x + C.
9. Tsin 5 x dx = cos x -f 1 cos 3 or i cos 5 x -f- C.
INTEGRATION
10. jcos 5 x dx = sin j f sin 3 jc -f * sin 5 .r -f C.
11. f s ) n5 ?/ d?/ = - 2Vcos ?/ ( 1 - = cos- v -f - cos 4 /A -f- C.
J V cos y * \ o '9 * /
12 . r COBUL d, = 3 sin s , /! i sin , , i Bin , A r
''N/riin 2 V 2 /
Work out each of the following integrals and verify your results by
differentiation.
13. fsin 3 2 6 dO. 18. /sin 1 ' m/ cos 2 wf <Jf.
/j ^
14. fcos 3 " d0. 19. J sin-' 1 r </j-.
15. J"sin 2 a- cos 2 a- dr. 20.J'coa 1 ( + bt)dt.
[ tan
16. J sin :! / cos :t < <. Z1 J V
17. cos^' sin*
Example II. To ^?irf I tan" u du or \ ctn n u (in.
These forms can be readily integrated, when n is an integer, on
somewhat the same plan as in the previous examples.
The method consists in writing, as the first step,
tan n u tan" ~ 2 u tan 2 u tan n - ?v (?ec* L> u 1) ;
or ctn n u = ctn r ' " a ?i ctn 1 ' w - ctn ri L> ?/ (esc- w 1 ) . By (2), Art. 2
The examples illustrate the subsequent steps.
ILLUSTRATIVE EXAMPLE 1 . Find / tan 1 r dx.
Solution. I tan 4 x dx = | tan'-' x(sec ? j- - 1 )djc
= ftan 2 a- sec 2 x (Lr - I tan 2 x do-
= f (tan x)- d(tan x) - \ (sec 2 J ~ l)dx
= ianiLH: _ tan x -f x -f- C. By (4) and (10)
3
ILLUSTRATIVE EXAMPLE 2. Prove
fctn' ! 2 x dx = - i ctn- 2 x - i In sin 2 x -f C.
Solution. Let 2 x = u. Then x = i w, dx = J dw. Substituting,
(4) fctn :i 2 x dx = U ctn :i u du.
216 INTEGRAL CALCULUS
Now / ctn a u du = I ctn u ctn 2 u du
I ctri w(csc 2 u l)du
= / ctn u esc 2 u du I ctn u du
= - \ ctn 2 u - In sin u + C. By (4) and (15)
Using this result in the right-hand member of (4) and substituting back u = 2 z,
we have the answer.
Example III. To find I sec^udu or I csc n udu.
These can be easily integrated when n is a positive even integer.
The first step is to write
n 2
sec r 'u = sec"" 2 u sec-w = (tan-w + 1) 2 sec 2 w- ;
n-2
or csc n w = csc"~ 2 ?/, csc 2 ^ = (ctn 2 w + 1) 2 esc 2 w. By (2), Art. 2
The example shows the subsequent steps.
ILLUSTRATIVE EXAMPLE ,'j. Prove f sec 4 \ x dx = * tan l ?, z + 2 tan 3 z + C.
Solution. Let \ x n. Then x 2 u, dx = 2 du. Substituting,
f- /
sec 4 .j j* r/j = 2 / sec 4 M dit.
//
sec 4 w d?/ ~ I sec 2 w . sec 2 u du
= I* (tan 2 M -f 1 ) sec 2 M d?z by (2), Art. 2
= I tan 2 w sec 2 u du -f I sec L> ?/dzt
- J tan ' u -f tan u + C. By (4) and (10)
Substituting back in the right-hand member of (5) and putting u = \ x gives
the answer.
EXERCISE. Set sec L w = 1 -f tan 2 w in the right-hand member of (5), square,
and follow Illustrative Example 1 above.
Example IV. To find I taw it sec n u du or I ctn m u csc n u du.
When 7? is a positive even integer we proceed as in Example III.
ILLUSTRATIVE EXAMPLE 4. Find f tan x sec*xdx.
Solution. ftanx sec 4 z dx = |\an fi x(tan 2 z -f 1) sec 2 x dx by (2), Art. 2
= I (tan x) 8 sec 2 xdx+ Ct&n*xsec 2 xdx
= ^ + ^ + C. By(4)
Here r = tan x, dv = sec 2 xdx, etc.
INTEGRATION 217
When m is odd we may proceed as in the following example.
ILLUSTRATIVE EXAMPLE 5. Find / tan* x sec ' x dr.
Solution. I tan 6 x sec 3 x dx = I tan 4 j sec- .r sec j tan x dx
= j (sec- j- - 1 )* sec- .r sec x tan .r dr by (2), Art. 2
= / (sec ft j 2 sec 4 j 4 sec- .r) sec .r tan J fir
= ^! _!** + **lr + Ci (4)
< o o
Here r = sec x, dr = sec x tan x dr, etc.
The methods used in the above examples are obviously limited in
their application. For example, they fail in the following case.
I sec 3 u du = I sec u sec- ?/ du
= I sec u tan 2 u (In + In (sec ?/ + t^n u).
For we cannot proceed further by the elementary standard forms.
Later other methods will be developed of more general use.
PROBLEMS
Work out the following integrals.
1. J tan 3 xdxl tan 2 x + In cos jr + C.
2. fctn 3 | dx = - ~ ctn- 1 - U In sin ^ 4- C.
Jo O '>
3. fctn 3 2 x esc 2 x dr - ^> esc 2 r - J, csc :i 2 T -f- C.
4. fcsc 4 | dx = - | ctn- { | - 4 ctn ^ -f r:.
5. Jtan 5 30d6 = T V tan 4 3 - i tan 2 30+ J In sec 3 f? + C.
cos 4
= tan 2 j + \ tan 3 2 x - \ ctn 2 x + C.
- -
sin 2 2 x cos 4 2 x 6 2
220 INTEGRAL CALCULUS
r A , 3 x mn 2 x , sin 4 x . n
3. / cos 4 x dx = -f 1 i" c -
5x sin_2jr sin* 2 x 3 sin 4 a:
16 4~~ 48 64
5 x sin 2 x sin 3 2 x . 3 sin 4 x
~
C.
5. /cos x dx = -:y -f
48
1
64
/x
sin 2 ax dx = -
L*
r . ,, x o x 7
7. / sin 2 - cos- - dx =
/ tL M
s sin or
.,
h C .
4a
x sin 2 x
8 ~ 16
+
r . . . 3 x sin 2 ax . sin 4 ax ^
8.J S1 n^arrL: = T + -|^- + C.
Z_ _i_ 8 ^ n:< 4 x __ sin 8 x
~~16 96 128
9.
10./(2
12.
C.
sn
. fsin2x
x cos 4
. n sin 2 .
4 cos h T.
cos 2 x cos 6 x
4 12
sin x sin 5 x
, T o o j sin :r sin t> x ^
13. I sin 3 x sin 2 x dx = ~ h C.
r . o j sin x . sin 7 x . ^
14. I cos 4 x cos 3 x dr = 7; 1 rr h C.
Work out each of the following integrals and verify your results by
differentiation.
15. J cos 2 ax dx.
16. / cos 4 ax dx.
17. I sin 2 ax cos 2 ax dx.
18. fsin 4 - cos 2 - dO.
J ^ ij
19. (sin 4 2 a cos 4 2 a da.
20. / sin 2 x cos 6 x dx.
-fcosx) 3 dx.
22. f( Vsin 2 6 - cos 2 0) 2 d0.
23. r(VSoT5 - 2 sin 0) 2 d0.
24. /(sin 2 x - sin 3 x) 2 dx.
25. J (sin x -f cos 2 x) 2 dx.
26. J(cos x + 2 cos 2 x) 2 dr.
INTEGRATION
221
135. Integration of expressions containing Va 2 - u 2 or Vu 2 fl 2 by
trigonometric substitution. In many cases the shortest method of in-
tegrating such expressions is to change the variable as follows.
When Va 2 u 2 occurs, let u = a sin z.
When Va 2 + w 2 occurs, let u a tan z.
When V?-/ 2 a- occurs, let ?/ = a sec 1 z.
These substitutions were used in Arts. 132-133. By them the radi-
cal sign is in each case eliminated. For
(1)
(2)
(3)
Va 2 a' 2 sir\~z =
Va 2 + a 2 tan 1 ':: = a Vl 4 tan- z a sec z ;
sin L> r = a cos z;
Va 2 sec 2 z a 2 a-Vsec 2 ^ ~ 1 = a tan z.
ILLUSTRATIVE EXAMPLE 1. Find f ^ -.
J ( fl l! _ M '.')!!
Solution. Let u a sin z ; then d</ cos c (/. and, using (1),
du _ r a cos z dz = j_ / ck _ J_ /'^..j.^,
a 2_ w 2)f J a :i cos^ fl-.' cos-r r/*.' '
For, since sin z - draw a right triangle and mark the
a
sides as in the figure. Then tan z
ILLUSTRATIVE EXAMPLE 2. Prove f =ii=r = - In -
' j-\ 4 j- } 9 l{
Solution. Here v'4 x 2 -f 9 \''u- -f </ J if = - J*. <i ~ 3.
Hence let 2 r = w ; then j = ,\ M, dx = ,1 (///. Substituting,
^ '
/* (/J* __ / _ jj_//^ ____ r JTJ^/
J y\ 4 j 2 -f 9 ~/ J u \ 7/* -f -' ^ wx'w 2 -I- a 2
Let M = a tan c. Then dn - a sec-r (/r, and, using (lij,
/rf?/ / a sec- c Jc __ 1 T sec z dz __ !_ r_dz_
uVu 2 -f a 2 ~^ a tan - ' w se(> - ~ ""' tan c ~ (/t ' sin
= - 1 - fcsc z dz - - In (esc z - ctn z) -f r.
o ./
Since tan c = - draw a right triangle and mark the sides as in the figure. Then
. f = -
u
/du 1
/ .. 9 = ~
u V u 2 -\- a o
In
Substituting back in (4), and setting, as above, u = 2 x, /\
a = 3, we have the answer.
222
INTEGRAL CALCULUS
PROBLEMS
Work out the following integrals.
dx
r dx _ x_
C dx __ x
i. I ... ,= ~r i-
r f 2 df _ _ I
r ir 2 dx x
a C ?/2(/ " - 1L
o. I " .; ,
J (9 _ W 2)& v9 - v/ 2
4 - /-' -f 2 arc sin ^-f r.
arc sin h C.
7.
9,
10.
.rVj' 2 -f 4 2 \2 4. V.r 2 -f 4
./*
5 \5 -f V25 - .r 2
r.
-f c.
.f ^ ^^~ 7 + r.
J // 2 V/y 2 7 ^ .'/
/ 2
/
Work out each of the following integrals and verify your results by
differentiation.
13 .
. r
J
A/ 2 - 9 <///
l*.f j -;
9 fa
VlOO - ?/ 2 du
- 2 - 5
INTEGRATION 223
136. Integration by parts. If u and r are functions of a single inde-
pendent variable, we have, from the formula for the differentiation of
a product (V), Art. 94,
F
or, transposing, 'u dr - d(uv) r du.
Integrating this, we get the inverse formula,
(A)
I u dv - uv - I v du y
called the formula for integration by parts. This formula makes the
integration of u dr, which we may not be able to integrate directly,
depend on the integration of dr and r du, which may be in such form
as to be readily integrable. This method of integration by parts is one
of the most useful in the integral calculus.
To apply this formula in any given case the given differential must
be separated into two factors, namely, ?/ and dr. No general direc-
tions can be given for choosing these factors, except that
(a) dx is always a part of dv;
(b) it must be possible to integrate dv; and
(c) when the expression to be integrated is the product of two fwnc-
tions, it is usually best to choose the 'most complicated-looking one that it
is possible to integrate as part of dv.
The following examples will show in detail how the formula is
applied :
ILLUSTRATIVE EXAMPLE 1. FindJ x cos xdx.
Solution. Let u = x and dv cos x dx ;
then du = dx and v \ cos xdx = sin x.
Substituting in (A),
u dv u v v du
f x cos x dx = x sin x ~ f * in xdx~ x sin x -f cos x -f C.
ILLUSTRATIVE EXAMPLE 2. Find j x In x dx.
Solution. Let u = \n x and dr xdx;
then du=: ^ and v
Substituting in (A),
;
x
dx
224 INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 3. Find j*xe**dx.
Solution. Let u = c aT and dv = xdx;
r x*
then du = c ai a dx and r = | x dx =
J
Substituting in (A),
Cxe tt 'dx =- c ax - ^ - f ~ < >ar a dx
But x 2 c aT dx is not as simple to integrate as xc ar dx, which fact indicates that we
did not choose our factors suitably, instead, let
u = x and dv e" r dx ;
/xioz
c ai dx --
d
Substituting in (A),
,. f,<ir i'(' (lx
I xc ar dx x - -- I dx
J a J a
It may be necessary to apply the formula for integration by parts
more than once, as in the following example.
ILLUSTRATIVE EXAMPLE 4. Find f x' 2 c" f dx.
Solution. Let u x~ and dv = c a *dx;
fc ax
c (lj dx =
Substituting in (4),
The integral in the last term may be found by applying formula (4) again,
which gives
a
Substituting this result in (1), we get
a a 2
ILLUSTRATIVE EXAMPLE 5. Prove
Csec*zdz - \ sec z tan z -f \ In (sec z + tan 2) + C.
Solution. Let u = sec z and dv = seQ 2 zdz;
then du sec z tan z dz and v = tan z.
INTEGRATION 225
Substituting in (A),
j sec^zdz = sec z tan z - f sec z tan 2 rdz.
In the new integral, substitute tan-'c = sec*,: - 1. Then we get
J sec :i zcfc = sec r tan c - | see- rdc + In ^sec r -f tan z) + C.
Transposing the integral in the right-hand member and dividing by 2, we have the
required result.
ILLUSTRATIVE EXAMPLE (>. Prove
f f' sin ru-dr - l!lL<L!LH ~ * < MJ) r
' (/' 4 n~
Solution. Let H = c ar and c/r = sin HJ- (/JT ;
then dw = w a 'd.r and r = - ^Ji.
/i
Substituting in formula (.4), the result is
(2) fc< sin nx dr = - ^j^nx a r . ( . ()s ^^
-' N ,/
Integrate the new integral by parts.
Let u e ax and t/r = cos nx djr ;
then dw = ac ar dx and r = ^LM.
n
Hence, by (v4),
(3 ) />* cos nx dx = I^'JJIIL^ _ 1! />, sin nj f / x>
/ n 11 J
Substituting in (2), we obtain
/C nf / \ (] ^ /*
c (i:r sin nx djr a sin nx n cos no* ) | c"- 1 sin nx dx.
n' 2 \ / w- J
The two integrals in (4) are the same. Transposing the one in the right-hand
member and solving, the result is as above.
Among the most important applications of the method of inte-
gration by parts is the integration of
(a) differentials involving products,
(b) differentials involving logarithms,
(c) differentials involving inverse circular Junctions.
PROBLEMS
Work out the following integrals.
1. I jr sin jr djr sin x x cos x -f C.
2. Jin x dx = x(ln x - 1) + C.
226
INTEGRAL CALCULUS
3. \x sin ^ dx 4 sin ~
J
A C
. I x
J
2 or cos ~ + C.
7
cos nx dx
cos nx
- -
n 2
sin nx
n
. ,,
h C.
5. / w sec 2 u du ~ u tan ?/ + In cos u + C.
6. J v sin 2 3 v dr = \ r' 2 ^ r sin 6 r ^ cos f> r + C.
7. > Bin n7/ rf*
COS
2 ?/ Rin
cos
C.
8.
9.
a' (Y^- - =-4-1 + r -
[ In a In 2 a j
In x <Lr -
-f
In
-j-
4- r.
10. f arc sin x dx ~ x arc sin x + "N/l x~ + r.
11. J arc tan .r dx x arc tan r I In (1 + j~~) + C.
12. / arc cot ?/ d/y = ?/ arc cot y + J In (1 + y 2 ) + r\
13. | arc cos 2 x dx ~ x arc cos 2 x -~ J Vl 4 .r 2 + C.
14. / arc sec ?/ r/?/ ?/ arc sec ?/ In < ?y + V//- 1 ) + C.
15. fare esc - dl t arc esc ^ + 2 In (/ + V/ 2 4) + C.
/x 2 + 1 ^
^ arc tan x dx - arc tan x - + c\
17. fare tan Vj" ^ (x + 1) arc tan W Vx + r.
18. \ JT-C x ^/.r = c '(2 + 2 x + .r L> ) + r.
19. Cc cos dO ~~ (sin + cos 0) + C'.
In x ~ In (x + 1) + C.
= 2 V [In (x + 1) - 21 + C.
INTEGRATION 227
Work out each of the following integrals and verify your results by
differentiation.
25. Cx sec 2 ^ dr. 36. fare tan V.r dr.
26. J JT cos 2 2 x <Lr. 37 f r
.r*
37. j .r { arc tan .r dr.
27. J jr~ cos JT dr. 38. f (< ' -f '2 .r}- dr.
28. J arc sin rw.r dr. 39. I (U 1 -h .r*)-' dr.
29. fare ctn ^ dr. 40. / r '' cos ^ (/fl.
./ - 7 !L
30. Jarc cos - dr. 41 j", ' sin 7r/ (//
31. fare sec - <///. 42. /<'' cos "' dr.
J // ' .^ >
32. J arc esc nt dt. 43. / c - cos \1 f ill.
33. Tare sin ^1^ dr. 44 /^J ( . ()S nf (ff
34. J .r' arc sin r dr. 45.fr 4 sin <//.
/.r arc sin .r dr r
/^ _ 2 46. / csc' ; dO.
137. Comments. Integration is, on the whole, a more difficult op-
eration than differentiation. In fact, so simple an integral (in
appearance) as r
\f'jr sin x dx
cannot be worked out ; that is, there is no clvmentary function whose
derivative is V? sin x. To assist in the technique of integration,
elaborate tables of integrals have been prepared. A short table is
given in Chapter XXVII in this book. The use of this table is ex-
plained below in Art. 176. At this point let it suffice to remark that
the methods thus far presented are adequate for many problems.
Other methods will be developed in later chapters.
228
INTEGRAL CALCULUS
MISCELLANEOUS PROBLEMS
Work out each of the following integrals and verify your results by
differentiation.
r 3xdx
?i r x3dx -
'J V5 - 2 x 2
n r 3 x dx
J Vl - x 2
on r^fo
J 5 - 2 x 2
rfojc -f ?;)dx
22. -
J x 1
T 4xdx
u. 1 '
J ~Vf 2 z 2
4. j x cos 2 x dx.
f(4x + 3)dx
^ Vl - 4 x 4
24. fc' 2t cos 3 / d.
nr r . r in
Jx 2 + 4x + 8
e r (4 x-f 3)dx
25. 1 sin 5 - cZ0.
f\n I ' .1 " 1/3
J Vx~ + 4 x + 8
7 r dx
26. 1 sin 1 - do.
J 5
nw r(t - csc' 2 2 t)dt
J (a*-x*)l
r dx
" ' J / 2 -h ctn 2 I
p.H f /arc sin x ,
J x 2 - ft j- + 9
o r ^
J V 1 - x 2 a '
T 5dr
J x* - 6 r -f 8
10. r dr
- 3 - J ,., _ x + 1
30 r 5dj *
J jr a 0/4- 10
11. f(r 2j --f 2r J )' J ^.
JU. I .
^ Vx 2 - x -f 1
rx
,,;5 o vrt i i Vi ^7 r*
12.JV 2 ' ~ 2x) 2 dx.
IQ r ^
13 ' J ^ - 4 r '
14. j sin 2 ax cos ax dx.
15. j sin 2 ax cos 2 ax dx.
IG.Jln (1- Vx)dx.
17.J(2tan2 ^-ct
19
'
20
r tfdx .
'^ Vx 2 + l
32. j (c* + sin x) 2 dx.
33. J"(x - ctn x) 2 dx.
34. y (1 -f tan x) 3 dx.
35. f (
J (1 cos 0)-'
(1 -f sin Q 3 dt
36.
r-
cos
Vx 2 - 1
37. y^-' sin 2 t dt.
38.ysin 2 6 cos 3 6 dd.
39. j sin sin 4 d</>.
40. J cos a cos 2 a da.
CHAPTER XIII
CONSTANT OF INTEGRATION
138. Determination of the constant of integration by means of initial
conditions. As was pointed out on page 1 90, the constant of integration
may be found in any given case when we know the value of the
integral for some value of the variable. In fact, it is necessary, in
order to be able to determine the constant of integration, to have
some data given in addition to the differential expression to be
integrated. Let us illustrate this by means of an example.
ILLUSTRATIVE EXAMPLE. Find a function whoso first derivative is 3 x' 2 2 x -f 6,
and which shall have the value 12 when .r -- 1.
Solution. (3 x' 2 2 .r -f 5k/.r is the differential expression to be integrated.
Thus
= .H - JT-
- -f T,
where C is the constant of integration. From the conditions of our problem this
result must equal 12 when .r = 1 ; that is,
12 = 1 - 1 -i 54- <\ or r = 7.
Hence x :$ x 2 +- 5 j* -f 7 is the required function.
139. Geometrical signification of the constant of integration. We
shall illustrate this by means of examples.
ILLUSTRATIVE EXAMPLE 1. Determine the equation
of the curve at every point of which the tangent line has
the slope 2 x.
Solution. Since the slope of the tangent to a curve at
any point is -^, we have, by hypothesis,
Integrating,
cfc '
dy = 2xdx.
y = 2 Cx dx, or
(1) y = x'* + C,
where C is the constant of integration. Now if we give to
C a series of values, say 6, 0, - 3, (1 ) yields the equations
y x 2 4- 6, y x' 2 , y - x' 2 - 3,
whose loci are parabolas with axes coinciding with the jy-axis and having 6, 0, 3
respectively as intercepts on the y-axis. ,
All the parabolas (1) have the same value of ^; that is, they have the same
direction (or slope) for the same value of x. It will also be noticed that the difference
229
230
INTEGRAL CALCULUS
in the lengths of their ordinates remains the same for all values of j. Hence all the
parabolas can be obtained by moving any one of them vertically up or down, the
value of r in this case not affecting the slope of the curve.
If in the above example we impose the additional condition that the curve shall
pass through the point (1, 4), then the coordinates of this point must satisfy (1),
fciving 4 ~ 1 -f r, or r ~ M.
Hence the particular curve required is the parabola y y' 2 -4- 3.
ILLUSTRATIVE EXAMPLE 2. Determine the equation of a curve such that the
slope of the tangent line to the curve at any point is the ^
ratio of the abscissa to the ordinate with sign changed. I
Solution. The condition of t he problem is expressed
by the equation ( {,i /
or, separating the variables,
?/ <///
dx.
Integrating,
or
This, we see, represents a family of con centric circles with their centers at the origin.
If, in addition, we impose the condition that the cur\<> must pas.^ through the
point (H, 4), then 9 4-1(5 ~ 2 r.
Hence the particular curve required is the circle .r-' -f //-' ^ 25.
PROBLEMS
The following expressions have been obtained by differentiating certain
functions. Find the function in each case for the given values of the vari-
able and the function.
Junction
1. IT - 3
2. 3 4- r - f> S-
O n n\ 1\ '2 ii
4. sin 4~ cos
M-J-h
6. sec 2 4- tan
* :r 2 4- (i 2
8. for 3 4- a.r 4- 4
9. Vt 4- r
Vt
10. ctn 6 - esc 2 6
11. 3fe 2 ' 2
of
vnlnc of junction
9
-20
i ^
o
sin
i
In (2
5
tan <
7T
1
a
- arc
2 a
a
b
10
4
j
3
4
Answer
\j- - 3 JT+ 13.
304 4- 3 J- 4- i .r 2 - f; jr\
\ .V 1 - 2 ^ L1 // 2 + 2 6 2 - 4.
sin cos 04-1.
tan </> 4- In sec </> 4- 5.
X 7T
a 4 a
CONSTANT OF INTEGRATION 231
Find the equation of the family of curves such that the slope of the
tangent at any point is as follows.
12. m. A us. Straight lines, // nur -f ('.
13. x. Parabolas, // - k j- 2 -f (\
14. - Parabolas, A //-' .r -f (\
x 2
15. Semicubical parabolas, \ //'-' J, .r l -f C
16. - L ^ Seinicubical parabolas, ; \ //* -. .J .r~ -f (\
17. tt x 2 . Cubical parabolas, \t -- .H -f- r.
18. Cubical parabolas, \ //< -- .r + r.
?/" ' '
JC
19. -- Equilateral hyperbolas, \r ~~ J"' 2 C.
20. - -- Equilateral hyperbolas, .r// = r.
21. - Hyperbolas, b-jc 2 a 12 // 1 -' r.
a 2 //
22. -^- Ellipses, //-'.r- 4 a'-'//- r.
a'-y
23. i-i^- Circles, j- -f //" -f 'J .r - i // = r.
In each of the following examples find the equation of the curve whose
slope at any point is the given function of the coordinates arid which
passes through the assigned particular point.
24. .r; (1,1). -4//.S-. '1 //----- jr* -f 1.
25. 4 //; (1, 1). In // - 4 jr - 4.
26. 2 JT//; (8, 1). In u ~ s~ - 9.
f''
27. -jry; (0, 2). y = 2 r ^ .
28. ^4; (0,1). (?/+ 1; ;! = U-f l) 2 -h 3.
?/ + 1
29. - ; (0, 0). ^ + y* - 2 f<s - ( 1 ky = 0.
y - k
30. 4,; (1, 1). x In // = r- 1.
x 2
31. ?yV7; (4, 1). 3 In // - 2(xV^ - 8).
32. 4^. (2fl) 4^-^-15.
4 j* 2 15
232 INTEGRAL CALCULUS
33. ; (1,4). 37.
2 y - 3 '
; (1, 9). 3g
35. ~~ Q
*3C __ _ _L__ /'I O \
" x 2 -f 4 ' ' 40. x cos 2 y ; (4, J TT).
41. Given dy (2 j* ~h l)<tr, ?/ = 7 when y = 1. Find the value of y
when x 3. A??.s. 17.
42. Given dA "N/2 />./ r/.r, A ~- when x = ~ Find the value of A
when .r = 2 7>. ' " ,4r?*. g ;? 2 .
43. Given dy j*Vl()0 ~ j' 2 r/^, ?y when .r ~ 0. Find the value of
?/ when x = 8.
44. Given dp cos 2 6 dO, p 6 when # J TT. Find the value of p
when = ij TT.
45. Given ^s = /V4 f -f 1 dt, K = when / = 0. Find the value of s
when t = 2.
46. At every point of a certain curve //" = .r. Find the equation of the
curve if it passes through the point (3,0) and has the slope ^ at that point.
.4w,s\ 6 y = j" 3 - 6 jc 9.
1
47. At every point of a certain curve y" ~ Find the equation of
the curve if it passes through the point (1, 0) and is tangent to the line
6 x + y 6 at that point. Ans. xy + 6 x = 6.
48. Find the equation of the curve at every point of which ?/" = /
and which passes through the point (1, 1) with an inclination of 45.
49. Find the equation of the curve at every point of which y" - and
which passes through the point (1, 0) with an inclination of 135'\
50. Find the equation of the curve whose subnormal is constant and
equal to 2 a. Aws. y' 2 4 ax -f C, a parabola.
HINT. From (4), Art. 43, subnormal = y-^--
ax
51. Find the curve whose subtangent is constant and equal to a
(see (3), Art. 43). Ans. a In y = x + C.
52. Find the curve whose subnormal equals the abscissa of the point
of contact. Ans. y 2 x 2 = 2 C, an equilateral hyperbola.
CONSTANT OF INTEGRATION 233
53. Find the curve whose normal is constant (= /?), assuming that
y = R when JT = 0. An*, a* 2 + ?r = A >2 , a circle.
HINT. From Art. 43, length of normal - ij^/l -f f^\", or
*
54. Determine the curves in \\hich the length of the subnormal is
proportional to the square of the ordinate. Ans. // = OK
55. Find the equation of the curve in which the angle between the
radius vector and the tangent is one half the vectorial angle.
An*, p ~ c(l cos 0).
56. Find the curves in which the angle between the radius vector and
the tangent at any point is n times the vectorial angle. .4 ;/s. p" = c sin ?/0.
140. Physical signification of the constant of integration. The fol-
lowing examples will illustrate what is meant,
ILLUSTRATIVE EXAMPLE 1. Find the laws governing the motion of a point
which moves in a straight line with constant acceleration.
Solution. Since the acceleration from (A), Art. 59 is constant, nay /,
we have * -I
<* _ ,
or dv=.fdt. Integrating,
(1) *=// + ('.
To determine C, suppose that the initial velocity is ?><>; that is, let v - p, when
1 = 0.
These values substituted in (1) give
p () =. -f C, or C = Vo.
Hence (1) becomes
(2) v=ft + v .
Since = ^ ^ c > Art " 51) we ^ ct from (2)
*=/ + *,
or dx = ft dt + vodl . Integrating,
(3) s = \ // 2 4 ?'o/4 r\
To determine C, suppose that the initial distance is ; that is, let =o when /=0.
These values substituted in (3; give
s = -h -f C, or C = 8 .
Hence (3) becomes
(4) B = J/J 2 + (,/ + *U.
234 INTEGRAL CALCULUS
By substituting the values / 0, vo = 0, 8 = 0, = h, in (2) and (4), we get the
laws of motion of a body falling from rest in a vacuum, namely,
?' = (//, and h ^ gl*.
Eliminating / between these equations gives r v t> <//?.
ILLUSTRATIVE EXAMPLE 2. Discuss the motion of a projectile having an initial
velocity ih\ inclined at an angle a. with the horizontal,
the resistance of the air being neglected.
Solution. Assume the A'OV-plane as the plane
of motion, OX as horizontal, and ()Y as vertical, and
let the projectile be thrown from the origin.
Suppose the projectile to be acted upon by ^
gravity alone. Then the acceleration in the hori-
zontal direction will be zero and in the vertical direction - g. Hence, from (F),
Art ' 84 ' dr dv
_* = 0. and ^ = -0.
Integrating, v f = fi and r u = yt -f C- 2 .
But 0o cos a ~- initial velocity in the horizontal direction,
and o sin a = initial velocity in the vertical direction.
Hence (\ ?>< cos a and (\> ?>o sin , giving
(6) 7v = ?'o c'os <.* and r u ~ gt + ?'o sin a.
But from (C) and (Z>), Art. 83, r, = ~ and r u ~ ^ ; therefore (5) gives
dx , </// . ,
~- = ?,' () cos a and -~ = yt -f- ? ( o sin a,
d/ d/
or dj' Vo cos a <:// and dy g\ dt + i\> sin a dL
Integrating, we get
((5) a: f) cos a / -f ( '.-i and y ~. ~ \ f// 1 -' -f ?;,, sin cv / 4- ^4.
To determine r 3 and (\ we obser\e that when / = 0, x and ^/ = 0.
Substituting these values in (G gives
( \ 4 - arid C , 0.
Hence
(7) x = 7>o cos a - /, and
(8) y \ f/'~ + ''o sin <x - t.
Eliminating t between (7) and (8), we obtain
(9) y = x tan a
'2 r<>- cos 2 a
which is the equation of the trajectory and shows that the projectile will move in a
parabola.
PROBLEMS
In the following problems the relation between r and / is given. Find
the relation between s and Mf s = 2 when / = 1.
Ans. s = a(t - 1) -f b(t 2 - 1) + 2.
3. r = ^ -f -
CONSTANT OF INTEGRATION 235
In the following problems the expression for the acceleration is given.
Find the relation between r and / if r 2 when t = 3.
4. 4 - f2. An*, r = 4 f - J /' - 1. 5. V/ -f 3.
In the following problems the expression for the acceleration is given.
Find the relation between .s- and / if .x = 0, v 'JO when t 0.
7. - 32. An*, s = 20 t - 16 /-'. 8. 4 - /. 9. - 16 cos 2 t.
10. With what velocity will a stone strike the ground if dropped from
the top of a building 120 ft. high? vl / = ;?'J.) An*. 87.61 ft. per second.
11. With what velocity will the stone of Problem 10 strike the ground
if thrown downward with a speed of 'JO ft. per second? if thrown upward
with a speed of 20 ft. per second? Ans. 89.89 ft. per second.
12. A stone dropped from a balloon \\hich was rising at the rate of
15 ft. per second reached the ground in 8 sec. How high was the balloon
when the stone was dropped? Ans. 904 ft.
13. In Problem 12, if the balloon bad been falling at the rate of 15 ft.
per second, how long \\ould the stone have taken to reach the ground?
.4//,s % . 7j' ( < sec.
14. A train leaving a railroad station has an acceleration of 0.5 -f- 0.02 / ft.
per second per second. Find how far it \vill move in 20 sec. AHS. 126.7 ft.
15. A particle sliding on a certain inclined plane is subject to an ac-
celeration downward of 4 ft. per second per second. If it is started up-
ward from the bottom of the plane with a velocity of 6 ft. per second,
find the distance moved after / sec. How far will it go before sliding
backward? An*. 4.5ft.
16. If the inclined plane in Problem 15 is 20 ft. long, find the necessary
initial speed in order that the particle may just reach the^tpp of the plane.
Aits. 4 VlO ft. per second.
17. A ball thrown upward from the ground reaches a height of 80 ft.
in 1 sec. Find how high the ball will go.
18. A projectile with an initial velocity of 160 ft. per second is fired at
a vertical wall 480 ft. from the point of projection.
(a) If a 45, find the height of the point struck on the wall.
An. 192 ft.
(b) Find a so that the projectile will strike the base, of the wall.
Anx. 18 or 72.
(c) Find a. so that the projectile will strike 80 ft. above the base.
AUK. 29 or 70.
(d) Find a for the maximum height on the wall and this height.
Ana. 59 ; 256 ft.
236 INTEGRAL CALCULUS
19. If the acceleration of a particle moving with a variable velocity r
is kv 2 , where k is a constant, and if r is the velocity when / = 0, show
that - = - + kt.
v r,,
20. The resistance of the air to an automobile, within certain limits
of speed, is proportional to the speed. Hence if F is the net force gen-
erated by the motor, we have M - = F AT. Express the velocity in
F *-
terms of /, knowing that r when / 0. Ans. v (1 c A/ ).
K
ADDITIONAL PROBLEMS
1. The temperature of a liquid in a room of temperature 20 is ob-
served to be 70", and after 5 min. to be 60 . Assuming the rate of cool-
ing to be proportional to the difference of the temperatures of the liquid
and the room, find the temperature of the liquid 30 min. after the first
observation. Ans. 33.1.
2. Find the equation of the curve whoso polar subtangent is n times
the length of the corresponding radius vector and which passes through
the point (, 0). 1
Ans. p = ac n
3. Find the equation of the curve whose polar subnormal is n times
the length of the corresponding radius vector, and which passes through
the point (a, 0). Ans. p = ac" e .
4. A particle moves in the .r //-plane so that the components of velocity
parallel to the .r-axis and the //-axis are ky and kx, respectively. Prove that
the path is an equilateral hyperbola.
5. A particle projected from the top) of a tower at an angle of 45 above
the horizontal plane strikes the ground in 5 sec. at a horizontal distance
from the foot, of the tower equal to its height. Find the height of the
tower (g = 32). AUK. 200 ft.
6. A particle starts from the origin of coordinates and in / sec. its
x-component of velocity is P 4 and its //-component is 4 t.
(a) Find the position of the particle after / sec.
Atis. .r- i / :i - 4 /, i/ = 2 / 2 .
(b) Find the distance traversed along the path. Ans. s ~ J t* + 4 /.
(c) Find the equation of the path. Ans. 72 .r 2 = ?/ M 48 ?/ 2 + 576 y.
7. Find an equation of a curve for which the length of the tangent
(Art. 43) is constant (= c).
SUGGESTION. Choose the minus sign in Problem 2 (a), p. 85, and assume y = c
when .r 0. / / A
Ans. jr = c\n ( c + VfL ~ ?/" ) - Vc 2 - y*.
\ V /
8. Find the equation of the curve for which (Art. 96) a 2 ds = p 3 dO t
and which passes through the point (a, 0). Ans. p 2 = a 2 sec 2 6.
CHAPTER XIV
THE DEFINITE INTEGRAL
141. Differential of the area under a curve. Consider the con-
tinuous function <t>(x), and let
y = 0(x)
be the equation of the curve An. Let CD be a fixed and MP a
variable ordinate, and let u be the measure of the area CMPD.
When x takes on a small increment Ax, ?/ takes on an increment
Aw (= area MNQP). Completing the rectangles 1\1\KI> and MNQS,
we see that
Area MNRP < are&MNQP < area MNQS,
or MP - Ax < Aw < NQ - Ax ;
and, dividing by Ax,
if*
-x
Now let Ax approach zero as a limit ; then since M P remains fixed
and NQ approaches MP as a limit (since y is a continuous function
of x), we get 1
or, using differentials, dn = y rfr.
Theorem. T/?e differential of the area bounded by any curve, the
x-axis, a fixed ordinate, and a variable ordtnate is equal to the product oj
the variable ordinate and the differential of the corresponding abscissa.
142. The definite integral. It follows from the theorem in the last
article that if the curve AB is the locus of
then du = ?/ dx, or
(1) dw = <f)(x)dx,
* In this figure MP is less than NQ ; if MP happens to be greater than NQ, simply
'erse the inequality signs,
237
reverse
238
INTEGRAL CALCULUS
where du is the differential of the area between the curve, the x-axis,
and two ordinates. Integrating, we get
Denote ( </>(/) ds byf(x) + C.
'
(2)
We determine r by observing that u =
when x = a. Substituting these values in (2), we get
o=/(o)-f r,
and hence r = /(a;.
Then (2) becomes
(3) w -/(.r) -/(a).
The required area CKF1) is the value of u in (3) when x = 6,
Hence we have
(A) Area CEFD = f(b) - f(a).
Theorem. The difference of the values of I yds for x = a and x = 6
gives the area bounded by the curve whose ordfnate is y, the x-axis, and
the ordinatex corresponding to x = a and, .r = b.
This difference is represented by the symbol*
(4)
S*h fh
I yd* or I 0(r)rf.r,
/a ^'u
and is read "the integral from a to b of //rf.r." The operation is
called integration between limits, a being the lower and 6 the wpper
limit.f
Since (4) always has a definite value, it is called a definite integral.
For, if
I c/)(.r)dr-/(.r) + T,
t/
then
or
r h
I 0(.rW.r-
Ja
/(r) + T = [/(&) + T] - [/(a) + C],
the constant of integration having disappeared.
* This notation is due to Joseph Fourier U 768- 1830).
f The word "limit " in this connection means merely the value of the variable at one end
of its range (end value), and should not be confused with the meaning of the word in the
Theory of Limits.
THE DEFINITE INTEGRAL 239
We may accordingly define the symbol
r*> /*.
j <t>(jc)djr or I n df
'a J
as the numerical measure of the area bounded by the curre // ~ </>(>),*
the x-ajris, and the ordinates of the curre at j ~ a and JT h. This
definition presupposes that these lines hound an area; that is, the curve
does not rise or fall to infinity and does not cross the .r-uj/.s, ami both
a and b are finite.
143. Calculation of a definite integral. The process may be sum-
marized as follows :
FIRST STEP. Integrate the (jircn differential expression.
SECOND STEP. Substitute in this indefinite inlcynrt first the upper
limit and then the loirer limit for the rar/ahle, and subttacf the last
result from the first.
It is not necessary to brinpc in the constant of integration, since
it always disappears in subtracting.
ILLUSTRATIVE FAAMPLE 1. Find j x-' dr.
Solution, f * JT~ djc - I ~ 1 4 - - - -- - 1 - 12 1 . \ ws.
J\ I 'Hi '* ^
/7T
ILLUSTRATIVE FXAMPLE 12. Find I MH .r d,r.
rn i I? 1 I 111
Solution. I sin j- djr ~ - cos / ( 1 ) ' " -- ^ nK -
Jo l J< I J i J
ILLUSTRATIVE FA AMPLE 3. Prove / . - -
,/ a- -| f- 4 <i
ion. / 7
Jo *
Solution '
: -f JC*
- arc tan -
a a
1 i 1 A /, 7T
- arc tan 1 - - arc tan
4 a
/o r j r ]
V~ Q "- ~ v> ' T1 f) ~ "" ' 134 -
] 4 J - .> 1
Solution. Comparing with (19) or (19 a), r - 2 r, ry - :j f r/?> ^- 2 c/j.
To decide between the use of (19) or (19 a), consider the limits. The values of
x increase from - 1 to 0.
Then v (= 2 x) increases from 2 to 0.
Hence v 2 - 4. But o 2 = 9. Therefore ?'~ ' -, and (19 a) must be used. Thus
r(} dr /*
Evaluating in (1) gives the answer. The result is negative because the curve and
the bounding ordinates lie below the j%axis.
* d)(j) is continuous und single-valued throughout the interval [a, b].
240 INTEGRAL CALCULUS
144. Change in limits corresponding to change in variable. When
integrating by the substitution of a new variable it is sometimes
rather troublesome to translate the result back into the original
variable. When integrating between limits, however, we may avoid
the process of restoring the original variable by changing the limits
to correspond with the new variable. This process will now be
illustrated by an example.
ILLUSTRATIVE KXAMPLK. Calculate \ x
Solution. Assume x z 1 .
Then dx - 4 z ' dz, x^ = z'-', x* = z. Also, to change the limits we observe that when
JT = 0, z - 0,
and when JT 16, z 2.
r* dx _ r 2 z 4 z'' dz
1 + X*
! -f 4 arc tan 2. Ans.
The relation between the old and the new variable should be such that to each
value of one within the limits of integration there is always one, and only one, finite
value of the other. When one is given as a many-valued function of the other, care
must be taken to choose the right values.
PROBLEMS
1. Prove that f V(.r)dr = - f a f(x)dx.
Ja ' Jl>
Work out the following integrals.
ra 04
2. / (a' 2 JC - x*)dx = ~- 9. - = 5.6094.
Jo 4 Jo x -f 1
0.3167.
-2x
# 9
- ~ ln 2>
12. f W ^/2 + 2 cos 6 d6 = 4.
g i * . i< -v _ ^ i^ o ^o
'() X + 1 3 7T
in 3 x cos 3 x dx =
THE DEFINITE INTEGRAL
Find the value of each of the following definite integrals.
20. I l jcc " dr.
241
21. / '"cos-
Jv
(3 J0.
22. I "sin- ~ cos | dO.
23.
145. Calculation of areas. On page 238 it was shown that the area
between a curve, the j-axis, and the ordinates
x = a and x == b is given by the formula
r b
(B) Area= I y dx,
Ja
where the value of y in terms of x is substituted
from the equation of the given curve.
ILLUSTRATIVE EXAMPLE 1. Find the area bounded by
the parabola y = x-, the /-axis, an4 the ordinates x =-- 2
and x 4.
Solution. Substituting in the formula,
Area ABDC =
ILLUSTRATIVE EXAMPLE 2. Find the area bounded by the circle x' 2 + y' 2 =25,
the x-axis, and the ordinates x - 3, x ~ 4.
Solution. Solving, y V'25 -r-. Hence
by (22)
= 6 4- arc sin | -f 6 - ^ arc sin (- g) = 31 .6. Am.
The answer should be compared with the area of the semicircle, which is
) =39.3.
ILLUSTRATIVE EXAMPLE 3. Area under a parabola wtwsc axis i parallel to ttw
y-axis. In the figure (p. 242) the point P' on the parabolic arc PP" is chosen so
that AO = OB. The ordinates of P, P', P" are, respectively, y, y', y" . Prove that
the area between the parabola, the z-axis, and the ordinates of P and P" equals
i h(y + 4 y' + y"} if 2 fe is the distance apart of the ordinates of P and P".
242
INTEGRAL CALCULUS
Solution. Take the ?/-axis along the ordinate of P', as in the figure. Then
AB = 2 k. The equation of a parabola with axis parallel to the #-axis is, by (7),
p. 4, (x h)' z 2 [>(y k). If this is solved for y,
the result takes the form
(1) // ~ as* + 2 bx + r.
The required area A /'/'"/? (= u) is, by (B),
(2) H- (" (ax- + 2 bx + r)ds = * ah' + 2 ch.
J h
if x - 0, ?/' - O/'' r ;
if j - /,, y/ " -_- /{/>" - r ///- -f 2 />// -f r.
Therefore ', //(// + 4 ?/' + ?/" ) -; /? ( -f 2 r// = M.
P"
y'
o
/^ ^\"
O.E.D.
146. Area when the equations of the curve are given in parametric
form. 4 Let the equations of the curve be Driven in the parametric form
jr -/(/). y/ = (/>(/).
We then have // -- </>(/), and djr J'(t)dt. Hence
[!> fh
Area-- / // <1 x = / </>(/)/'(/)</*,
./a .//,
r;, and / = i^ when j = b.
CD
where
when
ILLUSTKATIVK KXAMPLK. Find the area, of the ellipse whose parametric
equations (Art. SI ) are
.r a cos (/>, // ^ b sin c/>.
Solution. Here // b sin </^,
and ds = a sin </> c?</>.
When j- --= 0, </> =-- i IT ;
and when jr (7, </> 0.
Substituting these in (1\ above, we get
Area f" , f , . ., , , , irab
I // as I ah sin" <p ad) ~
4 Jo * Ji >7r 4
Hence the entire area eciuals TT?). .4ns.
PROBLEMS
1. Find by integration the area of the triangle bounded by the line
y = "2 .r, the .r-axis, and the ordinate .r = 4. Verify your result by finding
the area as half the product of the base and altitude.
2. Find by integration the area of the trapezoid bounded by the line
jr -f // = 10, the .r-axis, and the ordinates jr = 1 and x 8. Verify your
result by finding the area as half the product of the sum of the parallel
sides and the altitude.
* For a rigorous proof of this substitution the student is referred to more advanced
treatises on the calculus.
THE DEFINITE INTEGRAL 243
Find the area bounded by the given curve, the j-axis, and the given
ordinates.
3. y .r 3 ; .r = 0, jr = 4. Aw. 64.
4. ?/ = 9 -.r 2 ; .r = 0, .r = 3. ' 18.
5. y = r< + 3 .r- -f 2 .r ; .r = - 3, .r = 3. 54.
6. ?/ = j-2-f j-f 1; .r = 2, j- = 3. 9-;.
7. .r// = A- 2 ; jr = a, j- = fc. F In f~)-
9. // = , 10 ; .r = 0, j- - 5. 20.
Vj- 4-4
10. a// = .r V J - .r 1 ' ; a* = 0, .r - a. ; \ -.
11. ?/- + 4 .r = ; JT = - 1, .* = 0. 14. // = 4 .r - .r L> ; jr -- 1, .r = 3.
12. /r = 4 .r + 16 ; .r = - 2, JT - 0. 15. //-' - <> - .r ; .r - 0, jr = 8.
13. y = .r 2 + 4 jr ; .r = - 4, j- = - 2. 16. 2 // - .r { ; ./ - 0, .r -= 2.
Find the area l>ounded by the given curve, the //-axis, and the given lines.
17. ?r = 4 jr ; // = 0, // - 4. AH*. 5.\.
18. ?/ - 4 - j-- ; // = 0, // = 3. 4;|.
19. jc 9 ?/ - /y :{ ; y 0, ?/ 3. 21. // :i - rrV ; // - 0, // - a.
20. jry 8 ; y 1, // = 4. 22. <//r - .r^ ; // ~ 0, // - a.
Sketch each of the following curves and find the area of one arch.
23. = 2 cos.r. AUK. 4.
8
24. ?/ = 2 sin \ irx. -
25. ?/ cos 2 jr. 1
26. ?/ = sin \x. 4.
27. Find the area bounded by the coordinate axes and the parabola
Vx -f V?/ = Va.
28. Prove that the area of any segment of a parabola cut off by a chord
perpendicular to the axis of the parabola is two thirds of the circumscrib-
ing rectangle.
29. P and Q are any two points on an equilateral hyperbola xy = k.
Show that the area bounded by the arc PQ, the ordinates of P and Q,
and the x-axis is equal to the area bounded by PQ, the abscissas of P
and Q, and the 7/-axis. ^
30. Find the area bounded by the catenary y = %a(c n + c V, the
x-axis, and the lines x = a and x = a. ^ ns a ^( e _ 1\
\ e)
244
INTEGRAL CALCULUS
31. Find the area included between the two parabolas t/ 2 = 2 px and
x' 2 = 2 py. Am. f p 2 .
32. Find the area included between the two parabolas y 2 = ax and
x 2 = &#. Ans. J a6.
33. Find the area inclosed by the loop of the curve whose equation is
4 y' 2 = x' 2 (4 - x). Ans. *f-
34. Find the area bounded by the curve whose equation is
y 2 - x' 2 (jr' 2 - 1) and by the line x-2. Ans. 2\/3.
35. Find the area inclosed by the loop of the curve whose equation is
36. Find the area bounded by the curve whose equation is y 2 = j* 3 x 2
and by the line JT = 2. Ans. ff .
37. Find the area inclosed by the loop of the curve whose equation is
7/2 x(x 2)' 2 . Ans. ^|V2.
38. Find the area inclosed by the loop of the curve whose equation is
4 y~ r'(4 x). Ans. *.
39. Find the area bounded by the hyperbola x 2 - y 2 a 2 and the line
x = 2a. Ans. a 2 [2 V3 - In (2 + V3)].
40. Find the area bounded by the hyperbola x 2 4 y 2 4 and the
line x (>.
41. Find the area bounded by one arch of the cycloid x = a(0 sin 0),
y a(l cos 0), and the .r-axis. Ans. 3 ira 2 .
42. Find the area of the cardioid
x ~ a (2 cos t cos 2 t),
y = a(2 sin / sin 2 /). Ans. 6 ?ra 2 .
43. The locus in the figure is called the "companion to the cycloid."
Its equations are
x = aO, Tl
y = a (l COS 0).
Find the area of one arch.
Am. 2 ?ra 2 .
44. Find the area of the hypocycloid
x = a cos 3 0,
y = a sin 3 0,
c - 9
being the parameter. Ans. -
* that is, three eighths of the area
of the circumscribing circle.
147. Geometrical representation of an integral. In the preceding
articles the definite integral appeared as an area. This does not
necessarily mean that every definite integral is an area, for the
THE DEFINITE INTEGRAL
245
physical interpretation of the result depends on the nature of the
quantities represented by the abscissa and the ordinate. Thus, if x
and y are considered as simply the coordinates of a point, then the
integral in (B), Art. 145, is indeed an area. But suppose the ordinate
represents the speed of a moving point, and the corresponding ab-
scissa the time at which the point has that speed ; then the graph
is the speed curve of the motion, and the area under it and between
any two ordinates will represent the distance passed through in the
corresponding interval of time. That is, the number which denotes
the area equals the number which denotes the distance (or value of
the integral). Similarly, a definite integral standing for volume, sur-
face, mass, force, etc. may be represented geometrically by an area.
148. Approximate integration. Trapezoidal rule. We now prove two
rules for evaluating
Jo
approximately. These rules are useful when the integration in (1) is
difficult, or impossible in terms of elementary functions.
The exact numerical value of (1 ) is the measure of t he area bounded
by the curve
(2) y=f(x),
the x-axis, and the ordinates
x = a, x = b. This area may be
evaluated approximately by add-
ing together trapezoids, as follows.
Divide the segment b a on
OX into n equal parts, each of
length Ax. Let the successive
abscissas of the points of division
be x (= a), xi, x 2 , -, x n (= b). At these points erect the correspond-
ing ordinates of the curve (2). Let these be
2/0 =f(xo), y\ =/(i) 2/2 =f(' jr 2), - , y n =/O r n).
Join the extremities of consecutive ordinates by straight lines
(chords) forming trapezoids. Then, the area of a trapezoid being
one half the product of the sum of the parallel sides multiplied by
the altitude, we get
y + 2/1 ) Ax = area of first trapezoid,
yi + 2/2) Ax = area of second trapezoid,
i( 2/n _ 1 _j_ / n )Ax = area of nth trapezoid.
246
INTEGRAL CALCULUS
Adding, we get the trapezoidal rule,
(7) Area = (jy + JA + ?2 + - + y n - i -h |y n )Ax.
It is clear that the greater the number of intervals (that is, the
smaller Ax is), the closer will the sum of the areas of the trapezoids
approach the area under the curve.
c 12
ILLUSTRATIVE KXAMPLK 1. Calculate \ x~dx by the trapezoidal rule, dividing
x = 1 to x = 12 into eleven intervals. l
^ _ a ] f > _ ]
Solution. Here -^ , 1 = Aa*. The area in question is under the
, 12 in this equation, we
n 11
curve y ~- x~. Substituting the abscissas x 1, 2, 3,
get the ordinates // 1, 4, 9, -, 144. Hence, from (T),
Area = ( i +
4 + 9+16 + 25 + 36 +49+64 + 81 + 100 + 121 + J 144) 1 = 577?,.
By integration / ":r- dx - " = 575^. Hence, in this example, the trape-
M I '> h
zoidal rule is in error by less than one
third of 1 per cent.
ILLUSTRATIVE EXAMPLE 2. Find
the approximate valw of
i*~
^
/
/
7= / "\ 4 + jc^dx
M)
by (T), taking r? = 4.
Solution. Let 2 ^
t y y
y = X 4 + r;i - 2.000 - ?/0
XT \ A r - 5 2 - 0:11 = i/
Now A.r = 0.5. j L ,^ 6 =
Make a table of val- . r ^ r, 1r
1.5 J. It) = //. 1 H
ues of j- and // as 2 ;] 4{;4 =
shown. Applying (T),
I - (1 .000 + 2.03 1 + 2.236 + 2.71 G
+ 1 .732) x 0.5 = 4.858. Am.
If we take n 10, we obtain / =4.826,
a closer approximation.
O
X
PROBLEMS
Compute the approximate values of the following integrals by the
trapezoidal rule, using the values of n indicated. Check your results by
performing the integrations.
io/ r
1
rio/ r
.1 ; w = 7.
^3 X
2. f x V25 - x 2 dx; n = 10.
/o
THE DEFINITE INTEGRAL
247
Compute the approximate values of the following integrals by the
trapezoidal rule, using the values of n indicated.
5.^ 4 -^==; n = 4. Ans. 1.227. 10. {V^O + .r 4 dr; w = 5.
fcjjVl + r'dx; n = 4. 3.283. 11. jf Wl<;~H dr ; w = 4.
7. f 1 \/125-a' 2 (ir; n = 5. 44.17. 12. T vT^'TiT^ cir ; = 5.
/o Ji
8. r 6 Vl26-o^dr; n = 4. 34.78. 13. ( 4 --^L^ ; n = 6.
9.
= 6.
9.47.
VIO ~f .r 2
l\
149. Simpson's rule (parabolic rule). Instead of connecting the
extremities of successive ordinates by chords and forming trapezoids,
we can get a still closer approxi-
mation to the area by joining
them with arcs of parabolas and
summing up the areas under these
arcs. A parabola with a vertical
axis may be passed through any
three points on a curve, and a
series of such arcs will fit the curve
more closely than the broken line
of chords. In fact, the equation ~<; " ^ ]u 1 ~M.,""M y ~ji/
of such a parabola is of the form
(1) in Illustrative Example 3, Art. 14f>, and the values of the con-
stants a, b, c may be determined so that this parabola shall pass
through three given points. In the present investigation, however,
this is not necessary.
We now divide the interval from x a = OA/o to x = b OM n
into an even number (= n) of parts, each equal to Ax. Through
each successive set of three points /'<>, ]\, P 2 ; /^, /'n, /*4 ; etc., are
drawn arcs of parabolas with vertical axes. The ordinates of these
points are 2/0, y\, yz, , y*> as indicated in the figure. The area
MoP P2 P n M n is thus replaced by a set of "double parabolic
strips" such as MoPvP\P<2M< 2 , whose upper boundary is in every
case a parabolic arc (1) of Illustrative Example 3, Art. 145. The
area of each of these double strips is found by using the formula
of this example.
For the first one, h = Ax, y 2/0, y' = yi, y"
z. Hence
248 INTEGRAL CALCULUS
Area of first parabolic strip MoP G PiP 2 M2 = - (#o + 4 y\ +
Similarly, Second (double) strip = -rp (#2 + 4 2/3 + 1/4),
o
AT-
Third (double) strip = (z/ 4 + 4 y 5 + o),
Ax
Last (double) strip = (y n -z + 4 y n ~i + f/ n ).
Adding, we get Simpson's rule (n being even),
A*
(S) Area =(y + 4^ + 2
As in the case of the trapezoidal rule, the greater the number of
parts into which MuM n is divided, the closer will the result be to the
area under the curve.
rl()
ILLUSTRATIVE EXAMPLE 1. Calculate I jc :{ dx by Simpson's rule, taking ten
intervals. ^
Solution. Here > = = 1 = Ax. The area in question is under the
n 10
curve y - .r\ Substituting the abscissas x 0, 1, 2, -, 10 in y = x'\ we get the
ordinates y = 0, 1 , 8, 27, -, 1000. Hence, from (S),
Area = Jj(0 + 4 + 1 6 4 108 -f 128 + 500 + 4 32 + 1372 -f 1024 +2916 4- 1000) =2500.
By integration, f ^' { dr = I " 2500, so that in this example Simpson's rule
gives an exact result.
ILLUSTRATIVE. EXAMPLE 2. Find the approximate value of
1 \ "\/4 -f a-* dx
by (S), taking n = 4. Jo
Solution. The table of values is given in Illustrative Example 2 of the preced-
ing article. Hence
1 = (2.000 -f 8.124 + 4.472 4- 10.864 -f 3.464) x ^ = 4 . 82 i.
o
Compare this result with that given by (7) when n = 10, namely 4.826.
In this case formula (S) gives a better approximation than (T) when n = 4.
PROBLEMS
Compute the approximate values of the following integrals by Simp-
son's rule, using the values of n indicated. Check your results by perform-
ing tbe integrations.
_ . n = 6 .
Ja 4 -f x
2. jV25-r 2 dx; n = 4.
. f Vl6 + x 2 dx ; n = 6
*/4
THE DEFINITE INTEGRAL
249
Compute the approximate values of the following integrals by Simp-
son's rule, using the values of n indicated.
* f 4 dx
5. / ; n = 4.
Jo >/4 _j_ j-a
6. rVl + .r'dr
Jo
= 4.
7. f 5 Vl26 -.r'dr; n = 4.
Ans. 1.236.
3.239.
35.68.
9.49.
9. / V6 + X 2 dx; n = 4.
r5 ~ri A*
10.
'* VTT^'
n.
12.
,r * jr (Lr ; n 4 .
-
- V 5 -f j- {
Calculate the approximate values of the following integrals by both
the trapezoidal and Simpson's rules. If the indefinite integral can be
found, calculate also the exact value of the integral.
13. f 4 Vl6 - x' 2 dx\ n = 4.
18. | "r - dr; n = 4.
14.1 XV ID X*
ax ; n ^.
.A) v^ ^T" =
r7 xdx .
-n 4
7T
n = 6.
*< V64 - x 2
20. / ~ Vi> - cos 2 6 (W
Jo
ic f 7 dx
?l~4.
rt - r' 2 10 r/f?
gj i
4
1O. 1 . ,
Ja V64 - x 2
Jo Vl -f cos 2 ;5 TT(?
17 r 8 ***. . , t
n - fi.
22. r a V4 -;{sin 2 7T0^
0: n-
Jz V3 -f r 3 ^o
150. Interchange of limits. Since
f <fr(x)dx=f(b)-f(a),
/a
and C4>(x)dx = /(a) - /(&) = - [/(&) - /(a)],
Xb /
(/>(x)dx = - I c/)(x)cfx.
Jb
Theorem. Interchanging the limits is equivalent to changing the sign
of the definite integral.
250
INTEGRAL CALCULUS
151. Decomposition of the interval of integration of the definite
integral. Since
f Tl <l>(x)dx=f(xi) -/(a), (a < x l < 6)
Jtt
r h
and I <t>(x)dx=f(b)-J(xi),
Jx\
we get, by addition,
( "<KJK/j + f *<t>(x)dx =/(fc) -/(a).
Ja Jxi
r h
But I <t>(x)dx=f(b)-f(a)',
J a
therefore, by comparing the last two expressions, we obtain
Interpreting this theorem geometrically, as in Art. 142, we see
that the integral on the left-hand side
represents the whole area CEFL), the
first integral on the right-hand side the
area CM PL), and the second integral on
the right-hand side the area MEFP. The
truth of the theorem is therefore obvious.
Evidently the definite integral may be
decomposed into any number of separate definite integrals in this way.
152. The definite integral a function of its limits
From
r
I
'a
<p(jc}dx= /(?>)- /(a)
we see that the definite integral is a function of its limits. Thus
C b C b
I <f>(f)dz has precisely the same value as I (j)(x}dx.
J Of J O.
Theorem. A definite integral is a function of its limits.
153. Improper integrals. Infinite limits. So far the limits of the
integral have been assumed as finite. Even in elementary work,
however, it is sometimes desirable to remove this restriction and to
consider integrals with infinite limits. This is possible in certain
cases by making use of the following definitions.
THE DEFINITE INTEGRAL
When the upper limit is infinite,
= lim / <t>(x)dx,
and when the lower limit is infinite,
251
X
/t> /- b
4>(jc)dx = lim I <t>(x)dx,
oo <i -* - oc t / a
provided the limits exist.
ILLUSTRATIVE EXAMPLE 1. Find ( x
./, JP-
Solution. f = lim f fr ^= lim |-1]"= lim \-}+l\=\.
J\ x 2 b- -i xJi x 2 b , f oc I .r 1 1 /, . i oc L & .1
ILLUSTRATIVE EXAMPLE 2. Find /
J
Ans.
x '
r + 8a ;< Ar v r^ 8''f/j r \
i n - X ^TI^ = fc L X FTT^ = , l' m K [
= lim
4 a' 2 arc tan
r 1 f)
JU |o
4 a' 2 "2 7R7". .Arts.
Let us interpret this result geometri-
cally. The graph of our function is the
witch, the locus of
y
8a :{
x 2 + 4 a 2
. ,
= 4 a ? arc tan -
Then as the ordinate bQ moves indefinitely to the right, the area OPQb ap-
proaches a finite limit 2 Tra 1 -'.
/-f 00 ^3.
ILLUSTRATIVE EXAMPLE 3. Find I '
Jj x
Solution.
= Hm :c =: lim (ln
X b--foo
The limit of In b as b increases without limit does not exist ; hence the integral
has in this case no meaning.
154. Improper integrals. When y (/>(*) is discontinuous. Let us
now consider cases where the function to be integrated is discontinuous
for isolated values of the variable lying within the limits of integration.
Consider first the case where the function to be integrated is con-
tinuous for all values of x between the limits a and 6 except x = a.
If a < b and c is positive, we use the definition
(1)
Xb f*b
(t>(x)dx = \im I 4>(x}dx.
e->OJa+t
252 INTEGRAL CALCULUS
Likewise, when <fr(x) is continuous except at x = b, we define
r b r*>-<
(2) I 0(z)az = lim I <t>(x)dx,
Ja - J a
dx
provided the limits exist.
ILLUSTRATIVE EXAMPLE 1. Find
Solution. Here
o 2 x' 2
becomes infinite for x = a. Therefore, by (2),
.
\ a 2 -
= lim arc sin /I -- -\ 1 = arc sin 1 = ? Ans -
* - o L \ a/ J 2
ILLUSTRATIVE EXAMPLE 2. Find f -
J x 2
Solution. Here becomes infinite for x = 0. Therefore, by (1),
In this case there is no limit, and therefore the integral does not exist.
If c lies between a and b, and 0(j) is continuous except at x = c,
then, e and e' being positive numbers, the integral between a and 6
isdcjirwd by
(3) f 0Cr)d:r = lim f' 0(^)dj + lim f 0(o;)dx,
i / -0./ a e '_0,L/r + e'
provided each separate limit exists.
ILLUSTRATIVE EXAMPLE 3. Find
2 x rfx
Solution. Hefe the function to be integrated becomes discontinuous for x = a,
that is, for a value of x between the limits of integration and 3 a. Hence the above
definition (3) must be employed. Thus
= limf 3(z 2 - a 2 )*l a ' + lim \S(x* - a 2
*-oL Jo '-*oL
= lim [3S^(a c) 2 - a 2 + 3 a^J + lim
t - o *'-* o
= 3 o* -f 6 a* = 9 a'. Ans.
To interpret this geometrically, let us
plot the graph, that is, the locus, of
2x
v =
(x*-a*rt
and note that x = a is an asymptote.
--< 2xox
o + c') 2 - a 2 ]
THE DEFINITE INTEGRAL
253
Then as PE moves to the right toward the asymptote, that is, as c approaches
zero, the area OPE approaches 3 a^ as a limit.
Similarly,
2 J dx /
Area E'QRG -
a 4- e') a - a a
approaches 6 a 5 as a limit as QE' moves to the left toward the asymptote, that is,
as c' approaches zero. Adding these results, we get 9 <i*.
ILLUSTRATIVE EXAMPLE 4. Find \ f ( r
./o (.r - a^-
Solution. This function also becomes infinite between the limits of integration.
Hence, by (3), 8
t
^ rf_ = lim f V^ + lim f "-^-
o (z-a) J -.oJo (x-a)- '-,o-'a+t'(-r - ")-
=Um /i_iU, im /_i + i
e->0\6 a/ f'-0\ rt
In this case the limits do not exist and the inte-
gral has no meaning.
If we plot the graph of this function the condi-
tion of things appears very much the same as in the
last example. We see, however, that the limits do not exist, and therein lies the
difference.
That it is important to note whether or not the given function becomes infinite
within the limits of integration will appear at once if we apply our integration
formula without any investigation. Thus
dx \ 1 2 _ 2
a'
r^i
J
(x - a)' 2
result which is absurd in view of the above discussion.
PROBLEMS
Work out each of the following integrals.
C+ dx _ TT
Jo x 2 -hl"2
7T
r+ dx _ ^TT_
o a 2 4- 6 2 x 2 ~~ 2 a6
7.f
Jo
9.
10.
= V2.
1 z
4 ira
x 2 + 2x + 2
oo x dx _ 1
(1 +x 2 ) 2 ~4*
CHAPTER XV
INTEGRATION A PROCESS OF SUMMATION
155. Introduction. Thus far we have defined integration as the
inverse of differentiation. In a great many of the applications of the
integral calculus, however, it is preferable to define integration as
a process of summation. In fact, the integral calculus was invented
in the attempt to calculate the area bounded by curves, by supposing
the given area to be divided into an "infinite number of infinitesimal
parts called elements, the sum of all these elements being the area
required." Historically, the integral sign is merely the long S, used
by early writers to indicate "sum."
This new definition, as amplified in the next article, is of fun-
damental importance, and it is essential that the student should
thoroughly understand what is meant in order to be able to apply
the integral calculus to practical problems.
156. The Fundamental Theorem of integral calculus. If </>O) is the
derivative of /(, then it has been shown in Art. 142 that the value
of the definite integral
(1) f h <t>(x)dx=f(b)-f(a)
gives the area bounded by the curve
y = (/>(/), the j-axis, and the ordinates
erected at x a and .r = b.
Now let us make the following
construction in connection with this
area. Divide the interval from :r a to x = b into any number
n of equal subintervals, erect ordinates at these points of divi-
sion, and complete rectangles by drawing horizontal lines through
the extremities of the ordinates, as in the figure. It is clear that
the sum of the areas of these n rectangles (the shaded area) is an
approximate value for the area in question. It is further evident
that the limit of the sum of the areas of these rectangles when
their number n is indefinitely increased will equal the area under
the curve.
254
INTEGRATION A PROCESS OF SUMMATION
255
Let us now carry through the following more general construction.
Divide the interval into n subintervals, not necessarily equal, and
erect ordinates at the points of division. Choose a point within each
subdivision in any manner, erect ordi-
nates at these points, and through their
extremities draw horizontal lines to form
rectangles, as in the figure. Then, as
before, the sum of the areas of these
n rectangles (the shaded area) equals
approximately the area under the curve ;
and the limit of this SUM as n increases
without limit, and each subinterval ap-
proaches zero as a limit, is precisely the area under the curve. These
considerations show that the definite integral (1) may be regarded
as the limit of a sum. Let us now formulate this result.
(a) Denote the lengths of the successive subintervals by
(b) Denote the abscissas of the points chosen in the subintervals
*^ 'v_ v.v <->". . ... <"f 3
Then the ordinates of the curve at
these points are
(c) The areas of the successive rec-
tangles are obviously
Ail 1 ',;
\
(d) The area under the curve is therefore equal to
lim [<Kxi)Azi + 0fejAx 2 + 0fe)Ax ;{ H ----- h
-
r
But from (1) the area under the curve - / ^(x
Therefore our discussion gives "
(A) f (f>(x)dx = lim [^(xi) Axi + ^(
Ja n ^
This equation has been derived by making use of the notion of area.
Intuition has aided us in establishing the result. Let us now regard
(A) simply as a theorem in analysis, which may then be stated as
follows.
256 INTEGRAL CALCULUS
FUNDAMENTAL THEOREM OF THE INTEGRAL CALCULUS
Let (f)( r jr) be continuous for the interval x = a to x = b. Let this in-
terval be divided into n subintervals whose lengths are Aa*i, Aj 2 , ,
Az n , and points be chosen, one in each subinterval, their abscissas being
Xi, 2*2 1 * > fjr n respectively. Consider the sum
(2) 0(j] )Aj, + </)(r 2 )Ar 2 H ----- h </>(
Then the limiting value of this sum when n increases without limit,
and each subinterval approaches zero ax a limit, equals the value of the
definite integral
f
Ja
<t>(x)dx.
Equation (A) may be abbreviated as follows.
C h ^
(3) I (p(x)dx = lim 2,(/>0\)Aa%-.
J a n * oc t _ !
The importance of this theorem results from the fact that we are
able to calculate by integration a magnitude which is the limit of a sum
of the form (2).
It may be remarked that each term in the sum (2) is a differen-
tial expression, since the lengths AJI, A;r 2 , , Aj ri approach zero as
a limit. Each term is also called an element of the magnitude to be
calculated.
The following rule will be of service in applying this theorem to
practical problems.
FUNDAMENTAL THEOREM. RULE
FIRST STEP. Divide the required magnitude into similar parts such
that it is clear that the desired result will be found by taking the limit of
a sum of such parts.
SECOND STEP. Find expressions for the magnitudes of these parts
such that their sum will be of the form (2).
THIRD STEP. Having chosen the proper limits x = a and x = b f we
apply the Fundamental Theorem
it //>
lim ^j<t>(Xi)&Xt = I <t>(x)dx
-* i = 1 Ja
and integrate.
x
INTEGRATION A PROCESS OF SUMMATION 257
157. Analytical proof of the Fundamental Theorem. As in the last ar-
ticle, divide the interval from x = a to x = b into any number n of sub-
intervals, not necessarily equal, and denote
the abscissas of these points of division by
feit fe2, , fen- if and the lengths of the
subintervals by Axi, Ax 2 , , A.r n . Now,
however, we let x'\, jr' 2> , x' n denote ab-
scissas, one in each interval, determined by
the Theorem of Mean Value (Art. 116), erect
ordinates at these points, and through their
extremities draw horizontal lines to form
rectangles, as in the figure. Note that here
4>(x) takes the place of f'(x). Applying (5), Art. 116 to the first interval
(a = a, fe = fei, and JT'I lies between a and feO, we have
or, since
Also, /(7> 2 ) /(fei) = </>(.r' 2 )A.r 2 , for the second interval,
/(fea) ~/(fe2) = 0Cr'3)A.r:,, for the third interval,
/(fe) -/(fen-i) = 0(x' w )Ax B> for the wth interval.
Adding these, we get
(1) /(fe) -/(a) = 0(x'i)Axi + 0(.r' 2 )A.r a -f -h (/>(:r' n )A:r n .
But 0(x'i) Axi = area of the first rectangle,
<j>(jc f 2) Aj 2 = area of the second rectangle, etc.
Hence the sum on the right-hand side of (1) equals the sum of the
areas of the rectangles. But from (1), Art. 156, the left-hand side of (1)
equals the area between the curve y = c/>(j-), the ar-axis, and the ordinates
at x = a and x = b. Then the sum
(2)
fei a
fei a == A.TI,
equals this area. And while the corresponding sum
tT
(3) ^P (/>(>) Ax t (where x t is any abscissa of
1 = 1 the subinterval Ax,)
(formed as in last article) does not also give the area, nevertheless we
may show that the two sums (2) and (3) approach equality when n in-
creases without limit and each subinterval approaches zero as a limit.
For the difference </>(x\) 0(x t ) does not exceed in numerical value the
difference of the greatest and smallest ordinates in Ax,. And, furthermore,
it is always possible* to make all these differences less in numerical
* That such is the case is shown in advanced works on the calculus.
258
INTEGRAL CALCULUS
value than any assignable positive number c, however small, by continu-
ing the process of subdivision far enough, that is, by choosing n sufficiently
large. Hence for such a choice of n the difference of the sums (2) and (3)
is less in numerical value than t(h a), that is, less than any assignable
positive quantity, however small. Accordingly, as n increases without
limit, the sums (2) and (3) approach equality, and since (2) is always
equal to the area, the fundamental result follows that
/ f> n
I <t>(jr)djc = lim
'<i n+K*-
in which the" interval [a, l>\ is subdivided in any manner whatever and x t
is any abscissa in the corresponding subinterval.
158. Areas of plane curves; rectangular coordinates. As already
explained, the area between a curve, the j-axis, and the ordinates at
x a and x ~ b is given by the formula
Area
the value of // in terms of x being substi-
tuted from the equation of the curve.
Equation (B) is readily memorized by
observing that the element of the area is
a rectangle (as ('#) of base ds and altitude ?/. The required area
ABQT is the limit of the sum of all such rectangles (strips) between
the ordinates A I* and BQ.
Let us now apply the Fundamental
Theorem (Art. 150) to the calculation of
the area bounded by the curve .r = </>(//)
(AB in figure), the //-axis, and the hori-
zontal lines ?/ = c and // = (/.
FIRST STEP. Construct the n rectan-
gles as in the figure. The required area
is clearly the limit of the sum of the
areas of these rectangles as their number
increases without limit and the altitude of each one approaches
zero as a limit.
SECOND STEP. Denote the altitudes by A?/i, A# L >, etc. Take a
point in each interval at the upper extremity and denote their
ordinates by y } , y, etc. Then the bases are 0(?/i\ 0C//2), etc., and
the sum of the areas of the rectangles is
INTEGRATION A PROCESS OF SUMMATION
THIRD STEP. Applying the Fundamental Theorem gives
259
* /(/
h'm 2>U/<>.A//,== / <f>G
71 ' x ~ 1 Jc
Hence the area between a curve, the //-axis, and the horizontal
lines y = c and y = d is given by the formula
(C)
r
=
J c
xdy,
X
the value of x in terms of // being substi-
tuted from the equation of the curve.
Formula (C) is remembered as indicating
the limit of the sum of all horizontal strips
(rec tangles) within the required area, jr and </// being, respectively,
the base and altitude of any strip. The element of the area is one
of these rectangles.
Meaning of the negative sign before an area. In formula (B), a is
less than b. Since we now interpret the right-hand member as the
limit of the sum of ti terms resulting from //, A.r, by let ting i = 1 , 2, :*,
, ?/, then, if y is 'ticyatire, each term of this sum will be negative,
and (B) will give the area with a negative sign prefixed. This means
that the area lies below the .r-axis.
r
ILLUSTRATIVE EXAMPLE 1. Find the urea of
one arch of the sine curve y sin jr.
find
Solution. Placing y = and solving for .r, we
x 0, TT, 2 TT, etc.
Substituting in (B),
Also,
Area GAB = C 'ijdx - f \\nxdx = 2.
>'u >'()
Area BCD - f 'yds - | "**\nxdjr - - 2.
ILLUSTRATIVE EXAMPLE 2. Find the area bounded
by the semicubical parabola ay' 2 x'', the //-axis, and
the lines y = a and y = 2 a.
Solution. By (C) above, and the figure, the ele-
ment of area = xdy a V" dy f substituting the value
of x from the equation of the curve MN. Hence
Area BMNC = f % V dy
Ja
- I 2 (^2 - l)= 1.304 a 2 . Ans.
Note that a 2 = area OLMB.
r
r T
7
vN
?(*,y)
260
INTEGRAL CALCULUS
In the area given by (B) one boundary is the x-axis. In (C) one
boundary is the $/-axis. Consider now the area bounded by two curves.
ILLUSTRATIVE EXAMPLE 3. Find the area bounded by the parabola y 2 = 2 x
and the straight line x y = 4.
Solution. The curves intersect at A (2, 2), B(8, 4). Divide the area into
horizontal strips by a system of equidistant lines parallel to OA' drawn from the
parabola AOH to the line AH. Let
their common distance apart be dy.
Consider the strip in the figure whose
upper side has the extremities (xi, y),
(x<2, y}. From these points drop per-
pendiculars on the lower side. Then
a rectangle is formed and its area is
given by
(1) dA = (x 2 -xi)dy. (x- 2 > x,)
This is the element of area. For the
required area is obviously the limit of
the sum of all such rectangles. That
is, by the Fundamental Theorem,
(2) Area
r d
= j c to-*')
in which x^ and x\ are functions of y determined from the equations of the bounding
curves. Thus, in this example, from x y 4 we find x X* 4 + y ; from
y* 2 x we find x Ji = \ y' 2 . Then we have, by (1),
(3) dA = (4 + y - I y'^dy.
This formula will apply to the rectangle formed from any strip. The limits are
r = - 2 (at A), d = 4 (at B}. Hence
Area =
- i y' 2 )dy = 18. Ans.
In this example the area can be divided also into strips by a system of equidis-
tant lines parallel to OK. Let A.r be their common distance apart. The upper
end of each line will lie on the parabola
O7>. But the lower end will lie on the
parabola OA when drawn to the left of
A, but on the line AH when drawn to
the right of A. If (x, y> 2 ) is the upper
extremity, and (x, y\) the lower, the
rectangle whose area is
(4) dA = (y, - y,)dx (y, > y,}
is the element of area. But in this ex-
ample it is not possible to find by (4) a
single formula to represent the area of
every one of the rectangles. For while
y. 2 = V2 x, we have ?/i = V2 x or
?/! = x 4, according as the lower ver-
tex of dA is on the parabola or on AB. Thus from (4), we have two forms of dA
and two integrations are necessary.
X
INTEGRATION A PROCESS OF SUMMATION 261
In any problem, therefore, the strips should be constructed so that only a
single formula for the element of area is necessary. Formula (4) is used when these
strips are constructed by drawing lines parallel to the y-axis.
In the Fundamental Theorem some or all of the elements 0(x,) Ax,
may be negative. Hence the limit of their sum (the definite integral)
may be zero or negative. For example, if 0(x) = sin .r, a 0, 6 = 2 TT,
the definite integral (3), Art. 156, is zero. The interpretation of this
result using areas appears from Illustrative Example 1 above.
PROBLEMS
1. Find the area bounded by the hyperbola x\i = a 2 , the .r-axis, and
the ordinates x = a and x 2 a. Ans. a' 2 In 2.
2. Find the area bounded by the curve // = In x, the .r-axis, and the
line a- = 10. AUK. 14.026.
3. Find the area bounded by the curve // = xc r , the .r-axis, and the
line x = 4. An*. 164.8.
4. Find the area bounded by the parabola V.r + V# - Va, and the
coordinate axes. A u*. { a 2 .
5. Find the entire area of the hypocycloid x^ -f ?r' = a 5 . Ans. $ ira 2 .
Find the areas bounded by the following curves. In each case draw the
figure, showing the element of area.
6. y 2 - 6 .r, x' 2 = 6 y. A HK. 12. 10. ir = 2 .r, .r- -f ?/ 2 = 4 x.
7. y' 2 = 4 x, x 2 - 6 y. 8. 11. // = 6 .r - x~, y - x.
8. y' 2 - 4 JT, 2 x - y = 4. 9. 12. // = .r ' -- 8 x, /y = x.
9. y = 4 - x 2 , y - 4 - 4 x. lOg. 13. y- = 4x,x = 12 + 2 ?/ - // 2 .
14. Find the area bounded by the parabola y 6 -f 4 a- -r 2 and the
chord joining (- 2, - 6) and (4, 6). An*. 36.
15. Find the area bounded by the semicubical parabola y* = x 2 and
the chord joining (- 1, 1) and (8, 4). ^4n. 2.7.
16. Find an expression for the area bounded by the equilateral hyper-
bola x 2 y 2 = a 2 , the .r-axis, and a line drawn from the origin to any
point (x, y) on the curve. Am q^_ ^ ix + y\
"2 \ a )
17. Find the area bounded by the curve y - x(l Vx) and the line
r = 4. Aw. -4*.
18. Find the area bounded by the curve x' 2 y = x' 2 1 and the lines
y = 1, x = 1, and x = 4. ^s. f .
19. Find the area bounded by the curve y = x 3 - 9 x 2 + 24 x - 7, the
j/-axis, and the line y = 29. Ans. 108.
262
INTEGRAL CALCULUS
A square is formed by the coordinate axes and parallel lines through
the point (1, 1). Calculate the ratio of the larger to the smaller of the two
areas into which it is divided by each of the following curves.
20. y = x 2 .
21. y = x*.
22. y = x 4 .
23. y* = x 2 .
24. Vx +
25. x 8 + ir*
ATMS. 2.
3.
4.
5.
32-3 TT
3 7T
26. y = sin ~- -
27. ?/ = jrc J
28. = tan '
29.
Ans.
7T - 2
For each of the following curves calculate the area in the first quad-
rant lying under the arc which extends from the //-axis to the first inter-
cept on the .r-axis.
30. jr -f // + y' 2 = 2. An*. I}-.
31. ?/ = ^-8r 2 + 15 j. 15J.
32. y = c r sin .r. 12.07.
33. 2 = (4 - x)\
34. ?/ = e- cos 2 j\
x
35. ?/ 4 r - cos J T
36. y sin (j- + 1).
159. Areas of plane curves; polar coordinates. Let it be required
to find the area bounded by a curve and two of its radii vectores.
Assume the equation of the curve
to be p=/(0), \D
and let OP\ and OI) be the two radii.
Denote by a and /j the angles which the
radii make with the polar axis. Apply
the Fundamental Theorem, Art. 156.
FIRST STEP. The required area is o<
clearly the limit of the sum of circular
sectors constructed as in the figure.
SECOND STEP. Let the central angles of the successive sectors be
A0i, A0i>, etc., and their radii pi, p 2 , etc. Then the sum of the areas
of the sectors is
n
pr A0i + % p 2 2 A0 2 H h i Pn 2 A0 7t = V J Pl -' A0,.
For the area of a circular sector = J- radius X arc. Hence the area of
the first sector J PI p\ A0j = o pr' A0i, etc.
THIRD STEP. Applying the Fundamental Theorem,
lim
4 = f A p 2 rfff.
./a
INTEGRATION A PROCESS OF SUMMATION 263
Hence the area swept over by the radius vector of the curve in mov-
ing from the position OP\ to the position OD is given by the formula
(D)
the value of p in terms of 6 being substituted from the equation of
the curve.
The element of area for (D) is a circular sector with radius p and
central angle dO. Hence its area is i p" dO.
ILLUSTHATI\ E EXAMPLE, Find the entire area of the lemniscate p 2 a 2 cos 2 8.
Solution. Since the figure is symmetrical
with respect to both OX and OY, the whole
area = 4 times the area of OAB.
Since p = when 9 = j. we see that if
varies from to , the radius vector OP sweeps
4
over the area OAB. Hence, substituting in (D),
Entire area = 4 x area OA B = 4 if" p- dO = 2 '-' f* cos 2 <W = a' 2 ;
*- v/<i Jo
that is, the area of both loops equals the area of a square constructed on OA as
one side.
PROBLEMS
1. Find the area bounded by the circle p = a cos and the lines
= and = 60. An*. 0.37 a 2 .
2. Find the entire area of the curve p a sin 2 0. 4n,s. i Tra 2 .
Calculate the area inclosed by each of the following curves.
3. p 2 = 4 sin 2 0. An.s. 4. 10. p = 3 + cos 3 0.
4. p = a cos 3 0. i Tra 2 . 1 1. p - a cos -f 6 sin ^.
$. p = a(l - cos 0). 1 Tra 2 . o ^
12. p = 2 cos 2 -
6. p = 2 - cos 0. f TT. ^
7.p = sin4 ITT. 13. p - a sin n0.
8. p - i + cos 2 0. J TT. 14 - P = cos * ^ ~ cos fl -
9. p = 2 + sin 3 0. 1 TT. 15. p = cos 3 0-2 cos 0.
16. Find the area bounded by the parabola pfl + cos 0) = a and the
lines = and = 120. Am. 0.866 a 2 .
17. Find the area bounded by the hyperbola p 2 cos 2 = a 2 and the
lines = and = 30. Ans. 0.33 a 2 .
264 INTEGRAL CALCULUS
18. Prove that the area generated by the radius vector of the spiral
p = e e equals one fourth of the area of the square described on the radius
vector.
a
19. Find the area of that part of the parabola p a sec 2 - which is
intercepted between the curve and the latus rectum. ~ _
a 2 .
20. Show that the area bounded by any two radii vectors of the hyper-
bolic spiral pO a is proportional to the difference between the lengths
of these radii.
o.' 2 b' 2
21. Find the area of the ellipse p' 2 = - : - : Ans. irab.
a 2 sin 2 6 -f b' 2 cos 2 6
22. Find the entire area of the curve p = afsin 20 + cos 2 0).
Ans. ira' 2 .
c\
23. Find the area below OA' within the curve p = a sin-^ -
o
&(io 7r + 27V3)a 2 .
24. Find the area bounded by p' 2 a 2 sin 4 0. Ans. a 2 .
Find the area bounded by the following curves and the given lines.
25. p = tan 0; = 0, = \ TT.
26. p = <*"; 0=1 TT, 6=1 TT.
27. p = sec + tan ; = 0, 6 = .] TT.
28. p - a sin + ft cos ; (9 - 0, = | ?r.
Calculate the area which the curves in each of the following pairs have
in common.
29. p = X cos 0, p 1 -f cos 0. Ans. TT.
30. p - 1 -f cos 0, p - 1.
- .
31. p 1 cos 0, p sin 0. | TT 1.
32. p 2 = 2 cos 2 0, p = 1. i TT + 2 - Vs.
33. p 2 = cos 2 0, p 2 = sin 2 0. 1 - J V2.
34. p = V6 cos 0, p 2 = 9 cos 2 0. i(?r + 9 - 3V).
35. p = V2 sin 0, p 2 = cos 2 0. ( TT -f 3 - 3V).
36. p = V2 cos 0, p 2 - V3 sin 2 0.
37. 3 p = V3 cos 0, p - cos 2 0.
38. 3 p = V6 sin 2 0, p 2 = cos 2 0.
39. Find the area of the inside loop of the trisectrix p = a(l 2 cos 0).
For figure, see limayon, Chapter XXVI. Ans. \ a 2 (2 TT - sVs).
INTEGRATION A PROCESS OF SUMMATION
265
160. Volumes of solids of revolution. Let V denote the volume of
the solid generated by revolving the plane surface A BCD about the
x-axis, the equation of the plane
curve DC being
FIRST STEP. Divide the seg-
ment AB into n parts of lengths
Axi, Ax 2 , *, Ax u , and pass a
plane perpendicular to the x-axis
through each point of division.
These planes will divide the solid
into n circular plates. If rectangles
are constructed with bases Axi,
Ax L >, , Ax n within the area
A BCD, then each rectangle gen-
erates a cylinder of revolution
when area ABCD is revolved. Thus a cylinder is formed correspond-
ing to each of the circular plates. (In the figure n = 4 and two cyl-
inders are shown.) The limit of the sum of these n cylinders (n > oo )
is the required volume.
SECOND STEP. Let the ordinates of the curve DC at the points
of division on the x-axis be ?/i, ?/i>, , //. Then the volume of the
cylinder generated by the rectangle AEFD will be iry\~ Axi, and the
sum of the volumes of all such cylinders is
iry 2- Ax 2
ry n - Ax rt = V
V
THIRD STEP. Applying the Fundamental Theorem (using limits
OA = a and OB = b),
lim
pi>
, -. I
J a
Hence the volume generated by revolving about the x-axis the
area bounded by the curve, the x-axis, and the ordinates x a and
x b is given by the formula
(E) V x =n f y* dx,
Ja
where the value of y in terms of x must be substituted from the
equation of the given curve.
This formula is easily remembered if we consider a thin slice or
disk of the solid between two planes perpendicular to the axis of
revolution and regard this circular plate as, approximately, a cylin-
266
INTEGRAL CALCULUS
der of altitude dx with a base of area Try 2 and hence of volume Try 2 dx.
This cylinder is the element of volume.
Similarly, when OY is the axis of revolution, we use the formula
(F) V v - TT jf x* dy,
where the value of x in terms of y must be substituted from the
equation of the given curve.
ILLUSTRATIVE EXAMPLE 1. Find the volume generated by revolving the ellipse
~ 4- #_ i about the x-axis.
a 2 b 2
Solution. Since y' 2 = ^ (a 2 - x 2 ), and the
required volume is twice the volume gen-
erated by GAB, we get, substituting in (E),
V /vj /'! J,2
^ = 7T/ y' 2 dx = TT ^(a'-
L Jo Jo a-
2 irab*
To check this result, let b - a. Then V x =
4 ira :
, the volume of a sphere, which
is only a special case of the ellipsoid. When the ellipse is revolved about its major
axis, the solid generated is called a prolate spheroid ; when about its minor axis, an
oblate spheroid.
ILLUSTRATIVE EXAMPLE 2. The area bounded by the semicubical parabola
(1) ay* = x\
the ?/-axis, and the line AB (y a) is revolved about AB.
Find the volume of the solid of revolution generated.
Solution. In the figure, OPAB is the area revolved.
Divide the segment AB into n equal parts each of
length AJ. In the figure, NM is one of these parts. The
rectangle NMPQ when revolved about AB generates a
cylinder of revolution, whose volume is an element of
the required volume. Hence
Element of volume = irr-h TT(G i/) 2 Ax,
since r = PM = RM - RP = a - y
and h = NM = Ax.
Then, by the Fundamental Theorem,
4(a,a)
(2)
ra pa
Volume of solid = V = TT / (a - y)~ dx = ir I (a 2 - 2 ay +
Jo Jo
for the limits are x = and x = AB = a. Substituting for y its value given by (l) f
the answer is V = 0.45 ira\ Ans.
This should be compared with the volume of the cone of revolution with
altitude AB (= a) and base of radius OB (= a). Volume of cone = J 7ra :? .
INTEGRATION A PROCESS OF SUMMATION
267
If the equations of the curve CD in the figure on page 265 are given
in parametric form . _ /,,\ , /A
y = </>(/),
then in () substitute y = <f>(t\ dx=f(t)dt, and change the limits
to /i and 2, if
t = /, when x = a, t = / L . when x = b.
Volume of a hollow solid of revolution. When a plane area is re-
volved about an axis not crossing the area, a hollow solid of revolu-
tion is formed. Consider the solid ob-
tained by revolving about the rr-axis the
area ACBDA of the figure. Pass through
the solid a system of equidistant planes
perpendicular to the axis of revolution
OX. Let Ax be their common distance
apart. Then the solid is divided into
hollow circular plates each of thickness
Ax. If one of the planes dividing the solid passes through M, the
hollow circular plate with one base in this plane is approximately a
hollow circular cylinder whose inner and outer radii are respec-
tively MPi (=y\) and MP> 2 (=2/2). Its altitude is A.r. Hence its
volume is ir(y> 2 - yr)Ax. Let there be w hollow cylinders, where
b a = n-Ax. The limit of the sum of these v hollow cylinders
when ft cc is the volume of the hollow solid of revolution. Hence
(3)
r b
V; = TT
Ja
The element of volume in (3) is a hollow circular cylinder with
inner radius y it outer radius // L >, and altitude Ax. The radii y { and
y>> are functions of x (= OM) found from the equations of the curves
(or the equation of the curve) bounding the area revolved.
ILLUSTRATIVE EXAMPIK 3. Find the volume of the ring solid (anclwr ring or
torus) obtained by revolving a circle of radius n about an external axis in its plane
b units from its center (b > a).
Solution. Let the equation of the circle be
x* + (y- &) 2 = a\
and let the z-axis be the axis of revolution.
Solving for y, we have
.-. dV - ir(y 2 2 - 2/1 2 ) Ax = 4 irb Va 2 - x 2 Ax.
By (3),
V x = 4 irb
x 2 dx = 2 7T 2 a 2 6. Aw.
268
INTEGRAL CALCULUS
A solid of revolution may be divided into cylindrical shells by
passing through it a system of circular cylinders whose common axis
is the axis of revolution. If the area
AC ED of the figure be revolved about
the ?/-axis, it may be shown that
(4;
r b
= 2ir \ (y>> -y } )xdx,
Ja
where OM = x, MP\ y\, MP 2 = 2/2-
The element dV is now a cylindrical shell
of radius r, altitude y? y\, and thickness Ax. Illustrative Ex-
ample 3 may be solved by (4).
PROBLEMS
1. Find the volume of the sphere generated by revolving the circle
x' 2 -f y' 2 r' 2 about a diameter. Ans. f Trr*.
2. Find by integration the volume of the truncated cone generated by
revolving the area bounded by y = 6 x, y = 0, jr = 0, j* = 4 about OA~.
Verify geometrically.
3. Find the volume of the paraboloid of revolution whose surface is
generated by revolving about its axis the arc of the parabola y' 2 = 2 px
between the origin and the point (j"i, ?/]).
Ans. irpxi' 2 \ Tr\i\ i x\ ; that is, one half of the volume
of the circumscribing cylinder.
4. Find the volume of the figure generated by revolving the arc in
Problem 3 about OY.
Ans. I 7r.ri 2 ?/i ; that is, one fifth of the cylinder of
altitude y\ and radius of base x\.
Find the volume generated by revolving about OX the areas bounded
by the following loci.
5. ?/ - jr\ u = 0, .r = 2.
6. ay' 2 .r\ ?/ = 0, .r = a.
7. The parabola Vj- -f V// = Vo, x = 0, y = 0.
8. The hypocycloid x$ -f y* = cA
9. One arch of y = sin x.
10. One arch of ?/ = cos 2 x.
11. y = e-*, y = Q, x ^ 0, x = 5.
Ans.
TTO 3
7T(1 -
INTEGRATION A PROCESS OF SUMMATION
An$. 48 TT.
12. 9 x 2 + 16 y 2 = 144.
13. y = jrf, y = 0, x= 1.
14. The witch (jr 2 -f 4 a 2 )?/ = 8 a 3 , ?/ = 0.
- 1).
4 ?T 2 a 3 .
16. ?/ 2 (2 a - x) = JT*, y = 0, jr = a. 0.2115 wa\
17. y = a- 2 - 6 x, // = 0.
18. j/ 2 = (2 - .r) 3 , ?/ = 0, jc = 0, j- = 1.
19. i/ 2 (4 + .r 2 ) = 1, // = 0, jc = 0, a* = oo.
20. Cr - 1 )y = 2, ?/ = 0, j- = 2, r = 5.
Find the volume generated by revolving about OY the areas bounded
by the following loci.
21. y = .r : ', ?/ = 0, x = 2. Aw.s. V TT.
22. 2 y 2 = x : \ ?/ = 0, T = 2. TT-
23. ?/ = < >T , // = 0, x 0. 2 TT.
24. 9 JC 2 -f 16 y' 2 - 144. 64 TT.
25. (-
26. y 2 = 9 - a% a- = 0.
27. x 2 = 16 - y, y = 0.
28. ?/ 2 = GLT, ?/ = 0, j- = a.
29. The equation of the curve OA in the figure is ?/ 2 = .r 3 . Find the
volume generated when the area
(a) OAtt is revolved about OA'.
(b) GAB is revolved about AB.
(c) OAB is revolved about CA.
(d) O.4 # is revolved about OY.
(e) OAC is revolved about 07.
(f) OAC is revolved about CA.
(g) OAC is revolved about AB.
(h) OAC is revolved about OX.
y
Arts. 64 TT.
Z ^ J 7T.
192 TT.
(U.H)
(4.8)
(4,0)
30. Find the volume of the oblate spheroid generated by revolving
the area bounded by the ellipse b 2 x 2 + a 2 ?/ 2 = a z b' 2 about the t/-axis.
An.s. | ?ra 2 6.
31. A segment of one base of thickness h is cut from a sphere of radius r.
Show by integration that its volume is * l ' --
270 INTEGRAL CALCULUS
Calculate the volume generated by revolving about each of the follow-
ing lines the area which it cuts from the corresponding curve.
32. = 3; y = 4 x - x 2 . Ann. jf TT.
33. x = 4 ; ?/ 2 = y*. *%$* *
34. y - - 4 ; T/ = 4 + 6 x - 2 x 2 . -i^p TT.
35. y = r; y = j- a . gig ;rV2.
36. = JT ; # = 4 j - x 2 . 54 TT V2.
37. ?/ = .r + 7 ; ?/ = 9 - x 2 . 4 TT Vi
38. s -f ?/ = 1 ; V7 + \/T/ = 1. ^ 7r\/2.
39. r -f ?/ = 7 ; .n/ = 6.
40. Find the volume generated by revolving one arch of the cycloid
x r arc vers ^ \/2 ry
about OX, its base.
r
HINT. Substitute dx - V :, and limits // = 0, y - 2 r, in (), Art. 160.
V2 r ^ ~ ""' Aw. 5 7T 2 r :i .
(I T \
c" 4- f "/
about the .r-axis from x = to r 6. "
4 7T^' 5 ( ? ?) _L ^ 2 ^
A w. V<' " - f (/ / -f-
42. Find the volume of the solid generated by revolving the cissoid
jT r<
y/ 2 - - about its asymptote jr 2 (/. .4//,s. 2 TrV.
L rt jc
43. Given the slope of the tangent to the tractrix - -- ?/ find
the volume of the solid generated by revolving it about OX. An*. ?ra : '.
44. Show that the volume of a conical cap of height a cut from the
solid generated by revolving the rectangular hyperbola .r j //- = a 2 about
OX equals the volume of a sphere of radius a.
45. Using the parametric equations of the hypocycloid
f i* rr (j COS :i 0,
} y = a sin : * 6,
find the volume of the solid generated by revolving it about OX.
Ans. -ffa Tra 3 .
46. Find the volume generated by revolving one arch of the cycloid
x- a(0 - sin 0),
= ad cos 6}
about its base OA". Ans. 5 ?r 2 a 3 .
Show that if the arch be revolved about ())', the volume generated
is 6 7r 3 a 3 .
INTEGRATION A PROCESS OF SUMMATION 271
47. Find the volume generated if the area bounded by the curve
y = sec \ TTJC, the .r-axis, and the lines jr = } is revolved about the .r-axis.
Ans. 4.
48. The area under the curve // = r 1 sin .r from x to a* TT is re-
volved about the .r-axis. Find the volume generated.
49. Given the curve x / 2 , ?/ = 4 t - K Find (a) the area of the loop
and (b) the volume generated by the area inside the loop when revolved
about the x-axis. Ans. (a) vV ; (b) 67.02.
50. Revolve the area bounded by the two parabolas ir = 4 x and
y* = 5 x about each axis and calculate the respective volumes.
.4 wx. OX : 10 TT; OY : *%* IT.
51. Revolve about the polar axis the part of the cardioid p 4 -f 4 cos
between the lines 6 = and = ^ and compute the volume.
Ans. 160 TT.
161. Length of a curve. By the length of a straight line we com-
monly mean the number of times we can superpose upon it, another
straight line employed as a unit of length,
as when the carpenter measures the length
of a board by making end-to-end applica-
tions of his foot rule.
Since it is impossible to make a straight
line coincide with an arc* of a curve, we
cannot measure curves in the same manner as we measure straight
lines. We proceed, then, as follows.
Divide the curve fas AB) into any number of parts in any manner
whatever (as at r, I), E] and connect the adjacent points of division,
forming chords fas AC, r/>, ])K, EB}.
The length of the curve is defined as the limit of the sum of the chords
as the number of points of division increases without limit in such a way
that at the same time each chord separately approaches zero as a limit.
Since this limit will also be the measure of the length of some
straight line, the finding of the length of a curve is also called "the
rectification of the curve."
The student has already made use of this definition for the length
of a curve in his geometry. Thus the circumference of a circle is
defined as the limit of the perimeter of the inscribed for circum-
scribed) regular polygon when the number of sides increases without
limit.
272
INTEGRAL CALCULUS
The method of the next article for finding the length of a plane
curve is based on the above definition, and the student should note
very carefully how it is applied.
162. Lengths of plane curves; rectangular coordinates. We shall now
proceed to express, in analytical form, the definition of the last article,
making use of the Fundamental Theorem.
Given the curve
and the points P'(a, c), Q(b, d) on it; to find
the length of the arc P'Q.
FIRST STEP. Take any number n of points
on the curve between P' and Q and draw the
chords joining the adjacent points, as in the
// c
figure. The required length of arc P'Q is evidently the limit of the
sum of the lengths of such chords.
SECOND STEP. Consider any one of these chords, P'P" for ex-
ample, and let the coordinates of P' and P" be / /"
P'(x f , y f ) and I>"(x' + Ax', //' + Ay').
Then, as in Art. 95,
or
P'P" = V(A:r')-
p'p" = 1 +
^
V
1
<\x' >
1
a;j
a:'* Ax'
[Dividing inside the radical by (A.r') z and multiplying outside by Ax'.]
But from the Theorem of Mean Value (Art. 116) (if Ay' is denoted
by /(ft) /(a) and A?' by b - a), we get
f^' = /W (x f < x { < x' + Ax')
Xi being the abscissa of a point Pi on the curve between P' and P" at
which the tangent is parallel to the chord.
Substituting, P'P" = [1 +f'(xi ) 2 ]*A:r' = length of first chord.
Similarly, p"p"' = [1 +f'(xz) 2 ]*Ax" = length of second chord,
p()Q - [i +/'(j M )2]*Aj fw) = length of nth chord.
The length of the inscribed broken line joining P' and Q (sum of
the chords) is then the sum of these expressions, namely,
1=1
INTEGRATION A PROCESS OF SUMMATION 2755
THIRD STEP. Applying the Fundamental Theorem,
lim V [1 +/'(j i )2)4A.ri" = f "[1 + f\r)*$ dx.
w-",^l Ja
Hence, denoting the length of arc P'Q by s, we have the
formula for the length of the arc
= f
Ja
(G)
where y f = -^ must be found in terms of x from the equation of the
given curve.
Sometimes it is more convenient to use y as the independent vari-
able. To derive a formula to cover this case, we know from Art. 39
that
= -7- ; hence dx ~ x' dy.
dx or
dy
Substituting these values in (G), and noting that the correspond-
ing y limits are c and d, we get the formula for the length of the arc,
(H) *= rV+l]dy,
J c
where x' = must be found in terms of y from the equation of the
dy
given curve.
Formula (G) may be derived in another way. In Art. 95,
formula (D),
(1) ds = (I + y' 2 )* dx
gives the differential of the arc of a curve. If we proceed from (1)
as in Art. 142, we obtain (G). Also, (H) follows from (E) in Art. 95.
Finally, if the curve is defined by parametric equations
(2) *=/(*), =
it is convenient to use
(3) 8
since, from (2), dx=f'(t)dt, dy = 4>'(t}dt.
274
INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 1. Find the length of the circumference of the circle
Solution. Differentiating, *-& -
(Is y
Substituting in (G),
Arc BA =
dx.
Jo I y
[Substituting y' 2 = r' 2 jr' 2 from the equation of the i
Lcirele in order to get everything in tcrrm of jr. \
.-. arc BA = r f' -p^L^ =- ~ (See Illustrative Example 1, Art. 154.)
./o vr' - jr 2 ~
Hence the total length equals 2 irr. Ans.
ILLUSTRATIVE EXAMPLE 2. P'ind the length of arc of one arch of the cycloid
y -_ a(0 sin 6), y a(\ - cos 0).
See Illustrative Example 2, Art. 81.
Solution. r/j = a(] cos #)(/#, (/// = a sin f?r/^.
Then dx' 2 -f <-///-' 2 -'( 1 cos ^ir/fl : ' 4 <-/- sin j I 6 (W 2 . By (5), Art. 2
Using (,'}), ,s . j 2 (/ sin ^, ^r/fy 8 . Ana.
'o
The limits are the values of at and 1) see figure, Illustrative Example 2,
Art. SI ), that is, = and = 2 TT.
ILLUSTRATIVE EXAMPLE ,'i. Find the length of the arc- of the curve 25 y 2 - x 5
from x to x ~- 2.
Solution. Differentiating, y' = \ j2. Hence, by (G),
(4) = f (1 f j jc*)*djr = \ I "(4 -f a-'M'^dr.
The integral in (4) was evaluated approximately in Illustrative Example 2,
Art. 148, by the trapezoidal rule, and also in Illustrative Example 2, Art. 149, by
Simpson's rule. Taking the latter value, s ?, (4.821) - 2.41 linear units. .4/i.s.
163. Lengths of plane curves ; polar coordinates. From (/), Art. 96,
by proceeding as in Art. 142, we get the formula Nj c >
for the length of the arc,
where p and -^ in terms of must be substi-
ctu
tuted from the equation of the given curve.
In case it is more convenient to use p as the independent variable,
and the equation is in the form
then
dp
INTEGRATION A PROCESS OF SUMMATION
Substituting this in [p 2 d6- + dp 2 ]*
276
gives
Hence if pi and p 2 are the corresponding limits of the independent
variable p, we get the formula for the length of the arc,
\dp.
where -r~ in terms of p must be substituted from the equation of the
dp
given curve.
ILLUSTRATIVE EXAMPLE. Find the perimeter of the cardioid p u([ 4- cos 0).
Solution. Here -fl <i sin 0.
uu
If we let vary from to TT, the point P will gen-
erate one half of the curve. Substituting in (/),
- - C r [a J (l -f cos O)' 2 + a- sin* 0]* dO
2 ./o
= a r ('1 -f 12 cos 19 >2 dO = 2 a f "cos - </0 = 4 a.
,/o ,/(i li
.% = 8 a. Ans.
PROBLEMS
1. Find the length of the curve whose equation is ?/ :{ = jr~ between the
points (0, 0) and (8, 4). AUK. 9.07.
2. Find the length of the arc of the semicubical parabola ay* x :i from
the origin to the ordinate x 5 a. A IW5 a
A n .S . -
y3 1
3. Find the length of the curve whose equation is if -f --- from
6 2 x
'the point where x ~ 1 to the point where jr = 3. Ar/,s. - 1 ,*-.
4. Find the length of the arc of the parabola // 2 = 2 ;>.r from the vertex
to one extremity of the latus rectum. ^ c . ?>\/2 , P i., M . x/o'i
/-I A/ ,S . I -p 111 \1 *7~ V ^j y
2 2
5. Find the length of arc of the curve y 2 = x 3 from the point where
x to the point where x \. Ans. .]|.
6. Find the length of arc of the parabola 6 y = x 2 from the origin to
the point (4, |). Ans. 4.98.
7. Approximate by Simpson's rule the length of arc of the curve
3 y = x 3 from the origin to the point (1, -g). ^s. 1.09,
276 INTEGRAL CALCULUS
8. Find the length of arc of the curve y log sec x from the origin to
the point (~ In 2\ Am. In (2 + Vs).
9. Find the length of arc of the hyperbola x 2 - y 2 = 9 from (3, 0)
to (5, 4). Ans. 4.56.
10. Find the length of the arch of the parabola y 4 x x 2 which
lies above the x-axis. Ans. 9.29.
11. Find the entire length of the hypocycloid x* + y* = a*. Ans. 6 a.
12. Rectify the catenary y = - (c u + e a ) from x = to the point (x, y).
' 2 ^
13. Find the length of one complete arch of the cycloid
x = r arc vers - V2 ry - y' 1 . Ans. 8 r.
HINT. Use (//), Art. 162. Here = 9 V _ = '
14. Rectify the curve 9 a?/ 2 = x(x 3 a) 2 from x = to x = 3 a.
Ans. 2 a VS.
? 5J
15. Find the length in one quadrant of the curve (-J -f (r) = !
a 2 -f ab + b' 2
Ans.
a -f
c* 4- 1
16. Find the length between x = a and x = 6 of the curve f jy = -~-
x 1
C 2h 1
Ans. log + a b.
H a 1
17. The equations of the involute of a circle are
(x = a (cos + 0sin 0),
\ y = a (sin cos B).
Find the length of the arc from 6 = to = ft. Ans. -J aft 2 .
18. Find the length of arc of curve !* = c * sm f from = to = -
I // << cos 2
Ans. V2(e2 - 1).
Find the length of arc of each of the following curves.
19. y = In (1 - .r 2 ) from .r = to x - \.
x 2 1
20. y = - In x from x = 1 to x ~ 2.
21. // = In esc x from x = ^ to x =
22. 3 x 2 = ?/ M from y = 1 to ?/ = 20.
23. One arch of the curve # = sin x.
INTEGRATION A PROCESS OF SUMMATION
277
24. Find the length of the spiral of Archimedes, p = a6, from the
origin to the end of the first revolution.
Am?. TraVl -f 4 TT- + ~ In (2 w + Vl -f 4 ?r 2 ).
25. Rectify the spiral p c a9 from the origin to the point (p, 0).
HINT. Use (/).
n
26. Find the length of the curve p = a sec 2 -
from = to =
Aw*. [V2 + ln (V2 + i)] a .
27. Find the length of arc of the parabola
p = from = to = -7-
1 -f cos 2
V + ln (V2 + 1).
28. Find the length of the hyperbolic spiral
p0 = a from (p lf 0i) to (p 2 , 02).
- a In
' (a +
.
4- Va'-' -f p, 1 -)
29. Show that the entire length of the curve p = a sirr* ~ is -> Show
that OA, AB, BC (see figure) are in arithmetical progression.
30. Find the length of arc of the cissoid p = 2 a tan sin from =
to e = f '
4
31. Approximate the perimeter of one leaf of the curve p = sin 2 0.
164. Areas of surfaces of revolution. A surface of revolution is
generated by revolving the arc CD
of the curve
y=J(x)
about the axis of A'.
It is desired to measure the area
of this surface by making use of the
Fundamental Theorem.
FIRST STEP. As before, divide
the interval AB into subintervals
Aari, Ax2, etc., and erect ordinates
at the points of division. Draw
the chords CE, EF, etc. of the
curve. When the curve is revolved, each chord generates the lateral
surface of a frustum of a cone of revolution. The area of the required
278
INTEGRAL CALCULUS
surface of revolution is defined as the limit of the sum of the lateral
areas of these frustums.
SECOND STEP. For the sake of clearness let us draw the first
frustum on a larger scale. Let M be the middle point of the chord
CE. Then
(1) Lateral area = 2 ir\ 7 M CE.*
In order to apply the Fundamental Theorem
it is necessary to express this product as a
function of the abscissa of some point in the
interval AJI. As in Art. 162, we get, using the
Theorem of Mean Value, the length of the chord
(2)
A/,,
I*- Axr
where Ji is the abscissa of the point P\(r\, y\)
on the arc CE where the tangent is parallel to
the chord CE. Let the horizontal line through
M intersect QP\ (the ordinate of 7'j) at /?, and denote RP\ by
Then
(3) /V A/ = //!-,.
Substituting (2) and (3) in (1 ), we get
2 TT(//, - 6, )[1 + /'(*] )~J> A.ri = lateral area of first frustum.
Similarly,
2 irtite ej)[l + /'(A)-|*A.r 1 > = lateral area of second frustum,
2 ir(y n c n )[l + /'u,,)~J*A.r,, = lateral area of last frustum.
Hence
rr
2 2 ir(y t c,)[l +f'(jr l Y 2 \* AJ, = sum of lateral areas of frustums.
This may be written
(4) 2 2 iry^
-
IT
1=1
* The lateral area of the frustum of a cone of revolution is equal to the circumference of
the middle section multiplied by the slant height.
t The student will observe that as A.TI approaches zero as a limit, ci also approaches the
limit zero.
INTEGRATION A PROCESS OF SUMMATION 279
THIRD STEP. Applying the Fundamental Theorem to the first
sum (using the limits OA = a and OB = 6), we get
lim V 2 7n/,[l +f'(x i r~] A.T, = iry[l + /W1* dx.
The limit of the second sum of (4) when n *oo is zero.* Hence the
area of the surface of revolution generated by revolving the arc CD
about OX is given by the formula
(TT\ Q 9 .
(A; OJT *
where S z denotes the required area. Or we may write the formula
in the form >b
(I) S = 2 TT ( y ds.
J a
Similarly, when OY is the axis of revolution we use the formula
(M) S
- 2 TT f x ds.
Jc
In (L) and (M) ds will have one of the three forms (C), (Z)),
of Art. 95, namely,
v~> i i
depending upon the choice of the independent variable. The last
form must be used when the given curve is defined by parametric
equations. In using (L) or (M), calculate ds first.
The formula (L) is easily remembered if we consider a narrow
band of the surface included between two planes perpendicular to the
axis of revolution, and regard it approximately as the convex surface
of a frustum of a cone of revolution of slant height r/s, with a middle
section whose circumference equals 2 Try, and hence of area 2 iry ds.
ILLUSTRATIVE EXAMPLE 1. The arc of the cubical parabola
(5) a'-y = * :<
between x = and x = a is revolved about OA', Find the area of the surface of
revolution generated.
* This is easily seen as follows. Denote the second sum by S n If equals the largest of
the positive numbers f d |, | 62 I, ,'!, then
7?
S n ^ t^[l+f'(r
1=1
The sum on the right is, by Art. 162, equal to the sum of the chords CE, EF, etc. Let
this sum be In Then S n = d n Since lim c = 0, S n is an infinitesimal, and therefore lim S n = 0.
280
INTEGRAL CALCULUS
3 X 2
Solution. From (5), y' = - Hence
a 2
= (1 +
: = (a 4 -f 9
dx.
O
i
Then the element of area = 2 iryds = (a 4 -f 9 x 4 ) 2 x 3 (
a*
Therefore, by (),
27
-I)" 1 ' = 3.6 a 2 .
ILLUSTRATIVE EXAMPLE 2. Find the area of the ellipsoid of revolution gener-
ated by revolving the ellipse whose parametric equations are (see (3), Art. 81)
x '-- a cos 0, y = 6 sin about OX.
Solution. W 7 e have
dx a sin d0, dy b cos d0,
and d.s - (dx' 2 4- dy' 2 )% (a' 2 sin 2 -f b 2 co;,- 0)i d0.
Hence the element of area = 2 iry ds = 2 7r6(a 2 sin 2 4- 6~ cos-' 0)i sin d0.
r
(6) .'. i &, = 2 TT& / 2 (a 2 sin 2 -f 6 2 cos- 0)i sin d0.
To integrate, let u cos 0. Then du = sin d0. Also,
o' J sin- 4 b' 2 cos 2 = a' 2 (I - cos 2 0) 4- b 2 cos 2 = a' 2 - (a 2 - 6 2 )w 2 .
Hence, using the new limits u = 1, w = 0, and interchanging the u limits (Art.
150), the result is
- (a 2 -
du.
(a>6)
Working this out by (22), we get
i . ,
^Sj 2 TO- H
2 irab
. . .
arc sin f , where c eccentricity
Ans.
ILLUSTRATIVE EXAMPLE 3. Find the area of the surface of revolution generated
by revolving the hypocycloid x<* 4- ?/ s ^ about the x-axis.
Solution. Here -^ = ^-, w = (^Ti ja)3
dx V A
1 .
= 1 4-
,
d.r =-.
.r 4- V fl
- T~- -
X* / X*
- dx.
Substituting in (L), noting that the arc B A
generates only one half of the surface, we get
f = 27ro^ rV
^ Jo
This is an improper integral, since the func-
tion to be integrated is discontinuous (becomes
infinite) when x = 0. Using the definition (1),
Art. 154, the result is _ 6 va < 2
12
INTEGRATION A PROCESS OF SUMMATION 281
PROBLEMS
1. Find by integration the area of the surface of the sphere generated
by revolving the circle x 2 + y 2 = r 2 about a diameter. Ans. 4 irr 2 .
2. Find by integration the area of the surface of the cone generated
by revolving about OA~ the line joining the origin to the point (a, 6).
AUK. TrhVa 2 -f b' 2 .
3. Find by integration the area of the surface of the cone generated
by revolving the line y = 2x from x = to x = 2 (a) about OX ; (b) about
O Y. Verify your results geometrically.
4. P'ind by integration the lateral area of the frustum of a cone gen-
erated by revolving about OX the line 2 // = .r - 4 from .r = to jr = 5.
Verify your result geometrically.
5. Find the area of the surface generated by revolving about OY the
arc of the parabola // = x 2 from ?/ = to y 2. AUK. \? TT.
6. Find the area of the surface generated by revolving about O.V the
arc of the parabola y = x 2 from (0, 0) to (2, 4).
7. Find the area of the surface generated by revolving about OX the
arc of the parabola ?/ 2 = 4 - jr which lies in the first quadrant, AUK. 3G.18.
8. Find the area of the surface generated by revolving about OX the
arc of the parabola ?/ 2 = 2 px from x = to .r 4 p. AUK. : v* irp 2 .
9. Find the area of the surface generated by revolving about OY the
arc of // = x* from (0, 0) to (2, 8).
Find the area of the surface generated by revolving each of the follow-
ing curves about OX.
10. 9 y = x*, from y = to x = 2. Am. jj J TT.
11. y 2 = 9 jr, from x to x 4. 49 TT.
12. // 2 = 24 - 4 jr, from x = 3 to .r = 6. '$ w.
13. 6 // - j-2, from .r = to .r = 4. f820 - 81 In 3)^
72
14. y = c~ r , from j- - to r = oo. 7r[V2 -f In ( 1 -f \/2 )].
15. The loop of 9 ay 2 .r(3 a x) 2 . 3 ira 2 .
16. 6 a V?/ x 4 -f 3 a 4 , from x = a to x 2 a. -} ( 7 ; ira 2 .
17. One loop of 8 a 2 ?/ 2 = a 2 x 2 - x 4 . j vrrz 2 .
18. ?/ 2 + 4 jr = 2 log y t from ?y = 1 to y - 2. -\? TT.
19. The cycloid \ X = a ff "" sin J-
, y = a(l cos 6).
20. The cardioid / x = a(2 cos J ~ cos 2 *>
i y = a(2sin 6 - sin 2 (9).
282 INTEGRAL CALCULUS
21. y' 2 = 4 r, from a- = to x = 3.
22. x 2 + ?/ 2 = 4, from x = 1 to x = 2.
23. x 2 + 4 ?/ 2 = 36.
24. 9 x 2 + 4 ?/ 2 = 36.
Find the area of the surface generated by revolving each of the fol-
lowing curves about OY.
25. x = y'\ from y = to y = 3. Ans. ^ 7r[(730)* - 1].
26. y = x 3 , from // = to ?/ = 3.
27. 6 a 2 .r// jr + 3 a 1 , from jr = a to x = 3 a. (20 + In 3)?ra 2 .
28. 4 ?/ = .r 2 - 2 In .r, from x = 1 to a- = 4. 24 TT.
29. 2 ?/ - .rvV' - 1 + log (V - V.r 2 - 1 ), from x = 2 to j- - 5.
Am. 78 TT.
30. ?/ 2 = jr { , from x to x 8. 713.
31. 4 // = .r-, from // = to y = 4. 33. 4 .r 2 -f /y 2 64.
32. jr* -f 4 if = 16. 34. 9 jr = // :j , from // = to // = 3.
Find the area of the surface generated by revolving each of the fol-
lowing curves.
b
A About OY
35. The ellipse ^ + f! = 1. 2 vra 2 -f ^lnl
a- r>- r 1
HINT, c = eccentricity of ellipse
36. The catenary ?/ = - \r" -f r "/>
^ >
TTfl
trom a- = tox = a. ^j-(^ 2 4- 4-^~ 2 ). 2 ?ra 2 (l -^^ 1
(Figure, [). 532)
37. r 4 -h3 = 6x//,fromx^ltoa* = 2. H w. 7r(V ? -hln2).
40. The slope of the tractrix at any point of the curve in the first
quadrant is given by -2 = ~ y . Show that the surface generated
Of. Vc 2 - //-'
by revolving about OX the arc joining the points (xj, ?/i) and (x 2t 1/2) on
the tractrix is 2 TJT(//I // L ). (Figure, p. 537)
41. The area in the first quadrant bounded by the curves whose equa-
tions are y x* and y = 4 .r is revolved about OA'. Find the total surface
of the solid generated. Ar?s. 410.3.
INTEGRATION A PROCESS OF SUMMATION
283
42. The area bounded by the ?/-axis and the curves whose equations
are x 2 = 4 y and x 2 y -f 4 = is revolved about OY. Find the total
surface of the solid generated. Ans. 141.5.
43. Find the surface generated by revolving about OX the arc of the
X 3 1
curve whose equation is y = + from .r = 1 to .r = 8. Am.
* IT
44. Find the entire surface of the solid generated by revolving about
OX the area bounded by the two parabolas y' 2 = 4 x and // 2 .r -f 3.
An*, i 7r(17Vl7 -f 32 V2 - 17) = 51.53.
45. Find the area of the surface generated by revolving about OA' one
arch of the curve y = sin jr. Ans. 14.42.
165. Solids with known parallel cross sections. In Art. 160 we dis-
cussed the volume of a solid of revolution, such as is shown in the
accompanying figure. All cross sec-
tions in planes perpendicular to the
.r-axis are circles. If OA1 = x, MC = y,
then
(1) Area cross section
ACBD=iry* = TT[ </>(*) ] 2 ,
if 2/ = (/>(x) is the equation of the
generating curve OCG. Herice the area
of the cross section in any plane perpen-
dicular to OX is a function of its perpendicular distance (~ x) from
the point O.
We shall now discuss the calculation of volumes of solids that are
not solids of revolution when it is possible to express the area of any
plane section of the solid which is
perpendicular to a fixed line (as OX)
as a function of its distance from a
fixed point (as O).
Divide the solid into n slices by
equidistant sections perpendicular to
OX, each of thickness AJ.
Let FDE be one face of such a
slice, and let ON = x. Then, by hypothesis,
L)
(2)
Area FDE = A(x).
The volume of this slice is equal, approximately, to
(3) Area FDE X A:r = A(x)Ax (base X altitude).
24 INTEGRAL CALCULUS
n
Then ^ A(x l )Ax l = sum of volumes of all such prisms. It is evi-
i = i
dent that the required volume is the limit of this sum ; hence, by the
Fundamental Theorem,
n
lim Y A(x l )Ax l = I A(x)dx,
n-^ t:= , J
and we have the formula
(N) V
were A(x) is defined in (2).
The element of volume is a prism (in some cases a cylinder)
whose altitude is dx and whose base has the area A(x). That is,
civ = A(x)dx.
ILLUSTRATIVE EXAMPLE 1. The base of
a solid is a circle of radius r. All sections
perpendicular to a fixed diameter of the
base are squares. Find the volume of the
solid.
Solution. Take the circle x- + y~ r-
iri the AT-plane as base, and OA" as the
fixed diameter. Then the section PQRS
perpendicular to OX is a square of area
4 y~, if 1>Q 2 //. (In the figure, the por-
(ion of the solid on the right of the section
1'QRS is omitted.)
Hence /\ (.r) - 4 //'-' - 4(r 2 - .r--), and,
by (/V),
Volume 4 I (r-' j'~)djr = V r ;j . Am.
ILLUSTRATIVE EXAMPLE 2. Find the
volume of a right conoid with circular
base, the radius of the base being r and the altitude a.
Solution. Placing the conoid as shown in the figure, consider a section PQR
perpendicular to OA". This section is an isosceles triangle ; and, since
RM = V2 rx - x~
(found by solving .r 2 + ^ - 2 r.r, the equation of the
circle ORAQ, for //) and
MP = a,
the area of the section is a
Substituting in (N),
a L.J. J4-E~^
This is one half the volume of the cylinder of the same base and altitude.
INTEGRATION A PROCESS OF SUMMATION 285
ILLUSTRATIVE EXAMPLE 3. Calculate the volume of the ellipsoid
by a single integration.
Solution. Consider a section of the ellipsoid perpendicular to OX, as ARCD,
with semiaxes b' and c'. The equation of the ellipse HKJG in the YOV-plane is
Solving this for y (= b') in terms of
x (= OM) gives
Similarly, from the equation of the
ellipse EFGI in the A"OZ-plane we get
Hence the area of the ellipse (section) ARC!) is
, , , irbc , ., ,,. . , N
-n-b'c' - (a 2 - j-') -- A(x}.
Substituting in (W),
, r TTbc C* " , ,
\ - / (a- jr~
a 2 J,
)(/.r - TTabc. Ana.
o
PROBLEMS
1. A solid has a circular base of radius r. The line A H is a diameter
of the base. Find the volume of the solid if every plane section perpen-
dicular to AB is
(a) an equilateral triangle ; Am. % r :i V5.
(b) an isosceles right triangle with its hypotenuse in the plane of the
base ; AUK. ^ r :{ .
(c) an isosceles right triangle with one leg in the plane of the base ;
An.s. $ r\
(d) an isosceles triangle with its altitude equal to 20 in. ; Ann. 10 ?rr 2 .
(e) an isosceles triangle with its altitude equal to its base. Am. r 3 .
2. A solid has a base in the form of an ellipse with major axis 20 in.
long and minor axis 10 in. long. Find the volume of the solid if every
section perpendicular to the major axis is
(a) a square ;
(b) an equilateral triangle ;
(c) an isosceles triangle with altitude 10 in.
An*. 1,333 cu. in.
577.3 cu. in.
785.4 cu. in.
286 INTEGRAL CALCULUS
3. The base of a solid is a segment of a parabola cut off by a chord
perpendicular to its axis. The chord has a length of 16 in. and is distant
8 in. from the vertex of the parabola. Find the volume of the solid if
every section perpendicular to the axis of the base is
(a) a square ; Arts. 1024 cu. in.
(b) an equilateral triangle; 443.4 cu. in.
(c) an isosceles triangle with altitude 10 in. 426. 7 cu. in.
4. A football is 16 in. long, and a plane section containing a seam is an
ellipse, the shorter diameter of which is 8 in. Find the volume (a) if the
leather is so stiff that every cross section is a square; (b) if the cross
section is a circle. Anx. (a) 341 : \ cu. in. ; (b) 535.9 cu. in.
5. A wedge is cut from a cylinder of radius 5 in. by two planes, one
perpendicular to the axis of the cylinder and the other passing through a
diameter of the section made by the first plane and inclined to this plane
at an angle of 45". Find the volume of the wedge. Ans. *% Q cu. in.
6. Two cylinders of equal radius r have their axes meeting at right
angles. Find the volume of the common part. Ans. -^ r a .
7. A circle of radius a moves with its center on the circumference of
an equal circle, and keeps parallel to a given plane which is perpendicular
to the plane of the given circle. Find the volume of the solid it will
generate. Ans. I a ri (3 TT + 8).
8. A variable equilateral triangle moves with its plane perpendicular
to the .r-axis and the ends of its base on the points on the curves y' 2 = 16 as
and \r = 4 cue, respectively, above the .r-axis. Find the volume generated
by the triangle as it moves from the origin to the points whose abscissa
ifl- Ans. ^V5a 3 .
9. A rectangle moves from a fixed point, one side being always equal
to the distance from this point, and the other equal to the square of this
distance. What is the volume generated while the rectangle moves a
distance of 2 ft.? Am. 4 cu. ft.
10. On the double ordinates of the ellipse + ^-=1, isosceles tri-
a'~ b' 2
angles of vertical angle 90 are described in planes perpendicular to that
of the ellipse. Find the volume of the solid generated by supposing such
a variable triangle moving from one extremity to the other of the major
axis of the ellipse. Am. $ a/A'.
Calculate the volumes bounded by the following quadric surfaces and
the given planes.
11. z = .r' 2 + 4 y* ; z = 1. Ans. \ TT.
12. 4 x' 2 + 9 c 2 + ?/ = ; # + 1 = 0. & TT.
13. a- 2 + 4 y 2 = 1 + * 2 ; z + 1 = ; z - I = 0. i TT.
14. 25 ir + 9 z 2 = 1 + .r- ; .r = ; .r = 2. ft TT.
INTEGRATION A PROCESS OF SUMMATION 287
15. .r 2 -f 4 ?/ 2 + 9 z 2 = 1. Ans. $ w.
16. z* = jr~ + 9 f/2; c -f 1 = 0. -i TT.
17. Given the parabola z - 4 - jr 2 in the A'Z-plane and the circle
0-2 4- 1/2 = 4 in the A' V-plane. From each point on the parabola lying
above the circle two lines are drawn parallel to the r#-plane to meet the
circle. Calculate the volume of the wedge-shaped solid thus formed.
Ans. 6 TT.
18. Find the volume of the solid bounded by the hyperboloid of one
" " T ~ 11 ^
sheet ^ = - f- -f- 1 and the planes jr = 0, j- = a. AUK. $ Trafor.
r* a- b 2 tj
19. A solid is bounded by one nappe of the hyperboloid of two sheets
^ - ~ - ~ - 1 and the plane jr = 2 a. Find the volume. Ans. % irabr.
20. Find the volume of the solid bounded by the surfac
'~^ 4- TT; -f -7, 1 . Ann. > irahc.
a* 0- r"
ADDITIONAL PROBLEMS
1. Find the area of the loop of the curve
y* - (jr -f 4)(.r L ' - jr -f 2 y - 4). AUK. vV'-
2. A point moves along a parabola in such a way that the radius
joining it to the focus generates area at a constant rate. If the point moves
from the vertex to one end of the latus rectum in 1 sec., what will be its
position at the end of the next 8 sec. ?
Ans. Distance from focus latus rectum.
3. Find the perimeter of the figure bounded by the lino y =_l and the
curve 4 y = e' 2r -f c~' 2r . Ans. V.3 -f In (2 -f V.'J ) 3.05.
4. The arc OP of the curve xy = x - y joins the origin to the point
P(jr\, 7/1 ), and bounds with the x-axis and the line .r = jr\ an area A. The
same arc bounds with the ?/-axis and the line y = ?/, an area />'. Prove that
the volumes obtained by revolving A about the .r-axis and tt about the
?/-axis are equal.
5. The area bounded by the curve 16 y' 2 (x -f 4) :i and its tangent
at the point (12, 16) is revolved about the r-axis. Find the volume
inclosed. Am. 1J) # 4 *.
6. The base of a solid is the area bounded by the parabola y 2 = 2 px
and its latus rectum. Every section of the solid made by a plane at right
angles to the latus rectum is a rectangle whose altitude is equal to the
distance of the section from the axis of the parabola. Find the volume of
the solid. Ans. j p 3 .
288
INTEGRAL CALCULUS
7. Given the ellipse 9 s' 2 + 25 y 2 = 225. A solid is formed about this
curve in such a way that all plane sections perpendicular to the :r-axis are
ellipses whose foci are on the given ellipse. The major and minor axes of
each section are proportional to those of the given ellipse. Find the
volume of the solid. Ans. *-& TT.
8. Let (r, y) be a point on the curve of Art. 159, O being the origin
and <)A the x-axis. Show that (D) may be written
Area = \ j (x dy - y ds),
by using the transformation (5), p. 4. The limits are determined by the
coordinates of the extremities of the curve.
9. Derive the formula of the preceding problem directly from a fig-
ure, making use of (B) and (C), Art. 158.
The formula (1) of Problem 8 is useful for parametric equations. Find
the following areas by (1).
10. The area between the involute of a circle
.r r cos + rO sin 6, y r sin rO cos 6
and the .r-axis produced to the left, in the figure of Chapter XXVI,
p. 537.
11. The entire area of the hypocycloid
of three cusps
j j' = 2 r cos -f r cos 2 6,
1 ?/ = 2 r sin - r sin 2 6.
(See figure.) Ans. 2 ?rr 2 .
12. A straight uniform wire attracts a
particle P according to the law of gravita-
tion. The particle is in the line of the wire
but not in the wire. Prove that the wire
attracts the particle as if the mass of the
wire were concentrated at a point of the
wire whoso distance from P is the mean
proportional of the distances from P to the ends of the wire.
13. Find the area of the loop of the folium of Descartes, r
HINT. Let y = tx; then .r - :
1 +/<
* at \, and dx = \ ~ 2t ;\ 3 a dt.
(1 -f
The limits for / are and
CHAPTER XVI
FORMAL INTEGRATION BY VARIOUS DEVICES
166. Introduction. Formal integration depends ultimately upon
the use of a table of integrals. If, in a given case, no formula is
found in the table resembling the given integral, it is often possible
to transform the latter so as to make it depend upon formulas in the
tables. The devices which may be used are
(a) integration by parts (Art. 136),
(b) application of the theory of rational fractions,
(c) use of a suitable substitution.
We proceed to discuss (b) and (c).
167. Integration of rational fractions. A rational fraction is a frac-
tion the numerator and denominator of which are integral rational
functions, that is, the variable is not affected with negative or frac-
tional exponents. If the degree of the numerator is equal to or greater
than that of the denominator, the fraction may be reduced to a
mixed quantity by dividing the numerator by the denominator. For
example, 40* r
* + e * J ' == j2 _|_ x __ 3 _|_ T + <* .
The last term is a fraction reduced to its lowest terms, having
the degree of the numerator less than that of the denominator. It
readily appears that the other terms are at once integrable, and hence
we need consider only the fraction.
hi order to integrate a differential expression involving such a
fraction, it is often necessary to resolve it into simpler partial frac-
tions, that is, to replace it by the algebraic sum of fractions of forms
such that we can complete the integration. That this is always pos-
sible when the denominator can be broken up into its real prime
factors is shown in algebra.*
Case I. When the factors of the denominators are all of the first
degree and none are repeated.
*See Chapter XX in Hawkes's "Advanced Algebra" (Ginn and Company, Bonton).
289
290 INTEGRAL CALCULUS
To each nonrepeated linear factor, such as x a, there corre-
sponds a partial fraction of the form
A
x a
where A is a constant. The given fraction can be expressed as a
sum of fractions of this form. The examples show the method.
ILLUSTRATIVE KXAMPLK. Find / -v X .. ' H f
J x ' 4 x 2 - 2 x
Solution. The factors of the denominator being x, x 1, x 4 2, we assume*
v x(x - 1)U 4 2) x x - 1 x42
where A, K, r are constants to be determined.
Clearing ( 1 ) of fractions, we get
(2) 2x 4 3 = A(JC - l,(x 4 2 i f- B(JC 4 2)x 4 C(x - l)x,
2 x -I 3 = 1/1 4 n 4 Ox- 4 (A 4 2 7? - Dor - 2 A.
Since this equation is an identity, we equate the coefficients of the like powers
of x in the two members according to the method of Undetermined Coefficients,
and obtain three simultaneous equations
r A 4 7? + T = 0,
cn -; A 4 2 /? - r -2,
Solving equations (in, we get
Substituting. these values in U),
2x4 a 3 5 I
.r(,r-lHx42) 2x .S(x-l) 6(.r42)
_ i r_^L_
l 6./X42
",/ x(x- DU42) 2. x ;K
= - In x 4 rl In (.r - I) - J In (j 4 2) 4 In c
, r(x-l) >; t
= In -~ - ~ A HS.
.r'^Cr 42)"
A shorter method of finding the values of A, B, and f from (2) is the following:
Let factor x ~ ; then 3 = - 2 A, or A - - .
Let factor x - -I = 0, or x = I ; then 5 = 3 #, or B = .
Let factor x 4 2 = 0, or x = - 2 ; then - I - 6 C, or C = - /; .
In every example in rational fractions the number of constants to
be determined is equal to the degree of the denominator.
* In the process of decomposing the fractional part of the given differential neither
the integral sign nor dx enters,
FORMAL INTEGRATION BY VARIOUS DEVICES 291
Case II. When the factors of the denominator are all of the first
degree and some are repeated.
To every w-fold linear factor, such as (x a) n t there will corre-
spond the sum of n partial fractions,
(x-a)" (x-a)"" 1 x-a
in which A, B, , L are constants. These partial fractions are
readily integrated. For example,
C X 1 * + 1
ILLUSTRATIVE EXAMPLE. Find J _ 4 ctr.
Solution. Since x 1 occurs three times as a factor, we assume
4.
cr-1) 2 ar-1
Clearing of fractions,
x< + 1 = A(x- 1) ;< + B;r + Or(j-- 1) + />j(j- I) 2 .
x' + 1 = (A +/.*' + (-3^4 4-r-2/x 2 + (3.4 -f /? - T + /jr - A.
Equating the coefficients of like powers of j, we get the simultaneous equations
A + />= 1,
- 3 v4 4- C - L* /> - 0,
3 A -f B - T + D -- 0,
- A = 1.
Solving, A = - 1, B = 2, C = 1, /> = 2, and
y(jr 1)" x (x
' ,. ln LH^L a +
Cr-l)-'
PROBLEMS
Work out the following integrals.
JT* X
2.
Q / (4 x -r o;ax _ J- i ^ i r*
' r 3 -f 8 x 2 -f 3 x ~~ 2 (2 x + 1 N /0 " ' v
292 INTEGRAL CALCULUS
r(lr* + 2.r 2
J 4 rr 1 -
l )n (2 x + l)(2or- I) 2 { c
2 r 2
. -
J ?/ J + 2 ?/< 2 ?/
4 3
" f -t i v o c
J] x* -f x^ 3 <:
10.
12 /" (a^ + 7 J )i._ = ln 4 = 2877-
Ju (.' + l)(j- + 2)U + :i) 3
^ a -3)^ = i n 1 _ = _ o.4K!9.
JLCllk = 5 i n n _ 4 = 1.4930.
Work out each of the following integrals.
n5.r~ + 14 j- -h 10)dr
i f( 5j ' 2 " 9)( ^ 2^
lb 'J ^-9.r "
17 f ^ <] - + 7v/ ~ ___ 24
'
10
n f
y './ (2
(2 j- + 8)(4 .r^- 1 - 1)
/ 4 ~ 3 y*)dy
20.
01 r(^ 2 - J - 5)(ir OR f (2 / 4 -f 3 / - 20 / - 28)d/
J^x. I ' "" * <wO. I *
Case III. W r /w7i //^' denominator contains factors of the second degree
but none are repeated.
FORMAL INTEGRATION BY VARIOUS DEVICES 293
To every nonrepeated quadratic factor, such as x 2 + px -f q t there
corresponds a partial fraction of the form
Ax + B
x 2 + px + q
The method of integration of this term is explained on page 209
[Illustrative Example 2).
If p is not zero, we complete the square in the denominator,
Let x + \ p = u. Then x = u \ p, dx = du. Substituting these
values, the new integral in terms of the variable u is readily integrated.
ILLUSTRATIVE EXAMPLE 1. Find |
/ X -f~ 4 X
4 A . Ex -f C
Solution. Assume J) = * + ~xMT "
Clearing of fractions, 4 = ,4(x 2 4- 4) + x(Bx 4 C 1 ) = M 4 ft)x* 4- Cx + 4 A.
Equating the coefficients of like powers of x, we get
A + B - 0, C = 0, 4 A = 4.
= 0,so that
/- 4 rfar _ rdx _ r xdx
"J x(x 2 -f 4) ~J x J x 2 -f 4
= In r - - In Or 1 -' 4 4) 4- In c = In C * Ans.
2 Vx 2 4 4
ILLUSTRATIVE EXAMPLE 2. Prove
r_&E--i n ^^ 2)2 4-^-V^arctan^
J x a + 8 ~ 24 x 2 - 2 x -f 4 ^ 12 V3
Solution. Factoring, x < -I- 8 = (x + 2) (x 2 - 2 x -f 4).
1 Ax -f B C
x :< + 8 ~~ x 2 - 2 x + 4 x 4- 2 '
1 = (Ax -f B)(x 4- 2) -f C(x 2 -2x4-4),
1 = (A 4- Ox 2 + (2 A + B - 2 C)x -f 2 B -f 4 C.
Then A = - ^ B = J, C = ^7-
Now x 2 - 2 x + 4 = (x - I) 2 -f 3 = w 2 4- 3, if x - 1 = w.
Then x = u + l,dx = du, and
Substituting back u = x - 1, using (4), and reducing, we have the answer.
294 INTEGRAL CALCULUS
Case IV. When the denominator contains factors of the second degree
some of which are repeated.
To every n-fold quadratic factor, such as (x 2 + px + q) n , there
will correspond the sum of n partial fractions,
A * + B I Cx + D I . . Lx+M _
, \n ' (~2 J_ /n^ _L n \ n ~ 1 ^2
(x 2 + px + q) n (x' 2 + px + q) n ~ l x 2 + px + q
To carry out the integration, the "reduction formula"
C du 1 I u
(0) J (u' 2 + a 2 r
proved in the next chapter, is necessary. If n > 2, repeated applica-
tions of (5) are necessary. If p is not zero, we complete the square,
jc 2 + px + q = (x + p) 2 + \ (4 q - p 2 ) = u 2 + a 2 , etc., as before.
ILLUSTRATIVE FA AMPLE. Prove
Solution. Since x 2 + 1 occurs twice as a factor, we assume
2 x-* + x 4- X __ Ax 4- 7* , O 4- />
(x- -f \)' z ~ (^ + I) 2 x 2 + 1 *
Clearing of fractions,
L J :< 4x-f3 = >lxfB4- (Car + D)(x 2 -f 1).
Kcjuuting the coefficients of like powers of x arid solving, we get
,4^-1, 7? = 3, r = L\ 7^ = 0.
The first of these two integrals is worked out by the power formula (4), the
second by (5) above, with u - x, a - 1, n = 2. Thus we obtain
Reducing, we have the answer.
Conclusion. Since a rational function may always be reduced to
the quotient of two integral rational functions, that is, to a rational
fraction, it follows from the above discussion that any rational
function whose denominator can be broken up into real quadratic
and linear factors may be expressed as the algebraic sum of integral
rational functions and partial fractions. The terms of this sum have
forms all of which we have shown how to integrate. Hence the
FORMAL INTEGRATION BY VARIOUS DEVICES 295
Theorem. The integral of every rational function whose denominator
can be broken up into real quadratic and linear factors may be found,
and is expressible in terms of algebraic, logarithmic, and inverse trigono-
metric functions, that is, in terms of the elementary functions.
PROBLEMS
Work out the following integrals.
J ^ + 3.r "'" N " "'
L f (*"* + ^^ = In (j- - 1) -h arc tan j- + C.
2 f 2 - 8 / - 8)rf/ 4> , / 2 -f 4 ,
_ .., . .^, - in -+ ( .
4 r(^ + ^-iQ)^ = i ln fii + arctan
J (2x- 3)(jT-H-4) 12 2J--3 12
5 - = In + arc tan -f C.
4^-f9o- ^ L 6 :J
(2?/: '? + ^+t 2)% - ln v + >2) + arc lan
7 r__^_ - _ 1 __ arc tan z + C.
J z 4 -f z' 2 z
-h
^ tl
I) 2
_
15 . fA^r = In 2-^| - 2 arc tan x + C.
J or 1 - 1 x + 1
=2 , (z + 2 ) _ arc tan (z + 1) + C.
296 INTEGRAL CALCULUS
* = arc tan (t + 2> ~
r
r-' 1 + 4 jr
19. f ' 14^ = l n I + * = 0.667.
Jo U -f 2) Or 2 -f 1) 9 4
u *> T^ ^ i j ;n.< i ^ j /i ___ rt $ 7 1
r o U" -f 1)(^ 2 -h 1 j 4
22.
23.
Work out each of the following integrals.
A f(6 -r 2 + ?** + 4 )dr 9q f(4 J a -h3j 2
4 - J ^T^ 29 - J ^
I) I 1 : t . 30. I ' i
J (z + l)(z 2 -h 1) ^o (2 y -h 1)(4 i/ 2 -f
r(3 r' J 4- 3 a- + l)(Lr 01 T 1 (2 jr :t 4)dj
'I ; o ^ 31- I / v , IN/ r~r^*
J .r 4 + 3 J- 2 ./o (x 2 H- l)(j -f I) 2
f(3 ^ -f jc' 2 -h 3)<ijr 32 r 3 (x -f 10)dj
J j* 4 -f3^ ' *Ji x 3 -h2r 2 -h 5x"
f(5 ^ 2 -f 12 .r + 9)dr QQ f a (2 x 3 -f 18
. I '" ^ 5O. I
168. Integration by substitution of a new variable ; rationalization.
In the last article it was shown that all rational functions whose
denominators can be broken up into real quadratic and linear factors
may be integrated. Of algebraic functions which are not rational, that
is, such as contain radicals, only a small number, relatively speaking,
can be integrated in terms of elementary functions. By substituting
a new variable, however, these functions can in some cases be trans-
formed into equivalent functions that are either in the list of standard
forms (Art. 128) or else rational. The method of integrating a func-
tion that is not rational by substituting for the old variable such
a function of a new variable that the result is a rational function is
sometimes called integration by rationalization. This is a very im-
portant artifice in integration, and we shall now take up some of the
more important cases coming under this head.
FORMAL INTEGRATION BY VARIOUS DEVICES 297
Differentials containing fractional powers of x only. Suck an expres-
sion can be transformed into a rational form by m.eans of the substitution
x = z n ,
where n is the least common denominator of the fractional exponents ofx.
For x, dx, and each radical can then be expressed rationally in
terms of z.
ILLUSTRATIVE EXAMPLE 1 . Prove f J *** = | x* - 1 In U + x% ) + C.
J 1 + .r* 6 6
Solution. Here n - 4. Hence let x - ?.
Then x* = z 2 , x* = z-<, da- = 4 z' A dz.
=!/(''- rfp)" = l''- 1 '-in-rt + r.
Substituting back 2 = x*", we have the answer.
The general form of the irrational expression here treated is then
where R denotes a rational function of x' 1 .
Differentials containing fractional powers of a + bx only. Such an
expression can be transformed into a rational form by mea,m of the
substitution a + bx = z",
where n is the least common denominator of the fractional exponents of
the expression a + bx.
For x, dx, and each radical can then be expressed rationally in
terms of z.
fdx
- _ - -.
- (1 -fx)' 2 -f (1 +x) 2
Solution. Assume 1 4- x z 2 .
Then dx = 2 z dz, (1 -f x)* = z\ and (1 -f xY 1 = z.
2zdz _ 2
= 2 arc tan z + C = 2 arc tan (1 -I- x) 2 -f C
when we substitute back the value of z in terms of x.
The general integral treated here has then the form
R[x, (a + bx) v ]dx,
where R denotes a rational function.
298
INTEGRAL CALCULUS
PROBLEMS
Work out the following integrals.
2 s
l r <*
J
V7 - 1 V3
r
3. / ij
J x x
+ C.
1 -x
= I? x " " B * 2 + c -
- ^ 6 r^ + 6 x + 1 c
) ;; 12(4 j+1)'
= ~f- 4- 2 In ~-^l -f 4 arc tan x 8 + C.
-v. f v. M ' *
do* 2(2 a -f
'/^
(a -f
^'a -f />.r
8. /v'aH- // (/// = t -~ (4 ?/ - 3 )(a -f //)" -h C.
9.
__
/.( + 1 - 1
*frrjk
10. / ^ = ~ u -f )' ! - 3(x -f ) : ' -f 3 In (1 + Vx 4- a) -f C.
~ 2V/ + 2 + \/2 arc tan
= = 2 arc tan 2 -
1 -f V .r -f a
(/ + 5k//
(/ +
12. f !
./() ( r 4-
13 ./- 4 _^r 4 -21n;j.
Jo l _j_ V.r
14 r 4 y w = L^ V5.
''i V2 -f 4 // 2
16
*
r ! ^^7r^4
* Jo a- -f- 1 2 3'
15.
d/
3 9 arc tan
1 r^
r l7 - 1
2V? + V/
- = 5.31.
(.r - 2) s +3
FORMAL INTEGRATION BY VARIOUS DEVICES 299
Work out each of the following integrals.
19. I -^~ 23. f - .
20 -/^fe' 4.
(.r + 5)Vr + 4
22. / '-" t . M j(^-^rr+Tt:
27. Find the area bounded by the curve ?/ jr -f V.r -f 1, the .r-axis,
and the ordinates JT ~ 3 and .r = 8. AN*. 40 / ; .
28. Find the volume generated by revolving about the .r-axis the area
of the preceding problem.
29. Find the volume generated by revolving about the .r-uxis the area
in the first quadrant bounded by the coordinate axes and each of the
following curves.
(a) y 2 Vx. (c) // a VJIr.
(b) y-2- \/jr. (d) ?/ = 4 - A
30. Find the area bounded by the curves // - 2 .r -f V'J j -f 1 and
y JT v2 j* -f 1 and the ordinates .r 4 and j % ~ 12.
31. Find the area bounded by the curve
and the ordinates x 3 and x = 8.
V5 4- In
4 - V2
169. Binomial differentials. A differential of the form
(1) x m (a + bx n r dx,
where a and b are any constants and the exponents m, n, p are rational
numbers, is called a binomial differential.
Let x = z a ; then dx az a ~ ' rfc,
and x m (a + bx n ) p dx = az m " ' ff - ] fa + bz"") f> dz.
If an integer a be chosen such that ma and w* are also integers,*
we see that the given differential is equivalent to another of the same
form where m and n have been replaced by integers. Also,
(2) x(a + bx n ) p dx = x mJf ni '(ax~ n + b) l) dx
* It is always possible to choose a so that ma and not are int^ors, for we can take a an
the L C.M. of the denominators of m and n.
300 INTEGRAL CALCULUS
transforms the given differential into another of the same form where
the exponent n of x has been replaced by n. Therefore, no matter
what the algebraic sign of n may be, in one of the two differentials
the exponent of x inside the parentheses will surely be positive.
When p is a positive integer the binomial may be expanded and
the differential integrated termwise. In what follows p is regarded
as a fraction ; hence we replace it by -> where r and s are integers.*
s
We may then make the following statement.
Every binomial differential may be reduced to the form
r
x m (a+ bx n )*dx,
where m, n, r, s are integers and n is positive.
In the next section we prove that (1) can be rationalized under the
following conditions.
CASE I. When m = an integer or zero, by assuming
n
a + bx n == z*.
CASE II. When - -- h - = an inteqer or zero, lyy assuming
n s
a + bx n = z*x n .
ILLUSTRATIVE EXAMPLE 1. f - - = fx*(a
J (a + bx' 2 )l J
_ 1 2 a 4- bx' 2 c
~~ b * \ a + fa-2
Solution. m = 3,n = 2,r = 3, 8 = 2; and here m "*" 2, an integer. Hence
this comes under Case I, and we assume n
a + bx 2 = z 2 ; whence x = (-f- ' dx = 1 - . and (a -f
b
r x*dx _ rlz* - a\ zdz
' J (a -f bx*)'* ""' I &
= p f (1 ~ az ~
= (z
_ 1 2 a -f bx 2 c
~ V \ a + bx*
ILLUSTRATIVE EXAMPLE 2. f f x * x ~~ \ -
Solution, m = - 4, n = 2, ^ = - \ ; and here m "^ + - = - 2, an integer.
82 n s
* The case where p is an integer is not excluded, but appears as a special case, namely,
r = p, = 1.
FORMAL INTEGRATION BY VARIOUS DEVICES 301
Hence this comes under Case II, and we assume
(\
whence x 2 =
also x = -
i
; and
u~ - D*
1
-
PROBLEMS
Work out the following integrals.
._2(3r<-2)(
1. /VVl + x*dx=
J
r r/o , -
3.J j- 5 (8 4- x j
( a _|_ f^)* 3 f>-va -
rfjr (1 4-
4 C.
4 r.
fi r djr - - (1 + ' r3) ^ 4- r
6 - 1 r 7 ~~ > j-
J J : H1 4- x 3 ) 7 " ^
J -r2/1 4. r 4 ^
4 C.
rfr
(1 4-r")
1 4- 3 J 3
(^-1^ ( ;*_ }) \
(2 -r' J - I'd 4 .r 2 )
i__L_i__ 4 r.
loTT
_, ,,
h C.
-4 C.
4- C.
302 INTEGRAL CALCULUS
Work out each of the following integrals.
11. jV'Vl - x 3 dx. IP f xdx_ m 13
170. Conditions of rationalization of the binomial differential
r
(A) x m (a + bx n ) ; dx.
CASE I. Assume a + bs 71 = z*.
1 r
Then (a + bx j = 2, and (a + fxr" ) ; = sf ;
1
, /z* a\" Am /z* a\ Tl
also =( ^ ) and r m ~( - ) ;
j^ j
hence dx = 7^- 2* ~ J ( . ) dz.
on \ o '
Substituting in 04), we get
m + 1 1
f *
hn ~ V fe '
The second member of this expression is rational when
m + 1
n
is an integer or zero.
CASE II. Assume a + fe.r" zx n .
Then x n = ^ r and a + bx n zx" ~ ^- r -
^ fe s- b
r r _ r
Hence (a + fex n V = a 8 (^ fe) *z r ;
also x a"(z* 6) ", x m = a rl (z s fe) n ;
A J 8 ^ U M-- 1 ^
and dx a^*" 1 (z 8 fe) n dz.
n
Substituting in (A), we get
x m (a + bx n )*dx = - - a~ + *(z* - fe)'^"^" 4 "^ 1 )^^*-
n
FORMAL INTEGRATION BY VARIOUS DEVICES 303
The second member of this expression is rational when 2?L f- -
us
is an integer or zero.
Hence the binomial differential
r
x m (a + bx n )*dx
can be rationalized in the cases given in the preceding article.
171. Transformation of trigonometric differentials.
Theorem. A trigonometric differential involving sin u and cos u
rationally only can be transformed />// means of the substitution
(1) tan | = z,
or, what is the same thing, by the substitutions
2 z 1 z 2 2 dz
(2) sin u = -> cos u ~> du ~ ~~-Z-
~ ~ ~
into another differential expression which is rational hi z.
Proof. From the formula for the tangent of half an angle in (5),
Art. 2, after squaring both members, we have
, o 1 1 cos ?/
tan- -u ^T
2 1 + cos ?/
Substituting tan \ u = z, and solving for cos u,
-i ^2
(3) cos u- *. 2 > i-z*
one of the formulas (2). The right triangle in the figure shows the
relation (3) and gives also sin u as in (2). Finally, from (1),
u 2 arc tan z,
2 dz
and hence du .. : 2 -
Thus the relations (2) are proved.
It is evident that if a trigonometric differential involves tan u,
ctn u, sec u, esc u rationally only, it will be included in the above
theorem, since these four functions can be expressed rationally in
terms of sin u, or cos ?/, or both. It follows, therefore, that any rational
trigonometric differential can be integrated, provided the transformed
differential in terms of z can be separated into partial fractions (see
Art. 167).
304 INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE. Prove f -- - = - arc tan / 5tanx + 4 \ + C.
J 5 -f 4 sin 2 x 3 \ 3 /
Solution. Let 2 x = M. Then .r = ^ ?<, d jr = i c/?^ Substituting these values, and
then using (2), we have
/' _ jfr _ = I f f/ " = i
/544sin^j- 2 ./ -f 4 sin M ~2
8 ; ~. 5 2- + 82 + 5
Substituting back z = tan i // = tan j- gives the above result.
PROBLEMS
Work out th< 4 following integrals.
, / df) / n
l - = ln 1 + tan -
. __..
. 4 4- 5 cosr 3 tanf-3/
5. /^-7'- > - - - -: arc- tan (- tan ~ ) -f C.
*> 3 -f- cos <v v '3 V Vu ~ '
6. r_- - ^ - - arc tan ( 1 4 t> tan -') -f- C.
' J sin .r cos j* ~f 3 \ 2'
r cos ^ <1Q 0.5 . / . 0\ ,
7. 7 --- - - 7, - - - 4- ^ arc tan 2 tan - -f C.
; o :? cos ^ 3 6 ^ 2/
c.
'.(.4 -3 0080 V?' 11. f 2 ^ :
.'o 2 + sin x
10 _ < - In - 19
1U ' ~ n 12 '
_
o 1'J 4- 13 cos ~ r 2 ',o 3 -f 5 sin a - 4
FORMAL INTEGRATION BY VARIOUS DEVICES 305
Work out each of the following integrals
n.f ^ 16 r j
J 1 -f sm x cos x .'13 cos / - 5
14 C d t* r dx
'J cot 4- esc 6 \l 2 cos x -f 1*
15. f ^ 18 /'_jrf<L_.
J 13 - 5 cos < * J i> -f sin a
19. r f. . 21. /v-^ 23. r" ~
J I -f 2 sm x J 5 sec f - 4 .' 3 -f 2 cos
0. f sin ^ ^ og r ~ * djr 24 T 2 rf<y
*/ 5 -f 4 sin Vo T> -f 3 cos x "'o 2 -f cos tv
172. Miscellaneous substitutions. So far the substitutions con-
sidered have rationalized the given differential expression. In a
great number of cases, however, integrations may be effected by
means of substitutions which do not rationalize the given differential,
but no general rule can be given, and the experience gained in work-
ing out a large number of problems must be our guide.
A very useful substitution is
1 dz
x ~~ z z' 2 '
called the reciprocal substitution. Let us use this substitution in the
next example.
ILLUSTRATIVE EXAMPLE. Find f , dx.
Solution. Making the substitution x = - . dx = - ~- , we get
/v Q a
I
(a 2 -
^- j , .,-_ 3 fl2
PROBLEMS
Work out the following integrals.
I./ d ^ = In (__^===).
Let x = -
2
- x -f
2 V2 \Vx 2 - T -f 2 -f x -f
Let j2- r-f 2 = 2 - x.
3 f rfj = 2 arc tan (x + Vx^T 2 x - 1 )+ C.
' xVx 2 + 2 * - 1 T*t x/^TTTT
Let Vx^ -f2x-l=z-x.
306 INTEGRAL CALCULUS
4 C dx - * / V2 4- 2 x - V2 - x\
"J xV2 4- x - x 2 ~~ V2 W2 4- 2 x 4- V2 - x/
C.
Let V2 4- x - x 2 = (x -f 1)2.
/
x- 6-
dx
Let V5 x - 6 - x 2 = (x - 2)z.
xV3 x 2 - 2 x - 1
-dx
' ^ xVl 4-4x4-5x 2
.r . d -
J xVx 2 4- 4 x - 4
8.f y dx =-.
J x 2 Vl 4- 2 x 4- 3 x 2
= - arc sin
14- C.
-f 2 x 4- Vl -f 4 x -f 5 x 2
C.
rV2/
.-_ Vl -f 2 x -f 3 x 2
In
Vl -f 2 x -f 3 x 2
Let x =
Let x = -
Let x = -
2
-f C.
Let x = -
z
10. r^
J -r2x/ < >7 -r^
V27 x 2 -f 6 x - 1 r
3 arc sin ( f ^i -f C.
\ 6 x /
11. f 1 <*-?**" =6.
J X^
Let x = -
z
12. f - = arc tan e
./o r r -f r~ r 4
18. f
/0
14.
ax x 2
/ 2 d/ = V3-^ In (2
Let x = a sin 2 ir.
Let / 4- 1 = z.
Work out each of the following integrals.
15>
xVx 2 -2x4-3
4 x-dx
" J (x 2 - 2 x 4- 3)*
2dx
V5 x - 6 - x 2
2xdx
V5 x - 6 - x 2
Let Vx 2 -2x4-3 z - x.
Let Vx 2 -2x4-3 = z - x.
Let V5 x - 6 - x 2 = (x - 2)2.
Let V5x - 6 - x 2 =r (x - 2)2
CHAPTER XVII
REDUCTION FORMULAS. USE OF TABLE OF INTEGRALS
173. Introduction. In this chapter formal integration is completed.
The aim is eventually to lay down directions for using a table of in-
tegrals. Methods of deriving certain general formulas, called reduc-
tion formula's, given in all tables are developed, since these methods
are typical in problems of this sort.
174. Reduction formulas for binomial differentials. When the bi-
nomial differential cannot be integral ed readily by any of the methods
shown so far, it is customary to employ reduction formulas deduced
by the method of integration by parts. By means of these reduction
formulas the given differential is expressed as the sum of two terms,
one of them not affected by the sign of integration, and the other
an integral of the same form as the original expression, but one
which is easier to integrate. The following are the four principal
reduction formulas.
'/
ym n
x m (a + bx n ) f) dx =
(np+m-i l)b
(m-n +
(np + m + l)b,
I x
J
n (a-f bx n )t>dx.
np + m+
/ x
+ - - C X m (Q + fort
lj
b r , B( fr
J
/ x m+ i / a 4. bx")'' 4 1
*~ (a + wy dx = - x n [ p + +1) >
l r x m (fl + bx ny
J
n(/> -f l)a
307
308 INTEGRAL CALCULUS
While it is not desirable for the student to memorize these for-
'mulas, he should know what each one will do and when each one fails.
Thus :
Formula (A) diminisfos m by n. (A) fails when np + m -f 1 = 0.
Formula (B) diminishes pbyl. (B) fails when np + ra + 1 = 0.
Formula (C) increases w by n. (C) fails when w + 1 = 0.
Formula (D) increases p by 1 . (D) fails when p + 1 = 0.
I. To derive formula (A). The formula for integration by parts is
(1) (u dv - uv - f r du. (A), Art, 136
We may apply this formula in the integration of
r r "(a + bs n ) J> dx
by placing u = z m ~ n + ' * and dv = (a + ftr n ; 7> r n ~ J da* ;
then du - (m n + 1 )/ m ~ n dx and v = ' , \
?/.&(/> + l y
Substituting in (1),
/, r m n i \ ( n _[_ /, r n \ /' -f 1
~m(n j JL
J (tt -f- 0.
C
J
But
Cx m ' n (a + bx n )^ [ dx= r.r m - n
= a I x m " n (a + bx n )*dx
Substituting this in (2), we get
m~ n f |
/
o
nb(p
Uar m-
) J
Transposing the last term to the first member, combining, and
solving for I x m (a + bjr n )"dx, we obtain (A).
* In order to intoRrate r/r by the power formula it is necessary that x outside the
parenthesis shall have the exponent w - 1. Subtracting n 1 from m leaves m n + 1 f or
the exponent of j in u,
REDUCTION FORMULAS 309
It is seen by formula (A) that the integration of r m (a + bx n ) p dx is
made to depend upon the integration of another differential of the
same form in which m is replaced by m n. By repeated applications
of formula (-4), m may be diminished by any multiple of n.
When np + m + 1 = 0, formula (A) evidently fails (the denomi-
nator vanishing). But in that case
hence we can apply the method of Art. 169, and the formula is not
needed.
II. To derive formula (B). Separating the factors, we may write
(3) Cx m (a + bx n ) " dx = jx m (a + bx n ) "~ ' (a + bx) dx
= a I x m (a + bx") p ~ l dx
+ b j x m * n (a + bx n ) p ~ l dx.
Now let us apply formula (^4) to the last term of (3) by substi-
tuting in the formula m + n for rn, and p I for p. This gives
/
x^
np
Substituting this in (3), and combining like terms, we get (B).
Each application of formula (B) diminishes p by unity. Formula
(B) fails for the same case as (4).
III. To derive formula (C). Solving formula (A) for
and substituting m + n for m, we get (C).
Therefore each time we apply (C), m is replaced by m + n. When
m + 1 0, formula (C) fails, but then the differential expression can
be rationalized by the method of Art. 169, and the formula is not
needed.
IV. To derive formula (D). Solving formula (B) for
(a+bx n ) J '- l dx,
and substituting p + 1 for p, we get (D).
310 INTEGRAL CALCULUS
Each application of (D) increases p by unity. Evidently (D) fails
when p + 1 = 0, but then p = 1 and the expression is rational.
Formula (5) of Case IV, Art. 167, is a special case of (Z)), when
m = 0, p = n, n = 2, a = a 2 , 6 = 1.
ILLUSTRATIVE EXAMPLE 1. f ?===== = - - CH -f 2)(1 - z*)* + C
J V 1 z* 3
Solution. Here ra = 3, n = 2, /> = J, a = 1, 6 = i.
We apply reduction formula (4) in this case because the integration of the dif-
ferential would then depend on the integration of Cx(\ - x' 2 )~ *dx, which comes
under the power formula. Hence, substituting in (A), we obtain
(
2. f ^^ - - /I T ; + i j a ^\ Va 2 -
; (a a -a- a 4 ^ 4 8 /
ILLUSTRATIVE EXAMPLE
; (a a -a- a )4
+ -a 4 arc sin- -f C.
cS a
HINT. Apply (A) twice.
ILLUSTRATIVE EXAMPLE 3. C (a 2
-f a-dz = \ a 1 ' -f x*
-f ^ In (x -f V^T
Him 1 . Here w = 0, n = 2, /> = .J, a = -', b = 1. Apply (B) once.
ILLUSTRATIVE EXAMPLE 4. f
^ a-' 1
HINT. Apply (C) once.
PROBLEMS
Work out each of the following integrals.
/^3 ,] T 1 J _
/ = (a- 2 - 2 a 2 ) Va 2 -f x 2 -f C.
Va 2 + .r 2 *
/ r 5 r / r 1
t- - = -7r(3x 4 + 4r 2 + 8) VT^ + r.
V 1 .r 2 l^
. f^ 2 Va 2 - x* df = ^ (2 .r 2 - a 2 ) Va 2 - .r 2 -f ^ a r<* sin - + C.
J o 8 a
* / / o , J .no = ^ .,/ T . ^ 4- TT^T arc tan - -f C.
J(a- + Qc 2 Y 2 a-(a 2 -f x 2 ) 2 a- { a
REDUCTION FORMULAS
311
g - Va 2
j3 Va 2 - x 2 2 a 2 x 2 2 a ; <
x 3 dx x 2 -f 2 a 2
-f C.
(a 2 -f x 2 )
(a 1
"
Q C dx _ x(3 a 2 - 2 x 2 )
O- / 5 3 " *- '
J (a 2 -x 2 )* 3a 4 (a 2 -x 2 ) 2
9. f(x 2 4 a 2 )^ dx = J x(2 x 2 4 5 a 2 )Vx- 4 a 2 4 3 a' 1 In (x 4 Vx 2 4a 2 ) 4 C.
10. fx 2 Vx 2 4-a 2 dx = J x(2 x 2 4 a 2 )Vx 2 4a 2 - a' In (x 4 Vx 2 4 a 2 ) 4- C.
C x 2 dx
V ^
11
(x 4- 3 a)V2 ax x 2 . 3 a 2 /- x\ . ,,
1 arc cos 1 4- ( .
2 2 V a)
i twice.
V2 ax - x 2 2
HINT, f . yl> rf>C =r fx ? (2 a - .r) idr. Apply
J V 2 ax - x 2 '
l2.f^M= = - r 0/ L> + 5 ?/ + 30)V4//- ^ 4 20 arc cos (l - 1 A -f C.
r ds K 3_^ 3 - 4- r
'^ (a 2 -f s 2 ) : * 4 a 2 (a 2 4- ^' 2 ) 2 8 a l (~ 4- s '") 8 a- 1 a
14 -/vfirp = ~ 5 - wV9 - 4 ?/a + S arp sin + c -
'J Vl 4- 4 / 24
16. f?/ 2 V4 - 9 //- d?/ = ?/(9 ?/ 2 - 2)V4 - 9 //- 4 ~; arc sin ^ 4 C.
J ' 3o - / ^
17 '^(1+ 9/ 2 ) 1 81 Vl 49/ 2
18. 2 Vl 4- 4 / 2 r// = T /2 /(I -h 8 / 2 )Vl +4/ 2 - ^ In (2 / + Vl-f4/ 2 ) + C.
Work out each of the following integrals.
/x 2 dx o r X K dx
7~^ ^TiT *** / /A r'
(a 2 x 2 ) 2 J V4 x h
dx
25.
.s 7 ds
20. f
J
x 2 (l +x 2 ) 2
28.
23. r Va ' 2 + y ig-
24 fa-*'
V x
29.
26
27.
l d|/.
'/;
(a-f ?>,s
dx
312 INTEGRAL CALCULUS
175. Reduction formulas for trigonometric differentials. The method
of the last article, which makes the given integral depend on another
integral of the same form, is called successive reduction.
We shall now apply the same method to trigonometric differen-
tials by deriving and illustrating the use of the following trigonometric
reduction formulas :
(E) I sin m x cos n x dx =
sin"
m -f n
n - 1 C .
/ si
m-h nj
sin m xcos n ~ 2 ;tcte.
f . _, sin m ~
(F) I sin m x cos n xdx =
m -f n
i /*
-f I sin m 2 xcos n xdx.
m-j- n J
, _. . . , sin m f l x cos n j * x
(G) I sin m jtcos n xdx=
/
fi+ 1
sin m f * x cos n f 1 x
m-f 1
m+ 1
Here the student should note that
Formula (E) diminishes n by 2. (E) fails when m + n 0.
Formula (F) diminishes m by 2. (F) /a?"fe wtoi m + r^ = 0.
Formula (G) increases n by 2. (G) /a?k when n + 1 = 0.
Formula (H) increases m by 2. (H) fails when m + 1 = 0.
To derive these we apply, as before, the formula for integration
by parts, namely,
(1 ) f?/ dv = uv - Cv du. (A), Art. 136
Let u = cos"" 1 x, and dv sin" 7 x cos x dx ;
,. , / 1 x > ; j sin 7 "^^
then dw = (n l)cos ri ~-x sin jdx, and v =
m + 1
Substituting in (1), we get
^ i /*
+ T~T I sm m+2 xcos n ~-
?/i + 1 J
REDUCTION FORMULAS 313
In the same way, if we let
u = sin m ~ l x, and dv = cos n x sin x dx,
we obtain
/ClTl"* 1 '/ f O^" "^ ^ X
sin m x cos n x dx = - -~^
+ ^f S in-,o a
n+lj
But | sin m + -x cos n ~-x dx I sin"'/ (1 cos 2 /) cos" 2 x dx
C C
I sin' J '/cos r '~~xdx I sin"'/cos"xdx.
Substituting this in (2), combining like terms, and solving for
I sin m x cos"xdx, we get ().
** Making a similar substitution in (3), we get (F).
Solving formula (E) for the integral on the right-hand side, and
increasing ?/ by 2, we get (G).
Jn the same way we get (H) from formula (F).
Formulas (E) and (F) fail when m + n = 0, formula (G) when
n + 1 = 0, and formula (H) when m + 1 = 0. But in such cases we
may integrate by methods which have been explained previously.
It is clear that when m and n are integers, the integral
sin m x cos 7J xdx
may be made to depend, by using one of the above reduction for-
mulas, upon one of the following integrals :
fdx, fsin x dx, /cos x dx, / sin x cos x dx, I 4^ = j esc x dx,
' C-d? = fsec x dx, f . f tan x dx, I ctn / dx,
J cos x J J cos x sin x J J
all of which we have learned how to integrate.
ILLUSTRATIVE EXAMPLE 1. Prove
r?* cos' r dx = - ** ! '* + **SS* + (sin * cos x + x) + G
Solution. First applying formula (F), we get
/i sin .r cos r> x 1 r,, ria 4
sm 2 * cos 4 xc/x = + g J cos
[Here w = 2, w = 4.]
314 INTEGRAL CALCULUS
Applying formula () to the integral in the second member of (4), we get
/c\ C 4 j & m x cos3 x i 3 C ., ,
(5) I cos 1 x dx - - -- 1- - I cos- x dx.
[Here m = 0, n - 4.]
Applying formula (E) to the second member of (5) gives
,, C -j j
(6) J cos 2 x dx =
sin -r cos x , x
- - - 4 -
Now substitute the result (6) in (5), and then this result in (4). This gives the
answer as above.
ILLUSTRATIVE EXAMPLE 2. Prove
Aan^2jr dj I sef . 2 T tan 2 x - - In (sec 2 j + tan 2 jr) + C.
J cos 2x4 4
. .. tan j 2 x sin- 2 JP 1 sin" 2 x
Solution. - - = - 7- --- -
cos ^J jc cos* Z x cos 2 x cos 1 ' J j*
Let 2 x = u. Then x \ u, djr \ du, arid
(7) / sin- 2 x cos ' 2 r dx J / sin 2 M cos * ?/ c?//.
Apply (G) to the new integral in (7), with rw = 2, ?? 3, rej)lacirig x by u.
(8) Ann' M cos-' M rfw = - A<< ^ S l + Tsiri^' u cos ' M c/u.
Apply (F) to the new integral in (8), with m = 2, ?i -- 1.
(9) / sin 2 /v cos J u du = sin M + / cos J u du = sin // + In (sec w -f tan ?0.
Substituting from (9) and (8) into (7), reducing, and sotting M = 2 x, we have
the answer.
PROBLEMS
Verify the following integrations.
1. I sin 4 jc cos 2 x djc sin jc cos jc\ - sin' 1 jc - s i n2 J* 7 -f- - h C.
J [ 6 24 16 16
2. ftan-' 5 f dr = 4 ^ tan 2 f -f 3 In cos + r.
%/ AJ O t>
/ p f n ;{ /9
ctn 4 OdO=- ^~ + ctnO + + C.
4. Tsec :{ / r// - \ sec / tan / -f J In (sec / + tan /) 4- C.
5. f csc 3 .r djc I esc x ctn x -f J In (esc .r ctn j-) -f C.
6. J*csc 6 OdO = - SC g 4 ctn ^ ('csc- ^ -f j}) + | In (csc (9 - ctn 0) + C.
7. C sin- cos- (t> d<t> = i sin cos (2 sin 2 1) -f J -f r.
REDUCTION FORMULAS 315
8. f ctn2 2 B f e = - 7 ctn 2 esc 26-] In (esc 2 - ctn 2 (9) + C.
J sin 2 4 4
9 r dj? - _ cosx - 2cos - r + c
'.' sin 4 * 3 sin 3 x 3 sin .r
10. (Wfldfl = [ 8 cos'fl + 10 cos 2 4-15] + - + C.
11. f 2 sin<
J o
I o
12. f "cos 4 jcdx = ^- 14.
Jo o
o 128
TT n
13. r'sin2 fld = |f- 15. ^2^ = 5 _ 3_7r
Jo 32 Jn sm-x 4 8
4
Work out each of the following integrals :
16. fsin2 BdO. 18.
J
cos-v
17. fcsc< c/ff. 19. r-- 21.
J 2 J sin jl cos J Jo
ZL ?
22. f a sin 3 6 cos : < c/0. 23. ('"(! + sin 6) 4 dO.
Jo Jo
176. Use of a table of integrals. The methods of integration de-
veloped in Chapters XII, XVI, and XVII have been directed to re-
ducing a given integral to one or more of the Standard Elementary
Forms in Art. 128. Various devices have been elaborated to this
end, such as
integration by parts (Art. 136) ;
integration by partial fractions (Art. 167) ;
integration by substitution of a H.CW variable (Arts. 168-172) ;
use oj 'reduction formulas (Arts. 174 175).
When, however, a more or less extensive table of integrals is
available, the first step in any problem in formal integration is to
search for a formula in the table by which the problem can be solved
without the use of any of these devices. Such a table is given in
Chapter XXVII. Some examples will now be given.
ILLUSTRATIVE EXAMPLE 1. Prove, by the Table of Integrals,
Solution. Use 14, with a = 2, 6 = 1, and u = x.
This example, without the table, would be worked out as in Case II, Art. 167.
316 INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 2. Verify, by the Table of Integrals,
dx__ = J_ , /_i_ v , r
9 + 4:rO 18 \9+4:r7" 1 " "
Solution. Use 22, with a = 3, b = 2, and ?/ = /.
This example, without the table, is solved as in Case III, Art. 167.
ILLUSTRATIVE EXAMPLE 3. Verify, by the Table of Integrals,
r dx _ 1 V4 + 3 JT - 2 ,,
I / -- ^ in . ....... r -- -f ' .
JxV4-f3x 2 N/4+3J-+2
Solution. Use 31, with a = 4, 6 = 3, and a x.
This example, without the table, is worked out, by the substitution 4 + 3 x = z 2 ,
as shown in Art. 1 68.
ILLUSTRATIVE EXAMPLE 4. Verify, by the Table of Integrals,
f
./
xdx V3 J r" + 4x-7 2
V 3 x* -}- 4 j - 7 3 3V3
Solution. Use 113, with a 7, 6 = 4, c 3, and u = x.
Without the table the example would be solved by completing the square as
in Illustrative Example 2, p. 206.
ILLUSTRATIVE EXAMPLE 5. Verify, by the Table of Integrals,
cos 2 x dx = g '^ 2 sin 2 J -f 3 cos 2 x) ^
13 ^
Solution. Use 154, with a 3, n 2, u x.
Without the table the example would be solved by integration by parts. See
Illustrative Example 6, Art. 136.
In many problems the given integral cannot be identified with
one in the table as easily as in the preceding examples. In such cases
we search for a formula in the table similar to the given integral,
and such that the latter can be transformed into the former by a
simple change of variable. This method has been used constantly
in Chapter XII and in all integration problems hitherto.
ILLUSTRATIVE EXAMPLE 6. Verify, by the Table of Integrals,
1 O r
;=iln- ==l= + r.
zV4 x~ -f 9 3 3 + V4 or 2 + 9
Solution. Formula 47 is similar. Let u = 2 x. Then x - \ u, dx = J du, and,
substituting the values in the given integral, we obtain
C <fr = r jdM = r du
J xV4^ 2 +9 J \ wVw 2 -f 9 J u^/u' 2 +
9
Hence, applying 47, with a = 3, and substituting back v = 2 x, a = 3, we have
f
J x
Without tables we should proceed as in Illustrative Example 2, Art. 135.
REDUCTION FORMULAS
317
ILLUSTRATIVE EXAMPLE 7. Verify, by the Table of Integrals,
r
J
\ 9 J - 4
.
2 (9 x - 4
-
Solution. Formula 84 is similar. Let u = 2 x. Then x = t n, dx = | dt*.
Substituting, we obtain
r\ '
J
'9ar-4
^ rVjj u -" 2 (/K
J J M 3 2
du.
This is now 84 with a = j ;. Hence, applying 84, and substituting back
u = 2 x, we get the required result.
If no formula from the table can be applied as in the preceding
two cases, there remains the possibility that the use of one or more
of the devices mentioned at the beginning of this article will lead
to new integrals solvable by the table. No general directions can be
given other than the rules already developed in the text for the
employment of these devices.
The student should study the arrangement of the table. He will
find that the Standard Forms of Art. 1 28 appear in their proper places.
The reduction formulas of Art. 174 are given, with modifications, by
96-104. Also, the reduction formulas of Art. 175, with additional
ones for various cases, are numbered 157-174. Increased power
in the technique of integration will come from familiarity with the
table and practice in using it.
PROBLEMS
Work out the following integrals.
-4) + C.
1. Vz 2 + 5 dx = A (3 x 1 - 10 )(x' 2 + 5) + C.
= ^V9 x 2 - 4 -f - In (3 x -f V9 x
7. Ce< sin 2 ^ dt = \ e<(2 - sin t - cos rt -f C.
J Z 4
318 INTEGRAL CALCULUS
g Csn26jd0 ^ 2 In (1 + cos 0) - 2 cos -f C.
J 1 4- cos
9 -/2 + 2 t + ^ = arCtan(j:+1) + C -
10. f x :< sin x 2 dx = \ sin x 2 - \ x 2 cos x 2 -f C.
dx
'J V(x- 1)(2 -x)
= 2 arc sin Vx 1 -f (;.
C
Work out each of the following integrals.
4 -f sin 2
Evaluate each of the following definite integrals.
37.
38.
.
Jo (1 + J") 2
dT
35. /
36.
x 2 V25 - 9 x 2 25
dx _ 2
(4x 2 -h 9) >:j
39. f l<L^>du=ira.
J~a \ a -f w
REDUCTION FORMULAS 319
0. r f x2dX - = 1-3:
J\ V9 2 x' 2
40. = 1.338. 42.
44.
43. / /V ' <// = 0.1605.
dx
.jf 2 d " 2 46. jf-^ , - 48. jf 3
r~ *i
45. f 2 cos 2 sin 4 6 dO. 47. / e i cos A TT/ dt. 49 f ^ C os A d> dd>.
Jo Jo Vo
ADDITIONAL PROBLEMS
1. Verify the following results.
2. A parabola with its axis parallel to the //-axis passes through the
origin and the point (1, 2). Find its equation if the area between the
parabola and the x-axis is a maximum or a minimum.
Anx. // = 6 x 4 s' 2 gives a minimum.
3. Sketch the curve y V!r = In jr. Find the volume of the solid of
revolution formed by revolving about the .r-axis the area bounded by the
curve, the .r-axis, and two ordinates, one through the maximum point
and the other through the point of inflection. AUK. -^-^ TT.
4. A solid right circular cone of metal is formed so that the density at
any point P is 20(5 r) Ib. per cubic foot, where r is the distance in feet
of the point P from the axis of the cone. Find the weight of the cone if
its altitude and the radius of the base are each 3 ft. Am. 630 TT Ib.
NOTE. The weight of an element of uniform density is its volume times its
density.
5. A hollow metal sphere has an inside radius of 6 in. arid an outside
radius of 10 in. The density of the metal at any point varies inversely as
the distance of the point from the center of the sphere, and at the outside
surface the density is 2 oz. per cubic inch. Find the weight of the sphere.
Am. 2560 TT oz.
6. If n is an even integer, show that
7T TT
r 2 j r 2 7 ( n - D(n - 3) (!) TT
I sm 71 x dx = I cos" x dx = ^ ^r TT
Jo JQ n (n 2) (Z)
7. If n is an odd integer, find the value of
/*2
/ sin" x dx.
Jo
CHAPTER XVIII
CENTROIDS, FLUID PRESSURE, AND OTHER APPLICATIONS
177. Moment of area; centroids. The centroid of a plane area is
defined in the following manner.
A piece of stiff, flat cardboard will balance in a horizontal position
if supported at a point directly under its center of gravity. This
point of support is the centroid of the area of the flat surface of the
cardboard.
For certain areas considered in elementary geometry the cen-
troids are obvious. For a rectangle or a circle t he centroid coincides
with the geometrical center. In fact, if a plane figure possesses a
center of symmetry, that point is the centroid. Furthermore, if a
plane figure has an axis of symmetry, the centroid will lie on that axis.
The following considerations lead to the determination of the
centroid by mathematical means. It is beyond the purpose of
this book to justify the argument by
mechanics.
Consider the area AMPNB of the
figure. Divide it into n rectangles, each
with base A.r, as heretofore. The figure
shows one of these rectangles. Let d A be
its area, and C(h, k) its centroid. Then
(l)dA =
x, k
The moment of area of this elementary
rectangle about OA' (or O Y) is the product of its area by the per-
pendicular distance of its centroid from OX (or OY). If these
moments are, respectively, dM x and dM u , then
(A)
dM x =kdA, dM v =hdA.
The moment of area for the figure AMPNB is obtained by applying
the Fundamental Theorem (Art. 156) to the sum of the moments of
area of the n elementary rectangles. Thus we obtain
1 x = Ck dA, M v = f
320
hdA.
CENTROIDS, FLUID PRESSURE
321
Finally, if (x, y) is the centroid of the area AMPNB, and A its
area, then the relations between the moments of area (B) and x
and y are given by
(C) Ax = M Vt Ay = M x .
To calculate (x, y), find the moments of area M x and M v . From
(1) and (), these are, for the above figure,
(2)
ri> s*b
x = % y 2 dx, M v = /
*Ja Ju
xydx,
in which the value of y in terms of jr must be substituted from the
equation of the curve MPN.
If the area A is known, we have, from (C),
(3)
x =
If A is not known, it may be found by integration, as in Art. 145.
ILLUSTRATIVE EXAMPLE 1. Find the centroid of the area under one arch of the
sine curve
(4)
y = sm x.
Solution. Constructing an elementary rectan-
gle, we have
(5) dA = ydx sinx dx,
dM x = k dA = I y' 2 dx = \ sin 2 x dx, dM v = h dA = xy dx x sin x dx.
The limits are x 0, x = TT. Hence
(6) A = f^sinzdz = 2, M x = \ f "sin 2 * do- = \ IT, M V =C zsinzdz = 7T.
J *' ^
Then, from (3), x = \ TT, y = I TT. Aws.
The value of x might have been anticipated, since the line x = } TT is an axis of
symmetry.
ILLUSTRATIVE EXAMPLE 2. In the figure the curve OP A is an arc of the parabola
y' 2 2 px. Find the centroid of the area OPAB.
Solution. Draw an elementary rectangle, as in
the figure, and mark its centroid (h, k). Then
dA = x dy, h = \ x, k = y.
Using (4), dM z = kdAxy dy,
<?A (a, 6)
Finding x in terms of y from y 2 = 2 px, and
integrating between the limits y = 0, y = 6, we find
Hence x = . y = 7 b. But x = a, y = 6 satisfy the equation i/ = 2 px.
20 p 4
Hence 6 2 = 2 pa, and x = r 3 ^ a. The centroid is therefore ( T ' { a, J b). Am.
322 INTEGRAL CALCULUS
PROBLEMS
Find the centroid of each of the areas bounded by the following curves.
1. y 2 = 2 px, x = h. Ans. (f h, 0).
2. y = x'\ jr = 2, y = 0. (, - 1 ^).
3. y /*, y 4 x. (First quadrant.) (j|, f^).
4. .r 4 ?/ ?/*, ?/ j. (^, ij).
5. y 2 = 4x, 2.r-0 = 4. (, 1).
8. j/ = .f 2 f i/ = 2.r + 3. (!,)
7. ?/ = x 2 - 2 x - 3, // = 6 jr - j 2 - 3. (2, 1).
8. y = /*, ?/ = 8, .r = 0.
9. // = 6 x .r 2 , // jr.
10. ?/ = 4 jr ~ x' 2 , y = 2x- 3.
11. ?/ j- H 3 jr, y r. (First quadrant.)
12. y 2 a- cur, jr = 0, y = 0. (First quadrant.)
13. =^ - ^ = 1, // = 0, x = 2 a. (First quadrant.)
a- fr
14. Find the_centroid of the area bounded by the coordinate axes and
the parabola V.r -f V// = Va. /b?x. 7 ]/ = I a.
15. Find the centroid of the area bounded by the loop of the curve
y* 4 .r 2 r*. 4/f.x. J = 1 /' 7/ ^-
16. Find the centroid of the j ortion in the first quadrant of the ellipse
jr 2 . y' 2 1 _ 4 a - 4 b
; + ~~l. Aws. x = - // = o
O" /> 2 t5 7T J 7T
17. Find the centroid of the area bounded by the parabola y 2 2 ;xr
and the line ?/ = rax. , - 4 /> - p
Ans. x = *- ?/
5 m 2 m
18. Find the centroid of the area included by the parabolas y 2 = ax
and .r- = ?>?/. 4 - R k - __ 9 J/i
/i AiS . J 2 o i ^ " // 20 ^ ^
19. Find the centroid of the area bounded by the cissoid // 2 (2 a x) x*
and its asymptote x 2 a. .4 MS. 7 = !; a, // = 0.
20. Find the centroid of the area bounded by the witch x 2 y = 4 a 2 (2 a ?/)
and the .r-axis. Ans. x = 0, ]y ^ a.
21. Find the distance from the center of the circle to the centroid of
the area of a circular sector of angle 2 6. . ^ 2 r sin 6
30
22. Find the distance from the center of the circle to the centroid of
the area of a circular segment the chord of which subtends a central
an le261 ' Am 2 r sin-^
3(6 - sin flcos 6)'
CENTROIDS, FLUID PRESSURE
323
23. Find the centroid of the area bounded by the cardioid
p = a(l + cos 6). An$. jc = a, // = 0.
24. Find the centroid of the area bounded by one loop of the curve
p a cos 2 8.
Ans. Distance from origin =
128aV2
105 TT
25. Find the centroid of the area bounded by one loop of the curve
p = a cos 3 0.
AHS. Distance from origin -
81
80 TT
178. Centroid of a solid of revolution. The center of gravity of
a homogeneous solid is identical with the centroid of that body
considered as a geometrical solid.
The centroid will lie in any plane
of symmetry which the solid may
possess.
To achieve a mathematical def-
inition of the centroid of a solid
of revolution, it is necessary to
modify the discussion of the pre-
ceding article only in the details.
Let OX be the geometrical axis
of the solid. The centroid will
then lie on this axis. Let dV be an
element of volume, that is, a cylin-
der of revolution with altitude Ax
and radius y. Then dV iry- Ax.
The moment of volume of this cylinder with respect to the plane
through OY perpendicular to OX is
(1) dM v = x dV = 7TX// 2 Ax.
The moment of volume for the solid is then found by the Funda-
mental Theorem, and x is given from
(2)
Vx = M y I irxy- dx.
ILLUSTRATIVE EXAMPLE. Find the centroid
of a solid cone of revolution.
Solution. The equation of the element OB is
y = AB ^r
x OA h'
or !, = --
Hence M v = f TTX ^~ dx = - irr 2 h 2 .
Jo h* 4
Since V = 3 Trr' 2 h, x \ h. Ans.
324
INTEGRAL CALCULUS
Ans. jr jj a.
PROBLEMS
Find the centroid for each of the following solids.
1. Hemisphere. (See
figure.) Ans. x = I r.
2. Paraboloid of rev-
olution. (See figure.)
Ans. ~x%h.
The area bounded by
OX and each curve given
below is revolved about
OX. Find the centroid of
the solid of revolution
generated.
3. x' 2 y' 2 a' 2 , x 2 a.
4. 2 xy = a' 2 , x - J a, x - 2 a.
5. ay = x' 2 , x = a.
6. i/ 2 = 4 x, x = 1, x = 4.
7. x 2 4- 2 = 4, x = 0, x = 1. ? = f J.
8. ?/ = a sin x, x = ^ IT.
The area bounded by OY and each of the curves given below is re-
Tolved about OY. Find the centroid of the solid of revolution generated.
9. y' 2 = 4 ax, y b. Ans. y = | 6.
10. jr 2 - ir = 1, y = 0, ?/ = 1. y = A-
11. a// 2 = /', ?/ a.
12. The radii of the upper and lower bases of a frustum of a cone of
revolution are, respectively, 3 in. and 6 in., and the altitude is 8 in.
Locate the centroid.
13. Find the centroid of the solid formed by revolving about the ?/-axis
the area in the first quadrant bounded by the lines ?/ = 0, x a, and the
parabola y* = 4 ax. Ans. y = f a.
14. Find the centroid of the solid formed by revolving about the r-axis
that part of the area of the ellipse 2 + |r, = 1 which lies in the first
quadrant. a Ans. x = a.
15. Find the centroid of the solid formed by revolving about the z-axis
the area in the first quadrant bounded by the lines y 0, x 2 a, and
x' 2 t/ 2
the hyperbola ~ = 1 .
CENTROIDS, FLUID PRESSURE
325
16. Find the centroid of the solid formed by revolving about the .r-axis
the area bounded by the lines x 0, JT = a, y 0, and the hyperbola
17. Find the centroid of the solid formed by revolving about the .r-axis
the area bounded by the lines // = 0, x = , and the curve // sin 2 x.
18. Find the centroid of the solid formed by revolving about the x-axis
the area bounded by the lines .r = 0, .r = a, // = 0, and the curve // e r .
19. The area bounded by a parabola, its axis, and its iatus rectum is
revolved about the Iatus rectum. Find the centroid of the solid generated.
Am. Distance from focus fa of Iatus rectum.
179. Fluid pressure. We will now take up the study of fluid pressure
and learn how to calculate the pressure
of a fluid on a vertical wall.
Let ABDC represent part of the
area of the vertical surface of one wall
of a reservoir. It is desired to deter-
mine the total fluid pressure on this area.
Draw the axes as in the figure, the
//-axis lying in the surface of the fluid.
Divide AR into n subintervals and
construct horizontal rectangles within
the area. Then the area of one rec-
tangle (as EP) is y Ax. If this rectangle
was horizontal at the depth x, the fluid pressure on it would be
W xy A j,
[The pressure of a fluid on any given horizontal surface equals the weight!
of a column of the fluid standing on that surface as a base and of height
equal to the distance of this surface below the surface of the fluid. J
where W = the weight of a unit volume of the fluid. Since fluid pres-
sure'is the same in all directions, it follows that Wxy A.r will be ap-
proximately the pressure on the rectangle EP in its vertical position.
Hence the sum n
D
represents approximately the pressure on all the rectangles. The
pressure on the area ABDC is evidently the limit of this sum. Hence,
by the Fundamental Theorem,
lim
n x>
im V Wxflt Ax, = / Wxy dx.
-oo i = 1 J
326
INTEGRAL CALCULUS
Hence the fluid pressure on a vertical submerged surface bounded
by a curve, the /-axis, and the two horizontal lines x = a and x = b
is given by the formula b
(D) Fluid pressure - W \ yx dx,
J a
where the value of y in terms of x must be substituted from the equa-
tion of the given curve.
We shall assume 62 Ib. ( W) as the weight of a cubic foot of water.
ILLUSTRATIVE KXAMPLE 1. A circular water main
6 ft. in diameter is half full of water. Find the pres-
sure on the gate that closes the main.
Solution. The equation of the circle is x' 2 + y' 2 9.
Hence y= \ 9 - x~,
and the limits are from .r to x = II. Substituting
in (>), we get the pressure on the right of the
x-axis to be
Pressure = 62 C'\ 9 - x* x dx = [- <. (9 - x u ;a]jj = 558.
'o
Hence Total pressure = 2 x 558 1 1 16 Ib. Ans.
The essential part of the above reasoning is that the pressure
( dP) on an elementary horizontal strip is equal (approximately)
to the product of the area of the strip (= dA) by its depth (= h)
and the weight (= W) of unit volume of the fluid. That is,
() dP = WhdA.
With this in mind, the axes of coordinates may be chosen in any
convenient- position.
JLLI 'STRATI VK Kx AMPLE !2. A trapezoidal gate in a dam is shown in the figure.
Find the pressure on the gate when the v
surface of the water is 4 ft. above the
top of the gate.
Solution. Choosing axes OX and OY
as shown, and drawing an elementary
horizontal strip, we have, using (),
dA 2 .r (///,
WATER
The equation of AB is y = 2 x 8.
Solving this equation for .r, and substi-
tuting, the result is
dP W(% y)(y +
Integrating with limits y and y 4, we obtain
P = W f 4 (64 - y~)dy = ^^ W = 14,549 Ib. Ans.
t/O
CENTROIDS, FLUID PRESSURE 327
PROBLEMS
In the following problems the //-axis is directed vertically upward, and
the .r-axis is at the surface level of a liquid. Denoting the weight of a
cubic unit of the liquid by ir, calculate the pressure on the areas formed
by joining with straight lines each set of points in the order given.
1. (0, 0), (3, 0), (0, - 6), (0, 0). Ans. 18 \V.
2. (0, 0), (3, - 6), (0, - 6), ^0, 0). 36 W.
3. (0, 0), (2, - 2), (0, - 4), i- 2, - 2), (0, 0). 16 W.
4. Calculate the pressure on the lower half of an ellipse whose semi-
axes are 2 and 3 units respectively, (a) when the major axis lies in the
surface of the liquid; (b) when the minor axis lies in the surface.
AH*, (a) 8 If; (b) 12 \V.
5. Each end of a horizontal oil tank is an ellipse of which the horizontal
axis is 12 ft. long and the vertical axis 6 ft. long. Calculate the pressure
on one end when the tank is half full of oil weighing 60 lb. per cubic foot.
An^. 2160 lb.
6. The vertical end of a vat is a segment of a parabola (with vertex
at the bottom) 8 ft. across the top and 16 ft. deep. Calculate the pressure
on this end when the vat is full of a liquid weighing 70 lb. per cubic foot.
/b/K. 38,22911).
7. The vertical end of a water trough is an isosceles
right triangle of which each leg is 8 ft. Calculate the
pressure on the end when the trough is full of water
(W = 62.5). AUK. 3771 lb.
8. The vertical end of a water trough is an isosceles triangle 5 ft.
across the top and 5 ft. deep). Calculate the pressure on the end when the
trough is full of water. A//,s. 1302 lb.
9. A horizontal cylindrical tank of diameter 8 ft. is half full of oil
weighing 60 lb. per cubic foot. Calculate the pressure on one end.
Ans. 2560 lb.
10. Calculate the pressure on one end if the tank of Problem 9 is full.
11. A rectangular gate in a vertical dam is 10 ft. wide and 6 ft. deep.
Find (a) the pressure when the level of the water ( W 62.5) is 8 ft. above
the top of the gate ; (b) how much higher the water must rise to double
the pressure found in (a). Am. (a) 41,250 lb. ; (b) 11 ft.
12. Show that the pressure on any vertical surface is the product of
the weight of a cubic unit of the liquid, the area of the surface, and the
depth of the centroid of the area.
13. A vertical cylindrical tank, of diameter 30 ft. and height 50 ft., is
full of water. Find the pressure on the curved surface. Ans. 3682 tons.
328
INTEGRAL CALCULUS
180. Work. In mechanics the work done by a constant force F
causing a displacement d is the product Fd. When F is variable, this
definition leads to an integral. Two
examples will be considered here.
Work done in pumping out a tank.
Let us now consider the problem of
finding the work done in emptying
reservoirs of the form of solids of rev-
olution with their axes vertical. It
is convenient to assume the r-axis of
the revolved curve as vertical, and the
?/-axis as on a level with the top of
the reservoir.
Consider a reservoir such as the one
shown ; we wish to calculate the work
done in emptying it of a fluid from the depth a to the depth b.
Divide AB into w subintervals, pass planes perpendicular to the
axis of revolution through these points of division, and construct
cylinders of revolution, as in Art. J60. The volume of any such
cylinder will be 7n/~A:r and its weight M'7r// 2 A.r, where W = weight
of a cubic unit of the fluid. The work done in lifting this cylinder
of the fluid out of the reservoir (through the height jr) will be
Wiry 2
fWork done in lifting equals the weight multiplied by the vertical height.]
The work done in lifting all such cylinders to the top is the sum
The work done in emptying that part of the reservoir will evi-
dently be the limit of this sum. Hence, by the Fundamental Theorem,
lim
r
c l = / Wiry 2 xdx.
J
Therefore the work done in emptying a reservoir in the form of a
solid of revolution from the depth a to the depth 6 is given by the
formula
(F) Work= WIT f y 2 xdx,
J a
where the value of y in terms of x must be substituted from the
equation of the revolved curve.
CENTROIDS, FLUID PRESSURE
329
ILLUSTRATIVE EXAMPLE 1. Calculate the work done in pumping out the water
filling a hemispherical reservoir 10 ft. deep.
Solution. The equation of the circle is
& + 2/- = 100.
Hence y* = 100 - *-',
W = 62,
and the limits are from x = to r = 10.
Substituting in (F), we get
/> 10
Work = 62 TT I 000 - x'^rdr = 155,000 TT ft.-lb.
Jo
The essential principle in the above reasoning is that the element
of work (= dw} done in lifting an elementary volume ( dV) through
a height (=A) is dw=KhdV,
where W = weight of unit volume of the fluid. With this in mind,
the axes of coordinates may be chosen in any convenient manner.
ILLUSTRATIVE EXAMPLE 2. A conical cistern is 20 ft. across the top and 15 ft.
deep. If the surface of the water is 5 ft. be-
low the top, find the work done in pump-
ing the water to the top of the cistern.
Solution. Take axes OX and OY as in
the figure. Then
dV = TTX 2 dy,
h = 15-y,
dw VT(15 y}irx~ dy.
1(10, 15)
The equation of the element OA is
x = y. Substituting,
dw = 7rW(15 - |/)J u z <ly = i 7rH T (15 y' 2 - y*)dy.
The limits are y = and y = 10, since the water is 10 ft. deep. Integrating,
w = $ irW f 10 n5 it' 2 - y"')dy = 216,421 ft.-lb. An*.
' Jo
Work done by an expanding gas. If a gas in a cylinder expands
against a piston head from volume v cu. ft. to vi cu. ft, the external
work done in foot-pounds is
(G)
Work
r*>i
= I p
JvQ
where p = pressure in pounds per square foot.
Proof. Let the volume increase from v to v + dv.
Let c area of cross section of the cylinder.
Then = distance the piston moves.
c
330 INTEGRAL CALCULUS
Since pc = force causing the expansion dv,
Element of work done = pc - = p dv.
c
Then (G) follows by the Fundamental Theorem. To use (G), the
relation between /; and v during the expansion must be known. This
relation has the form
0) pi? = constant,
the exponent n being a constant.
Isothermal expansion occurs when the temperature remains con-
stant. Then n 1, and the pressure-volume relation is
(2) pv por () = piVi.
If a graph of (1) is made (pressure-volume diagram), plotting
volumes as abscissas and pressures as ordinates, the area under this
curve gives, numerically, the work done, as calculated by (G). In
isothermal expansion the graph of (2) is a rectangular (equilateral)
hyperbola.
PROBLEMS
1. A vertical cylindrical cistern of diameter 16 ft. and depth 20 ft. is
full of water (VT 62.5). Calculate the work necessary to pump the
water to the top of the cistern. Am. 800,000 TT ft. Ib.
2. Jf the cistern of Problem 1 is half full, calculate the work necessary
to pump the water to the top.
3. A conical cistern 130 ft. across the top and 20 ft. deep is full of water
(IV rr 62.5). Calculate the work necessary to pump the water to a height
of 15 ft. above the top of the cistern. , . 2,500,000 TT ,,
J\ / / 8 . ~ 1 1 . ID.
4. A hemispherical tank of diameter 10 ft. is full of oil weighing 60 Ib.
per cubic foot. Calculate the work necessary to pump the oil to the top
of the tank. Am. 9375 TT ft. Ib.
5. A hemispherical tank of diameter 20 ft. is full of oil weighing 60 Ib.
per cubic foot. The oil is pumped to a height of 10 ft. above the top of
the tank by an engine of } H.P. (that is, the engine can do work at the
rate of 16,500 ft. II). per minute). How long will it take the engine to
empty the tank ?
6. Find the work done in pumping out a semi-elliptical reservoir full
of water (W 62). The top is a circle of diameter 6 ft., and the depth
is 5ft. Ans. 3487 \ TT ft. Ib.
7. A conical reservoir 12 ft. deep is filled with a liquid weighing 80 Ib.
per cubic foot. The top of the reservoir is a circle 8 ft. in diameter. Cal-
culate the work necessary to pump the liquid to the top of the reservoir.
-4ns. 15,360 TT ft. Ib.
CENTROIDS, FLUID PRESSURE 331
8. A water tank is in the form of a hemisphere, 124 ft. in diameter, sur-
mounted by a cylinder of the same diameter and 10 ft. high. Find the
work done in pumping it out when it is filled within 2 ft. of the top.
9. A bucket of weight M is to be lifted from the bottom of a shaft
h ft. deep. The weight of the rope used to hoist it is in Ib. per foot. Find
the work done.
10. A quantity of air with an initial volume of 200 cu. ft. and pres-
sure of 15 Ib. per square inch is compressed to 80 Ib. per square inch.
Determine the final volume and the \\ork done if the isothermal law holds,
that is, pr = C. -4//s. :i7.f> on. ft. ; 72:5,000 ft. Ib.
11. Determine the final volume and work done in Problem 10 if the
adiabatic law holds, that is pr 7 ' (\ assuming n ~ 1.4.
An*. 60 cu. ft.; 648,000 ft. Ib.
12. Air at pressure of 15 Ib. per square inch is compressed from
200 cu. ft. to 50 cu. ft. Determine the final pressure and the work done if
the law is pv C. A//s. 60 11). per square inch ; 599,000 ft. Ib.
13. Solve Problem 12 if the law is pr" - <\ assuming // 1.4.
AHS. 104.5 Ib. per square inch ; 801,000 ft. Ib.
14. A quantity of gas with an initial volume of 16 on. ft. and pressure
of 60 Ib. per square inch expands until the pressure is ;}() Ib. per square
inch. Determine the final volume and the \\ork done by the gas if the
law is pr = C. An*. :*2 on. ft. ; 95,800 ft. Ib.
15. Solve Problem 14 if the law is pr" (\ assuming n 1.2.
An*. 2K.f> cu. ft.; 75,600 ft. Ib.
16. A quantity of air with an initial volume of 200 cu. ft. and pressure
of 15 Ib. per square inch is compressed to 150 cu. ft. Determine the final
pressure and the work done if the law is pr ~- - (\
17. Solve Problem 16 if the law is pr 7 ' = (\ assuming r? = 1.4.
18. A gas expands from an initial pressure of 80 11>. per square inch and
volume of 2.5 cu. ft. to a volume of 9 cu. ft. Find the work done if the
law-is pr n 0, assuming n 1.0646.
19. Solve Problem 18 if n = 1.131.
20. Determine the amount of attraction exerted by a thin, straight,
homogeneous rod of uniform thickness, of length /, and of mass M upon
a material point P of mass m situated at a distance of a from one end of
the rod in its line of direction.
Solution. Suppose the rod to be divided into equal infinitesimal portions (ele-
ments) of length dx. M
= mass of a unit length of rod ;
hence ~ dx = mass of any element.
332 INTEGRAL CALCULUS
Newton's law for measuring the attraction between any two masses is t
Force of attraction = Produrt of masses
'distance between them;*
therefore the force of attraction between the particle at P and an element of the rod is
which is then an clement of the force of attraction required. The total attraction
between the particle at V and the rod being the limit of the sum of all such elements
between x = and j /, we have
/-./M ,
r'-r-mdx
- I J - = ^ C l
-'o (j + aj- 2 f J
Force of attraction l dx M
21. Determine the amount of attraction in the last example if P lies
in the perpendicular bisector of the rod at the distance a from it.
Am. 2mM 2 -
22. A vessel in the form of a right circular cone is filled with water.
If // is its height and r the radius of the base, what time will it require
to empty itself through an orifice of area a at the vertex?
Solution. Neglecting all hurtful resistances, it is known that the velocity of dis-
charge through an orifice is that acquired by a body falling freely from a height
equal to the depth of the water. If, then, x denotes the depth
of the water, -
v = \ 2 gx.
Denote by dQ the volume of water discharged in time
dt, and by dx the corresponding fall of surface. The volume
of water discharged through the orifice in a unit of time is
a \
being measured_as_ a right cylinder of area of base a and
altitude r (= \ 2 r/.r). Therefore in time dt
(1)
Denoting by S the area of the surface of the water when the depth is x, we have,
from geometry,
'' >
, or
, ' VI
h~
But the volume of water discharged in time dt may also be considered as the
volume of a cylinder AB of area of base S and altitude dx; hence
Equating (1) and (2) and solving for dt,
Therefore t = JL = ^. Ans.
2 gx 5o\ 2 g
CENTROIDS, FLUID PRESSURE .
333
181. Mean value of a function,
value) of n numbers y\, y< 2 , -,#
(i) y = \
The arithmetic mean (or average
s
+ y<2 + - + //).
We proceed to establish the formula
r b
I 6(x)dx
(H) Mean value of 4>(x) 1 = J a T
from x = a to x = b j b a
The figure shows the graph of
(2) y = </>(/).
The mean value (= y) of the ordinates of the arc PQ is to be de-
fined. Divide AB into n equal parts each equal to A.r and let //i,
yz, ' * 2/n, be the ordinates at the w points of division. Then (1 ) will
give an approximate value for the mean value required. Multiply
numerator and denominator of the right-hand member of (1) by Ax.
Then, since n Ax = b a, we get
(3) y (approximately) = ^ , __
But the numerator in (3) is, approximately, the area APRQB.
The average value of y (or (/>(j)) is defined as the limit of the right-
hand member in (3) when n > oo. This gives (//).
In the figure the mean value of <j>(x) equals CR if area rectangle
ABML = area ABQRP.
Taking y as the function (dependent variable), then (H) becomes
f
Ja
ydx
Given
ILLUSTRATIVE EXAMPLE.
the circle
(4) x 2 + y 2 = r 2 .
Find the average value of the or-
dinates in the first quadrant
(a) when y is expressed as a func-
tion of the abscissa x ;
(b) when y is expressed as a func-
tion of the angle = Z MOP.
Solution, (a) Since y = \V 2 - x 2 ,
the numerator in (/) is
X
C
Jo
Vr 2 - x 2 dx = I Trr 2 . Then y = \ irr = 0.785 r. Ans.
334 INTEGRAL CALCULUS
(b) Since y r sin 0, and the limits are 6 = = a, 8 \ IT b, the numerator
in (/) is
A * 2 r
I r sin 6 d6 = r. Since b - a = \ TT, we have y = = 0.637 r. Ans.
JO 7T
Thus we have quite different values of y, depending upon the independent
variable with respect to which the mean value is taken.
As shown in the above example, the average value of a given
function y will depend upon the variable chosen as the independent
variable. For this reason, we write (/) in the form
C h
I ydx
x b a
in order to indicate explicitly the variable with respect to which y
is averaged.
Thus, in the Illustrative Example, we have ~y x 0.785 r, and y ft = 0.637 r.
PROBLEMS
1. Find the average value of ?/ = s~ from jr = to x 10. Ans. 33f 3 .
2. Find the average value of the or di nates of y 2 = 4 r from (0, 0) to
(4, 4) taken uniformly along the .r-axis. Ans. 2.
3. Find the average value of the abscissas of y' 2 = 4 jc from (0, 0) to
(4, 4) when uniformly distributed along the ?/-axis. Ans. 1^.
4. Find the average value of sin .r between .r and .r = TT. Ans.
IT
5. Find the average value of sin 2 .r between jr = and *c = TT. (This
average value is frequently used in the theory of alternating currents.)
Ans. %.
6. If a particle in a vacuum were thrown downward with an initial
velocity of v (} ft. per second, the velocity after t sec. would be given by
/- N i /
The velocity after falling s ft. would be given by
(2) r - Vz\r + 2 gs. (Take g = 32)
Find the average value of v
(a) during the first 5 sec., starting from rest ; Ans. 80 ft. per second.
(b) during the first 5 sec., starting with an initial velocity of 36 ft.
per second ; Am. 116 ft. per second.
(c) during the first 2-J sec., starting from rest; Ans. 40 ft. per second.
(d) during the first 100 ft., starting from rest; Ans. 53 J ft. per sec.
(e) during the first 100 ft., starting with an initial velocity of 60 ft.
per second. Ans. 81 ft. per second.
CENTROIDS, FLUID PRESSURE 335
7. In simple harmonic motion s = a COP nt. Find the average value
of the velocity during one quarter of a period (a) as to the time ; (b) as
to the distance.
8. Show that in simple harmonic motion the average kinetic energy
with respect to the time for any multiple of a quarter period is half the
maximum kinetic energy.
9. A point is taken at random on a straight line of length a. Prove
(a) that the average area of the rectangle whose sides are the two seg-
ments is a 2 ; (b) that the average value of the sum of the squares on
the two segments is fa 2 .
10. If a point moves with constant acceleration, the average as to the
time of the square of the velocity is J(r - -f r ( ,rj + r, 2 ), where r,> is the
initial and i\ the final velocity.
11. Show that the average horizontal range of a particle projected
with a given velocity at an arbitrary elevation is 0.6366 of the maximum
horizontal range.
HINT. Take a = in the formula of Problem 35, p. 1 14.
The formulas ~ ~
I x ds I y ds
(6) * '
where (x, y) is any point on a curve for which ds is the element of arc,
define the centroid of the arc. They give, respectively, the average
values of the abscissas and ordinates of points on the curve when
distributed uniformly along it. (Compare Art. 177.)
12. Show that the area of the curved surface generated by revolving
an arc of a plane curve about a line in its plane not cutting the arc equals
the length of the arc times the circumference of the circle described by
its centroid (6). (Theorem of Pappus. Compare Art. 250.)
HINT. Use (L), Art. 164.
13. Find the centroid of the arc of the parabola y 2 4 x from (0, 0)
to (4, 4). Am. 7 = 1.64, y = 2.29.
14. Find the centroid of an arc of the circle p a between arid -f 6.
A ns.
15. Find the centroid of the perimeter of the cardioid p = a(l -f cos 0).
Ans. ~jc = I a, y == 0.
16. Find by the Theorem of Pappus the centroid of the arc of the circle
x' 2 + 3/2 - r 2 w hich lies in the first quadrant, ^^ x = T =
336 INTEGRAL CALCULUS
17. Find by the Theorem of Pappus the surface of the torus generated
by revolving the circle (x b') 2 -f y 2 = a 2 (b > a) about the |/-axis.
18. A rectangle is revolved about an axis which lies in its plane and is
perpendicular to a diagonal at its extremity. Find the area of the surface
generated.
ADDITIONAL PROBLEMS
1. An area is bounded by the lines y = x 2 , x + y = 6, y 0, and x = 3.
Find its centroid. Ans. x = f , y = f .
2. The abscissa of the centroid of the area bounded by the curve
2 // = x' 2 and a certain line through the origin is 1. Find the ordinate of
the centroid. An.s. f .
3. Find the centroid of the area bounded by y = x n (n > 0), the x-axis,
and x 1. Discuss the locus of the centroid as n varies.
_ n-hl- n+ 1
AUK. x =
n + 2 y 2(2 n-f 1)
4. Find the equation of the locus of the centroid of the area bounded
by the .r-axis and the parabola y = ex x' 2 when c varies. Ans. 5 y = 2 x' 2 .
5. Given the parabola x' 2 = 2 py and any oblique line y mx + b meet-
ing the parabola in the points A and B. Through C, the midpoint of
A /*, draw a line parallel to the axis of the curve meeting the parabola at
D. Prove that (a) the tangent to the parabola at /> is parallel to the line
.4 H ; (b) the centroid of the area ACIU) lies on the line CD.
6. Let P be a point on the parabola // x 2 , and let C be the centroid
of the area bounded by the parabola, the .r-axis, and the ordinate through
P. Find the position of P so that the angle OPC is a maximum.
4ns. Ordinate = -f%.
7. A cistern has the form of a solid generated by revolving about its
vertical axis a parabolic segment cut off by a chord 8 ft. long, perpendicu-
lar to the axis and at a distance of 8 ft. from the vertex. The cistern is
filled with water weighing 62.5 Ib. per cubic foot. Find the amount of
work required to pump over the top of the cistern one half the volume of
water it contains. AnSm 16,000 (\/2 - 1) f = 6937 ft.-lb.
o
8. A hemispherical cistern of radius r is full of water. Two men, A
and B, are to pump it out, each doing half the work. If A starts first, what
will be the depth d of the water when he has finished his share of the work ?
CENTROIDS, FLUID PRESSURE
337
9. A tank in the shape of an inverted circular cone is full of water.
Two men are to pump the water to the top of the tank, each doing half
the work. When the first man has finished his share of the work, let s
denote the ratio of the depth of water left in the tank to the original depth.
Show that z is determined by the equation 6 r 4 8 r :{ 4- 1 = 0. Calculate
the value of z to two decimals. An*. 0.61.
10. A well is 100 ft. deep. A bucket, weighing 3 lb., has a volume of
2 cu. ft. The bucket is filled with water at the bottom of the well and is
then raised at a constant rate of 5 ft. per second to the top. Neglecting
the weight of the rope, find the work done in raising the bucket if it is
discovered that the water is leaking out at a constant rate of 0.01 cu. ft.
per second. (A cubic foot of water weighs 62.4 lb.) An*. 12,156 ft.-lb.
11. The area OAB is divided into elements such as OPQ by lines from
O. Show that the area A and the moments of area
M x and M v are given by Y j
r
\\ (xy r y)dx,
= 3 I ^
My
r
M x = JJ y(jcy'
n/ y)djr.
(The centroid of a triangle is on any median at
two thirds of the distance from the vertex to the - }
opposite side.)
12. Find the centroid of the hyperbolic sector bounded by the equi-
lateral hyperbola x a sec 6, y a tan and radii from the origin to the
points (a, 0) and (x, y).
2 tan - _ 2 sec
An*. x - a
V 3 in (sec + tan 0)
CHAPTER XIX
SERIES
182. Definitions. A sequence is a succession of terms formed ac-
cording to some fixed rule or law.
For example, 1, 4, 9, 16, 25
, -, X 2 X* X* X r '
and i,_ 3, _,__.,_,__
are sequences.
A scries is the indicated sum of the terms of a sequence. Thus
from the above sequences we obtain the series
1+4 + 9 + 16 + 25
, ! , X 2 X* . X* X*
and 1 _ :r + .__._ + T -~.
When the number of terms is limited, the sequence or series is
said to be finite. When the number of terms is unlimited, the sequence
or series is called an infinite sequence or infinite series.
The ycH-cral term, or r/th term, is an expression which indicates
the law of formation of the terms.
ILLUSTRATIVE EXAMPLE 1. In the first example given above, the general term,
or nth term, is n 2 . The first term is obtained by setting n = l, the tenth term by
setting u = 10, etc.
ILLUSTRATIVE EXAMPLE 2. In the second example given above, the nth term,
except for n = 1, is ^ : -
?il
If the sequence is infinite, this fact is indicated by the use of dots, as
1, 4, 9, - -, r/-, - -.
Factorial numbers. An expression which occurs frequently in con-
nection with series is a product of successive integers, beginning
with 1. Thus, 1x2x3x4x5 is called 5 factorial and is indicated
by [5 or 5 !
In general, [w = lx2x3x---x(?i-l)xw
is called n factorial. It is understood that n is a positive integer. The
expression [n has no meaning if n is not a positive integer.
338
SERIES 339
183. The geometric series. For the geometric series of n terms,
(1) S n = a + ar + ar 2 + + ar n ~\
it is shown in elementary algebra that
or
or
n _ ^ r) _____
the first form being generally used if \r\ < 1, and the second form
if \r\> 1.
If \r\ < 1, then r n decreases in numerical value as // increases and
lim (r n ) = 0.
n - ao
From formula (2) we see, therefore, that (Art. 16)
(3) limS^T-^--
n - oo l T
Hence if \r\< 1 the sum S n of a geometric series approaches a
limit as the number of terms is increased indefinitely. In this case
the series is said to be convergent.
If \r\> 1, then r n will become infinite as // increases indefinitely
(Art. 18). Hence, from the second formula in (2), the sum S n will
become infinite. In this case the series is said to be divergent.
A peculiar situation presents itself if r 1. The series then
becomes
(4) a a + r; a -f a a - - .
If t] is even the sum is zero. If n is odd the sum is a. As //, increases
indefinitely the sum does not increase indefinitely and it does not
approach a limit. Such a series is called an oscillating series.
ILLUSTRATIVE EXAMPLE. Consider the geometric series with
(5) S n = 1
We find, by (2), that S n =
Then
(6) lim S n = 'A which agrees with (3), when a = 1, r = \.
n -*oo
It is interesting to dis- 1- +; *
cuss (5) geometrically. To 6l ' Sj! - 5 ^
do this, lay off successive values of S n on a straight line, as in the figure.
n 1 2 3
Sn 1 1-2- ij
Each point thus determined bisects the segment between the preceding point and
the point 2. Hence (6) is obvious.
340 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
In each of the following series (a) discover by inspection the law of
formation ; (b) write three more terms ; (c) find the nth, or general, term.
1. 2 + 4 + 8 + 16 + -. Arts, nth term = 2 n .
3. - i + + J + jj + -jj +
6 * 2 ~ f 2.4~ f 2-4.f>~ f 2.4.6.8~ f *"'
6 ^_l'j_^ 5L 1 + . . . y **'
3579 2 n + 1
Write the first four terms of the series whose nth, or general, term is
given below.
7.^- Ans. l + -4- + 4- + -7= + "*-
V2 V3 V4
8 JLA. 3 + 1 + ^6
o. rt ., o-r-t--f-r.
O O i
n " . 1 . ? , 3 , _4_ ,
^7i J -^Q^o^97^
O O C/ i-j <
,fH 1 'y v2 /j3
i o - 1 + -^ + - I- -
1U. / J- i^ / 1^ , i^ i I
V^ V^ V3 V4
- - ( 1 ) ' X X ' . X'* X ,
llt . __}___{ ..
12.
n -M
> f/^"""^ r?y ~l~ n^^ w ^ *^2
[w ' 2"[n ' Vn~T2
184. Convergent and divergent series. In the series
S n = Ui + U 2 + Us + + Uny
the variable S n is a function of n. If we now let the number of terms
(= n) increase without limit, one of two things may happen.
CASE I. S n approaches a limit, say u, indicated by
(1) lim S n = u.
SERIES 341
The infinite series is now said to be convergent and to converge to the
value u, or to have the value u.
CASE II. S n approaches no limit. The infinite series is now said
to be divergent.
Examples of divergent series are
1 + 2 + 3 + 4 + 5H ---- ,
1-1 + 1-1 + ....
As stated above, in a convergent series the value of the series is the
number u (sometimes called the sum) defined by (1). No value is
assigned to a divergent series.
In the applications of infinite series, convergent series are of major
importance. Thus it is essential to have means of testing a given
series for convergence or divergence.
185. General theorems. Before developing special methods for
testing series, attention is called to the following theorems, of which
proofs are omitted.
Theorem I. // S n is a variable that always increases as n increases
but never exceeds some definite jired nuwher A, then as n increases
without limit, S n will approach a limit n which is not (jreater than A.
The figure illustrates the statement. The points determined by
the values Si, Sa, S, etc. approach the point w, where
and u is less than or equal to A.
ILLUSTRATIVE EXAMPLE. Show that the infinite series
a) 1 + 1+ r L 2 + rnh) + --- + [i + ---
is convergent.
Solution. Neglect the first term, and write
(2) S n = 1 + 2 + JT^l + ' * * + 1 -2-3---n"
Consider the variable s n defined by
(3) *n = l+7,+~ + '-+~
in which we have replaced all integers in the denominators of (2), except 1, by 2.
Obviously S B < Also, in (3j we have a geometric series with r = I and .<> < 2
no matter how large n may be (see Art. 1H3). Hence S w as defined by (2) is a vari-
able which always increases as n increases but remains less than 2. Hence S n ap-
proaches a limit as n becomes infinite, and this limit is less than 2. Therefore the
infinite series (1) is convergent, and its value is less than 3.
We shall see later that the value of (1; is the constant e = 2.71828 , the
natural base (Art. 61).
342 DIFFERENTIAL AND INTEGRAL CALCULUS
Theorem II. // S n is a variable that always decreases as n increases
but is never less than sonw definite fixed number B, then as n increases
without limit, S n will approach a limit which is not less than B.
Consider now a convergent series
for which lim S', t = u.
n oo
Ijet the points determined by the values S'i, .S\>, Sz, etc. be plotted
on a directed line. Then these points, as n increases, will approach
the point determined by ?/ (terms in S n all of the same sign; or
cluster about this point. Thus it is evident that
(A) 1imM M = 0.
it oo
That is, in a convergent series, the terms must approach zero as a limit.
On the other hand, if the general (or //th) term of a series does
not approach zero as // becomes infinite, we know at once that the
series is divergent.
(A) is not, however, a suflicient condition for convergence; that
is, even if the //th term does approach zero, we cannot state posi-
tively that t he series is convergent . For, consider the harmonic series
1 + 2 + a + 4 + ' " ' + w'
Hero
lim (//) ~ lim (-) = 0;
.* V W'
that is, condition (A) is fulfilled. Yet we shall show in Art. 186 that
the series is divergent.
We shall now proceed to deduce special tests which, as a rule, are
easier to apply than the above theorems.
186. Comparison tests. In many cases it is easy to determine
whether or not a given series is convergent by comparing it term
by term with another series whose character is known.
Test for convergence. Let
(1) Mi + Wj + U;\+ ' ' '
be a scries of posit ire terms irhich it is desired to test for convergence. If
a series of positive terms already kmnm to be convergent, namely,
(2) fli +fl- + :i + ,
raw be found whose terms are never less than the corresponding terms in
the series (1) to be tested, then (1) is a convergent series and its value
does not exceed that of (2).
SERIES 343
Proof. Let s n = MI + w 2 + MS + + u> n ,
and Sn = fli + <io -f a.< + -f a n ,
and suppose that lim N fl == A.
- oo
Then, since S n < A and s n ^ tf llf
it follows that s n < A. Hence, by Theorem I, Art. 185, s n approaches
a limit and the series (1) is convergent and its value is not greater
than .4.
ILLUSTRATIVE EXAMPLE 1. Test the series
(3) 1 + -;
Solution. Compare with the geometric series
(4) 1 + | + r* + if- + ? "'
which is known to be convergent. The terms of (4) are never less than the corre-
sponding terms of (3). Hence (3) also is convergent.
Following a line of reasoning similar to that applied to (1) and (2),
we may prove the
Test for divergence. Let
(5) '//i + W- -f W:i +
be a series of positive terms to he tested, which are never less than the
corresponding terms of a series of positive terms, namely,
(6) fci +&i. + b:i-4 ,
known to be divergent. Then (5) is a divergent series.
ILLUSTRATIVE EXAMPLE 2. Show that the harmonic series
(7) 1 + i + .4 + 1 +
is divergent.
Solution. Rewrite (7) as below and compare with the series written under it.
The square brackets are introduced to aid in the comparison.
(8) 1 + ^ + [J + 11 + [' + i + 7 -4- Jl + [ J + + f',,1 + -.
(9) 4 + i -Kd + 1] + [ 1 + t + t + 11 + Li'e + ' ' ' + A] + -.
We observe the following facts. The terms in (8) are never less than the
corresponding terms in (9).
But (9) is divergent. For the sum of the terms in each square bracket is 4,
and S n will increase indefinitely as n becomes infinite.
Hence (8) is divergent.
ILLUSTRATIVE EXAMPLE 3. Test the series
1+ i +-L + -L + ...
\/2 V3 V4
344
DIFFERENTIAL AND INTEGRAL CALCULUS
Solution. This series is divergent, since its terms are greater than the cor-
responding terms of the harmonic series (7) which is divergent.
The "p series/'
(10; 1-
is useful in applying the comparison lest.
Theorem. The p series is convergent when p > 1, and divergent for
other values of p.
Proof. Rewrite (10) as below and compare with the series written
under it. The square brackets are used to aid in the comparison.
(11) M
(12) 1 +
J^
2"
!
2"
i
If p > 1, the terms in (12) are never less than the corresponding
terms in (11 ). Hut, in (12), the sums within the brackets are
1 . 1 _ 2. _ . .]_
/> ' ;) 2'' " ~ 2 ; '
and so on. Hence, to test (12) for convergence, we may consider
the series
<" '+^+(rO'+(F
When p> 1, series (13) is a geometric series with the common
ratio less than unity, and is therefore convergent. Therefore (10) is
also convergent. When p 1, series (10) is the harmonic series and
is divergent,. When p ^ 1, the terms of series (10) will, after the
first, be greater than the corresponding terms of the harmonic series ;
hence (10) is now divergent. Q.E.D.
ILLUSTRATIVE KXAMPLK 4. Show that the series
4
(14)
-f
2-3-4
is convergent.
Solution. In (14),
, __
3-4-54 5-6
2n
(n
or -u n < ;
that is, .], u n is less than the general term of the /) series when p = 2. Hence the
series each of whose terms is half the corresponding term in (14) is convergent
and therefore (14) also is convergent.
SERIES 345
PROBLEMS
Test each of the following series.
L i + JU + -W *= + **
Convergent.
' * \/2 V3 V 7 ??
Divergent .
900 2
Convergent.
C^onvcTgent.
4 " 1 2 *" 2 3 h 3 4 M(H -f 1)
4 , 8 , 12 . 4 H
Divergent.
2 -3 3 -4 4 5 ' (H + 1 UN + 2)
, 3 ,57 2 W +1 { ...
Convergent.
2-3-4 3-4-5 4 5 f> ' ( ti -f 1 > ( w -f 2 ) ( H -f 3 )
'i+f,^-^-
Divergent.
Divergent.
Convergent.
Q 1 1 1 I _1_ 1- . ,
4 10 28 3" + l
' ^ V3 v'3 \/3
Divergent.
11 l + l + l f ! '
Convergent.
Convergent.
Divergent.
1 1 1 j_ 1
i o 1 i " L 1 ~~ ~h " " "
2-2-3 2-3-4 2-4-5 ' 2(w + l)(r/ + 2)
14. 1 -f i -f i 4- i J 2 + a V> + aV) + 4*2 + - - -.
Convergent.
01 11
1 c 2 , - - , -' , - i . . . 17. U .
4 r-^-T + ' ' .
126
187. Cauchy's test-ratio test. In the infinite geometric series
a + ar + ar~ + + ar" + or 1 ' M + -,
the ratio of the consecutive general terms ar" and ar" f ' is the com-
mon ratio r. Moreover, we know that the series is convergent when
i r \ < 1 and divergent for other values. We now explain a ratio test
which may be applied to any series.
346 DIFFERENTIAL AND INTEGRAL CALCULUS
Theorem. Let
(]) U\+ Wli + U-A + ' ' ' +U n + Mn + 1 + ' ' '
be an infinite scries of positive terms. Consider consecutive general
terms ?/ n and ?/ n i i > ^^rf /orrw the test ratio.
Test ratio = ^2LLi .
/"/'A/// JAtf //w?'/ o/ /r/,s tes/ ratio when n becomes infinite. Let this be
l- W<n 4 1
p lim --
n - cr W'n
I. H7/cw p -' 1, the scries is convergent.
II. M ; //r/? p > 1, ///r scries is divergent.
III. H/7/r// p = 1, ///r test foils.
Proof. I. H7/(v/ p 1. My the definition of a limit (Art. 14) we
can choose u so lar^e, say // ~~ m, (hat when n ^ m the ratio ^-^ will
diller from p by as little as we please, and therefore be less than a
proper fraction r. Hence
?/, f i ' ?/,,,r ; //, , - '" //, , \r < n ,,,r' 2 ; w m , ;? <- w,,,r* ;
and so on. Therefore, after the term ?/ each term of the series (1)
is less than the corresponding term of the geometric series
(2) u , n r + y/,,,r- + //,,,r' +
J^ut since r ^ 1, the series (2), and therefore also the series (1),
is convergent (Art. 1S(>).
II. \\'hen p ^ 1 (or p --- oo). Following the same line of reasoning
as in I, the series (1 ) may be shown to be divergent.
III. When p = 1 , t he series may be eit her convergent or divergent ;
that is, the test fails. For, consider the p series, namely,
The test,
st, ratio is ^ = (-^-rY= (l - -^)" ;
n n \H. + 17 \ n + I/
and lim l'^ = lim (l -- ~-- (1)" - 1 (= p).
n * \ W H 7 w -. x \ W + 1 7
Hence p = 1, no matter what value p may have. But in Art. 186
we showed that
when p > 1, the series converges, and
when p ^ 1, the series diverges.
SERIES 347
Thus it appears that p can equal unity both for convergent and for
divergent series. There are other tests to apply in cases like this,
but the scope of our book does not admit of their consideration.
For convergence it is not enough that the test ratio is less than
unity for all values of n. This test requires that the limit of the test
ratio shall be less than unity. For instance, in the harmonic series
the test ratio is always less than unity. The limit, however, equals
unity.
The rejection of a group of terms at the beginning of a series
will affect the value but not- the existence of the limit.
188. Alternating series. This is the name given to a series whose
terms are alternately positive and negative.
Theorem. // MI M L > + M : < M } ~f
is an alternating series in which each term is numerically less than the.
one which precedes it, and if ]j m u ___ ^
n * >
then the series is convergent.
Proof. When n is even, S n may be written in the two forms
(1) S n = (MI - M L .) + (M.I - MI) -f -f (n n , - ?/),
(2) S tt MI (M L > ~ M.O (M, t 2 M,, i ) M, t .
Each expression in parentheses is positive. Hence when n in-
creases through even values, d ) shows that S tl increases and (2) shows
that S H is always less than ?/j ; therefore, by Theorem I, Art. 185,
S n approaches a limit /. But S' n t j also approaches this limit /,
since S H + \ ~ X n + '"*} i and lirn u n n = 0. Hence when n increases
through all integral values, <S' n > / and the series is convergent.
ILLUSTRATIVE EXAMPLE. Test the alternating scries 1 - i -f i - } +
Solution. Each term is less in numerical value than tl
Hfcnce lim (w n ) 0, and the series is convergent.
Solution. Each term is less in numerical value than the preceding one, and u n -~ -
An important consequence of the above proof is the statement :
The error made by breaking a convergent alternating series off at any
term does not exceed numerically the value of the first of the terms discarded.
Thus, the sum of ten terms in the above Illustrative Example is 0.646, and the
value of the series differs from this by less than one eleventh.
In the above statement it is assumed, however, that the series has
been carried far enough so that the terms are decreasing numerically.
348 DIFFERENTIAL AND INTEGRAL CALCULUS
189. Absolute convergence. A series is said to be absolutely or un-
conditionally convergent when the series formed from it by making
all its terms positive is convergent. Other convergent series are said
to be conditionally convergent.
For example, the series 1 9^ + 0^ 77 + FI ~~ ' ' '
is absolutely convergent, since the series (3), Art. 186, is convergent.
The alternating series
! 1,1 1,1 ____
1-9+3-4+5
is conditionally convergent, since the harmonic series is divergent.
A series with some positive and some negative terms is convergent if
Ike series deduced from it by waking all the signs positive is convergent.
The proof of this theorem is omitted.
190. Summary. Assuming that the test-ratio test of Art. 187 holds
without placing any restriction on the signs of the terms, we may sum-
marize our results in the following
General directions for testing the series
U\ 4" U2 + H3 + H4 + ' + l/n + U n + l + ' ' '.
When it is an alternating series whose terms never increase in numer-
ical value, and 'if lim w w =- 0,
Tl '00
then, the series is convergent.
In any series in which the above conditions are not satisfied, we deter-
mine the form of ?/ n and u n 1 i, Jorm the test ratio, and calculate
lim
= p.
I. When | p | < 1, the series is absolutely convergent.
II. When | p i > 1, the series is divergent.
III. When \ p = 1, the test faifo, and we compare the series with
some series which we know to be convergent, as
a + ar + ar~ + ar* + ; 0" < 1) (geometric series)
1 + ~~ + -~~ + ^ + J (P > 1) (P series)
*j O ~r
or compare the given series with some series which is known to be
divergent, as 1 + l + l + l+... ; (harmonic series)
1+ + r~ + + (p<l) (P series)
SERIES 349
ILLUSTRATIVE EXAMPLE 1. Test the series
Solution. Here M n = , - n +i=r
Mn+lsJ"- 1 = 1
u n [n n
p = lim - = 0,
n-oc K
and the series is convergent.
ILLUSTRATIVE EXAMPLE 2. Test the series
Ji + .yL + Ji +
Solution. Here
10" n + 1
T7 = ~io~ -
P= lim , -"'
n -* oo J U
and the series is divergent.
ILLUSTRATIVE EXAMPLE 3. Test the series
1 } 1 j 1 , ...
Solution. Here = (2 n _* 1)2 w * M ~' = J2"n + i,( 2 n + 2)*
7i ni i _ (2 r? - 1 ) 2 n _ 4 r? 2 - 2 n ^
w n ~ (2ri -f lK2n-f 2) "" 4 n 2 -f 6 n -f 2*
,. 4 n 2 2 n 1
p lim - = 1,
77 , oc 4 w- -h 6 w 4- 2
by the rule in Art. 18. Hence the test-ratio test fails.
But if we compare the given series with the p series, when p = 2, namely,
we S ee that it must be convergent, since its terms are less than the corresponding
terms of this ;; series, which was proved convergent.
PROBLEMS
Test each of the following series.
..... Ans. Convergent.
2 3 ,_l + A + .-H . Convergent.
*' 2 2 2 2 3 2 4
350 DIFFERENTIAL AND INTEGRAL CALCULUS
3 3 2 3 ; * 3 4
3. - -f ~-~ 2 + ~~ + -~-^ + . An,s. Divergent.
4
1
1 -
2
1
^ j . i ''
3
3-
5
3
' 5 ' 7 3 - 5 7 (2 n -f 1 ;
Convergent.
1 ,
1
3
l 1
3-5 l .:j.5--- (2 H-I)
1 '
1-
4
1 1
-4-7 1 4 7 .-. (3 n - 2)
Convergent.
1
1 .
<>
i
'{ 1 'i 4
~L
1 '
1
3
~T
3 -5 1 3 -5-7
Lx on vergent.
r
r ?
r'i
"4
7.
H
+
l
^"
Convergent.
1
|2
l ; f
M
8.
Q
?, +
t
.4
b H
h ~ -f '
Divergent.
[i
212 1^13 2 :i |4 -i 4 . h
10. 1 -f -p- 4- -77- -f -rb 1 + . 12. ^ -h - -f ~- -f ~ -f .
f> 10 \t ;> ;j- 5- ! ;V
13. ;; -f ^ + .7~ir~i7 -f jj.^.j^.jo +
191. Power series. A series whose terms are monomials in ascend-
ing positive integral powers of a variable, say j, of the form
(1 ) flu + fli J + u L >.r 2 + a ;i r :{ + -,
where the coefficients uo, i, LS ure independent of a*, is called a
power scries in x. Such series are "of prime importance in the study of
calculus.
A power series in j may converge for all values of .r, or for no value
except x = ; or it may converge for some values of x different from
and be divergent for other values.
We shall examine (1) only for the case when the coefficients are
such that . .
Mm (^A = L,
where L is a definite number. To see the reason for this, form the
test ratio (Art. 187) for (\], omitting the first term. Then we have
i<n a, t x n a n
Hence for any fixed value of x,
i- A* * I \ ,. /dni 1
p = hm I-* i j) =xhm \- JL - L ) =xL.
SERIES 351
We have two cases :
I. If L = 0, the series (P will converge for all values of j, since
p = 0.
II. If L is not zero, the series will converge when rL(= p) is nu-
merically less than 1, that is, when x lies in the interval
and will diverge for values of .r outside this interval.
The end points of the interval, called the interval of convergence,
must be examined separately. In any given series the test ratio
should be formed and the interval of convergence determined by
Art. 187.
ILLUSTRATIVE EXAMPLE 1. Find the interval of convergence for the series
Solution. The test ratio here is
Hn .1 ft'' A i i- n' 2
_-L-_L = ~- jc. Also, lim = 1,
U n (/< + lr n . oc (n i- })' 2
by Art. 18. Hence p j*. and the series converges when JT is numerically less
than 1 and diverges when .r is numerically greater than 1 .
Now examine the end points. Substituting jr ~ 1 in (2), we get
1 - r^ + -; - ^ + -,
which is an alternating series that converges.
Substituting x = 1 in (L! i, we get
which is convergent by comparison with the p series (p > 1).
The series in the above example has [ 1, 1] as the interval of convergence. This
may be written 1 - x r 1, or indicated graphically as follows:
ILLUSTRATIVE EXAMPLE 2. Determine the interval of convergence for the series
Solution. Omitting the first term, the test ratio is
1
Also lim r- = 0. Hence the series converges for all values of x.
- (2w + l)(2n + 2)
352 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
For what values of the variable are the following Graphical representations
BCries convergent? of intervals of convergence*
1. 1 + x + z 2 + x 3 + . Ann. - 1 < x < 1.
3. x -f x 4 4- x 9 + x 16 + -. Am. - 1< x < 1.
4. z + -V -f -~= + -. A?i. - 1 < x < 1.
V2 V3 ~
5. l-hx+|~-fr H . Arcs. All valuesof x. ^ L
l~_ UL o
02 04 ^r> -
6. 1 r~-+ r~ rrrH . Ans. All values of 0. "^ L
7. 0-+ ,~ , + . Anx. All valuesof 0.
1 - 3 z 5 1 3 5
- 4 6 '
s. - 1 <z< 1.
8v _i_ - 1 ^ -1_ - -i- J * I I
. X -f- ~r T ~~ ~r ^ :; :; "^ ~r *. . |
2 3 2-4 5 2 4 (> < ITJ 5
9. x -f 2 x 2 + 3 x 3 -f 4 x 4 H . Ans. - 1 < x < 1.
10. l_:r + i!-^ + .... -1 gjr gl.
^. r 2 r .T r 4
1 i x -* i _ ^ i . . . _ 9 <^ r <
J>1 ' -I O O . 02 ' J . 03 4 . >4 ~ ' - \ .* - -.
Jlfcrf 4-1 ^ O- T*rf
O r 2 'J Wi /I r 4
12. x 4- + + + All values.
z 2 . l-3x 4 , l-3-5z
2 2 2
1 Q 1 4_ ** _L i i . . .
2 4 2 4 2 4 - 6 2 '
* End points that are not included in the interval of convergence have circles drawn
about them.
SERIES 353
ld - l+ l + f^ + f^ + '"- (a>0) AM. -agjr<a.
92 ->. 32 r 2 42 r3
16. 1 + ^ + ^ -f i^ 4- - All values.
Lfi l Li
17 ' 3 + 2^3^ + 22 . 33 + 03 . 34 H - 6 < x < 6.
18. x + 4 J 2 -f 9 x 3 -f 16 x 4 + - .
19 * , 2 * 2 , 3 J3 4 x*
2 -<
1 2 2 2 3 2 2 3 4 2-< 4 5
.r |2 x 2 13
-
10 i | 100 j | 1000 j* ; 23. r + ^' + ^ + ^
192. The binomial series. This important series is
/iv .. , , w(?w 1) o , in(m \}(m 2) ...
(1) 1 + mx + ^ 9 J- + ^ l 9 ; 3 - ^ H ----
i w ( w 1 ) (w ~ 2 ) (?w n + 1 ) w ,
+ ^ +
where m is a constant.
If m is a positive integer, (1 ) is a finite series of m -f 1 terms, since
all terms following that containing x m have the factor m m in the
numerator and vanish. In this case (1) is the result obtained by
raising ! + / to the wth power. If m is not a positive integer, the
series is an infinite series.
Now test (1) for convergence. We have
- l)(m-2) - (m - n + 2) , .
1 -2-3- (w-1) '
, m(m l)(m 2) - - (m n + 2)(m ri + 1) n
and ^^ J ~
Hence ^ = m-n + 1 x = tml
u n n V n
Then, since lim (^ - -- 1)= 1, we see that p = x,
n -. oo \ tl /
and the series is convergent if x is numerically less than 1, and di-
vergent when x is numerically greater than 1.
354 DIFFERENTIAL AND INTEGRAL CALCULUS
Maclaurin's Series, Art. 194, implies the following statement.
Assuming that ra is not a positive integer and that \x\ < 1, the
value of the binomial series is precisely the value of (1 + x) m . That is,
(2) (1 + xr = 1 + mx
If m is a positive integer, the series is finite and equals the value
of the left-hand member for all values of x.
Equation (2) expresses the special binomial theorem. We may also
write
(3) (a + b) m = a r "(l + x) w , if x = --
a
Thus the left-hand member of (3) may also be expressed as a
power series.
Examples of approximate computation by the binomial series are
given below.
ILLUSTRATIVE EXAMPLE. Kind v (;w approximately, using the binomial series.
Solution. The perfect square nearest to G30 is 625. Hence we write
\ GIJO = \ (52~5~T5 = 25( 1 + ] I -;)*
Now write out (2) with ?w J. The result is
(1 -f ar)i = 1 4- ?, .r - jj, .r-' f ,V -r' ( - T ? ft x 4 + %
In this example, j = 1 J r> ~ 0.008. Hence
(1 -f 1 .\ f| )i - 1 -f ().()04 ~ O.OOOOOS -f ().0000()00:r2 H ---- .
(4) 25(1 -f T ^2 = 25 -f O.i - 0.0002 4 0.0000008 =25.099S01 (to the nearest
figure in the sixth decimal place). Ans.
The series in (4) is an alternating series, and the error in the answer is less
than 0.0000008.
193. Another type of power series. We shall frequently use series
of the form
(1) 60 + MJT - a) + 6 2 (.r - a) 2 H ----- h b,,(x - a) n H ----
in which a and the coefficients &o, 6 lf - -, 6 n , - - are constants,
Such a series is called a power series in (x a).
Let us apply the test-ratio test to (1), as in Art. 191. Then, if
lim ^-i = M,
SERIES 355
we shall have, for any fixed value of j,
= lim ^-^ = (x a)M.
We have two cases :
I. If .17 = 0, series (1) is convergent for all values of x.
II. If M is not zero, series (1) will converge for the interval
A convergent power series in .r is adapt ed for computation when x
is near zero. Series (1), if convergent, is useful when x is near the
fixed value a, given in advance.
ILLUSTRATIVE EXAMPLE. Test the infinite series
1 /, ] , f ( - r - 1>" (*~- 1"' , . . .
i - tor i M ^ - \-
for convergence.
Solution. Neglecting the first term, we have
n + 1
Also, lim ( - )= 1.
w - oc \ N -f 1 /
Hence |p| = \x - 1 1 , and the seri(\s will converge when x lies between and 2.
The end point jr '2 may he included.
PROBLEMS
1. Using the binomial series, show that
1 + jr '
Verify the answer by direct division.
Using the binomial series, find approximately the values of the fol-
lowing numbers.
2. V98. 5. Yii5. 8. -rp 11.
V990
^ 1 1
4 --7" -x
4.^/630. V412 V30 \16
356 DIFFERENTIAL AND INTEGRAL CALCULUS
For what values of the variable are the following series convergent ?
/~ t 1 \ 2 /-.. i i \a /,.- i 1 ^4
-. Ans. 2 < x < 0.
15. (jr- 1) +
16. 2(2.
23 4
(T- I) 2 j fr-_l) 3 { (x-l)
X/2 V3 V4
^T < 2.
All values.
_
17. 1^(^-2)^ + ^
18. 1 -2(2jr- U) 4-3(2 jr - H) 2 ~ 4 (2 ^ - 3 )- { H .
19.
jr - 3 , (.r - 3)' 2
i - a
; ''./.. i ..'.- ' ' i
2 - 3- 3 3 :t
(x - 3) 4
4 - 3 4
CHAPTER XX
EXPANSION OF FUNCTIONS
194. Maclaurin's series. In this chapter the question of represent-
ing a function by a power series, or, otherwise expressed, developing
(expanding) the function in a power series, will be discussed.
A convergent power series in jc is obviously a function of j for all
values in the interval of convergence. Thus we may write
(1) /(*") = 00 + J J + a * r ~ + + a,,/" + - -.
If, then, a function is represented by a power series, what must be
the form of the coefficients a , fli, -, (in, etc.? To answer this ques-
tion we proceed thus :
Set x = in (1). Then we must have
(2) /(()) - oo.
Hence the first coefficient a in (1) is determined. Now assume that
the series in (1) may be differentiated term by term, and that this
differentiation may be continued. Then we shall have
( f(jr) = a, + 2 a 2 r + 3 a ; <:r 2 H h na n x n ~ l H
(3) J /"(j) = 2 a 2 + 6 flsr H h n(n - 1 )a r ,j n ~ 2 H
[f'"(x) = 6 a.-H h n(n -l)(n- 2)a n x n -* -\
etc.
Letting x = 0, the results are
(4) /'(O) = ai, /"(O) = 12 02, / ;// (0) = [3 a 3 , , / (n) (0) = lwo n .
Solving (4) for a\ t a, -, On, etc., and substituting in (1 ), we obtain
(4) /(x) - /(O) + /'(o) i + /- (o) | + - + / (w) (o) g -f - - -.
This formula expresses /(x) as a power series. We say, "the function
/Or) is developed (or expanded) in a power series in x." This is
Maclaurin's series (or formula).*
* Named after Colin Maclaurin (1698-1746), and first published in his " Treatise of
Fluxions" (Edinburgh, 1742). The series is really due to Stirling (1692-1770).
357
358 DIFFERENTIAL AND INTEGRAL CALCULUS
It is now necessary to examine (4) critically. For this purpose
refer to (G), Art. 124, and rewrite it, letting a = 0, 6 = x. The result is
(5) f(x) =/(0; +
IT
where # ^/' r "(^i) r (0 < jj < jj
The term R is called the remainder after n terms. The right-hand
member of (5; agrees with Maclaurin's series (A) up to n terms. If
we denote this sum by N/,, then (5) is
S n + K, or
Now assume that, for a fixed value x ~ xn, R approaches zero as a
limit when n. becomes infinite. Then S tl will approach /(.r ) as a limit
(Art. 14). That is, Maclaurin's series (4) will converge for x = x
and its value is/Oo). Thus we have the following result :
Theorem. /// order that the series (A) should converge arid represent
the function f(x) it is necessary and sufficient that
(G) lim 7? = 0.
n oc
It is usually easier to determine the interval of convergence (as
in the preceding chapter) than that for which ((>) holds. Hut in
simple cases the two are identical.
To represent a function /(.n by the power series (A), it is obviously
'necessary that the function and its derivatives of all orders should be
finite. This is, however, not sufficient.
Examples of functions that cannot be represented by a Maclaurin's
series are
In x and ctn x,
since both become infinite when .r is zero.
The student should not fail to note the importance of such an
expansion as (A). In all practical computations results correct to a
certain number of decimal places are sought, and since the process
in question replaces a function perhaps difficult to calculate by an
ordinary polynomial with constant coefficients, it is very useful in sim-
plifying such computations. Of course we must use terms enough to
give the desired degree of accuracy.
In the case of an alternating series (Art. 188) the error made by
Stopping at any term is numerically less than that term.
EXPANSION OF FUNCTIONS 359
ILLUSTRATIVE EXAMPLE 1. Expand cos x into an infinite power series and de-
termine for what values of J it converges.
Solution. Differentiating first and then placing x - 0, we get
/(jrt = cos x. /(O) = 1,
f'(x) = - sinx, /'(( ~-0,
f"(x] = - cos x, f'\0) = - I,
/'"(j) = sin J, /'"(O) = 0,
/iv(j-) = COST, /1V(( rr \ f
fv( x ) = _ s in jr, /v(0) = 0,
/Vl(j) zr: _ COS J, / vi (0) = - 1,
etc., etc.
Substituting in (4),
(7) C08 T=l-|-f^ - jj + --"
Comparing with Problem G, Art. 191, we sec that the series converges for all
values of J*.
In the same way for sin j,
(8) sin, -,-?! + - t - ?1 -.
which converges for all values of JT (Problem 7, Art. 191 .
In (7) and (8) it is not difficult to show that the remainder /v approaches zero
as a limit as n becomes infinite, when jc has aw// fixed value. Consider (7). Here we
may write the nth derivative in the form
,,
Hence A = cos x
Now cos(ji 4- ) never exceeds 1 in numerical value. Also, the second factor
of R is the nth term of the series
which is convergent for all values of x. Therefore it approaches zero as n become:-!
infinite (see (A), Art. 1S5). Hence (6) holds.
Jim - = 0.
,i .* [W
From the above example we see that
r n
Jim -
,i .* [W
Also, on the page preceding, we had
K= fn)(
Hence lim /? = if f (n] (x\) remains finite when n becomes infinite.
. (0< X, < X)
\n
360 DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 2. Using the series (8) found in the teat example, cal-
culate sin 1 correct to four decimal places.
Solution. Here x \ radian ; that is, the angle is expressed in circular measure.
Therefore, substituting x = 1 in (8) of the last example,
Summing up the positive and negative terms separately,
1 = 1.00000 - -J- = 0.16667
~ - 0.008,'W ,- - 0.00020
y? _ y; _
1.008U3 0.16687
Hence sin 1 - 1 .008W - 0.16687 = 0.84146
which is correct to five decimal places, since the error made must be less than ;
I?
that is, less than 0.000003. Obviously the value of sin 1 may be calculated to any
desired degree of accuracy by simply including a sufficient number of additional
terms.
PROBLEMS
Verify the following expansions of functions by Maelaurin's series and
determine for what values of the variable they are convergent.
1. r r - 1 + .r + ~ + !' + ...+ -1-1- -|- .... Ans. All values.
-
r-'l r- r > ( _ 1 \ " r
2. sin , = ,--+- ---- + 1_1J - :L - + . . .. A11 values _
.r-' .r ;t ,r
4. In U - j-) - - x - TT - ~ - j ----- ^j- ---- . - 1 ^ JT < 1.
:$ ' ^
5. arc sin jr = x
6. arc tan j- = r - -f ---- +
A n,s. All values.
8. In (a + J-) = In a -f- - - -^- 4- ^ -----
a 2 a- 3 a a
(n -
EXPANSION OF FUNCTIONS
Verify the following expansions.
361
( - 4
\4
12. tan ( - 4 + JT = I -f 2 jc + 2 x- +
13. arc tan - = ^ - x + - 4- -.
j* 2 3 5
or LS
16.1ncos J -=-^-^-g..,
Find three terms of the expansion in powers of JT of each of the fol-
lowing functions.
17. cos j* ~
18. sin (.r + 1).
19- e
20. J(r'- c
Compute the values of the following functions by substituting di-
rectly in the equivalent power series, taking terms enough to make the
results agree with those given below.
21. c = 2.7182 -.
Solution. Let x 1 in the series of Problem 1 ; then
First term =1.00000
Second term = 1 .00000
Third term = 0.50000
Fourth term = 0.16667
Fifth term =0.04167-
Sixth term = 0.00833
Seventh term = 0.00139
Eighth term = 0.00020
, etc.
Arts.
Adding, e = 2.71826
(Dividing third term by 3.)
(Dividing fourth term by 4.)
(Dividing fifth term by 5.)
(Dividing sixth term by 6.)
(Dividing seventh term by 7.)
22. arc tan (I) 0.1973 - ; use series in Problem 6.
23. cos 1 = 0.5403 ; use series in (7), Illustrative Example 1.
362 DIFFERENTIAL AND INTEGRAL CALCULUS
24. cos 10 = 0.9848 - ; use series in (7), Illustrative Example 1.
25. sin 0.1 = 0.0998 ; use series in Problem 1>.
26. arc sin 1 1.5708 - ; use series in Problem 5.
27 sin - ~ 0.7071 ; use series in Problem 2.
4
28. sin O.T> ~ 0.4794 ; use series in Problem 2.
29. <'-' = 1 + li f ^' 4 + - - - 7.3890.
30. V? = 1 4- + 7^7, -f oJTTj
-f . - l.b'487.
195. Operations with infinite series. One can carry out many of the
operations of algebra and the calculus with convergent series just as
one can with polynomials. The following statements are given with-
out proof.
],et f/o + fl,/4 <*#' + + <i>s" + '
and fro f- b\jr 4 b^jr- 4 * 4 b n s" -[
be convergent power series. Then we obtain new convergent power
series from them as follows :
1. fry adding (or subtracting) term, hi/ term..
(a< ( i bu) 4 (MI i ^1 )*" 4 4 ( a n b n )jr n 4 *
2. #// multiplication, and (jroupnnj terms.
ILLUSTRATIVE K\ AMPLE 1. Computation of logarithm*. From the series (Prob-
lems 3 and 4, Art.. 194)
In (1 4 oO = J-- } J-'-f - 1 ,-'" -i *' ^' ' '
In (1 - J*) - - J" - 1 - r ' J - a - r - \ - r * - ' '
we obt ain, by subt raet ion of corresponding terms, and using (2) , Art . 1 , the new series
(1) ln L, 2lJr+ .,. + U M!,. + -,
This series converges when ', j* ' ^ 1.
To transform (1) into a form better adapted to computation, lot A be a positive
number. Then, if we set
1 , 1 + JT __ A* -f 1
/2) x = - : whence ; r:
then | j | < 1 for all values of .V. Substituting in (l\ we get the formula
x- o 1 1__1__4-1_ i_ 4- 1
(3) In (A' + 1) = In .\ + ~ iTy"^ + 3 (2 A' + 1 )' 5 (2 :V 4- 1> 5 J
EXPANSION OF FUNCTIONS 363
This series converges for all positive values of A T , and is well adapted to com-
putation. For example, let A" = t. Then
In (A + 1) = In 2, r - = |
Substituting in (3), the result is In 2 = 0.69315.
Placing A* = 2 in (3\ we get
1,1 1
4
- 1.09861
It is only necessary to compute the logarithms of prime numbers in this way,
the logarithms of composite numbers being then found by using formulas (2),
Art. 1. Thus, l n 8 = In 2 ' = 3 In 2 = 2.07944
In 6 = In 3 + In 2 - 1.791 76- -.
All the above are Xapicrian, or natural, logarithms, that is, the bust* is
r 2.71828 . If we wish to find /ir/f/f/s's, or common, logarithms, where the base
10 is employed, all we need to do is to change the base by means of the formula
, In 11
'<>^
In the actual computation of a table of logarithms only a few of the tabulated
values are calculated from series, all the rest being found by employing theorems
in the theory of logarithms and various ingenious devices designed for the purpose
of saving work.
ILLUSTRATIVE FA AMPLE 2. Find the power series for C T sin x.
Solution. From the series
sin x = x - ^ -f --^ ---- Problem 2, Art. 1 94
and <*= * +* + ? + ? + + fSi ?robl " m ] ' AH ' * 94
we obtain, by multiplication,
c* sin x = x~f-x 2 -f^-~-f- terms in x' 5 etc. An.
3. By division. A special case is shown in the example below.
ILLUSTRATIVE EXAMPLE 3. Find the series for sec x from the series for cos x
(see (7), Art. 194).
X"' X* X
(4) cos x = 1 ~ n + |jj ~ -^ + ' '
Solution. From the formula sec x = , we see that we have to carry through
the division of 1 by the series (4). This is best done as follows.
Write (4) in the form cos x = 1 - z, where
r 2 r \ 7-
,r<. __ __ i _ __ ...
lo; " 2 24 720
Then
(6) sec x = -^ = 1 + 2 + z' 2 + 2 3 + ' ;
if ! 2 ! < 1 (Problem 1, Art. 193).
364 DIFFERENTIAL AND INTEGRAL CALCULUS
From (6), we have the series
T* x r>
z 4" ~ 24 "*"" terrns f higher degree,
Substituting in (6), the result is
sec x = 1 -f - x - + x 4 -f rr + . . .. .4ns.
PROBLEMS
Given In 2 - 0.69315, In 3 = 1.09861, calculate the following natural
logarithms by the method of the example above.
1. In 5 = 1.60944. 3. In 11 -2.39790
2. In 7 -1.94591. 4. In 13 - 2.56495.
Work out the following series.
5. c ' cos / = 1 / + ; { /'< _ ii /4 _j_ . . , f
6. ( ' X =1 + 2^-4- j-2 +? ^ + 55^
7 JL55JL 1-1 _ 1 .2 __ J_ 3 jilL 4
10. r' tan .r = JT + .r 2 + ~ j* -f - .r 4 -f . . .
-
1 1. f r sec j- 1 j- 4- j- 2 - " .r< -f I ^4 _j_ . . %
12. c 2 sin 2 / - i> / - / 2 - ~ /.'* -f- 5 /4 + . . B>
13. (1+ x) cos V.7 = 1 + i a- - 11 j.2 + ^i. ^ _ _11_
2 24 ^720 8064
14. (1 -f 2 j*) arc sin jc = x -f 2 r' 2 -f - j- a -f - .r 4 -f-
6 3 "
15. Vl - JT arc tan j- = j- - 1 ^ - 11 ^ -f A ^ + . . .
^4 48
16. Vl - tan x = l--x--x 2 -~^-^L.r^^
X ^ ^
EXPANSION OF FUNCTIONS 365
17. Vsecx = 1 + 7 J 2 4- ~ J* 4 + - .
4 96
. -
1+smx 2 b 12
20. V4 + sin </> = 2
16 2;>b 6144
'
For the following functions find all terms of the aeries which involve
pow r ers of x less than j* 5 .
-*- . 24. \/\\ -f c J .
21. c r> smx.
22.<*cosjVI 25.^jy^,
VI 4- x
sin JT
23.
x 26. V5 - cos x.
196. Differentiation and integration of power series. A convergent
power series
(1) oo + aiJ + a 2 x 2 + aa.r :{ -f + a n x n +
may be differentiated term by term for any value of x within the
interval of convergence, and the resulting series is also convergent.
For example, from the series
we obtain, by differentiation, the new series
Both series converge for all values of x (see Problems 6 and 7,
Art. 191).
Again, the series (1) may be integrated term by term if the limits
lie within the interval of convergence, and the resulting series will
converge.
ILLUSTRATIVE EXAMPLE 1. Find the series for In (1 + x) by integration.
Solution. Since -r- In (1 -f x) - - we have
dx 1 -f x
(2) ln(l -f x) = I r
/() L ~T X
NOW T-T- = 1 ~ X + * 2 ~ X * + ^ -
366 DIFFERENTIAL AND INTEGRAL CALCULUS
when |x|<l (Art. 192). Substituting in (2) and integrating the right-hand
member term by term, we obtain the result
In (1 -f ji = x - \ x' 2 -f 3 x :i - I .T 4 -f .
This series also converges when \x\ < I (see Problem 2, Art. 191).
ILLUSTRATIVE EXAMPLE 2. Find the power series for arc sin x by integration.
Solution. Since -- arc sin x . we have
dx \ i _ X 2
dx
r f dx
.'f, \l-j-l'
By the binomial series ((2j, Art. 192;, letting /// J, and replacing x by x 2 ,
we have
1 2-4-6 '
This series converges when \x\ < 1. Substituting in (3i and integrating term by
term, we get
an- sin , = T -, -,
This series converges also when \x\ - I (see Problem 8, Art. 191).
By this scries, the value of TT is readily computed. For, since the series con-
verges for values of / between 1 and + 1, we may let x = , giving
or TT = .'{.1415 -.
Evidently we might have used the series of Problem (>, Art. 194, instead. Both
of these series converge nit her slowly, but there are other series, found by more
elaborate methods, by means of which the correct value of TT to a large number of
decimal places may easily be calculated.
ILLUSTRATIVE EXAMPLE ,'{. Using series, find approximately the value of
I sin jr-dx.
Solution. Let r - x ? . Then
sin z - z - rr + fp - * * ' Problem 2, Art. 194
jtt flO
Hence sin x 2 = x' 2 - r~ + 'rr ' ' %
lii 1A
r ! , r 1 / r (i T U) \ , . ,
and / sin x-dx = I ( a* 2 r^ -f V ) dx, approximately,
= I ^1 _ _ 4. --I 1 = 0.3333 - 0.0238 -f 0.0008
= 0.3103. Aws.
EXPANSION OF FUNCTIONS 367
PROBLEMS
1. Find the series for arc tan x by integration.
2. Find the series for In (1 JT) by integration.
3. Find the series for sec 2 .r by differentiating the series for tan .r.
4. Find the series for In cos .r by integrating the series for tan .r.
Using series, find approximately the values of the following integrals.
i ~~~"' -4"- s '- 0.0295.
0.185. 10. f 1 c ''dx.
Jo
7. f Vln (1 -f jr)djr. 0.0(528. 11. fin U 4 V7 ) (/.r.
/o A)
i
8. f ( ',- 0.1815. 12. fVsin V7'</.r.
Jo -\/i _ j.'J Jo
13. fVl - .H <Lr. 14. ( 'r J cos Vr f/.r. 15. / \/2 - sirTx dr.
Jo ./o -A)
197. Approximate formulas derived from Maclaurin's series. By
using a few terms of the power series by which a function is repre-
sented, we obtain for the function an approximate formula which
possesses some degree of accuracy. Such approximate formulas are
widely used in applied mathematics.
For example, taking the binomial series ((2), Art. 192), we may
write down at once the following approximate formulas.
First <i}>j)}(>j'ini(ition Second dj>proriniatioH
(1 + x} m 1 + mx 1 + mx + \ m(w 1 )x- ;
_ = 1 mx = 1 mx + 2 'm(m l)x 2 .
In these \x\ is small and m is positive.
Again, consider the sine series
> r
X X'*
Then
(2) sin x x,
' ''___
sin x x ~
etc.
(3)
are approximate formulas. Let us examine the first.
368 DIFFERENTIAL AND INTEGRAL CALCULUS
In the series in (I), assume values of x such that the terms de-
crease. If the first term only is retained, the value of the remaining
series is numerically less than its first term j :{ (Art. 188). That is,
sin x = x, with | error |< | x*\.
We may inquire, For what range of values of x will (2) hold to three
decimal places? Then
" | < 0.0005,
that is, LH < V0.003 < 0.1443 rad.
We then conclude that formula (2) is correct to three decimal places
for values of x between 0.144.'* and +0.1443, or, in degrees, for
values between - 8.2 and + 8.2.
PROBLEMS
1. How accurate is the approximate formula sin x x when
(a)jr = 30? (b).r-60 ? (c).r~90? 6
Am. (a) Error < 0.00033; (b) error < 0.01 ; (c) error < 0.08.
2. How accurate is the approximate formula cos jr = 1 ~ when
(a) jc - 30? (h) JT = (JO ? (c) .r = 90? 2
Am. (a) Error < 0.0032; (}>) error < 0.05; (c) error < 0.25.
3. How accurate is the approximate formula e~ f = 1 jc when
(a) j- = 0.1? (b) .r = 0., r >?
4. How accurate is the approximate formula arc tan x = x ~ when
(a)jr=0.1? (h)j^ = 0.5? (c) jc - I ? 3
Am. (a) Error < 0.000002 ; (b) error < 0.006; (c) error < 0.2.
5. How many terms of the series sin .r .r j^- -f - must be
taken to give sin 45 correct to five decimals? LL L2- .477.8. Four.
jc~ x 4
6. How many terms of the series cos j- = 1 ~ irr + r '** mus t be
taken to give cos 60 correct to five decimals? '- I
7. How many terms of the series In (1 -f x} = x ~ -f- ^ - must
be taken to give log 1.2 correct to five decimals? ~ Ans. Six.
Verify the following approximate formulas:
10. r~" cos = 1 - 6 + ~- 14. fare sin xdx = C -f -f ~
o ' 2 24
11. cosVj(/j'=r+ x-~+- 15. (>sin0(/0= C -f -f
/ 4 ^2 */ 23
EXPANSION OF FUNCTIONS 3(59
198. Taylor's series. A convergent power series in x is well adapted
for calculating the value of the function which it represents for small
values of x (near zero). We now derive an expansion in powers of
x a (see Art. 193), where o is a fixed number. The series thus
obtained is adapted for calculation of the function represented by it
for values of x near o.
Assume that
(1) f( x ) = fr + bi(j - a) + b>2(Jc - a) 2 + + b n (x - a) n + ,
and that the series represents the function. The necessary form of
the coefficients & ( >, 61, etc. is obtained as in Art. 194. That is, we dif-
ferentiate (1) with respect to r, assuming that this is possible, and
continue the process. Thus we have
J'(x) = bi+2 6 L >U - a] H ---- + nb n (jr - a) n " ' + ,
/"(*) = 2 6,> + - + n(n - 1 )M-r - a)"~- + ,
etc.
Substituting x a in these equations and in (1 ), and solving for
bo, bi, bi>, , we obtain
6 =/(o), 6,=/'(d). k
Replacing these values in (1 ), the result is
(B) /(x) - /(a) + /'(a) ^ + /"(a) - +
The series is called Taylor's series (or formula).*
Let us now examine (B) critically. Referring to (G), Art. 124, and
letting b = x, the result may be written thus :
(2) /(or) =/(o)
where
The term R is called the remainder after n terms.
Now the series in the right-hand member of (2) agrees with
Taylor's series (B) up to n terms. Denote the sum of these terms
by S n . Then, from (2), we have
f(x) = S n + R, or
* Published by Dr. Brook Taylor (1685-1731) in his " Mpthoclus Incrementorum "
(London, 1715J.
370 DIFFERENTIAL AND INTEGRAL CALCULUS
Now assume that, for a fixed value x = XQ, the remainder R ap-
proaches zero as a limit when n becomes infinite. Then
(3) lim S n =/0o),
n - w
and (B) converges for x xo and its value is /(.TO).
Theorem. 77^ infinite series (B) represents th-e function for those
values of j, and those only, for which the remainder approaches zero as
tlie number of terms increases without limit.
If the series converges for values of x for which the remainder does
not approach zero as // increases without limit, then for such values
of x the series does not represent the function f(x).
It is usually easier to determine the interval of convergence of the
series than that for which the remainder approaches zero ; but in
simple cases the two intervals arc identical.
When the values of a function and its successive derivatives are
known and finite for some fixed value of the variable, as jr a, then
(B) is used for finding t he value of the function for values of x near a,
and (B) is also called the expansion off(x) in the neighborhood of x a.
iLLrsTKATIVK EXAMPLE 1. Expand In x in powers of (x 1 ).
Solution. f(jr ) = In jr, /( 1 ) - 0,
f'(x] =-, /'(D = l,
x
/'V) = -^. /"(1) = -1,
/"V)= J;. /"'(I) = 2,
etc., etc.
the ririnity
ec., ec.
Substituting in (B), In x - .r - 1 - J(r - 1 ) 2 + ^(x - 1)'' - . Ans.
Tliis converges for values of x between and 2 and is the expansion of In x in
ririnity of x 1. See Illustrative Example, Art. 198.
ILLUSTKATIX E EXAMPLE 2. Expand cos .r in powers of / x j j to four terms.
Solution. Here f(x) cos x and a ~ - Then we have
4
f(x) = cos .r, M~ | = =
\4/ \ 2
.
etc., etc.
EXPANSION OF FUNCTIONS 371
The series is, therefore,
v 2 N 2
The result may be written in the form
To check this result let us calculate cos 50 . Then x 5 expressed in radians,
> ., 4
or x _ 1L - 0.08727, (x - -\ - 0.00762, (x - ~Y~- (UHMMJG. With these values
4 \ 4 / \ 4 /
the series ahove gives cos 50 = 0. (54278. Five-place tables give cos 50 0.64279.
199. Another form of Taylor's series. In (), Art. 19S, if we re-
place a by x () and let .r - a = h, thut is, .r = a f- // = .r u + //, the result is
In this second form the new value of /(.r) when .r changes from j\> to
Jo + h is expanded in a power series in //, the increment of .r.
ILLUSTRATIVE EXAMPLE. Expand sin r in a power series in h when .r changes
from Jo to /o 4- h.
Solution. Here f(.r> - sin x, and f(x (} 4/0 " sin (j,, 4 h). Differentiate and ar-
range the work as below.
f(x) sin j, /(J ( >) ~ sin JT O ,
/'(jri = cos .r, f'(xu) cos .r () ,
/"(ji sin x, / "(J<>) = sin X ,
etc., etc.
Substituting in (C), we obtain
sin (xo -f h i = sin x 4- cos x ~ sin j ( , cos j ( 4- *. Am.
PROBLEMS
Verify the following series hy Taylor's formula.
1.
2. sin x = sin a 4- (x a) cos a ^ r; sin a ' r i q cos a 4-
3. cos a- = cos a (x a; sin a ' " -~ cos a 4- 7 sin a 4-
372 DIFFERENTIAL AND INTEGRAL CALCULUS
x 2 x 3
5. cos (a -f x) = cos a x sin a nr cos a + pr sin a + .
6. tan (x + h) = tan x -f- /* sec 2 x + /^ sec 2 x tan x + .
7. (x + A)" = ^ -f nx n -*h + n(n ~ 1} x"- 2 A 2
8. Expand sin x in powers of f x \ to four terms.
(-!)'
12
Ans. sin r = _
9. Expand tan x in powers of Lr - ~\ to three terms.
Am. tan .r = 1 + 2 a- - + 1> j - -f
10. Expand In r in powers of (r - 2) to four terms.
11. Expand <" in powers of (-r - 1) to five terms.
12. Expand sin f- -f jr \ in powers of x to four terms.
13. Expand ctn f ~ -f .r j in powers of x to three terms.
200. Approximate formulas derived from Taylor's series. Such
formulas are provided by using some terms of series (B) or (C).
For example, if /(.r) = sin j, we have (see Problem 2, Art. 199)
(1) sin y ~ sin a + cos a(x a]
as a first approximation.
A second approximation results by taking three terms of the series.
This is
f y _ Q \ 2
(2) sin x = sin a + cos a(x a) sin a -^
From (1), transposing sin a and dividing by x a, we get
rinar- S ina ==COBOj
j a
Since cos a is constant, this means that (approximately)
The change in the value of the sine is proportional to the change in
the angle for vahics of the angle near a.
Formula (3) illustrates interpolation by proportional parts.
EXPANSION OF FUNCTIONS 373
ILLUSTRATIVE EXAMPLE 1. For example, let a = 30 = 0.5236 radian, and sup-
pose it is required to calculate the sines of 31 and 32 by the approximate formula
(1). Then, since x - a = 1 = 0.01745 radian,
sin 31 = sin 30 + cos 30 (0.01745)
= 0.5000 + 0.8660 x 0.01745
= 0.5000 + 0.0151 =0.5151.
Similarly, sin 32 = sin 30 + cos 30 (0.03490) = 0.5302.
These values by (1) are correct to three decimal places only. If greater ac-
curacy is desired, we may use (2).
Then sin 31 = sin 30 + cos 30 (0.01745) - sm j* ' ^0.01745)2
= 0.50000 + 0.0151 1 - 0.00008 ~
= 0.51503.
sin 32 = sin 30 + cos 30 M (0.03490) - sin .j* (0.03490)'-
= 0.50000 + 0.03022 - 0.00030 "
= 0.52992.
These results are correct to four decimal places.
From (C) we derive approximate formulas for the increment of
/(j) when x changes from jr (} to Jo -f h- For, transposing the iirst
term of the right-hand member, we get
(4) /(J + h) -/(Jo) = /'(j u )// +/"(:ro) j~ + - -.
Ik
The left-hand member expresses the increment of /(j) as a power
series in the increment of JT (=//).
From (4) we derive, as a first approximation,
(5)
This formula was used in Art. 92. For the left-hand member is the
value of the differential of f(x) for x = J and AJ h.
As a second approximation, we have
(6) /(Jo + h) - /(J ) - /' (J )// + /"(J ) ~
<u
ILLUSTRATIVE EXAMPLE 2. Calculate the increment of tan x, approximately,
when x changes from 45 tJ to 46, by (5) and by (6).
Solution. From Problem 6, Art. 199, if x = x (] , we have
tan (x -f h} tan x 4- sec 2 x tt h -f sec 2 x<, tan x (} h 2 H ---- .
In this example x () = 45, and tan x () = 1 , sec 2 XG = 2.
Also, fe = 1 expressed in radians = 0.01745. Hence, by (5),
tan 46 - tan 45 = 2(0.01745) = 0.0349 ;
by (6), tan 46 - tan 45 = 0.0349 + 2(0.()1745) 2 = 0.0349 4- 0.0006 = 0.0355.
From the second approximation we get tan 46 = 1.0355, which is correct to four
places of decimals.
374 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
1. Verify the approximate formula
In (10 + *) = 2.303 +
Calculate the value of the function from this formula and compare
your result with the tables, when (a) x = 0.5 ; (b) x = 1.
Am. (a) Formula, 2.253 ; tables, 2.251.
(b) Formula, 2.203 ; tables, 2.197.
2. Verify the approximate formula
sin + x =0.5 + 0.8660 x.
\b /
Use the formula to calculate sin 27, sin 33, sin 40, and compare your
results with the tables.
3. Verify the approximate formula
tan K- + x ) = 1 + 2 x + 2 x 2 .
Use the formula to calculate tan 46, tan 50, and compare your
results with the tables.
4. Verify the approximate formula
cos x cos a (x a) sin a.
Given cos 30 = sin 60 = 0.8660,
cos 45 = sin 45 = 0.7071,
and cos 60 = sin 30 = 0.5,
use the formula to calculate cos 32, cos 47, cos 62, and compare your
results with the tables.
ADDITIONAL PROBLEMS
r\
1. Given the definite integral / x 5 In (1 + x)dx.
*) o
(a) Obtain its value by series correct to four decimal places. Ans. 0.0009.
(b) Obtain its value by direct computation and compare with the ap-
proximate value derived in (a).
(c) Prove that if n terms of the series are used in the computation the
1
error is less than
x
2. Given f(jr) c - cos -
7)
(a) Show that f ( *\x) = - } f(x).
(b) Expand f(x) in a Maclaurin series to six terms.
(c) What is the coefficient of x 12 in this series? Ans.
64 [12*
CHAPTER XXI
ORDINARY DIFFERENTIAL EQUATIONS *
201. Differential equations order and degree. A differential equa-
tion is an equation involving derivatives or differentials. Differential
equations have been frequently employed in this book. The illus-
trative examples of Art. 139 afford simple examples. Thus, from the
differential equation (Illustrative Example 1)
g-2i.
we found, by integrating,
(2) y = x~ + C.
Again (Illustrative Example 2 ) , integration of the differential equation
(3) - ? ~
v } dx y
led to the solution
(4) x 2 + y~ - 2 C.
Equations (1 ) and (3) are examples of ordinary differential equations
of the first order, and (2) and (4) are, respectively, the complete
solutions.
Another example is
(
This is a differential equation of the second order, so named from the
order of the derivative.
The order of a differential equation is the same as that of the deriva-
tive of highest order appearing in it.
The derivative of highest order appearing in a differential equa-
tion may be affected with exponents. The largest exponent gives the
degree of the differential equation.
Thus, the differential equation
(6) y" 2 = (1 + ?/' 2 ) 3 ,
* A few types only of differential equations are treated in this chapter, namely, such
types as the student is likely to encounter in elementary work in mechanics and physics.
375
376 DIFFERENTIAL AND INTEGRAL CALCULUS
where y 1 and y" are, respectively, the first and second derivatives of
y with respect to x, is of the second degree and second order.
202. Solutions of differential equations. Constants of integration. A
solution or integral of a differential equation is a relation between the
variables involved by which the equation is satisfied. Thus
( 1 ) y = a sin x
is a solution of the differential equation
For, differentiating (1;,
(3) -7^ = a sin x.
Now, if we substitute from (1) and (3) in (2), we get
a sin x + a sin x = 0,
and (2) is satisfied. Here a is an arbitrary constant. In the same
manner
(4 ) y = b cos x
may be shown to be a solution of (2) for any value of b. The relation
(5) // fj sin r+c>2 cos x
is a still more general solution of (2). In fact, by giving particular
values to c\ and 2 it is seen that the solution (5) includes the solu-
tions (1) and (4).
The arbitrary constants c\ and c> 2 appearing in (5) are called con-
stants of integration. A solution such as (5), which contains a num-
ber of arbitrary essential constants equal to the order of the equation
(in this case two), is called the complete or general solution.* Solutions
obtained therefrom by giving particular values to the constants are
called particular solutions. In practice, a particular solution is ob-
tained from the complete solution by given conditions to be satisfied
by the particular solution.
ILLUSTRATIVE EXAMPLE. The complete solution of the differential equation
is y = C] cos x + C'2 sin x (see (5) above).
Find a particular solution such that
(2) y 2, y' = 1, when x = 0.
* It is shown in works on differential equations that the general solution has n arbi-
trary constants when the differential equation is of the nth order.
ORDINARY DIFFERENTIAL EQUATIONS 377
Solution. From the complete solution
(3) y = c i cos x + r 2 sin x,
by differentiation, we obtain
(4) y' = c i sin x + c 2 cos x.
Substituting in (3) and (4) from (12), we find c\ = 2, r 2 1. Putting these
values in (3) gives the particular solution required, y = 2 cos r sin j. v4ns.
The solution of a differential equation is considered as having been
effected when it has been reduced to an expression involving integrals,
whether the actual integration can be effected or not.
203. Verification of the solutions of differential equations. Before
taking up the problem of solving differential equations, we show how
to verify a given solution.
ILLUSTRATIVE EXAMPLE 1. Show that
(1 ) y = c\x cos In x -f c?x sin In x -f x In x
is a solution of the differential equation
(2) x^-4^2 W = x,nx.
Solution. Differentiating (1), we get
(3) ^ = (r 2 - d) sin In x -f (r^ -f o ) cos In .r -f In JT -f 1,
dx
Substituting from (1), (3), (4), in (2), we find that the equation is identically
satisfied.
ILLUSTRATIVE EXAMPLE 2. Show that
(5) y 2 - 4 x =
is a particular solution of the differential equation
(6) *2/' 2 -1 = 0.
Solution. Differentiating (5), the result is
2
yy' 2 = 0, whence y' = -
Substituting this value of y' in (6) and reducing, we obtain 4 x - y 2 = 0, which
is true by (5).
378 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
Verify the following solutions of the corresponding differential equa-
tions.
Itifferentitil equations Solutions
(r jc
r <7r
' 4- r,
6 . f >l:Jl + ., 'Ht _ j.,, = o. ry = 2 f - 3 r '.
f/.T J (/J*
7. -- -f 4 .S r_r (). rr r, COS (2 t -f <* 3 ).
8 - (/ /X - r > '17 + 1:i J " = ;5;) - *" = r " cos - * + 3 -
/* a/
9. // y^ (
10. .r?/ ^ 4-
11. ^-^ '
<iv 1 -f r- 1 - rr
12. ''-- -f 4 ,s = 8 t. * = 2 sin 2 t + cos 2 / + 2 /!.
<//-
13 . <!lll -z^- * ll = r 2 '. // = r,r*' -f ce * - J f 2 '.
</.r* a.r
14. f ll _|_ <) j. 5 cos o /. j- = cos 2 / + 2 cos 3 / + 3 sin 3 /.
(// 2
15> 1^ 4. 9 j. = 3 C os 3 /. x = c, cos 3 / -f c 2 sin 3 / 4- \ t sin 3 /.
a/*
16. ^ -f .r// = J 3 // 3 - A - J-' 2 4- 1 -f- re"',
(/.r * /r
204. Differential equations of the first order and of the first degree.
Such an equation may be brought into the form
(A) Mdx
ORDINARY DIFFERENTIAL EQUATIONS 379
in which M and .V are functions of x and y. The more common dif-
ferential equations coming under this head may be divided into
four types.
Type I. Variables separable. When the terms of a differential
equation can be so arranged that it takes on the form
(1) /(J)rf.r+Fu/)rf// = 0,
where /(j) is a function of JT alone and F(y) is a function of \i alone,
the process is called separation of the variables, and the solution is
obtained by direct integration. Thus, integrating (1 ), we get the gen-
eral solution
where c is an arbitrary constant.
Equations which are not given in the simple form (1) may often
be brought into that form by means of the following rule for separat-
ing the variables.
FIRST STEP. Clear of fractions; and if the equation involves deriva-
tives, multiply through by the differential of the independent variable.
SECOND STEP. Collect all the terms containing the same differential
into a single term. If the equation, then takes on, the form
XYdx + -Y'y'(/i/ = 0,
where X, X' are functions ofx alone, and >', Y' are functions of y alone,
it may be brought to the form (1) by dividing through by X'Y.
THIRD STEP. Integrate each part separately, as in (2).
ILLUSTRATIVE EXAMPLE 1. Solve the equation
dy __ 1 4- j/' 2
dx (1+ x*)jry
Solution. First Step. (1 4- x' 2 )xydy =(14- y'*)dx.
Second Step. (1 -\- y' 2 )dx x(l 4- x' 2 )ydy ~ 0.
To separate the variables we now divide by o?(l 4- x 2 j(l 4- y 2 ), giving
dx ___ ydy _
x(\ 4-x*) 1 +y*
In x - J In (1 4- x 2 ) - \ In (1 4- y 2 ) = C,
In (1 4- x 2 )(l 4- y*) = 2 In x - 2 C.
380 DIFFERENTIAL ANT) INTEGRAL CALCULUS
This result may be written in more compact form if we replace 2 C by In f,
that is, give a new form to the arbitrary constant. Our solution then becomes
In (1 + T*)(\ + y' 2 ) = In j 2 4- In c,
In (1 -f r')fl -f y-) - In ex' 2 ,
(1 4 Jr-)(l -f r') - rj 2 . .4ns.
ILLI STKATIVE EXAMPLE 2. Solve the equation
Solution. Firat Step. as dy -f 2 a // djc sydy.
Second Step. 2 ay dx + jcla - y>dy = 0.
To separate the variables we divide by ry.
x y ~
Third Step. 2 f -\-aC^~ Cdy = C,
a In j -f a In // ?/ C,
r/ In j'^/ =.- r -f ?/,
By passing from logarithms to exponentials this result may be written in the
form
or
c
Denoting the constant f hy r, we get our solution in the form
y
x' 2 y = ce. Ans.
Type II. Homogeneous equations. The differential equation
(A) M dx + N dy =
is said to be homogeneous when 717 and .V are homogeneous functions
of * and // of the same degree.* Such differential equations may be
solved by making the substitution
(3) y = rx.
This will give a differential equation in r and x in which the vari-
ables are separable, and hence we may follow the rule under Type I.
* A function of x and y is said to bo homogeneous in the variables if the result of replac-
ing .r and // by \jc and Xj/ (X being arbitrary) reduces to the original function multiplied by
some power of X. This power of X is called the degree of the original function.
ORDINARY DIFFERENTIAL EQUATIONS 381
In fact, from (A) we obtain
--
Also, from (3),
dy dv .
< 5) l= r dr*+ v -
The right-hand member of (4) will become a function of v only when
the substitution (3) is used. Hence, by using (5) and (3), we shall
obtain from (4)
(6) *g + t>=/(p).
and the variables x and v may be separated.
ILLUSTRATIVE EXAMPLE. Solve the equation
'+x'^ = ^^.
dx dx
Solution. y 2 dx -f (x- - xy }dy = 0.
Here M y 2 , N = x' 2 xy, and both are homogeneous and of the second degree
in or and y. Also we have
ay _ y*
dx xy x 2
Substitute y = t?x. The result is
dx 1 v
or r dx -f x( 1 v)dr = 0.
To separate the variables, divide by rx. This givea
,
In x -f- In f - v T,
In rx = C -f ,
But v % Hence the complete solution is
X y
y = ce x .
PROBLEMS
Find the complete solution of each of the following differential equa-
tions.
1. (2 + y)dx - (3 - x)dy = 0. Ans. (2 -f y)C3 - x) = c.
2. xy dx - (1 -h ^ 2 )d?/ = 0. c?/ = 1 + x 2 .
3. x(x -f 3)dy - i/(2 x + 3)dx = 0. t/ = cx(x -f- 3).
382
DIFFERENTIAL AND INTEGRAL CALCULUS
4. Vl 4- f' 2 dy Vl y' 2 dx 0. Am. arc sin y = In c(x 4- Vl 4- x 2 ).
5. dp 4- p tan dO = 0. p = r cos 0.
6. (1 j-jd?/ y 2 ds 0. j/ In rfl .rj = 1.
7. (/- -f 2 y)(Lc + (2 jr - 3 ?/)</?/ = 0. x 2 4- 4 .n/ - 3 ?/ 2 - r.
8. (3 jr 4- 5 y)djr 4- (4 y 4- 6 ?/;r/?/ = 0. (jr 4- ?/j 2 (.r 4- 2 ?/) = r.
9. 2O 4- y)ds 4- ?/ r/?/ 0.
Aim. \ In (2 r 2 -f 2 r?/ 4- //*; - arc tan ( J ?/ ) = r.
10. (8 // 4-10 jr )ds 4- (5 // 4- 7 .r )r/// = 0. f.r 4- ?/ ) J (2 r 4- ?/)'* = r.
11. (2 *r 4- y)djr 4- <V -f 3 y)(h{ = 0. 2 ./* L> 4- 2 ./*// 4- 3 //-' = r.
12. Vl - 4 /-' dx 4- 2 Vl - ,s~ f// = 0. .,- Vr^lT' 4- 2 / Vl - .s- - c.
13. 2cf3 c-f I)////' 4- Cl - 2 /r)r/-: - 0. (2 // - 1 )(1 4- 3 c) 3 r^.
14. 2 y r/c - 2 : ds ~ VT^TT^ <r. 1 4- 4 r: - r->- = 0.
15. 0' -f 4 //K/.r 4- 2 jr (/// = 0. j- { 4- (> jr~y = r.
16. (2 j' L> -f //-' )(/.r 4- (2 sy 4- 3 /r )r/// ~ 0. 2 .r { + 3 .r//- 4- 3 //'< = r.
</w 1 4:^ L> r-hr
^-i/r'i r^' ^-rr^'
18. (3 f 2 //)./ J.r 4- (./' - 2)(/// = 0. 22. (3 s -f //)</./ 4- (.r 4- //)<7// ~ 0.
19. 2(1 4- // )(/j- ( 1 ~ ./)</// = 0. 23. jryi // 4- 2 )r/.r - ( // 4- 1 )r/// = 0.
20. ( 1 -f //).r r/.r - ( 1 -f .r )// (/// = 0. 24. f 1 4- /-V// - (1 - // L ')r/.r - 0.
21. (a.r 4- />)^/// //- djr ~ 0. 25. O 2 //>/.r (2 ^* 4-- //)(/// = 0.
26. (3 s 4- 2 //)</j- -f j-J// rr o.
27. 3(5 .r -f 3 i/)(ls 4- ( 1 1 ./* -f 5 /y )(/// = 0.
28. (.r- 4- // :> U/.r -f (2 .r// 4- // L ')</// = 0.
29. 2 // </.r ~ (2 .r ?/)r/// 0.
In each of the following problems find the particular solution which is
determined by the given values of .r and //.
//* jr* .r.
1 + 40-4.^=: o.
31. (.r- 4- // L 'W/.r 2 .r// dy ; .r = 1, // = 0.
32. .r dy - // J.r ~ Vr" 4- //- (Lr ; j- ~ ?,, // 0.
33. (1 + //-')r/// = // ( /.r ; .r = 2, // = 2.
34. Find the equation of the curve whose slope at any point is equal to
-(l4- -) and which passes through the point (2, 1). Anz. x 2 4- 2 nj 8.
35. Find the equation of the curve whose slope at any point is equal to
j, j
; and which passes through the point (1, 0). Am. y(\ 4- jr) = 1 .r.
ORDINARY DIFFERENTIAL EQUATIONS 383
Type III. Linear equations. The linear differential equation of the
first order in y is of the form
where P and Q are functions of .r alone, or constants.
Similarly, the equation
(C) ^ + Hx - J,
<*y
where // and .7 are functions of // or constants, is a linear differential
equation.
To integrate (#), let
(7) y = uz,
where z and u are functions of x to he determined. Differentiating (7),
( ,x, &rf= + ..!j.
or a.r </.r
Substituting from (S) and (7) in (#), the result Is
We now determine u by integrating
do, iJ7 + / '" = '
in which the variables jr and ?/ are separable. Using the value of u
thus obtained, we find z by solving
(11) u^ = Q,
df
in which j and z can be separated. Obviously, the values of u and z
thus found will satisfy (9), and the solution of (#) is then given by (1).
The following examples show the details.
ILLUSTRATIVE EXAMPLE 1. Solve the equation
Solution. This is evidently in the linear form (B), where
2 *
p = _ an( } Q=(x + l)*.
x -f 1
Let y = uz; then . . .
* a?y a2 aw
dx dx dx
384 DIFFERENTIAL AND INTEGRAL CALCULUS
Substituting in the given equation (12), we get
dz , du 2 uz , , s
' i ' \ i -t
dx \ dx 1
To determine M we place the coefficient of a equal to zero. This gives
du 2 M
dx, 1 -f x
du 2 dx
u 1 -far
Integrating, we get In if = 2 In (1 -f x) = In (1 -f x) 2 .
(M) .-.u = (1 -f j) a .*
Equation (KJ) now becomes, since the term in z drops out,
Replacing M by its value from (14),
=<r + 1,
dx
Integrating, we get dz = u -f 1 i
(15) s '
,
Sut>stituting from (15) and (14 ) in y ~ uz, we get the complete solution
y = **i + C (r + l,*. An*.
ILLUSTRATIVE EXAMPLK 2. Derive a formula for the complete solution of (5).
Solution. Solving (10), we have
In u -f f /' dr = In k,
v\ here In ^r is the constant of integration ;
whence u ke ' Pd *.
Substituting this value of // in (11 ), and separating the variables z and x, the
result is
fc = 2^ p ' e 'dr.
Integrating, and substituting in (7), we obtain
It should be observed that the constant k cancels out of the final result. For
this reason it is customary to omit the constant of integration in solving (10).
* For the sake of simplicity wt hu\ assumed the particular value zero for the constant
of integration. Otherwise we should have M = r(l -f j-)-. But in the work that follows
r finally cancels out (See Illustrative Example 2 >
ORDINARY DIFFERENTIAL EQUATIONS 385
Type IV. Equations reducible to the linear form. Some equations
that are not linear can be reduced to the linear form by means of a
suitable transformation. One type of such equations is
dx' *
where P and Q are functions of x alone or constants. Equation (D)
may be reduced to the linear form (B), Type 111, by means of the
substitution z = y~~ w + 1 . Such a reduction, however, is not necessary
if we employ the same method for finding the solution as that given
under Type III. Let us illustrate this by means of an example.
ILLUSTRATIVE EXAMPLE. Solve the equation
(16) ^ 4 & = fl In JT- y' 2 .
dx x
Solution. This is evidently in the form (/>), where
P^i, Q = a In x, n = 2.
j
Let y - uz-, then ^ - u ^ 4- z ^-
dx dx dx
Substituting in (16), we get
dz , du , uz , , ,,
M T" + - -7- -f ~- ~ In x u~z 2 ,
dx dx x
(17)
J" \r X
To determine u we place the coefficient of z equal to zero. This gives
,
dx x
du _ dx
u x
Integrating, we get In u In x In -,
x
(18) u = i-
Since the term in z drops out, equation (17) now becomes
u^- =a In x u'*z*
dx
= a \n x uz' z .
dx
Replacing u by its value from (18),
= a\ nx .*-,
dx x
386 DIFFERENTIAL AND INTEGRAL CALCULUS
1 a(m x) 2 . r
Integrating, we get
2
~ adnx^-f 2C
Substituting from (19) and (18) in y = uz, we get the complete solution
_ 1 2
* x f/(lnj)~' + 2r
or xy[a(\n t}- -f 2 f'J -f 2 = 0. An*.
PROBLEMS
Find the complete solution of each of the following differential equa-
tions.
I j. ( lli __ 2 // 2 r. An*. // -- cs' 2 - 2 j".
2. JT ^ - 2 /y ~ - J-. ?/ = .' + ^' 1> -
3. d i( __ 2 // - I - 2 r. ?/ = JT -f rr -".
4> j. ( J/ _ ;j y - - 2 nj. y = us -f r.r { .
5. ( !ii - ;/ rrr - 2 C r . ?/ = C r + C(" .
6. ~ - ctn / = 1 - (/ -f 2) ctn t. s = / + 2 + r sin /.
cx 2 u 4- 2 j-?/ -1=0.
djr ' jr '
8. 4 tan / = 2 / 4 / 2 tan /, s rr / L) 4 c cos /.
u/
( ili _ / _ 1 \ >r u c * -j- ^ r
(f.r j*
12. nx -f 12 // = w n + l . rj* 2 ?/" 4- A'/ n -
djr
13. -^ + i? = cos / - sin /. 8 cos t -f cc~
14. ^ - x ctn / = f r (l ~ ctn /). 8 = ^' ( + c sin ?
ORDINARY DIFFERENTIAL EQUATIONS 387
15. x ^3L _ 2 y 4 3 x = 0. 20. ~* - * ctn / 4 csc / = 0.
16. ^ 4 ?/ = 2 -f 2 .r. 21. 2 -^ -f // = (JT - l)r'.
17. .r -~ 4 V = (1 4 -rU". 22. r ~ /y = ^ cos .r sin x.
ajc * ax
18. ( -^ - y = 1 - L> j-. 23. N ^ - // -f (.r 2 -f 2 .r)?/" + 1 =
19. j- ^ -f ?/ -f j- 2 //- = 0. 24. ^ 4 * tan t = c ' (tan / - 1 ).
In each of the following problems find the particular solution which is
determined by the given values of x and //.
25. 7^ - ^ = /-V ; x = 1 . ?/ = 0. .4 //s. // =. jr' 2 (c' - e).
27. ~ -f ?/ tan jr sec ^ ; JT ~ 0, // 1 . ?/ sin .r cos jr.
djr
28. ^ - -^- = (.r + 1 )' ; jc = 0, y/ = 1. 2 ?/ = (r -fl ) 4 -f (jr -f I) 2 ,
dj* j" -f 1
29. Find the equation of the curve whose slope at any point is equal to
, // 4- .r -f- an( j ^.^^.^ p asses through the point (1, 0).
s Ann. 2 ?/ = tt.r 2 - 2-r- 1.
30. Find the equation of the curve whose slope at any point is equal to
y n f y ^^ which passes through the point (1 , 1 ). Anx. y(\ -f In x) = 1 .
205. Two special types of differential equations of higher order. The
differential equations discussed in this article occur frequently.
g-'-
where X is a function of x alone or a constant.
To integrate, first multiply both members by dx. Then, inte-
grating, we have
Repeat the process (r? 1) times. Then the complete solution con-
taining n arbitrary constants will be obtained.
388 DIFFERENTIAL AND INTEGRAL CALCULUS
d*i/
ILLUSTRATIVE EXAMPLE. Solve -rr = xe*.
dx A
Solution. Multiplying both members by dx and integrating,
=* + <:,.
g = ^-e* + c,
Repeating the process,
^ =Jxe* dx - JV dx + Jd dx + C 2 ,
or ^ = xr* - 2 <-* + T,x + 0V
j/ = Cxe* dx - C r* dx -f f r,x dx -f f C 2 dx + C 3
= xe 1 - c* -f ^TT- + r ^ x + c --
Hence y = xe z - 3 r 3 " -f r,x 2 4- r 2 x -f f:t. ^4ns.
A second type of much importance is
'
where }' is a function of y alone.
To integrate, proceed thus. Write the equation
d//'= Ydx,
and multiply both members by y'. The result is
y' dy' '= r//'^.
But ^/' dx = d//, and the preceding equation becomes
y'dy' = Ydy.
The variables |/ and //' are now separated. Integrating, the result is
The right-hand member is a function of y. Extract the square root,
separate the variables, jr and //, and integrate again. The following
example illustrates the method.
ILLUSTRATIVE EXAMPLE. Solve c ~~ f a*y 0.
Solution. Here C ~^- = ^ = a' 2 y, and hence the equation belongs to type (F).
ax ox*
Multiplying both members by y'dx and proceeding as above, we get
y ' dy' = - a' 2 ydy.
Integrating, \ y" 2 ~ - \ a~y' 2 -f C.
y' = \ 2 C - a'^ 2 .
ORDINARY DIFFERENTIAL EQUATIONS
389
setting 2 C C\ and taking the positive sign of the radical. Separating the vari-
ables, we get
d 'j = Ar _
Integrating, - arc sin == = x -f i
\ r.
arc sin -~^^= = or -f uC 2 .
This is the same as
Hence
-~= sin(j -f u("-)
^ C,
= sin ax cos a( \, -f ros ax sin d(\., (4), Art. 2
# = - - oos </( \, sin a.r 4 - ! sin ( '- cos ax.
a a
y c\ sin ax -f <'-j c-os ax. A us.
PROBLEMS
Find the complete solution of each of the following differential
equations.
3. - 4 sin 2 t.
x = - sin 2 / + nf + r 2 .
6.
Vas
as
,72 Z-
9- TT + =
j = ar\ V? x -f r 2 .
ro
. Find /, having given that s = a, ^ = 0, when / rr Q.
at
Ans. t
390 DIFFERENTIAL AND INTEGRAL CALCULUS
206. Linear differential equations of the second order with constant
coefficients. Equations of the form
where p and q are constants, are important in applied mathematics.
To obtain a particular solution of (G), let us try to determine the
value of the constant r so that (G) will be satisfied by
(1 ) y = e rr .
Differentiating (1 ), we obtain
Substituting from (1) and (2) in (G) and dividing out the factor
c rr , the result is
CJ) r 2 + pr + q=-(),
a quadratic equation whose roots are the values of r required.
Equation (3) is called the auxiliary equation for (G). If (3) has dis-
tinct roots r\ and r^ y then
(4) y = c r ^ and y = e 7 **
are distinct particular solutions of (G), and the complete solution is
(5) '// = c\e r * r + fiK >r -" r .
In fact, (5) contains two essential arbitrary constants, and (G) is
satisfied by this relation.
ILLUSTRATIVE EXAMPLE 1. Solve
(6 ) ^ - >> ( lM _ a y = o.
dx~ dx
Solution. The auxiliary equation is
(7) r a -2r-3 = 0.
Solving (7), the roots are 8 and 1, and by (5) the complete solution is
y nr tx -f C' 2 e *. Ans.
Check. Substituting this value of y in (6), the equation is satisfied.
Roots of (8) imaginary. If the roots of the auxiliary equation (3)
are imaginary, the exponents in (5) will also be imaginary. A real
complete solution may be found, however, by choosing imaginary
values of c\ and r, in (5). In fact, let
(8) r, = a + 6\/^l, r 2 = a - bV^T
be the pair of conjugate imaginary roots of (3). Then
ORDINARY DIFFERENTIAL EQUATIONS 391
Substituting these values in (f>\ we obtain
(10) y = f"(rif'" x ~ + c?c ''- rx ~~0.
In the algebra of imaginary numbers it is shown that*
^-i C os bx + V 1 sin bx, c ' trx ' = cos hx V^~T sin bx.
When these values are substituted in (10), the complete solution
may be written
(11) y = c" J (A cos bx + # sin M,
if the new arbitrary constants .4 and B are determined from c\ and r-
by .4 = ci + CL>, B (o r L >) V 1. That i-vwe now take for c, nm^r-
in ( 5 ) the imaginary values r i = A ( .4 H V 1 ^ , r- ,\ ( .4 -f 7> \/ 1 ) .
By giving to A and B in (11) the values 1 and 0, and and 1, in
turn, we see that
( 12 ) // = (> UJ cos bx and // c af sin bx
are real particular solutions of (G).
ILLUSTRATIVE EXAMPLE 2. Solve
(18) g 4 A--'// = 0.
Solution. The auxiliary equation (3) is now
r' 4- /r-' = 0. /. r- i AV^l.
Comparing with (8), we see that a 0, b k. Hence, by (11), the complete
solution is y = ^4 C()S kjr + />> s j n A:J ^
Check. When this value of y is substituted in (13), the equation is satisfied.
Compare this method with that used for the same example in Art. 205 (k a).
REMARK. A different form for the solution is obtained by setting A --- (' cos <*,
B = r sin o- in the above value of y. Then y - <' cos (A\r - ). (By (4), p. 3j.
* Let ? = V^~T, and assume that the scri(\s for c r in f'roblf-m 1, Art 191, rfprf.sonts
the function when J is replaced by ibs. Then, since ?- 1, /'* - /,, * 1 ~ I, etc.,
we have
. t , . ?/<>- .^J 1 , fcV , , IW
(ln flfc x =r ,4 lfcjr ._^ l _4 +l -^ -.--.
Also, replacing j by bj in (7> and ''S), Art 104,
b^j-l- 5^4 . 63j.J fofijfi
cos 6x = 1 - ^ + ^ ---- . Bin ft* = fer - + ---- .
Then, by Art. 195,
. . . , .. b'W bW . b*r* . b ; 's f >
(15) cos bx -f i sin bx - 1 -f- tkr - -- i -f 4- i - ,
assuming that the series represents the function. The right-hand members of (14) and (15)
are identical. Therefore c t>>x - cos bx + i sin bx
392 DIFFERENTIAL AND INTEGRAL CALCULUS
Roots of (3) real and equal. The roots of the auxiliary equation (3)
will be equal if p 2 = 4 q. Then (3) may be written, by substituting
q = \ P-,
04; r* + pr + lp 2 =(r+> 2 p)* = 0,
and r\ ~ r> 2 = 3 p. In this case
(15) y~e r>T and y - xe rtX
are distinct particular solutions, and
(16) // = ?'"(ri -f r,j)
is the complete solution.
To corroborate this statement, it is only necessary to prove that
the second equation in (15) gives a solution. But we have, by dif-
ferentiating,
(17) y - xe r >', ^ = <"'(! -f r,.r), ^ - <T"(2 r, -f rr'j).
Substituting from (17) into the left-hand member of (G), the
result is, after canceling e rijr ,
(18) (rr + />r, -j- q}jr + 2 r } + p.
This vanishes, since r\ satisfies (3) and equals \ p.
ILLUSTRATIVE EXAMPLE ;*. Solve
+ ^., = o.
Find the particular solution such that
c/s
s = 4 and ~ = 2 when / = 0.
cf/
Solution. The auxiliary ecjuation is
r-' -f L r -f 1 = 0, or (r -f 1 )* = 0.
Hence the roots are equal, r, = 1, and, hy (16),
(20) s = r-'(r, 4- r^/).
This is the complete solution.
To find the required particular solution, we must find values for the constants
d and C'2 such that the given conditions,
= 4 and -- = - 2 when t =
are satisfied. dt
Substituting in the complete solution (20) the given values s = 4, t = 0, we
have 4 = c\ t and hence
(21) s = e-'(
ORDINARY DIFFERENTIAL EQUATIONS 393
Now differentiate (21) with respect to t. We get
s'
By the given conditions, = 2 when / 0.
a/
Substituting, the result is 2 c -2 4, and hence c% = 2. Then the particular
solution required is 8 = e ~'(4 + 2 f). Arts.
PROBLEMS
Find the complete solution of each of the following differential equa-
tions.
1. ( ^~ -~-2x = Q. Atis. j- = r,r r + r 2 r a '.
2. 7^ - 4 ^ + 3 ?/ = 0. v = <v" -f r,r".
dr 2 dr
rj*
4. -7^7 -f 16 -r = 0. j* c\ cos 4 / -f ('-2 sin 4 i.
d/ J
6. ~ - 4 = 0. = or 2 ' -f r.r -'.
(I/~
_ f/ 2 ?/ , . dij f . . Ar
6. -7-^' -f 4 -f - 0. ?/ = r, 4 r 2 r 4x .
(/u- 2 djr
7. ^ -f 2 ~ + 2 .s- = 0. * =-- r f (n cos / -f r, sin /).
dt 2 dt
Q d~s r/K , r _ A ^' ~ <"'(o cos li / -f c* sin 2 0-
8 - ~ ~ + ' * ~ -
In the following problems find the particular solution which satisfies
the given conditions.
17. ^!f + 3 ^f 4. 2 = ; s = 0, ~ = 1, when f ^ 0. yin,s. = c
dt 2 dt dt
18. ^ -h n 2 x - ; x = a, ^ = 0, when / = 0. x - a cos
a< 2 ar
394 DIFFERENTIAL AND INTEGRAL CALCULUS
19. ~ - w V = () ; r = 2, ~ = 0, when / = 0. An*. x = c "' + c '.
ar" /
20. ~f 4 2 ^ - 8 // = ; y = 0, ^ = 24, when / = 0.
Arts. y = 4(> 2 ' r ~ 4 ').
21. ~ - 8 ~ 4 16 K = ; = 0, ^ = 1, when / = 0. 8 = tc 4t .
at- dt dt
22. -^~a~^(>; r = 0,^ = a,when/ = 0. a- = f-'-l.
<//- dt dt
23. ~) -f 8 ~ 4- 25 = ; .s- = 4, ^ = - 16, when / = 0.
Awx. ,s = 4 f 4f cos 3 /.
24. ^ - 6 ( jf ( -f 10 r = ; r = 1, ^ - 4, when / = 0.
Aw.s. J7 = e 3/ (cos/ -I- sin /).
25. 4 4 .s- = ; = 0, ~ - 4, when t = 0.
26. - 4 j- - ; jr - 10, - 0, when t = 0.
no " ^ - r . ^ /v . ^J" i /x
28. - --; - 4 4- 4 r = ; .r 2, ;>, when / = 0.
dt- <11 dt
To solve the differential equation
where /> and r/ are constants and X is a function of the independent
variable .r or a constant., three steps are necessary.
FIRST STEP. Sol re the equation (G). Let the complete solution be
(22) y = u.
Then u /,s called the complementary function for (H).
SKCOND STEP. Obtain a particular solution
(23) y = v
of (H) &// trial
THIRD STEP. 77/c complete solution of (H) is now
(24) = w + r .
ORDINARY DIFFERENTIAL EQUATIONS 395
In fact, when the value of // from (24) is substituted in (#), it is
seen that the equation is satisfied, and t,24) contains two essential
arbitrary constants.
To determine the particular solution (23), the following directions
are useful (see also Art. 208). In the formulas all letters except x,
the independent variable, are constants.
General case. If y = X is not a particular solution of (G), if
Form of X Form of v
X a + bx, assume // = r A f Hx ;
A" = ae t>jr , assume // = r ~ Ac'" ;
A' = a i cos b.r + a 2 sin bx, assume // = r = A \ cos bx -f A : sin />.r.
Special case. If ?/ = X is a particular solution of (G), assume for /?
the above form multiplied by ;r (the independent variable).
The method consists in substituting // r, as given above, in (//),
and determining the constants A, R, A\, A-^ so that (//) is satisfied.
iLLt'STRATIVK EXAMPLK 4. Solve
(25) rffB_ 2 rfK_;, v = 2x.
djr 2 djc
Solution. First Step. The complementary function u is found from the com-
plete solution of
S- 2 S---
By Illustrative Example 1, above, therefore
(27) y ~ u = rir' 7 ' f c->c x .
Second S/r;>. Since // = X 2 x is not a particular solution of (2(5), assume for
a particular solution of (25)
(28) y = r = A 4 Bx.
Substituting this value in (25) and collecting terms, the result is
(29 1 - 2 B - 3 A - :J Ilx = 2 x.
Equating coefficients of like powers of x, we get
-2B -XA=(), - U /* = 2.
Solving, A = |, B = - |, and substituting in (28), we obtain the particular
solution
(30) y = v = $-Hx.
Third Step. Then, from (27) and (30), the complete solution is
y = u -f v = c^ 1 -f c 2 e * -f j - !5
,396 DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 5. Solve
dx* " dx
Solution. First Step. The complementary function is (27), or
(32 j y - u = c l e'^ -f c 2 ('' x .
Second Step. Here y X = 2 e * is a particular solution of (26), for it is
obtained from the complete solution (32) by letting fi = 0, c 2 = 2. Hence assume
for a particular solution v of (31 1,
(33; y = v = Axe x .
Differentiating (33), we obtain
(34; 4r = Ar ' (1 ~ TI - S"' 4 '' ' (X ~~>-
Substituting from (33) and (34) in (31), we obtain
(35; Ac r (x 2; 2 Ae r (\ x) 3 Axe x = 2 c ~ T .
Simplifying, we get 4 Ac * 2 e *, and hence A ~-\. Substituting in (33)
we obtain
(30; y = r = - I xe *.
Third Step. The complete solution of (31 ) is, therefore,
y u -f r c\r u -f c 2 e J \ xe f . Ana.
ILLUSTRATIVE EXAMPLE (5. Determine the particular solution of
(37) ~ + 4 * = 2 cos 2 /,
such that s - and ~ = 2 when t = 0.
dt
Solution. Find the complete solution first.
First Step. Solving
(38) ~ -f- 4 .s = 0,
we find the complementary function
(39) u = r, cos 2 / -f c 2 sin 2 /.
Second Step. Considering the right-hand member in (37), we observe that
s = 2 "os 2 / is a particular solution of (38) resulting from (39) when = 2, c-2 0.
Hence for a particular solution s = v of (37) assume
(40) s - r = t(Ai cos 2 / -f A 2 sin 2 /).
Differentiating (40") , we obtain
f^' = 4! cos 2 / -f A - sin 2 / - 2 Mi sin 2 / - -4 2 cos 2 /).
a/
(41) |
' ^ = - 4 ^1, sin 2 / -f 4 .4, cos 2 / - 4 t( A } cos 2 / + .4 2 sin 2 /).
ORDINARY DIFFERENTIAL EQUATIONS 397
Substituting from (40) and (41 > in (37). and simplifying, the result is
(42) - 4 .4i sin 2 t + 4 A; cos 2 / = 2 cos 2 f.
This equation becomes an identity when A\ = 0, A a = J. Substituting in (40),
we get
(43) g = r = ? / sin 2 t.
Third Step. By (39) and (43) the complete solution of (37) is
(44) = r, cos 2 / + c-> sin 2 / 4 .\ / sin 2 t.
We must now determine d and r 2 so that
(45 ) s = and ^ = 2 when / = 0.
dt
Differentiating (44),
(46) = 2 Ci sin 2 t -f 2 r- cos 2 / -f - 1 , sin 2 / -f / cos 2 t.
at
Substituting the given conditions (45) in (44) and (16), the results are
= r,, 2 = 2 fa. .'.<, = 0, r, - 1.
Putting these values back in (44), the particular solution required is
(47) s = sin 2 / -f \ /sin 2 /. /I;i8.
PROBLEMS
Find the complete solution of each of the following diflVrenthil equa-
tions.
l. + jr = at + b. Ans. s ~ c\ cos / + c> 2 sin / + at -I It.
dt 2
2. ^~ -f x = ae ft/ . .f - O cos / -f r, sin / -f ^-j
3. ^ + f - 4 cos /,. s - r, cos / -f r- sin / f 2 / sin /.
dt-
4. + X = 4sln2t. JT = CI cos / -f c> sin / - '\ sin 2 /.
d/ 2
5 <* _ 4 s = at + b. - r,^' -f r,r ^ - \(at -f 6j.
d/ 2
e^-4 s = *>^ = r,r-' -f r 2 r 2; - 1 r'.
d/ 2
7. ^! _ 4 s = e ' 2t . R = o^' -f r 2 f" 2 ' -f J ^' 2t -
' d^ 2
8 ^! 4 s = 2 cos 2 t. 8 = fie 2 ' -h fzf' 2 ' - i cos 2 t.
dt 2
9 ^ 2 y _}_ 9 y 5 x 2 . ?/ = d cos 3 x 4- c 2 sin 3 x 4- ^ x 2 - if.
* dx 2
398 DIFFERENTIAL AND INTEGRAL CALCULUS
10. - ~ - 2 s - 4 /. Anx. s - r,f ' + r*--" -f 1 - 2 *.
11. 2'- -'~i '=- -r = r,<'' + r./r' + 8.
12. ^" 7 - 1 '' -{ :< = fj r~". * = r,r' -f r^r" - 6 c 2 '.
///- <//
13. |-~ f 2 j* -f 2 - 8 H'. s = r 'fr, cos / -f c> sin /) -f | r 2 '.
14. y/( - 4 -^ -i ;{ .r ^ 4 r'. r = r,r' -f r a r*' - li /r'.
~
-f /! cos / j :{ sin /.
16. -, -- '2 ~j- -f 5 // -_- U sin 12 /. // r'(r, cos 1* / -f rj sin 12 /)
-f ] s cos "2 f -f /V sin 2 /.
17. '-' ; f 9 .<; - :i cos 12 /. 22. 4 ^ -f * - 5 cos f -
21. f - .r - ^ - 2. 26. | + f, | + K, * = 5 sin o ^
In the following problems find the i>articular solution which satisfies
the tfiven conditions.
-
28.^ + 9,-9r^ ^=1, ^ = ^,when/ = 0.
J 3 / + r 1 ').
29. ~~ + 9 x = 5 cos 2 / ; x = 1 , ~ = 3, when / = 0.
30. |~| + 9 * = X cos 3 / ; = 0, ~ = 6, when / = 0.
us. s = sin 3 t -f cos 2 f.
= 0.
An*, x = 2 sin 3 / -f it sin 3 t.
g-j.. b ,. t -
Ans. or = i(e 3 ' + e-'-6 J + l).
ORDINARY DIFFERENTIAL EQUATIONS 399
32. ( ^~ - 6 ^ + 13 JT - 39 ; jr = 4, ~ =- 3, when / = 0.
(it- dt dt , .., , ,
, , t .4 us. .r r'' cos 2 / -f 3.
33. ~ -f 9 * = 4 - 3 / ; * = 0, ^ = (), when / = 0.
34. ^~ - 9 * = 6 / ; 8 = 0, ~ = 0, when / = ().
36. -f a- = Li cos 2 / ; jr = 0, ^~ :--- -J, when / = 0.
37. ^y + 4 ,s- = 1> cos L> / ; .s- = 0, y = "J, when / r- 0.
38. ^ - 2 ~ -f L> ^ - L> sin / ; .r -- 0, ~- - 0, when / = ().
dt 2 (It dt
39. ^ + 5 <& + 4 ,/ = 12 ,- // = 1, ? - 0, when .r = 0.
a.r- c/.r ' (/./
40. <%i + 4 ^ + 4 ,/ = 4 ,- ; = 0, ty ^ 2. when , = 0.
dj'- as ' (/.r
207. Applications. Compound-interest law. A sim]>k k application of
differential equations is afforded by problems in which the rate of
change of the function with respect to the variable (Art. f>0) for any
value of the variable is proportional to the corresponding value of
the function. That is, if // =/(.r),
-**
where k is a constant. In this differential equation the variables are
separable as in Type I, p. 379. Sohing, we get
(2) y - c ^ J t
where c is an arbitrary constant. Thus the function y is an expo-
nential function (Art. 62). Conversely, given (2), it is easily shown
by differentiation that y satisfies (1 ). The connection of (1 ) with the
name "compound-interest law" is shown as follows:
Let y = a sum of money in dollars arcumulatiri^ at compound interest;
i interest in dollars on one dollar for a year;
A/ = an interval of time measured in years ;
A?/ = the interest of y dollars for the interval of time At.
Then Ay = iy At. Therefore
Equation (3) states that the average rate of change of // (Art. 50) for the period
of time At is proportional to y itself. In business, interest is added to the principal
at stated times only, - - yearly, quarterly, etc. In other words, y changes dis-
400 DIFFERENTIAL AND INTEGRAL CALCULUS
continuously with /. But in nature, changes proceed on the whole in a continuous
manner. So that to adapt equation (3) to natural phenomena we must imagine
the sum // to accumulate continuously ; that is, assume the interval of time A/
to be infinitesimal. Then equation (3j becomes
dy _
~dt " ly >
and the rate of change of y is proportional to y, agreeing with (1) if k = i.
In ( 1 ) the function y is said to change according to the compound-
interest law.
A second example is afforded by the complete solution of the
equation
(4) S =T/1 ' tf + f '
where k and c are constants not equal to zero. For let c = ak. Then
(4) may be written
( r >) ~ 0/ + a) - k(y + a).
This equation states that the function y + a changes according to
the compound-interest law. The differential equation (4), or (5),
is solved as in Type I, p. 379. The solution is
(G) // = cc kj a.
ILLUSTKATIYIO EXAMPLE 1. The function // of x changes according to the
compound-interest law. When x 1, y - 4 ; when x - 2, y = 12. Find the law.
Solution. By (1) we have
(7) g = *,.
Separating the variables and integrating, we get
In // -- k.r 4 (\
We have to find the values of k and ('. Substitute the given values of x and y.
Then In 4 --- k + r, In 12 - 2 k + C.
Solving, k - In 12 - In 4 = In 3 - 1 .0980, (* = In 4 - In 3 = In J.
Therefore In // = 1.0986 JT -f In J, and y = J r 1 098Bj . .4n,s.
ILLUSTRATIVE KXAMPLK 2. Washing down a solution. W T ater is run into a
tank containing a saline (or acid) solution with the purpose of reducing its strength.
The volume r of the mixture in the tank is kept constant. If s = quantity of salt
(or acid) in the tank at any time, and j amount of water which has run through,
show that the rate of decrease of s with respect to x varies as s, and, in fact, that
Solution. Since s = quantity of salt in the mixture of total volume f, the
quantity of salt in any other volume u of the mixture is - u.
ORDINARY DIFFERENTIAL EQUATIONS 401
Suppose a volume Ax of the mixture is dipped out of the tank. The amount
of salt thus dipped out will be ^ A.r, and hence the change in the amount of salt
in the tank is given by v
(8) As = - * A.r.
r
Suppose now that a volume of water A.r is added to fill the tank to its original
volume r. Then from (8) the ratio of the amount of salt removed to the volume
of water added is given by \ s .
_ = _-.
When AJ- we obtain the instantaneous rule of change of s with respect
to j, namely, rfs s .
-~ - - /b<s.
dx r
Hence s changes according to the compound-interest law.
PROBLEMS
1. The rate of change of a function // with respect to .r equals - { //, and
y 4 when JT 1. Find the law connecting j- and //. A us. \i = 5.58 ci r .
2. The rate of change of a function // with respect to .r equals "2 //,
and y 8 when jc = 0. Find the law*. An*. // --= f> c ' -f 2.
3. In Illustrative Example li, if r 10, 000 gal., how much water must
be run through to wash down 50 per cent of the salt ? AUK. (>i),'U gal.
4. Neirton'K law of cooling. If the excess temperature of a body above
the temperature of the surrounding air is .r degrees, the time rate of de-
crease of jc is proportional to .r. If this excess temperature was at first
80 degrees, and after 1 min. is 70 degrees, what will it be after 2 min.?
In how many minutes will it decrease 20 degrees?
5. Atmospheric pressure ]> at points above the earth's surface as a
function of the altitude // above sea level changes according to the com-
pound-interest law. Assuming ;> 15 Ib. per sq. in. when // - 0, and 10 Ib.
when h = 10,000 ft., find p (a) when // = 5000 ft. ; (b) when // = 15,000 ft.
AUK. (a) lU.'Jlb.; (b) 8.1511).
6. The velocity of a chemical reaction in which / is the amount trans-
formed in time t is the time-rate of change of T,
Reaction of the first order. Let a = concentration at the beginning of
the experiment. Then ~ k(a r), since the rate of change is propor-
tional to the concentration at that instant. (Note that a r, the concen-
tration, changes according to the compound-interest law.)
Prove that k, the velocity constant, is equal to - In
/ a - f
7. In the inversion of raw sugar, the time-rate of change varies as the
amount of raw sugar remaining. If, after 10 hr., 1000 Ib. of raw sugar
have been reduced to 800 Ib., how much raw sugar will remain at the
expiration of 24 hr. ? Ans. 586 Ib.
402 DIFFERENTIAL AND INTEGRAL CALCULUS
8. In an electric circuit with given voltage K and current i (amperes),
the voltage A 1 is consumed in overcoming <]) the resistance ft (ohms) of
the circuit, and (2) the inductance L, the equation being
This process therefore comes under equation (4) above, h\ ft, L being
constants. Given L ~ f>40, ft = 250, ft 500, and / = when / = 0, show
that the current will approach 2 amperes as / increases. Also find in how
many seconds / will reach W) per cent of its maximum value. Ann. 5.9 sec.
9. In a condenser discharging electricity, the time-rate of change of
(lie voltage c is proportional to r, and c decreases with the time. Given
/; -- 40, find / if r decreases to 10 per cent of its original value. A nx. 92 sec.
10. llniUlimj n /> a saline (or acid) XO/M//OJ/ by adding salt (or acid),
maintaining constant volume, leads to the equation - (r //), where
as r ^
r -- the constant volume, // ~ salt (or acid) in the tank at any moment,
j' salt 'or acid) added from the beginning. Derive this result and com-
pare with Illustrative Example 2 above.
208. Applications to problems in mechanics. Many important prob-
lems in mechanics and physics are solved by the methods explained
in this chapter. For example, problems in rectilinear motion often
lead to differential equations of the first or second order, and the solu-
tion of the problems depends upon solving these equations.
Before giving illustrative examples it is to be recalled (Arts. 51
and :><)) that
where /' and a are, respectively, the velocity and acceleration at any
instant of time (- /), and N equals the distance of the moving point
at this time from a fixed origin on the linear path.
lu,rsTK-vn\ K !']\AMPLK 1. In u rectilinear motion the acceleration is inversely
proportional to the square of the distance x, and equals 1 when ,s' = 2. That is,
4
(2) Acceleration a -- -
Also, r 5, s- S, when / - 0.
(a) Find r when * 'J4.
Solution. From (2), using; the last form for a, we obtain
/o\ dr 4
(3) r ^
a,^ 6'-
Multiplying both members by </s and integrating, the result is
(4) = - + r, or f-" = ? + r.
ORDINARY DIFFERENTIAL EQUATIONS 403
Substituting in (4) the above conditions r = ft, 8 =- 8, we find (" ~- 1M. Hence
(4 > becomes
(5) r- 1
,S"
From this equation, if s = lM. r = \ \ 1M9 ~ 1.9;?. /I NX.
(hi Find the time which elapses when the point moves from s = S to ,s* = 24.
Solution. Solving (5j for r, we get
Separating the variables s and / and solving for / with limits as given, x -- 8,
s -4, we find, for the elapsed time,
(7) / =
NOTE. Using the first form in il for </, t~> is
^x _ 4
dl' " .s'- 1 '
which ib of the form (F), Art. 'J05. The method of integration here is the same as
in Art. 205.
An important type of rectilinear motion is that in which tlie ac-
celeration and the distance are in a constant ratio and differ in si^n.
Then we may write
(8) a = - Ks- f
where k~ = magnitude of (i ut unit distance.
Remembering that, a force and the acceleration caused by it differ
in magnitude only, we see in the above cast 4 that the effective force
is directed always toward the point ,s -- and is, in magnitude, di-
rectly proportional to the distance ,s-. The motion is called MM pic
harmonic vibration.
From (8), we have, using (1 ),
a linear equation in s and / of the second order with constant, coeffi-
cients. Integrating (see Illustrative Example 2, Art. 20(lj, we obtain
the complete solution,
(10 ) .s- = n cos ki + C'j sin kt.
From (10), by differentiation,
(ID r = k( C] sin /:/ + c- cos A-/).
It is easy to see that the motion defined by (10) is a periodic
oscillation between the extreme positions ,s- = />, and s = - fc, de-
termined by
(12)
404 DIFFERENTIAL AND INTEGRAL CALCULUS
In fact, we may replace the constants c\ and 2 in (10) by other
constants b and A, such that
(13j c\ = b sin A y 02 = b cos A.
Substituting these values in (10), it reduces to
(14) s=bsm(kt + A), by (4), Art. 2
and now the truth of the above statement is manifest.
In the following examples we give cases when the simple harmonic
motion is disturbed by other forces. In all cases the problem de-
penis upon the solution of an equation of the form (G) and (H),
discussed above.
ILLUSTRATIVE KXAMPLK 2. In a rectilinear motion
(15) a - - 'I x - v.
Also, v = 2, K =- 0, when / 0.
(aj Find the equation of motion (x in terms of /).
Solution. Using (1), we have, from (15),
nr\ rf2 ' s ' _L d$ ,5
df.) ^ + ^ + 4 S = '
an equation of the form (G). The roots of the auxiliary equation r 2 -f r + J = are
r, = - J + \ ITT , r 2 = - - vC"I.
Hence the complete solution of (16) is
(17) s = r" 2 '( r , cos / + r 2 sin /).
By the piven conditions, s when / = 0. Substituting these values in (17),
we find o = 0, and hence
(IS) s = c 2 e~
Diflerent.iuting to find r, we get
(19) r ~ r,f *'(- J sin / + cos /).
Substituting the given values r 2 when / = 0, we have 2 = 2.
With this value of r 2 , (18) becomes
(b) For what values of r will r = 0?
Solution. When r = 0, the expression in the parenthesis of the right-hand
member of (19) must vanish. Setting this equal to zero, we readily obtain
(21) tan < = 2.
For any value of / satisfying (21), r will vanish. These values are
(22) t = 1.10 + rnr. (n = an integer). Ans.
ORDINARY DIFFERENTIAL EQUATIONS
405
Successive values of t from (22) differ by the constant interval of time TT.
Discussion. This example illustrates damptd harmonic vibration. In fact, in
(15) the acceleration is the sum of two components
(23) fli = - J , ci 2 = - r.
The simple harmonic vibration corresponding to the component a\ is now dis-
turbed by a damping force with the acceleration a 2 , that is, by a force proportional
to the velocity and opposite to the direction of motion. The etTects of this dumping
force are twofold.
First, the interval of time between successive positions of the point where r =
is lengthened by the damping force. In fact, for the simple harmonic vibration
(24) i --- - ; *,
we have, by comparison with iS\ k - \\ f> - 1.12, and the half-period, by (12), is
0.89 TT. As we have seen above, for the damped harmonic vibration the correspond-
ing interval is TT.
Second, the values of * for the successive extreme positions where r ~ 0, instead
of being equal, now form a decreasing geometric progression. Proof is omitted.
ILLUSTRATIVE EXAMPLE IJ. In a rectilinear motion
(25) a - 4 s f 2 cos 2 t.
Also, s - 0, r = 2, when / = 0.
(a) Find the equation of motion.
Solution. By (1), we have, from (25),
(26)
44x^2 cos 2 t.
dt '
The particular solution required was found in Illustrative Example fi, Art. 206,
and is given by equation (47), p. 397. Hence
(27) R = sin 2 / + j / sin 2 /. Ans.
(b) For what values of / will r - 0?
Solution. Differentiating (27) to find r, and
setting the result equal to zero, we get
(28) (2 -f / icos 2 / -f \ sin 2 / - ;
or, dividing (28) through by cos 2 /,
(29) -J tan 2 / 4- 2 4 / = 0.
The roots of this equation may be found as ex-
plained in Arts. 87-89. The figure shows the curves
(see Art. 88)
(30) y = ] tan 2 /, y = - 2 - /,
and the abscissas of the points of intersection are, approximately,
/ = O.SS, 2..'H>, etc. Anx.
Discussion. This example illustrates forced harmonic vibrut'ion. In fact, in (25)
the acceleration is the sum of two components
(31) a, = 4 s, a- = 2 cos 2 t.
The simple harmonic vibration corresponding to the component GJ with the
period TT is now disturbed by a force with the acceleration a z , that is, by a periodic
406 DIFFERENTIAL AND INTEGRAL CALCULUS
force whose period (= TT) is the mme as the period of the undisturbed simple
harmonic vibration. The effects of this disturbing force are twofold.
FirHt, the interval of time between successive positions of the point where
v ~ is no longer constant, but decreases and approaches J TT. This is clear from
the above figure.
Second, the values of * for the successive extreme positions where r = now
increase arid eventually become, in numerical value, indefinitely great.
PROBLEMS
In each of the following problems the acceleration and initial conditions
are given. Find the equation of motion.
1. a = - A' L '.s ; s =.(),v= r,,, when / - 0. A nx. * = y' sin /;/.
2. a A--.S ; x ,s ()> r 0, w r hen / = 0. s = *,, cos kt.
3. a A'^.s ; .s ,s,,, r r () , \\ hen / 0.
4. a ft x ; .s- 0, r - 0, when / 0. = f> ( 1 cos /).
5. a - sin 2 / - s ; x - 0, r - 0, when / 0. s 5 sin / - \ sin 2 t.
6. a 2 cos / - x ; x 2, r 0, when t 0. * - 2 cos / -f / sin /.
7. a = - 2 r - 2 x ; .s = .'5, r - - ii, when / 0. ,s = 3 c ' cos /.
8. a = - A- L '.s- -f b ; x = 0, r = 0, when 1 = 0.
9. a - //r ; x 0, r ~ n, when / - 0.
10. a - 8 / - 4 x ; x = 0, r = 4, \\ hen / - 0.
11. a 4 sin / 4 x; x = 0, r = 0, \\hen / = 0.
12. a = 2 sin 2 / - 4 s ; x = 0, r 0, when / -- 0.
13. (/ - - 2 r 5 ,s- ; .s- 1 , r - 1 , when / 0.
14. Given a = 8 - -4 x, and r 0, s - 0, when / 0. Show that the
motion is a simple harmonic vibration \\ith the center at s 2, with an
amplitude 2 and a period TT.
15. The acceleration of a particle is given by
a 5 cos 2 / 9 s.
(a) If the particle starts from rest at the origin, find its equation of
motion. A H*. * = cos 2 / - cos 3 /.
What is the greatest distance from the origin reached by the particle?
(b) If the particle starts from the origin with velocity v = f>, find its
equation of motion. AHS. s cos 2 t + 2 sin 3 t cos 3 /.
What is the greatest distance from the origin reached by the particle?
16. Answer the questions of the preceding problem if the acceleration
is given by a - 3 cos 37-9 *.
Ans. (a) x = \t sin 3 / ; (b) s = \ t sin 3 / -f 2 sin 3 /.
ORDINARY DIFFERENTIAL EQUATIONS 407
17. A body falls from rest under the action of its weight and a small
resistance which varies as the velocity. Prove the following relations:
-.
= -^ (kt + < ; ' - 1).
*"
18. A body falls from rest a distance of 80 ft. Assuming n = 32 r,
find the time. AH*. 3. -17 sec.
19. A boat moving in still \\ater is subject to a retardation propor-
tional to its velocity at any instant. Show that the velocity /sec. after
the po\ver is shut off is given by the formula r a ' ', \\here c is the ve-
locity at the instant the power is shut off.
20. At a certain instant a boat drifting in still vater has a velocity of
4 mi. per hour. One minute later the velocity is 'J mi. per hour. Find
the distance moved.
21. Under certain conditions the equation defining the swing of a
galvanometer is
Show that it will not swing through the zero point if p. > k. Find the
complete solution if fj. < k.
209. Linear differential equations of the nth order with constant coef-
ficients. The solution of the linear differential equation
in which the coefficients p\, p>>, p,, are constants, will now be dis-
cussed.
The substitution of c rx for y in the first member gives
(r n + pir 71 " 1 + 7>>r n ~ 2 + + Pn)e".
This expression vanishes for all values of r which satisfy the
equation
(1 ) r n + p } r n ~i + p->r n ~' 2 + - + p n = ;
and therefore for each of these values of r, c rs is a solution of (7).
Equation (1 ) is called the auxiliary equation of (/). We observe that
the coefficients are the same in both, the exponents in (1 ) correspond-
ing to the order of the derivatives in (7), and y being replaced by 1.
408 DIFFERENTIAL AND INTEGRAL CALCULUS
From the roots of the equation we may write down particular
solutions of the differential equation (/). The results are those of
Art. 2(K> extended to cases when the order exceeds two, and are
proved in more advanced textbooks.
Rule to solve the equation (/)
FIRST STEP. Write down the corresponding auxiliary equation
(1 ) r" + p.r" - ' + jw n ~'- + + Pn = 0.
SECOND STEP. Solve completely the auxiliary equation.
THIRD STOP. From the roots oj the auxiliary equation write down the
corresponding particular solutions of the differential equation as follows:
ArxiLiAHv EQUATION DIFFERENTIAL EQUATION
(a) Kach dlstin-ct real \ , ,
f (jives a particular solution e T]X .
root r\ ) '
([}} Each distinct pair \ . ( two particular solutions c cos bx.
, . . , i ' <Jtves . ,
of nnaynuirtj roots a hi i * i e"- r sin b.r.
I 8 (or 2 s) particular solutions ob-
(c) A nndti}>h' root occur- \ . > \ taiwd by multiplying the par-
ritiy s thne-s J * ' ticular solutions (a) (or (b)) by
7, :r, .r 1 -', -, j-- 1 .
FOURTH STI:P. Multiply each of the n * independent solutions thus
found hy an arbitrary constant and add the results. This result set equal
to y gives the complete solution.
ILLUSTRATIVE EXAMPLE 1. Solve ( ~-~ ;i ( - - 4 ?/ 0.
CU' f (U*'
Solution. Follow tho above rule.
First S/r/>. r' - ;i t- -f 4 0, auxiliary cxiuation.
Second N/r/>. Solving, the roots arc 1, -, !2.
Third Sh'p. (ii\ The root - 1 gives the solution c ~*.
(r) The double root li givers the two solutions c- x , xe 2x .
Fourth Step. The complete solution is
^ = dc z -I- c>>c~ J -f riiJY-*. A?is.
ILLUSTRATU-E EXAMPLE 2. Solve ^-4^- / + 10^~12^-h5|/ = 0.
(/j >l dr 1 c/j- J dr
Solution. Follow the above rule.
First Step, r 4 4 r' -f 10 r- 11! r -f 5 = 0, auxiliary equation.
Second Step, Sohdng, the roots are 1 , 1 , 1 2 i.
* A chock on the accuracy of the work is found in the fact that the first three steps
must give n independent solutions.
ORDINARY DIFFERENTIAL EQUATIONS 409
Third Step, (bj The pair of imaginary roots 1 *J i* gives the two solutions
e* cos 2 x, < >J sin 2 x. (0 = 1,6 2)
(c) The double root 1 gives the two solutions r*. xe*.
Fourth Step. The complete solution is
y dc 1 -f ra-N'* -f c^c* cos 2 j -f <V' T sin 2 x,
or I/ = (<*i -f r 2 x 4- fa cos 2 x + <'i sin 2 .rir 1 . A?w.
The linear differential equation
in which pi, ;;_>, , p n are constants, and A' is a function of jc or a
constant, is solved by methods like those used in Art. 20(> for equa-
tion (//). The three steps described on page 394 are to be followed
here also. That is, we solve first the equation (/), obtaining
(2) ? / = w,
the complete solution of (/). Then u is the complementary function
for (/).
Next we find in any manner a particular solution of (/),
(3) y = v.
Then the complete solution of (7) is
(4) y = ?/ + v.
In finding (3), methods of trial may be used analogous to those
given on page 395 for // = 2. The rules given there for the general
case apply also for any value of ?/. In any case we may follow the
Rule to find a particular solution of (7)
FIRST STEP. Differentiate successively tht (liven equation (7) and
obtain, either directly or by elimination, a differential equation of a
higher order of Type (/).
SECOND STEP. Solving this new equation by the rule on page 408,
we get its complete solution.
y ~ u + v,
where the part u is the complementary function of (/) already found
in the first step,* and v is the sum of the additional terms found.
THIRD STEP. To find ihv values of the constants of integration in
the particular solution v, substitute
* From the method of derivation it is obvious that every solution of the original
equation must also be a solution of the derived equation.
410 DIFFERENTIAL AND INTEGRAL CALCULUS
and its derivatives in the yiven equation (/). In the resulting identity
equate the coefficients of Like terms, solve for the constants of integration,
and substitute their values back in
?/ = // + v,
(jiri)ty the complete solution of (J).
This method will now be illustrated by means of examples.
NOTK. The solution of the auxiliary equation of the new derived differential
equation is facilitated by observing that the left-hand member of that equation is
exactly divisible by the left-hand member of the auxiliary equation used in finding
the complementary function.
ILU'STRATIVH K\ AMPLE. Solve
Solution. We first find the complementary function u. Solving
I) y" ~ a \f 4 2 v = 0,
the result is
(7 ) y ?/ c\v~* 4 r-yff.
First ,SVr/>. We now dillerentiate (5), obtaining
(S) //'" \\ //" -f 2 //' = M* + e x .
Subtracting (5) from (S), the result is
(!) ) y'" - 4 //" -f 5 y' - '2 y e z .
Dill'-reritiating (9>, we obtain
(10) ?/ lv 4 //'" - 1 - t5 //" 13 ?/' = e x .
Subtracting (9) from (10), we get
(1 1 //' v - 5 //'" -f 9 y" - 1 ij' -f 2 // = 0,
an equation of Type (/).
iSYrow/ N/r/>. Solve (11). The auxiliary equation is
( 12) r 1 - 5 r< + 9 r-' - 7 r -f 2 = ().
The left-hand member must be divisible by r- H r + 2, since the auxiliary
equation for (t>> is r'- 1 8 r -f 2 = 0. In fact, we find that (12) may be written
(1M) (r-' - ;{ r -f 2>ir - 1 )- = 0.
'1 he roots are r = 1. 1, 1.2. Hence the complete solution of ( 11 ) is
(14) y = or 2 * 4- c J (c- 2 -f r ;i j- -f c&*}.
Third Step. Comparing (7) and (14 \ we see that
(15) y ~ r r'(r.{j -f r 4 J' J >
will be a particular solution of (5) for suitable values of the constants c.\ and c^.
Differentiating (15), we obtain
(10) y r = r'co 4- (r.< 4- 2 r 4 )x 4- r 4 .r 2 ),
I/" = p*(2(r 3 4- f 4) 4- (r 3 -f 4 r^or 4- r 4 x 2 ).
ORDINARY DIFFERENTIAL EQUATIONS 411
Substituting in (5) from (15) and (16\ dividing both members by c r and reduc-
ing, the result is
(17) 2<- 4 - c - 2r,j = J-.
Equating coefficients of like powers of x t we obtain 2r~ 1, '-d ra = 0,
whence we find c\ = 1, r, = J. Substituting these values in (15), the particular
solution is
(18) y = r = r"\- j- - .1 J J ),
and the complete solution is
PROBLEMS
Find tbe complete solution of each of the following differential equa-
tions.
I L-J. + } _if o A H$ t // C] -f ^ , (o s *j >r -|- c,\ sin 2 .r.
f/r { djr
<2 LJJ _|_ 4 1 d o^ y r , -)_ ^.^j- -f r , c() S 2 j* -}- c \ sin 2 s.
3. - ^ 0. // r, -f C'2( >r 4- r. ( r J 4- r i cos .r 4- o, sin u*.
(/j-- 1 as
// 4 ^ ^/ L \-
4. - ' 4- ;r 4 ,s = 0. ,s- - CiC* 4- r.r ' 4- r/, cos 2 / 4 r i sin 2 /.
,4 ?/,s. // = r, -I- cy" -f r. { cos JJ .r -f r i sin 3 .r.
-f r L ./) cos 4 J / -f (r :i -f r ,/) sin 2 /.
6 l + I L > + 8 ^ = 0. -4/**. a* - r L>r (r, + c*t -f r :i / a ).
J/ L> (//
;j/ t ^ WtSt .s- = nr'-r r-r ' + r.j cos / -f r i sin / - P - 3 ^.
9. i _ 4 = 2 ,-'. 7/ = r, + r,r-' + r.,r
rfr :l d.r
IQ. f HlL - ( l! = 4 . r . v = n + r,r + r : ,f
d.r* (Lr~
1L g _ 3 g + 2 = ". y = c,c- + r,f-' - 2 r" + i se>.
18.^-9^+ 20 = /V. A,. . = .,.' + ^'+ f: " (7 + V + 2<2)
(^/ L> a/ 4
13. ^j + 4 .s = / sin 2 /.
4 . . , i t cos 2 t t 2 sin 2
Am. s = cj cos 2 ^ + c 2 sin 2 / -f -- ---
412 DIFFERENTIAL AND INTEGRAL CALCULUS
"--- , 9 . + ,3 + 36, = 18-36.
,6. i; - 1:1 + 30 , = (I. 20. - 2 + g = '.
4 dt^ at 2 at
MISCELLANEOUS PROBLEMS
Find the complete solution of each of the following differential equations.
. Am. ?/=(/ + O\
4. (1 4- .!'-')</// =- Vl - y-dx.
// __ -r 4- r
Vi - //-'
5. (.r -f v/)r/.r ~ (.r //)(///. In (JT~ -f 'r ) 2 arc tan *- = c.
x
7. -4-f li = 0. = o<-' + r 2 ^'.
(//^ <//
8. ~ - 1 ( ^ -H 4 = 0. .s- = r,r-" 4- r,/r 2 '.
(//^ <//
9- ( ^' _ 4 ^ + 8 s = 0. A//K. = r j/ (r, cos 2 f + c> 2 sin 2 <).
(//^ dt
10. ^ - 2 ^ - 8 // = r-'. // - r,r" + ^^' r - \ e-<.
<//* (//
11. ~ + A--.r = / -f b. j- - d cos A-/ -f r,> sin A-/ -f ( ^
dr A"*
.r = n cos A'/ -f r.. sin A'/ + 2 -
!3^ ( l:- r _ / ; i> - a C os A-/. .r - c\c kt + cc~ u - -^ cos kt.
<//-' 2 A"^
14. ^4 -f A- 2 .r = a sin A*/. .r = c\ cos A'/ -f <" 2 sin kt -j t cos kt.
(if" n
15. (.r 2 2 /rW.r -f 2 .r// (/// 0. // 2 -f .r 2 In CJT = 0.
16. ^ -f 4ri- = .>l iv r ?vU' L) + I) 2 = arc tan x + c.
dr j- 4- 1 U'~ + 1)'
ORDINARY DIFFERENTIAL EQUATIONS 413
17. ~j - 5 ^ -f 4 * = 0. Ans. * = nc 1 4- c 2 e < 4- W 4- <V 2 '.
a/ 4 ar
Anx. s c\ cos / 4- r-j sin f 4- o cos 2 / 4- c* sin 2 f.
19. .n/ 2 (i?/ = (.r* 4- li'^djr. po [^ , * . __ /
20. (/// 4- j-z/d ~ -rV-'W-r = 0. (// "'
21. <-^ _ 8 - r 4- 25 .r - 0. 24. '- 4- 4 .r = 6 cos 3 t.
22. 4- 4 jr = 8 / 4- 2. 25. -^ 4- 4 .r = 2 cos 2 /.
df 2 (it-
Solve each of the following differential equations by making the trans-
formation suggested.
26. 1 ' ~ - 2 $t - *' = 0. Let ,s- = ~- A ns. ~ 4- ~ = r.
27. (/-' 4- 0(/.s- = (^' J 4- 2 x/ 4- .s 1 )^/. Let .s- = r/. ^ws. * = r/(l + t) - t.
28. C5 4- 2 K/)S r/^ = (3 - 2 .s7)/ (/.s\ Let x/ r.
29. (.r 4- ?/)' J -^ = 2 jr 4- 2 // 4- 5. Let .r 4- // = r.
ADDITIONAL PROBLEMS
1. For a certain curve the area bounded by the curve, the .r-axis, and
any two ordinates is k times the length of arc intercepted between the
ordinates, arid the curve passes through the point (0, /;). Show that the
curve must be a catenary.
2. The acceleration of a man dropping in a parachute from a stationary
balloon is 32 i r-' ft. per second per second, where ?> is the velocity in
feet per second. If he reaches the ground in one minute, prove that the
height of the balloon is a little more than 950 ft.
' 3. A point moving on the .r-axis is subject to an acceleration directed
toward the origin and proportional to its distance from the origin and to
a retardation proportional to its velocity. Given that the differential
equation for jr is of the form
where m and n are positive, and given the initial conditions x=W,
- 0, when / = 0, find in each of the following cases x and ~ and dis-
dt at
cuss the motion.
(a) m = 4, n = 5 ; (b) m - 4, n = 4 ; (c) m = 4, n = 3.
CHAPTER XXII
HYPERBOLIC FUNCTIONS
210. Hyperbolic sine and cosine. Certain simple expressions in-
volving exponential functions (Art. 02) occur frequently in applied
mathematics. They are called hyperbolic functions. The justifica-
tion for this name is brought out below in Art. 215. Two of these
functions, the hyperbolic sine and hyperbolic cosine of a variable r,
written, respectively, sinh v and cosh r, are defined by the equations
r C v C 1 ' ~4~ v
(4) sinh v = cosh v = !
where e is, as usual, the Napierian base (Art. (>1 ). These functions
are not, however, independent, for we have from (A)
(B) cosh 2 v sinh 2 v ^ 1.
r-' 1 4 '2 f e ' , f-' - 2 -f r -''
From (A), Hquarm^. cosh-' r : ~~~~ sum-' r = :
Hence, by suht ruction, cosh- r Mnh-' r - I.
From (A), by solving for the exponential functions, we get
(1) e r cosh r -f sinh r, e ' = cosh r sinh r.
ILLUSTRATIVE EXAMPLE. Show that the complete 1 solution of the differential
equation
/n\ rf'"'/ . f\
(2) Ji -'// ()
a.r-
may be written // A sinh as -4- />* cosh (/.r,
where .4 and /? are constants.
Solution. By Art. 2()(> the auxiliary equation for (2) is r~ a' 2 = 0, whose roots
are a and a. Therefore the complete solution of (2) is
\l rif"- r 4- or "'.
The values of r" z and r ~ nj> are found from (1) by taking r ax. Hence
r" x ~ cosh n.r -f- sinh u.r, r " x = cosh a.r sinh a.r,
y = n(eosh (i.r + sinh a.r} -f- r-j(cosh ax sinh ajr]
= (c\ -f- r\'Vosh ax -j (ri ^Osinh a.r.
Letting ri r-j = .4, TI -f c ~ /-?, we obtain the desired form.
The result should be compared with Illustrative Example 2, p. 391.
414
HYPERBOLIC FUNCTIONS 415
211. Other hyperbolic functions. The hyperbolic tangent, tanh r, is
defined by
sinh v e 1 e v
C tanh v -
^
r - -r"l
cosh v e 1 + e
The equations
define, respectively, the hyperbolic cotangent, hyperbolic secant, and
hyperbolic cosecant. The ratios used in (C) and (1) are the same as
those in (2), p. 2, for the corresponding trigonometric functions.
The following relations hold :
(2) 1 tanh 1 ' r = sech j r, ctnh-' r 1 = csch- r,
analogous to formulas in (2), p. 2. The proof of the first formula is
given below.
The following statements hold for the values of the hyperbolic
functions. They should be verified. ^
sinh r can have any value ; cosh r, any positive value not less
than 1 ; sech r, any positive value not exceeding 1 ; tanh r, any value
numerically less than 1 ; ctnh /', any value numerically greater than 1 ;
csch r, any value except zero. Also, from the definitions, we have
sinh (- r) =- - sinh r, csch (- r) = - csch v,
(3) cosh (- n = cosh r, sech (- r) = sech v,
tanh (- r) = tanh r, ctnh (- r) = - ctnh v.
II.USTRATIVK EXAMPLE. Given 1anh JT- \. Find the values of the other
hyperbolic functions.
Solution. In () divide each term by cosh 2 .r. Then we got
^" 1 >r ^
cosh*' J* cosh* x
Therefore 1 - tanh- .r - sech 2 x. By (C) and (1)
Since tanhjr= ?, this equation gives sech JT = J, the negative value being in-
admissible. Then
cosh jr = \ =~> bv (*)
sech x 3
sinh j = cosh x tanh 3 = ? by (C)
ctnh x = J, arid csch jr = 4 '. By (1)
212. Table of values ot the hyperbolic sine, cosine, and tangent.
Graphs. A table giving the values, to four significant figures, of
sinh r, cosh ?>, tanh r for values of r from to 5.9 is shown on p. 416.
For negative values of r, use the relations (3;, Art. 211.
416
DIFFERENTIAL AND INTEGRAL CALCULUS
HYPERBOLIC FUNCTIONS
V
Sinh v
CoHh i
Tanhr
p
Sinh v
Cosh r
Tanh r
r
Sinh r
Cosh r
Tanh r
00
0000
000
.0000
.50
.5211
1.128
.4621
1.0
1.175
1.543
.7616
01
0100
.000
.0100
.51
.5324
1.133
.4700
1.1
1.336
1.669
.8005
.02
0200
.000
.0200
.52
.5438
1 138
.4777
1 2
1 509
1.811
8337
03
.0300
000
.0300
.53
.5552
1 144
4X51
1 3
1 698
1 971
8617
04
0400
001
.0400
54
5666
1 149
4930
1 4
1 904
2 151
8854
.05
0500
001
0500
55
5782
1.155
5005
1 5
2 129
2 352
.9052
.06
0600
002
.0599
.56
5897
1.161
50X0
1 6
2 376
2 577
.9217
07
0701
002
0699
57
6014
1 167
5154
1 7
2 646
2.828
.9351
.08
.0801
003
0798
.58
6131
1 173
.5227
1 8
2 942
3.107
.9468
09
.0901
004
.0898
59
6218
1.179
5299
1.9
3.268
3.418
.9562
.10
1002
005
0997
.60
.6367
1 185
5370
2
3 627
3.762
9640
.11
1102
006
.1096
.61
.6485
1 1912
5441
2 1
4 022
4 144
.9705
12
1203
007
1191
.62
.6605
1 198
5511
22
4 457
4 568
.9757
.13
1301
008
.1293
.63
6725
1 205
55X1
2 3
4 937
5 037
9801
14
1405
010
1391
.64
6816
1 212
.5649
2.4
5.466
5 557
9837
15
1 506
Oil
11X9
65
.6967
1 219
57 1 7
2 5
6 050
6.132
.9866
16
1607
013
1 5X7
.66
.7090
1 226
57X1
2 6
6 695
6.769
9890
.17
1708
01 1
1 68 1
.67
.7213
1.233
5850
2 7
7.406
7.473
.9910
.18
.1810
016
1781
.68
.7336
1 210
.5915
2.8
8 192
8 253
9926
.10
.1911
018
1878
.69
.7461
1 248
5980
2.9
9.060
9.115
9940
.20
2013
020
1974
.70
.758 6
1 255
6011
3
1 02
10 07
9951
21
2115
022
2070
.71
.7712
1 263
6107
3 1
11 08
11.12
9960
''2
22 1 8
021
2165
.72
.7X.JX
1 271
6169
.'{ 2
12 25
12.29
.9967
.23
2320
027
2260
73
7966
1 27 X
623 1
3 3
13 51
13 57
9973
24
2123
029
2355
7 '
8091
1 287
6291
3.1
14 97
1 5 00
9978
r >
2526
031
2149
75
X22. () >
I 295
6:552
3 5
1 6 54
1 (i 57
99X2
26
2629
03 1
.2513
.76
.8353
1 303
.6111
3 6
1 h 29
1831
.9985
27
.2733
037
.2636
77
.8181
1.311
6169
3 7
2021
2021
99XX
28
.2837
039
2729
78
.8615
1 320
6527
3 X
22 34
22 36
.9990
29
2941
.042
2821
7!)
,X74X
1 329
(.584
3 9
24 69
21 71
9992
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3015
.045
29 1 3
XO
XXXI
1 MM7
6610
1
27 21'
27 31
9993
31
3 1 50
018
3001
81
<I015
1 316
6696
4 1
30 16
30 18
9995
32
3255
052
30 ( l5
82
9150
1 35. r
.6751
1 2
33 3 1
33 35
.9996
33
3360
055
318")
83
9286
1 365
6X05
1.3
36 X 1
36 X6
.9996
31
3166
058
3275
81
9123
1.371
6xr>x
4.1
40 72
40 73
9997
3f>
3572
062
3364
8,'>
9561
i :;x i
6911
1 5
45 00
45 01
9998
36
3(178
066
3152
.86
.9700
1 393
6963
4 6
49 74
49.75
.9998
37
3785
069
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.9810
1 4 03
.701 i
4 7
54 97
54.98
9998
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3892
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.7064
4 8
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027
1 133
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on
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82 01
82 01
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070
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100 17
100 17
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122 35
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182 52
0000
HYPERBOLIC FUNCTIONS
417
When r -* + oo, sinh r and cosh r become infinite, while tanh v
approaches unity as a limit.
Graphs of sinh x, cosh x, and tanh a* (Figs. 1, 2, 3) are easily
drawn by making use of the table.
(O.-l)
(0,1)
ZlL
tanh x
FIG. 1
Fie. ,'J
213. Hyperbolic functions of v -f w. Formulas for hyperbolic func-
tions, corresponding to two of (4), p. 3, are
(D) sinh (i; + w) = sinh v cosh w + cosh v sinh w,
() cosh (v + w) ~ cosh i; cosh w + sinh i> sinh i#.
Proof of (/)). From the definition (A), replacing r by r + w, we have
(1 ) sinh ( + w } - - ~~ >
(2)
cosh (v -f w)
The right-hand member of (1) is transformed as follows, making
use of (1), Art. 210.
e
e'e 11 e (>
2 2
(cosh ?? + sinh ?>)(cosh w + sinhw) (cosh ?* sinh ?>j(eosh7^--sinh
Multiplying out and reducing, we get (D). Formula (E) is proved
in the same way.
If we set w = v in (D) and (E), we have
(3) sinh 2 # = 2 sinh v cosh v,
(4) cosh 2 t? = cosh 2 ?; + sinh 2 v.
418 DIFFERENTIAL AND INTEGRAL CALCULUS
These are analogous to the formulas for sin 2 x and cos 2 x, re-
spectively, of (5), p. 3. From (B) and (4), we get results which
correspond to the formulas for sin- x and cos j x in (5), p. 3. These are
(5; sinh 2 v = | cosh 2 r \, cosh 1 -' r .V cosh 2 r + ?,.
Other relations for hyperbolic functions, which may be compared
with those on page 3 for trigonometric functions, are given in the
problems.
ILLUSTRATIVE Ex AMPLE. Derive the formula
sinh 2 ?'
(6) tanh
cosh 2 ?> -f 1
Solution. From (5), by division, we get
rush 2 r - 1
en;- h 2 r | 1
cosh 2 r 1 cosh "2 r i 1 cosh- 2 ?> 1
(7) tanh-' r -
Now
cosh 2 7' + 1 cosh 2 M 1 (cosh 2 /' -f 1 j-'
By (B), cosh- 2 v ~ ] ~- sinh* 2 r. Hence ('7,< becomes
i ;.,i, - > .,1
(8) tanh-'r-
and therefore tanh r - i
(cosh 2 /' + 1 )-'
sinh 2 r
>sh 2 r 4 1
The sign before the right-hand member must now be examined. From (3)
we have . ,
, 2 sinh r , , ,, . , , ,
smh 2 r ~~ , cosli- r 1 tanh r cosh- r.
cosh r
Therefore sinh 2 f and tanh r will always a^ree in sign. Also cosh 2 r -f- 1 is
always positive. Hence the positive sign must be used, and we get (f>>.
If r is replaced by \ r, (T>) becomes
sinh r
(9) tanh =
cosh r -f- 1
PROBLEMS
1. The value of one hyperbolic function is given. F"ind the values of
the others and check as far as possible by the table on page 416.
(a) cosh JT 1.25. (c) sinh .r = 10.
(b) csch .r 0.75. (d) ctnh s = 2.5.
Prove each of the formulas in Problems 27, and compare with the
corresponding formula (if any) in (2), (4)-(6), pp. 2, 3.
2. 1 ctnh 2 r = csch 2 r.
3. sinh (r ?r ) = sinh r cosh w cosh r sinh ir,
cosh (r w] = cosh r cosh ir sinh r sinh ?r.
4. tanh (r w)
HYPERBOLIC FUNCTIONS 419
tanh r tanh w
1 tanh r tanh w
, . i r . /cosh r 1 , r /cosh r -f 1
5. sinh - = A/ cosh - = -f A/ 1>
6. sinh r -f sinh // 2 sinh \(r 4 /r) cosh \(r w),
cosh r -f cosh /r = 2 cosh -\(v -f w) cosh J(r /r).
sinh r sinh //
7. tanh J (r w) =
cosh r -f cosh /r
8. Show that the equation of the catenary (figure, p. 532) may be
i'
written ?/ a cosh -
a
9. Solve the differential equation // in terms of hyperbolic
u.r-
functions, given that // -= 3 when .r 0, and // when tanh x J.
/l>/,s i . // rr ;{ cosh .r 3.75 sinh jr.
10. Show that sech ( s) = sech s. Draw the ^rapli and prove
lim sech .r 0.
j- oc
11. Show that ctnh ( .r) - - ctnh .r. Draw the graph arid prove
lim ctnh .r 1.
X-* i QC
12. Show that csch ( .r) = csch .r. Draw the graj)h and prove
lim csch s 0.
jr -> ex,
13. Prove (a) sinh 3 i< = 3 sinh // -f 4 sinh' 5 u ;
(b) cosh X // 4 coshr // 15 cosh //.
14. Show that (sinh .r -f cosh .r)" sinh ns 4 cosh tijr. (n any positive
integer.)
15. Prove that sinh- ./ sinh j y sinh O -f //) sinh O //).
16 . simplify C0 ; " + TOS , h * " A *. ctnh (u + 2 v).
sinh 2 // -f sinh 4 r
17. Parametric equations for the fmctrijr may be written
S = t a tanh - ?/ ~ a sech -
a ' a
The parameter is /, and a is a constant. Plot the curve when a 4. (The
tractrix is the curve for which the length of the tangent (Art. 43) is con-
stant and equal to a. Figure in Chapter XXV 7 !.)
18. Solve = n 2 (y -
ax 2
2 tfi,
Ans. y = A cosh nx -f B sinh nx 4- rax 2 H -- -
420 DIFFERENTIAL AND INTEGRAL CALCULUS
214. Derivatives. The formulas, in which v is a function of x, are as
follows.
XXVII 2- sinh v = cosh v ^ -
dx dx
XXVIII 4- cosh v = sinh v -
dx dx
XXIX 4- tanh z; - sech 2 y
dx dx
XXX ctnh v csch 2 f
dx dx
XXXI - sech v = sech v tanh v
dx dx
XXXII csch v csch v ctnh i; -^
Proof of XXVII. liy 01), sinh r - ( = ^
rru ' u
1 hen -7- sinh ?' ~
dx 2
lJL ''
~
i
= COSH /' -;- >
dx
^1).
Formula XXVIII is proved in a similar manner. The proof of
XXIX is analogous to that given in Art. 72 foi the derivative of
tan r. To prove XXX XXXII, differentiate the forms as given in (1),
Art. 211. The details are left as exercises.
215. Relations to the equilateral hyperbola. The curve for which
(1) x = a cosh r, y = a sinh r
are parametric equations is the equilateral hyperbola x- y 2 = a 2 .
For, eliminating the parameter r by squaring and subtracting, we
have X L> __ y - = a 2( c . os h- v ~ sinh'-' r) = a~. By (B)
Fig. 2, p. 421, shows a hyperbolic sector OA P\ bounded by the
arc A PI of (1), the semitransverse axis 0.4, and the radius vector
or i. At PI, ? = ri.
HYPERBOLIC FUNCTIONS
421
Theorem. The area of the hyperbolic sector OAI*i equals A- ai.
Proof. Let (p, 6) be the polar coordinates of any point on the arc
A PL Then the element of area is (Art. 159) dA = J p- dQ.
Or ,,?/,)
FIG. 1
But p~ = x 2 + y' 2 = a- (cosh- + sinh- r).
Also by (5), p. 4,
= arc tan = arc tan (tanh r).
Using (1)
By (1)
Therefore
(j[d _ sechr v
dv 1 + tanh- v
Using (C) and (1), Art. 211, we get
By XXII, p. 87, and XXIX
cosh 2 v + sinh- v
and therefore dA i a- r/?;.
The theorem follows by integration, since /? = at ^4. Q.E.D.
The parametric equations of the circle in Fig. 1 are
x r cos /, y = r sin /. Art. 81
The parameter / equals t\ at PI, and /j is the measure of the central
angle AOP\ in radians. Hence the area of the circular sector AOP\
is \ r-fi.
Let r = a = 1. Then in Fig. 1, for P(x, y),
x = cos t, y = sin , J t = area ^0/\
In Fig. 2, for P(x, ?/),
x = cosh v, ?/ = sinh ?;, i ?; = area AOP.
Hyperbolic functions, therefore, have the same relations to the
equilateral hyperbola as the trigonometric functions do to the circle.
422 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
1. Show that the element of length of arc for the catenary z/ = a cosh -
j. a
is given by J,x cosh - ds.
2. In the catenary of Problem 1 prove that the radius of curvature
i y 2
equals tL --
Verify the following expansions of functions by Maclaurin's series, and
determine for what values of the variable they are convergent.
3. sinh jr = jr 4- ~ -f + -f ,./*"_', 4- . Aws. All values
4. cosh x = 1 -f --: -f V -f -f h -A NX. All values.
\~ A
Verify the following expansions, using the series in Problems 15 and 4
and the methods explained in Art. 195.
5. sech .r = 1 - .1 .r-' -f .;:, jr* - 7 (> ,' / + .
6. tanh .r = jr - .', x'< + A r :> - .r, 7 :, ^ 7 + -
7. Test the function 5 cosh .r -f 4 sinh .r for maximum or minimum
values. .4 NX. Minimum value, .'5.
8. Test the function .4 sinh x -f /> cosh .r for maximum or minimum
values. .4 NX. If /*- > A 2 , a maximum value \//>- ,4 J if B < 0;
a minimum value -f \/ />*'-' .4* if B > 0.
9. Derive the series in Problems 15 and 4 from the series for c j and c T
by subtraction and addition. (Use (^1) and Art. 195.)
10. Let (/x ~ length of the element of arc ; let p ~ VV J -f ir ~ radius
vector of /'(.r, //) for the circle or equilateral hyperbola of Art. 215, and
take limits of integration for the arc .4 l\ in Figs. 1 and 2, p. 421. Prove
(a)
| ^ = /, for the circle ; (b) / ?', for the hyperbola.
J p J p
11. Prove lim (cosh .r sinh .r) = 0.
j- f oc
12. Evaluate each of the following indeterminate forms.
f ^ v sinh x /u\ v tanh jr
(a) lim (b) hm
An*, -
13. Given tan = sinh x. Prove - = sech x.
dx
HYPERBOLIC FUNCTIONS
423
14. Derive the expansion
arc tan (sinh x} .r J .r< -f ^ 4 x* Tj J| x
by integration as in Art. 196, using the result yj
in Problem 5.
15. Prove the following theorems for the
tractrix (see figure)
-f
jr = 1 a tanh
t
y = a sech -
(a) The parameter / equals the intercept of
the tangent on the .r-axis.
(b) The constant a equals the length of the
tangent (Art. 43).
(c) The evolute is the catenary fi = a cosh
/ a
(d ) The radius of curvature PC is a sinh -
M X
216. Inverse hyperbolic functions. The relation
(1 ) y = sinh v
is also written
(2) v = sinh l y,
and read " v equals the inverse hyperbolic sine of y." Therefore
sinh v and sinhr l y are inverse functions (Art. 39). The same nota-
tion and nomenclature are used for the other inverse hyperbolic
functions, cosh ' v (" inverse hyperbolic cosine of v "), etc.
The curves
(3) y = sinh x, y = cosh x, y = tanh x
are shown again on page 424. Assume now that y is given.
In Fig. 1, y may have any value, positive or negative, and then
the value of x is uniquely determined.
in Fig. 2, y may have any positive value not less than 1. When
y > 1, x has two values equal numerically and differing in sign.
In Fig. 3, y may have any value numerically less than 1, and then
the value of x is uniquely determined.
Summarized, the results are
The function sinh" 1 v is uniquely determined for any value of v.
Also sinh~ l ( v) = sinh~ ! v.
The function cosh" 1 v, when v > 1, has two values differing only
in sign. Also cosh" 1 1 = 0.
The function tanh" 1 v is uniquely determined when v-< 1. Also
tanh'K 0) = tann" 1 v.
424
DIFFERENTIAL AND INTEGRAL CALCULUS
FIG. 1
FIG. 2
FIG. 3
Hyperbolic functions were defined in Art. 210 in terms of ex-
ponential functions. The inverse hyperbolic functions are expressible
in terms of logarithmic functions. The relations are
sinh l x - - ln(jc
l).
(Any x)
(O
cosh l x =-- In (x i N// 2 l).
tanh 1 x = I In
/ 1 y
Jl A
Proof of (F). Let v sinh ] .r. Then
(4)
sinh r
By
To solve (4) for v, write it as follows :
<"' - 2 JT = 0, or
r
This is a quadratic equation in e v . Solving, r r = x Vj>' + 1.
Since r 1 is always positive, the negative sign before the radical
must be discarded. Hence, using Napierian logarithms, we have (F).
Proof of (G). Let v = cosh l x. Then
(5) jc = cosh /? = - ^
By
Clearing and reducing, we have c 2 '' 2 arc 1 ' + 10.
Solving, c r = x Vx 2 ~ 1.
Both values must be retained. Taking logarithms gives (G).
HYPERBOLIC FUNCTIONS 425
Proof of (//) Let r = tanh- l *. Then
(6) j = tanhr = -=^- By (C)
Clearing of fractions and simplifying, the result is
l+x
(x \)e' + (jc + l)e ' = 0. Hence t'- r =
Taking logarithms, we have (//).
1-z
ILLUSTRATIVK EXAMPLE. Transform
(7) 5 cosh .r -f 4 sinh jc
into the form (" cosh (jr -f a), where r and u are constants, and find C and a.
Solution. By (), Art. 213, we have
(8) C cosh (.r + a) T cosh .r cosh a -f (' sinh j* sinh a.
Hence (7) will hav^e the desired form if T and a satisfy the equations
(9) C cosh a - 5, C sinh a 4.
Squaring, subtracting, and using (/?), Art. 210, we get ("' = 9. Then C = -4 3,
since cosh must he positive. Also, by division, tanh a /,. Hence
H = tanh ' 0.8 - i In 9. By (//)
Therefore a = 1.099 and
(10) 5 cosh .r f 4 sinh JT ~~ 3 cosh (JT -f 1.099).
The graph of the function 5 cosh .r j 4 sinh .r may be obtained from the graph
of 5 cosh x by translating the //-axis to the new origin (1.099, 0). (Compare with
Illustrative Example 2, p. 391.)
When x is given, the values of sinh l .r, cosh ] j, or Umh ! jc can
be determined by the table on page 41 (> to not more than three signifi-
cant figures. For example, sinh ' 0.25 = 0.247 ; cosh ' IJ = 1.76.
For greater accuracy (F), (G), or (//) may be used if tables of
Napierian logarithms are at hand.*
217. Derivatives (continued). The formulas, in which v is a function
of x, are as follows. ^ v
XXXIII - sinh- 1 v - ,^ - - (Any v)
/i; 2 + 1
dv
XXXIV - cosh - 1 i; = d * (v > 1)
dv
/I rlY
XXXV
cfx 1 v 2
* The Smithsonian Mathematical Tables. "Hyperbolic Functions" (1909), give the
values of sinh u, cosh u, tanh 11, ctrih u to five significant figures Values of the correspond-
ing inverse functions to live significant figures may be found from these tables.
426 DIFFERENTIAL AND INTEGRAL CALCULUS
Proof of XXXIII. (Compare Art. 75.; Let
y = sintr l v;
then r = sinh y.
Differentiating with respect to y, by XXVII,
dv ,
-= cosh y ;
dy
therefore ^ = ~ By (C), Art. 39
dv cosh y J \ "
Since v is a function of x, this may be substituted in (A), Art. 38,
djr cosh // dx vV' + 1 d.
[cosh // = Vsinh'-' // 4 1 vV' -f 1, by
The proofs of XXXIV and XXXV are similar. Other formulas are
the following.
(/) ctnh J x - In (x 2 > 1)
(7) sech- 1 x = ln- --l- (0 < x g 1)
(K) csch 1 ^-!
\ A
_dv
ft rlv
XXXVI -,
_
XXXVII sech i i; = * (0 < i; < 1)
XXXVIII csch~i t; = - =- (^ 2 > 0)
Details of the proofs are called for in Problems 58 on the next
page.
HYPERBOLIC FUNCTIONS 427
PROBLEMS
1. Show that the two values of cosh ' .r in (G) differ only in sign.
2. Draw the graph of ij - \ sinh > jr. Check in the figure the values
of // and //' when s = 2. An*, tj = 0.72, //' = 0.21236.
3. Prove XXXIII directly by differentiating (f).
4. Draw the graph of each of the following and check in the figure
the values of ?/ and //' for the given value of jr.
la) ?/ = cosh l JT ; ./ 2.
(bl // = tanh ! .r; .r = - O.Tfi.
5. Prove XXXIV and XXXV.
6. Derive (7) and XXXVI.
7. Derive (/) and XXXVII.
8. Derive (K) and XXXVIII.
9. Derive the expansion
tanh ' j .' } -: + -- +
.{ ;>
by Art. 19 r>.
10. (riven sinh .r tan c/>. I 'rove
, </' /
(a) .r = In (sec -f tan 0) ; (b) ^-- - sec 0.
11. Show that csch ' r =- sinh ' 1- Derive XXXVIII from XXXIII,
using this relation.
12. Evaluate lim j- ctnh ' f. An *- 1 -
J X
13. Evaluate lim s csch ] j\
J - X
14. Derive the expansion
sinh '' = .'--J i - rl + ^lf ""
15. Evaluate lim fsinh ' f - In .0. ^w.s. In '2
4- X
16. Show that ctnh ' r = tanh ' y sech ' r ^ cosh J -. and verify
XXXVI and XXXVII from these relations.
d , . tanh a tan r _ s'nb a
17. Prove - tanh ^ fl + se( . ,. ~ j + cosh a cos r '
18. Draw the graphs off a) ?/= ctnh J r; rb)?y = sech ] j; (c) ?y=c8ch" 1 A
using the theorem of Problem 28, p. 41.
428 DIFFERENTIAL AND INTEGRAL CALCULUS
218. Telegraph line. Assume in a telegraph line that a " steady
state " of flow of electricity from A, the home end, to B, the receiving
end, has been established, with perfect A j ^ p f B
insulation and uniform linear leakage. P
is any intermediate point. We need to
consider :
the electromotive force (volts;, e.m.f., L\\ at A, En at B, E at P;
the current strength (amperes), 7 A at A, IB at B, 7 at P ;
the characteristic constants a and ro, whose values depend upon
the linear resistance and leakage. They are positive numbers.
Let x = A P. Then it is shown in books on electrical engineering
that E and 7 are functions of x such that
We wish to find the e.m.f. and current strength at P. They are
(8) E = E.\ cosh ax r () 7.i sinh ax,
(4) 7 7,1 cosh ax - sinh ax.
Proof. The complete solution of (1) is (Illustrative Example,
Art. 210)
(5; E A cosh ax + B sinh ax.
Substituting in (2), the result is
(0) r ( )7 A sinh ax B cosh ax.
But E -~ E i, 7 = J A when x 0. Therefore A ~ E A , B nJ A ,
and (5) and ((>) become (8) and (4) respectively.
For the solution in terms of the e.m.f. and current strength at
the receiving end, see Problem 2 below.
PROBLEMS
All refer to a telegraph line in a "steady state," and L AB.
1. Given E A = 200 volts, L = 500 kilometers, r - 4000 ohms, a = 0.0025,
In = 0. Find I A and En.
Ans. 1 1 = 0.05 tanh 1.25 = 0.04238 ampere ;
E B = 200 sech 1.25 = 105.8 volts = 0.53 EA.
HYPERBOLIC FUNCTIONS 429
2. If y = PB = distance of P from the receiving end, show that
E = EB cosh ay + r//? sinh a//, / I fi cosh ay + sinh ay.
ro
3. Given EA = 200 volts, I A = 0.04 ampere, r> = 4000 ohms, a = 0.0025.
Show that
E = 120 cosh (1.099 - 0.0025 jr}, I = 0.03 sinh (1.099 - 0.0025 x).
(See the Illustrative Example, Art. 216. Thus A 1 tends towards a minimum
value of 120 volts and / approaches as .r approaches 439.6.)
4. Given E.\ - 1GO volts, 7,i = 0.05 ampere, r n = 4000 ohms, cv = 0.0025,
Show that
E = 120 sinh (1.099 - 0.0025 x\ I = 0.03 cosh (1.099 - 0.0025 .r).
(See the Illustrative Example, Art. 216. Thus E approaches zero and 1
decreases to a minimum value of 0.03 ampere when .r approaches 439.6.)
d' 2 I
5. Prove that -pr, a*l = 0. (Thus /; and / are solutions of the same
dx 2
linear differential equation, which has the form //" a' 2 y 0.)
6. Given E,\ = r u /.4. Prove
(a) K= E A c af , I = IAC * ;
(b) E = r 7 ;
(c) E * when .r becomes infinite.
(Tor example, if TO = 4000, and the impressed e.m.f. at the home end
of the line is 4000 times the current strength, then at mr// point of the line
the e.m.f. is 4000 times the current strength and diminishes towards zero
as the length of the line is indefinitely increased.)
7. In Problem 6 show that the decrease in E at P at unit distance along
the line from P equals Ec~" t where e is the Napierian base.
8. Prove the following.
(a) If IK - 0, then E\ - r,,/.i ctnh aL.
(b) If En = 0, then E A = r ( J A tanh <*L.
ADDITIONAL PROBLEMS
Derive the following relations.
1. If E A > r n l A and T = tanh- 1 ^~A then
EA
E = E A sech r cosh (r ax), I I A csch T sinh (r ax).
2. If E A < r Q I A and T = tanh^ 1 -^X then
ro/A
E = EA csch T sinh (T ax), I = I A sech r cosh (r ax).
430 DIFFERENTIAL AND INTEGRAL CALCULUS
219. Integrals. A list of elementary integrals involving hyperbolic
functions and supplementing Art. 128 is given here.
(24) I sinh v dv = cosh v + C.
(25) / cosh vdv= sinh v + C.
(26) j tanh v dv = In cosh v + C.
(27) I ctnh vdv=la sinh v+C.
(28) I sech 2 vdv = tanh v+C.
(29) I csch 2 vdv=^~ ctnh v+C.
(30) I sech v tanh v dv = sech v+C.
(31) I csch v ctnh v dv csch v+C.
The proofs follow immediately from XXVII-XXXH, except for
(26) and (27). To prove (26), we have
f tanh v dv - C^LH rf r by (c)
J J cosh ?' J \ /
f(/(cosh r) , , . _,
i - - _ ] n eos h ? , _j_ ^
J cosh /'
The proof of (27) is similar.
ILLUSTRATIVE FA AMPLE. Derive the formulas
(1) I sech v dv = arc tan (sinh r) -f C;
(2) fcsch P dr = In tanh + C.
Solution. Since sech r = -J- = -^LL = cosh r
cosh r cosh- r l-fsmh 2 r J v 7
we have fsech r dr = f ^^ df - f ^ sinh , f)
J J 1 -f sinh- r J 1 -f sinh 2 z;
=J d[arc tan (sinh f)] = arc tan (sinh v) + C.
HYPERBOLIC FUNCTIONS 431
To derive (2) we have (compare Art. 131)
, , csch r ~\ ctnh r
csch r = csch ? ; -
csch r 4 ctnh r
csch- r 4- csch f ctnh /
/ csch r dv = I
= -f-
ctnh r 4- csch r
csch 2 r csch r ctnh r
ctnh r 4 csch r
'(/(ctnh r 4 csch M
dr
ctnh v 4- csch r
= In (ctnh r 4- csch r) 4- ('
= _l n /ilU + .__L_\ + r b v (n Art. 211
\suin r sinh rl
In ( cosh r -f 1 ) 4 In sinh r f- (" In ^ f (
cosh r 4 1
= In tanh + r. By (9), Art. 213
PROBLEMS
Work out the following integrals.
1. fsinh" r dr ~ \ sinh 'J r \ r 4- r.
2. Tcosh L ' r dr = 4 sinh 4 J r 4- i r 4- C.
3. | tanh' J r dv = r tanh r 4- C.
4. | ctnh 2 r dr ~ r ctnh r 4- r.
5. | sinh :< r dr = cosh :i 7 1 cosh ?' 4- C.
6. (cosh :? r d? 1 = fj sinh :? r 4- sinh 7 1 4- r.
7. | tanh 3 v dv In cosh r J tanh 2 ?' 4- r.
8. I tanh 4 r dr r tanh v > T tanh :< r 4- r.
9. Jcsch H r dr = I csch r ctnh ?' - \ In tanh ~ 4- C.
10. | x sinh x dx x cosh r sinh x 4- C.
1 1. fcos x sinh x dx = I (cos x cosh r 4- sin x sinh r) 4- C.
/I
sinh (rax) sinh (ns) = r fw sinh (nx) cosh (rax)
ra~ n z
n cosh (nx) sinh (rax)] + C.
432 DIFFERENTIAL AND INTEGRAL CALCULUS
Work out each of the following integrals.
13. J sinh 4 x dx. 15. Ann x cosh x dx.
14. J^sec-h 4 2 x dx. 16 -/- r tanh x djr '
17. j x 2 cosh x dx. 18. JV sinh a- dx. 19. /V" cosh x dx.
Work out each of the following, using the hyperbolic substitution in-
dicated. (Compare Art. 135.)
20. J Vx' 2 - 4dx; / = 2 cosh r. Am. \ x^/x' 2 - 4 - 2 cosh ' x + C.
/-
"/^TaTTs^" 2 " 1
23. The arc of the catenary // = a cosh - from (0, a) to (x, y) is revolved
about the //-axis. Find the area of the curved surface generated, using
hyperbolic functions.
24. Find the centroid of the hyperbolic sector OA P\ in Fig. 2, Art. 215.
(Compare Problem 12, p. 337.) . - __ 2 sinh r, - 2 cosh r, - 1
/r AtiS. .r (I ~ y en (j, .
'> ?'i * 3 ?i
220. Integrals (continued). From XXXIII XXXVIII we may
derive integrals. Some of them we have already met in Art. 128.
Their values are now expressible in terms of inverse hyperbolic
functions.
(32)
(33)
(34)
(35)
(36)
(37)
r dv
r dv
sinh~ 1 - + C.
a
cosh~ * - + C.
u
(v any value)
(v^)
(i> 2 > a 2 )
(0 < v < a)
(v any value)
J vV - a 2
/dv
v*-a 2
r dv
a a
eppV, 1 V 1 C
J v Va 2 - v 2
r dv
a a
I v
popVi ~1 ..1 /^
J vVi; 2 4-a 2
CZ CL
HYPERBOLIC FUNCTIONS 433
/_ _____ 2
Vi> 2 + a 2 dv = V7 2 + a 2 + |- sinh- 1 - + C.
(39) f V> - a 2 rfi; = Vi^~^ _ f*_ cosh -i + c
j * <& a
In (33) and (39) the positive value of the inverse hyj>erbolic
cosine must be used, and in (36) the positive value of the inverse
hyperbolic secant.
Proofs of (32) and (33). Let x = - in (F). Then
In (r + vV + <7~) In a.
a \a \ u,~ i
Hence
(1) In (v + Vr- + a~) = sinh ! --' + In a.
In the same way, from (G) we get
sinh- 1 - = In - + + 1
(2) In (v + Vr- - a 1 ') - cosh ' - 4 In a.
Using these results in the right-hand member of (21), p. 193, we
get (32) and (33).
Proofs of (34) and (35). Let x = - in (#). Then
/r^ 1 1 ft + /; i i , ?'
(3; 2 ln ^T7 = t;inh 'a'
Then (34) follows from (3) and (19 a), p. 192.
In the same way, from (I) and (19), p. 192, we get (35).
[in (19), In ^ = - In ^- I
I ' r + t> u J
Proof of (36). Since
d (-}
by XXXVII
v~
we
have I / = - sech ] - if the ijositive sign before the
J v V a 2 r- a a
radical is chosen. The proof of (37) is similar.
Formulas (38) and (39) follow from (23), p. 193, using (1) and (2).
REMARK. Since
,,,# 1,0 1,0 i,^ , , v .,.
ctnh" 1 - = tanh" 1 - seen' 1 - = cosh" 1 - csch l - = smh^ 1 - >
a v a v a v
the integrals (35) -(37) may also be expressed in terms of the functions most con-
venient for use of the table of Art. 212.
434
DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE. Derive (37) by means of the substitution v a csch z.
(Compare Art. 135.)
Solution. We have
By (2), Art. 211
Also
iv 2 + a' 2 - Va~ csch- z + a 2 = a ctrih z.
dr = -a csch z ctnh z dz.
By XXXII
t r dv C a csch z 01
Therefore / r- = I
J v vV -f a* J a (>sch z '
a csch 2 ctnh z dz 1 , ~
: z -f- C.
ctnh z a
Since 2 csch" 1 - we have (37).
PROBLEMS
1. In the figure the curve is the equilat-
eral hyperbola .r' 2 y' 2 = a-. Using Art. 142,
prove that
(a) area A MJ > triangle OMP - a' 2 cosh ' - ;
(b) sector OA I> - \ a 2 cosh l - = a' 2 r,
if x a cosh ?'. (Thus we have an alternative
proof of the theorem of Art. 215.)
2. Derive each of the following power
series by integration (Art. 19G).
(a) tanh
(b) sinh
= jr -f ~ .r< -f | .r r > 4-
<> .)
2 ;* 2-4 5
Work out the following integrals.
3. fsinh l JT djr JT sinh ' .r VI -f .r-' 4- C.
4. Mann ' .r dr. 5. (.rcosh" 1 ^
Work out, using hyperbolic functions.
'farrb-
9. Find the length of arc for the parabola .r 2 = 4 y from (0, 0) to (4, 4),
using hyperbolic functions. Ans. V20 + sinh J 2 = 5.92.
10. Find the area bounded by the catenary y a cosh - and the line
y = 2a.
11. The downward acceleration a of a falling body is given by
a = 32 - I r' 2 , and r = 0, s = 0, when / = 0. Find r and ,s.
A nx. r = 8 tanh 4 /, ,s = 2 In cosh 4 /.
HYPERBOLIC FUNCTIONS
435
221. The gudermannian. The function arc tan (sinh v), which occurs
frequently in mathematics (for example, in (1), Illustrative Example,
Art. 219), is called the gudermannian* of i. The symbol used is
gd r (read " gudermannian of v "). Thus
(1) gd r = arc tan (sinh r).
The derivative is
XXXIX 4- gd
. dv
sech v >
dx
assuming r to be a function of jr.
Proof. Differentiating (1), we get-
d
dv
cosh r
But
and
1 + sinh-' v
1 + sinh- r = cosh- v,
1
Then
cosh v
d
sech r.
dx
gd v sech r
dr
By XXII and XXVII
by (B)
By (1), Art. 211
By (A), Art. 38
From the definition (1 ) and Art. 77, we have
(2) gd(0)=0;gd (-r)=-z<lt>; gd ( f oo )= .1 TT ; gd (-00)=-- \ TT.
When r increases, gd /' increases (since sech r> 0).
Its value lies between J TT and + i K-
values are given in the accompanying table.
By (1), Art. 219,
(40) I cech vdv = gdv+ C.
To find the inverre function (Art. 39) let
(3) = arc tan (sinh v), ( i TT ^ <t> <
and solve for r. The result is
(4) v = sinh" l (tan </>).
From (3) we have tan (p sinh ?/. Since cosh 2 v = 1 + sinh' J w,
by (B), the trigonometric functions of <f>, when v > 0, can be read
off from the accompanying right triangle.
Thus sin </> = tanh v, cos = sech v, etc.
The inverse function j (4) may be written ^^^
?; = In (sec
7T)
V
gd v
0.5
0.480
1.0
0.864
1.5
1.132
2.0
1 .302
2.5
1.407
3.0
1.471
3.5
1.510
4.0
1.534
4.5
1.549
5.0
1.557
sinh v
(5)
+ tan 0).
* Named after the mathematician Gudermann. His papers were published in 1830.
t The symbol ^d '0 is used by some writers (r tfd '</>).
436
DIFFERENTIAL AND INTEGRAL CALCULUS
Proof. Replace x in (f), Art. 216, by tan <, and note that
1 -f tan 2 4> is equal to sec 2 0, by (2), p. 2.
Conversely, given (5), then = gd ?;.
Proof. Changing to exponentials, (5) becomes
sec + tan (/> = e r , or tan cf> e r = sec </>.
Squaring both members, substituting sec 2 </> = ! + tan 2 0, and
reducing, the result is
- 2 tan </> ^' + e 2 " = 1.
Solving for tan </>, we get
/>'' /> ~ '*
tan </> = - = sinh v.
Hence
2
(j> arc tan (sinh v) gd v.
By W)
Q.E.D.
ILLUSTRATIVE EXAMPLE. In the
tractrix let
a length of the tangent PT
(constant by definition) ;
/ - intercept of the tangent on
the x-axis ;
= angle between the tangent
line directed upward and the //-axis;
0-0 when / = 0.
Then #((), a) is on the curve.
To prove
(6) <f> = f
T'X
Proof. When / is given, </> is determined. Hence is a function of t. Let the
values of t and for the tangent line at Q t a point near P, be, respectively,
f 4- A/(=O7 V ) and </> 4 A</>. Draw 77 r perpendicular to QT 1 '. Let the tangent
lines at P and Q intersect at 8. Then in the right triangles UTT and STU we
have TV - !Tr cos LTr ; TV - 7\S sin TSU.
Therefore TT' cos [ T T7 1; = TS sin 7VS7 T .
But angle ITT 1 ' = </> + A</>, angle TSV = A0, rr = A/. Hence
A/ cos (0 4 A0) = rs sin A0.
Let (? move along the curve towards r, and let Ac/> 0. Then A/ and A</> are in-
finitesimals. Also S approaches P and TS . Hence, by the Replacement
Theorem, Art. 98, and (5), Art. 68, we have
dt cos </> = a d<f>, or cf - = sec d> dd>
a ^ v
Integrating, and remembering that 4> = when / = 0, we get
- In (sec </> -f tan </>).
Therefore, by (5),
Q.E.D.
HYPERBOLIC FUNCTIONS
437
PROBLEMS
1. The figure shows the circle .r 2 -f //- 1 and equilateral hyperbola
x 2 y' 1 = 1 in the first quadrant. From M, the foot of the ordinate MP
of any point P on the hyperbola, draw AIT
tangent to the circle. Let v = area of the hyper- Y>
bolic sector OAP (Art. 215), and ^ = angle AOT.
Prove </> = gd v.
2. Prove
(a) gd v = 2 arc tan e v TT;
(b) I sinh r tanh v dr = sinh v gd r -f C.
3. Draw the graph of y = gd .r. Calcu-
late y and y' when x = 1. See figure.
Am. # = 0.8ti, ?/-0.65.
4. In the Illustrative Example, p. 43f>, if
P is (x, #), prove that
x = a sin 0, ?/ = a cos </>.
From these and (6) derive the parametric equations
x = f a tanh -* ?/ = a sech -
a ' a
for the tractrix. Find the rectangular equation also.
5. Derive f sech 3 v dv | sech v tanh v+Jgdr-fC.
6. If the length of the tangent of a curve (Art. 43) is constant ( a),
/ x dy
(a) prove -r 1 = -- .
dx Va 2 -
y*
(b) Integrate by the hyperbolic substitution y = a sech - and the con-
dition x = when / = 0, and in this way derive the equations of the
tractrix in Problem 4.
- 7. Evaluate each of the following by differentiation.
gd :r x
K
r->0
(b) lim
X--.0
gd JT sin x
(a) - i; (b) ^.
8. Using the expansion of Problem 14, Art. 215, we have
gd x = x - \ x 3 -f & x* - sU?) x 7 -f -.
Calculate the value of gd 0.5 to four places of decimals. Am. 0.4804.
9. The equation (5), p. 435, may be written
v = In tan (\ TT + J (/>).
Prove this statement, making use of (2), (4), and (5), pp. 2, 3,
438
DIFFERENTIAL AND INTEGRAL CALCULUS
A r North Pole
222. Mercator's Chart. The figure shows a portion (one eighth)
of a sphere representing the earth. North Pole N, equator EF,
longitude 61 and latitude <t>\ of the point P\ are indicated. Q, with
longitude 0j + A0, latitude
0j + A</>, is a second point,
near P\ on the curve P\QV.
The meridians and parallels
through PI and Q are shown.
They form the quadrilateral
PiSQR. We seek expressions
for the circular arcs RQ and 7'j R.
Since O is the center of the
equal arcs RQ and 7'jtf, each
with central angle Ac/>, we have
(1 ) arc 7^Q = arc PiS = a Ac/>.
Also, r is the center of arc PiR, ^
with central angle A6. Hence
arc 7'i/t = CI'i A0. But, in the right triangle OPjT (right angle
at ('), ^P\ = a cos (/>,. Hence
arc 7' 1 7? = a cos 0i A0.
The line 7Ve' is tangent at 7^ to the parallel PiK. The line l\T
is tangent* at V\ to the curve 7'iQV. The angle at P\ between the
curve and the parallel is the angle R'P\ T. Then
(2)
tan
i T = sec fa -^
the value of the derivative being found from the equation
(3) 0=/0
satisfied by the latitude and longitude of each point of the curve
PiQV.
Proof of (2). By the Replacement Theorem, Art. 98, it can be
shown f that
(4) tan R'P 1 T = lim arc KQ
A0-o arc P\R
Substituting the values from (1), we get (2).
* Defined as in Art. 28 as the limiting position of the secant through Pi and Q when
Q approaches P\ along the curve P\Q.
t The details are indicated in the Additional Problems (p. 143) Note that arc RQ and
arc P\R are, respectively, opposite and adjacent to the angle at Pi in the curvilinear
triangle PiRQ.
HYPERBOLIC FUNCTIONS 439
On Mercator's* Chart of the earth's surface the point with
latitude <t>, longitude 6, is represented by the point (x, y) such that
(5) x = 0, y = In (sec </> + tan 0),
or, inversely,
(6) = ar t = gdy. By Art. 221
In (5) and (6), 6 and 4> are expressed in radians. Meridians (0 = con-
stant) are represented on the chart by lines parallel to the ?/-axis,
parallels (<f> = constant) by lines parallel to the j-axis. The curve
(3) is given by the parametric equations
(7) x = A(/>), y = In (sec c/> -f tan c/>).
Theorem. The angle between a curve on the sphere and an intersect-
ing parallel is unchanged by mapping.
Proof. Let (x\, //i) be the point on (7) where c/> = c/>i. The parallel
becomes the line y ~ y\ on the chart. Hence we have to prove that
the curve (7) is such that
From (7) and (3), we get
(\H , dx f , , . , (W
^sec</>, _=/(*):,_.
Then (8) follows from (4), Art, 81, and (C), Art, 39. Q.E.D.
Two important corollaries follow.
Cor. /. The angle at 7^ on the sphere formed by two curves,
P\QR and P\Q'R r , will equal the angle on the chart at (x\, y\)
formed by the corresponding curves. Hence angles remain un-
changed by mapping.
Cor. 77. A straight line on the chart with slope tan a corresponds
to a curve on the sphere cutting all parallels under the same angle a.
This curve is called a rhumb line (or loxodrome).
Along a rhumb line
(9) < = gd(0tan + &).
This follows from (6) and y = x tan a + b. The course of a ship
proceeding always in the same direction lies along a rhumb line. In
the representation (5), 0, and therefore x, has values from -K
to+ TT, inclusive. On the other hand,?/ may have any value (Art. 221).
Hence the entire surface of the earth is mapped on the strip of the
r?/-plane determined by the lines x = TT and x = + IT.
* Grrardus M creator fl 512-1 594), a noted cartographer, published his Chart of tho
World in lf>69 His name is the Latinized form of Gerhard Kremer.
440
DIFFERENTIAL AND INTEGRAL CALCULUS
By the table of Art. 221 we may find the latitude in degrees of the parallels
which are given on the chart by the lines y = constant.
y 0.5 1.0 1.5 2 3
lat. 0" 27 31' 49 36' 64" 51' 74" 35' 84 18'
A rhumb line is represented on the map by a series
of parallel line segments such as AA\, BB\ t CTi,
etc. in the figure, where BA\ t CBi, etc. are parallel
to the r-axis. The representation is "conformal " ;
that is, the K/M/K? of small areas is preserved. This
4
87 54'
y,
5
89 C 14'
Ci
follows from Cor. 1. For example, a triangular ^
O ^^
B\
figure * on the earth's surface bounded by rhumb , ^
lines will be a triangle on the map, and corre-
^
^*^
* X
A
sponding angles in the two figures will be equal. B
But the distortion of tin area on the earth's sur-
face by mapping depends upon its distance from
the equator. Problem 4, p. 44U, brings out this A
point.
Y
^
Al
223. Relations between trigonometric and hyperbolic functions. Let
the exponent v of the exponential function e 1 ' be a complex number
x -f /// (JT and // real numbers; / V l). Then we assume as a
definition
(1 ) c j * <" = (>*(>' = (''(cos y + i sin y).
If x = 0, we have (sec p. 391)
(2) e'" cos '// + i sin y.
Change y to y. Then (2) becomes
(3) (>-'" = cos // / sin ?/.
Solving (2) and (3) for sin y and cos y, the results are
c n < - c '"
(4)
sin y =
cos y
z / " 2
Thus the sine and cosine of a real variable are expressed in terms
of exponential functions with imaginary exponents.
Formulas (4) and (A) suggest definitions of the functions con-
cerned when the variable is any complex number z. These definitions
are
(>'' ** (>1Z -4_ W
(5) sin z = ' cos z =
sinh z =
cosh z
* The lines .r = - TT and j- = -f TT re-present the same meridian (180 W. or 180 E.).
It is assumed that this meridian does not cross the curvilinear triangle. In the figure, A\
and B represent the same point on the earth, us do also B\ and C.
HYPERBOLIC FUNCTIONS 441
The other trigonometric and hyperbolic functions of z are defined
by the same ratios as are used when the variable is a real number.
From (5) we may prove the following :
(L) sinh iz = i sin 2, cosh iz cos z.
[fit _ e it 1
sinh iz = = i sin z, using (5) ; etc. I
From (L), by division, we get
(6) tanh iz = i tan z.
The similarity of many formulas in this chapter to others for
trigonometric functions is explained by the relations (L) and (6)
(see Illustrative Example 2). The right -hand members of (f>) are
expressible as complex numbers whose real parts involve only trigo-
nometric and hyperbolic functions of real variables. This appears
below in Illustrative Example 1.
ILLUSTRATIVE EXAMPLE 1. Derive the formula
(7) sinh (jc + iy) = sinh x cos // -f i cosh x sin y.
Solution. By (5), if z - x + iy, we have
c * * "/ _ c r iv
(8) sinh (x + iy} = TJ
(9) = ^(cos y + i sin ?y) - f-*(ros y - i sin y) ^ By (1 )
By (1), Art. 210, if v = x, we have
c r cosh x + sinh x, c~ T = cosh x - sinh x.
Substitute these values in (9), and reduce. The result is (7).
Changing i to ?, (7) becomes
sinh (x iy} sinh x cos // i cosh j- sin y.
' The form of the right-hand member here and in (7) should be noticed.
ILLUSTRATIVE EXAMPLE 2. Prove directly by (5) the relations
sin 2 z + cos 2 2=1, cosh 2 z sinh-' 2=1.
, Solution. The details are the same as in the proof of (#), Art. 210.
The first relation may be derived from the second as follows:
Let z- iv. Then cosh 2 iv - sinh 2 iv = 1. But, by (L), cosh iv - cos r,
sinh iv i sin v. Hence cos 2 v -f- sin 2 v 1.
PROBLEMS
1. Using differentials, show that the distance apart on M creator's Chart
of the lines parallel to the z-axis which represent the parallels at latitudes
</>i and <i -f- Ac/>, respectively, varies as sec c/>i.
2. Along a rhumb line = gd (6 tan a + b). Prove by differentiation
j d<t>
that tan a = sec cp -f-
du
442 DEFERENTIAL AND INTEGRAL CALCULUS
3. The altitude h of the zone on the sphere bounded by the parallels
</> = fa, c/> = 0i (fa > <t>}) is " ( iri fa - sin fa) (see figure, p. 438). If
= ?/2 ,' = ?/, are "the corresponding parallels on the map, prove the
following.
(a) h aCtanh 7/2 tanh ?/i ) ;
i
(b) (li/ ~~ sec 2 </>i d//, if 0;> 0i 4~ cz0.
4. Using (b), Problem U, show that equal zones of small altitude whose
lower bases are parallels at latitudes 0, 30", 45', 60, respectively, map
into rectangles whose areas are as 3:4:6: 12. (The area of a zone equals
its altitude times the circumference of a great circle.)
5. Describe the direction of a curve on the sphere (a) if - ;
(V>) if becomes infinite.
dO
6. Derive each of the following formulas by the method used in Il-
lustrative Example 1.
(a) cosh (jr -f /'//) cosh jr cos y 4- ? sinh x sin y ;
(b) sin (x 4 ?'//) = sin x cosh ?/ -f / cos x sinh y ;
(c) cos (x -f /'//) = cos x cosh ?/ - / sin x sinh ?/.
From these write the values of cosh (x - ?'//), sin (x - ??/), cos (x - iy).
7. Prove (a) sinh ( ? J .r j = ? cosh x ;
(b) cosh ( i ~ x } = / sinh x,
8. Evaluate each of the following to two places of decimals.
(a) sinh (1.5 -f /) ; (b) cosh (1 - ?);
(c) cos (0.8 -f 0.5 /') ; (d) sin (0.5 + 0.8 0-
ANK. (a) 1.15 -h 1.98 i ; (c) 0.78 - 0.37 i.
ADDITIONAL PROBLEMS
1. In the figure of Art. 2l?2, P\M\ is drawn perpendicular to CR,
and therefore perpendicular to the plane of the meridian NQR. Then
triangle PiQMi is a right triangle (the chord PiQ is not shown), and
tan MiPiQ = -^^.- When M -> 0, the line Pi A/i (produced) approaches
Pi Mi
the tangent PiR', and angle MjPiQ approaches angle R'PiT. Therefore
(10) tan / Pir = lim^^-
v '
HYPERBOLIC FUNCTIONS
Compare with (4), Art. 222, and show that
(a) lim Pl **\ = I (see Fig. 1) ;
Afl-oarc P\It
(h) lim Afl( ^ = 1 (see Fig. 2, which shows
v . oarc KQ
the plane of the meridian
In triangle .V
show that Af i K is an infinitesimal of
higher order than QR when A6 and JK/> are of
the same order (Art. 99;. Then see Problem,
p. 148.
Using fa) and (h) and the Replacement Theo-
rem, Art. 98, (10) becomes (4), Art. 222.
2. If r/.s'i is the element of the length of arc for
a curve on the sphere of Art. 222, prove that
dx } ' 2 = a'-((l<f>' 2 -4- cos-' <W 2 ). (In the figure of
Art. 222, (chord I\Q} 2 = Pi MI" +
y j andlirn c - >nord/>l( ?^l.)
arc PiQ
3. If dx is the differential of the arc of a curve on M creator's Chart,
show that r/.s- sec 2 0u70 L ' -f cos 2 <10~). (Comparing with Problem 2,
we have dx\~ = a' 2 cos 2 (/.s- 2 .)
4. Find the length of a rhumb line between points whose difference
of latitude is A0. Ann. a esc a A0. (a = radius of the earth.)
5. Prove that the first four formulas in (4), p. 3, and (>), (), Art. 213,
hold when JT, y, r, /r are replaced by complex numbers. (Use the defi-
nitions (5).)
6. Prove the formulas of Problem 6, p. 442, by using the results in
Additional Problem 5 and (L).
~ n <^u ^ 4. u / i \ sm h 2 J ' + ' sm 2 ?/
7. Prove that tanh (.r 4- /?/) = : : ~*- m
cosh 2 r -I- cos 2 ?/
8. I>erive the formula for tan (.r -f- it/) from the result in the preceding
problem.
CHAPTER XXIII
PARTIAL DIFFERENTIATION
224. Functions of several variables. Continuity. The preceding
chapters have been devoted to applications of the calculus to func-
lions of one variable. We now turn to functions of more than one
independent variable. Simple examples of such functions are af-
forded by formulas from elementary mathematics. Thus, in the
relation for the volume v of a right circular cylinder,
(1 ) v = irjr'-y,
v is a function of the two independent variables x ( radius) and
?/ ( altitude). Again, in the formula for the area u of an oblique
plane triangle,
("2) u = ], yy sin otj
u is a function of the three independent variables r, ?/, and a, rep-
resenting, respectively, two sides and the included angle.
Obviously, in (1), as well as in (2), the values which can be as-
signed to the variables in the right-hand member are entirely in-
dependent of one another.
The relation
('*> z = f(f,y)
can be represented graphically by a surface, the locus of the equation
(I}) obtained by interpreting j, //, z as rectangular coordinates, as in
solid analytic geometry. This surface is the graph of the function
of two variables, f(jr, ?/).
A function /(.r, ;//) of two independent variables x and y is defined
as continuous for .r = a, y ~ b, when
04) lim/(x,y)=/(a, b),
X -*
*- >>
no matter in what way x and y approach their respective limits a
and b.
This definition is sometimes roughly summed up in the statement
that o very small change in one or both of the variables produces a
very small change in the value of the function.*
* This will be hotter understood if the student again reads over Art. 17 on continuous
functions of a single variable.
444
PARTIAL DIFFERENTIATION 445
We may illustrate this geometrically by considering the surface
represented by the equation
(3) z=f(x,y^.
Consider a fixed point P on the surface where :r = a and y b.
Denote by A.r and A// the increments of the variables x and
//, and by Ac the corresponding increment of the function z, the
coordinates of P' being
At P the value of the function is
z=f(a, b) = MP.
If the function is continuous at P, then how-
ever A.r and A?/ may approach zero as a limit , A::
will also approach zero as a limit. That is, 717 'P' will approach coin-
cidence with MP, the point P' approaching the point /' on the sur-
face from any direction whatever.
A similar definition holds for a continuous function of more than
two variables.
In what follows, only values of the variables are considered for
which a function is continuous.
225. Partial derivatives. In the relation
(1) z=/Cr, ?/),
we may hold y fast and let x alone vary. Then z becomes a function
of one variable x, and we may form its derivative in the usual manner.
The notation is
= partial derivative of z with respect to x (y remains constant).*
- c)x
Similarly,
= partial derivative of z with respect to y (x remains constant).*
c)y
Corresponding symbols are used for partial derivatives of functions
of three or more variables.
In order to avoid confusion the round b \ has been generally
adopted to indicate partial differentiation.
* The constant values are substituted in the function before differentiating.
t Introduced by Jacobi (1804-1851).
446
DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 1 . Find the partial derivativas of z = ax 2 + 2 bxy + cy 2 .
Solution. Y 1 = 2 ax -f 2 fr#, treating # as a constant,
i)"
-r- - 2 ore -f 2 n/, treating x as a constant.
ILLUSTRATIVE EXAMPLE 2. Find the partial derivatives of u sin (ax + by + cz).
Solution. ~ ~ a cos (ax -f by 4- re), treating # and z as constants,
~ fr cos (ax 4 by + cz), treating x and z as constants,
-f by -f r^), treating # and x as constants.
~ _ c cos
Referring to (I), we have, in the notations commonly used,
~ = T^/(y, y) = -^ =/ r (j, ?/) =/, = z x ;
Similar notations are used for functions of any number of variables.
Referring to Art. 24, we shall have
(2)
(3)
*) = Mm
AJ- .
/ ; ,(.r (l , y) = lirn ^^-
Aj;
226. Partial derivatives interpreted geometrically. Let the equation
of the surface shown in the figure be
z= /Or, y).
Pass a plane EFGI1 through the
point /' (where x ~ a and y ft) on
the surface parallel to the XOZ-
plane. Since the equation of this
plane is ,
y^b*
the equation of the curve of inter-
section JPK with the surface is
2=/Cr, b\
if we consider EF as the axis of Z and EH as the axis of X. In this
U ( ] :
plane TT- means the same as -r 1 ' and we have
fa dx
(1) |5 = tan MTP = slope of curve of intersection JX at P.
ox
PARTIAL DIFFERENTIATION 447
Similarly, if we pass the plane BCD through P parallel to the
YOZ-plane, its equation is j a
and for the curve of intersection DPI, means the same as j^- Hence
(2) ~ = tan MT'P slope of curve of intersection DI at P.
ILLUSTRATIVE EXAMPLE. Given the ellipsoid ~- 4 - ~- 4- ~ 1 ; find the slope
LI4 1 !_ (>
of the curve of intersection of the ellipsoid made (a > by the plane // 1 at the point
where x = 4 and z is positive ; (b) by the plane j = 2 at the point where j/ = H and
z is positive.
Solution. Considering y as constant,
24 6 to ~~ ' to ~ 4 z
?/ 4> "* ()~ ft" y
When J is constant, ~f + -rr 1 = 0, or - - - -f-
12 () c^/ _ f).v 2 ^
(a) When ?/ = 1 and r 4, 2; = x/h" * T. ~
1 <*)' *J
(b) When x - 2 and y = ,'J, 2 = = /. ~ - - - \ 2. A/ts.
PROBLEMS
Work out the following partial derivatives.
2. /(j, ?
a -f 2 to?/ -f
2 -f 2 f:jr// -f
_ A JT + /?// x ^ - M/>- /^') ? y
3. /(JT, ?y) - AMS - ": - -- 2 '
4. 7 < = /?/ -f z/z 4- ZJT. An.s. M T - // 4- z ; w v = s 4- 2 ; n, ~jr + y.
5. /(j, ?/) = Cr 4- ?y) sin fx - ?/).
Ans. //JT, i/) - sin (r - ?/) + (r + ?/) cos (x - ?/) ;
/ y (j- t T/) = sin (x - y) ~ (* 4- y) cos (r y).
6. p = sin 2 cos 3 </>. An. |^ = 2 cos 2 cos 3 ^ ;
^ = 3 sin 2 6 sin 3 (/>.
7< p = 8++ cos (j _ ^). Aws. ^ = e++ fcos (8-<j>)- sin (fl - </>)} ;
^P^ e 4c/>r cog ^_ 0) + sin (^_ 0^| e
448 DIFFERENTIAL AND INTEGRAL CALCULUS
Find the partial derivatives of the following functions.
8. f(s, y) = 3 jr 4 - 4 s*y + 6 jr*tf. 11. f(s, y) = (x + 2y) tan (2 x + I/).
Q y + ^ ?/
-~0 + 2/ 12. p = tan 2 (9 ctn 4 0.
10. z = fMn 2. 13. p = < cos^-
^"
14. If /(V, ?/) = 2 .r* - 3 jry -f 4 ?/*, show that / z (2, 3) = - l,/ v (2, 3) = 18.
15. If/O, //) = r ~j-show that/, (3, 1) = - i,/,,(3, lj = jj.
16. If/(r, //) = <' 'sin (JT + 2 ?/), show that /,((>, T W ~ 1, ^0, -W 0.
\ 4/ \ 4/
17. If w = ^1 r 4 -f 2 Rr 1 '// 2 -f ry , show that jr -f ?/ = 4 w.
r.r ' f'?/
18. If M = ~ ; ^-- , show that jr -f ?/ = 3 ?/.
^ 4- // r'x ' r'/y
19. If u - jr*u -f ifz -f- rV, show that -f + = (jr -f y + z) 2 .
fj- cy (Z
20. If u = Ajr l + 'ty*' ^ow that .r -f ?/ = (w - 2)w.
TJ- -f />//- rx r -, ;
21. The area of a triangle is given by the formula K o be sin /4.
Given b 10 in., r - 20 in., .4 = GO".
(a) Find the area.
(b) Find the rate of change of the area with respect to the side b if r
and A remain constant.
(c) Find the rate of change of the area with respect to the angle .4 if
/> and c remain constant.
(d) Using the rate found in (c), calculate approximately the change in
area if the angle is increased by one degree.
(e) Find the rate of change of c with respect to b if the area and the
angle remain constant.
22. The law of cosines for any triangle is a 2 = b 2 -f c 2 - 2 be cos A.
Given b = 10 in., c = 15 in., A = GO .
(a) Find a.
(b) Find the rate of change of a with respect to b if c and A remain
constant.
(c) Using the rate found in (b), calculate approximately the change in
a if b is decreased by 1 in.
(d) Find the rate of change of a with respect to A if b and r remain
constant.
(e) Find the rate of change of c with respect to A if a and b remain
constant.
PARTIAL DIFFERENTIATION 449
227. The total differential. We have already considered the dif-
ferential of a function of one variable in Art. 91. Thus, if
=/(*),
we defined and proved
(1) cfy=
We shall next consider a function of two variables. Consider the
function
(2) w=/(j, //).
Let Ax and A?/ be the increments of x and // respectively, and let
Aw be the corresponding increment of u. Then
(3) Aw = /(x + Ax, // + A//) - f(jr, //)
is called the total increment of u.
Adding and subtracting /U, y -f A-//) in the second member,
(4) Aw = f f(x + Ax, '// -f A-//) ~/{x, // + A//) ]
-H/Or, '// + A//) -/U, //)|.
Applying the Theorem of Mean Value (Z)), Art. 116, to each of
the two differences on the right-hand side of (4), we get, for the
first difference,
(5) /(x + Ax, y + A?/) - J(x, y + Ay/) = /, (x + 0, Ax, y/ + Ay/) Ax.
|fj = x, Aa ~ A/, arid since* j varies while n f A// remains!
'constant, \vt' ^ct the partial derivative \vith resj)ecl toj.l
For the second difference,
(6) /(x, v + Ay/) - /(x, //) - / /y (x, // + 6. A//) A?/.
I = //, A^/ = A?/, and since // varies while .r reinuins con- J
stant, we got the partial derivative with respect to //-I
Substituting from (5) and (6) in (4) gives
(7) A^ - f,(r + 0i Ax, y + Ay/) Ax + /,,(x, y + 0- A?/) A?/,
where 6\ and 02 are positive proper fractions.
Since J,(x, y) and f u (x, y) are continuous functions of x and y, the
coefficients of Ax and A?/ in (7) will approach f r (x 9 y) and f u (x, y),
respectively, as limits when Ax and &y approach zero as common
limits. Hence if e and f are infinitesimals such that
lim 6 = 0, lim e' = 0,
AJ- - A j -.
ly A?y -
we may write
(8) /,(z + 0i Ax, ?/ + Ay) - / x (x, |/) + ,
(9) f v (x, y + 6<2 A?/) = f y (x, y) + e',
450 DIFFERENTIAL AND INTEGRAL CALCULUS
and (1) will become
(10) &u=f f (x, y)kx+Sy(x, y)&y + *x + 'A?y.
We then define as the total differential (= du) of u
(11) <fa =/,(*, y)&x+f y (x, y)y.
The right-hand member in (11 ) is the "principal part" of the right-
hand member of (10), that is, du is a close approximate value of \u
for small values of \s and A// (compare Art. 92). Obviously, if
u = x, (11 ) becomes rfj = Ax. If w = //, (11 ) becomes dy = A//.
Substituting, then, in (11) for AJ and A// their corresponding dif-
ferentials, we obtain the important formula
(B) du - /,(*, y) dx + /(*, y) dy - j^ dx + ^ dy - ^ dx + ^ dy,
which should be compared with (1) at the beginning of this article.
If u is a function of three variables, its total differential is
(C) dv^dx + ^dy+'^dzi
(x (y (i
and so on for any number of variables.
A geometric interpretation of (B) is given in Art. 238.
ILLUSTRATIVE EXAMPLE 1. Compute An and du for the function
(12) n = 2 J-- -4 3 /r,
when x= 10, /y ~ 8, A>r = 0.2, Ji/y = 0.3, and compare the results.
Solution. Substitute in (12) for j-, //, /<, respectively, x + AJ*, // + A/y, M + A?J,
and proceed as below (compare Art. 27).
u + A// - 2(.r -r- A.D- 4- 3(/y -f A?y) J
^ 2 jr* -f 3 /y-' 4- 4 J- A.r -f 6 yy A/y 4- 2(A.r)^' + 3(A?y) 2 .
M = 2 .r- + 3 /y j
(13) AM = 4 j- A.r 4 G // A/y -f 2( A.r)"' + 3(A#) 2 .
Differentiating (12), we find
(111 . ()l(
= 4 .r, =6 y.
dx dy
Substituting in (B), the result is
(14) du = 4 x dx 4- 6 /y J/y.
Remembering that AJ = dr, A?y = <///, we see that the right-hand member in
(14) is the "principal part" of the right-hand member in (13), for the additional
terms are of the second degree in AJ* or A?y. This statement illustrates (10) and
(11) above (namely, e = 2 A.r, ' = 3 A/y).
Substituting the given values in (13) and (14), we get
(15) AM = 84- 14.4 -f 0.08 4- 0.27 = 22.75 ;
(16) du = 8 + 14.4 =22.4.
Then AM - du = 0.35 = 1.6% of AM. Ans.
PARTIAL DIFFERENTIATION 451
ILLUSTRATIVE EXAMPLE 2. Given u - arc tan ^, find du.
x
e i * ^" V <"'* .r
Solution.
(V .r* -f ir dij i' 2 +
Substituting in (5), (/ ^ ( ^ ~ ^. (J .4ns.
PROBLEMS
Find the total differential of each of the following functions.
1. z = 2 JL' :{ - 4 .n/-' + 3 // :l . ^ N,S-. d; - (fi .r-' - 4 ij-)djr -f (9 // 2 - 8 .r?/ )<///.
((V-f- />//)'-'
3. u = jryW. dn = ir^ djc 4 2 .r//c- ! dy -f .S .r//'-',: 1 -' (/c.
4. u .r- cos ^ //. 5. 6> = arc tan - 6. n (.r - //) In (.r -f //).
7. If ^ 4- >r + ^ - a^, show that d; - - J ' <*' +
8. Find ^ if 4 j-' - 9 //-' ~ 16 c j = 100.
9. Compute A// and </// for the function u .r- 3 .r// -f 12 // L> when
.r = 2, y = - 3, A.r = - ().:>>, A// =r 0.2. An*. A// = - 7.15, du = - 7.5.
10. Compute dn for the function // (j- -f //) V.r // when .r (J, // = 2,
J.r ~ \, dij - A. /1//X. 1.
11. Compute A// and r/w for the function M j*/y -f 2 r - 4 ?/ when
x - 2, t/ = 3, A.r = 0.4, A?/ = - 0.2.
12. Compute <7p for the function p = c* sin (0 </>) when 0,
<t> = J TT, cZO = 0.2, c/0 = - 0.2.
228. Approximation of the total increment. Small errors. For-
mulas (B) and (C) are used to calculate Au approximately. Also,
when the values of x and y are determined by measurement or ex-
periment, and hence subject to small errors AJ and A?/, a close ap-
proximation to the error in u can be found by (B). (Compare
Arts. 92, 93.)
ILLUSTRATIVE EXAMPLE 1 . Find, approximately, the volume of tin in a thin
cylindrical can without a top if the inside diameter and height are, respectively,
6 in. and 8 in., and the thickness is 1 in.
Solution. The volume v of a solid right circular cylinder with diameter x and
height y is
(1) v = \ vx*y.
Obviously, the exact volume of the can is the difference Ar between the volumes of
two solid cylinders for which x = 6J, y - 8, and x = 6, y = 8, respectively. Since
only an approximate value is required, we calculate dv instead of Av.
452 DIFFERENTIAL AND INTEGRAL CALCULUS
Differentiating (1), and using (B), we get
(2) df = \ Trxy dx + | irx' 2 dy.
Substituting in (2) x = 6, y = 8, dx = \, dy = 1, the result is
dv 7i TT = 22.4 cu. in. Ans.
The exact value is At? - 23.1 cu. in.
ILLUSTRATIVE KXAMI'LE 2. Two sides of an oblique plane triangle measured,
respectively, 63 ft. and 78 ft., arid the included angle measured 60'. These meas-
urements were subject to errors whose maximum values are 0.1 ft. in each length
and 1 in the angle. Find the approximate maximum error and the percentage
error made in calculating the 1 third side from these measurements.
Solution. Using the law of cosines ((7), Art. 2),
(3) it- = x 2 + y' 2 - 2 xy cos ,
where x, ij are the given sides, a the included angle, and u the third side. The given
data are
(4) r = 63, ?/-78, = 60 =, dx = d/y = 0.1, da = 0.01745 (radian).
o
Differentiating (3), we get
J ~~ y (l()s c * Of* y ~ x (><)s (y ^ u ~~ *y s ^ n a
dx H dy n Oa >'
Hence, using (C),
, (.r // '_cos (\')(1x -} (// .r (*os <v)dij -\- xij sin cy dcx
Substituting the values from (4), we find
/ 2.4 +4.65 + 74.25 , 10 . .
du ~ - = 1.13 ft. Aiis.
< 1. 1
The percentage error = 100 =. 1.6'\ . An*.
H
PROBLEMS
1. The legs of a right triangle measured 6 ft. and 8 ft. respectively,
with maximum errors in each of 0.1 ft. Find approximately the maximum
error and percentage error in calculating (a) the area, (b) the hypotenuse,
from these measurements. AHX. (a) 0.7 sq. ft., 2.9 ( \ ; (b) 0.14 ft., 1.4 < ;.
2. In the preceding problem find the approximate error in calculating
the angle opposite the longer side from the given dimensions, and the
approximate maximum error in that angle in radians and degrees.
3. The radii of the bases of a frustum of a right circular cone measure
5 in. and 11 in. respectively, and the slant height measures 12 in. The
maximum error in each measurement is 0.1 in. Find the approximate
error and percentage error in calculating from these measurements (a) the
altitude; (b) the volume (see (12), Art. 1).
-4?w. (a) 0.23 in., 2.2 <; ; (b) 32 TT cu. in., 4^ '~,
PARTIAL DIFFERENTIATION 453
4. One side of a triangle measures 2000 ft., and the adjacent angles
measure 30 and 60 respectively, with a maximum error in each angle
of 30'. The maximum error in the measurement of the side is 1 ft.
Find the approximate maximum error and percentage error in calculating
from these measurements (a) the altitude on the given side; (b) the area
of the triangle. AN*, (a) 17.88 ft., 2.1 %.
5. The diameter and altitude of a right circular cylinder are found by
measurement to be 12 in. and 8 in. respectively. If there is a probable
error of 0.2 in. in each measurement, what is approximately the greatest
possible error in the computed volume? Ans. 16.8 TT cu. in.
6. The dimensions of a box are found by measurement to be 6 ft.,
8 ft., 12 ft. If there is a probable error of 0.05 ft., (a) what is approxi-
mately the greatest possible error in the computed volume? (b) What is
the percentage error? Aw,s. (a) 10.8 cu. ft. ; (b) *%.
7. Given the surface z - ^-^- If, at the point where x - 4, y = 2,
s + //
x and y are each increased by A, what is the approximate change in 2?
P
8. The specific gravity of a solid is given by the formula s where
P is the weight in a vacuum and w is the weight of an equal volume of
water. How is the computed specific gravity affected by an error of
A in weighing P and ^ in weighing w, assuming P - 8 and w = 1
in the experiment, (a) if both errors are positive? (b) if one error is
negative? (c) What is approximately the largest, percentage error?
An*, (a) 0.3; (b) 0.5; (c) 6i%-
9. The diameter and slant height of a right circular cone are found by
measurement to be 10 in. and 20 in. respectively. If there is a probable
error of 0.2 in. in each measurement, what approximately is the greatest
possible error in the computed value_of (a) the volume? (b) the curved
surface? Af ^ (a) 37 7rVl5 = ^ cu Jn . (h) 8 ^ = ^ ^ m
18
10. Two sides of a triangle are found by measurement to be 63 ft. and
78 ft. and the included angle to be 60. If there is a probable error of
0.5 ft. in measuring the sides and of 2 in measuring the angle, what is
approximately the greatest possible error in the computed value of the
area? (See (7), Art. 2.) Am. 73.6 sq. ft.
^
11. If specific gravity is determined by the formula s = _ where
A is the weight in air and W the weight in water, what is (a) approxi-
mately the largest error in s if A can be read within 0.01 Ib. and W within
0.02 Ib., the actual readings being A = 9 Ib., W - 5 Ib.? (b) the largest
relative error? An. (a) 0.0144; (b) ?$fa.
454 DIFFERENTIAL AND INTEGRAL CALCULUS
jrt
12. The resistance of a circuit was found by using the formula C =
R
where C = current and / = electromotive force. If there is an error of
1*5 ampere in reading C and 3*5 volt in reading E, (a) what is the approxi-
mate error in It if the readings are C = 15 amperes and E 110 volts?
(b) What is the percentage error? Am. (a) 0.0522 ohms; (b) || f ;.
13. If the formula sin (x -f y) = sin jr cos // -f cos x sin y were used to
calculate sin (x + y), what approximate error would result if an error of
0.1 were made in measuring both x and ?/, the measurements of the two
acute angles giving sin x j? and sin y -f 3 ? Ans. 0.0018.
14. The acceleration of a particle down an inclined plane is given by
a = g sin i. If g varies by 0.1 ft. per second per second, and ?, which is
measured as 30, may be in error 1, what is the approximate error in the
computed value of a ? Take the normal value of g to be 32 ft. per second
per second. Ans. 0.534 ft. per second per second.
15. The period of a pendulum is P 2 ir\ - (a) What is the greatest
* g
approximate error in the period if there is an error of 0.1 ft. in measur-
ing a 10-foot suspension and (/, taken as 32 ft. per second per second, may
be in error by 0.05 ft. pe*r second per second? (b) What is the percentage
error? Ann. (a) 0.0204 sec. ; (b)gj r ;.
16. The dimensions of a cone are radius of base = 4 in., altitude = 6 in.
What is the approximate error in volume and in total surface if there is a
shortage of 0.01 in. per inch in the measure used?
Ans. dV - 3.0159 cu. in. ; dS = 2.818 sq. in.
17. The length I and the period P of a simple pendulum are connected
by the equation 4 ir' 2 l P' 2 g. If / is calculated assuming P = 1 sec. and
g = 32 ft. per second per second, what is approximately the error in / if
the true values are /' = 1.02 sec. and g ~ 32.01 ft. per second per second?
What is the percentage error?
18. A solid is in the form of a cylinder capped at each end with a
hemisphere of the same radius as the cylinder. Its measured dimensions
are diameter = 8 in. and total length 20 in. What is approximately the
error in volume and surface if the tape used in measuring has stretched
uniformly r ; beyond its proper length ?
19. Assuming the characteristic equation of a perfect gas to be rp Rt,
where r = volume, p = pressure, t = absolute temperature, and R a con-
stant, what is the relation between the differentials dr, dp, dt ?
Ans. v dp + p dv = R dt.
20. Using the result in the last example as applied to air, suppose that
in a given case we have found by actual experiment that t = 300 C.,
p = 2000 Ib. per square foot, v = 14.4 cu. ft. Find the change in p, assum-
ing it to be uniform, when / changes to 301 C., and v to 14.5 cu. ft.
R = %. Ans. 7.22 Ib. per square foot.
PARTIAL DIFFERENTIATION 455
229. Total derivatives. Rates. Turn now to the case where x and y in
CD M=/(*,0)
are not independent variables. Assume, for example, that both are
functions of a third variable /, namely,
(2) r = <t>(U, ?/=M)-
When these values are substituted in (1), M becomes a function of
one variable t, and its derivative may be found in the usual manner.
We now have
(3) <f = f *. ^ = f *. dv^dt.
Formula (B) was established with the assumption that x and y
were independent variables. We may easily show that it holds also
in the present case. To this end, return to (10), Art. 227, and divide
both members by A/. Changing the notation, this may be written
(A) r - i 4. 4. /
M ox M ry A/ \ Af A/
Now when A? > 0, A:r and A// -> 0. Hence (see Art. 227)
lim e = 0, lim e' = 0.
A/ - A/ -
Therefore, when A/ > 0, (4) becomes
rfu _ fiu c?x , f^u dy
( ' dt~~ ax dt ny dt '
Multiplying both members by dt and using (3), we obtain (B).
That is, (B) holds also when x and y are Junctions of a third variable t.
In the same way, if u /(x, y, z),
and x, ?/, z are all functions of /, we get,
du __ (hd dx_ , _ ^ , f
( ' dt ~ f>x dt cJydt ' <)z dt
and so on for any number of variables.
In (D) we may suppose t x ; then y is a function of x, and u is
really a function of the one variable x, giving
djc f)x r>y dx
In the same way, from (E) we have, when y and 2 are functions
of x
' du __ f du , r;u jy . <r)udz
dx 'ox by dx f)z dx
456 DIFFERENTIAL AND INTEGRAL CALCULUS
The student should observe that and -^ have quite different
ox dx
meanings. The partial derivative is formed on the supposition
'Ox
that the particular variable x alone varies, all other variables being
held fast. But , . A
(111 i / *_a 1JL \
= lim
dx AX .0 \Ax/
where AM is the total increment of u due to changes in all the variables
caused by the change Ax in the independent variable. In contra-
distinction to partial derivatives, ( -j> ^ are called total derivatives
with respect to / and x respectively.
It should be observed that has a perfectly definite value for any
point U, y), while -, depends not only on the point (x, y) but also on
the particular direction chosen to reach that point.
ILLUSTRATIVE EXAMPLE 1. Given M sin -, JT - <', y = t' 2 ; find ~
y dt
, A . du 1 x du x x d.r , dy .
Solution. - cos -, - cos - ; = c', - - 2 /.
dx // y dy // y dt dt
Substituting in (D), (/ 2) cos Am.
at r i 2
iLLUSTKATivK'Kx AMPLE 2. Given // = f" T (y z), y = a sin x, z cos x ; find ~-
ax
Solution. <u"' J (y - z}, =- r" r , - r - = - I M * ; - a cos x, - sin x.
dx ' dy dz dx dx
Substituting in (G),
M __ ac nr(y _-)-(- (ie nr cos x -I- r T sin x = f" T (a' 2 + 1) sin x. Ans.
NOTE. In examples like the above, n could, by substitution, be found explicitly
in terms of the independent variable and then differentiated directly ; but generally
this process would be longer and in many cases could not be used at all.
Formulas (D) and (E) are useful in all applications involving
time-rates of change of functions of two or more variables. The
process is practically the same as that outlined in the rule given in
Art. 52, except that, instead of differentiating with respect to t
(Third Step), we find the partial derivatives and substitute in (D)
or (E). Let us illustrate by an example.
PARTIAL DIFFERENTIATION 457
ILLUSTRATIVE EXAMPLE 3. The altitude of a circular cone is 100 in., and
decreases at the rate of 10 in. per second; and the radius of the base is 50 in.,
and increases at the rate of 5 in. per second. At what rate is the volume
changing ?
Solution. Let x = radius of base, y altitude; then
1 , , dn 'J dn I
u - 7r.r-?/ volume, -7- = :; Trrj/, T~ ~ flM".
o ox o ()y o
Substituting in (>), -- = ^ irxy ~ 4 \ 2 ^f-
But a- = 50, y = 100, Y - 5, ^ = -10.
/. ^-r = S TT 5000 5 - TT 2500 -10 15.15 cu. ft. per second, increasing. /Ins.
at o o
230. Change of variables. If the variables in
(1) M=M 0)
are changed by the transformation
(2) r ~~~ <p(r, s), // = ^(r, 8),
the partial derivatives of u with respect to the new variables r and s
can be obtained by (Z>). For, if we hold s fast, then x and y in (2)
are functions of r only. Hence we have
(3) ^M + ^4/,
^r r-'wT ( r ( // fr
all derivatives with respect to r now being partial.
In the same way,
(ju chi 'vx . 'cu (hi
(4) == r~ ~r 7*~'
r's r/x 6\s r''// rAs
In particular, let the transformation be
(5) x = z' + /A, ;-y = ?/' + &,
the new variables being x' and //', and h, k being constants. Then
rx - ex ^ ()y __ /v ^?y i
r? ~ ' (jy f ~ ' vx f ~ ' vy' ~~
Then we obtain, from (3) and (4),
(>IL !L '
fix 'ox 1 ' vy by'
Hence the transformation (5) leaves the value of the partial derivatives
unchanged.
If the values of x and y in (5) are substituted in (1), the result is
458 DIFFERENTIAL AND INTEGRAL CALCULUS
The results in (6) may now be written
(8) f x (x, y) = FA*', y'), /(*, y) = Mr', #')
In Art. 229 it was shown that (B) is true when x and y are func-
tions of a single independent variable t. We prove now that (B)
holds also when x and y are functions of two independent variables
r, s, as in (2). For by (B), when r and s are the independent variables,
we have ^ r/:r r ^
dx dr + ds, rf?/ = -^ rfr H ** ds.
r^r fs rr r;s
Substitute these values in the expression
(9) dx + dy
ox vy
and reduce by (3) and (4). The result is
/in '(''U j i '(''M> j
(10; dr + ds.
vr els
But, by (1 ) and (2), u becomes a function of the independent vari-
ables r and s. Then, by (B), (10) equals du. Hence (9) also equals du.
Therefore (B) holds when x and y are functions of one or of two
independent variables. In the same way, it can be shown that (C)
holds when x, y, z are functions of one, two, or three independent
variables.
231. Differentiation of implicit functions. The equation
(1) M ?/) = o
defines either x or y as an implicit function of the other. It repre-
sents an equation containing x and y when all its terms have been
transposed to the first member. Let
(2) M=/Cr,0);
,u du W , ?/ dy
then -7-^7- + ^* by (F )
dx ex f y dx J \ /
and y is an arbitrary function of x. Now let y be the function of x
satisfying (1). Then u = and du = 0, and hence
(3) 4^ + ^? = 0.
f)x eydx
Solving, we get
(H) ^ = _. W^
dx fl \dy
('y
PARTIAL DIFFERENTIATION 459
Thus we have a formula for differentiating implicit functions.
This formula in the form (3) is equivalent to the process employed
in Art. 41 for differentiating implicit functions, and all the examples
on pages 40 and 41 may be solved by it.
When the equation of a curve is in the form (1), formula (H)
affords an easy way of getting the slope.
ILLUSTRATIVE EXAMPLE 1. Given j-v + sin u = find ^-
dx
Solution. Let f(x, y) = x 2 y* + sin y.
V - O 2L A , , j.
-~ L xy\ -- 4 x-y* -f cos y.
dx dy
Therefore, from (//), ~ - - - , " An*.
ILLUSTRATIVE EXAMPLE 2. If jr increases at the rate of 2 in. per second as it
passes through the value x = 3 in., at what rate must // change when // ~ 1 in., in
order that the function 2 xy' 2 - 3 j-'y shall remain constant?
Solution. Let u = 2 xy- - 3 x'-y ; then, since u remains constant, -j: 0. Sub-
stituting this value in the loft -hand member of (/>), transposing, and solving for
dy
' we K pt du
___
dt ~~ du dt
dy
Also, -y- - 2 /y ? - 6 j#, -T = 4 xy - 3 x 2 .
Now, substituting in (4), "77 ~ " 4 * -^ - 17
But x = 3, //-I, ^"=2-
dy
Therefore, -r- 2j\ in. per second. ATIS.
In like manner, the equation
(5) F(x, y,z) = Q
defines z as an implicit function of the two independent variables x
and y. To find the partial derivatives of z with respect to x and to y,
proceed as follows.
Let u=f(x,y,z).
dF dF dF
Then du = - dx + T dy + -7 dz,
ox tiy dz
by (#), and this holds no matter what the independent variables are
(Art. 230). Now let z be chosen as that function of the independent
460 DIFFERENTIAL AND INTEGRAL CALCULUS
variables x and y which satisfies (5). Then u = 0, du 0, and we have
oF dF , cF
(6) dx + dy + dz = 0.
rx <">y rz
P>ut now dz = -r^ dx + -^ rfy. Ey (5)
Substituting this value in (6) and simplifying gives
(vF (F f)z\ I'cF cF rz\
( + }dx + ( + -- }dy - 0.
\r>x vz fx/ \f)y (Z ryj
Here dx (= Ax) and dy (= A?/) are independent increments. We
may therefore set dy = 0, dx ^ 0, divide through by dx, and solve
()Z
for T The result is
( z
Proceeding in a similar manner, we may prove
(J) = _-
( y f_l
( Z
Formulas (7) and (7) are to be interpreted as follows: In the
left-hand members z is the function of x and y satisfying (5). In
the right-hand members F is the function of three variables, x, y, z,
given in the left-hand member of (5).
The generalization of (//), (/), (J) to an implicit function u of
any number of variables is now obvious.
ILLUSTRATIVE EXAMPLE. By the (HI nation
S + 12 + *?- 1 = -
z is defined as an implicit function of .r and ?/. Find the partial derivatives of this
function.
Solution. F = ^ l + -\.
dF J dF y dF z
TlPTIfP _ __ . _ _
00 r'ir~12' fy~6' ~d~z ~ 3
Substituting in (7) and (7), we get
dz .r dz y .
a- = -T- ?-=-?r~- ^ws-
ar 4 z dy 2 z
(Compare with the Illustrative Example in Art. 226.)
PARTIAL DIFFERENTIATION 461
PROBLEMS
In Problems 1-5 find
at
1. u = x 2 3 xy -f 2 ?/ 2 ; a* = cos /, y = sin f.
Aws. = sin 2 f - 3 cos 2 /.
at
2.u=x + 4 VJv - 3 // ; .r = f 3 , u = \- ~ = 3 / 2 + 4 4- 75-
/ (// / J
3. K = f * sin y + < v{/ sin j- ; .r = J f, ?/ = 2 /.
Ans. = r J '( > sin 2/4-2 cos 2 /) 4- f "(2 sin J / + j cos J /)
4. w = 2 .r 2 j-?/ -f y' 2 ; j- = cos 2t,y sin /.
5. ?i = xy 4- ?/- -f ^ ; j- = -?/ = r', 2 = f "'.
In Problems 6-10 find ( -f-' by formula (/f).
a.r
6. Ajr 2 -f 2 7ij-?/ 4- Cy 2 -f 2 /Xr -f 2 /fy -h ^ = 0.
-f T// 4-
o r w i
8. e* sin // f j/ cos x 1.
au' r" cos JT r j cos //
9. x 4 - .ry - s' 2 4- 2 ?/ 2 = 8.
10. Ax 4 -f 2 Rr 2 // 2 -f r'// 4 = (r 2 4- ?/ 2 ) 2 .
In Problems 11 15 verify that the given values of x and ?/ satisfy the
equation, and find the corresponding value of -
11. x 2 -f 2 j-?/ 4- 2 /y = 22 ; jc = 2, // = 3. Am. ^ = - |-
12. j' 3 - // 3 4- 4 ^/y rr () ; j- = 2, y = - 2. - = 1.
13. Ax 4- By 4- O'" = C ; a- = 0, y = 0. ~
14. 2 x - V27^ 4- i/ = 4 ; .r = 2, y = 4.
15. e z cos ?y 4- ^ y sin x 1 ; x 0, ?/ = 0-
In Problems 16-20 find and ~
r^j <^?/
16. Ax>+B V > + Cz>=D. Ans. ^ = -*t;te = _*L
tfx Cz dy Cz
f)z Ay -i- Cz dz _ Ax 4- Bz
17. Axy 4- Byz + Czx = D. - = - - - ~ "
462 DIFFERENTIAL AND INTEGRAL CALCULUS
. o . o / - 1 n A ()Z V
18. x -f 2 // + z 2vxyz = 10. Ans. =
' ()Z V Z ~ -rtJZ CZ XZ -
.
Vxyz xy W Vxyz xy
19. x* + ?/< -f z' A - 3 axyz = 0.
20. ^^ -f /ty 2 -f Cz 2 -f 2 Dxy + 2 7/2 -f 2 Fz.r = G.
21. A point is moving on the curve of intersection of the sphere
x' 2 -f ir -f z 2 ~ 49 and the plane y = 2. When x is 6 and is increasing
4 units per second, find fa) the rate at which z is changing and (b) the
speed with which the point is moving.
Ans. (a; 8 units per second ; (b) 4 V5 units per second.
22. A point is moving on the curve of intersection of the surface
x' 2 -f xy -f ?/- z 1 and the plane x y + 2 = 0. When / is 3 and is
increasing 2 units per second, find (a) the rate at which // is changing,
(b) the rate at which z is changing, (c) the speed with which the point
is moving. Aws. (a) 2 units per second; (b) V units per second;
(c) 4.44 units per second.
23. The characteristic equation of a perfect gas is Rd pr, where 6
is the temperature, p the pressure, r the volume, and R a constant. At
a certain instant a given amount of gas has a volume of 15 cu. ft. and is
under a pressure of 25 Ib. per square inch. Assuming R 96, find the
temperature and the rate at which the temperature is changing if the
volume is increasing at the rate of \ cu. ft. per second and the pressure
is decreasing at the* rate of ,' Ib. per square inch per second.
Ans. Temperature is increasing at the rate of .^ degrees per second.
24. A triangle AHC is being transformed so that the angle A changes
at a uniform rate from O 1 to 90 L in 10 sec., while side AC decreases 1 in.
per second and side A /> increases 1 in. per second. If at the time of ob-
servation A GO 11 , AC = 1(1 in., and A /> - 10 in., (a) how fast is BC
changing? (b) how fast is the area of ABC 1 changing?
Atis. (a) 0.911 in. per second ; (b) 8.88 sq. in. per second.
232. Derivatives of higher order. If
(1) w =/(j, ?/ ),
then
/o\ f) u f / \ vn f f x
(2) - / r (r, ?/), - f v (x, y)
( 1 X f'lf
are themselves functions of ;r and ?/, and can, in turn, be differentiated.
Thus, taking the first function and differentiating, we have,
(3) "/,,Cr, ?/), 7r = /i;,(:r,?/).
tfx- fy fx
In the same manner, from the second function in (2), we obtain
/A\ ^"U f f \ 2 U f / x
(4) - /,(*, y), = f vu (x, y}.
)-
PARTIAL DIFFERENTIATION 463
In (3) and (4) there are apparently four derivatives of the second
order. It is shown below that
eycx exey
provided, merely, that the derivatives concerned are continuous.
That is, the order of differentiating vuccewirelH with respect to x and y is
immaterial. Thus /(x, //) has only three partial derivatives of the
second order, namely,
(5) fxjrfr, y\ &(*, '//) = /^(x, //), /, /?/ (x, y).
This may be easily extended to higher derivatives. For instance,
since (K) is true,
_r*H _ (>__ / (~u \ C''H __ e_- /( H\ e- /eu\ _ Ir^u
cx' 2 ( } y rxVxr/// rxetjfx <(it\(jc) eyex\cx/ ( ] y ex 2
Similar results hold for functions of three or more variables.
ILLUSTRATIVE EXAMPLE. Given // .r 1 // - ,'J .r 2 //-' ; verify ~~~ 7^-.
r')// oV d.r dy
Solution. -,-- *J f'"'y (> .r//'. T- :~ % J*" - IX -H/ 2 .
Hence the formula is verified.
Proof of (AT). Consider the expression
(6) F=f(jr+ Ax, // + A//) - /(x + Ax, y) - /(x, y + &y]
Introduce the function
(7) 0(w) =/(M, ?/ + A?/) /(M, ?/),
-where u is an auxiliary variable. Then
4>(x + Ax) =f(x + Ax, y + A?/) -/(x + Ax, ?/),
(8) c/>(x) = /(x, ?/ + Ay) - /(x, |/).
Hence (6) may be written
(9) F = <t>(x + Ax) -
Applying the Theorem of Mean Value (Z>), Art. 116,
(10) F = Az0'(x + 6li Ax). (
f /"(or) = 0(w), = a-, ^a = Ax.J
464 DIFFERENTIAL AND INTEGRAL CALCULUS
The value of 0'(z + 0i Ax) is obtained from (8) by taking the
partial derivative with respect to x and replacing x by x + 6\ Ax.
Thus (10) becomes
(11) F = Ax(/,(x + 61 Ax, y + A?/) -/,(* + 0, Ax, ?/)).
Now apply (/)), Art. 116, to / a (x + #j Ax, v), regarding v as the
independent variable. Then
(12) F = Ax &yf vf (x + 61 Ax, y + 0, A-//). (0 < 2 < 1)
If the second and third terms of the right-hand member of (6)
are interchanged, a similar procedure will give
(13) F = A?/ Ax/,,(x + ft. Ax, y + 4 Ay). (0< ft* < 1 , < 4 < 1 )
Hence from (12) and (13),
(14) f vf (x + Oi Ax, y + L , A?/) = /,(* + ft, Ax, y + 4 A/y).
Taking the limit of both sides as Ax and A?/ approach zero as
limits, we have
(15) f v ,(x,y)=Jr V (r,y),
since these functions are assumed to be continuous.
PROBLEMS
Find the second partial derivatives of each of the following functions.
1. /(x, //) = Ax 2 -f 2 lljry + O/ 2 -
An*, /^(x, ?/) = 2 .4 ; /, tf (x, ?y) = 2 It ; /^(x, 7y) = 2 C.
2. /(x, /y) = Ax :< + Rr 2 ?/ -f Tx// 2 -f /;//'.
. / (x, ?y) = 6 AJT + 2 Bij ; /^(x, ?/) = 2 Rr -f 2 r\v ; / (x, ?y) = 2 t\r + 6 /)?/.
3. /(x, ?/) - .4.r -f- KU -f rr'".
-4w. f 2f (x, y) = Cy' 2 c tv ; f* v =C(l
-
4 -
5. /(x, ?/) = x 2 cos ?/ 4- ?y 2 sin x.
6. If /(x, |/) = a- 3 -f 3 x 2 y 4- 6 x?y 2 - ?y 3 , show that
/ xx (2, 3) = 30, / XJ ,(2, 3) - 48, / yy (2, 3) = 6.
7. If /(x, //) = x 1 - 4 x a ?/ 4- 8 jrj/ 3 - 7/ 4 , show that
/(2, - 1) = 96, .^(2, - 1) = - 24, / w (2, - 1) = - 108.
8. If /(x, ?/) = 2 x' - 3 x 2 ?/ 2 -f /y 4 , find the values of
/(2, - 2), / T1/ (2, ~ 2), / vy (2, - 2).
PARTIAL DIFFERENTIATION 465
9. If u = Ax 4 4- Rjr*y -f O* 2 ?r -f />.n/ a 4- ?/ 4 , verify the following re-
sults. ,I3 p3
- = 24 .4r 4- 6 #?/. - fc Rr -f 4 O/,
2!L = 4 fj + 6 /; ?/ , ^1 = 6 Rr -f 24 //.
r'j ( } \f <"//
10. If w = (ax 2 -f ft?/ 2 + cz-}'\ show that
r'.r
= __!/_ - 2
L If M = ___, S how that j- 2
-f 2 xy -- 4- ?/ 2 = 0.
f>.r' 2 " r'j* r??/ <"/r
12. If M = In .H + /y 2 , show that 4- = 0.
2 *
ADDITIONAL PROBLEMS
1. A circular hill has a central vertical section in the form of the
curve whose equation is .r 2 + 160 y - 1600 = 0, where the unit is 1 yd.
The top is being cut down in horizontal layers at the constant rate of
100 cu. yd. per day. How fast is the area of the horizontal cross section
increasing when the top has been cut down a vertical distance of 4 yd.?
Am. 25 sq. yd. per day.
2 if w = -^~, show that + = Or -f ?/ - l)w.
c f + c y fa ry
3 If w = , where r = Vj- 2 + y' 2 + 2 2 , show that
7/ cr '()'*% 'f* ?/ 2
4. If z = x 2 arc tan ~ - y' 2 arc tan - show that - ^ 2 _^_ '
5. If u z arc tan -> show that - -f - 4- - = 0.
y c)x 2 ()y 2 (^
ffin
6. If u = In (c r 4- e" 4- r*), show that - : ~ - 2 ^
7. If u f(?, y) and x = r cos ^, y = r sin 0, show that
tfu n ()u sin 6 '(^u
_____
8. Let M = (/i 2 + x*' 2 + ---- h * n 2 )'. What values of * will satisfy the
^ ---+ = 0? An*. * = l-(n>2).
J?
CHAPTER XXIV
APPLICATIONS OF PARTIAL DERIVATIVES
233. Envelope of a family of curves. The equation of a curve gen-
erally involves, besides the variables x and //, certain constants upon
which the size, shape, and jxxsition of that particular curve depend.
For example, the locus of the equation
(x a) 2 + y 2 = r 2
is a circle whose center lies on the j-axis at a distance of a from the
origin, its size depending on the radius r. Suppose a to take on a
series of values while r is held fast ; then
we shall have a corresponding series of
circles of equal radius differing in their
distances from the origin, as shown in
the figure.
Any system of curves formed in this
way is called a family of curves, and the
quantity a, which is constant for any one curve, but changes in
passing from one curve to another, is called a variable parameter.
To indicate that a enters as a variable parameter it is usual to
insert it in the functional symbol, thus:
/Or, y, a) = 0.
The curves of a family may be tangent to the same curve or group
of curves, as in the above figure. In that case the name envelope of
the family is applied to the curve or group of curves. We shall now
explain a method for finding the equation of the envelope of a family
of curves. Suppose tha;, the curve whose parametric equations are
0) J- = </>(), y=\f,(a)
is tangent to each curve of the family
(2) /U, y, a) = 0,
the parameter a being the same in both cases. For any common
value of a equations (1) will satisfy (2). Hence, by (), Art. 229,
since u =f(x, y, or), du = df = 0, and z is replaced by a, we have
(3) fx(x, y, a)(f>'(a} + f y (x, y, a)\f/'(a) +/(>, y, a) = 0.
466
APPLICATIONS OF PARTIAL DERIVATIVES 467
The slope of (1) at any point is
g-*$.
and the slope of (2) at any point is
(5) = - - r ' ' <v - (tf), Art. 231
rfj- ./(*, //, a)
Hence if the curves (1) and (2) are tangent, the slopes at a point
of tangency will be equal, giving
__ fj(x, ?/, nr)
or
</> Uv) f u (x, //, a)
(6) /,(*, //, <v)</>'(aO + /(/, //, W() = 0.
Comparing (6) and (3) gives
(7) f a (x, y, a) = 0.
Therefore the coordinates of the point of tangency satisfy
(8) f(x, ?/, <v) - and / n (j, //, a) - ;
that is, the parametric equations of the envelope, in case an envelope
exists, may be found by solving these equations for x and y in terms
of the parameter a.
General rule for finding the parametric equations of the envelope
FIRST STEP. Write the equation of the family of curves in the form
/(x, y, a) and derive the equation f a (x, y, a) = 0.
SECOND STEP. Solve these two equations for x and. y in terms of the
parameter a.
The rectangular equation may be found either by eliminating a
-between the equations (8) or from the parametric equations (Art. 81).
ILLUSTRATIVE EXAMPLE 1. For the family of circles at the beginning of this
article, ^ (j> ^ a) = (j _ a) , + ^ _ r -z = .
Hence /(-r, 2/, J - (f - ) = 0.
Eliminating a, the result is /y 2 - r 2 - 0, or y = r, y - - r, and these are the equa-
tions of the lines .4K and CD in the figure.
ILLUSTRATIVE EXAMPLE 2. Find the envelope of the family of straight lines
x cos a -4- y sin a - ;>, a being the variable parameter.
Solution. (9) f(x, y, a] - x cos a + y sin a - p - 0.
468
DIFFERENTIAL AND INTEGRAL CALCULUS
First Step. Differentiating with respect to a,
(10) f a (x, y, a) - - x sin a 4- y cos a = 0.
Second Step. Multiplying (9) by cos a and (10) by sin a and subtracting, we get
x p cos a.
Similarly, eliminating x between (9) and (10),
y p sin a.
The parametric equations of the envelope are therefore
(11) P^Iina'
a being the parameter. Squaring equations (11) and
adding, we get ^ _^_ yi w^
the rectangular equation of the envelope, a circle.
ILLUSTRATIVE EXAMPLE 3. Find the envelope of a line of constant length a,
whose extremities move along two fixed rectangular axes.
Solution. Let AH = o in length, and let
(12) x cos a 4 y sin a p -
be its equation. Now as Aft moves, both a and p will vary. But p may be found
in terms of a. For AO Att cos a o cos a, and also
p = AO sin a = a sin cos a. Substituting in (12), we get
(13) x cos a -f y sin a a sin a cos a = 0,
where a is the variable parameter. This equation is in
the form f(x, y, a] 0. Differentiating with respect to
a, the equation fa(x, y, a) = is
(14) x sin a- -f y cos a -f a sin 2 a. a cos- a: = 0.
Solving (13) and (14) for x and y in terms of a, the
result is
\ x = a sin- 4 a,
] y a cos-' a,
(15)
the parametric equations of the envelope, a hypocycloid. The corresponding rec-
tangular equation is found from equations (15) by eliminating a: as follows:
Adding,
a*** = a d sin- a,
? 2
^3 a^ cos 2 a.
228
r j + ?/ a = fl-\ the rectangular equation of the hypocycloid.
Many problems occur in which it is convenient to use two param-
eters connected by an equation of condition. By using the latter,
one parameter may be eliminated from the equation of the family
of curves. It is, however, often better to proceed as in the following
xample.
APPLICATIONS OF PARTIAL DERIVATIVES
469
ILLUSTRATIVE EXAMPLE 4. Find the envelope of the family of ellipses whose
axes coincide and whose area is constant.
Solution. (16) + ^ = 1
o 2 b' 2
is the equation of the ellipse, where a and
b are the variable parameters connected
by the equation
(17)
irab A",
irab being the area of an ellipse whose
somiaxes are a and b. Differentiating,
regarding a and b as variables and .r and
y as constants, we have, using differentials,
= 0, from (16),
U" U"
and bda -f adb 0, from (17).
Transposing one term in each to the second member and dividing, the result is
Therefore, using (16),
whence
a = j\ '2 and ft ~ //% 2.
Substituting these values in (17), we get the envelope xy ~ , a pair of
conjugate rectangular hyperbolas (see figure). ~ w
234. The evolute of a given curve considered as the envelope of its
normals. Since the normals to a curve are all tangent to the evolute
(Art. 110), it is evident that the evolute of a curve
may be defined as the envelope of its normals. It
is also interesting to notice that if we find the
parametric equations of the envelope by the
method of the previous article, we get the coor-
dinates x and y of the center of curvature ; so
that we have here a second method for fiivditm the
coordinates of the center of curvature. If we elimi-
nate the variable parameter, we obtain the rec-
tangular equation of the evolute.
ILLUSTRATIVE EXAMPLE. Find the evolute of the parabola y~ = 4 px considered
as the envelope of its normals.
Solution. The equation of the normal at any point (x\ t y\) is
470 DIFFERENTIAL AND INTEGRAL CALCULUS
from (2), Art. 43. As we are considering the normals all along the curve, both Ji
and 2/1 will vary. Eliminating x\ by means of y\ 2 4 px\, we find the equation of
the normal to be
(1) y-Vi^--j 1 ' or x yi +2py - 2 pj/i - = 0.
o p & p ^ /'
Setting the partial derivative of the left-hand member with respect to the
parameter y\ equal to zero, and solving for x, we find
4 p
Substituting this value of i in (1) and .solving for y,
'
Equations (2) and (3) arc the coordinates of the center of curvature of the pa-
rabola. Taken together, they are the parametric equations of the evolute in terms
of the parameter y\. Eliminating ?/i, we obtain
27 pi/ 2 =-4(.r ~2 p)\
the rectangular equation of tho evolute of the parabola. This is the same result we
obtained in Illustrative Example 1, Art. 109, by the first method.
PROBLEMS
Find the envelopes of the following systems of straight lines and draw
the figures.
1. y rns -f nr. A ns. jr 2 -f- 4 y = 0.
2. // = -f nr. 27 x 2 = 4 ?y 3 .
w
3. // = m' 2 j' 2 m ;< . 27 ?/ = ^ :i .
4. ?/ = 2 mjr -f- ?w 4 . 16 ?/ : ^ -f 27 u* 1 = 0.
5. y tjr t' 2 . 6. // f-s -h /. 7. ?/ = m.r 2 m 2 .
Find tbe envelopes of the following systems of circles and draw the
figures.
8. (x c}' 2 4- ir = 4 r. Am. y 2 4 jr -f 4.
9. a- 2 -h (?/ - O 2 = 2 /. 10. (T - O 2 + (?/ + O 2 = / 2 -
Find the envelopes of the following systems of parabolas.
11. y* = c(x c). Ans. 2 y = jr.
12. cy' 2 = 1 - c 2 jc.
13. Find the evolute of the ellipse b~x 2 + a 2 y 2 a 2 5 2 , taking the equa-
tion of tbe normal in the form by = ax tan (a 2 6 2 ) sin </>, the eccen-
tric angle c/> being tbe parameter.
Am. x = a2 ~ ?>2 cos 3 0, ?/ = b2 ~ a " sin 3 ; or (ax) ? -f- (byY = (a 2 - 6 2 )l
'
APPLICATIONS OF PARTIAL DERIVATIVES
471
2 2 2
14. Find the evolute of the hypocycloid .r s -f ir = a 3 , the equation of
whose normal is
r being the parameter.
r cos r x sin r = a cos 2 r,
Arts, (.r -f '<
(JT- ?/) = 2 a.
15. Find the envelope of the circles which pass through the origin and
have their centers on the hyperbola x' 2 ?/- = r 2 .
Ans. The lemniscate (x' 2 -f ?/ 2 ) 2 = 4 r 2 (.r 2 ?/ 2 ).
16. Find the envelope of a line such that the sum of its intercepts on
the axes equals c. AnKf The parabola ,1 + y \ = r *.
17. Find the envelope of the family of ellipses b 2 x' 2 -f a 2 // 2 a 2 ?) 2 when
the sum of its semiaxes equals r. 4 ?/<s . f ^ e hypocycloid x** -f ifi = c^.
18. Projectiles are nred from a gun with an initial velocity r>. Sup-
posing the gun can be given any elevation and is kept always in the
same vertical plane, what is the envelope of all possible trajectories, the
resistance of the air being neglected?
HINT. The equation of any trajectory is
y = x tan ex - - ~ ,
2 r<r cos- a
(i being the variable parameter.
Ans. The parabola y = - ~^--
2g 2 i,r
19. If the family
has an envelope, show that it is
g 2 (x, y) L
, ?/) + /'Or, ?/) =
235. Tangent line and normal plane to a skew curve. The student
is already familiar with the parametric representation of a plane
curve (Art. 81). In order to extend
this notion to curves in space, let the
coordinates of any point P(x, y, z) on
a skew curve be given as functions of
some fourth variable which we shall
denote by t ; thus,
(1) * =
2=x(0-
The elimination of the parameter
t between these equations two by two
will give us the equations of the
projecting cylinders of the curve on the coordinate planes.
472 DIFFERENTIAL AND INTEGRAL CALCULUS
Let the point P(.r, y, z) correspond to the value t of the param-
eter, and the point P'(x + Az, y + Ay, z + Az) correspond to the
value t + A2 where Ax, A?/, Az are the increments of x, y, z due to the
increment AZ as found from equations (1). From analytic geometry
of three dimensions, we know that the direction cosines of the secant
(diagonal; PI*' are proportional to
Ax, A#, Az ;
or, dividing through by A/ and denoting the direction angles of the
secant by a', ft', y',
cos a' __ cos (3 f _ cos y'
( } Ax ~ Ag " Az
AZ At A*
Now let P' approach P along the curve. Then AJ, and therefore
also AJ, Ay, Az, will approach zero as a limit, and the secant PP' will
approach the tangent line to the curve at P as a limiting position.
Now lim ^ = ^ = </>'(0, etc.
A* - o A at
Hence, for the tangent line,
cos a _ cos P _ cos y
04) cfo (/y dz
dt dt dt
When the point of contact is P\(x\, y\, z\), we use the notation
(3)
= value of c -j- when x = .TI, y = y\, z = z\,
and similar notation for the other derivatives.
Hence, by (2) and (4), p. 5, we have the following result.
The equations of the tangent line to the curve whose equations are
(1) .r = <W), y = t(V, z
at the point PI(XI, y\, zO are
x xi_y y\ _ z z\
(B) \dx
\dt
\dt
I I
\dt\
The normal plane of a skew curve at a point P\(x\, y\, Zi) is the
plane which passes through PI and is perpendicular to the tangent
line at PI. The denominators in (B) are the direction numbers of
the tangent line at PI. Hence we have the following result.
APPLICATIONS OF PARTIAL DERIVATIVES 473
The equation of thf normal plane to the curve (1) at Pi (x\, y\, z\) is
(x-
\dt
(w -
ILLUSTRATIVE EXAMPLE. Find the equations of the tangent line and the equa-
tion of the normai plane to the circular helix (6 being the parameter)
( x = a cos 0,
- y fl sin 6,
(4)
(a) at any point (xi, 3/1, d) ; (b) when - 2 TT.
Solution. -^ = - a sin = - ?/, ^ = a cos = .r, -f- - b.
do du (<('
Substituting in (B) and (C), we get, at (.n, //,, z\ >,
(5) x ~ x - V ~ V 1 = T"^' tangent line,
and
normal plane.
When '2 TT, the point on the curve is
(a, 0, 2 &TT), giving
g_--ji _ ?y _ z - 2 bir
~ a ~ 6
or a* = a, &# = a^ 2 O^TF,
the equations of the tangent line, and
ny -f ?w - 2 ?> 2 7T = 0,
the equation of the normal plane.
REMARK. For the tangent line (5) we have by (2) and (4), p. 5,
fo 6
cos 7 =
: = a constant
-f yi 2 + b' 2 Va~ -f b 2
That is, the helix cuts all elements of the cylinder x 2 + y' 2 = a 2 under the same angle.
236. Length of arc of a skew curve. From the figure of the pre-
ceding article we have
/i x (Chord PP'Y _ /AxV
-
Let arc PP' = As. Proceeding as in Art. 95, we easily prove
(2) (*) 2 =
From this we obtain
(D) s =
where x 0(0, y = ^(0> z x(0 as in (1^ A*t. 23").
474 DIFFERENTIAL AND INTEGRAL CALCULUS
The direction cosines of the tangent line can now be given a simple
form. For, from (A) of the preceding article, and by the above
equation (2), using formulas in (2), p. 5, we have
(3 ) cos > ft ^ ~
ds " ds " ds
ILLUSTRATIVE EXAMPLE. Find the length of arc of the skew cubic
(4) x = t, y = } 1*, z = J f
between the points where / = and t = 4.
Solution. Differentiating (4), we obtain
dx = dt, dy = / (it, dz = t' 2 dt.
Substituting in (D), = f Vl + t' 2 + t* dt = 23.92,
Jo
approximately, by Simpson's Rule, letting n = 8.
PROBLEMS
Find the equations of the tangent line and the equation of the normal
plane to each of the following skew curves at the point indicated.
1. x = at, y = bt 2 , z = </; / = 1.
AUK. ^^ = ^-^ = ^p-f ; ax 4- 2 by -f 3 cz = a 2 4- 2 6 2 4- 3 c 2 .
3. a- = / 2 - 1, ?/ = / + 1, r = / :< ; / = 2.
4. j = P - 1, ?/ = r 2 -f /, ^ = 4 /3 - 3 / -hi ; / = 1.
5. a- = 2 / - S, ?/ = 5 - / 2 , c = - ; / = 2.
6. x = a cos /, // b sin /, z ~ / ; / = J TT.
7. x = /, ?/ = r', ^ = r ' ; / = 0.
8. x = cos /, z/ sin t, z = tan / ; f = 0.
9. Find the length of arc of the circular helix
x = a cos 0, y = a sin 6, z = b6
between the points where B = and 6 = 2 TT. Ans. 2 ?rVa 2 -f 6-.
APPLICATIONS OF PARTIAL DERIVATIVES 475
10. Find the length of arc of the curve
x = 3 cos 0, y = 3 6 sin 8, c = 4
between the points where 6 = and = 4. Ann. 26 + ^ In 5 = 32.70.
11. Find the length of arc of the curve
x = 2 t, y = t' 2 -2, ; = 1 - t 2
between the points where t = and / = 2.
12. Given the two curves
(5) x = t, y = 2t*, z = -y;
(6) x = 1 - 0, y = 2 cos 0, c = sin 6 - 1.
(a) Show that the two curves intersect at the point A(l, 2, 1).
(b) Find the direction cosines of the tangent line to (5) at A.
4 1 4 1
Ans. ;= p= 7=-
Vl8 Vl8 Vl8
(c) Find the direction cosines of the tangent line to (6) at A.
(d) Find the angle of intersection of the curves at A . Ans. 90.
13. Given the two curves
a- = 2-/, i/ = / 2 -4, 2 = f-8;
:r = sin 0, y = 0, z = 1 cos 0.
(a) Show that the two curves intersect at the origin O.
(b) Find the direction cosines of the tangent line to each curve at O.
(c) Find the angle of intersection of the curves at O.
14. (a) If OF, OE, ON in the first figure of Art. 222 are chosen as axes
of coordinates OX, OY, OZ, respectively, and if P(JC, ?/, z) is a point on the
sphere, prove that r = a cos $ sin 0, ?/ = a cos < cos 0, z = asin</>, if </> and
are, respectively, the latitude and longitude of P.
(b) Using (3), and (3) on page 5, find the angle u at P between a curve
on the sphere for which = /(</>) and the parallel through P.
Ans. tan a sec </> -777 , as in Art. 222.
au
237. Normal line and tangent plane to a surface. A straight line is
said to be tangent to a surface at a point P if it is the limiting position
of a secant line through P and a neighboring point P' on the surface
when P f is made to approach P along a curve on the surface. We
now proceed to establish a theorem of fundamental importance.
Theorem. All tangent lines to a surface at a given point lie in a plane.
Proof. Let
(1) F(x, y,z)=0
be the equation of the given surface, and let P(x, y, z) be the given
Doint on the surface. If now P' be made to approach P along a curve
476
DIFFERENTIAL AND INTEGRAL CALCULUS
C lying on the surface and passing through P and P', then evidently
the secant line PP' approaches the position of a tangent line to the
curve C at P. Now let the equations of the curve C be
Then the equation (1) must be satisfied identically by these
values. Hence, if u = F(x, y, z), then u = 0, du = 0, and by (), Art. 229,
tix dt 'dy dt cz dt ~
This equation (see (3), Art. 4) shows that the tangent line to (2),
whose direction cosines are proportional to
dx d/n dz
dt dt dt
is perpendicular to a line whose direction cosines are proportional to
(4) > > By (3), p. 5
(>x oy cz
Let PI(:TI, y\, z\) be a point on the surface and
/e\ < )F ? )F
(O) TT" > - >
vx i vy \
the values of the partial derivatives in (4) when x = Xi, y = y\ 9
z z\. The line passing through P\ whose direction numbers are
given by (5) is called the normal line to the surface at PI. Hence we
have the following result :
The equations of the normal line to the surface
H 1 F(x y ~] =
\i ) r \A., y j ** i \/
at PI('X\, '//i, zO are
_. y y\ __ z z\
x -
dF
dx
IdF
\9y
dF\
'dz\\
The preceding argument shows that all tangent lines to the sur-
face (1) at PI are perpendicular to the normal line at PI. Hence
they lie in a plane. Thus the theorem is proved.
This plane is called the tangent plane at (Pi).
We may now state the following result.
The equation of the tangent plane, to the surface (1) at the point oj
contact PI(XI, y\, z\) i-s
<*> i?i(.-
= 0.
APPLICATIONS OF PARTIAL DERIVATIVES
REMARK. If all the denominators in (E) vanish, the normal line and tangent
plane are indeterminate. Such points are called singular points and are excluded
here.
In case the equation of the surface is given in the form
z = /(z, y), let
(6) F(x, ?/, z) =/(j, //) - z = 0.
rp^ ?>F ?j C'Z ( ] F (f f } Z ( } F ,
i nen -r ~T~ TT~ -r ~r~ ~~ ~ == !
ra OT ex dy r } y ( } y cz
Hence, by (), we have the following result.
The equations of the normal line to the surface z /( J, y) at ( x\ , y\ , z\ )
are
(G)
2-
- 1
Also, from (F), we obtain
- xi) 4-
(y - yi) - (z - zi) =
which is then the formula for the equation, of a plane tangent at (x\ t y\, z\)
to a surface whose equation is given in the form z =f(x, y).
238. Geometric interpretation of the total differential. We are now
in a position to discuss formula (#), Art. 227, by geometry, in a man-
ner entirely analogous to that in Art. 91.
Consider the surface
(1)
and the point
A'hen
(2)
i, y\, z\) on it. Then the total differential of (1) is,
x = x\ t y~yi,
Ayy,
using (5), Art. 227, and replacing dx and dy by their equivalents, Ax
and A^/, respectively. Let us find the z-coordinate of the point in
the tangent plane at P\ where
x = xi + Ax, y = yi + A?/.
Substituting these values in (H) of Art. 237, we find
(3)
478 DIFFERENTIAL AND INTEGRAL CALCULUS
Comparing (2) and (3), we get dz = z z\. Hence the
Theorem. The total dif-
ferential of a function f(x, y)
corresponding to the incre-
ments A x and Ay equals the
corresponding increment of
the z-coordinate of the tan-
gent plane to the surface
Thus, in the figure, PP'
is the plane tangent to
surface PQ at P(x, y, z).
Let
and CD &y,
then dz = z - z\ = DP' - DE = EP'.
Notice also that Az = DQ - DE = EQ.
ILLUSTRATIVE EXAMPLE. Find the equation of the tangent plane and the equa-
tions of the normal line to the sphere x' 2 -f y 2 -f z 2 14 at the point (1, 2, 3).
then
Solution. Let F(x, y, z) = x 2 + y* + z 2 - 14 ;
dF
dx
Therefore
= 4,
vSubstituting in (F), 2(x - 1 ) + 4(y - 2) + 6(z - IJ) = 0,
or x + 2 y + 3 c = 14, the tangent plane.
y - 2 _ s - 3
4 " 6 '
Substituting in (),
giving z = 3 x and 2 s = 3 y, equations of the normal line. Arts.
PROBLEMS
Find the equation of the tangent plane and the equations of the nor-
mal line to each of the following surfaces at the point indicated.
1. x' 2 + y' 2 + z 2 = 49 ; (6, 2, 3).
An*. 6 x 4-
2. z =
8 -l; (2, 1,4).
x-6 = y-2 = z-3
J ' 6 ~ 2 3
- 1 2-4
APPLICATIONS OF PARTIAL DERIVATIVES 479
3. r 2 + xy 2 4- ?/ 3 -f z 4- 1 = ; (2, - 3, 4).
Ans. 13 .r -f 15 y + * + 15 = ; ^-^ = i^J = L=-l.
A o 1 1
4. x~ 4- 2 jry 4- ?/ 2 -f z - 7 = ; (1, - 2, 6).
5. j* 2 ?/ 2 -f .nr - 2 ?/< - 10 = ; (2 t 1, 4).
Ans. 4 .r + // -f z - 13 = ; ^-=-? = 2-^ = ^-
411
6. .r 2 - y 2 -z' 2 =: 1; (3, 2, 2).
7. ar 2 + 2 - z 2 = 25 ; (5, 5, 5).
8. 2 x 2 + 3 j/ 2 + 4z-' = 6; (1, 1, J).
9. y-f ?/- r 2 ^3; (3, 4, 2).
10. Find the equation of the tangent plane to the hyperboloid of two
sheets 1 - ^ - ! = i at (.r,, ?/ z,). .4**. l _ M _ l| = ! .
a- 6 2 r- 2 ?> 2 r 2
11. Find the equation of the tangent plane at the point (.r,, ?/ ]t z\) on
the surface ax 2 + 6// 2 -f re 2 + d ~ 0. ^^,s. a.ri.r -f ^?/i// -f rr,^ -f d = 0.
12. Show that the equation of the plane tangent to the sphere
x 2 + r -f z- -f 2 L.r -f 2 Af?/ -f 2 A^ + /> =
at the point (xi, y it z\) is
^i^ 4- 3/1 1/ + Ziz + L(J- + j-i) 4- M(/y -f ?/]) -h A/ T (- 4- Ci) 4- D = 0.
13. Find the equation of the tangent plane at any point of the surface
j% -f y* -f 2 rr a",
and show that the sum of the squares of the intercepts on the axes made
by the tangent plane is constant.
14. Prove that the tetrahedron formed by the coordinate planes and
any tangent plane to the surface -jryz a :< is of constant volume.
/2 4
15. The surface x 2 4 y 2 4 z = is cut by the curve x > y = -
/ _ 9 /2 J t
2 = - - - at the point (2, 2, 3). What is the angle of intersection?
Am. 90 - arc cos 1 / !L-. = 32 37'.
3V138
3
16. The surface x 2 + ?/ 2 -f 3 z 2 ~ 25 and the curve x = 2t, y = ->
t
z = 2 f 2 intersect at the point on the curve given by / = 1. What is
the angle of intersection ? A ^ 9Q o _ arc CQS _J9 = 3QO 16/>
7V29
17. The ellipsoid x 2 -h 2 y 2 4- 3 z 2 = 20 and the skew curve x - \(P 4- 1 ),
i/ = f 4 -f 1, z P meet at the point (3, 2, 1). Show that the curve cuts
the surface orthogonally.
480
DIFFERENTIAL AND INTEGRAL CALCULUS
239. Another form of the equations of the tangent line and normal
plane to a skew curve. If the curve in
question be the curve of intersection AB
of the two surfaces F (x, y, z) = and
G(x, y, z) = 0, the tangent line PT at
P(x\, y\ t Z}) is the intersection of the
tangent planes CD and CE at that,
point, for it is also tangent to both sur-
faces and hence must lie in both tangent
planes. The equations of the two tan-
gent planes at P are, from (F),
(1)
tix
f)G
IrJx
(x -
tiy
^ <*-">
vF
c)z
(z Zi) = 0.
Taken simultaneously, these equations are the equations of the
tangent line PT to the skew curve AB.
If A, B, C are direction numbers for the line of intersection of the
planes (1), then, by (6), Art. 4,
dG
(2)
c)JP
f)Z
C =
'dy
B =
bF\ VvG_
'c)z i cjy
8F
dx
dz
8y
c)F_
fry
c)G
Then the equations of the tangent line CPT are
(3)
A
_y y\ ^z
B C
The equation of the normal plane PHI is
(4) A(x - j,) + B(y - yi) + C(z - Z]) = 0.
ILLUSTRATIVE EXAMPLE 1. Find the equations of the tangent line and the
equation of the normal plane at (r, r, r N/2) to the curve of intersection of the sphere
and cylinder whose equations are, respectively, x 2 + y 2 H- z 2 = 4 r 2 , x 2 + j/ 2 = 2 rx.
Solution. Let F = a- 2 4 ?/ 2 4- z 2 - 4 r 2 and G = x 2 + y 2 - 2 nc.
aF
ax
= 0,
dG]
dy\
APPLICATIONS OF PARTIAL DERIVATIVES
Substituting in (2), we find
A = -4r 2 \/2, B = 0, C = 4r2.
Hence, by (3), we have
x r y r z rV2
481
or
- V2
^ = r, z
the equations of the tangent PT at P to
the curve of intersection.
Substituting in (4), we get the equa-
tion of the normal plane,
or
- r) 4- 0(y - r) + (z - r\ V =. 0.
\ 2 x - z 0.
ILLUSTRATIVE EXAMPLE 2. Find the angle of intersection of the surfaces in the
preceding example at the point given.
Solution. The angle of intersection equals the angle between the tangent planes
or normal lines. We have found direction numbers for these lines above in Illus-
trative Example 1 (see (), Art. 237).
These are a = 2 r, b = 2 r, r - 2 r\/2.
a' = 0, 6' = 2 r, c' = 0.
Hence, by (6), Art. 4
= t| = . = 60. Am.
PROBLEMS
Find the equations of the tangent line and the equation of the normal
plane to each of the following curves at the point indicated.
1. x 2 + y 2 + z 2 = 49, x* + y 2 13 ; (3, 2, - G).
Aw*. rLzJ = -=, 2+6 = 0; 2 s - 3 # = 0.
2. 2 = a- 2 + y 2 - 1, 3 ? 2 + 2 // 2 + z 2 = 30 ; (2, 1, 4).
Am.
5 -11
3 . x* + y* _ Z 2 = 16. x - + 4 ?/* + 4 z 2 = 84 ; (2, 4, 2).
x 2 _ y 4 _ 2 2
5
,4ns.
16 - 5 6
4. x a + ?y 2 + 3 z 2 = 32, 2 r 2 + ?/ 2 - z 2 = ; (2, 1, 3).
; 16 j-- 5 ?y + 6,7 = 24.
5. x 2 - y 2 - z 2 = 1, x 2 - y 2 + z 2 = 9 ; (3, 2, 2).
6. x 2 + 4 # 2 - 4 z 2 - 0, 2 x + ?/ + 2 - 24 = ; (8, 3, 5).
482 DIFFERENTIAL AND INTEGRAL CALCULUS
7. The equations of a helix (spiral) are
X 2 + y'2 _ f2>
iy = x tan -
r
Show that at the point O], ?/i, zj) the equations of the tangent line are
c(x - si) -f 1/1(2- ?]) = 0,
r(?y - ?y } ) - xi(z - Zj) = 0;
and the equation of the normal plane is
. 7/1 .r ~ x\y r(z ~ z\) = 0.
8. The surfaces x 2 ?/- + 2 j- -f z :< 16 and 3 jc 2 4- ?/ 2 2 z = 9 intersect
in a curve which passes through the point (2, 1, 2). What are the equa-
tions of the respective tangent planes to the two surfaces at this point?
Am. 3a- + 4i/ + 6c = 2'2; 6 x -f y - z ~ 11.
9. Show that the ellipsoid x~ -f 3 y' 2 + 2 z' 2 9 and the sphere
^2 _|_ ?y2 _|_ ~2 _ & r __ g ^ _ g _}_ 24 = are tangent to each other at the
point (2, 1, 1).
10. Show that the paraboloid 3 x~ -f 2 ?/ 2 22 = 1 and the sphere
x 2 -f y' 2 + z' 2 - 4 y - 2 z -f 2 = cut orthogonally at the point (1, 1, 2).
240. Law of the Mean. The applications of partial derivatives to be
given now depend upon the Law of the Mean for functions of several
variables. The result to be derived is based upon the discussion in
Art. 116. We proceed to establish the formula
(1) /fa, + h, yo + k) = /(a-o, 2/0) + V,(ro + Oh, ?/o + 0k)
+ kf tl (x*+0h,yv+0k). (0<0<1)
To this end let
(2) F(t)=f(ro + ht,yo + kt).
Apply (Z)), Art. 116, to F(t), with a == 0, and Aa = 1. Then we
have
(3) F(l) = F(0) + F'(0). (0 < 6 < 1)
But from (2), by (D), Art. 229, since x = xn + hi, y = yo + kt,
(4) f'(0 = fc/,(:ro + fa
Then, from (2), we get
(5) F(l) = /(Jo
and, from (4),
(6) F'tf) = hf x (x + eh, yo + 6k) + kf v (x + Oh, yo + Ok).
When these results are substituted in (3), we obtain (1).
APPLICATIONS OF PARTIAL DERIVATIVES 483
If we desire a formula analogous to (F), Art. 124, we must form
F"(t). Applying again (D), Art. 229, we get
x , kt) = hf tf (xo + ht, y (} + kt) + kf vr (x + hi, y + kt) ;
at
v-/y(aro + kt, .2/0 + kt) = hf,(x + hi, // + kt) + kf tw (x () + ht, y + kt).
at
Hence from (4), we have by differentiating with respect to t,
(7) F"(t) = k*f xf (x<> + ht, ?/ + kt) + 2 hkf xu (xo + kt. //o + kt)
+ A--/,,, (Jo + kt., ?/o + kt).
From (F), Art. 124, letting 6 = 1, a = 0, ^2 = 0, we get
(8) ni) = no) + r(0) + r|r'(0).
We may easily prove now the extended Law of the Mean for a
function of two variables by substituting in (8) from (5), (4), and (7).
Thus we get
(9) /(jr + h, ?/o + k) =/(TO, v/o) + A/r(:r<>, ?/o) + A/i/to), ?/o)
+ r| t^ 2 /xX^> + Oh, /A, + 0A-) + 2 Wn/ao + 0A, y () + Ok)
+ Ar 2 /^(y + 0//, ?/o + 0*)]- (0 < < 1).
There is no difficulty in establishing the corresponding formulas
for functions of more than two variables, nor in extending the laws
in a manner analogous to that at the end of Art. 124.
241. Maxima and minima of functions of several variables. In
Art. 46, and again in Art. 125, were derived necessary and sufficient
conditions for maximum and minimum values of a function of one
variable. We now take up this problem when several independent
variables are present.
The function /(.r, ?/) is said to be a maximum, at x = a, y = b
when /(a, b) is greater than f(x, y) for all values of x and y in the
neighborhood of a and b. Similarly, /(x, y) is said to be a minimum
at x a, y = b when /(a, 6) is less than f(x, y) for all values of x and y
in the neighborhood of a and 6.
These definitions may be stated in analytical form as follows:
If, for all values of h and k numerically less than some small
positive quantity,
Q) j( a + h f b + k) /(a, 6) = a negative number,
then /(a, b) is a maximum value of f(x, y). If
(2) /(a + A, 6 + k) /(a, 6) = a positive number,
then /(a, b) is a minimum value of /(x, #).
484
DIFFERENTIAL AND INTEGRAL CALCULUS
These statements may be interpreted geometrically as follows.
A point P on the surface
z = f(x,y)
l\ a maximum point when it is " higher" than all other points on the
surface in its neighborhood, the coordinate plane XOY being assumed
horizontal . Similarly,
P' is a minimum point
on the surface when
it is "lower" than all
other points on the
surface in its neigh-
borhood.
Hence if
21= /(a, 6)
is a maximum or mini-
mum, the tangent plane
at (a, b, zi) must be horizontal, that is, parallel to XOY. But the
tangent plane (H), Art. 237, is parallel to XOY when the coeffi-
cients of x and // are zero. Hence we have the following result.
A necessary condition that /(a, b) shall be a maximum or minimum
value off(x, y) is that the equations
(3)
c)x
8y'
shall be satisfied by x a, y b.
The conditions (3) may be obtained without use of the tangent
plane. For, when y ~ b, the function /(.r, b) can neither increase nor
decrease when x passes through a (see Art. 45). Hence follows the
first of equations (3). The same statement applies to the function
/(a, y}. Thus we have the second equation in (3).
The method just expounded applies to a function of three variables
f(x, y, z). That is, a necessary condition that /(a, 6, c) shall be a
maximum or a minimum value is that the equations
(4)
ftp
8z
=
shall have the common solution x = a, y b, z c.
For necessary and sufficient conditions the problem is much more
difficult (see below). But in many applied problems the existence of
a maximum or minimum value is known in advance, and no test is
necessary.
APPLICATIONS OF PARTIAL DERIVATIVES 485
ILLUSTRATIVE EXAMPLE 1. A long piece of tin 24 in. wide is to be made
into a trough by bending up two sides. Find the width
and inclination of each side if the carrying capacity is a
maximum.
Solution. The area of the cross section shown in the
figure must be a maximum. The cross section is a _
trapezoid of upper base 24 2 x -f 2 x cos a, lower r * 24 ~ 2a? *i
base 24 - 2 x, and altitude x sin a. The area A is given by
(5) A 24 x sin a - 2 x 2 sin a + x 2 sin a cos a.
By differentiation we have
-r = 24 sin a 4 x sin a + 2 jr sin a cos .
dx
dA
-r = 24 x cos a 2 .r- cos a -f a*" (cos-' a sin- a).
d
Setting the partial derivatives equal to zero, we have the two equations
2 sin (12 - 2 x + x cos a) = 0.
[24 cos a 2 .r cos a 4- x(cos 2 cv sin a )J = 0.
One solution of this system is a = 0, x = 0, which has no meaning in the
physical problem. Assuming a ^ 0, x & 0, and solving the e-quations, we get
cos a 2, x 8.
A consideration of the physical problem shows that there must exist a maxi-
mum value of the area. Hence this maximum value occurs when a = 60" and
x = 8".
We now establish a sufficient condition. Assuming that equa-
tions (3) hold, we obtain from (9), Art. 240, substituting x () = a,
yo = b, and transposing,
(6) /(a + M + A') -/(a, b) = r \h^f xx (x y y}+2 kkf fll (x, y)
^ +k*f w (x,y)\,
where we have set x = a + Oh, y b + 6k. By (1) and (2), /(a, b) will
be a maximum (or a minimum) if the right-hand member is negative
(or positive) for all values of h and k sufficiently small in numerical
value (zero excluded). Set
(7) A =/(*, y), B=:f xv (x, y), r=f yy (x, y),
and consider the identity
(8) Ah 2 + 2Bhk+ Ck* = [(Aft + Bt)2 + (AC - B*)k*].
The expression within the square brackets in the right-hand
member in (8) is always positive if
(9) AC - B 2 > 0,
486 DIFFERENTIAL AND INTEGRAL CALCULUS
and the left-hand member therefore has the same sign as A (or C,
since, by (9), A and C must agree in sign). The question now is,
therefore, to interpret the criterion (9; for the right-hand member in
(6), in which, as already stated, h and k are numerically small.
Assume that (9) holds when x = a, y = b. Then, the derivatives in
(1) being continuous, it will hold also for values of x, y near a, b.
Also, the sign of A (or C) will be the same as the sign of f xx (a, b)
(or f vv (a, b)). Thus we have established the following rule for find-
ing maximum and minimum values of a function / (x, y).
FIRST STEP. Solve the simultaneous equations
# = 0, # = 0.
c)x ( j y
SECOND STEP. Calculate for these values of x and y the value of
A^LM /J^LY\
()jc~ '( ] >\i~ \<rxvyj
THIRD STEP. The Junction will have
a maximum value if A > and ^ (or ~{ } < ;
()x* \ f)y 2 /
a minimum value if A > and -^ ( or ^-~ } > 0.
vx~ \ r)y-J
If A is negative, it is not difficult to see that /(or, y) will have
neither a maximum nor a minimum value.
The student should notice that this rule does not necessarily give
all maximum and minimum values. For a pair of values of x and y
determined by the First Step may cause A to vanish, and may lead
to a maximum or a minimum or neither. Further investigation is
therefore necessary for such values. The rule is, however, sufficient
for solving many important examples.
The question of maxima and minima of functions of three or more
independent variables must be left to more advanced treatises.
ILLUSTRATIVE EXAMPLE 2. Examine the function 3 axy x 3 y 5 for maxi-
mum and minimum values.
Solution. f(x, y] = 3 axy - x 3 - y 3 .
First Step. | = 3 ay - 3 z 2 - 0, | = 3 ax. - 3 y* = 0.
Solving these two simultaneous equations, we get
x = 0, x a,
y = 0, y = a.
APPLICATIONS OF PARTIAL DERIVATIVES 487
SecondStep. = - 6 x, = 3 a, = -6 V ;
\dxdyj -"
Third Step. When x = and {/ = (), A - - 9 a 2 , and there can be neither a
maximum nor a minimum at (0, OK ^
When x = a and y = a, A = 4 27 a 2 ; and since ~ = - (5 a, we have the con-
ditions for a maximum value of the function fulfilled at (a, </K Substituting u* = a,
y a in the given function, we get its maximum value equal to a :{ .
ILLUSTRATIVE EXAMPLE 3. Divide a into three parts such that their product
shall be a maximum.
Solution. Let x = first part, // = second part ; then a-(x + y)=a-x-y*=*
third part, and the function to be examined is
f(x t y} xy(a x ?/).
First Step. | = ay - 2 xy - y' 2 = 0, ^ = -r - 2 J?/ - x 2 = 0.
(7 _ a
Solving simultaneously, we get as one pair of values .r -' y - -
Second Step. ^ = - 2 ?/, ^ = a - 2 ar - 2 ?/, ~^T, = - 2 x ;
A = 4 xy - (a - 2 .r - 2 //)".
r>. When jr = " and y = ^ A = ^ ; and since ^ = - =~ . it is seen
that our product is a maximum when x = -' y^r^' Therefore the third .part is
also 7 , and the maximum value of the product is
PROBLEMS
Discuss for maxima and minima the following functions.
1. x 2 4- xy + y' 2 - 6 x 4- 2. Arts, r = 4, ?/ = - 2 gives min.
2 . 4 j. + 2 ?/ - r a 4 r// - ?/ 2 . ^ = J .'?, ?/ = S gives max.
3. 2 x 2 - 2 A// 4- /y 2 + 5 j- - 3 ?/. ^ = - L // = i gives min.
4 x >3 3 a^y 4- if- x = y a gives min.
5. sin x 4- sin y 4- sin (x 4- y). x = y = ~ gives max.
x = y -- gives min.
o
6. x 2 - xy 4- !/ 2 + oar 4- 60 4- c.
7 ^ , 2! + L
' ^ x y 2
8. Show that the maximum value of ^ . J 2 . ^ is a2 + fr2 "^ c2 '
488 DIFFERENTIAL AND INTEGRAL CALCULUS
9. Find the rectangular parallelepiped of maximum volume which
has three faces in the coordinate planes and one vertex in the plane
5 + 2+5 = 1. Ans. Volume = ~
a b c 27
10. Find the volume of the largest rectangular parallelepiped that can
be inscribed in the ellipsoid - + 2_ + = 1. Ans.
a 2 b 2 c 2
11. A pentagon is composed of a rectangle surmounted by an isosceles
triangle. If the perimeter of the pentagon has a given
value P, find the dimensions for maximum area.
P
AUK. a 30, 2 JT = >
24-2 sec a tan a y
p
y = .r(l + sec a).
12. Find the shortest distance between the lines x = ~ == - and
, j Z o
13. A manufacturer produces two lines of candy at constant average
costs of 50 cents and 60 cents per pound respectively. If the selling price
of the first line is jr cents per pound and of the second line is y cents per
pound, the number of pounds which can be sold each week is given by the
formulas
N ^ _ 25Q(?/ _ ^ ^ = 3 ^ ()00 + 25Q(x __ 2 y).
Show that for maximum profit the selling prices should be fixed at 89 cents
and 94 cents per pound respectively.
14. A manufacturer produces razors and blades at a constant average
cost of 40 cents per razor and 20 cents per dozen blades. If the razors
are sold at x cents each and the blades at y cents per dozen, the demand
, ., , . , , . 4,000,000 , 8,000,000 , ,, ,
of the market each week is - - - - razors and - - - - dozen blades.
.r?/ ry
Find the selling prices for maximum profit.
242. Taylor's theorem for functions of two or more variables. The
expansion of /(x, y) is found by using the methods and results of
Arts. 194 and 240. We consider
(1) F(t}=f(x + ht,y + kt),
and expand F(t) as in (5), Art. 194. The result is
' "
(2)
We obtain the values of F(0), F'(Q), F"(0), by substituting t =
in (2), (4), (7), Art. 240. By differentiating (7), and putting 1 = 0,
the expressions for F r/ '(0) etc. will result. These are omitted here.
Note, however, that F'"(0) is homogeneous and of the third degree
APPLICATIONS OF PARTIAL DERIVATIVES 489
in h and k. A similar property holds for higher derivatives. If these
values are substituted in (2), and we set / = 1, the result is
(3) /(:r + h, y + k) =/(j, y) + hf,(x. y) +
R.
The expression for R is complicated and will be omitted from this
point on.
In (3) write x = a, # 6, and then replace A by (x a) and k by
(?/ ~~ &) The result is Taylor's theorem for a function of two variables,
- a) +/ y (a,
- a)(y -
Finally, setting a = 6 = 0, we obtain an expansion corresponding
to Maclaurin's series (4), Art. 194,
The right-hand member in (7) may be written as the infinite series
fA\ I U l I W ii I
(4) Mo + [T + f|" + '"'
where uo = f(0, 0),
w 2 =/(0, 0)x 2 + 2/, w (O f
etc.
These terms in (4) are homogeneous polynomials in (x, y). The
degree of each is equal to the subscript. That is, by (7) the function
is expanded into a sum of polynomials homogeneous in (x, y) and of
ascending degree. Similarly, in (/) the terms in the expansion are
polynomials homogeneous in (x a, y b).
Formula (/) is called the expansion off(x, y} at the point (a, b).
Reference must be made to more advanced treatises for proof of
the problem of determining those values of O, y) for which the
expansions (/) and (/) hold.
By breaking off series (4) at any term, an approximate formula for
f(x, y) is obtained for values near (a, b} or (0, 0). Compare Art. 200.
490 DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE. Expand
xy 2 4- sin xy
at the point (1, \ IT) up to terms of the third degree.
Solution. Here a = 1, b = | IT,
and f(x, ?/) = *y* + si" *y
f y (x, y) =2xy + x cos xy,
= -y 2 smxy,
= 2 # + cos x|/ - xy sin xy,
Substituting x = 1, y = J TT, the results are
/(I, JT) = iir'
/,(!, - 2 TT) - 1 7T2,
/(!, 2 T) = T.
= 2
Substituting in (/), we get
xy 2 + sin xy = l + -J
: 1) +7r(2/ ITT)
Formulas for expanding a function of three variables f(x t y, z) are
readily derived, and are left as problems.
PROBLEMS
1. From (1) above, show that
dr
+ 3
+ 3
2. Verify the following expansion.
cos x cos y = 1 TTT *
4- 6
J4
_ x 6 + 15 J 4 ?/ 2 + 15 j-V 4- y 6 ....
3. Expand sin .r sin y in powers of x and y.
4. Verify the following expansion.
a' log (1 + 10=0+ U2^1oga-t/ 2 + x 2 i/log 2 a
5. Expand .r 3 -f x?/ 2 at the point (1,2).
6. Verify the following expansion.
- 3 x * y + 3
sin (x 4- !/) = x + y
.
+
Verify the following approximate formulas for small values of x and y.
7. r r sin T/ = ?/ -f .r//. (
8. P J In (1 -h /) = I' 4- ry
CHAPTER XXV
MULTIPLE INTEGRALS
243. Partial and successive integration. Corresponding to partial
differentiation in the differential calculus we have the inverse process
of partial integration in the integral calculus. As may be inferred
from the connection, partial integration means that, having given a
differential expression involving two or more independent variables,
we integrate it, considering first a ///(//< on.e only as varying and all
the rest constant. Then we integrate the result, considering another
one as varying and the others constant, and so on. Such integrals
are called double, triple, etc., according to the number of variables,
and are known as multiple integrals.
In the solution of this problem the only new feature is that the
constant of integration has a new form. We shall illustrate this by
means of examples. Thus, suppose we wish to find u, having given
Integrating this with respect to j, considering ?/ as constant, we
have
u - x 2 + .r// + 3 x + 0,
where c/> denotes the constant of integration. But since y was re-
garded as constant during this integration, c/> may involve y. We
shall then indicate this dependence of 4> on y by replacing (/> by the
symbol <t>(y). Hence the most general form of u is
where </>(?/) denotes an arbitrary function of y.
As another problem let us find
u^\ I (x 2
This means that we wish to find u, having given
dxdy
491
492 DIFFERENTIAL AND INTEGRAL CALCULUS
Integrating first with respect to y, regarding x as constant, we get
()U 0,2,
where \l/(x) is an arbitrary function of x.
Now integrating this result with respect to x, regarding y as con-
stant, we have
where <b(y) is an arbitrary function of y, and
244. Definite double integral. Geometric interpretation. Let/(j, y)
be a continuous and single- valued function of x and y. Geometrically,
(1) *=f(*,V)
is the equation of a surface, as KL. Take some area 5 in the A'OY-
plane and construct upon S as a base the right cylinder whose
elements are parallel to OZ. Let this cylinder inclose the area S'
on KL. Let us now find the volume V of the solid bounded by
S, S', and the cylindrical surface. We proceed as follows :
At equal distances apart (= Ar) in the area 5 draw a set of lines
parallel to OY, and then a second set parallel to OX at equal distances
apart (= Ay). Through these lines pass planes parallel to YOZ and
XOZ respectively. Then
within the areas S and S f
we have a network of
lines, as in the figure, that
in S being composed of
rectangles, each of area
AJ Ay. This construction
divides the cylinder into
a number of vertical col-
umns, such as MNPQ,
whose upper and lower
bases are corre-
sponding portions
of the networks
in S' and 5 re-
spectively. As the upper bases of these columns are curvilinear, we
of course cannot calculate the volume of the columns directly. Let
MULTIPLE INTEGRALS 493
us replace these columns by prisms whose upper bases are found thus :
each column is cut through by a plane parallel to XOY passed through
that vertex of the upper base for which x and y have the least
numerical values. Thus the column MNPQ is replaced by the right
prism MNPR, the upper base being in a plane through P parallel
to the X01 T -plane.
If the coordinates of P are (x, y, z), then MP = z = /(x, y), and
therefore
(2) Volume of M \PIt = /(x, ?/)A// Ax.
Calculating the volume of each of the other prisms formed in the
same way by replacing ;r and // in (2) by corresponding values, and
adding the results, we obtain a volume \ rf approximately equal to
V ; that is,
(3) y'-vv j( .
where the double summation sign ^^ indicates that values of
two variables x, y must be taken account of in the quantity to be
summed up.
If now in the figure we increase the number of divisions of the
network in S indefinitely by letting Ax and A// diminish indefinitely,
and calculate in each case the double sum (3), then obviously V will
approach V as a limit, and hence we have the fundamental result
(4) V = Hm
A./ (I
A/y - (I
We show now that this limit, can be found by successive integration.
The required volume may be found as follows: Consider any
one of the slices into which the solid is divided by two successive
planes parallel to YOZ ; for example, the slice whose faces are FIHG
and JTL'K'. The thickness of this slice is Ax. Now the values of z
along the curve HI are found by writing x ~ OD in the equation
z = /(x, y) ; that is, along ///
Hence Area FIHG - f /(OD, y)dy.
JltF
The volume of the slice under discussion is approximately equal
to that of a prism with base FIHG and altitude Ax ; that is, equal to
Ax - area FIHG = Ax f /(OD, y)dy.
JDF
494 DIFFERENTIAL AND INTEGRAL CALCULUS
The required volume of the whole solid is evidently the limit of
the sum of all prisms constructed in like manner, as x (= OD) varies
from OA to OB ; that is,
/ OB /^DG
(5) V = / dx I J(x, y)dy.
JOA JDF
Similarly, it may be shown that
/m r S*KV
(6) V=l dy f(x,y)dx.
Joe JKW
The integrals (5) and (6) are also written 'in the more compact fonn
no II nlHi nOV nEU
I I /(*, y)dy dx and I I f(y, y)dx dy.
JOA JDF Joe JEW
In (5) the limits DP and DO are functions of a% since they are
found by solving the equation of the boundary curve of the base of
the solid for y.
Similarly, in (6) the limits EW and EV are functions of y. Now
comparison of (4), (5), and (6) gives the result
(A) V=limf(x,y)&y. Ax-
A*-0
Av-o
'i r fl i
/ Hx.
. _2 J V 2
where v\ and r L > are, in general, functions of //, and ?/i and y/2 functions
of ;r. The second integral sign in each case applies to the first dif-
ferential.
Equation (A) is an extension of the Fundamental Theorem of
Art. 150 to double sums.
Our result may be stated in the following form.
The definite double integral
I ' / /Or, y)dy d2
Ja-2 J u-t
may be interpreted as that portion of the volume of a right cylinder which
is included between the plane, XOY and the surface
z^/0*, J/),
the base of the cylinder being the area in the XOY -plane bounded by the
/7,/r?v><?
CM/IM. |/ = 1*1, y = U2, x = ai, x a 2 .
A similar statement holds for the second integral.
It is instructive to look upon the above process of finding the vol-
ume of the solid as follows.
MULTIPLE INTEGRALS
495
Consider a column with rectangular base dy dx and of altitude z as
an element of the volume. Summing up all such elements from
y = DP to y = DG, .r in the meanwhile being constant (say = OD),
gives the volume of a thin slice having FGHI as one face. The
volume of the whole solid is then found by summing up all such
slices from x = OA to x = OB.
In successive integration involving two variables the order of
integration denotes that the limits on the second integral sign cor-
respond to the variable whose differential is written first, the differ-
entials of the variables and their corresponding limits being written
in the reverse order. Before attempting to apply successive integra-
tion to practical problems it is best that the student should acquire
by practice some facility in evaluating definite multiple integrals.
ILLUSTRATIVE EXAMPLE 1. Find the value of the definite double integral
/a /"A/a- 1 x~
I (.r -f y)dydx.
Jo Jo
Solution, J ( O -f y)dy dx
Jo Jo '
Jo I Jo
ir 1 )^
= I (x v a- - x' 2 -f -
Anx.
Interpreting this result geometrically, we have found the volume of the solid of
cylindrical shape* standing on OAB as base and bounded at the top by the surface
(plane) z = x + y.
The solid here stands on a base in the A'OV-plane bounded by
y (line OB)
/ , , . . . , I from y limits ;
y v a~ x~ (quadrant of circle AH] J
x- (line OA) ^
x^a (line BE) _
, .. ..
f rom x limits.
ILLUSTRATIVE EXAMPLE 2. Verify
Solution,
(a y}x 2 dydx
ra rZ^V ?v 2 la r2h/i~
| (a - y}x' 2 dydx= I \ay-- \ x 2 dx= ~
Jo J b L ^ lo Jft -
/ a / ~\/a 2 x 2 9 ^j.'i
ILLUSTRATIVE EXAMPLE 3. Verify / I _ xdydx -
JQ JVa^ x 2 "
/* /* " v/a " y2 , , /* a f "|"v/a 2 - a; 2
Solution. / I _ xdydx=l \xy\ _ dx
JQ J_V a ^ _ x 2 J L J-Vaa-x^
496 DIFFERENTIAL AND INTEGRAL CALCULUS
In successive integration involving three variables the order of
integration is denoted in the same way as for two variables; that
is, the order of the limits on the integral signs, reading from the
inside to the left, is the same as the order of the corresponding
variables whose differentials are read from the inside to the right.
c 3 r 2 c r> *} *~\
ILLUSTRATIVE FA AMPLE 4. Verify I I / ry* dz dy djr =
Ai -h J'2. **
nt> fr /->;> r> j nr^ ~ rjj /*2f ~15
" I j-y~ dz dy dj- I I " / .n/- dz dy dr = I ' I xy 2 z dy dx
J' ' -k '! \.'h ' *k J\ L J2
f\\ r'2. r<\\ r2 ~\
/ x u * dyds^'3 I xy* dy \dx
-~ J\ ->2 Ml 1
35
In Problems 1 10 in the following list the solid whose volume
equals the value of the integral should be described.
PROBLEMS
Work out the following definite integrals.
(.r + L> )</// d.r = 5. 6.
'
t -
^ /(/ -" '''' ^ l "* 8. f f "' (.r
t/ l./o
2/
4. r r " ^ <''/ = , 7 9 - ('" f (j ~" + " 2) ''-" ^
Ji /o ' ()
2 rr ? ri r r '-' -
\ //f///(/.r = ] - r '. 10. I / f dij(Lr= I.
- ., ./o ' ' ' - ; o At
11. f" PV' sin (JO dp - \(a { - /r)(cos /J - cos a').
.//, .V
12. / >7T /" Mt(>sf/ p s j n ^ d p dO = ,\ a 1 '.
Jo . 'ii
/^TT /^a(1 ) t-(Y*(h
13. / / p 2 sin ^ (/p ^/
A) .'o
15. ~ //-'c (/c c/// (^ = i a^ia-* - 6 3 ).
16.
MULTIPLE INTEGRALS
497
c " r^ c * ~ x
17. / I I jrdzdjrdii = 3*5.
Jo <Jtf 'o
is. r 1 r 1 " x f 1 "'zdzduds --
Jo Jo /o
/2 /? rr^ I
r /
Jo Jo
20.
. f ' f T f r + V ' " ' *d: <ln itr = i r 1 -
Jo Jo Jo
7
245. Value of a definite double integral taken over a region S. In the
last article the definite double integral appeared as a volume. This
(joes not necessarily mean that every definite double integral is a vol-
ume, for the physical interpretation of the result, depends on the
nature of the quantities represented by z, //, z. If x, y, z are the co-
ordinates of a point in space, then the result is indeed a volume. In
order to give the definite double integral
in question an interpretation not neces-
sarily involving the geometric concept of
volume, we observe that the variable z
does not occur explicitly in the integral,
and therefore we may confine ourselves
to the A'OV-plane. In fact, let us con-
sider simply a region S in the A'O V-plane,
and a given function /(z, y). Within this
region construct rectangular elements of
area by drawing a network of lines, as in Art. 244. Choose a point
(z, y) of the rectangular element, of area Az A?/, either with in the rec-
tangle or on its periwrtrr. Form the product
/(z, ?/)Az A//,
and similar products for all other rectangular elements. Sum up
these products. The result is
Finally let Ax - 0, and A?/ * 0.
We write the result
(1) lim y V/(x, ?y)Az A?y - ff/Yz, y}dx dy,
Aa'-O*""'^ .''
y->Q S
and call it the double integral of the function f(x, ?/) taken over the region S.
By (A) the value of the left-hand member in (1) was found by
successive integration when /(y, y) had no negative values for the
region S. The reasoning of Art. 244 will hold, however, if the portion
498 DIFFERENTIAL AND INTEGRAL CALCULUS
S' of the surface z =/(z, y) lies below the plane XOY. The limit of
the double sum will then be the volume with a negative sign. The
integrals in (A) will give the same negative number. Finally, if
/(x, y) is sometimes positive, sometimes negative for points of S, we
may divide S into subregions in which J(x, y) will be either always
positive or always negative. The reasoning will hold for each sub-
region and therefore for the combined region S. Hence the con-
clusion: the double integral in (1; may be evaluated in all cases by
successive integration .
It remains to explain the method of determining the limits of
integration. This is done in the next article.
246. Plane area as a definite double integral. Rectangular coordi-
nates. The problem of plane areas has been solved by single integra-
tion in Art-. 145. The discussion using double integration is useful
chiefly because the determination of limits for the general problem
of Art. 245 is made clear. To set up the desired double integral,
proceed as follows.
Draw a network of rectangles as before. Then, in the figure,
(1) Element of area = Ax A//.
If A is the entire area of the region S, obviously, by (1), Art. 245,
Ay - f fd
J J
OB) A = lim ]l;t Ay - dx dy.
Ax-0 ^"- f J J
Ay *0 S
Referring to the result stated in Art. 245, we may say :
The area of any region is the value of the double integral of the
function f(x, y) = 1 taken over that region.
Or, also : The area equals numerically the volume of a right cylinder
of unit height erected on the base S. (Art. 244.)
The examples show how the limits of integration are found.
ILLUSTRATIVE EXAMPLE 1. Calculate that portion of the area above OX which
is bounded by the semicubical parabola y' 2 x* and the straight line y = x.
Solution. The order of integration is indicated in the figure. Integrate first
with respect to jc. That is, sum up first the elements dr dy in a horizontal strip.
Then we have
rAC pAC
I djc dy = dy I dx area of a horizontal strip of altitude dy.
JAB JAB
Next, integrate this result with respect to y. This corresponds to summing up
all horizontal strips. In this way we obtain
rOD rAC
A = \ dxdy.
Jo JAB
MULTIPLE INTEGRALS
499
The limits AB and AC are found by solving each of the equations of the bound-
ing curves for x. Thus from the equation of the line, jr = AB - y, and from the
equation of the curve, .r = AC - ?/-" { . To determine ()/>,
solve the two equations simultaneously to find the point
of intersection E. This gives the point (1, 1); hence
OD = 1. Therefore A
Am.
Or we may begin by summing up the elements dr dy
in a vertical strip, and then sum up these strips. We
shall then have
'* ~ / / ^ IJ d ' r ~ I ^" r ~ jl> ^ r " i ~
In this example either order of integration may be chosen. This is not al-
ways the case, as the following example shows.
ILLUSTRATIVE EXAMPLE 2. Find the area in
the first quadrant bounded by the .r-axis arid the
curves j;2 + 2,2=10, tf = $ Xf
Solution. Here we first integrate with respect
to x to cover a horizontal strip, that is, from the
parabola to the circle. We then have, for the
entire area. i<{
A = I I d.r (///,
Jo .'IK;
since the point of intersection is (1, ,3). To find
}JG, solve y~ = 9 x for .r. Then
To find ///, solve x~ -f y 2 = 10 for x. We get
jc HI = + VlO ?/-'.
Hence
^ _ (' C y fly ,1*, \ li
i -|3
- ^- i/M =6.75.
27 |
Ans.
If we integrate first with respect to y, using ver-
tical strips, two integrals are necessary. Then
na'v'j /^ / Io /
dydx+ /
Ji Jo
The order of integration should be such that the
area is given by one integral, if this is possible.
The examples above show that we set
or A=ffdydx
500
DIFFERENTIAL AND INTEGRAL CALCULUS
according to the nature of the curves bounding the area. The figures
below illustrate, in a general way, the difference in the summation
processes indicated by the two integrals.
PROBLEMS
1. Find by double integration the area between the two parabolas
,'J \f l 25 jr and 5 jr'' 9 //, (a) by integrating first with respect to y ;
(b) by integrating first with respect to .r.
,<\ ~^LL ^5 /-V ^
AH*, (a) I I </// f/.r 5 ; (b) I \ ' (i.r dy 5.
*> 25
Calculate by double integration the finite area bounded by each of the
following pairs of curves.
2. y =: 4 .r - .r 1 ', y ~ jr.
3. y' 2 = 4 j-, 2 j- - // = 4.
4. ?/ ~ .r 2 , 2 .r - // + H = 0.
5. r' = 2 .r, .r 2 ^ (> //.
6. y' 2 4 .r, jr = 12 + 2 // ?/~.
7. // 2 2 .r, .r 2 + //'' 4 jc.
8. ?/ 2 = 9 + J, // 2 = 9 - :i s.
9. (j- 2 + 4 2 )// = 8 \ 2 // = .r, j- = 0.
10. jr^ + //* =a-', JT + // = a.
11. .r' + // r< = a-\ jr + // = a.
12. ?/ = j- :i - 2 j-, // = 6 .r -~ r : .
9.
4.
48.
a 2 (7r-l).
- 3 7r)a 2 .
16.
13. j = 6 ?/ - ?/ 2 , 2/ = .r. 16. .r 2 + ?/ 2 = 25, 27 ?/ 2 = 16 x?.
14. 4 // 2 = x'\ u - x. 17. (2 a - .r)// 2 = .r 3 , ?/ = or.
15. ?/ 2 = .r + 4, /y 2 = 4-2 .r. 18. ;r 2 - 7/ 2 = 14, x- + ?/ 2 = 36.
MULTIPLE INTEGRALS
501
247. Volume under a surface. In Art 244 we discussed the volume
of a solid bounded by a surface
(l) ; = M 0),
the A'Oy-plane, and a cylinder. The elements of the cylinder were
parallel to OZ, and its base was a region S in the A'OV-plane. The
volume of this solid is, by 04),
(2)
V =
c, y)dx dy.
The order of integration and the limits are the same as for the
area of the region S. The volume of a solid of this type is the "volume
under the surface (1)." The analogous problem for the plane, "area
under a curve," has been treated in Chapter XIV. As a special case
the volume may be bounded by the surface and the A'OF-plane itself.
Note that the element of volume in (2 ) is
a right prism with base d.r dy and altitude z.
ILLUSTRATIVE EXAMPLE 1. Find the volume
bounded by the elliptic paraboloid
(3) 4 z= 16 ~-4.r 2 - \r
and the AW -plane.
Solution. Solving (3) for z, we get
(4) 2 = 4- a >J - 1 //'.
Letting 2 = 0, we obtain
(5) 4 JT' J + //- = 16,
which is the equation of the perimeter of the base
of the solid in the AW -plane. Hence by (2), using
the value of z in (4),
(6) V -- 4 1 I (4 - x- - \ ij' 2 }dy dx = 16 TT. Am.
The limits are taken for the area OAB of the ellipse (5) lying in the first quadrant.
ILLUSTRATIVE K \AMPLE 2. Find the volume of the
solid bounded by the paraboloid of revolution
(7) -r* + ?r - nz,
the A'Oy-plane, and the cylinder
(8) .r- 4 !i 2 = 2 ax.
Solution. Solving (7) for 2, and finding the limits
for the area of the base of the cylinder (8) in the XOY-
plane, we get, using (2;,
Jo Jo
dy dx l - 7ra :< . Ans.
For the area ON A (see figure), MN = v2 ax - x' 2 , (solving (8) for #), and OA = 2 a.
These are the limits.
502 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
1. Find the volume under z = 4 x' 2 , above z 0, and within y' 1 = 4 x.
Ann. V = 2 f' 2 f 2 X Z (4 - x 2 )dy dx = 17.24.
Jo Jo
2. Find the volume under the plane x -f z = 2, above z 0, and within
x 2 -f ?y 2 = 4. r2 /-A- 4
An*. F =
r2 /-A- 4 x- 1
2/ / (2 - .r;c/?/ dr = 8 TT.
J- 2 J()
3. Find the volume bounded by the plane - -f y -f - 1 and the co-
ordinate planes. a ; r ^^ iabCu
4. Plnd the volume bounded above by j" -f z = 4, below by ^ = 0, and
laterally by // 2 = 4 .r. Ans. \ l .
5. Find the volume of the solid bounded above by //'-' = a 2 az, below
by 2 = 0, and within x' 2 -f //- a 2 . -4s. % ira*.
6. Find the volume under the elliptic paraboloid z 1 ] .r 2 J // 2
and above z 0. A/i,s\ 3 TV.
7. Find the volume under the plane x -f y -f z = 8, above ^ = 0, and
between the planes j: -f 2 // 8, x 2 // = 8. Ans. 170^.
8. Find the volume bounded by the cylindrical surface x 2 -f ac a 2
and the planes x -f // = a, // = 0, c = 0. yl/^s'. 4 { a ;i .
9. A solid is bounded by the surfaces ?/ 2 -f 2 == 4 ar, x = 3 c/, and
lies within // 2 = a.r. Find the volume. Ann. (6 TT 4- 9\/3)a :i .
10. Plnd the volume below the cylindrical surface ?/ 2 = a 2 a^, above
2 = 0, and wit?iin the cylindrical surface .r 2 + // 2 ax. Anx. % yra- 5 .
11. Find the volume below ,: 2 x -f , al>ove ^ = 0, and within
x 2 -f- // 2 = 2 ax. .4j/s. 3 ?ra :i .
12. Find the volume under // 2 + ^ = 4, above r = 0, and within the
cylindrical surfaces ?/ 2 2 .r 0, // 2 8 2 .r. Ans. -\ l .
13. A solid is bounded by the paraboloid .r 2 + ?/ 2 az, the cylindrical
surface ?/ 2 = a 2 a.r, and the planes .r 0, z = 0. Find the volume.
Ana. $ a 3 .
14. Find the volume under 4 z = 16 4 j- 2 ?/ 2 , above ^ = 0, and within
x* + y* = 2x. ' A?w?. ']! TT.
15. The axes of two circular cylindrical surfaces intersect at right angles
and their radii are equal (= r}. Find the common volume. .4??.s\ } ^ r 3 .
16. Find the volume of the closed surface jr :J + y ' + - y = a 3 . (The trace
on each coordinate plane is the astroid, Chapter XXVI.) Ans. $ ?ra 3 .
17. Find the volume common to ?/ 2 4- z 2 4 a.r and x 2 -f ?/ 2 = 2 a.r.
(2 TT + J #)a 3 .
MULTIPLE INTEGRALS
503
248. Directions for setting up a double integral. We shall now state
a rule for forming the double integral which will give a required
property. Applications are made in the following articles. For
single integration the corresponding rule is given in Art. 156.
FIRST STEP. Draw the curves 'which bound the region,, or area, con-
corned.
SECOND STOP. At any point I\.r, y) irithin the area construct the
rectangular element of area AJ A//.
THIRD STEP. Work out the function f(jr, //), which, 'when, multiplied
by AT A?/, gives the required property for the rectangular element of area.
FOURTH STEP. The required integral is
ffi
/(.r, y)djr <
taken over the given region, or area. The order of integration and limits
are determined in the same wanner a,s in finding the area, itself.
249. Moment of area and centroids. This problem is treated in
Art. 177 by single integration. Double integration is often more
convenient.
We follow the rule of the preceding article. The moments of area
for the rectangular element of urea are, respectively,
.r AJ A?/, with respect to OV,
y AJ A'//, with respect to () V.
Hence for the entire area, using the notation of Art. 177, we have
(C) M x =j*J*ydxdy, M y = ffxdxdy. *
The centroid of the area is given by
(D)
x =
area " area
In (C) the integrals give the values of the .
double integrals of the functions
f(x, y) = y an( l fte, y) = x >
respectively, taken over the area. (Art. 245.)
For an area bounded by a curve, the x-axis, and two ordinates
(the "area under a curve"), w ^ derive from (C)
(D
fft rv rb
I y dy dx - i I y' 2 dx,
J() Ja
r\> rv r^>
My == I I x c ^ dx= I xy dx.
Ja Jo Ja
504
DIFFERENTIAL AND INTEGRAL CALCULUS
These agree with (2), Art. 177. Note that y in (!) is the ordinate
of a point on the curve, and its value in terms of x must be found
from the equation of the curve and substituted in the integrand be-
fore integration.
ILLUSTRATIVE EXAMPLE. Find the centroid of the area in the first quadrant
bounded by the semicubical parabola y~ j ( and the
straight line y x.
Solution. The order and the limits of integration
were found in Illustrative Example 1, Art. 24(5. Ileuce,
using (C),
r 1 c r 1 ?j ., i
1 ~J() Jy * ' Jo
2
M v =11 y djc dy = i / (.v a y~)dy = fa.
Jo J y ,/o
Since A area i 1 ,,,
we have, from (D), ?=J-0.48, // = A> = 0.42. An*.
250, Theorem of Pappus. A useful relation between centroids and
volumes of solids of revolution is expressed in the following theorem.
// a plane area is reroltwl about an axis lying in its plane and not
crossing it, the volume of the solid of revolution, thus generated is equal
to the prod net of the plane area by the circumference described by its
centroid.
Proof. Let the area in the figure be
revolved about the x-axis. The rectangu-
lar element of area within the region X at
P will generate a hollow circular cylinder
whose volume A Y is given by
_\ y = TT(// + A//) L> AJ Try 2 A.r.
Factoring and simplifying, we get
AV-2 TTG/+ -2 A//) Ax A//.
Now, in (1), Art. 245, (x, y) in/(j, y) is a point " either within the
rectangle PQ or on its perimeter." But (x, y + \ A//) is a point on
the perimeter of PQ. Therefore let f(x, y > = 2 Try. Then A V has the
form fix, //) Ax A//, and, by (1), Art. 245, and (C),
ii) r x = 2
Finally, using (D), we get
(2) V, - 2 iry A,
MULTIPLE INTEGRALS
505
where A is the area of the region 5. The right-hand member is the
product of the area by the circumference described by its centroid.
Hence the theorem is proved. We write the result
(3) V = 2 iry - A.
If two of the quantities V, y, A are known, the other can be found
by (3).
ILLUSTRATIVE EXAMPLE. Find the centroid of the trapezoid OMPB of the
figure by the Theorem of Pappus.
Solution. Area OMPR = J(3 + 5^,8 = 32. Revolving the figure about OX, the
solid formed is a frustum of a cone of revolution. Hence, by (12), Art. 1, since
a = 8, R = 5, r = 3,
Hence, by (3), "y = ~-~ = f^ = 2.04.
2 7T-4 19J
Revolving the figure about OY, the volume generated
is the difference of the volumes of the cylinder generated
by OrPM and the cone generated by the triangle HCP.
M
Hence
V u = 320 7T -
12S TT
3
Hence, by the theorem, x
The centroid is (4j, 12.04). Arts.
_
3*
PROBLEMS
Find the centroid of the area bounded by each of the following curves:
1. y = .r 3 , y 4 jr. (Area in first quadrant.) Am. ([ Ji, !j \).
2. y = 6 x jr 2 , ?/ .r.
3. ?/ = 4 x - j'~, // = 2 jr - 3.
4. x 2 = 4 ?/, r - 2 ?/ -f 4 - 0.
5. ?/ = ^ 2 , 2 .r - y 4-3 = 0.
6. y = x 2 - 2 .r - 3, ?/ = 2 .r - 3.
7. y* = jr f jr 4- ?y = 2, ?y = 0. (First quadrant.)
8. ?/ 2 = .r, r -f ?/ = 2, x = 0.
9. ?/ 3 = J 2 , 2 // = r.
10. 4 ?/ - 3 j- 2 , 2 ?/ 2 = 9 x.
11. ?/ 2 = 2 x, ?/ = r ~ x 2 .
12. ?/ = 8 x, x + ?/ = 6,
13. y 2 = 4 .r, y 2 = 5 - x.
14. y = 6 x - x 2 , x 4- ?/ = 6.
(. 5).
(1, I)-
0. g)-
0. -'e 1 )-
(2,-f).
(I!. A)-
/ H 1 1 \
V2 T,' TOV'
/J 4 ON
V 3" dn)-
/ 9 2 7 \
Vii)' J'o>
(U- - H
(-4).
(.0).
506 DIFFERENTIAL AND INTEGRAL CALCULUS
Ans. (-V-, I).
15. x = 4 y y 2 , y = x.
16. ?/ = 4 x x 2 , # 5 - 2 a;.
17. ?/ 2 = 4 x, 2 x - y = 4.
18. ?/ = x 2 - 2 x - 3, y = 6 x - x 2 - 8.
19. * 2 + 2 = l, z + jy = l.
20. x 2 + ?/ 2 = 32, ?y 2 = 4 x.
21. i/ 2 = 4 x, 2 x + y = 4.
22. x 2 + ?y 2 - 10 x = 0, x 2 = y.
23. x 2 = ?y, 2 |/ = 6 x - x 2 .
24. x 3 -f y 3 = a 3 . (Area in first quadrant.)
25. x 1 * -f ?y^ = a*, x = 0, y = 0.
(3, I).
(2, 1).
(0.58.5. 0.585).
/25B(7 256 a \
\31f> Tr'ttlo 7T/
(-' -V
\5 5/
26. P'ind the controid of the area under one arch of the cycloid
x = a(9 sin 0), y = <j(l cos 0). * ^ / 5 o\
I 71 * ' 6 /
27. Using the Theorem of Pappus, find the centroid of a semicircle.
Ans. Distance from diameter = -
3 TT
28. Using the Theorem of Pappus, find the centroid of the area of the
ellipse \-~-~l which lies in the first quadrant. ..,,,,. , , ,
a 2 ft- \3 TT 3 7T/
29. Using the Theorem of Pappus find the volume of the torus gen-
erated by revolving the circle (x ft) 2 -f- ?y 2 = a 2 (ft > a) about the ?y-axis.
A M. 2 ?T 2 a 2 ft.
30. A rectangle is revolved about an axis which lies in its plane and is
perpendicular to a diagonal at its extremity. Find the volume of the
solid generated.
251. Center of fluid pressure. The
problem of calculating the pressure of
a fluid on a vertical wall was discussed
in Art. 179.
The pressures on the rectangular ele-
ments of the figure constitute a system
of parallel forces, since they are per-
pendicular to the plane of the area
XOY. The resultant of this system of
forces is the total fluid pressure P, given
by (Z)), Art. 179.
D
(D
r b
' = W I yxdx.
Ja
MULTIPLE INTEGRALS 507
The point of application of /' is called the center of fluid pressure.
We wish to find the .r -coordinate (~ .r u ) of this point.
To this end we use the principle of force nionieuts. This may be
stated thus :
The sum of the turning moments of a system of parallel forces
about an axis is equal to the turning moment of their resultant about
this axis.
Now the fluid pressure dP on the rectangular element EP is, by
Art. 179,
(2) dP = TT>// A:r.
The turning moment of this force about the axis OY is the product
of dP by its lever arm OK (== .r), or, using (2),
(3) Turning moment of dP = x dP = Wx 2 y Ax.
Hence we have, for the entire turning moment for the distributed
fluid pressure,
r' 1
(4) Total turning moment / Wx~{/dx.
But the turning moment of the resultant fluid pressure P is x^P.
Hence
(5) x ( )P W I x' 2 ydx.
Solving for /<> and using (1 ), we get the formula for the depth of the
center of pressure
f *x 2 dA
(6) a-o = -
/'
Jii
xdA
where dA element of area = ydx.
The denominator in (6) is the moment of area of ABCD with
respect to OY (see Art. 177). The numerator is an integral not met
with hitherto. It is called the moment of inertia of the area ABCD
about OY.
The letter / is commonly used for moment of inertia about an
axis, and a subscript is attached to designate the axis. Thus (6)
becomes
508 DIFFERENTIAL AND INTEGRAL CALCULUS
The usual notation for moment of inertia about an axis I is
(8)
in which
(9)
r = distance of the element dA from the axis I.
The problem of this article is one of many which lead to moments
of inertia. In the following section the calculation of moments of
inertia by double and single integration is explained. Applications
are also given.
252. Moment of inertia of an area. In mechanics the moment of
inertia of an area about an axis is an important concept. The calcu-
lation of moments of inertia will now be explained. We follow the
rule of Art. 248. A
For the elementary rectangle PQ at
P(j, y) the moment of inertia about OA'
is defined as
(1) '// 2 AxA*/,
and about the ?/-axis it is
(2)
Then, if I x and I y are the corresponding moments of inertia for the
entire area, we have (compare (8), Art. 251)
The radii of gyration r x and r v are given by
*U
2 _
2
area
area
In (E) the functions whose integrals are taken over the area are,
respectively, /(.r, y) = y 2 , and f(x, y) = .r 2 .
Formulas (E) become simple for an area "under a curve," that is,
an area bounded by a curve, the x-axis and two ordinates. Thus we
obtain r \, ry ~ b
I*= / V 2 dydx=l I y*dx,
Ja J() J u.
(3)
/*& rv rb
,= \ x 2 dydx= / yzydx
J a Jo J a
In these equations y is the ordinate of a point on the curve, and
its value in terms of x must be found from the equation of this curve
and substituted in the integrand.
MULTIPLE INTEGRALS
509
Formulas for moments of inertia / are written in the form
(G) I=Ar*,
where A = area and r radius of gyration. Solving (F) for I x and I v
will give this form.
Dimensions. If the linear unit is 1 in., the moment of inertia
has the dimensions in. 4 . By (F), r x and r v are lengths, in inches.
ILLUSTRATIVE EXAMPLE 1. Find /,, /, and the corresponding radii of gyration
for the area of Illustrative Example 1, Art. 246.
Solution. Using the same order of integration and
the same limits as before, we have, by (E),
Since A - area = 1 J n , we find, by (F),
r t = 0.48, r t = 0.53. Am.
ILLUSTRATIVE EXAMPLE 2. Find I s and I v for the parabolic segment BOC in
the figure.
Solution. With the axes of coordinates as drawn, the equation of the bounding
parabola is
(4) y' 2 = 2 px.
Since B(a, b} is a point on the curve, we get, by substitut-
ing x a, y = b in (4), b' 2 = 2 pa. Solving this equation for
2 p and substituting its value in (4), we obtain
2 &*jr bx*
y 2 = , or 2/ ~
a ^7
(5)
O
The moments of inertia for the area under the parab-
ola OPB in the first quadrant will be half the required
moments. Hence, using (3j, and substituting the value of
y from (5), we get
B(a,b)
15
For the area of the segment, we find
Hence, by (f), rS = = 6 s , and /, = A6 2 ,
The results are in the form (G). Ans.
510
DIFFERENTIAL AND INTEGRAL CALCULUS
In the figure on page 325 the axis Y lies in the surface of the fluid.
If we denote this axis in any figure by s, then the depth of the center
of pressure, is by (7), Art. 251
if r = radius of gyration about the axis s,
and h a = depth of centroid below the axis s.
ILLUSTRATIVE EXAMPLE 3. Find the depth of the center of pressure on the
trapezoidal water gate of the figure. Compare Illustrative Example 2, Art. 179.
Solution. Choose axes OX and OY as shown, and draw an elementary hori-
zontal strip. Let the distance of this strip from the axis s at the water level be r.
r = 8 - y, dA = 2 x dy.
Hence, by (8), Art. 251, and by the
definition of moment of area (Art. 177),
we have
(7)
(8) M, =
The equation of AB is y = 2 x - 8.
Solving this for x, substituting in (7) and (8), and integrating with limits y = 0,
y = 4, we obtain 4
/. = f (8 - i/) 2 (8 + y)dy = 1429 J,
A)
M. = f 4 (64 - y*)dy = 234.
Hence, by (7), Art. 251, x tt = 6.09. Ans.
253. Polar moment of inertia. The moment of inertia of the ele-
mentary rectangle PQ about the origin O is the product of the area
and OP 2 , that is,
(1) (^ + \
Hence, by Art. 248, for the entire area
WATEK LEVEL
1
1
1* 1
\
//MM)
\
/(^>2/)
\
/
; o J!(4.o)
L- To ^
y
(2)
ij 2 )dx dy.
We may, however, write the right-hand
member as the sum of two integrals, for
(2) is clearly the same as
(3) lo-jy^dxdy+JJy*
Hence we have the following theorem.
The moment of inertia of an area about the origin equals the sum oj
its moments of inertia about the x-axis and the y-axis.
/.
MULTIPLE INTEGRALS 511
PROBLEMS
Find /*, /, and / for each of the areas described below.
1. The semicircle which is to the right of the ?/-axis and which is
bounded by x 2 + y 2 = r 2 . . , __ . _ Ar 2
/ins. If iy -"~:
2. The isosceles triangle of height h and base a whose vertices are
/A m /i, a \ li a \ A j Aa' 2 T Ah' 2
(0> 0) 'r' 2/'\ ' ~2/ LS> '^"iSr v = ~'
3. The right triangle whose vertices are (0, 0), (b, a), (b, 0).
r - ^ a2 7 - Ab *
8. 1 x T7 ' - y Q *
4. The ellipse + | = 1. 4ns. /x = 4r~ ^ = '
a 2 b 2 44
5. The area in the first quadrant bounded by y' 2 4 .r, x = 4, ?/ = 0.
j ~ 5 ~ 7
J* 2 77 2
6. The area included between the ellipse - + *- = 1 and the circle
y* = 2y. ; 4 r 19 A 7 53 A
Am. y^-^-,/^-^.
j-2 ^i2 f<2
7. The area included between the ellipses -f TT 1 ano ^ x 2 + 4- = 1.
16 9 4
, r 5 A r 19 A
Arts. / x ^ , 7 V ^ |
8. The area included between the circle x 2 -f 2/ 2 16 and the circle
(y + 2)* = l. . _239 A _17A r
Aw*. /,- 6Q , /- 4
9. The area included between the circle x' 2 + ?/ 2 = 36 and the circle
( + 3)* = 4. 4ns j^ OA>
T 2 77 2
, 10. The area betwee'n the circle x 2 -f 7/ 2 = 4 and the ellipse -f f- = 1.
36 16
. r 23 A r 53 A
Aws. /, = , / tf = -y
11. The entire area bounded by xf -h y* = a^. Ans. I x = I v = Z_4.
64
12. Find the depth of the center of pressure on a triangular water
gate having its vertex below the base, which is horizontal and on a level
with the surface of the water.
13. Find the depth of the center of pressure on a rectangular water
gate 8 ft. wide and 4 ft. deep when the level of the water is 5 ft. above
the top of the gate. Ans. 7.19 ft. below the surface of the water.
512
DIFFERENTIAL AND INTEGRAL CALCULUS
14. Find the depth of the center of pressure on the end of a horizontal
cylindrical oil tank of diameter 5 ft. when the depth of oil is (a) 2.5 ft.;
(b) 4ft.; (c) 6ft.
(a)
==1 .47f t .. (b) approximately 2.4 ft. ;
= 3*95 ft.
254. Polar coordinates. Plane area. When the equations of the
curves bounding an area are given in polar coordinates, certain modi-
fications are necessary.
The area is now divided into elementary portions, as follows:
Draw arcs of circles with the common center with successive
radii differing by Ap. Thus, in Fig. 1, OP = p, OS = p + Ap. Then
-A r U
-x
FIG. 1
draw radial lines from O such that the angle between any two con-
secutive lines is the same and equal to A0. Thus, in Fig. 1, angle
FOR = A0.
The area will now contain a large number of rectangular portions,
such as PSQR in Fig. 1.
Let PSQR = AA. Now AA is the difference of the areas of the
circular sectors FOR and SOQ. Hence
(1) AA == i
^ Ap 2 A0.
The function f(x, ?/) of Art. 245 is to be replaced by a function
using polar coordinates. Let this be F(p, 6). Then, proceeding as
in Art. 245, we choose a point (p, 6) of AA, form the product
F(p, 0)AA
for each A A within the region S, add these products, and finally let
Ap and A0 > 0. It is shown in Art. 258 that the limiting value
of this double sum may be found by successive integration. We now
write (compare (1), Art. 245)
(2)
lim
Ap - '
AH -
AA
=f (V(p
J J
, d)pdpd6,
MULTIPLE INTEGRALS 513
and call it the double integral of the function F(p, 8} taken over the
region S.
Note in (2) that the value of A A in (1) has been replaced in the
integral by p dp dO.
The simplest case of (2) is that of finding the area of the region S.
We then have
(H) A= CCp dp dB =JTp dO dp.
These are easily remembered if we think of the elements (checks)
as being rectangles with dimensions p dd and dp, and hence of area
p dd dp.
The figures below illustrate, in a general way, the difference in
the processes indicated by the two integrals.
In the first, we integrate first with respect to p, since dp precedes
dd, keeping 6 constant. This process will cover the radial strip KGHL
in Fig. 2, p. 512. The limits for p are p = OG and p = OH, found
by solving the equation (or equations) of the bounding curve (or
curves) for p in terms of 0. Then integrate varying 0, the limits
being = Z JOX and 6 = Z IOX.
The second integral in (2) is worked out by integrating with re-
spect to 0, p remaining constant. This step covers the circular strip
ABCD in Fig. 1, p. 512, between two consecutive circular arcs.
Then integrate varying p.
When the area is bounded by a curve and two of its radii vectores
(area swept over by the radius vector), we obtain from the first form
r ft r? + r
= I pdpdd = %
Jot t/0 Jci
which agrees with (D), Art. 159.
Double integrals in polar coordinates have one of the forms
(3) ffF(p,0)pdpd0 or JjF(p,0)pdOdp.
514 DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 1. Find the limits for the double integral giving some
required property related to the ar^a inside the circle p = 2 r cos 6 and outside the
circle p ~ r.
Solution. The points of intersection are
A(T, ~j. and B(T, - |V Use the first form
in (3).
The limits for p are
p = O//-2r costf;
for 6 they are ~ and ^ - Ans.
o o
ILLUSTRATIVE EXAMPLE 2. Find the area inside the circle p = 2 r cos 6 and
outside the circle p r.
Solution. From Illustrative Example 1 above, we have
7L
r(4 r 2 cos 2 6 - r*)d6 = r 2 (i IT + J Vs) = 1 .91 r\ Ans.
255. Problems using polar coordinates. There should now be no
difficulty in establishing the following formulas:
(1) M x = ffp 2 sin 9 dp d6.
(2) M v = CCp*ca&8dpdO.
(3) /,= CCp*siY\ 2 6dpd8.
(4) /
(5)
The order of the differentials will have to be changed if integration
with respect to 6 is performed first.
ILLUSTRATIVE EXAMPLE 1. On account of important applications the moments
of inertia of a circle are now worked out.
Let a = radius. Then, by (5), the polar moment
of inertia with respect to the center is
(6) 7 =
where A = area of the circle. ~TV\~\ V ' / / / / / x
Also, since I, = I v , by symmetry , we have, by (3),
Art. 253,
MULTIPLE INTEGRALS
515
In words : The polar moment of inertia of a circle with respect to its center equal*
the product of half the area and the square of the radius; the polar moment of inertia
with respect to any diameter equate the product of one fourth the area and the square
of the radius.
ILLUSTRATIVE EXAMPLE 2. Find the centroid of a loop of the lemniscate
p* = a' 2 cos 2 6.
Solution. Since OA" is an axis of symmetry, we
have y = 0.
i r -V * r (i v f t >H 2 ft
\A^\< f pdpd.6
Jo Jo
X
p" f " Vros2 V'cos
Jo '(}
3 3 f * " (1 - 2 sin 2 0)* cos 9 dO
Jo
Hence z = = a v 2 = 0.55 a. Ana.
by (5), Art. 2
ILLUSTRATIVE EXAMPLE 3. Find 7 over the region bounded by the circle
p = 2r cos B.
Solution. Summing up for the elements in
the triangular-shaped strip OP, the p-limits
are zero and 2 r cos (found from the equation
of the circle).
Summing up for all such strips, the 0-limits
are - ^ and - Hence, by (5),
. Ann.
Or, summing up first for the elements in a circular strip (as QR), we have
PROBLEMS
1 Find the area inside the circle p = $ and to the right of the line
4 p cose = 3. " Arts . 3(4, -3Vi).
ID
2. Find the area which is inside the circle p = 3 cos and outside the
circle p = f .
A 718.
3(2 7T + 3V3)
516 DIFFERENTIAL AND INTEGRAL CALCULUS
3. Find the area which is inside the circle p = 3 cos 6 and outside the
circle p = cos 6. Ans. 2 IT.
4. Find the area inside the cardioid p = 1 + cos 6 and to the right of
the line 4 p cos 6 = 3. ^ 9\/3
Ans. g + -16-
5. Find the area which is inside the cardioid p = 1 -f cos 6 and outside
the circle p = 1. AnK I + 2 .
8. Find the area which is inside the circle p = 1 and outside the cardi-
oid p=a + cos0. AnSt 2 __]r t
4
7. Find the area which is inside the circle p = 3 cos 6 and outside the
cardioid p = 1 + cos 9. Ans. IT.
8. Find the area which is inside the circle p 1 and outside the parab-
ola p(l + cos 0) = 1. 4ws . !_?.
-in.s. 2 3
9. Find the area which is inside the cardioid p = 1 -f cos 6 and outside
the parabola p(l + cos 0) = 1. x_ 3jr,4.
43
10. Find the area which is inside the circle p = cos -f sin and out-
side the circle p 1. Ans. %.
11. Find the area which is inside the circle p = sin 6 and outside the
cardioid p = 1 cos 0. Ans. 1
4
12. Find the area which is inside the lemniscate p 2 2 a 2 cos 2 6 and
outside the circle p = a. Ans. 0.684 a 2 .
13. Find th area which is inside the cardioid p = 4(1 -f cos 6) and
outside the parabola p(l - cos 9) = 3. Ans. 5.504.
14. Find the area which is inside the circle p = 2 a cos 9 and outside
the circle p a. Find the centroid of the area and I f arid /.
-+ -
15. Find the centroid of the area bounded by the cardioid ^ g ^
p = a(l + cos0). Ans. x =
16. Find the centroid of the area bounded by a loop of the curve
= acos20.
105 ir
17. Find the centroid of the area bounded by a loop of the curve
= acos30. A - 81 Via.
Ans - x ~ SO IT
MULTIPLE INTEGRALS
51 1 !
18. Find / for the lemniscate p 2 - a 2 cos 2 0.
19. Find I r for the cardioid p = a(l -h cos 6}.
20. Find /, and I v for one loop of the curve p = a cos 2 0.
21. Prove from (1), Art. 254, that
Am. ~ (3 TT + 8)a 2 .
48
lim
mil
o p Ap SO
-
and therefore SA "differs from p Ap SO by an infinitesimal of higher
order" (Art. 99). Then SA in the left-hand member of (2), Art. 254, may
be replaced by p Ap SO. (Proof omitted.)
256. General method for finding the areas of curved surfaces. The
method given in Art. 164 applied only to the area of a surface of
revolution. We shall now give a more general method. Let
(1; z=f(jc,y)
be the equation of the surface KL in the figure, and suppose it is
required to calculate the area of the region S f lying on the surface.
Denote by tf the region on the A'Oi'-plane which is the orthogonal
projection of S f on that plane. Now pass planes parallel to YOZ and
XOZ at common distances
Ax and Sy respectively. As
in Art. 244, these planes form
truncated prisms (as PB)
bounded at the top by a
portion (as PQ) of the given
surface whose projection on
the A'O r -plane is a rectangle
of area Ax A// (as AB). This
rectangle also forms the lower
base of the prism. The coor-
dinates of P are (x, y, z).
Now consider the plane
tangent to the surface KL
at P. Evidently the same rectangle AB is the projection on the
A'OY-plane of that portion of the tangent plane (PR) which is in-
tercepted by the prism PB. Assuming 7 as the angle the tangent
plane makes with the XOV-plane, we have
Area AB = area PR cos 7,
["The projection of a plane area upon a second plane is equal to the area of the!
or
[portion projected multiplied by the cosine of the angle between the planes
Sy Sx = area PR - cos 7
518 DEFERENTIAL AND INTEGRAL CALCULUS
Now 7 is equal to the angle between OZ and a line from per-
pendicular to the tangent plane. Hence from (H), Art. 237, and (2)
and (3), Art. 4, we have
1
cos 7 =
L \f)x) W
Then Area PR = 1 1 + (-Y+ (\~ | J Ay Ax.
L \tixl \vy! J
This we take as the element of area of the region S'. We then define
the area of the region S' as
lim
Ax
^y -* o
the summation extending over the region S, as in Art. 245. Denoting
by A the area of the region N', we have
the limits of integration depending on the projection on the XOY-plane
of the region whose area we wish to calculate. Thus, for (/) we choose
our limits from the boundary curve or curves of the region S in the
.YOY-plane precisely as we have been doing in the previous sections.
Before integrating, the expression
must be reduced to a function of j and y only, by using the equation
of the curved surface on which the area lies.
If it is more convenient to project the required area on the XOZ-
plane, use the formula
(/) A-
' S
where the limits are found from the boundary of the region S, which
is now the projection of the required area on the A'OZ-plane.
Similarly, we may use
S
the limits being found from the projection of the required area on the
VOZ-plane.
MULTIPLE INTEGRALS
519
In some problems it is required to find the area of a portion of one
surface intercepted by a second surface. In such cases the partial
derivatives required for substitution in the formula should be found
from the equation of the surface whose partial area is wanted.
Since the limits are found by projecting the required area on one
of the coordinate planes, it should be remembered that
To find the projection of the area required on the XOY-plane, elimi-
nate z beticeen the equations of the surfaces whose inter sections form the
boundary of the area.
Similarly, ire eliminate ij to find the projection o)i, the XOZ-plane, and
x to find it on the YOZ-plane.
This area of a curved surface gives a further illustration of inte-
gration of a function over a given area. Thus in (/) we integrate the
function
over the projection of the required curved surface on the -YOV-plane.
As remarked above, (J) and (AT) must, be reduced to
I lf(jr, z)dz dx and I ff(y, z}dy dz,
respectively, by means of the equation of the surface on which the
required curved surface lies.
ILLUSTRATIVE EXAMPLE 1. Find the urea of the surface of the sphere
j>2 _|_ y'-' -|_ ~? _ r ?
by double integration.
Solution. Let ABC in the figure be one eighth of the surface of the sphere. Here
dx
and
(f s )'=
\dy/
. The projection of the area required on the A'OV-plane is AOB, a region bounded
by x = (= OB) ; y = (= OA ) ; & + ^- r- (= 7M ).
Integrating first with respect to y, we sum up
all the elements along a strip (as DEGF) which is
also projected on the A^OY-plane in a strip (as
MNGF) ; that is, our ^--limits are zero and MF
(= Vr 2 x 2 ). Then integrating with respect to x
sums up all such strips composing the surface ABC ;
that is, our x-limits are zero and OA ( r). Sub-
stituting in (7), we get
r d y dx _ irr 2
Vr- x' 2 y 2 %
A = 4 Trr 2 . Ans. 5
A = r p-'-
8 J Jo
JfU.
520
DIFFERENTIAL AND INTEGRAL CALCULUS
ILLUSTRATIVE EXAMPLE 2. The center of a sphere of radius r is on the surface
of a right cylinder, the radius of whose base is - Find the area of the surface of
the cylinder intercepted by the sphere.
Solution. Taking the origin at the center of the sphere, an element of the cylin-
der for the z-axis, and a diameter of a right section of the cylinder for the .r-axis, the
equation of the sphere is
x' 2 + y' 2 + z' 2 = r 2 ,
and of the cylinder x' 2 + y' 2 = rx.
ODATB is evidently one fourth of the cylindrical
surface required. Since this area projects into the
semicircular arc ODA on the A r Ol r -plane, there is
no region S from which to determine our limits
in this plane ; hence we shall project our area on,
say, the A'OZ-plane. Then the region N over
#hich we integrate is OACB, which is bounded
by z = (~ OA), x (- OB), and z 2 + r x = r' 2
(= ACTt), the last equation being found by elim- y/
inating y between the equations of the two sur-
faces. Integrating first with respect to z moans that we sum up all the elements in
a vertical strip (as 7V, the z-limits being zero and Vr- - rx. Then, on integrating
with respect to j, we sum up all such strips, the j-limits being zero and r.
Since the required surface lies on the cylinder, the partial derivatives required
for formula (J) must be found from the equation of the cylinder.
dy r 2 x dy _
Hence -^ ^ ?/ ' TH
Substituting in (7),
A = rr^^~
4 Jo Jo
Substituting the value of y in terms of / from the equation of the cylinder,
PROBLEMS
1. In the preceding example find the surface of the sphere intercepted
by the cylinder. A f rr r^'-** dy dx _ 9( 9 , 2
Am. 4rl / -j=======--6(Tr ^)r.
Jo Jo Vr 2 a- y-
2. The axes of two equal right circular cylinders, r being the radius
of their bases, intersect at right angles. Find the surface of one inter-
cepted by the other.
HINT. Take x 2 + z 2 = r 2 and x 2 + y 2 = r 2 as the equations of the cylinders.
A o C r f^r a -*2 dy dx
Ans. 8 r / , =
Jo Jo v r 2 _ x 2
o
r-.
MULTIPLE INTEGRALS 521
3. Find the area of that portion of the sphere .r 2 -f ?/ 2 -f z 2 = 2 a// cut
out by one nappe of the cone .r- 4- :'-' //-. An*. 2 Tra 2 .
4. Find the surface of the cylinder .r 2 -f //- r 2 included between the
plane z = nix and the A'OV-plane. An*. 4 r-m.
5. Find the area of that part of the plane - -f -- -f - 1 which is in-
tercepted by the coordinate planes. a ^_
An*. \Vlrc~ -f c-a 2 -f a 2 b' 2 .
6. Find the area of the portion of the sphere .r 2 -f ir 4- z' 2 2 ay which
lies within the paraboloid by f' 2 -f r 2 . An*. 2 irab.
7. In the preceding example find the area of the portion of the pa-
raboloid which lies within the sphere.
8. Find the area of the surface of the paraboloid // 2 -f z' 2 \ a.r in-
tercepted by the parabolic cylinder //'-' a.r and the plane JT = 3 a.
An*, -y Tra 2 .
9. In the preceding problem find the arra of the surface* of the cylinder
intercepted by the paraboloid and plane. A , /lo^/Hi i\ a2
/A )l*. ( 1 ) V 1 ) 1 ) ^r
Vii
10. Find the surface of the cylinder -: J -f (.r cos c\ -f // sin cv) 2 ~ r 2 which
is situated in the positive compartment of coordinates.
HINT. The axis of this cylinder is the line 2 - 0, jr cos a -f // sin a ; and the
radius of the base is r. r 2
An*.
sin (\ cos rv.
11. Find the area of that portion of the surface of the cylinder
y* -|~ - !5 a' ! bounded by a curve whose projection on the A' V-plane is
x% + y r * = a*. An*. V 5 a 2 .
12. P"ind by integration the area of that portion of the surface of the
sphere jr 2 + ?y 2 -f z- - 100 which lies between the parallel planers jc = 8
and x 6.
257. Volumes found by triple integration. In many cases the vol-
ume of a solid bounded by surfaces whose equations are given may
be calculated by means of three successive integrations, the process
being merely an extension of the methods employed in the preceding
articles of this chapter (see also Art. 247).
Suppose the solid in question is divided by planes parallel to the
coordinate planes into rectangular parallelepipeds having the dimen-
sions Az, AT/, Ax. The volume of one of these parallelepipeds is
Az A?/ Ax,
and we choose it as the element of volume.
Now sum up all such elements within the region R bounded by
the given surfaces by first summing up all the elements in a column
parallel to one of the coordinate axes, then summing up all such
522
DIFFERENTIAL AND INTEGRAL CALCULUS
columns in a slice parallel to one of the coordinate planes containing
that axis, and finally summing up all such slices within the region in
question. The volume V of the solid will then be the limit of this
triple sum as Az, A?/, Ax each approach zero as a limit. That is,
(1) V= lim Y V VA; A?/ A*,
R
the summations being extended over the entire region H bounded by
the given surfaces. This limit is denoted by
By extension of the principle of Art. 245, we speak of (L) as the
triple integral of the Junction f(x, y, z) = I throughout the region R.
Many problems require the integration of a variable function of x, y,
and z throughout a given region. The notation is
(2)
y>
dx >
which is, of course, the limit of a triple sum analogous to the double
sums we have already discussed. In more advanced treatises it is
shown that the triple integral (2) is evaluated by successive integra-
tion. The limits are found in the same manner as for (L).
Simple examples of (2) are afforded by the formulas for thecentroid
&> y> z) (center of gravity) of a homogeneous solid, namely,
Vx = f f
y dx dy dz, Vz = z dx dy dz.
They are obtained by reasoning as in Art. 249, using moments of
volume. In the integrands, (x, y, z) is an interior point. The centroid
will lie in any plane of symmetry.
ILLUSTRATIVE EXAMPLE 1. Find the volume of that portion of the ellipsoid
i I 1, ) ' . i
a* o- r*-
which lies in the first octant.
Solution. Let O-ABC be that portion of the
ellipsoid whose volume is required, the equations
of the bounding surfaces being
(3)
(4)
(5)
(6)
= i (= ABC),
c 2
2 = (= GAB),
y = (= OAC),
x = (= OBC) .
MULTIPLE INTEGRALS
323
PQ is an element, being one of the rectangular parallelepipeds with dimensions
Az, A?/, A.r into which the planes parallel to the coordinate planes have divided the
region.
Integrating first with respect to z, we sum up all such elements in a column
(as ftS), the z-limits being zero (from (4)) and TR = rWl - - f- (from (3) by
solving for z). V (/ ~ 6 *
Integrating next with respect to //, we sum up all such columns in a slice (as
I>EM\GF\ the /y-limits being zero (from (5)) and M<1 = b^J I - ~ ( from the
e(iuation of the curve AGF>, namely -f ^- 1, by solving for */).
a 2 b-
Lastly, integrating with respect to .r, we sum up all such slices within the entire
region O-ABC, the .r-limits being zero (from ((>)) and OA a.
Hence
dz dy dx
Therefore the volume of the entire ellipsoid is
ILLUSTRATIVE K\ -\MPLE 2. Find the volume of t he solid bounded by the surfaces
(7) z ~ 4 - .r~ - ', //-,
(8) 2 = 3 /-' + 1 //-.
Solution. The surfaces are the elliptic paraboloids of
the figure. Eliminating z between (7) and (8), we find
(9) 4 J a + 1 y- =4,
which is the equation of the cylinder AB(^I> (see figure)
that passes through the curve of intersection of (7) and
(8) and has its elements parallel to OZ.
We have
(10)
r = 4/ M /' 2V2(1 ' ) f 4 "V 1
J{) Jo J'.ljr- \ } y-
The limits are found as follows :
Integrating with respect to z, we sum up the elements of volume dz dy dx in
a column of base dy dx from the surface (8) to the surface (7) (M P to MQ in figure).
The limits for z are, then, #iven by the right-hand members in these equations.
Thus we find
(1 /'2 V^ (] - s- )
\ (4 - 4 x~ - \ y*)dy dx.
., Jo
The limits on this double integral are those for the region OAB, the portion of
the area of the base of the cylinder (9) which lies in the first quadrant. Working
out (11), we find V ~ 4 ?rV2 = 17.77 cubic units. Ans.
The problem given may be such that the first integration should
be performed with respect to x or ?/, and not with respect to z, as
above. The limits must then be determined in accordance with the
preceding discussion.
524
DIFFERENTIAL AND INTEGRAL CALCULUS
258. Volumes, using cylindrical coordinates. In many problems in-
volving integration the work is much simplified by employing cy-
lindrical coordinates (p, 0, z) as defined in (7), p. 6. The cylindrical
equation of any one of the bounding surfaces may often be written
down directly from its definition. In any case it may be found from
its rectangular equation by the substitution
(1)
x p cos 6, y ~ p sin 0.
Cylindrical coordinates are especially useful when a bounding
surface is a surface of revolution. For the equation of such a surface,
when the axis is OZ, will have the form z =f(p) ; that is, the coordi-
nate 6 will be absent.
Volume under a surface. Let
(2)
z = F(p, 0)
be the cylindrical equation of a surface, as KL in the figure. We
wish to find the volume of the solid bounded above by this surface,
below by the plane XO >',
and laterally by the cylin-
drical surface whose right
section by the plane XOY
is the region S. This cylin-
drical surface intercepts on
the surface (2) the region S'.
Divide the solid into ele-
ments of volume as follows :
Divide S into elements of
area AA by drawing radial
lines from and arcs of cir-
cles about 0, as in Art. 254.
Pass planes through the
radial lines and OZ. Pass
cylindrical surfaces of revo-
lution about OZ standing on the circular arcs within S. Then the
solid is divided into columns such as MNPQ, where area MN = AA,
and MP = z. The element of volume is then a right prism with base
AA and altitude z. Hence
R
(3)
V = z AA.
The volume V is found by summing up the prisms (3) whose bases
lie within S and finding the limit of this sum when the radial lines
MULTIPLE INTEGRALS 525
and circular arcs within 5 increase in number so that Ap > and
A0->0. That is,
AH-O
We now show that the double limit in (4) may be found by suc-
cessive integration. (Compare Art. 244.) This is done by finding
the volume, approximately, of a slice of the solid included between
two radial planes such as ROZ and XOZ, and then taking the limit
of the sum of these slices.
Let DEFG be the section of the solid in the plane ROZ. The
values of z along the curve GTF are given by (2) when (= angle XOR)
is held fast. In the plane ROZ take OR and OZ as rectangular axes
and (p, z) as coordinates. Let (p, z) be the centroid of area DEFG.
Then by (2) and (3), Art. 177,
p - area DEFG = f ' pz dp = $* pf(p, 0)dp.
The integral will be a function of 0.
Now revolve area DEFG about OZ. By Art. 250, the volume of the
solid of revolution thus generated is 2 ir~p area DEFG. The planes
ROZ and SOZ cut out a wedge from 1 his solid of revolution whose vol-
ume is A0p area DEFG, since angle ROS A0 (radians). Therefore
(5)
is equal, approximately, to the volume of the slice of the solid in-
cluded between the planes ROZ and $OZ. The limit of the sum of
the wedges (5) when A0 > is the exact volume.
Hence
(6) V
where a - Z XOA, = Z A'O/*, pi - O/> =/,(), P2 = OE =/->(0),
values to be found from the polar equations of the curves bounding S.
The element of the integral in (6j, namely,
F(p, e)p<lpd6 = zpdp(W,
may be thought of as the volume of a right prism of altitude z and
base of area p dp d6. Thus AA in (3) is replaced by p Ap A0, as in
Art. 254.
We have now the formula*
(M) V =//*P dp d6 =fff(p, fl)P ^P ^
\s s
*The orf/<T of integration is immaterial. Proof is omitted.
526
DIFFERENTIAL AND INTEGRAL CALCULUS
X
for the volume under the surface (2), and the limits are found as in
Art. 254 for the area of the region S.
From (M) and (4) we may derive (2), Art. 254.
ILLUSTRATIVE EXAMPLE 1. Show that the volume of the solid bounded by
the ellipsoid of revolution b-(x 2 -f y") -f a-z~ = Q"b~ and the cylindrical surface
x '2 + y* _ ax o j s given by
/) r\"K f'U COH
(7) V = *7,f i
o-Jo Jo
Evaluate this integral.
Solution. By (1) the cylindrical equa-
tion of the ellipsoid is b~p~ + a-z~ a-b-.
Hence
(8)
The polar equation of the circle
x -i _j_ y-2 _ ax 9 j n th e AT-plane
bounding S is, by (1),
(9) p - a cos 6.
For the semicircle the limits for p are
zero and o cos 9, when is held fast, and
for 0, zero and \ TT. Substituting in (M) the value of z from (8) and the above limits,
we get (7). Integrating,
V = a 2 &(3 TT - 4) - 1.20G a' 2 b.
Volume by triple integration. The clement of volume A V will now
be an clement of the right prism used above in (3), that is, a right
prism with base &A and altitude Az. The solid is divided into such
elements by passing through it the planes and cylindrical surfaces
used in the figure at the beginning of this article and also planes
parallel to the plane XOY at distances apart equal to Ac. We
now have
(10) AV = AzA4.
By summation and taking the limit when Az--0, Ap 0,
A0 * 0, we have
(N)
for AA may be replaced by p Ap M as before.
Formulas (3) in Art. 257 for the centroid become
Vx =fffp 2 cos 6 dz dp d6, Vy == fffp 2 sin 6 dz dp dO,
Vz^CCCpzdzdpdO,
when cylindrical coordinates are used.
MULTIPLE INTEGRALS
527
ILLUSTRATIVE EXAMPLE 2. Find the volume of the solid whose upper surface
is on the sphere
(in .r~-fr+2- = 8
and whose lower surface is on the
paraboloid of revolution
(12) .r-4 r' = --.
Solution. The figure shows
the sphere arid the paraboloid in
the first octant. The curve of in-
tersection All lies in the plane
z = 2. Its projection DK on the
AT-plane is the circle
(13) *-' + r' = 4.
The cylindrical equations
are, by (1) :
(14) p' 2 -f z 2 = 8 (the sphere
(ID);
(15) p' 2 - 2 2 (the paraboloid
(12));
(1G) p = 2 (the circle (1,T.
An element of area A/1 in the circle 1 (1(1) is drawn at A/(p, 0) in the figure. An
element of volume AT is shown at /'(/>, 0, z).
We have, by (AO,
r> /r r2 / v N - p-
I I p c/s (/p dO.
.. Jo J J, /( "
The limits are found as follows: Integrating with respect, to z (holding p and
6 fast), we sum up the elements of volume (10) in a column from the surface (15)
to the surface (14) (M/Vt^J/7'i in the figure). From (15), z = MP-> = J p 2 ;
from (14), 2 = M/'i = V8 - p~, the 2-limits. The limits for p and are those
for the area of the circle (16). Integrating with respect to p gives the sum of the
columns in the slice included between the plane passing through ()Z and OM and
the plane passing through OZ and ON. The final integration sums up these slices.
Integrating in (17),
AUK.
In the following problems, formulas (M) and (AO are to be used
when the equations of bounding surfaces are in cylindrical coordi-
nates (cylindrical equations;. If the corresponding rectangular equa-
tions are needed for drawing a figure, they may be found by the
transformation
(18) p 2 =
to which may be added
y' 2
6 = arc tan ->
x
(19)
sin c; =
cos =
528 DIFFERENTIAL AND INTEGRAL CALCULUS
PROBLEMS
1. Find the volume of the solid below the cylindrical surface x 2 -f z = 4,
above the plane jr -f z = 2, and included between the planes y = 0, y 3.
/!{ /~2 r4 - r 2
j. V / / / d2 dr dy = 13^ cubic units.
JO J \J2.~ jr
2. Work out Illustrative Example 2, Art. 247, using cylindrical co-
ordinates. rjTT /*2rtros0 p.'$ 3
An.s. I 7 = 2 / I i -~ dp d6 - Tra 3 .
Jo Jo a 2
3. Find the volume of the solid bounded above by the cylinder
2 = 4 x 2 and below by the elliptic paraboloid z = 3 .r 2 -f- 2/ 2 .
/"I /"2 x 1 - J -' /*4 - J 2
Ans. V 4 / / / dzdydx =
J() JO J3 j" -I y- 1
4. Two planes forming an angle <* radians with each other meet along
a diameter of a sphere of radius a. Find the volume of the spherical wedge
included between the planes arid the spherical surface, using cylindrical
coordinates. Ans. - aa 3 .
5. Find the volume below the plane z = .r and above the elliptic
paraboloid z = .r 2 4- y' 2 . Ans. ^ TT.
6. Work Problem 5 using cylindrical coordinates.
/?, 7T /^COS fl //) COS ^?
= 2/- / /
Jo Jo Jp^
7. Find the volume bounded by the sphere p 2 -f c 2 = a 2 within the
cylinder p = a cos ^. Aw.s-. g a^(w J).
8. Find the volume above c = 0, below the cone c- jc' 2 -f ?y 2 , and
within the cylinder jr' 2 + y' 2 2 aj-, using cylindrical coordinates.
A??,s. -^ a 3 .
9. Find the volume of the solid bounded by z ~ .r -f 1 and 2 z ~ x 2 -f y 2 .
Ans. | TT.
10. In Problem 3 show that integration with respect to z gives (with-
out further integration) V 4 A ~ 4 I v 1 M \vhere .4 is the area of the
ellipse 4 a* 2 -f y' 2 = 4, and 7 X and / tf are moments of inertia for this ellipse
as given by (E), Art. 252.
11. Find the volume below the plane 2 z 4 -f p cos 0, above z 0,
and within the cylinder p = 2 cos 0. Ans. f TT.
12. A solid is bounded by the paraboloid of revolution az = p' 2 and the
plane z c. Find the centroid. Ans. (0, 0, c).
13. A solid is bounded by the hyperboloid z 2 = a 2 + p 2 and the upper
nappe of the cone z 2 = 2 p 2 . Find the volume. Ans. 7ra 3 (V2 1).
14. Find the centroid of the solid in Problem 13.
Ans. (0,0, ia(V2+l)).
MULTIPLE INTEGRALS 529
15. Find the centroid of the solid in Problem 1. An*. (A, , V )
16. Find the centroid of the solid in Problem 2. Ans. ($ a, 0, V } a).
17. Find the centroid of the solid in Problem 8.
18. Find the volume of the solid bounded below by z 0, above by
the cone z = a p, and laterally by p = a cos 0. A NX. T <\. a ;< (9 TT 16).
19. Find the centroid of the solid in the preceding problem.
20. Find the volume of the solid below the spherical surface p L> -f z* = 25
and above the upper nappe of the conical surface z = p -f 1.
21. Compare Illustrative Example 3, Art, 165, and Illustrative Ex-
ample 1, Art. 257, and derive (N), Art. 165, from (L), Art. 257.
22. Derive formula (2), Art. 178, from the first formula in (;*), Art. 257.
ADDITIONAL PROBLEMS
1. Find the volume bounded above by the sphere p-' -f z- ~ r~, below
by the cone z = p ctn 0, and included between the planes ft,
6 ft -f A#, </> and jtf being acute angles. (The solid is part of a spherical
wedge, like O-.SQA' in the figure of Art. 222 when OQ is drawn.)
An*. ; \ r ! Atf(l - cos 0).
2. Find (without integration) the volume bounded by the sphere
p^ _|_ Z 2 r i> t the cones z - p ctn 0, ,: p ctn (0 -f A0), and the planes
ft, = ft + A^ f using the result in the preceding ])roblem. CTho solid
is like 0-Pi/iQS in the figure of Art. 222 when OR arid OQ are drawn.)
Xb/x. '] r> A/i sin (0 -f- -\ A0) sin \ A0.
3. Find (without integration) the volume bounded by ;' = p ctn 0,
2 p ctn (0 -f A0), 6 - rf, ft -f A/i and included l>etween the spheres
p'2 _|_ /: 2 _ f j > ^2 _j_ z *> (r -f Ar) 1 ', using the answ( r in Problem 2.
.4??.s. 2 A/i Ar sin (0 4- A A0) sin \ A0(r- -f r Ar -f ', Ar*).
(The solid is obtained from the figure in Art. 222 by producing each of the
radii OPi, OR, OQ, OS a distance Ar to /',', /'', Q', S f/ on the sphere
p* -f z 2 - (r 4- Ar)-. The cones intersect this sphere in the circular arcs
JYA" and Q'S', and the planes in the ar^s of great circles /Y,S", H'Q'- The
solid has the vertices PJtQS-Pi'H'Q'H'.)
4. The solid of Problem U is the element of volume AV 7 when spherical
coordinates (8), p. 6, are used. Replace ft by 0. Then one vertex P of
AV has the spherical coorclinat.es (r, 0, 0). Prove, from Problem 3,
AV
Ar->o r 2 sin Ar A0
v , o
A./. -
Therefore AT 7 differs from r 2 sin </> Ar A0 A0 by an infinitesimal of higher
order (Art, 99).
530 DIFFERENTIAL AND INTEGRAL CALCULUS
5. In the solid of the preceding problem prove that the edges of AV
meeting at any vertex are mutually perpendicular, and that the lengths
of those intersecting at (r, 0, 0) are, respectively, Ar, r A0, r sin A0.
6. Describe the three systems of surfaces (spheres, cones, planes)
which must be drawn to divide a solid R into elements of volume AV
(Problem 4) when spherical coordinates are used. Let (r, 0, 0) be any
point of AV. Then we write
^ V^/^'fr, 0, 0;A V =fff F ( r > 0J r ~ sin </> ^ r ^0 f/ #-
lim
In the left-hand member Al r may be replaced by r 1 -' sin <^ Ar A< A0
(see Problem 4), that is, by the product of the throe edges in Problem 5.
The right-hand member is calculated by successive integration. (Proof
omitted.)
7. Work out the integral in the preceding problem if F(r, </>, 0) = r,
and A' is the sphere r = "2 a cos (/;, that is, .r~ -f /r -f z- 'J a^.
2 7T /^ * TT /^2 <l COS
/ '-' /
./o .At
8. Work out the integral in Problem 6 if /^tr, 0, 0) = r-' cos and A'
is the region r 12 a cos 0. /I//*. I;/, 7r<z r '.
CHAPTER XXVI
CURVES FOR REFERENCE
For the convenience of the student a number of the more common
curves employed in the text are collected here.
CUBICAL PARABOLA
SEMICUBICAL PARABOLA
y 2 = ax 3 .
THE WITCH OF AGNESI
THE Cissoin OF DIOCLES
x 2 y = 4 a 2 (2 a y).
2/ 2 (2 a - x) = x 3 .
532 DIFFERENTIAL AND INTEGRAL CALCULUS
THE LEMNISCATE OF BERNOULLI THE CONCHOID OF NICOMEDES
(x 2 -+ y/ 2 ) 2
- = a- cos 2 0.
p a csc + 6.
(In the figure, b
CYCLOID, ORDINARY CASIO
~ a arc vers ~~ V2 a// - /r
:r = d(0 sn ^),
U ^ a(l cos 0).
CYCLOID, VERTEX AT ORIGIN
y
.r = a arc vers ^ + V2 a?/ y 2
a
- sin 0),
- cos 6).
PARABOLA
a cosh - -
a
CURVES FOR REFERENCE 533
HYPOCYCLOID OF Fouu CUSPS ASTROII^ KVOLUTK OF ELLIPSE
r
= a siv .
+ (ft//)" - (a-- ft 2 )
~~ a cos' 5 0,
/) sin ;< f?.
CARDIOIP
yr + ^^* =
p a(l cos 0).
3 axy = 0.
SINE CURVE
COSINE CURVE
y
' = sin x.
i = COS X.
534 DIFFERENTIAL AND INTEGRAL CALCULUS
STROPHOID
Y
p = fr a cos 0.
(in the figure, b - a.)
SPIRAL OF ARCHIMEDES
p = a0.
LOdARITHMIC OR EQUIANGULAR
SPIRAL
p = c' l( *, or
log p ad.
HYPERBOLIC OR RECIPROCAL
SPIRAL
= a.
LITUUS
= a 2 .
CURVES FOR REFERENCE 535
PARABOLIC SPIRAL LOGARITHMIC CURVE
r
(p- a)- = 4ac0.
y = log x.
EXPONENTIAL CURVE
PROBABILITY CURVE
v
SECANT CURVE
|37T II
y = sec
TANGENT CURVE
y = tan x.
53G DIFFERENTIAL AND INTEGRAL CALCULUS
THREE-LEAVED ROSE THREE-LEAVED ROSE
p -~ a sin 13 0.
p =: a cos 3 6.
FOUR- LEANED ROSE
FOUR-LEAVED ROSE
p = a sin 2
p = cos 2 0.
T\VO-LK\VEI> ROSE LEMNISCATE
r
EIGHT-LEAVED ROSE
= a 2 sin 2 5.
p = a sin 4 0.
CURVES FOR REFERENCE 537
PARABOLA EQUILATERAL HYPERBOLA
x
.r// = a.
INVOLUTE OF A CIRCLE
r cos -f rO sin 6,
r sin rO cos 0.
CHAPTER XXVII
TABLE OF INTEGRALS
Some Elementary Forms
l.fdf(x) =ff'(s)dx=f(s) + C.
2. fa du = aCdu.
u dv dw ) = Cdu C dv Cdw
Rational Forms containing a + bu
See also the Binomial Reduction Formulas 96-104.
U ~ ln (a + ''" )] + C
- f. It v . = r:\-rT- + '"(
J (a -f bw) 2 6^L a -f ^>w
= a + bu ~ r - 2 a in (a + 6zo -f c.
3 - X
- T-T-T ~ r
J (a -f ftw) 2 b 3 {_ a -f
r rfM i r i a
J (a -h 6w) 3 ^> 2 L a -f 6w ^ 2(a -h 6w) 2
C .
IA r <*M _ 1 , ^> i /a -f *>M\ , x,
14 - / "~77 , r x = --- 1 -- o In ( - ) -f C.
J u 2 (a 4- OM) aw a^ \ u I
15 f du = 1 - 1 In ^ a + ftM> \ 4- r
538
TABLE OF INTEGRALS
Rational Forms containing a 2
1 , bu
= arc tan + r -
539
2 ab \a -
C d = JL in ( bu ~ a \ 4- r
J b' 2 u 2 -a' 2 2ab \bu + a'
18. f *(a 2
'
(a 2
(a 2
20
(a 2 b' 2 u' 2 )'' b 2 (m 2 p -f l)(a' 2 b' 2 u' 2 } l1 ~ l
b'~(m 2 /> -f 1 ).' (<r b 2 u' 2 ) J>a
J (a' 2 b-u' 2 ) 1 ' 2 a' 2 (p 1 )(a~ b' 2 u 2 ) v l
_ _L1 L f _ i __ ,
22. f f ..*!".. ., = ^ In ( ..";., .,) + C.
' u((i" />*//-) J a*- \a- n*M*/
23. f (/ " !
u m (a' 2 b' 2 u' 2 )" a-(m ~ l)ir'- l (u' 2 6 2 M 2 )"" !
_ fe 2 (7w + 2 7) - 3) / du
a 2 (m -
/ _ u __
./ v rn - <2 (a' 2 b'W) p '
24 f _ - _ - _ -
J u r "(a' 2 b' 2 i( 12 )" 2 a' 2 (p - l)u r "- [ (a' 2 b*u*)"
m -f- 2 p 3
-f- p r a?/
2 a 2 (7> - 1) J u m (a' 2 b' 2 ^' 2 ) 1 '- 1 '
Forms containing Va -f- bu
The integrand may be rationalized by setting a + bu = v*. See also
the Binomial Reduction Formulas 96-104.
25. fu V^T^rf K = - 2(2 a ~ 3 6< jV fl + M *+ C.
*/ lo o 2
26.
27.
28. f MtfM = -
^ Va -f bu
28 a2 ~ 12
105 b 3
2
2(2 a - 6M)V(/ -f bu
3b' 2
540 DIFFERENTIAL AND INTEGRAL CALCULUS
u' 2 du __ 2(8 a 2 - 4 abu + 3 6 2 M 2 ) Va + bu
29 -/;
30. f-
>
31./'-
32. f~
/
oo. f ~"
J t
34 -/ :
u r "du 2 K m Va -f
2am
' + C.
u m ~ l du
/a + bu b(2m+l) b(2m+ 1) J \^~+~ l
- 4= In ( V "*" "" ~ v ^ ) 4- T, for a > 0.
/a 4- bu Va VVa 4- btt + Va
(/K
arc tan
4-
~ a
4- T, for < 0.
dw,
w \/a 4- />M ( ^ v ' w ~ 1 ^ ^ " ' 2 a ( r/?
fu + budu tl ,/~~TT~ , /' d?/
a 4-
2 Va 4- ow 4- a f-
(a -f-
u Va 4- ?>M
2 rr/
a 4-
a(m 1 )//"'
1 )
Forms containing Vu 2 a 2
In this group of formulas we may replace
36.
\n(u 4- V?/ 2 4- a-) by sinh" 1 -
u 4- Vw 2 a 2 ) by cosh" 1
a 4- Vu 2 4- a 2
)(j
)>y sinh" 1 -
?^
= ~ Vw 2 a 2 ~ln (w. 4- Vw^ia 2 ) 4- C.
?/ -f 1
4-
38. /'//CM 1 ' a 2 Y~dn ~ (* r ^ fl2 \" h C,
' ?/ 4- 2
// 4- m + 1
= ln (M 4- VM- a j ) 4- C.
TABLE OF INTEGRALS
541
4- In (M -f Vw 2 a 2 ) + r.
i'. _ i m n -f 1 J
n m ~' 2 du
49. f J
-}/ 2 f >/ -
u*(u*<i*)* a ' 2 "
50 */~; 7"^ 77 = ^aw'* + ^ ln ( q * v ^ + QW
arc sec - -f C.
c , r ^M V?/ 2 a 2 .1
51. / = -f - r
J ii(w 2 - a 2 )* - M - a
52.
C/M
1
a 2 (m -
-f n
2-1
O 2 a 2 ) 2
53./-
C/M
_ J^_-ZLzi:L f.
a 2 (m l)J
u
\
,
a 2 (n 2)?/ m " ! (
m -f n - 3 r
a 2 (n - 2)J "
- 1
a 2 ) 2
du
u m (u 2 a 2 ) 2
. A r(u 2 + a^du A /~r^ ^ , /a -f Vu 2 -f a 2 ^ , ^
54. I = V u 2 -f a 2 a In I / -f C.
J M V u /
KK r(u 2 a 2 )^du
55, / ^ L
t/ 24
542 DIFFERENTIAL AND INTEGRAL CALCULUS
66 rluiI
J u*
58.
a' 2 (w 1 )' n m ~ 2
(u*a*p , fl 2 w
(n~ TW -f 1 )" r " J w - w -h
Forms containing v a 2 i/ 2
59. f(a 2 - w^^r/w = ^ Vr/ 1 ' - M* -h if arc sin ^ -f C.
Tl
60. (,^ - *^ du = "(--' )5 + ( z -
61.
/ ., ., n j
62. J < - -)-d =
-f
63
^ = arc sin - + C.
(,/ 2 -,/^ w
r///
^ -f C.
.> } r>
C t /" // dn ((J 1 M*) " i r-
bO. rr h ( .
' IL N 2
66
arc sn
" a
67. r ?<2(/// ,= " -
^ fa 2 - ?/ 2 ) 3 V 2 - w 2
- arc sin - 4- C.
/u
a 2 (m 1) rj
2-1 m n + lJ
(a 2 - w 2 ) 2
AQ r_irdu_- -u m + l m-n + Zr <u^au
69 J n- 2-1 a 2 (n-2)J i l -i
(a 2 - w 2 ) 2 a 2 (n-2)(a 2 - i/ 2 ) 2 (a 2 - w 2 ) 2
TABLE OF INTEGRALS
543
70.
u(a 2 u' 2 fi a
, r (in V- u*
71 -J u w-*^~~*^~
72 . f & . = .
73 -/
(In
1 , , a . ^
cosh ~ ' - -f C.
a- {
1
a -(m 1 ) // '"
m 4- // '* /'
<r(m - 1 )-'
5 . r(
.'
2
-f-
<///
- 2) J . !
w m ( a ~ ""/"
> _ a ln
va j ?/ L ' a cosh" J - -f C.
w
76.
arc sln
a
77.
70
(a 2 f/ 2 ) 2
,
(n m -f 1 jM"'"" 1 n m +
Forms containing V2 au i/ 2
The Binomial Reduction Formulas 96-104 may be applied by writing
\/2 au u~ - y}(2 a u}*.
-2i 2 -f ~ arc
+ C.
80.
V2^
3 a 2 -f an - 2 w s
, a ' 3 /i w \ i /-
+ arc cos (1 ~-)+C>
544 DIFFERENTIAL AND INTEGRAL CALCULUS
. ,,^ 3
8L/>V2au-w 2 dw--
m -f-
m + 2
82. f M - M f M _ V2 aw - w 2 -f a arc cos ( 1 - -) -f C.
J u \ a/
/"\/ *) fj 1 }! Ij'^filJ '^ \/'^ fl 11 It '* / 1J\
- 2 - - ~ - arc cos 1 - - + C.
w 2 w \ a/
04 r^/2 aw K 2 dM _ __ (2 aw w 2 )^ ,,
f'V 2 a?/ w 2 dw (2 aw w 2 ) 7 w J /^"v 2 a?/
w m ~ a(2 ?w ,S)w'" a (2 w 3)J M m " !
86. f y ?/ === = arc cos ( 1 -
J V2 aw - u' 2 \ a
87.
- = In (w -f a -f V2 a?^ -f u 2 ) -f r.
V2 aw -f- u 2
88. CF(U, V2 a?^ 4- ?/ 2 ) r/// = f^fc a, V^ 2 <i' 2 ) dz, where 2
w 4- a.
89.
?/(/?/
. f f
^ V2 aw - w
QH f "*<*" =
J V2a?/-?/ 2
92./
93.
= - V^CIW-H* 4- a arc cos ( 1 - ^) + C.
2 v a/
94./
95. r
^
V2 aw - w 2
dw .
wV2 aw w 2
dw
7/ m \/2 aw w 2
(2 aw - w 2 )*
udu
m
V2 au w^
-f r.
V2 a?/ i/ 2
w 1 _ / a
a(2 m - 1 ).' M - ^V2
a 2 V2 aw - w 2
-f C.
(2 aw w 2 )* aV2 aw w 2
-f C.
96.
Cu m (a
J
-f
Binomial Reduction Formulas
ll m-<]+ 1 ( n i 7)7/';V' 4 *
dM = - - 1.
b(pq+ m -f 1)
m
TABLE OF INTEGRALS 545
97. / u m (a + bu q Ydu - ^ -~^-
J pq + m + 1
pq -f w -f 1 '
8i r tj "
v ?/ m (<i -f
a(m l)n tn ] (a -f
fo^w (/ -f jx/ 1 ]
99.
<i( w 1 ) ' M ni "( -f
(/;/ _ 1
u n '(d -f- bu' 1 ) 1 ' uq(}> 1 }u m l (n -f btcn*' l
~ ( / ~^ y )f y """ ^ / ' '^ w
,
100 .r d " == -Li n /-j^) +c .
c/ M( -f f>" <; > ar/ \a -f bu' 1 '
1A1 f (a 4- bit' 1 ) 1 ' da __ ( -}- />/rn ;> M
1U1. I
J u m
:
a ( m 1 ) // '" l
b(_m __~ </ ~- j>(] 1 i ,'(a -f
102 . r( a + bnWdu _ (// + bit")' 1
;>r/ m -f 1 ' //"
u m du //'" (/ ' '
103. f
104.
(a -f ^> M </)/' ?>(m - pq + 1 )(a 4- />// (/ ) ; '~ '
_ (m (/ -f 1 ) r H"' '' dn
b(m pq -f 1).' (a -f bu v )*
u m dn u 7 " 4 l
m 4- g - p(( H- 1
(iq(p 1) ' (a -f
Forms containing a -f bw ci/ 2 (c > 0)
The expression a + &M -f rw 2 may be reduced to a binomial by writing
w = z AC =
4 r 2
Then a + 6w -f r?^ 2 = r(z 2 - A").
i lien u f wo, -f r// - ( \z A;.
The expression a 4- bu cu 2 may be reduced to a binomial by writing
==2: " f "27' = 4r 2
Then a -f bu - cu 2 - c(k - z 2 ).
106 . f -^ . = 2 arc tan( 2 . CU + 6 )-f C,
J a -h 6w -f rw 2 V4 ac - b 2 \ V4 or - & 2 / w hei
when 52 < 4 ac<
C du = 1 In ( 2cu + b - ^ - 4 ac \ . c
J a + Jm -f cw 2 Vb 2 - 4 a \2 rw -f 6 -f Vfc 2 - 4 or/ '
540 DIFFERENTIAL AND INTEGRAL CALCULUS
106.^
107. L
[ > 2 4 4 ar 2 rw 4 b/
108. CLMJLJ^lll = M i n ( a + & M ra 2 } }
; a -I- 6w cu l 2 r
when b 2 > 4 ac.
* + 4 r + 2r M -^ + c<
(/W
109. fVa -f- bu -f r?/^/?i = ^-i- Va -f bn -f rw 2
./ 4 r
4 r
- ^"-iC ] n (9
a -f- bu
110.
~cv*du = 2 ^ ?/ ~ /; Va
4 r
/> 2 4- 4 (ic . / 2 r?
urc sln
'
- 6
'/;
= -Vln (2rw 4-6-f 2
Va -f bu -h rw 2 Vr
112. r- 7 =J^L== - 4= urc sin ( ~r^"~ b= }+ c -
tf Va 4- bu rw 2 Vr v Vh 2 + 4 ar/
// du
113. f
J
-f />// 4-
4 ow 4 r?/ 2 <'
~ln (2 ru 4 /) 4 2 \/rVa 4 few 4 <
-j- ^ ^
i r ^ " ?/ _ _ v 4- f>?/.
4. I . _
J Va 4 ww- rw 2 r
H ~ arc sin / -y >( j 4 C.
Other Algebraic Forms
117.
'
118.
it
_
-f (a b) log, ( Va H- w -f V?> -f M)+ C.
_
+ M arc sin > - + C.
b) arc sin
f \ T 1/ du Vl M 2 4 arc sin ?^ 4- C.
= 2arcsin x /^-^4C.
^w a)(b u)
TABLE OF INTEGRALS
Exponential and Logarithmic Forms
120. C c an dw = -f T.
J a
121. Cb au du = -^-r + C.
J a In o
122. Cue*" du = ^ ut - 1) -f C.
J a-
123. Cu n e au d?* = Cu" V" du.
J a aJ
124. /V'ft a " du = ^^ r-r f " ! ft' 1 " dw + C.
J (Mn ft (i In ftJ
/"ft " d?/ ft"' 11 a In ft rb"" du
J u" ( H 1 )//" l n IJ u" l
547
105
126. An u du = ?/ In u - u 4- r.
10 - T N i ;
127. I u" In ?/ d// M"
l
1)-'
4- C.
128. I u"' In" // d?/ = -- In" // ---- ; I u m In" ' /( d//.
J m -f 1 m -f IJ
r , i 7 r "" i n ?/ i r r "" ;
. I f" 3 " In u du --- I dn.
J a aJ n
inn
129
130.
?/ In ?/
rrln (In M) -h r.
Trigonometric Forms
In forms involving tan //, ctn ?/, sec M, esc //, which do not appear
below, first use the relations
1
sin u . cos u 1
tan // = ctn ?/ = -: sec ?/ =
cos // sin ?/ cos z/
131. I sin // d// cos u -f- C.
132. I cos ?/ d// = sin // -h (\
133. I tan ?y d?/ = In cos M + C = In sec ?/ -h C.
134. f ctn ?/ d?/ In sin ?/ -f r.
//" d ?^
sec ?/ dw / = In fsec w -f tan w) + C
J cos ?/ 7
esc u =
sin
136. /esc u du I . ( ?/ = In (esc w ctn w) -f C
J J sin w
= In tan - -f C.
648 DIFFERENTIAL AND INTEGRAL CALCULUS
137. /see 2 u du = tan u 4- C.
138. J esc 2 M dw = ctn u 4- C.
139. / sec w tan u dv = see w 4- r.
140. / esc u ctn w du esc ?/ 4- C.
141. j sin 2 u du ~ \ u \ sin 2 u + C.
142. / cos 2 u du = ^ M 4- ;J sin 2 ?/ 4- C.
143. /cos" u sin r< d?,* 4- r.
J w 4- 1
144. fsin" u cos w du = - -^ 4- c:.
/ n 4" 1
145. fsin mn sin WM rfw = - -" (w + " '" + s '^ (wt ~ n)u + c.
146. fcos m?/ cos n JH = sin (m + w)l< + sin (m ~ "Jf + r .
/ 2(m 4 w) 2(?w w)
147. / sin mw cos riw dw = ( QH w "^~ r/ ;/ _ cos ( m ?? ?/ _f_ (\
-/ 2(m 4- w) 2(rw ?o
du
2 esc a arc tan (tan ?, a tan J w) 4- r.
+ cos a cos
149.
1
cos a 4- cos u \1 - tan i a tan I ul
= 2 esc a tanh ] (tan i a tan A ?/ ) -f r i tan- J u < ctn 2 J
= 2 CS a arc tan (CSC fl tan i M + ctn fl) + r -
1. f - ^__ = cscaln / tan a- tan .^i/ - secax
J cos a 4- sin ?/ \tan a 4- tan J ?/ 4- sec a I
[ (ctn a tan // 4- esc a) 2 < 1]
= 2 esc a tanh J (ctn a tan -I u -f esc a } 4- r
[ (ctn a tan \ u -f esc a) 2 < 1]
152. f rf " . = 1 Bre tan (Itanj'N a
./ a 2 cos 2 ?/ 4- h 2 sin 2 ?/ a/> \ a /
163. ff sin rf = f " (a sin M " ~ M cos " M) + C .
? 2
a cos
a 1 4- w 2
155.J M sin w dw = sin ?/ u cos M -f C.
156.J w cos M du = cos w 4- u sin M 4- r.
154. /> cos d = '""
^
TABLE OF INTEGRALS 549
Trigonometric Reduction Formulas
/sin* 1 ~ " u cos ?/ it "~ 1 /*
sm n udu h i sin" ~' 2 u du.
n n J
ito r j cos"" 1 !/ sin M , r? 1 / _ 9 ,
168. / COS"M du = h I cos"~ 2 w du.
J n n J
159 C du = cos M j_ >? 2 r dj/
V sin" M (n 1 ) sin" ~ l u n 1 J sin" ~ 2 M
160 . f-rfJL- = smj/ + !L -^ r__clu
J cos" w (n 1 ) cos" ' u n 1 J cos" " 2 ?/
161. /cos m M sin"wdz/ = 1 : icos~' 2 u sin^wdu.
J m -f n m -f n J
162. fcos m u sin" M du = ^ c s -f ^-^ fcos m M sin" ~ 2 w du.
J m + n m -f n/
V cos m w sin n M (m IJsin"" 1 ?^ cos" 1 " 1 !/
,
"*"
m + n 2 r du
n r
- 1 J
B m -2 ,
m 1 ./ cos 7 " z u sin" M
164. f - -
^cos m usin n w (n 1) sin"~ ' u cos*" 1 u
m + n 2 / dw
n 1 J cos m i< sin" ~ 2 u
rcos m udu cos" 1 ^ 1 u m r? -f- 2 rcoR m udu
'J sin" u (n 1 ) sin" ~ ] M, n 1 / sin"~ 2 ii
IAA rcos m ?/ d/>< ___ cos*"" 1 ?/ m 1 rcos m ~ 2 u (
V sin"w ~" (m w) sin"" 1 it m nJ sin n u
- fl /sin n M d?i sin" 4 ] w n m + 2 /sin n i
m + rt
* W "J cos'"?/ (m l)cos m ~ 1 w m 1 ./ cos m ~ 2 w
16 r^Hllli-^ sin" ~ ] i/ n 1 rsin n ~ 2 u du
J cos m u (n m) cos" 1 " 1 ?/ n mJ cos m u
169. ftan"?/ dw = ^- ftan"~ 2 w d?>t.
./ n 1 J
170. fctn n w du = - ctnW "" lM _ r c tr\ n ~ 2 u du.
J n ~ 1 J
171. Ce au coB n udu = ^-^ ^ '
J a z + n z
. n(n 1) / _ t n o ,
+ ^T^/^ cos wrfM -
172./e-"8in M d M = -"n-'(asin - n COBU)
a + n
550 DIFFERENTIAL AND INTEGRAL CALCULUS
173. Cu m coa au du = (au sin au 4- m cos au)
v a
777(771 1 )/"*, o j
, L I u m ~ 2 cos au aw.
a J
174. / M m sin aw dw = ~ (m sin a?v au cos a?/)
*/ a
in ( m 1 ) r m . r, . ,
___ / M w ^ sln aw C / Wy-
Inverse Trigonometric Functions
175. J arc sin w dz/ = M arc sin u 4- Vl w 2 4 C.
176. J arc cos u du ?/ arc cos w Vl u* 4 C.
177. f arc tan ?/ r/// = ?> arc tan w In Vl -f u' 2 4- r.
178. / arc ctn u du ?/ arc ctn ?/ 4 In Vl 4 u' z 4 r.
179. / arc sec */ dw -- u arc sec // In ( u 4 vV-' 1 ) 4 C.
= u arc sec // cosh ] // 4 (*.
180. I arc esc u du ~ n arc esc ?/ 4- In ( // 4- Vw J 1 ) 4- C
u arc esc u 4- coh J ?/ 4 r.
Hyperbolic Functions
181. / sinh ?/ d?^ ~ cosh // 4- C\
182. I cosh ?/ d?/ = sinh tt -\- ('.
183. | tanli // du In cosh w -}- C.
184. j ctnh // d>'/ = In sinh u 4- (\
185. | sech M </// = arc tan (sinh ?/) -f C = gd M 4 C
186. /"csch M du = In tanh J M 4- C.
187.j"sech L> M du = tanh ?^ 4- C.
188. |csch' J ?f dw = ctnh ?/ 4- C.
189. | sech w tanh u du sech n 4- C.
TABLE OF INTEGRALS 551
190. I csch u ctnh u du = csch u -f C.
191. j sinh* u du \ sinh "2 n \ u -f C.
192. / cosh 2 u <ln = i sinh 2 M -f J u -f f.
193. / tanh- M <i?/ = M tanh u -f (\
194. I ctnh 2 u du u ctnh u -f C.
195. j ?/ sinh M du = u cosh // - sinh u -f f.
196. I n cosh M du u sinh >/ cosh // ~h r.
/> ___
sinh ' u du = u sinh ' u VlH- n~ + (\
198. j cosh * u du ?/ cosh ' // vV- 1 -f r\
199. f tanh J // c/// = M tanh ' // -f \ In (1 - u : ) -f C.
200. fctnh ' i/ du u ctnh ' // -f J In (1 ?/-) -f r.
201. I sech ' M <^/^ = n sech ' // -f g<i (tanh J w ) -f <"
u sech ' // -f arc sin u + C.
202. f csch J ?/ du ?/ csch ' ?/ -f sinh ' ?/ 4- r.
f . . . . _ sinh OH + ?V)M sinh Cm ?/)// . ., / > \
203. / sinh m?-< sinh nu du ^~ - - r -- ~r. -- ; --- h < . (m^n\
J Km + u) l(m-~ n) \ /
n ~ A ft, u ; sinh fw -f ?/)w . sinh (m u)u . ,, ( > \
204. / cosh mu cosh nu du = - - - -- --- -\- ( . [ m ^ n
J l l(m+ u) 'Km - n) \ ^ /
206
nn - r . , u 7 cosh (w-f ?/)M , f'osh (?w n ) u , ,, / > \
205. / sinh ?^^?/ cosh nu du ~ - } i -\ -- -- ' f (\ m ^ n
J 'Km + n] 2(w-r/) \ < /
. f : - r = 2 csch a tanh - ' (tanh .', ?/ tanh i a) -f r:.
J cosh a -f cosh u "
207.
. f - . - 2 esc a arc tan ftanh J ?/ tan i a) + r.
./ cos a -f cosh u
208. f - - - = 2 esc a tanh ] (tanh J ?/ tan i a) 4 r:
Jl + cosacosh, (tanh*i W <ctn*Ja)
209. f- sinh m/ du = - c<
J a j n 2
rt , rt r , , p a "(a cosh TIU n sinh nw) , ~,
210. I r a '^ cosh n?( du = - - - - - - -f C.
J a 2 n 2
INDEX
(The numbers refer to pages)
Absolute convergence, 348
Acceleration, curvilinear motion,
121 ; rectilinear motion, 83
Adiabatic law, 70
Agnesi, witch of, 531
Anchor ring, 267
Angle of intersection, of plane
curves, 43 ; polar form of, 126 ;
of skew curves, 475 ; of surfaces,
481
Approximate formulas, 367, 372,
490
Arc, centroid of, 335 ; differential
of, 142, 144, 473 ; length of, plane
curve, 271 ; skew curve, 473
Archimedes, 127, 128, 155, 277, 534
Area, of a curved surface, 517 ; mo-
ment of, 320, 503, 514 ; moment
of inertia of, 508, 514; plane, 241,
258, 498 ; in polar coordinates,
262, 512 ; of a surface of revolu-
tion, 277
Astroid, 119, 533
Auxiliary equation, 390
Bending, direction of, 75
Bernoulli, lemniscate of, 532
Binomial differentials, 299, 307
Binomial theorem, 1, 353
Boyle's law, 70
Calculation, of e, 361 ; of loga-
rithms, 362 ; of TT, 366
Cardioid, 117, 119, 125, 135, 145,
155, 244, 271, 275, 281, 323, 335,
516, 517, 533
Catenary, 152, 270, 276, 282, 423,
432, 434, 532
Cauchy, 345
Center of fluid pressure, 506
Centroid, of a homogeneous solid,
522, 526; of a plane area, 320,
337, 503 ; of a solid of revolution,
323
Change of variable, 166, 240, 457
Cissoid, 44, 46, 270, 277, 322, 531
Complementary function, 394
Complex number, 440
Compound-interest law, 399
Conchoid, 532
Conoid, 284
Constant, 7; absolute, 7; arbi-
trary, 7 ; of integration, 189, 229,
233, 376 ; numerical, 7
Continuity of functions, 12, 444
Convergence, 340
Coordinates, cylindrical, 6, 524-
527; polar, 123; spherical, 6,
529, 530
Cosine curve, 533
Critical values, 52
Cubic, skew, 474
Curvature, 149 ; center of, 157,
171; circle of, 153, 170; radius
of, 152
Curve-tracing, 81
Curvilinear motion, 120, 146
Cycloid, 116, 119, 144, 151, 161,
244, 270, 274, 276, 281, 532
Cylindrical coordinates, 6, 524-527
Derivative, definition, 21 ; inter-
pretation of, by geometry, 25,
446 ; partial, 445 ; as a rate, 64 ;
symbols for, 22, 445 ; total, 455 ;
transformation of, 166
Descartes, folium of, 46, 119, 288,
533
553
554
INDEX
Differential, 136; of arc, 142, 144,
473; of area, 237; formulas for,
140; geometric interpretation,
137, 477; as an infinitesimal,
146; total, 449
Differential coefficient, 21, 136
Differential equations, applications
to mechanics, 402 ; definitions,
375 ; first order, 378 ; higher
order, 387 ; homogeneous, 380 ;
linear, 383, 390, 407
Differentiation, 22 ; formulas for,
28, 29, 86, 87, 115, 119, -120, 425,
426, 435 ; general rule for, 23 ; of
implicit functions, -40, 73, 154,
458; logarithmic, 93; partial,
445, 462; successive, 73, 462
Diocles, cissoid of, 44, 46, 270, 277,
322, 531
Direction of a curve, 42
Ellipsoid, 280, 285, 522
Envelopes, 466
Equations, graphical solution, 128;
interpolation, 129 ; of motion,
120; Newton's method, 131
Errors, 138, 451; percentage, 138;
relative, 138
Evolute, 158, 469; of the cycloid,
161 ; of the ellipse, 160, 533; of
the parabola, 159, 470; proper-
ties of, 1 62
Exponential curve, 89, 535
Exponential function, 89
Factorial number, 338
Family of curves, 230, 466
Fluid pressure, 325 ; center of, 506
Fluxions, 19, 357
Folium of Descartes, 46, 119, 288,
533
Formulas, approximate, 367, 372,
490 ; for reference, 1-6
Fourier, 238
Function, complementary, 394 ;
continuity of, 12, 444 ; decreas-
ing, 50 ; definition of, 8, 444 ;
derived, 21; differentiate, 21,
23 ; discontinuous, example of,
108 ; exponential, 89 ; of a func-
tion, 37; graph of, 10, 444; hy-
perbolic, 414, 415; implicit, 39,
73,458; increasing, 50 ; inverse,
38 ; inverse hyperbolic, 423 ; in-
verse trigonometric, 105 ; loga-
rithmic, 89 ; mean value of, 333 ;
periodic, 97 ; of several variables,
444 ; sine, 97 ; table of hyper-
bolics, 416 ; transcendental, 86 ;
trigonometric, 99
Fundamental theorem of the in-
tegral calculus, 254-257
Graph of a function, 10, 444
Gravity, center of, 320, 323
Greek alphabet, 6
Gudermann, 435
Gudermannian, 435; inverse, 435
Gyration, radius of, 508
Harmonic vibration, 403, 405
Helix, circular, 473, 474, 482
Homer's method, 130
Hyperboloid, 528
Hypocycloid, 46, 119, 156, 244, 268,
270, 276, 280, 288, 468, 533
Increments, 19; approximation of,
137, 451
Indeterminate forms, 174
Inertia, moment of, 508, 514
Infinitesimals, 17; replacement
theorem, 147
Infinity, 13
Inflectional points, 79
Initial conditions, 229
Integrals, 188 ; change in limits,
240 ; decomposition of interval,
250 ; definite, 237 ; discontinu-
ous, 251 ; geometric representa-
tion, 244, 492; improper, 250
253 ; indefinite, 189 ; interchange
of limits, 249 ; multiple, 491 ;
table of, 538 ; use of table, 315
INDEX
555
Integrand, 195
Integration, 187 ; approximate, 245 ;
of binomial differentials, 299;
formulas for, 191 193, 430, 432,
433, 435 ; fundamental theorem
of, 254 ; by mlsceJhmpotrs subsU-
tutions, 221, 305, J32 ; by parts,
223 ; by raTional^fracIions, 289 ;
by rationalization, 221, 296; by
reciprocal substitution, 305; by
reduction formulas, 307, 312;
successive, 491 ; of trigonometric
forms, 213, 303, 312
Interpolation, 129, 372
Interval of a variable, 7
Involute, 163 ; of a circle, 156, 276,
288, 537
Isothermal expansion, 330
Jacobi, 445
Laplace, 19
Laws of the mean, 172, 182, 482
Leibnitz, 26
Lemniscate, 127, 155, 263, 515, 516,
517, 532
Length of arc, plane curves, 271 ;
in polar coordinates, 274 ; of
skew curves, 473
Limac.on, 534
Limit of a variable, 10
Limits, change in, 240 ; of an inte-
gral, 238; theorems on, 11, 17
Lituus, 534
Logarithmic curve, 89, 535
Logarithmic differentiation, 93
logarithmic function, 89
Logarithms, common, 88 ; natural,
87
Loxodrome, 439
Maclauriri's series, 357, 367
Maxima and minima, 47 ; analytic
treatment, 182 ; definitions, 52 ;
first method, 53 ; functions of two
variables, 483 ; second method,
76
Mean value, extended theorem of,
182 ; of a function, 333 ; theo-
rems of, 172, 482
Mechanics, 402
Mercator, 439 ; chart, 439
Moment, of area, 320, 503; of in-
ertia, 507, 508, 514; polar, 510,
514
Motion, curvilinear, 120 ; recti-
linear, 65, 83
Napierian logarithms, 88
Xewton, 19, 27, 131, 132, 133, 332,
401
Nicomedes, conchoid of, 532
Normal, to a plane curve, 43 ; plane
to a skew curve, 471, 480; to a
surface, 475
Osculating circle, 170
Pappus, theorems of, 335, 504
Parabola, 532, 537; cubical, 156,
531 ; semicubical, 266, 498, 531
Parabolic rule, 247
Paraboloid of revolution, 268, 521,
528
Parameter, 1 15, 466
Parametric equations, 115; first
derivative, 115; second deriva-
tive, 119
Point of inflection, 79
Polar coordinates, 123 ; moment of
inertia, 514; subnormal, 126;
subtangent, 126
Power rule, 32
Pressure, fluid, 325 ; center of,
506
Probability curve, 535
Projectile, 121, 234, 471
Quadratic equation, 1
Radius, of curvature, 152; of
gyration, 508
Railroad curves, 152
Rates, 64, 455
556
INDEX
Rectification, of plane curves, 271 ;
in polar coordinates, 274 ; of
skew curves, 473
Reduction formulas, 307, 312
Replacement theorem, 147
Rhumb line, 439
Rolle's theorem, 169
Roots of equations, 128
Rose-leaf curves, 536
Secant curve, 535
Sequence, 338
Series, 338 ; absolute convergence,
348; alternating, 347; approxi-
mate formulas from, 367, 372 ;
binomial, 353 ; Cauchy's test,
345 ; comparison tests, 342 ; con-
vergent , 340 ; differentiation and
integration of, 365 ; divergent,
340 ; geometric, 339 ; harmonic,
342; Maclaurin's, 357; opera-
tions with, 362 ; oscillating, 339 ;
p, 344; power, 350, 354;
Taylor's, 369, 488
Simpson's rule, 247
Sine curve, 533
Skew curves, 471, 480; length of,
473
Slope of a curve, 42 ; parametric
form, 115; polar form, 125
Solids of revolution, centroid of,
323 ; surface of, 277 ; volume of,
265, 267, 268
Speed, 121, 146
Spherical coordinates, 6, 529, 530
Spheroid, oblate, 266 ; prolate,
266
Spiral, of Archimedes, 127, 128, 155,
277, 534 ; hyperbolic or recip-
rocal, 119, 128, 264, 277, 534;
logarithmic or equiangular, 127,
128, 534 ; parabolic, 535
Stirling, 357
Strophoid, 534
Subnormal, 43 ; polar, 126
Subtangent, 43 ; polar, 126
Successive differentiation, 73, 462
Successive integration, 491
Table, of hyperbolic functions, 416 ;
of integrals, 538
Tangent, horizontal, 42, 117; to a
plane curve, 43 ; to a skew curve,
471, 480 ; plane to a surface, 475 ;
vertical, 42, 117
Tangent curve, 535
Taylor's theorem, 369, 488
Telegraph line, 428-429
Torus, 267
Tractrix, 85, 270, 282, 419, 423, 436,
537
Transformation of derivatives, 166
Transition curves, 152
Trapezoidal rule, 245
Triple integration, 521
Trisectrix, 155, 264
Variable, change of, 166, 457 ; defi-
nition, 7 ; dependent, 8 ; inde-
pendent, 8
Velocity, curvilinear motion, 120,
146; rectilinear motion, 65
Vibration, damped harmonic, 405 ;
forced harmonic, 405 ; simple
harmonic, 403
Volume, of a hollow solid of revo-
lution, 267 ; of a solid with
known cross section, 283 ; of a
solid of revolution, 265, 267;
under a surface, 501, 524; by
triple integration, 521, 526
Witch, 62, 251, 269, 322, 531
Work, 328
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