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Unformatted text preview: CarnegieMellon University UNIVERSITY
LIBRARIES
PRESENTED BY Person MATHEMATICAL TEXTS
FOR COLLEGES
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
and F. SMITH, PH.D. 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
AND WILLIAM RAYMOND LONGLEY F. SMITH ALL EIGHTS RESERVED
641.10 COPYBIGHT, 1904, 1911, COPYRIGHT, 1929, BY WILLIAM ANTHONY GRANVILLE AND PERCEY 1934, BY WILLIAM ANTHONY GRANVILLE, PERCEY
AND WILLIAM RAYMOND LONGLEY &tfienarum ffiregg
GINN AND COMPANY PROPBIETOKS BOSTON U.S.A. P. F. SMITH SMITH PREFACE
The
changes alterations in the text
in the details of some in this revision consist of verbal 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 completeness 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 selfreliance 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 wellnigh universal favor accorded Granville'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
Formulas
trigonometry, 2 Formulas from plane analytic geometry, 3
from solid analytic geometry, 4 Greek alphabet, 6 CHAPTER VARIABLES, FUNCTIONS, II. AND LIMITS 7 Variables and constants, 7 Interval of a variable, 7 Continuous variation, 7 Functions, 8 Independent and dependent variables. 8 Notation
Division by zero excluded, 9
of functions, 8
Graph of a function conTheotinuity, 10 Limit of a variable, 10 Limiting value of a function, 11
rems on limits, 1 1 Continuous and discontinuous functions, 12 Infinity (oo),
Theorems concerning infinitesimals and limits, 17
13
Infinitesimals, 17
; CHAPTER III. DIFFERENTIATION 19 Introduction, 19 Increments, 19 Comparison of increments, 20 Derivative of a function of one variable, 21
Symbols for derivatives, 22 Differentiable functions, 23 General rule for differentiation, 23 Interpretation
of the derivative by geometry, 25 CHAPTER IV.
FORMS RULES FOR DIFFERENTIATING ALGEBRAIC
28 Differentiation of a constant, 29
General Rule, 28
Importance
Differentiation of
Differentiation of a variable with respect to itself, 29
a sum, 30 Differentiation of the product of a constant and a function, 30
Differentiation of the
Differentiation of the product of two functions, 30
Differentiation of a
product of n functions, n being a fixed number, 31
Differentiation
function with a constant exponent. The Power Rule, 32
of a quotient, 32
Differentiation of a function of a function, 37 DifferenDifferentiation of
tiation of inverse functions, 38
Implicit functions, 39
implicit functions, 40
of the CHAPTER V. VARIOUS APPLICATIONS OF THE DERIVATIVE . Direction of a curve, 42 Equations of tangent and normal lengths of subMaximum and minimum values of a functangent and subnormal, 43
tion introduction, 47
Increasing and decreasing functions. Tests, 50
First
Maximum and minimum values of a function definitions, 52
method for examining a function for maximum and minimum values.
Maximum or minimum values when f'(x) becomes
Working rule, 53
Maximum and minimum values.
infinite and f(x) is continuous, 55
Derivative as the rate of change, 64
Velocity in
Applied problems, 57
rectilinear motion, 65
Related rates, 67
; ; ;   42 CONTENTS viii PAGE CHAPTER VI. SUCCESSIVE DIFFERENTIATION AND APPLICA TIONS ... . 73 Definition of successive derivatives, 73
Successive differentiation of imDirection of bending of a curve, 75 Second method for
Points of inflection, 79
testing for maximum and minimum values, 76
Curvetracing, 81 Acceleration in rectilinear motion, 83 plicit functions, 73 CHAPTER VII. DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS. APPLICATIONS 86 Formulas for derivatives; second list, 86 The number c. Natural logarithms, 87
Exponential and logarithmic functions, 89 Differentiation of
a logarithm, 90 Differentiation of the exponential function, 91 Differentiation of the general exponential function. Proof of the Power Rule, 92
The function sin j*, 97 Theorem, 98
Logarithmic differentiation, 93
The other trigonometric functions, 99 DifDifferentiation of sin v, 99
ferentiation of cos ?;, 100 Proofs of formulas
XIX, 100 Comments,
101
Inverse trigonometric functions, 105
Differentiation of arc sin r,
Differentiation of arc cos r, 106
Differentiation of arc tan v, 107
106
Differentiation of arc ctn v, 108 Differentiation of arc sec r and arc esc v,
108 Differentiation of arc vers r, 109 XV CHAPTER VIII. APPLICATIONS TO PARAMETRIC EQUATIONS,
POLAR EQUATIONS, AND ROOTS 115 Parametric equations.
Parametric equations of a curve. Slope, 115
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 subReal roots of equations. Graphical
Uingent and polar subnormal, 126
Second method for locating real roots, 130
Newton's
methods, 128
method, 131 CHAPTER IX. DIFFERENTIALS Introduction, 136 136 Definitions, 136 Approximation of increments by
Small errors, 138 Formulas for finding the
differentials of functions, 140 Differential of the arc in rectangular coordiDifferential of the arc in polar coordinates, 144
nates, 142
Velocity as
the timerate of change of arc, 146 Differentials as infinitesimals, 146
Order of infinitesimals. Differentials of higher order, 148 means of differentials, 137 CHAPTER
TURE X. CURVATURE. RADIUS AND CIRCLE OF CURVA. . . .149 Curvature, 149 Curvature of a circle, 149 Formulas for curvature; rectangular 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 curvature, 157 E volutes, 158 Properties of the evolute, 162 Involutes and
Transformation of derivatives, 166
their mechanical construction, 163
; CONTENTS ix PAGE CHAPTER XL THEOREM OF MEAN VALUE AND
TIONS . . ITS APPLICA169 . . Rolle's Theorem, 169 Osculating circle, 170 Limiting point of intersection of consecutive normals, 171
Theorems of Mean Value (Laws of
Indeterminate forms, 174
Evaluation of a function
the Mean), 172
Evaluation of the indeterminate
taking on an indeterminate form, 174 form > 175 Evaluation of the indeterminate form , 17$ Evaluation . indeterminate form
Evaluation of the indeterminate
QO, 178
Evaluation of the indeterminate forms 0, 1, 00,
form oc  oo, 179
180 The Extended Theorem of Mean Value, 182 Maxima and minima
treated analytically, 182
of the INTEGRAL CALCULUS
CHAPTER XII. INTEGRATION; RULES FOR INTEGRATING
STANDARD ELEMENTARY FORMS
. . . 187 Constant of integration. Indefinite integral, 189
Integration, 187
Rules for integrating standard elementary forms, 190 Formulas (3), (4),
Proofs of (0) and (7), 198 Proofs of (8) (17), 200 Proofs of
(5), 193
Proofs of (22) and (23), 211 Trigonometric differentials,
(18) (21 ), 203
2
?/
a 2 by
or V u'z
213
Integration of expressions containing Vutrigonometric substitution, 221 Integration by parts, 223 Comments, 227
' CHAPTER XIII. CONSTANT OF INTEGRATION 229 Determination of the constant of integration by means of initial condiGeometrical signification of the constant of integration, 229
tions, 229
Physical signification of the constant of integration, 233 CHAPTER XIV. THE DEFINITE INTEGRAL 237 The definite integral, 237
Differential of the area under a curve, 237
Calculation of a definite integral, 239 Change in limits corresponding to
change in variable, 240 Calculation of areas, 241 Area when the equaGeometrical repretions of the curve are given in parametric form, 242
trapezoidal rule,
sentation of an integral, 244 Approximate integration
245
Interchange of limits, 249
Simpson's rule (parabolic rule), 247
Decomposition of the interval of integration of the definite integral, 250
The definite integral a function of its limits, 250 Improper integrals. In<(x) is discontinuous, 251
finite limits, 250 Improper integrals. When //
; CHAPTER XV. INTEGRATION A PROCESS OF SUMMATION Introduction, 254 . of Integral Calculus, 254
257 Areas of plane curves . The Fundamental Theorem Analytical proof of the Fundamental Theorem,
rectangular coordinates, 258 Areas of plane curves ; polar coordinates,
Volumes of solids of revolution, 265 Length of a curve, 271
262
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 ; 254 CONTENTS x CHAPTER FORMAL INTEGRATION BY VARIOUS DEVICES XVI. PAGE
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 Transformation of trigonometric differentials, 303 Miscellaneous substitutions, 305
; CHAPTER REDUCTION FORMULAS. USE OF TABLE OF XVII. INTEGRALS 307
307 Reduction formulas for binomial differentials,
Reduction formulas for trigonometric differentials, 312 Use of a table of
integrals, 315
Introduction, 307 XVIII. CENTROIDS, FLUID PRESSURE, AND OTHER
APPLICATIONS CHAPTER Moment of area centroids, 320
Fluid pressure, 325 Work, 328
; * 320 Centroid of a solid of revolution, 323 Mean value of a function, 333 DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER XIX. SERIES 338 The geometric series, 339 Convergent and divergent
Definitions, 338
Comparison tests, 342 Cauchy's testseries, 340 General theorems, 341
Absolute convergence, 348
ratio test, 345
Alternating series, 347
Summary, 348 Power series, 350 The binomial series, 353 Another
type of power series, 354
  CHAPTER XX. EXPANSION OF FUNCTIONS 357 Operations with infinite series, 362 Differentiaseries,
tion and integration of power series, 365 Approximate formulas derived
Another form of
from Maclaurin's series, 367
Taylor's series, 369
Taylor's series, 371 Approximate formulas derived from Taylor's series, 372
Maclaurin's 357  ... CHAPTER XXL ORDINARY DIFFERENTIAL EQUATIONS 375 order and degree, 375 Solutions of differential
equations. Constants of integration, 376 Verification of the solutions of
Differential equations of the first order and of
differential equations, 377
the first degree, 378 Two special types of differential equations of higher
order, 387 Linear differential equations of the second order with constant
Differential equations coefficients, 390 Applications. Compoundinterest law, 399 Applications
Linear differential equations of the nth
to problems in mechanics, 402
 order with constant coefficients, 407 CHAPTER XXII. HYPERBOLIC FUNCTIONS 414 Other hyperbolic functions, 415 Table
of values of the hyperbolic sine, cosine, and tangent. Graphs, 415 HyperRelations to the equibolic functions of v + w, 417
Derivatives, 420
Derivatives
Inverse hyperbolic functions, 423
lateral hyperbola, 420
Integrals, 430
Integrals (conTelegraph line, 428
(continued), 425
tinued), 432 The gudermannian, 435 Mercator's Chart, 438 Relations
between trigonometric and hyperbolic functions, 440
Hyperbolic sine and cosine, 414  CONTENTS
CHAPTER XXIII. xi PARTIAL DIFFERENTIATION PAGE
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 derivatives. 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 considered 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 difAnother form of the equations of the tangent line and
ferential, 477
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 functions of two or more variables, 488
  CHAPTER XXV. MULTIPLE INTEGRALS 491 integration, 491 Definite double integral. Geomet492 Value of a definite double integral taken over a
region S, 497 Plane area as a definite double integral. Rectangular coordinates, 498 Volume under a surface, 501 Directions for setting up a double
Moment of area and centroids, 503 Theorem of Pappus,
integral, 503
504 Center of fluid pressure, 506 Moment of inertia of an area, 508
Polar coordinates. Plane area, 512
Polar moment of inertia, 510
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 Partial and successive  ric interpretation,
     CHAPTER XXVI. CURVES FOR REFERENCE
CHAPTER XXVII. TABLE OF INTEGRALS
INDEX 531 538
553 DIFFERENTIAL CALCULUS
CHAPTER I COLLECTION OF FORMULAS
1. Formulas from elementary algebra and geometry. For the convenience of the student we give in Arts. 14 the following lists of
formulas. We begin with algebra. (1) Solution. to zero, , 1. By and solve By 2. of J 2 AJT* Quadratic factoring B.r + C= Factor Ax'2 completing the square : Nature of the roots. 4 r, set each factor equal of half the coefficient of x, and _R
x =
The Ii expression called the di^crlmhuint. is 0. Bx Transpose r, divide by the coefficient add to both members the square the formula 4 for x. extract the square root.
3. By the formula real : + 2 4 The two and equal, or imaginary, according AC beneath the radical in
and unequal, roots are real as the discriminant is positive, zero, or negative.
(2) Logarithms = Jog a f log b.
= log a log b. log ab
Jog ^
(3) = n log a.
 log a.
'a
log v'a = 7
log a ' log 1=0. log a a = 1. Binomial theorem (n being a positive integer)  (4) Factorial numbers. n! = [n = l234(w l)w. In the following formulas from elementary geometry, r or
denotes radius, a altitude, B area of base, and s slant height.
(5) Circle. = 2 Trr. Area = Trr 2
= \ r 2 a, where a = central angle of the sector
Area Circumference (6) Circular sector.
measured in radians. R . 1 DIFFERENTIAL CALCULUS
Volume (7) Prism. = Ba. \ Ba.
Pyramid. Volume
(9) Right circular cylinder. Volume (8) Total surface
(10) Right
Total surface 2 irr(r
circular = Volume cone. Volume Frustum 2 Lateral J irr a. = :! Trr 3 Surface . = irs(R 2 Trra. surface = Trrs. Measurement Trr 2
. = 2 J ?ra(fi 2
f r f fir). f r). Many Formulas from plane trigonometry.
mulas will be found useful.
2. that ; of the following for There are two common methods
there are two unit angles. of angles. ing angular magnitude 4 Volume of a right circular cone. Lateral surface (1) = Lateral surface a). irr(r f s). (11) Sphere.
(12) + 2 ?rr a. of measur is, complete revolution and is Circular measure. The unit angle is an angle whose subtending arc
equal to the radius of that arc, and is called a radian.
The relation between the unit angles is given by the equation is Degree measure. The unit angle is of a l
;i ( called a degree. 180 degrees
the solution of
1 degree From TT radians 3.14159 (TT ), which gives = T^ = 0.0174 radian ; 1 radian 1 = 80 57.29 degrees. 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
1 ctn .r j* ; ctn (3) x f cos 2 x Formulas for = 1 ; 11 csc r ~
sin x JT cos x , .r cos x
sin 2 . cos sin x , tan sec x tan sin tan 2 x reducing angles = x sec 2 .r ; 1 f ctn 2 x = esc 2 x. COLLECTION OF FORMULAS
(4) Functions of (x f y) sin (x
sin cos tan (jr +  y} = sin
= cos
(x f
 #) = cos
/y) sin j ^L^^UL
 tan tan
jr cos + cos x sin y. cos #  cos .r cos //  sin .r cos // .r (.r Functions of 2 x and 2s = 2 sin y) sin x cos y #) 1 (5) (x y) f (.r cos and jc x sin f sin ^ tau . sin y. .r __ .r = } * //. sin ?/. 1 f  .r _lan // tan tan y .
tan y .r x  cos 2 ; { cos 2 .r  jc sin 2 x tan 2 ; .r  __r tan ^
1 sin' (6) j' cos ;   i =, fl^L'
   1 cos 2 .r COS ; L> .r  = tun : \ .. tan 2 x 2 + J cos 2 .1 j. Addition theorems
sin jr f sin .sin x sin // cos = '2 . 2 cos .1 co^ \ // cos .r f // cos.r cos// L* sin ^ i (.r \ (.r ?/). f //)sin ! (.r //). f (.> ( cos // f ./ //) ) cos \ (j ?/). ' i? yiu ', //)sin ^(.r (./  jy). (7) Relations for any triangle Law of sine*. sin Law Formulas for sin .4 a2 of cos hi ex.   ^~ = . /> C sin h f r 2 2 be cos A. area. sin (/> 4  a) (a T) where r), />)(.; ,s ?, (a f b f r). Formulas from plane analytic geometry. The more important formulas are given in the following list.
3. (1) Distance between two points /*i(xi, yi) and P2(x2, yz) Slope of P^Pz. Midpoint.
(2) x\ (x, + x 2 ), ?/ = 2 (2/i f 2/2). Angle between two lines
1 f (For parallel lines, wi = W2 ; mim2 for perpendicular lines, mint = 1.) DIFFERENTIAL CALCULUS
Equations of straight lines (3) y Pointslope form. = m(x = y\ mx + b.
~^
yz y Slopeintercept form.
"" y
x Twopoint form. '  x% x\  Intercept form. yi = ~ f xi). Xi 1. + By f C = Perpendicular distance from the line Ax (4) between rectangular and polar coordinates (5) Relations (6) Equation of the Center
(7) to PI(JCI, yi) circle (h, k). (JT  2 h)' +  (y = 2 k) r 2
. Equations of the parabola
y' = 2 px, focus (\ p, 0). = 2 Vertex the origin. 2 p(x h), axis // 2 p(y k}, axis x 2 2 py, focus x' Vertex (h, k). k) (y 2 h}' (x Axis ///r 2 ylj^ y yaxiii. 2 (0, i /;). = /c. /i. f C. Equations of other curves (8) Ellipse with center at the origin Hyperbola with center and with at the origin ^1
a2 _ 111 and ~ foci on the xaxis. (a > b) with foci on the xaxis. i b2 Equilateral hyperbola with center at the origin and with the coordinate
axes for asymptotes. xy
See also Chapter
4. = C. XXVI. Formulas from solid analytic geometry. Some of the more impor tant formulas are given.
(1) Distance between PI(XI, d= V(xi  x2 yi, z\)
2
) h (yi and  P2(*2, l/ 2 ) 2 + 1/2, (21 22)  z2 2
) . COLLECTION OF FORMULAS
(2) Straight line Direction cosines cos a, cos
Direction numbers: a, b, c.
: cos 7. (3, Then
a cos 2 6 cos 2 a f + 2
i cos cos For the line joining Two cos : Direction numbers: = 2 cO and _ cos = cos cos a + r2 h2 f ~+ 6 (j 2 , ?/ L
., .,) cos 7 cos <v', __ ft cos f W/ c/a' 42 + cos ft, 7 ; + b r2 ft cos /2 aa cos f ft' // "f cos 7'. 7 cos 7', r 2 2
f r' ^ a'
lincx. ft', f re' Va =~= Parallel lines. Perpendicular cos a', //, r'. lines, cos <V = t/ cos <rv, a, b, r; angle between the
cos (4) r2 2 b' r Vu or + 1. lines Direction cosines
If 2 = (jc\, ?/i, cos
(3) Va 7 = b  ft 7 a = a cos r cos 2 f ft 6' x r f 6/>' f rr' . ~ (). Equations of the straight line with direction numbers ing through (xi, z/i, 21) x  n _ y  a 7/1 _z b a, b c passt :\
t c For the plane Ax f Ky + Cz + J>() the coefficients A, B, C
numbers of a line perpendicular to the plane.
a plane passing throuyJi (>i, //j, ^ij and perpendicular to the
Equation of
line with direction numbers A, B, C.
(5) Plane. are the direction A(x
(6) Two  XL) + B(y  ?/,) f C(z  2,) = 0. planes Equations : Ax + By f Cz 4 D =
A'x+ B'y+ C'z+ D' = 0,
(). Direction numbers of the line of intersection EC'  CB' t CA'  AC', AB' :  BA'. DIFFERENTIAL CALCULUS
If 6 = angle between the planes, then AA' = cos + BB' + CC' B'2 The (7) Cylindrical coordinates.* the
its C /2
distance z of a point P(x, y, z) from 7plane and the polar coordinates (p, 6) of
projection (x, y, 0) on the X Tplane are called
A' 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 2 p cos  0, + (8) Spherical y
9 P sin 2 0, arc tan  x The coordinates.* radius vec between OP and
between the projecand the angle
on the ATplane and the .raxis are tor r of a point P, the angle the zaxis, OP tion of called the spherical coordinates of
the colatitude and 6 the longitude. coordinates of P are written (r, /'. (/>, is </> The called spherical 0). the rectangular coordinates of P are x, y, z,
then, from the definitions and the figure, we have
If =
f * r c/> cos =: ?/ = r sin 6 sin = arc tan > sin ^ $, 2 </> = r cos ; = arc tan LETTERS Z
T
T d NAMES Rho P p
s Sigma r Tau i> Upsilon
Phi <f> X * Psi a Omega of cylindrical and spherical coordinates see Smith, Gale, and
Analytic Geometry, Revised Edition" (Ginn and Company), pp. 320322. For a discussion "New Chi i/' 12 Neellry's x ^ CHAPTER
VARIABLES, FUNCTIONS,
6. Variables and constants. unlimited number of values Variables are denoted usually A quantity whose value is II AND LIMITS A variable is a quantity to which an
can be assigned in an investigation.
by the later letters of the alphabet.
any investigation is called fixed in a constant. Numerical or absolute constants retain the same values
lems; as 2, 5, V?, TT, in all prob 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
the intercepts, for
line, while the arbitrary constants a and b are
which definite values are assumed.
The numerical (or absolute) value of a constant a, as distinguished
= 2 = 2.
from its algebraic value, is represented by a. Thus,
2
value of a."
The symbol a is read "the numerical
7. Interval of a variable. Very often we confine ourselves to a portion 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.
 be included, or either or both excluded. We shall
employ the symbol [a, 6], a being less than b, to represent the numbers a, b, and all the numbers between them, unless otherwise stated.
u
This symbol [a, 6] is read the interval from a to b."
8. Continuous variation. A variable x is said to vary continuously
a to the
through an interval [a, 6] when x increases from the value
value b in such a manner as to assume all values between a and 6
= 6 to
in the order of their magnitudes, or when x decreases from x
values. This may be
x = a, assuming in succession all intermediate
illustrated geometrically by the diagram on page 8.
Also, a and b may DIFFERENTIAL CALCULUS 8 The
and B origin being at O, lay off corresponding to the on the straight numbers a and b. line the points A Also, let the point P correspond to a particular
value of the variable x. Evi^
^
g
dently the interval [a, b\ is represented by the segment AH. As x varies continuously through
the point P generates the segment AB when x
the interval [a,
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. H problems deal with quantities and relations of
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 Nearly this sort, all scientific and in Similarly, the distance a his strength, other things being equal. boy 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 can run may 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
variable, or junction. is given, is called the dependent we are considering two related variables, it is
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 exFrequently, when in our power to fix 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 function of J, and is read / of x. In order to distinguish between different
is changed, as F(.r), (/>(;r), /'(*), etc.
a functional symbol indicates the same
During any investigation
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 functions, the prefixed letter of the variable. Thus, if J(x) =r j2 then f(y) = y2  9 .r 9 y + 14,
+ 14. AND LIMITS VARIABLES, FUNCTIONS,
Also /(&+l) = (6 + l) a 9 + l) + 14 = & 2 7& + 6, 9ifc /(0)=0 9. 0+1414,
2 /( 1) /(3) (  1)2 0( = 3 93 + a + 14 4. 1) 14 The Division by zero excluded.
number x such that a = 12.
6 is = />.r. 24, quotient of two numbers a and
Obviously, division by zero is ruled out by this definition. For if
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
l> a
()' ()' are, therefore, meaningless. Care should he taken not to divide hy zero inadvertently.
is an illustration. The low ing fallacy Assume that a Subtracting ab
b(<i Factoring, Dividing by <i~. a'2 bb) />, a b. = ah Then, evidently, = b,  a therefore b 1 The ahsurd result is />). b. ; 2 b or + a l> But 2
b' . (a f />)(</ />, 2.  due to the we divided by fuel that a b PROBLEMS
20, show 3. =  5 sthat
= 12, /(5) = 0, /(())   2/C}), f^} ^ r)/( 1).
/(I)
 2 s f r find /(O ),/(!),/( l),/(2 ),/( 2).
Itfis) = 4
If F(0) = sin 2
+ cos 6, find /'(O), /M TT), F(TT). 4. Given 1. Given 1 jr* f(s) .r }  2. 1 , :( 5 ,r /(.r) 2 4 jr' /(/f 1)
5. Given 2 /(//) 6. Given /(j*) = f 7. Given //) f = > /(r) = f 3 ?'*> fU /
/O L> ?/ j,  f  20, show that ^  2 /+ 11 12. show that
r 2 y f 6, ?/ f(y = jr
;! 2 ?/ f f f> 12 O/ ~ Given 0(z) = 4* h2 f R show that  f(s) = Ms 2 + show that /(V f /O 1 ;// f 3 ^  /O)  2 show that <f>(z + 1)  <t>(z) =  . ^ J^ f   Jf 8. + 1 J/' 3 ~T" X/ 0(). 0. fol DIFFERENTIAL CALCULUS 10
9. If <f>(x) = a', 10. Given 0(x) 11. Given /(x) show that 0(y) = x 1 x, 13. and (6), p. of Graph = 0(y + 2). show that f(x f 2 h) HINT. Use </>(*) show that log = sin  f(f) = 2 cos (x f sin /?) /?,. .'}. a function; continuity. Consider the function x 2 , let y (1) = x'2 . This relation gives a value of y for any value of x
is, y is defined by (1) for all values of the independent variable. The locus of (1), a parabola (see
2
If x
figure), is called the yraph of the function x
= />, then
varies continuously (Art. 8) from x
a to x
a to y/ = &', and
y will vary continuously from y
; that . /'(x, ?/) will move continuously along the graph from the
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." the point point (a, Consider the function  and
x
y (2) = let  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 an equilateral hyperbola
tinuously through any interval [a, of (2), is then '// will will trace Then we
x A (see figure).
/>] If x increases con which does not include x decrease continuously from  to p the graph between the corresponding points
(a, )>
say, "the function  is continuous for all (?>, y) ^ values of x except 0." There is no point on the graph for x = 0.
These examples illustrate the concept of continuity
definition 0, and the point P(x, of a function. 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.
is AND LIMITS VARIABLES, FUNCTIONS, 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) increases 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
r
/
ultimately becomes and remains
than any preassigned positive number, however small. value of the difference The relation r ~ /. For convenience, we shall
"r approaches / as a limit," or, more
/, read,
"r approaches Z." (Some authors use the notation v == /.) defined use the notation
briefly, less is written lim r ILLUSTRATIVE EXAMPLE. Lot the valuer + 2 1, li i. t , without end. Then, obviously, lim we mark on a r  2, r .... 2 f 12 of or r be + 1, f U. straight line, as in Art. S, the point /, correspondon each side a length 6, however
\\ill ultimately all lie within
small, then the points determined by If ing to the limit and from L lay /, off /' the segment corresponding to the interval [/
e, / 1
e.
15. Limiting value of a function. In applications, the situation that
have a variable r, and a given function z
usually arises is this. We The independent variable r assumes values such that v
/.
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
of r. We lim
r read, "the limit of z, z = a, I as v approaches is I, a." 1G. Theorems on limits. In calculating the limiting value of a function, the following theorems may be applied. Proofs are given in Art. 20.
and w are functions of a variable j, and suppose that
Suppose u,
?', u lim
x  Then the following
(1) A, lim v B, x a lim
r  + v  w) = A
lim (uvw) = ABC. f B C. a x* a (3) a relations hold. lim (u jc (2) = a lim
jta u
"
V =A
B 9 if B is not zero. w= C\ DIFFERENTIAL CALCULUS 12 Briefly, in words, the limit of of a quotient an sum, of a product, or
algebraic sum, product,
provided, in the Last named, that the
algebraic same is equal, respectively, to the or quotient of the respective limits,
limit of the denominator is not zero.
If c is a constant (independent of rj and B is not zero, then, from the above,
lim (u (4) x + c) = A + lim cu c, a * lim cA, a JT* c
 c = Jj a V > a Consider some examples.
Prove lim 1. <JT'* f 4 a)  12. * JT Solution. The given function is the
ing values of these two functions. By
2
jr' lim
X By the answer (1), ~ ' Solution. 4 is ~ ~ 4, a = 4 lim x '.: 1 = x X f  8  article, where jc. We first find the limit x. 8. 12. t  U) it Hence, by 4. i^ :> 9) 5, we have '.'}), by (2) and (4). For the re<juired result. In Ex. of the pre 1 was shown that
lim
.r we observe 4 li Continuous and discontinuous functions. 17. ceding Thai ' since j : (Considering the numerator, lim the denominator, lim and .c r 9 : Prove lim 2. = ~ J" Hence, by of li lim 4 (4), sum
rji, (.r +4 .r) = 12, 1! that the answer is the ralue of the function for x 2. the limiting value of the function when ;r approaches 2 as a
equal to the value of the function for jc = 2. The function is is, limit 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
the limiting value of the function when .r approaches a as a limit
the value assigned to the function for .r = a. In symbols, if
lim
x then /(.r) is The
is not * if is /(*)=/ (a), a continuous for x
a.
is said to be discontinuous for x function = a if this condition satisfied. Attention
currence. is called to the following two cases of common oc VARIABLES, FUNCTIONS, AND LIMITS 13 CASE I. As an example illustrating a simple case of a function continuous for a particular value of the variable, consider the function
J For z = x ' *> X = /(I) = 3. Moreover, if x
1, f(x)
approaches
the function f(x) approaches 3 as a limit
(Art. 16).
function is continuous for x = 1. CASE 1 as a limit, Hence the The II. definition of a continuous function assumes that
already defined for xo. If this is not the case, however, 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. the function is Theorem. If f(x) is =
= B, not dcjiucd for x lim f(x) mil f(x) for x = be continuous for x if = a if tt is assumed as the value of a.
A 1> ~ Thus, the function
is not, and a x then f(x) a, defined for x T.ut for 2 (since then there would be division by zero).
every other value of :r, and lim (x therefore it 2) = 4 ; lim Although the function
sign to + is 2, it = 2, we arbitrarily as becomes continuous for this value. not defined for x the value 4 for x if A function f(x) is said to be coHtitiuoux in an niter ml whew
continuous for all valves of x in this in terra I.*
In the calculus we have of a function of a variable it is to calculate frequently the limiting value
when r approaches as a limit a value a ?' lying in an interval in which the function is continuous. This limiting 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. DIFFERENTIAL CALCULUS 14 however
values, we say large, it v becomes becomes positively becomes negatively
v = lim v = + oo lim , infinite; takes on only positive
negative values only, it If v
if The notation used infinite.
QO infinite. lim , for the three cases is
r = 00 . In these cases does not approach a limit as defined in Art. 14. The
"
v becomes infinite/'
notation lim v = QO, or v > oo, must be read
and not "v approaches infinity."*
v We may now write, for example, lim x meaning that  becomes
Referring to Art. 17, infinite
it = when x approaches appears that
lim/(.r)
x that
f(x) is, is A if becomes f(x) oo, *<)  = zero. if 00, a infinite as x approaches a as a discontinuous for x
a.
function may have a limiting value limit, then when the independent variable becomes infinite. For example,
lim i
X And, in general, when x > oo, if we oc = 0. X /O) approaches the constant value A as a limit use the notation of Art. 17 and write
lim /(*)
x  = A. oc Certain special limits occur frequently. These are given below.
r is not zero. The constant Written in the form, of limits
(1) lim   oo. = oo. = oo. V r (2) = Abbreviated form often used limczj =oo.
c oo = oo. r * oo /o\ (o) (4)
* V r Jim
r  oo  OO C lim " = oo. C = = 0. 0. 00 On account of the notation used and for the sake of uniformity, the expression
"
oo is
sometimes read P approaches the limit plus infinity." Similarly, r *
"
read "r approaches the limit minus infinity," and v
is read
r, in numerical value,
p * f oo is approaches the limit infinity."
This phraseology is convenient, but the student must not forget that infinity
limit, for infinity is not a number at all. is not a AND LIMITS VARIABLES, FUNCTIONS, 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
5 x   tS
.  x 7 x* Divide numerator and denominator by
either. Then we have Solution. present in J* The
first term
and (3) limit of each Hence, by (1) is step I>iride   J* x*\ _.
7 the highest power of x X* < numerator or denominator containing x is zero, by (4).
we obtain the answer. Jri any similar case the in of Art. H> therefore as follows. numerator and denominator by both the highest power of the variable occurring in cither. and u are functions of v if A is * if A, = x and and jr, lim u If lim a  a not zero, then u
r
lim r . r = 0, (i =: oo. tf This notation provides for the exceptional case of
= and A is not zero. See also Art. 20. (3), Art. 16, when B PROBLEMS
Prove each of the following statements, ^ lim Proof. = lim x
2
[Dividing both numerator and denominator hy x .] The
(4). limit of each Hence, by (1) term and in numerator and denominator containing x (3), Art. 16, 2. 3 4 t . 4. ,. lim = 2. + 3 + 2 = ""
+2 6 2  J obtain the answer. X lim r we ja / / r 2 /? r 3 4. x/z 2 _ f h*
, 2 1 3* =x is zero, by DIFFERENTIAL CALCULUS
2
3 h + 2 xh* + s'W
_
4j/ 3 _ _
 1
~
7
43*A2^ = 27
2i^ + 3*
+ =
ft'****' 4
=
8. Km
!
 **'' 1. 16 M
. ,. ,. 6 " ',! lim J 2 z(2z A.O "' 9 jj ', ft"7 . ' f
I h.r" ar
r
11. 11 in . . r : 4 (L'" , a, < 3. 7 = ao & 60 f  r
f ^^
 f r = 4 , , +
+ +
'+ Qi^" &./" m */*/*
2^* + 4z x,* /r; A 0. = cs* f ^^ 4 kr f
r
lim 7
fj (if f f'J"
f JS
' 1 , 12. l ^ ; lim a * . .s , oo. {/ ~ a  13. f . .1
j Hj_ _^Z_ rn nj,n i^ ( n positive integer) // ir. lc. 4 S* m i:
11 III ........ .7
. * S 5
(>_ ~* " .....  4 4 lim
16. r
i Proof. The obtain (Art. limiting value cannot" bo found lli) by substituting h We the indeterminate form  = Q, for a suitable manner as indicated below, namely, rationalize the numerator.
\ __x  h TT J Ten ce \_JT Given \ lim /(j) jr .r Given /(j) 19. Given = = x*, f \ \ r /z .r , a.r ,. lim
/*o\ f  ^ 2 * f 6.r f r, 1 j f W show that  show that h + *)
find lim /(* .r h
_  : show that 2 *~o
20. If f(x) //  = /(or) f // *> /j 18. J J JT +
  = ,. /* 17. \
\ f we then then transform the expression in == + \ x 2v
1 VARIABLES, FUNCTIONS,
19. Infinitesimals.
is an called A variable infinitesimal. This lim r is AND LIMITS 17 which approaches zero as a limit r written (Art. 14) = or r 0, and means that the numerical value of r ultimately becomes and
remains less than any preassigned positive number, however small.
lim If variable r = and /, its Conversely, then lim (r
/) = (); that
is an infinitesimal. the difference between is, a Inn it the different bit ween a variable if 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
I. An sum algebraic infinitesimals. of u infinitesimals is an infinitesimal, n being will become and remain less
becomes and a fixed number. For the numerical value
than e remains
II. less sum of the when the numerical value of each infinitesimal than n The product of n constant c by an infinitesimal is For the numerical value an infinitesimal. of the product will be less than the numerical value of the infinitesimal is less t han e when , kl
The product of n infinitesimals is an, infinitesimal, n being a
number.
fixed
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.
III. IV. If lim
i by v is also v = an /, and I is not zero, ihen the quotient of an infinitesimal 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. DIFFERENTIAL CALCULUS 18 Then i, j, k are functions of x, and each approaches zero as x
that is, they are infinitesimals (Art. 19). From equations (1),
obtain u (2) w f v The righthand member
Hence, by Art. (1), we have u = A
AH we get
ur (4) By k. an infinitesimal by theorem is lim (u h v transposing = i +j C) a\ we I above. 19, (3) From A+B ( > the above theorems I w) =A f B C. v=B+j. By { i,  AB = Aj f Bi multiplication and f ij. III the righthand member is an infinitesi mal, and hence
lim uv (5) ? The proof
Finally, is = AB. a readily extended to the product urw.
write we may f("\ 1* _ v 1 B~~ A i \ B+j _A_
~
B Bi Aj m B(B +j)' The numerator in (G) is an infinitesimal by theorems I and IT. By
and (4), lim /*(/*+./) = B'2 Hence, by theorem IV, the righthand member in (6) is an infinitesimal, and
(3) . U
V
hm  /r^
(7) j Hence the statements in Art.  v =A B 16 are proved. CHAPTER III DIFFERENTIATION
21. Introduction. We shall now proceed to investigate the manner 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 The increment modern mathematician. changing from one
numerical value to another is the difference found by subtracting the
An increment of x is denoted by the
first value from the second.
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,
22. Increments. of a variable in 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) dependent variable (or y). always to be reckoned from the definite
value of y corresponding to the arbitrarily fixed initial value
of x from which the increment Ax is reckoned. For instance, consider
the function The increment Ay is initial Newton Cl 6421 727), an Englishman, was a man of
He developed the science of the caloulas uml**r the name
Newton had discovered and made use of the new science 5ts early as
"
* Sir Isaac genius. the most extraordinary
of Fluxions. Although HJ70, his first published
Philosonhiao Naturalis Principia work in which it occurs is dated 1687. having the title
"
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 DIFFERENTIAL CALCULUS 20 Assuming x = value of y. initial 10 for the Suppose x increases to x initial = then y increases to y
x decreases to x
Suppose y decreases then =
to y = value of x fixes y 12, that Ax = is, Ay =
Ax = and 144, Ay = and that 9, is, 81, 2 = 100 as the ; 44.
1 ; 19. In the above example, ?/ increases when x increases and y decreases
The corresponding values of Ax and Ay have
decreases.
It may happen that // decreases as x increases, or the
like signs. when x
reverse
23. ; case in either Comparison Ax and A// will then = (1) have opposite signs. Consider the function of increments. 32. 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 + Ay/ (x + Ax) 2
y + Ay/ = x + 2 x Ax +
=x
2x Ax+
Ay/=
?/ or Subtracting
(2) we , 2 (Ax) 2
. 2 (1 ) // (Ax) 2 get the increment A// in terms of x and Ax.
find the ratio of the increments, divide both To members of (2) by Ax, giving V = 2 x
Ax
If the initial value of x is 4, it is + Ax.
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. DIFFERENTIATION 21 apparent that as Ax decreases, A?/ also diminishes, but their
on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01, It is ratio takes illustrating the fact that the value of '( 8 as we please by making A.r sufficiently small. lim
AJ
Az0 ^'
24. Derivative of a function = derivative* of ct function /,> Therefore 8. one variable. The of definition of the ditlerential calculus The can be brought as near to //,v is fundamental as follows. limit of the ratio of the increment of the function to the increment of the independent variable,
latter increment varies and approudtcx zero ax a limit. When the limit of this ratio exists, the function is when the said to be differ cntiable, or to possess a deriralire. The above may be given in a more compact form symGiven the function definition bolically as follows. 0) ?/ and consider x to have a = /U), fixed value. take on an increment A.r then the function y takes on an
Let
increment A?/, the new value of the function being
; (2) ?/ To find + A//  /(.r f A/). the increment of the function, subtract Av//(j (3) (1) from (2), giving + A.r)/M. Dividing both members by the increment of the variable, AJ,
(A\
( } The limit of the A# ="f(.r + A:r) f(jr)
A*
Ax
righthand member when AJ definition, the derivative of /(:r), or symbol
,,,
(A) ^ by (1 ), of ?/, and > is (} is, Therefore dy
dx /(x = r
hm
AXO + A*)/(*)
Ax defines the derivative of y [or f(n)] with respect
From (4) we get also dy
dx from the denoted by the to x. = r m Ay
lim
AX.O AJC * Also called the
differential coefficient or the derived function. 22 DIFFERENTIAL CALCULUS
u if Similarly, =
at The a function of is lim
A/ * A o = then t, derivative of u with respect to process of finding the derivative of a function is t. called differ entiation.
25. have Symbols Since for derivatives. Ay and Ax are always finite and 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
x
considered as a whole.
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 we may
which
of f, , !/=/(*), is ^ write = /'(j), read "the derivative of y with respect to x equals / prime x" The symbol ^
Tx when considered by itself is called the
any function written after indicates that respect to x. differentiating operator,
it is Thus, JL or T f(x) indicates the derivative of f(x) with respect to x
2
(2 x y' is and to be differentiated with y indicates the derivative of y with respect to x ; ; + 5) indicates the derivative of 2 x 2 + 5 with respect to x. an abbreviated form of The symbol Dc is dy
ax ? used by some writers instead of  ax If, then, DIFFERENTIATION
we may It write the identities must be emphasized that the variable, in the essential step of
Ax
0, is Ax, and not x. The value of the latter is assumed
from the start. To emphasize that x = xo throughout, we may
> letting
fixed 23 write From the Theory of Limits it is clear
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.
26. Differentiable functions. that if 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
for isolated values. most 27. ative General Rule for Differentiation. From the definition of a derivseen that the process of differentiating a function y = f(x) it is consists in taking the following distinct steps. GENERAL RULE FOR DIFFERENTIATION*
FIRST STEP. In the Junction replace x by x
new value of the function, y + Ay. SECOND STEP. + Ax, and cakulate Subtract the given value of the function from the value and thus find Ay This the new is the derivative required. 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 increment of the independent variable) varies and approaches zero as a limit.
(the 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 FourStep Rule. DIFFERENTIAL CALCULUS 24 ILLUSTRATIVE EXAMPLE x'2 4 5. Differentiate 3 1. we get, Solution. Applying the successive steps in the General Rule, = 3 4 5.
A^/ = 3(x 4 Ax) 4 5
= 3 x*' 4 x Ax 4y Firxf Step. y 4 2 ft Second Step, y 4 A?// A?/ '/Vnrd =
= 3 x fix Fir*/ tf/cyj. // y + A//  =
=
 Second Step. // 4 A// A// Fourth Step. 4 member = (3 x 4 5) 3(Ax)* x  ' Then by 0. 1 3 x 6 x. x'  1 2 x 4 7. f ( ( jr' ( ( 3 x  " 3 x 4  ! (x' Ax 3 x 3 x 4 2 x 4 Ax)
' 4 member  ( (Ax) let 3 x 7) 4  Ax = 2 ( Ax)' >  Then by 0. 2. // xA'JS/ Step. y f A//
(x Second Stej>. 4 A.r) a y 4~ A//
(x 4 Ax) (x 4 Ax) c , Third Step.
Fourth Step. ^=^Ax c . _ c Ax (2 x x j (x 4 x 4 Ax) 2 J__AL_.
.r'Cr In the righthand dr  Ax + Ax) member x\x) 5 let x <! Ax * 0. Then by 4 7.
4 7
4  2. Differentiatex 3. ~ Place Ax : j ILLUSTRATIVE KXAMPLE Ax 2 2 2x In the righthand y' (A)  2(x f Ax) 4 7
Ax f 3 x Ax) 4 (Ax)  2 x Ax 4 3 x Ax) J 4 Ax)  2 x  ' (x 4 A.r) 4 * 2 x 4 7. x 4 3
! Ax let Differentiate 2. = x   Or Solution. Ax Z,ZJ // Third Step. 5 4 6x43 Ax. ILLUSTRATIVE FA AMPLE
Place f 5. 3 (Ax) 4_5 In the righthand y' Solution. f 3( Ax) 4 ' S/f'/). Or Ax 6 x 3 x1 ~ Fourth Step. after placing x'' (A) 2 Ax 7 DIFFERENTIATION 25 PROBLEMS
Use the General Rule
1. 2.
3. 4.
5. 6. 7.
8. = 2 3 An*.
# = mx f 6.
= ax
2
s = 2t3
y = ex
 3
y = 3 J
2
M = 4 r f 2 r*
=x
r. i/ in differentiating =  3. //' 12. ?/ each of the following functions. =  4; 2 <ty . // t' . . j . 4 . // li_ 9.P^
; 10. y = 10,,
lo. // 4 I' 24. = 3 a  4  5.
8 = a/ +
+ r.
w = 2r*  3 r 2
 ar + bx' f r.r
p=(abOY
?/= (2  jr)(l ~ 2 25.  21.
22.
23. ar ?/ 26. x  27. j 2 2 :> )  ft/ s (r; // a . 2 + 2
t>x' d. ?/ 2 . j 29. of calculus to by geometry. We shall now
fundamental in all applications of the is It geometry. is of the necessary to recall the definition
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). = the derivative consider a theorem which
differential // a f 28. Interpretation Let and approach P />><
} ^ Q move along the curve
indefinitely. Then the secant will revolve about P, and
line at P. a s* i/= 20. f 8.T  2 / 19. j 3 = 11. s (.r 1 its limiting position is the tangent Let (1) y=f(?) be the equation of a curve AD.
(see figure). This curve is the graph of /(r) DIFFERENTIAL CALCULUS 26 Now differentiate (1) by the General Rule and interpret each step
from the figure (p. 25). We choose a point P(x, y)
geometrically
on the curve, and a second point Q(x + Ax, y + Af/) near P, also
on the curve. _ + Ay = f(x + Ax)
y + Ay = f(x + Ax) FIRST STEP. y SECOND STEP. =/(x) y A^
Ax THIRDSTEP. /(x = NQ
= NQ + Ax)  /(x) = Jgg. MN g
PR Ax = 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
r = inclination (/> Then lim </> = Assuming that tan r. Ar~ Art. 70), we of the secant line PQ, of the tangent line
is PT. a continuous function (see have, therefore, FOURTH STEP. ^~ /'(x) = lim ttX = tan <t> = tan r, Ar^O slope of the tangent line at P. Thus we have derived the important
Theorem. The value of
to the slope of the It was this tangent any point of a curve is equal
curve at that point. the derivative at tangent line to the problem that led Leibnitz* to the discovery of the differential calculus.
* Gottfried Wilhelm Leibnitz (16461716) was a native of Leipzig. His remarkable
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
abilities were shown by original investigations in several DIFFERENTIATION
ILLUSTRATIVE EXAMPLE. Find the slopes
and at the point where x = $ at the vertex of the tangents to the =2 3* parabola y = x* . we Differentiating by the General Rule (Art. 27), Solution.
(2) 27 x = slope To find slope of tangent of tangent line at any point at the vertex, substitute x giving = get on curve. (x, y) in (2) , dy dx~'
Therefore the tangent at the vertex has the slope zero that
it is parallel to the xaxis and in this case coincides with it.
To find the slope of the tangent at the point P, where x = i,
; is, substitute in (2), giving that is, , the tangent at the point P makes an angle of 45 with the xaxis. 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
2. 3. y y =
= 2, 2 x
4 =
X
6.  2 x' where x \ x ~, 2 Am. 1. where x , where x = 3. 4. 5. 2. y + 3 y = .r 3 ; 63 x \ where x x 3 x2 ;< 2 : , where x 26'.
1. 1. L = Find the point on the curve y the tangent line
7. = is 5 x 45. Find the points on the curve y parallel to tbe line y x* f 4 x. In each of the three following problems find x 2 where tbe inclination of
Anx. (2, 6).
x where the tangent line is
Ans. (1, 2), (
1,
2).
(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. =1 x
= x 2 1.
2
y = x
z2/ + 2 =
2 y Ans. Angle of intersection , = arc tan $ = 53 8'. y 9. 10. , 0. = x*3x,
2z + = 0. y ?/ Find the angle of intersection between the curves 9 y = x3 and
Ans. 21 21'.
6 + 8 x  x3 at the point (3, 3). 11. y = Newton were already in existence, and from these some claim Leibnitz got the
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 introwritten by new ideas. duced by Leibnitz. CHAPTER IV RULES FOR DIFFERENTIATING ALGEBRAIC FORMS
Importance of the General Rule. The General Rule for differentiation, 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
29. : 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 differentiate. FORMULAS FOR DIFFERENTIATION
dx dx
7dx II ...
III _ IV
_. v d
dx , (U +
, . V d
jdx W)
. . (ci>) = 1. = du
dx ^ c d dv dw + 3 Tdx
dx dv
dx du dv VI Via
v () =
dx\v/
28 du
dx __ u dv
_
dx which are RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 29 du VHa E(i?)=T' ^=^ VTTT
VIU dv
5U 5J 5x = IX " u em
*
.
k
S a function of
 ' J' y being a function of T, v. x. dy A 30. Differentiation of a constant. same value for every value
stant, and we may denote it by
the function that of the // = is known to independent variable is have
con c. As x takes on an increment AJ, the function does not change
value, that is, A?/ = 0, in and But ~ ~ oA.r dx lim
AX . ~~ 0. * dx The derivative of This result is a constant For the locus of y readily foreseen. OX, and line parallel to is zero. its is slope therefore zero. c is a straight 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.
FIRST STEP.
y + A?/ = x SECOND STEP.
THIRD STEP. FOURTH STEP. 27), we have + Ax. A?/ = A.r. ~ = 1. = 1. ( ~
dx !=
The derivative of a variable with respect to itself is unity. This result
unity. is readily foreseen. For the slope of the line y =x is DIFFERENTIAL CALCULUS 30 32. Differentiation of a sum Let y =u+ v w?. the General Rule, By FIRST STEP. + ky = u + Aw + v + A?;
Aw?.
A?/ = A?/ + Av y SECOND STEP. ^^ THIRD STEP. Now Ax
<to, Hence, by lim e/x A/  o ^!
Ax (1), Art. 16, ^ FOURTH STEP. c/x m
HI + Ax Ax
^___^. Ax d
.'. dx + v , ^, dx ^
dx . . (u == w) = im AX  o ^^.
Ax
dx ^
+ rfx _ ^_
dx ; du
dx + dv
dx A similar proof holds for the algebraic
functions. .  dwdx sum of any number sum of n functions is equal to the
of their derivatives, n being a fixed number. derivative of the algebraic algebraic sum of same product of a constant and a function 33. Differentiation of the Let By Aw. (Art. 24),
: The w y = cv. the General Rule, FIRST STEP. # SEC:OND STEP. A// = cA?\ ^c^*
Ax
Ax THIRD STEP.
Whence, by + A?/ = c(v + Ar) = cv + cAt?. (4), Art, 16, FOURTH STEP. ?=^:rax
ax ,.i(, = c.
The
the derivative of the product of a constant product of the constant and 34. Differentiation of the product of two functions y = uv. Let By and a function the General Rule, FIRST STEP. y is the derivative of the Junction. + Ay = (u + Aw) (v + Av). equal to RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 81 Multiplied out, this becomes y + + wAr + ?>Aw + AwAv.
rAw + A//Ar. ur A// SECOND STEP. Ay = wAr + THIRD STEP. ^ ^+
Ax
Ax Applying (2) w and A.  is ^= FOURTH STEP. = d . rfx + dx du dx dx and hence r dv ~ * 0, we have zero, ?/ ax
' ^ ^~ + Aw Ax
Ax Art. 16, noting that lim A// (4), that the limit of the product V r 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, number. When both sides of V are divided by
assumes the form
( A. llL ^!1 __ djr (, dx <\ djc UV
If, then, we may we have n being a fixed
this formula UP, U the product of V functions 71 write
"
flr ^ dx _ df Vn
ZJ dp
ax , f(r;^...,
ax
' Vi V'2 /'.'jr/i dv _
Multiplying both sides by viv<2  dux dx dx  , vn + (V\V2 ' , we dv:\ dx_ get ' * vn
( , n) DIFFERENTIAL CALCULUS 32 The
is equal to the sum of the the derivative of each function by Power Rule. functions. of VI "dx*"
x, this v all the other a function with a constant exponent. The
If the n factors in the above result are each equal to v, 36. Differentiation When n functions, n being a fixed number,
n products that can be formed by multiplying derivative of the product of dx becomes Via
dx We far proved VI only for the case when n is a positive
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 have so integer. 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 By y = > (t; v the General Rule, FIRST STEP. SECOND STEP. y u
u
+ Ay/ = v + ^
A?;
+
Av = ^M^   = **/*?;**.
+ Ac v
0(0 + A) THIRD STEP.
Applying il)(4), Art. FOURTH STEP.  A?/ Ax
16,  Ar
'(( A * + At') ^ 0) 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 from IV as follows also get VII a : du
d / u\
dx V c du
dx 1 ' The c dx
c" derivative of the quotient of a function
by a constant is equal to
function divided by the constant. the derivative of the PROBLEMS*
Differentiate the following functions.
1. y = x*. Solution.
/~ 2. y = ax 4 Solution. = ~r  bx 2 = x* +
& Solution. y = Ans. By VI
= a 3.] . a ax :i  4 (x = 7 bx 2 ) 1  =~
dx 6 ) 2 6 x. (rr*) f :r>. j ^=~
ax
dx (3 =
When x'2 . (ax 4
)  f dx (bx 2 by ) III by IV (x*') A ns. By VI a by (5) Ans. By VI III a and I ig^ + 8>^. Solution. * 3 5. =
4. ) ^ = f (ax
dx
dx
=4 y = <{ In = 3. (x x^ : y V 1 x ^) + I ~
ax (7 xJ + x~^) V f ~^ dx An * learning to differentiate, the student should have oral simple functions. by (8 x*) III By IV and VI
drill in a differentiating DIFFERENTIAL CALCULUS 34
5. y  (x 2  3) 5. & = 5(x Solution. 2  2  3) 3)< ax 5(x  2 2 x= 4 by VI 3) = x2  [p = (x f
ax = and n 3, 10 x(x 2  3) 5.]
4
. An*. We
p. 1), might have expanded this function by the Binomial Theorem ((3),
and then applied III, etc., hut the above process is to be preferred. = Solution. 7. y = (3 (a + r2  2 2)  * + l 5 ^ = & + 2) y Solution. (3 dx (<* jc  2
. + (1 5 a*)* 4 (1 =
= (3x 2  3 + 1 a' 2)i(l f x^i f
dx 5 and 2, 5x 2 f r f 5x 2 )' + _2 f5x 2 )4 (1 f g f 5 r((7 2  v x2 :r 2 ) f xfa 2 f  " 5x 2 1 f (a 2  10. 11. ~ .r 4 (4 + 4 (a/ dr
ar 6  4 (3 x 2 )^ 2 x 2 f 8) 3 x ~ 5  2 jfi) W3 =
) = 12 a 3  = 36 x2  fa 5 a/ 4 15 4 x. . 2
. by V 5x 2 )^Gx by VI etc. 5 x2 H T2 ) f Prove each of the following differentiations.
dx +2) )2 x [Multiplying both numerator and denominator by 9. x* 5 j'J )l] 5x + 6x(l 2 ;; dx (1 (1 )~^y _ =  (3 _ _ _
x 2 42H1 (3 V1 . + c/x [u .
r///
A
Solution. ~ by VI (a 2 hv VIT
DV v X1 RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 35  16. (2j 8 17.^U
i Q
18 ' d
d? i J f 4 ' = '
f. JT  X a ;=s.
s a f bx f <~.r~ 7 i v; _ 4V.r
20. ; =a+ bt + r/ 2 2V?
21. V a = p=Vaj 23. /(/) =  (2 3 :l /) .  (4
<f/?/ ^r~ , ^= 2
 J .2 /., 2  (2 dj: 9 .r __ (a 5  ~ I ^* _ 2 a h 3 bx ds< 2 ds _ a2 dy 30. s = <Va + 31. =o 2 x < z. Va 4f 2 f 2 a ^
2 DIFFERENTIAL CALCULUS 36 dy dx~~ = Va +
2 33. y X (a x 2 (a 36. = // 39. 6 Va 2 x2 f  x 2 )*'
2 10 r ^~
2 /a + (l+rx)Vr 2 x' 2 a 2r <// ' ^ Va'2 x'2 ,_ 3/2+in.
' (a 2 o 2 )Va 4
4 d,s \^TJn 77 2 d^ ~ 38. 2
) de~ ,__ 37. _6 dr * 2 a2 _ rfjy = x a2 d:r 35. r 2 _ d?/ ^ X 4 a 2x __ Ul a (2h30 (2~ do: ?/ 40. dx ay 41. Differentiate each of the following functions. = V2.r 43. 44. // // = = Va  = ft/ = V5 _ = // a 2 bx V 45. ,s 46. r 47. y 48. 42. /(.r) f . 50. 2/=  V^H3
In each of the following problems find the value of
value of .r.
52. y 53. // = = (x 2  o) .r ; VJ ; = x 3. = 64. JT ^ do for the given Am. 540. RULES FOR DIFFERENTIATING ALGEBRAIC FORMS
54. y = 55. y = Vif+771 ,_ + (2 jV* (2 .rV' Ii 7/ = Vl6 f 57. 58. ?/ = jr\/8  .r 59. // 60. .rVl // = (4 // 61 = =  + ; * = Arcs. 4. g. L\ .r 2 .r ; .*; .rV rri~7~> j ; .r ; = ' 37 ; r = .r = = .r ^ 0. 2. 20. 2. U. 63. 64. j= J. // 65.  // // = jN/.S f 2 . = = 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 function of x. In that case y is a function of jc through v and is called a
function of a function. For example, y if 2 v
= ^~ > r, and
then ?/ is a function of a function. P>y eliminating v we may express y
is not the best plan
JT, but in general this directly as a function of when we wish to find  dx = f(v) and v = is a function of # through 0. Hence,
<(JT), then
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 If y when we '// let functions amj FIRST STEP. y+&y=f(v SECOND STEP. ^ y+&y=f(v
y + A?;) /O)
Ay  /(^ + A.) /QQ
y =f(v THIRD C^TFP
IHIRDbTEP. AJH A?; At; = <f>(x + Ax) Ag_(a; + Ag)
AJ"~ Aar DIFFERENTIAL CALCULUS 38 The lefthand members show one form of the ratio of the increment of each function to the increment of the corresponding variable, and
the righthand members exhibit the same ratios in another form. Before passing to the limit let us form a product of these two ratios,
choosing the lefthand forms for this purpose.
J/ This gives . which equals _J?, Av Ax T* = Av Ax
*f'Ax
When Ax  0, then also Write this FOURTH STEP. r^ Ax Ai> * Passing to the 0. limit, *y
dx (A)
v ' This may JL.Q.
=d By (2), Art. 16 dv dx also be written (B) ^=/'00f. // y =/(?;) and v </>(x), tlie the product of the derivative of with respect derivative of y with respect to y with respect x equals by the derivative of v to x. Let y be given as a func 39. Differentiation of inverse functions. tion of x to v 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) When we wish to distinguish between
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
are said to be inverse functions. the two it is then the corresponding members in the second
be respectively the inverse functions. direct functions, column will = x 2 + 1,
x
y = a
y = sin x,
y , = \ly  1.
= loga y.
x = arc sin y. x ar RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 39 Let us now differentiate the inverse functions and y=J(x) = x 0(^/) _ simultaneously by the General Rule. FIRST STEP. + Ax = 4)(y + Ay).
x + Ax = (y + Ay) y+Ay=f(x + Ax) x SECOND STEP, y + Ay = f(x + Ax)
JL =f(x + Ax)f(x) THIRD STEP Ay^f(x+Ax}.f(x) Ax Taking the product Ax Ax
r AT ^ <(?/ + A?/)
Ay
Ay
we of the lefthand forms of these ratios,
AJ/ AT
Ax Ay get ' A// FOURTH STEP. When Ax  Ay  r = dx 0, then also, in general, by 0. Pass ing to the limit,
(C) ' dx (3), Art. 16 dy
or The derivative of the inverse function is equal to tlie reciprocal of the derivative of the direct function. by 40. Implicit functions. When a relation between x and y is given
means of an equation not solved for y, then y is called an implicit function of x. For example, the equation
x* (1) 4y= 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
. y be solved for
i x2
r
4 y, giving , 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
When y is defined as an
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
41. Differentiation of implicit functions. implicit function of s, it function of y).
We then follow the rule : Differentiate the terms of the equation as given, regarding y as a Junc and tion of x, 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 substituted in the derivative. Let us apply this rule
cu in
f ' finding + 2 .r'ty us  ix = ( ) I O / >' /V.M T *'} C I , . 7 '2. and 1 . T> ?/ .' __ /I ' /? ^* 7 The student should observe ^ "* // () \ rf//__ j 10. ^+^)^ Then both from 6aj e> G^ > 1 '// thai in general the result will contain v/. PROBLEMS
Find ' for each of the following functions.
An*. <^
dr 2. y = 2 =  15 y ?/ 2 + 5. 15 6. ,= V + \/ A .r , 5 ?( : ?/ < = + ja  3 y fi
. .r. = ^. V  1). RULES FOR DIFFERENTIATING ALGEBRAIC FORMS
7. y = 2 x 9. 6 + 2 2 8. 2 px. x2 ?/ o ^ 12. r2
2 2 + jc' f jry .r< ~ 21. V2~7 f 22. jc 20. 2  25. + 2 y' +
VlTjy =
2 3 xy' 2V7// Show  ir :< * ' f ?/ 18 ?/' 5 = 1 . ?/ 2 2 / ?/ .' ft3  ' 3 :l r;.r /> * 2 f r?/ .r// = 3 1. ' 0. each of the following curves of 2 2 .r ' = 3
j/ Find the slope
19. J ' 17. 3 arj/ 2 ' =r
= a.
f
2V^7/_+
+ a ^ r + =
4 3 ?/ . ?/ 3  JT 15 2 2 .r 14. . =a 6
V# = Va. f a Vx f 10.  2 3 13. 41 = 28
=
1 ; ; v. (2, 52 ; Awx. '*).  1 ). 23. (2, 3). ; the given point. at 24. (8, 2). ~ that the parabolas //' .r'  .r''  1 2 pj f =3a
=6 a.n/ f 3 a//.r \fxlj  and /> 2
' : /r' ~ 2 p' //  }.
J. :t ; (a, a). (4, 1). ; 2 ;>.r inter sect at right angles.
26. Show circle s f that the circle 2
if + 2 jr f it = ./' f //' 12 x 10 at the point (> f // 25 ~ is tangent to the (2, 1). 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)
fix) and rotate it
may be found as follows: construct the graph of
around the origin 9(T counterclockwise.
27. ?/ ADDITIONAL PROBLEMS
The vertex of the parabola y 2 = 2 px is the center of an ellipse.
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
2
Ans. 4 x 2 + 2 y 2
p .
equation of the ellipse.
1. The 2. A circle is drawn with that the circle cuts the ellinse its
2
b' center at (2 a, 0) and with a radius such
+ a 2 ?/ 2 = a 2 b 2 at right angles. Find the x2 r2 AUK. radius. From any point P on an ellipse lines are drawn to
these lines make equal acute angles with the normal 3. that 4.
2 b x 2 4 Prove that the line Bx f Ay
2 2
a 2 ?/ 2 = a 2 b 2 if, and only if, # a the 2
2
2(3 a f b ). foci., Prove at P. = AB is tangent to the ellipse 2 2
f ,4 />  A 2 B2 . m + n at
m n
any
Find the equation of the tangent to the curve x y = a
between the axes is divided
point. Prove that the portion of it intercepted
5. in the ratio n at the point of contact. Ans. myiix x\) + nx\(y #,) = 0. 2 2 a 2 ?/ 2 = a 2 b 2
the slope of a tangent to the hyperbola b x
 kx Va^k 2  b 2 and show that the locus of
prove that its equation is y
2
2
2
b2
the points of intersection of the perpendicular tangents is x f y = a 6. If A* is , , . 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), ~ = The then
slope of the line tangent
to the curve at P(x, y).
direction of a curve at any point is defined as the direction of the tangent line to the curve
Let r
inclination of the tangent line. Then the at that point. slope = tan ~ and r, = = tan r slope of the curve at any point P(x, y). At points such as D,
parallel to the xaxis r = ; and where the direction F, H, therefore ^
ax = 90; therefore ILLUSTRATIVE EXAMPLE
(a) (b)
(c) (d)
(e) 2x 3 Find
Find
Find
Find
Find
y = the inclination r
r when x = 3. = o infinite. Given the curve y when x I =*
3 x2 f 2 (see figure). 1. the points where the direction of the curve is parallel to OX.
the points where T = 45.
the points where the direction of the curve is parallel to the line 6 (line Solution. & becomes
1. is = 0. At points such as A, B, G, where the direction
of the curve is perpendicular to the xaxis and
the tangent line is vertical,
T of the curve the tangent line, is horizontal, AB). Differentiating, f ax =x 2 2 x 42 = tan T. VARIOUS APPLICATIONS OF THE DERIVATIVE When x=l,tanr=l2 = l; therefore r = 135. Ans.
When x = 3, tan 7 = 96 = 3; therefore r = 71 34'. Ans.
When r = 0, tan r =
therefore x'2  2 x = 0. Solving this equation, we (a) (b)
(c) get x x = ; = = y 0, Substituting in the equation of the curve, we find y = 2 when
f when x  2. Hence the tangent lines at 0(0, 2) and D(2, j) are or 2. horizontal. Ans. When (d) (or tangent) = r_= 45, tan r v 2 = 2.41 x= 1 get 43 is and 1 therefore r 2 ;  2 x = Solving this equation, we
of the curve 1. two points where the slope 0.41, giving unity. = therefore x'2  2 x = J.
Solving, we get
the abscissas of the points F and E where
0.29, giving
the direction of the given curve (or tangent^ is parallel to the line AB. Slope of the given line (e) x = VS = 2.29 1 and I  ; Since a curve at any point has the same direction as its tangent
between two curves at a common point
will be the angle between their tangent lines at that point.
line at that point, the angle ILLUSTRATIVE EXAMPLE
(A) Find the an^le of intersection of the
x 2 + y 2 4;r = l (B) x2 and W2 Then from
and from y*  mi (A), = 3, circle A at (or, R at (jr. y). = by Art. 41 jg By ~r
ax = 2, we have
mi =
slope
J
7H 2 =
3 = slope Art. 41 y of tangent to (A) at (3, 2).
of tangent to (/?) at (3, 2). between two for finding the angle = 7^
1 f tan 6 Substituting, 3 = ^ 1 The equation through the point
is (2), Art. 3 7W1///2 /. = 45". y _y1 and normal; lengths =m Arw.
2). of subtangent of a straight line passing (xi, yi) whose slopes are m\ ^ f ; lines also the angle of intersection at the point (1, 43. Equations of tangent m be j/), circle tan normal. 9. tangent to the is is = find the points of intersection to tangent to the = m^ (B), The formula This 2 y we 2). = slope of the
= slope of the Substituting x and m2 + Solving simultaneously, (1, Let mi and circles f Solution.
(3, 2) 2. and sub Y and having the slope (x . Xl ). (3), Art. 3 tangent to the curve AB at the
M N
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
If this line is DIFFERENTIAL CALCULUS 44 The normal being perpendicular to the tangent, its slope is the
And since it also passes
((2),
negative reciprocal
the point of contact Pi(r\, y\), we have for the equation
through Art. 3j. of rwi of the normal PiN,
y (2) /i =
m\ (x jfi). 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 raxis is called the length of the subtangent
} (= TM). Similarly, we have the length of rite,
normal (= P\K) and the length of the subnormal (= yi/A ).
i) y\ r /i TP\M In the triangle
therefore 9 r~m\ tan = (3) In the triangle 1\1P N, tan r
} (4) 71 / A"* f 7V17 ' _f length of sub tangent. = in\ = therefore = 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(a, a) on the cissoid tangent, subnormal, tangent, and normal, at the point ^a x Solution. dx
Substituting x a, ?/ = y(2 a
a,  x)~ we have
rr
 2 = slope of tangent. Substituting in (1) gives
U = 2x a, equation of tangent. 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,
If the subnormal extends to the right of M, we consider it positive, if to the negative,
Jeft, negative. VARIOUS APPLICATIONS OF THE DERIVATIVE
Substituting in (3) gives TM = ~ = length of subtangent. Substituting in (4) gives MX = 2 a = length 45 ( PT = V(TM)* + (MP^ = x + <2=\5 = length of tangent,
\ 4 Also, PN = V(MN) 2 + and Find the equations
2. y = ar  3 x; 3 9 T 4 = ^L1.
o
X T/ 4. 2x 2 xy + 6. ellipse 2
7/ + of the tangent fa = y 9  .r 7 16; Find the equations
2 2
2 2
b' x
a~b'2
f a = a\ 5 length of normal. //  j y 0, .r f 9 ?/ 9 = 0, .r + 7  20 = 0.  = 37 = 0. <// 16 (3, 2). = jrf 4 = and normal at the given point. .4//,s. (2, 2). ' \ 4 (2, 5). ?/4 2 (MP)*  1 3. 5. subnormal. of 0; 2). (1, and normal of the tangent at (x\, y\) to the . ?/ Ann. bfix h ay\y = a L> /> 2
, a~ii\x b~jr\y h 2 ). jr]y\((i~ 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.
8. JC\JT f 7/1 y = 2
r' , .r\y
' ~ ijix 0, ' * Show JT\. xi ~ 2 ;>.r is bisect (d at
that the subtangent to the parabola /r
is constant and equal to ;>. the vertex, and that the subnormal 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. 10. 2 x2 ay 11.
12. x  4 A us. (a, a). ; = 2
?y 9 f> (5, 2). ; = 72;
9x + 4
2
+ 2 = 0;
xy +
2 2 2/ ?y 2
.r  a, y jr 8 ?/ = x 9, 8 x 4 2 + // 5 ~
?/ 3 a, ^ 2 a. = 50, V, J. (2, 3).  (3, 2). 13. Find the area of the triangle formed by the jaxis 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
j at the point (5, 2).
and the normal to the curve // = 9 Find the angles
15.
16. y y = x 4=6 2 1, a* 2
, of intersection of x2
7 + x 2 y + 2 = y 2 each of the following pairs of curves. = 18. = x ,y*3 y = 2
z + 4 2 = 61, 2x y* = 41.
2 y jr. 2 2 ?/ 109 39 r
. 32. Ans. At
17. Am. 13. ( 2, 2), 5 54'; at ( 1, 5), 8 58'. DIFFERENTIAL CALCULUS 46 Find the points of contact of the horizontal and vertical tangents to
each of the following curves.
19. y = 20. 3 y'2
21. x2 5 x  +  x2  23. x2  24. 169 x 2 25. Show  6 y + 6 xy 22. 2 x2 + 8 xy 24 x?/ +  0. 25 y 2 25 y 2 + 10 Ans. Horizontal, . x Vertical,  = 16. x?/ ( Horizontal, ) 3, 1).  (3, Vertical, (5,  ( 1), 3, 1). 5, f ). ( ), 81. 2 = 25. 2 169 (, = t/ I t/ 144. that the hyperbola x 2  y 2 = 5 and the x2 + cissoid (2 a  ellipse 4 2 9 = 72 2 = x3 ?/ intersect at right angles.
26. Show (a) are that the circle 2 jr' + 2
?/ = 8 ax and the perpendicular at the origin;
an angle of 45 at two other points. (b) intersect at x)?/ (See figure in Chapter XXVI.)
27. Show that the tangents to the folium of Descartes x f if = 3 axy
= ax are parallel to the //axis.
: ^ at the points where it meets the parabola y 2
(See figure in Chapter XXVI.)
28. Find the equation of the normal to the parabola y
of 45 with the xaxis. f x 2 which = x 5 58 which makes an angle
29. Find the equations are parallel to the line 3x of the tangents to the circle
1 y
19. x2 + y 2 30. Find the equations of the normals to the hyperbola 4 x 2
which are parallel to the line 2 x 4 5 y = 4.  2 y' = 36 31. Find the equations of the two tangents to the ellipse 4 x 2 + 2 = 72
y'
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 tangent line at any point to the parabola x^
equal to a. (See figure in Chapter XXVI.) + ifi = a* of the constant and is 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 xaxis 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 building 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 supporting 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. Maximum and minimum values of a function introduction. In
many practical problems we have to deal with functions 44. ; a great 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
is ED. rectangle, as CD = x Let ; DE = VlOO  x, and then the area of the rectangle evidently A = j VlOO (1) That a rectangle
Let the base
the altitude become
zero ; of CD maximum (= Now L>
. area must exist may be seen as follows. x) increase to 10 inches (the diameter) DE = VlOO zero. :r let then the altitude x will decrease to zero ; and the area then
will the base decrease to
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
its area would be
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 rectangle becomes a square
greatest, ing the graph of (1), we observe that
(a) from the nature of the problem both be positive ; it is evident that x and and x range from zero to 10 inclusive.
a table of values and draw the graph, as in the
construct (b) the values of Now
figure A must on page
* There 48. may be more than one of each, aa illustrated on page 55, DIFFERENTIAL CALCULUS 48 What do we learn from the graph? '> 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, OA1 = 3 1\1 f> = 28. 6 sq. in. when x then A and when = ON ~~
A = A< Q ~
x then
(b) = There is in., 4\ ; in., about 39.8 sq. in. (found by measurement). one horizontal tangent (A\S ). The ordinate TH from
T is greater than any other ordinate. Hence this
Y contact its point, of 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.
cannot find observation : We this value (= 7/7") by measurement, but exactly very easy to
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
find by the calculus. it is We ; (1) = arVlOO  2
dA_ 100  2 x 2 100  2 x 2 = 0.
dx
VlOO  x 2 VlOO  x 2
.4 , x Solving, Substituting, = 5 V2.
DE = VlOO jc' we get x2 = 5 V& VARIOUS APPLICATIONS OF THE DERIVATIVE
Hence the rectangle of maximum 49 area inscribed in the circle is a square of area A = CD X DE = 5V2 X 5 V2 = The length of HT is therefore 50.
Take another example. A wooden box
108 cu. ft. It is to is 50 sq. to in. be built to contain 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 and y = length of side of square base in feet,
= height of box. Since the volume of the box
ever, y may Volume is given, terms of be found in  x'u  108 ; ..?/ = how Thus, jr. ^
.r~ We may now express the number
as a function of jr as follows.
required
of base = JT~ of four sides = 4 Area and
(2) Area M= ( sq. M) of square feet of ft., 432
.r// 2 x' lumber sq. ft. Hence + 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. is DIFFERENTIAL CALCULUS 50 What do we learn from the graph ? (a) If carefully drawn, we may measure the ordinate corresponding to any length (= x) of the side of the square base and so determine 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 obserOne of the boxes evidently takes less lumber than any of the
vation
others. In other words, we may infer that the function defined by
: (2) has a minimum point, Let us find this point on the graph ex value. actly, using the calculus. Differentiating (2) to get the slope at dM 9r 432
j ax At the lowest point T the that any we have is, when x = Substituting in <<" +' rr * x2 Hence slope will be zero. 6 the least amount of lumber will be needed.
(2), we see that this is M = 108 sq.
The fact that a least value of M exists ft. 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
varies from a large value, grows less, then increases 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 M least cost. We will now proceed
maxima and minima.
45. Increasing
is x
if to the treatment in detail of the subject of and decreasing functions.* Tests. A function y =f(x) said to be an increasing function if y increases (algebraically) when
increases.
function y
f(x) is said to be a decreasing function A y decreases (algebraically) as x increases.
The graph of a function indicates plainly whether or decreasing. For instance, consider the graph * The
proofs given here depend chiefly on geometric
and minima will be treated analytically in. Art. 125. it is increasing in Fig. a, p. 51. intuition. The subject of maxima VARIOUS APPLICATIONS OF THE DERIVATIVE
As we move along the curve from 51 left to right the curve is rising
as x increases the function (= y) increases. Obviously, Ay
and Ax agree in sign.
On the other hand, in the graph of that Fig. ; is, 6, we move along as the curve from to right the curve is falling that is,
as x increases, the function (== //) always
left ; decreases. Ay and Ax have Clearly, op posite signs. That a function maj be sometimes increasing and sometimes
decreasing is shown by the graph (Fig. c) of
r As we move along the curve from left to
right the curve rises until we reach the point
from A, falls B. A to B, and rises to the right of Hence
(a) from x increasing ;
(b) from x creasing oo = 1 to = 2 to to x = x = the function I 2 the function \ is is de ; from x (c) = x + oo the function is FIG. b increasing. At any point, such as C, where the function is increasing, the
tangent makes an acute angle with the jaxis. The slope is positive.
the other hand, at a point, such as D, where
the function is decreasing, the tangent makes
an obtuse angle with the jaxis, and the slope is On negative. Hence the following criterion : A function
positive, is increasing when its derivative
and decreasing when its derivative is
is negative. For example,
have
(2) g = f(x When x <
When 1<
When x >
These differentiating (1) above, 1, ) = 6 z2  18 * /'(*) is positive, + 12 and = we 6(z f(x) is  l)(z  2). increasing. x < 2, f'(x) is negative, and /(or) is decreasing.
is increasing.
2, f'(x) is positive, and f(x) results agree with the conclusions arrived at the graph. above from DIFFERENTIAL CALCULUS 52 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
(= 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 necessarily 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 MA If /OO is B that are greater than the maximum MA, and
that are less than the minimum NB.
an increasing function of jr when .r is slightly less than a, values to the right of
values to the left of A and a decreasing function of x when
is, if /'Or) changes sign from + to jr is slightly greater than a, that
as x increases through a, then maximum value when x
must vanish when j = a.
Thus, in the above example (Fig. f(x) has a a. Therefore, if continuous, }'(x) A, /'(.r) = ; at />, /'(:r) is r), at (\ J'(x) is negative. positive; at
/ 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 minimum value when r = a. Therefore, if continuous, /'Or) must vanish when r = a. Thus, in Fig. c, at D, J'(x) is negative; \
jj FlG ' at B, f(x) =  x c 0; at E, f f (x] is positive. We may then state the conditions in general for maximum and
minimum values of /Or).
= and /'(*) changes sign from + to
/(jc) is a maximum if /'(*)
= and/'(x) changes sign from to f.
f(x) is a minimum if /'(*)
. = The values of the variable satisfying the equation f'(x)
are
1 and x
2 are the
called critical values; thus, from (2), Art. 45, x
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
a value of the variable particular turning point, substitute in it, first, than the corresponding critical sliglitly less value, and then one VARIOUS APPLICATIONS OF THE DERIVATIVE + 53 If the first sign is
and the second
then the
function has a maximum value for the critical value considered.
and the second +, then the function has a
If the first sign is slightly greater. minimum , value. is the same in both cases, then the function has neither
nor a minimum value for the critical value considered.
For example, take the above function in (1), Art. 45, If a the sign maximum y (1) = f(jr) = 2 JT*  9 JT + 12 jc  3. Then, as we have seen,
(2) f'(jr) Setting /'(y)
first test x = 1. = 6(j l)(j2). = 0, we find the critical
We consider values of jc values y = 1, :r = 2. Let us
near this critical value to be substituted in the righthand member of
of the factors. (Compare Art. 45.) =
 and observe the signs
1 (2),  +.
x
1, f'(jr)
(+)()  .
Hence /(r) has a maximum value when x ~ 1. By the
= 2.
table, this value is ?/=/(!)
test y = 2. Proceed as before, taking values of
Next,
When
When x <
> the critical value 2. 1, /'(*) ()() < 2, /'(/)  (+)()  .
When x > 2, fix) = (+)(+.) = +.
minimum value when x 2. By W hen x now near f Hence /(y) has a this value We is shall y =/(2) = .r now summarize our results in a workruy method for examining a function
values. Working rule. 47. First mum the table above, 1. for ride. maximum and mini FIRST STEP. Find the first derivative of the function.
Set the first derivative equal to zero and solve the resulting
are the critical values of the variable.
equation for real roots. These roots
one critical value at a time, test the first
THIRD STEP. Considering
' SECOND STEP. derivative, first for than
the a value a trifle less * and then for a value a trifle greater + and then ,
the critical value. If the sign of the derivative is first
maximum value for that particular critical value of the
has a function then
variable; but if the reverse is true, it has a minimum value. If the has neither.
sign does not change, the function
" * In this connection the term "little less/' or trifle less," means any value between the
"
little
next smaller root (critical value) and the one under consideration and the term
means any value between the root under consideration and
greater," or "trifle greater,"
the next larger one.
; DIFFERENTIAL CALCULUS 54 In the Third Step,
factors, as in Art. 46. it often convenient to resolve is ILLUSTRATIVE Ex AMPLE 1. In the first problem worked out
of the graph of the function in Art. /'(a*) 44 into we showed by means A = xVlOO x
that the rectangle of
tained 50 8(4. in. This above maximum area inscribed in a circle of radius 5 in. conmay now be proved analytically as follows by applying the rule. Solution.
r, 2 , = a, /// First Step. x\ 100  x 2 = /(x) 100 N f'(x) Second Step. Setting 0, f'(x)  . 2 X2 VI 00 x 2
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 .
the function has a
Since the sign of the first derivative changes from + to
, , maximum value 5\ 2 /(5V2) ILLUSTRATIVE EXAMPLE
and minimum values. 5\ 2 = An*. 50. Examine the function 2. (x l) 2 (x f !)' for maxi mum Solution, = (xl) (xf IP.
= 2(x  l)(x +
+ 3(x  l)(x + 1) (5 x  1) = 0.
(x
2 /(x) First Step. f'(x) Second Step, Hence x 1 1, 1, =  D(x /'(x) Examine the critical value x first Therefore,
of 2 (x + 1) = (x l)(z + l) (5x 1).
2 2 are the critical values. \, Third Step. When
When l) )'< 2 x < 1, /'(x) x > 1, /'(x) when x 1 5(x f = 5()(+)
= 5(+ )(+) = 1 2
) 1 2 (f )
L
'(f) (x  (C in figure). J). = .
= +. the function has a minimum "
value /(I) = ( ordinate D.
Examine now the critical value x I (B in figure). U'(x) = 5()(+)
When x > I /'(x)  5()(+) W henx<
r Therefore,
nate of B). Examine when x \ the function has a Therefore,
value. when x 1 2 ()
(+) maximum = 1 (A in
1, /'(jO = 5(X~)
1, /'(x) = 5()(+ lastly the critical value When x < When x >  2 x K
. value /(J) =1.11 (= ordi figure).
2 2 ) =
 ()
() the function has neither a =
= +. +. maximum nor a minimum VARIOUS APPLICATIONS OF THE DERIVATIVE
48. Maximum is f(x) or continuous. minimum when values becomes /'(*) Consider the graph in the figure. 55 infinite At and B, or G, FIG. d f(x) is continuous and has a maximum value, but /'(a*) becomes insince the tangent line at B is parallel to the //axis. At E, f(x) finite, has a minimum cussion of all value, and /'(/) again possible becomes In our dis infinite. values of /(j), we
values also those values of x for maximum and minimum must therefore include as critical
which /'(x) becomes infinite, or, what the is same thing, values of x satisfying The Second Step must then of the rule of the preceding section (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. be modified as indicated by   r) (and f'(x] = ILLUSTRATIVE EXAMPLE. Examine the function a
minima. b(x for maxima and Solution. 2b
Since x
f(x) is X = C not = c is a ^^^1 ^ va u e for which infinite, let us test i the function for When x < c, f'(x) When = c = OM, oo), but maximum and minimum * Hence, when x  x > c, f'(x) =
= the function has a for values which when f . . maximum value /(c) = a = MP. DIFFERENTIAL CALCULUS 56 PROBLEMS
Examine each of the following functions for maximum and minimum values. r 1. + 10 2. + x2 6 { 12  .r 3 jr =
=
= Ans. x
x 9 x.  2 2 jr 1 . a 1, 3,
1, x
3. 2 4. /< 5. 2 :r + { + .r 3 j* .r f  2.r 2 2 2 12 4. .r No x4 7. r4 8. 3 4 .r 4 j' 4 a* . x .r. + 2  max. or min. = 0, = 1,  9. x* 10. 3 r  JT* 4
. 20 jr j2 12. 2, 13. r2 14 a* 2 3. .r 2
. == 0,
= 2,
x = 0,
x = 4, gives max.
gives min. =
= gives min. = a 1, = gives min. gives max. =  gives min. 32.
0. 256. a, 3 a2 T
17. x a4
a: + = a, x
x a 4 = a, x gives min. = 2 a2 gives min.
gives max. = = a, J. = gives min. = . 4 a*
rr " ^ + , 1Q /O
v^2/1
ID. ^~T~Uy ^JL o*^2 . I + x)
+ r(j* 19. (2 20. ?> 21. a . g ^+
2 5. 0. 15. 16 1. . ____,
2 ' = gives min. 3 2 rr*
^__. + 11. 0. x 12 x
x 5 = gives max. gives min.
1, 1. ^ 4 10. 15 j* 20. x
6. = 4.
= 0.
gives max. = 17.
2, gives min. =
gives max.
gives min. b(x 2 (l U^  x)*. 2 a) . cr. No a, max. or min. = 6. . 22. 23. VARIOUS APPLICATIONS OF THE DERIVATIVE
57
 x) 5
(2 + *)*(!
Ans. x = l, gives min. = 0.
x
1, gives max. = V4 = 1.6.
2
 x)*.
=  a, gives max. = 0.
(a +
(a
= \ a, gives min. = ~ jj a
~ a, gives max. = ^JJ a
. a r) .r ft .r . r> r j* 24. (2  a: 3* 25 + ^2 26 j + + ^ * + f 2 ^
~f~ 4 = x ' 2 ~
= J 4 gives neither. o, x
r
* a)*. 2 ' :r 27 a)*(j  . /, J a, = a, 0. gives max.
gives min. = i gives max.  gives
gives min. 1, a = j ^ 2 f r + 4
x 2 + 2 .r + 4 2, (.r  J. 5. ,'{. = gives max.
gives min. = *J, _
a J. =max. ~  .'i, u*  gives min. 4, T 1 gives neither. <*> 0, J a. .3. [;. _ . Drives f   max. 4 />  .  /> . > a a f b
. = gives max. on 30. (a
^ a 3 x)
_^_. x a
= 2 x . mm. =
. givers ., (a a 7 o 2
*? a . . x2 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. DIFFERENTIAL CALCULUS 58
(c) To (d) In practical problems maximum and a it is values. usually easy minimum which a which to tell value, so critical value it is not always the third step. apply
the graph of to necessary
(e) maximum and minimum finding (p. 53) for will give a single variable apply the above rule the resulting function of Draw The work the function in order check the work. to maximum and minimum of finding may values fre which
quently be simplified by the aid of the following principles,
follow at once from our discussion of the subject.
(a) The
must occur
(b) maximum and minimum values of a continuous Junction alternately. When mum for such c is a positive constant, cf(x) is a maximum, or a miniand such only, as makef(x) a maximum or a values of z, minimum.
Hence, in determining the When
and values of x and testing for maxmay be omitted. critical ima and minima, any constant factor maximum when a c is negative, cf(x) is f(x) is a minimum, conversely. (c) // c is mum a constant, f(x) and values for the same values + f(x) c have maximum and mini of x. Hence a constant term may be omitted when finding critical values
and testing. of x 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 ?
Let Solution. a then and volume
which is is x = side 2 x V= bottom p: ; of box, made maximum by a varying x. rule, p. 53, First Step. rtV
~ = (a Second Step. Solving a 2
is depth of box square 2 x) 2 j, (a the function to be Applying the It of small side of square forming  2 x) 2 8 ax  + 4 x(a 12 evident from the figure that x or = 2   = 2 x} = a2  8 ax + 12 x 2 gives critical values x must give a minimum, for tin would be cut away, leaving no material out of which to make a box.
test, x is found to give a maximum volume  Hence the side jj be cut out is one sixth of the side of the given square. .  and  then By of the all the the usual square to VARIOUS APPLICATIONS OF THE DERIVATIVE
The drawing
left to 2. of the of the function in this graph 59 and the following problems is the student. Assuming that the strength beam with rectangular of a cross sec and as the square of the depth, what
beam that can be sawed
a round log whose diameter is d! tion varies directly as the breadth
are the dimensions of the strongest out of Solution. have From = x If maximum breadth and y = depth, then the beam will
when the function xy is a maximum.
= d~ x'2 hence we should test the function strength the figure, 2 y' ; First Step. Second Step, d 2
Therefore, if =~ /'Or) the = 3 x2 beam is 0. .. the 3. will What have maximum \
j{ \ If x2  2 d' critical  3 2 x' . value which gives a maximum. \ of diameter of log, of diameter of log, strength. the width of the rectangle of maximum area that can be
segment OAA of a parabola?
Or = h, BC = hx and PP' = 2 y thereis f inscribed in a given HINT.  d'2 = ~ = x = Breadth beam + cut so that Depth
and 2 x ; fore the area of rectangle PDD'P' is 2(hx)y.
But since be tested P lies on the parabola y 2 is f(x) = 2px, the function to = 2(hx)\f2px.
AUK. Width 4. Find the altitude of the cone inscribed in a sphere of radius HINT. Volume
X2 of cone = BC  h. maximum volume of = J irxy. But
CD = y(2 r  y} ; is A\ TT f(y) ^ y*(2 r y). Ans. Altitude of cone
5. Find the altitude of the cylinder of
inscribed in a given right cone. HINT. Let But from AC r and BC = h. Volume of
ABC and DBG, = $ to be tested cylinder = irx 2 y. is 2
2
f(y)=y y(hy} T. maximum volume similar triangles Hence the function that can be r. x therefore the function to bo tested = . Ans. Altitude =  h. that can be DIFFERENTIAL CALCULUS 60 three sides of a trapezoid are each 10
if the area is a maximum? 6. If in. long, how fourth side be
7. It is divide the it long must the
Ans. 20 in. required to inclose a rectangular field by a fence, and then to
two lots by a fence parallel to one of the sides. If the area of into field is given, find the ratio of the sides so that the total length of fence minimum. shall be a Ans. 2 A 3. to be laid out along a neighbor's lot and is
rectangular garden
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
8. is of inclosing A 9. it minimum? be a may Ans. 18 radio manufacturer finds that he can 3 p.
375
at p dollars each, where 5 x
x 2 dollars. Show that the
(500 4 15 x f
?, } when the production sell The .r rd. instruments per week cost of production maximum profit is is obtained about 30 instruments per week. is In Problem 9 suppose the relation between x and 10. x 24 rd. /; is JT= 1001: Show that the manufacturer should produce only about 25 instruments per week for 11. In maximum
Problem profit. 9 suppose the relation T2 between x and p is = 2500  20 p. How many instruments should be produced each week for maximum profit?
The 2 producing x articles per week is (ax
dollars and the price (p dollars) at which each can be sold is /) =
Show that the output for maximum profit is
12. total cost of _ Va x 2 f 3aqi3 a b)  f
/3 bx f c) a x2 . 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 determines the output and price under the new conditions.
(a) (b) tax for
(c) Show that the price increases by a little less than half the tax.
Express the receipts from the tax in terms of / and determine the maximum return.
When the tax determined in (b) is imposed, show that the price is increased by about 33 per cent.
total cost of producing x articles per week is (ax 2 f bx f c)
which is added a tax of
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.
14. The dollars, to NOTE. In applications t in economics a, b, c, a, are positive numbers. VARIOUS APPLICATIONS OF THE DERIVATIVE
A 15. steel plant is  capable of producing x tons per day of a lowgrade and y tons per day steel 61 of a highgrade steel, where = // If the x 10 market price of lowgrade steel is half that of highgrade steel, show
that about 5 tons of lowgrade steel are produced per day for maximum
fixed profit. A 16. ment if telephone company finds there is a net profit of an exchange has 1000 subscribers or If less. if 15
per instruthere are over 1000 subscribers, the profits per instrument decrease lc for each subscriber
above that number. How many subscribers would give the maximum net AUK. profit? 11250. manufacturing a given article is p dollars and the number 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. The 17. cost of tlV
* * A What 18. should be the diameter of a and requiring the
if (b) least amount 19. The if 1 can holding 1 qt. (58 cu. in.)
the can is open at the top? = vV
(a) v \ 5.29 in. ; = ; (b) \ 7T lateral surface of a right circular cylinder the cylinder
the cylinder.
a of tin (a) n the can has a cover? Ans. is tin //,s\ 4 is TT 4.20 in. 7T sq. ft. From cut a hemisphere whose diameter equals the diameter of
Find the dimensions of the cylinder if the remaining volume is maximum or minimum. minimum. .4//s. Determine whether it is a
1 ft., altitude = 2
Radius maximum
ft. or a 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
2
Ans. 16.
12  s and 6 //  s  12.
parabolas 3
// 21. are on The other two Two vertices of a rectangle are on the .raxis.
2 jr arid 3
the lines whose equations are y value of y will the area of the rectangle be a jc f y maximum? 30. vertices For what Am. y 6. an isosceles trapezoid is a diameter of a circle of radius
22. One
and the ends of the other base lie on the circumference of the circle.
a,
Am. a.
Find the length of the other base if the area is a maximum.
base of 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
23. 1 of the largest rectangle to the area of the segment is /' beam varies as the product of the
Find the dimensions of the strongbreadth and the square of the depth.
est beam that can be cut from a log whose cross section is an ellipse of
Ans. Breadth = 2 />\/A depth = 2 en/Isemiaxes a and b.
24. The strength of a rectangular ; DIFFERENTIAL CALCULUS 62 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. a X 4ns.
26. The equation of the path of a ball is y  mx  m " x aVi.
where ", 800 m the origin is taken at the point from which the ball is thrown and
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
Ans. (a) 1 (b) $.
greatest height on a vertical wall 300 ft. away?
; 27. A window of perimeter p ft. is in the form of a rectangle surmounted 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 by an triangle.
28. The sum of the surfaces of a sphere and a cube being given, show
sum of their volumes will be least when the diameter of the sphere that the equal to the edge of the cube. is greatest
29. Find the dimensions the ellipse ~}~ in a its
is When will the sum of the volumes be ? ^~ = of the largest rectangle which can be inscribed
Ans. a\/2 X &V5. 1. 6* 30. Find the area of the largest rectangle which can be drawn with
base on the .raxis and with two vertices on the witch w hose equation
r ?/ 8ff> = ., ir* f * 4 a2 (See figure in Chapter Ans. 4 a 2 XXVI.) . 31. Find the ratio of the area of the smallest ellipse that can be circumscribed about a rectangle to the area of the rectangle. The area of an
Ans.  TTellipse is irab, where a and b are the semiaxes. The two lower vertices of an isosceles trapezoid are the points
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.
32. ( 6, 0) Ans. 64.
33. The distance between the respectively, is c. centers of Find from what point P two spheres
on the of radii a line of centers and AB b, 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
35. Given a sphere of radius 6
the following solids 63 Calculate the altitude of each of in. : maximum volume inscribed right circular cylinder of (a) ; (b) inscribed right circular cylinder of maximum total surface
(c) circumscribed right cone of minimum volume, Ans.
36. Prove that a conical tent (a) 4 V3 in. ; (b) 6.31 in. ; 24 (c) ; in. given capacity will require the least
V2 times the radius of the base.
flat it will be a circle with a sector
canvas would be required for a tent 10 ft.
Ans. 272 sq. ft.
of a amount of canvas when the height is
Show that when the canvas is laid out
of 152 cut out. 9' How much high? 2
2 px at a distance a
37. Given a point on the axis of the parabola
from the vertex find the abscissa of the point on the curve nearest to it.
?/ ; A HK.
38. Find the point on the curve 2 y = x2 which is JT = a p. nearest to the point Am. (4, 1). (2, 2). 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.
a"
where
40. A formula for the efficiency of a screw is E =
h f tan
39. If PQ is ~ is the angle of friction and h efficiency. is > the pitch of the screw. Find // for
4ns. h = sec maximum
tan 6. The distance between two sources of heat A and /?, with intensia and b respectively, is I. The total intensity of heat at a point P
between A and B is given by the formula
41. ties ~ where x is P the distance of \" / from A. For what position perature be lowest ? of P will Ans. x the tem = T a ij  of an isosceles trapezoid is the major axis of an
the ends of the upper base are points on the ellipse. Show that
ellipse
the maximum trapezoid of this type has the length of its upper base
half that of the lower. The lower base 42. ; 43.
ellipse An isosceles triangle with vertex at (0, b) is to be inscribed in the
2 2
Find the equation of the base if the area of
a2b2
a' y 2 2
b' x' the triangle + is . a Ans. 27/46 maximum. 44. Find the base and altitude
which circumscribes the ellipse parallel to the zaxis. = 0. minimum area of the isosceles triangle of a 2 ?/ 2 = a 2 6 2 and whose base is
Ans. Altitude = 36, base = 2 aV. b 2 x 2 4 , DIFFERENTIAL CALCULUS 64 and b) be a point in the first quadrant of a set of rectangular
a line through P cutting the positive ends of the axes at A
Calculate the intercepts of this line on OX and OY in each of the Let P(a, 45. Draw axes.
B. following cases.
(a) (b)
(cj (d) when
when
when
when the area OAK is a minimum;
the length AH is a minimum;
the sum of the intercepts is a minimum;
the perpendicular distance from O to AB An*, (a) 2 a, 2 b
(c) 50. Derivative a 4 ; V^, (b) a
fe as the rate of change. + <6\ b + VS; (d) is + a maximum. cPb* ; 2l^,
a 2!^!.
6 In Art. 23 the functional relation y (1) = r2 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
when jr increases from x
4 to x = 4.5. equals 8.5 In general, the ratio (A) ^X average rate of change of y with respect to x
A.r.
from x to x when x changes + Constant rate of change. When
y (4) we have r& Ax = ax + b, = a. 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 (B) change of y with respect
f = to x. Hence, by Art. 24, instantaneous rate of change of y with respect to x for a
definite value of x. VARIOUS APPLICATIONS OF THE DERIVATIVE
For example, from above, (1) (o) , When x = 4, 65 ^ Z x. the instantaneous rate of change of y is 8 units per
is often dropped in (B). unit change in x. The word "instantaneous"
Geometric interpretation. Let the graph of #=/(*) (6) be drawn, as in the figure. When x into OA7 then y increases
creases from OM MP from to , NQ. The average rate of change of y with respect to x equals the
slope of the secant line PQ. The instantaneous rate when x = OM equals the slope of the tangent Hence the instantaneous rate of the constant rate of When
is f'(xo). x XD, x If at P(x, y) change of y
change of y along the tangent line PT. equal to at P. the instantaneous rate of change of //, or /(.r) in (6)
increases from .TO to xo
A:r, the exact change in
r + now is not equal to/'(jo)A.r, unless f'(x) is constant, as in (4).
see later, however, that this product is equal to A//, nearly,
is sufficiently small. y 51. Velocity in We shall when Ax motion. Important, applications arise
variable in a rate is the time. The rate is rectilinear when the independent then called a timerate. Velocity
in rectilinear motion affords a
. line is  ^ 2 < A o
4 , r ^
B f i> 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 position of P, and let t be the corresponding elapsed time. To each value
of corresponds a position of
s will be a function of t Hence P and
/, therefore a distance (or space)
may write s. and we s=f(t). Now let t take on an increment As, and n
v
of
If A = ; then s takes on an increment the average velocity Az P when the point moves from P to P', during the time
P moves with uniform motion (constant velocity), the will at )
' A have the same value any instant. for every interval of time and is interval above A. ratio the velocity DIFFERENTIAL CALCULUS 66 For the general case of any kind of motion, uniform or not, we
define the velocity (timerate 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 with respect When v is positive, P and the point
s is any instant is distance (= space)
timerate of change of the distance. is the derivative of the to the time, or the the distance s moving in a decreasing function of /, is an increasing function of t, AB. When v is negative,
moving in the direction BA. the direction and P is (Art. 45.) To show that this definition agrees with the conception we already 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
s (2) where s = distance of fall General Rule (Art. 27) to  in feet, 16.1 /, time in seconds. Apply the t (2). + As=lfi.l(J + A/) = 6.1 2 + 32.2 AZ+16.1(A/) 2
SECOND STEP. A* = 32.2 tAt + 16.
As
1
THIRD STEP.
r = 32.2 1+ 16.1 At = average velocity throughout
FIRST STEP. Placing t 2 s 1 Z . the time interval At. = 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
; , ( during the corresponding small interval of time. But what
the velocity at the end of two seconds is the limit of
that is, the vethe average velocity when At diminishes toward zero
at the end of two seconds is, from (3), 64.4 feet per second.
locity
Thus even the everyday notion of velocity which we get from experience involves the idea of a limit, or, in our notation,
velocity we do mean by ; v = / As\
lim
)= 64.4
AioVAt/ ft. per second. VARIOUS APPLICATIONS OF THE DERIVATIVE 67 52. Related rates. In many problems several variables are involved 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 timerates 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,
which vary with the time.
SECOND STEP. Obtain a relation between the variables involved which p, z, etr. the quantities will hold true at instant. any 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 A man 1. 80 ft. Solution. y is walking at the rate of 5 mi. per hour toward the foot of
is he approaching the top when he is high. At what rate
from the foot of the tower? a tower 60 ft. Apply the above rule. First Step. Draw the figure. Let x
distance of the
his distance from the top, of the tower at any instant. = man from Second Step. Since we have a right triangle,
y y = x2 + 3600. Third Step. Differentiating, we get **%=**%'<* m
U 4y '  dt = xdx
y dt' This means that at any instant whatever
(Rale of change ofy] Fourth Step. x () times (rate of change ofx). ~ 80, at =
=_ ? 5 mi. an 5 x 5280 hour
ft. dt Fifth Step. Substituting in (1), 80
& = _ JjL x 5 x 5280
f dt = ft. per hour
100
4 mi. per hour. Ans. an hour. the foot, and DIFFERENTIAL CALCULUS 68 = x 2 in such a way that when
point moves on the parabola 6 y
6 the abscissa is increasing at the rate of 2 ft. per second. At what
x
rate is the ordinate increasing at that instant?
A 2. Solution. Plot the parabola. First Step. 6 y Second Step. = x'2 . 6^ = 2x^.or Third Step. .. ) dt~3 dt This moans that at any point on the parabola
(Rate of ckanac of ordinate) x Fourth Step. = liws [r] ~2 6. ( rafe ft. f change of abscissa). per second. dt Fifth Step. Substituting in (2), ^=~ x 2=4
o ft. at From the first result we note tliat at per second. Ans. the point / J (G, 6) the ordinate changes twice as rapidly as the abscissa.
If we consider the point /"(
t>, 6) instead, the result
the minus sign indicating that the ordinate is decreasing = is 4 ft P^r second, as the abscissa increases. A circular plate of metal expands by heat so
radius increases at the rate of 0.01 in. per
second. At what rate is the surface increasing
when the radius is 2 in. ?
3. that its Solution. Let x = radius and
y = 7r.r~.
du rt area of plate. y Then dx That is, at any instant the area of the plate is increasing in square inches 2
times as fast as the radius is increasing in linear inches. Substituting in ?nr (U), = 2 TT x 2 x 0.01 = 0.04 TT sq. in. per second. Arts. dt 4. An arc light is walk on which a boy 5 hung 12 ft. ft. tall is directly walking. above a straight horizontal How fast is the boy's shadow VARIOUS APPLICATIONS OF THE DERIVATIVE
when he lengthening is 69 walking away from the light at the rate of 168 ft. per minute?
Let x = distance of the boy from a point directly under the light L,
length of boy's shadow. From the figure, Solution. and y y : y =5
y = 8 + x or : 12, x. dx t/v 5 d/ ..
,.
T^.Differentiating, 7 d/ ' is, the shadow is lengthening
walking, or 120ft. per minute. that v> as fast as the boy is = A 12 jc in such a way that its
5.
point moves along the parabola //'
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 ^is which the rate
12 jc 2 + 45 .r  of change of 13 zero. /b/.s\ 3 and 5. A barge whose deck is 12 ft. below the level of a dock is drawn up
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.
7. to it by means A fastened to a rope which is wound about a windlass 20 ft.
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?
8. boat above the is level at 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
fast is
pulled away from the wall at the rate of 3 ft. per minute, fa) How the top of the ladder descending when its foot
(b) When will the top and bottom of the ladder
(c) When is is 14 move ft. from the wall? at the same rate? 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. DIFFERENTIAL CALCULUS 70
11. The side of an equilateral triangle is o in. long, and is increasing at
in. per hour.
fast is the area increasing? How the rate of k An.s.
12. The edges \ afcV3 of a regular tetrahedron are 10 in. long at the rate of 0.1 in. per minute. Find the rate sq. in. per hour. and are increasing 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 that
the area is changing at the rate an f bm.
r 14. At a piped are 6 certain instant the three dimensions of a rectangular parallele8 in., 10 in., and these are increasing at the respective rates in., How of 0.2 in. per second, 0.3 in. per second, 0.1 in. per second. fast is the volume increasing?
15. The period (Psec.) of a complete oscillation of a pendulum of
0.324
Find the rate of change
given by the formula P VL in. is length
of the period with respect to the length when / = 9 in.
result approximate the change in /' due to a change in
/ AJIK. By means
/ of this from 9 to 9.2 in. 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 increasing at the rate of 1 in. per minute, what change in the altitude will keep the volume constant? The radius A?IK. Decreasing 4 in. per minute. 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.
17. of the 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.
is A stone dropped. rate of ty ft. is dropped down Show that, a' deep shaft and after t sec. another stone
the distance between the stones increases at the per second. 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
Ans. 10 cu. ft. per hour.
Law pv = c.)
20. : The adiabatic law for the expansion of air is PV IA = C. If at a
time the volume is observed to be 10 cu. ft. and the pressure is
given
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.
21. VARIOUS APPLICATIONS OF THE DERIVATIVE
22. If = y r 3 and x how second, find x 4 .r is increasing steadily at the rate of A unit per
graph is changing at the instant when fast the slope of the = 2.
23. An*. Water 71 Decreasing 4 units per second, flows from a faucet into a hemispherical basin of diameter
in. per second.
How fast is the water rising 14 in. at the rate of 2 cu. (a) when the water is halfway to the top?
volume of a spherical segment is irr h f (b) just as >2 ^ ?r/r
, ! , it where // runs over? (The
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. denotes the radius of a sphere, 25. If r ,. , x, prove the relation tf the surface, and V the volume, dV = rdS
dt 2 dt A railroad track crosses a highway at an angle of (10". A locomotive
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 rate of change of the
distance between them?
Ans. Increasing 15 mi. per hour or IftV.'J mi. per hour.
26. is 500 ft. 1 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
Aws. I ft. per minute.
is 2 ft. deep?
28. to In Problem 27, at what rate must water be poured into the trough
the level rise I ft. per minute when the water is 3 ft. deep ? make 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
30.
if is 2 ft. deep?
is water being drawn from the trough
per minute when the water is 3 ft. deep ? In Problem 29, at what rate the level is falling 0.1 ft. ^intercept of the tangent line to the positive branch of the
xy = 4 is increasing 3 units per second. Let the //intercept be
hyperbola
OB. Find the velocity of B at the end of 5 sec., the xintercept starting at
31. The Am. the origin. A f unit per second. x so that its abscissa in32.
point P moves along the parabola y
creases at the constant rate of k units per second. The projection of P on
the .raxis is M. At what rate is the area of triangle
changing when
2 OMP P is at the point where x = a? Ann. 2 fcVa units per second. DIFFERENTIAL CALCULUS 72 ADDITIONAL PROBLEMS
2 =
16 x
1. Rectangles inscribed in the area hounded by the parabola y
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 .raxis. Find the volume of Ans. Aj*j&Vb the largest parallelepiped. = 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 Am. minimum. k 2 jr 2 + 2 h 2k 2 2
h' y 2 . 3 /// + //'* =
has a loop in the first quadrant sym3. The curve j
metric with respect to the line // = jr. Isosceles triangles having a common
~ a are inscribed in this
vertex at the origin and bases along the line j f y
loop. Find the value of a if the area of the triangle is a maximum.
:< Am. J(l + Vl3) =2.303. 2
jr
a tangent
the first quadrant on the curve // = 7
is drawn, meeting the coordinate axes at A and B. Find the position of P
Am. Ordinate */.
which makes A B a minimum. 4. At a point /' in 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
Am. 17.
the investment?
6. For a certain article the increase in the number of pounds consumed
proportional to the decrease in the tax on each pound. If the consumption is in Ib. when there is no tax and n Ib. when the tax is / dollars per is pound, find the tax which should be imposed on each pound to bring in the most revenue.
2
7. A triangle ABC is formed by a chord BC of the parabola y ~ kx
and the tangents AB and AC at each extremity of the chord. If BC remains 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. A tank of radius 10 in. has a hole of radius 1 in.
with which the water contained runs out of the
velocity
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 ?
8. vertical cylindrical in its base. The .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 new function We of successive derivatives. may have seen that the general also a function of x. This
also be diil'erentiable, in which case the derivative derivative of a function of x is in 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.  V Thus, if 3 r>, <LM ._ 19
i^
, ax ~3
x , dx\ drv/.r Notation. for the successive derivatives are usually The symbols abbreviated as follows.
rf If 2
?/ y =/(j), the successive derivatives are also denoted by = , _
 . .  . , , . dxn = 3 x 4 y' = 12 x 3
In the example given above, the notation y
= 36 x 2 T/'" = 72 r, ?y w = 72 is most convenient.
illustrate the
54. Successive differentiation of implicit functions. To
, I/'' , process
(1) we shall find ^ rom f b 2x 2 the equation of the hyperbola a 2 ?/ 2
73 = a2 62 . , DIFFERENTIAL CALCULUS 74 Differentiating with respect to x (Art. 41), 2 b*x  2 a'2 y ^=
^L
dx (9) Differentiating again, ^ its ., ^  ay remembering that y d~y
dx*
Substituting for 0, _ ^ value from
,i, a function of is dx (2), /fcVN a 4 '//2
Hut, from the given equation, bx r/ ay~ = 4 ;{ v/ a6 2 . PROBLEMS
Prove each = 2. of the following differentiations. ^~ >/" + W. "'* 3_ \LJ.y^. y /~~
Va_ 4. // d 2 //
*j< . r; 2
f r .
; d~u
fi\~
dl 4 __
__ dx'2 bx (i iL^!___. 8(af6/)^
b. (a // = dy + (i dx'2 r = 6. '2 7. f(s) 8. y = t h 1 r< *? =
.rfl 2 a =
(a' 5. a?> _ 2
(a + r 2 a' + x) 3 x, SUCCESSIVE DIFFERENTIATION
9. j* = 2 2
f */ AND APPLICATIONS r 2.
<ir 2 10. y = 2 2 4 d' ru*. y y3
__ _ 4 a2 dx'2
11. fc.r 2 a f 75 2 = a' b 2 2 // //* 2
. a 4 ?/'
12. or' f 2 Jixy f fy/ 2 1.
f by)*' (//.r ^ JT* + ?/ 14. .r + 2 T2^2 13. :{ 1. djr = 4
a V . 2 </.r In Problems 15 25 find the values of and //' //'' for the given values of the variables. 16. ;// 17. ?/ ; 18. j x2 ?/ ?/= 21. ?/ 22. ?/ 23. ?/ j* ; 2 19. 20. = V25  3 x j = 3.
= x Vx 2 + 9
= 4.
4 2=9
= 5, = 2.
f 4 xy 4x = 2,
f 3 =
; j ?/ 2 z/ (3 a: 2 ) 4 = Vl
=
= 4 2 ; ; x ; 2 ?/ = 1. =
= 2.
x = j J; 4 = j ; 2 24. ?/ 25. .r' { +
 2 .r/y 2 = f .r/y 1 G //" ; = = .r 8 x ; ^ y 3, 2, y 2. = 2. In each of the following problems find
29. y
2 30.
27. ?/a: 28. T/ 2 H a = V2  3 bending varies. :i 4 xy 3 ajry f 2 . :{ ?y = ?> 3
. y
\ P(x, y) traces a curve, the
slope of the tangent line at P x  x
 If). a:. When the point 55. Direction of of a curve. // 31. 2 jVr/"' When the tan / X x FIG. 6
FIG. a
gent line is below the curve
the arc is concave upward; when above the curve (Fig. 6),
(Fig. a),
the arc is concave downward. In Fig. a the slope increases when DIFFERENTIAL CALCULUS 76 P describes the arc AP'. On the other hand, in Fig. and decreases, J'(jr) is Hence f'(x) is an increasing function of x.
b, when P describes the arc QB, the slope a decreasing function. In the first case, therethe second case, negative (Art. 45). Hence fore, f"(x) is positive; in 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
with respect to x is positive, and concave downward if this derivative
y
is negative. 56. At A Second method for testing for maximum and minimum values. a of the preceding section, the arc is concave upward,
and the ordinate has a minimum value. That is, /'(x) = and /"(/)
is in Fig. At B positive. and f"(x) in Fig. b, f'(x) is negative. We may
minimum
j(x) is then state the sufficient conditions for maximum and
values of f(x) for critical values of the variable as follows
a /(jt) is a maximum if f'(x)
minimum if f'(x) = Following
for is and/"(*) a negative number, and/"(*) a positive number. the corresponding working rule for applying this test maximum and minimum values. FIRST STEP. Find the first derivative of the function.
Set the first derivative equal to zero and solve the resulting equation for real roots in order to find the critical values of the variable. SECOND STEP. THIRD STEP. Find the second derivative.
FOURTH STEP. Substitute each critical value
second mum derivative'. If the result is for the variable in, the maxifunction is a negative, then the function is a for that critical value; if the result is positive, the minimum.
or does not exist, the above process fails, al/"(/) =
there may even then be a maximum or a minimum
in that
though
case the first method given in Art. 47 holds, being fundamental. When ; 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
the function 1 . Let us now apply M = x~ in the example worked out on page Solution. /(;r)=.r found f x 49.
2 +
439 First Step. the above rule to test analytically , f(x)=2x~> AND APPLICATIONS SUCCESSIVE DIFFERENTIATION
Second Step. ; 77 ^=.
x = /"(6) =
= 6, critical value. Third Step.
Fourth Step. Hence /(6) ILLUSTRATIVE EXAMPLE f . 108, P'xamine
minima. Use the second method.
2. Solution. 6 values are critical Fourth Step. /"( 1)
1) ../( =4  3 x2 Second Step. /"(3)  3 x value.  9 x 4 5 for maxima and = **  3 x 9x45.
2 6 x  9.
/'(*) = 3 x
x  9 =
x =
1 and 3.
/(*) First Step. hence the minimum r* 12. /. /(3) ; =  12.
= 10 (ordinate of /I) = maximum
= 22 (ordinate of #) minimum value. value. PROBLEMS
of the following functions for Examine each
1. x 3 3 x 4  2 An*. 2. maxima and minima.
 JT x
2. 3. x3  3 x  2 x3 3 ox 2 4 a 3 4. 2 4 12 x 5.  3 x x
x 3 x2 4 2 x  2 (a .  2 x3 ; 0) = x
x + 4. ~ = 6. 3 x 7. x4   4 x  3 4 x2 4 4 2. ax 9. x3 4 10. 12 x 11. 2
4 9 x x 2 (x4) 2 . 27 x  4 9. 4 x3 2. a, 2, 2.
6. , = 2. a3 , gives min. = 0.
gives max. = 22.
=
1, gives min.
gives max. = \.
=
 gives min.
gives max. = 2.
=
1, gives min.
, 5. $. = 0,
=
3.
= 2, gives min.
30.
4.
x = 0, gives max.
min. = 0.
x =
V2, gives
x = a, gives max. = .
= \.
x
a, gives min. x
x
x f 4. 9 x2 12 x 2 gives max. \, jr 4 gives max.
gives min. 0, x  1, 1 x x = . gives max.
gives min. 2,
0, + 12. * 13. x . DIFFERENTIAL CALCULUS 78 A rectangular 14. Find the volume box with a square base and an open top is to be made.
box that can be made from 1200 sq. ft. of
Ans. 4000 cu. ft. of the largest material. A water tank is to be constructed with a square base and an open
to hold 125 cu. yd. If the cost of the sides is $2 a square yard,
top,
and of the bottom $4 a square yard, what are the dimensions when the
Ans. A cube of edge 5 yd,
cost is a minimum?
15. and A is rectangular flower bed is to contain 800 sq. ft.
It is to be surwide 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
Ans. 20 ft. x 40 ft.
dimensions of the flower bed? 16. rounded by a walk which is 3 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 to be is made up U\o edges so as to give
piece is 14 in., ing capacity how deep should may be a is 15 amount ft. the trough be in order that its carry Ans. of a rectangle Find in perimeter. of light. made maximum? A window composed 19. triangle mum of a long rectangular piece of tin by bending
a rectangular cross section. If the width of the its surmounted by an equilateral dimensions An*. Rectangle is 3.5 in. 3.51 ft. if it admits the maxiwide and 2.23 ft. high. What is the weight of the
20. A solid wooden sphere weighs ?/> Ib.
heaviest right circular cylinder which can be cut from the sphere?
Ans. The 21. slant height of a right circular cone Find the altitude if the volume is a is r^lb. V3 a given constant a. maximum. a A ri/io. 7^ V3
An oil can
Show that 22. is to be made in the shape of a cylinder surmounted by a for a given capacity the least material is required
altitude of the cylinder is equal to the altitude of the cone. cone. if the 2
23. Ciiven the parabola // = 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). A given isosceles triangle has a base of 20 ft. and an altitude of 8 ft.
are the dimensions of the maximum inscribed parallelogram, one side
coinciding with the base of the triangle, if the acute angle of the parallelo24. What
gram is arc tan ^? A Ans. 5 ft. X 10 ft. 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.
25. AND APPLICATIONS SUCCESSIVE DIFFERENTIATION
26. A 79 sheet of paper for a poster is to contain 16sq. ft. The margins
and the bottom are to be (\ in., and those on the sides 4 in. at the top What are the dimensions the printed area if is to be a
.4 27. An electric current flows the center of the by F coil _,
2 (r of the coil. + .r and perpendicular to w here 1.90ft. through a coil of radius
the axis of which is on a line F on a small magnet force maximum? MS. .r is is a its plane. r x 3.27 ft, and exerts a drawn through This force is given the distance to the magnet from the center 2 )* Show that F 57. Points of inflection. maximum A for .r \ r. 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 At points of (1) 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 values 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 inand the signs obtained determine if the flection, 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 curve evidently crosses the curve. Following
equation is y is a rule = f(x). for finding points of inflection of the curve whose
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)
trifle greater than each root sign, we have a point for values of x first a found in of inflection. trifle less the second step. and then a If f"(x) changes DIFFERENTIAL CALCULUS 80 When f"(x) = +, = Whenf"(x) upward + .*
concave downward ^^^. the curve is concave
the curve is , 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 x3 1. Solution. /(x) First Step. /"(x) 36 Second Step.
.  2 J' x 24 x =3 = 36
~ 0. and x 3 Third Step. x4 /"(x)  x2 4  x'< 4 1. 24 x. tl are the roots. =36T(x When x <
When 3 >x> I). =
= 0,/"(x)
0,/"(x) Therefore the curve is concave upward to the
the right of x
(A in figure). When ()<x <
When x > 3,/"(x)
l,f"(x) +.
. left =
= and concave downward to .
f. Then^fore the curve is concave downward to the left and concave upward to
the right of x =
(/ 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
rj downward between ,4(0, 1) and (, A}), and concave upward everywhere to the right of B.
2. (y2}*= (jr4). Solution. y Second Step. When =2 x = f 4, (x  both When 4)*. first and second derivatives become < 4, W hen x > Third Step. 4, x T * This
it is
( ) may easily concave upward
water. +. ^= . dx 2
dr 2 if we say that a vessel
shaped like the curve where
hold (+) water, and where it is concave downward will spill be remembered
will = infinite. SUCCESSIVE DIFFERENTIATION AND APPLICATIONS
We may therefore conclude that the tangent at xaxis, that to the left of (4, 2) the curve
(4, 2) it is
3. 4.
5. is = x2
y= 5
3
y= x Ans. . is a point of inflection. Concave upward everywhere. x2. 2 x perpendicular to the concave upward, and to the right of concave downward. Therefore (4,2) y (4, 1^ is 81 Concave downward everywhere.
Concave downward to the left and
concave upward to the right of . (0, 0). 6.
7. y =
y = x4 Concave upward everywhere. . 2 x3  3 x2  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. Curvetracing. 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 corresponding values of y (or r), plot the respective points, and draw a smooth
curve through them, the result being an approximation to the required curve. This process is laborious at, best, and in case the equation 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
; 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
is in his work the student may follow the Rule for tracing curves, using rectangular coordinates FIRST STEP. Find
to the first derivative; place find the abscissas of the it to zero; solve points. Test these equal maximum and minimum values. SECOND STEP. Find the second derivative; place it equal to zero;
Test these values. solve to find the abscissas of the points of inflection. THIRD STEP. Calculate the corrcspomling ordinates of the points
whose abscissas were found in the first two steps. Calculate as many DIFFERENTIAL CALCULUS 82 more points as may be necessary to (jive a good idea of the shape of the 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 curve. 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 plotting 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.
the e([iiations of the tangent and normal at each Also find point of inflection.
1. y rr .r :{ Solution.  9 jr 2 4 1M x Use the above Ft rat Step.  7. rule.  3 x'~  18 x
 0,
x = 2, 4.
y" = G x  18,
 18 = 0,
x  3.
y'  (lx lS.r Second Step.
G.r Third Step. f f 24, 24 Plotting the points and sketching in the curve, Fourth Step. we get the figure shown. To /V3,
tangent and 3 use formulas (1), (2\ Art. 43.
x
30 for the normal. 11), ?/ 2. ;] // = An*. .r<  3 x'2 Max. 1, ( 9 x f 11.
J tangent, 4
3. tangent and normal to the curve at the point of
20 for the
This gives 3 x + y find the equations of the inflection 6 y = VI  24 x A//S.  Max. /); min.
jc + 15 :r ( y
2 1, (3, 4  2 j\
^) min.
: point of inflection,
4 //
1 = 0.
normal, .r \f)
; ( ; 4,  5) ; (1, 0) ; point of inflection, AND APPLICATIONS SUCCESSIVE DIFFERENTIATION
4. = x4  y 8 x2 . Am. Max. (0, 0) min. ; x 5 y  2, ( 16) points of inflection, ; Y>. (?,Vi*,
5. 83 J*. Am. Max. (1, min. ( 4); 1,  4) point of inflection, ; (0, 0). *?Ti(V3, V3)
(3, ij), (0, Am. Max. =
= 14. +6
4 + 3
= 4 .r<  18 r 2 +
3
= O  a) f b.
=  I)  24
12
= r (9 =2  5 2
= 3  5 .r\ 15. ?/ 7. ?/ 8. ?/ 9. /< .T ; min. ( ~ V3,  2 18. . ) points of inflection, ; ^ (I If  + .r .r<. .r ?/ 19 ''// 20. 15 x. a'// ^ .'"' ^ .r< ~ f f :* 10. ?/ 1 11. (.r // 2 12. r .r> ?/ (.r 1)2. ^,. L> // 13. ). .r 21. . = //   "' ., .r 1 a 1 .r> // = .r  5
= jr(s r> 16. y .r 1 22. .  ?/ (.r 4) ''  f <i) 2
. 23. 4 59. Acceleration in rectilinear motion. 1 .r '//  (.r' { 1 )'. In Art. 51 velocity in recof change of the distance. motion was defined as the timerate tilinear We now
That V3 0), (3, H). define acceleration as the timerute of climiye of the velocity. is, Acceleration (A) From (C), Art. 51, we = dv
a
(h obtain also, since v * Referring to Arts. 45, 47, and 56, we have the following
which apply to a definite instant / =
/ If a If a If a If a
a If >
<
>
< = criteria : 0, v is increasing (algebraically). 0, v is decreasing (algebraically). and v = 0, s has a minimum value.
and v = 0, s has a maximum value.
and changes sign from + to
(from passes through to, then v has a maximum (a to +) when
minimum) value when t DIFFERENTIAL CALCULUS 84 In uniformly accelerated rectilinear motion, a is constant. Thus
in the case of a body falling freely under the action of gravity only,
32.2 ft. per second per second. Namely, from (2), Art. 51,
a = = 8 16.1 1 2 v , = = 32.2 = <* t, = 32.2. PROBLEMS
1. in a By experiment it has been found that a hody falling freely from rest
vacuum near the earth's surface follows approximately the law ~ 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.
s = , ; Solution. (1) s = 16.1/ 2 ~ = 32.2 . ds
(a) Differentiating,
or, from (2) Differentiating again,
or, from (A) above, which t, (3) v  32.2 r. (C), Art. 51, = 32.2, a / 32.2 per second. ft. per ft. (sec.) 2
, 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.
and </ at the end of the first second, substitute /
1 in (2) and (3).
(b) To find
tells ; Then
(c) 32.2 T To find rand Then v end a at the = 161 per second, ft. 32.2 ft. per (sec. of the fifth second, substitute per second, ft. = a a = 32.2 ft. t 2
) . = 2
per CsecJ 5 in (2) and (3). . Given the following equations of rectilinear motion; find the position,
velocity, and acceleration at the instant indicated.
2.
3. .s 4 1 * 4.
5. .r // =  = ~ 120 { 32
(J (>   /  /' 8
2 t lf> = = { t' / 2. s 1. y ; / ; t ; 4. .  7. / 8. // =
= 9. * = V5/ f = 4 10. * 16 2
t' 100 \  4 20
/  8 / ; 2 ; / / 2
/ ; ^ = V5f 7 3 4; f / * = t' /' AUK. 2. f ; = 2. / = 2.
= 3.
5. * 4, r 10, a = 8.   8, a = = 32, r = 0, a =  16.
= 4, r = 6, a = 0.
= 224, r 32. SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 85
In the following problems find the acceleration at the instant indicated. = 8032/:f = 0.
= 4 !()/; = 2.
r
2 12. = 16  64 + 04.
*= 120 / 16
* = 3 c*t A ball thrown directly 14. * / 2 of rectilinear
first comes 16. motion * it will 19. If rise; = 80 and velocity how (c) far + upward moves according
,s (a) its position 20 Pt
5/4 ,s /'. 18. it will / 1 . 1 find the position ; AHS. . 17. + to rest. / / ^_., =
/ J 15. high 18r = 6. / / Given the following equations
and acceleration when the particle Find 32. An.. ll.r = s 0, a = 32. ~. 1 to the law If) /. and after 2 sec. move after 3 sec. (b) ; how in the fourth second. f 1, show that the
the equation of a rectilinear motion is =
is negative and proportional to the cube of the velocity. V ,s / acceleration 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 Find a formula j gt~. for the greatest height reached by the body. ~ 1(>0, g 32. Find (a) the
preceding problem suppose r\
sec.; tb) the distance
velocity at the end of 4 sec. and at the end of
moved during the fourth second and during the sixth second.
21. In the (5 A car makes a trip in 10 min. and moves according to the law
in feet. Ca) How far
250 f2
t\ where is measured in minutes and
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.
22. ,s = / :\ ADDITIONAL PROBLEMS
2 x  s 2 arid find the equations
Trace the curve (4  2 x + s)tj
at each point of inflection.
of the tangent and normal
2 y =
Ans. Max. (1, \). Point of inflection CO, 0): tangent, x
2 =
0. Point of inflection (2, 0)
tangent, x + 2 y
normal, 2 x + y
1. , ; : normal, 2 x
2. y 4 = A certain curve (the tractrix) is point of contact Pur, y} to
c). Show that
the constant r (A P from ( 3. its \ 4lL V ; 0. such that the length of every tangent
its intersection A with the xaxis is x / d 2 ij _ c 2y Determine k so that the normals at the points
= k(x 2 3) 2 will pass through the origin. curve y of inflection of the ... _ri/lc>. 7. K> _ 1
. 4V2 CHAPTER VII DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS.
APPLICATIONS We consider now functions such as
sin 2 3% jr, log + s~), (1 called transcendental Junctions, as distinguished from the algebraic functions hitherto discussed. The following formulas,
60. Formulas for derivatives; second list.
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. X d = ^ (lnl dv
dx 1 T~ ITx Xa
^(a")
dx = a"lna^dx f <"> = ?
dx
dx XI a 4.
dx XII XIII XIV = <!") d , ^ (cos , v) 1*1" 1^ + inu.u*^.
dx
dx cos v (sin v) XV = dv
sin v XVI (ctn y) ^
dx ^ (sec r)  sec 2 v (tan v) dx c?x XVII On^log^) dx v XI dv = esc 2 v ^~ = sec 86 dv
i> tan i; . clx TRANSCENDENTAL FUNCTIONS
XVin (esc v) dx & / \ (vers v) = 87 esc v ctn v dx = sin v dv
.  dv XX
rf* XXI dx (arc sin i;) (arc cos v) ^n v) i;) " 4 (a1
dx XXIII XXIV ^ ._ Vl  ~ i; _ ^/i (arc ctn XXII = (arc sec v) vz = 7^1 f ^ =  7^..
dv
dx = dv XXV ~  XXVI 61. The number (arc csc ") (arc vers v) = = Natural logarithms. e. One of the most important limits in the calculus is Km (1) x (1 + ar = 2.71828 . > is denoted by c. To prove rigorously that such a limit e
beyond the scope of this book. For the present we shall This limit
exists is content ourselves with plotting the locus of the equation
(2) y=(l + xf the function (1 + x) x (= y)
and showing graphically that as x
and therefore
takes on values in the near neighborhood of 2.718
+ e = 2.718 approximately. DIFFERENTIAL CALCULUS 88 from the left, y decreases and approaches e as a
from the right, y increases and also approaches e as a As x
As x * The
As
?/ fact expressed in (1)
> jr f oo, approaches // used is The lines ?/
we shall show how XX In Chapter
of ; and as .r
and f = from the 1 > right, are asymptotes.
to calculate the value of c to any increases without, limit. number limit. in Art. 63. as a limit 1 limit. 1 1 decimal places. Natural, or Napierian, logarithms are those which have the number
base These logarithms play a very important role in mathe e for To distinguish between natural logarithms and common
logarithms when the base is not explicitly stated, the following nomatics. tation will be used. Natural logarithm of Common logarithm of e (3) If .r
j, ==
_._ A =
r 0, _ oo^ 1, and (h en The student is j  A;
In 1 T ,\  0, 0. and r>. T number A  In A
If x = 1, AT = e,
we write In = that  = In r.
= log c) (base 10) r the natural logarithm of a By definition, x in the equation If (base r is the exponent 7 is, .r . and In e = 1. oo. familiar with the use of tables of common loga rithms, where the base is 10. The common logarithm of a number
is the exponent // in the equation
= log N.
10" = A
or
//
(4) N T , Let us find the relation between
In (3), In A and
7 log A 7
. take logarithms of both members to the base 10. Then we have, from (2), p. 1, x log (5) Solving for x, which equals c = In \, log by A' . (3), _ log W
~~
log e we get the desired relation TRANSCENDENTAL FUNCTIONS
That
its we obtain is, 89 any number by dividing the natural logarithm of common logarithm by log e.
Equation (A) may be written = log .V (6) Hence the common logarithm
its natural logarithm by log (= ji/) common of By log c  In .Y. number of a is obtained by multiplying
is called the modulus This multiplier e. logarithms.  tables, log e 0.4343, i = and 2.303. log e Equation may now (4) be written  In ;V (7) 2.303 log A T
. Tables of natural logarithms should be at hand. and logarithmic 62. Exponential fined The function of jr de by (1)
is functions. = ij e jr (c Its graph is shown
an increasing function for all values of
is everywhere continuous. called an exponential function. The function
see later, From and
(1), is
it we = 2.718 in the figure.
j, as have, by definition,
y (2) The functions e f = and ) we shall y In y.
In y are inverse functions (Art. 39). Interchanging x and (3) y  // in (2), we have , In y, 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
in values of x That is > 0, and (Art. 17), for is everywhere continuous.
of x greater any value a than zero
lim In x (4) = In a. x*a When x > asymptote The
e r and 0, then y * as remarked above. The ?yaxis is an log,, x (a > 0) have the same properties as
and graphs similar to the above. functions a x and In x oo, in the graph. DIFFERENTIAL CALCULUS 90 63. Differentiation of a logarithm. Let = y In v. (v Differentiating by the General Rule (Art. 27 j, considering
independent variable, we have FIRST STEP. SECOND STEP. y
Ay/ = + A// = In (r f + In (v Jn Ar) r > 0) as the Arj.
?' By(2),p.l ^= iin A A?' THIRD STEP. A/? \ + ) r / We cannot evaluate the limit of the righthand member as it
stands by Art. 16, since the denominator Ar approaches zero as a
limit. But we can rewrite the equation as follows: M ull iplymtr I l>y  r L J P,v(2),p.l The
ber of expression following (2), Art. 61, with the form of the righthand In is in mem = .r r ~ FOURTH STEP.  in ^ r = r a." /; r When Ar*0,
(1), Art. 61.  Honco lim 0. fl Ar(A r FsiiiK (1), Art. (>2, +
' V }? / wo havc the = r, by] 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_ _ Substituting the value of we (ir dy_ dv <7,r di ^ from the result of the Fourth Step, get v
X d
dx ,. . (In v) = dv
dx
v 1
=  dv v dx TRANSCENDENTAL FUNCTIONS
The 91 a function derivative of the natural logarithm of is equal to the derivative of the function divided by the function (or the reciprocal of the
function times its derivative}. Since log log e In v we have r, at once (IV, Art. 29) v
Xa
dx v dx 64. Differentiation of the exponential function Let // Taking the logarithm =a 1 (a . of both sides to the base
In we 0) get v In a, y
r e, > = lM = L.ln V
In a In . a Differentiating with respect to y by formula X, ^^_L .1.
a
In (/// From y (C), Art. 39, relating to inverse functions, we get *="*
or ^ = lnaa". (1) tfV Since ?? is respect to j% r
required to differentiate a with
use formula (A), Art. 38. Thus we get a function of we .r and dii
~ = In a
i /. ax When The a e, In a ., a' dv
y ax aa* XI it is (a^)lna^ ax = In e = 1,  and XI becomes 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. DIFFERENTIAL CALCULUS 92 65. Differentiation of the general exponential function. Proof of the Power Rule
Let y Taking = (n u''. the logarithm of both sides to the base
In '// or y
Differentiating = v In u
= e'^". by formula XI > 0) e, t By (3), Art. 61 a, ^ = <'"" 4 ("ton)
ax
ax
to + nw.rfr\
l u byVandX (Is/ (Ijr <' s+""s1 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
the exponent as eonstant, and ayam differentiating by XI, regarding
7Vzt' function Let v (is = constant.
n, any constant f then XII reduces to ; = (,,") WM " i^.
rf^ rf.r Thus we have shown that the Power Rule VI holds true
value of the constant for any n. ILLUSTRATIVE EXAMPLE 1. In Differentiate y x2 + 2
(:r 4 a). Arts. a 2 x
' ILLUSTRATIVE EXAMPLE
Solution. By (2), p. 1,
2, Then
lhen 2. Differentiate y we may = log 2 write this r ^=
dx log  log
o 2 j dx j. _ (1 + J 2 ). EJL l+x 2 (i <te + a2) bylllandXfl TRANSCENDENTAL FUNCTIONS
ILLUSTRATIVE EXAMPLE &= Solution. =
ILLUSTRATIVE EXAMPLE &= Solution. In a = ILLUSTRATIVE EXAMPLE
Solution. 3^
dx = = { (3 or 2 by XI ) Ans. a**'. be^ Differentiate y 4. b ~ a*** 6 x In a dx = a *\ Differentiate y 3. r 4dx be fy * * * (c fz'' . by IV
J ) +**(c*+x ' 4 z2 2 2 dx + (.r) by XL ) = Differentiate y 5. 93 T
.r' T . 1 z In r' a dx (r by XII ) J (^x^ l In f x' <' 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 whenever this is possible. ILLUSTRATIVE EXAMPLE
Solution. follows By using Differentiate y 1. (2), p. 1, we may = In Vl x2 . write this in a form free from radicals, as : ILLUSTRATIVE EXAMPLE
Solution. Differentiate y Simplifying by means of
y Then 2. i*
dx = i [In (1 = J + ' In (2), p. 1, *2 )  In (1  ; A
I by III and X In differentiating an exponential function, especially a variable
with a variable exponent, the best plan is first to take the natural DIFFERENTIAL CALCULUS 94 logarithm of the function and then to differentiate. Thus Illustrative
Example 5, Art. 65, is solved more elegantly as follows
: ILLUSTRATIVE EXAMPLE
Solution. 3. Taking the natural logarithm
In y Now x fT Differentiate y = both of sides, e r In x. differentiate both sides wit h respect to or . By (2), p. 1 .r. dx ILLUSTRATIVE EXAMPLE
Solution. 4. Differentiate Taking the natural logarithm
In y   (2 4 vV~' 5 = ( 2 f V^^5) L4 j f x'' (4 7) V * both sides, of xj In (4 ) Differentiating both sides with respect to rfj:  ?/  7 ). .r, l n (4 r2  7) 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
Solution.
In y 5. Differentiate y Taking the natural logarithm = i[ln (J  1) + In (x  2)  of = ir_i_
2[arl , _! j2 both In (x Differentiating both sides with respect to i^/
ydx = \/ nor 2)
V (j  sides, 3)  In (x j,
i x3 L x~4  4)]. TRANSCENDENTAL FUNCTIONS 95 PROBLEMS
Differentiate each of the following functions.
1. 2. 3. 4. ?/ + In (ax ?, = ?/ = ?/ = In ax n In + ln(ax dli dx + (a.r> " b}. 'A' 6). dx ax ft = _JL_.
ox +
2 ^ ft)'. h  + ax dx ft 2fl . dx
5. 6. 7. 8. y = .] .. In x :? . y=\ri*x\=
ij = In ^ (2 (Inx)  :>> x ( 3 2 lli
. = In 10 ln ^ + n2 x x dy 6 x(x __ </.r~2.r ^ //log.
?/ ~^ dx  3^ 2 :{ >r : d?/. V9  dx
^> .r x(l h ~ a 11. ?/  In (a.rVc7T7). 12. /O)  r 13. /(.r) = . > 17. y j/ = /'(. r ) .r. VfTT (j f /u r __ i^
in 15. f(x)
16. In (// ^ = .r 2 <>"*. In x 2
. 19. 2 2 ). /'(^)
a,L = +
=
1 In #. Vl
aft r  a2 (// = 2 y(l
^ = w . f'(jc) = 10 ft _ e = .
J / 2 In x). ^s ft 2
''. In 10. . ^ = _1.
dt = xe r dx e^ <? 21. r 2 2 f ^ = w 10 ^ 20 . dx y ?/ ) 2 J dx
. 2 ^=
92 x
^= 2a+3y
^ 2 r(a + x) 2 dx i f 4" 2 s2 In 1) dc J" 9. x dx e
. f 2^/1' ^=2
d// z ft 2" In ft. DIFFERENTIAL CALCULUS 96
22. u = 23. v = se". dv  24. /, 25. /y 26. ?/ 27. 7/ 28. ^
= In (.rV) = // 1) e"(u du u ds
dy . JT _ 2 c j ~~ry = .rY = ~(c' g= '.  1 "(2 ,**). c a= t + (,' U" fir  2 30.*^.
3Lf(j )== in^il^:.
V /'(,)= . J f 1 f < :)  JT 7 4 InJ ^
== Vx 2 f 1 HINT. First rationalize the denominator.
r
32. y == s 35. ?/ = .r ^^ s ' . \/3 .r + = r r ' ( a
3 s 36. y = V4 + 37. = .r" // // >r + 2 a 1 ^^a , 2 x J mb In Problems 38 47 find the value of ^ for the given value of x.
Ans. y' = %.
38. y  In (jc39.
40. 41.
. = log (4  3);
In V.r f 3
y 2r
= J.
y = xeJT ; \ rt 42. .r /y 7/ = In .r .r 2 y = 4.
4 ; .r = 0.3474.
= 1.4319.
y'
= 0.
' J
r 7/' t/' = 0.0483. TRANSCENDENTAL FUNCTIONS 97 = 43. y = log V25  4 x
= 10 x7 x4. 44. y
45. n x = .r'V.r a 5. 2 + 9 = ,
4. \20 ; ~ Find ; for each of the following functions. djc'
\i 49. In 48. ?/ cjc. 52. = f". 50. 2/ x In // 51. = = In
.r a*. 53. y r
<' // = f a ~
 '. Differentiate each of the following functions.
:
58. (^
In 54. In In
55. Va V 59. 10' log/.
2 jc 2 60. (ttf)'". 61. 2 2 s
,s . ' 57. In 6$ Vl> / f 3 The function 67. The graph sin x. y = (1)
is of shown in the figure. an angle Any of sin x value of x is assumed to be the measure in radians (Art. 2). n
T i y Thus for x = 1, y = of y is l = sin is repeated. = 0.841. of x The
It is a periodic function with the period = 2 when the value 57 18' is continuous for all values of x. For
is, = sin (1 radian) function sin x is defined and
important to note that sin x That 2 TT) is increased sin x. by a period, the value DIFFERENTIAL CALCULUS 98 The property
graph on page 97 of periodicity has the following interpretation in the
The portion of the curve for values of x from Oto2w
: OX OQBRC in the figure) may he displaced parallel to
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.
(arc 68. Theorem. Before differentiating sin x (Art. G9) it is necessary to prove that hm
,. (B) x sin x =  1. % > This limit cannot be found by Art. We 16. proceed by geometry and trigonometry.
Let O be the center of a circle whose radius
is unity. Let x angle AOM measured Since the radius in radian?. unity, arc AM = x, also.
Lay off arc AM = arc AM and draw
the circle at M and M' respectively. From
MM' < arc MAM < MT
is f ~\ by trigonometry,
2 sin x < 2 x < Dividing through by 2 sin x, we M'T. get 1 Replacing each term by
signs, we have 1 Therefore when x But when x = * 0, is 2 tan x. 1 sin x cos x reciprocal ^ sin x ^ > > = cos = and reversing the inequality cos x. small, the value of lim cos x (see Art. 17). < its 1 x tangent to geometry, 1 Or, M'T and 71/7' 1, sin x
lies between since cos x Thus we have proved is 1 and cos continuous for (B). interesting to note the behavior of this function from
graph, the locus of equation
oiri x
It x. is its x The function not defined for x = 0. Let us, however, assign
for x = 0. Then the function is defined and is is the value 1 to it
continuous for all values of x (see Art. 17). TRANSCENDENTAL FUNCTIONS 99 69. Differentiation of sin v Let By y = sin v.
the General Rule, Art. 27, considering v as the independent we have variable, FIRST STEP. y + SECOND STEP. =
Ay = A// sin (v
sin (v + Ar).
+ A?0 sin r. The righthand 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 setting A  sin B = 2 cos \ (A + B) sin ] (A  B),
A = v + Ar, B = 0.
A  B) = A?;.
A + #) = r + J A/?,
" Then ?, ( ?> ( Substituting, + A/0 sin Hence = sin v ~ A// 2 cos
\ / 2 cos ( ?' + (v + *> Ar) sin A?;. \n \ 7^ j ?, sin sin THIRD STEP. Av
,2 FOURTH STEP. = cos v. p
dv /sinf\
 Since lim I Ar Ar .Ol 1, (M '
f Art. 68, and lim cos
V
A,. .0 by I (
~ in
(4), Art. 38,
Substituting this value of a r/v/
~
ax XIII /. The statement ) we  coi get t/ = cos dv
z; ax cos v (sin v) of the corresponding rules will now be left to the student.
70. fined The other trigonometric functions. The function cos x is deand is continuous for any value of x. It is periodic, with the period 2
is TT. The graph of y _ cos x obtained from the graph of Art. 67 of sin x by taking the line
TT as the #axis. x=i DIFFERENTIAL CALCULUS 100 The graph number = = y tan x shows that the function tan x (see figure)
infinite when x of (n + is discontinuous for an of values of the independent variable x; namely, where n denotes any positive or negative ^)TT, integer. when x
\ TT, tan x becomes infinite. Hut from the relation tan (TT + x) = tan x, we see that
In fact, > the function has the period TT, and
the values x = (n + ])ir differ from
the period.
2 TT by a multiple of The
TT. for function ctn x has the period
defined and is continuous It is
all values of x except x
nir, n being any integer as before.
x becomes infinite. Finally, secx and esc x P'or these values ctn are periodic, each with the period 2 TT. The former is discontinuous
mr. The values
(n
only when x
2)^, the latter only when x
of x for which these functions become infinite determine vertical + asymptotes in the graphs. 71. Differentiation of cos v Let = y may P>y (3), p. 2, this Differentiating cos be written by formula
du
Jl = XIV .'. Since cos d
dx f . (cos u) XIII, c =
I v. sin r\= (^ = sin r, by (3), p. 2. 1 du . sin v dx formulas XVXIX. These formulas are readily derived by expressing the function concerned in terms of other functions
whose derivatives have been found, and differentiating.
72. Proofs of Proof of XV. Let By (2), p. 2, this y may = tan v. be written
cost? _ TRANSCENDENTAL FUNCTIONS
by formula Differentiating VII, cos r _ sin (sin r) dx rf^_
dx r r 1 (cos r)
y do* cos cos 2 101 + r 7 dx sin 2 v v r~ dx dr  ^  sec 2
cos XV . VM . dx ^ To prove XVIXIX, , ^  ; fdx r .  Using (2), p. 2 dx differentiate the form as given for each of the functions below. XVI. ctn v = ^ XVII. sec tan XIX. The /?  ^cos ?> versine r = vers r = XVIII. esc v cos 1 = ^
sin v /> v. details are left as exercises. Comments. In the derivation of formulas IXIX it was necessary to apply the General Rule, Art. 27, only for the following.
73. TTT
III
TT V d f
 (u
dx +v
d
rdr (I'M" 1 + d.r
as
, ^ , <]P </'// ^ w)
(uv) T T dv =w 7 ax + dx
dvx , ' dn VII ()=
v dx <!& ~ IX ,> , , Product, dr> 2 = r  <to dx dv ax * v' = <!&. ax ax 1 Quotient. ' \ VIII T" Al Algebraic sum.  ax Function of a function.
Inverse functions, 7 a ?/
cfe X
XIII all 4 HOK
ax 1 = Logarithm.
?; jLfein^^eos,,^.
dx
dx Sine, Not only do all the other formulas deduced depend on these, but
we shall deduce hereafter depend on them as well. Hence we see DIFFERENTIAL CALCULUS 102 that the derivation of the fundamental formulas for differentiation
involves the calculation of only two limits of any difficulty, namely, = lim 5!5?
 t by Art. 68 i V I and + v) = e.
v lim (1 By Art. 61 PROBLEMS
Differentiate the following functions.
1. y = sin u.r 2 . & = cos Solution. oar 2 by XIII f (ax*)
dx dx [va^.] = 2 ax cos ax' 2 2. Vl tan ?/ \ 2
3. y 1 x~ 1 sec' \ = sec = x \ 1  This may 1) x also be written dx :{ (cos.r)
!i ( . dx = cos.v and
= 3 cos T  sin ?u: sin" Solution. 3.J ?? ( = by VI cos J^TC cosy) [r = x)~ *( 1 ( by XV cos x. y^= ?/ \  x)* (1 3 Solution. 4. Ans. x. ^ Solution. . sin by XIV J) 3 sin x cos 2 x. Ans. jc. ^
Cw* = sin nx ~ (sin x) n 4 sin" r fix [?< = sin = sin nx n 4 sin" ??T and (sin a:)"" x cos nx sin" r
1 r ax r dx ~ (sin wx) ax
a.] (s i n x) (nx) = w sin nx sin"" 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
1 ! by V by VI and XIII TRANSCENDENTAL FUNCTIONS
y 6. y = s = tan 3 y 3 cos 2 7. =2 = r sin ax. 5. ?/' t. = dn ctn  a cos or. = s' .r. 6 sin 2 x.
3 sec 3 =  2 = f y 4 sec 4
(76 p' sin ?/'  c/.s _ dp
^? 18. y = sin 2 19. ?/ = In sin ax. 20. ?/ = In Vcos 2 x. (jj) 21. y e a<f sin
' cos 2 f 23. y 24 ?/ = ' = sin 2 27. p = 28. i  tan 3 (cos x) + (IT = x**.
= /. sin 26. /(x) = 0. tan 2
cos / sin 0 = = c"
= x x sin x. x. bx (a sin 4 '(2nm 2 c sin 2 fo / cos
f ?>x). cos 2 0 X
A
9 X
ctn  sec 2  i
. =  x). tan f 6. a). . f'(0) = = 2 sin = tan 4 0. p' =* CLX . tan 2
r !'(*) a) cos (6  x. x ^r
x .r a ctn ax. y' 1 sin r0 S 2 cos 2 x cos T/ . \ ^ x sin _
"" / = i/ .H cos x 7/' 25. f(0) 29. L> / y' In tan  y sin 2 .r. ?/' x. frx. 22. s . cos To
x cos x. esc bO ctn bO. V sec /'(0)  tan 4 .r (tan30) 7/' P .r SOC dx 17. /. o ?'
esc  dv d 103 V' cos 2 Hin *(?^x
\ = (9. X ?/On cos x (TT  x) cos 4. C os  x x tan In x). (TT x)/  x). DIFFERENTIAL CALCULUS 104 Find the second derivative = sin kx. p = \ u = 30. ?/ 31. 32. of each of the following functions. Arts. d2y
~
dx 2 k 2 sin kx. ^= cos 20. cos 2 (9. d$~
tan = v. ' 2 sec 1 tan r. j* r .r (/?" 33. =x ?/ _ *>/i J^ 35. sin L
c/ '//
i r d
 =  cos j. 2 x cos
 r 2 sin x x2  x { sin .r
. ' s* r' cos /. a/ 36. _ dx 2 x = 2 sin ~^(
<ir * ?/ .s cos x. ' r sin 2 ~ 2 ^= /. r' sin /. fc r '(3 sin 2/44 cos 2 /). <// 37. c y/ nr sin ^ ftj. r c" \ (a  />'') sin bx 4 2 a?> cos </j*" Find
38. 7, from each of the following equations.   cos (.r sin (x <Lr 40. cos // 4 In (x dn
dx //). In Problems 4150 find the value of cos ("' I 4 ~ (x 1 //) 4 (.r 4 //) 1
//) sin // for the given value of x (in radians).
41. // =x cos 42. y = x sin 43. y  In cos 44. y  ; 45. 46.
47.
48. x .r x ; j = = Ans. I. .r j :r ?/ .r ; ; t/ 49. y = 5 f 2 s in 50. = 10 c" y 10 ^ ; sin 3 = .r x ; 2. j = = 0.546. = 3.639.
1.754. ?/' 1. ?/' =
=
= ?/' TT. .r ; 1.381. ?/ 0.5. = ?/' 0.5. = sin cos 2
1.
= In Vtan
\
# = r* sin x x  2.
= 10c'cos TTX; x = 1.
?/ 1.841. y'
//' 2.  x ; = x ; = 21.35. y' = 27.00. y' ?/ 1. 3.643.
3.679. TRANSCENDENTAL FUNCTIONS
From 74. Inverse trigonometric functions.
(1) ?/ 105 the equation = 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
x (2) arc sin x equals an arc whose sine
we obtain
read //, " arc sin y (3) and arc x sin is Interchanging x and y in is //." (2), .r, called the inverse xine function of It is defined :r. of x numerically less than or equal to 1. From (1)
appears that sin x and arc sin // are inverse functions any value for and (2), it (Art. 39). Equation (,'J) is often written sin // read "the inverse sine of x." ' .r, This notation is inconvenient, for the reason that sin1.
might be read as sin s with the exponent Consider the value of y determined by x
y  arc sin (4) One value A is of// satisfying (4) .r, as thus written, in (3), \ that since sin J ~ TT r second value is y \\ TT, since sin any multi])le of
number of vain ex of these solutions Hence the function arc sin x The graph When well. is 2 = OM, y=MI> lt 150 ?,. may be added shows ,17/V, 30 = \.
To each of or subtracted. this property Ml**, , MQi, For most purposes in the calculus it is allowable
and advisable to select o'tie of the many values of y. We
TT and \ TT, that is,
select, then, the value between
?, the smallest numerical value.
(5) arc sin The J = TT, function arc sin x (6) y = arc Thus, for example, arc sin
is sin x, now
then = 0, by sin (4) ix without limit.
// satisfying
"
to be
then said
multiplevalued/' of arc sin x (see figure) x sin TT
; ; /r is, \. I TT, // = ' arc sin singlevalued, ( 1) and = \ TT. if \ir~y~\ir. In the graph we confine ourselves to the arc QOP. / The DIFFERENTIAL CALCULUS 106 manner each of the other inverse trigonometric
made singlevalued. Thus, for arc cos x, if
^ y ~ TT.
y = arc cos x, then In the same
tions be may (7) func As examples,
arc cos From = \ and (6) arc cos ( TT, ?, (1) = ) we now have arc cos \ TT, = 1) TT. rj TT. we In the graph of arc cos x (see figure),
selves to the arc QP\P.
the ( the identical relation + arc cos x = arc sin x (8) \ confine our Definitions establishing a single value for each of
other inverse trigonometric functions are given below.
75. Differentiation of arc sin u Let =
= y
v then Differentiating with respect to dv // r is a function of .r, r \ TT ( ; i y ^ by os y \ 1 XX
y // is By ^~ may this be substituted (C), Art. dx
arc sin arc QP P, and equals :r, \f of the figure.
1 1 rH , the positive si^n of tho radical bpinpf) all (arc sin v) = values of y between = =  and inclusive. . Vl  v2 The graph is the The slope becomes infinite at Q and
The function increases (y' > 0) at 0. throughout the interval x =~ 1 to or = 1. 76. Differentiation of arc cos u Let
then arc cos v; y = cos ?/. (0 g y 39 in (A), Art. 38, dx '// positive for .', = \ sin'// taken, since cos If cos TT) XIII, _
dx \ ?y. = ~ therefore Since arc sin
sin ^ TT) TRANSCENDENTAL FUNCTIONS
Differentiating with respect to y by XIV, But since a function of is r Art. 38, giving l rf?/ // = \ 1  ince sin y 1 to = \ 1 r', positive for all the plus
values of 1 (arc PQ sitfn of the radical being taken,
 // between 1 then to TT < (//' y
v (2) TT inclusive.  x increases from on page 0). Let 77. Differentiation of arc tan v.
(1) When y' of the first figure 106). y decreases from and dy
dx . , dx + j "~ (arc cos in .. = arc cos J, substituted in 04), dx sin cos"// is XXI y rfr may be this .r, . 'dx" If Art. 39 By (O, therefore iin 107 = arc tan
= tan r; then ?/. The function (1) becomes singlevalued if
we choose the hast mmcrical value of //, that
 1 TT and i TT, corresponding to arc
is, a value between
the figure. Also, when v >  oo, yy '  2 *" I when Or, symbolically,
(3) arc tan (+ oo)  \ TT, arc tan (00) Differentiating (2) with respect to y
sec~ y + v ""' = \ ' ?y of
""' 2 IT. by XV,
, dy and
Since
giving dv
v is a function of dy
c/x x, this sec
[sec XXII may
<fo 1 ^
"~ 2 By sec 2 y be substituted = _j_jg.. 2
1 + v' dx
dx
= 1 + tan'' y = 1 + '.]
dv ?y 2 *' ?/ L (C), Art. 39 in (A), Art. 38, DIFFERENTIAL CALCULUS 108
If y = arc tan x, then ?/' = 1 and the function , an increasing is function for all values of x. The function arc tan  furnishes a good
jr. example one branch fining ourselves to of the graph of arc tan  y we Con of a discontinuous function. approaches zero from the left, ?/ approaches
approaches
approaches zero from the right,
a limit. Hence the function is discontinuous when x
(Art.
can be assigned at pleasure.
value for x
see that as limit, and as jr. :r ;// 78. Differentiation of arc ctn v. TT as a 17). as
Jts Following we the method of the last article, \ f i ?r get dv ~ XXIII (arc ctn <** = i;) 1 f i^ The function is singlevalued if,
when
arc ctn v,
< < TT,
// // corresponding to the arc
()
if
Also, if v > + oo, // AB > ; arc ctn (f oc  r of the figure.  . = ) ; 79. Differentiation of arc sec v is
1. To make v is positive, choose when r is negative choose if + v if oo, ^^> Solving (1), ? arc ctn jr. That
oo; i and arc esc is, = symbolically, TT. Let v. r. defined for when Also, , // arc sec y (1) This function
tween  1 and + oo, = > // between // // ^> between TT 1 // sec all values of r
except those lying bethe function single valued (see figure), and  7T. ?/. Differentiating with respect to y
dr
sec vy tan y
T by XVII, = ; therefore . sec y tan y By
y v
(C),
'' TT (arc and  ; TT J Art Ot7
u 39 1 AB) TT (arc ; CD). TRANSCENDENTAL FUNCTIONS
Since a function of v is be substituted may z, this 109
in (A), Art. 38, giving ^
= and tan y I sec y I since tan y r, is 1 dr dx sec y tan y dx = V sec 2 positive for all y = 1 \ r' (h 1 _ v~ 1 the plus sign of the radical being taken,"!
1, values of y between and ^ and between  7T and ~ dv XXIV /.  = (arc sec v) Differentiation of arc esc
2/ then = arc esc
esc v Let v. v ; //. Differentiating with respect to y by XVIII
and following the method of the last section, we get dv XXV
The  arc
( dx function y lying between v) = = arc esc
1 and +  defined for r is
1, 1 and is all values of except those v manyvalued. To make the func tion singlevalued (see figure above), when v is positive, choose y when v is negative, choose y between between and \ TT TT and (arc AB) \TT (arc CD).
} 80. Differentiation of arc vers v Let
then : 27T y = v = vers y. arc vers Differentiating with respect to y v *
; by XIX, dv a v \
therefore By
dv (C), Art. 39 sin y * Defined only for values of r between
and 2 inclusive, and manyvalued. 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. DIFFERENTIAL CALCULUS 110
Since v is a function of K ivin dji 1 ffjin y ~ V'l  COR // ~ V]  [being taken, since sin y dv sin y dx =
~ Ax < t V2 all in (4), Art. 38. dv 1 = 2 c  versv/^ = V'J 1 positive for is be substituted may x, this r' dx r
r a the plus sign of the radical!
arid TT inclusive.J
values of y between
, dv XXVI = (arc vers v) /. PROBLEMS
Differentiate the following functions.
1. arc tan ax'2 y . <((ax \
" } dx . Solution. dx 1 f byXXn (  [r Ans. . 2. ?/ = arc sin (3 4 X H jr ).  by XX Solution. \ 1  9 .r' f 24 x  4 16 x r ' vl x' 2 ~
3. y = arc sec
*" JL ./* by XXIV Solution. x 4 ix 1 i x*  1)2 + 1\ .r^'l j I (r= 4 O2 a; __ __
4. 5. i/ y = arc cos dx a = arc sec 2! a = __!!_. Ans.
x2 + 1 a2 lu a dx xVx 2 c
. _. a2 TRANSCENDENTAL FUNCTIONS
6. 7. 8. y y ?/ =
= = As. 22 = arc ctn a ax dy arc sec  dx
</?/ arc esc 2 x. ^4a* f .r 1
__
~
Vl  <k* 111 2 .r 1 x'2 .rV4 2 ^^ 2Vx  1 1 9. arc sin y 10. 6 Vx. ~~ d.r = arc vers p 2 . N? .r = </p 11. y x arc sin 2 .r. V2 <iy
dy
 o
arc sin 1 </x 2 2 p 2  x , I Vl . 4 x2 DIFFERENTIAL CALCULUS 112 Differentiate each of the following functions. Vx. 28. arc sin 30. 34. e* arc cos x. x arc cos  01 arc ctn 2
31. 35. In arc x 0/, versa*). V / 36. x
32. arc Vx. 33. arc sec 2
29. arc tan x tan
: arc cos 37 x. arc sin 2 x. V^ Vx PROBLEMS
Sketch the following curves, and find the slope at each point where
the curve crosses the axes of coordinates.
1. 2. y In u 3. ?/ =r =
= log 5. Show 6. // ?/ 8. // 9. 10 =
=
 that = At (3, 0), (JT cos j%
j*, ?/ 3), ?/ ?/ = 0, ; . = a(r"' f c J, a
), of intersection of
1 ), xjy sin at x m = 1.
m = 0.434.
m=1
m =  J. 2 x' if +
+ (.r tan y
?/ In
In (1, 0), JT). Find the angles 7. At _
 V4  In // (1, 0), jr. In (4 4. AUK. At .r. y
y =
= In (7 In (5 then ?/" = $2
a* each of the following pairs of curves.  2 x).
 x'2 Ans. 127 53'. 109 28'. ).  cos j.
= ctn
= sin 2 x. 53 jr. 8'. Find the maximum, minimum, and inflectional points on the following
curves and draw their graphs.
ll.y = Ans. x\njr. Min. (>  V \e _ 10 In
13. 14. Min. J = In y = xe x y = (8 . ; inflectional point, (e 2 , J e 2 ). x ?/ e) (e, e) x  x 2 ). Max. (4, Min. ( In 16).
1, ); inflectional point, (
15. 2, V TRANSCENDENTAL FUNCTIONS
16. A 113 submarine telegraph cable consists of a core of copper wires
made of nonconducting material. If u* denotes the ratio with a covering of the radius of the core to the thickness of the covering, Show the speed of signaling varies as .rln it is known that that the greatest speed J is attained 17. 18. of y = when What is = JT 1 p Vf minimum the value of ac kj // + Find the maximum point and the points
J
~, and draw the curve. Ax be Anx. 2Vob. ? of inflection of the f~ Am. Max. (0, 1) points of inflection, ; ~. ( Show that maximum the p V2 Vf \ 19. graph f rectangle with one side on the .raxis
in Problem 18 has two of its which can be inscribed under the curve
vertices at the points of inflection. Find the maximum, minimum, and inflectional points
and sketch the following curves. for the range indicated,
20. y I x sin x to 2 (0 ; Am. TT). Min. (\ TT,  0.3424); max. inflectional points, (0, 0), 21. y = 2 x tan x (0 ; to (jf (TT, \ Alax. 1 ( , TT, 0.571); min. (] inflectional points, (0, 0),
22. 4 x (0 ; to Min. 1 f; , TT,  2.457); max. inflectional points, (0, 0),
i/ = 3 sin x 4 cos x TT, TT). TT, 2 (TT, 5.712);
TT). TT). AW.S. 23. (2 TT), ?r). An*. y = tan x 3.4840); TT, (J  TT, 4 (TI, 10.11); TT). (0 to 2 TT). ; Max. AUK. (2.498, 5); min. (5.G40,  5); inflectional points, (0.927, 0), (4.069, 0).
24. y
25.
26. ?/ 0= 27. y 28. y
29. = x f
= sin + cos TTJ Show (0 that the ; ; to TT). (0 sin2.r;
:r TT). (0 to 2). ; to TT). (0 to 2). maximum value of y = a sin x 4 b cos x is Va 2 4 6 2. effect of a ship's rudder is shown theoretically to be
where 6 is the angle the rudder makes with the keel, and
is the rudder most effective ?
a constant. For what value of
An*. About 55. 30. is TTJC to (0 ; = x  2 cos 2
= TTx + sin irx The turning k cos 6 sin 2
k J.r cos 2 x 6, DIFFERENTIAL CALCULUS 114
31. Into a and generating angle a conical wineglass of depth a full carefully dropped a sphere of such size as to cause the greatest
overflow. Show that the radius of the sphere is there is a a sin
sin + a. u cos 2 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.) Problem 32 if the convex surface
using the same parameter. 34. A body of weight W is dragged along a horizontal
P whose line of action makes an angle with of a force JT magnitude* of the force m denotes the coefficient of
when tan x = m.
35. If a projectile is from fired makes a constant angle
by the formula with cv (> jc ?> and 2 36. y are constants cos jr Show that the pull least is so as to strike an inclined plane which
horizontal at (>, the range is given the*
. ,,, ( and maximum giving the + friction. O[ , is (Q  cos (/ value of plane by meant
the plane. The ^ _
  A
where a given by the equation is m sin
where be of the cylinder is to 33. Solve* maximum, (v) cv the angle of elevation. Calculate the
TT +
CY.
up the plane. AHX. 9
} range* I and angle For a squareheaded screw with pitch
is given by the formula of friction </> the efficiency tan __ p tan (0 where /
is a is a constant. known constant + 0) Find the value +/ of for maximum when efficiency angle. ADDITIONAL PROBLEMS
1. The curves ?/ = jr In .r and jr // In (1 jc} intersect at the origin at another point A. Find the angle of intersection at
2. and
30'. Sketch the following curves on the same axes and find their angle of intersection. , ?/ 3. Ans. 103 .4. The line = AB at A and of AB is a minimum. is  { In ('  x
1 ), ?/ = ( In (:* .r  2
'  x
1 ). tangent to the curve whose equation
Find the coordinates of A crosses the .raxis at B. Am.
is
if 32 28'. y = e + 1
the length Ans. x (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 point on the curve.
of the curve. If we .r of // and determines a Equations (1) are called parametric equations
eliminate / from equations (1), we obtain the For example, rectangular equation of the curve. [_// and a value = rsin/ are parametric equations of the circle in the
t
being the parameter. For if we elim figure, inate t have by squaring and adding the
.>.,,.> JN
=
sin3. f , ?/ r(cos / + /) results, we 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 dydt by '' dx dt rfx
' dj J_.
dt' dx
y (A), Art. 38 by (C), Art. 39 dt that is, 04)
dt By this formula we may find the slope of a curve whose parametric
equations are given.
115 DIFFERENTIAL CALCULUS 116 ILLUSTRATIVE EXAMPLE 1. Find the equations of the tangent and normal,
and the lengths of the subtangent and subnormal to the ellipse* =a x at the point where $ = (/>, /J dx sm 77
acp COS a = # Substituting in (A),
Substituting < 45. The parameter being Solution. cos = a sin = ff =  ctn a 45" in the given equations (13),
and the slope m becomes (/> 77
a<p </>, we & cos slope at get xi = any point
\ a\ 2, y\ = m.
^ b\'2 as the point of contact, a a
in Substituting f and 1) bx Substituting in
\ b^ i a2 by) and (3) &V2I = i ILLUSTRATIVE EXAMPLE 2 = a/ \ get Art. 43, (4), \ / we V2 ab .equation of the tangent,
 b' = equation of the normal. ay f V2(ax  Art. 43, and reducing, (2), x  2 J ~ V " 2 a = length of subtangent, length of subnormal. Given the equations 2. of the cycloid f in parametric form,
f
v 4 x j
)
' l# <i( , = cos v), being the variable parameter; find the lengths
and normal at the point (.n, //i) whore = 6 of the subtangent, subnormal, } . * 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 and . ., l/  or cos 4> = a cos /> = AP = OD = Or sin
= cos = and <jf> sin = 6 sin <, </>. j Now squaring
o2 f and adding, we get % cos 2 </) + sin 2 (/> = 1, ft the rectangular equation of the ellipse.
times called the eccentric angle of the point
<j> is some P of the ellipse. described by a point on the circumference of a circle which rolls without
line is called a cycloid. Let the radius of the rolling circle be a, P
the point of contact with the fixed line X, which is called the
the tracing point, and fThe path sliding on a fixed straight M PARAMETRIC AND POLAR EQUATIONS
Solution. = Differentiating, du  a (I cos 117 ~ = a sin0. t?), du Substituting in (A), Art. 81, dy
dx When = 0i, y = y\ = sin =  1 = m = slope
. cos cos 0i), a (I , at any polnt
sin 0t m = m\ = cos 0i 1 Following Art. 43, we find (see figure at foot of tnis page) cos 7W = subtangent = 0i J NM = subnormal = a sin ) sn MP = length of normal = a\
a sin 6\
In the figure, PA
the construction for the normal (if PM 2(1  =2 cos 0,) a sin \ 0,. By ft. (5), p. 3. NM = as above. Hence
0,) = the subnormal
and tangent Ptt is as indicated. Horizontal and vertical tangents. From (4), and referring to Art. 42,
see that the values of the parameter / for the points of contact of
these tangent lines are determined thus we : Horizontal tangents
Vertical tangents solve ~ : = solve & : = f or t. for \ t. ILLUSTRATIVE EXAMPLE 3. Find the points of
contact of the horizontal arid vertical tangents to
the cardioid (see figure) '%
^  a cos = y dx
du Solution. 73 a sin = sin a( Horizontal tangent*.
cos 20 = ' 2 cos 1 + i Then sin 2 0) = cos = ;  and solving, = OM PM a J ^ 0, P  a(cos 20 = cos solving, we get
Then  sin 6 f sin in longth, then
If arc
equals
by 0,
have, denoting the angle base. We a cos 2
a sin 2 0. 01, and Vertical tangents.
cos 0,
2 sin sin 2  J
\  0, 20 = 0. cos 2 0). Substituting (using
Substituting (using 0. y = ON  NP = will touch at if the circle OM  NM = aO  a ain = a(0  sin 0),
fl/C the parametric equations of the cycloid, the angle 6 through which
the radius of the rolling circle turns = 2 ira
being the parameter. OD
is called the base of one arch of the
cycloid, and the point V is called
the vertex. Eliminating 6,
the rectangular equation we get _ AC  a (5), p. 3) 60, 180, 300. 7TM x (5), p. 3) 120", 240".  a cos  a(l  cos 6), is rolled to the left. DIFFERENTIAL CALCULUS 118
The common
nominator
(5), = root = y =
Substituting the other values in x For both numerator and de should he rejected. become zero, and the slope is indeterminate
when = 0. The point is called a cusp. in (A) From (see Art. 12). the results are as follows: (5;, 7 Horizontal tangents: points of contact ( \ a, _J a\ 3).
2 a,
Vertical tangents: points of contact (\ a, i \ </\/3 ), ( Two vertical tangents coincide, forming a
These results agree with the figure. 0). "double tangent" line. 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. \ormal 7\inynit = ,7/ = 2/f t1. Am.
=1
x = f\ y = 3 1. j 2. / 2 1 / ; / +2=
  2 = 0, .r?/ ; .r . :r (), .r ?/ yr2 = Q, x = sin
=J
=2=
rr /a,
x
=
=
3
y =
= 2 + 3  0.
=6 = f 3 t; = 1.
r = r, = cos 2
6. 0, //  3. y 35 = 0, 6,  G.r  J.  / 1 / ; 8. 9. 10. j t'\ j / '1 /'' ; 12. r / ?/ = 1 = 2/; = !.
/ p?/ m j = 15. :{ , .r 14. t ; jr 13. J!. / / ?/ /'*,  1, = ctn
=4
= 2 r' = 0.
 5 sin
a=J
= cos
 J = tan
= r=ln(/2),3?/ 11. . 4 7. 3, 9. TT. ; // X. 3, = 0,
5. Subn. Subt. + ?/4 = 0,
f y f 4  0, j* 0, // r ,'{ TT. ; ', // 3 cos n', ?/ sin 2 0, / ; a ?y ; TT. TT. ; = /; = 3.
/ In each of the following problems plot the curves and find the points
of contact of the horizontal
16. 17. JT x =3 t /\ ?/ = 3  4 sin = +
/ 0, // and vertical tangents. Horizontal tangents, none Ann. !. tangents, =4+3 cos (2, 2), ( 6. Ann. Horizontal tangents, 1), (3, vertical tangents, (7, 4),
18. 19. (a) = t~  2
=
s = + r cos 0, jc /, 1i // .y r A 12 + vertical ; 2, 0). 20. x 21. /. r siri 0. .r = sin
= cos 1, 7);
4). = sin
= sin 4 0.
0, y 2
4 (3, ( f, y t. In the following curves (figures in Chapter XXVI) find lengths of
subtangent, (b) subnormal, (c) tangent, (d) normal, at any point.
rt _ 22. m
The
, l a = curve a (cos f f .
t 4r?s. ?/ (a) 11 ctn sin / '
/), . a (sin ; /, / / (b) cos n. ?y tan /, (c) ^sin / , (d) cos f 23. PARAMETRIC AND POLAR EQUATIONS
x = a c OR
/ The hypocvcloid ' (astroid) =f
4 I// .4?js. _,
24. . r The 119 (a)  . ctn // t, ?/ ( fx~ rcos/,
= r sin /
[^/
= </( cos/
f x
\
= a (2 sin
^ , circle (Figure, p. 533) ''f sin*/. ^ tan t, (c) ~
sin (d) / ^
cos t . /. nc 25. rm. The j j cardioid . { . / // 26. J r The hyperbolic ..
P 11 (Figure, p. 533) J^jT^  COS /, sin <. ^
spiral (Figure,
h /), ./' 7., 2/= I 27. . "1+P i The folium cos2 sm J () (Figure, p. 534) ! ?/ i = L y
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) To V' = find the second derivative y", use this formula (A) again, re placing y by Then we have y'. dt ifx=/(0,asin (1), Art. 81. ILLUSTRATIVE EXAMPLE. Find y" for the cycloid
Art. 81) fx = Solution. We found y' a(d
^y a(l =
l sin (see Illustrative ^~  ^"^'
sin 2 cos 6) cos Substituting in (), lcos^) 6), and = a(1 "
^
cos __  cos ^ __L__.
(lcos(9) Ans. negative, and the curve therefore
figure for the cycloid, p. 117.
is ,j_. (lcoc0J 15  Note that y" 2, cos 0). Also, differentiating,
c^_' __ (1 Example is concave downward, as in the DIFFERENTIAL CALCULUS 120 PROBLEMS
In each of the following examples find ~ and 1. ax = tl,y = (a)x t* + _
 /<ix
; v o
L  __ . 7 y , i, P
T _
6 x
(f) x
(K) x
(h) x
(e) <_!. 2
(_ . 2 = = 2 terms of ^ /, = a cos
= b sin
= 2(1 sin 0, = 4
= sin
= sin 2
= cos 2 y = sin
/, '> t. 2. _._ _ 3 = y >)J = Ans. l. .
/^ T
^c; x 7^ in ax 2 f. / cos *. < ?/ ?. /, 2. Show 3. In each of the following examples plot the curve and find the maxi that the curve r mum, minimum, and
(a) x = 2 a ctn x = tan 0, y y /, ^, ?/ inflectional points Arts.
(b) sec Ans. Max. = 2 a sin Max.
siri ; points of inflection, => ( "V" /. min. (1, j); : 6. (0, 2 a) cos t 2 tan ^ has no point of inflection. 1, ( p>oints of inflection, ( \}\ V3,  V
4 / ,
(0, 0),
, V /V3,
, 4 / \ motion. Velocity. When the parameter / in the
(1), Art. 81, is the time, and the functions /(/) 83. Curvilineai parametric equations and </>(/) are continuous,
trace the curve or path. if /, varies continuously the point P(jr, ?/) will
then have a curvilinear motion, and We 0=0(0 *=/(0, (1) are called the equations of motion.
The velocity v of the moving point P(x, ?/) at any instant
termined by its horizontal and vertical components. The horizontal of the projection component M of P, and Hence, from (C), Art. 51, In the same way when Lay de equal to the velocity along OX
therefore the time rate of change of x. v x is s is replaced by .r, we get the vertical component vv, or time rate of change of m is is ?/, is *J.'
off rectangle, the vectors vx and v v from P as in the figure, complete the
and draw the diagonal from P. This is the required vector PARAMETRIC AND POLAR EQUATIONS
velocity v. From the figure,
by the formulas 121 magnitude and direction are given its . vx dx
~dt (4), Art. 81, we see that tan r equals the slope
at P. Therefore the direction of v lies along the tangent
path
line at P. The magnitude of the vector
velocity is called the Comparing with of the speed. 84. Curvilinear motion. Component accelerations. In treatises on it is shown that in curvilinear motion the vector acceleranot, like the vector velocity, directed along the tangent, but
toward the concave side of the path of motion. It may be resolved mechanics tion a is a into a tangential component,
at and a normal component, a n where t , _dv , a _?/2 t "~jT ~dt' 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
parallel to OX and OY, (R is /m (F) Also, , if a rectangle is = dvXm av ,  dv,,  constructed with vertex a vy then a. is the diagonal from P. P and sides ax and Hence a (G) which gives the magnitude (always positive) of the vector acceleration at any instant.
In Problem 1 below
projectile, which we make use illustrate equations of motion of a
and the preceding article. of the very well this PROBLEMS
1. Neglecting the resistance of the air, the equations of motion for a projectile are = vi cos
16.1 2
y = 0i sin
where v\ = initial velocity, = angle of projection with
horizon, and = time of flight in seconds, x and y being
x <t> t, <t> t 1 ; <f> t "of measured in feet. Find the component velocities, component accelerations, velocity, and acceleration (a) at any instant
the end of the first second, having given Vi = 100 ft. per second, ; </> (b) at = 30. DIFFERENTIAL CALCULUS 122 Find (c) the direction of motion at the end of the
rectangular equation of the path. From Solution. (C) and Also, From and (G), (F) (b) Substituting 1, t vv ; 64.4 av ; 100, t'i </>  rr = a., cos PI =v from (), = 86.6 ft.
32.2
< = vy 17.8 ft. per sec. v (speed; = 88.4 ft. per sec. When (dj v\ = 100, /, '  11" 37' the result 50 /V3, is 2. Show Problem /. 2
t =  y 50 ^ y downward. we get 32.2 ft. ft. per per (sec.) (sec.; 2
. 2
. angle of direction of motion with of motion become  16.1 /  . 32.2, direction 32.2 6 x~, a <5 \ 3 in ; 30, the equations 4> x
Eliminating ^ 8(j 1036.8 a a arc tan v* I </> 32.2 30' in these results, per sec. =  sin /*>, <p sin v\ ax = 0.
a, = r, = arc tan
(c) r
the horizontal. second; (d) the (Z>), = TJ (a) first /. parabola. that the rectangular equation of the path of the projectile
1 is = y .r  tan ^
ir ^ (1 + tan 2 2 0).r . 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 motion at the same instants.
113.1
Ans. (a) When t
2, ?v tion of (b) when
when
when / /
/ ft. per sec., v = 48.7 ft. per sec.,
= 113.1 ft. per sec., r = 15.7 ft. per sec.
r = 123.1 ft. per sec., r = 23 18',
r = 114.2 ft. per sec., r = 172 6'.
y 4, r x = 2,
= 4, v ; as in Problem 3 find the greatest height reached by
the projectile strikes the ground at the same horizontal
level from which it started, find the time of flight and the angle of impact.
4. With the data the projectile. If A 5.
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 Ans. this height?
6. If a point referred to rectangular coordinates x show that its a cos t + b and y = a sin velocity has a constant magnitude. t moves so that
f c, = 59. PARAMETRIC AND POLAR EQUATIONS
7. If the path of a moving point = .r
I the sine curve is at, b sin a/, u \ 123 show (a) that the .rcomponent 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 motion of .r = t\ // (t equation
path
rectangular coordinates,
the velocity and acceleration vectors for /  A, / (b) of the in values of the time
its speed 10 is minimum? the speed a
per second ? ft. is 1 Ans. Parabola, (a) .r + i/~ ~ 1, 1)'. Draw 1 ; (c) ( / is = a ) Find the the path with 2. / Where (d) =  (c) For what the point ; when (d) (16, 9). In uniform motion (speed constant) in a circle, show that the acceleration at any point P is constant in magnitude and directed along
the radius from P toward the center of the circle.
9. The equations of a curvilinear motion are j ~ cos 2
3 cos
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.
10. '2 /, // /. (a) 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.
, 14. =
x =
x =
x = 15. x 2 t,y= 11. 12.
13. x t 2 2 , 3 ?/ ? /'; 2 t' /, 17. /, 18. x t, 19. x 20. sin 2 t 1 = f /; a sin / y 2 cos / y /, = 2 sin
x = tan t; line. CD / ; 2 cos / y y / 3. = ; / ; / ; t ; TT. j TT. J t ; f 0. = f = cos 2
=
= ctn y ; 85. Polar coordinates. gent ; / x = a cos
x = 4 sin 16. ; ; , = 2t; = 2.
= /<; t= 1.
=
= 2.
y
3
=
y =
y /, P, , ; =
= / \ w. 1, 1 T. TV. Angle between the radius vector and the tanof a curve in polar coordinates p, 6, be Let the equation p DIFFERENTIAL CALCULUS 124 We proceed to prove the Theorem. // \f/ angle between the radius vector is the OP and the tangent line at P, then $= tan (H)
where . dd Through P and a Proof. 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
and OR = p cos A0. Also, PR
PR ~
RQ OQ OR " (2) POQ = A0, p sin __
~~ p + Ap PR = p sin A0, A0
A0 p cos 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 . , tan ^
Y (3)
v } p sin =v
hm
o p + Ap A0
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 2
p sin 1 A0
A0
A0 sin p. A0 [Dividing both numerator and denominator by A0 and factoring.] PARAMETRIC AND POLAR EQUATIONS
When A0  0, then, by Art. 68, 125 sin = lim and 1, =L lim \f\ = 0, lim sin Also, Hence the limits of numerator and denominator
p and p'. Thus (H) is proved. are, respectively, 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 p cos 3 (4) 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), = ? du  ' p cos p sin Slope of tangent
Formula OPT, ""' tan = tan (0 Then tan = 7 tan > \L r + 4 P and reducing, we have p cos 0, cos p sin 6 on page 124.
Uin
f ^) =
tan
1 figure* COS0 For, from 1he triangle Substituting tan ^ (7). P ILLUSTRATIVE EXAMPLE
slope for the cardioid p =  Solution. and p' cos the 4* p' sin p' sin = tan r (7) is easily verified for = T = &
du 0, dd a sin p' 1. Find tan ^ and the
cos 0). a(\
6. Substituting in (77) (7), tan P i (i(l
^ = \I/
t a P = tan
tan r = 2 a sin si J ((5), p. 3) + a sin 2 a sin cos \ J 0. "(1 cos 6 cos cos 2 sin 2   a(l ; cos 0) cos  cos = tanl!0. i ) sin ((5), (6), p. 3) sin 6 = angle XO7\ If the tangent
P in the figure, ^ = angle OPT =
PT is produced to cross the axis OX, forming with it the angle T, we have
 angle OPT + r.
angle XOP = 180
 180, and tan r = tan 0, as above ((3), p. 3).
Therefore r = i
At J line for the figure on page 124. In each probthe relations between the angles i/s T, and 6 should be determined by examinlem,
and drawing a figure.
ing the signs of their trigonometric functions NOTE. Formula (H) has been derived DIFFERENTIAL CALCULUS 126 To find the angle of intersection two curves C and C" whose
we may proceed as follows of (/> equations are given in polar coordinates, Angle TPT'
or <j) / n * JL tan^ (7)
' =
= : OPT  angle OPT,
1 angle  tan = Hence \[/. \//' ^r' + tan 1 tan t/r zp tan t/r' r/r and tan \{/
by
\[/'
from the equations of the curves and
evaluated for the point of intersection.
where tan are calculated (//) ILLUSTRATIVE EXAMPLE
intersection of the curves p Find the angle 2. = a sin 20, of a cos 2 0. p Solving the two equations simultaneously, Solution.
intersection, = tan 2 From the From 1, = 2(9 = 45, > i>L we get, at the point of J. the second curve, first curve, using (H),
\[/ = tan 2 J cot 20 = = for J, 22.^, ?,, for = 22\. in (/), tan The curves i ^ =  tan
Substituting = f tan are = </> shown * = f $. /. </> = arc tan Chapter XXVI. in 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
86. OT ON and length of polar subtangent, length of polar subnormal, of the curve at P. In the triangle OPT, tani// = Therefore
P (1) OT = p tan^ In the triangle
(2)
* When  p 2 ~ = length OPN, ON = tani/' = ON = dp tan increases with of polar subtangent.* Therefore length of polar subnormal. \[/ p, do
 is positive and i/' is an acute angle, as in the above figure. tip Then the subtangent OT is and looking along OP. positive When to the left of the observer.  dp is and is measured to the right negative, th^ subtangent is of an observer placed negative and is at O measured 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.
subnormal to the lemniscate = a 2 cos 2 of the polar (figure in subtangont and polai
Chapter XX YJ). Differentiating the equation of the curve, regarding p as an implicit Solution. function of Find the lengths
p~ , 6, dO
Substituting in (1) and (2), Length we dO get sub tangent of polar  = , 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
a ctn 2 Veos~2~0.
length of the polar subtangent = PROBLEMS
In the circle p 1. a sin 0, find \p and r in terms of #. A//*. In the parabola p 2. =a sec  show that r f ^  t=0,T = 20. TT. 3. Show that t/' is constant in the logarithmic spiral p
e<
Since the
tangent makes a constant angle with the radius vector, this curve is also
called the equiangular spiral. (Figure, p. 534)
l . 4. Show values of ty that tan \p =2 when =
TT in and 4 the spiral
TT. of (Figure, p.
4 Find the slopes
5. 6.
7.
8.
9. w.s'. ^^ Am. 1.
3. ; 0, 1, oo ; a sin 3 ; 11. 13. p ; = , 0, origin. 1 1. , oo 0, ; p Find 80 57' and 85 27'. 2 p = aO. 5.*J4) of the following curves at the points designated.  cos 0)', = ~p = a(l
2 a.
p = a sec
p
= a sin 4
origin.
p
2
2
origin.
p = a sin 4 = a cos 3
= acos20;
p 10. Archimedes, p 1. V3, , V3. origin.
origin. 14. p ad; 6 = ~ DIFFERENTIAL CALCULUS 128
Find the angle = 17. p cos 6
18. 19. = p a sin = p sin =
=
= 20. p 21. p
22. p 2 a, p
6, (5 =
= 25. p a(\ Show p 2(1 + = cos 2 = 28. p  a0, 29. p  p cos 0), = 0, p = 2 sin 30. p 2 sin 32. two other
y4ws. 8, 0). 16 sin 2  6(1 + cos 2 = p 60.
30. = = 2 sec = 0, 2 cos p 60. 0. cos 0). 1. 0. p0 + cos
20 = a a sec 0. y 0),
2 p , p = p cos = /> 20 = a(l cos
62 0). . esc'2  Find the lengths of the polar subtarigent, subnormal, tangent, and
of the spiral of Archimedes, p = aO. Subtangent subnormal
Find the lengths of the polar = = a, > tangent normal =  = Va
is Va +
2 2 p 2
, 2 f p
constant. . subtangent, subnormal, tangent, and
a ft
. Subtangent subnormal Show = the fact that the subnormal in the logarithmic spiral p A NS. 34. ^ a. The student should note
33. . and arc tan Ann. normal 45 if. _
3V3. points, arc tan 0). p 4, p' a(l normal at arc tan that the following pairs of curves intersect at right angles. 27. p 31. p 0; 0. + 26. p 2 sin 2 cos
cos p = sin 2 4(1  of curves. Am. 0. n
a sec . 0, 0, 23. p'2 sin 2
24. p = =
= 0, cos sin a sin 5 p = a sin 2 0.
j4w. At origin, 2 a, p 4 cos between the following pairs of intersection that the reciprocal spiral = ~> tangent = p^/lf = p In a, normal = p Vl f In
p0 = a has a constant polar tj 2 a. sub tangent.
87. Real roots of equations. Graphical methods.
the equation A value of x which satisfies
(1) /(J)=0 A called a root of the equation (or a root o//0)).
root of (1) may
be a real number or an imaginary (complex) number. Methods of
is determining real roots approximately will now be developed. PARAMETRIC AND POLAR EQUATIONS
Location and number of the FIRST METHOD. If 129 roots. the graph of /fr), that is, the locus of 0=/fr) (2) constructed, following Art. 58, the intercepts or? the xaxis are the
From the figure, therefore, we know at once the number of
roots.
roots and their approximate values.
is ILLUSTRATIVE EXAMPLE.
!'< (3)  9 j2 f 24 JT  Locate
7 = The graph has been Solution. all real roots of 0. con Problem 1. It
crosses the axis of x between
and 1.
Hence there is one real root between
these values, and then are no other real roots.
The table gives the values of /(()) and /(I showing a change of sign. structed in Art. 58, 4 ), The table of values of x plotting the graph namely,
of x. if y If not, = may and // used in locate a root exactly, for some value
the values of for X
a '// two successive values x = a,
b may have opposite signs.
x y = XQ
b The Kb) corresponding
points
P(a,f(a)), Q(b,f(b)) are, therefore, on opposite sides of the raxis, and
the graph of (2) joining these points will cross this axis. That is, a root between a and b.
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.
XQ will lie An exact 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 interpolation. This amounts to determining the intercept on the xaxis of the
chord PQ. That is, the portion of the graph joining
placed, as an approximation, by the chord.
_z
ILLUSTRATIVE EXAMPLE (CONTINUED). The root
0.4
between and 1 may be located by calculation more
closely be this between 0.3 and 0.4. See table. Let 0.3 +
root. Then, by interpolation (proportion), __ =
332 = and Q is re f(x) =y E224 + z(root) 0.3
DiflF. Q32  0.583 0.1 1.807 1.807' 0.1 Hence x 583 0.3 * P is a second approximation. This of the line that joins the points Q(0.4, 1.224) is the intercept on the xaxis and P(0.3,  0.583), which He DIFFERENTIAL CALCULUS 130
of on the graph NQ = 1.224, N are 0.3 MP =  0.583, In the figure, (3;. drawn The to scale. MC = 0.4 respectively. Also,
sides of the similar triangles and homologous
PQIt give the above proportion. M and
and the
MPC and abscissas of
z, For an algebraic equation, of which (3) is
an example, Homer's method is best adapted
to calculating a numerical root to any desired
in degree of accuracy, as explained
88. textbooks on algebra. for locating real roots. Second method *" The method of Art. 58 is of f(jr). By this graph
well adapted to constructing quickly the graph
are located and their number determined. In many exthe roots
however, the same result is attained more quickly by drawing amples, certain intersecting curves.
is following example shows The done. '" ' 10 20" I ' Id '10
1 50 ItiO' 7U M)"( )
c
H Rad ion ILLUSTRATIVE EXAMPLE. Determine the num U ber of real roots in radians) of the equation ctnjj (1) and locate the smallest
Solution.
(2)
If y
will (1) thus : the curves
ctn x on the same axes,
tersection 0, Transpose and write
x.
ctri x we draw (3) = root. be and y x the abscissas of Ihc points of in roots of (1\ For, obviously, eliminating y from (3) gives equation (1), from
which the values of x of the points of intersection are to be obtained. I'JO 200 'X. how this PARAMETRIC AND POLAR EQUATIONS
In plotting it is 131 well to lay off carefully both scales (degrees and radians ^ on OX.
.r consists of an infinite number of branches Number of solutions. The curve y  ctn
congruent to AQB the figure of 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 may locate the smallest
as we shown
find x in the table. = 0.860. of degrees, root By more we closely interpolation Ann. The Second Method may be
Transpose certain described as follows. selected terms of f(.r] so that it becomes /iWMO. (4) Plot the curves y=fiW, (5) 2/=/u(ar) on the same axes, choosing suitable scales (wot necessarily the same on
both axes). The number of points
ber of real roots of f(x) The terms of intersection of these curres equate the numand the abscissas of these points are the roots. = 0, selected in (4) can often be chosen so that one or both of the curves in (5) are standard curves.
real roots of For example, to locate the
: < .r 4 r =
{ write the equation The curves + in (5) are now
y x
r>  5 = 0, 4 x. the standard curves = A
JC' , y ~ 4 x, f> a cubical parabola and a straight line.
As a second example, consider
2 sin Write this in the Then the curves form
in (5) sin 2x41 x =
2 x = IO z 2 0. 1 ) are the standard curve = sin 2 x
= \ (** ~ l
y
y and the parabola ) method
Newton's method. Having located a root, Newton's
value.
affords a procedure to calculate its approximate
The figure shows two points
89. DIFFERENTIAL CALCULUS 132 on the graph of /(x) on opposite sides of the xaxis. Let PT be
the tangent line at P (Fig. a). The intercept a' of this line on the xo, / X x^a'\
j [ = &' =& X (,
FIG. 6 FIG. a xaxis 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 xintercept of PT. xi We find
= a,
'//i this intercept a' as follows. The coordinates of P are
is mi =f'(a). Hence the equation
/(a). The slope of PT of PT'is ((1), Art. 43) yf(a)^f'(a)(xa). 0)
Putting ?/ =
approximation and solving for x(= gives Newton's formula for a') ,aM.
Having found a' by (K), we may substitute
hand member, obtaining
,,
f a' for a in the right , <<$&
The as a second approximation.
a sequence of values process might be continued, giving , 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
x ctn x = 1 = by Newton's method.
Solution. Here /(*) = ctn x /'(x) = j,
J csc' x  2  ctn 2 x. PARAMETRIC AND POLAR EQUATIONS
the illustrative example of Art. 88, we take a By Art 88
' ' Afl>=
/'(a) = 0.855 f Also, Hence, by (K),
If we used a' = 6 = = ^0.869)* ~~ = Then, by the table 0.855. in 0.014.
2.76. Ans. 0.860. 0.873 in (K), then
b'  0.873 we found x = By interpolation
places of decimals. From  2 = 133  ~^~  The above 0.860. the figures on page 132 0.861. results are valid to three 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 This statement is, however, subject to the reservation
has no root between a and b, that is, that there is no
point of inflection on the arc PQ.
interpolation. = that f"(x) 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. .r'  3. 4. .r 3 a
.r 2 6.
7. .r 3 8< j.3 9. x 12. r4 1 3 x' 2  4 .r + 9 j2 _ 5 .r  I 4
4 14 .r 2 f 2  7  12 = 0.
^  6 x + 20
 6 x'  20
:r< 0. JT 2 2 2.21, 0.54, 1.07.  2.44,
1.53, 0.00, 3.10.
0.35, 1.88. 0.88, 1.35, 2.53. 1.49. = 0.
8 = 0.
x f 5 = +
 .07.  .f 10. x* f 8 11. A //*. 8  2 x* 4 jr .r  ^
^+ 5.  = 0.
4
+ 2 = 0.
8 x  5 = 0.
 1 = 0.
3
3 jr* + 3 = 0.
3 x 2  10 = 0. + .r x f 9 = 0.
 23 = 0. 1.71, 1.14, 3.57.  2.70, 1 .30, 2.12. 0.51, 0.71, 0.80. 2.30, 1.22.  2.10, 0.41, 2.41, 4.16. 4.00, 2.00. Determine graphically the number of real roots of each of the following
the smallest root (different from zero), using both
equations. Calculate
interpolation and Newton's formula.
13. cos
14. x +
x 0. x tan x 15. cos 2 Ans. x 0.  x = 0. One root Infinite One ; x = number root ; x = 0.739. of roots. 0.515. DIFFERENTIAL CALCULUS 134  = 0.
2 sin x
x  0.
cos x  2 x = 0.
ctn x f x = 0.
x 16. 3 sin
17. Ans. Three roots x Two
Two 2 2 18. 2 19. 21. sin x 22. cos x f cos 23. c' 24. tan x 25. ( ' f j* 3 J = 77 A f x 1.237. 0.511.
0. number of roots ; x 1.29. roots ; x 3.65. One = root ; x = 0.792. = 0.314.
= 0.517.
x = 1.44. Infinite number of roots ; x 2x^0. Infinite number of roots ; .r 0. 2 c* = 3.032. number of Infinite number of roots ; .r inner radius 0. and outer radius (R) (r) steamer transmitting N revolutions per minute satisfy
= 2500, N = 160, r = 6, find R.
30. ; ; Infinite 0. cos .r root = roots Infinite 0. steel driving shaft of a of One 0. root; x = x ; One 0. cos 2 x The 29.  x 27. 2 sin = log x .r 26. sin 3 28. tan 1 = x ; = 2.279.
= 1.404.
x
x = 0.635.
x number of Three roots 1 x ; roots Infinite 2 x  x
0.
+ x  = 0. 20. 2 sin roots ; 77 in inches of the hollow
horse power at a speed the relation 7i 4 r4 = ^ Nir " If cylindrical shell with a hemispherical end has a diameter d in.,
cu. in. The length of the cylindrical part is // in. Show and contains V
that (P + 3 hd'2 = ^^  Given // = 20, V = 800, find d. Ans. d = 6.77 7T 31. The quantity width 7> ft. where 77 is Given Q is of water Q cu. ft. per second flowing over a weir of
given by Francis's formula the height of the water (the head) above the crest of the weir.
12.5, H= 3, find (Solve the formula for the factor 77^ and 77. then plot.)
32. If perature V
7" Ans.
cu. F. ft. is the volume of P and pressure V=
Given
33. V= 2,8, The chord T= r of 420, an arc .s Ib. 1 Ib. of per square inch, 0.6490 ^~~4?.
P
PS in a circle of radius r is given approximately by the formula
24
If r = 4 ft., c = 5.60 ft., find 2 s. Am. * A find P.  s II = 1.23. superheated steam at a tem = 6.23 ft. PARAMETRIC AND POLAR EQUATIONS
The 34. area u of a circular segment whose arc s subtends the central
2
is u = % r' (x
sin x)
Find the value of x if r = 8 in. angle x (in radians)
and u = 64 sq. in. Ans. x The volume V 35. of 135 of a spherical CD = h is
V= ir(rh'2  i/? 3
= 4 ft., V = 150 cu. ft. = 2.554 radians. segment one base of height ). Find h if r Ans. h =4.32 The volume V 36. R and radius Derive this of a spherical shell of thickness result. If ft. / is R= 4 ft. and V is one half the volume of a solid sphere of equal
Ans. t = 0.827 ft,
radius, find t. A 37.
in wooden sphere solid water to a depth (See Problem 35.) of specific gravity Let x h. Find x for a S and diameter d sinks  and show that 2 maple ball for which S :{ \\ .r x f 0. tf 0.786. Ans. 0.702.
cos
for which the curves p
38. Find the smallest positive value of
e~ e intersect. Find the angle of intersection at this point.
and p
= 1. 29 radians; 29
Ans. = (> . ADDITIONAL PROBLEMS
1. Find the angle of intersection of the curves p = 2 cos and p at the point of intersection farthest from the origin.
0.54 radian
Ans. Point of intersection is = 2. Show that the curve p = a sin 4 \ cuts itself 75 ; ~ e 56'. at right angles. radius vector of the cardioid p = a(l + cos 0) is OP. From the
C of the circle p = a cos a radius of the circle CQ is drawn parallel
center
normal to the cardioid.
to OP and in the same direction. Prove that PQ is
3. Any A square, one of whose diagonals
a(l
cumscribed about the cardioid p
4. fi(2
5. = + V3)o 2 along the polar axis, is cirShow that its area is
0). cos . The path moves lies  of a particle is the ellipse p = gp
i. *~ C The particle COS u time rate.
so that the radius vector p describes area at a constant
the major axis.
of the velocities of the particle at the ends of Find the ratio A
Ans. 1  1 + e CHAPTER IX
DIFFERENTIALS
90. Introduction. Thus
y=f(x) by the notation far we have represented the ^f'fcr)
J(X) dx~ We have taken derivative of ' 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 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. as 91. Definitions. value of JT, and Ax differential of J(x), is the derivative of f(x} for a particular
an arbitrarily chosen increment of x, then the
denoted by the symbol df(s), is defined by the If f'(x)
is equation d/(x)= (A)
If now /(;r) = :r, then f'(x) =
dx 1, = and (4) reduces to
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)
* On account of the position
called the differential coefficient dy=f'(x)dx*=^dx.
dx
which the derivative 136 f'(x) here occupies, it is sometimes DIFFERENTIALS
differential of a function equals
differential of the independent variable. The its 137 derivative multiplied by the Let us illustrate what this means geometrically. Draw the curve y =/(x).
Let/'(.r) be the value of the derivative at P. Take dx = ~4 PQ, then dy = f'(x)dx = tan r  PQ  ^
y Therefore dy, or df(x), is the increment (=
tangent corresponding to PQ = QT. t QT) of the ordinal e of the dr. This gives the following interpretation of the derivative as a
fraction. // an arbitrarily chosen increment of the independent variable j/or
P (x, y) on the curve y = f('jc) be denoted by dx, then in the a point derivative dy denotes the corresponding increment of the ordinate of the tangent line at P. especially that the differential (= dy)
the increment (= A?/) of the function corresponding to the same
and
value of d'jr(=A'jr) are not in general equal. For, in the figure, The student should note = QT, but = QP'.
differentials. From
Approximation of increments by means of
Art. 91 it is clear that Ay(= QP' in the figure; and r///( QT) are
When only an apapproximately equal when dx(= PQ) is small.
value of the increment of a function is desired, it is usually
proximate
easier to calculate the value of the corresponding differential and use dy A'# 92. this value. ILLUSTRATIVE EXAMPLE 1 Find the volume approximately
in.
diameter 10 in. and thickness
. Solution. of a spherical shell ^ of outside y The volume of a sphere of diameter x is V=l*x\ (1, is the difference AV between the
Obviously, the exact
Since
volumes of two solid spheres with diameters 10 in. and 9J in. respectively.
an approximate value of AV is required, we find dV. From (1) and (),
only volume dV =\ of the shell 7TX 2 dx, since = 2 ^ 2
 DIFFERENTIAL CALCULUS 138  J, we obtain dV  19.63 cu. in., approximately,
Substituting x = 10, dx =
neglecting the sign, which merely means that V decreases as x decreases. The
exact value is AVr ~ 19.4 cu. in. Note that the approximation is close, for dx is
relatively Kmall that is, small a* compared with x (= 10*. The method would be
t worthless otherwise. ILLUSTRATIVE EXAMPLE 2.
given tan 45 = 1, sec 45 Calculate tan 4G', approximately, using differen\ 7/2. 1 = 0.01745 radians. = tials, Let y tan x. x changes to x + Solution. When  substitute x
y f \ (45"), IT ~ dx Then, by (J3), dx, y will change to y = Then dy 0.0175. = f In
tan 45 dy, approximately. 0.0350. Since y  (1), = 1, approximately. Anx.
(Fourplace tables give tan 46 = 1.0355.)
1.0350 dy tan 46 ", r 93. when Small errors. A second application of differentials
small errors in calculation are to be determined. is afforded The diameter of a circle is found by measurement
1.
with a maximum error of 0.05 in. Find the approximate maximum
the area when calculated by the formula ILLUSTRATIVE EXAMPLE
to be 5.2
error in in., A (1) \ TTX". (x diameter) 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
Solution. dA Relative i irxdx  and percentage
u = 100 (3) may relative error x 5.2 x 0.05  0.41 sq. errors.
= (2)
v
' The I TT If di( is in. Ans. the error in u, then the ratio the relative error ; the percentage error. be found directly by logarithmic differen tiation (Art. 66). ILLUSTRATIVE EXAMPLE Find the relative and percentage errors in the 2. preceding example.
Solution. Taking natural logarithms
log ~.~ ,. 1 ..  5.2, dx Relative error in The log
2 .4 in (1),
\ TT
, an = 0.05, we
= 0.0192
; f 2 log x.  dA = 2dx
T~
A
x find percentage error = errors in calculation considered here are lfifa %. Ans. due to small errors upon which the calculation is based. The latter may
from lack of precision in the measurements or from other causes. in the data
arise dA "TT"^"'
A dx, x Differentiating, Substituting x A DIFFERENTIALS 139 PROBLEMS
1. If A the area of a square of side
and A A. is Draw dA. x, find 2. What for the area of a circular ring of
the exact formula ? is = Ans. dA of = Find an approximate formula radius r and width dr. 3. a figure show2 x dx. Aus. dA ing the square, dA, What is the approximate edge 6 2 error in the TIT dr ; AA = ir(2 r volume and surface + Ar)Ar. of a cube an error of 0.02 in. is made in measuring the edge?
Ans. Volume,
2.16 cu. in.
1.44
surface, in. if sq. in. ; and volume of a sphere are S = 4 rrr 2
3
If the radius is found to be 3 in. by measuring, (a) what
and V
Trr
J
is the approximate maximum error in S and V if measurements are accurate 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.;
4. The formulas for the surface . ; (b) s, I
5. Show by means
x ; <;'< v, 1 ;,. of differentials that L (approximately).
x2 x f djc Find an approximate formula for the volume of a thin cylindrical
with open ends if the radius is r, the length h, and the thickness L
Ans. 2 irrht. 6. shell A box is to be constructed in the form of a cube to hold 1000 cu. ft.
accurately must the inner edge be made so that the volume will
be correct to within 3 cu. ft.?
Ans. Error ^ 0.01 ft.
7. How 8. If what is ?/ = a: 5 and the possible error in measuring x the possible error in the value of Use ?/? differentials to find 0.9 Am. proximate values of (27.9)^ and (26.1)". Use is an approximate value when x = 27, this result to obtain ap 0.2; 9.2; 8.8. of each of the following expressions.
9. V66. 11. ^120. 15. \/35. 13. T^. A 1 __
/ 10. V98. 17. If In 12. 10 \^1010. 14. 2.303, approximate In 10.2 7='   V51 by means of differentials. Ans. 2.323.
18. If e 2 = Given 7.39, approximate e2  1 by means = = of differentials. and Ans. 8.13. = 0.01745 radians,
0.5,
0.86603, cos 60
use differentials to compute the values of each of the following functions
to four decimals
(d) cos 58.
(b) cos 61
(c) sin 59
(a) sin 62
19. sin 60 : ; Ans. (a) 0.8835; 1 ; ; fb) 0.4849; fc) 0.8573; fd) 0.5302. DIFFERENTIAL CALCULUS 140 The time 20.
t* , where t of
is one vibration measured in of a pendulum seconds, g = is given by the formula and 32.2, /, the length of the V pendulum, is measured 0.01 ft. (c) ; Find in feet. the length of a (a) ; vibrat (b) the ; ; How 21. pendulum change in / if the pendulum in (a) is lengthened
how much a clock with this error would lose or gain in a day.
 2 min. 12 sec.
An*, (a) 3.26 ft.
(c)
(b) 0.00153 sec.
must the diameter of a circle be measured in order
exactly ing once a second Arts. Error ^
that the area shall be correct to within 1 per cent?
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.
' ; . that the relative error in the ?rth power of a number is
n times the relative error in the number.
 times
24. Show that the relative error in the nth root of a number is Show 23. the relative error in the number. When block of metal is heated, each edge increases
increase in temperature. Show that the surface
r^ per cent per degree
increases fa per cent per degree, and that the volume increases f\ per cent
25. a cubical per degree. Formulas for finding" the differentials of functions. Since the
differential of a function is its derivative multiplied by the differential of the independent variable, it follows at once that the formulas
94. for finding differentials are the given in Arts. 29 and 60,
This gives if d(c ) I II d(j) d(u III f v same as those for finding derivatives we multiply each one by  IV d(r/) V d('ur) 0. dy. + dv du ?r) = u dv d(v Via d(jr") = wn ~
= nj? n  rf(") = ) dw. fd/>. VI n dx. +
} } v du. dv. dx. vn
Vila dp X
XI
XI
XII
XIII yv a = a In a dv.
= e*dv.
= vu ~ du + In u
= cos vdv.
r ! d(a ') d(e*) r d(w.") d(sin r) } ur dv. DIFFERENTIALS
XIV d(cos d(tan = sin vdv.
= sec rdv. Etc.
= ^L==. Etc. r) XV r) XX 2 d(arcsinr) The term 141 "differentiation" also includes the operation of finding differentials. In finding differentials the easiest way is to find the derivative as
and then multiply the result by dx. usual, ILLUSTRATIVE EXAMPLE Solution. dy Find the 1. +* }
+ 3/ X = d( a \ar
2 ( x' + 3 )dx + 3 )2 (x
. (x 2 + (te ( + x* (  3 3) 2 h' 2 6 2 x dx Find 2. ~ x2  : G j:  a~y~
2 2 a' y dy 3. 2 P'ind p~ 2 p Solution. dp j 4 Ans. a/.
y dp from =
= op =
, .". . 0. <*' ILLUSTRATIVE EXAMPLE ) 2 2 a' b' bjc , /. aij J from </// ~
~ 1 . 3) (x f ILLUSTRATIVE EXAMPLE
Solution. .r differential of a'2 cos 2
2 a' #. sin 202 2 a' sin 2 , <i#. n
(/. >inK. P ILLUSTRATIVE EXAMPLE
Solution. d[arc sin (3 4. /  Find d[arc sin
4 (3 /  4 /')_). :< / ) \ 1  (3 /  \lr 4 /')' 4 PROBLEMS
Find each
1. y of the following differentials. x* 3 x. ax 4/00
3. T/ Vax y = xv a f 6. f 6 , . 2 x 7 2 . ay _
= C/I2
'*
' __ O A
r2 tt a2 ds abc ht i j
aw = dv 6. s = ae
u = In rr. 7. p sin ad. dp = a cos aO d6. 5. bt . dt. v DIFFERENTIAL CALCULUS 142 = y 8. cos = 10. ,4ns. In sin x. p 9. r f cos 10
lo. 19. If = VV // 2 Find
20. 2 + 21. x 22. x 23. Vx ft. ment :{ ?r 6 x/r Vx// 4 = .r?/ V?/ Trsin TT/ of each of the following functions. show that a, .r, ?/ = and dx from each //, = 2 f 4 f/?/ 0.1 ft. of the following equations. __ 20. 4 2 = 10.
= a.
4
= Va.
//" x: // 12 24. 25. x ft. ' f  /' = ?/ 4 3 ?/)dx
3x48//
jr r j4 //) ^.  cos (x are found by measurement The maximum respectively. (4 = a\ 26. sin (x legs of a right triangle and 21.4 is r'(cos 0sin 8)(W. f 1. terms of f 3 f 4 The 27. 14.5 f in JT'~  r==r
/
Va~ ~ x ,,, jr ctn (In TT/. Find the differential 12. v = </?/ f/p (cos 0. 4 ?/). to be error in euch measure maximum Find the error in degrees in calculating the
using the formula for the tangent of angle opposite the smaller side by
that angle. Let s be
95. Differential of the arc in rectangular coordinates.
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 apAy arc From
(1) (Chord PQ) ^ the figure,
2 = (A*) 2 + 2 (A?/) . 2
Multiplying and dhdding by (As) in the lefthand
2
we get
dividing both members by (Ax)
, (2) /Chord
V
As 2
2 2
\ /A8\ _
PQV / As V_
~ / VAx/ member and DIFFERENTIALS Now let Q approach P 143 as a limiting position then ; Ax > and we have
(3) \dx Multiplying both members by dr%
(C) Or, by = ds 2 we if dx 2 we + dy get the result 2
. extract the square root in (3) and multiply both dr, members i From we may (C), show readily also that
i All these From forms are useful. (>), since IT = sec T
we obtain
we may easily prove
r/,s assuming the angle r to be acute. djc, dx
(F) sec*"7% = ds cos r,
^ tan du
~
ds = T COB r sin r. = sin r. f/8 For later reference, we add Hence I the formulas, setting y f =
J 1 i COST (G) j>
2 sinr y
2 = 2 (1 (1 f y' )" + y p 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 accompanying figure, I'Q = A:r = rfy,
tangent at P, and r is acute. Angle In the FT is
PQT is a right angle. Therefore Then Q T = tan
/> 7 1 r rfj = W/r + = r/y/. 2 rf?/ By Art.  rfs. By 91 (C) The figure will help in memorizing the relations above.
The assumption made at the beginning of this article
in Art. 99. is proved DIFFERENTIAL CALCULUS 144 From 96. Differential of the arc in polar coordinates. = j (1) p cos y 0, = p sin 6 between the rectangular and polar coordinates
by V, XIII, XIV, of Art. 94,
dx (2) = cos p sin dp dy d6, the relations = sin of a point, we + p cos dO. dp obtain, Substituting in (C), Art. 95, reducing, and extracting the square
we obtain the result root, = ds (H)
This may Vd/> 2 + p2 d0 2 . be written
/^*\2 r o dO. The figure is drawn so that the angle
OP and the tan \f/ between the radius vector PT gent line is and Sp (= OP'
the for p acute (Art. 85). Also p, SO,
 op) are positive. Take But Then
PQT, take variable. independent Sp = dp. In the right
PQ = dp. Then QT = triangle tan
/ tan ^ ^ = r/p. p 7 P = Therefore PT = hence ILLUSTRATIVE EXAMPLE
x2 + y* = 1. pp
dp
Vrfp + Find the differential aj: find To find ds in terms of y rfs in terms of we a* ample  2, the of arc of the circle sin 0), y = a(l y substitute in (/)), giving we ILLUSTRATIVE EXAMPLE
a (6 rfs. ^ = _. Differentiating, To = = ra. Solution. x L>  substitute in (), giving 2. Find the differential cos 0), in terms of and of the arc of the cycloid
(See Illustrative Ex dd. Art. 81.) Solution. Differentiating, dx = a(l  cos 6)d6, dy = a sin dd, DIFFERENTIALS 145 Substituting in (C)  a" (I Art. 2, ds'2 From (5),  cos 6}~d6" f a
1  cos 6 = 2 sin ILLUSTRATIVE EXAMPLE
p = a(l 3. cos 0) in terms of  sin.\ dO* 0. Hence 2 a(l ds Find the differential =  cos 0}dQ'2 2 a sin J . A?ks. <J0. the arc of the cardioid of 6. du
( Solution. 2 Differentiating, = a sin 6. Substituting in (/) gives
cte = LaHlcos0j + a 2 sin 0]adt? = a^ PROBLEMS
For each
1. 7/ = 2 y 2. = 2 of the following curves find
.r 2 px. f a2 c
6 x?y = a* 5. ?y ~ 6. a 2 // >, 4. 2 7. a?/
8. 4  a 26 2 o , ; ds jr. For each = = (^
 = f = sec j 2 ?/ // = sin 11. Va. dr.  \
\ 10. . dr. j* dr. ^ 9. V/y f  d^ f 3. :? a* and jr = Vl + ds . // =  jr djc. J cos 2 x JT\ 2 V.r 2
// In sec = of ds Ans. . x2 3. & terms in ds (/.s 2 of the following curves find ds in c f . jr. terms of ?/ and dy. , :j?y J 14. x^ 42
15. a ?y 16. t 2
?y ?y^ =  x3 2 19. =2
x = 3 x a*. f 3, f / 2
, ^J~ d?/. .  .r <is = 3 ?/ = 17. 2 x// 2 0.  y of the following curves find ds, sin r, For each
and dt.
18. = J ?y y = t 2 2 P. 2. 20. x 21. x 2 0. and cos a sin = 4 = 4 cos /, f, y
?/ r in terms of a cos J. 3 sin /. DIFFERENTIAL CALCULUS 146
For each of the following curves find dx in a cos 22. p 23. p ~ 24. p = 1 25. 27. = + 1 p^sec 2 28. p 29. p =
=  2 2 4 a dO. dn <& 0. sin 0. 4 cos 0. = ds 12 sin 4 3 sin p 26. p  A tin. 0. 5 cos terms of 6 and dO. = V2 a cos 30. p
31. p 0. ^Z = 32. p cos 13 d8. = 0. 3 sin 33. p 34. p 0. e/0. 0. { 4 sin f4 +
' 1
1 cos 2 sin  cos i 4
: 3 cos i 4 =
1  3 cos 6 97. Velocity as the timerate of change of arc. In the discussion of
curvilinear motion in Art. 83, the velocity, or, more correctly, the speed v was given by (),
2 (1) By (C) = r,  fl and (D) in Art. 83, rx *+7y>.
> a/ Substituting in (1), r, = d
%a/ using differentials and (C), Art. 95, the result Extracting the square root, taking the positive sign,
V = is we have ds 3T Hence, in curvilinear motion the speed of the moving point is the
timerate of change of the length of arc of the path. This statement should be compared with the definition of velocity
motion as the timerate of change of distance (Art. 51). in rectilinear 98. Differentials as infinitesimals. In applied mathematics differis, as vari entials are often treated as infinitesimals (Art. 20), that ables approaching zero as a limit. Conversely, relations between
which these are replaced
"
involved here is
"principle of replacement infinitesimals are frequently established in by differentials. very useful. The DIFFERENTIALS
If x is the independent variable, 147 we have seen that = dr, and
any equation.
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 thus AJ may be replaced by in djc A.r If A.r > lim^l. (1) JLXO W?/ Proof. \im=& Since AX ^ we may o =/'(?), A.r ^= write /'(.r) AT + /, lim if i = 0. .0 Aj Clearing of fractions, and using (B),
A// = Dividing both members by A//, _ dy dy + i and transposing, the 1 i lim
Ar We now state, , ^= 1 , result is AJ  ^ _^ A?/ Hence A.r. A// or also A'// ^= lim
Ax0"?/ 1 Q. E. D. without proof, the Replacement Theorem. //? problems involving only the ratios of infinitesimals 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 theorem is simple above in application. ILLUSTRATIVE EXAMPLE 1. By (5), p. 3, if x  \ i, 1  cos i = 2 sin 2 J i. Let
2
{ be an infinitesimal.
Then, by (B), Art. 68, sin i may be replaced by i, sin J i
2
cos i sin i) may be replaced
Also tan i (
cos i by J i 2
by ^ i and therefore 1
. , by i. ILLUSTRATIVE Kx AMPLE
ultimately, since A.r Then In (1), equation By is all quantities are infinitesimals homogeneous, each term being of the the theorem,
arc ; 2 (1) Art. 95, we may replace the infinitesimals as follows
 As, and As by ds A// by ety and Ax by dx.
PQ
PQ by
becomes ds 2  dx + d?y that is, (C). second degree. Chord 2. 0. The , ; : DIFFERENTIAL CALCULUS 148 Order of 99. Differentials of higher order. Let i and
which simultaneously approach zero, and let infinitesimals. ; be infinitesimals lim 4 = L. i If
If
If L is not zero, and j are said to be of the same order.
L = 0, j is said to be of higher order than i.
L becomes infinite, j is said to be of lower order than i.
? Let L Then 1. i is j of higher order than =
=
[hm(^) Iim^l) hmj The converse
from also is In this case (L true. 1 i. =
O.j = 1), j is said to differ by an infinitesimal of higher order. i For example, dy and AJ are of the same order if f'(jr] neither vanishes nor
 dy is of higher
becomes infinite. Then A// and AJ are of the same order, but A/y
A.r. For this reason dij is called the "principal part of A?/." Obviously
order than
i.
powers of an infinitesimal are of higher order than
i of Art. 95, ILLUSTRATIVE EXAMPLE. Prove the assumption Solution. In the figure chord PQ Therefore, by 1 < <' arc we have, by geometry, PQ < PT + division, " Now PQ chord chord chord PQ = PQ
sec PQ
PT = sec chord
</> A.r, r AJ, PT and hence
lim TQ. chord PQ chord sec <' ^limj Differentiate of higher order. \
^ Let = PQ / . Jim =/(j). // dy, Ax PQ TQ /_Z^_\ = lf TQ = Ay  dy
A// \ ^rc =L The equation defines the second differential of //. If //" neither vanishes nor becomes
2
dy is of the same order as Ax and therefore of higher order infinite, than dy. In a similar manner d*y,   , d ny may be defined. PROBLEM
In triangle ABC the sides approach zero, and r is of a, b, c are infinitesimals which simultaneously higher order than b. Prove lim r = ! CHAPTER X
CURVATURE. RADIUS AND CIRCLE OF CURVATURE
In Art. 55 the direction of bending of a curve 100. Curvature. was discussed. The shape of a curve at a point (its ilatness or sharpness) depends upon the rate of change of direction. This rate is called
the curvature at the point and is denoted by K. Let us find the
mathematical expression for K.
In the figure, P is a second point on
a curve near P. When the point of conf 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 We now set
gent line.
ing definitions.
^r down the follow = 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 K= (A) lim
AS. = = oAs In formal terms the curvature
respect to the arc curvature at P. ds
is the rate of change of the inclination (compare Art. 50). with
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 same at
of the radius, and is therefore
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
CP and CP'. Hence
As AT ~
As angle As _. = 1, ~As~ R 149 the reciprocal DIFFERENTIAL CALCULUS 150 PCP' since the angle measured in radians. is That curvature of the arc PP' is equal to a constant.
have the result stated in the theorem. From the average is, Letting As > we 0, the standpoint of curvature, the circle is the simplest curve,
Obviously, the curvature of a since a circle bends at a uniform rate. is everywhere zero.
Formulas for curvature rectangular coordinates straight line
102. ; When Theorem. the equation of a curve is given in rectangular co ordinates, then = /; (B) where y' and with respect //" to __,, w>/w/?/W//, the are, and second Jirst derivative* of y jr.
( Proof. Since arc tun r ^)
dx' (// //, \ differentiating, we have
'y _/>!_ (Lc (1 1  Y (1 l ) P>y XXII, Art. . 60 //'" Hut
() Dividing by (1) EXKIUUSK. A //''' l>.v (o>, Art. 95 gives (B). (2) If ?/ is h Q.E.D. the indo])endent variable, show that (O "*"  A' , 2
(1 4 *' )* j and jc" are, respectively, the first and second derivatives
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' bewhere of j .r comes infinite, that is, when the tangent
.r'  at K and P is vertical. Then in (C) .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 negative according as the curve is concave upward or concave downward.
ILLUSTKATIVIO EXAMPLE 1. Find the
the point (1, '2); (Iri at the vertex. curvature of the parabola y 2 =4x (a) at > Solution. ii' = < v ' " = (a)
" = When
8 \2~ x = 1 and 0.177. // ^ 2, Hence f1 /'>\ ^(:) = '7 at then
(1, 2) //' = !, ,/' .rJK.. dx\ij/ // y y" the curve 2 =L
is Substituting in 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 //' " I ' ( ILLUSTRATIVE EXAMPLE
x
Solution. Example 2 for the cycloid (see Art. 81) 2, (! cos 0). // we found Art. 81,
sin _. , (C). 1 2 sin 0], a(0 In Illustrative K Find 2. r 1 I Jll
2 dy ' 2 Hence use infinite. cos 6 .1 >) Hence +?/' 1 cos T
1 Example, Art. Also, in the Illustrative was shown that 82, it _ 1 ~ ' c(} (>(>s ~ 0) Substituting in (B),
1 2 a\ 1 4 a sin 2 cos 2 \ From equation 103. Special formula for parametric equations.
Art. 81, we have, by differentiation,
djr dy
dtdf2 f , dy (lj ~j dtdl'2 "^^ ' (ft) Whence, using
reducing, (B), Art. 82, and substituting in (B), Art. 102, and we obtain where the accents indicate derivatives with respect to
, 2 dx '?' . w* Ti d x x ,, di/ , II
'* t ** lj' ,, 11
'
i dt 2 dt /, ; that is, ~ dy
~
7i'> dt~ di Formula (D) is convenient, but it is often better to proceed as
as in Art. 81, '//" as
Example 2, Art. 102, finding
Art. 82, and substituting directly in (B).
Illustrative 104. Formula Theorem.
(E) //' for curvature When ; in in polar coordinates the equation of a curve is given in polar coordinates, K= p2
P ^^
. 2 P' 2 PP"
*f, (P 2 + P' 2 2 )" DIFFERENTIAL CALCULUS 152
where and p" p' with respect By Proof. are, respectively, the first and second derivatives of p to 0. Art. 85, (/), r = + ^. Hence <> s'+Sby Also, ^= Art. 85, (#), arc tan Hence
Then, by Dividing
at (1), (2) by gives (). (3) ILLUSTRATIVE EXAMPLE.
any point.  ( t~ Solution. Substituting in p' = Q.E.D. Find the curvature of the logarithmic spiral p = ac at) ap ^= ; </>"" p" = = e" 9 a p. (),
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 (),
( Jf (F) =I == (l2l.
y ILLUSTRATIVE EXAMPLE. Find the radius
_ / catenary y = \r a f r 2 / . ' ?/ = I\
2 1 fl/' 2 ^ 1 (figure in  / a
Solution. of curvature at any point of the i\ , ( ._ ,, \
uj +7\f"f Chapter XXVI). . y / " / = J_ \
2 a =Mfa+f 106. Railroad or transition curves. r ffl ^_ tf \ a) = JL.
a =  Ans. a In laying out the curves on a
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
railroad it will RADIUS AND CIRCLE OF CURVATURE
circular track. 153 This curve should have zero curvature at its point
and the curvature of the circular
joins the latter. Arcs of cubical parabolas are generally of junction with the straight track track where it employed as transition curves.
ILLUSTRATIVE EXAMPLE. The transition curve on a railway track has the shape
an arc of the cubical parabola // = J ,r
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)?
l of . ^= ^ = ^.
dx Solution. A" Substituting in (B), =
(1 (a) At (3, 9), 2x. dx 2 f radian per mile K= = 28' per mile. radian per mile = 3 ATM. (812)2 (b) At (2, 5), K = ~ Ans. 16' per mile. (17)3
(c) At (1 , I \ K = ^ = U radian
(2)2 per mile  40" 30' per mile. Am. V2 Consider any point /' on the curve (\
drawn to the curve at P has the same slope as the 107. Circle of curvature. The tangent line curve at itself P (Art. 42). In an analogous 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 /0 / ', To do this, proceed as follows. Draw
point.
the normal to the curve at P on the concave
side of the curve. Lay off on this normal the distance PC = radius of curvature the circle passing through P. (= K) at P. With The curvature \ c as a center draw of this circle is then **
which also equals the curvature
so constructed is of the curve itself at P. The called the circle of curvature for the point circle 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. DIFFERENTIAL CALCULUS 154 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
in given Art. 28. ILLUSTRATIVE EXAMPLE. Find the
equilateral hyperbola xy dy
ax * i Solution. For W. K . (U, 4 12, ) circle of curvature. corresponding ~ ra on the
and draw the dius of curvature at the point d^y
~~l
dx  4). ix 2 v
*.  jr* dx'2 .'J 9 Jl^lLi*j=5..v
4
M The curvature crosses the curve at two points. circle of iLLi'STHATivE K.VAMi'LE Find 2. at (2, /i jj
Solution. I )iiferent iiitui^, regarding
2 Differentiating again, regarding .r 1 4 4 //' }  Substituting the given values Hence, by 2 // 1 ij and  2 JTIJ' //' .r ?  //' hyperbola
10. // \/5.  we ////'  1,  //)//' we find get (). we as implicit functions of x, f 2(.r 2, 2 us an implicit function of x, // .//"' for the )  4 /// f 2 U  (F), The method f get 0. ?/'  2, l y" . .\.s. and //' as implicit functions
example (namely, regarding
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.
of this // PROBLEMS
Find the radius of curvature for each of the following curves at the
)raw the curve and the corresponding circle of curvature. point indicated.
1. 2 2. 6 // // L> 3. ?/ 4. .r = 7. 2 .r r i(), An*. 0). (2, ; ; .',). 7i ft =
= 1.
g Vs. ; (1, 1). ; .r j ; ^ TT, 1). (0, 1). ; 4
=9
= ^f8;
2 // 2
// j ; sin // 6. .r =
~ .r // 5.  I ; (5, 2). (1,8). 8. y 9. y = 2 sin 2 x
= tanx; (
; ( J TT, TT, 1). 2). RADIUS AND CIRCLE OF CURVATURE
Calculate the radius of curvature at any point
the following curves. = ^. 10.,, 155
on each y\] (.r\, of K= A,:,. (> = 2 11. jr' 12. Wx*  13. bjr f 14. + ,* JU.
* aW  a*b*.
2 2
a' b' . ay ** = *. /, = ^' +/>'
a* +
=r 15. j3 16. x 17. ?y 2/ = aI A* V2 arc vers  rz/ 8 x A* the point of contact of the tangent line moves along the curve a distance As approximately, will the tangent line turn?
19. Use inclination of the curve 27 A.s sec { at jv to the parabola
through what angle, (Use differentials.
.r // ~ n/i. the point ) 1) is .\(IJ, 45. approximately the inclination of the curve at the
on the curve such that the distance along the curve from A to B H = 0.2 units. The circle p = a sin Archimedes p The spiral of 22. The cardioid p = a (I 23. 24. = The parabola p
The curve p 26. m
The = 3 J cos 2 a sec = a sin tnsectnx p  =a The lemniscate p 2 25. An*. 0. 2 R of \ a. equilateral hyperbola p oo TU
The conic p 1 e 0. K= (Fig., p. 5:M) #= (Fig., p. 5153;
(Fig., p. A' 5.'52) e2)
' cos = ^ _^ 3^ a 2 ^ Pi a2 ^^ f>:J7) J ^= (Fig., p. R = 2a sec 3 0. 2 a cos  ad. 0). J 0. The =  a(l  cos 2 27.  (M on each (pi, 3 21. 28. 2\/2 (2, 4) at
().l, Calculate the radius of curvature at any point
the following curves.
20. 3 (an //,)*. differentials to find point
is The = /i z/~. In sec x. 18. If
?/* S J 0^ sin  4 0j. r>
^ _flf54cos0,)
7 a. ( 2 2 = a2 cos 2 p fi .3
. aft
fl  ' cos { L ) j DIFFERENTIAL CALCULUS 156 Find the radius of curvature for each of the following curves at the
Draw the curve and the corresponding circle of curvature. point indicated.
29. =
=
=
= jr 30. jr 31. .r 32. r 33. ~ jr 34. .r 35. 2  x = ?/ t // , 2 r ', // a cos /, ?/ cos /, ?/ ; =
= ' t 1 / = a cos / ; / tf 1. 72 / ; 1 =
= 1 / of Find the radius of jr = 2 sin . 37. x  tan !. 38. r / :! /, // =
~ // (f aloe* / a (sin + / sin 42. a cos (t = cos 2 =
/, = ctn // /i ) ft = / ; / / a. t\) t ;  = / TT. / ; on the hypocy;i a sin /i cos /i. on the involute ?/ = An*, ft Find the points on the curve 3 y ~ .r at\. where the curvature r' An$. s Show TT. IT. \ cos 1 /). ! 2 is a 0.347. x where the curvature maximum. 43. . 2 V2. /;, t Find the point on the curve / maximum.
is ;// ,4//,s. /. curvature at any point
.r /, sin curvature at any point a sin 4>/2. ft t\. 36. ; =
=
= ft 0. . of the circle 41. = 2 sin /;// // = t ; / A MS. 1. < a sin  1 /, : ' =
1 = / ;  2 f  /  3 ?/ /,, =// 2
t =
= 2 Find the radius cloid j
40. 3 /, / f 1, 4 39. 2 ,4^/.^. jr 0.931. that the radius of curvature becomes infinito at a point of inflection. Given the curve
3 jr
.r\
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
44. // (a) maximum
(c) of curvature. Find to the nearest hundredth maximum of a unit the abscissa of the point An*, curvature. .r 1.01. Find the radius of curvature at each maximum and minimum
4
2 .r
Draw the curve and the circles of curpoint on the curve // ~ x
vature. Find the points on the curve where the radius of curvature is a
45. L> . minimum.
46. Show curve y = that at a point of f(x] minimum radius of curvature on the we have '^Vl
,dx) \ that the curvature of the cubical parabola 3 a 2 ?/ = x3 increases from zero to a maximum value when x increases from zero to
47.
4 I Show _____ aV125. Find the minimum value of the radius of curvature. An*. 0.983 a. RADIUS AND CIRCLE OF CURVATURE 157 The tan pent line at P(r, y) has the
that x, y, and y have the same values at P for the tangent
property
line and the curve. The circle of curvature at P has a similar prop108. Center of curvature.
f namely, x, y, y', and //" have the same values at
curvature and the curve. erty; P for the circle of DEFINITION. The center of curvature (a, /j) for a point P(x,
a curve is the center of the circle of curvature. The coordinates Theorem. (a, /j) y) on center of curvature for of the P(x, y) are y The equation Proof. is Cra^+fojCO^/i', (1) \\here y of the circle of curvature K is given by (F). Differentiating (1), a x / (2) ,, ?/"
" yP R~ =(y From
of R the second of these equations, after substituting the value
from (F), we obtain From the , ' first y*
of equations (2), (3) for Solving in I /3, in (4) for a, have (G). Work 1. out using (G), Art. 95.
f R cos EXERCISE we (a =x di (G) rectly from the accompanying /3 get, using (3), Q.E.D. EXERCISE =y we figure, R sin r, r, etc.) If x' 2. and x" are, re spectively, the first and second derivatives of x with respect to y, derive (G) in the form a Formulas (H) ~ x" x" may be used when ferentiation with respect to y is y' simpler. becomes infinite, or if dif DIFFERENTIAL CALCULUS 158 ILLUSTRATIVE EXAMPLE. Find the coordinates of the center of curvature of
2
4 px corresponding (a) to any point on the curve; (b) to the
y' the parabola
vertex. "' Use Solution. Hence 2 a=x Therefore (a) Then (//). y' ( 3 x + f = x' 4 . x" = 1 j = 2 p, + 3 x ~~~^ }
4 p/
t V 2 p, Ls the center of 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) = 0; and from (F),
Therefore, by (B), Art. 102, the curvature
Art. 105, and (G), Art. 108, we see that in general a, 0, and R increase without limit as the second derivative approaches zero, unless K the tangent line is vertical. That is, if we supP with its tangent to move along the curve pose to P', at the point of inflection Q the curvature
is zero, the rotation of the tangent is momen and as the direction of rotation
the center of curvature moves out inchanges,
definitely and the radius of curvature becomes
tarily arrested, 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 curvature 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 eliminate x and y between the two expressions and the equation of the
given curve. No general process of elimination can be given that 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 will the following steps. General directions for finding the equation of the evolute in rec tangular coordinates. FIRST STEP. Find a and /3from (G) SECOND STEP.
of a and or (H), Art. 108.
Solve the two resulting equations for y and y in terms /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 Find the equation 1. of the evolute of the parabola V=4p*.
Solution. First Step.  ,
, dx a= 3 ar C,,l dx y + 2 p, = 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. DIFFERENTIAL CALCULUS 160 ILLUSTRATIVE EXAMPLE c
Solution.
. Find the equation of the evolute of the 2. = aw. ^2 X 2 4. n iyi
..  =  TT . a 2 ?/ 3 a 2 !/ dc /" p / ei
tm Mtfp. a= Second Step, x ellipse 1 ./ "\a 2 <*  \*
z b' ) (au; + (/^j = (a 2  6 2 )% the
c?quation of tho evolute KHK'H' of the ellipse
are the centers of curvature corresponding to the points
A^/l'/^'.
/, A ', ^/', //
v
("" correspond to the points P, P', P".
A, A', K, It', on the curve, and (\ (
1 , ILLUSTRATIVE EXAMPLE a. The
/ paranietric equations of a curve are
2 4 1 : (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 responding where / = 1, and draw the cor circle of curvature. X By Substituting in (G) and reducing gives
(2) ^^^1 f 3 t (A), Art. 81 By Solution. (B), Art, 82 and calculate x, ^ from (1) and tabulate the (2), Now plot a, j8 from results. the curve and evolute. its point (J, 0) is common to the
given curve and its evolute. The given
curve (semicubical parabola) lies entirely to the right and the evolute
entirely to the left of x
J.
The circle of curvature at A (
) The , Ji , where
A'(
dius = f 1, have will Jl center at its on the evolute and
1,}
AA'. To verify our work ra k, = From the radius of curvature at A.
(F), Art. 105, we find get This should equal the distance AA' =
ILLUSTRATIVE EXAMPLE " By Art. 3 Find the parametric equations of the evolute of the 4. cycloid
r.r
V (1), = </</ ; COS/). Solution. As in the Illustrative _ d// sin dx Example
d' y
dx~ / cos/ 1 of Art. 2 cSl?, a ait  (4)
1 jg 4 t/d sin get 1 a(\ cos/) Substituting these results in formulas (G), Art. 108,
f we we 2 get /),  cos /). Ann. NOTE. If we eliminate between equations (4), there results the rectangular
The cocirdinates
equation of the evolute OO'Q* referred to the axes O'a and O'fi.
with respect to these axes are (~ mi,  2 a). Let us transform equations (4) to
of
the new set of axes OX and OY. Then
/ a x ?ra, Also, let t ft = 2 a. y TT. t' Substituting in (4) and reducing, the
equations of the evolute become x
(5) y = a(t'  sin
= a(l cos Since (5) and
form, (3) ^ /'),
/') are identical in o we have: The evolute of a cycloid is itself a cycloid given cycloid. whose generating circle equals that of the DIFFERENTIAL CALCULUS 162 evolute has two interesting The 110. Properties of the evolute. properties. Theorem The normal 1. at (Ij ft line (See figures in the From the figure,
Rsin T,
a=x Proof. P, given curve is tangent to C(a, p) for P. article.) preceding The to the P(x, y) the evolute at the center of curvature PC = + R COS T. y along the normal at lies and the
(2) Slope of PC  a x tan r slope of normal at P. We show now that the slope
Note that of the evolute equals the slope of = Slope of evolute
since a arid fi PC. j/j da
any point on the are the rectangular coordinates of evolute.
l^et us choose as independent variable the length of arc on the given curve then
(1) with respect to x, ; s R, //, d'jr as ds rf/i j (4) = ( = cos Substituting in dy ~~ . ^ sm ds ^ = and ds sin r,  (4), sin r rfs Theorem dr rf/2 , from Art. 95 : and =  dR
~ = COSTT dR rf/J r/.s* =  ctn T  (5)  by the = tan r The length of an first gives slope of PC. R increases or decreases. Q.E.D. arc of the evolute is equal to the dif which are tangent
along the arc of the given curvt fer erne between the radii of curvature of the given curve
to this arc at its extremities, . R and reducing, we obtain da.
2. J ds T T" H cos r ~T~'
ds
ds Dividing the second equation in
(6) Differentiating ds da = (5) as (3) T, rf.s 8. ) j ds But ~r are functions of a,  # cos r dr __ sin T dR da
(8) T, gives provided tfiat RADIUS AND CIRCLE OF CURVATURE
Proof. Squaring equations and adding, we get (5) (DXiNff <"
But = if s' length of arc of the evolute, (C), Art. 95, if s = = x s', a, y = ^T (8) + dfi, da ds'~ by 163 =
or ' Hence /i that dff = i (7) asserts ds Confining ourselves to an arc on the given curve for which the
righthand member does not change sign, we may write = ^ w
dR~ 41 or (9) That is, or i s __ i R ,~ ' Jfee _ =  dtf R rate of change of the arc of the evolute with respect to
Hence, by Art. 50, corresponding increments of s' and i. are numerically equal.
s' (10) That  159) Arc is, =
CCi = (R s' flu),  PC). or (first figure, p. at Thus the theorem is proved.
In Illustrative Example 4, Art.109, we observe that at
P v R = 4 a. Hence arc O'QQ V = 4 a. (PiCi O', R= ; , Tfce length of one arch of the cycloid (as length of the radius of the general ing OO'Q V
) / eight times the circle. and their mechanical construction. Let a flexible
form of the curve r,r.,, the evolute of the curve
and suppose a string of length
PI Po,
Ru, with one end fastened at Co, to
be stretched along the ruler (or
curve). It is clear from the results
111. Involutes ruler be bent in the of the article that when
unwound and kept taut, last the the
string
free end will describe the curve PjP<>.
is 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. DIFFERENTIAL CALCULUS 164 The /Y/V, /Y'/V involutes PiP*, 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
evolutes.
pages 159, 160 may be constructed in this way from their 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 = + 2 2. x'
3. 2 x'
2 4 y' = 6;
= <"; 5. // 6. y 7. ?/ 8. 10. 11.
12. 13. =
=
= 25 19 ; lnr; (0, 0). TT, (J 2). + ^?/ =
(3,
= x 4; (0, 2).
2?/
xy = x' + 2; (2, 3).
(J, 1).
y = sin
= tan2o;
J).
(/ f 6) ~
TT, ( 2 :i 3 (i (1, 0). 2 sin 2.r; ) ( ~ 3 ) (0, 1). ; H) 5?) (?i (0, 1). cosr ~ (W, (3, 2). ; (0, p). (iVb (3, 2). ; (2, 3). 4.x?/ 9. = r i/ '=
j Am. (0, 0). ; ; ( 3). 2)  ) 13, 8). 2 2 TT.T ; (J ?, ?/ TT, of the center of Find the coordinates curvature at any point (x, of each of the following curves.
2 = ?/ 15. ^ = a^. 2
16. 6 2 x  3 <r f 2 A 2 14. a 2 ?/ 2 ' 4 a= a
=a 2 52 . f 15 + y = a. ,. . a=
2 17. j.8 4
?/
 2 a=x 4 3 +
x?/, P= a 4 ?/  9
4 y) RADIUS AND CIRCLE OF CURVATURE
18. points
ters. 165 4 at the
Find the radii and centers of curvature for the curve xy
Draw the arc of the evolute between these cen(1, 4) and (2, 2). What is its length ? __ Ans. At
at
A>, (1, 4), A>, (2, 2), A> 2  ft, = = V Vl7, a = V,
= 2 \/2, a = 4, = = V ; 4; 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. = 3/
20. x = 3
21. x = 6 =2
22.
23. x = 4
=924.
19. , t, a 26. j* =
= J f , a cos f, /' , jj. . . 2 / f 4 . ' 4 = ?> 4 <. , / . r
?/ = /*), f 3 = ?/  2 f. ?/ t' :l f / f. 2 2 2 2 =3 =J
=  2 P.
=43
t*
=3 2
a   2 /*,
=  /\ 0=11 + 3 2
=  2 P.
=73
27 + 4
 12/ + 9 J y + %(l a =2
=  2.
= 3 4 2
=2
y , i/ 2*, . f 2 /, a Am. a = :i 2 #
* f, 25.* = 3f?
=  6. 2 x ^= / 4 el t /^2 a sin L .> fo'2\ cos :i /, z _ a COS ^
= a sin
x = a(cos + sin /),
28.
cos /).
y = a(sin
=2
=429. x
= 16 =2
30.
= i ^.
31. x =
= 1 cos y = 32. x
4
= sin
33. x = cos
= b tan
= a sec
34. x
35. x = cos
y=
x = 6 sin
36.
y = 3 cos
= 4 ctn
37. x = 3 esc
38. x = a(t + sin 0= a(l cos/)39. x = 2 cos + cos 2
y = 2 sin + sin 2 27^ a 3 x :i f. ?y t t / / JT 2 /. ?/ , 2 / ?/ /, ot / . 2 / , ?y 4 ?/ /, , sin t f, /. ?/ f. . t, J. t, /, J. i/ 7/ t ? i. ^. /. a =
= cos* '* " cos 2 a cos / /, +
/ 3 a cos
sin = / + / sin 2 a sin a sin t. , :{ iL DIFFERENTIAL CALCULUS 166
40. + (3 41. Show that in the parabola x* = X(x + H y = a we 2 2 have the relation y). of the equilateral Given the equation hyperbola 2 xy  that =a 2
, show * 2 From this derive the equation of the evolute (+ 0)1  (a 0)3 112. Transformation of derivatives. = 2 a*. 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 ?/' , x' 2/" =
= dx
, = ,,
x" = dx' !
etc., f, d 2x , dy dy By = dy . etc. z IX, Art. 29, XT Now
Using 'f
~~ // y" (/), we Ret = dx 7
x' ^~
x' (/// >
2 dy"
Again, y'" = *f =
dx A,tt Using (/), ? = dy ,y.r~rrf And
y'> x/ ~
x' v X '"
v
' ..y'" *3L. = i Q , 4 O v
*x " ' Y '5 so on for higher derivatives. y", '//'"> etc. By these formulas equations in
can be transformed into equations in x , x", x'", eta
1 RADIUS AND CIRCLE OF CURVATURE
ILLUSTRATIVE EXAMPLE. Transform
Solution. Using (/) 167 (B), Art. 102, into (C) in that article. and (/) above,
1L x" Transformation from rectangular m ' Given the polar coordinates. to relations = x (1) 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 parametric equations for that curve,
being the parameter.
variable is 0, and x', .r", y', y", p', p"
NOTATION. The independent
denote successive derivatives of these variables with respect to 0.
Differentiating (1),
(2)
(3) = p sin + p' cos 0,
z" = 2p'sin0 + (p"p)cos0, formulas By (1), (2), y' in j, equations (3), transformed into equations in Solution.
(2), (3), _ y> x " = p + 2 p' 2  PP" ?/, x', 104, directly Taking numerator and denominator
and reducing, we obtain the results
X 'y p cos ; ?/, r", ?/" may be p". p, 0, p', ILLUSTRATIVE EXAMPLE. Derive (), Art, from = + p' sin
<//"2p'cos0 + (p"p)sin0. x' ; ^ /1! from (/)), Art. 10,'J. in (D) separately, substituting + //" 2 = P" + P /2
 Putting these values in (D) gives (). PROBLEMS
variables,
In Problems 15 interchange the dependent and independent
4 Ans. z d 2*
rf 4 2 /dx\ =0.
A ?/(7)
V(]f DIFFERENTIAL CALCULUS 168 dy
6. ~
by assuming x Transform = p cos 6, = y p sin 6. P A~<. 7. Transform the equation 8. Transform the equation s 2 y~   ~ ( ~ ~? + 2 ~^ f
2
dr
.c cLr  f   ^7 y
' =  = 2 by assuming by assuming .r = .r / A Ans. d'2 ij
7*7 4~ v a v . = A 0. ADDITIONAL PROBLEMS
3 cos f 4 cos 3 t, y = 3 sin t
sin 3 t. Find the
1. Given the curve or
parametric equations of the evolute. Find the center of curvature for
/ =
arid show that it coincides with the corresponding point on the given
= 6 sin t 4 2 sin 3 t.
Ans. a = 6 cos i  2 cos 3 t,
curve.
of curvature at any point of the ellipse
the perpendicular distance from the origin to
the tangent drawn at this point, prove that RD* = a 2 b 2
2. If It b 2 jc 2 f a*// 2 is = the 2 2 a' b' radius and D . Find the equations of the evolute of the parabola ?/ 2 = 4 x using x
as a parameter. Find the points of the parabola for which the correspond3. t ing centers of curvature are also points of the parabola. Hence find the
length of the part of the evolute inside the parabola.
 l).
Ans. (2,
2V2); 4(V7 At every point (jc, y} of a certain curve,
and the curve passes through the point (2, 4. (a)
?/ its slope 0). is equal to Verify that the V5 jc
equation of the curve is log (1 f y) = 1
(b) Find the curvature of the curve at this point and draw a small
portion of the curve near it.
Ans. K = oVVH.
Ans. cv = f ft = .
(c) Draw the circle of curvature at this point.
2 . , 5. The given by slope of the =  where tangent to a certain curve
.s jr is C at any point P the length of arc (measured from some is fixed Depoint) and a is a constant. The center of curvature of C at P is P
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
1 '. R= a2 CHAPTER XI THEOREM OF MEAN VALUE AND
113. Rolle's Theorem. A theorem which the theoretical development of the ITS APPLICATIONS
lies at the foundation of calculus will now be explained. Let y=J(x) be a singlevalued function of x, continuous throughout the interval [a, 6] (Art. 7) and
vanishing at its extremities
(/(a) = 0, f(b) = Sup 0). pose also that /(x) has a
derivative/' (2) at each interior point (a < x < 6) of
the interval. The function will then be represented graphically by
at
a continuous curve as in the figure. Geometric intuition shows us
b the tangent is paronce that for at least one value ofx between a and
that is, the slope is zero. This illustrates
allel to the xaxis (as at P)
; continuous throughout the interval [a, b]
at every
and vanishes at its extremities, and if it has a derivative /'(x)
at least one
the interval, then f'(x) must vanish for
interior point of
value of x between a and b.
Rolle's Theorem, ///(x) is or negative in some
is simple. For/(x) must be positive
in this
interval if it does not vanish at all points. But
parts of the
that /(x)
case the theorem is obviously true. Suppose, then,
special
Then /(x) will have a maxiis positive in a part of the interval.
if f(x)
value at some point within the interval. Similarly, The proof mum value. But if f(X) is a maxinegative, it will have a minimum
< b), then f'(X) = 0. Otherwise, /(x)
or minimum (a <
would increase or decrease as x passes is X mum through X (Art. 51). The figure illustrates a
Theorem does not hold f(x) case in which Rolle's continuous throughnot
out the interval [a, 6]. f'(x), however, does
At no point
exist for x = c, but becomes infinite.
xaxis.
of the graph is the tangent parallel to the
169
; is DIFFERENTIAL CALCULUS 170 We give first two applications 114. Osculating circle. Theorem of Rolle's to geometry. a circle be If drawn through three neighboring points
Po, Pi, P'2 on a curve, and if Pj and P 2 be made to approach P along the
curve as a limiting position, then this
f, circle will in general in approach 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. Let the equation of the curve be Proof. U) ?/=/(*); and let (a', fi ) xi, x>> be the abscissas of the points Po, PI, Po respectively,
the coordinates of the center, and R the radius of the circle
passing through the three points. Then the equation of the circle is x ( , f f and since the coordinates
equation, we have of the points P , must PI, P2 satisfy this (2) Now consider the function of x defined by
F(*) in = (x~ which y is defined by
Then from equations
) a') 2 +  (y f i3 ) 2  R' 2 , (1). we (2) = 0, get F(X!) = 0, F(x 2 ) = 0. Hence, by Rolle's Theorem (Art. 113), F'(x) must vanish for at
two values of x, one lying between xo and x\, say x and the
other lying between xi and x^ say x" that is,
f least , ; = 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"(z3 ) = 0. THEOREM OF MEAN VALUE AND
Therefore the elements must points Po, PI, P2 F(zo) Now let the points R ITS APPLICATIONS 171 f of the circle passing through the
the three equations
satisfy = a', 0', 0, = 0, F'(x') F"(.r.i) = 0. PL approach P as a limiting position
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"(x )=0;
then xi, ?2 , x', PI and x", x3 will > ; all () f or, dropping the subscripts, by  a) 2 + (y  0) 2 =
 a) + (y  0,
(x
2
 J3)y" = 0,
1 + if + (y
(x (3) (4) ff',
8)'//' (5) Solving (4) and
x (6) (5) a a and x f or = differentiating (3).
differentiating (4).
0, ?/ > get ~ rf 7/
?/ we (?/" ^ 0), ^ , '// 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. P was defined as the limiting posidrawn through P and a neighboring point Q In Art. 28 the tangent line at
tion of a secant line on the curve. We now see P may that the circle of curvature at defined as the limiting position of a
other points Q, R on the curve. circle be drawn through P and two 115. Limiting point of intersection of consecutive normals Theorem. The center of curvature (^ for a point limiting position of the intersection of the normal a neighboring normal.
Proof.
(1) Let the equation curve be f is the P with /*j c[ '^
<tt =/(*) ^ If a curve curve at
C(a of a The equations of the normals to the
curve at two neighboring points Po and
D are
^
Oo  y) + ('//o  ?/)/'Oo) must P on to the the normals intersect at C'(a',
satisfy both equations, giving /3'), = A'.^o'l/o) 0, the coordinates of this point DIFFERENTIAL CALCULUS 172 Now consider the function of x defined = (x  <t>(x) which y in is defined by Then equations (2) a') + , now p')y' f (1). = ) 0(xi) 0, = 0. But then, by Rolle's Theorem (Art.
some value of x between XQ and x\, say
determined by the two equations If (y show that </>(z x by PI approaches P must vanish for
Therefore a' and /3' are 113), <'(j)
x'. as a limiting position, then x' approaches giving and f"(a', 0') will on the normal at approach as a limiting position a point C(a, ]8)
the subscripts and accents, the last /V Dropping two equations are + (y &)y' = 0,
1 + y' 2 + (y  fry" = 0.
(xa) a and /3, the results are identical with (G), Art.
Theorems of Mean Value (Laws of the Mean). For Solving for
116. cations we need 108. Q.E.D.
later appli 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 </>() = </>(?>) =
be applied. Differentiating, 0, and This must vanish for a value x ~ /(a) r*/ Rolle's xi Theorem, Art. 113, between a and 6. may THEOREM OF MEAN VALUE AND ITS APPLICATIONS 173 Dividing through by F'(XI) (remembering that F'(XI) does not
vanish),
If and transposing, the
= x, (A) becomes result Q.E.D. is (4). F(x) b  J a 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, OD = b, * CA=/(a),
DB=f(b). Hence
r/_\ of chord a b Now f'(xi) AB. 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 tangent 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 /(&)=/(<*) (C)
6 = a r Aa Let
between a and 6, ; then we may a b + = (b  <0/' (*0 Aa, and since x\ is + a number lying =a + 6 Aa, where 6 is a positive proper fraction. Substituting
another form of the Theorem of Mean Value,
f(a form write
Xl (D) in the Aa) /(a) = Aa/' (a + Aa). in (C), (0 we get < < 1) PROBLEMS
finding the values of x for which f(x)
Verify Rolle's
vanish in each of the following cases.
and/'fa)
 cos TTX.
= x 3  3 x.
(e^ f(x) = sin irx Theorem by 1. (a) f(x) (b) f(x)
(c) f(x) (d) f(x) = 6 x2  x3 = a + bx f
= sin x. (f) . ex 2 . f(x) (g) f(x) (h) /(x) = tan x  x.
= x In x.
= xe*. DIFFERENTIAL CALCULUS 174 and /(ir) = 0. Does Rolle's
tan x. Then /(O) =
Theorem justify the conclusion that f'(jc) vanishes for some value of x
and TT? Explain your answer.
between = 2. f(x) 3. x Given Given (y f l) =+ 1. Does some value for 4. :i = Rolle's 1 and
when x =
Then y =
x2
Theorem justify the conclusion that between of x  + and 1 = y' when vanishes Explain your answer. 1 ? In each of the following cases find x\ such that (dj /(x) f b) = /() +
= x, a 1, b 2.
= Vx, a = 1, b = 4.
= 6", a = 0, b = 1.
 , = 1, 6 = 2. (e) /(x) = (ba)f'(jci). ( (a) f(x) (b) /(x)
(c) /(x) 5. // . Given x, =
= Xi Ans. = In Xi 1.5. 2.25.
(<?  1) = 0.54. r; In x, a = f(x)  0.5, b  a x 1.5. = 1,6 = /() + (b )/'(x!)?
6. Given /(x) = x^, a =
/(ft) =/(*;) f (fcaWi)? For what value of Xi, if any, will For what value 1. of x,, if any, will f(b) 1. /> 1, When, for a particular value of the
a function takes on one of the forms
independent variable,
117. Indeterminate forms. L\ it is 2?, X said to be jndetcrvmuite,  oo oo, 0, oo, and the function value of the independent variable I oc", is 'not t defined for that the given analytical expression. by For example, suppose we have where for some value of the variable, as x
/(a) For this value of x fore assign to it gone before (Case II, possible to do F(a) 0, our function any value we is make it = 0. not defined please. Art. 17) that function a value that will
it is  = a,
and we may there evident from what has
desirable to assign to the It is it is continuous when x = a whenever so. 118. Evaluation of a function taking on an indeterminate form.
the function f(x) assumes an indeterminate form when x = a, then If
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 transformations, as the following examples show.
^~
ILLUSTRATIVE EXAMPLE 1. Given /(j)
^. Prove lim f(x] =4.
Solution.
fix} =x+ indeterminate. But, dividing numerator
by denominator is /(2) Then lim 2. z  (x + ILLUSTRATIVE EXAMPLE
Solution. /(*) sec x ir>  2) = 4. 2 Given 2. indeterminate tan x = L^J^LZ =
fos Hence the limit (*> /(x)  = sec x tan j\ . cos  ^ = 0. TT Transform as follows: L^L Lin* j Prove lim f(x)
.r oc). .r 1 f sin cor x __ ^ .r 4 1 sin x is 0. See also Art, 18. General methods for evaluating the indeterminate forms of Art. 117 depend upon the calculus.
119. Evaluation of the indeterminate the form ff.r) j~~ such that /(a) indeterminate when x We shall Proof. that /(a) a. It is form  and F(a) (iiven a function of ~ 0. The function is then required to find prove the equation Referring to (4), Art. 116, and setting 6 = F(a) = 0, = jr, remembering we have m=m <<<'> >
If x
a, so also .TI > a. Hence, if the righthand member of (1)
approaches a limit when r\ > a, then the lefthand member will approach the same limit. Thus (E) is proved.
From (), if /'(a) and F'(a) are not both zero, we shall have W
(2) ii 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. DIFFERENTIAL CALCULUS 176
In case it happens that /'(a) derivatives also vanish for x = a, = = 0, and F'(a) then () that is, the first can be applied to the ratio F'(x) and the rule will give us ij m xaF(x)
It may F"(a) be necessary to repeat the process several times. The student is warned against the very careless but common mistake
whole expression as a fraction by VII. of differentiating the
If a = oo, the substitution Thus = jc z of the limit for z reduces the problem to the evaluation 0. = lim lim = m
li Therefore the rule holds in this case
ILLUSTKATIVK EXAMPLE F^ove lim 1. ^H!^ = x .0 Solution. = l^tf(x)=^Kinnx F(x)
l Hm /(l = lim
1 ILLUSTRATIVE EXAMPLE
Solution. =0, Let /(.r)   3 x n , JT Hl = Hm ncoanr = + 2, = 0. Then/(0) =0, F(0) 3 ? *"* Prove lim 2. ^ jr. also. + F(.r) w _ 2 jr^x + ' Therefore, by (),
b 3 2 1 =** x*  x + 1. Then /(I) = F(0) = Therefore, by (), ILLUSTRATIVE EXAMPLE
Solution. Let /(x) Therefore, by (), = e*  3. Prove lim e~ r   2 x, F(x) _
t = x  sin x. = 2.
Then /(O) = 0, THEOREM OF MEAN VALUE AND ITS APPLICATIONS 177 PROBLEMS
Evaluate each of the following indeterminate forms by differentiation.* ~ ~2
1. lim 2 :4X
jra x
z4
2. lim c 3. . 4. 1C x20 f M lim  ,. e ,. lim
x . 1 na" a" In x n 1. e~ o
2. : sin o .r 2. _ox
. 6. v hm  sin
In sin JT  7. lim
,. 8. hm ~  2 x  ^I= ln r   1 sin ev  arc sin ^ 0*o + i 4 i
* 6 sin y 1 15
15. hm
lim tan 6 + sec ^ _^
e ,/> 16. v hm
. r sin ^ + 1 l ; 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
and P and the line through A and O.
through
Find the limiting position of B as P approaches M A as a limiting position. An*. OB = 2 r. * After differentiating, the student should in every case reduc the resulting expression
form before substituting the value of the variable. to its simplest possible DIFFERENTIAL CALCULUS 178 form ~. 120. Evaluation of the indeterminate In order to find lim
x^a when both f(x) and F(x) become infinite when x
a, we follow the
in Art. 119 for evaluating the indeterminate
same rule as that given
> form . Hence Rule for evaluating the indeterminate form ^. Differentiate the
numerator for a new 'numerator mid the denominator for a new denomi The value of nator. 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 n/r
 _
  ILLUSTRATIVE KXAMPLE. Prove lim C.C .r = I*t /(j) Solution. osc f(jr) lim
, $L o f  (jc) J
im f'(jf)
f~ ,.
)  ,<)/<'(>) j , I Then, by /I^
(A), j. o 121. Evaluation
f(y) cos " I'
lim jr o cos , J j;  </>(**') F(Q)   oo . 'j /I <l 0. If oo. oo for x T7 T^ Q.E.D. x sin Hence, L lim the indeterminate form of , siri .. ' : o   i* takes on the indeterminate form  Bi JT  /(O) J ctn c'..c (,  " hm
1 j the Then .r  " rni 0. .r. .. = hm of this book. JT I ,. j = In r, F(JT) by the rule, beyond the scope is a function
a, we write 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
X Solution. Since sec J TT o sec 3 = oo
, c .r + cos J cos 5 J ~ IT  Let/(j) cos 5 or, F(x)  cos 3 = 1 cos = (sec 3 jr. cos 5 J* = .r) . 7T /, 7 0, we write c
cos 5 x = Then/( C()S 5 X cos 3 3 .r TT) = 0, j" F(\ TT) 3sm3.r = 0. Herice, by (), THEOREM OF MEAN VALUE AND APPLICATIONS ITS 179 oo.
form oo
It is possible in
to transform the expression into a fraction which will assume
general 122. Evaluation of the indetei ruinate form  or ~ either the ILLUSTRATIVE EXAMPLE. Prove lim
X Solution. ^
Bv We have sec o y (2), p. 2, sec =  I \ * tan jr sin tan TT .r F(rt = cos J (sec tan x] .r j* j. 0. JT T = x ~ x
 cos
jc, t, sni v 1 = l  ~~ sin 1 cos x indeterminate. .. . *" cos r Then/(\ TT) = 0, F(\ TT)  0. Hence, by (E), p ,. hm (.r) __ 0; ^ X '.TT ^H  ~ Q.E.D. ' s in PRO3LEMS
Evaluate each of the following indeterminate forms. 2. , o ,. 3. hm lim  lim ctn x lim
j 12. 13. 0.
Ans. A (1 tan 0) sec 2 0. Ann. 4. ctn 2 r ^
tan 3  tan 6 ^ _ 4. lim 5. lim
x , .r. . 6. hm 7. lim s c : In i x ctn x
i 2 x In sin ,o In sin
8. lim .r In sin jc jr. JT rt 9. 10. lim  . tan 7T0
  lim r sin x 11. 7T T ^ . oo lim (TT 2 .r)tan x. 22. lim (sec 5 1. <? tan 6.) DIFFERENTIAL CALCULUS 180 7T
' TT
" 23. lim 1 t\f* _ r* *( i ' . 4 ^* limf4 24. 28 123. Evaluation of the indeterminate tion of the form f(x)+ . forms 0, 00. Given a func 1, (T) In order that the function shall take on one of the above three we must forms, have, for a certain value of x, /(x)=0, 0Cr)=0, = !,
= oo, = oo,
= 0, or /(*) or /(j) 0tr)
(f)(x) Let giving giving I Invy ; (f) both of = 000 00 giving 00. y=f(x)+ Taking the natural logarithm 0; . sides, In /(a). In any of the above cases the natural logarithm of y (the function)
take on the indeterminate form will oo. this Evaluating by the process illustrated in Art. 121 gives the
This being equal to the loga limit of the logarithm of the function. rithm of the limit of the function, the limit of the function
For if limit In //
a, then lim //  e".
ILLUSTRATIVE EXAMPLE Prove lim x z 1. j The function assumes Solution. y then
r> = In y
* u. 1 ni By By Art, 120, In lim
 r Therefore lim In y
' *  0, 1. x1
a* j ?/= =
1 and when x = 0. ; In In x i Art, 121, known. . the indeterminate form Let is =  when x oo, oo = 0. whenz = , 0. lim
 J lim
X () y lim x*
 X = e = 1. O.E.D. THEOREM OF MEAN VALUE AND
ILLUSTRATIVE EXAMPLE Prove lim 2. j) 181 2 i  (2 APPLICATIONS ITS
tan 5 WJr = f * jrl The Solution. function assumes the indeterminate form Let y
In By  (2 tan .r) Art. 119, ln (2 lim ctn .T.I  In (2 ~ ^ = hm
.ri   j) ~ L> i TTJ oo ' 31 lim '
y
x In sin x Art. 120, lim In y 0, sinj ^ = r2 " im and = i. when x ~ 00 = j oc esc ilH = LL. = ^ nm
T .r = Hinj " lim (ctn .r) = c  2. tanx Ans. . ' i[ z lim + m
. r e*. lim fl
x V h lim Hm COS \ V. a c . lim x f x) (cos ^
? 11. lim (r^ ctnT
. r. 12. + 2 lim (x x
. 4 e l) /  (1 + ntY. . /I \sina; 2
. 13. lim '.
j 1 lim x)^ ctnx x0 ( 7. . T. o (1 f sin x) (c V
y/ (COB 6. / Hm J y/ i/* xO 9 10. lY. Slimx^. 5. g I / 4. 1.  c\ 14. 0. lim
J  . cos'*' .r .<> Evaluate each of the following indeterminate forms.
lim (sinx) = j* csc x ctn y 0. whon x , oo _ _
lirn 0. when j = , PROBLEMS x " ; j: j^o OED
v 77 sinj ^^ = = = ^ .r) r csc lim l!L*!L
csc J
x *o (2 .1 (ctn j) In ctn  1. . TT tan  the indeterminate form  Therefore 1 *o y= (ctnx)
In y By 1. when j = 0, Trr .^ " Let Art. 121, lim
j. Prove lim 3. The function assumes then ._ .1 j By . i and 7T ILLUSTRATIVE EXAMPLE
Solution. = r csc TT \ o lim In y  , Therefore jr ; 7r.r i when l ^ tttn * ,n^! Art. 121, By i/ =
= lnr
(I f r) . 1. Q.E.D. DIFFERENTIAL CALCULUS 182 124. The Extended Theorem of
be defined by the equation
J(b) f(o.) (1) Mean (b a)}' (a)  (&  a^R = Let F(y) be a function formed by replacing
of (1)
that is, member b by x R  in the lefthand ; = f(x) J(a) and from
F(b) = F(x) (2) From (1), ; (x  (2), a)/' (a) F(a)  = Theorem (Art. 113), at least one value of x
will cause F'(x) to vanish. Hence, since we Let the constant Value. ;  a)*R. therefore, by between a and F'(x)=J'(x)f'(<i)(xa)R,
 (x  a)R
F'(Xi) =f'(xi) ~J'(o) get \(x l = b, Rolle's say x h 0. 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
Since F f (x\ ) F"(x) =f"(jr)  R\ tlierefore F"(jrJ and ), /f (^) R= 0, R=f"(xJ. Substituting this result in (1 =f we get ' a) By 2 "
/"(^). (a continuing this process we get the general result, (G) Equation (G) is called the Extended Law of the Mean.
Maxima and minima Theorem of Mean Value, or the Extended
125. Art. 116 and the discussion of By making use of
we can now give a general treated analytically. results of the last section maxima and minima of functions of a single independent variable. Let h be a positive number as small as
/(.r).
then the definitions given in Art. 46 may be stated as Given the function we please follows. ; THEOREM OF MEAN VALUE AND
for all values of If, /(x) (1) then /(fl) a +H h, a negative number, = /(a) /(.r) the interval a in = said to be a majriwinn whew x
on the other hand, (2) a. a positive number, said to be a minimum when x == a.
begin with an analytical proof of the criterion on page 51. /(.r) is We
A = 183 f(jc) is If, then from a different :r ITS APPLICATIONS increasing when the derivative is function ing when and is positive, decreas the derivative is negative. When y=f(x). For, let A;/ is and the small, numerically,
A.r derivative is Therefore f(jr) is ative is The agree in sign (Art. 24). will /'(.r) when Ax positive, so is and when A?/, A increasing. For, through a is if a. maximum or /'(a) 5^ 0, f(x) But then f(a) minimum is is neither a now Let /'(a) = 0, and/"(a)^ /(a) Since /"(a) ^ 0, /(a) and /"Or) f (a) = 0. nor a minimum.
maxima and minima. 0. (f), Art. 124, replacing 6 (3) easily deduced. or decrease as x increases seek general sufficient conditions for
Consider the following cases. I. deriv maximum We From Then, 0. when the value of f(x), then would increase > jet /'(/) similar proof holds negative.
truth of the following statement ///(a) I A.r is negative, so is A?/. by n and transposing = *~ a}2 f"(.r<2). /(a), ( is + h\ (a < x>> < x) assumed as continuous, we may
so small that /"(/) will have the choose our interval \a same Therefore
Also, (/
a)' does not change sign.
of (3) will not change sign, and the difference sign as /"(a). the second member h, a 2 /Or)
will have the same sign /(a) for all values of x in the interval [a same as the sign o//"(a).
fore follows from our definitions (1) and (2) that and, moreover, (4) /(a) is a (5) /(a) is a this sign mil be the maximum iff'(a)
minimum if f'((i) = and f"(a)
and j"(a) These conditions are the same as those h, a + h], It there = a negative number;
= a positive number. in Art. 56. DIFFERENTIAL CALCULUS 184 Let II. From  0, =/"(a; /'(a) (G), Art. 124, putting /'"(a) ^ n = 3, and replacing b 0. by r, and trans posing /(a),  /(a) = ,4 f(x) (6) te  a) /"W 3 (a < a* < a) 12 As have the same sign as f'"(a). But (x
a)
to f as x increases through a. Therefore :i before, /'"(or) will changes its sign from the difference must change
III. Let and sign, /'(a) is /(a)  =/"(a) fi
 f(a) . ., neither a maximum nor a minimum. =0, and/ (n) (a)^ =/<"'>(a) 0. continuing the process as illustrated in I and II, it is seen that
the derivatives of f(x) which does not vanish f or x = a
of even order (= ri), then By if is the first of /(a) (/) is a /(a) (H) is a maximum
minimum if if = a negative number;
(d) = a positive number.* / (n) (a)
f (n) If the first derivative of f(x) which does not vanish for x = a
odd order, then /(a) will be neither a maximum nor a minimum. ILLUSTRATIVE EXAMPLE minimum 1. Examine x*  9 2 x' f 24 x  7 for is of maximum and values. Solution. /(x)
/'Or) = z<  9
=3 { a; 2 x'2 + 18 z 24 x +  7. 24. 24 =
= 4. /./'(2) = 0, and /'(4) = 0.
~ 6 x 18.
Differentiating again,
/"(*)
 6, we know, from (//), that/(2) = 13 is a maximum.
Since /"(2) =
Since /"(4) = f 6, we know, from (/), that /(4) = 9 is a minimum.
3 x2 Solving, gives the critical values x =2 ILLUSTRATIVE EXAMPLE mum 18x4 and z 2. Examine e* + 2 cos x 4 e~ x for maximum and mini values. Solution. = ex f 2 cos f f*,
=  2 sin x  e 1 = 0,
/'(j)
x
2 cos x f <?"* = 0,
f"(x} = e
x
/"'(z) = g* f 2 sin x  e~ = 0,
= e' f 2 cos x <r T = 4,
/"(
(/), /(O) = 4 is a minimum.
/(r) .r <?* f Hence, from = for x for = 0,t x 0, = 0,
x = 0. for x for * As in Art. 46, a critical value x
a is found by placing the
zero and solving the resulting equation for real roots.
is the only root of the equation e*
2 sin r
e~ x
0.
tx = first derivative equal to THEOREM OF MEAN VALUE AND ITS APPLICATIONS 186 PROBLEMS
Examine each
1. + 4 x3 x4 maximum and minimum of the following functions for
of the last section. method values, using the 5. = 0,
= 3,
x= x = 0, Arts, jc x 2.
3. + 3 x 2 + 3 x,
3
x (x  2)
x3 2 . =L
= 2, x
x
4.
5. x(xl) 2 (x+l) 3
6 Investigate 4 x gives neither,
gives min. = 1, 22. gives neither. gives neither, gives max.
gives min. = 1.11,
= 0. .  15 x4 + 20 x 3  10 x 2 at x = 1. 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.
6. = ADDITIONAL PROBLEMS
1. If y = ex f e~ x find dx in terms of
, and ?/ Ans. dx dy. = ^
V;// 2. (3 x + 2 + V9 x2 f 12 x) = . . Vi) jf f 12 Prove that Vx4. 4 Prove that
T In 3. 2 Show that the curve x = t* + 2 * 2
, ?/ = 3 f 1 / 2 = i +4 f x 2 Vx 2 f 1. has no point of inflection.
5. y = cos x are points of inflection of the x sin x and
?/
Sketch both curves on of the curves 2 Prove that the points of intersection
first curve. the same axes.
6. Given the damped harmonic motion
s =r ae~ ht sin ct, and c are positive constants prove that the successive values
which v = form an arithmetic progression and that the corresponding values of s form a decreasing geometric progression.
 ax 2 is increas7. The abscissa of a point P moving on the parabola //
at the rate of one unit per second. Let O denote the origin and let T
where of t a, 6, ; for ing be the intersection of the xaxis with the tangent to the parabola at P.
Show that the rate of increase of the arc length OP is numerically equal
to the ratio TP DIFFERENTIAL CALCULUS 186 MP be the ordinate at any point P on the catenary (see page
line MA is drawn perpendicular to the tangent at P. Prove
that the length of MA
constant and equal to a.
8. Let The 532). is 9. The curve x 2 y of inflection. f 12 y = 144 has one maximum point and two points
formed by the tangents to the F"ind the area of the triangle curve at these three points.
10. Given In 6 = Arts. 1.792 and In 7 = and second by differentials.
between the two approximations. interpolation value lies 1. Calculate In 6.15 first by
Show graphically that the true 1.946. 2 2
2 2
2 2
ellipse b jr + a ?/ = a' b' find the length of the shortest
Ans. a 4 b.
intercepted between the coordinate axes.
tangent 11. Given the , 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
*
and one vertex on the curve ?/ = c~ r \ Find the area
along the line x =
. of the largest of these rectangles.
14. Find the A/?s. maximum and minimum values of X y ae a Given x 2 minimum 3 xy h 2 y 2
values of y.
f  2r> = 0.7788. y, if X 3 x 2 ac Ans. Max.
15. e~ 5 x 6 y = h 5 . a ; min. 0, find the Ans. Max. a(l 3 log 2). maximum and
1 ; min. = 5. INTEGRAL CALCULUS
CHAPTER
INTEGRATION ; XII RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS
The student is already familiar with the mutuinverse operations of addition and subtraction, multiplication
ally
and division, raising to a power and extracting roots. In the examples which follow, the second members of one column are, respecother column.
tively, the inverse of the second members of the
126. Integration. =
y = y a*, x sin x, x = log,,
= 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
namely operation, To of the integral calculus depend on the inverse : derivative
find a function f(x) whose
/'(*) (1)  4>W is given. Or, since
culus, it is we may (2) customary to use dj(x) = f'(x)dx = and state the problem as follows Having differentials in the integral cal write
cf>(x)dx : to find the function
given the differential of a function, 187 itself. INTEGRAL CALCULUS 188 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 of the given differential expression
(3) ; * integral sign ff'(x)dx=f(x } /in front ^ thus , 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 (b) If /(x) = sin a
, then/'(x)dx x, = 2 3 then/'(x)dx x' dx, = cos xdx, and
= sin cosxcfx
If /(x) (c) and x. /r = arc tan x, then/'(x)rfx = .,, 1 "T~
 arc tan x. + x~
Let us now emphasize what and X" 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 [=
(5) tf/(x)] from get fdf(x)=f(x).
i Therefore, considered as symbols of operation, 3 and dx inverse to each other When d
I is is /*
 / J by by
d, I they annul each other, as in as in dx are are using differentials, d and r followed followed we if or, ; inverse to each other. when we (2) in (3), (5), / J (4), are but 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
127. Constant of integration. ing section it 189 Indefinite integral. Prom 3 = = x3 3 x 2 dx, we have f 3 .r~dr since d(x* + 2) = 3 x~dx, we have f 3 xdx = since d(x 3  we have / snce rf(:r ) 7) = In fact, since where C 3 xdjr, + d(x* = c) f3 2 a rfjp 3 3 x 2 dx x3 = x3 ; +2 ; 7. .rrfjr, we have any arbitrary constant, is the preced follows that = a 3 + C. 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,
r
constants. Hence it has infinitely + Jf'(x)dx=f(x)
and since C is unknown and indefinite, many differing only by C; the expression +C /Or)
f is called the indefinite integral of f (x)dx. evident that if </>(#) is a function the derivative of which is
then <f>(x) + C, where C is any constant whatever, is likewise
f(x),
a function the derivative of which is /(a;). Hence the
It is Theorem. // two functions differ by a constant, they have the same derivative. 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
^
It is, _j_ where C is any constant. In other words, there remains to be proved the
Converse theorem. // two functions have
a constant. the same derivative, their difference is Proof.
tive /(x). Let <t>(x) and be two functions having the same deriva Place
F(x) (1) \l/(x) = F *(x) = </>(x) \l/(x) ; then, by hypothesis, INTEGRAL CALCULUS 190 But from the Theorem
F(x [Since by (1) Art. 116, F(x) the derivative of F(x) and (Z>), = Ax F'(x +
F(x + Ax)  F(x) = 0, + Ax)
.. Mean Value of  F(x is zero for all Ax). we have
(0 < 6 < 1) values of x.J + Ax) = F(x). This means that the function
F(x) = all when x takes on the increment Ax,
only by a constant.
\f/(x)
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 present we shall content ourselves with first learning how to find the does not change
that is, <t>(x) and in value at
differ In what folindefinite integrals of given differential expressions.
lows we shall assume that every continuous function has an indefinite a statement the rigorous proof of which is beyond the scope
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 verifying the results is that the differential of the integral must be equal integral, of this book. to the given differential expression. Rules for integrating standard elementary forms. The differencalculus furnished us with a General Rule for differentiation 128.
tial (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 differentiated, will yield the given differential expression ? Integration, then, essentially a tentative process, is and to expedite known integrals are formed called standard forms.
any integration we compare the given differential expression the work, tables of To effect 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
*
it of which employ Even though the may artifices which can be suggested by practice integral of a given differential expression may be known to
not be possible for us actually to find it in terms of known functions. exist, yet INTEGRATION 191 only. Accordingly a large portion of our text will be devoted to the
explanation of methods for integrating those functions which frequently 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 expressions to standard forms.
The (a) any integral of Proof. v, w du + I dv dw, I being functions of a single variable, du
/. / (1) (b) of differential expressions of the integrals of these expressions Differentiating the expression
/ u, sum algebraic sum equals the same algebraic
taken separately. A (du + dv constant factor dw) may we get + dv  dw. By = (du + Cdv  III, Art. 94 Cdw. be written either before or after the inte gral sign. Proof. Differentiating the expression By adv. gives
.'. (2) I adv= a I IV, Art. 94 dv. account of their importance we shall write the above two rules
M
as formulas at the head of the following list of Standard Elementary
Forms." On STANDARD ELEMENTARY FORMS
(1) i(du (2) I (3) Cdx / + dv dw) = idu +
=a ad v I =x+ V " dV = I dv  Cdw. dv. C. vn + l ^+l +C ' (*!) INTEGRAL CALCULUS 192 Cdv
(5) T = lnv+C J = In v + In c = In cv. (6) (7) (8) (9) C
/ sin v </v = I cos v dv = sin v + C. cos v + C. _/ (10) (11) I f sec 2 esc 2 i; i; c/v = tan v + C.
=  ctn v + C. rfi; (12) I sec i; tan v dv = sec v+C. (13) I esc i; ctn v dv = (14) / tan v dv (15) I ctn v dv (16) / sec v dv = ln (sec v + tan v) + C. (17) I esc v dv = In (esc *"
(18) Jf v 2 + a2 fffv
Va x 2 = esc v+C. In cos v In sin v + In sec v C. ctn v) v = iarc tan+ C.
a v 2 + C. G v
= arc sin  + C. + C. INTEGRATION Formulas 129. Proof of we (3). d(x = dx =x+ I dr, II, 7 ,n + \ 1 + /
This holds true for V Cj
n dv = C. Since (5). VI, Art. 94 tcdr, W+ = ? ,nH T
1 + (7. values of n except all 1, (4) involves division by zero.
The case when n =
1 comes under Proof of Art. 94 Since (4). (jjj we C) + get Proof of n These are easily proved. (3), (4), (5). Since 193 n = For when 1. (5). ,/., d(lnr+D =f
v
f
J get X, Art. 94 = In w + r. v The results we get from (5) may be put in more compact form
we denote the constant of integration by logc c. Thus
C^l J = n v v + i n c = i n if CVf Formula (5) states that if the expression under the integral sign is a
whose numerator is the differential of the denominator, then the
fraction integral is the natural logarithm of the denominator. ILLUSTRATIVE EXAMPLES* Work out the following integrations / 1. Cx^dx J 2.J
* = 6
776
j o +1
r + 1 + C r
= * + C, by : (4), where v = x and n = 6. / Vxdz =fx*dx = 2L+C = z*+C, by (4), where = When functions. learning to integrate, the student should have oral drill in a: and n = J. integrating simple INTEGRAL CALCULUS 194
3 Cdx
' = J y* r . = 4./ax*dx
3 5./(2x x 3dx J ^ + c = _J_ + Ciby
X = = a/Vdx where (4), r =x and and (4) +4 dx by (1) 3/xdx + 4/dx by (2) By <^+C. (2)  5x 2 3x /5 x x = 2/x 3 dx  5/x 2 2 d(x dr /3
 xrfx we
NOTE. Although each separate integration requires an arbitrary constant,
their algebraic sum.
a single constant denoting
write down
only  2 =f2 ax'^dx fbx~' dx f/3
= 2 a fx~*dx  bfx~ dx + 3
2 = 2a .26.
= 4 aVx f ./ + rx^dx by (1) cfx$dx by (2) by (4) + 3r.^ + C
5
'
1 rx
5 4 C. HINT. First expand. into the form (4). For
and
after the integral sign before x dx,
Those operations
its reciprocal fo/orr the integral sign.
balance each other by (2).
Solution. may be brought This insert the factor 2 2
b' : L =
5 ( fl % /
2 +
3 NOTE. The student
from one side is (a + 6x)l(26*zd I & 2 X 2)1 ^ Comparison with (4).
2
n= 2
a'2 + b' x'\
dv = 2b*xdx.
 v > = + C,
)[=^ /*i* ^5
5 by
(4)] "^ warned against transferring any function of the variable
would change the value of the integral sign to the other, since that of the integral. By <2) INTEGRATION
This resembles (5). we If insert the factor 2 c 2 after the integral sign and its reciprocal before
the value of the expression will not be changed. Hence 3 a C x dx
7
J 62 + c2x2
I ~ 2
3 a
= ^7 C2c xdx
r^
2 oj 6^ + cjc
I First divide the Solution. x / + 3 dx = x I 9 Solution. f Dividing, ^ _  The function to be
Example 1, Illustrative 2 Cdv
"~
I rj x 1 In (2  3. 3 a
= ^r~, L Comparison with
v it, = 62 3 a
= ;r^ +c , ln p 2 c2 v 2 * 2 dv
, /rv
b y ^ 5>
, +^
, , J r = JT 1 + 1 and integrating gives the answer.  2 f T.  f 3) ^  Substitute and use (1) etc.  integrated is called the integrand.
p. 193, the integrand is x
(] . Verify the following integrations. fx* dx = ^ + C. 6. a?/ 2 d?/ J*3 = a?/ 3 4 l* = _? +c
t" 3. / .r 5 dx = = + C. *  " . ~
8 / = r c:. t 5 . l
j PROBLEMS 1. (5).
c 2 x dx. =2 numerator by the denominator. Then ~ 2 f
J 2 x [ 2 Substituting in the integral, using (1),
11. 195 V2.r + r. Thus in INTEGRAL CALCULUS
= ~ 6x
2Va x
/* 2 4 x' V Vx
r h I) 2 cfcr ^r 4 5 In x 4 C. by 4 ~ "^ 7 +c  _
~ dx 8
6 z dz
:i = 4 4 V(532r 2 1
2
) SSz^"*" 4 C\ r / 2 rf/ __ "J (a 4 6/ 3 ) 2 4 C. INTEGRATION
36 = 3" /f*
x* '/ sin 2 = x cos x dx I +C a"
2 (sin x) 197 ' cos x dx = " ^QlTl 1*^3
l HINT. Use
oo making (4), r
j
ax cos ax ax
/ sin
J
 38. 39. 40. J = sin x, = dt> sin 2 aj^ /,
C . , h 2 a =  CQS cos x dx, n 2 r
' f C. ^ ftan " = 2 x cos 2 2 x dx Jsin see  V6 + sin = dx 1 cosaxdr / 4i v ax tan 2 ~ f r. = 2 V6 4 sin a.r ^ ^
a rf
42. tan = 44 Cr'<i* V2+ . 5> f a + r) ln(2 3 :! d< f
J ^+1^ _
~ In (a +M z
) .
26 = In In (^ + 3 + ?/) (;/" 4 L. > ' lie . + Ar . 49> be e  In
~ +
b r sin x dx
cos x i (a i / In (1 j 1 be 9 ) +
. N cos x) +
, /t
T. ^ =li n (a + b tan ^ 4 C.
f^/
o
tan
^ +
(2
= 2 x  In (x + 2) f C.
51. f
2
J
x +^
=  + 3 In (X + 1) +
62
50. f J a s ( h o ?/ )rf:r . . /Lt* f d 55. 6 = 2 In ; o / C. = 2. = Rin*^ f
^^
T" f C o C, INTEGRAL CALCULUS 198 Work out each of the following integrals and verify your results by differentiation. r
56. 2xdx 177===;.
V 6  5 x2 > Solution. /6  5 x* d [ ft Verification, (6  5 x*)l + ('[ = ft 3S(6  5 x 2 )4( 10 x)dx 2 x dx 57. O'* f 3 / j x 2 )dx. 58. 66. X(/J / 67, v 68 C
J 71. Vi^^ ' 5 ' <lt ITFT1 sin (iC (10
f_EL___
a^? + b J cos esc 2 72 (&fr 130. Proofs of (6) and (7). These follow at once from the corre
spending formulas for differentiation, XI and XI a, Art. 94. /^
ba'2T Solution. Cba* r dx This resembles (6). = b fa 2 ' Let r = dx = x * h By dx. 2 x ; then d?? = 2 dx. 2 before dx and the factor J before the integral sign, If we then (2) insert the factor we have = . + C. By (6) INTEGRATION 199 PROBLEMS
Work
1. 2. f6 out the following integrals. e** J JV dr dx = = + 2 c* r nr " + 7. C. 5. C. r(
Vf a ~ xZ dx f<' Bin x = 13. V? ) I c r2 (// dx 2 n/
%\c + dx cos x dx 11. /Vanfl sec 12.  "\ n r </// r" 'V f = + ^^ + C. C. C C. = r sin x = 2 VP  (i + C i&nf) ^ f/^ c a ^
) T. + C. f T. =
1 f In Work r^In 10 JV* = a(r  2 *ffC
10. = J fU" + f ~)(ir 8.J f 10' dx 4. a + C. out each of the following integrals and verify your results by differentiation. INTEGRAL CALCULUS 200 131. 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 dv r = f'sin cos v C sin r dv COS 7' _ r r) COS J V +C In cos v =
 In COM In r [Since Proof of (15). fctn v wc d,  In 1 f**^ = Jf J sec v sin v In sin v = sec v + Tsec r By (5) By (5) ^^
sin v C. + sec v ~ (5) r. v sec v tan /, In sec f
+ tan v
v + sec 2 sec sec r dr By In see r f (5) J Proof of (16). Since by r. r = J + In sec r tan tan + ?' ?> ?; sec 2 v , dv ! I J + tan r
(sec r + tan r)
sec + tan r
(sec v + tan r) +
sec* v ?> = In Proof of (17). Since r ctn r esc r esc v ctn r = esc v esc esc v ctn /, = f esc r 1 7 + esc ctn esc r esc v ay ctn ?> An alternative form of (17)
esc v dv v " v
+ esc 2M dv
! esc r ctn ctn v v) esc
/(/(esc r = 2 v
2 ?' I J f. ctn v In (esc v ctn r) + C. is = In tan %v+C (see Problem 4, p. 201 ) INTEGRATION
ILLUSTRATIVE EXAMPLE Prove the following integration. 1. This resembles Solution. fsin 2 ax c?x  i fsin 2
2 aj J = let r = [ +r.
then dr ; fsi
sin r eft'   f (tan  1  (tan 2 x we now  ~  cos r 4 T, we by get (8) (i Prove the following integration. 2. 2 x If . 2 iLLfSTRATH'E EXAMPLE 2 a dr. before the integral sign,  <*J L 2 a Solution. 2 ax the" factor 2 a dr aj flx fl For (8). ds and co * = Jsin2a*dj
insert the factor 2 o before 201 =
= )* 1 tan 2 2 = dx )'' x tan 2 J tan' 2 x sec 2 x + In x  cos 2 2 tan 2 s
1 \ f ( + 1 . By . (2), Art. 2 Hence, substituting,
 (tan 2 = sec* 2 x ds x dx = 2
\ J Then x. 2 (sec''  = ftan 2 ds 1 ) let r Now
I F .S 2 tan 2 *? Using (10) and 2 ds. d?' sec 2 d(2 ft an 2 x d(2 = x) ' x) sec I I 1 dr r  tan \ v dv .', = ? ^ n cos r <ur) 1 + tan 2 ds r tan bs ds = sec as 9. f  In (sec as
a djc esc v dv fsec 3 / fcsc 2 3 fctn
2 fs' I tan j: ds dr = / a?/ :i =sin x f// = ?, J ctn x ^ . sec 3 ctn 3 2 In sin  tan f ^ + tan x
f C. :i j f C. /  esc J \ ( Y J r f c/// = sec 2 J dx 10. tan In esc ay ctn 8.  In sec bs /I 4. f 7. sin rns h C. /^ 8./U 5. m ay + + C. C.
H C. C. x. I Verify the following integrations. / d. (14), the steps are as follows. PROBLEMS 2. tan 2 2 sec" 2 x dx /  x I ^V/x C. In cos 2 x. INTEGRAL CALCULUS 202 f^ =
X 11. tan x J COS12. 13. I 14 HINT. f (tan J ctn f tan (sec 2
<f)) tan ~ 2 (tan d$ ctn  = + jr + ctn 6 1 C . + sec 0) C. esc J f C. OOB.T f 1 r. Or dO = LL
'J + Multiply both numerator and denominator by cos x 1 and reduce before integrating. = f 15. J sin f 1 Sin **
16. ( = . ./ ./ 1 I j cos f 1 tan r  sec jr + C. ) + j' In (1 cos f . . 18. tan f
 cos jr J ^' djc .r)(/.r J >/4 L> f (\ cos 2 i(.r J f sin jr / ; .r) . 1 +r sin r) _ j^ J^L^L^ =
V f C. j> rn + cos^dj = }n(x + 21. 22. sin r ^l^^L, = 2V4COS., + C
_ cos /" 20. I 2 19. f(.r 4~ sin 2 Work C. ,s Vl f 2 tan f r. fl 2 tan 6 f out each of the following integrals and verify your results differentiation.
23. /"sin 24. ~
( Jcos r
esc*
25.J ftw 27. r
I esc 29. dr. (a </> y  . ctn a</>
1 y J J , 7 33. (/0. x ctn f rfj. f// tan 5 ^ J ctn Q
MJ (Lr. dr. 7 ?/ /"sin V.T . r 2 or Jtan 
f ^i
01 brtdjr. /> 28. 30. b f fljV/x. 2 sec  y^ c/j* by INTEGRATION O.T f (1  43. 35. 37. 203 sin 2 3  f a rfjr
J cos" bj" 38. I  (sec 2 45. 0 I (tan </> 1 f 0? Vu ' </>) rcsc
csc
40. 4 s f(tan 41. f(ctn.r 42. I (sec 'J ctn^  I) 2 I) 2 / 49 dr. 5 j dy. //) .r 3 f cos "./ csc^ f sec (V^
sin J 47. 39. 2 esc J t / !2 .r f ( . fo sin c
ctn f (10 4 csc r csc\r(/.r
cscj\r V  ctn .r (//. Formulas (18) and (20) follow
132. Proofs of (18)(21).
from the corresponding formulas for differentiation.
Proof of (18). Since , r d( arc tan a
\a we + rW ~
a ./ 1 ^^  r^:, by XXII, Art. 60 / r
get \  (II rf' 1 i
" arc ian "
?? n
+ " "= 4 ?'~ Proofs of (19) and (19 a ' ^
+r
i We prove (19) first. By algebra, we have a).
1 1 a v r f a2 a LI. _J 1 1 Hence a /) r f a \ by(1) Then = 7T ln
^1 d = _L
2a To prove easily (19 a), by i ( ??  a )7r ln
Ll (I + n C i algebra, 2 a 1 a +v a v a ' By (2), Art. 1 INTEGRAL CALCULUS 204 The rest of the proof proceeds as above. NOTE. The and (19 integrals in (19) r dv
r2 J a) satisfy the relation C __
~  dv
'  2 J '' a' 2 v' 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 *(\a = (j )' v
arc sin a
. we ^ Va
/dr get 2 2 Proof of Assume (21). differentiating, r a sec" dr V?' ./ dv = __ +a 2 2 = ;
a V?' 2 a Placing where seesec 2 z dz tan (tan z 2 + z __ a~ In (sec z + new a z is variable ; Vtan a secsera
C sec 2 z dz
J Vtan 2 z + + tan z) + +1) + r. In a r 1 by (16) + c. By (2), Art. 2 In a + /ir ln(*+
\ r, = In
I + Vr + a~)
2 we
(v ^ a~ get + +a +
2 Vtr' In the same manner, by assuming we . Hence, by substitution, sec z dz In (r r= z, ,,
( . + hence, P*+= = J a tan z dz.
<J  In
z = r C
a
J Va 2
I But tan r' by XX, Art. 94 => >K a, ) v a sec C. z, dv = a sec z tan z dz, get V
/dv
i r 1 ra
a 1 sec a tan z dz r \ , \ J va'sec^z = In
= In (sec a (sec z a L + tan z) + c
+ Vsec z 2 sec z ar J by
1) +c by (2), (16) Art. 2 205 INTEGRATION
ILLUSTRATIVE EXAMPLE. Work out the following integration. f_*_ = l arc taii^ + r.
3
4
+9 6
2 J
Solution. dv = 2dt, and This resembles
a = 3. 2, r_j_ir
9
4 z~ + 2J if we '
= 9 then r = 2 x
For, let f = 4 r* and a*
the numerator by 2 and divide in front
; (18). Hence by of the integral sign J x' we multiply get 2d*
(2 x) 2 + 2 (3) r=!f_*l_:=Larctan2+r. By (18)1
a
J
2j L r2 + 2 a =  arc tan
o <> PROBLEMS
Work out the following integrals.
1 r^_ = larctanf + C.
3 J x2 f 3 9 dg f , 2 6 f J  = + ( V.s^  16) 6 dx Vl6  In =
9 x2  arc sin
^ + ' C. r_lL = arc tan f + C. 9 _ ln EL +C . or ' i f ax dx
J x4 f 6 4
f. J (  = a
2 &2 ^
2) arc tan ! 4.
2
?> =2 f 9 3 arc tan ( \ C . T^
o  ) + C. f C. INTEGRAL CALCULUS 206
15. 18. Work f
J Vl
f
J V4 dy = i In
(ay
a + aV 2 <**  (u =
+ 3j + Vl + ay) ("*\ +
2
y
\ 4 arc sin 2 C C. . out each of the following integrals and verify your results by differentiation. The standard formulas (18)(21) involve quadratic expressions
with two terms only. If an integral involves a
quadratic expression containing three terms, the latter may be reduced to one with two terms by completing the square, as shown in (7^'ia, a L> ?; ) the following examples.
ILLUSTRATIVE EXAMPLE Verify the following: 1. =
+
Solution. x'' .
" This is in r
J jr 2 2 x +2 f + 5 + x 5 ^
f 1 arc tan = x f 2 j the form (18). J (r For  + 4 f 1 (\
(x f I) 2 f 4. ^ = r 2 j ^4_1 + 5 + let r I) 2 =x +2
+ 1, and
2 a = Then 2. dt; = dx. Hence the above becomes = r2
/; + ; a* arc tan a a /9+x /fv. V2 Solution. Now
Let v This
2 =x is f I, a in the x a* ?. 2 form =2 Then  C~ f  arc tan ~t 2 ^ y C. 2
9 2 arc sin 1 y. + :1:i * 2 x' C. 2
(20), since the coefficient of x' is negative.
2 (x'
dt> x f J) 4 1 = S (x 2 J) . = dr. dr
f_2dr
=2 r
^ v  (j ^v2+xx = 2 J)' 2 f d; JVa'r 2 = 2 arcsin 2 +
a c> by (20) INTEGRATION
ILLUSTRATIVE EXAMPLE f 3. +4x7 = 3(x 3z 2 Solution. <* f .
' 2
'J 3 x by form (19), r if = x + T = 4 x f a , ~ 3 x2 J + 2 + 1 x 4 x  C
J 7
*j, ~ In 3 x = 3 dx since also dr =  3 J) = J \ C
dr
3j r 3  a' Then wo have (ir. out the following integrals. 4. f J ' = dj<
. Va x  J o x2
2 jc*  = arc sin (2 s  3) f C. 2 2 x dx
/2 x  arc sin (x 1) f C. x2 = = ln
/2 as + + 3 y 2
s' + ^
lfxfx 2 ~
1 = A V5
2 V3 " \2 arc tan + C. (or PROBLEMS
Work 3 =3
+ J * + I  V)
= 3[(:r+{jVVl. j) 3[(x f ~~ 3x47 10 7  207 I/ f /2
\ + V5
+ l\
V3 /
3 INTEGRAL CALCULUS 208 16 . f
J4x Jfa
44x45 = J 1 arc tan
4 = (2^tl) +
2
\
/ 17. Work *
f
J V2
.  3 x  =
4 x2 =
Vll arc tan Vll

2 C. \ (*Z*\ + C.
\ V41 / arc sin out each of the following integrals and verify your results by differentiation.
jc 10 ' 4(^
V Vx  4 x f 13 09 f 2 .,
31 . _ r
J Vi> 2
I . a  Vl   4x41* r* 4 2 w2 r 36 42
x J 9 j.o 2 _ 4 4 S8./ V9 x 2 w dfx ^ _ i" t  : *'
28 f
J r2 _ 2 r  39
3 4 dx
 r ^ V4 x 2 y 4 l" 3 37. 15 P 2 t t 2 12 x f 8 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* 209 INTEGRATION
ILLUSTRATIVE EXAMPLE 3x " 1 r 1. Prove the following = V4^49lln (2 Multiply through by dr and apply Solution. C 3xl
J V4 x 2 f Then by (4) and (21) we ILLUSTRATIVE EXAMPLE ^
9 = r
> 3jdr
V4 x a h x 9) 4 dJ
r
J ^4 jr 2 f 9 obtain the answer.
2. 3x 2 +4z7=r 3[U f S V  V'l, by Illustrative Example
= x + 1. Then x = v  ?, and dr = dr. Substituting,
3
r
2 x  3
C<r^
dr _
^ T2 r J ^^
j3x^4x~7 ~J air^V)
 1 r 2 r (/r _ 12 r ^'~
~aj f  v 9 J r  V'
and (19), and substituting back = + 3, we have the above Solution. Let v C. (1).
__ 9 : ( "i 3, p. 207. 1 ) ,, _
a 5 Using (5) t> jr PROBLEMS
Work out the following integrals.
f 2 ~ 1U
10 f(2x 'Jx 2 + *)<** = _ VF^T^  arc sin j f C. result. INTEGRAL CALCULUS 210 9 r2 ln(1 6 4 = Vx 2 + 13. 14. V4 x  B. f 'J f ' .  6 x + C.
+ V2
+ 1 + Vx 2 + 2 2 In (x + x) C. r. f =  V27 f 6 x  + x2 sn
si 3 arc x2 fy+ ~')</x _
Vl9  5 x + x 2 V4 x2  x 4 ,. V In (r  Vl9  f 5 x h .r 2
) + C. f 5 : ~ .. h 1 9:r2) rjyizjjd J 1* f ,_ arc sin I 17 a 6 x2 15. /27 2 _ \3 x I  6x
!^J 12. ^ 'J ^J ^
Vl2x~4 x ^^'r
2   = In (2 x 2 Vl2 x  + V4 1 ~ x2 4 x2  + 4 x 5) f C. 5 5 Q /9 r X 3\ + ~arcsinr f C.
) out each of the following integrals and verify your results by Work differentiation. 19 /^T^' 20
21 2g jCi^4V/x. 28 .  V3 x 4  2
+ 2)^
a (5^
f
J Vx + 2 2 a 4 . 5 ' 'jT:ij 2 /yo^' 23 r(
^ 24.  lw r
V3 + 5r4  r  . r^4X J
25 (3 27.  r(i4J^. ^ V3j'x 2 30 f(8a
J x
' r QI  3 x)<ir
+x+1 (J" + 4)dr +
+ 2 x h 1
(3
o, f (3a +
33
J 9 x*  3 x  1
 r)(fa
(6
34 f.
2 V4x 12 x +
32. (2r 2 x2 )riJ ' 7 INTEGRATION
133. Proofs of (22) and To prove (23). 211 (22), substitute = a sin z.
dv = a cos z dz,
= Va 2 a 2 sin 2 z =
t' Then Va2 and v2 a cos z. Hence fVa 2 z; 2 <fo = a 2 f cos 2 z dz = To
z we obtain the result in terms of r, = arc sin  2 sin z cos z and a Substituting, we Proof of (23).
readily that
I (1) V?' 2 By Since tan z = it is Art. 2 = "~ a 2 a ?? a and a tan we show z, = a2 I (see Art. 132) sec* r dz. proved that l sec z tan z
sec z + I = we + a + ~ In
2 V? C'=C^ In a. Hence tan I In (sec z f a fx/V + a 2 d  (3) (5), have, from above, a sec z dz a sec z I = r substituting + a 2 dr = f sec 3 zdz = (2) sin 2 z by l)dz obtain (22). In a later section where f (cos 2z+ is (23) z) + r. derive, from (1) (r + v/r' and + a) + C' (2), 9 proved when the positive sign holds. By substituting v =
(4) I vV a 2 dv 2 = Comparing But we f Vfl 2
sec z z, a tan we obtain
z a sec z (see Art. 132) tan z dz = a2 (4) a2 = with
dtf = tan 2 z sec z dz
j  = (5) a sec a2 ( (2), a2 sec 3 2 dz sec z dz.
J we have y sec z tan and hence tan z z  ^ In (sec z + tan + C.
z) = obtain (23) for the negative sign before a Substituting in
2 . (5), INTEGRAL CALCULUS 212 ILLUSTRATIVE EXAMPLE f \/4  9 x 2 dx Compare with Solution. f \/4  Prove the following: 1. dx and setting (22), = P ILLUSTRATIVE EXAMPLE f \/3 x = 2 4 x f 1(3 x
1  Solution. 2)\ By v = x 4 <j , = a 2 (23), + = 3 dx  f = we have 4, C. v =3  a* 2 t> Hence dx.
d0. / <> 3 x, a 2 + Then dp 3 x. 3 J 4z7 25^2
18
Example  x f 4 Then . /} /. Using 2 ^ the answer. 2. Illustrative
3 x if arc sin = 4, r = a2 let 1 dx 7 / 3x* f 9 x 2 4 = 7 f V4  9x ,/ Using  \/4 and (22), x'2 9 =7 = 7 V9 f x + 12  12 x 2l) f C. I) 3 ( r2  fl2 ) dr. d?? C\'\lx* + 4 x and setting 2  V] = 2 f 4 207, 3, p. 3[(x x In (3 =x r  7dx = V^ f\ 4 = , . we  2
t' a 2 dr. obtain the answer. PROBLEMS
Verify the following integrations
1. 2. ^ Vl  4 = ? Vl 9 J 2 4 1 In (3
D fVl 9.r 2 dx 4 / 3. 4. f li /\ /~ \ 4 1 fV25  9 x 2 dx cij: = fV4 :r 2 .r 4 Vx 4 2 7 J In 4  V25  arc sin 2 x 4 C. 1 (a 24
7 9 2 x' + 2t / 5. : =
JVl 4x*dx 2 4 9 dx =  V4 ./ 2x x 7. TV3 8. rV5 2x4x 2 dx = x2 9 4 21l 4 4 a V.r  L> ~ arc sin 2 x 4 9 ) 4 C. 4) 4 C. ^ 4 C.  4 x2 4 V4 x2 f 9) 4 2 arc sin ^ = ^^ V5  2 x 4 x 2
 1 f V52x4x
4 2 In (x
arc sin (x  1)
2 x  x2 +
x 2 dx =
2 2 .r o t> 7 In (2 x V3  Vl 4 C. f C. rf.r 2 10.JVlO4x44x dx ~
= 2J 1 VlO  2 ) 4 C. 4 C. 4 x f 4 x 2 4 + 5 In (2 x  1 4 VlO4x44x 2 ) 4 C. INTEGRATION
Work 213 out each of the following integrals and verify your results by differentiation. 11.JV169 x 2 dx. 12. JV4 + 25 x 13. JV9 r 14. JV8 is. f V5 2  2 16. dx. 17. 3 .r f 2 .r 2 JVs 18. 1 rfr.  fV5  (Vr 4 x  4 2 JT +7*  8 .r J a dr. + dr. 7 dr. dr. 19. f V4  dr. 20. rVJ^^TTTn dr. 2 .r  ., dr. We shall now consider some trigo 134. Trigonometric differentials. 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
may be, this integration may be performed by means of simple other transformations and formula (4), it For example, if m is odd,
m ?/ sin we T write = sn m u sin u. m 1 is even, the first term of the righthand memThen, since
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
(1) integral takes the form
I (sum of terms involving cos ?/) sin u du. Since sin u du
r/(cos M), each term to be integrated is of the
n
v dv if v = cos u.
"
1
u cos u, and use the
Similarly, if n is odd, write cos" u = cos
2
2
sin u. Then the integral becomes
substitution cos ?/ = 1 form 7 (2) I (sum of terms involving sin u) cos u du. INTEGRAL CALCULUS 214 ILLUSTRATIVE EXAMPLE
Solution. I sin 2 1 Find . x COB* x dx =
= I  fsin 2 x(l
2 (sin = f (sin
sin' sin 2 x) 2 cos by 2 2 (2), Art. cos x dx
4 x) I (sin cos x dx + (sin x)  fi cos x dx x fl 2 sin x dx f sin" x) cos x dx 2 x)' x cos 5 x dx. x cos 4 x cos x dx 2 sin 4 x x x 1 sin 2 I 2 J sin By(4)  5 Here r =  cos x sin x, dr ILLUSTRATIVE EXAMPLE
Solution, l^t x J = \ x dx ?/ sin /sin' f sin' I and 6 respectively. 2, 4, sin' = x dx J w du sin 2 2 sin cos I Then x = 2u, dx = 2du. ?/. (3) Now Prove 2. ~ and n dx,  x :i J 2 cos x f C. Substituting, w du. :! j  ?i dw I cos 2 I  (1 cos 2 w) sin d?^ ?* j ~
Using this result sin u / du member the righthand in sin w d?/ ?/ = cos u f \ PROBLEMS
out the following integrals. dx = cos 1. fsin 2. fsin 2 B cos 8 dO :< j cos 2 3. sin ?, (/>  6. <f fl j* r^~^ d.r sin
4 esc x = ^T =  :< 1 6 sin' C. f r\ jr cos 4 2 J \ csc + $ :} j + fl + C. C. JT ^ 51 f
J COS 2 cos i d</> f C. JT ^ 4 r\ J sin^ fcos^ 2 8 sin 2 6 cos JT .r f/.r J
7. = ' cos B 4. Tsin'^ 5. : rf0 8. fcos 4 J sin 3 9. Tsin 5 x dx =
jc = sec d.r + = cos x + cos \ cos 5 f 1 JT cos 3 cos r< ?^ and substituting back u of (3) we have the answer. Work J C. +
or 7
\ cos x 5
i cos +
x C. f C. 4 C. x, INTEGRATION
10. 5 s n5
11. f J = x dx jcos ?/ d?/ ) V cos y = 2Vcos ?/
*  1 ( = . 3
sin s
, 2 i /!
V sin 5 * f jc f C. .r = cos v f  cos 4
/A
/
o '9 \ 12 r COBUL d,
''N/riin Work 3
f sin j sin A i sin , Bin , * , 2 f C. r / out each of the following integrals and verify your results by differentiation.
13. fsin 3 2 6 dO. 18. /sin " fcos 3 19. d0. r 1 sin' J </j. + 1 2 15. cos 2 a a ( dr. 20.J'coa J"sin
16. 17. J sin :! / cos :t Z1 <. < bt)dt. tan [ JV sin* cos^' Example <Jf. ^ /j 14. m/ cos 2 wf ' 1 II. To u du tan" I ^?irf 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
ctn or n u tan" = ctn The examples r' ~ 2
" a I  '
1 ctn w ?i illustrate the ILLUSTRATIVE EXAMPLE
Solution. tan n u tan 2 u tan 4 x dx 1 = Find . tan''  ri ctn u L> ?v L> (?ec* ; w 1 ) . By (2), By (esc ?/ 1) (4) Art. 2 subsequent steps.
tan r dx.
1 / x(sec ?  j 1 )djc  = ftan 2 = f (tan a sec 2 x (Lr   x) d(tan x) = ianiLH: _ tan x x f I tan 2 x \ (sec 2 do J ~ l)dx C. f 3 ILLUSTRATIVE EXAMPLE 2. fctn' 2 x dx
! Solution. Let 2 x = u. Prove = Then x = ctn 2 x
i fctn 2 x dx
:i (4) i w, = dx U  = i In sin 2 J dw. ctn :i x f C. Substituting, u du. and (10) INTEGRAL CALCULUS 216 Now / ctn a u du = I ctn 2 u du ctn u I ctri 2 w(csc u l)du = ctn u esc u du
=  \ ctn u  In sin
2 / Using member this result in the righthand ctn u du I 2 + C. u By (4) and and substituting back u of (4) (15) = 2 z, we have the answer. Example To III. These can be The first is step find sec^udu I or csc I when n easily integrated
to write n udu.
a positive even integer. is n = r sec 'u or csc n = csc"~ w u secw sec"" 2 = csc 2 ^ = (ctn 2
?/, The example shows Let \ x ,'j. f = 2 4 j* r/j sec 4 w d?/ .j 2 ~ I =
I* =
Substituting back in 2 2 w 2 ; w. By (2), Art. 2 =2 dx u, x dx \ = * du. tan z l ?, + 2 tan 3 z + C. Substituting, / //
sec 2 Prove f sec 4 Then x n. + 1) sec 2
n2
w + 1)
esc the subsequent steps. ILLUSTRATIVE EXAMPLE
Solution. 2 (tanw I / sec 4 sec 2 w dit. sec 2 u . 2
(tan M f 1 ) du sec 2 M d?z tan 2 w sec 2 u du tan u
' J M the righthand f tan u member (2), Art. 2 sec ?/dzt
L> f + by I By C. of (5) and putting u (4) = and
\ (10) x gives the answer. EXERCISE. Set sec L w = 1 f tan 2 w in the righthand member
and follow Illustrative Example 1 above. Example IV. When is 7? To find taw it ftanx 4. sec 4 z dx = =
tan x, u du or dv = sec 2 xdx, I ctn mu we proceed 2
z
\an x(tan
fi I Here n csc as in n u du. Example III. Find f tan x sec*xdx. = r = sec a positive even integer ILLUSTRATIVE EXAMPLE
Solution. I of (5), square, etc. (tan x) 8 sec 2 x dx sec 2 xdx+ ^^
+ f 1) + C. by (2), Art. 2 Ct&n*xsec 2 xdx By(4) INTEGRATION
When m is odd we may proceed as ILLUSTRATIVE EXAMPLE
Solution. I Find 5. = tan 6 x sec 3 x dx / = in the following example. tan* x sec x dr.
' tan 4 j sec I 217 (sec  j sec j tan x dx .r sec 1 )* .r sec x tan .r dr by j =
= (sec / 2 sec 4 j 4 sec j ft ^! _!** + **lr +
o < Here r = sec x, dr .r) = sec x tan x sec .r tan J Ci (4) dr, etc. in the above examples are obviously limited
For example, they fail in the following case. sec3 u du = I sec u sec = I 2 fir o The methods used
their application. (2), Art. I sec ?/ in du 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
1. 2. J out the following integrals. tan 3 xdxl tan Jo fctn 3  dx = 2 fctn 3 2 x esc 2 x dr 4. fcsc 4 5. 5 Jtan dx + ~ ctn1
O 3.  x =  ctn  30d6 = TV  ^> In cos  C.  i ^
'> J, 4 ctn 
tan 4 3 + U In sin esc 2 r  { jr ^ csc f tan 2 C. 4 :i 2 T f C. r:. 30+ J In sec 3 f? + C. cos4 sin 2 2 x cos 4 2 x = tan 2 j + 3
\ tan 2 x
6  \ ctn 2 x
2 + C. INTEGRAL CALCULUS 220
r
3. / A , cos 4 x dx = mn 3 x 2 x 5x /cos x dx = sin_2jr 4~~ x 5 /x
sin 2 :y ax dx sin 2 r
/ / x
o x
sin 2  cos M
tL
. r . ,, . . S1 n^arrL: or =x~ 3 x = x 16 ax sin 2 T +
. . , sn T
I 14. I r sin 4 ax o . j ^
C. ^ +
x __ sin 8 x C. 128
sin 2 n . h T. cos 6 x 4 12 cos 4 cos 4 x cos 3 x dr Work out + 96 4 cos o
j
x sin 2 x dx o sin 3 . cos 2 x x
fsin2x 13. 3 sin 4 x . _i_ ~~16
. 12. x 64 8 ^ n:< 4 Z_ 9. 10./(2 C. ., h C sin 2 8 8.J ~ a: 64 sin 3 2 x 4a 7 3 sin 4 48 1 sin dx x  48 L* 7. sin* 2 f =s n
c . i" 1 16
5. x sin 4
, f sin t> x
5 ~
= sin x :r = sin x
7; . 1 ^
h C. sin 7 x ^ . h C. rr each of the following integrals and verify your results by differentiation. 15. 3
fcosx) dx. cos 2 ax dx.
J 16. / cos 4 ax dx. 22. 17. I sin 2 ax cos 2 ax dx. 23.  cos 2  dO. 24. 18. fsin J 4 ^ 19. (sin 4 2 20. /
ij a cos 4 2 a sin 2 x cos6 x dx. da. 25. 26. f( Vsin 2 6 r(VSoT5 /(sin J (sin J(cos 2 x   cos 2 0) 2 d0. 2 2 sin 0) d0. sin 3 x) 2 dx. x f cos 2 x) 2 dx. x + 2 cos 2 x) 2 dr. INTEGRATION 221 Va  Vu 2
2
2
fl
u 2 or
135. Integration of expressions containing
by
trigonometric substitution. In many cases the shortest method of in tegrating such expressions change the variable as follows. to is When Va 2 u 2
When Va 2 + w 2
When V?/ 2 a occurs, 2 _ du 2_ Find f
J 1. a sin z Let u Solution. a then ; r a cos z dz
J a cos^ =  Here v'4 x 2
r = w ; Let M f a tan 9 = 2 = f ~^ a2 u / fcsc \ a 9 = ,1 dx _ = if jj_//^ z dz  '  w  ____
' <i J*. __
~1  In (esc z ~ 3. r
JTJ^/
^ wx'w 2 I a 2 and, using se(>   l{ Substituting, (///. f  In 9 } T ""'  (lij, sec z dz tan ctn __
~ !_ c z) f r_dz_ (/t ' sin r. = f = and mark the sides as in the figure.  u 1
~ In o Substituting back
we have the answer. = 3, J
</ right triangle in (4), a j ./ draw a .. u2 / a tan . /du
V and mark the =ii=r = a sec c Jc 1  o = r/*.' ~/ J u \ 7/*
 a secr (/r, 9 Then dn c. M, ,\ (/J* /rf?/ f
uVu c . j\ 4 \''u f / Since tan ; ' cosr Prove f 2. then j J y\ 4 j 2 = a sec z = a tan z. 1 z __ /*
' ^ cos z; _ J_ /'^..j.^, ' let 2 the radi M '.')!! right triangle Then tan ILLUSTRATIVE EXAMPLE Hence By them a sides as in the figure. Solution. 1 cos c (/. and, using (1), ck fl.' draw a ~ 2 ^
_ l! fl ( d</
/ j_ :i w 2)f For, since sin z ^ tan z ':: ILLUSTRATIVE EXAMPLE z. =a sin L> r 1 (3) z. a sec For 2 (2) z. a tan = ?/ 132133. in Arts. each case eliminated. 2
a' sir\~z =
Va
Va 2 + a 2 tan = a Vl 4
aVsec
Va 2 sec 2 z a (1) a sin u let occurs, let These substitutions were used
cal sign is in = occurs, let u and setting, as above, u = 2 x, /\ Then INTEGRAL CALCULUS 222 PROBLEMS
Work out the following integrals.
r dx _ x_ C dx __ x I i. ... f2 df r r a
o. ir 2 __  " .; (9 /' i f 2 arc sin ^f r. x I _  4 dx ?/2(/ " C
J I ,= ~r 1L v9  arc sin , W 2)& h C. 2 v/ r. 7. 2 .rVj' 2 f 4 \2 4. V.r 2 f 4 ./* 5 ^ .f
J 9, 2
// V/y 2 7 \5 ^^~
^ f 7 V25  + f c.
.r 2 r. .'/ 10. / Work 2 / out each of the following integrals and verify your results differentiation. 13 . r
J
A/ l*.f 2  9 j </// ; 9 fa
 VlOO  2
?/ du 2  5 by INTEGRATION 223 136. Integration by parts. If u and r are functions of a single independent variable, we have, from the formula for the differentiation of
a product (V), Art. 94,
F
or, transposing, 'u Integrating this, we  dr r d(uv) du. get the inverse formula, (A) u dv I   uv v I 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 directions can be given for choosing these factors, except that
(a) (b)
(c) tions, dx is always a part of dv;
must be possible to integrate dv; and
when the expression to be integrated is it it is usually best the product of two fwncchoose the 'most complicatedlooking one that it to is possible to integrate as part of dv. The
applied following examples will show in detail how the formula : ILLUSTRATIVE EXAMPLE
Solution. u Let du then
Substituting FindJ =x
= x cos xdx. and dv and dx v cos x dx \ cos ; xdx = sin x. in (A), u fx u dv
cos x dx v = x sin ILLUSTRATIVE EXAMPLE
Solution. 1. 2. u Let x Find = du=: then v du ~ f * in xdx~ \n ^ j x x In x sin x x dx. and dr and v xdx; Substituting in (A), ; dx
x f cos x f C. is INTEGRAL CALCULUS 224 ILLUSTRATIVE EXAMPLE Find 3. u Let then c du Solution. = = c ai j*xe**dx. aT and dv and a dx r = xdx;
r = x dx = x*  J in (A), Substituting = c ax Cxe 'dx
tt  ^ f ~ >ar a dx < But x 2 c aT dx is not as simple to integrate as xc ar dx, which fact indicates that
did not choose our factors suitably, instead, let =x u and e" r dv dx we ; dx
d
/xioz
c ai Substituting in (A), It may be more than xc ar dx x  (lx i'(' dx I a a necessary to apply the formula for integration by parts once, as in the following example. ILLUSTRATIVE EXAMPLE
Solution.  J
f,<ir ,.
I J Let 2 Find 4. u x' c" f dx. f and x~ dv = c a *dx; fc
c (lj dx = ax Substituting in (4), The term integral in the last may be found by applying formula (4) again, which gives
a Substituting this result in (1), ILLUSTRATIVE EXAMPLE 5. Csec*zdz then Let get a2 a Solution. we u
du  Prove
\ sec z = sec z tan z f \ In (sec z and sec z tan z dz = seQ
v = tan dv and + tan
2 2) zdz;
z. + C. INTEGRATION 225 Substituting in (A), j
In the new sec^zdz = = integral, substitute tan'c J sec :i zcfc = sec tan r  c  Transposing the integral f sec
 sec*,: + see rdc the righthand in  sec z tan z tan 2 rdz. z Then we 1. In ^sec r member and get tan f + z) dividing by C. 2, we have the required result. ILLUSTRATIVE EXAMPLE f f' Prove (>. ~
 l!lL<L!LH sin rudr ' (/' H then = c ar dw Let Solution. = w c/r = r and a = and 'd.r * MJ) < r n~ 4 sin HJ (/JT ; ^Ji.  /i Substituting in formula nx dr fc< sin new integral (2) the result (.4), = ^j^nx a r Let u then dw e = by . . ( N ' Integrate the is ^^ ()s ,/ parts. ax and ac ar dx t/r and r = cos nx djr ; = ^LM.
n Hence, by
(3 (v4), />* ) cos nx dx = I^'JJIIL^ /C Substituting in
c (i:r (2), sin we _ n / 1!
11 />, sin J nj f/ x> obtain
nf nx djr / a sin
2 n' nx n cos \ no* \ (] ^ ) / /*
 w c" 1 sin J nx dx. The two
member and integrals in (4) are the same. Transposing the one in the righthand
solving, the result is as above. Among the most important applications of the method of inteparts is the integration of gration by (a) differentials involving products,
(b) differentials involving logarithms,
(c) differentials involving inverse circular Junctions. PROBLEMS
Work out the following integrals.
1. 2. I jr Jin sin jr x dx djr sin = x(ln x x  x cos x
1) + C. f C. INTEGRAL CALCULUS 226
3.
A 4 sin ~
2
\x sin ^ dx
J
C
cos nx
x cos nx dx
J
n2  7 . 5. 6. I 2
/ w sec u J v u tan = sin 2 3 v dr > 7. ~ du a' 8. (Y^In a [ In 9. x <Lr 2  =41 + r
In 2 a 11.
12.
13. J
/  14. / arc tan
arc cot .r ?/ ~ ~ arc cos 2 x dx
arc sec fare esc  dl arc sec t fare tan Vj" 18. \ 19. Cc cos JTC x ^/.r = dO +  ^ 2 (x c '(2 ~~ +  +
+ (sin In = + In (1 J 2 V ~ In + 2 In +
+ + ?y y tan W 2 x + .r L> ) cos 0) ~ In (x + + [In (x + ) + r\ 2 + .r 1 2 + C. ) 4) Vx + r. + 1) C.  C. + C. + C. c\ C'. 1) + + V// ^
 1) arc 2 4 + V/ (/ r. j~~) Vl J arc tan x 1 +
x < + x~ "N/l I In (1 ?/ arc esc ^ ^ arc tan x dx 17. x arc cot y /x 15. sin x arc cos 2 x ?/ ?/ r/?/ C. 4 r. x arc tan r
?/ C. j x arc = + r f> cos In dx
d/y ^ cos j  x dx C. Rin ?/ f 10. f arc sin + r sin 6 r COS Bin n7/ rf* ,, . u In cos C. h C. n ^ 2
\ r' + nx sin + ?/ cos ~ or 21 + C. r. INTEGRATION
Work out 227 each of the following integrals and verify your results by differentiation. sec 2 25. Cx 26. J JT 27. J jr~ 28. J dr. ^ 36. fare tan V.r dr.
.r* 29. cos 2 2 x
cos { dr. JT arc sin r
37
37. f.r arc tan
j <Lr. 38. dr. rw.r 39. fare ctn ^ dr.
 Jarc J h .r*)' cos ^
!L '' dr. (/fl. 41 sin 7r/ (// "' 42. /<'' cos
.^ <///.
' dr.
> // arc esc nt 33. Tare sin 34. 1 j", J J (U dr. .r} '2 dr. ' cos  dr. 31. fare sec  32. I f (< 40. / r
7 ./ 30. ' f .r .r' ^1^ 43. /^ _ / 44 dr. arc sin r dr.
arc sin /.r dt. /^J 45.fr dr .r c  . ( cos ()S 4 f ill. \1 nf (ff sin <//. r
46. 2 / csc' ; dO. 137. Comments. Integration is, on the whole, a more difficult opIn fact, so simple an integral (in
eration than differentiation.
as
appearance)
r
\f'jr sin x dx 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 explained below in Art. 176. At this point let it suffice to remark that
the methods thus far presented are adequate for many problems. cannot be worked out Other methods will ; be developed in later chapters. INTEGRAL CALCULUS 228 MISCELLANEOUS PROBLEMS
out each of the following integrals and verify your results by Work differentiation. ~ 2x) 2 2' 12.JV
IQ
13 ' 14. 15. r
J ^
j j dx. ^  32. j ax cos ax dx. 33.  34. ax cos 2 ax dx. sin 2 (1 f tan x) 3 dx. (1 cos
f sin f J (1 Vx)dx. ^ct
37. ry^' 38.ysin tfdx 0)' Q3 dt cos
sin 2 t dt. 2 6 cos 3 6 dd. . ' '^ 20 y ( 35. 17.J(2tan2 r ctn x) 2 dx. J"(x 36. 19 sin x) 2 dx. (1 ' 4 r sin 2 IG.Jln + (c* Vx + l Vx  1 2 2 39. 40. sin 4 sin d</>. j J cos a cos 2 a da. CHAPTER XIII CONSTANT OF INTEGRATION
138. Determination of the constant of integration by means of initial
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
conditions. order to be able to determine the constant of integration, to have
in addition to the differential expression to be
integrated. Let us illustrate this by means of an example. some data given ILLUSTRATIVE EXAMPLE. Find a function whoso
and which shall have the value 12 when .r  1.
2
(3 x' Solution. 2 5k/.r .r f first the differential is = Thus .H  derivative expression = 12 Hence x x2 :$ + 5 j* 2 x f 6, be integrated. JT f T,
of our problem this ;  1 3 x'2  where C is the constant of integration. From the conditions
that is,
result must equal 12 when .r =
1 to is 1 i 54 <\ r= or 7. the required function. f 7 is 139. Geometrical signification of the constant of integration.
by means of examples. We shall illustrate this ILLUSTRATIVE EXAMPLE 1. Determine the equation
curve at every point of which the tangent line has of the the slope 2 x. Solution. any point is Since the slope of the tangent to a curve at
^, we have, by hypothesis,
' cfc dy = 2xdx. Integrating, y = 2 (1) y = x'* where C is C a series Cx dx, or + C, of values, say 6, 0, y x2 3, (1 ) Now if we give to
yields the equations the constant of integration. 4 6, x'2 , y y  x'2  3, 3
with the jyaxis and having 6, 0,
parabolas with axes coinciding
as intercepts on the yaxis.
respectively
that is, they have the same
All the parabolas (1) have the same value of whose loci are , ^; direction (or slope) for the same value of x. 229 It will also be noticed that the difference INTEGRAL CALCULUS 230 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),
in fciving 4 ~ f 1 or r, Hence the particular curve required is r~ M.
y'2 the parabola y 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 The Solution. condition of by the equation ( {,i t he problem is expressed / separating the variables, or, ?/ </// 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 9 41(5 ~ 2 r. then (H, 4), 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 variable and the function.
of  1. IT 3 2. O n n\ 9  3 4 r
1\ '2 f> Answer vnlnc of junction Junction S 20 \j  304 4 ii 1 \ 4. sin 4~ cos  .V 3 JT+ 13.
3
2 J ^ 4 i
2 L1 // 2 .r + f; 2 62  cos 041. sin MJh
6. sec 2 * :r 2 8. for 9. 43 Vt 4 (i tan 4 a.r 4 4 r Vt
10. ctn 6
11. 3fe 2 '2  </> 4 In sec X
a 2 4 tan esc 2 6 </> 7T 4 a 4 5. jr\ 4. 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. A us. m. 13. x. 14. Straight lines,
// Parabolas,  A nur //  Parabolas, k j 2 f f .r //' f ('. (\
(\ x2
Semicubical parabolas, 15. 16. L
 17. x2 Cubical parabolas, . //* ;  .H \t Cubical parabolas, 18. //'' \ Seinicubical parabolas, ^ tt \ . ' .r l f C f (\ .r~
.J r. f +  .r //< \ ' J, r. ?/" 19. 20. 21. JC
   Equilateral hyperbolas, a 2 // Hyperbolas, ^ 22. Ellipses, //'.r bjc 4 2 a C. J"' = .r// 1 12 2 ~~ Equilateral hyperbolas, \r r. r. ' // r. a''// a'y 23. ii^ Circles, j f //" f .r 'J  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. 26. 2 4//.S. (1,1). .r; 25. 4 //; (8, 1). JT//; '1 In (1, 1). jr* // ~ // In u 4 f 1.
jr  s~  4.
9. f'' 27. jry; 28. ^4;
+
y   30. 4,;
2 4 ^ 2 r
;! . = Uf (?/+ (0, 0). ^ + y*  2 1; l) 2 h 3. ; 2  ( 1 ky In // = r 3 In //  2(xV^  x (4, 1). 4^.
15 j* f<s k ?yV7; 32. = (0,1). (1, 1). x 31. y 1 ?/ 29. (0, 2). (2fl) 1. 4^^15. 8). = 0. INTEGRAL CALCULUS 232
33. ; (1, 9). ; 35. *3C
" .r = when x 46. curve \ ' x cos 2 y 40. j* ~h l)<tr, ?/ = 7 when y = "N/2 />./ ~ A r/.r, when x = 1. ; (4, J TT). Find the value = of y ~ 17. Find the value 2 ,4r?*. 7>. j*Vl()0 ~ j' 2 r/^, A of " ' 2 = when ?y ~ .r 0. g ;? Find the value . of 8. ij cos 2 6 dO, p 6 when # TT. Find the value 0. Find the value of J of p TT. Given ^s 45. when O /'I
' A??.s. Given dp 44. when ' 3g f 4 (2 Given dy 43.
?/ 3 3. Given dA 42.  2 y Q ____L__ x2 Given dy 41. when x
when ~~ 37. (1,4). = /V4 f f 1 dt, K = when / = s = 2. t of a certain curve //" = .r. Find the equation of the
passes through the point (3,0) and has the slope ^ at that point.
.4w,s\ 6 y = j" 3
6 jc
9. At every point
if it  47. At every point the curve
6 x +
48. if it of a certain curve passes through the point ~
1 y"
(1, 0) and Find the equation of
is 6 at that point. y Find the equation of the curve at every point of and which passes through the point (1, 1) tangent to the line
Ans. xy + 6 x = 6. which ?/" = with an inclination of 45. 49. Find the equation of the curve at every point of which y"
which passes through the point (1, 0) with an inclination of 135'\
50. Find the equation equal to 2 a. HINT. From
51. (4), Art. 43,  and curve whose subnormal is constant and
4 ax f C, a parabola.
Aws. y'2 subnormal = y^ ax Find the curve whose subtangent (see (3), Art. 43).
52. of the / is constant and equal to a
Ans. a In y = x + C. Find the curve whose subnormal equals the abscissa of the point
Ans. y 2
x 2 = 2 C, an equilateral hyperbola. of contact. CONSTANT OF INTEGRATION
53. y =R Find the curve whose normal when JT = is 0. HINT. From Art. 43, length constant (=
An*, a* 2 normal  ij^/l of * f 233
assuming that /?), + = A>2 ?r , a circle. f^\", or 54. Determine the curves in \\hich the length of the subnormal
proportional to the square of the ordinate.
Ans. // = is 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. The 140. Physical signification of the constant of integration.
will illustrate what is meant, fol lowing examples ILLUSTRATIVE EXAMPLE 1. Find the laws governing the motion
which moves in a straight line with constant acceleration.
Solution. from Since the acceleration we have _ <* or To determine f ^ ^c > v dx or s (3) + These values substituted = ft dt + vodl
= \ // 4 ?'o/4
. 2 in (3; = s (4) that is, let v  p, when (2) *, To determine C, suppose that the (3) ?><>; . Art" 51) we ^ ct from *=/ Hence is C = Vo. or C, v=ft + = Integrating, becomes (2) Since /, in (1) give p () =. (1) constant, nay ('. C, suppose that the initial velocity These values substituted is , *=// + = 0. Hence 59
I dv=.fdt.
(1) 1 (A), Art. * of a point initial Integrating,
r\ distance f h or C, becomes
B = J/J is ; give 2 + (,/ + *U. C=8 . that is, let =o when /=0. INTEGRAL CALCULUS 234 By substituting the values / 0, vo = 0, 8 = 0, = h, in (2) and
laws of motion of a body falling from rest in a vacuum, namely, = ?' Eliminating / and (//, h we get the ^ gl*. between these equations gives ILLUSTRATIVE EXAMPLE (4), v r t> <//?. Discuss the motion of a projectile having an
velocity ih\ inclined at an angle a. with the horizontal,
the resistance of the air being neglected.
Solution.
of
let 2. Assume the A'OVplane OX as the plane and ()Y as vertical, and
the projectile be thrown from the origin.
Suppose the projectile to be acted upon by motion, as horizontal, ^ gravity alone. Then the acceleration in the horizontal direction will be zero and in the vertical direction
Art 84
dr
dv
_*
' ' = 0o cos a o sin a = and Hence 7v (6) (C) and and initial dx
~ a and c'os <.* and ru Art. 83, ~ = </// ~ and a: ((5) we
f) ~ ru = giving , + gt and , a cos ?,' () r, sin ?>o ~ ^
. sin a. ?'o therefore (5) gives ; , sin a, ( yt ? f o d/ dj' Integrating, yt f C2 . (\> ?'o = (F), velocity in the vertical direction. cos d/ or = ru ?>< (Z>), Hence, from g. velocity in the horizontal direction, initial =  ^ = 0. and 0. fi ~ (\ But from = vf Integrating, But initial Vo cos a and <:// g\ dt dy + sin a dL cv i\> / get cos a f / ( and '.i To determine r 3 and (\ we obser\e
Substituting these values
4 = in (G  x ( \ ~. y ~ that 1 ' \ f// when f
/ ?;,, = sin 0, 4 ^4. and x ^/ = 0. gives C arid 0. , Hence
(7) 7>o \ y (8) Eliminating t between (7) y (9) cos = a and /, + f/'~ (8), x tan and  ''o we is <x  t. obtain a
'2 which sin r<> cos 2 a the equation of the trajectory and shows that the projectile will move in a parabola. PROBLEMS
In the following problems the relation between
the relation between s and Mf Ans. s = s a(t =  2 when 1) f / =
2 b(t r and / is given. Find 1.  1) + 2.
3. r =^ f  CONSTANT OF INTEGRATION 235 In the following problems the expression for the acceleration
Find the relation between r and if r
2 when = 3.
/  4 4. An*, f2. r = 4 f  J /'  1. /  7. An*, 32. With what 10. = s 20 t  .x 16 12. 15 f 3. 'JO when  ft. when 0. t 9. /.  16 cos 2 t. velocity will a stone strike the ground if dropped from
ft. high? vl = ;?'J.)
An*. 87.61 ft. per second.
/ 11. With what velocity will the stone of Problem 10
thrown downward with a speed of 'JO ft. per second? with a speed of 20 given. for the acceleration is given. 8. 4 /'. the top of a building 120 if V/ 5. In the following problems the expression
Find the relation between and if = 0, v
.s is t per second? ft. strike the
if ground
thrown upward Ans. 89.89 ft. per second. A stone dropped from a balloon \\hich was rising at the rate of
per second reached the ground in 8 sec. How high was the balloon
the stone was dropped?
Ans. 904 ft.
In Problem 12, 13. per second, how long if the balloon bad been falling at the rate of 15 \\ould the stone have taken to reach the ft. 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. A 15.
particle sliding on a certain inclined plane is subject to an acceleration downward of 4 ft. per second per second. If it is started upward from the bottom of the plane with a velocity of 6 ft. per second, find the distance moved after / sec. How far will it backward?
16. If
initial the inclined plane in Problem 15 is 20 ft. long, find the necessary
in order that the particle may just reach the^tpp of the plane. speed Aits.
17. A in 1 sec.
18. go before sliding
An*. 4.5ft. A 4 VlO ft. per second. ball thrown upward from the ground reaches a height of 80
Find how high the ball will go. projectile with an initial velocity of 160 ft. per second
ft. from the point of projection. ft. fired at is a vertical wall 480
(a) If a 45, find the height of the point struck on the wall. An. 192 ft.
the base, of the wall.
Anx. 18 or 72.
so that the projectile will strike 80 ft. above the base.
AUK. 29 or 70.
for the maximum height on the wall and this height. (b) Find a so that the (c) Find (d) Find a a. projectile will strike Ana. 59 ; 256 ft. INTEGRAL CALCULUS 236 the acceleration of a particle moving with a variable velocity r
is a constant, and if r
is the velocity when / = 0, show 19. If kv 2 is that  where k , =+ v 20. The kt. r,, of speed, is resistance of the air to an automobile, within certain limits
proportional to the speed. Hence if F is the net force gen erated by the motor, terms of /, we have knowing that M when r  = F Express the velocity AT. Ans. 0. / F v K in *c (1 A/
). ADDITIONAL PROBLEMS
The temperature of a liquid in a room of temperature 20 is observed to be 70", and after 5 min. to be 60
Assuming the rate of cooling 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
1. . Ans. 33.1. observation. curve whoso polar subtangent is n times
length of the corresponding radius vector and which passes through
1
point (, 0).
Ans. p = ac n
3. Find the equation of the curve whose polar subnormal is n times
length of the corresponding radius vector, and which passes through
Ans. p = ac" e
point (a, 0).
2. the
the the
the Find the equation of the . A particle moves in the .r //plane so that the components of velocity
parallel to the .raxis and the //axis are ky and kx, respectively. Prove that
4. the path A an equilateral hyperbola.
particle projected from the top)
is 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.
5. A particle starts
xcomponent of velocity
6. (a) Find the position from the origin of coordinates and
P 4 and its //component is 4 t. of the particle after (c) .r /  4 /, = 2 / 2
Find the distance traversed along the path.
Ans. s ~ J t* + 4 /.
Find the equation of the path.
Ans. 72 .r 2 = M
48 2 + 576 y.
i SUGGESTION. Choose the minus sign when .r :i for in Problem 2 (a), p. 85, jr = c\n + VfL ~
/ / Ans. c for
(a, of the which
0). ?/"
) V  Vc 2  =c
y*. / 2
(Art. 96) a ds Ans. tangent and assume y
A ( \
8. Find the equation of the curve
and which passes through the point ?/ which the length 0. . i/ ?/ Find an equation of a curve
43) is constant (= c). 7. (Art. sec. its / sec. / Atis. (b) in is p 2 = =
2 3
p dO a sec 2 t 6. CHAPTER XIV
THE DEFINITE INTEGRAL
141. Differential of the area under a curve. tinuous function <t>(x), and Consider the con let = y 0(x) MP be the equation of the curve An. Let CD be a fixed and
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 (=
we area MNQP). Completing the rectangles 1\1\KI> and MNQS, see that Area MNRP < MP or are&MNQP < area MNQS,
Ax < Aw < NQ Ax
  ; and, dividing by Ax, if*
x Now let Ax approach zero as a limit then since M P remains fixed
NQ approaches MP as a limit (since y is a continuous function
; and of x), or, we get 1 using differentials, = dn y rfr. Theorem. T/?e differential of the area bounded by any curve, the
to the product oj
xaxis, a fixed ordinate, and a variable ordtnate is equal
the variable ordinate and the differential of the corresponding abscissa.
142. The article that It follows definite integral.
if the curve AB is =
dw = du then
(1) MP * In this figure
is less than
'erse
reverse the inequality signs, from the theorem in the last the locus of NQ ; ?/ dx, or <f)(x)dx,
if MP 237 happens to be greater than NQ, simply INTEGRAL CALCULUS 238 where du is the differential of the area between the curve, the
and two ordinates. Integrating, we get Denote ( </>(/) + ds byf(x) xaxis, C. ' (2) We r by observing that u =
Substituting these values in determine when x = a. we get o=/(o)f r,
r = /(a;. and hence Then (2), becomes (2) w /(.r) /(a). (3) The required area CKF1)
Hence we have
Area (A) Theorem. The the value of u in (3) the curve corresponding This difference is to = x yds I for x whose ordfnate a and, = 6, .r = = a and x is y, the xaxis, = 6 and b. represented by the symbol*
S*h (4) when x CEFD = f(b)  f(a). difference of the values of bounded by gives the area
the ordinatex is yd* I or fh
I 0(r)rf.r, ^'u /a is read "the integral from a to b of //rf.r."
called integration between limits, a being the lower and The operation
and 6 is the wpper limit.f Since
For, always has a (4) definite value, it is called a definite integral. if
I c/)(.r)dr/(.r) + T, t/ r
then h I 0(.rW.r /(r) + T = [/(&) + T]  [/(a) + C], Ja
or the constant of integration having disappeared.
* This notation f The word is "limit due to Joseph Fourier " of its range (end value), Theory of Limits. U 768 1830). means merely the value of the variable at one end
and should not be confused with the meaning of the word in the in this connection THE DEFINITE INTEGRAL We may 239 accordingly define the symbol
r*> /*. or <t>(jc)djr j n df I J 'a as the numerical measure of the area bounded by the curre
and the ordinates of the curre at j ~ a and JT // ~ the xajris, 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 .ruj/.s, ami both a and b are finite. The process may be sum 143. Calculation of a definite integral. marized as follows : FIRST STEP. Integrate the (jircn differential expression.
Substitute in this indefinite inlcynrt first the upper
then the loirer limit for the rar/ahle, and subttacf the last SECOND STEP.
limit and result from the first. not necessary to brinpc in the constant of integration, since
always disappears in subtracting.
It is it ILLUSTRATIVE FAAMPLE Find 1. x' dr. j
* Solution, f JT~ djc  I J\ I ~ 4
1 'Hi   1    12 1 \ . ws. ^ '* /7T ILLUSTRATIVE FXAMPLE
rn
Solution. Find i sin I 12. j djr ~ Jo 7 / *
: f JC* Prove f  Evaluating in (1) gives the * d)(j) is lie <i arc tan  i 1  1 A  arc tan 7T /, 4 a V~
4 J r
" (19) or (19 a), r a), ~ Q
.>  decide between the use of (19) or (19
x increase from  1 to 0.
2 to 0.
Then v (= 2 x) increases from
2
Hence v 2  4. But o = 9. Therefore ?'~ the bounding ordinates   4 1 To dr ^ nK  J . a rj r(} i a a Comparing with " ' / /o
Solution. 111
) J  arc tan  ' Jo 1 I cos / ,/ ion.
Solution d,r. I 1 3. .r ( I?
J< l ILLUSTRATIVE FA AMPLE MH I  ~ v> ' T1 f) ""
' 2 r, ry  :j f r/?> ^ 2 consider the limits. '
, 134  1 and (19 a) c/j. The values of must be used. Thus /* answer. below the The result is negative because the curve and j%axis. continuous und singlevalued throughout the interval [a, b]. INTEGRAL CALCULUS 240 144. 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 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
variable. by an example. illustrated ILLUSTRATIVE KXAMPLK. Calculate
Solution. Assume x Then dx  4 z = x^ ' dz, z 1
. x* z'', = z. Also, to change the limits = and when JT _ r2 4 z z'' 0, z 16, JT r* dx x
\  z we observe that when 0, 2. dz + X* 1 ! Ans. 4 arc tan 2. f 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 manyvalued function of the other, care
must be taken to choose the right values. PROBLEMS
1. Prove that f V(.r)dr
' Ja Work
Jo a f f(x)dx.
Jl> out the following integrals. ra
2. / = 2 (a'  JC 04
= ~ x*)dx =  9. 4 Jo x 5.6094. f 1 0.3167. 2x
# ~ ln 9
2>
 W
g i< *
i '() v . X + 1 _ ^
3 12. i^ o f ^/2 ^o + 2 cos 6 d6 = 4. 7T in 3 x cos 3 x dx = THE DEFINITE INTEGRAL
Find the value of 241 each of the following definite integrals.
l 20. jcc " dr. I 21. / '"cos (3 J0. Jv
22. I "sin ~ cos dO.  23. 145. Calculation of areas. On page 238 it was shown that the area between a curve, the jaxis, and
x = a and x == b is given by the formula the ordinates rb Area= (B) I y dx, Ja where the value of y in terms of x is substituted
from the equation of the given curve.
1. Find the area bounded by
the /axis, an4 the ordinates x = 2 ILLUSTRATIVE EXAMPLE
the parabola y and x = x, 4. Solution. Area Substituting in the formula, ABDC = ILLUSTRATIVE EXAMPLE 2. Find the area bounded by the
 3, x ~ 4.
the xaxis, and the ordinates x
Solution. V'25 Solving, y r. 4 arc sin  f 6  ^ arc sin ( g) 2 y' =25, (22) = 31 .6. Am. The answer should be compared with the
) + Hence by =6 circle x'2 area of the semicircle, which is =39.3. 3. Area under a parabola wtwsc axis i parallel to ttw
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 zaxis, and the ordinates of P and P" equals
4 y' + y"} if 2 fe is the distance apart of the ordinates of P and P".
i h(y + ILLUSTRATIVE EXAMPLE yaxis. In the figure (p. . INTEGRAL CALCULUS 242 Take the ?/axis along the ordinate of P',
The equation of a parabola with axis parallel Solution. AB = 2 k. z p. 4, (x 2 [>(y h)' k). this If solved for is as in the figure. Then
to the #axis is, by (7), y, the result takes the form
~ (1) The H if x
j 146. form. 4 0,  ?/' /,, r)ds //(// + 4 then have + ?/'  // = ah' * (B), + 2 ch. ; r /// f 2 ?/" 2 f /? ) ; o f r. />// ( /(/). and </>(/), Area r// = M. /^ ^\" O.E.D. ) Here When j and when jr = I Jo Hence the , //
* as ^ b sin // = and Find the </> (7, = </> in b sin </>(/)/'(/)</*, / area, =
of i^ when j = b. the ellipse whose parametric c/>. </^, a sin 0, Substituting these
f" // (/>, ds and / are a cos .r Solution. SI Hence J'(t)dt. fh = x (/>(/). .//, r;, KXAMPLK. ILLUSTKATIVK
equations (Art. djr // <1 /
./a = y/ [!> when where 4 by is, Area when the equations of the curve are given in parametric
Let the equations of the curve be Driven in the parametric form CD Area r  /{/>" _ jr We u) O/'' y/ " ', + 2 bx  Therefore r. y'  if + (ax J("h + 2 bx A /'/'"/? (= required area (2) P" + as* // </> IT i c?</>. ; 0. (1\ above, f
I . , ., ah sin" , <p we
, , ad) get ~ irab
4 Ji>7r entire area eciuals TT?). .4ns. PROBLEMS
1. Find by integration the area of the triangle bounded by the line
y = "2 .r, the .raxis, and the ordinate .r = 4. Verify your result by finding
the area as half the product of the base and altitude. Find by integration the area of the trapezoid bounded by the line
= 10, the .raxis, and the ordinates jr = 1 and x 8. Verify your 2.
jr f // by finding the area
and the altitude. result sides as half the product of the * For a
rigorous proof of this substitution the student
on the calculus. treatises is sum of the parallel referred to more advanced THE DEFINITE INTEGRAL
Find the area bounded by the given curve, the 243
and the given jaxis, ordinates.
3.
4. 3 y .r .r y .r ; j .r .r// j jr ; 10 = // 2 A 18.
3, .r = 54. 3. 3. 9;. F In fc. = .r ; , 64. ' .r .r ; ?/ 9. Aw. 4. = 0, = 3.
= 9 .r
= = r< + 3 .r f 2
= j2f jf 1; = 2, =
= a, =
= ?/ 5. = jr 0, 2 6.
7. = .r ; j  5. = 0, .r 0, f~) 20. Vj 44 = 10. a//
11. +
=
/r
y = ?/ 12. 13. .r V 4 =
+ .r 4 .r .r + 2  J ' .r 16
jr a. 14.
15. JT 17. ?r 18. 19. ?/ =4
4 jr 9 jc 20. jry  ?/ 8 24. ?/ 25. =
=
= y ; 0,
// .r <>  //  .r .r //axis, ; jr ;  .r ;  { .r  1, L> ./ 0, 0, = ?/ \ = .r = = 3. 8. 2. and the given lines.
5.\. 3. 4;.
:i  rrV 21. // 22. 3. 4. <//r and  .r^ ; 0, // ~ find the area of  // 0, // ; // a. a. one arch.
AUK. 2 cos.r.
2 sin .r jr AH*. of the following curves cos 2 ?/ = 4  4. // 0, 1, y ; // = :{ /y Sketch each
23. ; 0, // // ; j  //' Find the area l>ounded by the given curve, the  = // 16. 2 j = . \ ; .* .r ;  =  1, = 0.
=  2,  0.
=  4, =  2. .r ; a* ; JT ; 4 1 4. 8
 jr. sin irx. 1 4. 26. ?/ 27. Find the area bounded by the coordinate axes and the parabola Vx f 28. V?/ = \x. Va. Prove that the area of any segment perpendicular to the axis of the parabola of a is parabola cut off by a chord
of the circumscrib two thirds ing rectangle. = k. any two points on an equilateral hyperbola xy
Show that the area bounded by the arc PQ, the ordinates of P and
and the xaxis is equal to the area bounded by PQ, the abscissas of
29. and Q, 30. xaxis, P and Q and the are = %a(c n ^ ns + P ^ 7/axis. Find the area bounded by the catenary y
a.
and the lines x = a and x = Q, c V, the a ^( e
\ _ 1\
e) INTEGRAL CALCULUS 244
31.
2 x' = Find the area included between the two parabolas 2
t/ = 2 px
Am. 2 py. and
f p 2
. 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. *f32. 34. y 2  Find x'2 (jr'2 35. the  1) area bounded and by the curve the by whose equation x2. line is Ans. 2\/3. Find the area inclosed by the loop of the curve whose equation is 2
36. Find the area bounded by the curve whose equation is y = j* 3
x2
Ans. ff
and by the line JT = 2.
37. Find the area inclosed by the loop of the curve whose equation is
. x(x 7/2 38.
4 y~ 2 2)' Ans. ^V2. . Find the area inclosed by the loop
r'(4 of the x). curve whose equation
Ans. Find the area bounded by the hyperbola x 2  y 2
2a.
Ans. a 2 [2 V3 a 2 and the line 39. x = 2
40. Find the area bounded by the hyperbola x
(>.
line x Find the area bounded by one arch
cos 0), and the .raxis.
a(l 41. y 42. Find the area x The of the cycloid  In (2 + V3)]. 4 y2 4 and the = a(0 sin 0), x Ans. 3 ira 2 . Ans. 6 ?ra of the cardioid y
43. is *. ~
= a (2 cos t cos 2 a(2 sin / sin 2 locus in the figure is called the t),
/). "companion 2
. to the cycloid." Its equations are = aO,
y = a (l Tl x Find the area of COS 0). one arch. Am.
44. Find the area 2 ?ra 2 . of the hypocycloid = a cos
3
y = a sin 3 x 0, 0, 9 being the parameter. Ans. c  * that is, three eighths of the area of the circumscribing circle. an integral. In the preceding
appeared as an area. This does not
necessarily mean that every definite integral is an area, for the
147. Geometrical representation of articles the definite integral 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 abscissa 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 That is, the number which denotes
the area equals the number which denotes the distance (or value of corresponding interval of time. the integral). Similarly, a definite integral standing for volume, surmass, force, etc. may be represented geometrically by an area. face, 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
y=f(x), (2) the xaxis, and the ordinates
x = a, x = b. This area may be
evaluated approximately by adding together trapezoids, as follows. Divide the segment OX into n equal a on b parts, each of Let the successive
length Ax.
abscissas of the points of division
be x (= a), xi, x2 , , x n (= ing ordinates of the curve
2/0 =f(xo), y\ b). (2). =/(i) At these points erect the correspond Let these be
2/2 jr
=f(' 2),  , yn =/Or 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
the altitude, we parallel sides multiplied get yi +
+ i( 2/n _ 1 _j_ y Ax
Ax
2/2)
2/1 ) /n )Ax = area of first trapezoid,
= area of second trapezoid, = area of nth trapezoid. by INTEGRAL CALCULUS 246 Adding, we get the trapezoidal
Area (7)
It is = rule, + JA + ?2 + (jy + yn   i h y n )Ax. clear that the greater the number of intervals (that is, the
is), the closer will the sum of the areas of the trapezoids Ax smaller approach the area under the curve.
ILLUSTRATIVE KXAMPLK
x = = to x 1 Calculate 1. Here Solution. ^ _
n a _
^11
f > 1 , c 12 x~dx by the trapezoidal \ rule, dividing l 12 into eleven intervals. = Aa*. The area in question curve y ~ x~. Substituting the abscissas x
1, 2, 3,
144. Hence, from (T),
1, 4, 9,
,
get the ordinates , is under the 12 in this equation, we // Area = By ( + i 4 + 9+16 integration zoidal rule third of 1 is in + 25 + 36 ":r/ " dx  M I error by +49+64 + '> = 81 575^. + 100 Hence, + 121 in this + J 144) 1 = 577?,. example, the trape h than one less per cent. If we take n
10, we obtain
a closer approximation. / =4.826, 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.
rio/ r
io/
1 .1 ^3
2. ; f x V25 /o w = 7. X  x 2 dx; n = 10. THE DEFINITE INTEGRAL 247 Compute the approximate values of the following integrals
trapezoidal rule, using the values of n indicated.
4 fcjjVl + = 4. n ^==; 5.^ r'dx; n Ans. 1.227. 3.283. {V^O + 11. = 4. 2
f \/125a' (ir; n = 5. 12. 44.17. /o .r 4 dr; w Wl<;~H dr jf 1 7. 10. by the T vT^'TiT^ cir ; ; =
w 5. = 4. = 5. Ji 6
8. r Vl26o^dr; n 9. 4 = 4. = 6. ^L^ 13. ( 34.78. n ; = 6. 9.47. VIO ~f .r 2 149. Simpson's rule (parabolic rule).
Instead of connecting the
extremities of successive ordinates by chords and forming trapezoids,
we can get a still closer approximation to the area by joining them with arcs of parabolas and
summing up the areas under these
arcs. A axis may parabola with a vertical
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 ~<;
of such a parabola is of the form
(1) in Illustrative stants a, b, c We now " ^ ~M.,""M y
u 1 ~ji/ 3, Art. 14f>, and the values of the conbe determined so that this parabola shall pass Example may through three given points.
this is not necessary.
into l\ In the present investigation, however, divide the interval from x an even number (= a = OA/o to x =b n) of parts, each equal to Ax. OM n Through each successive set of three points /'<>, ]\, P 2
/^, /'n, /*4
etc., are
drawn arcs of parabolas with vertical axes. The ordinates of these
y*> as indicated in the figure. The area
points are 2/0, y\, yz,
P n n is thus replaced by a set of "double parabolic
MoP P2
; ; , M 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 INTEGRAL CALCULUS 248 Area MoPG PiP 2 M2 =  of first parabolic strip Second (double) Similarly, = rp strip = Third (double) strip +4 + 4 y5 + AT = Ax Last (double) strip + (#2 (z/ 4 o + 4 y\ + (#o 2/3 1/4), o), + 4 y n ~i + (y n z f/ n ). Adding, we get Simpson's rule (n being even),
Area (S) As A*
=(y + 4^ + 2 in the case of the trapezoidal rule, the greater the MuM n is divided, number of the closer will the result be to the parts into which
area under the curve. rl() ILLUSTRATIVE EXAMPLE Calculate 1. :{ I jc ^ intervals. Here Solution. = > = 1 = The area Ax. in 10 n rule, dx by Simpson's curve y  .r\ Substituting the abscissas x
0, 1, 2,
ordinates y = 0, 1 8, 27,
Hence, from (S),
, 1000. , taking ten is under the question = x'\ we 10 in y get the , Area = By +4 + Jj(0 1 f dr = EXAMPLE 2. taking n Solution. ing article.
1 ^' = + 500 + 4 32 +
" I 1372 f 1024 +2916 4 1000) =2500. 2500, so that in this example Simpson's rule Find the approximate value of
1 \ 4. (2.000 "\/4 f a* dx Jo The table
Hence = { 128 result. ILLUSTRATIVE. (S), f integration, gives an exact by 4 108 6 of values f 8.124 is given in Illustrative + 4.472 4 10.864 f Example 3.464) x ^= 2 of the preced 4 82 i.
. o 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 Compare n = 4. PROBLEMS
Compute the approximate values of the following integrals by SimpCheck your results by perform son's rule, using the values of n indicated. ing tbe integrations.
_
Ja
2. 4 f . n =6 . x jV25r 2 dx; n = 4. . f Vl6
*/4 + x 2 dx ; n =6 THE DEFINITE INTEGRAL
Compute 249 the approximate values of the following integrals by Simp son's rule, using the values of n indicated.
* 5. f 4 ; Jo
6. dx / >/4 n = 4. Ans. 1.236. _j_ ja rVl + .r'dr = 4. 3.239. Jo
5 7. = 4. f Vl26 .r'dr; n 35.68. 9.49. 9. V6 + / r5 ~ri X 2 dx; n = 4. n. ,r A* VTT^' '* jr (Lr ; n 4 .  12. 10. *  V5 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.
4 13. f Jz Vl6  V3 f 2 x' dx\ n = 18. 4. r "r  dr; n = 4. ^o 3 150. Interchange of limits. f  Since <fr(x)dx=f(b)f(a), /a and C4>(x)dx Xb (/>(x)dx = /(a)  /(&) =  = Theorem. Interchanging
of the definite integral. [/(&)  /(a)], /
I c/)(x)cfx. Jb
the limits is equivalent to changing the sign INTEGRAL CALCULUS 250 151. Decomposition of the interval of integration of the definite Since integral. Tl <l>(x)dx=f(xi) /(a), f
Jtt
r and (a < xl < 6) h <t>(x)dx=f(b)J(xi), I Jx\ we get, by addition, ( "<KJK/j
Ja /(a). rh
<t>(x)dx=f(b)f(a)',
Ja But
therefore, + f *<t>(x)dx =/(fc)
Jxi I by comparing the last two expressions, we obtain Interpreting this theorem geometrically, as in Art. 142,
that the integral on the lefthand side
represents the whole area CEFL), the
first integral area CM PL), we see on the righthand side the
and the second integral on the righthand 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
152. The definite integral a function of its limits this way. r From I <p(jc}dx= /(?>) /(a) 'a we see that the definite integral Cb
I
J Of <f>(f)dz is a function of has precisely the same value its limits. Thus Cb
as
(j)(x}dx.
J
I O. Theorem. A definite integral is a function of its limits. Improper integrals. Infinite limits. So far the limits of the
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.
153. integral THE DEFINITE INTEGRAL
When the upper limit is infinite, = X
and when the lower limit lim / lim I <t>(x)dx, is infinite,
/ /t> = 4>(jc)dx oo provided the limits f fr = lim x2 b i r+
 <t>(x)dx, /a JP ^= lim
b 8a Ar
;< 2. .r I /, 1 1 . K oc \}+l\=\.
&
L [ i Ans. .1 ' / J = X ^TI^ Find lim 1]"= f oc , x n t ( f
xJi x 2 ILLUSTRATIVE EXAMPLE i  oc x Find 1. ./, J\ * <i b exist. ILLUSTRATIVE EXAMPLE Solution. 251 fc = r^ 8''f/j v
L X FTT^ lim 4 2 a' r =
, m l' arc tan r 1 \ 4 JU
2 a' "2 7R7". f) o .Arts. Let us interpret this result geometriThe graph of our function is the cally. witch, the locus of 8a
y x 2 + :{ 4 a2 =4 a
. Then as the ordinate bQ
finite limit 2 Tra ? arc tan , moves indefinitely to the right, the area ap '. /f ILLUSTRATIVE EXAMPLE 3. Find 00 ^3.
' I x Jj = Hm Solution. The OPQb 1 proaches a :c lim =: X (ln bfoo 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
If a < b and c is positive, we use the definition
(1) Xb (t>(x)dx = \im f*b
I e>OJa+t 4>(x}dx. = a. INTEGRAL CALCULUS 252 when Likewise, continuous except at x <fr(x) is rb
(2) = lim 0(z)az I provided the limits I becomes Here 2
x' lim
* ILLUSTRATIVE EXAMPLE no \ 1 = arc sin J \ by (2), 1 = ? Ans  2  x2 x infinite for = Therefore, 0. and therefore the limit, Therefore, by (1), integral does not exist. b, and 0(j) is continuous except at x = c,
being positive numbers, the integral between a and 6 between a and If c lies
e is a. a/  o L Find f
J 2. becomes In this case there then,  =  arc sin /I . = x infinite for \ a2 Here define dx Find 1. o2 Solution. we exist. ILLUSTRATIVE EXAMPLE
Solution. b, <t>(x)dx, Ja  Ja = r*>< and e' isdcjirwd by f (3) 0Cr)d:r = f' lim 0./ i/ provided each separate limit
ILLUSTRATIVE EXAMPLE
Solution. that is, 3. 0(^)dj + lim f a e '_0,L/r 0(o;)dx, + e' exists.
2 x Find rfx Hefe the function to be integrated becomes discontinuous for x = a,
between the limits of integration and 3 a. Hence the above
must be employed. Thus for a value of x definition (3) = limf 3(z 2  a = lim
t =  ' + a 2 )*l *oL lim \S(x* Jo [3S^(a c) 2  a2 + 3 a^J o =9 a'. To Ans. interpret this geometrically, let us
plot the graph, that is, the locus, of
v 2x = (x*a*rt
and note that x = a is + lim
*'* 3 o* f 6 a* an asymptote. < 2xox  a2 '*oL o o + c') 2  a 2] THE DEFINITE INTEGRAL
Then PE moves to the right toward the asymptote,
OPE approaches 3 a^ as a limit. as 253 that as is, c approaches zero, the area Similarly, Area 2 J dx E'QRG  / 5
approaches 6 a as a limit as QE' moves to the as c' approaches zero. Adding these ILLUSTRATIVE EXAMPLE results,
( Find 4. \ ./o This function also becomes Solution. Hence, by ^ (za) (3), rf_ = lim
J + lim (xa) .oJo o  left we  aa toward the asymptote, that get 9 is, <i*. r a^ infinite between the limits of integration.
8 V^ f f (.r a 4 e') a f "^ ',o'a+t'(r t ") =Um /i_iU, im /_i + i
a/ e>0\6 f'0\ rt In this case the limits do not exist and the integral has no meaning.
If we plot the graph of this function the condition 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
r^i
dx
\ J
result which is (x  12 _ 2 2
a' a)' absurd in view of the above discussion. PROBLEMS
Work
C+
Jo out each of the following integrals. dx _ TT
x 2 hl"2 7.f
Jo
7T = V2.
1 ira 9. r+
o _ ^TT_
~~ dx
a2 2
4 6 x 2 4 10. x2 2 a6 + 2x + 2
_1 x dx oo (1 +x 2 2
) ~4* z CHAPTER XV
INTEGRATION A PROCESS OF SUMMATION
155. Introduction. Thus inverse of differentiation. far we have In a great defined integration as the
of the applications of the many 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 fundamental 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
h (1) f <t>(x)dx=f(b)f(a) gives the area bounded by the curve
(/>(/), the jaxis, and the ordinates y = erected at x Now let a and
us .r = make b. the following connection with this
Divide the interval from :r
a to x = b into any number
area.
n of equal subintervals, erect ordinates at these points of diviconstruction sion, in 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 the curve.
254 under 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 ordinates 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 SUM as limit of this n increases and each subinterval apzero as a limit, is precisely the area under the curve. These
proaches
considerations show that the definite integral (1) may be regarded
without limit, Let us now formulate as the limit of a sum. this result. (a) Denote the lengths of the successive subintervals by (b) Denote the abscissas *^ 'v_ v.v ... <>". . Then the ordinates of the points chosen in the subintervals
3 <"f of the curve at these points are 1 (c) The Ail ',; areas of the successive rec \ tangles are obviously (d) The area under
lim [<Kxi)Azi the curve is therefore equal to + 0fejAx + 0fe)Ax
2 ;{ H h  But from (1) the area under the curve f Ja (f>(x)dx = lim [^(xi) Axi n^ / ^(x " Therefore our discussion gives
(A)  r + ^( 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
then be stated as
(A) simply as a theorem in analysis, which may
follows. INTEGRAL CALCULUS 256 FUNDAMENTAL THEOREM OF THE INTEGRAL CALCULUS
r Let (f)( jr) be continuous for the interval x Az n and
, 2*2 1 > fjr n Then
and each x = b. Let this in, respectively. 0(j] )Aj, (2) to subintervals whose lengths are Aa*i, Aj 2
points be chosen, one in each subinterval, their abscissas being * Xi, =a n terval be divided into + </)(r 2 sum Consider the )Ar 2 Hh </>( sum when n the limiting value of this , increases without limit, subinterval approaches zero ax a limit, equals the value of the definite integral f <t>(x)dx. Ja Equation (A) may be abbreviated as follows. C
Ja (3) The importance ^ h I (p(x)dx = lim 2,(/>0\)Aa%.
oc
_ n of this * t ! theorem results from the fact that we are
magnitude which is the limit of a sum able to calculate by integration a
of the form (2).
It may be remarked that each term in the sum (2) is a differen Aj approach zero as
expression, since the lengths AJI, A;r 2
a limit. Each term is also called an element of the magnitude to be tial , , ri calculated. The following rule will be of service in applying this theorem to practical problems. FUNDAMENTAL THEOREM. RULE
FIRST STEP. Divide
that it the required magnitude into similar parts such
found by taking the limit of is clear that the desired result will be sum of such parts.
SECOND STEP. Find expressions for the magnitudes of these
such that their sum will be of the form (2).
THIRD STEP. Having chosen the proper limits x = a and x =
a apply the Fundamental Theorem
//> it lim ^j<t>(Xi)&Xt
*
=1
i and integrate. = I Ja <t>(x)dx parts
bf we INTEGRATION A PROCESS OF SUMMATION
157. Analytical proof of the
ticle, Fundamental Theorem. As divide the interval from x = a to x = b into 257 in the last ar any number n of sub intervals, not necessarily equal, and denote
the abscissas of these points of division
by
feit fe2, fen , and the lengths if Ax 2 subintervals by Axi, , A.r n , of the
. Now, x'n denote abhowever, we let x'\, jr' 2>
scissas, one in each interval, determined by
the Theorem of Mean Value (Art. 116), erect
, ordinates at these points, and through their
draw horizontal lines to form extremities x 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 and fei, lies JT'I /(7> 2 ) /(fei) /(fea) ~/(fe2) =
= /(feni) = Adding these, (1) /(fe) But we </>(.r' 2 )A.r 2 , for the second interval, 0Cr'3)A.r:,, for the third interval, 0(x' w )AxB> for the wth interval. get = 0(x'i)Axi + 0(.r' 2 )A.r
h
= area of the first rectangle,
Aj 2 = area of the second rectangle, etc. /(a) 0(x'i)
f <j>(jc we have a == A.TI, fei /(fe) feO, a fei or, since Also, between a and 2) a f (/>(:r'n )A:rn . Axi Hence the sum on the righthand side of (1) equals the sum of the
areas of the rectangles. But from (1), Art. 156, the lefthand side of (1)
equals the area between the curve y = c/>(j), the araxis, and the ordinates
at x = a and x = b. Then the sum
(2) equals this area. And while the corresponding sum tT (3) ^P
1 (/>(>) =1 Ax t (where x is any abscissa of
the subinterval Ax,)
t (formed as in last article) does not also give the area, nevertheless we
that the two sums (2) and (3) approach equality when n increases without limit and each subinterval approaches zero as a limit.
For the difference </>(x\)
0(x ) 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 may show t * That such is the case is shown in advanced works on the calculus. INTEGRAL CALCULUS 258 value than any assignable positive number c, however small, by continuing 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,
limit, the sums however small. Accordingly, as n increases without
and (3) approach equality, and since (2) is always
the fundamental result follows that (2) equal to the area, /
I f> <t>(jr)djc '<i in
is = n lim n+K* which the" interval [a, l>\ is subdivided in any manner whatever and x
any abscissa in the corresponding subinterval.
158. Areas of plane curves; rectangular coordinates. explained, the area between a curve, the jaxis, x ~ a and x b is t As already and the ordinates at given by the formula Area
// in terms of x being substituted from the equation of the curve. the value of 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 Let us Theorem A I* and BQ. now apply the Fundamental (Art. 150) to the calculation of the area bounded by the curve .r = </>(//)
(AB in figure), the //axis, and the horizontal lines ?/ = c and // = (/. FIRST STEP. Construct the n rectangles 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
point in each interval at altitudes by A?/i, A# L >, etc. Take a 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 /(/ h'm
71 259 /
x 2>U/<>.A//,== Jc
~ 1 <f>G Hence the area between a curve, the //axis, and the horizontal
= c and y = d is given by the formula lines y = r (C) xdy, Jc the value of x in terms of // from the equation of tuted Formula (C) sum of horizontal strips all within the required area, the base and altitude of any strip.
of these rectangles. Meaning
than less X remembered as indicating is the limit of the
(rec tangles) being substithe curve. and jr being, respectively,
of the area is one </// The element of the negative sign before an area. we now Since b. In formula (B), a interpret the righthand member is as the terms resulting from //, A.r, by let ting i = 1 2, :*,
each term of this sum will be negative,
?/, then,
y
and (B) will give the area with a negative sign prefixed. This means
that the area lies below the .raxis.
limit of the sum of ti , is 'ticyatire, if , r ILLUSTRATIVE EXAMPLE 1. one arch of the sine curve y
Solution. Placing y = Find the urea
sin of jr. and solving for .r, we find x 0, TT, 2 TT, etc. Substituting in (B), Area GAB = C 'ijdx  f \\nxdx = >'u Area Also, BCD  f 'yds  "**\nxdjr ILLUSTRATIVE EXAMPLE 2. Find the area bounded
x'', the //axis, and
by the semicubical parabola ay'2
the lines y = a and y = 2 a.
Solution. By (C) above, = xdy ment and the figure, the ele of area
a V" dy substituting the value
x from the equation of the curve MN. Hence of Area BMNC = f
Ja Note that a 2 f %V
2 I dy (^2  l)= = area OLMB. 2. >'() 1.304 a 2 . Ans.   2. r
r T 7vN
?(*,y) INTEGRAL CALCULUS 260 In the area given by (B) one boundary is the xaxis. In (C) one
is the $/axis. Consider now the area bounded by two curves. boundary ILLUSTRATIVE EXAMPLE 3. Find the area bounded by the parabola y 2
and the straight line x
y = 4. = 2 x Divide the area into
Solution. The curves intersect at A (2,
2), B(8, 4).
horizontal strips by a system of equidistant lines parallel to OA' drawn from the
to the line AH. Let
parabola AOH common distance apart be dy.
Consider the strip in the figure whose
upper side has the extremities (xi, y),
their From these points drop per(x<2, y}.
pendiculars on the lower side. Then
a rectangle is formed and its area is
given by
dA = (1) (x 2 xi)dy. (x2 > x,) the element of area. For the
required area is obviously the limit of This is sum of all such rectangles. That
by the Fundamental Theorem, the
is, (2) Area d
= r to*')
j
c in which curves.
y* 2 x and
Thus, x^ we x\ are functions of y determined from the equations of the bounding
from
4 + y
4 we find x
X*
example, from x
y in this find x Ji ; = 2 \ y' . Then we have, by dA = (3) (4 + y  I (1), y'^dy. This formula will apply to the rectangle formed from any
r =  2 (at A), d = 4 (at B}. Hence Area  = 2 i y' )dy = 18. strip. The limits are Ans. In this example the area can be divided also into strips by a system of equidistant lines parallel to OK. Let A.r be their common distance apart. The upper
end of each line will lie on the parabola 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
O7>. (4) dA = (y,  y,)dx (y, > y,} X the element of area. But in this example it is not possible to find by (4) a is single formula to represent the area of
every one of the rectangles. For while
= V2 x, we have ?/i = V2 x or
y.2
= x 4, according as the lower ver?/!
tex of dA is on the parabola or on AB. Thus from and two integrations are necessary. (4), we have two forms of dA INTEGRATION A PROCESS OF SUMMATION 261 In any problem, therefore, the strips should be constructed so that only a
formula for the element of area is necessary. Formula (4) is used when these
strips are constructed by drawing lines parallel to the yaxis.
single In the Fundamental Theorem some or all of the elements 0(x,) Ax,
the limit of their sum (the definite integral) may be negative. Hence
may be zero or negative. For example, = sin 0(x) if a .r, 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
Find the area bounded by the hyperbola
2 a.
the ordinates x = a and x
1. Find the area bounded by the curve 2. line a = x = = 2 a' AUK.
// xc r
, the 14.026. and the .raxis, An*. 4. Find the entire area of the hypocycloid x^ f In 2. and the In x, the .raxis, = and .raxis, , Ans. 4. Find the area bounded by the parabola V.r
coordinate axes.
5. a 2 the 10. Find the area bounded by the curve 3. line // = x\i 164.8. + V#  Va, and the
A u*. { a 2
= a 5 Ans. $ ira 2
?r'
. . . Find the areas bounded by the following curves. In each case draw the
figure, showing the element of area.
6.
7.
8.
9.  6 x' = 6 y. A HK.
= 4 x, x 2  6 y.
y'
2 4
2 x  y = 4.
y'
 x 2 y  4  4 x.
y=4
2 y 2 .r, 2 JT, , = 2 .r f 2 = 4 x.
= 6  x~, y  x.
=  8 x, = x.
y = 4x,x = 12 + 2 12. 10. ir 8. 11. // 9. 12. // lOg. 13. .r, ?/ .r ' .r /y ?/ 6 f 4 Find the area bounded by the parabola y
chord joining ( 2,  6) and (4, 6).
14. Find an expression point (x, y
y) 2 = a2 , the for the area .raxis, and a = y = 1, 19.
j/axis, y* = x 2 and
^4n. 2.7. bounded by the equilateral hyperdrawn from the origin to any Am
"2 ^ on the curve. q^_  Find the area bounded by the curve
x = 1, and x = 4.
Find the area bounded by the curve y
and the line y = 29. 2 x' y ix + y\ \ a ) Vx) and the line
Aw. 4*. x(l 4. 18. and the line 17. Find the area bounded by the curve y r 2 An*. 36. Find the area bounded by the semicubical parabola
the chord joining ( 1, 1) and (8, 4).
16. . r a 15. bola x 2 2
// = x'2 1 and the lines ^s. f
 7, the
. = x 3  9 x2 + 24 x
Ans. 108. INTEGRAL CALCULUS 262 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 x2 ATMS. . 24. = x*.
4
y = x
2
y* = x
Vx + 25. x8 2. 26. y 21. y
22. + Ans.  7T 27. 4. . 23. = sin ~ 3. 28. jrc = ?/ . = tan f jr // ' 5. 323
ir* + 3 = 2. 2 y' = ^8r +
= c r sin
y
= (4  x)\
2 31. ?/ 32. 2 J TT 29. 7T For each of the following curves calculate the area in the
rant lying under the arc which extends from the //axis to the
cept on the .raxis.
30.  An*. = first quad first inter e cos 2 j\ I}. 34. ?/ 15J. 35. 4 r  ?/ 12.07. 36. y sin (j x 15 j. .r. cos J + 1). T 2 33. 159. Areas polar coordinates. Let it be required
of its radii vectores. of plane curves; bounded by a curve and two to find the area Assume the equation
to be of the curve \D p=/(0), 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
pi, p 2 etc. Then the sum of the areas A0i, A0i>, etc., and their radii
of the sectors is , n pr
For the area
the first A0i + % p 2 2 A0 2 H h i Pn of a circular sector sector = J 2 radius A0
X 7t = arc. = VJ lim p\ 4 = f
./a A p 2 rfff. A0,. Hence the area o pr' A0i, etc.
A0j
THIRD STEP. Applying the Fundamental Theorem, J PI ' Pl of INTEGRATION A PROCESS OF SUMMATION 263 Hence the area swept over by the radius vector of the curve in moving 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. Since the figure
Solution.
symmetrical
with respect to both OX and OY, the whole
area = 4 times the area of OAB.
is Since p
varies = from when
to , 9 = we j. see that the radius vector 4 over the area OAB. Hence, Entire area = 4 OP sweeps substituting in (D), x area OA B = 4 if" p* that one is, if dO = 2 '' f* <W cos 2 = a'2 ; Jo v/<i OA the area of both loops equals the area of a square constructed on as side. PROBLEMS
1. =
2. Find the area bounded by the
and = 60. circle p = and the a cos Find the entire area of the curve p a sin 2 4n,s. 0. lines 0.37 a 2 An*. i Tra . 2
. Calculate the area inclosed by each of the following curves.
3. 4. $.
6. p 2 p
p
p 7.p
8. p 9. p = 4 sin 2 0. An.s.
= a cos 3 0.
= a(l  cos 0).
= 2  cos 0. = sin4
 i + cos 2
= 2 + sin 3 16. lines =
 3 + f p p =o
2 13. 1 Tra 1 1. 12. i Tra p p  a sin n0.
cos * ^ 2
. a cos . TT. ITT. 17. 0. f 6 sin ^. ^ 2 cos 2 ^ ~ J TT. 14 P = 0.
0. 1 TT. 15. = cos 3 02  p + Find the area bounded by the parabola pfl
= and = 120. cos cos 0) Find the area bounded by the hyperbola p cos 2
= and = 30. fl  cos = Am.
2 lines cos 3 10. 4. 0. a and the
0.866 a 2
. = a2 and the
Ans. 0.33 a 2
. INTEGRAL CALCULUS 264 Prove that the area generated by the radius vector of the spiral
equals one fourth of the area of the square described on the radius 18. p = e e vector.
a a sec 2  which Find the area of that part of the parabola p
intercepted between the curve and the latus rectum.
19. 20. Show bolic spiral 22. _  a2 . that the area bounded by any two radii vectors of the hypera is proportional to the difference between the lengths pO of these radii.
21. is ~ of the ellipse p'2 Find the area =  2 o.' 2
b' : a 2 sin 2 6 Find the entire area of the curve p = 2 f b' 20 + afsin Ans. : irab. cos 2 6 cos 2 0). Ans. 2 ira' . c\ 23. Find the area below OA' within the curve p = a sin^ o 24. Find the area bounded by 2 a sin 4 p' 7r + 27V3)a 2 Ans. a &(io
2  2 0. Find the area bounded by the following curves and the given
25. 26.
27.
28. = tan 0; = 0, = \
= <*"; 0=1 6=1
p
= 0, 6 =
+ tan
p = sec
 0, =
+ cos
p  a sin
p . . lines. TT. TT. TT, ; .] ft TT. (9 ;  ?r. Calculate the area which the curves in each of the following pairs have
in common.
29. p 30. p
31. p =
 X cos 0, p 1 1 f cos 0, 1 cos 0,  p
p f cos Ans. 0. TT.  1. sin 0. = 2 cos 2 0, p = 1.
2 =
cos 2 0, p 2 = sin 2 0.
p
2
p = V6 cos 0, p = 9 cos 2  Vs.
 J V2.
TT + 2 i(?r + 9 i 33. 1
0. 38. = V2 sin 0, p = cos 2 0.
p = V2 cos 0, p  V3 sin 2 0.
3 p = V3 cos 0, p  cos 2 0.
3 p = V6 sin 2 0, p 2 = cos 2 0. 39. Find the area 35. p 2 36. 1.  TT 32. p 2 34. .  3V).
 3V). 2 37. For ( of the inside loop of the trisectrix p figure, see limayon, Chapter XXVI. Ans. = TT 2 cos 0). a(l
2 \ f 3 a (2 TT  sVs). INTEGRATION A PROCESS OF SUMMATION
160. Volumes of solids of revolution. Let V 265 denote the volume of
A BCD about the the solid generated by revolving the plane surface
xaxis, the equation of the plane curve DC being FIRST STEP. ment AB Divide the seg into n parts of lengths Ax u and pass a
*,
Axi, Ax 2
plane perpendicular to the xaxis
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
A BCD, then each rectangle
>, , area
gen a cylinder of revolution
when area ABCD is revolved. Thus a cylinder is formed corresponding to each of the circular plates. (In the figure n = 4 and two cyl erates The limit of the sum of these n cylinders (n > oo )
the required volume.
SECOND STEP. Let the ordinates of the curve DC at the points
of division on the xaxis be ?/i, ?/i>,
//. Then the volume of the
cylinder generated by the rectangle AEFD will be iry\~ Axi, and the
inders are shown.) is , sum of the volumes of all such cylinders iry 2 Ax 2 ry n is
 Ax =
rt V THIRD STEP. Applying the Fundamental Theorem OA = a and OB = (using limits b), . pi> lim , JIa Hence the volume generated by revolving about the xaxis the
a and
area bounded by the curve, the xaxis, and the ordinates x
x
b is given by the formula
(E) Vx =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 INTEGRAL CALCULUS 266 2
der of altitude dx with a base of area Try 2 and hence of volume Try dx.
This cylinder is the element of volume.
Similarly, when OY is the axis of revolution, we use the formula  Vv (F) x* dy, TT jf in terms of y must be substituted from the
equation of the given curve. where the value of x ILLUSTRATIVE EXAMPLE ~
a2 #_
4b2 about the i Since Solution. 2 y' Find the volume generated by revolving the 1. ellipse xaxis. =^ 2 (a  x 2 ), and the required volume is twice the volume generated by GAB, we get, substituting in (E), V /vj ^ = 7T/
L
Jo 2 y' dx = /'! J,2 TT Jo ^(a'a 2 irab* To check this result, let b  a. Then Vx = 4 ira :
, the volume of a sphere, which is only a special case of the ellipsoid. When the ellipse is revolved about its major
when about its minor axis, an
axis, the solid generated is called a prolate spheroid
oblate spheroid.
; ILLUSTRATIVE EXAMPLE
ay* (1) 2. = x\ and the line AB (y
Find the volume of the solid the The area bounded by the semicubical parabola ?/axis, a) is revolved about AB.
of revolution generated. Solution. In the figure, OPAB is the area revolved.
Divide the segment AB into n equal parts each of
is one of these parts. The
length AJ. In the figure,
when revolved about AB generates a
rectangle
cylinder of revolution, whose volume is an element of
the required volume. Hence NM NMPQ Element of r since and = irrh TT(G
= PM = RM  RP =
h = NM = Ax. volume 2
i/) a  4(a,a) Ax,
y Then, by the Fundamental Theorem,
(2) Volume of solid =V= ra
TT / Jo (a  y)~ dx = pa
ir I Jo (a 2  2 ay + = a. Substituting for y its value given by (l)
x = and x =
the answer is V = 0.45 ira\ Ans.
This should be compared with the volume of the cone of revolution with
altitude
(= a) and base of radius OB (= a). Volume of cone = J 7ra
for the limits are AB AB f :? . INTEGRATION A PROCESS OF SUMMATION
in If the equations of the curve CD
_ /,,\
parametric form
. () then in
to /i and 2, substitute y = <f>(t\ in the figure y = , 267 on page 265 are given /A </>(/), dx=f(t)dt, and change the limits if
t = /, when x = a, = t / L. when x = b. Volume of a hollow solid of revolution. When a plane area is revolved about an axis not crossing the area, a hollow solid of revolution is formed. Consider the solid obtained by revolving about the rraxis the
of the figure. Pass through area ACBDA the solid a system of equidistant planes
perpendicular to the axis of revolution OX. Let apart. Ax be common their Then the distance divided into solid is hollow
Ax. If
hollow
hollow circular plates each of thickness
one of the planes dividing the solid passes through M, the
circular plate with one base in this plane is approximately a tively MPi whose inner and outer radii are respec(=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 = nAx. The limit of the sum of these v hollow cylinders
cc is the volume of the hollow solid of revolution.
when ft
Hence
circular cylinder V; (3) = rb
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 Find the volume of the ring solid (anclwr ring or
about an external axis in its plane 3. torus) obtained by revolving a circle of radius n
b units from its center (b
a). > Solution. Let the equation of the
x* and + (y be &) 2 = a\ let the zaxis be the axis of revolution.
Solving for y, we have .. By circle (3), dV  ir(y 2 Vx = 4 irb 2  2
2/1 ) Ax = 4 irb x 2 dx Va  x =2 2 7T 2 a 2 6. 2 Ax. Aw. INTEGRAL CALCULUS 268 A may be divided into cylindrical shells by
a system of circular cylinders whose common axis
If the area
is the axis of revolution.
AC ED of the figure be revolved about
the ?/axis, it may be shown that
solid of revolution passing through = (4; it rb 2ir \ (y>> y } )xdx, Ja where OM = x,
The element dV
of radius r, may ample 3 MP\ MP =
2 y\, 2/2 now a cylindrical shell
altitude y?
y\, and thickness Ax.
is be solved by Illustrative Ex (4). PROBLEMS
1.
2 x' 2 f y' Find the volume of the sphere generated by revolving the circle
2
r' about a diameter.
Ans. f Trr*. 2. Find by integration the volume of the truncated cone generated by
x, y = 0, jr = 0, j* = 4 about OA~.
revolving the area bounded by y = 6 Verify geometrically.
3. Find the volume of the paraboloid of revolution whose surface is
2
generated by revolving about its axis the arc of the parabola y' = 2 px
between the origin and the point (j"i, ?/]).
2 Ans. \ irpxi' i
Tr\i\ x\ ; that one half of the volume is, of the circumscribing cylinder.
4. Find the volume
Problem 3 about OY. of the figure
2 Ans. I 7r.ri ?/i generated by revolving the arc in is, one
and radius that ; altitude y\ Find the volume generated by revolving about
by the following loci.
5. ?/  8.
9. 10.
11. = 0, .r = The parabola Vj f .r\ ?/ = 0, The hypocycloid x$ OX x\. the areas bounded Ans.
TTO 3 a. = Vo,
f y* = cA
V// = sin x.
One arch of = cos 2 x.
y = e*, y = Q, x ^ 0, x =
One arch base 2. .r 6. ay'
7. jr\ u = 2 of the cylinder of fifth of x = 0, y = 0. of y ?/ 5. 7T(1  INTEGRATION A PROCESS OF SUMMATION
12. 9
13. y = 16 y 2 + x2 = 144. = 0, x= jrf, y 14. The witch 16. ?/ 2 17.
18.
19. 2  = 2
)?/ a 2 20. Cr 1 = 2
.r ) = )y by the following
y = 2, jc ?/ 1, = // = ?/ 23. ?/ 24. 9 = ?/ = 0, ?/ = x\ < 2 : = >T
// , f JC 1). . 0, 0, j 1. j jc = =
2, 0, r = a* = oo. 5. OY the areas bounded loci. = : .r ', 22. 2 y 2 a 0.2115 wa\ Find the volume generated by revolving about 21. ?T 3 // +  4 0. jr , (4 i/ = ?/ , 2 2 3 j/ 8 a 3  x) = JT*, y = 0, = a.
 6 x, = 0.
 .r)
= 0, = 0, =
(2 (2 a y =
2 = TT. 1. a f 4 (jr An$. 48 16 0, 0, x  2 y' = x Aw.s. 2. V TT. = 2. T TT 2 0. TT. 64 144. TT. 25. (26. y 2
27. x 28. 2 ?/ 2 =
=
=  a% = 0.
16  y, y = 0. 9 GLT, a ?/ = 0, j = a. The equation of the curve OA
volume generated when the area
29. (a) (b)
(c) OAtt GAB
OAB (d) O.4
(e)
(f) (g) (h) # OAC
OAC
OAC
OAC is revolved about OA'. is revolved about AB. is . Find the revolved about OX. (U.H) revolved about CA. is 3 revolved about 07. is .r revolved about OY. is = y revolved about CA. is 2
?/ revolved about AB. is in the figure is 64 Arts. TT. Z J
^ 7T. 192 (4.8) TT. (4,0) Find the volume of the oblate spheroid generated by revolving
z 2
2 2
2 2
the area bounded by the ellipse b x + a ?/ = a b' about the t/axis.
An.s.  ?ra 2 6.
30. 31. A segment of one base of thickness h is cut from a sphere of radius Show by integration that its volume *
is l '  r. INTEGRAL CALCULUS 270 Calculate the volume generated by revolving about each of the followit cuts from the corresponding curve. ing lines the area which 39. = 3; y = 4 x  x
= y*.
x = 4
4
= 4 + 6 x  2 x2
y
= r; y =
y
=
# = 4 j  x
= +7
=9x
=
s f
V7 + \/T/ = 1.
= 7 .n/ = 6.
r f 40. Find the volume generated by revolving one arch 32.
33.
34.
35. 36.
37.
38. 2 Ann. . jf 2 ; ; T/ j JT i^p . a . gig 54 . 2 ; 1 ?/ ?/ 4 . ?/ ^ ; r arc vers ^ V2. TT Vi 7r\/2. its of the cycloid \/2 ry r HINT. TT ; x
about OX, TT. ;rV2. 2 ; .r ?/ TT. *%$* * ?/ base. Substitute dx  V V2 ~ r^ and limits :, // = 0, y  2 r, (), Art. in Aw. ""' 5 160.
2 7T r :i
. T about the .raxis from x = " to r  4 f "/ 6.
4
A w. 42. c"
(I \ 7T^' 5 ( ?" V<' f ?)
/
(/ _L ^ 2 ^ f Find the volume of the solid generated by revolving the cissoid
r< jT  2
y/ L
43. about its asymptote 2 jr 2 .4//,s. (/. TrV. jc rt Given the slope  of the tangent to the tractrix the volume of the solid generated by revolving it  ?/ find about OX. An*. 44. Show 45. ?ra : '. Using the parametric equations of the hypocycloid that the volume 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.
of a conical f } volume find the 46. i* y rr (j = a sin of the solid generated a(0 = ad
its Show
is 6 7r 3 :i : 0,
* 6, by revolving it about OX.
Ans. Find the volume generated by revolving one arch xabout COS a3 .  3
ffa Tra . of the cycloid sin 0), cos 6} base OA".
that if the arch be revolved about Ans. 5
())', ?r 2 a3 . the volume generated INTEGRATION A PROCESS OF SUMMATION
47. y = Find the volume generated if the area bounded by the curve
\ TTJC, the .raxis, and the lines jr =
} is revolved about the .raxis.
Ans. 4. sec 48. The area under the curve volved about the = sin .r from x
Find the volume generated. .raxis. // r 1 2
=4 K
49. Given the curve x
and (b) the volume generated by the area
about the xaxis.
/ 50. y* = 271 5 , Find t ?/ is TT a* re the area of the loop
when revolved (a) inside the loop Ans. (a) vV (b) 67.02. ; Revolve the area bounded by the two parabolas ir = 4 x and
x about each axis and calculate the respective volumes.
.4 51. to Revolve about the polar axis the part between the lines 6 = and =^ wx. OX : 10 OY TT; of the cardioid p : 4 f 4 Length of a curve. By IT. cos and compute the volume.
Ans. 161. *%* the length of a straight line monly mean the number of times we can superpose upon
straight line employed as a unit of length,
as when the carpenter measures the length
of a board by making endtoend applica 160 TT. we com it, another we measure straight tions of his foot rule. impossible to make a straight
with an arc* of a curve, we
cannot measure curves in the same manner as
Since it is line coincide We proceed, then, as follows.
Divide the curve fas AB) into any lines. whatever (as at r, forming chords fas I), number of parts in any manner E] and connect the adjacent points of division, AC, r/>, ])K, EB}. The length of the curve is defined as the limit of the sum of the chords
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.
as the 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 circumscribed) regular polygon when the number of sides increases without
limit. INTEGRAL CALCULUS 272 of the next article for finding the length of a plane
based on the above definition, and the student should note The method
curve is 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. // c 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
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
f P'(x f
, y ) and I>"(x' Then, as in Art. P'P"
or = p'p" = + Ax', //' + Ay'). 95, V(A:r')1 + a;j [Dividing inside the radical by (A.r') z and multiplying outside by But from the Theorem of Mean Value
 a), we get
/(a) and A?' by b
by /(ft) =
f^'
Xi (Art. 116) /W f (x a:'* Ax'.] (if Ay' < x { Ax' < is denoted x' + being the abscissa of a point Pi on the curve between P' and
is parallel to the chord. Ax') P" at which the tangent = [1 +f'(xi 2 ]*A:r' = length of first chord.
p"p"' = [1 +f'(xz) 2 ]*Ax" = length of second chord, Substituting,
Similarly, P'P" )  +/'(j M )2]*Aj = length of nth chord.
The length of the inscribed broken line joining P' and Q (sum
the chords) is then the sum of these expressions, namely, p()Q 1=1 fw) [i of INTEGRATION A PROCESS OF SUMMATION 2755 THIRD STEP. Applying the Fundamental Theorem,
lim V [1 +/'(j i )2)4A.ri" w",^l = f "[1 +
Ja Hence, denoting the length of arc P'Q by
formula for the length of the arc s, f\r)*$ dx. we have the = f
Ja (G) where y = f ^ must be found in terms of x from the equation of the given curve.
it is more convenient to use y as the
independent variderive a formula to cover this case, we know from Art. 39 Sometimes To able. that dx = 7 hence dx ; or ~ x' dy. dy
Substituting these values in (G), and noting that the corresponding y limits are c and d, we get the formula for the length of the arc, *= rV+l]dy,
J (H) c where x' = must be found in dy terms of y from the equation of the given curve. Formula (G) may be derived in another way. In Art. 95, formula (D),
ds (1) = (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 by parametric equations *=/(*), (2)
it is defined = convenient to use
8 (3) since, from (2), dx=f'(t)dt, dy = 4>'(t}dt. INTEGRAL CALCULUS 274 ILLUSTRATIVE EXAMPLE Find the length of the circumference of the 1. *& Differentiating,  (Is Solution. circle y Substituting in (G), BA = Arc Jo dx. y I 2 = r'2
2 from the
jr'
equation of the
[Substituting y'
Lcirele in order to get everything in tcrrm of jr. .. arc BA = r f'
./o Hence the p^L^
vr' jr y a(0 See Illustrative Example
Solution. r/j dx'2 Using f = ,s (,'}), 1, Art. 154.) the length of arc of one arch of the cycloid sin 6), y  a(\ cos 0). Art. 81. 2, cos #)(/#, a(]
2 <///' Example Ans. irr. P'ind 2. _ (See Illustrative ~ total length equals 2 ILLUSTRATIVE EXAMPLE Then ~ = 2 i \ '( cos 1 2 . (/ sin j (/// = a sin ' ^ir/fl : 4
8 ^r/fy ^, </ . f?r/^. sin j I 6 (W 2 By (5), . Art. 2 Ana. 'o The limits are the values of Art. SI ), that is, = ILLUSTRATIVE EXAMPLE
to x ~ 2.
from x
Solution. The 2 =
f (1 see figure, Illustrative 1) = f Example 2,  x5 TT. Find the length ,'i. Differentiating, y' (4) and at = and j \ j2. of the arc of the Hence, by jc*)*djr = \ I "(4 curve 25 y 2 (G),
f a'M'^dr. was evaluated approximately in Illustrative Example 2,
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.
integral in (4) Art. 148, ?, 163. Lengths of plane curves by proceeding as in Art. 142, ; polar coordinates. we get the formula From (/), Art. 96, Nj > c for the length of the arc, where p and ^
ctu in terms of must be substi 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
2
[p d6 Substituting this in 276 + dp 2 ]* gives Hence if variable p, pi we and p 2 are the corresponding limits of the independent
get the formula for the length of the arc, \dp. where terms of p must be substituted from the equation of the r~ in dp
given curve.
ILLUSTRATIVE EXAMPLE. Find the perimeter
Solution.
If we Here TT, the point
Substituting in to erate one half of the curve.   2 P will gen (/), r C 2
J
[a (l f cos O)' + a sin* 0]* dO ./o = r a ('1 f 12 cos 19 >2 dO = 2 a ,/o = .% cos 0). 4 uu vary from let u([ sin 0. <i fl of the cardioid p 8 a. f "cos ,/(i </0 =4 a. li Ans. PROBLEMS
Find the length of the curve whose equation
points (0, 0) and (8, 4).
2. Find the length of the arc the origin to the ordinate x 3. Find the length 'the point where x ~ 1 of the :{ is 1. ?/ 5 a. where jr = 2
// = Find the length 7. = 1 2 x 6 from 2 ;>.r
, p 1 ,*. from the vertex
P i.,
111 M \1 . *7~ 2 x/o'i
V ^j y x 3 from the point where
Ans. .]. = x 2 from the origin to
Ans. 4.98. Approximate by Simpson's rule the length
x 3 from the origin to the point (1, g). of arc of the curve the point 3 y \. =  .S . f if ?>\/2 I 2 6. :i Ar/,s. c.
A/ ,S . of the curve y 2 x from
IW5 a y3
is 3. Find the length of the arc of the parabola
^
to one extremity of the latus rectum.
/I
Find the length of arc
to the point where x between the
AUK. 9.07. A
An curve whose equation to the point 5. jr~ of the semicubical parabola ay* 4. x = of arc of the parabola 6 y (4, ). ^s. 1.09, INTEGRAL CALCULUS 276
8. Find the length the point
(~
9. to In log sec x from the origin to curve y of arc of the Am. 2\  Find the length of arc of the hyperbola x 2 y In (2 = 2 from
Ans. 9 (5, 4). Find the length of the arch of the parabola y
above the xaxis. lies 11. Find the entire length of the hypocycloid 12. Rectify the catenary y =  u (c + a e = y* = from x ) + x* 4.56. a. to the point (x, y). Find the length of = r 2 ^ one complete arch of the cycloid x HINT. Use
14. 0) (3, Ans. 6 a*. ' 13. Vs). x 2 which
Ans. 9.29. 4 x 10. + (//), = 2 a?/ ry V = Here Art. 162. Rectify the curve 9 V2 arc vers  = _ 9  1 y' r. ' 3 a) 2 from x x(x Ans. 8 . = to x = 3 a. Ans. 2 a VS.
? 15. Find the length one quadrant of the curve
(J in 5J f = (r) a Find the length between x = a and x = 6 of the curve + a 2 f ab Ans.
16. ! fjy 2
b' f ~= c* 4 1
x
1 Ans. log
17. The equations 1 Ha 1 (x =
= + 0sin Find the length of arc = = ft.
to
= c sm
!* of curve I Ans. cos << // f from = = 20. y = 21. = 23. x2 of each of the following curves. In x (1 2  2
) from .r 1
 In x from x In esc = .r M
?/ One arch x from x = =
= ^ from y = of the curve # 1 to to x
1  \. to x ~ 2. to x = = 20.
= sin x. ?/ to J aft 2
. =2 Ans. Find the length of arc 22. 3 b. B). a (sin * // a 0), cos a (cos Find the length of the arc from 6 19. y + of the involute of a circle are \y 18. C 2h V2(e2  1). INTEGRATION A PROCESS OF SUMMATION
Find the length
end of the 24. origin to the of the spiral of
first HINT. Use c p a6, from the revolution. Am?.
25. Rectify the spiral = Archimedes, p 277 TraVl a9 4 f + TT ~ w+ In (2 from the origin to the point Vl f 4 ?r 2 ). (p, 0). (/). n Find the length of the curve p 26. = from = to [V2 + Aw*. Find the length 27. p = (V2 + ln of arc of the = from to = V
= . parabola 7 + ln (V2 + Find the length of the hyperbolic
a from (p lf 0i) to (p 2 02). 28. p0 i)] a 2 cos 1 f = a sec 2  1). spiral ,  a In ' (a +
. 4 29. Show that the entire length of the curve p BC that OA, AB,
30. to e 31. sirr* ~ is > Show of arc of the cissoid p = 2 a tan sin from = ' Approximate the perimeter 164. a (see figure) are in arithmetical progression. Find the length =f
4 = Va'' f p, 1 ) of one leaf of the Areas of surfaces of revolution. generated by revolving the arc
of the curve A curve p = sin 2 0. surface of revolution is CD y=J(x)
about the axis of
It is desired to A'. measure the area by making use of the
Fundamental Theorem.
FIRST STEP. As before, divide
of this surface AB the interval into subintervals Ax2, etc., and erect ordinates
at the points of division. Draw
Aari, the chords
curve. CE, EF, When etc. the curve of the revolved, each chord generates the lateral
surface of a frustum of a cone of revolution. The area of the required
is 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
be the middle point of the chord
frustum on a larger scale. Let M Then CE. Lateral area (1) = 2 ir\ In order to apply the 7 M CE.* Fundamental Theorem necessary to express this product as a
function of the abscissa of some point in the
is it As interval AJI. Theorem of we get, using the
Value, the length of the chord in Art. 162, Mean A/,, (2) where Ji is the abscissa of the point P\(r\, y\)
on the arc CE where the tangent is parallel to
I* Axr
the chord CE. Let the horizontal line through
intersect QP\ (the ordinate of 7'j) at /?, and denote RP\ by M Then
/V A/ (3) Substituting (2) and 2 TT(//,  6, )[1 = (3) in (1 + /'(*] //!,. we ), )~J> A.ri = get lateral area of first frustum. Similarly, 2 irtite ej)[l + /'(A)*A.r = lateral area of 2 ir(y n c n )[l + /'u,,)~J*A.r,, = lateral area of last > 1 second frustum, frustum. Hence
rr 2 2 ir(y This (4) c,)[l t may 2
+f'(jr Y
l \* AJ, = sum of lateral areas of frustums. be written 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
THIRD STEP. Applying the Fundamental Theorem
sum (using the limits OA = a and OB = 6), we get
lim The V2 +f'(x 7n/,[l i sum limit of the second A.T, r~] of (4) = iry[l when *oo n 279 to the first + /W1* dx.
is zero.* Hence the area of the surface of revolution generated by revolving the arc
about OX is given by the formula
Q (TT\ (A; 9
* OJT CD . where Sz denotes the required area. Or we may write the formula
in the form
>b S (I) Similarly, OY when is TT y J(a ds. the axis of revolution S (M) =2  2 TT f we use the formula x ds. Jc In (L) and (M) ds will have one of the three forms (C),
of Art. 95, namely,
v~> i (Z)), 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
and regard it approximately as the convex surface
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. axis of revolution,
of a ILLUSTRATIVE EXAMPLE 1. The arc of the cubical parabola
a'y (5) between x = = and x a is =* :< revolved about OA', Find the area of the surface of revolution generated.
* This is the positive easily seen as follows. numbers f d ,  62 Denote the second sum by S n I, ,'!, If equals the largest of then 7? Sn ^ t^[l+f'(r
1=1 The sum on the right is, by Art. 162, equal to
this sum be In Then S n = d n Since lim c = 0, S n the
is sum of the chords CE, EF, etc. an infinitesimal, and therefore lim S n Let = 0. INTEGRAL CALCULUS 280 = (1 (5), y' From Solution. = (a X2
 =3 + : Then the element Hence a2
4 =2 of area 9 f dx. O = iryds i (a 4 9 f x4 ) 2 x 3 ( a* Therefore, by (), I)" 1 =
' 27 3.6 a 2 . ILLUSTRATIVE EXAMPLE 2. Find the area of the ellipsoid of revolution generated by revolving the ellipse whose parametric equations are
(see (3), Art. 81)
x ' = a cos 0, y about OX. 6 sin We
7 Solution. have
dx and d.s (6) .'. integrate,
o'J let 4 dy' )% = 2 iry ds i &, = r
2 TT& 2 = cos 2 (a' = sin  a'2 (I = sin cos 2 0) = 1, w this out
i . , by we (22), 2 irab = 0, = d0. Also, = cos 2 2 a'  2 (a  and interchanging the u
2  6 2 )w 2 . limits (Art. d.r 4 w ^ s
?/ =  4 V T~/ fl dx. X* Substituting in (L), noting that the arc B A
generates only one half of the surface, we get rV
f = 27ro^ Jo
^
an improper integral, since the funcis discontinuous (becomes
when x = 0. Using the definition (1), is tion to be integrated Art. 154, the result is _ < 6 (a>6) va 2 . of the surface of revolution about the ja)3 (^Ti .r =. du. . VA , . 1 4 x<* X* infinite) d0. eccentricity Find the area 3. ^, dx This 0)i sin 0)i sin get
, Here ^ 1 2 . by revolving the hypocycloid = (a arc sin f where c ILLUSTRATIVE EXAMPLE Solution. 4 6~ cos' d0. 4 b 0)i d0. is 2 TO H ^Sj d0, 2
f b co;, 2
f 6 cos Working 2 2 7r6(a 2 sin 2 2 (a 2 sin 2 Then du cos 0. b' / Hence, using the new limits u
150), the result 2 b cos dy d0, 2 (dx' u 4 sin a sin
2 of area Hence the element To  12 xaxis. Ans.
generated INTEGRATION A PROCESS OF SUMMATION 281 PROBLEMS
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
1. . 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).
2. AUK. TrhVa 2 f b'2
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.
. 3. ; 4. P'ind by integration the lateral area of the
frustum of a cone generated 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 = arc of the parabola y x 2 from to (0, 0) by revolving about O.V the (2, 4). Find the area of the surface generated by
the
revolving about
arc of the parabola ?/ 2
4
jr which lies in the first
quadrant, AUK. 3G.18.
7. OX =  8. Find the area of the surface generated
2 px from x
to .r = 2 arc of the parabola ?/ = by revolving about
4 p. AUK. Find the area of the surface generated by revolving about
arc of // = x* from (0, 0) to (2, 8).
9. Find the area
ing curves about
10. 9 = x*, y 11. y 12. =
= 2 2
// 13. 6 // 9  = to x 4 4. from x jr, from j2, = 2. to x from x  by revolving each : v* the
irp OY 2 .r = = to 3 to = .r Am. = .r 6. J of the follow TT. 49 TT. jj '$ w.  f820 4. 81 In 3)^ 72 = c~ r from 14. y 15. The loop , 17. One loop 18. ?/ 19. The 2 + 4 of 9 ay 2 x4 16. 6 a V?/ jr The oo. a =  x = 1 , 2 2
?/ 2 log y t 2 a x from 2 ?y 2 3 . to x 2 a. \ cardioid . to y = a "" sin
ff
= a(l cos Jy
6).
= a(2 cos ~ cos 2
x
J
y = a(2sin 6  sin 2 /
i ira
7 } ( ; j  2. *>
(9). 2
. ira 4 X cycloid f In 7r[V2 x) =a of 8 a = = to r
.r(3 a 4 from x f 3 , 20.  j vrrz 2
. 2 \? TT. . . the OX. from y jr, 24 of the surface generated OX ( 1 f \/2 )]. INTEGRAL CALCULUS 282 =
+ 21. y'2
22. x2 23. 2 x = 2
?/ + 4 ?/ 4 = a 4, = to x from x = 1 3. = to x 2. = 36.
2 =
36. 2 + x2 24. 9 4 r, from ?/ Find the area of the surface generated by revolving each of the
lowing curves about OY.
25.
26. =
y = x x 3 from , a 2 .r// 27. 6 =
= from y y'\ + jr  28. 4 ?/ = .r 29. 2 ?/  .rvV' 2 // 3 a 2 In  ^ 7r[(730)*  Ans. 1 jr , a. 24 a + log (V  V.r 2  1 ), + (20 = from x 2 to In 3)?ra 2 31. 4 32. = 2
?/ // jr* jr = from x { , from .r, f 4 if = to x //  j 5. = 4. 78 8. to y = 33. 4 16. 34. 9 .r jr 2 f = 2 64. /y
:j // , from // = to About The ellipse ^ + f! =
a 2 1. vra 2 f r> HINT, 36. c = // trom a = (Figure,
37. r 40. = ^ ^lnl
1 tox [). \r" f r "/>
> TTfl = a. The
is 2 ^j(^ 4 4^~ 2 ). ^^ 2 ?ra 2 (l 1 532) H w. ? 7r(V hln2). slope of the tractrix at any point of the curve in the first
~y
2 =
Show that the surface generated
given by
. Vc 2  Of. //' by revolving about OX the arc joining the points
the tractrix is 2 TJT(//I
// L ).
(Figure, p. 537)
41. fol OY r h3 = 6x//,fromx^ltoa* = 2. quadrant = 3. eccentricity of ellipse The catenary ?/ 4 TT. 713. Find the area of the surface generated by revolving each of the
lowing curves.
b
A
35. . TT. Am.
30. 1]. ?/ .r, 1 = 3.
= 3.
to
from = a to x = 3
from x = 1 to = 4.
to y fol The area in the first x* and y
tions are y
of the solid generated. = 4 .r (xj, ?/i) and (x 2t 1/2) on quadrant bounded by the curves whose equais revolved about OA'.
Find the total surface
Ar?s. 410.3. INTEGRATION A PROCESS OF SUMMATION
The 42. area bounded by the =4 2 y f 4 =
y and x
surface of the solid generated. are x 2 283 and the curves whose equations
revolved about OY. Find the total
Ans. 141.5. ?/axis
is Find the surface generated by revolving about
X3
1
curve whose equation is y =
from .r = 1 to .r =
+
43. OX the arc of the
* IT Am. 8. Find the entire surface of the solid generated by revolving about
the area bounded by the two parabolas y'2 = 4 x and // 2
.r f 3. 44. OX An*, 7r(17Vl7 i f 32 V2  17) = 51.53. Find the area of the surface generated by revolving about OA' one
arch of the curve y = sin jr.
Ans. 14.42.
45. 165. Solids with known In Art. 160 parallel cross sections. cussed the volume of a solid of revolution, such as accompanying figure. is we shown dis in the All cross sec tions in planes perpendicular to the
.raxis are circles. If OA1 = x, MC = y, then
(1) Area cross section ACBD=iry* = 2
TT[ </>(*) ] , = if 2/
is the equation of the
(/>(x)
generating curve OCG. Herice the area of the cross section in any plane perpendicular to
is a function of its perpendicular distance (~ x) OX 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
slice, FDE and let be one face of such a ON = x. Then, by hypothesis, Area (2) The volume
(3) L) of this slice Area FDE X A:r FDE = is = A(x). equal, approximately, to A(x)Ax (base X altitude). 24 INTEGRAL CALCULUS
n ^ A(x )Ax Then l i = = sum l volumes of of all such prisms. It is evi i dent that the required volume the limit of this is sum ; hence, by the Fundamental Theorem,
n Y A(x )Ax
n^
lim l t:= , l = I J A(x)dx, and we have the formula V (N) 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
ILLUSTRATIVE EXAMPLE
a solid is a circle of radius 1. The base A(x)dx.
of All sections r. perpendicular to a fixed diameter of the
base are squares. Find the volume of the
solid. Solution. Take the circle x + y~
rthe ATplane as base, and OA" as the
fixed diameter.
Then the section PQRS
iri 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) by  4 //'' 4(r 2  and, .r), (/V), Volume 4 I j'~)djr (r' =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. Placing the conoid as shown in the figure, consider a section
perpendicular to OA". This section is an isosceles triangle and, since
Solution. ; (found by
circle RM = V2 rx  x~
+ ^  2
solving
the
.r ORAQ, for //) 2 r.r, equation of the and MP = a,
the area of the section is Substituting in (N), This is a a L.J. J4E~^ one half the volume of the cylinder of the same base and altitude. PQR INTEGRATION A PROCESS OF SUMMATION
ILLUSTRATIVE EXAMPLE by a 285 Calculate the volume of the ellipsoid 3. single integration. Consider a section Solution. with semiaxes and b' c'. Solving this for y (= x (= OM) HKJG ARCD, YOVplane is terms of b') in gives Similarly,
ellipse of the ellipsoid
perpendicular to OX, as
of the ellipse
in the The equation EFGI from the equation of the
the A"OZplane we get in Hence the area ARC!) of the ellipse (section)
, , , nb'c'  irbc , (a .,
2  j') is . ,,.  , N A(x}. Substituting in (W),
, r \ TTbc
 C* "
, (a / a 2 J,  , jr~ )(/.r TTabc. Ana. o PROBLEMS
1. A solid has a circular base of radius r. Find the volume of the if of the base. dicular to
(a) (b) base AB solid The line A H is a diameter
every plane section perpen is an equilateral triangle
an isosceles right triangle with Am. ; its hypotenuse in the :i
% r V5. plane of the
AUK. ^ r
:{ . ; (c) an isosceles right triangle with one leg in the plane of the base
An.s. (d)
(e) Ann. 10
an isosceles triangle with its altitude equal to 20 in.
an isosceles triangle with its altitude equal to its base. Am.
; ; $ r\
?rr 2
. r3 . 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 An*. (a) a square (b) an equilateral triangle
an isosceles triangle with altitude 10 (c) ; ; in. 1,333 cu. in.
577.3 cu. in.
785.4 cu. in. INTEGRAL CALCULUS 286 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.
; an equilateral triangle;
an isosceles triangle with altitude 10 (b)
(c) 443.4 cu. in. 426. 7 cu. in. in. A football is 16 in. long, and a plane section containing a seam is an
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
4. ellipse, section a circle. is A wedge Anx. (a) 341 \ : cu. in. (b) 535.9 cu. in. ; 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
5. is Find the volume at an angle of 45". of the Ans. wedge. *% Q cu. in. Two cylinders of equal radius r have their axes meeting at right
Find the volume of the common part.
Ans. ^ ra
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
Ans. I a (3 TT + 8).
generate.
6. angles. . ri 8. A to the .raxis and \r moves with variable equilateral triangle = 4 its plane perpendicular = and the ends of its base on the points on the curves y'2
16 as
respectively, above the .raxis. Find the volume generated cue, by the triangle as it moves from the origin to the points whose abscissa ifl Ans. ^V5a 3
. 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
9. What distance. is the volume generated while the rectangle moves a Am. distance of 2 ft.?
10. On the double ordinates of the ellipse
a'~ + ^=1,
b'
2 4 cu. ft. isosceles tri 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 = 12. 4 x'2
2 2 .r' + + 13. a 14. 25 ir + 4 y* 9 c2 4 y2 + + = 9 z2 = ; z ?/ = 1 = + *2 1 + 1.
; ; Ans. =
# +
z + 1 = .r ; .r = ; ; \ TT. & 0. 1 z
.r  = I
2. = 0. i TT. TT. ft TT. INTEGRATION A PROCESS OF SUMMATION
15. .r 2 16. z* 2 f 4 = jr~ 9 + 9 f/2; z 2 = = 4 above the the in c f 1 = 0.
z  4  $ w.
TT. i jr 2 the A'Zplane and the circle in Vplane. From each point on the parabola lying
lines are drawn parallel to the r#plane to meet the A' two circle Ans. 1. Given the parabola 17.
02 4 1/2 Calculate the volume of the wedgeshaped solid thus formed. circle. Find the volume of the 18. "" T~ r* a ^= sheet A 19.  solid ~~ ^ + ?/ 287 ^ 11 f2 f and the planes 1 b is
1 jr = bounded by one nappe
and the plane Find the volume of the 20. Ans. 6 TT.
bounded by the hyperboloid of one solid '~^ jr = 2 = AUK. a. of the hyperboloid of Find the volume. a. $ Trafor.
tj two sheets Ans. % irabr. Ann. > irahc. AUK. vV' bounded by the surfac solid 4 TT; f a* j 0, 1 7, . r" 0 ADDITIONAL PROBLEMS
Find the area of the loop of the curve 1. y* A 2.  (jr point moves along f 4)(.r '
L  jr f 2 y  4). a parabola in such a way that the radius 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. ? joining it Ans. Distance from focus Find the perimeter of the figure bounded by the
curve 4 y = e'2r f c~'2r
Ans. V.3 f In
3. . latus rectum.
lino y =_l
(2 f V.'J ) and the
3.05.  arc OP of the curve xy = x
y joins the origin to the point
and bounds with the xaxis 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 .raxis and tt about the The 4. P(jr\, 7/1 ), />'. ?/axis are equal. 5. The area bounded by at the point (12,
inclosed. The base 16) is the curve 16 y'2
and its tangent
(x f 4)
revolved about the raxis. Find the volume
:i Am. #4 1J) 2 *. = bounded by the parabola y
2 px
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
6. and its of a solid is the area latus rectum. the solid. Ans. j p 3
. INTEGRAL CALCULUS 288
Given the 7. curve such a in + 2
ellipse 9 s' way that = 25 y 2 225. A solid is formed about this plane sections perpendicular to the :raxis 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.
all 8. Let (r, y) be a point on the curve of Art.
159, O being the origin
and <)A the xaxis. Show that (D) may be written = Area (x j \ dy  y ds), by using the transformation (5), p. 4. The limits are determined by the
coordinates of the extremities of the curve. Derive the formula of the preceding problem directly from a 9. ure, making use The formula and of (B) (1) of Problem the following areas by 8 useful for parametric equations. is and the .raxis + cos r .r rO sin of a circle
r sin y 6, rO cos 6 to the left, in the figure of produced Chapter XXVI, p. 537.
11. The entire area of the hypocycloid of three cusps
j' = 2 r cos f r cos 2 ?/ = 2 r sin  sin 2 6. 6, j 1 12. A r Ans. 2 (See figure.) particle ?rr 2
. straight uniform wire attracts a
according to the law of gravita P 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
tion. wire whoso distance from P is the
proportional of the distances from
13. Find the area HINT. Let y = tx;
\, then .r : and dx = +/< ~
\
(1 f The limits for / are mean
P to the ends of the wire. of the loop of the folium of Descartes, 1 * at Find (1). The area between the involute 10. fig (C), Art. 158. and 2t
;\ 3 a dt. r 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. tion the A rational fraction numerator and denominator of which are functions, that the variable is, is is a frac integral rational 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,
r
* + * J
T + <*
== j2 __ x __ 3 __ 40*
' e . 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 fractions, that is, to replace it by the algebraic sum of fractions of forms
such that we can complete the integration. That this is always possible when the denominator can be broken up into its real prime factors is shown in algebra.* Case I. When the factors of
and none are repeated. the denominators are all of the first degree *See Chapter XX in Hawkes's "Advanced Algebra" (Ginn and Company, Bonton).
289 INTEGRAL CALCULUS 290 To each nonrepeated linear factor, such as x
sponds a partial fraction of the form there corre a, A
x where A sum The given fraction can be expressed
The examples show the method. a constant. is a
as a of fractions of this form. ILLUSTRATIVE KXAMPLK. Find X / The  1)U 4 x(x 2 x denominator being factors of the v f  .. ' Solution. H ' J v4 x 2
x  x x 2) x x, 1, x 4 2, we assume* x42 1 where A, K, r are constants to be determined.
Clearing ( 1 ) 2x 4 3 = 2 x (2) 3 = I Since this equation
of x in the we of fractions, A(JC
1/1 is get  l,(x 4 n 42
Ox 4 4 i f B(JC 4 2)x 4 2 7? (A  4 C(x Dor   l)x, 2 A. an identity, we equate the coefficients of the like powers two members according to the method Undetermined of Coefficients, and obtain three simultaneous equations
r cn ; Solving equations we (in, A 4
A 4 2 7? +T= /?  r 2, 0, get Substituting. these values in U), 2x4 a
",/ 3 5 I .r(,rlHx42) 2x .S(xl) 6(.r42) DU42) 2. x(x _ = x ;K 4  In x rl In l r(xl)~
= In ~
, .r'^Cr A shorter  (.r Let factor x factor x I = Let factor x 4 2 = ; then 3 0, or x 0, or x = =
=I 2 In c A HS.
t ; and f from 2 A, or then 5 ; 4 2) 42)" of finding the values of A, B, Let r_^L_ 6./X42
 J In (j 4 >; method ~ I) i then =
 3 #, or
I  6 C, or the following: (2) is A B=
C=  . . . /; In every example in rational fractions the number of constants
be determined is equal to the degree of the denominator. to * In the
process of decomposing the fractional part of the given differential neither
the integral sign nor dx enters, FORMAL INTEGRATION BY VARIOUS DEVICES
Case II. When the factors of
and some are repeated. denominator are the 291 all of the first degree To every wfold
spond the sum of n linear factor, such as (x n there will corre t partial fractions, (xa)"" (xa)"
in a) which A, B, L , xa 1 These are constants. partial fractions are readily integrated. For example, Solution. Since x X* C ILLUSTRATIVE EXAMPLE. Find 1 J +
_ 1 ctr.
4 we assume occurs three times as a factor, 1 4. ar1 cr1) 2 Clearing of fractions,
x< + + = A(x 1) + B;r + Or(j 1) + />j(j I)
= (A +/.*' + (3^4 4r2/x + (3.4 1 x' 1 ;< Equating the coefficients of like powers of j, we A + />=
 3 v4 4 C  /> 3 A f B  T + D  A =
L* Solving, 2 2 A = B = 2, C = 1, x 1)" y(jr 1, /> = 2, . f /?  T + /jr  1, 0,
0,
1. and (x LH^L a '
,. ln + Crl)' PROBLEMS
Work out the following integrals. 2. X JT* Q (4 /
' r 3 x f r o;ax 8 x 2 f 3 x _
~~ J 2 i ^ (2 x A. get the simultaneous equations + 1 N i /0 " ' v r* INTEGRAL CALCULUS 292
r(lr*
J + + 2 2.r 2 l  + 2 1 4 rr x (2 )n l)(2orr2 I) 2
{ c  . J J ?/ 2 ?/< ?/ 4 " f t J] o v i 3
c 3 x* f x^ <: 10. (a^ + 7 J )i._ =
l)(j + 2)U + /" 12 Ju + (.' ^ 3)^
a = i = JLCllk Work 4 n 1 _ 5 n n i = 2877 =_ o.4K!9. ln :i) 3 _ 4 = 1.4930. out each of the following integrals. n5.r~ i
lb 17 f( 5j
'J ' " 2 9)( ^ ^<] f  + 7v/ f n
y './ j (2 + 1 8)(4 14 .r^ ~  ___ h j 10)dr 10 2^ " ^9.r ' + 24 1)
4
/ ~ 3 y*)dy 20.
01 r(^
J^x.
' I 2
""  J  5)(ir
* OR f (2
<wO. 4
/ f 3 /  20 /  28)d/ * I r
Case III. W /w7i //^' denominator contains factors of
but none are repeated. the second degree FORMAL INTEGRATION BY VARIOUS DEVICES
2
quadratic factor, such as x
corresponds a partial fraction of the form To every nonrepeated + px f q 293
t there +B
+ px + q Ax
x2 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 values, the new u. Then x ILLUSTRATIVE EXAMPLE Solution. =u \ p, dx = du. Substituting these
integral in terms of the variable u is readily integrated. = p 1. Find X  / Assume 4 X f~ = A + Ex
* 4 . C f " ~xMT J) M 2
=
Clearing of fractions, 4 = ,4(x 4 4) + x(Bx 4 C )
the coefficients of like powers of x, we get
Equating
1 A +B C= 0, 4 0, A = 4 ft)x* 4 Cx +4 A. 4. = 0,so that
4 / "J x(x 2 rfar f _ rdx _ r xdx
J x 2 f 4
4) ~J x = In r ILLUSTRATIVE EXAMPLE 2.  In Or
2 1 ' 4 4) 4 In c = C* Ans. In Vx 4
2 4 Prove 2)2
~
r_&Ei n x ^^ x 4 ^ 12
4^V^arctan^
2
V3
+ 8 24
2 J xa
Solution. Factoring, x < I 8 1 x :< + f 2
(x + 2) (x
Ax f B
~~
x2  2 x + 4 =
8 2 f 4). C
' x 4 2 = (Ax f B)(x 4 2) f C(x 2x44),
= (A 4 Ox + (2 A + B  2 C)x f 2 B f 4 C.
B = J, C = ^7A=2 1 2 1 Then x ^ x 2  2 x + 4 = (x  I) 2 f 3
Then x = u + l,dx = du, and Now Substituting back u =x 1, = w2 using (4), 4 3, if x  1 = w. and reducing, we have the answer. INTEGRAL CALCULUS 294 Case IV. When the denominator contains factors of the second degree
some of which are repeated.
n
To every nfold quadratic factor, such as (x 2 + px + q) there
will correspond the sum of n partial fractions,
, +B
+ px + q) n (x , I 2
(~2 J_ /n^ _L n n ~
px
q) ' \n (x' C
(0) J proved in du ^2
x2 _ + px + q 1 the next chapter, + px + q = u I + a2r 2 (u' (x + 2 p) is p If necessary. If tions of (5) are necessary.
2 Lx+M . . I l
1 \ carry out the integration, the "reduction formula" To jc +D
+ +
Cx A*
2 is not zero, > n repeated applicathe square, 2, we complete + \ (4 q  p 2 = u 2 + a 2
) , etc., as before. ILLUSTRATIVE FA AMPLE. Prove + 2
Solution. Since x 2 1 x* occurs twice as a factor, + x 4 X __ Ax
~ 4 7* (^ + z (x f \)' I) , 2 O
x2 we assume
4 />
* + 1 Clearing of fractions,
L J :< 4xf3 = >lxfB4 (Car + D)(x 2 f 1). Kcjuuting the coefficients of like powers of x arid solving, ,4^1, The 7? these two integrals
above, with u  x, a first of second by (5) Reducing, we have = is  r= 3, L\ 7^ = we get 0. worked out by the power formula
1, n = 2. (4), the Thus we obtain the answer. Conclusion. Since a rational function may always be reduced to
the quotient of two integral rational functions, that is, to a rational
rational
fraction, it follows from the above discussion that any
real quadratic
function whose denominator can be broken up into
and linear factors may be expressed as the algebraic sum of integral
rational functions forms all of which and partial fractions. we have shown how The terms
to integrate. of this sum have Hence the FORMAL INTEGRATION BY VARIOUS DEVICES 295 integral of every rational function whose denominator
broken up into real quadratic and linear factors may be found,
can be
and is expressible in terms of algebraic, logarithmic, and inverse trigono Theorem. The metric functions, that is, in terms of the elementary functions. PROBLEMS
Work out the following integrals. f L "'" ^ + 3.r J (*"* 2 2 f +  8 _ = ^^
/
..,  '? +
7 z 4 f 2
z' ln
12 In 1 i = ^L __ f 4 j + ( . fii + arctan 2J3 + arc tan
6 ln arc tan z + C. , + 2)%  ^+t r__^_  _
J / , tan 1) h arc  in 4^f9o(2?/: 2 = 5 (j 4> .^, 3)(jTH4)
(2x ^iQ)^
 J "' " In 8)rf/ . r(^ + 4 N f 12 C. :J v+ >2) + arc lan C. z ^2 h tl I) _ 15 . fA^r = In 2^  2 arc tan x +
1
x + 1 J C. 1 or =2 , (z + 2) _ arc tan (z + 1) + C. INTEGRAL CALCULUS 296 *= arc tan (t + ~ 2> r
1 r' f 19. 4 jr U f 14^
2) Or Jo u
r + ' *> T^ 2 ^ j ;n.< i U" f 1)(^ o =
f 1) 2 h 1 j l n I
9
i ^ + *= 0.667. 4
/i j ___ rt 71 $ 4 22. 23. Work
A  f(6
4 r out each of the following integrals. + 2 1 I J (z r(3 l)(z r' J 4 'I J .r f(3 ^
j* f(5 ^ I
. '" 2 ;
4 f 2 3 a 30. . + 2 l)(Lr 01
31 2 h 3)<ijr
' f3^
f 12
+ 32 T 1 I r 3 QQ f
5O. :t (2
/ (x v
2 , H IN/ l)(j (x f a (2 x 3 2
i/ f 4)dj
r~r^*
2 jr f I) 10)dj x3 h2r 2 *Ji 9)dr i ' (2 y h 1)(4 ./o 4 .r ^ I ^o ^ o 3 J h3j 2 J h 1) +
jc' a
9q f(4 J
29
 t : + J 4 )dr ^T^ J I) + ?** f h 5x" 18 I ^ 168. Integration by substitution of a
In the last article it was shown that new variable ; rationalization. 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 transformed into equivalent functions that are either in the list of standard
forms (Art. 128) or else rational. The method of integrating a function 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 imall rational functions 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 expression can be transformed into a rational form by m.eans of the substitution =z x
where n n
, common denominator is the least of the fractional exponents ofx.
in For x, dx, and each radical can then be expressed rationally
terms of z.
J ILLUSTRATIVE EXAMPLE
Solution. Here n  Prove f
J 1 . Hence 4. x* Then 1 = z =!/(''Substituting back = = + .r*   x*
6 1 In
6 U + x% + C.
) ?. rfp)" = da z<, = 4 A z' dz. +
l'' 1 'inrt r. have the answer. general form of the irrational expression here treated The where = x*", we 2 x* ,  x let 2 *** R denotes a rational function of x' is then 1
. + bx only. Such an
Differentials containing fractional powers of a
can be transformed into a rational form by mea,m of the
expression
substitution where n a is the least the expression a + + bx = z", common denominator
bx. of the fractional exponents of  For x, dx, and each radical can then be expressed rationally
terms of z. fdx . _  2 (1 Solution. Then Assume
dx =2 1 4 z dz, x
(1 z2
f fx)' f (1 x)* = = The z\ and 2 arc tan substitute back the value of z in terms (1 z f xY1 = z. 2 +C= 2 arc tan of x. general integral treated here has then the form R[x, (a where +x) 2 . 2zdz _ when we R in + bx) ]dx, denotes a rational function. v (1 I x) 2 f C INTEGRAL CALCULUS 298 PROBLEMS
Work
r
J l out the following integrals.
2 <* s V7 r
3. /
J x x + ij =
6 r^ = ~fv. f M v. 2(2 a do* (a f 1 /.( ^
*frrjk
V
f .r +
+ (/ 12. f ./() 4 13 ./Jo r 14 ''i 4 ~^l f 4 f + C. (4 />.r  ?/ 3 )(a f ~ u f ! )'  //)" h : 3(x f == 2 + C. ) ' f 3 In (1 + Vx 4 a) f C. \/2 arc tan 2 arc tan 2  _j_ y V2 4 21n;j. V.r w = f 4 L^ V5.
16 2 r!
Jo 3 (.r 9 arc tan  2) s +3 a f 1 2 3' r^ 1 r ^^7r^4 * // d/
15. arc tan x 8 r 4 _^r
l c j+1)' ~ 2V/ + ! ( 1  f a 5k// (/ +c 1 = / 1 ~
t __ + 9. 10. = // (/// 2 f ^'a /v'aH + x 6 2 In 4 B* * ' '/^
8. + 12(4 ) " x" I? ;; C. x 1 ^ V3 1 l7  1  2V? + V/ = 5.31. FORMAL INTEGRATION BY VARIOUS DEVICES
Work out each of the following integrals.
19. 20 I ^~ 23. /^fe'  f 22. '" + 5)Vr + t 27. Find the area bounded by the curve
and the ordinates JT ~ 3 and .r = 8. V.r f jr ?/ f 1, the .raxis, AN*. Find the volume generated by revolving about the of the preceding 4 M j(^^rr+Tt: . / . 4. (.r 28. 299 .raxis 40 / . ; the area problem. Find the volume generated by revolving about the .ruxis the area
the first quadrant bounded by the coordinate axes and each of the
29. in following curves.
(a) (b) 30. y JT 31. Vx. 2 y y2 (c) (d) \/jr. = ?/ Find the area bounded by the curves
v2 j* f 1 and the ordinates .r 4 and VJIr. a // // j % 4  ~ A 2 .r f V'J j f 1 and 12. Find the area bounded by the curve and the ordinates x = and x 3 V5 4 8. In 4 169. Binomial differentials. xm (1) A
(a differential of the  V2 form + bx r dx,
n where a and b are numbers, called a binomial differential. is any constants and the exponents m, Let x
xm and
If we + bx za
n p
) az a dx then ; dx = az m " ' ff an integer a be chosen such that see that the given differential form where m is bx n p
) dx = fa ma p are rational ~
' rfc, + bz"") f> dz. and w* are also integers,* equivalent to another of the same x mJf always possible to choose a so that
of the denominators of m and n. L C.M.  and n have been replaced by x(a + (2)
* It is the (a = n, ni integers.
n '(ax~ ma and + b) not are l) Also, dx int^ors, for we can take a an INTEGRAL CALCULUS 300 transforms the given differential into another of the same form where
n. Therefore, no matter
the exponent n of x has been replaced by
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 a positive integer the binomial is as a fraction We may hence we replace ; make then by it may be expanded and
is regarded In what follows p the differential integrated termwise. where > r and s are integers.* s the following statement. Every binomial differential be reduced to the may form r x m (a+ bx n )*dx,
where m, n, and n s are integers r, In the next section we prove is positive. that can be rationalized under the (1) following conditions. CASE m When I. = an n integer or zero, by +
h  = an
a CASE When II. n inteqer or zero, ILLUSTRATIVE EXAMPLE + bx f 1. J (a n  + =
= 2
bx' )l _ ~~ z*x . fx*(a
J
2a 1 2
4 bx' 3, 8 = 2; and here
3,n = 2,r =
comes under Case I, and we assume a + bx 2 m= = z2 ; whence x = (f r
J rlz*  a\
""' I
& (a f bx*)'* = ILLUSTRATIVE EXAMPLE m= 4, case where p is Solution,
*
p, The = 1. n ~ p f (1 _
~ = = 1 2a f V \ a + = 2, f bx  an integer is \ m "*" 2, n  = . and an integer. Hence (a f (z 2 c bx*
* x ~~ \ f =
82 ^ c fa2 1 ~ az x
2. + zdz _ x*dx ' r dx ' b assuming n b* \ a this lyy s a Solution. assuming bx n == z*. ; and here  m "^ +  =  2,
s
n an integer. not excluded, but appears as a special case, namely, FORMAL INTEGRATION BY VARIOUS DEVICES
Hence this comes under Case II, and we assume
(\ also x2 = x whence = i and ;  D* u~
1 (^1^  (2 ;*_ })
 I'd 4
\ ( J r' .r 2
) PROBLEMS
Work out
1. the following integrals. ._2(3r<2)(
/VVl + x*dx= J i__L_i__ r 5
r/o
j (8
3.J a __ _, xj f^)* 3 f>va
(1 C.  4 1
J J J H1 4 r7
x 7 r2/1 4. r4 : ~~ (1 4 r3) r. r C. ^ ' >
j" ^ 3 ) ^
(1 rfr + 4 4 djr fi  4r") 1 4 3 J3 ,, h C. loTT rfjr 6 r r.  , 4 4
( 4 4 C. 4 C. 301 INTEGRAL CALCULUS 302 Work
11. out each of the following integrals. jV'Vl  x 3 dx. IP xdx_ f 13 m 170. Conditions of rationalization of the binomial differential
r x m (a (A) CASE Assume I. a + bs = 71 + bx n) ; dx. z*.
r 1 (a + Then bx = j 1 /z* , rm and ) ^ /z* hence on we Substituting in 04), ( \ sf ; a\ Tl  ; j dz. J . ) ) j^ ~
dx = 7^ 2* = ; fxr" Am ~( a\" =( also (a + and 2, )
' o get m + 1
1
* f ~ hn The second member V of this expression m+ ' fe rational is when 1 n
is an integer or zero. CASE II. Then Assume
xn = a + ^ ^ zx n fe.r" and r a (a x also A
and bx n zx" ~ ^r 6) a"(z*
8 Substituting in (A), n we _ (^ xm ", ^ 8 = a r *z fe) fe) ; n (z n 1 r s rl M a^*" U 8
(z  b s r + fex n V = a J
dx x m (a + fe r Hence . fe) 1 ; ^ dz. get + bx n )*dx =  n a~ +
*(z*  4 fe)'^"^" "^ 1 )^^* FORMAL INTEGRATION BY VARIOUS DEVICES
The second member
is an integer or of this expression is rational when 2?L zero. Hence the binomial 303 us
 f differential
r x m (a can be rationalized + bx n )*dx in the cases given in the preceding article. 171. Transformation of trigonometric differentials. A Theorem. trigonometric differential involving sin u and cos
means of the substitution
/>// u rationally only can be transformed tan (1) or, what is the sin (2) same
u =
 z, thing, by the substitutions 2 = z z 1 cos u >
~ into another differential expression which From Proof. o tan 1 cos Substituting tan \ u = z, and rational hi ~ z. one of the formulas (5), we have solving for cos u, ^2 u cos *. iz* > 2 The right triangle in the figure shows the
also sin u as in (2). Finally, from (1), (2). and gives u and hence Thus the 2 dz
~~Z ?/ i relation (3) ~ ?/ + cos 1 ^T
1 u
2 (3) is du the formula for the tangent of half an angle in Art. 2, after squaring both members,
, 2 ~>
~ du 2 arc tan z, 2 dz
.. : 2 relations (2) are proved. 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 It is evident that the transformed
trigonometric differential can be integrated, provided
into partial fractions (see
differential in terms of z can be separated Art. 167). INTEGRAL CALCULUS 304   ILLUSTRATIVE EXAMPLE. Prove f
J 5
Let 2 x Solution. then using /' = Then M. = .r f ?<, ^ 4 sin 2 x d jr = =  arc tan 5tanx + 4 \ + C.
/ 3 3 \ Substituting these values, and i c/?^ we have (2), _ _
jfr = /544sin^j ./ = z f/ I f
2 Substituting back f tan " =i ~2 4 sin M // i = tan 8 gives the j ~. ; above 5 2 + 82 + 5 result. PROBLEMS
Work out th< following integrals.
4 df) / , cosr ln 1 + tan  __.. . 4 4 5 n / = l . 3 tanf3/
  > 5.  /^7'3 f cos <v r_ 6. arc : ^ v  *> J sin ' r
7. cos .r  cos
 ^ 7 <1Q 7, o ; :? j* V  arc tan 0.5 ~ (
\ 4 ^ arc tan
3
6
. ~ tan Vu 3 ~f   cos ^ tan ( '3 1 4 ) C. f ' tan t> ') f C. 2'
/
^ 0\
2 tan 2/ , . f C. c. 3 '.(.4 0080 _
' o 1'J 4 < 13 cos ^ 2 V?' 11. f
.'o 10
1U / ~ In n
r 2 19
12 2 + sin : x ' ',o 3 f 5 sin a 4 FORMAL INTEGRATION BY VARIOUS DEVICES
Work out each of the following integrals n.f 1
J ^ sm x f 'J cot Jf 13
19. j
16 r
.'13 cos cos x d 14 C 15. 305 t* 4  5 cos f sin ^ 5 dx r 18 /'_jrf<L_.
i> f sin a * J < f.
2 sm  \l 2 cos x f 1* ^ r
J I esc 6 / 21. . x ^ /v^  J 5 sec og r 0. f
*/ 5 f 4 sin ~ * Vo 172. Miscellaneous 23.
4 f f 3 f 2 cos 2
24 T djr
T> ~ r"
.' 3 cos x "'o rf<y 2 f cos tv So far the substitutions con substitutions. 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 working out a large number of problems must be our guide.
A very useful substitution is
1 dz z x z' ~~ ' called the reciprocal substitution. 2 Let us use this substitution in the next example.
ILLUSTRATIVE EXAMPLE. Find f
Solution. Qa /v substitution x Making the ^ dx.
, =  . dx ~ , we get
(a 2 .,_ , 3 j I =  fl2 PROBLEMS
Work out the following integrals. d I./
2 = In ^
 x f V2 2 \Vx 2  T f 3 f
' xVx 2 + 2 *  =
1 2 arc tan (x = 2 f x f Let
rfj Let x (__^===).
j2 rf 2 =2 x. + Vx^T 2 x  1 )+ C.
T*t x/^TTTT
Let Vx^ f2xl=zx. INTEGRAL CALCULUS 306  dx 4 C xV2 "J  x 4 x V2 4 2 x W2 * ~~ 4 2 x / V2 2  V2  x\
C.
 x/
4 V2
Let V2 4 x  x 2 = (x f 1)2. x 6Let dx = V5 x  6  x2 = (x  2)z. 2
dx x2  x arc sin .r
J = =  Let x 14 C. =  1
f ' ^ Let x Let x / xV3 2 x 4 Vl 4 x f 5 x 2 f xVl 44x45x 2
d. xVx 2 4 4 x  C. dx 8.f y
J x 2 Vl 4 rV2/ 4 ._ Vl
=. 2 x 4 f 2 x f 2 3 x 2 3 x2 Vl In f 2 x f 3 x 2
f C. Let x
10. < z r^ J r2x/ >7 = V27 x 2 f 6 x  1 f r 3 arc sin ( \ r^ ^i f C. 6 x / Let x = z 1 11. 12. f r ./o 18.  <*?**" =6.
X^ f J r f r~ r 4 = Let 4 1 a sin 2 ir. x2 ax 14. / 2 d/ = V3^ In (2 / = z. out each of the following integrals. Let 15> " arc tan e Let x f /0 Work = Vx 2 2x43  x. Let Vx 2 2x43 = z  x. Let V5 = (x Let V5x  6  x 2 =r (x xVx 2 2x43
4 xdx J
(x 2  2 x 4 3)* 2dx V5 x  6  6  x2 x  6  x2  2)2.  2)2 x2 2xdx V5 x  z CHAPTER XVII
REDUCTION FORMULAS. USE OF TABLE OF INTEGRALS
In this chapter formal integration is completed.
eventually to lay down directions for using a table of inMethods of deriving certain general formulas, called reduc 173. Introduction. The aim is tegrals.
tion formula's, given in all tables are developed, since these
are typical in problems of this sort. methods Reduction formulas for binomial differentials. When the bicannot be integral ed readily by any of the methods
nomial
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
174. differential which easier to integrate. is The following are the four principal reduction formulas.
x m (a '/ + bx n ) dx
f) = ym n (np+mi (mn +
+m+ (np + l)b
I J  m+
np + n
(af bx )t>dx. C X m (Q + fort lj x / b / n x l)b, *~ (a + wy dx =  x m+ i /a n [p +
4. bx")''
> 4 r
J , B( 1 +1)
l n(/> f l)a
307 r x m (fl + J bx ny fr INTEGRAL CALCULUS 308 While not desirable for the student to memorize these forknow what each one will do and when each one fails. it is 'mulas, he should Thus : Formula
Formula
Formula
Formula
I. To m (A) diminisfos w (C) increases by (D) increases p by derive (u We may 1 . dv  formula this apply n. The formula formula (A). (1) = zm ~n+ ' * m~ n bs n = /,
du  n (m Substituting in + 1 )/ = 0.
= 0. by parts (A), Art, is 136 the integration of + and for integration  f r du. dv r by placing u uv in r "(a then np + m f 1
when np + ra + 1
(B) fails
(C) fails when w + 1 = 0.
(D) fails when p + 1 = 0.
(A) fails when by n.
diminishes pbyl.
(B) J>
) dx + (a and dx ftr v n 7> ; = rn ~ J da* ' , ?/.&(/> ; +\ l y (1), ~m(n
J (tt j r m n i \ ( n _[_ /, rn \ /' f 1 JL f 0. C J
But Cx m 'n + bxn )^ dx=
[ (a r.r =a Substituting this in (2), we mn I + bxn )*dx get m~ / "
x m n (a n f  o Uar
nb(p Transposing the
m
for I x solving (a + last
bjr n term to the )"dx, first we obtain m J ) member, combining, and (A). * In order to intoRrate r/r by the
power formula it
 1. Subtracting n
parenthesis shall have the exponent w
the exponent of j in u, is necessary that x outside the
n + 1 f or
from
leaves 1 m m 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
n. By repeated applications
same form in which m is replaced by m
of formula (4), m may be diminished by any multiple of n.
When np + m + 1 = 0, formula (A) evidently fails (the denomi But nator vanishing). in that case hence we can apply the method of Art. 169, and the formula
needed.
II. To derive formula (B). Separating the factors, we Cx m (a (3) + bx n "
) dx = = m (a jx
a x m (a I +b
Now let us apply formula tuting in the formula xm + bx") p ~
* n (a l rn, write dx + bxn and p not + bx) dx ' ) p ~ l dx. to the last term of (3) (^4) m + n for / x^ j bx n ) "~ (a + may is I for p. by substi This gives np (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). Substituting this in To III. derive and substituting formula (C). Solving formula (A) for 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 )
and substituting p + 1 for p, we ' J l get (D). dx, INTEGRAL CALCULUS 310 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 Here Solution. ra = f ?===== V J = 2, n 3, We apply reduction 1. = /> formula (4) = 3 a J, = CH 1, z* 1 6 under the power formula. Hence, substituting 2)(1 =  z*)* +C i. because the integration of the in this case would then depend on the integration ferential f  of Cx(\ 2 x' )~ *dx, which comes we obtain in (A), ( ILLUSTRATIVE EXAMPLE ^^ f
2. ; (a a  /I T 4
a a )4 ^ 4 + i a ^\ Va 8
j ; 2 / + a 4 arc sin f C.
a cS HINT. Apply (A) twice. ILLUSTRATIVE EXAMPLE C (a 2 3. f adz = ' \ a1
f w = = 2, ILLUSTRATIVE EXAMPLE 4. Him Here 1 . n 0, = /> .J, = a ', b = f ^ In x* (x f V^T Apply (B) once. 1. f ^ 1 a' HINT. Apply (C) once. PROBLEMS
Work out each of the following integrals. ,] T /
/^3 +
Va
2 r5 r/ t /V
. =
.r 2  1 .r  x* df J
/ o + Va 2 , Qc x2 f = 7r(3x4 + 4r 2 +
2
l^
=^
o J(a _ J 2 a2)  f^ 2 Va 2 / 2
(a f C. * 1 r J
* 1 .no
2 Y = (2 ^ .,/ T
2 a(a 2 . f .r 2 ^ x2) a 2) 4 8) VT^ + Va 2  .r 2 f r. ^a
8 r<*  f C.
TT^T arc tan
2 aa
{ dif sin a + C. REDUCTION FORMULAS
g Va  2 a x x2 2 j3 2 x2 x3 dx
2 Q
O (a
9. 10. f(x _
2 3a )* 2 dx )^ = ax (a = Jf =r . V 2 ax  x  r =~ ' 5 4 4 V4  9 '^(1+ 9/ 2
2 Vl Work 4 2 7~^
2
(a x2 a' In (x 3 a2 . ?  (2 a + L> 0/ 5 idr. .r) // d?/ wV9  4 ?/(9 ?/  / =
3o 81 Vl 1
) 4 8 a ) + ?/a l (~ C. 4 Vx 4 a 2 C. 2 2 / x\
a) 1 V 3 8  a A 1 2 r// = T/2  2)V4  2 C. r 4  9 // 4 ~; arc sin
 / ^ ^ 4 C. 49/ 2
/(I h 8 / 2 2 )Vl +4/ r
o
*** dx
2 2
x ^TiT
) ^ In (2 / + Vlf4/ 2 +
) X K dx
/A .s 7 ds 25.
r'
h V4 x
y
r Va + ig 26 fa*' 27. J/ (af ?>,s ' 2 23. x 2 (l f a 1 +c S arp sin 4 ,,
( . . 4  ^ 4 20 arc cos (l s'") 4 ) twice. i 3_^
2 2
^' 4 Vx 4a 2 ) 4 arc cos Apply + 30)V4// ?/ In (x 2 dx
Jf 2 1 out each of the following integrals. /x
20. 3 a' 24 / 17 18. 4a  K /vfirp J fx
' 2 4 a 2 (a 2 4 :* 14 16. f?/ 2 a 2 )Vx 2 4 2 2 ds Vl 4a 4 x2 '^ (a 2 f s 2 ) 'J 5 a 2 )Vx 1 l2.f^M= =
r ' a)V2 ax (x 4 3  * 2
) 2
x(2 x J " 3
2 4 2
x(2 x J yl> rf>C HINT, 2 x2) x 2 4 x 2 dx V^
V2  x(3 a 2 fx 2 Vx 2 4a 2 dx
C 11 4a 2 2 a2 f 5 x 2
1 f C. ) dx C
J/ 2 a  Va 2 ;< " x2 f (a 2 311 +x 2 2 '/; ) 24 V x
l 28. 29. d/. dx C. INTEGRAL CALCULUS 312 175. 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
tials shall now apply the same method to trigonometric differenby deriving and illustrating the use of the following trigonometric reduction formulas
I sin (E) : m x cos n x dx = sin" m f n n  1
C m
~
sin xcos n 2 ;tcte.
si
mh nj
. / sin m f sin m x cos n xdx =
. (F) _, ~ I m f f , _. sin m jtcos n
. . (G) I sin m , xdx= f x cos n l mf m+
Here the student should note by derive these x 1 f x 1 that when m + n
(F) /a?"fe wtoi m + r^
(G) /a?k when n + 1
(H) fails when m + 1
(E) fails 0. = 0.
= 0.
= 0. apply, as before, the formula for integration parts, namely,
(1 f?/ dv ) u Let
,. we * 1 Formula (E) diminishes n by 2.
Formula (F) diminishes m by 2.
Formula (G) increases n by 2.
Formula (H) increases m by 2. To j xcos n xdx. 1 x cos n f * 2 J fi+
sin m / sin m I mj n n /* i , then dw = = cos""
/ (n Substituting in 1 and 1 x, x = uv  Cv du.
sin" x cos
7 dv > ; l)cos ~x sin jdx,
(1), ri we (A), Art. j and v x dx = ; sin m+ get + ^ i T~T ?/i /* +1J I "^^ 7 sm m+2 xcos n ~ 1 136 REDUCTION FORMULAS
In the same way,
u we = we if m~
sin l 313 let and x, = cosn x sin x dx, dv obtain = 1 f O^" "^ X
^ /ClTl"* ~^
sin m x cos n x dx ^f
n+lj +
sin m + x cos But n ~x dx '/ S in,o a
cos 2 /) cos" sin"'/ (1 I 2 x dx  C
sin' I Substituting this in
I sin mx J r I sin"'/cos"xdx. and solving like terms, combining (2), C '/cos '~~xdx cos"xdx, we get (). similar substitution in (3), we get (F).
Solving formula (E) for the integral on the righthand side, ** for Making a and by 2, we get (G).
increasing
the same way we get (H) from formula (F).
Jn
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.
when m and n are integers, the integral It is clear that sin m x cos xdx
7J be made to depend, by using one of the above reduction
mulas, upon one of the following integrals may for : fdx,
' Cd? J
all fsin x dx, cos x of which = /cos x dx, fsec x dx, J we have f cos x sin x
. 1. how tan x dx, I ctn / dx, J to integrate. ! '* + **SS* + First applying formula (F), /i
sm 2 * 4^ = j esc x dx, Prove **
r?* cos' r dx = Solution. f I J J learned ILLUSTRATIVE EXAMPLE x cos x dx, / sin cos 4 xc/x = [Here we sin w = 2, * cos x get
cos x
r> .r (sin w = 4.] + 1 g r,, ria 4 J cos + x) + G INTEGRAL CALCULUS 314
Applying formula () /c\
(5)   C cos 4 x dx
j
1 & I m xcos3 x m= [Here C cos 2 x dx
j
J ,,
(6) 4.] of (5) gives cos x 4 x
 r and then substitute the result (6) in (5), get I  sin = j we ., 1  n 0, of (4), 3
,
 C cos x dx. i Applying formula (E) to the second member Now member to the integral in the second , This gives the this result in (4). answer as above. ILLUSTRATIVE EXAMPLE 2x4 Aan^2jr dj
J cos
. cos = Let 2 x Then x u. / (7) Apply (G)
(8) Ann' Apply (F) to the (9) / sin 2 /v cos new
u J Substituting from
the answer. (9) ' = = 1 ' J C. j* sin 2 M cos / J ^ A<< sin M and + jr) du, arid \ 2 r dx S l + / cos (8) into (7), = + ?/ c?//. 3, rej)lacirig u cos Tsiri^' ?i 2,  = du u J * ?? 2, m = integral in (8), with du tan 2 sin" 2 x x cos 2 x integral in (7), with rw M cos' M rfw cos JP cos* Z \ u, djr new  +  sin 2 x cos to the  In
(sec 2 j
4 2
1
= sin7 jc ^J  T tan 2 x sef 2
. 2
tan j  x .. Solution. Prove 2. I ' x by 1. sin // + In (sec w f reducing, and sotting M = tan ?0. we have 2 x, PROBLEMS
Verify the following integrations.
1. I sin 4 jc cos 2 x sin djc jc cos J jc\
[ 2. ftan' 5 %/ = f dr 2
^ tan f
O 4 f  sin'
6 ;{ :{ Tsec 5. fcsc 3 / r// 6 6. .r  \ sec djc OdO I = / Csin / x ctn x 4 cos (t> d<t> f  16 16 h C. r. + f J In (sec C. = i sin / + tan f J In (esc SC g ctn ^ J*csc
7. tan esc 7 J* /9 4 4. s i n2 24 t> p n
^~ + ctnO +
/ctn OdO=f jc + 3 In cos AJ  1 ('csc cos .r ^ f
j}) (2 sin +
2 /) 4 C. ctn  j) u. M c/u. f C. In (csc (9 1) f J  ctn 0) f r. + C. REDUCTION FORMULAS
ctn2 2 B
8. Jf 9 r
'.' 10. e
f =  7 ctn 2
4 sin 2 cosx _ dj? sin 4  3 sin 3 x * 2cos
3 sin (Wfldfl = 26] In
4 esc [8 r  315  (esc 2 ctn 2 (9) + C. + c .r cos'fl + 10 cos 2 415] +  + C. 2 11. f sin< Jo I = f "cos 4 jcdx 12. Jo o ^o 14. 13. 128 o
n TT r'sin2 fld Jo = f32 ^2^ = 15. Jn smx 5 _ 4 3_7r 8 4 Work
16. out each of the following integrals 17. fcsc<
J BdO. 18. c/ff. Jfsin2 19. : cosv 2 r J sin jl 21. cos J Jo
? ZL a 22. 3
f sin 6 cos : < 23. ('"(! c/0. + sin 6) 4 dO. Jo Jo The methods of integration deXVI, and XVII have been directed to reChapters XII,
veloped
ducing a given integral to one or more of the Standard Elementary
Forms in Art. 128. Various devices have been elaborated to this
176. Use of a table of integrals. in end, such as
integration by parts (Art. 136) ; integration by partial fractions (Art. 167)
integration by substitution of a H.CW variable (Arts. 168172)
; ; use oj'reduction formulas (Arts. 174 175). When, however, a more or less extensive table of integrals is formal integration is to
available, the first step in any problem
which the problem can be solved
search for a formula in the table by
without the use of any of these devices. Such a table is given in
in Chapter XXVII. Some examples
ILLUSTRATIVE EXAMPLE Solution. Use 14, with a 1. will now be given. Prove, by the Table of Integrals, and u = x.
would be worked out as = 2, 6 = 1, This example, without the table, in Case II, Art. 167. INTEGRAL CALCULUS 316 ILLUSTRATIVE EXAMPLE 2. by the Table Verify, dx__ = J_ + 4:rO
a = 3, b = Solution. Use 22, with This example, without the table, ILLUSTRATIVE EXAMPLE r  / _ as shown Use in Art. 1 V3 xdx f V 3 x* } 4 j J r" ?/ Case " III, Art. 167. of Integrals,  2
V4 + 3 rN/4+3J+2
. ....... and 3, a worked 4. Verify, ,, f ' . x. by the substitution 4 out, by the Table + 4x7 +3 x = z2, 7 of Integrals, 2 3 3V3
6 = 4, Use 113, with a
c
7,
3, and u = x.
Without the table the example would be solved by completing the square as
Illustrative Example 2, p. 206.
Solution. in r JT ^ in
2 is " 68. ILLUSTRATIVE EXAMPLE ./ 1 JxV4f3x
31, with a = 4, 6 = This example, without the table, in , 1 /. and by the Table Verify, dx I Solution. 3. = solved as 2,
is v \9+4:r7" 18 9 of Integrals, /_i_ , ILLUSTRATIVE EXAMPLE 5. Verify, = cos 2 x dx by the Table g '^ 2 sin 2 J of Integrals, 3 cos 2 x) f 13
Solution. Use 154, with a n 3, 2, u ^^ x. Without the table the example would be solved by integration by
Illustrative In Example 6, parts. See Art. 136. many problems the given integral cannot be identified with
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 one in the table as easily as in the This method has been used constantly simple change of variable.
in Chapter XII and in all integration ILLUSTRATIVE EXAMPLE 6. Verify, by the Table
O 1 zV4 x~
Solution. Formula 47 is f ;=iln3 9 similar. problems hitherto. ==l= + r. + V4
u = 2 x. 3 Let of Integrals, r or 2 +9 Then x  \ u, dx = J du, and, substituting the values in the given integral, we obtain
du
jdM
= r <fr
C
= r
J xV4^ 2 +9 J \ wVw 2 Hence, applying 47, with a = 3, f 9 J u^/u'2 +9
=2 and substituting back v x, a = 3, we have f
J x
Without tables we should proceed as in Illustrative Example 2, Art. 135. REDUCTION FORMULAS
ILLUSTRATIVE EXAMPLE
r\ 9 J 7. Verify, 4 by the Table  2 . Solution. Substituting, Formula 84 we is r\ '9ar4
' This is = 2 x, we = u of Integrals, 4 2 x. Then x = dx t n, =  dt*. obtain ^ rVjj J u Let similar.  x (9 J 317 J
now 84 with a = " u J M j 2 (/K 3 du. 2 Hence, applying ;. 84, and substituting back get the required result. no formula from the table can be applied as in the preceding
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
If two cases, there 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
96104. Also, the reduction formulas of Art. 175, with additional
ones for various cases, are numbered 157174. 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. Vz +
2 1. 5 dx = 7. Ce< sin 2 J ^
Z dt = A (3 ^V9 x =\
4 e<(2  x1 2  sin  4 t 10 )(x'2 f   cos + 5) In (3 rt f x C. + f C. V9 x 4) + C. INTEGRAL CALCULUS 318 9 ^ Csn26jd0 g J + /2 10. fx :< 2 In (1 cos 0)  2 cos f C. t + ^ = arCtan(j:+1) + C  2 = x 2 dx sin \ sin dx
'J + cos 1 4 V(x x2 =2 1)(2  \ x 2 cos x 2 arc sin Vx f C. 1 f (;. x) C Work out each of the following integrals. 4 f sin 2 Evaluate each of the following definite integrals.
37. . Jo (1 + 2
J") dT
35. / x 2 V25  _ dx
36. (4x 2 h 9) 25 9 x2 >:j 38. 2
39. f \l<L^>du=ira.
w
a J~a f REDUCTION FORMULAS r 40.
0. J\ x2dX
f V9  = 1.338.
13: 42. 2 x'2 43.
2 d" 44.
.jf /V / ' <// = 0.1605.
3 46.
2 r~ 45. 319 jf^ , dx 48.  jf *i f 2 cos 2 sin 4 6 dO. 47. Jo /
Jo e i cos A 49 f dt. TT/ Vo ^ C os A d> dd>. ADDITIONAL PROBLEMS
1. Verify the following results. 2. A parabola with its axis parallel to the //axis passes through the and the point (1, 2). Find its equation if the area between the
parabola and the xaxis is a maximum or a minimum. origin Anx. // = 6 x 4 s'2 gives a minimum. Sketch the curve y V!r = In jr. Find the volume of the solid of
revolution formed by revolving about the .raxis the area bounded by the
curve, the .raxis, and two ordinates, one through the maximum point
AUK. ^^ TT.
and the other through the point of inflection.
3. A is formed so that the density at
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
Am. 630 TT Ib.
its altitude and the radius of the base are each 3 ft. 4. cone of metal solid right circular any point NOTE. P is 20(5 The weight r) Ib. of an element of uniform density is its volume times its density. A hollow metal sphere has an inside radius of 6 in. arid an outside
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.
5. radius of 10 6. If n an even integer, show that is TT 7T j
r 2 sm x dx
71 I Jo
7. If n is = r 2 cos" x dx
7 I = ( n  D(n ^
n (n JQ an odd integer, find the value
/*2
/ Jo sin" x dx. of 2) 3) (!) ^r
(Z) TT TT 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
This
if supported at a point directly under its center of gravity. point of support is the centroid of the area of the flat surface of the
cardboard.
For certain areas considered in elementary geometry the centroids 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.
of the
Consider the area
figure. Divide it into n rectangles, each
with base A.r, as heretofore. The figure
shows one of these rectangles. Let d A be AMPNB its area, and C(h, (l)dA =
The moment x, (A) k of area of this elementary rectangle about OA' (or
pendicular distance of moments Then k) its centroid. O Y)
its are, respectively, is the product of
centroid from OX dM x and dM u , its area by the perOY). If these (or then dM =kdA, dM =hdA.
x v AMPNB is obtained by applying
The moment of area for the figure
the Fundamental Theorem (Art. 156) to the sum of the moments of
area of the n elementary rectangles. Thus we obtain 1x = Ck dA, M 320 v = f hdA. CENTROIDS, FLUID PRESSURE 321 and A its
Finally, if (x, y) is the centroid of the area
area, then the relations between the moments of area (B) and x AMPNB, and y are given by To
(1) calculate find the (x, y), and (), these M = Ax (C) ri> = x 2 y dx, % M = moments above are, for the (2) Ay Vt x. M of area x and M From v. figure,
s*b M = xydx, / v Ju *Ja 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), If = x (3) A is not known, be found by integration, as may it ILLUSTRATIVE EXAMPLE Find the centroid 1. of the area in Art. 145. under one arch of the sine curve y (4) Constructing an elementary rectan Solution.
gle, = sm x. we have
(5) dA = ydx sinx dx,
dMx = k dA = I y' dx = \ sin x dx, dM v =
2 2 The limits are x 0, x = 2,
(6) A = f^sinzdz
J
Then, from (3), x = \
The value of = TT. M =\ x h dA = xy dx \ IT, x sin x dx. Hence = f "sin 2 * do*' M =C
^
V = I TT. Aws.
TT, y
x might have been anticipated, since the line x zsinzdz = } TT is = 7T. an axis of symmetry.
2 y' ILLUSTRATIVE EXAMPLE 2. In the figure the curve OP A
2 px. Find the centroid of the area OPAB. Draw an elementary Solution. the figure, and mark dA =
Using (4), its x dy, centroid h = \ x, is an arc of the parabola rectangle, as in (h, k). k Then <?A (a, 6) = y. dMz = kdAxy dy, 2
Finding x in terms of y from y = 2 px, and
between the limits y = 0, y = 6, we find
integrating Hence x
Hence 6 2 = . 20 p y =7 = 2 pa, and x = b. But x = a, y = 6 satisfy the equation i/ 4 ' 3
r ^ a. The centroid is therefore { ( T a, J b). Am. = 2 px. INTEGRAL CALCULUS 322 PROBLEMS
Find the centroid
2 = 2 px, 1. y 2. y = x'\ jr y 3. y /*, 4. .r 4 5. y 8. j/ 2 = 8. 9. // 2 f 2 ?/ 10. ?/ ?/*, ?/ 12. y 13. (f h, 0).
1 (,  ^). 0. (First quadrant.) (j, f^). j. ?/ (^,
(, // = ?/ 4 jr 8, .r ~ j 3 a = x'2 y , = 2xr. y jr, cur, ^= 1, //  jr j2  3. (2, 1). jr. // , 6 0. 2 a  .r ij). 1). (!,) i/ 6 x 2 =^ y = 2.r0 = 4.
= 2.r + 3.
 2 x  3, = H 11. Ans. h. 4 x. ?/ .f =
= 2, bounded by the following curves. 4x, =
=x
= /*,
y 7. = x = of each of the areas 3. (First quadrant.) = 0, y = 0.
= 0, x = 2 a. jr fr (First quadrant.)
(First quadrant.) 14. Find the_centroid of the area bounded by the coordinate axes and
= I a.
/b?x. 7
the parabola V.r f V// = Va./ Find the centroid 15. 4 y* .r 2
; bounded by the loop of the j = curve 1 /' ^ 7/ ortion in the first quadrant of the ellipse _ + ~~l.
2
y' of the J 4/f.x. 2 . O" of the area r*. Find the centroid 16.
jr 2 1 Aws. x 4
= a  = // 4 b
o J 7T t5 /> 7T Find the centroid of the area bounded by the parabola y 2
2 ;xr
4 * and the line = rax.
/>
p
Ans. x =
?/
2
17. , ?/ m 5 18. and .r Find the centroid = of the area included
4 ?>?/. /i AiS . m by the parabolas y 2 = ax
 __ 9
R
J/i
J
//
2o ^ "
20 ^ ^ k i 19. and its 20. Find the centroid of the area bounded by the cissoid // 2 (2 a
2 a.
.4 MS.
7 =
asymptote x !; Find the centroid and the .raxis. of the area bounded by the witch x 2 y x* x)
a, // = 4 a (2 a
= 0, = 0. 2 Ans. x
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
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
2 r sin^
22. ' Am 3(6  sin flcos 6)' CENTROIDS, FLUID PRESSURE
23. Find the centroid of the area p
24. a cos 2 p 25. p Find the centroid a(l bounded by the cardioid + of the area 8. cos 6). of the area 0. An$. jc bounded by one loop Ans. Find the centroid = a cos 3 = 323 = Distance from origin bounded by one loop AHS. a, // = 0. of the curve = 128aV2
105 of the Distance from origin  TT curve 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, is it necessary to
of the pre modify the discussion ceding article only in the details.
Let OX be the geometrical axis
of solid. The centroid on this axis. Let the then lie dV will be an element of volume, that is, a cylinder of revolution with altitude Ax and radius OY through
(1) Then dV iry Ax.
volume of this cylinder with respect to the plane
of
perpendicular to OX is
dM v = x dV = 7TX// 2 Ax. y. The moment The moment of volume for the solid
mental Theorem, and x is given from Vx = (2) M I irxy y ILLUSTRATIVE EXAMPLE. is dx. Find the centroid of a solid cone of revolution. Solution. The equation of the y = AB ^r OA x Hence
Since V= M
3 v = f
Jo
2 Trr' h, TTX x or element
!, OB =  h' ^~ dx = 4
h*
\ h. Ans. irr 2 h 2
. is then found by the Funda INTEGRAL CALCULUS 324 PROBLEMS
for each of the following solids. Find the centroid
1. figure.)
2. Hemisphere.
Ans. x (See = I r. 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 revolution of solid generated. 4.
5. 6. 7.
8. 2 2 2 3. x' a' y' 2 xy = 2 a' x , x ,  2 a.
J x a,  2
ay = x' x = a.
2 =
4 x, x = 1, x = 4.
2
x 4 2 = 4, x = 0, x =
= a sin x, x = ^ 2 a. Ans. , jr jj a. i/ ? 1. = f J. IT. ?/ The area bounded by OY and each of the curves given below is reTolved about OY. Find the centroid of the solid of revolution generated. =
 2 9. y' 10. jr 2 11. a//
12. 2 4 ax, y
ir = The = /', 1, Ans. b. = y 0, ?/ = 1. y a. ?/ radii of the upper and lower bases revolution are, respectively, 3 in. and 6
Locate the centroid.
13. parabola y* of a frustum of a and the altitude cone of
is 8 in. of the solid formed by revolving about the ?/axis
a, and the
quadrant bounded by the lines ?/ = 0, x
Ans. y = f a. = first 4 ax. Find the centroid of the solid formed by revolving about the raxis that part of the area of the ellipse quadrant.
15. in., Find the centroid the area in the 14. =  6.
= A y first
2 the hyperbola x' + = 1 which lies in the r, Ans. x = first
a. of the solid formed by revolving about the zaxis
2 a, and
0, x
quadrant bounded by the lines y Find the centroid the area in the a2 2 ~= t/ 1 . CENTROIDS, FLUID PRESSURE 325 16. Find the centroid of the solid formed by
revolving about the .raxis
the area bounded by the lines x
0, JT = a, y
0, and the hyperbola 17. Find the centroid of the solid the area bounded by the lines // = formed by revolving about the 0, x = , and the curve // .raxis sin 2 x. 18. Find the centroid of the solid formed by revolving about the xaxis
r
the area bounded by the lines .r = 0, .r = a, // = 0, and the curve //
e
. 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.
19. 179. Fluid pressure. We will now take up the study of fluid pressure and learn how
of a fluid on to calculate the pressure
a vertical wall. ABDC Let
represent part of the
area of the vertical surface of one wall
of a reservoir. It is desired to deter mine the total Draw
//axis fluid pressure on this area.
the axes as in the figure, the
lying in the surface of the fluid. Divide AR into n subintervals and
construct horizontal rectangles within
the area. Then the area of one rectangle (as EP) is y Ax. If this rectangle was horizontal at the depth x, D the fluid pressure on it would be W xy A j,
pressure of a fluid on any given horizontal surface equals the weight!
of the fluid standing on that surface as a base and of height
[The to the distance of this surface below the surface of the fluid.
equal
J
of a where column W = the weight of a unit volume of the fluid. sure'is the same in all directions, it follows that proximately the pressure on the rectangle Hence the sum EP Since fluid pres Wxy A.r will be ap in its vertical position. n represents approximately the pressure on all the rectangles. The
is evidently the limit of this sum. Hence,
pressure on the area
the Fundamental Theorem,
by ABDC
n im
lim V Wxflt Ax, =J oo i = 1 x>
/ Wxy dx. INTEGRAL CALCULUS 326 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
Fluid pressure (D) where the value of y  W yx dx, \ Ja terms of x must be substituted from the equa in tion of the given curve. We shall assume 62 Ib. ILLUSTRATIVE KXAMPLE as the weight of a cubic foot of water. W) ( A circular water main 1. 6 ft. in diameter is half full of water. Find the pressure on the gate that closes the main.
Solution. The equation y= Hence and the \ 9  2 is x' + 2 9. y' x~, .r
to x = II. Substituting
pressure on the right of the from limits are (>), we get
xaxis to be in of the circle the Pressure = C'\ 62  9 x dx x* = [ <. (9  x u ;a]jj = 558. 'o Total pressure Hence = essential part of the The 2 x 558 1 1 Ans. 16 Ib. above reasoning that the pressure is 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 A !2. trapezoidal gate in a Find the pressure on the gate when the
surface of the water is 4 ft. above the dam shown is v WATER top of the gate. OX OY and
Solution. Choosing axes
as shown, and drawing an elementary
horizontal strip, we have, using (), dA The equation 2 .r (///, of AB is Solving this equation for
tuting, the result is y
.r, dP = W(% W 4 f t/O 8. y)(y and y Integrating with limits y P= 2 x and substi (64  y~)dy +
4, we obtain = ^^ W = 14,549 Ib. Ans. in the figure. CENTROIDS, FLUID PRESSURE 327 PROBLEMS
In the following problems the //axis is directed vertically upward, and
.raxis 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. the 1. (0, 0), (3, 0), 2. (0, 0), (3,
3. (0, 0), (2,  (0,  6), (0, 2), (0, Ans. 6), (0, 0).  4), i 2, 18 \V. 36 W. 6), ^0, 0).  16 W. 2), (0, 0). 4. Calculate the pressure on the lower half of an
ellipse whose semiaxes 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*,
5. 12 8 If; (b) 12 \V. Each end is (a) of a horizontal oil tank is an ellipse of which the horizontal
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. axis ft. An^. 2160
6. The end of a vat is a segment of a parabola (with vertex
across the top and 16 ft. deep. Calculate the pressure
the vat is full of a liquid weighing 70 lb. per cubic foot. vertical at the bottom) 8 on this end when ft. /b/K. The lb. 38,22911). end of a water trough is an isosceles
of which each leg is 8 ft. Calculate the
right triangle
pressure on the end when the trough is full of water
7. (W = vertical AUK. 3771 62.5). 8. The lb. end of a water trough vertical across the top and 5 ft. deep).
trough is full of water. is an isosceles triangle 5 ft.
Calculate the pressure on the end when the 1302 lb. horizontal cylindrical tank of diameter 8 ft. is half full of
weighing 60 lb. per cubic foot. Calculate the pressure on one end. oil Ans. 2560 lb. 9. 10. A//,s. A Calculate the pressure on one end if the tank of Problem 9 is full. A rectangular gate in a vertical dam is 10 ft. wide and 6 ft. deep.
the pressure when the level of the water (
62.5) is 8 ft. above
the top of the gate
how much higher the water must rise to double
(b)
Am. (a) 41,250 lb. (b) 11 ft.
the pressure found in (a).
11. Find W (a) ; ; 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.
full of A vertical cylindrical tank, of diameter 30 ft. and height 50 ft., is
Ans. 3682 tons.
water. Find the pressure on the curved surface. INTEGRAL CALCULUS 328 180. Work. In mechanics the work done by a constant force F
causing a displacement d is the product Fd. When F is variable, this Two an integral.
be considered here. definition leads to examples will 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 revolution with their axes vertical. It
is convenient to assume the raxis 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.
into w subintervals, pass planes perpendicular to the
axis of revolution through these points of division, and construct Divide AB The volume cylinders of revolution, as in Art. J60. cylinder will be 7n/~A:r and
of a cubic unit of the fluid. its 2 weight M'7r// The work done A.r,
in fWork done in lifting The work done The work done jr) will cylinder be 2 equals the weight multiplied by the vertical height.] in lifting all in W = weight lifting this of the fluid out of the reservoir (through the height Wiry any such of where such cylinders to the top is emptying that part of the reservoir the sum will evi dently be the limit of this sum. Hence, by the Fundamental Theorem,
lim c l = r
/ J 2
Wiry xdx. Therefore the work done
solid of revolution in emptying a reservoir in the form of a
from the depth a to the depth 6 is given by the formula
(F) where the value Work= WIT f
Ja 2 y xdx, of y in terms of x must be substituted from the
equation of the revolved curve. CENTROIDS, FLUID PRESSURE
ILLUSTRATIVE EXAMPLE
filling 1. Calculate the work done in
a hemispherical reservoir 10 ft. deep. The equation Solution. &+
=
W= Hence y* of the circle = 2/ 100 is  *', 62, to r
limits are from x =
Substituting in (F), we get
/> Work = 62 TT 10 I Jo = 10. 000  x'^rdr = 155,000 TT ft.lb. above reasoning is that the element
an elementary volume ( dV) through essential principle in the work (= dw} done
a height (=A) is in lifting of where pumping out the water 100. and the The 329 W = weight dw=KhdV,
volume of unit may the axes of coordinates
ILLUSTRATIVE EXAMPLE A 2. With this in mind,
convenient manner.
any of the fluid. be chosen in conical cistern 20 is across the top and 15 ft. deep. If the surface of the water is 5 ft. below the top, find the work done in pumping the water to the top of the cistern. Take axes OX and
Then Solution. the figure. dV =
h y. VT(15 y}irx~ dy. of the element OA is Substituting, dw = 7rW(15 The as in TTX 2 dy, The equation = 1(10, 15) = 15y, dw
x OY limits are y = w= Work done and y = /)J ' irW f = z <ly n5 2
it'  y"')dy = is 10 2 y' ft. 216,421  y*)dy. deep. ft.lb. Work = v If a gas cu. ft. to in
vi r*>i
I p = pressure in pounds per square foot.
Let the volume increase from v to v Proof. + dv. area of cross section of the cylinder. c = distance the piston Then
c An*. a cylinder expands
cu. ft, the external JvQ where p Integrating, Jo by an expanding gas. (G) T 7rH (15 i water 10, since the 10 $ u against a piston head from volume
work done in footpounds is Let ft. moves. INTEGRAL CALCULUS 330
Since pc = force causing the expansion dv, = Element of work done pc = p dv.  c Then (G)
relation by the Fundamental Theorem. To use (G), the
This
/; and v during the expansion must be known. follows between relation has the form 0) pi? = constant, the exponent n being a constant.
Isothermal expansion occurs when the temperature remains constant. Then n
1, and the pressurevolume relation is pv (2) a graph of If (1) is por made () = piVi. (pressurevolume diagram), plotting volumes as abscissas and pressures as ordinates, the area under
curve gives, numerically, the work done, as calculated by (G).
isothermal expansion the graph of (2) is this In a rectangular (equilateral) hyperbola. PROBLEMS A 1. vertical cylindrical cistern of diameter 16 and depth 20 ft. ft. is Calculate the work necessary to pump the
Am. 800,000 TT ft. Ib.
water to the top of the cistern. full water (VT of 2. Jf to 62.5). the cistern of Problem pump the water to the
3. A conical cistern 130 1 is half full, calculate the ft. across the top and 20 (IV rr 62.5). Calculate the work necessary to
of 15 ft. above the top of the cistern. , is full deep ft. pump A hemispherical tank of diameter 10 ft. is
cubic foot. Calculate the work necessary to
per
of the tank.
4. A work necessary top.
of water the water to a height
,,
2,500,000 TT 1
~
1
. J\ // 8. . ID. weighing 60 Ib.
the oil to the top full of oil pump Am. 9375 TT ft. Ib. 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
5. hemispherical tank of diameter 20 rate of 16,500 ft. II). per minute). How ft. is full long will it take the engine to empty the tank ?
6. Find the work done in pumping out a semielliptical reservoir full
62). The top is a circle of diameter 6 ft., and the depth
of water (W
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. Calculate the work necessary to pump the liquid to the top of the reservoir. 4ns. 15,360 TT ft. Ib. CENTROIDS, FLUID PRESSURE
8. A water tank is in the form of a hemisphere, 124 ft. mounted by a cylinder of the same diameter and 10
work done in pumping it out when it is filled within 2 in diameter, sur Find the ft. high. ft. of the top. bucket of weight M is to be lifted from the bottom of a shaft
deep. The weight of the rope used to hoist it is in Ib. per foot. Find 9. h 331 ft. A the work done. A quantity of air with an initial volume of 200 cu. ft. and presIb. per square inch is compressed to 80 Ib. per square inch.
Determine the final volume and the \\ork done if the isothermal law holds,
10. sure of 15 that pr is, 11. = C. 4//s. Determine the volume and work done final adiabatic law holds, that :i7.f> is pr 7' (\ on.
in ft. 72:5,000 ; Problem 10 ft. Ib.
if the ~ 1.4.
assuming n
An*. 60 cu. 648,000 ft.; 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
A//s. 60 11). per square inch
599,000 ft. Ib.
C.
the law is pv
; 13. Solve Problem 12 the law if AHS. is 104.5 pr"
Ib.  <\ assuming 1.4. // per square inch 801,000 ; ft. Ib. 14. A quantity of gas with an initial volume of 16 on. ft. and pressure
60 Ib. per square inch expands until the pressure is ;}() Ib. per square
of
inch. Determine the final volume and the \\ork done by the gas if the
An*. :*2 on. ft.
95,800 ft. Ib.
law is pr = C.
; 15. Solve Problem 14 if the law is (\ pr" An*. assuming 1.2. n 2K.f> cu. ft.; 75,600 ft. Ib. quantity of air with an initial volume of 200 cu. ft. and pressure
Determine the final
is compressed to 150 cu. ft.
if the law is pr ~ (\
pressure and the work done
16. A of 15 Ib. per square inch  Solve Problem 16 17. if the law is pr 7' = (\ assuming r? = 1.4. inch and
gas expands from an initial pressure of 80 11>. per square
Find the work done if the
ft. to a volume of 9 cu. ft.
n
1.0646.
lawis pr
0, assuming n
18. A volume
19. of 2.5 cu. Solve Problem 18 if n = 1.131. 20. Determine the amount of attraction exerted by a thin, straight,
of mass M upon
homogeneous rod of uniform thickness, of length /, and
m situated at a distance of a from one end of
a material point P of mass the rod in its line of direction. Solution. Suppose the rod to be divided into equal infinitesimal portions ments) of length dx. hence M = mass of a unit length ~ dx = mass of any element. of rod ; (ele INTEGRAL CALCULUS 332
Newton's law measuring the attraction between any two masses
Produrt of masses
Force of attraction =
'distance between them;*
for is t P and an element of the rod is therefore the force of attraction between the particle at 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
= and j /, we have
between x ^ /./M
r'rmdx
, Force of attraction  J I 'o 21. (j = + Determine the amount 2 aj f C M dx l J of attraction in the last example if
from it. P lies in the perpendicular bisector of the rod at the distance a 2mM Am.  2 22. A If // to empty is its form of a right circular cone vessel in the height and
itself is filled with water. the radius of the base, what time will
through an orifice of area a at the vertex?
r it require Solution. Neglecting all hurtful resistances, it is known that the
velocity of discharge 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 dt, = \ 2 gx. Denote by dQ the volume of water discharged in time
and by dx the corresponding fall of surface. The volume of water discharged through the
a orifice in a unit of time is \ being measured_as_ a right cylinder of area of base a and
r (= \ 2 r/.r). Therefore in time dt altitude
(1) Denoting by S the area
from geometry, of the surface of the water when the depth
'' , '
, is x, we have, > VI
or 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 Therefore (1) and (2) t and solving = for dt, JL =
2 gx ^. 5o\ 2 g Ans. CENTROIDS, FLUID PRESSURE
Mean 181. value) of =\ y (i) We The value of a function, n numbers + y<2 arithmetic mean (or average s ,# y\, y<2 , 333 . +  + //). proceed to establish the formula rb
6(x)dx
T I Mean (H) value of from x The =a figure shows to x = = Ja 1 4>(x) bj b a the graph of = y (2) </>(/). y) of the ordinates of the arc PQ
into n equal parts each equal to A.r The mean value (= is to be de and let //i,
Divide AB
be the ordinates at the w points of division. Then (1 will
2/n,
yz,
give an approximate value for the mean value required. Multiply
numerator and denominator of the righthand member of (1) by Ax. fined.
' * ) = Then, since n Ax b we a, =^ y (approximately) (3) get
, __ But the numerator in (3) is, approximately, the area APRQB.
The average value of y (or (/>(j)) is defined as the limit of the rightoo. This gives (//).
hand member in (3) when n
> In the figure the mean
= area ABQRP. value of equals <j>(x) CR if area rectangle ABML Taking y as the function (dependent f variable), then (H) becomes ydx Ja ILLUSTRATIVE EXAMPLE. Given the circle + x2 (4) y 2 = r2 . Find the average value of the
dinates in the first quadrant when or X expressed as a function of the abscissa x
(b) when y is expressed as a func(a) y is ; tion of the angle
Solution, = Z MOP.
Since y = \V  x 2
2 (a) the numerator in C Vr 2 Jo , (/) is  x 2 dx = I Trr 2
. Then y = \ irr = 0.785 r. Ans. INTEGRAL CALCULUS 334
(b) Since y
in (/) r sin 0, and the limits are 6 = = 8 a, \ IT the numerator b, is A*
sin 6 r I d6 = r.  Since b a = TT, \ we have y JO = 2 r = 0.637 Ans. r. 7T Thus we have quite different values of y, depending
variable with respect to which the mean value is taken. upon the independent 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 Ch
I ydx x a b
in
is order to indicate explicitly the variable with respect to which y averaged.
Thus, the Illustrative Example, in we have 0.785 ~y x r, and y ft = 0.637 r. PROBLEMS
1. Find the average value of ?/ = s~ from = jr to x Find the average value of the or di nates of y 2
4) taken uniformly along the .raxis.
2. (4, 3. (4, Find the average value when uniformly 4)
4. of the abscissas of y'2 distributed along the Find the average value of sin .r = 10.
4 Ans. 33f 3 r from Ans. = 4 from jc and .r = .r TT. *c = 2.
to (0, 0) Ans. ?/axis. between (0, . 0) to 1^. Ans.
IT 5. Find the average value average value is of sin 2 .r between jr = and TT. (This frequently used in the theory of alternating currents.) Ans.
6. If a particle in a velocity of
/ The v (} ft. vacuum were thrown downward with an per second, the velocity after N i velocity after falling s ft. r (2) Find the average value t sec. %. initial would be given by / would be given by
Vz\r + 2 gs.  (Take g = 32) of v Ans. 80 ft. per second.
during the first 5 sec., starting from rest
during the first 5 sec., starting with an initial velocity of 36 ft.
Am. 116 ft. per second.
per second
(c) during the first 2J sec., starting from rest; Ans. 40 ft. per second.
Ans. 53 J ft. per sec.
(d) during the first 100 ft., starting from rest;
(e) during the first 100 ft., starting with an initial velocity of 60 ft.
(a) ; (b) ; per second. Ans. 81 ft. per second. CENTROIDS, FLUID PRESSURE
7. In simple harmonic motion s = a COP 335 Find the average value nt. of the velocity during one quarter of a period (a) as to the time
to the distance. (b) as ; 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
that the average area of the rectangle whose sides are the two segments is a 2
(b) that the average value of the sum of the squares on (a) ; the two segments is fa 2 . 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.
10. If ( 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, The formulas ~
I where (x, y) is 1 14. ~ x ds I y ds ' * (6) p. 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
12. an arc Show it. (Compare Art. 177.) that the area of the curved surface generated by revolving
about a line in its plane not cutting the arc equals of a plane curve the length of the arc times the circumference of the circle described by
its centroid (6).
(Theorem of Pappus. Compare Art. 250.)
HINT. Use
13. to (L), Art. 164. Find the centroid 2
of the arc of the parabola y Am. (4, 4). 14. Find the centroid of an arc of the circle p 4 x from 7 = 1.64, y a between 0)
2.29. (0, = arid f 6. A ns.
15. Find the centroid of the perimeter of the cardioid p Ans.
16. 2 x' Find by the Theorem + 3/2  r w hich lies in
2 the of Pappus the centroid first quadrant, =
~jc a(l = f I a, cos y 0).
== 0. of the arc of the circle ^^ x = T = INTEGRAL CALCULUS 336 17. Find by the Theorem of Pappus the surface of the torus generated
2
2
2
f y = a (b > a) about the /axis.
by revolving the circle (x
b') 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
18. generated. ADDITIONAL PROBLEMS An 1. Find its // = 2 x' is bounded by the lines y = x2 , + x y abscissa of the centroid of the area and a certain = 6, Ans. centroid. The 2. 2 area line through the origin y
x 0, = and x
f , 3. 4. = 3.
f An.s. Find the centroid of the area bounded by y = x n (n >
1.
Discuss the locus of the centroid as n varies.
_
nhlAUK. x =
y Find the equation of the locus
and the parabola y = ex .raxis + 2 0), 2 when c varies. f . the xaxis, n+ 1 2(2 nf 1) of the centroid of the area
x' . bounded by the curve
Find the ordinate of n by the = is 1. the centroid. and x y bounded Ans. 5 y = 2 x'2 . = 2 py and any oblique line y
mx + b meet5. Given the parabola x'
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
2 D.
.4 H Prove that (a) the tangent to the parabola at /> is parallel to the line
the centroid of the area ACIU) lies on the line CD.
(b) ; P x 2 and let C be the centroid
be a point on the parabola //
the parabola, the .raxis, and the ordinate through
of the area bounded by
P. Find the position of P so that the angle OPC is a maximum.
4ns. Ordinate = f%.
6. Let 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, perpendicular 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
to pump over the top of the cistern one half the volume of work required
water it contains. A AnSm 16,000 (\/2  f = 6937 ft.lb.
o
water. Two men, A
If A starts first, what
1) hemispherical cistern of radius r is full of
pump it out, each doing half the work.
will be the depth d of the water when he has finished his share of the work ?
8. and B, are to CENTROIDS, FLUID PRESSURE A 337 tank in the shape of an inverted circular cone is full of water.
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.
8 r 4 1 = 0. Calculate
Show that z is determined by the equation 6 r4
An*. 0.61.
the value of z to two decimals.
9. Two men pump are to :{ A bucket, weighing 3 lb., has a volume of
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.
An*. 12,156 ft.lb.
per second. (A cubic foot of water weighs 62.4 lb.)
10. 2 cu. A ft. well is 100 deep. ft. The bucket is filled The area OAB is divided into elements such
Show that the area A and the moments of area 11. O. M x M and v r by are given M r \\ (xy y)dx, My = 3 I ^ n/ x = Y by lines from j r
JJ y(jcy' y)djr. of a triangle is on any median at
of the distance from the vertex to the (The centroid two thirds OPQ as } opposite side.)
12. Find the centroid of the hyperbolic sector bounded by the equiand radii from the origin to the
a tan
a sec 6, y
lateral hyperbola x
points (a, 0) and (x, y). An*. x  2 a tan V _ 2
3 sec
in (sec + tan 0) CHAPTER XIX
SERIES A 182. Definitions. some cording to sequence a succession of terms formed ac is fixed rule or law. For example, 1, 4, 9, , and , i,_ 25 16, X2
3, r
' X
X* X*
_,__.,_,__ are sequences. A scries is sum the indicated of the terms of a sequence. from the above sequences we obtain the Thus series 1+4 + 9 + 16 + 25
X2
X*
X*
X*
1 _
+ .__._ + ~.
. ! , and , :r When the number The called is of terms limited, the sequence or series is When the number of terms is unlimited, said to be finite. or series T an is the sequence infinite sequence or infinite series. or r/th term,
the law of formation of the terms.
ycHcral term, is an expression which indicates ILLUSTRATIVE EXAMPLE 1. In the first example given above, the general term,
2
is n
The first term is obtained by setting n = l, the tenth term by or nth term, setting u = . 10, etc. ILLUSTRATIVE EXAMPLE
except for n
If the = 1, is ^ sequence 2. In the second example given above, the nth term,  : ?il
is infinite, this fact is indicated 1, 4, 9, Factorial numbers. nection with series
with 1. Thus, is [5 or 5 , r/,  by the use of dots, as . An expression which occurs frequently in cona product of successive integers, beginning 1x2x3x4x5 by  is called 5 factorial and is indicated ! In general, [w = lx2x3xx(?il)xw n factorial. It is understood that n is a positive integer. The
expression [n has no meaning if n is not a positive integer. is called 338 SERIES
The geometric 183. shown in if series of n terms, ar n ~\ elementary algebra that _ or ^ form being generally used first _____ r) < if \r\ and the second form 1, 1. \r\>
If + 2 n the For the geometric series. = a + ar + ar + Sn (1)
it is 339 \r\ < then 1, r n decreases in numerical value as
lim
n From formula we (2)  n (r ) = // increases and 0. ao see, therefore, that (Art. 16) limS^T^l
T (3) n Hence oo sum S n the \r\< 1 if  of a geometric series limit as the number the series approaches a said to be convergent. If \r\> (Art. 18). become A is 1, then r of terms n become will infinite as // Hence, from the second formula
In this case the series infinite. In this case increased indefinitely. is peculiar situation presents is in increases indefinitely
(2), the sum S n will said to be divergent.
1.
The series then itself if r becomes
a (4) + a a f a r; a   . even the sum is zero. If n is odd the sum is a. As increases
sum does not increase indefinitely and it does not
approach a limit. Such a series is called an oscillating series.
If t] is //, indefinitely the ILLUSTRATIVE EXAMPLE. Consider the geometric Sn (5) We find, by (2), that = Sn series with 1 = Then
(6) lim
n Sn = 'A which agrees with (3), = when a 1, r = \. *oo It is interesting to discuss (5) geometrically. To
do this, lay off successive values of 1 +; 6l S n on a straight n 1 2 1 12 Sj! *
5 ^ as in the figure. 3 Sn line, ' ij Each point thus determined bisects the segment between the preceding point and
the point 2. Hence (6) is obvious. DIFFERENTIAL AND INTEGRAL CALCULUS 340 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 3.  i + 6 ~ * 2 + 8 + 16 J + + + jj f 2.4~ 24.f>~ 2.4.6.8~ 3579 Write the first . + f f ^_l'j_^ 6 + jj = 2n nth term Arts, . f *"' 1 5L + **' y
. . . + 2 n four terms of the series whose nth, or 1 term general, is given below. 7.^8
o. Ans. 4 V2 + 4 + 7= +
V3 V4 "* 3 + 1 + ^6
ortfr.
O
O JLA.
rt + l ., i n " ,fH o
1U. i   llt 1 . ^7i
O . 1 1
J/ V^
' 1 ) + i^ ij +V^ V3
1^ / X'* . n
f/^"""^ r?y ~l~ n^^ w , < , /j3
Ii^ i V4
, .. M ' Vn~T2 2"[n [w Convergent and divergent = I *^2 ^ ' Sn _4_ v2 'y ^ ' 12. , X
X
__}___{ X
. > 184. 3 , C/  ( ? ^Q^o^97^
O J Ui series. + U 2 + Us + In the series + Uny the variable Sn is a function of n. If we now let the number of terms
(= n) increase without limit, one of two things may happen. CASE
(1) I. Sn approaches a limit, say lim S n u, indicated = u. by SERIES
The infinite series is value CASE II. now said have the value u, or to 341 to be convergent and to converge to the u. S n approaches no The limit. infinite series is now said to be divergent.
Examples of divergent series are + 2 + 3 + 4 + 5H11 + 11 + .... 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
nuwher A, then as n increases but never exceeds some definite jired without limit, S n will approach a limit n which The S, Si, Sa, etc. approach the point than or equal to A.
ILLUSTRATIVE EXAMPLE. Show that and u not (jreater than A. points determined by where w, is less the infinite series + 1+ rL2 + rnh) + 1 a)
is The figure illustrates the statement. the values is  + +  [i convergent.
Solution.
(2) Neglect the Sn = 1 Consider the variable
(3) first +
sn *n term, and write 2 + JT^l defined = + ' * * +
1 23n" by l+7,+~ + '+~ 2.
in which we have replaced all integers in the denominators of (2), except 1, by
= I and
< 2
Also, in (3j we have a geometric series with r
Obviously S B <
no matter how large n may be (see Art. 1H3). Hence S w as defined by (2) is a variHence S n apable which always increases as n increases but remains less than 2.
and this limit is less than 2. Therefore the
proaches a limit as n becomes infinite,
less than 3.
infinite series (1) is convergent, and its value is
the
We shall see later that the value of (1; is the constant e = 2.71828
.<> , natural base (Art. 61). DIFFERENTIAL AND INTEGRAL CALCULUS 342 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 for now a convergent which lim
n S', t = series u. 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 1imM M = (A) That is, in 0. oo it a convergent series, the terms must approach zero as a limit. On the other hand, if the general (or
not approach zero as // becomes infinite,
series is (A)
is, not,
if tively that the
t he series is Hero convergent + 2 + + a For, consider the harmonic series . 4 + condition (A) the series We is shall is " ' + ' ~ lim ()
W' lim (//) = w' 0; V .* is, of a series does at once that the however, a suflicient condition for convergence; that
//th term does approach zero, we cannot state posi 1 that term divergent. is even //th) we know Yet we shall show fulfilled. in Art. 186 that divergent. now proceed easier to apply than the to deduce special
above theorems. tests which, as a rule, are 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.
(1) Let
Mi + Wj be a scries of posit ire terms irhich a series of positive terms already
(2) fli +fl +
it does not exceed that of (2). is + ' ' to be + :i to test for convergence. If convergent, namely,
, than the corresponding terms in
a convergent series and its value less (1) is ' desired kmnm raw be found whose terms are never
the series (1) to be tested, then U;\+ SERIES =
Sn = Proof. Let sn and + w + MS +
+ f + MI  < u> n , an fl oo ^ sn tf llf I, convergent and (1) is , A. A. Hence, by Theorem Art. 185, s n approaches value its is not greater .4. ILLUSTRATIVE EXAMPLE 1. 1 + (3) Solution. ; ++ 1 is Test the series Compare with the geometric (4) which == S n < A and sn a limit and the series than a.< lim N Then, since
follows that +
f 2 <io fli and suppose that it 343 known Hence (3). + r* +
if ? The terms to be convergent. sponding terms of series (3) also is "' never of (4) are less than the corre convergent. Following a line of reasoning similar to that applied to we may prove (1) and (2), the Test for divergence. Let
(5) + '//i W f W:i + be a series of positive terms to he tested, which are never
corresponding terms of a series of positive terms, namely,
fci (6) known ILLUSTRATIVE EXAMPLE 1 is is (5) Show 2. (7) + +&i. Then to be divergent. + i b:i4 less than the , a divergent series. that the harmonic series + + .4 1 + divergent.
Solution. Rewrite The square brackets
(8) 1 (9) 4 (7) as below and compare with the series written under it. are introduced to aid in the comparison. + ^ + [J +
+ i Kd + We observe the
corresponding terms in 11 + [' + i 1] + t [ 1 + +
+ following facts. 7 4 Jl t + 11 +
+ [ J +
+ The terms in +
+
+ A] +
f',,1 ' ' Li'e (8) ' . . are never less than the (9). But (9) is divergent. For the sum of the terms in each square bracket
and S n will increase indefinitely as n becomes infinite.
Hence (8) is divergent.
ILLUSTRATIVE EXAMPLE 3. 1+ Test the series
i \/2 +L + L + ...
V3 V4 is 4, DIFFERENTIAL AND INTEGRAL CALCULUS 344 Solution. This series is divergent, since its terms are greater than the corresponding terms of the harmonic series (7) which is divergent. The "p series/' 1 (10;
is useful in applying the Theorem. The p comparison series is lest. convergent when p > 1, and divergent for other values of p. Rewrite (10) as below and compare with the series written
are used to aid in the comparison. Proof. under The square brackets it. (11) M (12) 1 p > If terms
1 . ! + 2" the terms 1, in (11 i J^ 2" 1_ _
~ 2. . ; are never less than the corresponding
sums within the brackets are the .]_ 2 " in (12) in (12), Hut, ). ' ;) /> 2'' ' Hence, to test (12) for convergence, we and so on. may consider geometric series with the common the series <" '+^+(rO'+(F When p> 1, series (13) is a ratio less than unity, and is When p
When p ^ 1, also convergent. therefore convergent. Therefore (10) is
1, series (10) is the harmonic series and the terms of series (10) will, after the
be greater than the corresponding terms of the harmonic series
hence (10) is now divergent.
Q.E.D.
is divergent,. first, ; ILLUSTRATIVE KXAMPLK
(14)
is 234 f 4. Show that the series __
3454
4 2n , 56 (n convergent.
Solution. that u n < ; u n is less than the general term of the /) series when p =
each of whose terms is half the corresponding term in (14) is, series or In (14), .], and therefore (14) also is convergent. 2. Hence the is convergent 345 SERIES
PROBLEMS
Test each of the following series. 126 187. Cauchy's testratio a test. + ar + ar~ + In the infinite + ar" + or 1' M + geometric series
, is the comar" and ar"
the ratio of the consecutive general terms
we know that the series is convergent when
mon ratio r. Moreover,
f r
i \ < which 1 and divergent may for other values. be applied to any series. We now ' explain a ratio test DIFFERENTIAL AND INTEGRAL CALCULUS 346 Theorem. Let U\+ (]) be an terms + Wli + UA +U n + Mn + + ' ' ' and n ?/ n ^^rf /orrw the test ratio. i > i = ^2LLi Test ratio
//w?'/ /"/'A/// JAtf lim ' H7/cw p
p > 1, ///r III. H/7/r// p = 1, ///r test M Proof. H7/(v/ I. 4 1 W'n cr the scries is convergent. 1, //r/? II. ;  p scries is divergent. foils. definition of a limit (Art. 14) My the 1. ~~ can choose u so lar^e, say // m, (hat when n ^m from p by as little as we please, and therefore be
proper fraction r. Hence
' f ?/,,,r i  //, ; '" , //, , \r < n ,,,r'2 ; wm <, ;? less w,,,r* each term of the
and so on. Therefore, after the term
is less than the corresponding term of the geometric series
?/ u (2) J^ut since r
is ^ 1 as in I, III. that is, The
and the series When p = r (1
1 t , ratio + and therefore also the series (1),  ^ = (^rY=
\H. + 17 nn l'^ =
WH (2), //,,,r' ; test,
st, * \ + series (1) For, consider the p series, namely, test fails. lim y/,,,r ; oo).
Following the same line of reasoning
be shown to be divergent.
may
he series may be eit her convergent or divergent the n + than a 1S(>). (or p
) is n the series 1, convergent (Art.
II. \\'hen p ^ , we ^^ will the ratio diller ?/, Let this be infinite.  W<n l n . when n becomes tes/ ratio /r/,s o/ p
I. ' Consider consecutive general infinite scries of positive terms.
?/ ' ' 1 7 lim (l
. x
\ w (l
\ ~W + 1 ^)"
n + I/
(1)"  1 ; (= p). 7 Hence p = 1, no matter what value p may have. But
we showed that
when p > 1, the series converges, and when p ^ 1, the series diverges. in Art. 186 SERIES 347 Thus it appears that p can equal unity both for convergent and for
There are other tests to apply in cases like this,
our book does not admit of their consideration. series. divergent
but the scope of For convergence
unity for all not enough that the test ratio is less than
This test requires that the limit of the test is it values of n. 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
the value but not the existence of the limit. will affect 188. Alternating series. This is the name given to a series whose
terms are alternately positive and negative. Theorem. //
is ML MI an alternating > and it, if
j n then the series is (1) Sn (2) S tt = : M < } ~f m ___ u is less numerically than the. ^ > * convergent. When Proof. M which each term series in one which precedes + n (MI is  even, S n
M L.)
(M L MI > + ~ (M.I may be  written in the two forms MI) f M.O (n f (M, t 2  n , M,, i ?/),
M, ) t . Each expression in parentheses is positive. Hence when n increases through even values, d shows that S increases and (2) shows
that S H is always less than ?/j
therefore, by Theorem I, Art. 185,
) tl ; also approaches this limit /,
S n approaches a limit /. But S' n
~ X n + '"*} and lirn u n n = 0. Hence when n increases
SH
/ and the series is convergent.
through all integral values, n
t since + j i \ <S' > ILLUSTRATIVE EXAMPLE. Test the alternating
Solution. Hfcnce An
The Each term lim (w n ) is less in 0, scries 1  f i  } + numerical value than tl preceding one, and un
the and the ~  series is convergent. important consequence of the above proof
error i made by breaking a convergent is the statement 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
value of the series differs from this by less than one eleventh. is 0.646, and the In the above statement it is assumed, however, that the series has
been carried far enough so that the terms are decreasing numerically. DIFFERENTIAL AND INTEGRAL CALCULUS 348 A 189. Absolute convergence. series is said to be absolutely or un formed from it by making
Other convergent series are said
convergent. convergent when the conditionally series terms positive is
be conditionally convergent.
to all its For example, the
is series 1 + 0^ 9^ + FI ~~ 77 ' ' absolutely convergent, since the series (3), Art. 186,
series ' is convergent. The alternating 1,1 ! 1,1 ____ 19+34+5 is harmonic conditionally convergent, since the A series with Ike series The some proof of this and some negative terms positive deduced from it series is divergent. by waking theorem all the is signs positive convergent if
is convergent. omitted. is Summary. Assuming that the testratio test of Art. 187 holds
without placing any restriction on the signs of the terms, we may sum190. marize our results in the following
General directions for testing the series
U\ 4" U2 When H3 + an alternating
and 'if it ical value, + H4 series is + ' l/n + Un + + ' l ' '. whose terms never increase in numer lim w w
Tl then, the series is + = 0, '00 convergent. In any series in which the above conditions are not satisfied, we determine the form of ?/ n and u n i, Jorm the test ratio, and calculate
1 = lim I. II. When
When
When   p
p <
>  i 1, the series is
1, p. absolutely convergent. the series is divergent. p = 1, the test faifo, and we compare the series with
some series which we know to be convergent, as
III. \ a + ar + ar~ + ar* + 1 + ~~ + O + ^ +
~~ compare series with the given divergent, as
1 + 1+ l J 0" (P < > 1) (geometric series) (P series) 1) ~r *j or ; some + l + l+... + r~ + + series which is known (harmonic ; (p<l) to be series) (P series) SERIES
ILLUSTRATIVE EXAMPLE 1. Test the series Mn = Here Solution.  un 1 series = = 1 = lim noc K 0, convergent. ILLUSTRATIVE EXAMPLE Test the series 2. Ji + Ji + .yL + Here Solution. 10" T7 P= nlim J
, * oo and the +i=r n [n p and the n , Mn+lsJ" is 349 U = + n 1
 ~io~ "' series is divergent. ILLUSTRATIVE EXAMPLE Test the series 3.
1 j = Here Solution. 7i ni i _
~ wn _* n (2 (2 r? (2ri f lim
77 , 1)2  4 w oc w 1 ) 2 , M ~' * n lK2nf
4 ,. p ... 1 1
} =
J2"n 4 _ "" n2 2 n h 6 w 4  2 + i,( 2 n +
2n = n2 f 6 n f 2* 1 1, 2 by the rule in Art. 18. Hence the testratio test fails.
But if we compare the given series with the p series, when p we that S ee terms it of this 2)* ^ 4 2) r? must be convergent, since its terms are
series, which was proved convergent. less = 2, namely, than the corresponding ;; PROBLEMS
Test each of the following series. ..... 2
*' 3 ,_l 2 22 +A+
3
2 .H
4
2 . Ans. Convergent. Convergent. DIFFERENTIAL AND INTEGRAL CALCULUS 350
3. 10. 3
 1 2 f f + 2 212
pf> 3
~~ + 34 ;* 3
~~ 1^13
7710 4 2
f ~^ + :i + 12. . \t series. ^ 4
 h ^ f A + series .7~ir~i7 .
~ f h f ~ f ;j ;> ;; Power i 4 rb1 13. 191. An,s. . 5 ! Divergent. . ;V f jj.^.j^.jo + whose terms are monomials in ascendj, of the form ing positive integral powers of a variable, say
(1 flu ) where the
power + fli coefficients uo, Such scries in x. +u J >.r
L 2 +a r + :{ , ure independent of LS i, ;i series are "of prime importance in a*, is called a the study of calculus. A power series in j may converge for all values of .r, or for no value or it may converge for some values of x different from
except x =
and be divergent for other values.
; We shall examine (1) only for the case such that Mm
where L is when the (^A = L,
To a definite number. see the reason for this, test ratio (Art. 187) for (\], omitting the first term. a, i<n Hence for any t xn an fixed value of x, = hm
i p A*
I* * i
I coefficients are . . \ j) =xhm
,. /dni L
JL 1
\ ) =xL. form the Then we have SERIES We If I. p have two cases = 0, L : the series (P = 0.
II. If merically and L is less 351 will converge for all values of j, since not zero, the series will converge when rL(= p)
than 1, that is, when x lies in the interval is nu diverge for values of .r outside this interval.
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
will The end Art. 187. ILLUSTRATIVE EXAMPLE Solution. The test ratio here Hn .1
_L_L = by Art. 18. Hence p
and diverges when
Now examine the end
1 is ft'' Un than Find the interval of convergence for the series 1. (/< ~ A jc. + lr n'2 i i lim Also,
n . oc (n =
2 i 1, })' and the series converges when JT is numerically
numerically greater than
~
in (2), we get
points. Substituting jr
j*. .r is 1 less . 1  1 r^ + ;  ^+ , is an alternating series that converges.
we get
1 in (L!
Substituting x = which i, convergent by comparison with the p series (p > 1).
above example has [
1, 1] as the interval of convergence. This
1  x r 1, or indicated
be written
graphically as follows: which is The may series in the ILLUSTRATIVE EXAMPLE Solution. Omitting the 2. first Determine the interval term, the test ratio of convergence for the series is 1 Also lim
(2w + l)(2n + r2) = 0. Hence the series converges for all values of x. DIFFERENTIAL AND INTEGRAL CALCULUS 352 PROBLEMS
For what values of the variable are the following
BCries convergent? 1. 1 + x 3. x f x4 4. z + V
V2 5. + + z2 + x 16 ~= f V3 lhx+~fr
02 7. 04
r~ r~+ . ^ 1  _i_
f 1_ ~r 2 < 1. Am.  1< . x < 1. <
~ x < 1. A?i. . H 10. x 3 2 x2 l_:r +  1 + . 5 z
~~
5 3
T 3 5 i 41 O x Q 1 4_ r2 4 1 :; :; 6 4  End . I "^ ~r (>  *. < <z< 1 03 l3x , i  1 1
~
i . . . _ ' + 4  5 1. Ans. 4 . >4
T*rf r4 /I . ITJ . r
^4 O Wi H r .T
. L "^ I ^
2 .... + _L
2 _
J 'J z2
2 ' + ** 2 * 02
^ . 0. *
' ~r 3 x 3 f 4 x i!^ + O L
o Anx. All valuesof 0. J
1
i 4 * O
Jlfcrf I 12. + ^  Ans. All values of . , 24 r2 ^.
x i 1
J>1 ' f 1 ^r> s. 9.  Arcs. All valuesof x. . rrrH 0+ ,~ X
8v x 1 UL l~_ 6. 1 +  < Ann. . + x9 4 + x3 Graphical representations
of intervals of convergence* 9
 < x 1. gjr gl.
<^
\ r .* <
 . All values. l35z i . . .
' 2 4 24 2 4  6 2 points that are not included in the interval of convergence have circles about them. < drawn SERIES
ld l+ l + f^ + f^ +  + 16. 1 ^
92 ^
32 >. + Lfi + 2^3^ + x + ' 18. 22 4 J 2 f 9 * 19 f 2 * x3 AM. agjr<a. (a>0) r3 i^ All values.  4 Li 33 . + 03 2 6< H  3 J3
22 3 100 j
 4 x* 3 < 4 5 4 2< 1000 j*
 /iv .. 1 +
i 1) , w w 1 ( ) 9 (w 2 .r ~ o 2 m is a constant.
m is a positive integer, +
, J \}(m
9 3 in(m
^
l n (?w ) ;  + 1 ) w 13 is 2) ^ x2 ^+^ + ^' + r This important series series. + mx + w(?w
^
, 23. ; The binomial (1) 6. .  192. < , 2 2 10 i x 34 . + 16 x4 f , 2 1 42 r2 l 3 17 '" 353 ^...H +
, where is a finite series of m f 1 terms, since
(1
m
in the
that containing x have the factor
all terms following
and vanish. In this case (1) is the result obtained by
numerator If +/ raising !
series is Now an ) m wth power. to the test (1) for convergence. 1
, m(m
^^ ~ is not a positive integer, the  23 l)(m We have
(m  n + 2) 2)   , .
' (w1)
n (m + 2)(m ri + 1) n J Hence ^ = mn + un Then, since lim (^
tl
oo \
n
. and the m infinite series.  l)(m2)
and If m 1 n x = tml
n V  1)= 1, we see that p convergent if x is numerically
vergent when x is numerically greater than 1.
series is = x, / less than 1, and di DIFFERENTIAL AND INTEGRAL CALCULUS 354 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) + xr = 1 + mx (1 m is a positive integer, the series is finite
If
of the lefthand member for all values of x. Equation and equals the value We may also expresses the special binomial theorem. (2) write
(3) (a Thus the + + x) w r a "(l member lefthand power
Examples = m b) if , x a
also be expressed as a may of (3) =  series. approximate computation by the binomial of series are given below. v (;w ILLUSTRATIVE EXAMPLE. Kind
Solution. The = \ GIJO Now write out (2) with
(1 f (4) The =
1 1 .\ f )i \ (52~5~T5 ?w 1 4 ~ J r> 1 The J. ?, .r  1 f ,V + I ;)* is r' ~ O.OOOOOS T ? ft x4 + % f ().0000()00:r2 H . 0.0002 4 0.0000008 =25.099S01 (to the nearest Ans. an alternating series in (4) is 25( series. Hence we write 625. Hence 0.008.  is ( .r' f ().()04 ^2 = result jj, = 25 f O.i
25(1 f
the sixth decimal place).
T figure in = f ar)i In this example, j
(1 approximately, using the binomial perfect square nearest to G30 series, and the error the answer in is less than 0.0000008.
193. of the
(1) Another type
form
60 + MJT  of a) power +6 2 (.r series.  We shall frequently use series a) 2 Hh b,,(x  n a) H which a and the coefficients &o, 6 lf
6n
are constants,
,
Such a series is called a power series in (x
a).
Let us apply the testratio test to (1), as in Art. 191. Then, if in   , lim ^i = M,  355 SERIES
we shall have, for fixed value of j, any = We have two cases
If .17 = II. If M is I. ^^ = lim a)M. (x : is convergent for all values of x.
not zero, series (1) will converge for the interval 0, series (1) A convergent power series in .r is adapt ed for computation when x
near zero. Series (1), if convergent, is useful when x is near the
fixed value a, given in advance.
is ILLUSTRATIVE EXAMPLE. Test the  1
i /,
tor i Mf (  infinite series  r 1>" (*~ 1"'
 , , . . . \ ^ for convergence. Solution. Neglecting the first term, we have n
 lim Also, w  ( oc \ N f )=
1 1 1 1. / and the seri(\s
Hence p = \x The end point jr
may he included.
1 + , will converge when x lies between and 2. '2 PROBLEMS
1. Using the binomial + 1 series, show that
' jr Verify the answer by direct division. Using the binomial
lowing numbers.
2. V98. series, 5. Yii5. find approximately the values of the fol 8. rp 11. V990
^ 4 4.^/630. 1 7"
V412 1 V30 x
\16 DIFFERENTIAL AND INTEGRAL CALCULUS 356 For what values
/~ of the variable are the following series 23
1 t + (T 15. (jr 1) \ I) 2 /.. i i \a /,. 3
j (xl) fr_l) { V3 V4 2(2. 17. 1^(^2)^ _ 1
jr 19.
i 2(2jr U) 43(2
 3)'2
 3
(.r
,  a x < 0. < 2. 2 All values. 16. 18. < ^T Ans. . 4 2 X/2 convergent ? 1 ^4 i ''./.. ; 2  3 i +
jr  ^
2 H)
' ..'. ~ '
i 3 3 :t 4 (2 ^ (x  4  3) 34 3
4 { ) H . CHAPTER XX
EXPANSION OF FUNCTIONS
In this chapter the question of representa power series, or, otherwise expressed, developing
ing a function by
(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
194. Maclaurin's series. /(*") (1) = 00 + J J + a,,/" + + a *r +
~ by a power then, a function is represented
the form of the coefficients a fli,
If, . what must be series, To answer etc.? (in, , ,  this ques tion we proceed thus
in (1). Then we must have
Set x =
: /(()) (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
( (3) J f(jr) /"(j) [f'"(x) =
=
= +2a r+3a
2 a + 6 flsr H
a, 2 2
; <:r h h n(n 2 6 a.H H l)(n h n(n na n x n
1 )a r,j 2)a n x ~ l H n~ 2 H n *
\ etc. = 0, the results are
= 12 02,
/'(O) = ai, /"(O) Letting x
(4) Solving
(4) /(x) (4) for a\ t a, , / On, etc.,  /(O) + /'(o) i + / (o) = [3 a 3 ;// (0) , , and substituting in  +  + / (w) (o) g / (n) (0)
(1 ), f  = lwo n . we obtain
 . This formula expresses /(x) as a power series. We say, "the function
This is
is developed (or expanded) in a power series in x."
Maclaurin's series (or formula).* /Or) " * Named after Colin Maclaurin (16981746), and first published in his
Treatise of
Fluxions" (Edinburgh, 1742). The series is really due to Stirling (16921770). 357 DIFFERENTIAL AND INTEGRAL CALCULUS 358
It is now necessary to examine (4) critically. For this purpose
and rewrite it, letting a = 0, 6 = x. The result is refer to (G), Art. 124, (5) f(x) + =/(0; # ^/' where IT r "(^i) < (0 r jj < jj The term R is called the remainder after n terms. The righthand
member of (5; agrees with Maclaurin's series (A) up to n terms. If
we denote this sum by N/,, then (5) is
Sn + or K, assume that, for a fixed value x ~ xn, R approaches zero as a
limit when n. becomes infinite. Then S 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 Now ) tl : Theorem. /// the function f(x) order that the series (A) should converge arid represent
it is necessary and sufficient that
lim (G) n = 0. 7? oc It is usually easier to determine the interval of convergence (as
the preceding chapter) than that for which ((>) holds. Hut in
simple cases the two are identical.
in To represent a function /(.n by the power series (A), it is obviously
and its derivatives of all orders should be 'necessary that the function
finite. This is, however, not sufficient. Examples of functions that cannot be represented by a Maclaurin's
series are
In x and become infinite when
The student should not fail since both expansion as (A).
certain
in number In .r is ctn x, zero. to note the importance of such an practical computations results correct to a
of decimal places are sought, and since the process
all question replaces a function perhaps difficult to calculate by an ordinary polynomial with constant coefficients, it is very useful in simplifying 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
Stopping at any term is numerically less than that term. made by EXPANSION OF FUNCTIONS
ILLUSTRATIVE EXAMPLE 1. termine for what values of J
Solution. Differentiating it Expand cos x into an infinite and then placing x  first /(O) = series and de get
1, ~0, f'\0) =  /'"(O)
/1V(( /v(0) jr, / vi zr: (0) = I, 0, rr \ f = 0,
= 1, etc. in (4), C08 (7)  T=lf^ Comparing with Problem
values of J*.
In the same we 0, /'(( etc., Substituting power converges. = cos x.
/(jrt
 sinx,
f'(x) =
=  cos x,
f"(x]
/'"(j) = sin J,
= COST,
/iv(j)
x = _ s in
fv(
_ COS J,
/Vl(j)
) 359 way G, Art. 191, we + "
jj sec that the series converges for all for sin j, sin, (8) +   ,?! ?1 . t 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
Here we
as a limit as n becomes infinite, when jc has aw// fixed value. Consider (7).
the nth derivative in the form
may write
. ,,
A = Hence Now
of R is which cos(ji 4 ) never exceeds x cos 1 in numerical value. Also, the second factor the nth term of the series is convergent for all values of infinite (see (A), Art. 1S5). From Hence the above example x. Therefore it approaches zero as n become:! (6) holds. we see that rn
Jim .* [W = 0. ,i Also, on the page preceding, we had K= fn)( . (0< X, < X) \n Hence lim /? = if f (n] (x\) remains finite when n becomes infinite. DIFFERENTIAL AND INTEGRAL CALCULUS 360 ILLUSTRATIVE EXAMPLE
culate sin 1 Using the series 2. (8) found Here x Solution. radian \ Therefore, substituting x = that ; the angle is expressed in circular measure.
the last example, is, in (8) of 1 _ _
= 1.00000 ~ 0.008,'W 1 y?  J , = 0.16667  0.00020 y; 1.008U3 Hence
is the teat example, cal the positive and negative terms separately, Summing up which in correct to four decimal places. sin  0.16687 1 1 .008W  0.16687 = 0.84146 correct to five decimal places, since the error made must be less than ; I? than 0.000003. Obviously the value of sin 1 may be calculated to any
is,
desired degree of accuracy by simply including a sufficient number of additional
that less 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 !' + ...+ 11 r 2. sin 4. In , = U  r'l j) x arc sin jr = 6. arc tan j =r 8. In (a J)  .r' TT  _
+ 1_1J .r 1 ( ~;t ,r j \ " Ans. All values. .... r :L  + . . A11 values ..   . ^j 1 ^ <
JT ^ ' :$ 5. + r > ,+  x = In a  f f a  ^ (n 2 a  + 4 ^3 aa A n,s. All values. _ 1. EXPANSION OF FUNCTIONS 361 Verify the following expansions. 12. tan ( + JT = 4 2 I f jc + 2 x + \4 13. arc tan j* =^2 x + 3 4 or
16.1ncos J . 5 LS =^^g.., Find three terms of the expansion
lowing functions.
17. cos 18. sin j* (.r Compute in powers ~ + of JT of each of the 19 e 20. 1). J(r' c the values of the following functions by substituting dienough to make the rectly in the equivalent power series, taking terms
results agree with those given below.
21. c = 2.7182 Solution. Let x . 1 in the series of Problem 1 ; then (Dividing
(Dividing
(Dividing
(Dividing
(Dividing 22. arc tan 23. cos 1 fol = (I) 0.1973 0.5403 ;  ; third term 3.) term by 5.)
term by 6.)
seventh term by 7.)
fifth sixth use series in Problem use series in by fourth term by 4.) 6. (7), Illustrative Example 1. DIFFERENTIAL AND INTEGRAL CALCULUS 362 = 24. cos 10 ; 26. arc sin 1.5708 1 ~ use series in Problem ; ; 0.7071 use series in Problem ; 1. 1>. use series in Problem  Example (7), Illustrative use series in  = 0.0998 25. sin 0.1 27 sin  0.9848 5. 2. 4 ~ 28. sin O.T> 29. = <'' 1 + 0.4794 f li use series in Problem ; + 4    2. 7.3890. ^' V? = 30. + 4 1  . f f l.b'487. oJTTj 7^7, 195. Operations with infinite series. One can carry out many of the
with convergent series just as
operations of algebra and the calculus one can with polynomials. The following statements are given without proof.,et + fl,/4 f/o and <*#' 4 b^jr 4 fro b\jr f be convergent power (a< ( i 4 bu) 2. #// multiplication, (1 In we obt ain, by (1 ln )*" [ 4 4 ( an b J'f   1 r ,'"  '  J * terms. 1 } 4 n n )jr From Computation of logarithm*. 4 oO = J J*)   J" subt raet ion of (1) 1. s" n hi/ term.. and (jroupnnj ILLUSTRATIVE K\ AMPLE
lems 3 and 4, Art.. 194)
In i ^1 (MI b ' + : term,
fry adding (or subtracting) 1. 4 * <i>s" Then we obtain new convergent power series. from them as follows series + + a i
 ^' *'
 r  \ * r the series (Prob ' '  ' ' Art
corresponding terms, and using (2) . , L, U M!,. 2lJr+ .,. + 1 the new series A be a positive , , + This series converges when j* ^ 1.
To transform (1) into a form better adapted to computation,
number. Then, if we set
' ', x /2) then j
 (3)  < 1 = 1  for all values of .V. In (A' + 1) = In x .\ + o
~ 1 , whence : Substituting
1 iTy"^ + + JT __ A* f lot 1 r:
; in (l\ we get the formula 1__1__41_
3 (2 A' + 1 )' i_ 5 (2 :V 4 1> 5 4 1
J EXPANSION OF FUNCTIONS
T This series converges for all positive values
= t. Then
putation. For example, let A" + (A In = 1) Substituting in (3), the result
Placing A* = 2 in (3\ we get In 2, is , well adapted to com =  r = In 2 is A and of 363  0.69315. 1,1  1.09861 1 4 in this way,
It is only necessary to compute the logarithms of prime numbers
the logarithms of composite numbers being then found by using formulas (2),
= 3 In 2 = 2.07944
Art. 1. Thus,
l n 8 = In 2
' In 6 = In 3 +  In 2 1.791 76 . the above are Xapicrian, or natural, logarithms, that is, the bust* is
If we wish to find /ir/f/f/s's, or common, logarithms, where the base
2.71828
10 is employed, all we need to do is to change the base by means of the formula
All r . 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
and various ingenious devices designed for the purpose
of
in the
of logarithms
theory
saving work. ILLUSTRATIVE
Solution. FA AMPLE From = x <*= and
obtain, by By A division.  * x f ^  = x~fx 2 Problem From the division of 1 (4) in = 1 ~ X"' n is in x' r2 __ r 2 " sec x (6) < 1 94 ' AH * 94 ' etc. (Problem 1, = An. the example below. in X* + X ~ + ' ' ^ jj , we we have see that 24 \ __ ... 7 __ i _ 720 Then ! 5 by the series (4). This is best done as follows.
the form cos x = 1  z, where ,r<. 2 1 fSi terms shown = the formula sec x lo; ! Art. Find the series for sec x from the series for cos x 3. cos x (4) if 2, ?robl " m + f^~f (see (7), Art. 194). Write sin x. ^ +* + ? + ? + special case ILLUSTRATIVE EXAMPLE Solution. T multiplication,
c* sin 3. C the series
sin x we Find the power series for 2. ^ Art. 193). = 1 +2+ 2 z' + 23 + ' ; to carry through DIFFERENTIAL AND INTEGRAL CALCULUS 364 From we have (6), the series
T* z x ~ Substituting in r> the result f higher degree, is = + (6), sec x terrns "*"" 24 4"  x f 1 x4 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 = 2. In 7 1.94591. Work 1.60944. out the following
' 5. c ' cos = 1 /  13 2.56495. series. + ; _ /'< { ii . /4 _j_ . , f X ( =1 + 2^4 6. 7 / 2.39790 3. In 11 4. In 11 JL55JL 10. r' tan 1 1. f r = .r _ +? ^ + 55^ 1 + ~ j* 4 j  / .r 2 J_ .2 __ j + JT j2 f  .r 3 4 f jilL . . 4 .  12. c 13. 2 sec sin 2 (1+ 16. /  i> / x) cos V.7 14. (1 f 2 15. 1 j Vl Vl  j*) JT tan x  =1+ arc sin arc tan 2 2 jc j =x .r< ~ i /.'*  a f 5 f 11 2 I 2 r'2 1 ^4 /4 + j.2 + 24 f = j "  f 6 _j_ . . . B> ^i. ^720
j a ^  11 ^
^4 f % .  ^_ _11_
8064 4 .r
" f 3
f A^+ . 48 = lxx ~^^L.r^^
X
^
^
2 . . EXPANSION OF FUNCTIONS
Vsecx = 17. + 1 2
7J
4 4 ~ 1+smx 2 . V4 + 20. sin = </> than j* 6144 2;>b
' 2 For the following functions
of x less 12 b 16 r   . pow ers + 4 J* 96 find all terms . 22.<*cosjVI
sin 25.^jy^,
VI 4 x x
196. Differentiation 26. and integration of V5  power cos x. series. A convergent series oo (1) + aiJ + a 2 x2 + aa.r +a :{ f n xn + be differentiated term by term interval of convergence, for any value
and the resulting series is For example, from the we . JT 23. may J 24. \/\\ f c . smx. r> which involve of the aeries 5 * 21. c power 365 obtain, Both by series of x within the also convergent. series new differentiation, the converge for all series values of x (see Problems 6 and 7, Art. 191).
lie Again, the series (1) may be integrated term by term if the limits
within the interval of convergence, and the resulting series will converge.
ILLUSTRATIVE EXAMPLE
Solution. (2) Since r dx 1. Find the In (1 f x)  we have  1 f ln(l series for In (1 x f x) = I /() NOW r
L ~T X TT = 1 ~ X + * ~ X + ^ 2 * + x) by integration. DIFFERENTIAL AND INTEGRAL CALCULUS 366 when x<l (Art. member term by 192). we term,
In (1 = ji f This series also converges when ILLUSTRATIVE EXAMPLE  x 2 \ x' < \x\ f (see I . dx \ x  :i .T I Problem 4 f . Art. 191). 2, we have _ i X2
r f dx \ljl' .'f, By 3 Find the power series for arc sin x by integration. 2. Since  arc sin x Solution. and integrating the righthand Substituting in (2)
obtain the result the binomial series ((2j, Art. 192;, letting /// J, and replacing x by x2 , we have
1 This series converges when
term, we < \x\ 246 Substituting 1. (3i in ' and integrating term by get = T an sin , , , This series converges also when
By this scries, the value of
verges for values of / between or TT =  \x\ (see I Problem 8, is and Art. 191). For, since the series con readily computed. 1 TT .'{.1415 + 1, we may let x = , giving . Evidently we might have used the series of Problem (>, Art. 194, instead. Both
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.
of these series ILLUSTRATIVE EXAMPLE Using ,'{. series, find approximately the value of sin jrdx.
I Solution. Let r  x? sin z Hence  x2 = sin Then . z  rr  x'2  + jtt r~ + r
/ ! , sin xdx = r 1 /
a* I ' = I ^1 0.3103. % flO r 2 _ ' Problem ' 1A _ T U) \ (i r^ ( = * 'rr lii and * fp f V ) 1 4. , . , dx, approximately, I = 0.3333  0.0238 f 0.0008 Aws. 2, Art. 194 EXPANSION OF FUNCTIONS 367 PROBLEMS
1. Find the series for arc tan x by integration. 2. Find the series for In (1 JT) 3. Find the series for sec 2 4. Find the series for In cos series, find Using by integration. by differentiating the .r .r series for tan by integrating the series for tan .r. .r. approximately the values of the following integrals.
i ~~~"' 4"s' 0.0295. 1 0.185. f Vln 7. 0.0(528. (1 f jr)djr. 10. 11. f ''dx. c Jo fin U 4 fVsin V7'</.r. V7 ) (/.r. A) /o
i ( f 8. Jo 13. ',\/i _ 0.1815. 12. Jo j.'J fVl  .H J 14. ( 'r <Lr. Jo cos Vr 15. / f/.r. \/2  sirTx dr. A) ./o 197. Approximate formulas derived from Maclaurin's series. By
using a few terms of the power series by which a function is represented, 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 (1 + x} _
In these \x\ is m 1 = Second dj>proriniatioH <i}>j)}(>j'ini(ition + \ m(w + 2 'm(m m Let us examine the first. 1 mx = 1 small and + mx + mx 1 is mx positive. Again, consider the sine series X > r X'* Then
(2) sin x (3) sin ' x, ''___
x ~ x etc. are approximate formulas. 1 )x l)x ; 2
. DIFFERENTIAL AND INTEGRAL CALCULUS 368 In the series in (I), assume values of x such that the terms deIf the first term only is retained, the value of the remaining crease. series is numerically less sin x than = its first with x,  term error < j :{ inquire, For what range of values of x
decimal places? Then
" that We LH is, < < V0.003 < then conclude that formula for values of x between
values between  8.2 and hold to three will (2) 0.0005, 0.1443 rad. (2) is 0.144.'* is, x*\.  We may  That (Art. 188). correct to three decimal places and +0.1443, or, in degrees, for + 8.2.
PROBLEMS 1. (a)jr How accurate is
= 30? (b).r60
Am. (a) the approximate formula sin x (c).r~90?
Error < 0.00033; (b) x error < 0.01 (c) ; when 6 ? < error 0.08. ~ = How accurate is the approximate formula cos
1
when
2
 30? (h) = (JO ? (c) = 90?
Am. (a) Error < 0.0032; (}>) error < 0.05; (c) error < 0.25.
accurate is the approximate formula e~ f = 1
3. How
when
= 0.1? (b) = 0.,r>?
(a)
~ when
4. How accurate is the approximate formula arc tan x = x
3
= 0.5? (c)  ?
(a)jr=0.1? (h)j^
Am. (a) Error < 0.000002 (b) error < 0.006; (c) error < 0.2.
2. (a) jr JT jc .r jc .r j jc I ; 5. How many taken to give sin terms of the series sin .r
.r
45 correct to five decimals? 6. How many terms of the series cos j = 1
taken to give cos 60 correct to five decimals? j^ ~ must be  f LL L2 jc~ x4 irr ' 7. How many terms of the series In (1 f x} = x
be taken to give log 1.2 correct to five decimals? + .477.8. '** r Four. mus t be I ~ ^
~  f must Ans. Six. Verify the following approximate formulas: 10. r~" cos = 1  6 cosVj(/j'=r+ 11.
/ + ~o 14. fare sin xdx = C f x~+4 ^2 15. (>sin0(/0= C */ f 2 ' f ~
24 23
f EXPANSION OF FUNCTIONS 3(59 A convergent power series in x is well adapted
the value of the function which it represents for small
for calculating
values of x (near zero). We now derive an expansion in powers of
a (see Art. 193), where o is a fixed number. The series thus
x
obtained is adapted for calculation of the function represented by it
198. Taylor's series. 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 + , series represents the function. The necessary form of
the coefficients & 61, etc. is obtained as in Art. 194. That is, we differentiate (1) with respect to r, assuming that this is possible, and and that the >, ( Thus we have continue the process.
J'(x) /"(*) = bi+2
=
2  6 L>U
6,> + a] H +  n(n +  nb n (jr
1  a) )Mr  n" ' a)"~ +
+ , , etc. Substituting x
bo, bi, bi>, 6 , we a in these equations Replacing these values
(B) /(x) The in in (1  /(a) + /'(a) ), the result (2) /(or) and solving ^  + /"(a) for + called Taylor's series (or formula).*
critically. Referring to (G), Art. 124,
x, the result may be written thus : series = ), is is Let us now examine (B)
letting b (1 k 6,=/'(d). =/(o), and obtain and =/(o) where The term R is called the remainder after n terms.
the series in the righthand member of (2) agrees with
Taylor's series (B) up to n terms. Denote the sum of these terms Now by Sn . Then, from (2), f(x)
* = we have
Sn + R, or Published by Dr. Brook Taylor (16851731)
(London, 1715J. in his " Mpthoclus Incrementorum " DIFFERENTIAL AND INTEGRAL CALCULUS 370 Now assume that, for a fixed value x = XQ, the remainder
proaches zero as a limit when n becomes infinite. Then ap Sn =/0o), lim
 w (3) R n and (B) converges x for xo and value its is /(.TO). Theorem. 77^ infinite series (B) represents the function for those
and those only, for which the remainder approaches zero as
tlie number of terms increases without limit.
values of j, 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 used for finding he value of the function for values of x near
and (B) is also called the expansion off(x) in the neighborhood of x
(B) is t iLLrsTKATIVK EXAMPLE 1. Solution. f(jr Expand = In ) In x f'(x] /( =, /'V) = Tliis  ) /"(1) ^. /"'(I)  = of (x 1 a. ). 0, l, = 1,
= 2, ec.
etc. ec.,
etc.,
In x powers
1 /'(D x /"V)= J;.
Substituting in (B), in jr, a, .r  1  J(r  1 ) 2 + ^(x  1)''  . Ans. and 2 and is the expansion of In x in
converges for values of x between
See Illustrative Example, Art. 198.
x
1. the ririnity of ILLUSTKATIX E
Solution. EXAMPLE Here f(x) 2. cos x Expand
and a cos .r ~ in powers of x
/ Then we have 4
f(x) = cos .r, M~ =
 \4/ =
\ 2 . etc., etc. j to four terms.
j EXPANSION OF FUNCTIONS
The series is, therefore, v2
The result To check
or x _ 1L may N 2 be written in the form this result let us calculate cos 50  0.08727, (x  Another form by x and let () 0.00762, (x / .r = Jo + is expanded thut Here f(.r> range the work as below. /'(jri sin Expand r cos /(J / i = when .r 19S, + .r u //, 4 sin JT O cos = "(J<>) when .r x 4 cos x ~ sin j ( , cos j , , 4 ( *. Am. PROBLEMS
Verify the following series hy Taylor's formula.
1. 2. sin x = sin 3. cos = a a 4 (x ^ a) cos a r; ' sin a q r
i cos a 4 " cos a (x a; sin a ' ~ cos a 4 to changes we obtain
sin j\> is Differentiate and ar h). X re .r. , .r () sin we changes from sin >) if the result the increment of (j,, 0.64279. etc. etc., sin (xo f h = series in h ~
( (), Art.
f // power f'(xu) .r, sin x, /"(ji //, in a (} sin j, = /(.r) in and f(x 4/0 " sin x, f(x) Substituting in (C), In =a .r is, a power series ILLUSTRATIVE EXAMPLE.
from Jo to /o 4 h. With these values (UHMMJG. / of Taylor's series.  a = h, in Solution. 4 Fiveplace tables give cos 50 0. (54278. In this second form the new value of
h  ~Y~ \ expressed in radians, 5 4 ., \ 4 \ the series ahove gives cos 50 199. Then x . > 4 place a 371 7 sin a 4 DIFFERENTIAL AND INTEGRAL CALCULUS 372
5. cos (a f x) = cos a 6. tan (x + = 7. (x 8. h) + A)" = Expand Ans. tan x ^ f nx sin sec 2 x f /* n + *h n(n x in powers of sin r x2 = f x3 + pr sin nr cos a x sin a + /^ sec x tan
~ 1}
2 2
2 x + a + . . x" A x to four terms. \ (!)' _ 12
9. Expand tan x in powers of Lr Am.
10. Expand In r in 11. Expand <" in 12. Expand sin 13. Expand ctn powers tan
of (r powers of f f jr ~ f .r f  (r .r =   ~\ to three terms. + 1 2 a  + 1>  f 2) to four terms. 1) to five terms. in powers of x to four terms. in \ j powers of x to three terms. j from Taylor's series. Such
some terms of series (B) or (C).
by using
= sin j, we have (see Problem 2, Art. 199)
/(.r) 200. Approximate formulas derived formulas are provided For example, if sin (1) as a first y ~ sin a + cos a(x approximation. A second approximation results by
This a] taking three terms of the series. is sin (2) From (1), x = f sin a + cos a(x a) sin a transposing sin a and dividing by x rinar Sina
j ^ y _ Q\2 a, we get ==COBOj a means that (approximately) Since cos a is The change in the value of the sine is proportional to the change in constant, this the angle for vahics of the angle near a. Formula (3) illustrates interpolation by proportional parts. EXPANSION OF FUNCTIONS 373 1. For example, let a = 30 = 0.5236 radian, and
suprequired to calculate the sines of 31 and 32 by the approximate formula
Then, since x  a = 1 = 0.01745 radian, ILLUSTRATIVE EXAMPLE
pose
(1). it is = sin 30
= 0.5000
= 0.5000
= sin 30 sin 31 sin 32 Similarly, These values by
curacy is (1) Then cos 30 (0.01745)
0.8660 x 0.01745
0.0151 =0.5151. + cos 30 = (0.03490) 0.5302. are correct to three decimal places only. we may use desired, +
+
+ If greater ac (2). = sin 30 + cos 30 (0.01745)  sm j*
= 0.50000 + 0.0151  0.00008 ~
= 0.51503.
= sin 30 + cos 30 M (0.03490)  sin
= 0.50000 + 0.03022  0.00030 "
= 0.52992. sin 31 ' ^0.01745)2 1 sin 32 These .j* (0.03490)' results are correct to four decimal places. From (C) we derive approximate formulas
when x changes from
to Jo f h For, /(j) for the increment of transposing the jr (} iirst term of the righthand member, we get
/(J (4) + h) /(Jo) = /'(j u +/"(:ro) j~ )// +  . Ik The lefthand series in the From member expresses the increment of /(j) as a power increment of (4) we JT (=//). derive, as a first approximation, (5) This formula was used in Art. 92. For the lefthand member
value of the differential of f(x) for x = J and AJ
h.
As a second approximation, we have
/(Jo (6) + h)  /(J  /' (J + /"(J )// ) ) is the ~
<u ILLUSTRATIVE EXAMPLE 2. Calculate the increment
when x changes from 45 to 46, by (5) and by (6). of tan x, approximately, tJ From Problem Solution. tan (x f h} In this example x () 6, Art. 199, tan x = 45, 4 sec 2 and tan x if x x h f tt () = 1 = x (] sec
2 , , 2 we have
x<, sec XG tan x h 2 H . = (} 2. = 0.01745. Hence, by (5),
1
Also,
 tan 45 = 2(0.01745) = 0.0349
tan 46
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
fe = expressed in radians ; places of decimals. DIFFERENTIAL AND INTEGRAL CALCULUS 374 PROBLEMS
1. Verify the approximate formula + In (10 *) = 2.303 + Calculate the value of the function from this formula and compare
0.5
1.
(b) x =
your result with the tables, when (a) x =
Am. (a) Formula, 2.253 tables, 2.251.
(b) Formula, 2.203
tables, 2.197.
2. Verify the approximate formula
; ; ; sin \b + x =0.5 + 0.8660 x. / Use the formula to calculate sin 27, 33, sin sin 40, and compare your results with the tables. Verify the approximate formula 3. tan K + x = ) 1 + 2 x + 2 x2 . Use the formula to calculate tan 46, tan 50, and compare your
results with the tables. Verify the approximate formula 4. cos x
cos 30 Given cos 45 and cos 60 cos a =
=
= use the formula to calculate cos
results with the tables. sin 60 sin 45 sin 30 a) sin a. (x =
=
= 0.8660,
0.7071,
0.5, 32, cos 47, cos 62, and compare your ADDITIONAL PROBLEMS
Given the definite integral 1. r\
/ x 5 In *) o Obtain
Obtain (1 + x)dx. value by series correct to four decimal places. Ans. 0.0009.
(a)
its value by direct computation and compare with the ap(b)
proximate value derived in (a).
(c) Prove that if n terms of the series are used in the computation the
error is less its 1 than 7)
2. Given f(jr) c  x
cos  (b) Show that f *\x) =  } f(x).
Expand f(x) in a Maclaurin series (c) What ( (a) is to six terms.
the coefficient of x 12 in this series? Ans. 64 [12* CHAPTER XXI
ORDINARY DIFFERENTIAL EQUATIONS *
order and degree. A differential equaan equation involving derivatives or differentials. Differential
tion
equations have been frequently employed in this book. The illustrative examples of Art. 139 afford simple examples. Thus, from the
differential equation (Illustrative Example 1)
201. Differential equations
is g2i.
we found, by integrating, y (2) Again (Illustrative Example 2 C. integration of the differential equation ~
y ?
dx  v }
(3) ) , = x~ + led to the solution x2 (4) Equations (1 ) and of the first order, (3) + y~  2 C. are examples of ordinary differential equations
(2) and (4) are, respectively, the complete and solutions. Another example is ( This is a differential equation of the second
order of the derivative. The order of a differential equation is order, so named from the 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
(6) differential equation y" 2 = (1 + 2 3
?/' ) , * A few types only of differential equations are treated in this chapter, namely, such
work in mechanics and physics.
types as the student is likely to encounter in elementary 375 DIFFERENTIAL AND INTEGRAL CALCULUS 376
1 where y and y" are, respectively, the first y with respect to x, is of and second derivatives 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
( 1 is by which the equation = a sin y ) Thus satisfied. is x a solution of the differential equation For, differentiating (1;, Now, if we a sin x. substitute from (1) and (3) in + a sin x and = 7^ (3) (2) is satisfied. Here a is a sin x = (2), we get 0, an arbitrary constant. In the same manner
(4 may y ) = b be shown to be a solution of (5) // fj sin cos x (2) for any value of b. The relation r+c>2 cos x more general solution of (2). In fact, by giving particular
c\ and
2 it is seen that the solution (5) includes the solutions (1) and (4).
The arbitrary constants c\ and c>2 appearing in (5) are called constants of integration. A solution such as (5), which contains a numis a still values to 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 obby given conditions to be satisfied tained from the complete solution
by the particular solution. ILLUSTRATIVE EXAMPLE. The complete solution of the is y differential equation = C] cos x + C'2 sin x (see (5) above).
Find a particular solution such that
y (2) * It is 2, y' = 1, when x = 0. shown in works on differential equations that the general
when the differential equation is of the nth order. trary constants solution has n arbi ORDINARY DIFFERENTIAL EQUATIONS
Solution. From the complete solution
y (3) by differentiation, 377 we =c cos x + r 2 sin x, c i sin i x + c2 obtain
y' (4) = cos x. Substituting in (3) and (4) from (12), we find c\ = 2, r 2
values in (3) gives the particular solution required, y = 2 cos r Putting these 1. v4ns. sin j. The solution of a differential equation is considered as having been
when it has been reduced to an expression involving integrals, effected whether the actual integration can be effected or not.
203. Verification of the solutions of differential equations. taking up the problem of solving differential equations,
to verify a given solution.
ILLUSTRATIVE EXAMPLE
(1 is y ) = 1. Show Before we show how that c\x cos In x f c?x sin In x x f In x a solution of the differential equation x^4^2 (2) Solution. Differentiating (1), (3) ^=
dx (r 2 Substituting from  d) we sin In x = x,nx. W get f (r^ f o we (1), (3), (4), in (2), ) cos In .r f In JT f 1, find that the equation is identically satisfied. ILLUSTRATIVE EXAMPLE 2. Show
y (5)
is 4 x = a particular solution of the differential equation
*2/' (6) Solution. 2 1 = 0. Differentiating (5), the result
yy' is 2 that 2 = 0, is whence y' Substituting this value of y' in (6) and reducing,
true by (5). 2
= we obtain 4 x  y 2 = 0, which DIFFERENTIAL AND INTEGRAL CALCULUS 378 PROBLEMS
Verify the following solutions of the corresponding differential equations. (r Solutions equations Itifferentitil jc r <7r ' 6 f + >l:Jl . f/.T 7.   9. // 4  /X
/* _ 'Ht ., j.,, = o. ry =2f 3 r '. (/J* f (/ 8 J 4 r, .S r_r + r '17
>
a/ y^ rr r, (). 1:i J " = ;5;)  *" COS (2 = r" t <* * cos  f + 3 3 ).  ( 10. .r?/ 11. ^^ 4 ^ ' <iv f r 1 12. '' 4 f ,s 1  rr * = 2 sin 2 r 2 '. // = r,r*' f 5 cos o = /. j = cos 2 3 C os 3 /. x = c, 8 t. + t cos 2 + / 2 /!. <// <!lll z^ * </.r* 13 a.r . f
14. ll __ <) j. 1^ 4. 9 j. (// 15> 2 = ll = / cos 3 ce +
/ *  J f 2 cos 3 / f c 2 sin 3 2 '. +
/ 3 sin 3 4 \ t /. sin 3 /. a/*
16. ^ f .r//
* = J 3 // 3 A/r  (/.r 2
J' 4 1 f re"', 204. Differential equations of the first order and of the first degree.
may be brought into the form Such an equation
(A) Mdx ORDINARY DIFFERENTIAL EQUATIONS
in which M and .V are functions of x and ferential equations coming under The more common y. head this may 379
dif be divided into four types. Type I. When Variables separable. equation can be so arranged that the terms of a differential takes on the form it = 0, /(J)rf.r+Fu/)rf// (1) 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 general solution 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 where ing the variables. FIRST STEP. Clear of fractions; and if the equation involves derivamultiply through by the differential of the independent variable.
SECOND STEP. Collect all the terms containing the same differential tives, into a single term. If the equation, then takes XYdx + the on, form = 0, Y'y'(/i/ where X, X' are functions ofx alone, and >', Y' are functions of y alone,
may be brought to the form (1) by dividing through by X'Y.
THIRD STEP. Integrate each part separately, as in (2). it ILLUSTRATIVE EXAMPLE 1. Solve the equation dy
dx Second Step. To (1+ 2
(1 4 x' )xydy Solution. First Step.
(1 2
\ separate the variables y' 4 1 __ )dx x*)jry =(142
4 x' x(l we now divide dx  J In (1 4 x2 )  y'*)dx. )ydy by o?(l ___ +y*
ydy x(\ 4x*) In x 2
j/' \ ~
4 0. x 2 j(l 4 y 2
), giving _ 1 In (1 4 y 2 )
2 In (1 4 x )(l 4 y*) = C,
= 2 In x  2 C. DIFFERENTIAL ANT) INTEGRAL CALCULUS 380 This result
that may be written in more compact form if we replace
2 C by
to the arbitrary constant. Our solution then becomes
In (1 + T*)(\ In (1 f r')fl f y) (1 ILLI 4 Jr)(l f r') STKATIVE EXAMPLE f, + 2 y' as dy Solution. Firat Step. 2 ay dx separate the variables = In j 4 In
 In ex'
 rj .4ns.
2 ) c, 2 , 2 . Solve the equation 2. Second Step. To In new form give a is, we + f 2 a //  jcla djc y>dy sydy. = 0. divide by ry. ~
x
Third Step. 2 y \aC^~ Cdy = f a In j
r/ f a In In j'^/ =. // r ?/ f C, C, ?/, passing from logarithms to exponentials this result By 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 Type Homogeneous II. is equations. Ans. The differential equation M dx + N dy = (A)
said to be of * = ce. and // homogeneous when 717 and .V are homogeneous functions
same degree.* Such differential equations may be of the solved by making the substitution y (3) = rx. This will give a differential equation in r and x in which the variables are separable, and hence we may follow the rule under Type I.
* ing .r A function of x and y and // some power by \jc of X. is said to bo homogeneous in the variables if the result of replacand Xj/ (X being arbitrary) reduces to the original function multiplied by
This power of X is called the degree of the original function. ORDINARY DIFFERENTIAL EQUATIONS
we In fact, from (A) 381 obtain Also, from (3), dv dy
< . l= r dr*+ 5) The righthand member
the substitution (3)
obtain from (4) of (4) will used. is v become a function of v only when
(5) and (3), we shall Hence, by using *g + t>=/(p). (6) and the variables x and v may be separated. ILLUSTRATIVE EXAMPLE. Solve the equation '+x'^ = ^^.
dx
dx
2
y dx Solution. M in Here
y N = x'
Also we have
or and y.
2 2 xy, , = Substitute y f (x  xy }dy = 0. and both are homogeneous and of the second degree
y*
ay _
dx
x2
xy t?x. The result is dx
or r To dx f x( 1 v 1 v)dr separate the variables, divide by = 0. This givea rx. , In x f In f  v In rx But %
X v T, =C Hence the complete solution f , is y y = cex . PROBLEMS
Find the complete solution of each of the following differential equa tions.
1. 2.
3. (2 + y)dx xy dx
x(x f   (1 h 3)dy   (3 ^ x)dy 2 )d?/ i/(2 x = = 0. 0. + 3)dx = 0. Ans.  x) = c. (2 f y)C3 c?/
t/ = =1+
cx(x x2 f . 3). DIFFERENTIAL AND INTEGRAL CALCULUS 382 Vl 4. 2 Vl dy 4 f' dO = p tan 5. dp 6. (1 jjd?/ (/ 2 y)(Lc 7. 4 f 8. (3 9. 2 + 0.  jr 3 4 jr 5 y)djr 4 (4 y 4 6 y)ds 4 =r x = ?/;r/?/ 0. cos 2  4 4 .n/
4 (jr = .rj 2 x2 ). = r.  c. 1.
2 3 ?/ 2 4 (.r ?/j 4 0. In rfl j/ = 0. ?/)</?/ = In c(x 4 Vl arc sin y p 4 2O Am. 0. 0. 2
y ds (2 dx y' ?/)  r.
= r. 0. ?/ r/?/ Aim. r2 In (2 \ f 2  r?/ 4 //*; J arc tan ?/ ( ) 10. (8 410 // 11. (2 *r 4 (5 Vl  13. 2cf3 cf 4 /'  dx 15. 0' f 4 //K/.r ~ 4 L> 16. (2 f j' 2 //' )(/.r 2 = 0. .r )r/// = 2
0.  /r)r/: = jr (/// { )r/// ~ 4 2 { 0. 2 19. 2(1 4 20. ( //)./ f //).r 1 21. (a.r 4 J.r 4 // )(/j r/.r  ( ~
1 = 2)(/// = .r )//
(/// = ~ 27. 3(5 .r 2 3 c) = .s r^. 0. r. .r// 4 3 = //'< r. //)</j // </.r :> (2 U/.r f // ~ (2 // 4 1 f f O 4 2  4 //)<7// (.r  )r/.r /V//
2 4 //)</./ ( (1  1 )r/// L
// (2 //>/.r 4 // ')r/.r ^* 4 ~ 0.
= 0.
 0.
= 0. //)(/// f jJ// rr o. f 3 i/)(ls 4 (.r 4 29. 2 24. 0. 25. 4 s 23. jryi 0. s 26. (3 28. 22. (3 0. 0. ./)</// f // djr />)^///  (./' 1 ( 3 r> = //' / ^rr^' ^i/r'i r^'
f 3 3
2 rhr 4:^ 1  = r.
= r.
Vl  ?/)'* 4 )(1 4 1 jr~y + L> </w 18. (3 .r (> 4 4 ./*// 4 r: j 4 3 /r sy 4 (2  // 4 1 0. 2 4 ./* r (2 ) Vr^lT' (2 VT^TT^ <r. ?/ L> ., 0. J 4 f.r = 0. ,s~ f//  2 I)////' 4 Cl ds r/c  2 Vl 4 : 14. 2 4 7 // 4 y)djr 4 <V f 3 y)(h{ 12. y )ds jr .r ( 1 1 4 .r// f 5 ./* /y L
// ')</// )(/// = = 0. 0. 0. ?/)r/// In each of the following problems find the particular solution which
determined by the given values of .r and //. 31. (.r 4 32. .r dy 33. (1 34. (l435.
j, j
; + L
// 2 'W/.r // J.r //')r/// .r// ~ Vr" = // ( /.r Find the equation
) ; dy
4.r ; .r // = = (Lr 2, of the 1,
; j // = =
~ // 0. ?,, //* 0. // of the jr* .r. + 404.^=: o. 2. curve whose slope and which passes through the point Find the equation 1 is at (2, 1). any point
Anz. x 2 curve whose slope at any point and which passes through the point (1, 0). Am. y(\ is 4 is 4 jr) equal to
2 nj 8. equal to = 1 .r. ORDINARY DIFFERENTIAL EQUATIONS
The III. Linear equations.
order in y is of the form Type
first where P and Q are functions of 383 linear differential equation of the alone, or constants. .r Similarly, the equation ^ + Hx  (C) J, <*y where // and .7 are functions of equation.
To integrate (#), y ( z and ,x, or and (S) determine which the variables and jr find z by ..!j.
</.r the result (7) in (#), Is +
?/ / '" = ' are separable. Using the value of u u^ = Q,
df which j and thus found z can be separated. Obviously, the values of u and z
and the solution of (#) is then given by (1). will satisfy (9), show the following examples ILLUSTRATIVE EXAMPLE Solution. This is = uz; then 1. details. Solve the equation evidently in the linear form (B), where p=
Let y
* (7), solving (11) The uz, a.r iJ7 we a linear differential by integrating u do, thus obtained, is he determined. Differentiating &rf= + We now in = u are functions of x to Substituting from in or constants, let (7) where // 2 _ x f
. * an(} Q=(x + 1
. . a?y a2 aw dx dx dx l)*. DIFFERENTIAL AND INTEGRAL CALCULUS 384 we Substituting in the given equation (12),
dz ' ' i M we 2 uz i \ dx To determine du , 1 f 1 x du
u
In get (M) 2 dx (KJ) now becomes, Replacing M by its =
= if we 2 In
(1 x) f (1 f j) a = In (1 f x) 2 . .* since the term in z drops out, value from (14), =<r + dx
Integrating, far 1 ..u Equation This gives 2 M dx, we s place the coefficient of a equal to zero. du Integrating, , t dx \ get
, = u dz get 1, f 1 i ' s (15) Sut>stituting from (15)
y and = ILLUSTRATIVE EXAMPLK
Solution. Solving (10), (14 , ) here In ^r is (r + get the complete solution l,*. An*. we have
f f /' dr = the constant of integration // in (11 ), In k, ; ke u Substituting this value of Pd ' *. and separating the variables z and x, the is fc Integrating, and substituting in It we uz, Derive a formula for the complete solution of (5). 2. whence result ~ **i + C
In u v\ in y = 2^
(7), p we 'e 'dr. obtain should be observed that the constant k cancels out of the final result. For
it is customary to omit the constant of
integration in solving (10). this reason * For the sake of
simplicity wt hu\
of integration.
r finally assumed the particular value zero for the constant
= r(l f j). But in the work that follows Otherwise we should have M 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' P and Q are functions of x alone or constants. Equation (D)
be reduced to the linear form (B), Type 111, by means of the
may
w
Such a reduction, however, is not necessary
substitution z = y~~ +
where 1 . we employ the same method for finding the solution as that given
under Type III. Let us illustrate this by means of an example.
if ILLUSTRATIVE EXAMPLE. Solve the equation ^4 (16) dx Solution. This is P^i,
 uz, we du
dx ,  7 f \r J" we , T" +
dx (17) u uz
~ get z ~ , In _ ,, , dx
x
In = x In , x i drops out, equation (17) =a
= u'*z* a \n x uz'z . value from (18), =
dx now becomes x In dx
its , u~z 2 X In u u^dx Replacing u by x , u Since the term in ^dx x u (18) 2. x du we = place the coefficient of z equal to zero. dx Integrating, n get dz
M . where (/>), 4 z dx Substituting in (16), 2 y' a In x, ^u^
dx then To determine In JT fl Q= j Let y = the form in evidently &
x a\ nx .*,
x This gives DIFFERENTIAL AND INTEGRAL CALCULUS 386 we Integrating, a(m 1 x) 2 r . get 2 ~ adnx^f 2C Substituting from (19) and (18) in y _ * or = uz, x xy[a(\n t} we get the complete solution 2 1 f f/(lnj)~' 2 f'J f + 2r = 0. 2 An*. PROBLEMS
Find the complete solution of each of the following differential equa tions.
( I j. 2. JT 3. lli __ di( j. __ J/ ( 5. 6. 2 /y !ii // _ ;j  rrr
;/ ~ ctn ?/ = .' 2 r. ?/ = JT  y 2 nj. y = us = C / 2 r C = . ?/  1 f 2) (/ ctn u/ tan / _ = 2 4 / _ / nx f 12 = // djr 14. 1>
 f r.r r + 4 2 . C(" + 2 / { j?/ . r sin 1 \ 2 tan /, >r s rr / u / c L) 4 c * cos /. ^ r j + ^ i? = x ctn cos / / = w
r n + l sin f (l rj* . ~ ?/" 4 A'/ cos 8 /. ctn 2 /). 8 = ( ^' + t n f /. 1=0. j* (f.r ^ ^' j". f rr ". = +
' 4 ili 13. + 2 jr ( 12. s cx 2 u ' djr 8. t.   cs'2  I   // J. ~   2 An*. 2 r. // ^( 4> 2 cc~ c sin ? ORDINARY DIFFERENTIAL EQUATIONS
^3L 15. x 16. ^4 17. .r ~ _ ?/ 2 y = 4 2 f 2 4 V = y = 1 * 3 x (1 = 20. ~* 0. 21. 2 .r. 4 22. rU". r  ^ ( 19. j ^ f  = 2 ?/ 23. N j. L> f j // ctn * 0. 24. 4 / csc  // ~
^ = (JT /y f ^ ax ajc 18.  = ^ cos // f (.r 2 ^ 4 * tan = = / 0. l)r'. sin x. .r + 2 f 1 .r)?/" ' c t 387 (tan /  =
1 ). In each of the following problems find the particular solution which
determined by the given values of x and //.
25.  7^ ~ 27. f ^ = /V
?/ tan jr ; x = sec ^ ; 1 . JT ?/ = 0. ~ 0, .4 //s. 1 // // . =. 2 jr' (c' sin ?/  is e). cos .r jr. djr ^  ^ = 28. dj* // 4 .r f (.r + 30. 1 )' ; jc = 0, =
y/ 2 1. ?/ = (r fl 4 f ) of the curve whose slope at any point is 2 ?/ = tt.r 2 Find the equation of the curve whose slope at any point
y f I) 2
, ^^ which passes through the point (1 equal to (1, 0). Ann. nf (jr 1 an( j ^.^^.^ p asses through the point s y f Find the equation 29.
, j" , 1 ). Anx. y(\  2ris 1. equal to f In x) 205. Two special types of differential equations of higher order.
differential equations discussed in this article occur frequently. = 1 . The g'where To X a function of x alone or a constant.
integrate, first multiply both members by dx. Then, inte grating, is we have 1) times. Then the complete solution conRepeat the process (r?
constants will be obtained.
taining n arbitrary DIFFERENTIAL AND INTEGRAL CALCULUS 388 d*i/ ILLUSTRATIVE EXAMPLE. Solve rr
dx A
Solution. = xe*. Multiplying both members by dx and integrating, =* + <:,. g = ^e* + c,
Repeating the process, ^ =Jxe* dx  JV dx + Jd dx + C
^ = xr*  2 + T,x + 0V or 2, <* j/ = Cxe* dx = xe 1 Hence A y = xe z second type of
' where To }' is   C
c* f r* 3 r f r,x 2 f f r,x dx f f C 2 dx + C3 + r^ x + c  ^TT " 3 dx 4 r 2 x f much importance ^4ns. f:t. is a function of y alone.
Write the equation integrate, proceed thus. d//'= Ydx, and multiply both members by y'. The
'= r//'^.
y' dy' But ^/' dx = d//, The / is and the preceding equation becomes = y'dy' variables result and //' now are The righthand member Ydy. a function of is Integrating, the result separated. Extract the square root, y. separate the variables, jr and //, and integrate again.
example illustrates the method. ~~ Solution. Here ~^ ax = ^
ox* = a'2 y, f a*y ' y dy'
2 \ following 0. and hence the equation belongs to type Multiplying both members by y'dx and proceeding as above, Integrating, The c ILLUSTRATIVE EXAMPLE. Solve
C is y" y' =  a' ydy.
~  a~y' f C.
\
= \ 2 C  a'^
2 2 2 . we get (F). ORDINARY DIFFERENTIAL EQUATIONS
C setting 2
ables, we C\ and taking the positive sign of the radical. 389 Separating the vari get d 'j =  arc sin Integrating, == Ar = x = or = sin =  _ f i \ r. arc sin ~^^= This is the same ~= as ^ C, # Hence y f uC 2 sin(j . f u(") ax cos
 oos a( \, f ros ax sin d(\., c\ sin  4 sin a.r </( \, a ! sin Art. 2 (4), ( ' cos ax. a ax f <'j cos A us. ax. PROBLEMS
Find the complete solution of each of the following differential equations.  3. 4 sin 2 x t. = sin 2 / + nf + r2 . 6. Vas
as j ,72 9 TT + Z = . Find /, Ans. having given that
t s = = ar\ V? x f r 2. ro a, ^ = 0,
at when / rr Q. DIFFERENTIAL AND INTEGRAL CALCULUS 390 206. Linear differential equations of the second order with constant
Equations of the form coefficients. where p and q are constants, are important in applied mathematics.
obtain a particular solution of (G), let us try to determine the
value of the constant r so that (G) will be satisfied by To
(1 (1 Differentiating c Substituting from
the result is rr
, = y ) ), we rr
. obtain and (1) e r2 CJ) (G) and dividing out the factor (2) in + + q=(), pr a quadratic equation whose roots are the values of
Equation (3) is called the auxiliary equation for (G). If
tinct roots r\ and r^ y = c r ^ and = y (5) '// In fact, (5) contains by = two c\e * r + fiK is >r " r
. essential arbitrary constants, and (G) is Solve 1. ^ (6 ) >> dx~ The auxiliary equation (7), nr lM
dx 1, tx = f and by
C'2 e Check. Substituting this value of y in Roots of a y o. 2r3 = 0. the roots are 8 and
y _ ( is ra (7) Solving r 7
e ** this relation. ILLUSTRATIVE EXAMPLE Solution. has dis and the complete solution are distinct particular solutions of (G), satisfied required. (3) then y (4) r *. (6), (5) the complete solution is Ans. the equation is satisfied. the roots of the auxiliary equation (3)
(8) imaginary.
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) If r, =a+ 6\/^l, r2 = a  bV^T be the pair of conjugate imaginary roots of (3). Then ORDINARY DIFFERENTIAL EQUATIONS
Substituting these values in
(10) = y we obtain (f>\
x f"(rif'" ~ + c?c In the algebra of imaginary numbers ^i V + C os bx if in new the by
( 5 ) c" CL>, B the imaginary values giving to A and
turn, we see that
12 ) = // rx ~~0. shown that* it is trx B UJ = ' V^~T cos hx sin bx. r i = M, sin and B are determined from
1. That ivwe now take for .4 V
A HV .4 ( and ^ 1 in (11) the values cos bx (> +# (A cos bx r L>) (o By
( J arbitrary constants = ci + .4 c '' are substituted in (10), the complete solution = y (11) ' 1 sin bx, When these values
may be written 391 c // , r and 1 af nm^r \/ and and 0, and c, f 7> .4 ( ,\ c\ 1 r ) . 1, in sin bx are real particular solutions of (G).
ILLUSTRATIVE EXAMPLE Solve 2. g (18) The Solution. Comparing with y When Check. REMARK. A
sin * Let o in is Then y. = . flfc x =r Also, replacing j cos 6x = 1 by bj  lfcjr in (7> b^jl and 1 , ''S), 5^4 + Art assuming that the
are identical. f i . . sin bx  .. , 1 f tkr  , + , IW Art 1, 1, +l (By ). a).
('  cos <*, (4), p. 3j. 191, rfprf.sonts  /'* satisfied. (k /,, * 1 ~ I, etc., and (15) .. ^ 104,
Bin  b'W series represents the function.
i sin bx
cos bx
c t>>x Therefore is 205 in Art.  (A\r ? fcV . . the equation f'roblfm in l ^ ^ . cos bx r cos Then, since ft* Then, by Art. 195,
(15) (13), obtained by setting A <' scri(\s for c complete A:J ^ in is  y ?/<> , n j same example ._^ .^J
_4 . ,4
t />> for the solution the above value of (ln + (11), the Hence, by k.
s substituted V^~T, and assume that the
the function when J is replaced by ibs.
we have
? b 0, that used for the form different r i AV^l. ^4 C()S kjr this value of y method with this = 0. now is /. 0. see that a is Compare B= r we (8), = /r' = A'// auxiliary equation (3)
r' 4 solution 4 = bW
i fer . f  63j.J b*r* + fofijfi b ;'s f > . 4 i  .  The righthand members , of (14) DIFFERENTIAL AND INTEGRAL CALCULUS 392 Roots of
q = \ p 2 The roots of the auxiliary equation (3)
Then (3) may be written, by substituting equal. = 4 q. r* if + pr + lp 2 =(r+>
2 P, 04;
and and (3) real be equal will ~ r\ r>
2 = = 0, p)* In this case 3 p. y~e (15) and r>T y  xertX are distinct particular solutions, and
(16)
is = // ?'"(ri f r,j) the complete solution.
To corroborate this statement, the second equation is it only necessary to prove that
But we have, by dif (15) gives a solution. in ferentiating, y (17)  xe r >', ^= <"'(! ^ f r,.r), <T"(2 Substituting from (17) into the lefthand
rijr
is, after canceling e result r, f member rr'j). of (G), the , (rr (18) + + />r, j q}jr This vanishes, since r\ satisfies (3) ILLUSTRATIVE EXAMPLE ;*. 2 r + } p. and equals Solve ^., = + \ p. o. Find the particular solution such that
s = 4 and ~= c/s 2 when or (r f / = 0. )* = cf/ Solution. The auxiliary ecjuation
r' Hence the f L r f roots are equal, d is = 0, =
s (20) This r, 1 1, = 1 0. and, hy (16), r'(r, 4 r^/). the complete solution.
To find the required particular solution,
and C'2 such that the given conditions,
is =4 and  = we must 2 when t find values for the constants = dt are satisfied. Substituting in the complete solution (20) the given values have 4 = c\ (21) t and hence
s = e'( s = 4, t = 0, we ORDINARY DIFFERENTIAL EQUATIONS
Now differentiate (21) with respect to s' By the given conditions, = 2 We t. when / 393 get 0. a/ Substituting, the result is
solution required is 8 = e ~'(4 2 + c 2 2 f). 4, and hence c% = Then the 2. particular Arts. PROBLEMS
Find the complete solution of each of the following differential equa tions.
( 1. 2. ^~ 7^
dr 2 ~2x =
 4 ^+3
dr rj*
4. 7^7 f
J 16 r = = 0. ?/ 0. j = r,r v Atis. Q. = <v" j* r c\ + f r 2 r a '. r,r". cos 4 / f ('2 sin 4 i. d/ 6.
~  4 = 0. = or 2 ' f r.r '. (I/~ _ f/ 2
?/ .
, 6. 7^' f 4
(/u 2 dij f  7. ^ f 2 ~+
dt 2 Q d~s ~ r/K r 8 dt 2  ?/ = r, = r f. 0. . 4 r 2r Ar
4x
. djr ~ +
, .s 0. * _
~ A ^' * ' = ~ f (n cos <"'(o cos li / / f r, sin /). f c* sin 2 0  In the following problems find the particular solution which satisfies
the given conditions.
17. ^!f dt 2 18. ^
a< 2 + 3 ^f 4. 2 = ; s = h n 2x  ; x 0, ~= 1, when f ^ 0. =c yin,s. dt dt = a, ^ = 0, when = 0.
/ ar x  a cos DIFFERENTIAL AND INTEGRAL CALCULUS 394 ~ 19. w V= () r ; ar" ~ 21. 2 8 at 22. ^  8 = // ~) f ^ 8  "' '. 26.  4 f "  ; =  ^ ; = .s ; 10 r = ; = 4, r= ^ . = y ~ ' r 4
'). . ^ = f'l.  jr r = /v  ^J" . ; .r = 0.
,s = 4 f 4f cos 3 /. / when 0, / Aw.s.
t 10, when  4, when = 0. 0, 4 4 a Awx. ^ 1, 16, = ~  4, when =
; r 4 dt = 4(> 2
8 = tc 4t Arts. dt ;  j = 0. / ( .s / = 0,^ = a,when/ = 0. r 25 4 jf
4 where ~ 6 4 To =c +c dt ( 25. no
28. ^ = 24, when = 0, = 0, ^ = 1, when = 0. dt <// 24. y ; ~ 4 16 K = ^~a~^(>;
dt 23. An*. x / / ~f 4 20. = 2, ~ = 0, when = 0. t ;>, = e 3/ (cos/ I sin /). 0. = 0. when = 0.
/x i 2, J7 / dt <11 solve the differential equation /> variable X and
.r is a function of the
r/ are constants and
independent
or a constant., three steps are necessary. FIRST STEP. Sol re the equation (G). (22) y Then u /,s called the of (H) = v trial THIRD STEP.
(24) a particular solution y
&// u. complementary function for (H). SKCOND STEP. Obtain
(23) = Let the complete solution be 77/c complete solution of = w +r . (H) is now ORDINARY DIFFERENTIAL EQUATIONS
In fact, when the value seen that the equation
arbitrary constants. of from (24) // and is satisfied, is t,24) 395 substituted in (#), it is
contains two essential 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.
Form X
A"
A' of =X y is ae not a particular solution of (G), X Form + bx, a =
= If assume + b.r a 2 sin bx, // assume , a i cos v = r A f Hx
= r ~ Ac'"
= r = A cos bx f A // assume t>jr of // if ; ; \ : 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,
iLLt'STRATIVK EXAMPLK A\, A^ so that (//) = rffB_ 2 rfK_;, v
2 The complementary First Step. 2x. djc djr Solution. satisfied. Solve 4. (25) is function u is found from the com plete solution of SBy Illustrative Example Second ~ y
Since S/r;>. // S above, therefore 1, (27) 2 =X = u 2 x is x ' rir' 7 f c>c . 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
(29  1 Equating 2 coefficients of like B powers 2B
Solving, A = , B= , 3 A  :J Ilx = 2 x. we XA=(), get  of x, is U /* and substituting in = 2. (28), we obtain solution y (30) Third Step. Then, from
y = u (27) f v = v = $Hx. and = c^ 1 (30), the
f c 2 e * complete solution f j !5 is the particular DIFFERENTIAL AND INTEGRAL CALCULUS ,396 ILLUSTRATIVE EXAMPLE Solve 5. " dx dx* (32 The complementary First Step. Solution.  y j = u function (27), or is x
c e'^ f c 2 ('' .
l X = 2 e * is a particular solution of (26), for it is
Second Step. Here y
obtained from the complete solution (32) by letting fi = 0, c 2 = 2. Hence assume
for a particular solution v of (31 1, Differentiating (33),
(34; = = v ~ y (33; TI = Ar ' (1 (35; we Simplifying,
obtain we (x 2 2;
* Ac 4 get Third Step. The complete solution
y ILLUSTRATIVE EXAMPLE (5. (37) ~+ such that . Substituting in (33) ~\. therefore, is, ) T *. J f c 2 e A ~ 2 c \ f xe Ana. . Determine the particular solution of  s xe I of (31 u
c\r u f r and hence =r= y (30; *, = x Axe 3 x) (\ 2 e ~~> (X '' we obtain r Ae ' S"'4  Substituting from (33) and (34) in (31),
r . we obtain 4r Ac x Axe = 4 * and 2 cos 2 /, ~ = 2 when t = 0. dt Find the complete solution Solution.
First Step. Solving ~ (38) we find the u Second Step. = 4 = .s 0, = cos 2 r, / f c 2 sin 2 /. member Considering the righthand in / we observe
= 2,
when (37), a particular solution of (38) resulting from (39)
v of (37) assume
for a particular solution s 2 "os 2 Hence f complementary function (39) s first. is c2 = s (40) Differentiating f^' = (40") , r= we 4! cos 2 / t(Ai cos 2 / f A 2 sin 2 /). obtain f A  sin 2 / a/  2 Mi sin 2 /  4 2 cos 2 /). (41)
'
 ^= 4 ^1, sin 2 / f 4 .4, cos 2 /  4 t( A cos 2
} / + .4 2 sin 2 /). that
0. ORDINARY DIFFERENTIAL EQUATIONS
Substituting from (40) and (41  (42) 4 2 .4i sin and simplifying, the in (37). > + t A; 4 cos 2 we 2 cos 2 Aa = 0, result is f. Substituting in (40), J. get and (43) the = By Third Step. cos 2 (39) (44) r, must now determine d and
(45 =r= g (43) We = / = This equation becomes an identity when A\ 397 s ) = r 2 so and ^
dt sin 2 / t. complete solution of (37) + / ? c> sin 2 / 4 sin 2 / .\ is t. that = 2 when = / 0. Differentiating (44), = (46) 2 sin 2 Ci t f 1 2 r cos 2 / f  sin 2 , / f cos 2 / t. at Substituting the given conditions (45) = 2 r,, Putting these values back
s (47) = = 2 fa. and in (44) .'.<, = 0, r, the results are (16),  1. in (44), the particular solution required is sin 2 / f \ /sin 2 /I;i8. /. PROBLEMS
Find the complete solution of each of the following diflVrenthil equa tions. l. dt 2 2. jr = at x = ae + f  4 cos + X = 4sln2t. + ^~ f ^ 3. dt4. d/ 2 <* _ 5 + cos / + c>2 sin / + at O cos / f r, sin / f ^j r, cos / f r sin / = CI cos / f c> sin /  r,^' f r,r = r,r' f r 2 r 2;  R ft/ = o^' f r 2 f" 2 2t
f J ^'  8 = h .f . = at s = 4 s = e =2 s /,. JT *>^ 4 s d/ 2 4 s e^4 + ~ c\ Ans. s b. b.  ^  f 2 '\ I / It. sin sin 2 /. /. \(at f 6j. 1 r'. d/ 2 7.
' ^! _ 8 ^!2
dt 9 * ' 2t
. ' d^ 2 ^ 2y
dx 2 _}_ 9 y cos 2 5 x2 . t. ?/ fie =d 2 ' fzf' 2 '  i cos 2 t. 2
cos 3 x 4 c 2 sin 3 x 4 ^ x  if. DIFFERENTIAL AND INTEGRAL CALCULUS 398  10. 11. ~2s4 2' '~i
 7 1 /// 13. ~
14.  r,f r = r,<'' '= ''
12. ^" Anx. s /. { f = :< 2 r~". * = r,r' 8 H'. s = 4 r'. r f 2  4  r ^ i ;{ .r ^ , f '2 5 // U sin _ 12 /. f 9 f  21. 6 c 2 '. .<;  .r cos 12 cos sin  /. f 5 cos  + f,  /) 12 /V sin 2 f * f /. rj sin f
f "2 + 26. / 1* ^ /r'. li j cos /.  / 22. 4  ^  2. :i 2
f c> sin /) f  r '. / :{ cos /! s f ;  *. 8. f r a r*' r'(r, // ~j 17. + cos 'fr, j* f '' f r./r' 2 <// ~
y/( 16.  r*" f 1 r^r" + = r,r' fj + ' 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 29. ~~ + 9 x = 5 cos 2 / ; x = 1 , ~ = 3, when / = 30. ~ + 9 * = X cos 3 / ; = 0, 6, when
An*, / =
x s or = = + r 1 '). = sin 3 t f cos 2 f. 0. 2 sin 3 gj..
Ans. / 0. us. ~= 3 b i(e 3' / f it sin
 3 t. ,. t + e'6 +
J l). ORDINARY DIFFERENTIAL EQUATIONS
 ( 32. ^~
(it 6 ^ + 13 34. 38. 39. = jr ; ~ 4, = 3, when dt = / f 9 * = 4   9 * = 6 / ^~ f a = Li ^
^+
dt 2 + = cos 1> L> ^ jr ; L> /  L> =
.s ; 0, sin : y= / ; 5 <& .., ~ 4 r'' .r ,
cos 2 , / f 3. 0. (). = / / 0. r 0.  0, 0, when when / = (). dt + 4 ,/ = 12 , = // ' ? 1, c/.r dj' = when "J,  0, .r / = / when J, ^~ 0, when (), when 0, 0, = = ^ ~= 0, / 8 ; = * ; (It a.r 40. <%i / cos 2 =
^y + 4
 2 ~ f
,s 3 0. , .4 us. 36. 37. 39 , t 33.  JT dt ~
, 399 .r = 0. (/./ ^+
as 4 ,/ = 4 , ; =
' 0, ty ^ when 2. , = 0. (/.r Compoundinterest law. A sim]>k application of
is afforded by problems in which the rate of
of the function with respect to the variable (Art. f>0) for any
change
value of the variable is proportional to the corresponding value of
207. Applications. k differential equations the function. That is, if // =/(.r), **
where k is a constant. separable as in Type I, In this differential equation the variables are
p. 379. Sohing, we get (2) where y  c^ an arbitrary constant. J
t Thus the function y is an expoConversely, given (2), it is easily shown
by differentiation that y satisfies (1 ). The connection of (1 with the
name "compoundinterest law" is shown as follows: c is nential function (Art. 62). ) Let y = a sum money of in dollars arcumulatiri^ at compound interest; interest in dollars on one dollar for a year; i = an interval of time measured in years
= the interest of y dollars for the interval
Then Ay = iy At. Therefore
A/ ; A?/ of time At. (3) states that the average rate of change of // (Art. 50) for the period
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 Equation of time At is  DIFFERENTIAL AND INTEGRAL CALCULUS 400 /.
But in nature, changes proceed on the whole in a continuous
So that to adapt equation (3) to natural phenomena we must imagine
that is, assume the interval of time A/
the sum // to accumulate continuously
to be infinitesimal. Then equation (3j becomes continuously with manner. ; _
" dy ly > ~dt and the rate
In ( 1 ) of change of y proportional to is the function y agreeing with (1) y, if said to change according to the is k = i. compound interest law. A second example afforded by the complete solution of the is equation + ' S (4) =T/1 tf f' where k and c are constants not equal to
(4) may be written ~ r
( >) 0/ +  a) zero. + k(y For let c = ak. Then a). This equation states that the function y + a changes according to
the compoundinterest law. The differential equation (4), or (5),
is solved as in Type I, The 379. p. (G) = // ILLUSTKATIYIO EXAMPLE 1.
compoundinterest law. When
Solution. By (1) The cc 4 y 1, is a. function
 x solution kj ; of // x changes according to the
Find the law.
2, y = 12. when x  we have g = *,. (7) Separating the variables and integrating, we get
In We have to find the values of k Then
Solving, Therefore In 4 k  In 12
In // =  In 4 1.0986  =
JT k + In 3
f //  and In 1 J, 4 (\ Substitute the given values of x and y. In 12 r,  k.r
('. .0980, and  2 k (* y = = + C. In 4  In 3 1 098Bj
J r . = In J. .4n,s. ILLUSTRATIVE KXAMPLK 2. Washing down a solution. W 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
amount of water which has run through,
(or acid) in the tank at any time, and j
show that the rate of decrease of s with respect to x varies as s, and, in fact, that
T Solution. Since quantity of salt in s = quantity of any other volume salt in the u of the mixture of total volume mixture is  u. f, the ORDINARY DIFFERENTIAL EQUATIONS
Suppose a volume Ax
of salt thus tank in the dipped
is of the mixture out will be ^
v The amount dipped out of the tank. is and hence the change A.r, 401 amount in the of salt given by As (8) * =  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 we AJ obtain the instantaneous rule of change of to j, namely, ~ s
  dx Hence s with respect . rfs r /b<s. changes according to the compoundinterest law. s PROBLEMS
The rate of change of a function with respect to .r equals
and
4 when JT
1. Find the law connecting
and
A us. = 5.58 ci
y
2. The rate of change of a function
with respect to .r equals
and y
8 when jc = 0. Find the law*.
c
An*.
f 2.
3. In Illustrative Example
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.
1.  // { //, r j //. . \i "2 // = // //, ' f> li, If the excess temperature of a body above
4. Neirton'K law of cooling.
the temperature of the surrounding air is .r degrees, the time rate of decrease of jc is proportional to .r. If this excess temperature was at first 80 degrees, and after 1 min. is 70 degrees, what will
In how many minutes will it decrease 20 degrees? it be after 2 min.? 5. Atmospheric pressure ]> at points above the earth's surface as a
above sea level changes according to the comfunction of the altitude
15 Ib. per sq. in. when //  0, and 10 Ib.
poundinterest law. Assuming ;>
// when = h 6. 10,000 The formed in find p (a) ft., when // = 5000 ft.
AUK. ; when (b) (a) lU.'Jlb.; velocity of a chemical reaction in which /
t is the timerate of change of T, is the = 15,000 ft. (b) 8.1511). amount trans time Reaction of the the experiment. first order. Then ~ Let a
k(a = concentration at the beginning of r), since the rate of change (Note that a
changes according to the compoundinterest law.) tional to the concentration at that instant.
tration, // Prove that k, the velocity constant, is equal to In
/ a  r, is propor the concen f 7. In the inversion of raw sugar, the timerate 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. DIFFERENTIAL AND INTEGRAL CALCULUS 402 8. In an electric circuit with given voltage K and current
the voltage A is consumed in overcoming <]) the resistance
the circuit, and (2) the inductance L, the equation being i (amperes), 1 ft (ohms) of This process therefore comes under equation (4) above, h\ ft, L being
= when / = 0, show
constants. Given L ~ f>40, ft = 250, ft
500, and
that the current will approach 2 amperes as
increases. Also find in how
/ / many seconds
(lie
/; / will reach W) per cent of maximum its sec.  In a 10. llniUlimj n /> a saline (or acid) XO/M//OJ/ by adding salt (or acid),  (r
//), where
r
as
the constant volume,
salt (or acid) in the tank at any moment,
salt 'or acid) added from the beginning.
Derive this result and com maintaining constant volume, leads to the equation
r Ann. 5.9 value. condenser discharging electricity, the timerate of change of
voltage c is proportional to r, and c decreases with the time. Given
A nx. 92 sec.
40, find / if r decreases to 10 per cent of its original value.
9. ^ ~  // j' pare with Illustrative Example 2 above. Many important probmechanics 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 solu208. Applications to problems in mechanics. lems in 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 instant of time and and acceleration (equals the distance of the
at this time from a fixed origin on the linear path. /), N any
moving point lu,rsTKvn\ K !']\AMPLK 1. In u rectilinear motion the acceleration
=
1 when
proportional to the square of the distance x, and equals
,s' Acceleration (2) Also, r
(a) Find Solution. 5,
r s when From S, when *  a 'J4. / (2), using; 0. the last form for
dr /o\
(3) r ^ a,^ Multiplying both members by
(4)  4 =  + </s r, a, we obtain 4
6' and integrating, the
or f" = ? + r. result is at is 2. inversely That is, ORDINARY DIFFERENTIAL EQUATIONS
Substituting in (4) the above conditions
(4 > = r ft, 8 = 8, we 403
Hence find (" ~ 1M. becomes
1 r (5) ,S" 1M9 ~ 1.9;?. /I NX.
Find the time which elapses when the point moves from From
(hi this equation, if s = Solution. Solving (5j for lM. r we r, = \ \ and solving for
Separating the variables s and
4, we find, for the elapsed time,
/ (7) NOTE. Using the S to ,s* = 24. get / s = s with limits as given, / x  8, = first form for in il ^x t~> is </, 4 _ " '
1 dl' which of the ib form (F), Art. .s' The method 'J05. of integration here is same the as in Art. 205. An important type of rectilinear motion and celeration where a is of magnitude Remembering
in that, (i = Ks rectly proportional to the distance f 4 ,s and  ,s. is, in The motion magnitude,
is called di MM pic vibration.
(8), we have, using a linear equation in s and / (1 ), of the second order with constant, coeffi Integrating (see Illustrative cients. ac a force and the acceleration caused by it differ
see in the above cast that the effective force magnitude only,
directed always toward the point From tlie differ in si^n. ut unit distance. we harmonic and write (8) = which that in the distance are in a constant ratio Then we may
k~ is Example 2, Art. 20(lj, we obtain the complete solution,
(10 .s) From (10), (ID
It is cos ki + C'j sin kt. differentiation,
r = k( C] sin /:/ + c cos A/). easy to see that the motion defined by (10)
= />, and
between the extreme positions oscillation termined by
(12) by =n ,s is s a periodic = fc, de DIFFERENTIAL AND INTEGRAL CALCULUS 404 we may In fact, constants replace the constants and c\ 2 in (10) by other and A, such that b (13j = c\ A b sin = 02 y Substituting these values in (10), b cos A. reduces to it s=bsm(kt + (14) by A), (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 depenis upon the solution of an equation of the form (G) and (H),
discussed above.
ILLUSTRATIVE KXAMPLK In a rectilinear 2.   a (15) motion
v. 'I Also, v = 2, K = 0, when 0. / Find the equation of motion (aj  x Solution. Using (1), we terms of (x in have, from
' (15), d$ ^ + ^ +,5
rf2 s nr\
df.) ' _L /). S = ' 4 an equation of the form (G). The roots of the auxiliary equation = r, Hence the complete By J s the piven conditions, o = find 0, ITT \ = r" 2 + J = are  vC"I. is cos '( r , when s = r2 , solution of (16) (17) we + r 2 f r + / = / sin r2 /). Substituting these values in (17), 0. and hence
s (IS) Diflerent.iuting to find r, r (19) we ~ = c 2 e~ get r,f J sin / when / *'( 2
Substituting the given values r
With this value of r 2 (18) becomes + cos = 0, /). we have 2 = 2. , (b) For what values of Solution. member When of (19) r = 0, r will r = the expression in the parenthesis of the righthand
this equal to zero, we readily obtain must vanish. Setting tan (21) For any value
(22) 0? of / < = 2. satisfying (21), r will vanish.
t = 1.10 + rnr. These values are
(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 disturbed 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 =
In fact, for the simple harmonic vibration
is lengthened by the damping force. we  1.12, and the halfperiod, by (12), is
\\
by comparison with iS\ k
As we have seen above, for the damped harmonic vibration the correspond TT. ing interval is TT.
Second, the values of the successive extreme positions where r
a decreasing geometric progression. Proof ILLUSTRATIVE EXAMPLE IJ.  a  0, r = when 2, Find the equation Solution. By (1), we = / 4 s 2 cos 2 f 44x^2 ' cos 2 t. dt R (27) = sin 2 / + j For what values / of sin 2 (28) (2 f / icos 2 / r  0? sin 2 f \ / dividing (28) through by cos 2
(29) J tan 2 / 4 2 4 =
/ Example in Illustrative Ans. /. will / Solution. Differentiating (27) to find
setting the result equal to zero, we get or, Art. 206, of motion. particular solution required was found
given by equation (47), p. 397. Hence (b) fi, t. have, from (25), The
is instead
omitted. 0, 0. (26) and is ~ In a rectilinear motion (25) Also, s f> * for now form of being equal, (a) ; *, i  have, 0.89   (24)  and r, ; /, 0. roots of this equation may be found as exthe curves
plained in Arts. 8789. The figure shows
(see Art. 88) The (30) and the abscissas = y tan 2 y /, = 2 of the points of intersection are,
/ = O.SS, 2..'H>, etc.  /, approximately, Anx. Discussion. This example illustrates forced harmonic
the acceleration is the sum of two components
(31) a, = 4 s, a = 2 cos 2 vibrut'ion. In fact, in (25) t. The simple harmonic vibration corresponding to the component GJ with the
with the acceleration a z that is, by a periodic
period TT is now disturbed by a force
, DIFFERENTIAL AND INTEGRAL CALCULUS 406 force whose period (= TT) is the mme as the period of the undisturbed simple
harmonic vibration. The effects of this disturbing force are twofold.
of the point where
FirHt, the interval of time between successive positions
~ is no longer constant, but decreases and approaches J TT. This is clear from
v
the above figure.
= now
Second, the values of * for the successive extreme positions where r increase arid eventually become, in numerical value, indefinitely great. PROBLEMS
In each of the following problems the acceleration are given.
1. a 2. a 4. a L a 3. = 5. 8.  0, ; .s ,s,,, r a A 4 sin = 13.   (/ amplitude f ; 2 / 0. b / =  = .'5, ~ 0, r = x n, r \\ hen 8 ;  4 x, , r  0, and 1 , / cos f> 1 ( *  ,s =3c 0. / 5 cos sin s / 2 cos / ' / f / cos sin t. /. /.  0. 0. /  s is 5 cos 2 a sin 2 \ 0. / when 0, acceleration of a particle /).  0.  0.
= when r *,, 0. \\hen 0, ; 1 when  / kt. = 0. t 1 when /;/. 0. / when = 0, sin when 0, Show 0. / a simple harmonic vibration \\ith the center at s
2 and a period TT. The (a) If when
ii, = = s 0. / 0,   * 0. / when 0, r .s ,s conditions initial y' 0, = 4,
4 x; x = 0, r =
 4 s x = 0, r
; / hen \\ = 0, r = x ; r 2, r .s ; w hen when 0, 0, r x ; 2 x 5 r a x  x ; x 4 x 2 sin 2 Given
is  '.s //r  / L /  s /  0, r .s ; 2 r 8  12. a = / r,,, A nx. 0. r () , sin 2 = 11. a 15. r x = 10. a 14. ,s ()> 2 cos a motion x ft a 9. ; A'^.s a 7. when  s A.S a 6. =.(),v= ; A' '.s and of motion. Find the equation that the 2, with an given by
/ 9 s. the particle starts from rest at the origin, find A H*. motion. * = its equation of cos 2 /  cos 3 /. the greatest distance from the origin reached by the particle?
= f>, find its
(b) If the particle starts from the origin with velocity v
cos 3 /.
AHS. s
cos 2 t + 2 sin 3 t
equation of motion.
What is the greatest distance from the origin reached by the particle? What 16.
is is Answer the questions given by a Ans. of the  (a) x preceding problem 3 cos = 379 \t sin 3 / if the acceleration *. ; (b) s = \ t sin 3 / f 2 sin 3 /. ORDINARY DIFFERENTIAL EQUATIONS
17. A body falls from 407 under the action of its weight and a small
Prove the following relations: rest resistance which varies as the velocity. .
= ^ *" A body 18. (kt + ; ' < from rest a distance falls  1). of 80 A 19. n Assuming ft. AH*. find the time. boat moving in \\ater still is = 32 r, 3. 17 sec. subject to a retardation propor any instant. Show that the velocity /sec. after
a
\\here c is the vethe po\ver is shut off is given by the formula r
at the instant the power is shut off.
locity tional to its velocity at ' ', 20. At a certain instant a boat drifting in still vater has a velocity of
mi. per hour. One minute later the velocity is 'J mi. per hour. Find
the distance moved.
4 Under 21. conditions the equation defining the swing of a certain galvanometer is Show will not that it complete solution if fj. swing through the zero point < if p. > k. Find the k. 209. Linear differential equations of the nth order with constant coefThe solution of the linear differential equation
ficients. in which the coefficients p\, p>>, are constants, will p,, now be dis cussed. The substitution of
n (r c rx + pir for y in the first
71 " 1 + 7>>r This expression vanishes for n ~2 all member gives + Pn)e". + values of r which satisfy the equation
(1 r ) and therefore
Equation (1 ) n +p } n
r ~i + p>r n ~'2 +  for each of these values of
is + pn =
r, c called the auxiliary equation of rs is (/). ; a solution of We observe (7). that the coefficients are the same in both, the exponents in (1 corresponding to the order of the derivatives in (7), and y being replaced by 1.
) DIFFERENTIAL AND INTEGRAL CALCULUS 408 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 Rule more advanced textbooks. to solve the equation (/) Write down the corresponding auxiliary equation FIRST STEP.
(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:
DIFFERENTIAL EQUATION ArxiLiAHv EQUATION
(a) root real dlstinct . , , \ a particular solution e T]X (jives f ([}}
, Kach r\ ' Each distinct pair . ( . \ two particular solutions c <Jtves ' i , of nnaynuirtj roots a . * hi e" i i r 2 8 (or A nndti}>h' root occur b.r. particular solutions ob s) taiwd by multiplying .
\ \ > cos bx. , sin I (c) . ) the par ' * ritiy s thnes ticular solutions (a) (or (b)) by J 7, FOURTH STI:P. the n * Multiply each of found hy an arbitrary constant and add
to y gives the ( 1. Solve , ', Follow tho above
r'  ;i t f ^ 4 . ;i f   4 0. ?/ (U*' rule.
0, = dc ILLUSTRATUE EXAMPLE auxiliary cxiuation. 2. z
I Solve Follow the above
r4 4 r' f 10 r J c>>c~ f riiJY*. c x
, xe 2x . A?is. ^4^ + 10^~12^h5/ =
dr
dr
/ (/j First Step, 1 independent solutions thus
This result set equal 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
Fourth Step. The complete solution is Solution. j the results. ( ~~ CU' First S/r/>. 1 complete solution. ILLUSTRATIVE EXAMPLE
Solution. .r :r, >l 1 c/j J 0. rule. 5 11! r f Second Step, Sohdng, the roots are 1 * A chock on the
accuracy of the work
must give n independent solutions. , is 1 =
, 0, 1 found auxiliary equation.
2 i.
in the fact that the first three steps ORDINARY DIFFERENTIAL EQUATIONS
Third Step, The (bj pair of imaginary roots
e* cos The double (c) The complete Fourth Step. dc 1 y or I/ The in = (<*i root 2 x, 1 f raN'* f c^c* 1 *J gives the i* sin 2 x. gives the solution f r 2 x 4 fa >J
< 409
two
(0 two solutions solutions = 2) 1,6 r*. xe*. is cos 2 j + cos 2 x f <V' T sin 2 x, sin 2 .rir 1 <'i . A?w. linear differential equation which constant,
tion (//). here also. p n are constants, and A' is a function of jc or a
like those used in Art. 20(> for equaThe three steps described on page 394 are to be followed pi,
is ;;_>, , by methods solved That is, we solve (2) first
?/ the complete solution of = the equation obtaining w, Then (/). (/), u is the complementary function for (/). Next we find in any manner a particular solution
y (3) Then the complete = v. solution of (7) y (4) = of (/), ?/ + is
v. (3), methods of trial may be used analogous to those
= 2. The rules given there for the general
on page 395 for
given
case apply also for any value of ?/. In any case we may follow the In finding // Rule to find a particular solution of (7) FIRST STEP. Differentiate successively tht (liven equation (7) and
a obtain, either directly or by elimination, a differential equation of higher order of Type SECOND STEP.
we get its (/). Solving this complete solution. y where the part u is the first step,* the particular solution v, ~ u + the rule on page 408, v, complementary function of (/) already found and v is
THIRD STEP. To find in the new equation by the sum of the additional terms found. ihv values of the constants of integration in 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. DIFFERENTIAL AND INTEGRAL CALCULUS 410 and In derivatives in the yiven equation (/). its the resulting identity equate the coefficients of Like terms, solve for the constants of integration, and substitute their values back in = ?/ (jiri)ty + // v, the complete solution of (J). This method now be will by means illustrated of examples. NOTK. The solution of the auxiliary equation of the new derived differential
equation is facilitated by observing that the lefthand member of that equation is member exactly divisible by the lefthand
the complementary function. of the auxiliary equation used in finding ILU'STRATIVH K\ AMPLE. Solve We Solution. the complementary function first find y" ~ I) the result
(7 y We now ,SVr/>. (5) from (9) (10), from we  v of Type 5 r 1 1 The  + (1M)
1. (14) Third Step. z
. = ex ?/'  1 f ij' 2 = // y = or (7) y 9 7 In fact,
2>ir r f ;{  r' r f . 0, 2 * 4 c and ~ J
(c2 (14 \ r (). H r
by rwe find that  f r we is = 2 1 ) = ;i j f + 2, (12) since the auxiliary may be written 0. Hence the complete 1.2. 1, Comparing (15)
will  (r' = 13 //" divisible r' r e y '2 auxiliary equation 5 r< member must be
is
8 r f 2 = 0. he roots are  get 1 '1 t5 9 y" lefthand
(t>> ex . (/). 12) equation for //' 5 y' //" f //'" //'" f Solve (11). iSYrow/ N/r/>. 4 4 //' an equation = M* + is  lv ?/ 1 ryff. we obtain (10) The 2 \\ y'" Dill'reritiating (9>, ( //" f the result (S), ) Subtracting 4 c\v~* ?/ //'" Subtracting (1 Solving 0, dillerentiate (5), obtaining (S) (!) 2 v u. is ) First 4 a \f = solution of ( 11 ) c&*}. see that J
r'(r.{j f r 4 J' > be a particular solution of (5) for suitable values of the constants
we obtain Differentiating (15), = r'co 4I/" = p*(2(r 3
r (10) y (r.< 4 is 2 r 4 )x 2 4 r 4 .r ), 4 f 4) 4 (r 3 f 4 r^or 2
4 r 4 x ). c.\ and c^. ORDINARY DIFFERENTIAL EQUATIONS
Substituting in
ing, the result is from (5) (15) and (16\ dividing both members by 2< 4 (17) of like Equating coefficients
whence we find c\ =
solution  2r,j =  c of x powers = 1, r, 411
and reduc J. 2r~ we obtain t r c 'd 1, ra = 0, Substituting these values in (15), the particular J. is y (18) and the complete solution = r = r"\  j J J ), .1 is PROBLEMS
Find tbe complete solution of each of the following differential equa tions. I LJ.
{ f/r <2 LJJ +
__ 4 d 1 C] f ^ y o^ ^ // 4 '^  (o s // t r ^.^j , )_ *j >r  f r r, f C'2( // L ^/ \ 4 , >r 0. as 1 (/j 4. A H$ o c,\ ;r 4 = 0. ,s ,s ,4  ?/,s. , 2 c() S J 4 r. r
( 4 r // = r, I 4 cy" r/, cos f r. { cos J l 6 J/ + L> I IQ. _ f
HlL = 4 :l 2 ^ WtSt t 4/**. 0. a* = nr'r .s ' rr +  .r r i f r sin 2 i + L>r (r, u*. /. sin 3 f (r :i f r ,/) sin / r cos r.j 7/ = v ,'.  ( .r. 2 c*t f r :i / /. a
). = c,c + r,f'  f r i sin / P 3 ^. = n + r,r + r ,f y / r, + r,r' + r.,r l! g_ 3 = 4 . r. : (Lr~ g+2 = ". 18.^9^+ 20 = /V.
a/
(^/ 13. = JJ 4 o, sin d.r d.r* 1L 8 ^ / .r 4 (// ;j/ rfr + > L sin 2 s. \ cos i cos 2 ' CiC* 4 r.r f r L./) i .r. c j* } 4 9. sin 2 djr  3. _if } A,. . r" + 2 i = .,.' + ^'+ se>.
f: " (7 + L> ^j + 4 .s = / sin 2
4 Am. V
4 + 2<2) /. s = cj . cos 2 ^ + c 2 sin 2 i .
, / f cos
 t 2 t t 2 sin 2 DIFFERENTIAL AND INTEGRAL CALCULUS 412 " ,6. + 1:1 i;
4 , 30 , = 9 + . 2 20. (I. + 36, = 1836. ,3 + at 2 dt^ g=
at '. MISCELLANEOUS PROBLEMS
Find the complete solution of each of the following differential equations. Am. . 4. (1 4 5. (.r f v/)r/.r (//^ ~( Vi In //)(///. = 0. = 0. ^
^+ H 4 .s ^' _ 4 ^~+ 8 2 ^8 r
(
!3^ l: = r,r" s = A//K. r j/ (r, = // r'. // ^4
(if" = c. 4 r,/r 2
'. + f c>2 sin 2 <). r ^^'  \ e<. A.r = / f b. j d ^
( cos A/ f r,> sin A/ f A"* _ /; i>  a C os .r A/. =n
 cos c\c kt A'/ f r.. sin A'/ + + cc~ u  ^
2  2 cos kt. A"^ f 2 A 2 .r = a sin 2 /rW.r f 2 (.r ^
dr * x r 2 ^'. cos 2  r,r" + <//' 16. 2 arc tan ) (// dr 15. 'r dt .r 14. 4 r //' (JT~ f = o<' + 0. = li r __ // ydx. <// <//* 11. (.r ( 1 (//^ 10. Vl  + O\ <// (//^ 9 ~ 4f 7. 8. = .!'')</// ?/=(/ f 4ri =
4 1 j A*/. .r// (/// .r 0. =
2 // c\ f
L) .>l iv r
+ 1)' U'~ ?vU' cos .r + 2 A'/ In
I) 2 f CJT = <" = 2 sin kt j
n 0. arc tan x + c. t cos kt. ORDINARY DIFFERENTIAL EQUATIONS
~j  17. ^
ar 5 a/ 4 = 4 * f Ans. 0. Anx.
2 19. .n/ 20. (i?/ 22. (.r* 4 ~ 4 jz/d (/// _ <^ 21. = df2 r 8 4 4 jr c\ cos 4 /  0. 8 / .r 1 sin 4 f (// 0. o cos 2 * . __ 4 .r = 4 4 .r =2 , 4 ^
(it 25. 4 2. 4 / 2 <V 4 4 c* sin 2 '. f. / "' ' 24. W < 4 c 2 e po [^ = 25 = nc rj li'^djr. rV'Wr 4 = s * 413 6 cos 3 cos 2 t. /. Solve each of the following differential equations by making the transformation suggested. ~2 ' 26. 1 27. 4 (/' $t 0(/.s = 28. C5 4 2 K/)S 29. J
?/)' (.r 4  ^ J
(^' r/^ = = *' 2 4 =
2 x/ 4  (3 jr Let 0. 4 2 2 ,s 1 .s )^/. .s7)/ // = (/.s\ 4 5. A ns. ~Let .s Let
Let .r = x/ 4 ^ws. r/. * ~ 4 = r/(l + ~= r.  t. t) r. = // r. ADDITIONAL PROBLEMS
1. For a certain curve the area bounded by the curve, the .raxis, and
any two ordinates is k times the length of arc intercepted between the ordinates, arid the curve passes through the point
curve must be a catenary. (0, Show /;). that the man dropping in a parachute from a stationary
is the velocity in
second per second, where
per
i
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.
2. The balloon is acceleration of a 32 r' ft. ?> 3. A point moving on the .raxis 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  0, m and when n
/ = are positive, and given the initial conditions
0, find in each at dt cuss the motion.
(a) m = 4, n = 5 ; (b) x=W, ~ and
of the following cases x and m  4, n =4 ; (c) m= 4, n = 3. dis CHAPTER XXII
HYPERBOLIC FUNCTIONS
210. Hyperbolic sine and cosine. Certain simple expressions involving exponential functions (Art. 02) occur frequently in applied
mathematics. They are called hyperbolic functions. The justification 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 sinh v (4) r = v C cosh v C = 1 v ' ~4~
! e is, as usual, the Napierian base (Art. (>1 ).
These functions
are not, however, independent, for we have from (A) where sinh 2 v cosh 2 v (B) ^ 1.  ' r' From (A), Hquarm^. cosh' by solving (A),
e (1) r cosh 4 '2 r e
~~~~ f , sum' : Hence, by suht ruction, cosh From 1 Mnh' r  r r = f' 2 f I. we for the exponential functions, r f sinh e r, = ' '' r : cosh sinh r get r. ILLUSTRATIVE EXAMPLE. Show that the complete solution of the differential 1 equation
/n\ rf'"'/
Ji (2) f\ . () '// a.r may be written where .4 and Solution.
are a and /? A // sinh as 4 values of r" z r" and r ~ nj> rif" or " f rj = result should .4, TI f c a'2 r~ = 0, whose roots "'. ~ cosh n.r sinh u.r, r x = cosh a.r
+ sinh a.r} rj(cosh ax
y = n(eosh
= (c\ r\'Vosh ax (ri ^Osinh a.r.
x (i.r ri 4 are found from (1) by taking r f f Letting (/.r, Art. 2()(> the auxiliary equation for (2) is
Therefore the complete solution of (2) is
r The cosh By
a. \l The />* are constants. ax. Hence sinh a.r,
sinh ajr] j ~ /?, we obtain the desired form. be compared with Illustrative Example
414 2, p. 391. HYPERBOLIC FUNCTIONS
hyperbolic tangent, tanh The 211. Other hyperbolic functions. defined by tanh v  ^
C sinh v  r cosh v 415 e e 1 r"l e + 1 r, is v e The equations and
define, respectively, the hyperbolic cotangent, hyperbolic secant,
cosecant. The ratios used in (C) and (1) are the same as
hyperbolic
the corresponding trigonometric functions. (2), p. 2, for those in following relations hold The tanh 1 (2) 1' = r : j sech ctnh' r, r = 1 csch r, of the first formula is
analogous to formulas in (2), p. 2. The proof
given below.
The following statements hold for the values of the hyperbolic
^
functions. They should be verified.
cosh r, any positive value not less
sinh r can have any value
tanh r, any value
than 1 sech r, any positive value not exceeding 1
value numerically greater than 1
ctnh
1
any
numerically less than
we have
csch r, any value except zero. Also, from the definitions,
; ; ; ; /', ; sinh ( r) = cosh ( n = tanh ( (3)  = r) sinh cosh csch ( r, sech ( r, tanh = r) = ctnh ( r, Given 1anh JT II.USTRATIVK EXAMPLE. r) sech =  v, v, ctnh v. Find the values of the other \. 2
by cosh r) csch .r. hyperbolic functions.
Solution. In () divide each term Then we got 1 ^" Therefore
Since 1 tanhjr= ?,
Then  >r cosh*' J* tanh .r ^ cosh* x  this equation gives sech sech 2
JT = By x. J, (C) and (1) the negative value being in admissible. cosh jr sinh j ctnh x =
=
= \
sech x bv (C) By (1) 3 cosh x tanh 3
J, (*) by =~> arid csch jr =
= ?
'. 4 and tangent.
212. Table of values ot the hyperbolic sine, cosine,
the values, to four significant figures, of
table giving Graphs. A
sinh r, cosh ?>, tanh r for values of For negative values of r, r from use the relations to 5.9
(3;, is shown on Art. 211. p. 416. 416 DIFFERENTIAL AND INTEGRAL CALCULUS
HYPERBOLIC FUNCTIONS HYPERBOLIC FUNCTIONS
When r * + oo, sinh and cosh r become r approaches unity as a limit.
Graphs of sinh x, cosh x, and tanh drawn by making use a* 417
while tanh infinite, (Figs. 1, v are easily 2, 3) of the table. (0,1) ZlL
(O.l) FIG. Fie. 1 tanh x ,'J 213. Hyperbolic functions of v f w. Formulas for hyperbolic functions, corresponding to two of (4), p. 3, are + w) = sinh v cosh w + cosh v sinh w,
+ w) ~ cosh cosh w + sinh sinh
From the definition (A), replacing by r + w, we have (D) sinh () cosh (v Proof of (/)). (i; r ) sinh (2) cosh (1 ( +w  (1), ~~  } of (1) is transformed as follows, making Art. 210.
e'e e 11 (cosh e (> 2 2
?? > w) (v f The righthand member
use of i#. i> i; + sinh ?>)(cosh w + sinhw) (cosh sinh ?>j(eosh7^sinh ?* Multiplying out and reducing, we get (D). Formula (E)
in same way.
we set w = v the
If in (D) and (E), (3) sinh 2 # (4) cosh 2 t? we have = 2 sinh v cosh
= cosh 2 + sinh v, ?; 2 v. is proved DIFFERENTIAL AND INTEGRAL CALCULUS 418 These are analogous to the formulas for sin 2 x and cos 2 x, reFrom (B) and (4), we get results which
spectively, of (5), p. 3.
j
to the formulas for sin x and cos x in (5), p. 3. These are
correspond = sinh 2 v (5; cosh 2  r cosh \, 1' r cosh 2 .V 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 tanh (6) cosh 2 From Solution. (5), by we division, cosh 2 Now
By (B), cosh 2 v 7' 1 ~ ~] + cosh 1 r  r  r "2 i (cosh 2 1 ('7,< i 1 cosh 2 1 M Hence r. 1 r h 2 cosh 2 sinh* 2 1 get en; cosh 2 f rush 2 tanh' r  (7) ?' ?> 1 ?> /' 1 f j' becomes ;.,i, >  .,1 tanh'r (8) (cosh 2 tanh and therefore sinh 2  i r + /' 4 >sh 2 r The sign
we have 2 r ~~ 2 sinh r f and tanh , ,, , , , tanh r will always a^ree Hence the positive sign must be
replaced by \ r, (T>) becomes = tanh (9) be examined. From (3) r , cosh r. r always positive.
If r is . 1 cosli r , cosh Therefore sinh 2 1 , . smh )' member must now before the righthand
, 1 r sinh cosh r in Also cosh 2 sign. used, and we get r f 1 is (f>>. r
f 1 PROBLEMS
F"ind the values of
1. The value of one hyperbolic function is given.
the others and check as far as possible by the table on page 416. cosh JT (b) csch .r (a) (c) 0.75. sinh (d) 1.25. ctnh s .r = 10.
= 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
3. ctnh 2 sinh (r cosh (r = csch
= sinh r cosh w
w] = cosh r cosh
r 2 r. ?r ) ir cosh r sinh ir, sinh r sinh ?r. HYPERBOLIC FUNCTIONS
4. tanh tanh r
tanh w
tanh r tanh w w) (r 419 1
, r
sinh . 5. i = 7. 8. sinh r f sinh r f cosh tanh J (r w) Show = sinh r sinh A/ f r f r f cosh 1 1> w),
/r). // cosh = /cosh = , 2 sinh \(r 4 /r) cosh \(r
2 cosh \(v f w) cosh J(r // /r r
cosh  1 A/ cosh 6. r /cosh . /r that the equation of the catenary (figure, p. 532) may be i' written a cosh ?/ a
9. Solve the differential equation terms of hyperbolic in // u.r functions, given that = // 3 when and 0, .r when tanh x //
i /l>/,s 10. Show lim sech sech = s) ( sech // rr Draw s. cosh ;{ the J. 3.75 sinh .r jr. and ^rapli prove prove 0. .r oc j 11. Show lim ctnh
X* i ( that csch (  ctnh .r) .r)  Draw the graph arid Draw .r. the graj)h and prove QC 12. > ctnh that
1. .r Show lim csch s
jr that . = csch .r. 0. ex, 13. Prove (a) sinh 3 (b)
14. Show i< cosh X that (sinh f .r = 3 sinh // coshr // 4 cosh .r)" f 4 sinh' 4 sinh ns u cosh 15 // 5 ; //. cosh tijr. (n any positive integer.)
15. Prove that sinh 16 simplify
. 17. C0 " + TOS h * // f ,
sinh 4 O f //) O
*. " ;
sinh 2 sinh sinh A sinh j y ./ //). r Parametric equations for the = S t fmctrijr a tanh a ?/
' ~ may ctnh (u + 2 v). be written a sech a 4. (The
is /, and a is a constant. Plot the curve when a
the curve for which the length of the tangent (Art. 43) is con The parameter
tractrix is stant and equal to
18. Solve =
ax 2 a. Figure in Chapter n 2 (y XXV !.)
7 Ans. y =A cosh nx f B 2
 sinh nx 4 rax 2 H tfi, DIFFERENTIAL AND INTEGRAL CALCULUS 420 The formulas, 214. Derivatives. in which v is a function of x, are as follows. XXVII 2 sinh v
dx = cosh XXVIII 4 cosh
dx = XXIX 4 tanh
dx XXX ^
dx sinh v   dx  z; sech 2 y dx csch 2 f ctnh v dx XXXI v v dx  sech v = sech v tanh v dx XXXII Proof of XXVII. rru
1 csch v ctnh csch v ' hen 7 dx liy 01), sinh u
sinh ?'  r i; dx ^ = ( ^ ~
2 lJL~ '' = COSH
i /' ; > dx ^1).
is proved in a similar manner.
The proof of
analogous to that given in Art. 72 foi the derivative of
To prove XXX XXXII, differentiate the forms as given in (1), Formula XXVIII XXIX
tan r. is Art. 211. The details are left as exercises. 215. Relations to the equilateral hyperbola. x (1) = a cosh are parametric equations arc X L> __  y = = a sinh a 2( c os h. i. At PI, ? = ri. which r y 2 = a2 . by squaring and subtracting, we r
v ~ sinh'' r) Fig. 2, p. 421, shows a hyperbolic sector
A PI of (1), the semitransverse axis 0.4, or for the equilateral hyperbola x is For, eliminating the parameter have y r, The curve = By (B) OA P\ bounded by the a~. and the radius vector HYPERBOLIC FUNCTIONS
Theorem.
Let the hyperbolic sector OAI*i equals A ai. be the polar coordinates of any point on the arc
Then the element of area is (Art. 159) dA = J p dQ. Proof. A PL The area of 421 (p, 6) Or ,,?/,) FIG. But = x 2 + y'2 = p~ Also by 1 + sinh a (cosh Using (1) By r). (1) (5), p. 4, = = arc tan (j[d Therefore sechr _ dv arc tan (tanh 1 v + tanh v XXII, p. 87, and By
Using (C) and (1), Art. 211, we dA i a + sinh v v r/?;. The theorem follows by integration, since =
The parametric equations of the circle in Fig. at /? x r cos XXIX get cosh 2 and therefore r). y /, = r sin The parameter equals t\ at PI, and
angle AOP\ in radians. Hence the area
/ /j is 1 Q.E.D. ^4. are /. Art. 81 the measure of the central of the circular sector AOP\ is \ rfi. Let r = a = x
In Fig. 2, for x Then 1. = cos P(x, t, in Fig. 1, for P(x, y), y = sin J , t = area ^0/\ ?/), = 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. DIFFERENTIAL AND INTEGRAL CALCULUS 422 PROBLEMS
1. Show that the element of length of arc for the catenary z/ = a cosh  a j. is given by
2. cosh  ds. J,x Problem In the catenary of prove that the radius of curvature 1 2 i equals y 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 4. cosh x = 1 f : f V f f h Aws. All values A NX. . All values. A \~ Verify the following expansions, using the series in Problems
in Art. 195. 15 and 4 and the methods explained
5. sech 6. tanh 7. .r .r = = 1 .r' .1 f x'< jr .', .;:, + A  jr* r :> Test the function 5 cosh .r / + (> 7 ,'
7
.r, ^ :, 7 + sinh f 4 .  for .r values.
8. maximum or minimum
Minimum value, .4 NX. Test the function values. NX. .4 .4 sinh x > A2
minimum If /* a cosh f /> maximum a , value f for .r maximum value \/ />*'' .'5. \//> .4* if Derive the series in Problems 15 and 4 from the
by subtraction and addition. (Use (^1) and Art. 195.)
9. B> or
,4 J minimum
if B < 0; 0. series for c j and c T ~ radius
~
J
f ir
length of the element of arc let p
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
10. Let vector of (a) 11. (/x VV ;  ^= for the circle /, ; (b) J p Prove lim (cosh
j 12. ~ /'(.r, f .r sinh .r) = / J p ?', for the hyperbola. 0. oc Evaluate each
f of the following
^ (a) v
lim indeterminate forms. sinh x /u\ v
(b) hm tanh jr An*, 13. Given tan = sinh x. Prove dx = sech x.  HYPERBOLIC FUNCTIONS
14. Derive the expansion
arc tan (sinh x} by
in J .r< f .r x* ^4 Tj integration as in Art. 196, using the result Problem x J f yj 5. Prove the following theorems 15. 423 for the tractrix (see figure)
jr = t a tanh 1 y = a sech  (a) The parameter
equals the intercept of
the tangent on the .raxis.
(b) The constant a equals the length of the
/ tangent (Art. 43).
(c) The evolute
(d ) The is the catenary radius of curvature fi =a cosh
a 216. Inverse hyperbolic functions.
(1
is y = sinh v ) /
 PC is a sinh = The sinh M X relation v also written
(2) l y, " Therefore
v equals the inverse hyperbolic sine of y."
and read
sinh v and sinhr y are inverse functions (Art. 39). The same notation and nomenclature are used for the other inverse hyperbolic
l functions, cosh ' v (" inverse hyperbolic cosine of v "), etc. The curves
y (3) = sinh shown again on page are y x, 424. = cosh x, y = tanh x Assume now that y is given. have any value, positive or negative, and then 1, y may
the value of x is uniquely determined. In Fig. may have any positive value not less than 1. When
x has two values equal numerically and differing in sign.
y
In Fig. 3, y may have any value numerically less than 1, and then
the value of x is uniquely determined.
in > Fig. 2, y 1, Summarized, the results are The function sinh" 1 v is = uniquely determined for any value of v. ! sinh~ v.
Also sinh~ ( v)
The function cosh" v, when
1
in sign. Also cosh" 1 = 0.
l 1 The function tanh"
tanh'K 0) = tann" 1 v is
1 v. v > 1, has two values differing only uniquely determined when v< 1. Also DIFFERENTIAL AND INTEGRAL CALCULUS 424 FIG. FIG. 2 1 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 cosh l tanh (O l 1 x
x  ln(jc l). i In (x = x = In I
/ 2 N// sinh v To solve (4) for write v,  <"' 2 = JT it Then .r. sinh (4) l). y
A 1
Jl Proof of (F). Let (Any x) By r as follows : or 0, r a quadratic equation in e v Solving, r r = x
Vj>' + 1.
Since r is always positive, the negative sign before the radical
must be discarded. Hence, using Napierian logarithms, we have (F). This is . 1 Proof of (G). Let
(5) v =
jc cosh = Clearing and reducing,
Solving, c r l cosh Then x. /? = we have = x c  2 ^
'' Vx ~ Both values must be retained. 2 2 By
arc 1 ' + 10. 1. Taking logarithms gives (G). HYPERBOLIC FUNCTIONS
Let Proof of (//) = r = j (6) tanh l Then *. = tanhr 425 By =^ Clearing of fractions and simplifying, the result
(x \)e' + + (jc = ' l)e Taking logarithms, we have 0. Hence (C) is = l+x r t' 1z (//). ILLUSTRATIVK EXAMPLE. Transform
5 cosh (7) into the form (" (7) will hav^e .r we have
T cosh .r the desired form C (9) 4 sinh f where r and (jr f a), By (), Art. 213,
C cosh (.r + a) Solution.
(8) Hence cosh cosh a  if 5, jc cosh a sinh f (' T and a satisfy
C sinh a
4. j* Therefore a = 5 cosh The graph tanh .r f 4 sinh i a. sinh a. ("'
/,. = Then C 9. = 4 3, Hence By In 9. (//) Example ~~
JT 3 cosh (JT f 1.099). be obtained from the graph
to the new origin (1.099, 0). (Compare with of the function 5 cosh x by translating the Illustrative  0.8 ' C and 1.099 and (10) of 5 cosh = find the equations Squaring, subtracting, and using (/?), Art. 210, we get
must he positive. Also, by division, tanh a
since cosh
H and u are constants, .r //axis 4 sinh j .r may 2, p. 391.) jc can
.r, cosh
j, or Umh
given, the values of sinh
be determined by the table on page 41 (> to not more than three signifi0.25
0.247 cosh
IJ
1.76.
cant figures. For example, sinh When x = ' For greater accuracy (F), (G), or
Napierian logarithms are at hand.*
217. Derivatives (continued). (//) ! = ' ; may be The formulas, of x, are as follows. XXXIII l is in used which if tables of v is a function ^v  sinh 1 v  ,^
+ /i; XXXIV  cosh  1
i; 2   (Any v) > 1) 1 dv
d* = (v dv
rlY /I XXXV
cfx 1 v2 * 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 corresponding inverse functions to live significant figures may be found from these tables. DIFFERENTIAL AND INTEGRAL CALCULUS 426 Proof of XXXIII. (Compare Art. = sintr
r = sinh y.
l y then Let 75.;
v; Differentiating with respect to y, by XXVII,
dv , = cosh y ; dy ~
^ = cosh y
dv therefore Since v is a function of cosh djr [cosh The proofs of
the following. // = may x, this dx // Vsinh'' // 4 ctnh J sech 1 (7) (K) csch 1  x x be substituted vV'
vV' 1 XXXIV and XXXV (/) By
J +
f 1 1, Art. 39 " \
(C), in (A), Art. 38, d. by are similar. Other formulas are In 2 > 1) x g 1) i; < 1) > 0) (x l = ln (0 < (0 < ^!
\A _dv
rlv ft XXXVI , _
XXXVII sech XXXVIII csch~i i; t; = * = i  = Details of the proofs are called for in Problems page. (^ 58 2 on the next 427 HYPERBOLIC FUNCTIONS
PROBLEMS
1. 2. of // Show two values Draw and //' that the the graph of
when s = 2.  ij of cosh sinh \ ' .r in (G) differ only in sign. Check > jr. 3. Prove XXXIII directly by differentiating 4. Draw // 0.72, //' = 0.21236. (f). 2. l (bl = tj the graph of each of the following and check in the figure
the given value of jr.
//' for the values of ?/ and
JT
la) ?/ = cosh = the figure the values in An*, tanh ./ ; ! = .r .r;  O.Tfi. XXXIV and XXXV. 5. Prove 6. Derive 7. Derive (/) and XXXVII. 8. Derive (K) and XXXVIII. 9. Derive the expansion and XXXVI. (7) tanh ' j .' + : } by Art.  + ;> .{ 19 r>. 10. (riven sinh = tan .r In (sec (a) .r 11. Show f </'  (b) ; / sec 0. ^ r = ' 'rove
, tan 0) csch that I c/>. Derive XXXVIII from XXXIII, 1 ' sinh using this relation.
12. Evaluate lim 13. ctnh ' Evaluate lim s csch
J 14. j An * f.  j\ Derive the expansion = '' rl  .'J + i Show that ctnh d
 tanh
, 18. Prove Draw f  = tanh ""
^w.s. In .0. ' r ' In '2 of sech y ' r ^ cosh J . and verify from these relations.
. ^ tanh a tan r
fl + se( Problem _
~ . j ,. the graphs off a) ?/= ctnh using the theorem ^lf X XXXVI and XXXVII
17. ' Evaluate lim fsinh
4 16.  X sinh 15. 1 X J J + s'nb a
cosh a cos r r; rb)?y 28, p. 41. = sech ' j; (c) ?y=c8ch" 1 A DIFFERENTIAL AND INTEGRAL CALCULUS 428 "
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
and uniform linear leakage. P A p ^ j B f insulation
is any intermediate consider point. We need to : 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. = A P. Then is shown in
E and 7 are functions of x such Let x
that We They are books on it positive numbers. electrical engineering that wish to find the e.m.f. and current strength at P. E= (8) 7 (4) Proof. E.\ 7,1 The complete cosh ax
cosh r () 7.i ax  solution of They are sinh ax, sinh ax. (1) is (Illustrative Example, Art. 210) E (5; Substituting in r )7 (0) But (2),
( E ~ E i, 7 = JA A cosh ax
the result A is sinh when x + B sinh ax. ax 0. B cosh ax. Therefore A ~ EA 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,"
1. In = and L AB. Given E A = 200 volts, L = 500 kilometers, r  4000 ohms, a = 0.0025,
0. Find I A and En.
Ans. 1 = 0.05 tanh 1.25 = 0.04238 ampere
1 EB = ; 200 sech 1.25 = 105.8 volts = 0.53 EA. HYPERBOLIC FUNCTIONS
2. If y = PB = distance E = EB cosh + ay P from of r//? sinh 429 the receiving end, show that
I fi / a//, cosh ay + sinh ay.
ro 3. Show Given EA = 200 volts, IA = 0.04 ampere, r> = 4000 ohms, a = 0.0025. that E=  120 cosh (1.099 0.0025 = I jr}, 0.03 sinh (1.099  0.0025 x). (See the Illustrative Example, Art. 216. Thus A tends towards a
/ approaches
as .r approaches 439.6.)
1 minimum value of 120 volts and
4. Show Given E.\  1GO volts, 7,i = 0.05 ampere, rn = 4000 ohms, cv = 0.0025, 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.)
5. Prove that 2
d' I
pr,
2 = a*l dx linear differential equation,
6. Given E,\ = (b)
(c) the e.m.f. is K=
E=
E which has the form are solutions of the is P //" a' y 0.) , I = when .r becomes EA c
r 7
* af * IAC ; ; infinite. indefinitely increased.) show that the decrease in E at P at unit distance along
equals Ec~" where e is the Napierian base.
t Prove the following.
(a) If IK (b) If En  0,
= 0, then
then E\  r,,/.i EA = r J A
( ctnh aL. tanh <*L. ADDITIONAL PROBLEMS
Derive the following relations.
1. If EA > rn l A E = EA
2. If same 2 In Problem 6 the line from
8. / TO = 4000, as the length of the line
7. and if
and the impressed e.m.f. at the home end
4000 times the current strength, then at mr// point of the line
4000 times the current strength and diminishes towards zero (Tor example,
of the line is /; Prove ru /.4. (a) (Thus 0. EA < and T = tanh ^~A then EA sech r cosh (r rQ I A 1 and T I ax), = tanh^ ^X
1 I A csch T sinh (r ax). then ro/A E = EA csch T sinh (T ax), I = IA sech r cosh (r ax). DIFFERENTIAL AND INTEGRAL CALCULUS 430 A 219. Integrals. of list elementary integrals involving hyperbolic functions and supplementing Art. 128 cosh v + C. vdv= sinh v + C. I sinh v dv (25) / cosh (27) vdv=la ctnh I I sech 2 (28) = In cosh v + tanh v dv j given here. = (24) (26) is vdv = v+C. tanh vdv=^~ v+C. sinh (29) I csch 2 (30) I sech v tanh v dv (31) I csch v ctnh v dv The
(26) ctnh v+C. = sech v+C. csch v+C. proofs follow immediately from and To prove (27). f tanh v dv  C^LH The proof of (27) is by
J (c)
\
/ ?' r) f(/(cosh i cosh J for rf r cosh J XXVIIXXXH, except we have (26), J C. _ , , n eos h . ?, _j_ ^
_, /' similar. ILLUSTRATIVE FA AMPLE. Derive the formulas we have v dv = arc tan (sinh P dr = In I (2) Solution. sech fcsch (1) Since sech fsech J r r dr + tanh r) f C; C. = J = ^LL = cosh r
cosh r
cosh r
lfsmh = f
J =J ^^ df 1 f sinh r  f ^ sinh,f)
J 1 f sinh 2 d[arc tan (sinh f)] = 2 J r z; arc tan (sinh v) + C. v 7 HYPERBOLIC FUNCTIONS
To derive (2) we have (compare = , csch r csch , csch r r csch r 4 / r csch r ctnh r r dr I ctnh
'(/(ctnh r = f = ctnh 4 csch r 4 v 4 r M csch csch r In (ctnh r 4 csch r) 4 = _l n /ilU +
In + tanh In 1 ) 4 (' +r sinh rl r f cosh ( .__L_\ r \suin = r r ctnh f csch r 4 csch 2 = dv 4 ; ctnh
csch ctnh
ctnh ~\ ? csch / Art. 131) r csch 431 In sinh b v (n
r f (" ^ In cosh r. f
4 r By PROBLEMS
Work out the following integrals.
r dr ~ \ sinh Tcosh L r dr = 4 sinh J tanh'J r dv = r tanh r 4 C. ctnh 2 r dr ~ r ctnh r 4 r. 1. fsinh" 2. ' 3. r 4 r. r 4 i r 4 C. r 'J \ 4  4.
 sinh 5. :< r = dr cosh :i sinh :? 7 cosh 1 ?' 4 C. 4 r.  6. (cosh :? r d? 1 = fj r 4 tanh 3 7. v dv In cosh r tanh 4 r dr r r dr J  8.
I 9. H Jcsch tanh = I csch sinh v r tanh 2
> T ctnh /I
x sinh x dx 10. 1 7 tanh
?'  sinh x x cosh r r. ?' 4 :< r 4 \ In r. tanh ~ 4 C. 4 C.  1 1. fcos x sinh x dx = I sinh (rax) sinh (ns) (cos x cosh r 4 sin = ra~ x sinh r) 4 C. r fw sinh (nx) cosh (rax)
nz
n cosh (nx) sinh (rax)] + C. Art. 211 ( 1 (9), Art. 213 DIFFERENTIAL AND INTEGRAL CALCULUS 432 Work out each of the following integrals.
13. J sinh 4 x dx.
4 14. 15. 2 x dx. 16 J^sech
18. j Work
dicated.
20. J r / x 2 cosh x dx. 17. Ann x cosh x JV sinh a tanh x dx. djr ' dx. 19. /V" cosh x dx. out each of the following, using the hyperbolic substitution in (Compare
Vx'2 Art. 135.)  4dx; / = 2 cosh r. Am. \ x^/x'2  4  2 cosh ' x + C. /"/^TaTTs^" 2 "
23. The about the arc of the catenary
//axis. 1 // = a cosh  from (0, a) to (x, y) is revolved 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.
 __ 2 sinh r, .
12, p. 337.)
2
cosh r,  1
~
/r AtiS. .r
en
(I (Compare Problem y 220. Integrals derive integrals.
Their values are
functions. ?'i . (j, * '> 3 ?i From XXXIII XXXVIII we may
(continued).
Some of them we have already met in Art. 128.
now expressible in terms of inverse hyperbolic HYPERBOLIC FUNCTIONS
_____ Vi>2 /_ 2 + a2 dv = V7 + a2 +  sinh2 f V>  (39) 433 a2 = Vi^~^ _
* rfi; j f*_ 1 cosh i <&  a + C.
+c In (33) and (39) the positive value of the inverse hyj>erbolic
must be used, and in (36) the positive value of the inverse cosine hyperbolic secant. = Proofs of (32) and (33). Let x sinh  1 =  In a + \a \ u,~  in
(F). +1 In (r Then + vV + <7~) In a. i Hence
In (v (1) In the + + a~) = Vr same way, from (G) we + In (v (2)  Vr a sinh ! ' + In a. 4 In a. get  cosh 1
') '  Using these results in the righthand member of
get (32) and (21), p. 193, we (33). Proofs of (34) and (35). Let x
1 /r^ 1 2  in
(#). + =
^T7
ft ln (3; = /; i Then ?' i , 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 I ^
r = +  In ^t> u I J Proof of (36). Since
d (} by XXXVII
v~ we have
radical I J is / v Va =
2 chosen. r  sech
a The proof  the ijositive sign before the if a of (37) is Formulas (38) and (39) follow from similar.
(23), p. 193, using (1) and (2). REMARK. Since ,,,# = tanh" 1,0 ctnh" 1 a 1 v 1,0 = seen' a
1 i,^
cosh" 1 v , csch , l v
 a .,.
= smh^ 1  > v the integrals (35) (37) may also be expressed in terms of the functions most convenient for use of the table of Art. 212. DIFFERENTIAL AND INTEGRAL CALCULUS 434 ILLUSTRATIVE EXAMPLE. Derive (37) by means
Solution. We a csch z. have + iv 2  Va~ csch z + a = a
= a csch z ctnh z dz. a'2 2 dr Also r t Therefore v csch" 2 = C dv
r / J
Since of the substitution v Art. 135.) (Compare 1  a* f J we have 1 : a (>sch z ctnh ' (2), Art. 211 By XXXII a csch z 01
2 ctnh z dz I vV By ctrih z. a z z , f ~ C. (37). PROBLEMS
In the figure the curve 1. 2 eral hyperbola
prove that
area (a) A MJ (b) sector
if x > OA a cosh = 2 .r' y' triangle I>  is OMP 2
\ a cosh the equilat Using Art. 142, a. l   2 cosh a' = ' ; 2 a' r, (Thus we have an alternative ?'. proof of the theorem of Art. 215.) Derive each of the following power 2. series by integration (a) = tanh jr (Art. 19G).
f ~ .r< f <> (b) sinh Work 2 .r r> 4 .) ;* 24 5 out the following integrals. 3. fsinh 4. Mann Work  l JT ' .r djr JT sinh ' .r VI f 4 .r' C. dr. 5. (.rcosh" 1 ^ out, using hyperbolic functions. 'farrb9. Find the length of arc for the parabola .r 2 = Ans. using hyperbolic functions. 4 y V20 + Find the area bounded by the catenary y
2a. 10. y = a = 11. 32 The downward acceleration a
 I r'2 and r = 0, s = 0, when
, / a falling of = A nx. 0. r from (0, sinh 0) to (4, 4),
J 2 = 5.92. a cosh  and the line body Find r and
= 8 tanh 4 /, is given by ,s. ,s = 2 In cosh 4 /. HYPERBOLIC FUNCTIONS
The gudermannian. The function arc tan 221. frequently in mathematics Art. 219), called the gd r is " (read gudermannian* of
r gd The derivative Proof. of v = arc i. tan (sinh dv
dx . sech v to be a function of we Differentiating (1), r). But cosh + 1 r By XXII and XXVII sinh' v + sinh = cosh 1 > get dv
r by (B) v, 1 and sech cosh r the definition (1 v and Art. ) r 77, gd gd(0)=0;gd (r)=z<lt>; value (1), dr gd dx When
Its sech By r. Art. 211 v d Then (2) is jr. d From Example, The symbol used Thus "). 4 gd r which occurs v), is XXXIX
assuming (sinh example, in (1), Illustrative (for gudermannian (1) 435 ( 38 (A), Art. By we have
f oo )= .1 TT ; gd (00)= \ TT. gd increases (since sech r> 0).
Kbetween
J TT and + i increases,
lies /' values are given in the accompanying table.
By (1), Art. 219,
(40) To I = gdv+ vdv cech C. find the inverre function (Art. 39) let = arc (3) and solve for The r. v (4) From (3) tan (sinh = v), result sinh" l <t> < 7T) is (tan we have tan ^ TT i ( </>). sinh (p ?/. 2
Since cosh by (B), the trigonometric functions of <f>, when
off from the accompanying right triangle. Thus The
(5) sin </> = tanh v, cos inverse function
?; = In (sec = j (4) + sech may v, v > v = 0, 1 + sinh v ^^^ tan 0). Named after the mathematician Gudermann. His papers were published
tfd
'</>).
'0 is used by some writers (r
t The symbol ^d * w, can be read etc. be written J sinh' in 1830. DIFFERENTIAL AND INTEGRAL CALCULUS 436 Replace x Proof.
1 f tan 2 in Conversely, given
Proof. + = gd tan (/> = e r becomes tan or , e cf> r Squaring both members, substituting sec 2
reducing, the result Solving for tan is </>,  2 tan +e =
2" ^' </> we get
tan Hence /> '* =
</> sec = ! + tan 2 0, and By W) v. 2
arc tan (sinh </>. 1. = sinh  (j> ILLUSTRATIVE EXAMPLE. ~ />'' = </> and note that <, ?;. to exponentials, (5) Changing
sec by (2), p. 2. then (5), by tan Art. 216, (f), 2
equal to sec 0, 4> is gd v) v. Q.E.D. In the tractrix let
a the xaxis = PT the tangent of length (constant by definition)
/ intercept of the ; tangent on ; between the tangent angle 00 upward and the
when = 0. Then #((), a) is line directed //axis; / on the curve. To prove
(6) Proof.
values of = <f> When T'X f is given,
is determined.
Hence
is a function of t.
Let the
and
for the tangent line at Q a
point near P, be, respectively,
V
4 A/(=O7 ) and
4 A</>. Draw 77 perpendicular to QT '. 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.
/
</> t t f 1 r </> ; Therefore But angle
Let (? = TS sin 7VS7
+ A</>, angle TSV = A0, rr = A/.
A/ cos (0 4 A0) = rs sin A0.
TT' ITT =
1' move along finitesimals. 98, </> = a d<f>, and remembering that
 Therefore, T7 by 1; T . (5), or 4> In (sec cf =
</> f . a Hence 0. Then A/ and A</> are inHence, by the Replacement let Ac/> S approaches P and TS
and (5), Art. 68, we have
dt cos Integrating, T [ the curve towards r, and Also Theorem, Art. </> cos = sec d> dd>
^ v when
tan / = 0, we get </>). Q.E.D. HYPERBOLIC FUNCTIONS 437 PROBLEMS
The shows the 2 circle .r f //1
and equilateral hyperbola
quadrant. From M, the foot of the ordinate
of any point P on the hyperbola, draw AIT
Y>
tangent to the circle. Let v = area of the hyper1. x2 = 1 y' figure
in 1 OAP bolic sector Prove </> = the gd MP first = angle AOT. Prove 2. gd (a) (b) I =2 v arc tan e v sinh r tanh v dr Draw 3. and late y y' TT; = sinh v gd = # 0.8ti, In the Illustrative Example,
prove that 4. r f C. the graph of y = gd .r. Calcuwhen x = 1. See figure. Am.
P and ^ (Art. 215), v. ?/0.65.
if p. 43f>, is (x, #), x From = a sin 0, ?/ = a cos </>. these and (6) derive the parametric equations = x a tanh f * ?/
' a
for the tractrix.
5. a sech a Find the rectangular equation Derive f sech 3 6. If = dv v  sech v tanh also. v+JgdrfC. the length of the tangent of a curve (Art. 43)  (a) dy
1
prove r
dx (b) Integrate by the hyperbolic substitution y / x dition x = tractrix in
 7. = Va  when = / Evaluate each constant a), y* and 0, a sech  and the con derive the equations of the way in this = 4. of the following K
gd :r by differentiation. x
(b) lim gd sin JT x X.0 r>0 (a)
8. ( . 2 Problem is  i; (b) ^. Using the expansion of Problem 14, Art. 215, we have
3
.
x*  sU?) x 7 f
gd x = x  \ x f & Calculate the value of gd 0.5 to four places of decimals.
9. The equation (5), p. v Prove this statement, 435, may = In tan making Am. be written
(\ TT + J use of (2), (4), (/>). and (5), pp. 2, 3, 0.4804. DIFFERENTIAL AND INTEGRAL CALCULUS 438 222. 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
latitude
0j
A0,
longitude + + A</>, A r second point,
near P\ on the curve P\QV.
The meridians and parallels
through PI and Q are shown.
They form the quadrilateral
0j is We PiSQR. a seek for the circular arcs North Pole expressions
7'j R. RQ and 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/>. r is the center of arc PiR,
with central angle A6. Hence
arc 7'i/t = CI'i A0. But, in the right triangle
at ('), ^P\ = a cos (/>,. Hence ^ Also, arc The line 7Ve' is = 7' 7?
1 OPjT (right angle a cos 0i A0. tangent at 7^ to the parallel PiK. The line l\T 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
is tan (2) i T= sec fa ^ the value of the derivative being found from the equation 0=/0 (3) satisfied by the latitude and longitude of each point of the curve PiQV.
Proof of shown f (2). By the Replacement Theorem, Art. 98, Substituting the values from 1 T = (1), arc can be KQ A0o arc tan R'P (4) Q it that P\R lim we get (2). * Defined as in Art. 28 as the
limiting position of the secant through Pi and
approaches P\ along the curve P\Q.
Note that arc
t The details are indicated in the Additional Problems (p. 143) arc P\R triangle are, respectively, opposite PiRQ. and adjacent to the angle at Pi Q when RQ and in the curvilinear HYPERBOLIC FUNCTIONS
On 439 Mercator's* Chart of the earth's surface the point with
<t>, longitude 6, is represented by the point (x, y) such that
x = 0, y = In (sec </> + tan 0), latitude
(5) or, inversely, = (6) In (5) and (6), 6 and 4> = ar t By gdy. are expressed in radians. Meridians Art. 221
(0 = con stant) are represented on the chart by lines parallel to the ?/axis,
constant) by lines parallel to the jaxis. The curve
parallels (<f> = given by the parametric equations (3) is = A(/>), x (7) y = In (sec f c/> Theorem. The angle between a curve on
ing parallel is unchanged by mapping.
Proof. Let becomes the the curve (7) From (7) be the point on
on the chart. (x\, //i) ~ line y
is y\ tan c/>). the sphere and an intersect = c/>i. The parallel
where
Hence we have to prove that (7) c/> such that and we (3),
(\H get
, ^sec</>, dx f, , . , (W _=/(*):,_. Then Q.E.D.
(8) follows from (4), Art, 81, and (C), Art, 39.
important corollaries follow.
Cor. /. The angle at 7^ on the sphere formed by two curves,
P\QR and P\Q'R will equal the angle on the chart at (x\, y\)
formed by the corresponding curves. Hence angles remain un Two r , changed by mapping. A Cor. 77.
straight line on the chart with slope tan
to a curve on the sphere cutting all parallels under the
This curve is called a rhumb line (or loxodrome). Along a rhumb a corresponds
same angle a. line = gd(0tan + &).
and y = x tan a + b. The course of a ship
(6)
proceeding always in the same direction lies along a rhumb line. In
K
the representation (5), 0, and therefore x, has values from
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
TT and x = + IT.
r?/plane determined by the lines x =
(9) < This follows from * Grrardus World in lf>69 M creator
His name fl is 5121 594), a noted cartographer, published his Chart of tho
the Latinized form of Gerhard Kremer. DIFFERENTIAL AND INTEGRAL CALCULUS 440 By the table of Art. 221 we may which are given on the chart by the 223. Relations find the latitude in degrees of the parallels lines y = constant. between trigonometric and hyperbolic functions. Let
be a complex number
Then we assume as a
real numbers; /
l).
' the exponent x f /// and (JT of the exponential function e v 1 V // definition
(1 If c ) x = j <" * we have 0, = to Change y cos Then y. (>'" (3) Solving (2) and
sin (4) Thus the sine y (''(cos + i sin y). (sec p. 391)
e'" (2) = (>*(>' = '// (2) cos y = n< z and cosine becomes
/ // (3) for sin
c + i sin y.
sin ?/. y and cos y, the results are '" c cos "
y 2 / of a real variable are expressed in of exponential functions with terms imaginary exponents. and (A) suggest definitions of the functions concerned when the variable is any complex number z. These definitions
Formulas (4) are
sin z = sinh z = (5) *
It is and ' cos z cosh = (>1Z 4_ W z and j = f TT represent the same meridian (180 W. or 180 E.).
meridian does not cross the curvilinear triangle. In the figure, A\
represent the same point on the earth, us do also B\ and C. The lines .r =  assumed that B ** (>'' this TT 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
: sinh (L) iz = =
[fit
sinh From by (L), e it we (6) sin i using z, (5) z.
etc. ; 1
I iz = tan i z. formulas in this chapter to others for
explained by the relations (L) and (6)
The right hand members of (f>) are
2). many similarity of trigonometric functions is Example
numbers whose (see Illustrative cos iz get tanh The = iz division, cosh sin 2, i _ involve only trigo expressible as complex real parts nometric and hyperbolic functions of
below in Illustrative Example 1. real variables. ILLUSTRATIVE EXAMPLE
sinh (7) By Solution. // (5), if sinh (x (8) + Derive the formula 1. + iy) = sinh x cos
z  x + iy, we have
r iv
* "/ _
c
c (jc = ^(cos y +
(1), Art. 210,
c if r i v = x, we
cosh x + sin i ?y)  sinh (x The form c~ T sinh x, ILLUSTRATIVE EXAMPLE The Solution. The
Let first = i sin v. + cosh x The i sin y) i // details are the same cosh 2 z  sinh result cosh here and in Prove directly by 2=1, relation z iv. sinh iv 2. member cos 2 sin 2 z
, sinh x cos iy} of the righthand  f*(ros y By ^ (5) j (7) is (7). sin y. should be noticed. the relations sinh' 2=1. as in the proof of (#), Art. 210. be derived from the second as follows:
cosh 2 iv  sinh 2 iv = 1.
But, by (L), cos 2 v f ) x. may Then
Hence (1 have Substitute these values in (9), and reduce.
?, (7) becomes
Changing i to ' cosh x sin y. TJ (9) By f * = iy} This appears sin 2 v cosh iv  cos r, 1. PROBLEMS M creator's Chart
show that the distance apart on
to the zaxis which represent the parallels at latitudes
of the lines parallel
varies as sec c/>i.
</>i and <i f Ac/>, respectively,
1. Using 2. Along a rhumb that tan differentials, a= sec j d<t> cp f du line = gd (6 tan a + b). Prove by differentiation DEFERENTIAL AND INTEGRAL CALCULUS 442
3.
</> = = =
= fa, > " is fa iri (  (see sin fa) parallels p. 438). figure, <t>})
0i (fa
are "the corresponding parallels on the
?/, c/> ?/2 ,' bounded by the altitude h of the zone on the sphere The If map, prove the following. aCtanh h (a) tanh 7/2 ?/i ) ; i ~~ (b) 4. Using (b), (li/ Problem sec 2 0i d//, if 0;> </>i 4~ cz0. show that equal zones U, of small altitude whose lower bases are parallels at latitudes 0, 30", 45', 60, respectively, map
area of a zone equals
into rectangles whose areas are as 3:4:6: 12. (The
of a great circle.)
its altitude times the circumference
5. (V>) Describe the direction of a curve on the sphere becomes if (a)  if ; infinite. dO
6. of the following Derive each lustrative Example cosh (jr f (b) sin (x 4 ?'//) (c) cos (x f /'//) (a) From 8. cosh /'//) =
= cos y 4 jr sin x cosh ?/ cos x cosh ?/  these write the values of cosh (x
(a) Prove Evaluate each sinh (b) 7. formulas by the method used in Il 1. cosh ( ( J ? i = .r /) (c) cos (0.8 f 0.5  ?'//), ? sinh x sin y
cos x sinh y
/ sin x sinh ?/. ? ; / ; sin (x cosh x  ??/),  iy). ; ~ x = / sinh x, } two places of decimals. (b) cosh ; /') cos (x j of the following to (a) sinh (1.5 f f (1 (d) sin (0.5 ; ANK. (a) 1.15 h 1.98 i ;  ?);
+ 0.8 (c) 0.78 0  0.37 i. ADDITIONAL PROBLEMS
1. In the figure of Art. 2l?2, P\M\ drawn perpendicular to CR,
meridian NQR. Then
chord PiQ is not shown), and is and therefore perpendicular to the plane
triangle PiQMi is a right triangle (the of the M > 0, the line Pi A/i (produced) approaches
tan MiPiQ = ^^. When
Pi Mi
the tangent PiR', and angle MjPiQ approaches angle R'PiT. Therefore
(10)
v
' tan / Pir = lim^^ HYPERBOLIC FUNCTIONS
Compare with Art. 222, (4), = Pl lim
**\
Afloarc P\It (a) Afl(^ (h) lim
v oarc . KQ = I 1 and show that (see Fig. 1) ; which shows (see Fig. 2, In triangle
the plane of the meridian
show that Af K is an infinitesimal of
higher order than QR when A6 and JK/> are of
the same order (Art. 99;. Then see Problem, .V i p. 148. Using fa) and (h) and the Replacement Theorem, Art. 98, (10) becomes (4), Art. 222.
is the element of the length of arc for
2. If
a curve on the sphere of Art. 222, prove that
<W2 ). (In the figure of
dx 2 = a'((l<f>'2 4 cos'
r/.s'i ' } Art. 222, (chord I\Q} 2
dx
that 3. If show
we have = Pi MI" j
c >nord/>l(
y andlirn ?^l.) + arc PiQ M the differential of the arc of a curve on
creator's Chart,
sec 2 0u70 L f cos 2
r/.s<10~).
(Comparing with Problem 2,
is dx\~ ' = a'2 2 cos 2 (/.s .) Find the length of a rhumb line between points whose difference
Ann. a esc a A0. (a = radius of the earth.)
of latitude is A0.
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 defi4. nitions (5).)
6. Prove the formulas of Problem
Additional Problem 5 and (L). ~ 7. 8. u
n
that
Prove <^u ^ tanh
4. / (.r i 4 \ /?/) = sm h I>erive the formula for tan problem. : 6, 2 J p. ' (.r f + sm
' : cosh 2 r
it/) 442, I by using the 2 ~*?/
cos 2 ?/ results in m from the result in the preceding CHAPTER XXIII PARTIAL DIFFERENTIATION
The preceding
224. Functions of several variables. Continuity.
chapters have been devoted to applications of the calculus to funclions of one variable. We now turn to functions of more than one
independent variable. Simple examples of such functions are afforded by formulas from elementary mathematics. Thus, in the
relation for the volume v of a right circular cylinder,
(1 v ) = irjr'y, a function of the two independent variables x ( radius) and
altitude). Again, in the formula for the area u of an oblique
(
?/
plane triangle,
v is u ("2) =
, yy sin otj a function of the three independent variables r, ?/, and a, representing, respectively, two sides and the included angle.
Obviously, in (1), as well as in (2), the values which can be assigned to the variables in the righthand member are entirely independent of one another. u is 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 A variables, function f(jr, ?/). /(.r, as continuous for .r ;//) = of
a, two independent variables x and y
y ~ lim/(x,y)=/(a, 04) X and in defined b), * * no matter b, is when >> what way x and y approach their respective limits a 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
functions of a single variable. if the student again reads over Art. 17 on continuous 444 PARTIAL DIFFERENTIATION We may
represented illustrate this geometrically by considering the surface by the equation
z=f(x,y^. (3) P Consider a fixed point
//, 445 = a and y b. of the variables x and on the surface where Denote by A.r and A// the increments
and by Ac the corresponding increment :r of the function z, the coordinates of P' being At P the value of the function z=f(a,
the function If ever A.r and A?/ is may b) is = MP. continuous at P, then howapproach zero as a limit A::
, approach zero as a limit. That is, 717 'P' will approach coincidence with MP, the point P' approaching the point /' on the sur will also face from A any direction whatever. similar definition holds for a continuous function of two variables.
In what follows, only values
which a function is continuous.
225. Partial derivatives. of the variables are considered for In the relation z=/Cr, (1) more than ?/), 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. we may hold 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 of functions
Corresponding symbols are used for partial derivatives of three or more variables. In order to avoid confusion the round b adopted to indicate partial
* The \ has been generally differentiation. constant values are substituted in the function before differentiating. t Introduced by Jacobi (18041851). DIFFERENTIAL AND INTEGRAL CALCULUS 446 ILLUSTRATIVE EXAMPLE
1 Solution. Y i)"
r = 1 Find the partial derivativas of . 2 ax f 2  2 ore f fr#, z = ax + 2
2 bxy + cy 2 treating # as a constant, 2 n/, treating x as a constant. ILLUSTRATIVE EXAMPLE 2. ~ ~ a cos (ax f by 4 re), treating # and z as constants, Solution. ~
~ Find the partial derivatives sin (ax _ cos (ax 4 by + cz), c cos f by f r^), + by + cz). treating x and z as constants, treating # and x as constants. we have, (I), ~ = T^/(y, y) in the notations =
^ =/ r (j, ?/) commonly =/, Similar notations are used for functions of any
Referring to Art. 24, we shall have
*) (2) / ,(.r
; (l , y) = zx used, ; number of variables. = Mm
AJ (3) of u fr Referring to = . lirn ^^ Aj; 226. Partial derivatives interpreted geometrically.
of the surface shown z= /Or, in the figure Let the equation be y). Pass a plane EFGI1 through the
ft) on
point /' (where x ~ a and y
the surface parallel to the XOZplane. Since the equation of this
is plane , y^b* the equation of the curve of intersection JPK with the surface is 2=/Cr, b\
if . we consider EF as the axis of U
plane
(1) TT fa means the same
5 = tan
ox (] : as r1 dx ' Z and EH as the axis of X. In this and we have MTP = slope of curve of intersection JX at P. PARTIAL DIFFERENTIATION
its YOZplane, we if Similarly, BCD pass the plane
is equation j ~= tan MT'P means the same as ~ 4 ~ ellipsoid LI4 made of the curve of intersection of the ellipsoid =4 and z is positive (a by the plane j (b) ; parallel to the ~ 4 DI 1 = at P. find the slope ; (> 1 !_ by the plane > Hence j^ slope of curve of intersection ILLUSTRATIVE EXAMPLE. Given the where x P through a and for the curve of intersection DPI,
(2) 447 1 // 2 at the point at the point
H and
j/ = where z is positive. Considering y as constant, Solution. ~~ 24 When J is
(a) (b) "* ?/ ~f +
12 constant, When ?/ When x =
 1 2 and r 4, = and y ,'J, 2; 2 to ()~
rr () ~ ' to 6
4> = 1 c^/ or 0, = x/h" T. ~ <*)' = /. f2 ^ ~ * 1 y f).v    _ = 4 z ft"   *J \ 2. A/ts. PROBLEMS
Work out the following partial derivatives. 2. /(j, ? a 2 _A +
JT 3. /(JT, 4. 7< = ?y) x
AMS /?// ^": = Cr 4 ?y) An.s. sin fx //JT, i/) / y (j t T/) 6. p = sin 2 cos 3 2 MT  // 4 z  sin
= sin (r
(x ; ' wv  ?/) +
 y) ~ (r p = 8++ cos (j _ ^). + 2 ?/) n, ; ~jr cos (x + y.  ?/) (* 4 y) cos (r = An. </>. = s 4 2 cos 2 Aws. ^ = e++ fcos cos 3 ^P^ e 4c/>r cog 3 sin 2 6 sin 3 (8<j>) sin ^_ 0) + sin (fl  (^_ ; y). ^ ^=
7< f
f f:jr// /^') ?y   ?/). Ans. 2 to?/
2   /?/ f z/z 4 ZJT. 5. /(j, ?/) M/> f
f ^ ; (/>. </>)} 0^ ; e DIFFERENTIAL AND INTEGRAL CALCULUS 448 Find the partial derivatives
8. f(s, y) y Q ~0 =3  4 s*y + of the following functions. 6 jr*tf. 11. f(s, y) +^
+ 2/
?/ = fMn 10. z jr 4 = (x 12. = tan 2 13. 2. p p = < (9 + 2y) tan (2 + I/). x ctn 4 0. cos^ ^" 14. If /(V, ?/) = 2 .r*  If/O, //) =r 16. If/(r, //) = 15. = r 17. If w 4 ^1 3 jry f 4 ~jshow
'sin <' 2 f Rr 1 (JT 2 '// ?/*, show that /z (2, that/, (3, 1) = = ~ 19. If  u ^ ; ^ 4 , W~ f ry show that , show that jr
f jr f = ?/
' r'x ifz f = + Ajr
l TJ The 21. Given rV, show that ^ow that in., r  20 .4 is = \ W
4/ 0. w. f'?/ = + = ?/ f (jr y + 2 z) . (Z cy rx in., ^0, 1, ?/. f f .r f />// area of a triangle
10 b 'ty*' 3 jj. =4 ?/
' = 18. r'/y fj 20. If u = + 2 ?/), show that /,((>, T
\
4/ // jr*u f l,/v (2, 3) i,/,,(3, lj r.r 18. If M = 3)  (w 2)w. r ,; K given by the formula be sin o /4. GO". 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.
(a) /> 22. Given
(a) (b) The law
b = 10 of cosines for in., c = 15 in., any triangle A = GO is a2 = b 2 f c 2  2 be cos A. . Find a.
Find the rate of change of a with respect to b if c and A remain constant. Using the rate found (c) a if b in (b), calculate approximately the change decreased by 1 in.
Find the rate of change of a with respect to in is (d) A if b and r remain constant.
(e) Find the rate of change constant. of c with respect to A if a and b remain PARTIAL DIFFERENTIATION
227. The 449 We have already considered the
one variable in Art. 91. Thus, if total differential. ferential of a function of dif =/(*), we defined and proved cfy= (1) We shall next consider a function of two variables. Consider the function w=/(j, (2) Let Ax and Aw be
is be the increments of x and A?/ the corresponding increment of Aw = /(x + (3) //). Ax, + // respectively, // and let Then u.  A//) f(jr, //) called the total increment of u. Adding and subtracting /U, y Aw = (4) + f(x f Ax, H/Or, '// the second f A//) in f '// ~/{x, A//) + /U, A//) + // A//) member, //). Applying the Theorem of Mean Value (Z)), Art. 116, to each of
the two differences on the righthand side of (4), we get, for the
first difference, (5) /(x + + Ax, y  J(x, A?/) = x, Aa ~ A/,
fj
'constant, \vt' y + = /, (x + Ay/) arid since* j varies while n f ^ct the partial derivative \vith Ax, 0, y/ + Ay/) Ax. A// remains!
resj)ecl toj.l For the second difference, + /(x, v (6) =
I //, A^/ stant, = A^  f,(r where 6\ and / /y (x, + // 6. A//) A?/. while .r reinuins con// varies
J
partial derivative with respect to //I (5) and 0i Ax, y + //) and since A?/, we got the Substituting from
(7)  /(x, Ay/) (6) in (4) gives + Ay/) Ax + /,,(x, y + 0 A?/) A?/, 02 are positive proper fractions. Since J,(x, y) and fu (x, y) are continuous functions of x and y, the
coefficients of Ax and A?/ in (7) will approach f r (x y) and fu (x, y),
as limits when Ax and &y approach zero as common
9 respectively,
limits. Hence if e and
lim f 6 are infinitesimals such that = lim 0, AJ  ly we may
(8) (9) Aj
A?y e' = 0, .
 write
/,(z + 0i Ax, fv (x, ?/ y + + Ay)  / x (x, /) 6<2 A?/) = fy (x, y) +
+ , e', DIFFERENTIAL AND INTEGRAL CALCULUS 450 and (1) will &u=f (10) We become
f y)kx+Sy(x, y)&y (x, then define as the
<fa (11) + *x + total differential 'A?y. du) of u (= y)&x+fy (x, y)y. =/,(*, The righthand member in (11 is the "principal part" of the righthand 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 = (11 x, becomes ) rfj = w = //,
AJ and A// Ax. If Substituting, then, in (11) for we ferentials, dx u is for any number of variables. geometric interpretation of (B) ILLUSTRATIVE EXAMPLE n
10, Solution. and proceed
u /y ~ = 8, A>r 0.2, Ji/y + for the function = 2 J 3 /r,
= 0.3, and compare
4 Substitute in (12) for
as given in Art. 238. is Compute An and du 1. (12) x= its total differential is dv^dx + ^dy+'^dzi
(i
(x
(y and so on when A//. their corresponding dif the beginning of this article. (1) at a function of three variables, (C) A = + /(*, y) dy  j^ dx + ^ dy  ^ dx + ^ dy, which should be compared with
If becomes dy ) obtain the important formula  /,(*, y) du (B) (11 j, /<, //, the results. respectively, x + AJ*, // + A/y, M + A?J, below (compare Art. 27).  2(.r A.D^ 2 jr* f 3
M = 2 .r + 3
AM = 4 A.r 4 G A// 4 3(/y f A?y) r 4 /y' 4 J A.r f 6 J yy + A/y 4 2(A.r)^' 3(A?y) 2
. j /y (13) j Differentiating (12), we f A/y // 2( A.r)"' + 3(A#) 2
. find
(111 = ()l( . 4 .r, dx =6 y. dy Substituting in (B), the result is du = (14) Remembering that
(14) is the "principal AJ = dr, part" A?y (15) AM = 84 (16) du = 0.35 = 4 6 /y J/y. we <///, in AJ* or A?y.
' Substituting the given values Then AM  du = see that the righthand member in
of the righthand member in (13), for the additional terms are of the second degree
(11) above (namely, e = 2 A.r, = 4 x dx = This statement illustrates (10) and 3 A/y). in (13) and (14), 14.4 f 0.08 4 0.27 8 + 14.4 1.6% of AM. Ans. =22.4. we = get 22.75 ; PARTIAL DIFFERENTIATION
ILLUSTRATIVE EXAMPLE Given 2. u  451 arc tan ^, find du. x ^" e
*
Solution.
i V (V .r* f ^
in (5), Substituting ( (/ .r <"'* ir ~ ^ i' dij + 2 (J .4ns. ^. PROBLEMS
Find the total
1. z = 2  :{ JL' differential of each of the following functions.
.n/' +3 ^ :l // . d;  = ir^  (fi .r' dn 4 N,S. 4 ij)djr f (9 ((Vf3. = u jryW. cos ^ 7. If ^ 8. Find 9.
.r .r Compute u 4. = 10.
J.r ~  2, 12.
<t> = ^ J if 6> arc tan  = TT, cZO = n J <*' ! dy .r//c ' f  .S (/c. .r//'',: + ' (.r //) In and = for the function </// ().:>>, A// =r 0.2. u An*.
(j // f //) 3 .r A// = V.r // .r// f 7.15, (.r when L> 12 du .r f //). // when =(J, A. 3, A.r Compute />//)'' j A// Compute
t/ .r?/ )<///. //' Compute dn for the function
 8 1 4 2 6. + ^  a^, show that d;  ~ 16 c = 100.
4 j'  9 3, A.r \, dij 11. x 4 >r = 2, y 5. //. = djc  2
// // 7.5. = /1//X. A// = and <7p 0.2, c/0 = the for = 228. Approximation for r/w 0.4, A?/ the function M j*/y f 2 r  4 ?/ 2,
1. when 0.2. function p = c* sin (0 </>) when 0, 0.2. the of 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 experiment, and hence subject to small errors AJ and A?/, a close approximation 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
and height are, respectively, cylindrical can without a top if the inside diameter
6 in. and 8 in., and the thickness is 1 in. Solution. height y
(1) The volume v of a solid right circular cylinder with diameter x and is v = Obviously, the exact volume of the can
two solid cylinders for which x = 6J, y
only an approximate value is required, \ vx*y. the difference Ar between the volumes of
and x = 6, y = 8, respectively. Since
we calculate dv instead of Av. is  8, DIFFERENTIAL AND INTEGRAL CALCULUS 452 Differentiating (1), and using (B), we get = \ Trxy dx +  irx' dy.
y = 8, dx = \, dy = 1, the
7i TT = 22.4 cu. in. Ans.
2 df (2) Substituting in (2) = x 6, dv The exact value is At?  result is 23.1 cu. in. ILLUSTRATIVE KXAMI'LE 2. Two sides of an oblique plane triangle measured,
ft. and 78 ft., arid the included angle measured 60'.
These measurements were subject to errors whose maximum values are 0.1 ft. in each length
in the angle.
and
Find the approximate maximum error and the percentage
error made in calculating the third side from these measurements.
respectively, 63 1 1 Using the law Solution. where of cosines ((7), Art. 2), = it (3)
x, ij are the given sides, + x2 2 y'  2 xy cos a the included angle, , and u the third side. The given data are
(4) r = = ?/78, 63, Differentiating (3),
J ~~ we Hence, using x d/y = da = 0.01745 0.1, ^u (><)s (y ~~ *y Oa n dy (radian). s^ n a >' (C),
(.r // '_cos (\')(1x Substituting the values from du ~ y Of* H , / = dx get (l()s c * y dx =,
o 60 ~ 2.4 < The percentage + +4.65
 error = 1. } (4), .r (// we 74.25 = (*os <v)dij \ xij sin cy dcx find 10
1.13
, . ft. . Aiis. 1 100 =. 1.6'\ . An*. H PROBLEMS
1. with The legs of a right triangle
errors in each of 0.1 maximum error and percentage error
from these measurements. measured 6 ft. and 8 ft. respectively,
Find approximately the maximum ft. in calculating (a) AHX. the area, (b) the hypotenuse, (a) 0.7 sq. ft., 2.9 (
\ ; (b) 0.14 ft., 1.4 < ;. In the preceding problem find the approximate error in
calculating
the angle opposite the longer side from the given dimensions, and the
2. approximate The maximum error in that angle in radians and degrees. radii of the bases of a frustum of a right circular cone measure
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).
3. 5 in. and 11 in. respectively, 4?w. (a) 0.23 in., 2.2 <; ; (b) 32 TT cu. in., 4^ '~, PARTIAL DIFFERENTIATION
One 4. side of a triangle measures 2000 and the adjacent angles ft., respectively, with a maximum error in each angle
1 ft.
maximum error in the measurement of the side is and 60 measure 30 The of 30'. 453 Find the approximate maximum error and percentage error in calculating
from these measurements (a) the altitude on the given side; (b) the area
AN*, (a) 17.88 ft., 2.1 %.
of the triangle.
5. The diameter and measurement altitude of a right circular cylinder are found by
and 8 in. respectively. If there is a probable to be 12 in. what
computed volume? error of 0.2 in. in each measurement, possible error in the
6. The dimensions is approximately the greatest
Ans. 16.8 TT cu. in. box are found by measurement to be 6 of a ft., a probable error of 0.05 ft., (a) what is approxift.,
mately the greatest possible error in the computed volume? (b) What is
Aw,s. (a) 10.8 cu. ft.
the percentage error?
(b)
12 8 ft. If there is ; 7. z  ^^ Given the surface s x and y are each increased by 8. The + If,  4, at the point where x *%.
y = 2, // A, what is the approximate change in 2? specific gravity of a solid is given by the formula P
s where 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 P is in weighing w, assuming P  8 and w = 1
weighing P and
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% A ^ in 9. The diameter and measurement error of 0.2 in. in each possible error in surface? 10. 78 ft. Two slant height of a right circular cone are found by
and 20 in. respectively. If there is a probable to be 10 in. measurement, what approximately the computed value_of (a) the volume?
37 7rVl5 = Af ^ (a) 18 ^ cu Jn . (h) 8 is the greatest (b) the ^ = curved ^^m sides of a triangle are found by measurement to be 63 ft. and
to be 60. If there is a probable error of and the included angle in measuring the angle, what is
ft. in measuring the sides and of 2
of the
approximately the greatest possible error in the computed value
Am. 73.6 sq. ft.
Art. 2.)
area? (See (7), 0.5 11. If specific gravity is determined by the formula s = ^ where _ W the weight in water, what is (a) approxithe weight in air and
within
the largest error in s if A can be read within 0.01 Ib. and
mately
 5 Ib.? (b) the largest
0.02 Ib., the actual readings being A = 9 Ib.,
An. (a) 0.0144; (b) ?$fa.
relative error? A is W W DIFFERENTIAL AND INTEGRAL CALCULUS 454
12. The resistance of a circuit was found by using the formula C = jrt R = current and / = electromotive force. If there is an error of
ampere in reading C and 3*5 volt in reading E, (a) what is the approximate error in It if the readings are C = 15 amperes and E
110 volts?
Am. (a) 0.0522 ohms; (b)  ;.
(b) What is the percentage error?
where C
1*5 f the formula sin (x 13. If f y) = sin jr cos // cos x sin y were used to f + 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
and sin y
Ans. 0.0018.
f3 ? calculate sin (x j? 14. a = The g sin i. acceleration of a particle down an inclined plane
If g varies by 0.1 ft. per second per second, and measured as 30, computed value may
a of ? per second.
15. The period of a is
?, given by which is be in error 1, what is the approximate error in the
Take the normal value of g to be 32 ft. per second
Ans. 0.534 ft. per second per second. pendulum 2 ir\ (a) What is the greatest
* g
0.1 ft. in measurthere is an error of is P approximate error in the period if
ing a 10foot 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
Ann. error?
16. What The dimensions
is of a 0.0204 sec. (a) cone are radius of base = 4 in., ; (b)gj altitude the approximate error in volume and in total surface
in. per inch in the measure used? if = r
;. 6 in. there is a shortage of 0.01 dV  3.0159 cu. in. dS = 2.818 sq. in.
and the period P of a simple pendulum are connected
2
P'2 g. If
is calculated assuming P = 1 sec. and
ir'
Ans. 17. The length by the equation
g = 32 ft. I 4 is / l per second per second, what the true values are What ; /' = 1.02 sec. and g is ~ approximately the error in / if
ft. per second per second? 32.01 the percentage error? 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
20 in. What is approximately the
are diameter = 8 in. and total length
error in volume and surface if the tape used in measuring has stretched
18. uniformly r
; beyond its proper length ? Assuming the characteristic equation of a perfect gas to be rp Rt,
where r = volume, p = pressure, t = absolute temperature, and R a con19. stant, what is the relation between the differentials dr, dp, dt ? Ans. v dp +p dv =R dt. Using the result in the last example as applied to air, suppose that
we have found by actual experiment that t = 300 C.,
= 2000 Ib. per square foot, v = 14.4 cu. ft. Find the change in p, assump
C., and v to 14.5 cu. ft.
ing it to be uniform, when / changes to 301
7.22 Ib. per square foot.
Ans.
R = %.
20. in a given case PARTIAL DIFFERENTIATION Turn now to the case where x and y in 229. Total derivatives. Rates. CD 455 M=/(*,0) are not independent variables. Assume, for example, that both are
functions of a third variable /, namely, r (2) = ?/=M) <t>(U, 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
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
were independent variables. r (A) M Now when A?  i ox
> lim when A/ e > ' Hence lim
A/ = e' (see Art. 227) 0.  becomes 0, (4) _ fiu c?x rfu
( A// > 0. 0,  / A/ \ and = 4. Af ry A/ 0, A:r A/ Therefore, 4. M f^u dy
ny dt , dt~~ ax dt ' Multiplying both members by dt and using (3), we obtain (B).
is, (B) holds also when x and y are Junctions of a third variable t. That In the and x, ( ' ?/, same way,
z are all if u functions of du __
~ dt (hd
f>x /(x, y, z), we /, dx_ _ , dt get, ^ , ' f <)z cJydt dt so on for any number of variables.
x then y is a function of x,
In (D) we may suppose t
really a function of the one variable x, giving and ; djc In the same way, from (E)
of x
' f)x we dx __ du have, 'ox r;u
, jy
by dx is dx r>y when y and f du and u . <r)udz
f)z dx 2 are functions DIFFERENTIAL AND INTEGRAL CALCULUS 456 and ^ have quite
dx The student should observe that
ox
meanings. The partial derivative different formed on the supposition is 'Ox that the particular variable x alone varies,
held fast. But , other variables being all A
/ *_a 1JL \
. = (111 dx i lim
AX \Ax/ .0 AM is the total increment of u due to changes in all
caused by the change Ax in the independent variable.
where the variables In contra ( ^ are distinction to partial derivatives,
j> with respect to / and x respectively.
has a perfectly definite value for any should be observed that It called total derivatives point U, y), while depends not only on the point , (x, but also on y) the particular direction chosen to reach that point.
ILLUSTRATIVE EXAMPLE 1. Given sin , M  JT y <', = 2 t' ; ~ find dt y
, A du . 1
 cos x
,
//
y Solution. dx
Substituting in (D), <u"' dx
Substituting x
cos y x  dy // 2) (/ J (y
'  2. Given // = r" z}, = r
, dy f" = dy  2
 , c', Am. 2 T (y = y z),  . /. dt dt i r
dz  d.r
; cos r at iLLUSTKATivK'Kx AMPLE
Solution. du M*
I ; = a sin x, dx z cos x ; find ~ax  a cos x, sin x. dx in (G), M __ acnr(y _)( nr cos (ie x I r T sin x = f" T 2 (a' + 1) sin x. Ans. NOTE. In examples like the above, n could, by substitution, be found explicitly
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.
in ; Formulas (D) and (E) are useful in all applications involving
timerates 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 ?
Let x Solution.
1 o u But a = =S 230. Change 5000 ox = = 100,  5 o dn 'J T~ Trrj/, :; o ~4  Y I flM". o
2 ^f \ ^ = 10. 5, 2500 10 TT ~ ()y irxy ^ y = 7  in (>), /. ^r TT dn volume, 50, altitude; then radius of base, y
, 7r.r?/ Substituting at = , 15.15 cu. ft. o of variables. the variables in If M=M (1) per second, increasing. /Ins. 0) are changed by the transformation r (2) ~~~ <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. (3) ^M Hence we have ^r all derivatives with respect to
In the same way,
(ju (4) == x new = variables being x' rx r? ~ ( // now being r 'cu (hi . r~ ~r
r/x partial. 6\s r''// 7*~'
rAs the transformation be let (5) the + ^4/,
fr r chi 'vx r's In particular, ( r'wT z' /A, and ex  + ' ~ //', = and and &, k being constants.
^?y ~~ /v Then i ' vx (3) h, ~ ' (jy + ?/' ()y __ ^ f Then we obtain, from ;y f vy' (4),
' (>IL !L '
1 fix Hence 'ox vy by' 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 DIFFERENTIAL AND INTEGRAL CALCULUS 458 The results in (6) fx (x, y) (8) In Art. 229 it may now = FA*', be written
/(*, y) y'), was shown that true is (B) = Mr', #') when x and y are func We tions of a single independent variable t.
prove now that (B)
holds also when x and y are functions of two independent variables
For by (B), when r and s are the independent variables,
r, s, as in (2). we have ^ dx dr + = r/:r ds, rf?/ r
^ ^
rfr rr fs r^r H ** ds. r;s Substitute these values in the expression dx (9) ox and reduce by and (3) /in
(10; '(''U dy
vy The (4). + j dr result j
ds. '(''M> +
i vr is els u becomes a function of the independent variThen, 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)
But, by (1 ables r and and s. holds when ) x, y, z (2), are functions of one, two, or three independent variables.
231. Differentiation of implicit functions. M (1) ?/) = The equation o defines either x or y as an implicit function of the other. It represents an equation containing x and y when all its terms have been transposed to the first member. Let M=/Cr,0); (2) W + ^*
?/ dy du ,u then and y , 7^7ex
dx
is Now an arbitrary function of x.
Then u = and du = satisfying (1). 4^
f)x (3) Solving, (H) by \ )
J (F / y dx f 0, let y be the function of x and hence + ^? = 0.
eydx we get ^ = _.
dx fl ('y W^
\dy 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, on pages 40 and 41 and ILLUSTRATIVE EXAMPLE
Solution. Let = f(x, y) x + y* the examples formula (H) u = find ^dx sin y. V  O xy\
L ~ ~ (//), + sin Given jv 1. 2 2L
dy dx
Therefore, from all (1), be solved by it.
When the equation of a curve is in the form
affords an easy way of getting the slope. may , A
4 xy* j.
f , cos y. "  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 = stituting this value dy  2 xy in the 3 ; then, since u remains constant, hand member of loft we K pt ' x'y 0. j: Sub transposing, and solving for (/>), du ___
~~
du dt dt dy  Also, Now, y substituting in /y ~ (4), 6 j#, " "77 But x = xy  3 x 2 4 . * ^ //I, 3, 2j\ r T 4 = dy
Therefore,  ? 2 in.  17 ^"=2 per second. ATIS. In like manner, the equation
F(x, y,z) (5) =Q defines z as an implicit function of the two independent variables x
y. To find the partial derivatives of z with respect to x and to y, and proceed as follows.
Let du Then
by (#), and (Art. 230). this holds Now let z = u=f(x,y,z).
dF
dF  ox dx +T
tiy dy + dF
7 dz dz, no matter what the independent variables are
be chosen as that function of the independent DIFFERENTIAL AND INTEGRAL CALCULUS 460 oF
(6) P>ut dx rx dF (vF + (F + = therefore set dy f)z\ dx 0, and we have and simplifying gives dz ^ + ( A?/)
0, Ey rfy. cF
 rz\
}dy
(Z ryj \f)y ^ 0. rz I'cF + }dx
vz fx/ Here dx (= Ax) and dy (= may dx (6) Substituting this value in \r>x + r^ = dz = 0, dy + <">y now ( , = 0, du Then u
cF variables x and y which satisfies (5).  (5) 0. are independent increments. We
divide through by dx, and solve ()Z for T The result is z ( Proceeding in a similar manner, we ( Formulas (7) y and (7) are prove _ = (J) may f_l
( Z be interpreted as follows: to In the members z is the function of x and y satisfying (5). In
the righthand members F is the function of three variables, x, y, z,
given in the lefthand member of (5).
The generalization of (//), (/), (J) to an implicit function u of
any number of variables is now obvious.
lefthand ILLUSTRATIVE EXAMPLE. By the S+
z is (HI nation +
12 defined as an implicit function of .r *? and ?/. 1 =  Find the partial derivatives of this function. F= Solution. l ^ dF
dF
_
_
r'ir~12'
J __ TlPTIfP
00 + \. y . fy~6' Substituting in (7) and (7), dz
aar (Compare with the we
.r z 3 ~d~z get = T4 z Illustrative dF
_~ dz y ?=?r~2 z . ^ws dy Example in Art. 226.) PARTIAL DIFFERENTIATION 461 PROBLEMS
In Problems 15 find
at
1. = u x2 2 3 xy f 2 ?/ = a* ; cos = sin y /, f. = Aws. sin 2  f 3 cos 2 /. at 2.u=x + 4 VJv  3
*
3. K = f sin y +
sin // v{/ j < = Ans.
4.
5. =2
= xy w .r ?i 2 4 f ?/ ^ = ; j j ; = ( In Problems 610 find sin > '( 2 f y' j?/ 3
f .r ; J r =
= .r ; J f, 2 4 = 4 75J
/ f "(2 sin J /) 4 sin 2t,y
r', 2 + /. 2/42 cos 2 = 2
/ (// = ?/ by formula f' ~=3 \/ cos ?/ = u , / + cos J j /) /. f "'. (/f). a.r
2 6. Ajr 2 f Cy 2 4 7ij?/ 2 /Xr f 2 /fy f h ^ = 0. f T// 4 o r sin 8. e* f // w i cos x j/ 1. au'
9. 10.  .ry  x4 Ax 4 Rr 2 2 f 2 s' 4 2 2 ?/ f r'// // =
= 2 4 r" cos (r 2 4 3 f  12. j' 13. Ax 14. 2
15. e z 2 j?/ 3
// x 4 4 By 4 2 ?y = /y 4 ^/y rr 4 O'"  V27^ cos 4 y
4 ^ sin i/ () 22 ; = j =C
=4 x jc ; // . ) ?/ satisfy the 2, 2, y = // = Am. 3. ^=
 2. = 0, y = 0.
= 2, y = 4.
= 0x
0, = 1. ~ .r ; 1 = a ; ; ?/ and In Problems 1620 find
r^j 16. cos find the corresponding value of  equation, and x2 j 2 2
?/ In Problems 11 15 verify that the given values of x and 11. r JT 8. Ax>+B V + Cz>=D.
> ~
<^?/ Ans. ^ = *t;te = _*L
Cz
Cz dy
 =  Ay Cz dz _ ~ Ax
tfx 17. Axy 4 Byz + Czx = f)z D. i 4 Bz
" DIFFERENTIAL AND INTEGRAL CALCULUS 462
18. x . f 19. x* o
2 + 20. 2 ^^ f 21.
x' A f z'
2 /ty point f ir f z  A ?/< 2 ~  / 3 axyz Cz 2 f f 2 1 n Ans.
A = VZ ()Z 10. 23. The CZ xy W Dxy +2 7/2 f 2 Fz.r = XZ  Vxyz xy G. moving on the curve of intersection of the sphere is = 49 and the plane y When 2. which speed with which the point is moving.
Ans. (a; 8 units per second A rtJZ . Vxyz = 0. 4 units per second, find fa) the rate at 22. ~ ' 2vxyz =
o +z
. // x is (b) 4 ; and 6 is increasing
(b) the changing and z is V5 units per second. moving on the curve of intersection of the surface
point
1
2
z
and the plane x
x' f xy f ?/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
Aws. (a) 2 units per second; (b) V units per second;
is moving.
is (c) 4.44 units per second. characteristic equation of a perfect gas Rd is pr, where 6 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
and the rate at which the temperature is changing if the
temperature
is is increasing at the rate of \ cu. ft. per second and the pressure
decreasing at the* rate of ,' Ib. per square inch per second.
Ans. Temperature is increasing at the rate of .^ degrees per second. volume
is A 24.
triangle
at a uniform rate AHC is from O 1 being transformed so that the angle A changes
L
to 90 in 10 sec., while side AC decreases 1 in. per second and side A /> increases 1 in. per second.
GO
AC = 1(1 in., and A />  10 in.,
servation A
11 , changing? (b) how fast is ABC the area of Atis. (a) 0.911 in. how fast is BC 1 ; sq. in. per second. If w =/(j, (1) (a) changing?
(b) 8.88
per second 232. Derivatives of higher order. the time of ob If at ?/ ), then
/o\
(2) u f)
1 ( X f
/ /
r vn \ (r, ?/), f
 fv (x,
f x y) f'lf ;r and ?/, and can, in turn, be differentiated.
function and differentiating, we have, are themselves functions of Thus, taking the
(3) first "/,,Cr, 7r ?/), fy fx tfx = /i;,(:r,?/). In the same manner, from the second function in
/A\
(4) ^"U f
 /,(*,
f 2
\ y), U ) f
= fvu (x,
/ x y}. (2), we obtain PARTIAL DIFFERENTIATION
In and (4) there are apparently four derivatives
shown below that (3) order. 463
of the second It is eycx exey provided, merely, that the derivatives concerned are continuous. That is, the order of differentiating vuccewirelH with respect to Thus immaterial. /(x, x and y is
has only three partial derivatives of the //) second order, namely,
(5) fxjrfr, This may since (K) _ _r*H
cx'2 ( is } (>__ y\ '//) = /^(x, y). /,/?/ (x, //), be easily extended to higher derivatives. For instance, true,
/ (~u C''H \ rxVxr/// y &(*, __ rxetjfx /( e_ H\ /eu\ _ e eyex\cx/ <(it\(jc) Ir^u
( 2
y ex Similar results hold for functions of three or more variables.
ILLUSTRATIVE EXAMPLE. Given  1 .r // ,'J // .r 2
//' ; verify ~~~
r')// Solution. *J , Hence the formula Proof of
(6) (AT). F=f(jr+ is (> f'"'y T .r//'. :~ % J*"  IX 7^. oV d.r 2 H/ . verified. Consider the expression
Ax, // + A//)  /(x + Ax, y)  /(x, y + &y] Introduce the function
0(w) =/(M, (7) where u is (8) (9) + A?/) /(M, + c/>(x) (6) + Ax) =f(x may be
F = /(x, ?/ + A?/) /(x + Ax,
+ Ay)  /(x, Ax, y ?/), /). written = <t>(x + Ax)  Applying the Theorem of Mean Value
(10) ?/), an auxiliary variable. Then
4>(x Hence ?/ F = f /"(or) (Z>), Az0'(x + 6li Ax).
= 0(w), = ^a = Ax.J
a, Art. 116,
( dy DIFFERENTIAL AND INTEGRAL CALCULUS 464 + The value of 0'(z
0i Ax) is obtained from (8) by taking the
derivative with respect to x and replacing x by x
6\ Ax.
partial + Thus becomes (10) F (11) = + Ax(/,(x + Ax, y 61 Now apply (/)), Art. 116, to /a (x
independent variable. Then
F = Ax &yfvf (x (12) + 61 Ax, y + /,(* A?/) + + #j 0, Ax, Ax, 0, v), F= Hence from
(14) + A?/ Ax/,,(x fvf (x and (12) + Ax, y Oi ft. + Ax, y (0 A//). 2 member
< ft* < 1 < , as the < 1) of (6) 4 < 1 ) (13), + = /,(* + A?/) , L ft, + Ax, y Ax and of both sides as Taking the limit
limits, we have (0< Ay). 4 v regarding If the second and third terms of the righthand
are interchanged, a similar procedure will give (13) ?/)). A?/ 4 A/y). approach zero as fv ,(x,y)=Jr V (r,y), (15) assumed since these functions are to be continuous. PROBLEMS
Find the second partial derivatives of each of the following functions.
1. /(x, //) = Ax 2 2 f lljry + An*, /^(x,
2. /(x, / . 3. (x, /(x, O/ 2
?/)  = 2 .4 /, (x, ; tf = ?y) = Ax + Rr f Tx// f /;//'.
= 6 AJT + 2 Bij /^(x, = 2 Rr f 2 r\v
 .4.r KU f rr'".
4w. f2f (x, y) = Cy'2 c tv f* v =C(l
2 :< 2 It ; /^(x, 7y) ?y) ?/) ; ; / (x, ?y) =2 t\r f ?/) ; 4 /(x, ?/) 6. If /(x, = x 2 cos /) = /(2,
8. If /(x, //) ?/) = x 1) = ?/ 3)  1 =
2 4 2 sin x. ?y 2
f 3 x y 3 a /xx (2,
7. If /(x, 2 C. ?/ /y)  5. = 2 x' =
4 x 4 a 4 ?/ 3 x /(2,  2 2
?/ 2), 8  3
jrj/  96, .^(2,   6 x?y 2 30, /XJ ,(2, 3) f 1)
/y , , show that 48,
4 7/ = 4 /T1/ (2, 3
?y  , /yy (2, 3) = 6. 1) = show that 24, /w (2,  find the values of ~ 2), /vy (2,  2). 108. + 6 /)?/. PARTIAL DIFFERENTIATION
9. If u = Ax4 sults. ,I3 O* 2 ?r 4 Rjr*y f  = 2!L 4 4 ?/  4 6 #?/. fj 4 a p3 24 .4r = f />.n/ + 6 ^1 = /; ?/ , r'j ( \f verify the following re , Rr fc Rr 6 465 4 O/, f 24 f //. } 10. If w = (ax 2 <"// + 2 f ft?/ show that cz}'\ r'.r L If M = __!/_ S how that
___,  j 2  2 xy
" f
2 12. If M = In .H + show that 2
/y , 4 0. <"/r = 4 = 2
?/ r'j* r??/ f>.r' 0. * 2 ADDITIONAL PROBLEMS
1. A form of the
where the unit is 1 yd. circular hill has a central vertical section in the
2 +  160 y 1600 = curve whose equation
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
distance of 4 yd.?
increasing when the top has been cut down a vertical
Am. 25 sq. yd. per day.
is 2 if w = ^~,
f
c + cy w = 4. If z = 3 If 5. If , z fa where r = arc tan > Vj 2  2 y' + u 7. If = In (c u r f(?, y) Let M = ^ 2 (/i and x + 22  ^,  f 2 2 y 2 +  ffin
: = n ()u h * n 2 )'. + = 0? 4 ()y  l)w. sin 6  2 'f* '()'*% ^ ?/
' 2 = _^_ 0. (^ ~2^ r sin 0, _____ x*' ?/ show that , show that = r cos Or f show that  cr
arc tan  c)x tfu 8. + 2 y' show that 4 e" 4 r*), 0, ry y
6. If = + show that 7/
x 2 arc tan ~ u .r show that '(^u What values of * will satisfy the An*. * = l(n>2). CHAPTER XXIV
APPLICATIONS OF PARTIAL DERIVATIVES
The equation 233. Envelope of a family of curves. erally involves, besides the variables x and of a curve gen certain constants upon
which the size, shape, and jxxsition of that particular curve depend.
For example, the locus of the equation
a) (x
is a circle whose center origin, its size y 2 = r2 on the jaxis at a distance of a from the
r.
Suppose a to take on a
held fast
then lies ; have a corresponding shall + depending on the radius series of values while r is we 2 //, series of circles of equal radius differing in J? their distances from the origin, as shown in
the figure. Any system
way is formed in this
and the of curves called a family of curves, 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) The curves of a family may above = 0. be tangent to the same curve or group 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, as in the of curves. Suppose 0)
is tha;,
J figure. the curve whose parametric equations are = y=\f,(a) </>(), tangent to each curve of the family /U, (2) y, a) = 0, a being the same in both cases. For any common
a equations (1) will satisfy (2). Hence, by (), Art. 229,
u =f(x, y, or), du = df = 0, and z is replaced by a, we have the parameter value of
since (3) fx(x, y, a)(f>'(a} + fy (x, y, a)\f/'(a) 466 +/(>, y, a) = 0. APPLICATIONS OF PARTIAL DERIVATIVES
The slope of (1) at any point 467 is g*$.
and the slope of (2) at any point = (5) Hence r ' ' ./(*, rfj of is //,  <v
 (tf), Art. 231 a) (1) and (2) are tangent, the slopes at a point
be equal, giving the curves if will tangency fj(x, ?/, nr) fu (x, //, a) + /(/, //, __
</> (6) /,(*, Comparing (6) //, Uv) <v)</>'(aO and (3) W() = 0. gives
fa (x, (7) or y, = a) 0. Therefore the coordinates of the point of tangency satisfy
f(x, (8) ?/, <v)  and /n (j, a) //,  ; 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.
that is, General rule for finding the parametric equations of the envelope FIRST STEP. Write the equation of the family of curves in
and derive the equation fa (x, y, a) = 0.
/(x, y, a) SECOND STEP. the form 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
article, 1. For the family ^ (j> ^ /(r, Hence a) 2/, J = (j  (f _
 a) ,
) 2 r 2  0, or
Eliminating a, the result is /y
tions of the lines .4K and CD in the figure. ILLUSTRATIVE EXAMPLE
x cos a 4 y sin Solution. a  (9) ;>, 2. a being + of circles at the ^ = 0.
y = r, _ y Find the envelope r z =  r, and these are the equa of the family of straight lines the variable parameter. f(x, y, a]  x cos a + y sin beginning of this . a  p  0. DIFFERENTIAL AND INTEGRAL CALCULUS 468 Differentiating with respect to a, First Step. fa (x, (10) Second Step. Multiplying
x   a) y, (9) x sin 4 by cos a and a= y cos (10) by sin 0. a and subtracting, we get p cos a. and Similarly, eliminating x between (9) p y The parametric a (10), sin a. equations of the envelope are therefore (11) P^Iina' a being the parameter. Squaring equations (11) and
w^
adding, we get
_^_ yi ^ the rectangular equation of the envelope, a circle. ILLUSTRATIVE EXAMPLE 3. Find the envelope of a line
whose extremities move along two fixed rectangular axes.
Solution. Let AH = and o in length, x cos (12) a 4 of constant length a, let y sin p  a equation. Now as Aft moves, both a and p will vary. But
o cos a, and also
in terms of a. For AO
Att cos a
cos a. Substituting in (12), we get
p = AO sin a = a sin
a sin a cos a = 0,
x cos a f y sin a
(13) be its where a is the variable parameter. This equation is
0.
the form f(x, y, a]
Differentiating with respect
a, the equation fa(x, y, a)
(14) x sin a f y cos = a p may be found in to is 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 = (15) y a sin a,
a cos' a,
4 the parametric equations of the envelope, a hypocycloid. The corresponding rectangular equation is found from equations (15) by eliminating a: as follows:
a*** = a d sin a,
2 ? a^ cos 2 a. ^3 228
+ rj Adding, a ?/ = fl\ the rectangular equation of the hypocycloid. problems occur in which it is convenient to use two paramby an equation of condition. By using the latter,
one parameter may be eliminated from the equation of the family Many eters connected of curves. xample. It is, however, often better to proceed as in the following APPLICATIONS OF PARTIAL DERIVATIVES
ILLUSTRATIVE EXAMPLE 4. axes coincide and whose area
Solution. is (16) o2 ^= + is Find the envelope of the family of 469 ellipses whose constant. 1 2
b' ellipse, where a and
parameters connected the equation of the b are the variable by the equation
irab (17) A", being the area of an ellipse whose
somiaxes are a and b. Differentiating,
regarding a and b as variables and .r and
irab y as constants, we have, using differentials, =
U" bda and 0, from (16), 0, from (17). U"
f adb 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
conjugate rectangular hyperbolas (see figure). ~
~ , w a pair of 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
be defined as the envelope of its normals. It
also interesting to notice that if we find the
parametric equations of the envelope by the may
is of the previous article, we get the coordinates x and y of the center of curvature so
that we have here a second method for fiivditm the method ; coordinates of the center of curvature. If we eliminate the variable parameter, we obtain the rec tangular equation of the evolute.
ILLUSTRATIVE EXAMPLE. Find the evolute of the parabola y~ as the envelope of its normals.
Solution. The equation of the normal at any point (x\ t y\) is = 4 px considered DIFFERENTIAL AND INTEGRAL CALCULUS 470 from (2), Art. 43. As we are considering the normals
4
and 2/1 will vary. Eliminating x\ by means of y\ 2
the normal to be
1
yVi^j p
&
o p (1) +2py  x yi or ' 2 all px\, pj/i along the curve, both Ji
we find the equation of  =
^ 0. /' Setting the partial derivative of the lefthand member with respect to the
y\ equal to zero, and solving for x, we find parameter 4 p Substituting this value of i in
' (1) and .solving for y, and (3) arc the coordinates of the center of curvature of the patogether, they are the parametric equations of the evolute in terms
of the parameter y\. Eliminating ?/i, we obtain
Equations (2) Taken rabola. 27 2 pi/ =4(.r ~2 p)\ same the rectangular equation of tho evolute of the parabola. This is the
obtained in Illustrative Example 1, Art. 109, by the first method. result we PROBLEMS
of the following Find the envelopes systems of straight lines and draw the figures.
1. rns y 2. // 3. // 4. ?/ 5. = =
= y A ns. nr. f w f 2 2 mjr m
?w f . 16 . t' 6. . Find tbe envelopes fs h // of the following 7. /. systems of ?/ = circles =4 = ?/ : = 4 y f 27 ;< 4 2 tjr 2 27 x 2 nr. m'2 j' jr ^ ?/ ^ f m.r 0.
3
. ?y :i . 27 1 u* 2 m = 0. 2
. and draw the figures.
8.
9. 2 a 2 h = 4 ir c}' (x  O =
2 (?/ Find the envelopes = c(x
cy' = 1 11. y*
12. 2 4 2 r. 10. (T /. of the following Am. y 2 4
 O2 + + O = 2 c / (?/ f 4.  systems of parabolas.
Ans. 2 y c).
2 jr 2 = jr. jc. a 2 5 2 taking the equaFind the evolute of the ellipse b~x 2 + a 2 y 2
2
= ax tan
6 2 ) sin </>, the eccention of tbe normal in the form by
(a
13. tric angle , c/> Am. x = being tbe parameter.
a2 ~ ?>2 cos 3 0, ?/
' = b2 ~ a" sin 3 ; or (ax) ? f (byY = (a 2  6 2 )l APPLICATIONS OF PARTIAL DERIVATIVES
2 2 Find the evolute of the hypocycloid
whose normal is
=
cos r
14. r being the parameter. f = ir a cos 2 x sin r r s 471 2 a3 the equation of r, .r Arts, f (.r , (JT '< ?/) = 2 a. 15. Find the envelope of the circles which pass through the origin and
2
have their centers on the hyperbola x'2
?/ = r
2
Ans. The lemniscate (x'2 f 2 ) 2 = 4 r 2 (.r 2
?/ ).
. ?/ 16. Find the envelope the axes equals
17. such that the AnKf Find the envelope sum the of a line c. sum of its intercepts = r *. a 2 ?) 2 when x** f ifi = c^. ,1 of the family of ellipses b 2 x'2 f a 2 // 2 of its semiaxes equals r. 4 ?/<s . ^ + y f e hypocycloid on \ The parabola 18. Projectiles are nred from a gun with an initial velocity r>. Supposing 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
y
(i = x tan ex  ~ 2 r<r cos is , a being the variable parameter. Ans.
19. If The parabola y  ~^ =
2g 2 i,r the family
, has an envelope, show that ?/) + /'Or, ?/) = it is 2 L g (x, y) 235. Tangent line and normal plane to a skew curve. The student
already familiar with the parametric representation of a plane
curve (Art. 81). In order to extend
is 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
denote by t thus, shall ; (1) t * = 2=x(0 The elimination of the parameter
between these equations two by two will give us the equations of projecting cylinders of the curve the on the coordinate planes. DIFFERENTIAL AND INTEGRAL CALCULUS 472 Let the point correspond to the value t of the paramy + 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
P(.r, y, z) + Az, and the point P'(x eter, (diagonal; PI*' are proportional to Ax,
or, dividing through
secant by a', ft', y',
} ~ Ax
let direction angles of the _ cos y'
"
Az At P P' approach f (3 Ag AZ Now ; by A/ and denoting the
cos a' __ cos ( Az A#, A* Then along the curve. AJ, and therefore approach zero as a limit, and the secant PP' will
approach the tangent line to the curve at P as a limiting position.
also AJ, Ay, Az, will Now lim
A*  o ^=^=
A </>'(0, etc. at Hence, for the tangent line, cos a _ cos P_ cos cfo When dt dt the point of contact (3) and (/y = y dz dt 04) value of c j is we use the P\(x\, y\, z\), when x = .TI, = y y\, z notation = z\, similar notation for the other derivatives. Hence, by (2) The equations
(1) and (4), p. 5, we have the following result. of the tangent line to the curve
.r at the point PI(XI, y\, = <W), y = t(V, whose equations are z zO are
x (B) xi_y \dx
\dt y\ _ z z\ I \dt I \dt\ The normal plane of a skew curve at a point P\(x\, y\, Zi) is the
plane which passes through PI and is perpendicular to the tangent
The denominators in (B) are the direction numbers of
line at PI.
the tangent line at PI. Hence we have the following result. APPLICATIONS OF PARTIAL DERIVATIVES 473 of thf normal plane to the curve (1) at Pi (x\, y\, z\) is The equation (x (w  \dt 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)
(
 (a) at any point Solution. (xi, d) 3/1, = ^
do = a sin x ~ (5) x ~ V V 1 = fl T"^' sin 6,  2 TT. = ^ = a cos
du ?/, we get, Substituting in (B) and (C), a cos 0, when (b) ; = x
y (4) at f b. (<(' //,, z\ >, (.n, tangent .r, line, and
normal plane. When
(a, 0, 2 '2 &TT), the TT, on point g_ji _
~ _z~ ?y a
or a* = a, &# = 2 O^TF, a^ line, of the and
f ?w cos 7 line (5) = ?> 2 7T = 0, cuts all we have by (2) 6
: 2 + 2 Va~ f elements of the cylinder x 2 + and = (4), p. 5, a constant b2 236. Length of arc of a
article 2 fo f yi That is, the helix  normal plane. REMARK. For the tangent /i is 2 bir ny ceding curve 6 the equations of the tangent the equation the giving skew b' curve. 2 y' From =a 2 under the same angle. the figure of the pre we have
(Chord PP'Y x Let arc PP' = As. _ /AxV
 Proceeding as in Art. 95, we easily prove
2 (2) From (*)
this we obtain
s (D) where x = 0(0, y = = ^(0> z x(0 as in (1^ A*t. 23"). DIFFERENTIAL AND INTEGRAL CALCULUS 474 The equation
(3 now be given a simple direction cosines of the tangent line can For, from (A) of the preceding article, form. using formulas in (2), cos ) > ^ ft " ds (4) between the points where = Solution. = / Substituting skew cubic f J t = we obtain = dy dt, = f Vl + in (D), = z } 1*, t, ds of arc of the y =
and = 4. Differentiating (4), dx ~ " ds ILLUSTRATIVE EXAMPLE. Find the length
x and by the above we have (2), p. 5, / + 2 t' dz (it, = dt 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 AUK.  3. a = / 4. j = P  5. a = 2 6. x 7.
8. 9. =
x =
x = 2 / = ?/ cos 2 = z , 1, ?/ = + 1, ?/ =  =
/, </; = / 1. ^^ = ^^ = ^pf a cos
/, bt S, /, ?/ z/ = ^ = , c r2 f = /,  5 2
/ b sin // r', 1, r / ^ = sin Find the length
x ' r / ; t, z = 4 = = / ; ~ z /, = :< /  /3  / ; / 4 2 by f 3 cz = a2 4 2 6 2 4 3 c 2 . 2. 3 / hi ; / = 1. = 2. = / ; ax ; J TT. 0. tan / = f ; 0. of arc of the circular helix = a cos between the points where B = 0, y = and a sin 6 = 2 6,
TT. z = b6 Ans. 2 ?rVa 2 f 6. APPLICATIONS OF PARTIAL DERIVATIVES
10. Find the length of arc
x = 3 cos = between the points where 6
11. Find the length
x 475 of the curve = y 0, 3 6 sin c 8, = 4. and =4
+ ^ In 5 = 32.70. Ann. 26 of arc of the curve = 2 between the points where t,
t = 12. x = 2 2, t' and = / =1 ; t 2 2. Given the two curves (5) y =
y =
 0,
x = 1
t, z 2t*, = y;  1.
y = 2 cos 0, c = sin 6
1).
(a) Show that the two curves intersect at the point A(l, 2,
(b) Find the direction cosines of the tangent line to (5) at A.
(6) 1 4 1 ;= 4
Ans. p= 7= Vl8 Vl8 Vl8
(d) Find the direction cosines of the tangent line to
Find the angle of intersection of the curves at A 13. A. Given the two curves (c) a:r = 2/,
= sin 0, i/ y = 4,
= 0, z =
/ 2 2 (6) at Ans. 90. . = f8;
cos 1 0. Show that the two curves intersect at the origin O.
Find the direction cosines of the tangent line to each curve at O.
(b)
(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
= a cos cos 0, z = asin</>, if and
sphere, prove that r = a cos $ sin 0, ?/
and longitude of P.
are, respectively, the latitude
at P between a curve
(b) Using (3), and (3) on page 5, find the angle u
(a) < on the sphere for which = /(</>) and the </> parallel Ans. tan a through P.
sec </> 777 au as in Art. 222. , A straight line is
237. Normal line and tangent plane to a surface.
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 is made to approach P along a curve on the surface.
now proceed to establish a theorem of fundamental importance.
f Theorem. All tangent a surface lines to at a given point lie We in a plane. Proof. Let
(1) F(x, y,z)=0 be the equation of the given surface, and let P(x, y,
Doint on the surface. If now P' be made to approach z) be the given P along a curve DIFFERENTIAL AND INTEGRAL CALCULUS 476 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
if u = F(x, y, z), then u = 0, du = 0, and by (), Art. 229, values. Hence, ~
tix dt cz dt 'dy dt This equation (see (3), Art. 4) shows that the tangent
whose direction cosines are proportional to
dx
dz
d/n
dt dt
is By > > (4)
(>x PI(:TI, y\, z\) dt whose direction cosines are proportional to perpendicular to a line Let line to (2), cz oy (3), p. 5 be a point on the surface and
?
)F <)F /e\
(O) TT" vx > vy i >
\ y = y\
z
The line passing through P\ whose direction numbers are
z\.
given by (5) is called the normal line to the surface at PI. Hence we
have the following result
the values of the partial derivatives in (4) when x = Xi, 9 : The equations of H \i the normal F(x y
r \A., y 1 ) at PI('X\, line to the surface '//i, zO are x  _. j y ~] ** i = \/ y\ __ z z\ dF IdF dF\ dx \9y '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?i(. is = 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
z case = /(z, the equation F(x, (6)
rp^ i the of surface given is in F form the , y), let nen ?>F ?/, =/(j, z) C'Z ~T~ ra OT ( TT~ ?j r F z= dy ( ~ } } y ( Hence, by (), we have the following The equations } f Z
~~ ~r~
r 0. } (f r ex //) == ! cz y result. of the normal line to the surface z /( J, y) at ( x\ , y\ , z\ ) are 21 (G) Also, from (F), we  obtain
xi) 4 (y  yi)  (z  zi) = which
to is then the formula for the equation, of a plane tangent at
a surface whose equation is given in the form z =f(x, y). (x\ t y\, z\) 238. Geometric interpretation of the total differential. We are now
a position to discuss formula (#), Art. 227, by geometry, in a manner entirely analogous to that in Art. 91.
Consider the surface
in (1) and the point i, y\, z\) on
x A'hen Then the it. = x\ t total differential of (1) is, y~yi,
Ayy, (2) using (5), Art. 227, and replacing dx and dy by their equivalents, Ax
and A^/, respectively. Let us find the zcoordinate of the point in
the tangent plane at P\ where
x = xi + Ax, y = yi + A?/. Substituting these values in (H) of Art. 237,
(3) we find DIFFERENTIAL AND INTEGRAL CALCULUS 478 Comparing (2) Theorem. The and we (3), get dz =z Hence the z\. total dif ferential of a function f(x, y) corresponding to the increments A x and Ay equals the
corresponding increment of
the zcoordinate of the tan gent to plane Thus, the surface PP' in the figure, the plane tangent to
surface PQ at P(x, y, z).
is Let CD and &y,
dz then Notice also that = z  z\ = DP'  DE = EP'.
Az = DQ  DE = EQ. ILLUSTRATIVE EXAMPLE. Find the equation of the tangent plane and the equa14 at the point (1, 2, 3).
normal line to the sphere x'2 f y 2 f z 2 tions of the Let F(x, Solution. y, z) = + x2 y* + z2  14 ; dF then dx = 4, Therefore vSubstituting in (F), x or 2(x + Substituting in (),
z giving =3x  + 4(y  2) + 6(z + 3 c = 14, the tangent
2_s3
y
"
1 ) 2 y IJ) = 0, plane. ' 6 4 and 2 s =3 y, equations of the normal line. Arts. PROBLEMS
Find the equation of the tangent plane and the equations of the normal line to each of the following surfaces at the point indicated.
2 1. x' + y'2 + z2 = 49 ; (6, 2, 3). An*.
2. z = 8 l; (2, 6 x 4 1,4). J ' x6 = y2 = z3
~
6 3 2
 1 24 APPLICATIONS OF PARTIAL DERIVATIVES + r2 3. xy 2 3 4 = f z 4 1 ?/ Ans. 13  (2, ; 3, 4). +*+ 15 y f .r 15 = ^^ = i^J = L=l.
o ; A 4. x~ 4 2 5. j* 2 2  2  f z ?/ f .nr ?/ 2 jry 4 ?/<  = 7 10 = Ans. 4
.r  2 6. 2 + 2 =: z' 2 y 2  7. ar 8. 2 9. yf ?/ + x2 3 z
2 + j/ r2 1; = 2 25 1 sheets  ^6
2 a !
r = = Show 12. + ^i^ 4 i at = 2^ = ^ 6; (1, 1, J). tangent plane to the hyperboloid of two of the
(.r,, M_ _ l .4**. z,). ?/ 2 ?> 2 of the f re 6// + r 2 + tangent plane at the point ~ d ^^,s. 0. 2 L.r f 2 Af?/ f z f (.r,, a.ri.r f ^?/i// f = l
r2 2 = d . on z\) ?/ ]t rr,^ f ! 0. f A^ + 2 /> = (xi, y it z\) is 3/1 1/ + Ziz + L(J Find the equation 13. 411 ^=?
; that the equation of the plane tangent to the sphere x2
at the point = (3, 4, 2). Find the equation 11. the surface ax 2 13 (5, 5, 5). ; 4z' ^3; 1 (3, 2, 2). Find the equation 10.  f z // 1 2, 6). (2 t 1, 4). ; + .r  (1, ; 479 + D= T M(/y f ji) 4 of the j% A/ ( 4 Ci) 4 ?/]) h 0. tangent plane at any point of the surface
rr a", y* f 2 f and show that the sum of the squares
by the tangent plane is constant. of the intercepts on the axes made 14. Prove that the tetrahedron formed by the coordinate planes and
a is of constant volume.
any tangent plane to the surface jryz
:< The 15. =  / _ 9
 /2 4 y2 surface x 2 4 z =  16. at the point The surface x2 (2, 2, + 2
?/ f What 3). 3 is Am. 2 25 > y = J
the angle of intersection? 90 ~ z2 /2 cut by the curve x is  / t = 32 37'. 2t, y 3
= > 1 arc cos 4
 !L. 3V138
and the curve x = t z = 2 f 2 intersect at the point the angle of intersection 17.
i/ = f 4 The on the curve given by 2
2
2
ellipsoid x h 2 y 4 3 z f 1, z P meet A^ ? at the point the surface orthogonally. = 9Q o _ arc CQS / = 1. _J9
7V29 What = 3QO is 16/> 20 and the skew curve x  \(P 4 1 ),
Show that the curve cuts
1). (3, 2, DIFFERENTIAL AND INTEGRAL CALCULUS 480 form of the equations
skew curve. If the curve of the tangent line 239. Another plane to a and normal in question be the curve of intersection AB
and
two surfaces F (x, y, z) = of the G(x, y, = 0, Z}) is z) P(x\, y\ tangent t planes point, for
faces and planes. the tangent line PT at
the intersection of the it is CD CE and at that, also tangent to both sur hence must lie in both tangent
The equations of the two tan P are, (x gent planes at  from (F), vF tix c)z tiy (1)
f)G <*"> ^ IrJx (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 Art. 4, (6), B= c)JP f)Z VvG_ bF\
'c)z 'dy 8F dG cjy dx dz i (2) C=
8y Then the equations fry of the tangent line _y
(3) A(x (4) normal plane of the  j,) + ^z y\ B A The equation c)G c)F_ B(y PHI  yi) + CPT are C
is  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 aF
ax = 0, dG]
dy\ r2 and G=x +
2 y 2  2 nc. APPLICATIONS OF PARTIAL DERIVATIVES
Substituting in A = 4r
Hence, by 2 find B = 0, C = 4r2. \/2, we have (3), x we (2), 481 r rV2 z r y  V2
or = ^ z r, PT the equations of the tangent
the curve of intersection.
Substituting in we (4), P at to get the equa tion of the normal plane, 0(y  r) \ 2 x   z r) 4 or +  (z V r\ =. 0. 0. ILLUSTRATIVE EXAMPLE 2. Find the angle of intersection of the surfaces in the
preceding example at the point given. The angle Solution. or normal lines. We trative 1 Example of intersection equals the angle between the tangent planes
for these lines above in Illus have found direction numbers
(see (), Art. 237). These are =2
= 0, a
a' Hence, by =2 b r, 6' r  c' = 0. r, = 2 r, 2 r\/2. Art. 4 (6), = 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. x2 + y 2 + z2 = 49, + x* y 2 13 = a 2 + y 2  1, + 3 ?2 + 2 2 // Am.
. + y* _ Z2 = 16. x + 4 ?/* x 2 ,4ns. +
_ 16
4. xa + 5. x2  6. x 2 + 2
?y y + 3 z2  z2 2 4 # 2  = = 32, 2 1, 4 z2 x2   = G). =, 2+6 =
30 0; 2 s  3 # = 0. (2, 1, 4). ; 11 5 3 x* z2  (3, 2, rLzJ = Aw*.
2. 2 ; r2 y 2 0, 2 x + 2
?/ 4 z2 = y 4 5
 z2 = + z2 = 9
+ +2 ; ?/ 84 _ ; (2, 4, 2). 2 2 ; 16 j 5 6
; (2, 1, 3). (3, 2, 2). 24 = ; (8, 3, 5). ?y + 6,7 = 24. DIFFERENTIAL AND INTEGRAL CALCULUS 482 The equations 7. of a helix (spiral) are + X2 _
= y'2
iy Show that at the point O],  c(x
r(?y and the equation The 8.  ?y } ) ~ 7/1 .r . surfaces x 2 ?/ + 2  Zj) =
= ~ z\) = si) f 1/1(2 ?]) xi(z normal plane of the r x tan the equations of the tangent line are zj) ?/i, f2> f z j 16 and 3 :< 0; is r(z x\y 0, 0.
2 jc 4 2 = 2 z ?/ 9 intersect curve which passes through the point (2, 1, 2). What are the equations of the respective tangent planes to the two surfaces at this point?
Am. 3a + 4i/ + 6c = 2'2; 6 x f y  z ~ 11. in a Show 9. ^2 __ ?y2 __ point x the
__ ellipsoid g ^ _ g _}_ 24 = + 2 x~ f 3 y' 2
z' 2 and 9 the sphere are tangent to each other at the (2, 1, 1). 22 = 1 and the sphere
that the paraboloid 3 x~ f 2 2
4 y  2 z f 2 =
cut orthogonally at the point (1, 1, 2). Show 10.
2 that
& r _ ~2 2 f y' + 2
z' ?/  Law of the Mean. The applications of partial derivatives to be
now depend upon the Law of the Mean for functions of several
variables. The result to be derived is based upon the discussion in
240. given We proceed to establish the formula
= /(ao,
+ V,(ro + Oh,
(1) /fa, + h, yo + k) Art. 116. 2/0) + kf
To this end F(t)=f(ro + 0k) (0<0<1) + ht,yo + kt).
and Aa Art. 116, to F(t), with a == 0, (Z)), ?/o (x*+0h,yv+0k). let (2) Apply tl = Then we 1. have
F(l) (3) But from
(4) (2), = f'(0 Then, from and, from When (2), F(0) + F'(0). (0 (D), Art. 229, since x fc/,(:ro F(l) (5) (6) by = we = xn + hi, y = < yo 6 < + kt, + fa
get = /(Jo (4), F'tf) = hfx (x + eh, yo + 6k) these results are substituted in + kf (3), v (x we + Oh, obtain yo (1). + Ok). 1) APPLICATIONS OF PARTIAL DERIVATIVES a formula analogous to (F), Art. 124, we must form
Applying again (D), Art. 229, we get we If F"(t). 483 desire x = hftf (xo + ht, y + kt) + kfvr (x + hi, y + kt) kt) , (} at
v/y(aro
at + kt, .2/0 Hence from + kt) = hf,(x + hi, (7) F"(t) we have by (4), = k*fxf (x<> + ht, + kt) + kf tw (x + ht, y + kt).
() // differentiating with respect to (F), Art. 124, letting ni) = no) (8) We may now easily prove kt) //o + r(0) + kt., ?/o rr'(0). the extended function of two variables by substituting Thus we Law in (8) of the from Mean (5), (4), for a and (7). get (9) /(jr + A/i/to),
+
+ h, + k) =/(TO,
+ r t^ 2/xX^> + Oh, + 0A) + 2 Wn/ao + 0A, y + Ok)
?/o) A/r(:r<>, ?/o) v/o) ?/o () /A, +
There t, + 2 hkfxu (xo + kt. + kt)
+ kt).
+
(Jo +
6 = 1, a = 0, ^2 = 0, we get + ?/ A/,,, From ; is + 2 /^(y Ar no 0//, ?/o + 0*)] (0 difficulty in establishing the < < 1). corresponding formulas 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
for functions of more than two Art. 46, and again in Art. 125, were derived necessary and sufficient
conditions for maximum and minimum values of a function of one
now take up this problem when several independent
variable. We variables are present.
The function /(.r, is said to be a maximum, at x = a, y = b
than f(x, y) for all values of x and y in the
greater
/(a, b)
neighborhood of a and b. Similarly, /(x, y) is said to be a minimum
of x and y
at x
a, y = b when /(a, 6) is less than f(x, y) for all values when in the neighborhood
These definitions
If, ?/) is of a and may for all values of h 6. be stated in analytical and k numerically less form as follows:
than some small positive quantity, Q)
then /(a,
(2) then +h j( a
b) is a /(a /(a, b) is f b + k) maximum + A, 6 + k) /(a, 6) =a value of f(x,
/(a, 6) a minimum value =a negative number,
y). If positive of /(x, #). number, DIFFERENTIAL AND INTEGRAL CALCULUS 484 These statements may be interpreted geometrically as
P on the surface A point = f(x,y) z a l\ maximum surface in its horizontal . when follows. " higher" than all other points on the
neighborhood, the coordinate plane XOY being assumed
point it is Similarly, a minimum point
on the surface when
it is "lower" than all
other points on the
P' is surface in its neigh borhood. Hence if 21= /(a, 6) a maximum or minimum, the tangent plane is But the
(a, b, zi) must be horizontal, that is, parallel to XOY.
tangent plane (H), Art. 237, is parallel to XOY when the coefficients of x and // are zero. Hence we have the following result.
at A necessary condition that /(a, b) shall be a
value off(x, y) is that the equations
(3) c)x shall be satisfied by The conditions x a, (3) y may maximum or minimum 8y'
b. be obtained without use of the tangent ~ b, the function /(.r, b) can neither increase nor
plane. For, when y
decrease when x passes through a (see Art. 45). Hence follows the
The same statement applies to the function
(3).
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
first of equations /(a, y}. maximum or a (4) minimum value is that the equations =
ftp 8z c.
have the common solution x = a, y
b, z
For necessary and sufficient conditions the problem shall difficult (see a maximum necessary. is much more below). But in many applied problems the existence of
or minimum value is known in advance, and no test is APPLICATIONS OF PARTIAL DERIVATIVES
ILLUSTRATIVE EXAMPLE 1. A long piece of tin 24
into a trough by bending up two sides. Find the width
and inclination of each side if the carrying capacity is a wide in. is 485 to be made maximum.
The Solution. area of the cross section shown in the
The cross section is a
2 x f 2 x cos a, lower
upper base 24
and altitude x sin a. The area A is given by must be a maximum. figure trapezoid of
base 24  2 x, A (5) By differentiation = r dx dA =
r
d 24 x sin a a + x2 sin 2 a 2 x 2 sin a 24 sin + x sin a 4 a 24 x cos 2 sin (12 [24 cos One A solution of this 2 cos .r a jr sin  2 x + f a*" (cos' 2 cos a .r system ^ a 4 a= is Assuming a = x cos a) 0, x a . sin a). we have the two equations 0. x(cos & x 0, cos 2 sin a cv )J = 0, which has no meaning in the
and solving the equations, we get 0, establish a sufficient condition. yo = b, (6) we obtain from (9), 0. = consideration of the physical problem shows that there
of the area. Hence this maximum value occurs We now *i cos a. mum value
x = 8". tions (3) hold, 24 ~ 2a? we have Setting the partial derivatives equal to zero, physical problem.
cos a
8.
2, x a _ * r must exist a when a = maxi 60" and Assuming that equa Art. 240, substituting x () = a, and transposing,
/(a + M+ A') /(a, b) = r ^ \h^fxx (x y y}+2 kkf (x, y)
+k*fw (x,y)\,
fll 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 righthand member is negative
(or positive) for all values of h and k sufficiently small in numerical
value (zero excluded).
(7) A =/(*, Set B=:fxv (x, y), y), r=fyy (x, y), Bt)2 + (AC  B*)k*]. and consider the identity
(8) Ah 2 + 2Bhk+ Ck* = [(Aft + The expression within the square
member in (8) is always positive if
(9) AC  B 2 > brackets in the righthand 0, DIFFERENTIAL AND INTEGRAL CALCULUS 486 and the lefthand member therefore has the same sign as A (or
by (9), A and C must agree in sign). The question now since, C,
is, therefore, to interpret the criterion (9; for the righthand member in
(6), in which, as already stated, h and k are numerically small. Assume that when x = = b.
Then, the derivatives in
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 fxx (a, b)
(or fvv (a, b)). Thus we have established the following rule for finding maximum and minimum values of a function / (x, y). holds (9) y a, (1) FIRST STEP. Solve the simultaneous equations #= #= 0, c)x 0. y j ( and y Calculate for these values of x SECOND STEP. A^LM the value of /J^LY\ ()jc~ THIRD STEP. The Junction
a maximum value if '( \<rxvyj >\i~ will have A> and ^ (or ()x* \ a minimum value If A is it negative, is if not A > ^
vx~ and ( ; f)y or ^~ } > 0. r)yJ \ that difficult to see ~{ } <
2
/ /(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 for solving many The question Further investigation is
is, however, sufficient or neither. The therefore necessary for such values. rule important examples. of of functions of three or maxima and minima independent variables must be
ILLUSTRATIVE EXAMPLE
and minimum values. 2. Solution. f(x, y] left more advanced to Examine the function 3 axy mum First Step.  =3 ay  3 z2 = 3  0,  axy x3 =
 Solving these two simultaneous equations,  3 we x = 0, x a, y = 0, y = a. y ax. 3
.  get 3 y* = 0. more treatises. x3 y 5 for maxi APPLICATIONS OF PARTIAL DERIVATIVES
= SecondStep. =3 6 x, = 6 V a, 487 ; "
\dxdyj
and =
Third Step. When x =
maximum nor a minimum at (0, OK
When x = a and y = a, A = 4 27 a
{/ ditions for a maximum a in the given function, we
3. shall ; and there can be neither a , ^=~ Substituting
value equal to a = Let x Solution. = a, . Divide a into three parts such that their product first part, = // second part and the function to be examined third part, f(x
First Step. =   ay x xy(a y} t  2 xy = 2 y' a(x + y)=axy*=* then ; is ?/). = ^ 0, r   2 J?/
(7 Solving simultaneously, Second Step. we = ^ 2 get as one pair of values When r>. that our product jr " 4 xy and y  = ^ and the maximum value of the a  2  ar 2 ' y .r x2 _
 = ?/, = 0. a
 2 x ; ~^T, (a maximum when a is = = ^ ?/, A= , u* :{ maximum. be a
also 7 we have the con a, (5 fulfilled at (a, </K maximum its get 9 a2 and since 2 value of the function ILLUSTRATIVE EXAMPLE y A   (),  2 A =
x =  .r ^ ' ; 2 //)". and since ^ =  =~ . it is seen Therefore the third .part y^r^' is is product PROBLEMS
maxima and minima Discuss for
x2 1. 2. 4
3. xy + y'  j.  + 2  r 4 r//
 2 A// 4 2 + 5
?/ 2 /y 3 a^y >3 the following functions.
Arts, 6 x 4 2. a 2 x 4 x
5. 4 2 j 2  3 4 if x
x2 7 ^
^ ' 8. , xy 4 2! 2 + Show x !/ + oar 4 60 ?/ // ?/. sin x 4 sin y 4 sin (x 4 y). 6. ?/ .'?, . ?/ = 4, =  2 gives min.
= S gives max.
^ =J
= i gives min.
^ =  L
a gives min.
x = y
x = y = ~ gives max.
r  =y o gives min. 4 c. L
y that the 2 maximum value of ^ . J 2 is
. ^ a2 + fr2 "^ c2 ' DIFFERENTIAL AND INTEGRAL CALCULUS 488 9. Find the rectangular parallelepiped of maximum volume which
has three faces in the coordinate planes and one vertex in the plane 5
a + 2+5 =
c
b Find the volume 10. be inscribed A = Volume Ans. 1. 27 of the largest rectangular parallelepiped that in the ellipsoid +  a2 2_ + b2 c2 = ~ can Ans. 1. is composed of a rectangle surmounted by an isosceles
the perimeter of the pentagon has a given
value P, find the dimensions for maximum area. 11. pentagon
If triangle. AUK. a 30, =p y Find the shortest 12. , 2 JT P = > 242 + .r(l sec a tan a y sec a). between distance the lines x = ~
Z j ==  and o A manufacturer produces two lines of candy at constant average
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
= 3 ^ ()00 + 25Q(x __ 2 y).
N _ 25Q(?/ _
13. costs of 50 cents ^ ^ ^ Show that for maximum profit the selling prices should be fixed at 89 cents and 94 cents per pound respectively. A manufacturer produces razors and blades at a constant average
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
14. cost of 40 cents per razor ., ,
of the , . , , market each week . is  4,000,000
  , razors and  8,000,000
  Find the selling prices for , , blades.
dozen ,, ry .r?/ 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 F(t}=f(x (1) and expand + ht,y + kt), 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 =
= 0, By differentiating (7), and putting 1
r/
the expressions for F '(0) etc. will result. These are omitted here.
Note, however, that F'"(0) is homogeneous and of the third degree
in (2), (4), (7), Art. 240. APPLICATIONS OF PARTIAL DERIVATIVES 489 and k. A similar property holds for higher derivatives.
values are substituted in (2), and we set / = 1, the result is in h (3) /(:r + h, y + k) =/(j, y) If these + hf,(x. y) +
R. expression for R is complicated and will be omitted from this
on.
point
In (3) write x
6, and then replace A by (x
a, #
a) and k by
~~
The result is Taylor's theorem for a function of two variables,
(?/
&) The =  a) +/y (a,
 Finally, setting a =6= 0, a)(y  we obtain an expansion corresponding to Maclaurin's series (4), Art. 194, The righthand member
fA\
(4) Mo in (7) +
I U l [T where uo = f(0, may be written +
I W ii +
I as the infinite series '"' f" 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
a, y
b).
polynomials homogeneous in (x
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. DIFFERENTIAL AND INTEGRAL CALCULUS 490 ILLUSTRATIVE EXAMPLE. Expand xy 2
at the point up to terms IT) (1, \ xy = 1, b =  IT,
= *y* + si" *y a Here Solution. and 4 sin of the third degree. f(x, ?/) fy (x, y) =2xy + x cos xy,
= y smxy,
2 # + cos x/  xy sin xy,
2 = = 1, Substituting x = y the results are TT, J  /,(!, 2 /(!, Substituting in xy 2 + sin xy = l + Formulas we (/), TT) T) T. 2 iir'
1 7T2, get
ITT) 1) +7r(2/ : J for 2 =
=
= JT) /(I, of three variables f(x t y, z) are expanding a function readily derived, and are left as problems. PROBLEMS
1. From above, show that (1) 2. +3 +3 dr Verify the following expansion. = cos x cos y 4 1 TTT 6 + 15 * J4 _
sin sin y in x6 + powers 15 J 4 ?/ 2 of x 3. Expand 4. and jV 4 y 6 .... y. Verify the following expansion. a' log (1 .r + 10=0+ U2^1ogat/ 2 + x
3 2 2 i/log 2 a at the point (1,2). 5. Expand 6. Verify the following expansion. .r f x?/ sin (x 4 !/) =x+y  3 x *y + 3 +
. for small values of x and
Verify the following approximate formulas
7. r
8. P = r sin J In (1 h T/ ?/ /) f .r//. = I' ( 4 ry y. 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
partial integration means that, having given a
differential expression involving two or more independent variables, from the connection, 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 Integrating this with respect to have u  x2 + .r// j, + considering
3 x + ?/ u, having given as constant, we 0, denotes the constant of integration. But since y was remay involve y. We
garded as constant during this integration,
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 c/> c/> where </>(?/) denotes an arbitrary function As another problem let of y. us find u^\ I (x 2 This means that we wish to find dxdy
491 u, having given DIFFERENTIAL AND INTEGRAL CALCULUS 492 Integrating first with respect to
()U where y, regarding x as constant, we get 0,2, an arbitrary function of x.
integrating this result with respect to x, regarding y as con \l/(x) is Now
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, *=f(*,V) (1) the equation of a surface, as KL. Take some area 5 in the A'OYplane 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
is : 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
we have a network of
f lines, in as in the figure, that S being composed of each of area
AJ Ay. This construction
divides the cylinder into
a number of vertical columns, such as MNPQ, rectangles, whose upper and lower
bases are sponding
of
in the
S' corre portions networks 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
is replaced by the right
numerical values. Thus the column
MNPR, the upper base being in a plane through P parallel
prism
T
to the X01 plane.
= z = /(x, y), and
If the coordinates of P are (x, y, z), then
: MNPQ MP therefore Volume (2) 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 \ approximately equal to
// rf V ; that is, y'vv j( (3) . where the double summation sign ^^ indicates that values of y must be taken account of two variables x, summed up.
If now in in the quantity to be 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 V = Hm (4) A./ A/y (I  (I We show now that this limit, can be found by successive integration.
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 The ; and JTL'K'. The thickness
along the curve
that
z = /(x, y)
; HI
is, of this slice are found by is Now the values of z
~ OD in the equation Ax. writing x along /// Area Hence FIHG  f /(OD, y)dy. JltF The volume of the slice to that of a prism with base Ax  area under discussion is approximately equal
FIHG and altitude Ax that is, equal to
; FIHG = Ax f
JDF /(OD, y)dy. DIFFERENTIAL AND INTEGRAL CALCULUS 494 The required volume of the whole solid is evidently the limit of
sum of all prisms constructed in like manner, as x (= OD) varies the OA from to OB that ; is, OB / V= (5) /^DG dx / JOA may it Similarly, /m V=l
(5) (6) S*KV are also written 'in the JDF more compact fonn nOV nEU
and y)dy dx /(*, I I f(x,y)dx. JKW nlHi II JOA and r dy Joe no J(x, y)dy. JDF be shown that (6) The integrals I I I JEW Joe f(y, y)dx dy. 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. EW EV and
are functions of y.
Similarly, in (6) the limits
comparison of (4), (5), and (6) gives the result V=limf(x,y)&y. AxA*0 (A) Avo r 'i fl / . and where v\ of The second ;r. Now >
r L are, in _2 JV2 general, functions of //, i Hx.
and ?/i and y/2 functions integral sign in each case applies to the first dif ferential. Equation (A) is an extension of the
Art. 150 to double sums. Our
The result may be stated Fundamental Theorem of the following form. in definite double integral
' I
Ja2 /Or, J/ y)dy d2 ut be interpreted as that portion of the volume of a right cylinder which
and the surface
included between the plane, may
is XOY z^/0*,
the base of the cylinder being the
/7,/r?v><?
CM/IM.
/ A = 1*1, y = U2, XOY plane
= ai, x a 2
x bounded by the . similar statement holds for the second integral.
look upon the above process of finding the volof the solid as follows. It is instructive to ume J/), area in the 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 correspond to the variable whose differential is written first, the differentials of the variables and their corresponding limits being written
Before attempting to apply successive integrait is best that the student should acquire in the reverse order. tion to practical problems by practice some facility in evaluating definite multiple integrals. ILLUSTRATIVE EXAMPLE
/a /"A/a x~ I Jo Jo
Jo = I y)dy dx
' Jo (x I y)dydx. f O ( J f (.r Jo Jo
Solution, Find the value of the definite double integral 1. 1 v a  x' 2  f 1 ir )^ Anx.
Interpreting this result geometrically, we have found the volume of the solid of
as base and bounded at the top by the surface
on OAB cylindrical shape* standing (plane) z The = x + y. solid here stands on a base y (line the A'OVplane bounded by in OB) / v a~ y (line x^a (line ILLUSTRATIVE EXAMPLE ra  2. ILLUSTRATIVE EXAMPLE
" /*
/ JQ /*
I from y limits J OA)
BE) ,
f .. rom x .. limits. _ (a Verify rZ^V y}x
?v 2 la 2 dydx _ v/a " J_V a _
^ y2 x2 3. / ~\/a 2 dydx= JI
Verify / JQ
, , /* a xdydx=l
J ~ r2h/i~ \ay lo x dx= Jft
^
L / a 2 y}x' Jo Solution. I , b (a  . AH] ^ x Solution, . . , , x~ (quadrant of circle f I _
_ JVa^
2 ""v/a \xy\
L 2 \  9 x2 xdydx x2 a; 2 JVaax^ dx ^j.'i
 " ; DIFFERENTIAL AND INTEGRAL CALCULUS 496 In successive integration involving three variables the order of
integration is denoted in the same way as for two variables; that the order of the limits on the integral signs, reading from the
left, is the same as the order of the corresponding
variables whose differentials are read from the inside to the right.
is, inside to the c3 r2 c ILLUSTRATIVE FA AMPLE 4. Verify I I h Ai fr " nt> I r>
" />;> dz dy jy~ ' dj I ry* dz dy djr *~\ ** ~ nr^ j dz dy dr .n/ / ' /*2f ' dyds^'3 I Ml >2 2 dy dx z L J2 r2 r<\\ xu* ~15 xy I I *k J\ r'2.
/ f\\ J\
~ rjj = \.'h '! *} = J'2. I k J' r> / ~\ xy* dy \dx
1 35 10 in the following list the solid whose volume
1
the value of the integral should be described.
equals In Problems PROBLEMS
Work out the following definite integrals.
(.r + L> )<///
' = d.r 5. 6.  t ^ /(/ " '''' ^ "' "* l f 8. f (.r l./o t/
2/ rr ^ 4. " ' Ji 2 rr ?
\  ., <''/ = 9 7  ('" , /o //f///(/.r = ri r  ' ' 10. '. I ' ./o  11. PV' sin f"  Mt(>sf/ >7T /" p s n ^ d p dO
j / Jo
/^TT 13. dp { \(a . = /^a(1 / ) r r '' (/p dij(Lr= /r)(cos /J  a 1'. ^/ .'o //'c (/c c/// (^ = i ^  f / t(Y*(h 2
p sin ^ / ~ 16. ,\ " 2) ''" At 'ii A) 15.  + ~" (j () .V .//, 12. (JO ;o f a^ia*  6 3 ). cos a'). I. MULTIPLE INTEGRALS
c
17. r^ c " I / Jo x ~ * jrdzdjrdii I 3*5. 'o <Jtf
1 = 497 " 1 x 1 r f Jo Jo /o r is. /? rr^ r
/2 Jo  I /
Jo r + T ' f f f 20.
. "'zdzduds V " ' ' *d: <ln itr = i r 1  Jo Jo Jo 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 volume, for the physical interpretation of the result, depends on the
nature of the quantities represented by z, //, z. If x, y, z are the coordinates 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 necesof
sarily involving the geometric concept volume, we observe that the variable z
does not occur explicitly in the integral, 7 and therefore we may confine ourselves
to the A'OVplane. In fact, let us conin the A'O Vplane, sider simply a region S and a given function /(z, y). Within this of
region construct rectangular elements
area by drawing a network of lines, as in Art. 244. Choose a point
of area Az A?/, either with in the rec(z, y) of the rectangular element, tangle or on its periwrtrr. Form the product /(z, ?/)Az A//, and similar products The these products. for all other rectangular elements. result  0, and A?/
Finally let Ax
write the result Sum up is * 0. We lim (1) y V/(x, ?y)Az A?y .'' y}dx dy,
ff/Yz, Aa'O*""'^
y>Q S over the region S.
double integral of the function f(x, ?/) taken
lefthand member in (1) was found by
By (A) the value of the
no negative values for the
successive integration when /(y, y) had
of Art. 244 will hold, however, if the portion
region S. The reasoning and call it the DIFFERENTIAL AND INTEGRAL CALCULUS 498 =/(z, y) lies below the plane XOY. The limit of
then be the volume with a negative sign. The
in (A) will give the same negative number.
Finally, if
integrals
is sometimes positive, sometimes negative for points of S, we
/(x, y)
may divide S into subregions in which J(x, y) will be either always
positive or always negative. The reasoning will hold for each subS' of the surface z sum the double will region and therefore for the combined region S. Hence the conclusion: the double integral in (1; may be evaluated in all cases by
successive integration . 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 coordinates. The problem of plane areas has been solved by single integration 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,
It proceed as follows.
Draw a network of rectangles as before. Then, in the
Element of area = Ax A//.
(1)
If A is the entire area of the region S, obviously, by A OB) = lim
Ax0 dxl;t Ay  f fd
^"f
JJ Ay *0 of function f(x, y) any region = (1), Art. 245, dy. S Referring to the result stated in Art. 245, The area figure, we may say : the value of the double integral of the is 1 taken over that region. Or, also The area equals numerically the volume of a right cylinder
unit height erected on the base S. (Art. 244.)
of
: The examples show how the limits of integration are found. ILLUSTRATIVE EXAMPLE 1. Calculate that portion of the area above OX which
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
is rAC
I JAB djc dy = dy pAC dx I JAB area of a horizontal strip of altitude dy. Next, integrate this result with respect to
horizontal strips. In this way we obtain y. This corresponds to all A= rOD rAC
Jo \ JAB dxdy. summing up MULTIPLE INTEGRALS
The 499 AB and AC are found by solving each of the equations of the boundfor x. Thus from the equation of the
= AB  y, and from the
line, limits ing curves jr 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.
in Or we may begin by summing up the elements dr dy
a vertical strip, and then sum
up these strips. We
have shall then ~ '* ^ IJ d r
' / / ~
I ^"r ~ jl> ^ r In this example either order of integration
case, as the following example shows. " may i ~ be chosen. This is not al ways the ILLUSTRATIVE EXAMPLE Find the area 2. the first quadrant bounded by the
curves
=$
j;2 + 2,2=10, Solution. Here we tf .raxis arid in the Xf integrate with respect
to x to cover a horizontal strip, that is, from the
then have, for the
parabola to the circle.
first We entire area. A i<{ = I d.r I Jo (///, .'IK; since the point of intersection
}JG, solve y~ = 9 x for .r. Then To find ///, solve x~ y f 2 = 10 for x. HI = + VlO jc To is (1, ,3). We find get ?/'. Hence ^_ (' y C i fly ,1*, \  ^ li 27 If we integrate tical strips, two first with respect to integrals are necessary. y, using ver Then / /^ Io na'v'j dydx+ / / Jo Ji The order of integration should be such that
is given by one integral, if this is possible. area The examples above show that we
or set A=ffdydx the 3 i/M
 =6.75. Ans. DIFFERENTIAL AND INTEGRAL CALCULUS 500 according to the nature of the curves bounding the area. The figures
illustrate, in a general way, the difference in the summation
processes indicated by the two integrals. below PROBLEMS
1.
l
,'J \f (b) Find by double integration the area between the two parabolas
9 //, (a) by integrating first with respect to y
25 jr and 5 jr''
; by integrating with respect to first ~^LL ,<\ AH*, (a) .r. I 5 </// f/.r I ^5 /V
(b) ; I ^
' (i.r \ dy 5. 25 *> Calculate by double integration the finite area bounded by each of the
following pairs of curves.
2. =: 4 y y' =
~ 3. 2 4. 4 2 .r  2 .r, .r .r, jr 2 .r, .r 9 + .r 2
// ?/ = (j 10. jr^ 2
, + 2 9.  j, 2 8. ~ y ', j = r' 2 1 .r 2 6. y'
7.  4 ?/ 5. .r //* 2 J, = )// + = // =a', 2 9  \ + // ?/~. // 4 8
JT 0.
4. //'' 2 9. //. 12 + = + (> = 2 4 + ^ 4. H // // 2 jr. = jc. 2 = 48. s. :i // = .r, = a 2 (7rl). 0. a. + = a\ + = a.
= 2
= 6 ~ r
12.
 2 =
13. j = 6
 x.
2 =
14. 4
x'\ u  r< 11. .r' jr // // :i j ?/ j, ?/ .r // ?/ , 2/ : 16. .r. 2
?/ = .r + 4, 2
/y = 42 2
.r 18. ;r + = 25,
2 =
a = 14,
2 ?/ 17. (2
.r. 3 7r)a 2 16. . // 15. j 2 .r)// 2 7/ .r ?/ 3 x = 16 x?.
?/ = or.
+ = 36.
2 27
, 2 ?/ . MULTIPLE INTEGRALS
247. 501 Volume under a surface. In Art 244 we discussed the volume
bounded by a surface of a solid ; (l) = M 0), the A'Oyplane, and a cylinder. The elements of the cylinder were
The
parallel to OZ, and its base was a region S in the A'OVplane. volume of this solid by is, 04), V= (2) c, y)dx dy. and the limits are the same as for the
The volume of a solid of this type is the "volume
area of the region
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'OFplane itself.
Note that the element of volume in (2 is The order of integration
S. ) a right prism with base dy and altitude d.r z. ILLUSTRATIVE EXAMPLE 1. Find the volume
bounded by the elliptic paraboloid z= 4 (3) and the AW Solution. (3) for Solving
2 (4) = = 4 a >J 2 \r 0, (5) we z, 1 we J get //'. obtain 4 Letting  16 ~4.r 2 plane. JT' + // = 16, the equation of the perimeter of the base
which
of the solid in the
plane. Hence by (2), using
the value of z in (4),
is AW (6) V The limits are taken for the area  4 1 (4 I  x  2
ij' \ }dy dx OAB of the ellipse ILLUSTRATIVE K \AMPLE 2. Find the volume
solid bounded by the paraboloid of revolution
r* (7) + ?r  = 16 TT. Am.
the first quadrant. (solving (8) for #), and OA = 2 a. (5) lying in of the nz, the A'Oyplane, and the cylinder
2 =
2 ax.
.r 4 !i
(8) and finding the limits for the area of the base of the cylinder (8) in the XOY Solving Solution. plane, we get, using (7) for 2, (2;,  l dy dx
Jo 7ra :< . Ans. Jo For the area ON A (see
These are the limits. figure), MN = v2 ax  x'2 , DIFFERENTIAL AND INTEGRAL CALCULUS 502 PROBLEMS
Find the volume under 1. z =4 2 x' above , V= Ann. 2 0,
X 2 and within y' = 4 x. (4 z 2  = 17.24. Z f f' 1 x 2 )dy dx Jo Jo Find the volume under the plane x 2. x2 f 2
?y = = f z 2, 4 x Find the volume bounded by the plane  Find the volume
and within x'2 5. by 2 = 0, f a2 // f z j" = y
; 4, ^^ below by = a 2 0, z paraboloid elliptic 1 .r // = f z above 8, below 2 2 J
// 3 0, Ans. // . % ira*. = ^ and
l A/i,s\ 7. Find the volume under the plane x f y
= 8.
2
between the planes j: f 2
8, x co \ az, 4s. . TT. iabCu = ^ Ans.
//'' 8 and the 1 r 0. z = dr .r;c/?/  f bounded above by of the solid Find the volume under the 6. and above f a ordinate planes.
4. Plnd the volume bounded above by
2
laterally by // = 4 .r. and within 0,  (2 / J 2 J()
3. z 1 F = 2/ An*. above /A r2 4. TV. and 170^.
a2 Find the volume bounded by the cylindrical surface x 2 f ac
and the planes x f // = a, // = 0, c = 0.
yl/^s'.
8. A 9. 0, and = a.r. 2
// f ?/ =2 A surface f 2 al>ove , = ?/ a = a2 9\/3)a 2 a^, = ^ 2 2 ?/ 2
// + ^ 2 0, .r = above 4, 2 8 // r = + 0, yra % 3 ?ra Find the volume under
y* = 2x. 14. x* + 15. and The axes of two 4 z = 16 4 j ?/
' , . az, l . the cylindrical
3
$ a above = ^ 0, . and within
A?w?. ']! TT. circular cylindrical surfaces intersect at right angles their radii are equal (= r}. Find the common volume. y
16. Find the volume of the closed surface jr + y + on each coordinate plane is the astroid, Chapter XXVI.)
:J 17. . :i \ Ana.
2 5 Find the volume. 0. 2 . and within the 0, 2
?/ :i above Ans. bounded by the paraboloid .r
0, z =
a.r, and the planes .r . and within .r. 2 ;i and c/, Anx. ax. // 3 .4j/s. solid is
2 ?/ 2 x ,: + 2 2 a 4 TT ax. Find the volume under 12. .r (6 { = 4 ar, x == Ann. wit?iin the cylindrical surface cylindrical surfaces
13. 2 Find the volume. Find the volume below 11.
f // 2 bounded by the surfaces is Plnd the volume below the cylindrical surface 10. = x2 2 within lies 2 solid 4 Find the volume common to 2
?/ 4 z 2 ' 4 a.r =a and x } ^ .4??.s\ r3 . 3 (The trace
3
Ans.
$ ?ra
. . 2 f 2
?/ (2 =
TT 2 + a.r.
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 bound the curves 'which the region,, or area, con corned. SECOND STOP. At any point AJ rectangular element of area by irithin the I\.r, y) THIRD STEP. Work out the function f(jr,
AT A?/, gives the required property for the
FOURTH STEP. The required integral is
y)djr /(.r, ffi taken over the given region, or area.
are determined in the same wanner
249. Moment Art. 177 by area construct the A//.
//), which, 'when, multiplied
rectangular element of area. < of integration and limits
in finding the area, itself. The order
a,s This problem
Double integration and centroids. of area single integration. is
is treated
often in more convenient. We The moments follow the rule of the preceding article. of area for the rectangular element of urea are, respectively, AJ
AJ .r y Hence A?/, with respect to OV, A'//, with respect to V. for the entire area, using the notation of Art. 177, M = M The centroid of the area is given x x y =j*J*ydxdy, =
area " we have * (C) (D) () ffxdxdy.
by area In (C) the integrals give the values of the
of the functions . double integrals f(x, y) = an(l y respectively, taken over the area. = fte, y) x > (Art. 245.) For an area bounded by a curve, the xaxis, and two ordinates
(the "area under a curve"), w ^ derive from (C)
rv (D I fft J()
r\> My I Ja Jo  rb
2 i y' I dx, Ja rv ==
I y dy dx r^> x ^ dx= Ja c I xy dx. DIFFERENTIAL AND INTEGRAL CALCULUS 504 These agree with (2), Art. 177. Note that y in (!) is the ordinate
on the curve, and its value in terms of x must be found
from the equation of the curve and substituted in the integrand beof a point fore integration. ILLUSTRATIVE EXAMPLE. Find the centroid of the area
j and the
bounded by the semicubical parabola y~ in the first quadrant ( straight line y were found x. The order and the limits of integration
Illustrative Example 1, Art. 24(5. Ileuce, Solution.
in using (C), r 1 c r 1 1
?j i ., *
' Jo ~J() Jy
2 M =11
Jo J y v djc dy = a
i A Since we have, from
250, = fa. 1 area i ,,, ?=J0.48, (D), Theorem volumes y~)dy (.v / ,/o y of Pappus. A of solids of revolution // = A> = 0.42. An*. useful relation between centroids and in the following theorem. is expressed // 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 xaxis. The rectangular element of area within the region X at P will generate a hollow circular cylinder
A Y is given by whose volume
_\ y = TT(// + L> A//) AJ 2 Try A.r. we get
Ax A//. Factoring and simplifying, AV2 TTG/+ 2 A//) " Now, in (1), Art. 245, (x, y) in/(j, y) is a point either within the
\ A//) is a point on
rectangle PQ or on its perimeter." But (x, y
the perimeter of PQ. Therefore let f(x, y = 2 Try. Then A V has the + > form fix, ii) //) Ax A//, and, by rx = Finally, using (D),
(2) we V,  (1), Art. 245, and (C), 2 get 2 iry A, MULTIPLE INTEGRALS 505 where A is the area of the region 5. The righthand member is the
product of the area by the circumference described by its centroid.
Hence the theorem is proved. We write the result If by = V (3) two of the quantities V, y, 2 iry A  A. are known, the other can be found (3). ILLUSTRATIVE EXAMPLE. Find the centroid
by the Theorem of Pappus. of the trapezoid OMPB of the figure a = formed 8, R = Hence, by is = 5, r (3), OMPR = J(3 + 5^,8 = 32. Revolving the figure about OX, the
a frustum of a cone of revolution. Hence, by (12), Art. 1, since Area Solution.
solid "y 3, = ~~ = f^ =
19J
2 7T4 2.04. Revolving the figure about OY, the volume generated
the difference of the volumes of the cylinder generated
and the cone generated by the triangle HCP.
by M is OrPM Vu =
Hence, by the theorem, The centroid 7T  12S TT 3 _ x 3* (4j, 12.04). is 320 Hence Arts. PROBLEMS
Find the centroid
1. 2.
3.
4.
5. 6.
7. 8.
9. 10.
11. 12.
13. 14. of the area bounded by each 4
y
(Area in first quadrant.)
y =
2
=6x
y
= 2  3.
=4x2 = 4
r  2
x
f 4  0.
= ^ 2  y 43 = 0.
 2  3, = 2  3.
y = x
= 2, = 0. (First quadrant.)
4y* =
= r f = 2, x = 0.
3 =
= r.
J2 2
 3 2 2 2 = 9 x.
4
= 2 x, = r ~ x 2
= 6,
?/ = 8 x, x +
2
 x.
y = 4
y = 5
= 6 x  x 2 x 4 = 6.
y
.r 3 jr. , jr // 2 jr ?/ ?/, ?/ .r. ?/ j'~, ?/ , .r , 2 .r jr jr f .r ?/ ?y ?y 2 .r, ?/ ?/ ?/ // , ?/ j ?/ , 2 . ?/ ?/ ?/ 2 .r, , ?/ of the following curves: Am. ([ Ji, (. \). !j 5). (1, I) 0. g) 0. 'e 1
) (2,f).
(I!.
/ H
T,' 3" 1 \ TOV'
4 /J
V A)1 V2 ON dn) / 9 2 7\ Vii)' J'o>  H
(4). (U (.0). DIFFERENTIAL AND INTEGRAL CALCULUS 506
15.
16.
17.
18.
19. 20. 21.
22. 23. 24. = 4 y y 2 y = x.
= 4 x x2 # 5  2
2 =
4 x, 2 x  y = 4.
= x  2 x  3, y = 6 x  x 2  8.
2
= l.
* + 2 = l, z +
x 2 + 2 = 32, 2 = 4 x.
2 =
4 x, 2 x + y = 4.
2
x + 2  10 x = 0, x = y.
= 6 x  x2
2
x2 =
x f y = a
(Area in first quadrant.)
x , ?/ , (V, I). (3, I). 2 (2, 1). ?/ (0.58.5. 0.585). jy ?y ?/ i/ 2 ?y ?y, ?y^ 256 a \ /25B(7 3 . = a*, x = 0, the controid 26. P'ind a(9 . / 3 3 1 = a;. ?/ 25. x * f x Ans. sin 0), y = <j(l y = (' V
\5 5/ 0. the area under of cos \31f> Tr'ttlo 7T/ one arch of the cycloid 0). / * ^ 5 o\ * 71 ' 6 I 27. Ans. Distance from diameter = 3 28.
ellipse 29. erated / Using the Theorem of Pappus, find the centroid of a semicircle. Using the Theorem of Pappus,
a2 \~~l which lies in the TT find the centroid of the area of the first quadrant. ft ..,,,,. , , \3 , TT 3 7T/ Using the Theorem of Pappus find the volume of the torus gen2
2 =
a 2 (ft > a) about the ?yaxis.
f ?y
ft)
by revolving the circle (x
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. The 251. Center of fluid pressure. 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 perpendicular 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 ' =W r
I Ja b yxdx. D MULTIPLE INTEGRALS
The point of application of /' is 507 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, dP = (2) of TT>// A:r. The turning moment of this force about the axis OY
its lever arm OK (== .r), or, using (2), is the product dP by Turning moment (3) Hence we have, of dP = x = Wx 2 y Ax. dP for the entire turning moment for the distributed fluid pressure, But the turning moment
Hence
( Solving for /<> and using / 1 Wx~{/dx. of the resultant fluid pressure x )P (5) r' moment Total turning (4) (1 ), W
we 2 I x' P is x^P. ydx. get the formula for the depth of the center of pressure f *x 2 dA
=ao
xdA (6) /'
Jii element of area = ydx.
where dA
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.
about It is called the moment of inertia of the area ABCD OY. The letter / is commonly used axis, and a subscript
becomes is for moment of inertia attached to designate the axis. about an Thus (6) DIFFERENTIAL AND INTEGRAL CALCULUS 508 The moment usual notation for of inertia about an axis I is (8) in which
r (9) = distance The problem dA from the axis I. is one of many which lead to moments
In the following section the calculation of moments of of inertia. by double and inertia of the element of this article single integration is explained. Applications are also given. Moment an area. In mechanics the moment of
is an important concept. The calculation of moments of inertia will now be explained. We follow the
252. inertia of of inertia of an area about an axis rule of Art. 248. A For the elementary rectangle PQ at
P(j, y) the moment of inertia about OA'
is defined as
2 (1) '// and about the AxA*/, ?/axis it is (2) Then, if Ix entire area, The and I y are the corresponding moments of inertia for the
(8), Art. 251) we have (compare and radii of gyration r x
2 rv _ are given by
*U 2 area area In (E) the functions whose integrals are taken over the area are,
= y 2 and f(x, y) = .r 2
/(.r, y)
Formulas (E) become simple for an area "under a curve," that is,
an area bounded by a curve, the xaxis and two ordinates. Thus we
~b
obtain
r ry
respectively, . , \, I*=
Ja
(3) /*& ,= /
J() V 2 dydx=l rv
\ J a Jo y*dx, I J u. rb x 2 dydx= / Ja yzydx In these equations y is the ordinate of a point on the curve, and
value in terms of x must be found from the equation of this curve
and substituted in the integrand.
its MULTIPLE INTEGRALS
Formulas moments for 509 of inertia / are written in the form I=Ar*, (G) A= radius of gyration. Solving (F) for Ix and I v
area and r
where
will give this form.
Dimensions. If the linear unit is 1 in., the moment of inertia has the dimensions in. 4 ILLUSTRATIVE EXAMPLE
for the area of Illustrative Solution. By . (F), r x Find 1. Example /,,
1, and rv are lengths, in inches. /, and the corresponding radii of gyration Art. 246. Using the same order of integration and
we have, by (E), the same limits as before, Since A  = area J 1 n , we by find, rt ILLUSTRATIVE EXAMPLE 2. (F), = 0.48, rt = 0.53. Am. Find I s and I v for the parabolic segment BOC in the figure.
Solution. parabola With the axes of coordinates as drawn, the equation of the bounding is = 2 y' (4) B(a,b) 2 px. a point on the curve, we get, by substitutSince B(a,
2 =
2 pa. Solving this equation for
a, y = b in (4), b'
ing x
2 p and substituting its value in (4), we obtain
b} is y (5) 2 = &*jr
, or ~ bx* O 2/ a ^7 under the parabquadrant will be half the required
moments. Hence, using (3j, and substituting the value of
y from (5), we get The moments ola OPB of inertia for the area in the first 15 For the area segment, we find = = results are in the form Hence, by The of the (f), rS 6s , (G). and Ans. /, = A6 2 , DIFFERENTIAL AND INTEGRAL CALCULUS 510 Y In the figure on page 325 the axis
If we denote of pressure, this axis in is by if r and ha (7), =
= any figure radius of gyration about the axis s,
depth of centroid below the axis s. ILLUSTRATIVE EXAMPLE Find the depth 3. = OX Choose axes Compare OY and as of the center of pressure  8 Hence, by
definition of dA = y, (8), Example 2, Art. 179.
shown, and draw an elementary horis at the water level be r. = 0, 2 x dy. Art. 251, moment on the Illustrative Let the distance of this strip from the axis zontal strip.
r the surface of the fluid.
then the depth of the center lies in s, Art. 251 trapezoidal water gate of the figure.
Solution. by and by the of area (Art. 177), we have
(7) (8) M, = is y = 2 x  8.
The equation of
Solving this for x, substituting in (7) and
y = 4, we obtain AB /. = 4 f (8  2
i/) (8 (8), + A) M. = f
Hence, by Art. 251, x (7), 253. Polar moment PQ mentary rectangle
2
and OP that is, tt 4 = (64  6.09. of inertia. y*)dy and integrating with limits y y)dy = 1429 J, = 234. Ans. The moment about the origin O is of inertia of the ele the product of the area , (1) (^ + y
\ Hence, by Art. 248, for the entire area
(2) We 2
ij )dx dy. may, however, write the righthand
as the sum of two integrals, for
is clearly the same as member
(2) (3) lojy^dxdy+JJy* /. Hence we have the following theorem.
The moment of inertia of an area about the origin equals
its moments of inertia about the xaxis and the yaxis. the sum oj MULTIPLE INTEGRALS 511 PROBLEMS
Find /, and / for each /*, The 1. semicircle which bounded by x 2 The 2. /A
(0> m
0) /i, The y 2 =r is to the right of the ?/axis 2 . . ' A whose + The ellipse 5. The area in the a2 2
b = first iy "~: vertices are Ah'
= ~'
2 Aa'2 T v '^"iSr (0, 0), (b, a), (b, 0).  r
1x 8. 4. If j LS> ~2/ right triangle and which is
2
__ . _ Ar and base a whose vertices are a\ li , /ins. isosceles triangle of height h a\ 'r' 2/'\ 3. + of the areas described below. ^ a2 2 quadrant bounded by y' 44 5
6. y* 7. The = area included between the ellipse +  2y. The 2 77 ;
Am. f (y 9.
( , 10. + 2)* = 3)* = The area f Aw*.
circle x'2 4. + 4ns betwee'n the circle x 2 f 2
7/ = r
/x . Aws.
11. The entire area bounded by xf h y* = ^ 2 a^. circle 53 7 A ^
ano x 2 5 A + 4 = 1.
4
f<2 r
7V , 19 ^ 2
?/ = 36 , _17A
/ 4 and the /, OA>
T2 ellipse = r circle j^ r A  16 and the circle 2/ and the 4 A _239 A
/,6Q . The area included between the + x2 circle l. 1 TT
9 16 7 and the 19 r , The area included between the 1 y^^,/^^. Arts. 8. = ^i2 j2 area included between the ellipses 2 *4 ' ?/ .r, j
J* * Q y  = 4r~ ^ =
= 0.
4
x = 4,
~
~ 4ns. /x 1.  Ab * 7
' T7 36 23 A Ans. I x , r
/ = 77 f 2 f = 1. 16 tf = Iv 53 A y = Z_4.
64 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
12. 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
Ans. 7.19 ft. below the surface of the water.
the top of the gate. DIFFERENTIAL AND INTEGRAL CALCULUS 512 Find the depth of the center of pressure on the end of a horizontal
ft. when the depth of oil is (a) 2.5 ft.; 14. cylindrical oil tank of diameter 5
(b) 4ft.; (c) 6ft. ==1 .47f t .. (b) approximately 2.4 (a) = 3*95 ft. ; ft. 254. Polar coordinates. Plane area. When the equations of the
curves bounding an area are given in polar coordinates, certain modifications are necessary.
The area is now divided into elementary portions, as follows: Draw arcs of circles with the radii differing by Ap. Thus, common in Fig. 1, r draw FOR = is from O such that the angle between any two conthe same and equal to A0. Thus, in Fig. 1, angle A0. The area will now contain a
such as PSQR in Fig.
Let PSQR = AA.
circular sectors
(1) x 1 radial lines secutive lines p, with successive OS = p + Ap. Then U A FIG. center OP = AA FOR large number of rectangular portions, 1. Now AA is the difference of the areas of the and SOQ. Hence == i ^ Ap 2 A0. f(x, ?/) of Art. 245 is to be replaced by a function
coordinates. Let this be F(p, 6). Then, proceeding as
using polar
in Art. 245, we choose a point (p, 6) of AA, form the product The function F(p,
for each AA 0)AA within the region S, add these products, and finally let
> 0.
It is shown in Art. 258 that the limiting value and A0 Ap of this double sum may be found by write (compare
(2) (1), lim
Ap AH  ' successive integration. Art. 245) AA =f (V(p
JJ , d)pdpd6, We now MULTIPLE INTEGRALS
and call it the 513 double integral of the function F(p, 8} taken over the region S. Note The We in (2) that the value of by p dp integral AA in (1) has been replaced in the dO. simplest case of (2) is that of finding the area of the region S. then have A= (H) CCp dp dB =JTp dO dp. if we think of the elements (checks)
with dimensions p dd and dp, and hence of area
as being rectangles These are easily remembered p dd dp. The figures below a general way, the difference in illustrate, in the processes indicated by the two integrals. In the first, we first integrate with respect to p, since dp precedes KGHL dd, keeping 6 constant. This process will cover the radial strip
OG and p OH, found
in Fig. 2, p. 512. The limits for p are p
by solving the equation (or equations) of the bounding curve (or = curves) for p in terms of = Z JOX being The second
spect to ABCD
Then 0, and 6 Then integrate varying 0, the limits
IOX.
(2) is worked out by integrating with re0. = Z integral in p remaining constant. in Fig. 1, p. = This step covers the circular strip 512, between two consecutive circular integrate varying p.
the area is bounded When arcs. by a curve and two of its radii vectores
we obtain from the first form (area swept over by the radius vector), = r ft r?
I Jot =% r
+ pdpdd Jci t/0 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. DIFFERENTIAL AND INTEGRAL CALCULUS 514 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. The Solution. A(T, ~j. points of intersection are  and B(T, Use the V first form in (3). The limits for p are p
for 6 they are = O//2r
~ and ^
o o costf; Ans.  ILLUSTRATIVE EXAMPLE
outside the circle p
Solution. From Find the area inside the 2. circle p =2 r and cos 6 r. Illustrative Example 1
above, we have 7L 2
2
r(4 r cos 6 255.  r*)d6 = r 2 (i Problems using polar coordinates. IT + J Vs) = 1 .91 r\ Ans. There should now be no difficulty in establishing the following formulas:
(1) M (2) M x v = ffp = CCp*ca&8dpdO. (3) /,= (4) 2 sin 9 dp d6. 2 / CCp*siY\ 6dpd8. (5) The order of the differentials will have to be changed
with respect to 6 is performed first.
1. On account
now worked out. ILLUSTRATIVE EXAMPLE
of inertia of a circle are Let a
of inertia
(6) where 7 = radius. Then, by (5), the polar
with respect to the center is of if integration important applications the moments moment = A = area ' of the circle. Also, since I,
Art. 253, = Iv , ~TV\~\ V by symmetry we have, by
, (3), / / / / / x MULTIPLE INTEGRALS 515 In words The polar moment of inertia of a circle with respect to its center equal*
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
: the of the radius. ILLUSTRATIVE EXAMPLE
p* = 2 Solution. Since OA" = have y is Find the centroid 2. cos 2 a' of a loop of the lemniscate 6. an axis of symmetry, we 0. r i V \A^\<
Jo * r (i t >H 2 X ft pdpd.6 f Jo " p" Jo Vros2 f V'cos '(} * 3 3 vf f " (1  2 sin 2 0)* cos 9 dO by Jo Hence z = = ILLUSTRATIVE EXAMPLE
p = 2r cos = av 2 0.55 a. Find 7 3. (5), Art. 2 Ana.
over the region bounded by the circle B. Solution. Summing up for the elements in
the triangularshaped strip OP, the plimits
are zero and 2 r cos
(found from the equation
of the circle).
Summing up for all such strips, the 0limits are  ^ and  Hence, by (5), . Or, summing up first for Ann. the elements in a circular strip (as QR), we have PROBLEMS
Find the area inside the
4 p cose = 3.
1 2. circle Find the area which
p =f . is circle p = $
" and to the right inside the circle Arts p = 3 . of the line 3(4, 3Vi).
ID cos and outside the A 718. 3(2 7T + 3V3) DIFFERENTIAL AND INTEGRAL CALCULUS 516 Find the area which 3. circle p = cos = inside the circle p is 3 cos 6 and outside the
Ans. 2 6. 4. Find the area inside the cardioid p
the line 4 p cos 6 = 3. = 1 + cos 6 and to the right of Ans. ^ 9\/3 + 16 g
5. Find the area which the circle p
8. oid = f cos 6 and outside 1. I+ AnK Find the area which p=a + =1 inside the cardioid p is inside the circle p is IT. =1 and outside the cos0. 2. cardi AnSt 2 __]r t 4
7. Find the area which cardioid p
8. ola p(l =1+ cos Find the area which + cos 0) = inside the circle p is = 3 cos 6 9.
is inside the circle p and outside the
Ans. IT. and outside the parab 1 1. !_?. 4ws
in.s.
. 3 2
9. Find the area which the parabola p(l
10. = cos 0) Find the area which side the circle p
11. + = 1 =1 f cos 6 and outside x_ 3jr,4. 43 1. inside the circle p is = cos f sin cos and outAns. 1. Find the area which cardioid p inside the cardioid p is inside the circle p is = sin 6 %. and outside the 0. Ans. 1 4
12. Find the area which outside the circle p
13. Find th = is inside the lemniscate p 2 2 a 2 cos 2 6 Ans. 0.684 a 2 a. area which outside the parabola p(l and  is inside the cardioid p cos 9) = = 4(1 f . cos 6) and Ans. 5.504. 3. Find the area which is inside the circle p = 2 a cos 9 and outside
a. Find the centroid of the area and I f arid /.
the circle p
14. + 15. Find the centroid of the area bounded by the cardioid p = a(l +
16. Find the centroid  ^ Ans. x cos0). = g ^ bounded by a loop of the curve the area bounded by a loop of the curve of the area = acos20.
105
17. Find the centroid of = acos30. A
Ans  x ir ~ 81 Via.
SO IT MULTIPLE INTEGRALS
18. Find for the lemniscate p 2 / 19. Find I r for the cardioid 20. Find /, 21. Prove from and I v for
(1), h of the cos o 256. General 48 (3 TT + 8)a 2 . = a cos 2 0. p Ap SO from p Ap SO by an infinitesimal of higher "differs member of (2), Art. 254, may for finding the areas of curved surfaces. The
164 applied only to the area of a surface of
shall now give a more general method. Let method in Art. We revolution. ~ ! 6}. curve p order" (Art. 99). Then SA in the lefthand
be replaced by p Ap SO. (Proof omitted.) method given Am. 0. 1 Art. 254, that  SA a 2 cos 2 = a(l p one loop lim
mil and therefore  51 z=f(jc,y) (1; be the equation of the surface KL in the figure, and suppose it is
required to calculate the area of the region S lying on the surface.
Denote by tf the region on the A'Oi'plane which is the orthogonal
projection of S on that plane. Now pass planes parallel to YOZ and
f f XOZ at common distances
Ax and Sy respectively. As
in Art. 244, these planes truncated prisms form 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 (as AB). This rectangle also forms the lower base of the prism. The coordinates of P are (x, y, z). Now consider the plane
tangent to the surface
at P. Evidently the same rectangle AB is the projection on the
A'OYplane of that portion of the tangent plane (PR) which is in KL tercepted by the prism PB. Assuming 7 as the angle the tangent
plane makes with the XOVplane, we have Area AB = area PR cos 7, of a plane area upon a second plane is equal to the area of the!
["The projection
[portion projected multiplied by the cosine of the angle between the planes or Sy Sx = area PR  cos 7 DEFERENTIAL AND INTEGRAL CALCULUS 518 Now 7 is and (3), we have Art. 4, cos 7 1 =
\f)x) L Then OZ and a line from perHence from (H), Art. 237, and (2) equal to the angle between pendicular to the tangent plane. Area PR = 1 1 W + (Y+ (\~
\tixl
J
 L J Ay Ax. \vy! This we take as the element of area of the region
the area of the region S' as S'. We then define 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 XOYplane
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
.YOYplane precisely as we have been doing in the previous sections.
Before integrating, the expression to a function of j and y only, by using the equation
curved surface on which the area lies.
If it is more convenient to project the required area on the XOZplane, use the formula must be reduced
of the (/) 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'OZplane.
Similarly, we may use S
the limits being found from the projection of the required area on the VOZplane. 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 XOYplane, eliminate z beticeen the equations of the surfaces whose inter sections
form the
boundary of the area. x to find it eliminate ire Similarly, ij to find the projection o)i, the XOZplane, and on the YOZplane. This area of a curved surface gives a further illustration of integration of a function over a given area. Thus in (/) we integrate the
function over the projection of the required curved surface on the YOVplane.
As remarked above, (J) and (AT) must, be reduced to
I ILLUSTRATIVE EXAMPLE z}dy dz, of the equation of the surface required curved surface on which the lies. Find the urea 1. j>2 by double I ff(y, by means respectively, and z)dz dx lf(jr, __ y'' _ ~? of the surface of the sphere _ r? integration. Solution. Let ABC in the figure be one eighth of the surface of the sphere. Here dx
s and
. (f )'=
\dy/
The by x = projection of the area required (= OB) ; y = (= OA ) on the A'OVplane & + ^ r (= 7M ; is AOB, ). Integrating first with respect to y, we sum up
the elements along a strip (as DEGF) which is
also projected on the A^OYplane in a strip (as
MNGF) that is, our ^limits are zero and
all MF ; x 2 ). Then integrating with respect to x
(= Vr 2
sums up all such strips composing the surface ABC
; that is, our xlimits are zero and stituting in (7), A=
8 we J Jo
Trr 2 . ( r). Sub get r p' A=4 OA r VrAns. _ d y dx
2 x' y 2 irr 2 %
JfU. 5 a region bounded DIFFERENTIAL AND INTEGRAL CALCULUS 520 ILLUSTRATIVE EXAMPLE 2. The center of a sphere of radius whose base of a right cylinder, the radius of is  r is Find the area on the surface of the surface of the cylinder intercepted by the sphere.
Solution. Taking the origin at the center of the sphere, an element of the cylinder for the zaxis, and a diameter of a right section of the cylinder for the .raxis, the
equation of the sphere is + x'2 and 2 y'
2 of the cylinder x' + z'2 = r
+ y' = rx.
2 , 2 ODATB is evidently one fourth of the cylindrical
surface required. Since this area projects into the
r
r
on the A Ol plane, there is
semicircular arc ODA no region S from which to determine our limits
hence we shall project our area on,
in this plane
Then the region N over
say, the A'OZplane.
#hich we integrate is OACB, which is bounded
2
2
( OB), and z + r x = r'
by z = (~ OA), x
the last equation being found by elim(= ACTt),
inating y between the equations of the two sur; y/ with respect to z moans that we sum up all the elements in
the zlimits being zero and Vr  rx. Then, on integrating
with respect to j, we sum up all such strips, the jlimits being zero and r.
Since the required surface lies on the cylinder, the partial derivatives required
faces. Integrating
a vertical strip (as for first 7V, formula (J) must be found from the equation of the cylinder. dy
^ Hence 2 x r dy _ ' ^ TH ?/ Substituting in (7), rr^^~ A=
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
rr r^'**
the cylinder.
dy dx
_ 9(
9 2
by
A
Am. 4rl /
j=======6(Tr ^)r.
, f Vr 2 Jo Jo a y right circular cylinders, r being the radius
of their bases, intersect at right angles. Find the surface of one intercepted by the other.
2. The axes of HINT. Take x 2 + two equal z2 = r2 and x 2 + y 2 = r2 as the equations of the cylinders. A
o
Ans. 8 r r
C f^r a *2
/ Jo Jo dy dx v r2 _ = , x2 o r. MULTIPLE INTEGRALS
3. Find the area of that portion of the
sphere
out by one nappe of the cone .r 4//. 521 .r 2 2 f Find the surface of the cylinder
plane z = nix and the A'OVplane.
4. Find the area 5. .r f r // 2 lies 2 f' 2
. f  1 which f ca 2 is in a of the portion of the sphere within the paraboloid by Tra included between the  ^_ \Vlrc~ An*. Find the area cut a// An*. 4 rm. of that part of the plane  f tercepted by the coordinate planes.
6. 2 An*. 2 :'' 2 = 2
f z ?/ f r .r 2 f ir 4 2
z' An*. 2 . . which 2 ay 2 a 2 b'2 f irab. In the preceding example find the area of the portion of the paraboloid which lies within the sphere.
7. 8. Find the area of the surface of the paraboloid // 2
a.r and the plane
tercepted by the parabolic cylinder
//'' f 2
z' = JT \ a.r in 3 a.
2
y Tra An*, . In the preceding problem find the arra of the surface* of the cylinder
intercepted by the paraboloid and plane.
A
/lo^/Hi
i\ a2
9. , /A )l*. ( 1 ) V 1 1 ) ^r ) Vii
J
10. Find the surface of the cylinder
f (.r cos c\ f // sin
situated in the positive compartment of coordinates.
: is HINT. The axis
radius of the base of this cylinder is the line 2  0, jr cos a f // cv) sin 2 ~ a ; y* x% Find the area An*.  !5 a' ~ + y r
* = jr 2 that portion of bounded by a curve whose projection on the ! + which and the (\ 2 cos rv. the surface of the cylinder
A' Vplane An*. a*. 12. P"ind sphere
and x of 2 r is r. sin
11. r V 5 is a2 . by integration the area of that portion of the surface of the
2
8
f z  100 which lies between the parallel planers jc =
?y 6. Volumes found by triple integration. In many cases the volbounded 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
257. ume of a solid 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 dimensions 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 DIFFERENTIAL AND INTEGRAL CALCULUS 522 one of the coordinate planes containing
summing up all such slices within the region in
The volume V of the solid will then be the limit of this
question.
triple sum as Az, A?/, Ax each approach zero as a limit. That is, columns in a slice parallel to that axis, and finally V= (1) lim Y V VA; A?/ A*,
R the summations being extended over the entire region
the given surfaces. This limit is denoted by By we speak extension of the principle of Art. 245, triple integral of the Junction f(x, y, z) = I H bounded by of (L) as the throughout the region R. z problems require the integration of a variable function of
throughout a given region. The notation is (2) y> Many
and which of course, the limit of a triple is, sums we have already discussed.
shown that the triple integral (2)
tion. The limits are y> z) found in by the formulas for thecentroid
homogeneous solid, namely,
y dx dy They are obtained by reasoning
volume. In the integrands,
of any plane I
i lies in Solution. the first Let 1, ' 1. ) (x, y, z) is Find the volume r* octant. (4)
(5) (6) =
c2 =
y =
x =
2 = dx dy dz. moments of z an interior point. The centroid be that portion of the ellipsoid whose volume is required, the equations
of the bounding surfaces being
(3) Vz as in Art. 249, using . i o OABC dz, symmetry. ILLUSTRATIVE EXAMPLE
a* to the double In Vx = f f which sum analogous more advanced treatises it is
is evaluated by successive integrathe same manner as for (L). (center of gravity) of a will lie in > of (2) are afforded Simple examples &> dx x, y, i (= ABC), (= GAB),
(= OAC),
(= OBC)
. of that portion of the ellipsoid 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 with respect to first (as ftS), the zlimits solving for we sum up z, being zero (from and (4)) all such elements in a column  TR = rWl V (/ ~ f (from (3)
6* by z). Integrating next with respect to I>EM\GF\ //, we sum up all such columns the /ylimits being zero (from (5)) and M<1 e(iuation of the curve AGF>, namely ^ f a2 1, = by solving b^J
for I  in a slice (as ~ ( from the */). b Lastly, integrating with respect to .r, we sum up all such slices within the entire
a.
OABC, the .rlimits being zero (from ((>)) and OA region dz dy dx Hence Therefore the volume of the entire ellipsoid ILLUSTRATIVE K\ \MPLE
z (7) ~ = 2 (8) The Solution. the figure. Find the volume of he solid bounded by the surfaces 2. 4  3 /' is t .r~  //, ', + //. 1 surfaces are the elliptic paraboloids of Eliminating z between + 4 Ja (9) (7) and (8), we find =4, y 1 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. which We have
r (10) = 2V2(1 M 4/
J{) ' /'
Jo ) f 4 J'.ljr "V 1 \ } y limits are found as follows
Integrating with respect to z, we sum up the elements of volume dz dy dx in
in figure).
a column of base dy dx from the surface (8) to the surface (7) (M P to
The limits for z are, then, #iven by the righthand members in these equations. The : MQ Thus we find
/'2 (1 Jo
\ V^ (]  s ) (4  4 x~  \ y*)dy dx. ., 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.
first integration should
with respect to x or ?/, and not with respect to z, as
be performed
above. The limits must then be determined in accordance with the The problem given may be such that the preceding discussion. DIFFERENTIAL AND INTEGRAL CALCULUS 524 258. Volumes, using cylindrical coordinates. In many problems in volving integration the work is much simplified by employing cylindrical coordinates (p, 0, z) as defined in (7), p. 6. The cylindrical equation of any one of the bounding surfaces
down directly from its definition. In any case
its rectangular equation by the substitution
x (1) p cos 6, y ~ p sin may often be written
it may be found from 0. Cylindrical coordinates are especially useful when a bounding
surface is a surface of revolution. For the equation of such a surface, when the is OZ, will have the form
be absent.
Volume under a surface. Let nate 6 axis z =f(p) ; that is, the coordi will z (2) = 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 whose
by the plane drical surface right section XOY is the region S. This cylin drical surface intercepts on the region S'. the surface (2) Divide the solid into ele ments of volume as follows : Divide S into elements of
area AA by drawing radial
from and arcs of circles about 0, as in Art. 254.
Pass planes through the
radial lines and OZ. Pass
lines R cylindrical surfaces of revolution about OZ standing on the circular arcs within S. and MP = AA and
(3) z. Then the MN = AA,
columns such as MNPQ, where area
The element of volume is then a right prism with base solid is divided into altitude z. Hence V = z AA. The volume V is found by summing up the prisms (3) whose bases
within S and finding the limit of this sum when the radial lines lie MULTIPLE INTEGRALS
and 5 increase circular arcs within A0>0. That in 525 number so that Ap and > is, AHO We now show that the double limit in
(Compare Art. 244.) (4) may be found by suc 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
This cessive integration. of the sum is of these slices. DEFG be the section of the solid in the plane ROZ. The
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. Let values of z Then by and (2) p  Art. 177, (3), area ' DEFG = f dp pz The integral will be a function of Now revolve area DEFG about ROZ and SOZ cut
ume DEFG, is out a wedge from A0p area pf(p, 0)dp. $* 0. OZ. By Art. 250, the volume of the solid of revolution thus generated is = 2 ir~p DEFG. The area his solid of revolution 1 since angle ROS A0 planes whose vol (radians). Therefore (5) equal, approximately, to the volume of the slice of the solid included between the planes ROZ and $OZ. The limit of the sum of
is the exact volume.
the wedges (5) when A0 is > Hence V (6) 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, where a =Z  Z XOA, = zpdp(W,
volume of a right prism of altitude z and
Thus AA in (3) is replaced by p Ap A0, as in F(p, e)p<lpd6
of as the be thought
base of area p dp d6.
Art. 254. may We  A'O/*, pi have now the formula* V (M) =//*P dp d6 =fff(p, *The orf/<T of fl)P ^P ^ s \s
integration is immaterial. Proof is omitted. DIFFERENTIAL AND INTEGRAL CALCULUS 526 volume under the surface for the ILLUSTRATIVE EXAMPLE 1. Show the ellipsoid of revolution b(x 2
o j s given by
x + y* _ ax and the (2), Art. 254 for the area of the region S.
From (M) and (4) we may derive (2), limits are found as in Art. 254. that the volume of the solid bounded by
= Q"b~ and the cylindrical surface y") f az~ f '2 r\"K f'U
V = *7,f i
oJo
Jo
/) (7) COH Evaluate this integral.
Solution. By the cylindrical equa (1) tion of the ellipsoid is + az~ b~p~ ab. the circle Hence
(8) The
x polar _ i _j_ y2 ax bounding S by is, p (9) of equation
9
jn  th e ATplane X (1), a cos 6. For the semicircle the limits for p are
is held fast, and
zero and o cos 9, when
for 0, zero and \ TT. Substituting in (M) the value we Integrating, get (7). V= a 2 &(3 TT   4) of z from (8) and the above limits, 2 1.20G a' 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 AV = AzA4. (10) By summation and
A0 * 0, taking the limit when Az0, Ap we have (N)
for AA may be replaced
Formulas Vx (3) in Art.
2 =fffp by p Ap M as before. 257 for the centroid become cos 6 dz dp d6, Vy == 2 fffp Vz^CCCpzdzdpdO,
when cylindrical coordinates are used. sin 6 dz dp dO, 0, MULTIPLE INTEGRALS
is EXAMPLE ILLUSTRATIVE
on the sphere Find the volume 2. of the solid 527
whose upper surface (in .r~fr+2 = 8
and whose lower surface is on the
paraboloid of revolution
(12) = .r4 r' The Solution. . shows figure the sphere arid the paraboloid in
the first octant. The curve of intersection All lies in the plane
z = 2. Its projection DK on the
ATplane is the circle
*' (13) are, + r' = 4. The cylindrical
by (1) equations : 2 (14) p' 2 (15) p' f z 2 = (ID);
 2 2 8 (the sphere (the paraboloid (12)); p (1G) An = 2 (the circle (1,T. element of area A/1 in the element of volume We AT is shown circle 1 / r2 r>
.. vN at A/(p, 0) in the figure. An  p p I I Jo J "
J, The drawn have, by (AO,
/r 6 fast), (1(1) is at /'(/>, 0, z). c/s (/p dO. /( Integrating with respect, to z (holding p and
the elements of volume (10) in a column from the surface (15) limits are found as follows: we sum up = MP> = J p
(M/Vt^J/7'i in the figure). From (15), z
are those
M/'i = V8  p~, the 2limits. The limits for p and
for the area of the circle (16). Integrating with respect to p gives the sum of the
and
columns in the slice included between the plane passing through ()Z and
the plane passing through OZ and ON. The final integration sums up these slices.
Integrating in (17),
2 to the surface (14) from (14), 2 ; = OM AUK. In the following problems, formulas (M) and (AO are to be used
when the equations of bounding surfaces are in cylindrical coordinates (cylindrical equations;. If the corresponding rectangular equations are needed for drawing a figure, they may be found by the
transformation
p (18) to which
(19) may 2 = 2 y' 6 = arc tan >
x be added
sin c; = cos = DIFFERENTIAL AND INTEGRAL CALCULUS 528 PROBLEMS
2
1. Find the volume of the solid below the cylindrical surface x f z
above the plane jr f z = 2, and included between the planes y = 0, y V j. Work 2. /!{
/ / 13^ cubic units. /*2rtros0 rjTT =2 p.'$
~
i / I Jo 3
 dp d6 a Tra 3 . 2 Find the volume of the solid bounded above by the cylinder
x 2 and below by the elliptic paraboloid z = 3 .r 2 f 2/ 2 3. = 3. jr Jo 2 4, Art. 247, using cylindrical co 2, 7
I = d2 dr dy \J2.~ out Illustrative Example
An.s. r2 / JO J ordinates.  r4 /~2 = 4 . V Ans. /"2 x /"I 4  1 J /*4 '  J2 dzdydx / / / JO J() J3 = 1 j" y I 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 Ans. coordinates. Find the volume below the plane 5. z paraboloid = .r 2 = ^ . 5 using cylindrical coordinates.
/?, = 2/Jo 7T /^COS //) COS fl / = a cos cylinder p Jo Jp^
2
f c ^. within the cylinder = a 2 within the Aw.s. Find the volume above 8. ^? / Find the volume bounded by the sphere p 2 7. 2 jr' + 2 2 y' c = aj, 0, below the cone a^(w g
2 c f jc' Find the volume , and using cylindrical coordinates. bounded by of the solid J). 2
?y A??,s. 9. . and above the elliptic
Ans.
TT. .r 2 4 y' Work Problem 6. z aa3  ~ z .r f 1 and 2 3
^ a ~ x 2 f z Ans.  y . 2
. TT. 10. In Problem 3 show that integration with respect to z gives (with4 A ~ 4 Iv
1 M \vhere .4 is the area of the
out further integration) V
2 ellipse 4 a* as given by 2 f y' = 4, and 7X and / tf are moments 11. Find the volume below the plane
and within the cylinder p = 2 cos 0. 12. plane A
z is 4 f p cos 0, 2 z bounded by the paraboloid
Find the centroid. solid
c. of revolution az . of the solid in = 2 p' Ans. bounded by the hyperboloid z 2
13. A solid
Find the volume.
of the cone z 2 = 2 p 2
nappe
Find the centroid above z Ans. f is 14. of inertia for this ellipse (E), Art. 252. Problem = a2 + 0,
TT. and the (0, 0, c). 2 p and the upper Ans. 7ra 3 (V2 1). 13. Ans. (0,0, ia(V2+l)). MULTIPLE INTEGRALS
Find the centroid 15. Problem of the solid in 529
An*. 1. Problem Ans. 16. Find the centroid 17. Find the centroid of the 18. Find the volume of the solid bounded below by
= a cos 0.
A NX.
z = a
p, and laterally by p of the solid in Problem solid in 2. Find the centroid z Compare
1, Example Art, 165, 3, and derive (N), Art. 165, from Derive formula 22. ) a). a (9
;< 16). TT of the solid in the preceding problem. Illustrative Art. 257, } above by 0, T<\. Find the volume of the solid below the spherical surface p
and above the upper nappe of the conical surface z = p f 1.
21. V ($ a, 0, 20. ample V , 8. the cone
19. (A, Art. 178, from the (2), first and L> f z* Illustrative = 25 Ex (L), Art. 257. formula in (;*), Art. 257. ADDITIONAL PROBLEMS ~ r~, below
Find the volume bounded above by the sphere p' f z= p ctn 0, and included between the planes
ft,
by the cone z
and
6
being acute angles. (The solid is part of a spherical
ft f A#,
wedge, like O.SQA' in the figure of Art. 222 when OQ is drawn.)
An*. \ r Atf(l  cos 0).
1. jtf </> ! ; 2. p^ __ Z2 ft, is like Find (without integration) the volume bounded by the sphere
r
the cones z  p ctn 0,
p ctn (0 f A0), and the planes
= ft + A^ using the result in the preceding ])roblem. CTho solid
0Pi/iQS in the figure of Art. 222 when OR arid OQ are drawn.)
r> A/i sin (0 fXb/x.
A0) sin \ A0.
i> ,: t f \ '] = p ctn 0,
3. Find (without integration) the volume bounded by
l>etween the spheres
ctn (0 f A0), 6  rf,
ft f A/i and included
p
j
2 _
(r f Ar)
f
using the answ( r in Problem 2.
__
^2 _j_ z
;' 2 *> p'2 /: 1 ', > A0) sin \ A0(r f r Ar f Ar*).
Art. 222 by producing each of the
(The
f/
radii OPi, OR, OQ, OS a distance Ar to /',', /'', Q', S on the sphere
2
intersect this sphere in the circular arcs
The cones
(r 4 Ar).
p* f z
JYA" and Q'S', and the planes in the ar^s of great circles /Y,S", H'Q' The
.4??.s. solid is obtained 2 A/i Ar sin (0 A 4 from the figure ', in  solid has the vertices PJtQSPi'H'Q'H'.)
4. The solid of Problem volume AV7 when spherical
by 0. Then one vertex P of
Prove, from Problem 3, U is the element of coordinates (8), p. 6, are used. Replace ft
has the spherical coorclinat.es (r, 0, 0). AV AV
Ar Ar>o r 2 sin
v , A0 Ar A0 A0 o A./.
2
7
Therefore AT differs from r sin
order (Art, 99). </> by an infinitesimal of higher DIFFERENTIAL AND INTEGRAL CALCULUS 530 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. 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
6. lim ^ V^/^'fr, 0, 0;A V =fff F 0J r (r > ~ sin ^r </> r ^0 f/ # member Al may be replaced by r sin <^ Ar A< A0
4), that is, by the product of the throe edges in Problem 5.
The righthand member is calculated by successive integration. (Proof
omitted.)
In the lefthand (see 7. and 1 ' Problem Work A' is out the integral in the preceding problem
the sphere r = "2 a cos (/;, that is, .r~ f /r f z2 7T /^ * TT /^2 <l if F(r, 'J = r' </>, 0) = r, a^. COS '' / ./o 8.
is Work the region / .At out the integral in Problem 6
12 a cos
0. r if /^tr, 0, 0) cos /I//*. and
I;/, 7r<z A'
r
'. CHAPTER XXVI
CURVES FOR REFERENCE
For the convenience of the student a number of the more
curves employed in the text are collected here.
CUBICAL PARABOLA SEMICUBICAL PARABOLA y THE WITCH OF AGNESI x 2y = 4 a 2 (2 a common THE 2 = ax 3 . Cissoin OF DIOCLES y). 2
2/ (2 a  x) = x3 . DIFFERENTIAL AND INTEGRAL CALCULUS 532 THE LEMNISCATE OF BERNOULLI (x 2 + THE CONCHOID OF NICOMEDES 2 2
y/ )  = a cos 2 p 0. + a csc 6. (In the figure, b CYCLOID, VERTEX AT ORIGIN CYCLOID, ORDINARY CASIO y ~
:r U =
^ a arc vers d(0 sn a(l cos ~~ V2 a//  /r .r =a arc vers ^
a + V2 a?/  0). sin 0),  ^), cos 6). PARABOLA a cosh a  y 2 CURVES FOR REFERENCE
HYPOCYCLOID OF Fouu CUSPS ASTROII^ 533 KVOLUTK OF ELLIPSE
r +
~~ = a siv . (ft//)" a cos'
/) sin  (a 5 0, ;< f?. CARDIOIP yr p + ^^* = a(l 3 axy
cos = 0. 0). COSINE CURVE SINE CURVE y ' = sin x. i = COS X. ft 2
) 534 DIFFERENTIAL AND INTEGRAL CALCULUS
STROPHOID
Y a cos 0.
p = fr
the figure, b
a.)
(in
 SPIRAL OF ARCHIMEDES p= LOdARITHMIC OR EQUIANGULAR
SPIRAL p
a0. HYPERBOLIC OR RECIPROCAL log p = l( c' *, or ad. LITUUS SPIRAL = a. = a2 . CURVES FOR REFERENCE 535 LOGARITHMIC CURVE PARABOLIC SPIRAL r (p a) = 4ac0. y = log x. PROBABILITY CURVE EXPONENTIAL CURVE v TANGENT CURVE SECANT CURVE 37T y = sec II y = tan x. 53G DIFFERENTIAL AND INTEGRAL CALCULUS
THREELEAVED ROSE p ~ a sin 13 0. FOUR LEANED ROSE p = a sin 2 THREELEAVED ROSE p cos 3 6. FOURLEAVED ROSE p T\VOLK\VEI> ROSE LEMNISCATE =: a = cos 2 0. EIGHTLEAVED ROSE r = a 2 sin 2 5. p = a sin 4 0. CURVES FOR REFERENCE 537 EQUILATERAL HYPERBOLA PARABOLA x .r// INVOLUTE OF A CIRCLE r cos r sin f rO sin 6, rO cos 0. = a. CHAPTER XXVII
TABLE OF INTEGRALS l.fdf(x)
2. Some Elementary Forms
C.
=ff'(s)dx=f(s) + = fa du
u aCdu. dw dv = ) Cdu Cdw C dv + bu Forms containing a Rational See also the Binomial Reduction Formulas 96104. U ~ ln (a   f.
J (a .
2 f f ^>w = a + bu 6w) 3 ^> 2 L ~
a i r rfM h b 3 {_ i a f 6w  r IA r
14 / "~77
2
15 J u f <*M
, r x _
= du 1
  ^> ,
1 aw (a 4 OM) = 1 + f a ^ 2(a h 2 6w) . f *>M\
x,
In /a
f C. i o 2 a in (a  C  ''" )] It v = r:\rT + '"(
bw)
6^L a J TTT 2
(a f ftw) r
J (a + ( a^ \  1 In
538 u a
^ , ) I + ftM>
\ 4 r 6zo
X f c. +C TABLE OF INTEGRALS 539 Rational Forms containing a 2
1 = bu ~
(
\bu + 2 ab 18. a'2 2
2
b' u +r  (a \a = JL in d C
J bu , arc tan 2ab a r 4 \ (a a' 2 2 f *(a 2
' 20 2 2 2 b' u' (a b 2 (m )'' J 2 bu'2 ) (a' 1 ' 2 2 a' (p ., ' (/ 23. J " f u m (a'2 24 f ^ = ..*!"..
f u((i"
f
/>*//) 2 2 __
 2
b' i( fe 2 (7w a 2 (m  a'2 2 )" m  (p f 2 a 2 (7>  l , (u'  _ __ 6 2 M 2 )"" 2 l 3) /
./ 2 [ du
u v rn  (a' a?/
' J u m (a'2 1) Va The integrand may be rationalized by
Formulas 96104. f 2 b' ^'2 ) 1 fu V^T^rf K = 2(2 a ~ setting a fl + o2
lo jV 28 26. 3 6< a2 ~ 105
2
27. MtfM
28. ^f Va f =bu 2(2 a  M *+ 12 6M)V(/
3b'2 f bu b3 ' 1 bu + the Binomial Reduction */ ' b'W) p 2 <2 b*u*)" (a' 3 r Forms containing 25. !  " r 2 p J>a C.  l)u l b2 u'2 ) ! l)ir'+ 2 7) ~ __ i + ( ..";., .,)
\an*M*/ ~ v ) _ l1 } (<r 2 u f _
_ 12 2 L a(m b' u' )" J u r"(a'2 In a* 1 ).' b' )(a~ _L1 _ 22. 1 f /> 2 b' u' f l)(a' 2 b'~(m 2 2 2 p C. bu = v*. See also DIFFERENTIAL AND INTEGRAL CALCULUS 540
29 2
u' du 2
2(8 a __  3 6 2 M 2 ) Va + 4 abu + bu + ' /; 30. f> 2 Km /a + bu  4= 31./' /a In Va bu 4 f~ 1) J ^
+ Va 4 btt l du \^~+~
l v 4 > T, for a 4 ) arc tan ~ 4 "*" VVa (/K 32. V ( ~ "" ~ um 2am
b(2m+ Va f
b(2m+l) u r"du C. T, for < 0. 0. a / oo. w
t 34 dw, ~"
f J
: \/a fu (^ v ' 4 />M + budu w tl 1 ^ ,/~~TT~
ow 2 Va / ~ , 4 (a " ^ f Va
u 1 \n(u 4 we may V?/ 2 u 4 Vw a 4 Vu 2 ) a2 replace 4 a) by sinh" 1 a 2) Vu 2 4 a 4 rr/ 1 )//"' Forms containing
In this group of formulas a 4 ?>M 2 f a(m r/? ( d?/ /' a 4 2 a ' by cosh"  1 2 4 a 2 )(j
sinh" )>y 1 ?^ = 36. ~ Vw ?/ ' 38. /'//CM 1 a 2 Y~dn a2 2 f 1 ~ r^
(* ' ?/ // = ln (M 4 ~ln (w. 4 4 fl2 h C, \" 4 2 4 m+ VM aj 1 ) 4 C. Vw^ia 2
) 4 C. TABLE OF INTEGRALS 4 In Vw (M f 541 2 a2 ) + r.
nm i'. 49. c , i n f 1 J ' f >/ 77 7"^ ^M r
J a2" 2
u*(u*<i*)* */~; 51. m J f }/ 50 _ / ii(w 2  =
= a 2 ^aw'*
V?/ 2 ^ + .1
 a2 f  )* C/M *v ( arc sec  r  a M q ln ^ + QW C. f 1 52.  a 2 (m O 21
2 a 2)2 f n
_ J^_ZLzi:L f.
l)J a 2 (m u
C/M \ 53./  a 2 (n
, m f 2)?/ n a 2 (n
. A
54. r(u 2 I J KK / ^
r(u 55, t/ + a^du =
M 2 a 2 )^du
L
24 A /~r^
V u f a
2 m " (  3 r"
 2)J du u m (u 2 ^
2 , a In 1 a2 )2 ! /a
I V f a2)2 Vu
u 2 f a2^
/ / , f ^
C. ~'2 du DIFFERENTIAL AND INTEGRAL CALCULUS 542
66 rluiI
J u* a'2 (w 1 n )' (u*a*p 58. (n~ TW f )" 1  2 w^^r/w = *^ du f(a =  ' 1
^ Vr/ ~ 2 fl
, r" w J 2  w v a2 Forms containing
59. m h i/ M* h if arc sin
^ w 2 f C. Tl 60.  (,^ "(' )5 z + (  61. / 62. ., < J  ., n j )d = f ^ 63 = ,/^ 2
(,/ arc sin w + C. ^ r/// f C. } Ct
bO. /" // .> 1 dn rr M*) ((J r>
"
i h ' r( . 2 N IL arc 66 sn
a " 67.  fa 2 2
?/  " ?<2(/// r
^ ,=
) 3 V 2  arc sin  a 2 (m /u
AQ
69 21 (a  m 1) n + rj lJ
(a r_irdu_nJ
2 4 C. w2 w 2 2
) u m+ a 2 (n2)(a 2 mn + Zr l  21
i/ 2 2
) 2  w2 2
) <u^au a 2 (n2)J
(a 2  l
i w2 )2 i TABLE OF INTEGRALS 543 70. u(a
r , 71 J 2 a 2 u' fi V (in u u* w*^~~*^~
& 72 f
. . = . 1 ~ , cosh 73 a
 , ' . f ^ C. { a1 /
a (m m 4 1 <r(m '"
) // '* //  /' 1 )' (In </// f  2 5 r( > . 2) J _ wm a . ~ ! ""/" ( a ln .' va j ' L a cosh" ?/ J f C.
w arc sln 76. a 77. (a 70 2 2
f/ 2
)
, m f (n 1 Forms containing m+ n 1 jM"'"" V2 au i/ 2 The Binomial Reduction Formulas 96104 may be applied by writing
\/2 au u~  y}(2 a u}*. 2i 2
3 a 2 f an 80. a3  ' +
, arc cos /i (1 f 2 ws ~ arc V2^ w\ i / ~)+C> + C. DIFFERENTIAL AND INTEGRAL CALCULUS 544 3 ,,^ . 8L/>V2auw dw2 J m+  M 82. f M  *) 1 2 K 2 dM w 2 dw y V2 aw V2 aw = === f . QH f
J  /  arc cos (2 w2 w2 aw
?w ~ 4 )^ w
w 7
) 2 /^"v J C. a?/ Mm 3)J a (2 " !  1 ( a In (w f ?/ 2 = r/// ) a V2 f u2 a?^ f f^fc ) f r. V^ 2 a, a arc cos 4 2 <i' 1 where dz, )  ^)
a/ ( v + 2 C. = aw  m w2 94./
(2 aw (2  V2 . w^
f V2 a?/ w
_
m1 2 i/ 1 a(2 V2 w 2 )* a2 aV2 aw  f C. w2 udu r
^ r. w2 w 2 )* aw au w2 dw
m \/2 aw
7/ f C. w2 aw  Binomial Reduction Formulas Cu m (a J + a/ ,, ,S)w'" =  V^CIWH* dw 96. 1J\
  1
\ w2 "*<*" 92./ wV2 aw 95. '* C. f )
a/ V2a?/?/ 2 V2 93. It aw (2 __ arc cos = V2 a?^ f
^f V2 aw  1 ( u2 ?/(/?/ 89. a arc cos f fl 11 \  CF(U, _ 2
u' 87. 88. ~ \/'^ a(2 wm
?/ f J w 2 2 w 2 a?/ f'V aw '^  Ij'^filJ fj }! w  f \ aw
04 r^/2 86. _ V2 M f u /"\/ m f dM = ll
 m<]+ 1 ( n i b(pq+ m m 7)7/';V' 1.
f 1) 4 * a / ).' M  ^V2 w 4 a. TABLE OF INTEGRALS
97. J/ u m (a +  bu q Ydu ^ ~^+ m+ 1 pq
pq
tj r 8i v m
?/ (<i a(m n bu' f 1 1 d" w 1A1 f (a
1U1.
J 4 bit'1 1 ' l 1 __ "( f w . ;> M '" 1 l // ) ~ + bnWdu _ '^ / : m ( l ' ^ ' />/rn } ( a b(_m __~ a
102 r( ni (n f btcn*'
""" y )fy f ar/ da M ) /j^) +c
\a
bu' Li n <; ) ~^ ( / um I }u 1 uq(}> > 1 ' 1 m ' ) == f f>" (a f 1 , M( f jx/ (/ ~ c/ tn l)n _ (/;/ 100 .r ' f 1 " f <i( u '(d w f fo^w 99. 545 + (// 1 j>(] </ ,'(a f i 1 bit")' . u 103. f (a m du f ^> ' m ;>r/ f //" 1
' M </)/'  ?>(m + pq (m
b(m _ u 7" 1 f (/ u m dn )(a 4 1 pq
4 ' (/ //'" />// r ) ; (/ '~ ' ) H"' dn '' bu v )* f 1).' (a f
l 104. m  4 g p(( H 1
' Forms containing a
The expression a
w = z AC + f &M (a f 1) (iq(p rw 2 f
+ wo,
6w f r?^
r// f bw ci/ 2 (c > 0) be reduced to a binomial by writing may =
4 r2 Then
i lien
The expression a = ==2: "f u
a
4 Then
. a ^ J a h 6w cu 2 may =
 r(z
\z
( 2  A;.
A"). be reduced to a binomial by writing 4r 2 "27' 106 f bu 2 f . rw 2 f bu  cu 2 2 =
V4 ac  b2  c(k  arc tan(
\ z 2 ). 2 CU
. V4 or +6
 & 2 )f C,
/
when
whei 52 < 4 ac< DIFFERENTIAL AND INTEGRAL CALCULUS 540
C du 106.^ a
J + Jm =
cw 2 f +b 2cu 1 In Vb 2  (
\2 rw 4 a ^
Vfc f 6 f  2 4 ac \ when +
4 * 107. L
[ > 108. CLMJLJ^lll =
a
cu l
6w ; I M i n ( + a 2 r 2 + 2r M 4 ar 2 rw > b2 b/ } } (/W 109. fVa bu f = f r?/^/?i ^i Va bn f  ^"iC
~cv*du = 110. a n (9 ^ 2 rw 2 f 4 r ./ ~ ?/ /; f bu Va 4 r
2
/> 4 4 (ic . urc sln 2 /  6 4 r? few ' '/;
Va
112. bu f h rw 2 (2rw 46f
sin 2 du // f
J 4 f 4 ow r?/ r
I J ^ " ?/
_ . Va 4 b= ~r^"~
Vh + 4
2 }+ c __ v (2 ru 4j 4 ^ 4 2 \/rVa ~ arc sin / y >( j Other Algebraic Forms b) log, f (a ( Va C. _ w f V?> f _ H 4 M)+  arc sin > 117. + C. b) +M
' arc sin it f \ T 1/ du Vl =
^w a)(b 4 ^ f>?/. /) r H 118.  ar/ <' rw 2 ww (
v 4 />// 2 ~ln
i
4. 2 Vr r7=J^L==  4= urc
Vr
Va 4 bu rw tf 113. = Vln u) M2 4 arc sin ?^ 4 C. 2arcsin x /^^4C. c ' 4 ^ + c<
4 2 ra &M r 4 . 4 or/ C. < ac. TABLE OF INTEGRALS 547 Exponential and Logarithmic Forms
120. = C can dw
J au
121. Cb = ^r + du J C. ^  a In o = du 122. Cue*" J a"
124. /V'ft = d?* du J 127. u T u"
N i I H ( = du
?/ 1 ! I u"' In" ft rb"" inn
129 . 4 C. f r f"
J ,
3 " I i In u r "" n i In  //  7
du rrln
?/ In" 1)' m 1 ; I u m In" ' /( d//. IJ f "" i rr
dn.
aJ n ?/ a 130. l u 4 r. m = d?/ J + C. du IJ u" M" d// // dw n l 128. " ft' a In
l )//"  In u ?/ ; In 1 " rr f
In ftJ (i ft"' u" An V" du. 11 d?/ /"ft J 10  ^^
(Mn = ft " 126. 1) f C. Cu"
aJ a J
105 ut a Cu n e au 123. T. f a ; I (In M) h r. ?/ Trigonometric Forms
In forms involving tan //, ctn
first use the relations ?/, sec M, esc //, which do not appear below, tan // = sin u cos . ctn cos u I sin // d// 132. I cos ?/ d// = 133. I tan ?y d?/ = 134. f ctn ?/ d?/ sin dw J du f In sin d ?^ ( ?/ ?/ = 1 esc u cos z/ C. ?/ . In sec ?/ h C. In fsec w f tan w) ?/ J sin w +C= f r. = / I sec ?/ h (\ // J cos 136. /esc u cos u
: In cos M //"
?/ = sin 131. sec ?/ // + C f C 7 =
= In (esc w
In tan  ctn w)
f C. = 1 sin 648 DIFFERENTIAL AND INTEGRAL CALCULUS 137. /see 2 u du = 2 M dw = 138. J esc tan u 4 C. ctn u = w tan u dv 139. / sec 140. / see w esc u ctn w du ~ \ cos 2 u du = ^ M 4 j 142. / 143. /cos" u sin u + sin 2 ;J 4 C. WM sin mw sin / 4 1 ^ =n = JH 4 c:. 4" 1 + (m sin + (w =  " rfw 146. fcos m?/ cos n
/
147. 4 r. w u cos w du mn ?/ C. r< d?,* / 145. fsin C. 4 ?/ sin 2 u \ J 144. fsin" r. 4 esc du sin 2 u 141. 4 C. w)l< 2(m 4 = cos riw dw ( 2(m + s + "^~ r/ ;/ ~ (wt n)u '^ ~ (m sin "Jf 2(?w w) w QH / " '" c. + r . w) cos _ + ( m ?? ?/ _f_ 4 w) (\ ?o 2(rw du + 2 esc a arc tan (tan cos a cos 149. cos a 4 cos u 1 \1  tan = 2 esc a tanh 1. f ^__ J cos a 4 sin =
= 2 CS i (tan a tan
i cscaln / a tan tan \tan a 4 tan fl A tan 2 esc a tanh J 152. f
./ 163. a 2 cos 2 ff = . ?/ 2
2
4 h sin sin rf = f ?/ " (a 1
a/> J ?/ />
^ sin M " ~ 155.J 156.J d = '"" a1 dw = sin w cos M du = cos w M sin w ?/ 4 M cos " M) a cos
w2 u cos M f C. 4 u sin M 4 r. + i tan ctn sec a J u +r fl) < ctn 2 J  secax
I (ctn a tan
I (ctn a tan a +C . esc a) 2 < 1] u f esc a) 2 < 1] // u f esc a Bre tan (Itanj'N
a
/
\
2 cos M r 4 (ctn a tan ? 154. f ) .^i/ [ rf " ?/ i [ = w) 4 r. J ul I a tan a arc tan (CSC ?/ a tan ?, } \ 4 4 r TABLE OF INTEGRALS 549 Trigonometric Reduction Formulas /sin* ~ 1 u cos " sm n udu j
ito r COS"M du
168. /
J du
159 C V cos"" = sin M !/ 1 (n = 1 cos" ) = "*" 164. f ^cos m usin n w ^
1 ^ c  fl * W 16 V 1 169. ftan"?/ 1 sin" 4 (m dw 2 = w du 171. Ceau coB n
J ^ 1 =  ctnW
~
n udu = lM sin" M ~ n(n r? n m
m ?/ it m~ w _ 2 w u 2 rcoR m udu ( sinn u 2 rt n i
/sin
~ cos m 2 w
~
1
rsin n 2 u du
mJ cos m u n
?/ 2 /
sin"~ 2 ii
m~2u
rcos 1 n 1 Jftan"~ f m+ i/ " ~ 1 1 m 1 u nJ n w
~ sin" i< M, 1 1 dw 1 ./ d?>t. r c tr\ n ~ 2 u du.
J 1 a M dM n/ 1 u cos*" m ^^
z
. ' u 1 n +
172./e"8in m
fcos M m f !/ / J cos m m) cos" "" n 1 sin" l)cos
(n ./ 170. fctn
J ) sin" r^Hllli^
cos m u " ^^ 2 u
z ,
sin" 7 ./ w) sin"" (m J 1 nJ f du B "
J cos m 1 1 cos*"" ___
~" cos'"?/ cos" ?^ n cos" ^ /sin n M d?i "J ?/  (n d/>< f 2 r 1) sin"~ 1 sin"w 2 icos~'2 u sin^wdu. m s IJsin"" m+n
m n IAA " n (n rcos m udu rcos m ?/ r__clu
J cos"
1 m+  sin" u dj/ J sin" ~ 2 M n m+
'J , : (m
, 1 n m f = cos m w sin n M 2 r n
!L u du. 9
cos"~ 2 w du. j_ + ~'
2 _ /
I J n u l u ' J 162. fcos m u sin" M du
J sin" >? ~ smj/
(n 161. /cos m M sin"wdz/ V 1 r?
, h sin" ) /*
i J n cos M = . h n sin" M 160 frfJLJ cos" w 1 1 "~ it ?/ n 1) / +
_t ^T^/^ ^ ' nz cosn = "n'(asin a + n o , wrfM  n COBU) ~2 w du. DIFFERENTIAL AND INTEGRAL CALCULUS 550
173. Cu m coa au du = (au sin au a v 777(771 )/"*, ~ 2 cos au j
o
um
aw. 1 L , /
*/ M m sin aw dw = ~
in m ( au cos sin a?v (m a I J a
174. m cos au) 4 1 ___ ) r
/ m Mw . . r,
^ sln a?/) aw ,
C / Wy Inverse Trigonometric Functions
175. J arc 176. = sin w dz/ M arc 177. f arc tan r/// ?/ 178. / / arc sec dw Vl w2 arc cos w Vl ?> arc tan w In Vl ?/ = arc ctn u du 179. 4 ?/ arc cos u du J u sin arc ctn In Vl 4  u ?/ 4 u* 180. I // In u arc sec // cosh ~ arc esc u du arc sec = */ n arc esc ?/ 4 In u arc esc u 4 ( ( u // coh 4 C. 4 C. 2
f u' 4 4 4 u'z 4
Vw
4 r. J J ?/ ) 4 C. 1 ) 4 C (*. // 4 r. 1 vV' r. Hyperbolic Functions
?/ d?^ ~ cosh // cosh ?/ d?/ = sinh tt tanli // du ctnh // d>'/ = In sinh u 4 (\ sech M </// = arc tan (sinh M du = In 181. / sinh 182. I 183. 184.
185. 186. ('. In cosh w } C. j  /"csch
L> 189. \  187.j"sech
188. 4 C\ J csch'
 M du = ?f dw = tanh tanh ?^ ctnh sech w tanh u du J ?/) f C M 4 C. 4 C. ?/ 4 C. sech n 4 C. = gd M 4 C TABLE OF INTEGRALS
190. 191. I j = csch u ctnh u du
sinh* u du sinh \ csch u
n "2 551 C. f u f C. \ 192. / cosh 2 u <ln = i sinh 2 M f J u f f. 193. / tanh M <i?/ = M tanh 194. I ctnh 2 u du u ctnh u 195. = sinh M du ?/ cosh u u f (\ C. f  sinh u f f. cosh // ~h r. j 196. I /> n cosh M du j u u cosh tanh du J 200. fctnh ' 199. f 201. I sech = c/// * ' M >/ ' cosh M = tanh csch f f
sinh
J/ n~A
204. nn  205. ?/ . . 203. J . m?< // ' ft, mu cosh
u
cosh / r
/ J . ?^^?/ f
./ 208. 209. rt , rt 210.   : . . u
7
cosh nu du , sinh f f r
cosh u  cos a . f =  J  f sinh m/ du r a
r
cosh
J
, '^ I arc sin u + ?/ f sinh = ' ?/ ?V)M , n?( du = . f'osh
 , \  ' 2 esc a tanh (tanh J (tanh " ?/
.', ?/  n)u
' (?w tanh tan J ?/ a cosh TIU
 a2 i tan . \ . ( ,, ( . a) i c< n sinh nw) , f ~, C. [ f
, ,,
(\ /
\ a) f i + a) > \ (m^n\
\
\ m>
^
^ n >
m^
< n \ / \ / r:. r. 4 r: (tanh*i W n2
n2 / ., < n)  p "(a h 2(wr/) 2 esc a arc tan ftanh = . / sinh (m
  u)u
 n] 2 csch a tanh ; n) 'Km u) } + aj = l(m~ cosh (wf ?/)M
i 'Km ?/)// ~r. f ?/)w
 l(m+ ) C. Cm
 sinh    sinh fw
l ~ f <" 4 r. ' cosh u Jl + cosacosh, w f + ; r. // Km + u)r nu du J C. ?/) f (tanh f ) f g<i sinh OH ^~ _ . : //
' csch sinh nu du u f J In (1 ' J 206 f
J cosh a
207. ?/ f (\ f r\  In (1 \ + n~ 1 // n sech du vV ' u ctnh = VlH ' u sech 202. u // u sinh ?/ <^/^ // ___ du i/ ' du // sinh 198. u sinh <ctn*Ja) INDEX
(The numbers refer to pages) Absolute convergence, 348 Center of Acceleration, curvilinear motion,
121
rectilinear motion, 83
Adiabatic law, 70
; Agnesi, witch Anchor
Angle of, 531 ring, 267
of intersection, of plane
curves, 43
polar form of, 126
of surfaces,
of skew curves, 475
481
; ; ; Approximate formulas, 367, 372,
490
Arc, centroid of, 142, 144, 473 335 differential ; length of, plane
skew curve, 473
curve, 271
Archimedes, 127, 128, 155, 277, 534
Area, of a curved surface, 517 moment of, 320, 503, 514 moment
of, ; ; ; ; of inertia of, 508, 514; plane, 241,
in polar coordinates,
258, 498
; 262, 512
tion, ; of a surface of revolu 277 fluid pressure, 506
Centroid, of a homogeneous solid,
522, 526; of a plane area, 320,
of a solid of revolution,
337, 503
323
; Change of variable, 166, 240, 457
Cissoid, 44, 46, 270, 277, 322, 531 Complementary function, 394
Complex number, 440
Compoundinterest law, 399
Conchoid, 532
Conoid, 284
Constant, 7; absolute, 7;
trary, 7 ; 233, 376
Continuity of functions, 12, 444
Convergence, 340
Coordinates, cylindrical, 6, 524527; polar, 123; spherical, 6,
529, 530
Cosine curve, 533
; Critical values, 52 Astroid, 119, 533 Cubic, skew, 474
center of, 157,
Curvature, 149
171; circle of, 153, 170; radius
of, 152 Auxiliary equation, 390 Bending, direction of, ; 75 Bernoulli, lemniscate of, 532 Curvetracing, 81
Curvilinear motion, 120, 146
Cycloid, 116, 119, 144, 151,
244, 270, 274, 276, 281, 532 Binomial differentials, 299, 307
Binomial theorem, 1, 353
Boyle's law, 70 Cylindrical coordinates, Calculation, arbi of integration, 189, 229,
numerical, 7 of 361
of TT, 366
e, ; of 524527 loga rithms, 362
Cardioid, 117, 119, 125, 135, 145,
155, 244, 271, 275, 281, 323, 335,
516, 517, 533
; 6, 161, Derivative,
pretation 446 21 definition,
of, ; inter by geometry, 25, 445 as a rate, 64
symbols for, 22, 445 total, 455
transformation of, 166
; partial, ; ; Catenary, 152, 270, 276, 282, 423,
432, 434, 532
Cauchy, 345 Descartes, folium
533
553 of, 46, ; ; 119, 288, INDEX 554 Differential, 136; of arc, 142, 144,
473; of area, 237; formulas for, 140; geometric interpretation,
477; as an infinitesimal,
146; total, 449
Differential coefficient, 21, 136
137, Differential equations, applications
to mechanics, 402
definitions,
; 375 ; first order, 387 378
higher
homogeneous, 380 order,
; ; ; linear, 383, 390, 407
formulas for,
Differentiation, 22
; 28, 29, 86, 87, 115, 119, 120, 425,
of
426, 435
general rule for, 23
; ; functions, implicit 73, 40, 154, 21; differentiate, 21,
discontinuous, example of,
of a funcexponential, 89 derived, 23
108
; ; ; 37; graph of, 10, 444; hyperbolic, 414, 415; implicit, 39,
73,458; increasing, 50 inverse,
38 inverse hyperbolic, 423 inverse trigonometric, 105
logamean value of, 333
rithmic, 89
of several variables,
periodic, 97
444
table of hypersine, 97
bolics, 416
transcendental, 86
trigonometric, 99
Fundamental theorem of the intion, ; ; ; ; ; ; ; ; ; ; ; 254257 tegral calculus, 458; logarithmic, 93;
partial,
445, 462; successive, 73, 462 Diocles, cissoid of, 44, 46, 270, 277,
322, 531 of a function, 10, 444
Gravity, center of, 320, 323
Greek alphabet, 6 Graph Direction of a curve, 42 Gudermann, 435
Gudermannian, 435; Ellipsoid, 280, 285, 522 Gyration, radius Envelopes, 466
Equations, graphical solution, 128;
of motion,
interpolation, 129
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
; of, inverse, 435 508 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 ; the parabola, 159, 470;
ties of, 1 62 proper Exponential curve, 89, 535
Exponential function, 89 Increments, 19; approximation
137, 451
Indeterminate forms, 174
Inertia, moment of, 508, 514
Infinitesimals, 17; of, replacement theorem, 147 Factorial number, 338
Family of curves, 230, 466
Fluid pressure, 325
center of, 506
Fluxions, 19, 357 Infinity, 13
Inflectional points, 79 Folium of Descartes, 46, 119, 288, Integrals, ; 533 Formulas, approximate, 367, 372,
490 for reference, 16
Fourier, 238
394
Function,
complementary,
decreascontinuity of, 12, 444
definition of, 8, 444
ing, 50
; ; ; ; ; 229 Initial conditions, 240
250 ; ; 188 change in limits,
decomposition of interval,
discontinudefinite, 237 ous, 251 ; ; geometric representation, 244, 492; improper, 250
253 indefinite, 189 interchange
of limits, 249
multiple, 491
use of table, 315
table of, 538
; ; ; ; ; ; INDEX
Integrand, 195
Integration, 187 approximate, 245
of binomial differentials, 299;
formulas for, 191 193, 430, 432, ; ; fundamental theorem
of, 254
by mlsceJhmpotrs subsUtutions, 221, 305,J32
by parts,
223
by raTional^fracIions, 289
by rationalization, 221, 296; by
reciprocal substitution, 305; by
433, 435 ; ; ; ; ; reduction formulas, 307, 312;
of trigonometric
successive, 491
forms, 213, 303, 312
; Interpolation, 129, 372
Interval of a variable, 7 555 Mean
182 ; value, extended theorem of,
of a function, 333
theo rems ; of, 172, 482 Mechanics, 402
Mercator, 439 chart, 439
Moment, of area, 320, 503; of inertia, 507, 508, 514; polar, 510,
; 514 Motion, 120 curvilinear, linear, 65, recti ; 83 Napierian logarithms, 88 Xewton, 19, 27, 131, 132, 133, 332, 401 Isothermal expansion, 330 Nicomedes, conchoid of, 532
Normal, to a plane curve, 43 plane
to a skew curve, 471, 480; to a
surface, 475 Jacobi, 445 Osculating Laplace, 19
Laws of the mean, 172, 182, 482
Leibnitz, 26 Pappus, theorems of, 335, 504
Parabola, 532, 537; cubical, 156,
531
semicubical, 266, 498, 531
Parabolic rule, 247 Involute, 163 ; of a circle, 156, 276, 288, 537 in of arc, plane curves, 271
of
polar coordinates, 274
; ; skew curves, 473
Limac.on, 534 Limit of a variable, 10
of an inteLimits, change in, 240
gral, 238; theorems on, 11, 17
; Lituus, 534 Logarithmic
Logarithmic
logarithmic
Logarithms,
87 circle, 170 ; Lemniscate, 127, 155, 263, 515, 516,
517, 532 Length ; Paraboloid of revolution, 268, 521,
528
Parameter, 1 15, 466
Parametric equations, 115; first
derivative, 115; second derivative, 119
Point of inflection, 79
Polar coordinates, 123 moment of
; 514;
subnormal, 126;
subtangent, 126
Power rule, 32
center of,
Pressure, fluid, 325
506
Probability curve, 535
Projectile, 121, 234, 471
inertia, curve, 89, 535
differentiation, 93 function, 89 common, 88 ; natural, ; Loxodrome, 439
Maclauriri's series, 357, 367 Maxima and minima,
treatment, 182 47 ; ; Radius, of curvature, method, 53 functions of two
second method,
variables, 483 gyration, 508
Railroad curves, 152 76 Rates, 64, 455 first ; ; 1 analytic definitions, 52 ; Quadratic equation, 152; of INDEX 556 Rectification, of plane curves, 271
of
in polar coordinates, 274 ;
; skew curves, 473
Reduction formulas, 307, 312
Replacement theorem, 147
Rhumb line, 439
Rolle's theorem, 169
Roots of equations, 128
Roseleaf curves, 536 ; Table, of hyperbolic functions, 416
of integrals, 538 ; ; ; ; ; , ; ; ; ; ; 350, power, 344; p, 354; rule, ; 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, 428429
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 Taylor's, 369, 488 Simpson's 357 ; Secant curve, 535
Sequence, 338
absolute convergence,
Series, 338
348; alternating, 347; approximate formulas from, 367, 372
binomial, 353
Cauchy's test,
345 comparison tests, 342 condifferentiation and
vergent 340
divergent,
integration of, 365
340
harmonic,
geometric, 339
342; Maclaurin's, 357; operations with, 362
oscillating, 339
; Stirling, Strophoid, 534
Subnormal, 43 polar, 126
Subtangent, 43 polar, 126
Successive differentiation, 73, 462
Successive integration, 491 247
Variable, change of, 166, 457 Sine curve, 533
Skew curves, 471, 480; of, nition, parametric
Slope of a curve, 42
form, 115; polar form, 125 7 dependent, 8 defi ; inde Velocity, length 473 ; pendent, 8
; Solids 323 ; revolution, centroid
volume
surface of, 277 of
; ; of,
of, 265, 267, 268
Speed, 121, 146
Spherical coordinates, 6, 529, 530
prolate,
Spheroid, oblate, 266
266
; Spiral, of curvilinear motion, 120,
146; rectilinear motion, 65
Vibration, damped harmonic, 405
forced harmonic, 405
simple
harmonic, 403
Volume, of a hollow solid of revoof a solid with
lution, 267
known cross section, 283 of a
; ; ; ; solid of revolution, 265, under a surface, 501, 524; Archimedes, 127, 128, 155,
277, 534
hyperbolic or reciprocal, 119, 128, 264, 277, 534; triple integration, 521, 526 logarithmic or equiangular, 127,
128, 534
parabolic, 535 Witch, 62, 251, 269, 322, 531
Work, 328 ; ; PRINTED IN THE UNITED STATES OF AMERICA 267; by DATE DUE Can 01113 <4fiM<4 51? G?6ea University Libraries CarnegieMellon University
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