elements of the differential and integral calculus

Carnegie-Mellon University

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MATHEMATICAL TEXTS

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EDITED BY

PERCEY F. SMITH, PH.D.PROFESSOR OF MATHEMATICS IN THE SHEFFIELD

SCIENTIFIC SCHOOL OF YALE UNIVERSITY

ELEMENTS OF THE

DIFFERENTIAL AND INTEGRAL

CALCULUS

Revised Edition

By WILLIAM ANTHONY GRANVILLE, PH. D., LL. D.

Formerly President of Gettysburg College

PERCEY F. SMITH, PH.D.

and WILLIAM RAYMOND LONGLEY, PH.D.

Professors of Mathematics, Yale University

GINN AND COMPANYBOSTON NEW YORK CHICAGO LONDON ATLANTA DALLAS COLUMBUS SAN FRANCISCO

COPYRIGHT, 1941, BY WILLIAM ANTHONY GRANVILLE, PERCEY F. SMITHAND WILLIAM RAYMOND LONGLEY

ALL EIGHTS RESERVED

641.10

COPYBIGHT, 1904, 1911, BY WILLIAM ANTHONY GRANVILLE AND PERCEY F. SMITH

COPYRIGHT, 1929, 1934, BY WILLIAM ANTHONY GRANVILLE, PERCEY P. SMITHAND WILLIAM RAYMOND LONGLEY

&tfienarum ffiregg

GINN AND COMPANY PRO-PBIETOKS BOSTON U.S.A.

PREFACE

The alterations in the text in this revision consist of verbal

changes in the details of some of the proofs and discussions, and in

the addition of a chapter on Hyperbolic Functions. This chapterhas been written with the same attention to clearness and com-pleteness that marks all other sections of the book. Also, cylindricalcoordinates 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. Someapplications 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 thestudent greater self-reliance in checking his work. Teachers whodesire answers to the other problems should communicate with thepublishers.

The labor of the authors will be amply repaid if this revised editionmeets with the generous and well-nigh universal favor accorded Gran-ville's Calculus since its first appearance.

PERCEY F. SMITHWILLIAM R. LONGLEY

CONTENTS

DIFFERENTIAL CALCULUSPAGE

CHAPTER I. COLLECTION OF FORMULAS 1

Formulas from elementary algebra and geometry, 1 Formulas from planetrigonometry, 2 Formulas from plane analytic geometry, 3 Formulasfrom solid analytic geometry, 4 Greek alphabet, 6

CHAPTER II. VARIABLES, FUNCTIONS, AND LIMITS 7

Variables and constants, 7 Interval of a variable, 7 Continuous vari-

ation, 7 Functions, 8 Independent and dependent variables. 8 Notationof functions, 8 Division by zero excluded, 9 Graph of a function ; con-

tinuity, 10 Limit of a variable, 10 Limiting value of a function, 11 Theo-rems on limits, 1 1 Continuous and discontinuous functions, 12 Infinity (oo),

13 Infinitesimals, 17 Theorems concerning infinitesimals and limits, 17

CHAPTER III. DIFFERENTIATION 19

Introduction, 19 Increments, 19 Comparison of increments, 20 Deriva-

tive of a function of one variable, 21 Symbols for derivatives, 22 Dif-

ferentiable functions, 23 General rule for differentiation, 23 Interpretationof the derivative by geometry, 25

CHAPTER IV. RULES FOR DIFFERENTIATING ALGEBRAICFORMS 28

Importance of the General Rule, 28 Differentiation of a constant, 29

Differentiation of a variable with respect to itself, 29 Differentiation of

a sum, 30 Differentiation of the product of a constant and a function, 30

Differentiation of the product of two functions, 30 Differentiation of the

product of n functions, n being a fixed number, 31 Differentiation of a

function with a constant exponent. The Power Rule, 32 Differentiation

of a quotient, 32 Differentiation of a function of a function, 37 Differen-

tiation of inverse functions, 38 Implicit functions, 39 Differentiation of

implicit functions, 40

CHAPTER V. VARIOUS APPLICATIONS OF THE DERIVATIVE . 42

Direction of a curve, 42 Equations of tangent and normal ; lengths of sub-

tangent and subnormal, 43 Maximum and minimum values of a func-

tion; introduction, 47 Increasing and decreasing functions. Tests, 50

Maximum and minimum values of a function ; definitions, 52 First

method for examining a function for maximum and minimum values.

Working rule, 53 - Maximum or minimum values when f'(x) becomesinfinite and f(x) is continuous, 55 Maximum and minimum values.

Applied problems, 57 Derivative as the rate of change, 64 -

Velocity in

rectilinear motion, 65 Related rates, 67

viii CONTENTSPAGE

CHAPTER VI. SUCCESSIVE DIFFERENTIATION AND APPLICA-TIONS . ... 73

Definition of successive derivatives, 73 Successive differentiation of im-

plicit functions, 73 Direction of bending of a curve, 75 Second method for

testing for maximum and minimum values, 76 Points of inflection, 79

Curve-tracing, 81 Acceleration in rectilinear motion, 83

CHAPTER VII. DIFFERENTIATION OF TRANSCENDENTALFUNCTIONS. APPLICATIONS 86

Formulas for derivatives; second list, 86 The number c. Natural loga-

rithms, 87 Exponential and logarithmic functions, 89 Differentiation of

a logarithm, 90 Differentiation of the exponential function, 91 Differen-

tiation of the general exponential function. Proof of the Power Rule, 92

Logarithmic differentiation, 93 The function sin j*, 97 Theorem, 98

Differentiation of sin v, 99 The other trigonometric functions, 99 Dif-

ferentiation of cos ?;, 100 Proofs of formulas XV XIX, 100 Comments,101 Inverse trigonometric functions, 105 Differentiation of arc sin r,

106 Differentiation of arc cos r, 106 Differentiation of arc tan v, 107

Differentiation of arc ctn v, 108 Differentiation of arc sec r and arc esc v,

108 Differentiation of arc vers r, 109

CHAPTER VIII. APPLICATIONS TO PARAMETRIC EQUATIONS,POLAR EQUATIONS, AND ROOTS 115

Parametric equations of a curve. Slope, 115 Parametric equations.Second derivative, 119 Curvilinear motion. Velocity, 120 Curvilinear

motion. Component accelerations, 121 Polar coordinates. Angle betweenthe radius vector and the tangent line, 123 Lengths of polar sub-

Uingent and polar subnormal, 126 Real roots of equations. Graphicalmethods, 128 Second method for locating real roots, 130 Newton'smethod, 131

CHAPTER IX. DIFFERENTIALS 136

Introduction, 136 Definitions, 136 Approximation of increments bymeans of differentials, 137 Small errors, 138 Formulas for finding the

differentials of functions, 140 Differential of the arc in rectangular coordi-

nates, 142 Differential of the arc in polar coordinates, 144 Velocity as

the time-rate of change of arc, 146 Differentials as infinitesimals, 146

Order of infinitesimals. Differentials of higher order, 148

CHAPTER X. CURVATURE. RADIUS AND CIRCLE OF CURVA-TURE . . . .149

Curvature, 149 Curvature of a circle, 149 Formulas for curvature; rec-

tangular coordinates, 150 Special formula for parametric equations, 151

Formula for curvature ; polar coordinates, 1 51 Radius of curvature, 1 52

Railroad or transition curves, 152 Circle of curvature, 153 Center of cur-

vature, 157 Evolutes, 158 Properties of the evolute, 162 Involutes andtheir mechanical construction, 163 Transformation of derivatives, 166

CONTENTS ix

PAGE

CHAPTER XL THEOREM OF MEAN VALUE AND ITS APPLICA-TIONS . . . . 169

Rolle's Theorem, 169 Osculating circle, 170 Limiting point of intersec-

tion of consecutive normals, 171 Theorems of Mean Value (Laws of

the Mean), 172 Indeterminate forms, 174 Evaluation of a function

taking on an indeterminate form, 174 Evaluation of the indeterminate

form -> 175 Evaluation of the indeterminate form , 17$ . Evaluation

of the indeterminate form QO, 178 Evaluation of the indeterminateform oc - oo, 179 Evaluation of the indeterminate forms 0, 1, 00,180 The Extended Theorem of Mean Value, 182 Maxima and minimatreated analytically, 182

INTEGRAL CALCULUS

CHAPTER XII. INTEGRATION; RULES FOR INTEGRATINGSTANDARD ELEMENTARY FORMS . . . 187

Integration, 187 Constant of integration. Indefinite integral, 189

Rules for integrating standard elementary forms, 190 Formulas (3), (4),

(5), 193 Proofs of (0) and (7), 198 Proofs of (8) (17), 200 Proofs of

(18) (21 ), 203 Proofs of (22) and (23), 211 Trigonometric differentials,

213 Integration of expressions containing Vu- ?/2 or V'

u'z a 2 bytrigonometric substitution, 221 Integration by parts, 223 Comments, 227

CHAPTER XIII. CONSTANT OF INTEGRATION 229

Determination of the constant of integration by means of initial condi-

tions, 229 Geometrical signification of the constant of integration, 229

Physical signification of the constant of integration, 233

CHAPTER XIV. THE DEFINITE INTEGRAL 237

Differential of the area under a curve, 237 The definite integral, 237

Calculation of a definite integral, 239 Change in limits corresponding to

change in variable, 240 Calculation of areas, 241 Area when the equa-

tions of the curve are given in parametric form, 242 Geometrical repre-

sentation of an integral, 244 Approximate integration ; trapezoidal rule,

245 Simpson's rule (parabolic rule), 247 Interchange of limits, 249

Decomposition of the interval of integration of the definite integral, 250

The definite integral a function of its limits, 250 Improper integrals. In-

finite limits, 250 Improper integrals. When // <(x) is discontinuous, 251

CHAPTER XV. INTEGRATION A PROCESS OF SUMMATION . . 254

Introduction, 254 The Fundamental Theorem of Integral Calculus, 254

Analytical proof of the Fundamental Theorem, 257 Areas of plane curves ;

rectangular coordinates, 258 Areas of plane curves ; polar coordinates,

262 Volumes of solids of revolution, 265 Length of a curve, 271

Lengths of plane curves ; rectangular coordinates, 272 Lengths of plane

curves; polar coordinates, 274 Areas of surfaces of revolution, 277 Solids

with known parallel cross sections, 283

x CONTENTSPAGE

CHAPTER XVI. FORMAL INTEGRATION BY VARIOUS DEVICES 289

Introduction, 289 Integration of rational fractions, 289 Integration bysubstitution of a new variable ; rationalization, 296 Binomial differentials,

299 Conditionsofrationalization ofthe binomial differential, 302- Transfor-

mation of trigonometric differentials, 303 Miscellaneous substitutions, 305

CHAPTER XVII. REDUCTION FORMULAS. USE OF TABLE OFINTEGRALS 307

Introduction, 307 Reduction formulas for binomial differentials, 307

Reduction formulas for trigonometric differentials, 312 Use of a table of

integrals, 315

CHAPTER XVIII. CENTROIDS, FLUID PRESSURE, AND OTHERAPPLICATIONS 320

Moment of area ; centroids, 320 * Centroid of a solid of revolution, 323

Fluid pressure, 325 Work, 328 Mean value of a function, 333

DIFFERENTIAL AND INTEGRAL CALCULUS

CHAPTER XIX. SERIES 338

Definitions, 338 The geometric series, 339 - Convergent and divergent

series, 340 General theorems, 341 Comparison tests, 342 Cauchy's test-

ratio test, 345 Alternating series, 347 Absolute convergence, 348

Summary, 348 Power series, 350 - The binomial series, 353 Another

type of power series, 354

CHAPTER XX. EXPANSION OF FUNCTIONS 357

Maclaurin's series, 357 Operations with infinite series, 362 Differentia-

tion and integration of power series, 365 Approximate formulas derived

from Maclaurin's series, 367 Taylor's series, 369 - Another form of

Taylor's series, 371 Approximate formulas derived from Taylor's series, 372

CHAPTER XXL ORDINARY DIFFERENTIAL EQUATIONS ... 375

Differential equations order and degree, 375 Solutions of differential

equations. Constants of integration, 376 Verification of the solutions of

differential equations, 377 Differential equations of the first order and of

the first degree, 378 Two special types of differential equations of higher

order, 387 Linear differential equations of the second order with constant

coefficients, 390 Applications. Compound-interest law, 399 - Applications

to problems in mechanics, 402 Linear differential equations of the nth

order with constant coefficients, 407

CHAPTER XXII. HYPERBOLIC FUNCTIONS 414

Hyperbolic sine and cosine, 414 Other hyperbolic functions, 415 - Table

of values of the hyperbolic sine, cosine, and tangent. Graphs, 415 Hyper-bolic functions of v + w, 417 Derivatives, 420 Relations to the equi-

lateral hyperbola, 420 Inverse hyperbolic functions, 423 Derivatives

(continued), 425 Telegraph line, 428 Integrals, 430 Integrals (con-

tinued), 432 The gudermannian, 435 Mercator's Chart, 438 Relations

between trigonometric and hyperbolic functions, 440

CONTENTS xi

PAGECHAPTER XXIII. PARTIAL DIFFERENTIATION 444

Functions of several variables. Continuity, 444 * Partial derivatives, 445

Partial derivatives interpreted geometrically, 446 The total differential,

449 Approximation of the total increment. Small errors, 451 Total de-

rivatives. Rates, 455 - Change of variables, 457 - Differentiation of implicit

functions, 458 Derivatives of higher order, 462

CHAPTER XXIV. APPLICATIONS OF PARTIAL DERIVATIVES 466

Envelope of a family of curves, 466 The evolute of a given curve con-

sidered as the envelope of its normals, 469 Tangent line and normal planeto a skew curve, 471 Length of arc of a skew curve, 473 Normal line and

tangent plane to a surface, 475 Geometric interpretation of the total dif-

ferential, 477 - Another form of the equations of the tangent line andnormal plane to a skew curve, 480 Law of the Mean, 482 Maxima andminima of functions of several variables, 483 - Taylor's theorem for func-

tions of two or more variables, 488

CHAPTER XXV. MULTIPLE INTEGRALS 491

Partial and successive integration, 491 - Definite double integral. Geomet-ric interpretation, 492 Value of a definite double integral taken over a

region S, 497 - Plane area as a definite double integral. Rectangular coordi-

nates, 498 Volume under a surface, 501 Directions for setting up a double

integral, 503 Moment of area and centroids, 503 - Theorem of Pappus,504 - Center of fluid pressure, 506 - Moment of inertia of an area, 508

Polar moment of inertia, 510 Polar coordinates. Plane area, 512

Problems using polar coordinates, 514 General method for finding the

areas of curved surfaces, 517 - Volumes found by triple integration, 521

Volumes using cylindrical coordinates, 524

CHAPTER XXVI. CURVES FOR REFERENCE 531

CHAPTER XXVII. TABLE OF INTEGRALS 538

INDEX 553

DIFFERENTIAL CALCULUS

CHAPTER I

COLLECTION OF FORMULAS

1. Formulas from elementary algebra and geometry. For the con-

venience of the student we give in Arts. 1-4 the following lists of

formulas. We begin with algebra.

(1) Quadratic AJT* + B.r + C = 0.

Solution. 1. By factoring : Factor Ax'2 4- Bx 4- r, set each factor equalto zero, and solve for x.

2. By completing the square : Transpose r, divide by the coefficient

of J 2, add to both members the square of half the coefficient of x, and

extract the square root. _ R3. By the formula x =

Nature of the roots. The expression Ii2 4 AC beneath the radical in

the formula is called the di^crlmhuint. The two roots are real and unequal,real and equal, or imaginary, according as the discriminant is positive,

zero, or negative.

(2) Logarithms

log ab = Jog a -f log b. log a7 ' = n log a. log 1=0.

'a = -

(3) Binomial theorem (n being a positive integer)

Jog ^= log a log b. log v'a = -

log a. loga a = 1.

|

(4) Factorial numbers. n! = [n= l-2-3-4---(w- l)w.

In the following formulas from elementary geometry, r or Rdenotes radius, a altitude, B area of base, and s slant height.

(5) Circle. Circumference = 2 Trr. Area = Trr 2 .

(6) Circular sector. Area = \ r 2a, where a = central angle of the sector

measured in radians.

1


(7) Prism. Volume = Ba.

(8) Pyramid. Volume \ Ba.

(9) Right circular cylinder. Volume ?rr2a. Lateral surface = 2 Trra.

Total surface 2 irr(r + a).

(10) Right circular cone. Volume = J irr2a. Lateral surface = Trrs.

Total surface irr(r -f s).

(11) Sphere. Volume = :! Trr3

. Surface = 4 Trr2

.

(12) Frustum of a right circular cone. Volume = J ?ra(fi2-f r 2 -f fir).

Lateral surface irs(R -f r).

2. Formulas from plane trigonometry. Many of the following for-

mulas will be found useful.

(1) Measurement of angles. There are two common methods of measur-

ing angular magnitude ; that is, there are two unit angles.

Degree measure. The unit angle is;i (

l of a complete revolution and is

called a degree.

Circular measure. The unit angle is an angle whose subtending arc is

equal to the radius of that arc, and is called a radian.

The relation between the unit angles is given by the equation

180 degrees TT radians (TT 3.14159 ),

the solution of which gives

1 degree = T^ = 0.01741 80

radian ;1 radian = 57.29 degrees.

From the above definition we have

Number of radia.ns In an angle =subtend hiq ore

radius

These equations enable us to change from one measurement to another.

(2) Relations

ctn .r

1

tansec x . csc r

~

, sin x ,

tan j* ; ctn .r

cos JT

cos x

sin x

cos x sin x

sin 2 x -f cos 2 x = 1; 1-1- tan 2 x = sec 2

.r ; 1 -f ctn 2 x = esc 2 x.

(3) Formulas for reducing angles

COLLECTION OF FORMULAS {

(4) Functions of (x -f y) and (x y)

sin (x -f- y) sin x cos y + cos x sin y.

sin (.r #) = sin .r cos # - cos .r sin y.

cos (x -f /y)= cos .r cos //

- sin x sin //.

cos (.r-

#) = cos .r cos // -f sin .r sin ?/.

tan (jr + y} - ^L^^UL . tau ^ __} = _lan .r - tan y .

1 - tan jr tan //

*

1 -f tan .r tan y

(5) Functions of 2 x and| x

sin 2s = 2 sin j- cos jc ; cos 2 .r cos 2jc - sin 2 x ; tan 2 .r - __r tan ^

..

1 tan 2 x

fl^L-' :

\ -; cos =,

: tun =- - \ - 2

sin-' j' - i - 1 cos 2 .r ; COS L>

.r --J + .1 cos 2 j-.

(6) Addition theorems

sin jr -f sin //= '2 sin i (.r -f- //) cos \ (.r ?/).

.sin x sin //-. 2 cos .1 (.> -f //)sin ! (.r //).

cos .r -f cos // L* co^ \ ( ./ -f //) cos \ (j- ?/).

cos.r- cos// ^ i? yiu ', (./

'

//)sin ^(.r-

jy).

(7) Relations for any triangle

Law of sine*. -. = - - -^~sin .4 sin /> sin C

Law of cos hi ex. a 2 h- -f r 2 2 be cos A.

Formulas for area.

sin (/> 4- T)

- a) (a />)(.-; r), where ,s ?, (a -f b -f- r).

3. Formulas from plane analytic geometry. The more important for-

mulas are given in the following list.

(1) Distance between two points /*i(xi, yi) and P2(x2, yz)

Slope of P^Pz.

Midpoint. x-\ (x, + x2 ), ?/=

2 (2/i -f 2/2).

(2) Angle between two lines

1 -f mim2

(For parallel lines, wi = W2 ; for perpendicular lines, mint = 1.)


(3) Equations of straight lines

Point-slope form. y y\ = m(x xi).

Slope-intercept form. y mx + b.

Two-point form.y ""

yi = y '

z~ ^ -

x x\ x% Xi

Intercept form.-

-f ~ = 1.

(4) Perpendicular distance from the line Ax + By -f C = to PI(JCI, yi)

(5) Relations between rectangular and polar coordinates

(6) Equation of the circle

Center (h, k). (JT-

h)'2 + (y

- k)2 = r 2

.

(7) Equations of the parabola

Vertex the origin. y'2 = 2 px, focus (\ p, 0).

x'2 2 py, focus (0, i /;).

Vertex (h, k). (y k)2 = 2 p(x h), axis //

= /c.

(x h}'2 2 p(y k}, axis x /i.

Axis ///r y-axiii. y ylj^2-f C.

(8) Equations of other curves

Ellipse with center at the origin and with foci on the x-axis. (a > b)

Hyperbola with center at the origin and with foci on the x-axis.

^1 _ 111 i

a 2 b 2~

Equilateral hyperbola with center at the origin and with the coordinate

axes for asymptotes.

xy = C.

See also Chapter XXVI.

4. Formulas from solid analytic geometry. Some of the more impor-

tant formulas are given.

(1) Distance between PI(XI, yi, z\) and P2(*2, 1/2, 22)

d = V(xi - x2 )2-h (yi

-l/2 )

2 + (21- z2 )

2.

COLLECTION OF FORMULAS

(2) Straight line

Direction cosines : cos a, cos (3, cos 7.Direction numbers: a, b, c.

Thena 6 r

cos 2a- -f cos 2

ft -f cos 2 7 = 1.

acos a =

cos ft-

cos 7 =

i2 + b'

2 + r 2

b

Va 2-f h 2

r

Vu 2~+ 6- + r 2

For the line joining (jc\, ?/i, cO and (j-2 , ?/L., -.,)

cos or _ cos ft __ cos 7

(3) Two lines

Direction cosines : cos <rv, cos ft, cos 7 ;cos <v', cos ft', cos 7'.

Direction numbers: a, b, r; a', //, r'.

If = angle between the lines,

cos = cos a cos <V -f cos ft cos ft' -f cos 7 cos 7',

c/a' 4- W/ -f re'cos t/ =

2 + b- + r 2 Va /2"-f // 2

-f r' 2

Parallel lines. - = ~ = -^.

a' 6' rx

Perpendicular lincx. aa r

-f 6/>' -f rr' ~ ().

(4) Equations of the straight line with direction numbers a, btc pass-

ing through (xi, z/i, 21)x - n _ y

-7/1 _ z - :\

t

a b c

(5) Plane. For the plane Ax -f Ky + Cz + J>-() the coefficients A, B, Care the direction numbers of a line perpendicular to the plane.

Equation of a plane passing throuyJi (>i, //j, ^ij and perpendicular to the

line with direction numbers A, B, C.

A(x - XL) + B(y - ?/,) -f C(z-

2,) = 0.

(6) Two planes

Equations : Ax + By -f Cz 4- D = 0,

A'x+ B'y+ C'z+ D' = ().

Direction numbers of the line of intersection :

EC' - CB' t CA' - AC', AB' - BA'.

If 6 = angle between the planes, then

AA' + BB' + CC'cos =

B'2 C/2

(7) Cylindrical coordinates.* The distance z of a point P(x, y, z) from

the A' 7-plane and the polar coordinates (p, 6) of

its projection (x, y, 0) on the X T-plane are called

the cylindrical coordinates of P. The cylindrical

coordinates of P are written (p, 6, z).

If the rectangular coordinates of P are x, y, z,

then, from the definitions and the figure, we have

x = p cos 0,

p2 - +

y P sin 0, 2

9 arc tan -x

(8) Spherical coordinates.* The radius vec-

tor r of a point P, the angle between OP and

the z-axis, and the angle between the projec-

tion of OP on the AT-plane and the .r-axis are

called the spherical coordinates of /'.</> is called

the colatitude and 6 the longitude. The spherical

coordinates of P are written (r, (/>, 0).

If the rectangular coordinates of P are x, y, z,

then, from the definitions and the figure, we have

= r sin c/> cos

f * =:

?/= r sin sin $, 2 = r cos

;

6 = arc tan -></>= arc tan

^

LETTERS NAMES

P p RhoZ d s SigmaT r TauT i> Upsilon<f> Phi

X x Chi

^i/'

Psi

12 a Omega

* For a discussion of cylindrical and spherical coordinates see Smith, Gale, and

Neellry's "New Analytic Geometry, Revised Edition" (Ginn and Company), pp. 320-322.

CHAPTER II

VARIABLES, FUNCTIONS, AND LIMITS

6. Variables and constants. A variable is a quantity to which an

unlimited number of values can be assigned in an investigation.

Variables are denoted usually by the later letters of the alphabet.

A quantity whose value is fixed in any investigation is called a

constant.

Numerical or absolute constants retain the same values in all prob-

lems; as 2, 5, V?, TT, etc.

Arbitrary constants are constants to which numerical values maybe assigned, and they retain these assigned values throughout the

investigation. They are usually denoted by the earlier letters of

the alphabet.

Thus, in the equation of a straight line,

x and y are the variable coordinates of a point moving along the

line, while the arbitrary constants a and b are the intercepts, for

which definite values are assumed.

The numerical (or absolute) value of a constant a, as distinguished

from its algebraic value, is represented by |a|. Thus, | 2|= 2 = |2|.

The symbol |a| is read "the numerical value of a."

7. Interval of a variable. Very often we confine ourselves to a por-

tion only of the number system. For example, we may restrict our

variable so that it shall take on only values lying between a and b.

Also, a and b may be included, or either or both excluded. We shall

employ the symbol [a, 6], a being less than b, to represent the num-

bers a, b, and all the numbers between them, unless otherwise stated.

This symbol [a, 6] is readuthe interval from a to b."

8. Continuous variation. A variable x is said to vary continuously

through an interval [a, 6] when x increases from the value a to the

value b in such a manner as to assume all values between a and 6

in the order of their magnitudes, or when x decreases from x = 6 to

x = a, assuming in succession all intermediate values. This may be

illustrated geometrically by the diagram on page 8.

8 DIFFERENTIAL CALCULUS

The origin being at O, lay off on the straight line the points Aand B corresponding to the numbers a and b. Also, let the point

P correspond to a particular

value of the variable x. Evi-g ^

-^

dently the interval [a, b\ is rep-

resented by the segment AH. As x varies continuously through

the interval [a, H the point P generates the segment AB when x

increases, or the segment BA when x decreases.

9. Functions. When two variables are so related that the value of

the first variable is determined when the value of the second variable

is given, then the first variable is said to be a function of the second.

Nearly all scientific problems deal with quantities and relations of

this sort, and in the experiences of everyday life we are continually

meeting conditions illustrating the dependence of one quantity on

another. For instance, the weight a man is able to lift depends on

his strength, other things being equal. Similarly, the distance a boy

can run may be considered as depending on the time. Or we maysay that the area of a square is a function of the length of a side, and

the *o!ume of a sphere is a function of its diameter.

10. Independent and dependent variables. The second variable, to

which values may be assigned at pleasure within limits depending

on the particular problem, is called the independent variable, or

argument; and the first, variable, whose value is determined when

the value of the independent variable is given, is called the dependent

variable, or junction.

Frequently, when we are considering two related variables, it is

in our power to fix upon either as the independent variable; but

having once made the choice, IM> change of independent variable is

allowed without certain precautions and transformations. For ex-

ample, the area of a square is a function of the length of its side.

Conversely, the length of a side is a funrlJO'i of the area.

11. Notation of functions. The symbol /(:r) is used to denote a func-

tion of J, and is read / of x. In order to distinguish between different

functions, the prefixed letter is changed, as F(.r), (/>(;r), /'(*), etc.

During any investigation a functional symbol indicates the same

law of dependence of the function upon the variable. In the simpler

cases this law takes the form of a series of analytical operations uponthe variable. Hence, in such a case, the functional symbol will indicate

the same operations or series of operations applied to different values

of the variable. Thus, if

J(x) =r j2 - 9 .r + 14,

then f(y) = y2 9 y + 14.

VARIABLES, FUNCTIONS, AND LIMITS 9

Also /(&+l) = (6 + l)a-9ifc + l) + 14 = &2 -7& + 6,

/(0)=02 -9. 0+14-14,/(- 1) = (- 1)2 - 0(- 1) + 14 - 24,

/(3) = 3-- 9-3 + 14 --4.12. Division by zero excluded. The quotient of two numbers a and

6 is a number x such that a = />.r. Obviously, division by zero is

ruled out by this definition. For if l> o, and we recall that anynumber times zero equals zero, we see that .r (ioes not exist unless

a = 0. But, in this case, jc may be any number whatever. The forms

a

()' ()'

are, therefore, meaningless.

Care should he taken not to divide hy zero inadvertently. The fol-

low ing fallacy is an illustration.

Assume that a b.

Then, evidently, ah = <i~.

Subtracting />-, ab b- a'2 b'2 .

Factoring, b()(</ />).

Dividing by a b, l> a + b.

But a - b;

therefore b 2 />,

or 1 --- 2.

The ahsurd result is due to the fuel that we divided by a b 0.

PROBLEMS

1. Given f(s) = jr*- 5 s- - 1 .r -} 20, show that

/(I) = 12, /(5) = 0, /(())- - 2/C}), f^} ^

-

r)/(- 1).

2. Itfis) = 4 - 2 s- -f r1

,find /(O ),/(!),/(- l),/(2 ),/(-- 2).

3. If F(0) = sin 2 + cos 6, find /-'(O), /M TT), F(TT).

4. Given /(.r) ,r:( 5 jr'

2 4 jr -f 20, show that

/(/-f 1) = /;! - 2 ^ - 11 /+ 12.

5. Given /(//) ?/2 2 y -f 6, showr that

f(y -f //) = ?/L> - 2 ?/ -f f> -f 12 O/

~1 J/' + h 2

-

6. Given /(j*) = ?'*> -f 3 j-, show that

fU -f /O-

f(s) = Ms 2 + 1 ;// -f 3 ^ 2-f R

7. Given /(r) = -> show that /(V -f /O- /O) - -- -

^.

Jf J^ ~T" X/

8. Given 0(z) = 4* f show that <f>(z + 1)-

<t>(z)= 3 0().

9. If <f>(x)= a', show that 0(y) -

</>(*)= 0(y + 2).

1 x10. Given 0(x) = log show that

11. Given /(x) = sin x, show that

f(x -f 2 h) f(f) = 2 cos (x -f /?) sin /?,.

HINT. Use (6), p. .'}.

13. Graph of a function; continuity. Consider the function x2,

and let

(1) y = x'2 .

This relation gives a value of y for any value of x;

that is, y is defined by (1) for all values of the inde-

pendent variable. The locus of (1), a parabola (see

figure), is called the yraph of the function x 2. If x

varies continuously (Art. 8) from x a to x = />, then

y will vary continuously from y a- to y/=

&-', andthe point /'(x, ?/) will move continuously along the graph from the

point (a, a2) to (h, />-). Also, a and b may have any values. We then

say, "the function x- is continuous for all

values of x."

Consider the function - and letx

(2) y = --

This equation gives a value of y for anyvalue of x except x = (Art. 12). For x =the function is not defined. The graph, the locus

of (2), is an equilateral hyperbola (see figure). If x increases con-

tinuously through any interval [a, />] which does not include x 0,

then'//

will decrease continuously from - to p and the point P(x, y)

will trace the graph between the corresponding points (a, -)> (?>, ^-

Then we say, "the function - is continuous for all values of x except

x 0." There is no point on the graph for x = 0.

These examples illustrate the concept of continuity of a function.

A definition is given in Art. 17.

14. Limit of a variable. The idea of a variable approaching a limit

occurs in elementary geometry in establishing a formula for the area

of a circle. The area of a regular inscribed polygon with any numberof sides n is considered, and n is then assumed to increase indefinitely.


The variable area then approaches a limit, and this limit is defined

as the area of the circle. In this case the variable v (the area) in-

creases constantly, and the difference a r, where a is the area of

the circle, diminishes and ultimately becomes less than any preas-

signed number, however small.

The relation illustrated is made precise by the

DEFINITION. The variable r is said to approach the constant / as

a limit when the successive values of r are such that the numerical

value of the difference r / ultimately becomes and remains less

than any preassigned positive number, however small.

The relation defined is written lim r -~ /. For convenience, we shall

use the notation r /, read, "r approaches / as a limit," or, more

briefly, "r approaches Z." (Some authors use the notation v ==/.)

ILLUSTRATIVE EXAMPLE. Lot the valuer of r be

2 + 1, li -t i. 12 -f|,

.... 2 + 1, -

f

without end. Then, obviously, lim r - 2, or r U.

If we mark on a straight line, as in Art. S, the point /, correspond-

ing to the limit /, and from L lay off on each side a length 6, however

small, then the points determined by /' \\ill ultimately all lie within

the segment corresponding to the interval [/ e, / -1 e|.

15. Limiting value of a function. In applications, the situation that

usually arises is this. We have a variable r, and a given function z

of r. The independent variable r assumes values such that v /.

We then have to examine the values of the dependent variable z,

and, in particular, determine if z approaches a limit. If there is a

constant a such that lim z = a, then the relation described is written

lim z = a,r- I

read, "the limit of z, as v approaches I, is a."

1G. Theorems on limits. In calculating the limiting value of a func-

tion, the following theoremsmay be applied. Proofs are given in Art. 20.

Suppose u, ?', and w are functions of a variable j, and suppose that

lim u = A, lim v B, lim w = C\x - a x - a -r

- a

Then the following relations hold.

(1) lim (u + v - w) = A -f B - C.jc- a

(2) lim (uvw) = ABC.x-* a

u A(3) lim

" = -9 if B is not zero.

jt-a V B

Briefly, in words, the limit of an algebraic sum, of a product, or

of a quotient is equal, respectively, to the same algebraic sum, product,

or quotient of the respective limits, provided, in the Last named, that the

limit of the denominator is not zero.

If c is a constant (independent of rj and B is not zero, then, from

the above,c c

(4) lim (u + c) = A + c, lim cu cA, lim - =x * a JT-* a a- -> a V Jj

Consider some examples.

1. Prove lim <JT'* -f- 4 a-)-

12.JT *

Solution. The given function is the sum of .c- and 4 jc. We first find the limit-

ing values of these two functions. By rji,

lim jr'2 = 4, since j :

' = x x.X li

By (4), lim 4 a- = 4 lim x 8.

J" ~ X -li

Hence, by (1), the answer is 4 -f 8 -- 12.

~ :'

9 rt

2. Prove lim ~ ~ -

Solution. (Considering the numerator, lim i^ :> 9) 5, by (2) and (4). For

the denominator, lim '.: 1 U) - 4. Hence, by '.'}), we have the re<juired result.

17. Continuous and discontinuous functions. In Ex. 1 of the pre-

ceding article, where it was shown that

lim (.r- + 4 .r) = 12,.r 1!

we observe that the answer is the ralue of the function for x 2.

Thai is, the limiting value of the function when ;r approaches 2 as a

limit is equal to the value of the function for jc = 2. The function is

said to be continuous for .r = 2. The general definition is as follows.

DEFINITION. A function /(.r) is said to be continuous for r = a if

the limiting value of the function when .r approaches a as a limit is

the value assigned to the function for .r = a. In symbols, if

lim /(*)=/ (a),x * a

then /(.r) is continuous for x a.

The function is said to be discontinuous for x = a if this condition

is not satisfied.

Attention is called to the following two cases of common oc-

currence.


CASE I. As an example illustrating a simple case of a function con-tinuous for a particular value of the variable, consider the function

J x '

X *>

For z = 1, f(x) = /(I) = 3. Moreover, if x approaches 1 as a limit,the function f(x) approaches 3 as a limit (Art. 16). Hence thefunction is continuous for x = 1.

CASE II. The definition of a continuous function assumes thatthe function is already defined for x-o. If this is not the case, how-ever, it is sometimes possible to assign such a value to the functionfor x = a that the condition of continuity shall be satisfied. Thefollowing theorem covers these cases.

Theorem. If f(x) is not dcjiucd for x = a, and if

lim f(x) = B,x a

then f(x) mil be continuous for x = a if tt is assumed as the value of

f(x) for x = a.

1> A

Thus, the function ~

is not, defined for x 2 (since then there would be division by zero).T.ut for every other value of :r,

and lim (x + 2) = 4;

therefore lim

Although the function is not defined for x = 2, if we arbitrarily as-

sign to it the value 4 for x 2, it becomes continuous for this value.

A function f(x) is said to be coHtitiuoux in an niterml whew it is

continuous for all valves of x in this in terra I.*

In the calculus we have to calculate frequently the limiting value

of a function of a variable ?' when r approaches as a limit a value a

lying in an interval in which the function is continuous. This limit-

ing value is the value of the function for r = a.

18. Infinity (oo). If the numerical value of a variable v ultimatelybecomes 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 functionin question is actually continuous.

however large, we say v becomes infinite. If v takes on only positive

values, it becomes positively infinite; if negative values only, it

becomes negatively infinite. The notation used for the three cases is

lim v = QO, lim v = + oo

, lim r = 00 .

In these cases v does not approach a limit as defined in Art. 14. Thenotation lim v = QO, or v > oo, must be read

"v becomes infinite/'

and not "v approaches infinity."*

We may now write, for example,

lim - = oo,x *<)

meaning that - becomes infinite when x approaches zero.

Referring to Art. 17, it appears that if

lim/(.r) = 00,x - a

that is, if f(x) becomes infinite as x approaches a as a limit, then

f(x) is discontinuous for x a.

A function may have a limiting value when the independentvariable becomes infinite. For example,

lim i = 0.X oc X

And, in general, if /O) approaches the constant value A as a limit

when x > oo, we use the notation of Art. 17 and write

lim /(*) = A.x -- oc

Certain special limits occur frequently. These are given below. Theconstant r is not zero.

Written in the form, of limits Abbreviated form often used

(1) lim - = oo. =oo.r - V

(2) limczj = oo. c oo = oo.r -* oo

/o\ r V OO(o) Jim - = oo. = oo.

r -- oo C C

(4) lim - = 0. - = 0."00

* On account of the notation used and for the sake of uniformity, the expressionp * -f oo is sometimes read

"P approaches the limit plus infinity." Similarly, r * oo is

read "r approaches the limit minus infinity," and v is read"

r, in numerical value,

approaches the limit infinity."

This phraseology is convenient, but the student must not forget that infinity is not a

limit, for infinity is not a number at all.

These special limits are useful in finding the limiting value of the

quotient of two polynomials when the variable becomes infinite.

The following example will illustrate the method.

ILLUSTRATIVE EXAMPLE. Prove lim "L"*"~

tS.

- _.5 x - x- - 7 x* 7

Solution. Divide numerator and denominator by x*\ the highest power of xpresent in either. Then we have

J* - J* - < X-*

The limit of each term in numerator or denominator containing x is zero, by (4).

Hence, by (1) and (3) of Art. H> we obtain the answer. Jri any similar case thefirst step is therefore as follows.

I>iride both numerator and denominator by the highest power of the variable

occurring in cither.

If u and v are functions of jr, and if

lim u = A, lim r -= 0,x * a a- - (i

and if A is not zero, thenr ulim - =: oo.r . tf

This notation provides for the exceptional case of (3), Art. 16,

when B = and A is not zero. See also Art. 20.

PROBLEMS

Prove each of the following statements,

Proof. lim ^ = lim

x

[Dividing both numerator and denominator hy x2.]

The limit of each term in numerator and denominator containing x is zero, by(4). Hence, by (1) and (3), Art. 16, we obtain the answer.

2. limX = 2.

r 4 t2 + 3 / + 2 1

3 -

J ja + 2 /- 6

= ""

3*

. ,. r 2/? 4- 3 x/z

2-f h* x

4. lim r . _ , 2= -

16 DIFFERENTIAL CALCULUS.

,.3 h + 2 xh* + s'W 1 _

,. 4j/2 -3 _ _M 4-3*A-2^ = -

27 7',! 2i^ + 3*

~ "

6- limft'****'- 4

**'' = 1. 8. Km */-*/*+! = 3.

A-.O 2 z(2z - /r;J x-,* 2^* + 4z- 7

9 jjm ft"-7"'

-f Qi^" ] + + a,< = ao

', . ' &./" -I- h.r" '+ + & 60

-f

. r, . , (L'" 4 cs* -f

, r ar4-f ^^ -f r A

11. 11 in - ---:

-r- = 0.

, r ^^ 4- kr 1'

-f ^l

12. lim ------;

-7- = oo.

.1. fj (if -f f'J" -f JS -f {/

-

13. lim-------~.s . a

* a-

, ]j rn _^Z_ Hj_ nj,n i^ (n positive integer)//

S- (>_ 5.7

..... " ~** - 4 4

ir. i:m S* 4 S-lc. 11 III

........

.

i r16. lim

Proof. The limiting value cannot" bo found by substituting h = Q, for we then

obtain (Art. lli) the indeterminate form - We then transform the expression in

a suitable manner as indicated below, namely, rationalize the numerator.

\__x -f h - \_JT \ jr -f // -f \ J J -f ^ - .r ]_

\ JT

TT ,.\ .r + // \ r ,. 1 1

JTence lim-- ----- = lim :

----= =-/* *> /z /*-o\ j- -f h + \ x 2v

17. Given /(j) = .r2

,show that

/j -* W

18. Given /(j) = a.r2-f 6.r -f r, show that

19. Given /(or)- show that

*~o h

20. If f(x) = x*, find lim /(* + *)


19. Infinitesimals. A variable r which approaches zero as a limit

is called an infinitesimal. This is written (Art. 14)

lim r = or r 0,

and means that the numerical value of r ultimately becomes andremains less than any preassigned positive number, however small.

If lim r = /, then lim (r /) = (); that is, the difference between avariable and its Inn it is an infinitesimal.

Conversely, if the different bit ween a variable and a constant is an

infinitesimal, then the variable approaches the constant as a limit.

20. Theorems concerning infinitesimals and limits. In the followingconsiderations all variables are assumed to be functions of the sameindependent variable and to approach their respective limits as this

variable approaches a fixed value a. The constant t is a preassigned

positive number, as small as we please, but not zero.

We first prove four theorems on infinitesimals.

I. An algebraic sum of u infinitesimals is an infinitesimal, n beinga fixed number.

For the numerical value of the sum will become and remain less

than e when the numerical value of each infinitesimal becomes and

remains less than -n

II. The product of n constant c by an infinitesimal is an infinitesimal.

For the numerical value of the product will be less than e when

the numerical value of the infinitesimal is less t han ,---

klIII. The product of n infinitesimals is an, infinitesimal, n being a

fixed number.

For the numerical value of the product will become and remain

less than e when the numerical value of each infinitesimal becomesand remains less than the r/th root of e.

IV. If lim v = /, and I is not zero, ihen the quotient of an infinitesimal

i by v is also an infinitesimal.

For we can choose a positive number r, numerically less than /,

such that the numerical value of v ultimately becomes and remains

greater than r, and also such that the numerical value of i becomesand remains less than ce. Then the numerical value of the quotientwill become and remain less than .

Proofs of the theorems of Art. 16. Let

(1) u A i, v B = j, w C = k.


Then i, j, k are functions of x, and each approaches zero as x > a\

that is, they are infinitesimals (Art. 19). From equations (1), weobtain

(2) u -f v w (A + B C) = i + j k.

The right-hand member is an infinitesimal by theorem I above.

Hence, by Art. 19,

(3) lim (u -h v w) = A -f- B C.

From (1), we have u = A -{- i, v=B+j. By multiplication and

transposing AH we get

(4) ur - A B = Aj -f Bi -f ij.

By the above theorems I III the right-hand member is an infinitesi-

mal, and hence

(5) lim uv = AB.? a

The proof is readily extended to the product urw.

Finally, we may write

f("\ 1* _ 1 A -\- i _ A _ Bi Aj m

v B~~ B+j B~B(B +j)'

The numerator in (G) is an infinitesimal by theorems I and IT. By(3) and (4), lim /*(/*+./) = B'2 . Hence, by theorem IV, the right-hand member in (6) is an infinitesimal, and

/r^ V U A(7) hm - =

j- - v BHence the statements in Art. 16 are proved.

CHAPTER III

DIFFERENTIATION

21. Introduction. We shall now proceed to investigate the man-ner in which a function changes in value as the independent variable

changes. The fundamental problem of the differential calculus is to

establish a measure of this change in the function with mathematical

precision. It was while investigating problems of this sort, dealingwith continuously varying quantities, that Newton* was led to the

discovery of the fundamental principles of the calculus, the mostscientific and powerful tool of the modern mathematician.

22. Increments. The increment of a variable in changing from one

numerical value to another is the difference found by subtracting the

first value from the second. An increment of x is denoted by the

symbol Ax, read "delta x." The student is warned against readingthis symbol "delta times :r."

Evidently this increment may be either positive or negative t

according as the variable in changing increases or decreases. Similarly,

Ay denotes an increment of //,

A(f> denotes an increment of 0,

A/(J) denotes an increment of /(j), etc.

If in y f(x) the independent variable x takes on an increment Ax,then A?/ will denote the corresponding increment of the function f(x)

(or dependent variable y).

The increment Ay is always to be reckoned from the definite

initial value of y corresponding to the arbitrarily fixed initial value

of x from which the increment Ax is reckoned. For instance, consider

the function

* Sir Isaac Newton Cl 642-1 727), an Englishman, was a man of the most extraordinary

genius. He developed the science of the caloulas uml**r the name of Fluxions. AlthoughNewton had discovered and made use of the new science 5ts early as HJ70, his first published

work in which it occurs is dated 1687. having the title"Philosonhiao Naturalis Principia

Mathematical' This was Newton's principal work. Laplace said of it,"It will always remain

preeminent above all other productions of the human mind." See frontispiece.

t Some writers call a negative increment a decrement.

19


Assuming x = 10 for the initial value of x fixes y = 100 as the

initial value of y.

Suppose x increases to x = 12, that is, Ax = 2;

then y increases to y 144, and Ay = 44.

Suppose x decreases to x = 9, that is, Ax = 1;

then y decreases to y = 81, and Ay = 19.

In the above example, ?/ increases when x increases and y decreases

when x decreases. The corresponding values of Ax and Ay have

like signs. It may happen that // decreases as x increases, or the

reverse;

in either case Ax and A// will then have opposite signs.

23. Comparison of increments. Consider the function

(1) = 3-2.

Assuming a fixed initial value for x, let x take on an increment Ax.

Then y will take on a corresponding increment A?/, and we have

or

Subtracting (1 )

(2)

?/ + Ay/ (x + Ax) 2,

y + Ay/ = x 2 + 2 x Ax + (Ax)2

.

//= x 2

Ay/= 2x Ax+ (Ax) 2

we get the increment A// in terms of x and Ax.

To find the ratio of the increments, divide both members of (2)

by Ax, giving-V = 2 x + Ax.Ax

If the initial value of x is 4, it is evident (Art. 16) that

Let us carefully note the behavior of the ratio of the increments of

x and // as the increment of x diminishes.

DIFFERENTIATION 21

It is apparent that as Ax decreases, A?/ also diminishes, but their

ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01,

illustrating the fact that the value of -'( can be brought as near to

8 as we please by making A.r sufficiently small. Therefore

lim^-'

= 8.Az-0 AJ

24. Derivative of a function of one variable. The fundamental

definition of the ditlerential calculus is as follows.

The derivative* of ct function /,> //,v limit of the ratio of the increment

of the function to the increment of the independent variable, when the

latter increment varies and approudtcx zero ax a limit.

When the limit of this ratio exists, the function is said to be differ-

cntiable, or to possess a deriralire.

The above definition may be given in a more compact form sym-

bolically as follows. Given the function

0) ?/= /U),

and consider x to have a fixed value.

Let take on an increment A.r;then the function y takes on an

increment A?/, the new value of the function being

(2) ?/ + A// - /(.r -f A/).

To find the increment of the function, subtract (1) from (2), giving

(3) Av/-/(j + A.r)-/M.

Dividing both members by the increment of the variable, AJ,

(A\ A# = f(.r + A:r) f(jr)( } Ax

"A*

The limit of the right-hand member when AJ > (} is, from the

definition, the derivative of /(:r), or by (1 ), of ?/, and is denoted by the

symbol Therefore

,,,

^dy r /(x + A*)-/(*)

(A) = hmdx AX-O Ax

defines the derivative of y [or f(n)] with respect to x.

From (4) we get also

dy rm Ay- = lim -

dx AX-.O AJC

* Also called the differential coefficient or the derived function.


Similarly, if u is a function of t, then

= lim = derivative of u with respect to t.at A/ -* o A

The process of finding the derivative of a function is called differ-

entiation.

25. Symbols for derivatives. Since Ay and Ax are always finite andhave definite values, the expression

Axis really a fraction. The symbol

dv

ax

however, is to be regarded not as a fraction but as the limiting value ofa fraction. In many cases it will be seen that this symbol does possessfractional properties, and later on we shall show how meanings maybe attached to dy and dor, but for the present the symbol ^ is to beconsidered as a whole. x

Since the derivative of a function of x is in general also a function

of y, the symbol f'(x) is also used to denote the derivative of f(x).

Hence, if f , ,

!/=/(*),

we may write -^ = /'(j),

which is read "the derivative of y with respect to x equals / primeof x" The symbol ^

Tx

when considered by itself is called the differentiating operator, andindicates that any function written after it is to be differentiated with

respect to x. Thus,

-JL or y indicates the derivative of y with respect to x ;

-T- f(x) indicates the derivative of f(x) with respect to x ;

(2 x2 + 5) indicates the derivative of 2 x2 + 5 with respect to x.

dyy' is an abbreviated form of -?-ax

The symbol Dc is used by some writers instead of -If, then,

ax

DIFFERENTIATION 23

we may write the identities

It must be emphasized that the variable, in the essential step of

letting Ax > 0, is Ax, and not x. The value of the latter is assumedfixed from the start. To emphasize that x = xo throughout, we maywrite

26. Differentiable functions. From the Theory of Limits it is clear

that if the derivative of a function exists for a certain value of the

independent variable, the function itself must be continuous for that

value of the variable.

The converse, however, is not always true, functions having been

discovered that are continuous and yet possess no derivative. But

such functions do not occur often in applied mathematics, and in this

book only different!able functions arc considered, that is, functions that

possess a derivative for all values of the independent variable save at

most for isolated values.

27. General Rule for Differentiation. From the definition of a deriv-

ative it is seen that the process of differentiating a function y = f(x)

consists in taking the following distinct steps.

GENERAL RULE FOR DIFFERENTIATION*

FIRST STEP. In the Junction replace x by x + Ax, and cakulate the

new value of the function, y + Ay.SECOND STEP. Subtract the given value of the function from the new

value and thus find Ay (the increment of the function).

THIRD STEP. Divide the remainder Ay (the increment of the function)

by Ax (the increment of the independent variable).

FOURTH STEP. Find the limit of this quotient when Ax (the incre-

ment of the independent variable) varies and approaches zero as a limit.

This is the derivative required.

The student should become thoroughly familiar with this rule byapplying the process to a large number of examples. Three such

examples will now be worked out in detail. Note that the theorems

of Art. 16 are used in the Fourth Step, x being held constant.

called the Four-Step Rule.

ILLUSTRATIVE EXAMPLE 1. Differentiate 3 x'2 4- 5.

Solution. Applying the successive steps in the General Rule, we get, after placing

y = 3 x'-' 4- 5.

Firxf Step. y 4- A^/ = 3(x 4- Ax)2

4- 5

= 3 x*' 4- ft x Ax 4- 3( Ax)- 4- 5.

Second Step, y 4- A?/ - 3 x- -f 6 x Ax -f 3 (Ax)- 4- 5

]/= 3 x 1

'

4_5

A?/ = fix- Ax 4- 3(Ax)*

'/Vnrd S/f'/).-~- 6x4-3- Ax.

Fourth Step. In the right-hand member let Ax * 0. Then by (A)

Or y'-

(3 x- 4- 5) = 6 x.

ILLUSTRATIVE FAAMPLE 2. Differentiate x'1 - 2 x 4- 7.

Solution. Place y = x ' - 2 x 4 7.

Fir*/ tf/cyj. // + A// = (x 4- A.r) ' - 2(x -f Ax) 4- 7

--- x 1 4 3 x- Ax -f 3 x ( Ax)- 4- (Ax)f - 2 x - 2 Ax 4- 7.

Second Step. // 4- A// = x !

4- 3 jr-' Ax 4- 3 x ( Ax)J4- ( Ax)

( - 2 x - 2 - Ax 4- 7

// Z,ZJ -2x 4- 7

A// -- 3 x- : Ax 4- 3 x ( Ax)24- ( Ax)-'

- 2 Ax

Third Step.-- - 3 x

"

4- 3 x Ax 4- (Ax)' - 2.

Fourth Step. In the right-hand member let Ax > 0. Then by (A)

Or y'-

-j-(x'

! - 2 x 4 7) = 3 x- - 2.

ILLUSTRATIVE KXAMPLE 3. Differentiate--x

Solution. Place //

~x-

A'JS/ Step. y f A//(x 4 A.r)

a

Second Stej>. y 4~ A//(x 4- Ax)-

, c _ -c Ax (2 x 4-

(x 4- Ax)- x- x j(x 4- Ax)

2

Third Step. ^-=^- c . J__AL_.Ax .r-'Cr + Ax)-

5

Fourth Step. In the right-hand member let Ax * 0. Then by

dr x-\x)- x <!

DIFFERENTIATION 25

PROBLEMS

Use the General Rule in differentiating each of the following functions.

1. i/= 2 -3 -r. An*. //'

= - 3.

2. # = mx -f 6.

3. //= ax 2

.

4. s = 2t- t'2

.

5. y = ex3 .

6. y = 3 J - j-3

.

7. M = 4 r 2-f 2 r*

8. //= x 4

.

12. ?/= -

9.P^;

10. y =

11. s =

-1

3

-4;

<ty-

li_

(.rj

-f a :>

)

2

/ I'2

10,,lo. //- 8.T

19. ?/= 3 ar

a - 4 j-- 5.

20. 8 = a/ 2 + ft/ + r.

21. w = 2r* - 3 r 2 .

22. ?/- ars + bx'2 -f r.r

23. p=(a-bOY2.

24. ?/= (2-

jr)(l~ 2 j-

25. i/=

4 - s*

26. x - (r;

d.

27. //

-

29. //=

a + t>x'2

a -f

28. Interpretation of the derivative by geometry. We shall nowconsider a theorem which is fundamental in all applications of the

differential calculus to geometry. It is />><

necessary to recall the definition of the } ^

tangent line to a curve at a point P on

the curve. Let a secant be drawn throughP and a neighboring point Q on the curve

(see figure). Let Q move along the curve

and approach P indefinitely. Then the

secant will revolve about P, and its limiting position is the tangent

line at P. Let

(1) y=f(?)

be the equation of a curve AD. This curve is the graph of /(r)

(see figure).

Now differentiate (1) by the General Rule and interpret each step

geometrically from the figure (p. 25). We choose a point P(x, y)

on the curve, and a second point Q(x + Ax, y + Af/) near P, also

on the curve.

FIRST STEP. y + Ay = f(x + Ax) = NQSECOND STEP. y + Ay = f(x + Ax) = NQ

y =/(x)_THIRDSTEP. A^ /(x + Ax) - /(x) = Jgg.

|gAx Ax MN PR= tan Z #/JQ = tan </>

= slope of secant line PQ.

At this point, therefore, we see that the ratio of the increments Ayand 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. Thevalue of x is now regarded as fixed. Hence P is a fixed point on the

graph. Also, Ax is to vary and approach zero as a limit. Obviously,

therefore, the point Q is to move along the curve and approach P as a

limiting position. The secant line drawn through P and Q will then

turn about P and approach the tangent line at P as its limiting

position. In the figure,

(/>= inclination of the secant line PQ,

r = inclination of the tangent line PT.

Then lim </>= r. Assuming that tan is a continuous function (see

Ar~

Art. 70), we have, therefore,

FOURTH STEP. ^ ~ /'(x) = lim tan <t>= tan r,

ttX Ar-^O

= slope of the tangent line at P.

Thus we have derived the important

Theorem. The value of the derivative at any point of a curve is equalto the slope of the tangent line to the curve at that point.

It was this tangent problem that led Leibnitz* to the discovery of

the differential calculus.

* Gottfried Wilhelm Leibnitz (1646-1716) was a native of Leipzig. His remarkableabilities were shown by original investigations in several branches of learning. He wasfirst to publish his discoveries in calculus in a short essay appearing in the periodical ActaEruditorum at Leipzig in 1684. It is known, however, that manuscripts on Fluxions

DIFFERENTIATION 27

ILLUSTRATIVE EXAMPLE. Find the slopes of the tangents to the parabola y = x*

at the vertex and at the point where x = $ .

Solution. Differentiating by the General Rule (Art. 27), we get

(2) 3* = 2 x = slope of tangent line at any point (x, y) on curve.

To find slope of tangent at the vertex, substitute x = in (2) ,

giving dydx~'

Therefore the tangent at the vertex has the slope zero ; that

is, it is parallel to the x-axis and in this case coincides with it.

To find the slope of the tangent at the point P, where x = i,

substitute in (2), giving ,

that is, the tangent at the point P makes an angle of 45 with the x-axis.

PROBLEMS

Find by differentiation the slope and inclination of the tangent line to

each of the following curves at the point indicated. Verify the result bydrawing the curve arid the tangent line.

l.y = x'2 - 2, where x = 1. Am. 2;63 26'.

2. y = 2 x \ x 2, where x 3. 4. y 3 + 3 x x :

\ where x 1.

43. y = ~, where x = 2. 5. y = .r

;< 3 x 2,where x 1.

X -L

6. Find the point on the curve y = 5 x x 2 where tbe inclination of

the tangent line is 45. Anx. (2, 6).

7. Find the points on the curve y x* -f x where the tangent line is

parallel to tbe line y 4 x. Ans. (1, 2), ( 1, 2).

In each of the three following problems find (a) the points of intersection

of the given pair of curves ; (b) the slope and inclination of the tangentline to each curve, and the angle between tbe tangent lines, at each point

of intersection (see (2), p. 3).

8. y = 1 x 2,

Ans. Angle of intersection = arc tan $ = 53 8'.

y = x2 1.

9. y = x2,

10. y = x*-3x,z-2/ + 2 = 0. 2z + ?/

= 0.

11. Find the angle of intersection between the curves 9 y = x3 and

y = 6 + 8 x - x3 at the point (3, 3). Ans. 21 21'.

written by Newton were already in existence, and from these some claim Leibnitz got the

new ideas. The decision of modern times seems to be that both Newton and Leibnitz

invented the calculus independently of each other. The notation used today was intro-

duced by Leibnitz.

CHAPTER IV

RULES FOR DIFFERENTIATING ALGEBRAIC FORMS

29. Importance of the General Rule. The General Rule for differen-

tiation, given in the last chapter (Art. 27) is fundamental, being found

directly from the definition of a derivative, and it is very important

that the student should be thoroughly familiar with it. However, the

process of applying the rule to examples in general has been found

too tedious or difficult : consequently special rules have been derived

from the General Rule for differentiating certain standard forms of

frequent occurrence in order to facilitate the work.

It has been found convenient to express these special rules bymeans of formulas, a list of which follows. The student should not

only memorize each formula when deduced, but should be able to

stale the corresponding rule in words.

In these formulas v/, i\ and w denote functions of x which are

differentiate.

FORMULAS FOR DIFFERENTIATION

dx

dxII -7- = 1.

dx

... d , ,

. du dv dwIII (U + V W) = + 3-- -T--

dx dx dx dx

_ d . . dvIV -j- (ci>)

^ cdx dx

_. d dv duv

VI

Via

du dvv __ u_dx dx

-(-) =dx\v/28

RULES FOR DIFFERENTIATING ALGEBRAIC FORMS 29

du

VHaE(i?)=T'

VTTT ^ ^ dv u - * .VIU5x=5U

"

5J'

J' kemS a function of v.

IX = -T-, y being a function of x.

dy

30. Differentiation of a constant. A function that is known to havethe same value for every value of the independent variable is con-

stant, and we may denote it by//= c.

As x takes on an increment AJ, the function does not change in

value, that is, A?/ = 0, and

But lim ~ -~ 0.AX . oA.r dx

dx~~ *

The derivative of a constant is zero.

This result is readily foreseen. For the locus of y c is a straight

line parallel to OX, and its slope is therefore zero. But the slope is

the value of the derivative (Art. 28).

31. Differentiation of a variable with respect to itself

Let y = x.

Following the General Rule (Art. 27), we have

FIRST STEP. y + A?/ = x + Ax.

SECOND STEP. A?/ = A.r.

THIRD STEP. -~ = 1.

FOURTH STEP.(-~ = 1.dx

!=The derivative of a variable with respect to itself is unity.

This result is readily foreseen. For the slope of the line y = x is

unity.

32. Differentiation of a sum

Let y = u + v w?.

By the General Rule,

FIRST STEP. y + ky = u + Aw + v + A?; w Aw.

SECOND STEP. A?/ = A?/ + Av Aw?.

THIRD STEP. ^^ + ^___^.Ax Ax Ax AxNow (Art. 24),

: <to, lim ^! == ^, ]im ^^.e/x A/ - o Ax dx AX - o Ax dx

Hence, by (1), Art. 16,

FOURTH STEP. ^^ + ^ _ ^_;

.

c/x dx rfx dx

m d , . . du dv dwHI .'. (u + v - w) = + - -

dx dx dx dx

A similar proof holds for the algebraic sum of any number of

functions.

The derivative of the algebraic sum of n functions is equal to the same

algebraic sum of their derivatives, n being a fixed number.

33. Differentiation of the product of a constant and a function

Let y = cv.


FIRST STEP. # + A?/ = c(v + Ar) = cv + cAt?.

SEC:OND STEP. A// = cA?\

THIRD STEP. ^-c^*Ax AxWhence, by (4), Art, 16,

FOURTH STEP. ?=^:r-ax ax

,.i(, =c|.

The derivative of the product of a constant and a function is equal to

the product of the constant and the derivative of the Junction.

34. Differentiation of the product of two functions

Let y = uv.


FIRST STEP. y + Ay = (u + Aw) (v + Av).

Multiplied out, this becomes

y + A// ur + wAr + ?>Aw + AwAv.

SECOND STEP. Ay = wAr + rAw + A//Ar.

THIRD STEP. ^ - w ^ + r ^~ + Aw ^Ax Ax Ax Ax

Applying (2) and (4), Art. 16, noting that lim A// = 0, and hence

that the limit of the product A. -is zero, we have

FOURTH STEP. ^ = ?/ + rax rfx dx

V .d dv du

' *

dx~

dx dx

The derivative of the product of tiro functions is equal to the first

function titnes the derivative of the second, plus the second functiontimes the derivative of the first.

35. Differentiation of the product of n functions, n being a fixed

number. When both sides of V are divided by UP, this formulaassumes the form

A. (, <\(llL ^!1

djc __ djr dx

UV U V

If, then, we have the product of 71 functions

we may write

fl-r"

^ _ df dx

Vn

ZJ

dp f(r;^...,n )

ax,ax

Vi V'2 /'.'jr/i' ' * vn

dv dux dv:\ (

_ dx,dx

dx_ ,

Multiplying both sides by viv<2- - vn ,

we get

+ (V\V2'


The derivative of the product of n functions, n being a fixed number,is equal to the sum of the n products that can be formed by multiplying

the derivative of each function by all the other functions.

36. Differentiation of a function with a constant exponent. The

Power Rule. If the n factors in the above result are each equal to v,

VI "dx*" dx

When v x, this becomes

Viadx

We have so far proved VI only for the case when n is a positive

integer. In Art. 65, however, it will be shown that this formula

holds true for any value of n, and we shall make use of this general

result now.

The derivative of a function with a constant exponent is equal to the

product of the exponent, the function with the exponent diminished by

unity, and the derivative of the function.

This rule is called the Power Rule.

37. Differentiation of a quotient

Let y = ->(t; ^ 0)

v


FIRST STEP. y + Ay/ =u +

^u

-

yv + A?;

SECOND STEP. Av = ^M^ - - = **/*-?;**.+ Ac v 0(0 + A)

THIRD STEP.A?/ - Ar A*-

Ax '(( + At')

Applying il)-(4), Art. 16,

FOURTH STEP.

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 !(Hdx\c

We may also get VII a from IV as follows :

dud / u\ 1 du dx

dx V c'

c dx c"

The derivative of the quotient of a function by a constant is equal to

the derivative of the function divided by the constant.

PROBLEMS*

Differentiate the following functions.

1. y = x*.

Solution.-/~= ~r (x

<{

)= 3 x'2 . Ans. By VI a

In = 3.]

2. y = ax4 - bx 2.

Solution. ^ =-f (ax4 - bx2

)= ~ (ax 4

)-

-f (bx2) by IIIdx dx dx dx

= a (x1

) 6 (x*') by IV

= 4 ax :i - 2 6x. A ns. By VI a

3. y = x* + 5.

Solution. & = -7- (rr*) -f (5) by III

= j :r>. Ans. By VI a and I

4. y =ig-^ + 8>^.

Solution. ^ = ~ (3 x1

^)-~

(7 x~^) -f (8 x*) by IIIax dx ax dx

= :

Vy x^ + I x-J + V ~^- An*- By IV and VI a

* When learning to differentiate, the student should have oral drill in differentiatingsimple functions.

5. y-

(x2 - 3) 5

.

Solution. & = 5(x2 -

3)< -f (x2 - 3) by VI

ax ax

[p = x2 - 3, and n = 5.]

= 5(x2 -

3)4 2 x = 10 x(x2 - 3)

4. An*.

We might have expanded this function by the Binomial Theorem ((3),

p. 1), and then applied III, etc., hut the above process is to be preferred.

by VISolution. = (a2 - * - (<* -

7. y = (3 r 2 + 2) l + 5 jc2

.

Solution. ^ = (3 & + 2) y- (1 + 5 a*)* 4- (1 + 5 x^i -f(3 x* +2) by V

dx c/x dx

[u - 3 a-'-1 + 2, and r - (1 + 5 j'J )l]

= (3x 2-f 2)i(l -f 5x 2

)~^y-(1 -f5x 2)4 (1 -f 5x2)^Gx by VI etc.

= (3 x 2 4-2H1 -f 5x 2)' 2 5x + 6x(l -f 5x 2

)2-

gV 1 -f 5 x2 v 1 -f 5 x2

_. A . r/// __H_ hv VIT

Solution. -~ =-;;

-"- DV v X1

dx - x-

_ 2 r((7 2 - :r2

) -f xfa 2-f T 2

)

[Multiplying both numerator and denominator by (a2

(a2 - x 2 )^

Prove each of the following differentiations.

9. ~ (3 .r4 - 2 x 2

-f 8) = 12 a-3 - 4 x.

dx

10. 4 (4 + 3 x - 2 jfi)= 3-6 x 2

.

dr

11. 4 (a/6 ~ 5 W3

)= 5 a/4 - 15 fa 2 .

ar

16. -

(2j-J

-f 4

17.^U8 -a s;=s.

i Q d i a -f bx -f <~.r~18'

d? i 7

v;

' = f. JT' - X

20. ; =

21. V =

a + bt + r/ 2

-p=-Vaj-

23. /(/) = (2- 3 /-)

:l.

_4V.r

2V?

a

(4- 9 ,

<f/?/ __ .r

^r~ /., 2 . 2 J

(2 - 5

^ = (a - ~dj: -

2I ^*

30. s = <Va 2 + <z

.

dy _ 2 a -h 3 bx

ds< 2Va 4- ^ds _ a 2

-f 2 f2

31. = o x 2 a

33. y =

35. r =

36. //=

37.

38.

Va 2 +X

X

,__ /a2 + x'

2

~Va'2 -x'2

'

,_ 3/2+in.'- \^TJn

39.

40.

41.

77 -

dy __ 4 a 2x

dx~~ (a2 -x 2

)2

*

d?/ _ a 2

^ x 2Va 2-f x 2

rfjy _ a 2

d:r(a

2 - x 2)*'

dr _6 6 - 10 2

de~

d^ r

^~ (l+rx)Vr<// 2 a 2r

^ (a2 -o- 2)Va4

d,s- 4

Ul(2-h30 a (2~

do: ?/

dx a-y

Differentiate each of the following functions.

42. /(.r) = V2.r

43. //=

44. //=

47. y = a-2V5 _

45. ,s-

Va - bx

V -f ft/.

48. //=

50. 2/=

46. r = -

V^H-3 JT

In each of the following problems find the value of ^ for the givenvalue of .r.

52. y = (x2 -

o-) ; .r = 3.

do-

Am. 540.

53. //= VJ ; x = 64.

54. y = (2 j-V* + (2 .rV' ; .r = 4. Arcs. g.

55. y = Vif+771'

;j- = L\

,_ Vl6 -f Ii .r

57. 7/=

58. ?/= jr\/8 - .r

2;

.r = 2. 0.

59. //= .r-Vl + .*; .r = 2. 20.

60. //= (4

- .r-V ; .r = U. 63. //= j-N/.S -f 2 .

61- //=

rri~7~> ;*r ^ - 64. //

=-

j-= J. 65. //=

38. Differentiation of a function of a function. It sometimes hap-

pens that y, instead of being defined directly as a function of J, is

given as a function of another variable r, which is defined as a func-

tion of x. In that case y is a function of jc through v and is called a

function of a function.2 v

For example, if y = ^---

r,>

~

and

then ?/ is a function of a function. P>y eliminating v we may express y

directly as a function of JT, but in general this is not the best plan

when we wish to find -

dx

If y = f(v) and v = <(JT), then'//

is a function of # through 0. Hence,

when we let x take on an increment AJ, v will take on an increment Av

and y will also take on a corresponding increment A?y. Keeping this

in mind, let us apply the General Rule simultaneously to the two

functionsamj ^

FIRST STEP. y+&y=f(v

SECOND STEP. y+&y=f(vy

y =f(v+ A?;) -/O) At; = <f>(x+ Ax)

THIRD C^TFPAy /(^+ A.) -/QQ Ag_(a; + Ag)

IHIRDbTEP.AJH-

A?; AJ"~ Aar

The left-hand members show one form of the ratio of the increment

of each function to the increment of the corresponding variable, and

the right-hand members exhibit the same ratios in another form. Be-

fore passing to the limit let us form a product of these two ratios,

choosing the left-hand forms for this purpose.

This gives-J/ . _J?, which equals -r^-Av Ax Ax

Write this T* = *-f'-Ax Av Ax

FOURTH STEP. When Ax -0, then also Ai> * 0. Passing to the

limit,

(A)*y = dJL.Q. By (2), Art. 16

v 'dx dv dx

This may also be written

(B) -^=/'00-f.

// y =/(?;) and v </>(x), tlie derivative of y with respect to x equals

the product of the derivative of y with respect to v by the derivative of v

with respect to x.

39. Differentiation of inverse functions. Let y be given as a func-

tion of x by means of the relation

y =/(*).

It is often possible in the case of functions considered in this bookto solve this equation for x, giving

that is, we may also consider y as the independent and x as the

dependent variable. In that case

/(x) and <t>(y)

are said to be inverse functions. When we wish to distinguish between

the two it is customary to call the first one given the direct function

and the second one the inverse function. Thus, in the examples which

follow, if the second members in the first column are taken as the

direct functions, then the corresponding members in the second

column will be respectively the inverse functions.

y = x2 + 1, x = \ly - 1.

y = ax, ar = loga y.

y = sin x, x = arc sin y.

Let us now differentiate the inverse functions

y=J(x) and x = 0(^/)

simultaneously by the General Rule.

FIRST STEP. y+Ay=f(x+ Ax) x+ Ax = 4)(y+ Ay).

SECOND STEP, y+ Ay = f(x + Ax) x + Ax = (y + Ay)

JL _=f(x+ Ax)-f(x) Ax

THIRD STEP Ay^f(x+Ax}-.f(x) r

AT ^ <(?/+ A?/)

Ax Ax Ay AyTaking the product of the left-hand forms of these ratios, we get

AJ/ ATAx Ay

'

A//

FOURTH STEP. When Ax - 0, then also, in general, Ay - 0. Pass-

ing to the limit,

(C) r = ' by (3), Art. 16dx dx

dyor

The derivative of the inverse function is equal to tlie reciprocal of the

derivative of the direct function.

40. Implicit functions. When a relation between x and y is given

by means of an equation not solved for y, then y is called an implicit

function of x. For example, the equation

(1) x* - 4 y =

defines y as an implicit function of x. Evidently x is also defined bymeans of this equation as an implicit function of y.

It is sometimes possible to solve the equation defining an implicit

function for one of the variables and thus obtain an explicit function.

For instance, equation (1) may be solved for y, giving

. i r2y 4 x ,

showing y as an explicit function of x. In a given case, however,

such a solution may be either impossible or too complicated for

convenient use.

41. Differentiation of implicit functions. When y is defined as animplicit function of s, it was explained in the last article that it

might be inconvenient to solve for y in terms of y or x in terms of y(that is, to find y as an explicit function of x, or x as an explicitfunction of y).

We then follow the rule :

Differentiate the terms of the equation as given, regarding y as a Junc-

tion of x, and solve forax

This process will be justified in Art. 231. Only correspondingvalues of x and y which satisfy the given equation may be substi-tuted in the derivative.

Let us apply this rule in finding fromus

cu f ' + 2 .r'ty- ix = 10.

Then -^+^)-^I

() /V.M >' I C . '2. 7 ^. T> .' __ /I

/O -*'} T , () \ //"*

^*'

/? >

rf//__ ?/7 6aj-e> G^ 1

'//

The student should observe thai in general the result will containboth j

1 and v/.

PROBLEMS

Find -'- for each of the following functions.

2. y = 2 -?/

2, ?( = j-a - .r.

An*. <^ = ^.dr V

-1).

5. 15 .r = 15 y + 5 ?/:< + 3 y

fi.

6. ,= V + \/ A

7. y2 = 2 px. 13. .r

34- 3 .r

2?/ -f ?/

:< = r 1

.

8. x 2 + ?/2 - r 2

. 14. JT -f 2V^7/_+ ?/= a.

9. 6 2x 2-f a 2

?/2 = a 26 2

.

15 ' J' 2 + a^-r

.'/ + ?/

2 = ft3 -

10. Vx -f V# = Va.' *

' *'

o 317. r;.r

:l 3 />2.r// -f r?/

3 = 1.

12. ^ - 3 arj/ + j/3 = 0.

18 '

Find the slope of each of the following curves at the given point.

19. jc'2

-f jry + 2 y'2 = 28 ; v-. '*). Awx. -

}.

20. .r< ~ 3 xy'2 + ?/'

= 1 ; (2,-

1 ). -J.

21. V2~7 -f VlTjy = 5; (2, 3). 23. .r'

1 -a.n/ -f 3 a//- = 3 a :t

; (a, a).

22. jc2 - 2V7// -

ir = 52; (8, 2). 24. .r'-'

-.r \fxlj

- 2 /r' = 6; (4, 1).

25. Show that the parabolas //-'

~2 pj- f />- and //

:' ~

p'2 2 ;>.r inter-

sect at right angles.

26. Show that the circle ./-' -f //-' 12 x (> // -f 25 ~ is tangent to the

circle s- -f if2 + 2 jr -f it

= 10 at the point (2, 1).

27. At what angle does the line ?/= 2 .r cut- the curve ^ 2

xy -f 2 ?/2 = 28 ?

28. If f(x) and </>(//) are inverse functions, show that the graph of </>(.r)

may be found as follows: construct the graph of fix) and rotate it

around the origin 9(T counterclockwise.

ADDITIONAL PROBLEMS

1. The vertex of the parabola y2 = 2 px is the center of an ellipse.

The focus of the parabola is an end of one of the principal axes of the

ellipse, and the parabola and ellipse intersect at right angles. Find the

equation of the ellipse. Ans. 4 x 2 + 2 y2 p2

.

2. A circle is drawn with its center at (2 a, 0) and with a radius such

that the circle cuts the ellinse b'2x 2 + a 2

?/2 = a 2b 2 at right angles. Find the

radius. AUK. r 2 - 2(3 a 2-f b 2

).

3. From any point P on an ellipse lines are drawn to the foci., Prove

that these lines make equal acute angles with the normal at P.

4. Prove that the line Bx -f Ay = AB is tangent to the ellipse

b 2x24- a 2

?/2 = a 2b 2

if, and only if, # 2a 2-f ,4

2/>2 - A 2B 2

.

5. Find the equation of the tangent to the curve xmyn = am + n at any

point. Prove that the portion of it intercepted between the axes is divided

in the ratio at the point of contact. Ans. myiix x\) + nx\(y #,) = 0.

n

6. If A* is the slope of a tangent to the hyperbola b 2x 2 - a 2?/

2 = a 2b 2,

prove that its equation is y - kx Va^k 2 - b 2, and show that the locus of

the points of intersection of the perpendicular tangents is x 2-f y

2 = a 2 b 2.

CHAPTER V

VARIOUS APPLICATIONS OF THE DERIVATIVE

42. Direction of a curve. It was shown in Art. 28 that if

=/(*)

is the equation of a curve (see

figure), then

-~ = slope of the line tangentto the curve at P(x, y).

The direction of a curve at

any point is defined as the direction of the tangent line to the curve

at that point. Let r inclination of the tangent line. Then the

slope = tan r, and

-~ = tan r = slope of the curve at any point P(x, y).

At points such as D, F, H, where the direction of the curve is

parallel to the x-axis and the tangent line, is horizontal,

r = ; therefore-^ = 0.ax

At points such as A, B, G, where the direction

of the curve is perpendicular to the x-axis andthe tangent line is vertical, I o

T = 90; therefore-& becomes infinite.

ILLUSTRATIVE EXAMPLE 1. Given the curve y =*

x 2-f 2 (see figure).

3

(a) Find the inclination r when x = 1.

(b) Find r when x = 3.

(c) Find the points where the direction of the curve is parallel to OX.(d) Find the points where T = 45.

(e) Find the points where the direction of the curve is parallel to the line

2x -3 y = 6 (line AB).

Solution. Differentiating, -f-= x2 2 x = tan T.

ax

42

VARIOUS APPLICATIONS OF THE DERIVATIVE 43

(a) When x=l,tanr=l-2 = -l; therefore r = 135. Ans.

(b) When x = 3, tan 7 = 9-6 = 3; therefore r = 71 34'. Ans.

(c) When r = 0, tan r =; therefore x'2 - 2 x = 0. Solving this equation, we

get x = or 2. Substituting in the equation of the curve, we find y = 2 whenx = 0, y = f when x - 2. Hence the tangent lines at 0(0, 2) and D(2, j) arehorizontal. -Ans.

(d) When r_= 45, tan r = 1 ; therefore r 2 - 2 x = 1. Solving this equation, weget x= 1 v 2 = 2.41 and 0.41, giving two points where the slope of the curve

(or tangent) is unity.

(e) Slope of the given line = I ; therefore x'2 - 2 x = J. Solving, we getx = 1 VS = 2.29 and -

0.29, giving the abscissas of the points F and E wherethe direction of the given curve (or tangent^ is parallel to the line AB.

Since a curve at any point has the same direction as its tangentline at that point, the angle between two curves at a common pointwill be the angle between their tangent lines at that point.

ILLUSTRATIVE EXAMPLE 2. Find the an^le of intersection of the circles

(A) x2 + y2 -4;r = l f

(B) x 2 + y*- 2 y = 9.

Solution. Solving simultaneously, we find the points of intersection to be

(3, 2) and (1, -2).

Let mi = slope of the tangent to the circle A at (or, j/),

and W2 = slope of the tangent to the circle R at (jr. y).

Then from (A), mi =jg

=

and from (B), m^ ~rax

by Art. 41

By Art. 41

Substituting x = 3, y = 2, we have

mi = J slope of tangent to (A) at (3, 2).

7H 2 = 3 = slope of tangent to (/?) at (3, 2).

The formula for finding the angle between two lines whose slopes are m\and m2 is

tan = 7^-^-

(2), Art. 3f

1 -f 7W1///2

3Substituting, tan 6 = ^ 1 ; /. = 45". Arw.

This is also the angle of intersection at the point (1, 2).

43. Equations of tangent and normal; lengths of subtangent and sub-

normal. The equation of a straight line passing Ythrough the point (xi, yi) and having the slope

m isy -_y1

= m (x -. Xl ). (3), Art. 3

If this line is tangent to the curve AB at the

point PI(XI, y\), then m is equal to the slope of

the curve at (xi, y\). Denote this value of m by mi. Hence at the

point of contact PI(ZI, y\) the equation of the tangent line TPi is

M N


The normal being perpendicular to the tangent, its slope is the

negative reciprocal of rwi ((2), Art. 3j. And since it also passes

through the point of contact Pi(r\, y\), we have for the equation

of the normal PiN,

(2) y |/i= (x jfi).

m\

That portion of the tangent which is intercepted between the

point of contact and OX is called the Ivtnjth of ihe, tauywt (= TP} ),

and its projection on the r-axis is called the length of the subtangent

(= TM). Similarly, we have the length of rite,y\

normal (= P\K) and the length of the sub-

normal (= yi/Ar

)./i f i)

In the triangle TP\M 9tan r~m\

therefore7V17

'

_f

(3) = length of subtangent.

In the triangle 1\1P} N, tan r = in\ therefore

(4) 71 / A"* = = length of subnormal.

The length of the tangent (TPi) and the length of the normal

(PjAO may then be found directly from the figure, each being the

hypotenuse of a right triangle having two legs known.When the length of subtangent or subnormal at a point on a curve

is determined, the tangent and normal may easily be constructed.

PROBLEMS

1. Find the equations of tangent and normal and the lengths of sub-

tangent, subnormal, tangent, and normal, at the point (a, a) on the cissoid

^a- x

Solution.dx y(2 a - x)~

Substituting x a, ?/= a, we have

rr 2 = slope of tangent.-

a, equation of tangent.

Substituting in (1) gives

U = 2x-

Substituting in (2) gives

2 y -f x = 3 a, equation of normal.

* If the subtangent extends to the ritrht of 7\ we consider it positive; if to the left,

negative, If the subnormal extends to the right of M, we consider it positive, if to the

Jeft, negative.

Substituting in (3) gives TM = (~ = length of subtangent.

Substituting in (4) gives MX = 2 a = length of subnormal.

Also, PT = V(TM)* + (MP^ = x- + <|2=-\5 = length of tangent,\ 4

and PN = V(MN) 2 + (MP)* - \ 4'

-fa- = a \ 5 = length of normal.

Find the equations of the tangent and normal at the given point.

2. y = ar3 - 3 x; (2, 2). .4//,s. 9 .r

-//- 16 = 0, .r -f 9 ?/

- 20 = 0.

9 T 4- 1

3. T/= ^L1. (2, 5). 7 j-

-y -9 = 0, .r + 7

<//

- 37 = 0.o X

4. 2x 2 -xy + y- = 16; (3, 2).

5. 7/2 + 2 ?/-4 jr-f 4 = 0; (1, -2).

6. Find the equations of the tangent and normal at (x\, y\) to the

ellipse b'2x 2

-f a 2?/2 a~b'2 .

Ann. b-fix -h a-y\y = a L>/>2

, a~ii\x b~jr\y jr]y\((i~ h 2).

7. Find the equations of the tangent and normal, and the lengths of

the subtangent and subnormal, at the point u*i, ?/i ) on the circle s- -f ir T~.

Ans. JC\JT -f 7/1 y = r'2

, .r\y ijix ~ 0, * JT\.' '

xi

8. Show that the subtangent to the parabola /r~ 2 ;>.r is bisect (d at

the vertex, and that the subnormal is constant and equal to ;>.

Find the equations of the tangent and normal, and the lengths of the

subtangent and subnormal, to each of the following curves at the pointsindicated.

9. ay x 2; (a, a). A us. 2 jr y a, x 4- 2 //

~ 3 a, ^ 2 a.

10. x 2 - 4 ?y2 = 9

; (5, 2). f> .r- 8 ?/

= 9, 8 x + 5 ?/= 50, V, J.

11. 9x 2 + 4 2/2 = 72; (2, 3).

12. xy + ?y2 + 2 = 0; (3,

-2).

13. Find the area of the triangle formed by the j-axis and the tangent

and the normal to the curve y = 6 jr-

.r'2 at the point (5, 5). A us. Ap.

14. Find the area of the triangle formed by the //-axis and the tangent

and the normal to the curve //-= 9 j- at the point (5, 2).

Find the angles of intersection of each of the following pairs of curves.

15. y2 = x 4- 1, x 2 + y

2 = 13. Am. 109 39r

.

16. y = 6 - a*2

,7 x 2 + y

2 = 32.

Ans. At ( 2, 2), 5 54'; at ( 1, 5), 8 58'.

17. y = x 2,y*-3 y = 2 jr.

18. z2 + 4 ?/2 = 61, 2x 2 -y* = 41.


Find the points of contact of the horizontal and vertical tangents to

each of the following curves.

19. y = 5 x - 2 x 2. Ans. Horizontal, (|, )

20. 3 y'2 - 6 y - x - 0. Vertical, (- 3, 1).

21. x 2 + 6 xy + 25 y2 - 16. Horizontal, (3,

-1), (- 3, 1).

Vertical, (5,-

), (- 5, f).

22. x 2 - 8 xy + 25 y2 = 81.

23. x 2 - 24 x?/ + 169 t/2 = 25.

24. 169 x 2 + 10 x?/ -I- t/2 = 144.

25. Show that the hyperbola x 2 - y2 = 5 and the ellipse 4 x 2 + 9 ?/

2 = 72intersect at right angles.

26. Show that the circle jr'2 + ?/

2 = 8 ax and the cissoid (2 a - x)?/2 = x3

(a) are perpendicular at the origin;

(b) intersect at an angle of 45 at two other points. (See figure in

Chapter XXVI.)

27. Show that the tangents to the folium of Descartes x :^-f if = 3 axy

at the points where it meets the parabola y2 = ax are parallel to the //-axis.

(See figure in Chapter XXVI.)

28. Find the equation of the normal to the parabola y 5 x -f x 2 whichmakes an angle of 45 with the x-axis.

29. Find the equations of the tangents to the circle x 2 + y2 = 58 which

are parallel to the line 3x 1 y 19.

30. Find the equations of the normals to the hyperbola 4 x 2 -y'2 = 36

which are parallel to the line 2 x 4- 5 y = 4.

31. Find the equations of the two tangents to the ellipse 4 x 2 + y'2 = 72

which pass through the point (4, 4). Ans. 2 x + y = 12, 14 x + y = 60.

32. Show that the sum of the intercepts on the coordinate axes of the

tangent line at any point to the parabola x^ + ifi = a* is constant andequal to a. (See figure in Chapter XXVI.)

33. Show that for the hypocycloid x* + y$ = cfi the portion of the

tangent line at any point included between the coordinate axes is constantand equal to a. (See figure in Chapter XXVI.)

34. The equation of the path of a ball is y = x - -^-; the unit of

distance is 1 yd., the x-axis being horizontal, and the origin being thepoint 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 thestarting point? (c) If the ball falls on a horizontal roof 16yd. high, at


what angle will it strike the roof? (d) If thrown from the top of a build-

ing 24 yd. high, at what angle will the ball strike the ground ? (e) If

thrown from the top of a hill which slopes downward at an angle of 45,at what angle will the ball strike the ground ?b

35. The cable of a suspension bridge hangs in

the form of a parabola and is attached to support-ing pillars 200 ft apart. The lowest point of the cable is 40 ft. belowthe points of suspension. Find the angle between the cable and the sup-porting pillars.

44. Maximum and minimum values of a function; introduction. In

a great many practical problems we have to deal with functionswhich have a greatest (maximum) value or a least (minimum)value,* and it is important to know what particular value of thevariable gives such a value of the function.

For instance, suppose that it is required to find the dimensions of

the rectangle of greatest area that can be inscribed in a circle of

radius 5 inches. Consider the circle in the following figure. Inscribe

any rectangle, as ED.Let CD = x ; then DE = VlOO - x-, and the area of the rectangle

is evidently

(1) A = j VlOO -:r L>.

That a rectangle of maximum area must exist may be seen as follows.

Let the base CD (= x) increase to 10 inches (the diameter) ; then

the altitude DE = VlOO x- will decrease to zero and the area will

become zero. Now let the base decrease to

zero ; then the altitude will increase to 10

inches and the area will again become zero.

It is therefore evident by intuition that there

exists a greatest rectangle. By a careful studyof the figure we might suspect that when the

rectangle becomes a square its area would be

greatest, but this would be guesswork. Abetter way would evidently be to plot the

graph of the function (1) and note its behavior. To aid us in draw-

ing the graph of (1), we observe that

(a) from the nature of the problem it is evident that x and A mustboth be positive ;

and

(b) the values of x range from zero to 10 inclusive.

Now construct a table of values and draw the graph, as in the

figure on page 48.

* There may be more than one of each, aa illustrated on page 55,

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 lengthof the corresponding ordinate. Thus,

when x =- OA1 = 3 in.,

then A 1\1 f> = 28. 6 sq. in. ;

and when x = ON ~~ 4\ in.,

then A = A< Q ~ about 39.8 sq. in. (found by measurement).

(b) There is one horizontal tangent (A\SY

). The ordinate TH fromits point, of contact T is greater than any other ordinate. Hence this

observation : One of the inscribed rectangles has evidently a greater areathan any of the others. In other words, we may infer from this thatthe function defined by (1) has a maximum value. We cannot find

this value (= 7/7") exactly by measurement, but it is very easy tofind by the calculus. We observed that at T the tangent washorizontal ; hence the slope will be zero at that point (Art. 42). Tofind the abscissa of T we then find the derivative of A with respectto x from (1), place it equal to zero, and solve for x. Thus we have

(1) .4 = arVlOO - jc'2

,

100 - 2 x2 100 - 2 x2dA_dx VlOO - x2

= 0.VlOO - x2

Solving, x = 5 V2.

Substituting, we get DE = VlOO x2 = 5V&


Hence the rectangle of maximum area inscribed in the circle is a

square of area

A = CD X DE = 5V2 X 5V2 = 50 sq. in.

The length of HT is therefore 50.

Take another example. A wooden box is to be built to contain

108 cu. ft. It is to have an open top and a square base. What must

be its dimensions in order that the amount of material required shall

be a minimum ; that is, what dimensions

will make the cost the least ?

Let x = length of side of square base in feet,

and y = height of box.

Since the volume of the box is given, how-

ever, y may be found in terms of jr. Thus,

Volume - x-'u- 108

; .-.?/=

.r~

We may now express the number ( M) of square feet of lumber

required as a function of jr as follows.

Area of base = JT~ sq. ft.,

432and

(2)

Area of four sides = 4 .r//-

sq. ft. Hence

M = x'2 +

is a formula giving the number of square feet required in any such box

having a capacity of 108 cu. ft. Draw a graph of (2), as in the figure.

What do we learn from the graph ?

(a) If carefully drawn, we may measure the ordinate correspond-

ing to any length (= x) of the side of the square base and so deter-

mine the number of square feet of lumber required.

(b) There is one horizontal tangent (RS). The ordinate of its

point of contact T is less than any other ordinate. Hence this obser-

vation : One of the boxes evidently takes less lumber than any of the

others. In other words, we may infer that the function defined by(2) has a minimum value. Let us find this point on the graph ex-

actly, using the calculus. Differentiating (2) to get the slope at anypoint, we have

dM -9r 432j

<<-" +'--

rr*

ax x2

At the lowest point T the slope will be zero. Hence

that is, when x = 6 the least amount of lumber will be needed.

Substituting in (2), we see that this is

M = 108 sq. ft.

The fact that a least value of M exists is also shown by the follow-

ing reasoning. Let the base increase from a very small square to a

very large one. In the former case the height must be very great andtherefore the amount of lumber required will be large. In the latter

case, while the height is small, the base will take a great deal of

lumber. Hence M varies from a large value, grows less, then in-

creases again to another large value. It follows, then, that the graphmust have a "lowest" point corresponding to the dimensions which

require the least amount of lumber and therefore would involve the

least cost.

We will now proceed to the treatment in detail of the subject of

maxima and minima.

45. Increasing and decreasing functions.* Tests. A function y =f(x)is said to be an increasing function if y increases (algebraically) whenx increases. A function y f(x) is said to be a decreasing function

if y decreases (algebraically) as x increases.

The graph of a function indicates plainly whether it is increasingor decreasing. For instance, consider the graph in Fig. a, p. 51.

* The proofs given here depend chiefly on geometric intuition. The subject of maximaand minima will be treated analytically in. Art. 125.

\

As we move along the curve from left to right the curve is rising ;

that is, as x increases the function (= y) increases. Obviously, Ayand Ax agree in sign.

On the other hand, in the graph of

Fig. 6, as we move along the curve from

left to right the curve is falling ;that is,

as x increases, the function (== //) alwaysdecreases. Clearly, Ay and Ax have op-

posite signs.

That a function majr be sometimes increasing and sometimes

decreasing is shown by the graph (Fig. c) of

As we move along the curve from left to

right the curve rises until we reach the point

A, falls from A to B, and rises to the right of

B. Hence

(a) from x = oo to x = I the function is

increasing ;

(b) from x = 1 to x = 2 the function is de-

creasing ;

(c) from x = 2 to x + oo the function is

increasing.

At any point, such as C, where the function is increasing, the

tangent makes an acute angle with the j-axis. The slope is positive.

On the other hand, at a point, such as D, where

the function is decreasing, the tangent makes

an obtuse angle with the j-axis, and the slope is

negative. Hence the following criterion :

A function is increasing when its derivative is

positive, and decreasing when its derivative is

negative.

For example, differentiating (1) above, we

have

(2) g = f(x )= 6 z 2 - 18 * + 12 = 6(z

-l)(z

-2).

When x < 1, /'(*) is positive, and f(x) is increasing.

When 1< x < 2, f'(x) is negative, and /(or) is decreasing.

When x > 2, f'(x) is positive, and f(x) is increasing.

These results agree with the conclusions arrived at above from

the graph.

FIG. b


46. Maximum and minimum values of a function; definitions. Amaximum 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 amaximum value MA (= y = 2) when x = 1, and a minimum valueNB (=-y = 1) when x = 2.

The student should observe that a maximum value is not neces-

sarily the greatest possible value of a function nor a minimum valuethe least. For in Fig. r it is seen that the function (= y) will havevalues to the right of B that are greater than the maximum MA, andvalues to the left of A that are less than the minimum NB.

If /OO is an increasing function of jr when .r is slightly less than a,

and a decreasing function of x when jr is slightly greater than a, that

is, if /'Or) changes sign from + to as x increases through a, then

f(x) has a maximum value when x a. Therefore, if continuous,

}'(x) must vanish when j = a.

Thus, in the above example (Fig. r), at (\ J'(x) is positive; at

A, /'(.r)=

;at />, /'(:r) is negative. /

On the other hand, if /Or) is a decreasingfunction when j is slightly less than a, and an

increasing function when x is slightly greater

than a, that is, if /'Or) changes sign from to +as x increases through a, then /Or) has a mini-

mum value when r = a. Therefore, if contin-jj

\ x

uous, /'Or) must vanish when r = a. ' FlG - c

Thus, in Fig. c, at D, J'(x) is negative; at B, f(x) = 0; at E,

ff

(x] is positive.

We may then state the conditions in general for maximum andminimum values of /Or).

/(jc) is a maximum if /'(*) = and /'(*) changes sign from + to .

f(x) is a minimum if /'(*) = and/'(x) changes sign from to -f.

The values of the variable satisfying the equation f'(x) = are

called critical values; thus, from (2), Art. 45, x = 1 and x 2 are the

critical values of the variable for the function whose graph is shownin Fig. c. The critical values determine turning points where the

tangent is parallel to OA".

To determine the sign of the first derivative at points near a

particular turning point, substitute in it, first, a value of the variable

sliglitly less than the corresponding critical value, and then one

slightly greater. If the first sign is + and the second , then the

function has a maximum value for the critical value considered.

If the first sign is and the second +, then the function has a

minimum value.

If the sign is the same in both cases, then the function has neither

a maximum nor a minimum value for the critical value considered.

For example, take the above function in (1), Art. 45,

(1) y = f(jr)= 2 JT* - 9 JT- + 12 jc

- 3.

Then, as we have seen,

(2) f'(jr)= 6(j- l)(j-2).

Setting /'(y) = 0, we find the critical values y = 1, :r = 2. Let us

first test x = 1. We consider values of jc near this critical value to be

substituted in the right-hand member of (2), and observe 1 the signs

of the factors. (Compare Art. 45.)

When x < 1, /'(*) = (-)(-) - +.

When x > 1, f'(jr)- (+)(-) - -.

Hence /(r) has a maximum value when x ~ 1. By the

table, this value is ?/=/(!) = 2.

Next, test y = 2. Proceed as before, taking values of x now near

the critical value 2.

Wfhen .r < 2, /'(/) - (+)(-) - -.

When x > 2, fix) = (+)(+.) = +.

Hence /(y) has a minimum value when x 2. By the table above,

this value is y =/(2) = 1.

We shall now summarize our results in a workruy ride.

47. First method for examining a function for maximum and mini-

mum values. Working rule.

FIRST STEP. Find the first derivative of the function.' SECOND STEP. Set the first derivative equal to zero and solve the resulting

equation for real roots. These roots are the critical values of the variable.

THIRD STEP. Considering one critical value at a time, test the first

derivative, first for a value a trifle less* and then for a value a trifle greater

than the critical value. If the sign of the derivative is first + and then,

the function has a maximum value for that particular critical value of the

variable; but if the reverse is true, then it has a minimum value. If the

sign does not change, the function has neither.

* In this connection the term "little less/' or"trifle less," means any value between the

next smaller root (critical value) and the one under consideration ; and the term"

little

greater," or "trifle greater," means any value between the root under consideration and

the next larger one.

In the Third Step, it is often convenient to resolve /'(a*) into

factors, as in Art. 46.

ILLUSTRATIVE ExAMPLE 1. In the first problem worked out in Art. 44 we showed

by means of the graph of the function

A = xVlOO -x 2

that the rectangle of maximum area inscribed in a circle of radius 5 in. con-

tained 50 8(4. in. This may now be proved analytically as follows by applying the

above rule.

Solution. /(x) = x\ 100 - x 2.

r,- , a, /// N 100 - 2 X 2

First Step. f'(x) =VI 00 -x 2

Second Step. Setting f'(x) 0, we have

x^r 5 \ '2 = 7.07,

which is the critical value. Only the positive sign of the radical is taken, since,

from the nature of the problem, the negative sign has no meaning.

Third Slcp. When x < 5\"2, then 2 x-' < 100, and /'(x) is -K

When x > 5\ , then 2 x 2 > 100, and f (xj is -.

Since the sign of the first derivative changes from + to,the function has a

maximum value /(5V2) 5\ 2 5\ 2 = 50. An*.

ILLUSTRATIVE EXAMPLE 2. Examine the function (x l)2(x -f !)' for maxi-

mum and minimum values.

Solution, /(x) = (x-l) 2 (x-f IP.

First Step. f'(x) = 2(x - l)(x + 1 )'< + 3(x - l)2(x + 1)

2 = (x -l)(z + l)2(5x- 1).

Second Step, (x- l)(x + 1)

2(5 x - 1) = 0.

Hence x 1, 1, \, are the critical values.

Third Step. /'(x) = 5(x - D(x -f 1 )2(x - -J).

Examine first the critical value x = 1 (C in figure).

When x < 1, /'(x) = 5(-)(+)2(-f )

= -."

When x > 1, /'(x) = 5(+ )(+) L'(-f-)

= +.

Therefore, when x 1 the function has a minimum value /(I) = ( ordinate

of D.Examine now the critical value x I (B in figure).

Wrhenx< U'(x) = 5(-)(+) 2(-) = -K

When x > I /'(x) - 5(-)(+) 2(+) - -.

Therefore, when x \ the function has a maximum value /(J) =1.11 (= ordi-

nate of B).

Examine lastly the critical value x = 1 (A in figure).

When x < -1, /'(jO = 5(-X~) 2

(-) = +.

When x > -1, /'(x) = 5(-)(+ )

2(-) = +.

Therefore, when x 1 the function has neither a maximum nor a minimumvalue.

48. Maximum or minimum values when /'(*) becomes infinite and

f(x) is continuous. Consider the graph in the figure. At B, or G,

FIG. d

f(x) is continuous and has a maximum value, but /'(a*) becomes in-

finite, since the tangent line at B is parallel to the -//-axis. At E, f(x)

has a minimum value, and /'(/) again becomes infinite. In our dis-

cussion of all possible maximum and minimum values of /(j), we

must therefore include as critical values also those values of x for

which /'(x) becomes infinite, or, what is the same thing, values of x

satisfying

The Second Step of the rule of the preceding section must then

be modified as indicated by (1 ). The other steps are unchanged.

In Fig. d above, note that /'(/) also becomes infinite at A, but the

function is neither a maximum nor a minimum at A.

ILLUSTRATIVE EXAMPLE. Examine the function a - b(x - r) for maxima and

minima.

Solution.

2b

Since x = c is a ^^^1 va iue for which ^ - (and f'(x] = oo), but for which

f(x) is not infinite, let us test the function for maximum and minimum values when

X = C *

When x < c, f'(x) = -f .

When x > c, f'(x) = -.

Hence, when x = c = OM, the function has a maximum value /(c) = a = MP.

PROBLEMS

Examine each of the following functions for maximum and minimumvalues.

1. r { 6 x 2 + 9 x. Ans. x = 1, gives max. = 4.

x = 3, gives min. = 0.

2. 10 + 12 .r- 3 jr

2 - 2 jr 1. a- = 1, gives max. = 17.

x 2, gives min. = 10.

3. 2 :r{ + 3 j*

2-f 12 .r 4. No max. or min.

4. /< + 2.r 2 - 15 j*- 20.

5. 2 .r2

.r4

. a* = 0, gives min. = 0.

x 1, gives max. = 1.

6. x4 4 .r. x = 1, gives min. 3.

7. r4 -j'

2 + 1.

8. 3 .r4 - 4 ^ - 12 .r

2. a- = -

1, gives min. = - 5.

x = 0, gives max. = 0.

x = 2, gives min. 32.

9. x* 5 r4. x = 0, gives max. = 0.

x = 4, gives min. = 256.

10. 3 JT*- 20 jr

3.

2 rr*11. j-2 + ^__. x a, gives min. = 3 a 2

.

12. 2, -g-a4

13. r 2 + a: = a, gives min. = 2 a 2.

14 ____, x = a, gives min. = J.'

a*24- a2 x = a, gives max. = .

15.

16- ^+ a*

T 2 -4- rr"

17.x , +^

1Q /O I v^2/1 o*^2ID. ^~T~Uy ^JL U^ .

19. (2 + x)2(l-

x)*.2

20. ?> + r(j* a) . x = a, gives min. = 6.

21. a b(x cr. No max. or min.

22. (2 + *)*(! - x)5

. Ans. x = l, gives min. = 0.

x 1, gives max. = V4 = 1.6.

23. a- (a + -r)2(a-

x)*. .r = -a, gives max. = 0.

.r = \ a, gives min. = ~jj a ft

.

-r ~/, a, gives max. = ^JJ a r>

.

j* o, gives neither.

24. (2 a:-

a)*(j-

a)*. x - J a, gives max. = J a.

r = a, gives min. = 0.

* i <*> gives neither.

253* + 2 x ~ 0, gives max. -

J.'

:r2-f 2 j- + 4

'

J- = -4, gives min. = -

J.

26^ 2 + ^ + 4 T .'i, gives max. ~ - 5.

^ ~f~ 1 u* 1, gives min. ,'{.

27^ 2

-f r + 4 a- = -*J, gives max. =

.3.*

x 2 + 2 .r + 4 j- 2, gives min. =-[;.

(.r- ._

- _Drives max. -- ------ -

a -f /> 4 />

----->

. .

a -f b a

(a -.

gives max. =a

on (a x)3 a . . ., 7 o

30. ^ _^_. x = -givers mm. = *? a 2

.

a 2 x

.

x 2 x -f 1

49. Maximum and minimum values. Applied problems. In manyproblems 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.


(c) To the resulting function of a single variable apply the above rule

(p. 53) for finding maximum and minimum values.

(d) In practical problems it is usually easy to tell which critical value

will give a maximum and which a minimum value, so it is not always

necessary to apply the third step.

(e) Draw the graph of the function in order to check the work.

The work of finding maximum and minimum values may fre-

quently be simplified by the aid of the following principles, which

follow at once from our discussion of the subject.

(a) The maximum and minimum values of a continuous Junction

must occur alternately.

(b) When c is a positive constant, cf(x) is a maximum, or a mini-

mum for such values of z, and such only, as makef(x) a maximum or a

minimum.

Hence, in determining the critical values of x and testing for max-

ima and minima, any constant factor may be omitted.

When c is negative, cf(x) is a maximum when f(x) is a minimum,and conversely.

(c) // c is a constant, f(x) and c + f(x) have maximum and mini-

mum values for the same values of x.

Hence a constant term may be omitted when finding critical values

of x and testing.

PROBLEMS

1. It is desired to make an open top box of greatest possible volume

from a square piece of tin whose side is a, by cutting equal squares out

of the corners and then folding up the tin to form the sides. What should

be the length of a side of the squares cut out ?

Solution. Let x = side of small square depth of box ;

then a 2 x side of square forming bottom of box,

and volume is V = (a 2 x) 2j,

which is the function to be made a maximum by varying x.

Applying the rule, p. 53,

p:

rtVFirst Step.

-~- = (a- 2 x)

2 - 4 x(a - 2 x} = a 2 - 8 ax + 12 x2.

Second Step. Solving a 2 8 ax + 12 or2 = gives critical values x - and -

It is evident from the figure that x = - must give a minimum, for then all the

tin would be cut away, leaving no material out of which to make a box. By the usual

test, xjj

is found to give a maximum volume -- Hence the side of the square to

be cut out is one sixth of the side of the given square.


The drawing of the graph of the function in this and the following problems is

left to the student.

2. Assuming that the strength of a beam with rectangular cross sec-

tion varies directly as the breadth and as the square of the depth, whatare the dimensions of the strongest beam that can be sawedout of a round log whose diameter is d!

Solution. If x = breadth and y = depth, then the beam will

have maximum strength when the function xy- is a maximum.From the figure, y'

2 = d~ x'2 ; hence we should test the function

First Step. /'Or) = ~ 2 x- + d'2 - x2 - d'2 - 3 x'2 .

Second Step, d2 3 x 2 = 0. .-. x = -~ = critical value which gives a maximum.

Therefore, if the beam is cut so that

Depth = \j{

of diameter of log,

and Breadth \\ of diameter of log,

the beam will have maximum strength.

3. What is the width of the rectangle of maximum area that can be

inscribed in a given segment OAA f

of a parabola?

HINT. If Or = h, BC = h-x and PP' = 2 y ; there-

fore the area of rectangle PDD'P' is

2(h-x)y.

But since P lies on the parabola y2 = 2px, the function to

be tested is

f(x) = 2(h-x)\f2px.

AUK. Width = | h.

4. Find the altitude of the cone of maximum volume that can be

inscribed in a sphere of radius r.

HINT. Volume of cone = J irx-y. But

X 2 = BC x CD = y(2 r - y} ;

therefore the function to bo tested is

TT A\f(y) ^ y*(2 r y).

Ans. Altitude of cone = $ T.

5. Find the altitude of the cylinder of maximum volume that can be

inscribed in a given right cone.

HINT. Let AC r and BC = h. Volume of cylinder = irx2y.

But from similar triangles ABC and DBG,

Hence the function to be tested is

f(y)=y2 y(h-y}2

.

Ans. Altitude = | h.


6. If three sides of a trapezoid are each 10 in. long, how long must the

fourth side be if the area is a maximum? Ans. 20 in.

7. It is required to inclose a rectangular field by a fence, and then to

divide it into two lots by a fence parallel to one of the sides. If the area of

the field is given, find the ratio of the sides so that the total length of fence

shall be a minimum. Ans. 2 3.

8. A rectangular garden is to be laid out along a neighbor's lot and is

to contain 432 sq. rd. If the neighbor pays for half the dividing fence,

what should be the dimensions of the garden so that the cost to the owner

of inclosing it may be a minimum? Ans. 18 rd. x 24 rd.

9. A radio manufacturer finds that he can sell .r instruments per week

at p dollars each, where 5 x 375 3 p. The cost of production is

(500 4- 15 x -f?,x 2

} dollars. Show that the maximum profit is obtained

when the production is about 30 instruments per week.

10. In Problem 9 suppose the relation bet-ween x and /; is

JT= 100-1:

Show that the manufacturer should produce only about 25 instruments per

week for maximum profit.

11. In Problem 9 suppose the relation between x and p is

T2 = 2500 - 20 p.

How many instruments should be produced each week for maximum profit?

12. The total cost of producing x articles per week is (ax 2-f bx -f c)

dollars and the price (p dollars) at which each can be sold is /)=

/3 a x 2.

Show that the output for maximum profit is

x _ Va 2-f 3a-qi- b)

- a

3 a

13. In Problem 9 suppose a tax of t dollars per instrument is imposed

by the government. The manufacturer adds the tax to his cost and de-

termines the output and price under the new conditions.

(a) Show that the price increases by a little less than half the tax.

(b) Express the receipts from the tax in terms of / and determine the

tax for maximum return.

(c) When the tax determined in (b) is imposed, show that the price is

increased by about 33 per cent.

14. The total cost of producing x articles per week is (ax2-f bx -f c)

dollars, to which is added a tax of t dollars per article imposed by the

government, and the price (p dollars) at which each can be sold is p = p ax.

Show that the tax brings in the maximum return when t = J(/3 b) and

that the increase in price is always less than the tax.

NOTE. In applications in economics a, b, c, a, are positive numbers.


15. A steel plant is capable of producing x tons per day of a low-grade

steel and y tons per day of a high-grade steel, where //= - If the

10 x

fixed market price of low-grade steel is half that of high-grade steel, showthat about 5| tons of low-grade steel are produced per day for maximumprofit.

16. A telephone company finds there is a net profit of if 15 per instru-

ment if an exchange has 1000 subscribers or less. If there are over 1000

subscribers, the profits per instrument decrease lc for each subscriber

above that number. How many subscribers would give the maximum net

profit? AUK. 11250.

17. The cost of manufacturing a given article is p dollars and the num-ber which can be sold varies inversely as the >/th po\\er of the selling price.

Find the selling price which will yield the greatest total net profit.

* tlVA //,s\*

n 1

18. What should be the diameter of a tin can holding 1 qt. (58 cu. in.)

and requiring the least amount of tin (a) if the can is open at the top?

(b) if the can has a cover?

-vV = 5.29\ 7T

Ans. (a) -v = 5.29 in. ; (b) ; = 4.20 in.\ 7T \ 7T

19. The lateral surface of a right circular cylinder is 4 TT sq. ft. Fromthe cylinder is cut a hemisphere whose diameter equals the diameter of

the cylinder. Find the dimensions of the cylinder if the remaining volume

is a maximum or minimum. Determine whether it is a maximum or a

minimum. .4//s. Radius 1 ft., altitude = 2 ft.; maximum.

20. Find the area of the largest rectangle with sides parallel to the

coordinate axes which can be inscribed in the figure bounded by the two

parabolas 3 // 12 - s- and 6 //- s 2 - 12. Ans. 16.

21. Two vertices of a rectangle are on the .r-axis. The other two vertices

are on the lines whose equations are y 2 jr arid 3 jc -f y 30. For what

value of y will the area of the rectangle be a maximum? Am. y 6.

22. One base of an isosceles trapezoid is a diameter of a circle of radius

a, and the ends of the other base lie on the circumference of the circle.

Find the length of the other base if the area is a maximum. Am. a.

23. A rectangle is inscribed in a parabolic segment with one side of the

rectangle along the base of the segment. Show that the ratio of the area

1

of the largest rectangle to the area of the segment is /-'

24. The strength of a rectangular beam varies as the product of the

breadth and the square of the depth. Find the dimensions of the strong-

est beam that can be cut from a log whose cross section is an ellipse of

semiaxes a and b. Ans. Breadth = 2 />\/A ; depth = 2 en/I-


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 stiffest

beam that can be cut from a cylindrical log whose radius is a.

-4ns. a X aVi.

26. The equation of the path of a ball is y - mx - m " x", where

800the origin is taken at the point from which the ball is thrown and m is the

slope of the curve at the origin. For what value of m will the ball strike

(a) at the greatest distance along the same horizontal level? (b) at the

greatest height on a vertical wall 300 ft. away? Ans. (a) 1; (b) $.

27. A window of perimeter p ft. is in the form of a rectangle surmounted

by an isosceles right triangle. Show that the window will admit the most

light when the sides of the rectangle are equal to the sides of the right

triangle.

28. The sum of the surfaces of a sphere and a cube being given, showthat the sum of their volumes will be least when the diameter of the sphereis equal to the edge of the cube. When will the sum of the volumes be

greatest ?

29. Find the dimensions of the largest rectangle which can be inscribed

in the ellipse ~-}~ ^~ = 1. Ans. a\/2 X &V5.a- 6*

30. Find the area of the largest rectangle which can be drawn with

its base on the .r-axis and with two vertices on the witch wrhose equation

is ?/=

.,

8ff>*

(See figure in Chapter XXVI.) Ans. 4 a 2.

ir* -f 4 a 2

31. Find the ratio of the area of the smallest ellipse that can be cir-

cumscribed about a rectangle to the area of the rectangle. The area of an

ellipse is irab, where a and b are the semiaxes. Ans. | TT-

32. The two lower vertices of an isosceles trapezoid are the points

( 6, 0) and (6, 0). The two upper vertices lie on the curve jc2 + 4 y 36.

Find the area of the largest trapezoid which can be drawn in this way.Ans. 64.

33. The distance between the centers of two spheres of radii a and b,

respectively, is c. Find from what point P on the line of centers AB the

greatest amount of spherical surface is visible. (The area of the curved

surface of a zone of height h is 2 irrh, where r is the radius of the sphere.)

|Ans. -jp^ 5 units from A.

a* + b*

34. Find the dimensions of the largest rectangular parallelepiped with

a square base which can be cut from a solid sphere of radius r.

Ans. h = $ rV3.


35. Given a sphere of radius 6 in. Calculate the altitude of each of

the following solids :

(a) inscribed right circular cylinder of maximum volume ;

(b) inscribed right circular cylinder of maximum total surface ;

(c) circumscribed right cone of minimum volume,

Ans. (a) 4V3 in. ; (b) 6.31 in. ; (c) 24 in.

36. Prove that a conical tent of a given capacity will require the least

amount of canvas when the height is V2 times the radius of the base.

Show that when the canvas is laid out flat it will be a circle with a sector

of 152 9' cut out. How much canvas would be required for a tent 10 ft.

high? Ans. 272 sq. ft.

37. Given a point on the axis of the parabola ?/2 2 px at a distance a

from the vertex ;find the abscissa of the point on the curve nearest to it.

A HK. JT = a p.

38. Find the point on the curve 2 y = x 2 which is nearest to the point

(4, 1). Am. (2, 2).

39. If PQ is the longest or shortest line segment which can be drawn

from P(a, b) to the curve y f(x), prove that PQ is perpendicular to the

tangent to the curve at Q.

40. A formula for the efficiency of a screw is E = ~ a"> where

h -f tan

is the angle of friction and h is the pitch of the screw. Find // for maximum

efficiency. -4ns. h = sec tan 6.

41. The distance between two sources of heat A and /?, with intensi-

ties a and b respectively, is I. The total intensity of heat at a point Pbetween A and B is given by the formula

~ \" /

where x is the distance of P from A. For what position of P will the tem-

perature be lowest ? aijAns. x = T -

42. The lower base of an isosceles trapezoid is the major axis of an

ellipse ;the ends of the upper base are points on the ellipse. Show that

the maximum trapezoid of this type has the length of its upper base

half that of the lower.

43. An isosceles triangle with vertex at (0, b) is to be inscribed in the

ellipse b'2x'2 + a'2y

2 a 2b 2. Find the equation of the base if the area of

the triangle is a maximum. Ans. 27/4-6 = 0.

44. Find the base and altitude of the isosceles triangle of minimum area

which circumscribes the ellipse b 2x 24- a 2

?/2 = a 26 2

,and whose base is

parallel to the z-axis. Ans. Altitude = 36, base = 2 aV.

45. Let P(a, b) be a point in the first quadrant of a set of rectangular

axes. Draw a line through P cutting the positive ends of the axes at Aand B. Calculate the intercepts of this line on OX and OY in each of the

following cases.

(a) when the area OAK is a minimum;(b) when the length AH is a minimum;

(cj when the sum of the intercepts is a minimum;

(d) when the perpendicular distance from O to AB is a maximum.

An*, (a) 2 a, 2 b; (b) a + <6\ b + cPb* ;

(c) a 4- V^, fe + VS; (d) 2l^, 2!^!.a 6

50. Derivative as the rate of change. In Art. 23 the functional

relation

(1) y = r 2

gave as the ratio of corresponding increments

When x = 4 and Ar = 0.5, equation (2) becomes

(3. -.Then, we say, the average rate of change of y with respect to x

equals 8.5 when jr increases from x 4 to x = 4.5.

In general, the ratio

(A) -^- average rate of change of y with respect to x when x changesXfrom x to x + A.r.

Constant rate of change. When

(4) y = ax + b,

we have -r& = a.Ax

That is, the average rate of change of y with respect to x equals a,

the slope of the straight line (4), and is constant. In this case, andin this case only, the change in y (A?/), when x increases from anyvalue x to .r + A.r, equals the rate of change a times Ao*.

Instantaneous rate of change. If the interval from x to x + Axdecreases and A.r -> 0, then the average rate of change of y with

respect to x in this interval becomes at the limit the instantaneous

rate of change of y with respect to x. Hence, by Art. 24,

(B) -f- = instantaneous rate of change of y with respect to x for a

definite value of x.

For example, from (1) above,

(o) , ^ Z x.

When x = 4, the instantaneous rate of change of y is 8 units perunit change in x. The word "instantaneous" is often dropped in (B).

Geometric interpretation. Let the graph of

(6) #=/(*)

be drawn, as in the figure. When x in-

creases from OM to OA7

,then y increases

from MP to NQ. The average rate of

change of y with respect to x equals the

slope of the secant line PQ. The instanta-

neous rate when x = OM equals the slope of the tangent line PT.

Hence the instantaneous rate of change of y at P(x, y) is equal to

the constant rate of change of y along the tangent line at P.

When x XD, the instantaneous rate of change of //, or /(.r) r in (6)

is f'(xo). If x now increases from .TO to xo + A:r, the exact change in

y is not equal to/'(jo)A.r, unless f'(x) is constant, as in (4). We shall

see later, however, that this product is equal to A//, nearly, when Axis sufficiently small.

51. Velocity in rectilinear motion. Important, applications arise

when the independent variable in a rate is the time. The rate is

then called a time-rate. Velocity

in rectilinear motion affords a -4

< 2 ^ ^. ,

A o r i>f B

simple example.Consider the motion of a point P on the straight line AB. Let

s be the distance measured from some fixed point, as O, to any posi-

tion of P, and let t be the corresponding elapsed time. To each value

of t corresponds a position of P and therefore a distance (or space) s.

Hence s will be a function of /, and we may write

s=f(t).

Now let t take on an increment A ; then s takes on an increment

As, and An)

= the average velocityv ' Az

of P when the point moves from P to P', during the time interval A.If P moves with uniform motion (constant velocity), the above ratio

will have the same value for every interval of time and is the velocity

at any instant.

For the general case of any kind of motion, uniform or not, wedefine the velocity (time-rate of change of s) at any instant as the limit

of the average velocity as At approaches zero as a limit ; that is,

<o .-

The velocity at any instant is the derivative of the distance (= space)

with respect to the time, or the time-rate of change of the distance.

When v is positive, the distance s is an increasing function of t,

and the point P is moving in the direction AB. When v is negative,

s is a decreasing function of /, and P is moving in the direction BA.

(Art. 45.)

To show that this definition agrees with the conception we al-

ready have of velocity, let us find the velocity of a falling body at

the end of two seconds.

By experiment it has been found that a body falling freely from

rest in a vacuum near the earth's surface follows approximately the

law

(2) s - 16.1 /-,

where s = distance of fall in feet, t time in seconds. Apply the

General Rule (Art. 27) to (2).

FIRST STEP. s+ As=lfi.l(J+ A/)2 = 1 6.1 Z

2+ 32.2 AZ+16.1(A/) 2.

SECOND STEP. A* = 32.2 t-At + 16.

AsTHIRD STEP. -r1 = 32.2 1+ 16.1 At = average velocity throughout

the time interval At.

Placing t = 2,

As(3) = 64.4 + 16.1 At = average velocity throughout the

time interval At after two

seconds of falling.

Our notion of velocity tells at once that (3) does not give us the

actual velocity at the end of two seconds ; for even if we take At very

small, say , o (y or nuio of a second, (3) still gives only the average

velocity during the corresponding small interval of time. But whatwe do mean by the velocity at the end of two seconds is the limit of

the average velocity when At diminishes toward zero ;that is, the ve-

locity at the end of two seconds is, from (3), 64.4 feet per second.

Thus even the everyday notion of velocity which we get from experi-

ence involves the idea of a limit, or, in our notation,

/ As\v = lim )= 64.4 ft. per second.

Ai-oVAt/

52. Related rates. In many problems several variables are in-

volved each of which is a function of the time. Relations betweenthe variables are established by the conditions of the problem. Therelations between their time-rates of change are then found bydifferentiation.

As a guide in solving rate problems use the following rule.

FIRST STEP. Draw a figure illustrating the problem. Denote by x,

p, z, etr. the quantities which vary with the time.

SECOND STEP. Obtain a relation bet-ween the variables involved whichwill hold true at any instant.

THIRD STEP. Differentiate with respect to the time.

FOURTH STEP. Make a list of the given and required quantities.

FIFTH STEP. Substitute the known quantities in, the result found by

differentiating (third step), and solve for the unknown.

PROBLEMS

1. A man is walking at the rate of 5 mi. per hour toward the foot of

a tower 60 ft. high. At what rate is he approaching the top when he is

80 ft. from the foot of the tower?

Solution. Apply the above rule.

First Step. Draw the figure. Let x distance of the man from the foot, and

y = his distance from the top, of the tower at any instant.

Second Step. Since we have a right triangle,

yy = x2 + 3600.

Third Step. Differentiating, we get

**%=**%'<*

m 4y = xdxU '

dt y dt'

- This means that at any instant whatever

(Rale of change ofy] (-)times (rate of change ofx).

Fourth Step. x 80,-~ = 5 mi. an hourat= - 5 x 5280 ft. an hour.

_?

dt

Fifth Step. Substituting in (1),

f = 80

100= 4 mi. per hour. Ans.

& = _JjL x 5 x 5280 ft. per hour

dt 100


2. A point moves on the parabola 6 y = x 2 in such a way that when

x 6 the abscissa is increasing at the rate of 2 ft. per second. At what

rate is the ordinate increasing at that instant?

Solution. First Step. Plot the parabola.

Second Step. 6 y = x'2 .

Third Step. 6^ = 2x^.or

)-..

dt~3 dt

This moans that at any point on the parabola

(Rate of ckanac of ordinate) [r]liws (rafe f change of abscissa).

Fourth Step. x = 6.~ - 2 ft. per second.

dt

Fifth Step. Substituting in (2),

= 4 ft- P^r second,

^ = ~ x 2 = 4 ft. per second. Ans.at o

From the first result we note tliat at the point /J

(G, 6) the ordinate changes twice

as rapidly as the abscissa.

If we consider the point /"( t>, 6) instead, the result is

the minus sign indicating that the ordinate is decreasing

as the abscissa increases.

3. A circular plate of metal expands by heat so

that its radius increases at the rate of 0.01 in. per

second. At what rate is the surface increasing

when the radius is 2 in. ?

Solution. Let x = radius and y area of plate. Then

y = 7r.r~.

du rt dx

That is, at any instant the area of the plate is increasing in square inches 2 ?nr

times as fast as the radius is increasing in linear inches.

Substituting in (U),

= 2 TT x 2 x 0.01 = 0.04 TT sq. in. per second. Arts.dt

4. An arc light is hung 12 ft. directly above a straight horizontal

walk on which a boy 5 ft. tall is walking. How fast is the boy's shadow


lengthening when he is walking away from the light at the rate of 168 ft.

per minute?

Solution. Let x = distance of the boy from a point directly under the light L,

and y length of boy's shadow. From the figure,

y : y + x = 5 : 12,

or y = 8 x.

T^.- .. ,. t/v 5 dxDifferentiating,

d/ 7 d/'

that is, the shadow is lengthening v> as fast as the boy is

walking, or 120ft. per minute.

5. A point moves along the parabola //-'= 12 jc in such a way that its

abscissa increases uniformly at the rate of 2 in. per second. At what pointdo the abscissa and ordinate increase at the same rate? Ans. (3, 6).

6. Find the values of .r for which the rate of change of

^ - 12 jc2 + 45 .r

- 13

is zero. /b/.s\ 3 and 5.

7. A barge whose deck is 12 ft. below the level of a dock is drawn upto it by means of a cable attached to a ring in the floor of the dock, the

cable being hauled in by a windlass on deck at the rate of 8 ft. per minute.

How fast is the barge moving towards the dock when 10 ft. away?Ann. 10 ft. per minute.

8. A boat is fastened to a rope which is wound about a windlass 20 ft.

above the level at which the rope is attached to the boat. The boat is

drifting away at the rate of 8 ft. per second. How fast is it unwinding the

rope when 30 ft. from the point directly under the windlass?

Ann. ().Gf) ft. per second.

9. One end of a ladder 50 ft. long is leaning against a perpendicular

wall standing on a horizontal plane. Suppose the foot of the ladder to be

pulled away from the wall at the rate of 3 ft. per minute, fa) How fast is

the top of the ladder descending when its foot is 14 ft. from the wall?

(b) When will the top and bottom of the ladder move at the same rate?

(c) When is the top of the ladder descending at the rate_of 4 ft. per minute?

Ans. fa) I ft. per minute; fb) when 25V2 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.


11. The side of an equilateral triangle is o in. long, and is increasing at

the rate of k in. per hour. How fast is the area increasing?

An.s. \ afcV3 sq. in. per hour.

12. The edges of a regular tetrahedron are 10 in. long and are increasing

at the rate of 0.1 in. per minute. Find the rate of increase of the volume.

13. If at a certain instant the two dimensions of a rectangle are a and 6

and these dimensions are changing at the rates ra, n respectively, showr that

the area is changing at the rate an -f bm.

14. At a certain instant the three dimensions of a rectangular parallele-

piped are 6 in., 8 in., 10 in., and these are increasing at the respective rates

of 0.2 in. per second, 0.3 in. per second, 0.1 in. per second. How fast is the

volume increasing?

15. The period (Psec.) of a complete oscillation of a pendulum of

length / in. is given by the formula P 0.324VL Find the rate of changeof the period with respect to the length when / = 9 in. By means of this

result approximate the change in /' due to a change in / from 9 to 9.2 in.

AJIK. 0.054 sec. per inch; 0.0108 sec.

16. The diameter and altitude of a right circular cylinder are found at a

certain instant to he 10 in. arid 20 in. respectively. If the diameter is in-

creasing at the rate of 1 in. per minute, what change in the altitude will

keep the volume constant? A?IK. Decreasing 4 in. per minute.

17. The radius of the base of a certain cone is increasing at the rate of

3 in. per minute and the altitude is decreasing at the rate of 4 in. perminute. Find the rate of change of the total surface of the cone when the

radius is 7 in. and the altitude is 24 in.

An*. Increasing 96 TT sq. in. per minute.

18. A cylinder of radius r and altitude // has a hemisphere of radius r

attached to each end. If r is increasing at the rate of.]

in. per minute, find

the rate at which // must change to keep the volume of the solid fixed at

the instant when r is 10 in. and Jt is 20 in.

19. A stone is dropped down a' deep shaft and after t sec. another stone

is dropped. Show that, the distance between the stones increases at the

rate of ty ft. per second.

20. A gas holder contains 1000 cu. ft. of gas at a pressure of 5 Ib. per

square inch. If the pressure is decreasing at the rate of 0.05 Ib. per squareinch per hour, find the rate of increase of the volume. (Assume Boyle'sLaw : pv = c.) Ans. 10 cu. ft. per hour.

21. The adiabatic law for the expansion of air is PVIA = C. If at a

given time the volume is observed to be 10 cu. ft. and the pressure is

50 Ib. per square inch, at what rate is the pressure changing if the volumeis decreasing 1 cu. ft. per second?

Ans. Increasing 7 Ib. per square inch per second.


22. If y = 4 x r3 and .r is increasing steadily at the rate of A unit per

second, find how fast the slope of the graph is changing at the instant whenx = 2. An*. Decreasing 4 units per second,

23. Water flows from a faucet into a hemispherical basin of diameter

14 in. at the rate of 2 cu. in. per second. How fast is the water rising

(a) when the water is halfway to the top? (b) just as it runs over? (Thevolume of a spherical segment is

],irr

>2h -f ^ ?r/r!

, where // is the altitude of

the segment.)

24. Gas is escaping from a spherical balloon at the rate of 1000 cu. in.

per minute. At the instant when the radius is 10 in., (a) at uhat rate is the

radius decreasing? (b) at what rate is the surface decreasing?AUK. (b) 200 sq. in. per minute.

25. If r denotes the radius of a sphere, tf the surface, and V the volume,

x , , ,. dV rdSprove the relation = -

dt 2 dt

26. A railroad track crosses a highway at an angle of (10". A locomotive

is 500 ft. from the intersection and moving away from it at the rate of

60 mi. per hour. An automobile is 500 ft. from the intersection and movingtoward it at the rate of 30 mi. per hour. What is the 1 rate of change of the

distance between them?Ans. Increasing 15 mi. per hour or IftV.'J mi. per hour.

27. A horizontal trough 10 ft. long has a vertical section in the shape of

an isosceles right triangle. If water is poured into it at the rate of H cu. ft.

per minute, at what rate is the surface of the water rising when the water

is 2 ft. deep? Aws. I ft. per minute.

28. In Problem 27, at what rate must water be poured into the trough

to make the level rise I ft. per minute when the water is 3 ft. deep ?

29. A horizontal trough 12 ft. long has a vertical cross section in the

shape of a trapezoid, the bottom being 3 ft. wide and the sides inclined to

the vertical at an angle whose sine is;?.

Water is being poured into it at

the rate of 10 cu. ft. per minute. How fast is the water level rising when

the water is 2 ft. deep?

30. In Problem 29, at what rate is water being drawn from the trough

if the level is falling 0.1 ft. per minute when the water is 3 ft. deep ?

31. The ^-intercept of the tangent line to the positive branch of the

hyperbola xy = 4 is increasing 3 units per second. Let the -//-intercept be

OB. Find the velocity of B at the end of 5 sec., the x-intercept starting at

the origin. Am. f| unit per second.

32. A point P moves along the parabola y2 x so that its abscissa in-

creases at the constant rate of k units per second. The projection of P on

the .r-axis is M. At what rate is the area of triangle OMP changing when

P is at the point where x = a? Ann. 2 fcVa units per second.

ADDITIONAL PROBLEMS

1. Rectangles inscribed in the area hounded by the parabola y2 = 16 x

and its latus rectum and such that one side always lies along the latus

rectum serve as the bases of rectangular parallelepipeds whose altitudes

are always the same as the side parallel to the .r-axis. Find the volume of

the largest parallelepiped. Ans. Aj*j&Vb = 73.27.

2. An ellipse symmetrical with respect to the coordinate axes passes

through the fixed point (//, k). Find the equation of the ellipse if its area

is a minimum. Am. k 2jr

2 + h'2y2 2 h 2k 2

.

3. The curve j :< 3 /// + //'*= has a loop in the first quadrant sym-

metric with respect to the line //= jr. Isosceles triangles having a common

vertex at the origin and bases along the line j- -f y~ a are inscribed in this

loop. Find the value of a if the area of the triangle is a maximum.Am. J(l + Vl3) =2.303.

4. At a point /' in the first quadrant on the curve //= 7 jr

2 a tangentis drawn, meeting the coordinate axes at A and B. Find the position of Pwhich makes A B a minimum. Am. Ordinate */-.

5. The cost of erecting an office building is $50,000 for the first story,

$52,500 for the second, $55,000 for the third, and so on. Other expenses

(lot, plans, basement, etc.) are $350,000. The net annual income is $5000for each story. How many stories will give the greatest rate of interest on

the investment? Am. 17.

6. For a certain article the increase in the number of pounds consumedis proportional to the decrease in the tax on each pound. If the consump-tion is in Ib. when there is no tax and n Ib. when the tax is / dollars per

pound, find the tax which should be imposed on each pound to bring in the

most revenue.

7. A triangle ABC is formed by a chord BC of the parabola y ~ kx 2

and the tangents AB and AC at each extremity of the chord. If BC re-

mains perpendicular to the axis of the parabola and approaches the vertex

at the rate of 2 units per second, find the rate at which the area of the

triangle is changing when the chord BC is 4 units above the vertex.

8. A vertical cylindrical tank of radius 10 in. has a hole of radius 1 in.

in its base. The velocity with which the water contained runs out of the

tank is given by the formula v 2 2 gh, where h is the depth of the water

and g is the acceleration of gravity. How rapidly is the velocity changing ?

.4r?8. Decreasing y^ g ft. per second per second.

9. A light is 20 ft. from a wall and 10 ft. above the center of a pathwhich is perpendicular to the wall. A man 6 ft. tall is walking on the pathtoward the wall at the rate of 2 ft. per second. When he is 4 ft. from the

wall, how fast is the shadow of his head moving up the wall ?

Am. f ft. per second.

CHAPTER VI

SUCCESSIVE DIFFERENTIATION AND APPLICATIONS

53. Definition of successive derivatives. We have seen that the

derivative of a function of x is in general also a function of x. This

new function may also be diil'erentiable, in which case the derivative

of the first derivative is called the second derivatire of the original

function. Similarly, the derivative of the second derivative is called

the third derivative ; and so on to the ni\\ deriratire. Thus, if

V - 3 r>,

<LM ._ 19 ~3,

i ^ x,

ax

dx\ drv/.r

Notation. The symbols for the successive derivatives are usually

abbreviated as follows.

rf2?/

If y =/(j-), the successive derivatives are also denoted by

, _ . . . .= -, ,

dxn- .

In the example given above, the notation y = 3 x4, y'

= 12 x3,

I/''= 36 x 2

, T/'" = 72 r, ?yw = 72 is most convenient.

54. Successive differentiation of implicit functions. To illustrate the

process we shall find^ from the equation of the hyperbola

(1) b2x2 - a2?/2 = a262 .

73

Differentiating with respect to x (Art. 41),

2 b*x - 2 a'2y^ = 0,

(9) ^L ^ -

dx a-y

Differentiating again, remembering that y is a function of x,

d~y _ dx

dx* ^Substituting for ^ its value from (2),

., ,i, /fcVN

a 4'//- r/

4v/

;{

Hut, from the given equation, b-x2a-y~ = a-62

.

PROBLEMS

Prove each of the following differentiations.

2. = >/" + W. ^ ~ iL^!___."'*

8(a-f6/)^

d 2// __ 4 a?>-

3_ y\LJ-.y^. . *j< __(i bx dx'2 (a b.

_ /~~ ; r; d~u a 2

4. // Va- -f r 2.

- =

5. //=

6. =

fi\~dl(a-' + r

d-y _ 2 a'2

(i + r dx'2 (a + x)3

7. f(s) =

8. y =

'2 t -h 1

r-< *?

.r-fl

<ir 2 y3

d'2y __ _ 4 a 2

dx'2 //*

SUCCESSIVE DIFFERENTIATION AND APPLICATIONS 75

9. j*2

-f- */2 = r 2

.

10. y2 = 4 ru*.

11. fc-.r2-f a 2

//2 = a'

2b 2.

12. or-' -f 2 Jixy -f fy/2

1.

13. JT* + ?/

:{ ^ 1.

14. .r + 2 T 2^ 2 - - 4

a4?/'

(//.r -f by)*'

= a4.

djr-

V

</.r2

In Problems 15 25 find the values of //' and //'' for the given values ofthe variables.

16.;//= V25 - 3 x

; j = 3.

17. ?/= xVx2 + 9 ; j* = 4.

18. j-2 - 4 ?/

2 = 9 ; j- = 5, ?/= 2.

19. x 2-f 4 xy 4- z/

2-f 3 =

; x = 2, ?/= -

20. ?/= (3 -a:2)4

; j- = 1.

21. ?/= Vl 4- 2 J-; j- =

22. ?/=

23. ?/=

4; x = 2.

2; x =

In each of the following problems find

24. ?/2 + 2 .r/y

= 1 G ; .r = 3, y ^ 2.

25. .r'{ -

.r/y2

-f //"= 8

;x- 2, y = 2.

27. ?/-a:

2H- a 2

29. y j-Vr/"' - x 2.

30. //2 - 4 xy

-If).

31. x :i - 3 ajry -f ?y:{ = ?>

3.

28. T/= V2 - 3 a:.

55. Direction of bending yof a curve. When the point \

P(x, y) traces a curve, the

slope of the tangent line at

P varies. When the tan-

gent line is below the curve

(Fig. a), the arc is concave upward; when above the curve (Fig. 6),

the arc is concave downward. In Fig. a the slope increases when

/

FIG. a

X x

FIG. 6


P describes the arc AP'. Hence f'(x) is an increasing function of x.

On the other hand, in Fig. b, when P describes the arc QB, the slope

decreases, and J'(-jr) is a decreasing function. In the first case, there-

fore, f"(x) is positive; in the second case, negative (Art. 45). Hencewe have the following criterion for determining the direction of

bending at a point.

The graph of y = f(x) is concave upward if the second derivative of

y with respect to x is positive, and concave downward if this derivative

is negative.

56. Second method for testing for maximum and minimum values.

At A in Fig. a of the preceding section, the arc is concave upward,and the ordinate has a minimum value. That is, /'(x) = and /"(/)

is positive. At B in Fig. b, f'(x) and f"(x) is negative.

We may then state the sufficient conditions for maximum and

minimum values of f(x) for critical values of the variable as follows

j(x) is a maximum if f'(x) and/"(*) a negative number,

/(jt) is a minimum if f'(x) = and/"(*) a positive number.

Following is the corresponding working rule for applying this test

for maximum and minimum values.

FIRST STEP. Find the first derivative of the function.

SECOND STEP. Set the first derivative equal to zero and solve the result-

ing equation for real roots in order to find the critical values of the variable.

THIRD STEP. Find the second derivative.

FOURTH STEP. Substitute each critical value for the variable in, the

second derivative'. If the result is negative, then the function is a maxi-

mum for that critical value; if the result is positive, the function is a

minimum.

When /"(/) = or does not exist, the above process fails, al-

though there may even then be a maximum or a minimum; in that

case the first method given in Art. 47 holds, being fundamental.

Usually the second method does apply, and when the process of

finding the second derivative is not too long or tedious, it is generally

the shortest method.

ILLUSTRATIVE EXAMPLE 1 . Let us now apply the above rule to test analyticallythe function ,

M = x~ -fx

found in the example worked out on page 49.

Solution. /(;r)=.r2 +

439First Step. f(x)=2x-~>


Second Step.

Third Step.

Fourth Step.

Hence

;-^=.x = 6, critical value.

/"(6) = -f .

/(6) = 108, minimum value.

ILLUSTRATIVE EXAMPLE 2. P'xamine r* - 3 x- - 9 x 4 5 for maxima and

minima. Use the second method.

Solution. /(*) = ** - 3 x- -9x45.First Step. /'(*) = 3 x 2 - 6 x - 9.

Second Step. 3 x2 - 6 x - 9 = ;

hence the critical values are x = 1 and 3.

Fourth Step. /"(- 1) = - 12.

.-./( 1) = 10 (ordinate of /I) = maximum value.

/"(3) = 4 12. /. /(3) = 22 (ordinate of #) minimum value.

PROBLEMS

Examine each of the following functions for maxima and minima.

An*. JT - 2, gives max. 2.

x ~ 0, gives min. 2.

x 1, gives max. 6.


0) x 0, gives max. a3,

x = a, gives min. = 0.

x - 2, gives max. = 22.


x \, gives max. = \.

jr | , gives min. = $.

x = 0, gives max. = 2.


x = 2, gives min. 30.

1. x34- 3 x 2 - 2.

2. x3 - 3 x + 4.

3. 2 x3 - 3 ox 24- a3

. (a ;

4. 2 4- 12 x 4- 3 x 2 - 2 x3.

5. 3 x - 2 x 2 - -

6. 3 x4 - 4 x3 - 12 x 24- 2.

7. x4 - 4 x 2-f 4.

ax

x = 0, gives max. 4.

x = V2, gives min. = 0.

x = a, gives max. = |.

x a, gives min. = \.

9. x3 4- 9 x 24- 27 x 4- 9.

10. 12 x 4- 9 x 2 - 4 x3.

11. x 2(x-4) 2.

12. * +

13. x-


14. A rectangular box with a square base and an open top is to be made.

Find the volume of the largest box that can be made from 1200 sq. ft. of

material. Ans. 4000 cu. ft.

15. A water tank is to be constructed with a square base and an open

top, and is to hold 125 cu. yd. If the cost of the sides is $2 a square yard,

and of the bottom $4 a square yard, what are the dimensions when the

cost is a minimum? Ans. A cube of edge 5 yd,

16. A rectangular flower bed is to contain 800 sq. ft. It is to be sur-

rounded by a walk which is 3 ft. wide along the sides and 6 ft. wide across

the endtt. If the total area of the bed and walk is a minimum, what are the

dimensions of the flower bed? Ans. 20 ft. x 40 ft.

17. A rectangular field to contain a given area is to be fenced off along

the bank of a straight river. If no fence is needed along the river, show that

the least amount of fencing will be required if the field is twice as long as

it is wide.

18. A trough is to be made of a long rectangular piece of tin by bending

up U\o edges so as to give a rectangular cross section. If the width of the

piece is 14 in., how deep should the trough be made in order that its carry-

ing capacity may be a maximum? Ans. 3.5 in.

19. A window composed of a rectangle surmounted by an equilateral

triangle is 15 ft. in perimeter. Find its dimensions if it admits the maxi-

mum amount of light. An*. Rectangle is 3.51 ft. wide and 2.23 ft. high.

20. A solid wooden sphere weighs ?/> Ib. What is the weight of the

heaviest right circular cylinder which can be cut from the sphere?

Ans. r^lb.V3

21. The slant height of a right circular cone is a given constant a.

Find the altitude if the volume is a maximum. A a-ri/io. 7^

V322. An oil can is to be made in the shape of a cylinder surmounted by a

cone. Show that for a given capacity the least material is required if the

altitude of the cylinder is equal to the altitude of the cone.

23. Ciiven the parabola //2 = 8 jc and the point P(6, 0) on the axis, find

the coordinates of the points on the parabola nearest to P. Ans. (2, 4).

24. A given isosceles triangle has a base of 20 ft. and an altitude of 8 ft.

What are the dimensions of the maximum inscribed parallelogram, one side

coinciding with the base of the triangle, if the acute angle of the parallelo-

gram is arc tan ^? Ans. 5 ft. X 10 ft.

25. A miner wishes to dig a tunnel from a point A to a point B 200 ft.

below and 600 ft. to the east of A. Below the level of A it is bed rock and

above A is soft earth. If the cost of tunneling through earth is $5 and

through rock is $13 per linear foot, find the minimum cost of a tunnel.

Ans. $5400.


26. A sheet of paper for a poster is to contain 16sq. ft. The marginsat the top and the bottom are to be (\ in., and those on the sides 4 in.

What are the dimensions if the printed area is to be a maximum?.4 MS. 1.90ft. x 3.27 ft,

27. An electric current flows through a coil of radius r and exerts a

force F on a small magnet the axis of which is on a line drawn throughthe center of the coil and perpendicular to its plane. This force is given

by F _, where .r is the distance to the magnet from the center

(r2 + .r

2)*

of the coil. Show that F is a maximum for .r \ r .

57. Points of inflection. A point of inflection (or inflectional point)on a curve separates arcs having opposite directions of bending (see

Art. 55).

In the figure, B is a point of inflection. When the tracing point,

on a curve passes through such a point, the second derivative will

change sign, and if continuous must vanish at the point. Hence wemust have

(1) At points of inflection, /"(*) = 0.

Solving the equation resulting from (1) gives the abscissas of the

points of inflection. To determine the direction of bonding in the

vicinity of a point of inflection, test f"(jr) for val-

ues of x first a trifle less and then a trifle greater

than the abscissa at that point.

If /"(#) changes sign, we have a point of in-

flection, and the signs obtained determine if the

curve is concave upward or concave downward in the neighborhood.The student should observe that near a point where the curve is

concave upward (as at A) the curve lies above the tangent, and at

a point where the curve is concave downward (as at C) the curve

lies below the tangent. At a point of inflection (as at B) the tangent

evidently crosses the curve.

Following is a rule for finding points of inflection of the curve whose

equation is y = f(x). This rule includes also directions for examiningthe direction of bending.

FIRST STEP. Findf"(x).SECOND STEP. Setf"(x) = 0, and solve the resulting equation for real

roots.

THIRD STEP. Testf"(x) for values of x first a trifle less and then a

trifle greater than each root found in the second step. If f"(x) changes

sign, we have a point of inflection.

When f"(x) = +, the curve is concave upward + .*

Whenf"(x) = ,the curve is concave downward ^^^.

Before the Third Step it is sometimes convenient to factor f"(x).

It is assumed that f'(x) and J"(x) are continuous. The solution

of Problem 2, below, shows how to discuss a case where f'(x) and

f"(x) are both infinite.

PROBLEMS

Examine the following curves for points of inflection and direction of

bending.

1. 77= 3 j-

4 - 4 x3 + 1.

Solution.

First Step.

Second Step.

/(x) = 3 x 4 - 4 x'< 4- 1.

/"(x) = 36 x 2 - 24 x.

36 J'2 - 24 x ~ 0.

. x 3 and x are the roots.

Third Step.

tl/"(x) =36T(x- I).

When x < 0,/"(x) = +.

When 3 >x> 0,/"(x) = -.

Therefore the curve is concave upward to the left and concave downward to

the right of x (A in figure).

When ()<x < 3,/"(x) = -.

When x > l,f"(x) = -f.

Then^fore the curve is concave downward to the left and concave upward to

the right of x =rj

(/ in figure).

Hence the points .4(0, 1) and B($, \\] are points of inflection.

Th*> curve is evidently concave upward everywhere to the left of 4, concave

downward between ,4(0, 1) and (, A}), and concave upward everywhere to

the right of B.

2. (y-2}*= (jr-4).

Solution. y = 2 -f (x - 4)*.

Second Step. When x = 4, both first and second derivatives become infinite.

Third Step. When x < 4, = +.dx 2

WThen x > 4,^ = -.dr 2

* This may easily be remembered if we say that a vessel shaped like the curve whereit is concave upward will hold (+) water, and where it is concave downward will spill

( ) water.


We may therefore conclude that the tangent at (4, 1^ is perpendicular to thex-axis, that to the left of (4, 2) the curve is concave upward, and to the right of

(4, 2) it is concave downward. Therefore (4,2) is a point of inflection.

3. y = x2. Ans. Concave upward everywhere.

4. y = 5 2 x x2. Concave downward everywhere.

5. y = x3. Concave downward to the left and

concave upward to the right of

(0, 0).

6. y = x4 . Concave upward everywhere.

7. y = 2 x3 - 3 x2 - 36 x + 25. Concave downward to the left andconcave upward to the right of

jc~

\

8. 2/=24x2 -x4. 9. ?/^x +

1-.

*'

x

58. Curve-tracing. The elementary method of tracing (or plotting)a curve whose equation is given in rectangular coordinates, and onewith which the student is already familiar, is to solve its equation for

y (or x), assume arbitrary values of x (or //), calculate the correspond-

ing values of y (or r), plot the respective points, and draw a smoothcurve through them, the result being an approximation to the re-

quired curve. This process is laborious at, best, and in case the equa-tion of the curve is of a degree higher than the second, it may not be

possible to solve the equation for y or x. The general form of a curve

is usually all that is desired, and the calculus furnishes us with

powerful methods for determining the shape of a curve with verylittle computation.

The first derivative gives us the slope of the curve at any point ;

the second derivative determines the intervals within which the curve

is concave upward or concave downward, and the points of inflection

separate these intervals; the maximum points are the high points,

and the minimum points are the low points on the curve. As a guidein his work the student may follow the

Rule for tracing curves, using rectangular coordinates

FIRST STEP. Find the first derivative; place it equal to zero; solve

to find the abscissas of the maximum and minimum points. Test these

values.

SECOND STEP. Find the second derivative; place it equal to zero;

solve to find the abscissas of the points of inflection. Test these values.

THIRD STEP. Calculate the corrcspomling ordinates of the points

whose abscissas were found in the first two steps. Calculate as many

more points as may be necessary to (jive a good idea of the shape of the

curve. Make a table such as is shown in tlie problem worked out below.

FOURTH STEP. Plot the points determined and sketch in the curve to

correspond with the results shown in the table.

If the calculated values of the ordinatcs are large, it is best to

reduce the scale on the ?/-axis so that the general shape of the curve

will be shown within the limits of the paper used. Coordinate plot-

ting paper should be employed. Results should be tabulated as in

the problems solved. In this table the values of x should follow one

unother, increasing algebraically.

PROBLEMS

Truce the following curves, making use of the above rule. Also find

the e([iiations of the tangent and normal at each

point of inflection.

1. y rr .r:{ - 9 jr

24- 1M x - 7.

Solution. Use the above rule.

Ft rat Step. y' - 3 x'~ - 18 x -f 24,

(lx- - lS.r -f- 24 - 0,

x = 2, 4.

Second Step. y" = G x - 18,

G.r - 18 = 0,

Third Step. x - 3.

Fourth Step. Plotting the points and sketching in the curve, we get the figure

shown.To find the equations of the tangent and normal to the curve at the point of

inflection /V3, 11), use formulas (1), (2\ Art. 43. This gives 3 x + y 20 for the

tangent and 3 ?/ x 30 for the normal.

2. ;] //= .r<

- 3 x'2 - 9 x -f 11.

An*. Max. ( 1,J/); min. (3, -\f-) ; point of inflection, (1, 0) ;

tangent, 4 jc + y -4; normal, .r 4 // 1 = 0.

3. 6 y = VI - 24 x - 15 :r2 - 2 j-\

A//S. Max. ( 1, ^) : min. (- 4, - 5) ; point of inflection,

4. y = x4 - 8 x 2.

Am. Max. (0, 0) ; min. ( 2,- 16) ; points of inflection,

(?,Vi*, -Y>.5. y 5 x J*.

Am. Max. (1, 4); min. (- 1,- 4)

; point of inflection, (0, 0).

*-?Ti-Am. Max. (V3, V3)

; min. ( ~ V3, - V3 ); points of inflection,

(-3, -ij), (0, 0), (3, H).

7. ?/= /< + 6 .T

2. 18. (I If

^ .r- + -

8. ?/= 4 + 3 .r

- .r<.

9. 3 ?/= 4 .r< - 18 r 2 + 15 x. 19-

'-'//^ .'"' -f ~

10. ?/= O -

a): *

-f b.

11. 12 //=

(.r-

I)1 - 24 (.r -1)2. 20. a-'//

^ .r< -f^,.

12. //= r 2

(9-

-rL>

).

13. ?/= 2 .r>

- 5 .r2

. 21. //= - - -"-' -

.,-

.r- -1 1 a-14. //

= 3 .r> - 5 .r\

15. ?/= .r

r> - 5 .r1

. 22. ?/- - ''-- -

(.r f <i)-

16. y = jr(s- - 4)2

.

4 23. .r1

'//- (.r-' { 1 )'.

59. Acceleration in rectilinear motion. In Art. 51 velocity in rec-

tilinear motion was defined as the time-rate of change of the distance.

We now define acceleration as the time-rute of climiye of the velocity.

That is,

dv(A) Acceleration = a

(hFrom (C), Art. 51, we obtain also, since v --*

Referring to Arts. 45, 47, and 56, we have the following criteria

which apply to a definite instant / = / :

If a > 0, v is increasing (algebraically).

If a < 0, v is decreasing (algebraically).

If a > and v = 0, s has a minimum value.

If a < and v = 0, s has a maximum value.

If a = and changes sign from + to (from to +) when t

passes through to, then v has a maximum (a minimum) value when

In uniformly accelerated rectilinear motion, a is constant. Thus

in the case of a body falling freely under the action of gravity only,

a = 32.2 ft. per second per second. Namely, from (2), Art. 51,

8 = 16.1 12

,v = = 32.2 t, <* = = 32.2.

PROBLEMS

1. By experiment it has been found that a hody falling freely from rest

in a vacuum near the earth's surface follows approximately the law

s~

16.1 t2

,where = space (height) in feet, t = time in seconds. Find

the velocity and acceleration (a) at any instant; (b) at end of the first

second; (c) at end of the fifth second.

Solution. (1) s = 16.1/ 2.

ds(a) Differentiating,

~ = 32.2 t,

or, from (C), Art. 51, (2) v - 32.2 / ft. per second.

Differentiating again, -r. = 32.2,

or, from (A) above, (3) a 32.2 ft. per (sec.)2

,

which tells us that the acceleration of a falling body is constant ; in other words, the

velocity increases 32.2 ft. per second every second it keeps on falling.

(b) To find and </ at the end of the first second, substitute / 1 in (2) and (3).

Then T 32.2 ft. per second, a = 32.2 ft. per (sec. )

2.

(c) To find rand a at the end of the fifth second, substitute t = 5 in (2) and (3).

Then v = 161 ft. per second, a = 32.2 ft. per CsecJ 2.

Given the following equations of rectilinear motion; find the position,

velocity, and acceleration at the instant indicated.

2. * 4 1- ~ (> t;

f 2. AUK. * 4, r 10, a = 8.

3. .s--= 120 {

- lf> t'-; / = 4. * = 224, r - -

8, a = - 32.

4. .r- 32 /

- 8 /' ; / = 2. s = 32, r = 0, a = - 16.

5. //= (J /-'

- 2 t'{

;t = 1. y = 4, r = 6, a = 0.

.

7. / = 16 t'2 - 20 / -f 4; / = 2.

8. //= 100 - 4 /

- 8 /2

; / = 3.

9. * = V5/ -f ; ^ = 5.

V5f

10. * = \7

3 / 4- 2 ; / = 2.


In the following problems find the acceleration at the instant indicated.

ll.r = 80-32/:f = 0. An.. -32.18r = ^_., = 1 .

12. r = 4 /2 -!()/; / = 2. 6. / + 1

Given the following equations of rectilinear motion ; find the position

and acceleration when the particle first comes to rest.

14. * = 16 /2 - 64 / + 04. AHS. s = 0, a = 32.

15. *= 120 /- 16 /J

. Pt 2017. ,s- 5/4- ~.

16. * = 3 c*t - /'. * + 1

18. A ball thrown directly upward moves according to the law

,s- = 80 /- If) /-.

Find (a) its position and velocity after 2 sec. and after 3 sec.; (b) how

high it will rise; (c) how far it will move in the fourth second.

19. If the equation of a rectilinear motion is ,s = V / -f 1, show that the

acceleration is negative and proportional to the cube of the velocity.

20. The height (*ft.) reached in /sec. by a body projected vertically

upward with a velocity of ?M ft. per second is given by the formula

8 = r\t j gt~. Find a formula for the greatest height reached by the body.

21. In the preceding problem suppose r\~

1(>0, g 32. Find (a) the

velocity at the end of 4 sec. and at the end of (5 sec.; tb) the distance

moved during the fourth second and during the sixth second.

22. A car makes a trip in 10 min. and moves according to the law

,s- = 250 f2:\ t\ where / is measured in minutes and in feet. Ca) How far

does the car go? (b) What is its maximum speed? (c) How far has the

car moved when its maximum speed is reached?

Anx. (a) 12,500 ft.; (b) 1924 ft. per minute; (c) 0944 ft.

ADDITIONAL PROBLEMS

1. Trace the curve (4- 2 x + s-)tj 2 x - s 2

,arid find the equations

of the tangent and normal at each point of inflection.

Ans. Max. (1, \). Point of inflection CO, 0): tangent, x 2 y = ;

normal, 2 x + y 0. Point of inflection (2, 0) : tangent, x + 2 y 2 =;

normal, 2 x y 4 = 0.

2. A certain curve (the tractrix) is such that the length of every tangent

from its point of contact Pur, y} to its intersection A with the x-axis is

the constant r (A P c). Show that

(\ 4lL V

/|x d 2

ij _ c 2y

3. Determine k so that the normals at the points of inflection of the

curve y = k(x2

3)2 will pass through the origin. ... 7. _ 1

_ri/lc>. K> .

4V2

CHAPTER VII

DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS.APPLICATIONS

We consider now functions such as

sin 2 -jr, 3% log (1 + s~),

called transcendental Junctions, as distinguished from the algebraic

functions hitherto discussed.

60. Formulas for derivatives; second list. The following formulas,

grouped here for convenience of reference, will be derived in this

chapter, and, with the formulas of Art. 29, comprise all formulas for

derivatives used in this book.

dv

d dx 1 dvX ^ (lnl = T~ ITx On^log^)

Xav dx

XI -^-(a")= a"lna^-dx dx

XI a f <"> = ?dx dx

XII 4.<!") = 1*1" 1^ + inu.u*^.dx dx dx

XIII (sin v) cos v

d , ,dv

XIV ^ (cos v) = - sin v -

XV (tan v) sec2 vc?x dx

XVI (ctn y) = - esc 2 v^~-

^ ^ dvXVII (sec r) = sec i> tan i;

.

dx clx

86

TRANSCENDENTAL FUNCTIONS 87

XVin (esc v) = esc v ctn vdx dx

& / \ . dv(vers v) = sin v -

dv

XX (arc sin i;)= ._

rf* Vl -i;

XXI (arc cos v) ~dx-^/i _ vz

XXII 4- (a1" ^n v) = 7-^-dx 1 -f ^

XXIII (arc ctn i;) = - 7-^..

XXIV ^- (arc sec v) =

dv

dx

dv

XXV ~(arc csc ") = -

XXVI --(arc vers v) =

61. The number e. Natural logarithms. One of the most importantlimits in the calculus is

(1) Km (1 + ar = 2.71828 -.

x ->

This limit is denoted by c. To prove rigorously that such a limit e

exists is beyond the scope of this book. For the present we shall

content ourselves with plotting the locus of the equation

(2) y=(l + xf

and showing graphically that as x -+ the function (1 + x)x(= y)

takes on values in the near neighborhood of 2.718 and therefore

e = 2.718 approximately.


As x from the left, y decreases and approaches e as a limit.

As x * from the right, y increases and also approaches e as a limit.

The fact expressed in (1) is used in Art. 63.

As jr > -f oo, // approaches 1 as a limit;and as .r > 1 from the right,

?/ increases without, limit. The lines ?/ 1 and f = 1 are asymptotes.

In Chapter XX we shall show how to calculate the value of c to any

number of decimal places.

Natural, or Napierian, logarithms are those which have the number

e for base These logarithms play a very important role in mathe-

matics. To distinguish between natural logarithms and common

logarithms when the base is not explicitly stated, the following no-

tation will be used.

Natural logarithm of r (base c) = In r.

Common logarithm of r (base 10) = log r>.

By definition, the natural logarithm of a number AT

is the exponent

x in the equation

(3) ej - A; that is, .r - In A7

.

If .r == 0, Ar =

1, and In 1 - 0. If x = 1, AT = e, and In e = 1.

If j,_._ _

oo^ (hen ,\T

- 0, and we write In = oo.

The student is familiar with the use of tables of common loga-

rithms, where the base is 10. The common logarithm of a number Nis the exponent // in the equation

(4) 10" = AT

,or //

= log N.

Let us find the relation between In A7 and log A7.

In (3), take logarithms of both members to the base 10. Then we

have, from (2), p. 1,

(5) x log c = log A' .

Solving for x, which equals In \, by (3), we get the desired relation

_ log W~~

log e


That is, we obtain the natural logarithm of any number by dividingits common logarithm by log e.

Equation (A) may be written

(6) log .V = log c - In .Y.

Hence the common logarithm of a number is obtained by multiplyingits natural logarithm by log e. This multiplier is called the modulus

(= ji/) of common logarithms.

By tables, log e - 0.4343, and -i- = 2.303.log e

Equation (4) may now be written

(7) In ;V - 2.303 log AT

.

Tables of natural logarithms should be at hand.

62. Exponential and logarithmic functions. The function of jr de-

fined by

(1) ij= e

jr

(c = 2.718 )

is called an exponential function. Its graph is shown in the figure.

The function is an increasing function for all values of j, as we shall

see later, and it is everywhere continuous.

From (1), we have, by definition, y

(2) y = In y.

The functions ef and In y are inverse functions

(Art. 39). Interchanging x and // in (2), we have ,

(3) y - In y,

in which y is now a logarithmic function of x. The graph is shown in

the figure. The function is not defined for negative values of x, nor

for x = 0. It is an increasing function for all

values of x > 0, and is everywhere continuous.

That is (Art. 17), for any value a of x greater

than zero

(4) lim In x = In a.x-*a

When x > 0, then y * oo, as remarked above. The ?y-axis is an

asymptote in the graph.The functions ax and log,, x (a > 0) have the same properties as

er and In x and graphs similar to the above.

63. Differentiation of a logarithm.

Let y = In v. (v > 0)

Differentiating by the General Rule (Art. 27 j, considering r as the

independent variable, we have

FIRST STEP. y + A// = In (v + Arj.

SECOND STEP. Ay/ = In (r -f Ar) Jn ?'

By(2),p.l

THIRD STEP. ^ = -i-in A +-)

A?' A/? \ r /

We cannot evaluate the limit of the right-hand member as it

stands by Art. 16, since the denominator Ar approaches zero as a

limit. But we can rewrite the equation as follows:

I M ull iplymtr l>y-

L r J

P,v(2),p.l

The expression following In is in the form of the right-hand mem-

ber of (2), Art. 61, with .r =r

FOURTH STEP. ~ ^- - in r = -a." r /;

r

When Ar-*0, - 0. Honco lim fl + V }? =r, by]

r Ar-(A ' /

(1), Art. 61. FsiiiK (1), Art. (>2, wo havc the result. J

Since r is a function of / and it is required to differentiate In v

with respect to .r, we must use formula (A), Art.. 38, for differentiating

a function of a function, namely,

dy_ _ dy_(ir

<7,r dv di

Substituting the value of ^ from the result of the Fourth Step,

we getdv

v d,.

. dx 1 dvX (In v) = = -

dx v v dx


The derivative of the natural logarithm of a function is equal to the

derivative of the function divided by the function (or the reciprocal of the

function times its derivative}.

Since log v log e In r, we have at once (IV, Art. 29)

vXadx v dx

64. Differentiation of the exponential function

Let //= a 1

. (a > 0)

Taking the logarithm of both sides to the base e, we get

In y v In a,

r = lM = -L.ln V .

In a In a

Differentiating with respect to y by formula X,

^^_L .1.(/-// In a y

From (C), Art. 39, relating to inverse functions, we get

*="*or

(1) ^ = lna-a".tfV

Since ?? is a function of .r and it is required to differentiate ar with

respect to j% we use formula (A), Art. 38. Thus we get

dii i ., dv-~ = In a a' -y-aa* ax

XI /. (a^)-lna-^- -

ax ax

When a e, In a = In e = 1, and XI becomes

The derivative of a constant with a variable exponent is equal to the

product of the natural logarithm of the constant, the constant with the

variable exponent, a,nd the derivative of the exponent.

65. Differentiation of the general exponential function. Proof of the

Power Rule

Let y = u''. (n > 0)

Taking the logarithm of both sides to the base e,

In'//= v In u

t

or y = e'^". By (3), Art. 61

Differentiating by formula XI a,

^ = <'"" 4- ("ton)ax ax

to+ lnw.rfr\ byVandX

u (Ijr (Is/

<-'- 1

s+-""s-7Vzt' derivative of a function 'with a variable exponent is equal to the

sum of the two results obtained bij first differentiating by VI, regarding

the exponent as eonstant, and ayam differentiating by XI, regarding the

function (is constant.

Let v = n, any constant ;then XII reduces to

f (,,")= WM "

i^.rf.r rf^

Thus we have shown that the Power Rule VI holds true for any

value of the constant n.

ILLUSTRATIVE EXAMPLE 1. Differentiate y In (:r24- a).

Arts.x 2 + a

2 xILLUSTRATIVE EXAMPLE 2. Differentiate y log

'

Solution. By (2), p. 1, we may write this

2,= log 2 r - log (1 + J 2

).

Then ^ = oj._ EJL

(i + a-2) bylllandXfllhendx 2 j dx l+x2 <te

ILLUSTRATIVE EXAMPLE 3. Differentiate y = a-{*\

Solution. & = In a a*** ~ (3 or2) by XI

= 6 x In a a**'. Ans.

ILLUSTRATIVE EXAMPLE 4. Differentiate y be^ 4 z2.

Solution. & = b 4- (cr* * * 2

) by IVdx dx J

= be fy+**(c*+x 2) by XL a

ILLUSTRATIVE EXAMPLE 5. Differentiate y = .r'T

.

Solution. 3^ = fz'' ' --(.r) + r'

TIn z (r

1) by XII

dx dx dx

= (^x^ l

-f x'J

In a* <'

Ans.

66. Logarithmic differentiation. Instead of applying X and X a at

once in differentiating logarithmic functions, we may sometimes

simplify the work by first making use of one of the formulas of (2)

on page 1. It is important that these formulas should be used when-

ever this is possible.

ILLUSTRATIVE EXAMPLE 1. Differentiate y = In Vl x 2.

Solution. By using (2), p. 1, we may write this in a form free from radicals, as

follows :

--ILLUSTRATIVE EXAMPLE 2. Differentiate y In

'

Solution. Simplifying by means of (2), p. 1,

y = i [In (1 + * 2)- In (1

-;

AThen i*

= - -I

by III and Xdx J

In differentiating an exponential function, especially a variable

with a variable exponent, the best plan is first to take the natural


logarithm of the function and then to differentiate. Thus Illustrative

Example 5, Art. 65, is solved more elegantly as follows :

ILLUSTRATIVE EXAMPLE 3. Differentiate y xfT.

Solution. Taking the natural logarithm of both sides,

In y = er In x. By (2), p. 1

Now differentiate both sides wit h respect to .r.

ordx

ILLUSTRATIVE EXAMPLE 4. Differentiate ?/-

(4 x'-' 7)V *


In y - (2 4- vV~' - 5 ) In (4 x j - 7 ).

Differentiating both sides with respect to .r,

=( 2 f V^^5) -L- f l n (4 r2 - 7)

rfj: 4 j- - i

4 j--'- 7

In the case of a function consisting of a number of factors it is

sometimes convenient to take the natural logarithm and simplify

by (2), p. 1, before differentiating. Thus,

ILLUSTRATIVE EXAMPLE 5. Differentiate y = -\/V (j --nor -2)


In y = i[ln (J- 1) + In (x

- 2) - In (x-

3) - In (x - 4)].

Differentiating both sides with respect to j,

i^/ = ir_i_ , _! i Lydx 2[ar-l j-2 x-3 x~4

PROBLEMS


1. ?/- In (ax + b}." dli

6. y=\ri*x\= (Inx)3

].

7. ij= In (2 ^ -

:>> x 2 +

9. ?/= In

dx ax -h ft

2. ?,= In (a.r-> + 6). 'A' = _JL_.

dx ox 2 + ft

3. ?/= ln(ax + ft)'.

^- 2fl-

dx ax +

4. ?/= In ax n

.

dx.] ..

5. y = In x:?.

dx x

(lli

^ ^n2 x

dx~

x

dy __ 6 x(x 1)

</.r~2.r:{ - 3^ 2-f 4"

8. //-log-. ^ :

J" dc

>r2

d?/.

s 2 dx x(l -h a-2)

10 ln V9 - ^>.r

2 ^ = ~ 2 Jdx 9-2 x 2

11. ?/- In (a.rVc7T7). ^ = 2 a + 3 y

.^ 2 r(a + x)

12. /O) - r In .r. /'(.r )= 1 + In #.

13. /(.r) = In (j- -f- VfTT2). /'(^) = ]

Vli __ i^ /u -r (// a,Lr aft

. > in - ^ (// a 2 -ft2/2

15. f(x) = .r2 In x 2

. f'(jc) = 2 y(l -f 2 In x).

16. y = <>"*. ^ = w-.dx

17. j/= 10- ^ = w 10- In 10.

dx

. y e _ xer .

19. ?/= -. ^ = _1.

^J dx ez

20 <? e^ ^s -. f

dt 2^/1'

21. r = ft2

''. ^ = 2 ft2" In ft.

d//

22. u = se".

23. v = -u

24. /,-^

25. /yIn (.rV)

26. ?/=~~ry

27. 7/= .r-Y '.

28. //= ~(c'

1 -t

dv e"(u -1)du

= .

ds JT

dy _ 2 cj

g = c "(2 ,-**).

a =|(,'

+ <- :) -

fir U"

30.*^.2 - 4 InJ

3Lf(j .

)== in^il^:.V J-- -f 1 -f JT

HINT. First rationalize the denominator.

32. y == s r.

^^ s'- // = r'r

(

^

/'(,)= -7==-

Vx 2-f 1

35. ?/= .r \/3 .r + a

36. y =

37. //= .r"

V4 + >r2

3 s + a 2 x

1 ,J

^-^amb

In Problems 38 47 find the value of-^

for the given value of x.

Ans. y' = %.38. y - In (jc-

39. /y= log (4 .r

- 3); J

40. y - JT In V.r -f 3 ; -r

41. y = xe- 2r\

.r = J.

. rt In .r2

442. 7/

=-y- ;

.r = 4.

' = 0.3474.

y' = 1.4319.

7/'= 0.

t/'= - 0.0483.

43. y =

44. y = log V25 - 4 x;x = 5.

45. n = 10 x7; x-4.

Find -~ for each of the following functions.djc'

2

.r'V.r a + 9,= 4.

\20

52. //= In

48. \iIn cjc.

49. ?/= f".

50. //x In a*.

51. 2/= <'

r-'.


.r -f a

53. y =~-

54. In

55.In Va 2

jc2

58. (^ : In V

59. 10' log/.

60. (ttf)'".

61. 2 s,s2

.

57. In'

6$Vl> / -f 3

67. The function sin x. The graph of

(1) y = sin x

is shown in the figure. Any value of x is assumed to be the measure

of an angle in radians (Art. 2).

n

i T 2

y = - l

Thus for x = 1, y = sin (1 radian) = sin 57 18' = 0.841. The

function sin x is defined and is continuous for all values of x. It is

important to note that sin x is a periodic function with the period

For 2 TT)= sin x.

That is, when the value of x is increased by a period, the value

of y is repeated.

The property of periodicity has the following interpretation in the

graph on page 97 : The portion of the curve for values of x from Oto2w(arc OQBRC in the figure) may he displaced parallel to OX either to

the right or left a distance equal to any multiple of the period 2 TT, and

it will be part of the locus in its new position.

68. Theorem. Before differentiating sin x (Art. G9) it is necessary

to prove that

,. sin x -

(B) hm = 1.

x -> %

This limit cannot be found by Art. 16. Weproceed by geometry and trigonometry.

Let O be the center of a circle whose radius

is unity. Let x angle AOM measured in radian?. Since the radius

is unity, arc AM = x, also.

Lay off arc AM f = arc AM and draw 71/7' and M'T tangent to

the circle at M and M' respectively. From geometry,

MM' < arc MAM 1 < MT ~\- M'T.

Or, by trigonometry,2 sin x < 2 x < 2 tan x.

Dividing through by 2 sin x, we get

1 <1

sin x cos x

Replacing each term by its reciprocal and reversing the inequality

signs, we have1 ^ sin x ^1 > > cos x.

Therefore when x is small, the value ofsin x

lies between 1 and cos x.

But when x * 0, lim cos x = cos =1, since cos x is continuous for

x = (see Art. 17). Thus we have proved (B).

It is interesting to note the behavior of this function from its

graph, the locus of equationoiri x

x

The function is not defined for x = 0. Let us, however, assign

the value 1 to it for x = 0. Then the function is defined and is

continuous for all values of x (see Art. 17).


69. Differentiation of sin v

Let y = sin v.

By the General Rule, Art. 27, considering v as the independent

variable, we have

FIRST STEP. y + A// = sin (v + Ar).

SECOND STEP. Ay = sin (v + A?0 sin r.

The right-hand member must be transformed in order to evaluate

the limit in the Fourth Step. To this end, use the formula from

(6), p. 3,

sin A - sin B = 2 cos \ (A + B) sin ] (A-

B),

setting A = v + Ar, B = 0.

Then ?, (A + #) = r + J A/?,

"

(A - B) = ?> A?;.

Substituting,

sin + A/0 sin v = 2 cos (v + *> Ar) sin ?, A?;.

/ \ \ \ n

Hence A//~ 2 cos ( ?' + -7^- j

sin

sin

Av,2

THIRD STEP.

FOURTH STEP. -p = cos v.dv

/sinf\Since lim I

-] 1, by Art. 68, and lim cos (M f )

- coi

Ar .Ol Ar I A,. .0 V -'

Substituting this value of(-~ in (4), Art. 38, we geta t/

r/v/ dv-~ = cos z;

ax ax

XIII /. (sin v) cos v

The statement of the corresponding rules will now be left to the

student.

70. The other trigonometric functions. The function cos x is de-

fined and is continuous for any value of x. It is periodic, with the

period 2 TT. The graph ofy_ cos x

is obtained from the graph of Art. 67 of sin x by taking the line

x = i TT as the #-axis.


The graph of y = tan x

(see figure) shows that the function tan x is discontinuous for an

infinite number of values of the independent variable x; namely,

when x = (n + ^)TT, where n denotes any positive or negative integer.

In fact, when x >

\ TT, tan x be-

comes infinite. Hut from the rela-

tion tan (TT + x) = tan x, we see that

the function has the period TT, and

the values x = (n + ])ir differ from

2 TT by a multiple of the period.

The function ctn x has the period

TT. It is defined and is continuous

for all values of x except x nir, n being any integer as before.

P'or these values ctn x becomes infinite. Finally, secx and esc x

are periodic, each with the period 2 TT. The former is discontinuous

only when x (n + 2)^, the latter only when x mr. The values

of x for which these functions become infinite determine vertical

asymptotes in the graphs.

71. Differentiation of cos v

Let y = cos v.

P>y (3), p. 2, this may be written

Differentiating by formula XIII,

duJl = c

= sin

I Since cos (^ r\= sin r, by (3), p. 2. 1

XIVd

f. . du

.'. (cos u) = sin vdx dx

72. Proofs of formulas XV-XIX. These formulas are readily de-

rived by expressing the function concerned in terms of other functions

whose derivatives have been found, and differentiating.

Proof of XV. Let y = tan v.

By (2), p. 2, this may be written

cost?

Differentiating by formula VII,

cos r (sin r) sin r -r1 (cos r)rf^_ dx_do*

y

dx cos- v

cos2r -7- + sin2 v -r~

_ dx dx

dr

- -^- - sec 2f-

;

- Using (2), p. 2cos- r dx

XV . .

dx VM^ ,-^ .

dx

To prove XVI-XIX, differentiate the form as given for each of

the functions below.

XVI. ctn v = ^ XVII. sec /? - -^- XVIII. esc v = -^tan ?> cos /> sin v

XIX. versine r = vers r = 1 cos v.

The details are left as exercises.

73. Comments. In the derivation of formulas I-XIX it was neces-

sary to apply the General Rule, Art. 27, only for the following.

TTT d f ^ </'//,

<]P (I'M"1

Al 1III -- (u + v w) -T- + -T Algebraic sum.

dx as d.r dx

TT d , ^ dv,

dvx-,-> , ,V -r- (uv) = w -7- + '

T"* Product,

dr ax ax

dn dr>

VII -(-)= Quotient.dx \

v'

v'2

VIII <!& = <!&. <toFunction of a function.

ax dv dx

IX -~ = -r- - Inverse functions,ax ax

a7

?/

cfe

X 4- HOK 1 = Logarithm.ax ?;

XIII jLfein^^eos,,^. Sine,dx dx

Not only do all the other formulas deduced depend on these, but

all we shall deduce hereafter depend on them as well. Hence we see

that the derivation of the fundamental formulas for differentiation

involves the calculation of only two limits of any difficulty, namely,

lim 5!5-? = i by Art. 68t - V

I

and lim (1 + v)v = e. By Art. 61

PROBLEMS


1. y = sin u.r2

.

Solution. & = cos oar 2 -f (ax*) by XIIIdx dx

[v-a^.]= 2 ax cos ax'2 . Ans.

2. ?/ tan Vl x.

Solution. ^ - sec-' \ 1- x ~ (1

- x)* by XV

= sec- \ 1- x \ ( 1 x)~ *( 1)

2 \ 1 - x

3. y = cos3 x.

Solution. This may also be written

y- (cos.r):{

.

^ = !i ( cos J^-T-C cosy) by VIdx dx

[r = cos.v and ?? 3.J

= 3 cos- T (- sin J-) by XIV

= 3 sin x cos 2 x. Ans.

4. ?/= sin ?u: sin" jc.

Solution. -^= sin nx ~ (sin x)

n4- sin" r ~ (sin wx) by V

Cw* fix ax

[?<= sin ??T and r sin" a-.]

= sin nx n (sin a:)""1 -r- (sin x)ax

4- sin" x cos nx -r- (nx) by VI and XIIIdx

= w sin nx sin"" 1 x cos x + n sin" x cos nx

= w sin" -1x(sin nx cos x 4 cos nx sin x)

= n sin"~

! x sin (H 4- l)x

5. y sin ax.

6. y = 3 cos 2 .r.

7. s = tan 3 t.

= 2 ctn --

17. P -

18. y = sin 2 x cos x.

19. ?/= In sin ax.

20. ?/= In Vcos 2 x.

21. y ea<f sin frx.

22. s f

'

cos 2 /.

23. y In tan -

24 . ?/=

yr = a cos or.

?/'= 6 sin 2 x.

s' = 3 sec- 3 /.

dn o ?'- = esc- -dv 2

yf = 4 sec 4 .r tan 4 x.

p' (76 esc bO ctn bO.

?/' sin .r cos .r.

c/.s-- sin 2 /

dp _ SOC L>.H ^

d ^? (tan30)S

dx Vsec .r

7/' cos x x sin x.

/'(0) = tan 2 0.

(jj) _ cos sin

To""

0-

?/' 2 cos 2 x cos x sin 2 x sin x.

y' a ctn ax.

T/ = - tan 2 x.

y' = c" r (a sin bx 4- fo cos ?>x).

' = - c '(2nm 2 / -f cos 2 0-

/ i A X 9 X7/' . ctn - sec 2 -

./= .

\ 1 sin x

25. f(0) = sin r0 + a) cos (6 - a). f'(0) = cos 2 (9.

26. /(x) = sin 2(IT

-x). !'(*) = - 2 sin (TT - x) cos (TT

-x).

27. p = i tan 3 - tan -f 6. p' = tan 40.

28. y = x**. ^r = * Hin *(?^x 4. Cos x Inx)-CLX \ X /

29. i/= (cos x)

x.

CLX \ X

V' = ?/On cos x - x tan x).

Find the second derivative of each of the following functions.

d 2y30. ?/

= sin kx. Arts. -~ k 2 sin kx.dx 2

31. p = \ cos 20. ^ = - cos 2 (9.

d$~

32. u = tan v. = 2 sec 1'r tan r.

(/?"

33. ?/= x cos x. ~-^( 2 sin j* .r cos j.

<ir-

*>/i _ sin r c/L'// _ 2 sin x 2 x cos r x 2 sin .r

J^ ?/* i- --------- .

x dx 2x-

{

d -'s*

35. .s = r' cos /.---- ~ 2 r' sin /.

a/ fc

36. = r' sin 2 /. ^ = - r '(3 sin 2/4-4 cos 2 /).

<//-

37. y/ cnr sin ftj-. ^ -c"

r

\(a- - />'-') sin bx 4 2 a?> cos

</j*"

Find - from each of the following equations.

38. 7,- cos (.r

-sin (x //) 1

<Lr ("' cos (.r 4 //)

40. cos // In (x 4 //).

dn -1

dx I 4 (x 4 //) sin //

In Problems 41-50 find the value of ~ for the given value of x (in

radians).

41.-//= x cos .r ; x = I. Ans. y' 1.841.

42. y = x sin ; x = 2. //'= 1.381.

43. y - In cos j- ; x - 0.5. ?/'= 0.546.

44. y - -; x = - 0.5. ?/ = - 3.639.

45. ?/= sin j- cos 2 .r ; .r 1. ?/' 1.754.

46. ?/= In Vtan :r ; .r \ TT. y' = 1.

47. # = r* sin x ; x - 2. ?/ = 3.643.

48. t/= 10c-'cos TTX; x = 1. ?/'

= 3.679.

49. y = 5 f 2 sin -^ ; .r = 2. ?/'= - 21.35.

50. y = 10 c" 10 sin 3 x ; j- = 1. y' = - 27.00.


74. Inverse trigonometric functions. From the equation

(1) ?/= sin x.

we may read "x is the measure of an angle in radians whose sine

equals //." For a central angle in a circle with radius unity, x equalsalso the intercepted arc (see Art. 2). The statement in quotationmarks is then abbreviated thus

(2) x arc sin//,

read"x equals an arc whose sine is //." Interchanging x and y in (2),

we obtain

(3) y arc sin .r,

and arc sin x is called the inverse xine function of :r. It is defined

for any value of x numerically less than or equal to 1. From (1)

and (2), it appears that sin x and arc sin // are inverse functions

(Art. 39).

Equation (,'J) is often written // sin '

.r, read "the inverse sine of x."

This notation is inconvenient, for the reason that sin- '

.r, as thus written,

might be read as sin s with the exponent 1.

Consider the value of y determined by x = \ in (3), that is, by

(4) y - arc sin \.

One value of// satisfying (4) is// I TT, since sin J TT

~ sin 30 =\.

A second value is y \\ TT, since sin;

r

;TT sin 150 ?,. To each of

these solutions any multi])le of 2 /r may be added or subtracted.

Hence the number of vain ex of // satisfying (4) ix without limit. The

function arc sin x is then said to be"multiple-valued/'

The graph of arc sin x (see figure) shows this property

well. When x = OM, y=MI> lt ,17/V, Ml**, -, MQi,

For most purposes in the calculus it is allowable

and advisable to select o'tie of the many values of y. Weselect, then, the value between ?, TT and \ TT, that is,

the smallest numerical value. Thus, for example,

(5) arc sin J = TT, arc sin = 0, arc sin ( 1) = \ TT.

The function arc sin x is now single-valued, and if

(6) y = arc sin x, then \ir~y~\ir. /

In the graph we confine ourselves to the arc QOP.


In the same manner each of the other inverse trigonometric func-

tions may be made single-valued. Thus, for arc cos x, if

(7) y = arc cos x, then ^ y ~ TT.

As examples,

arc cos \=

?, TT, arc cos (- \ )=

\ TT, arc cos ( 1) = TT. rj

From (6) and (1) we now have the identical relation

(8) arc sin x + arc cos x = TT.

In the graph of arc cos x (see figure), we confine our-

selves to the arc QP\P.Definitions establishing a single value for each of

the other inverse trigonometric functions are given

below.

75. Differentiation of arc sin u

Let y = arc sin r; ( \ TT i y ^ \ TT)

then v = sin ?y.

Differentiating with respect to // by XIII,

dv =

therefore~ -

^~- By (C), Art. 39

Since r is a function of .r, this may be substituted in (A), Art. 38,

_dx cos

'//dx

-os y \ 1 sin-'// \ 1 rH,the positive si^n of tho radical bpinpf)

taken, since cos//

is positive for all values of y between and inclusive.

XX .', (arc sin v) =

If y = arc sin :r, \f = - =

.

dx Vl - v2

The graph is the

arc QP of the figure. The slope becomes infinite at Q and

P, and equals 1 at 0. The function increases (y' > 0)

throughout the interval x = ~ 1 to or = 1.

76. Differentiation of arc cos u

Let y arc cos v; (0 g y ^ TT)

then = cos ?/.

TRANSCENDENTAL FUNCTIONS

Differentiating with respect to y by XIV,

107

therefore By (O, Art. 39

But since r is a function of .r, this may be substituted in 04),

Art. 38, givingrf?/ l rfr j

'dx""~

sin

.

dx

iin //= \ 1 -- cos"// = \ 1 r-', the plus sitfn of the radical being taken,

|

ince sin y is positive for all values of // between and TT inclusive.|

XXI, .

.-. (arc cos indx

dydx

If y = arc cos J, then y'

1When x increases from

1 to + 1 (arc PQ of the first figure on page

106). y decreases from TT to (//' < 0).

77. Differentiation of arc tan v. Let

(1) y = arc tan r; then

(2)v = tan ?/.

The function (1) becomes single-valued if

we choose the hast mmcrical value of //, that

is, a value between - 1 TT and i TT, corresponding to arc

the figure. Also, when v -> - oo, yy-' -

2 *" Iwhen v ""' + ' ?y

""'2

Or, symbolically,

(3) arc tan (+ oo) - \ TT, arc tan (-00) = -\ IT.

Differentiating (2) with respect to y by XV,

of

sec~ y ,

dy

and By (C), Art. 39dv sec 2

y

Since v is a function of x, this may be substituted in (A), Art. 38,

giving dy ^ 1 <fo = _j_jg..c/x

"~sec2

?y dx 1 + v'2 dx

[sec2

?/= 1 + tan'-' y = 1 + *'

L'.]

dv

XXII

1If y = arc tan x, then ?/'

=, and the function is an increasing

function for all values of x.

The function arc tan - furnishes a goodjr.

example of a discontinuous function. Con-

fining ourselves to one branch of the graph of

y arc tan -

we see that as jr. approaches zero from the left, ?/ approaches \ TT as a

limit, and as :r approaches zero from the right, ;// approaches -f i ?r as

a limit. Hence the function is discontinuous when x (Art. 17). Jts

value for x can be assigned at pleasure.

78. Differentiation of arc ctn v. Following

the method of the last article, we get

dv

XXIII ~(arc ctn i;)

= <**

1 -f i^

The function is single-valued if,

when // arc ctn v, < // < TT,

corresponding to the arc AB of the figure.

Also, if v -> + oo, //- > ()

;if r - . -

oo, //, jr. That is, symbolically,

arc ctn (-f oc)=

; arc ctn i oo;= TT.

79. Differentiation of arc sec v and arc esc v. Let

(1) y arc sec r.

This function is defined for all values of r except those lying be-tween - 1 and + 1. To make the function single-valued (see figure),

when v is positive, choose // between and J TT (arc AB) ;

when r is negative choose // between - TT and - 1 TT (arc CD).

Also, if v + oo, //>1

TT;

^^> // ^>7T.

Solving (1), ? = sec ?/.

Differentiating with respect to y by XVII,

if

drT- = sec vy tan y ;

therefore .

sec y tan yBy (C), Art 39y v '' u Ot7


Since v is a function of z, this may be substituted in (A), Art. 38,

giving

^ 1 dr _ 1 (h

dx sec y tan y dx v~ 1

I sec y = r,

I since tan

= r, and tan y = V sec 2y 1 = \ r-' 1, the plus sign of the radical being taken,"!

y is positive for all values of y between and ^ and between - 7T and ~-

XXIV /. - (arc sec v) =

dv

Differentiation of arc esc v. Let

2/= arc esc v

;

then v esc //.

Differentiating with respect to y by XVIII

and following the method of the last section,

we getdv

XXV -(arc

dxv) = - 1

The function y = arc esc r is defined for all values of v except those

lying between 1 and + 1, and is many-valued. To make the func-

tion single-valued (see figure above),

when v is positive, choose y between and \ TT (arc AB) :

when v is negative, choose y between TT and \TT (arc CD).

80. Differentiation of arc vers v}

27T

Let y = arc vers v;

*

then v = vers y.

Differentiating with respect to y by XIX,

dv

\a v

thereforedv sin y

By (C), Art. 39

* Defined only for values of r between and 2 inclusive, and many-valued. To makethe 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.

Since v is a function of x, this may be substituted in (4), Art. 38.

Kivindji = 1 dv = 1 dv

t

Ax~

sin y dx V2 c r'2 dx

ffjin y~ V'l - COR- //

~ V] -< 1

- versv/^ = V'J r ra,the plus sign of the radical!

[being taken, since sin y is positive for all values of y between arid TT inclusive.J

XXVI /. (arc vers v) =

dv

PROBLEMS


1. y arc tan ax'2 .

.

Solution.

<( -\

dx(ax

"}

dx 1 -f (

[r -

. Ans.

2. ?/= arc sin (3 jr 4 XH

).

-

Solution.

byXXn

by XX

3. y = arc sec ~./*

*"JL

\ 1- 9 .r-' -f 24 x 4 - 16 x r' vl- x'2

Solution.x- 4- 1 ix- + 1\-

by XXIV

i .r^-'-l I

x* - 1)2 j- (r= 4- O2 a;

__ __ = _. __!!_. Ans.x 2 + 1

4. i/= arc cos--

a

5. y = arc sec 2! .

a

dx a 2

clu a

dx xVx2 a2

6. y = arc ctn -a

7. y = arc sec -

8. ?/= arc esc 2 x.

9. y arc sin Vx.

10. 6 = arc vers p2

.

11. y x arc sin 2 .r.

As. 22 = ^4-ax a* -f .r2

dy __- 1

dx~Vl -

.r2

</?/- 1

<k* .rV4 x'2 - 1

^^ 1

d.r~~2Vx .r

2

N? = 2

</p V2 p2

dy<iy - o ,--- arc sin 1 x -I .

</x Vl - 4 x 2



28. arc sin Vx. 33. arc sec Vx.2

29. arc tan - 34. e* arc cos x.x

30. x arc cos - 35. In arc tan x.

01 arc ctn 2 x 0/,/ :

-31. 36. Varc sin 2 x.

x

32. arc versa-*). 37arc cos V^

Vx

PROBLEMS

Sketch the following curves, and find the slope at each point wherethe curve crosses the axes of coordinates.

1. u In .r. AUK. At (1, 0), m = 1.

2. y =r log jr. At (1, 0), m = 0.434.

3. ?/= In (4 - JT). At (3, 0), m = - 1

;

at x = 0, m = -J._

4. //= In V4 - x'2 .

5. Show that if ?/=

J, a(r"' -f c a), then ?/" = $-

2a*

Find the angles of intersection of each of the following pairs of curves.

6. //= In (.r + 1 ), y = In (7

- 2 x). Ans. 127 53'.

7. ?/= In (JT + 3), y = In (5

- x'2 ).

8. //- sin xjy - cos j. 109 28'.

9. y tan j% ?/= ctn jr. 53 8'.

10 ?/= cos j*, ?/

= sin 2 x.

Find the maximum, minimum, and inflectional points on the following

curves and draw their graphs.

ll.y = x\njr. Ans. Min. (->- -V

\e e)

10 _ J Min. (e, e) ;

In x inflectional point, (e2

, J e 2 ).

13. ?/= In (8 x - x2

). Max. (4, In 16).

14. y = xex . Min. (- 1, --);

inflectional point, ( 2, -V

15. y =


16. A submarine telegraph cable consists of a core of copper wires

with a covering made of nonconducting material. If u* denotes the ratio

of the radius of the core to the thickness of the covering, it is known that

the speed of signaling varies as .r-ln-- Show that the greatest speed1

J

is attained when JT = -p-

Vf

17. What is the minimum value of // ackj + be Ax? Anx. 2Vob.

18. Find the maximum point and the points of inflection of the graphof y = f~ J

~, and draw the curve.f

Am. Max. (0, 1) ; points of inflection, (-~. p

\ V2 Vf19. Show that the maximum rectangle with one side on the .r-axis

which can be inscribed under the curve in Problem 18 has two of its

vertices at the points of inflection.

Find the maximum, minimum, and inflectional points for the range

indicated, and sketch the following curves.

20. y I x sin x; (0 to 2 TT).

Am. Min. (\ TT,- 0.3424); max. (jf TT, 3.4840);

inflectional points, (0, 0), (TT, \ TT), (2 TT, TT).

21. y = 2 x tan x; (0 to ?r).

An*. Alax. (

1

, TT, 0.571); min. (] TT, 5.712);

inflectional points, (0, 0), (TT, 2 TT).

22. y = tan x 4 x; (0 to TT).

AW.S-. Min. f;1

, TT,- 2.457); max. (J TT,

-10.11);

inflectional points, (0, 0), (TI, 4 TT).

23. i/= 3 sin x 4 cos x

; (0 to 2 TT).

AUK. Max. (2.498, 5); min. (5.G40,-

5);

inflectional points, (0.927, 0), (4.069, 0).

24. y = x -f cos 2 x; (0 to TT).

25. ?/= sin TTJC cos TTJ

; (0 to 2).

26. 0= J.r + sin2.r; (0 to TT).

27. y = x - 2 cos 2 :r; (0 to TT).

28. y = TTx + sin irx; (0 to 2).

29. Show that the maximum value of y = a sin x 4- b cos x is Va 24- 6 2

.

30. The turning effect of a ship's rudder is shown theoretically to be

k cos 6 sin 26, where 6 is the angle the rudder makes with the keel, and

k is a constant. For what value of is the rudder most effective ?

An*. About 55.

31. Into a full conical wineglass of depth a and generating angle athere is carefully dropped a sphere of such size as to cause the greatest

overflow. Show that the radius of the sphere is

a sin asin a. + cos 2 u

32. Find the dimensions 01 the cylinder of maximum volume which

can be inscribed in a sphere of radius 6 in. (Use the angle 6 subtended

by the radius of the base* of the inscribed cylinder as a parameter. Thenr = 6 sin 6, h = 12 cos 0.)

33. Solve* Problem 32 if the convex surface of the cylinder is to be a

maximum, using the same parameter.

34. A body of weight W is dragged along a horizontal plane by meantof a force P whose line of action makes an angle JT with the plane. The

magnitude* of the force is given by the equation

m sin jc + cos jr

where m denotes the coefficient of friction. Show that the pull is least

when tan x = m.

35. If a projectile is fired from (> so as to strike an inclined plane whichmakes a constant angle cv with the* horizontal at (>, the range is given

by the formula2 ,,, (

.O[, ^ (Q _ (v)

A -------(/ cos- cv

where ?> and y are constants and is the angle of elevation. Calculate the

value of giving the maximum range* up the plane. AHX. 9} TT + -I CY.

36. For a square-headed screw with pitch and angle of friction</> the

efficiency is given by the formula

p __ tan

tan (0 + 0) +/where / is a constant. Find the value of for maximum efficiency when

is a known constant angle.

ADDITIONAL PROBLEMS

1. The curves ?/= jr In .r and // jr In (1 jc} intersect at the origin and

at another point A. Find the angle of intersection at .4. Ans. 103 30'.

2. Sketch the following curves on the same axes and find their angleof intersection. , -

{ x ( 2 x

?/= In ('-

- 1), ?/

= In(:*

.r- ' - 1

).Am. 32 28'.

3. The line A B is tangent to the curve whose equation is y = ex + 1

at A and crosses the .r-axis at B. Find the coordinates of A if the lengthof AB is a minimum. Ans. (0, 2).

CHAPTER VIII

APPLICATIONS TO PARAMETRIC EQUATIONS, POLAREQUATIONS, AND ROOTS

81. Parametric equations of a curve. Slope. The coordinates jr andy of a point on a curve are often expressed as functions of a third

variable, or parameter, /, in the form

(1)

Each value of t gives a value of .r and a value of // and determines a

point on the curve. Equations (1) are called parametric equationsof the curve. If we eliminate / from equations (1), we obtain the

rectangular equation of the curve. For example,

[_//= rsin/

are parametric equations of the circle in the

figure, t being the parameter. For if we elim-

inate t by squaring and adding the results, wehave .

, .>.,,.> JN3- -f ?/-

= r-(cos- / + sin- /) r~,

the rectangular equation of the circle. It is evident that if t varies

from to 2 TT, the point 7'O, //) will describe a complete circumference.

Since, from (1), y is a function of /, and t is a function (inverse)

of x, we haveby (A), Art. 38

dxdydt

''

dt rfx

dj J_.'

dt' dxy

dt

by (C), Art. 39

that is,

04)

dt

By this formula we may find the slope of a curve whose parametric

equations are given.115

ILLUSTRATIVE EXAMPLE 1. Find the equations of the tangent and normal,

and the lengths of the subtangent and subnormal to the ellipse*

x = a cos <

at the point where $ = 45.

Solution. The parameter being (/>, -77= a sin </>, -77 & cos

acp a45" in the given equations (13), we get xi = \ a\ 2, y\ ^ b\'2

as the point of contact, and the slope m becomes

a a

Substituting in f 1) and (2), Art. 43, and reducing, we get

bx -f ay V2 ab -.equation of the tangent,

V2(ax -by) a 2 - b'

2 = equation of the normal.

Substituting in (3) and (4), Art. 43,

\ b^ / -\ = -2

x - = length of subtangent,

i &V2I -i= J

~ V "length of subnormal.

\ a/ 2 a

ILLUSTRATIVE EXAMPLE 2. Given the equations of the cycloid f in parametric

form,

f4 )jx <i( ,

v '

l# = cos v),

being the variable parameter; find the lengths of the subtangent, subnormal,

and normal at the point (.n, //i) whore = 6} .

* As in the figure, draw the major and minor auxiliary circles of the ellipse. Through

two points /* and C on the same radius draw 7>M parallel toOY'and DP parallel to OA'.

These lines will intersect in a point />

(j, //) on the ellipse, ybecause _ . .,

y = OH = OB cos 4>= a cos />

and l/= AP = OD = Or sin

<jf>= 6 sin <,

or - = cos andj-

= sin </>.

Now squaring and adding, we get

-f%- cos2 </) + sin 2

(/>= 1,

o 2 ft-

the rectangular equation of the ellipse. <j> is some-

times called the eccentric angle of the point P of the

ellipse.

fThe path described by a point on the circumference of a circle which rolls without

sliding on a fixed straight line is called a cycloid. Let the radius of the rolling circle be a, Pthe tracing point, and M the point of contact with the fixed line X, which is called the

PARAMETRIC AND POLAR EQUATIONS 117

Solution. Differentiating, = a (I- cos t?),

~ = a sin0.du du

Substituting in (A), Art. 81,

dy sin . ,

dx=

1 - cos= m = slope at any polnt

When = 0i, y = y\ = a (I cos 0i), m = m\ =sin 0t

1 cos 0i

Following Art. 43, we find (see figure at foot of tnis page)

cos 0i )J

7W = subtangent = snNM = subnormal = a sin ft.

MP = length of normal = a\ 2(1- cos 0,) = 2 a sin \ 0,. By (5), p. 3.

In the figure, PA a sin 6\ (if = 0,) = the subnormal NM as above. Hence

the construction for the normal PM and tangent Ptt is as indicated.

Horizontal and vertical tangents. From (4), and referring to Art. 42,

we see that the values of the parameter / for the points of contact of

these tangent lines are determined thus :

Horizontal tangents : solve -& = for t.

Vertical tangents : solve -~ = for t.

ILLUSTRATIVE EXAMPLE 3. Find the points of

contact of the horizontal arid vertical tangents to

the cardioid (see figure)

'% a cos -J a cos 2 -

J a

^ y = a sin -\ a sin 2 0.

\

dxSolution. -73

= a( sin i

du+ sin 2 0) ;

^|= a(cos - cos 2 0).

Horizontal tangent*. Then cos - cos 20 = 0. Substituting (using (5), p. 3)

cos 20 = 2 cos 1' 0-1, and solving, we get -- 0, 120", 240".

Vertical tangents. Then - sin 6 -f sin 20 = 0. Substituting (using (5), p. 3)

sin 2 = 2 sin cos 0, and solving, = 0, 60, 180, 300.

base. If arc PM equals OM in longth, then P will touch at if the circle is rolled to the left.

We have, denoting the angle 7TM by 0,

x = ON - OM - NM = aO - a ain = a(0 - sin 0),

y- NP = fl/C _ AC - a - a cos - a(l - cos 6),

the parametric equations of the cy-

cloid, the angle 6 through which

the radius of the rolling circle turns

being the parameter. OD = 2 ira

is called the base of one arch of the

cycloid, and the point V is called

the vertex. Eliminating 6, we get

the rectangular equation

The common root = should he rejected. For both numerator and de-

nominator in (A) become zero, and the slope is indeterminate (see Art. 12). From

(5), x = y = when = 0. The point is called a cusp.

Substituting the other values in (5;, the results are as follows:

Horizontal tangents: points of contact (- \ a,_-J

a\73).

Vertical tangents: points of contact (\ a, -i \ </\/3 ), ( 2 a, 0).

Two vertical tangents coincide, forming a "double tangent" line.

These results agree with the figure.

PROBLEMS

Find the equations of the tangent and normal, and the lengths of the

subtangent and subnormal, to each of the following curves at the point

indicated.7\inynit \ormal Subt. Subn.

1. j-= /2,7/= 2/-f 1

;t-1. Am. .r-?/ + 2 = (), :r + ?/-4 = 0, 3, X.

2. x = f\ y = 3 /;

/ = - 1 . .r-

?/- 2 = 0, .r -f y -f 4 - 0,

-3,

- 3.

y-r2 = Q, G.r- y- 35 = 0, -6, --J.

= 0,-

1,- 9.

5. x = cos 2 0, //= sin

;=

J TT.

6. x rr /a, //= 2 - /

;/ = 1 . 11. j* = tan 0, //

= ctn;= 4 TT.

7. 3 j- = t'\ '1 y = /'-' ; / = 4

J!. 12. jr = -,'{ r ', //

= 2 r';

/ = 0.

8. j- = 6 /-

/'*, ?/= 2 / + 3

;t - 0. 13. .r -m 3 cos n', ?/

- 5 sin a-;a = J TT.

9. ,r = r, ?/= /

:{

-f 3 t; 1 = 1. 14. j- = sin 2 0, ?y= cos

;-

J TT.

10. r =p?/

= 2/; / = -!. 15. r=-ln(/-2),3?/ = /; / = 3.

In each of the following problems plot the curves and find the points

of contact of the horizontal and vertical tangents.

16. JT = 3 t /-\ ?/= / + !. Ann. Horizontal tangents, none ; vertical

tangents, (2, 2), (- 2, 0).

17. x = 3 - 4 sin 0, //= 4 + 3 cos 6.

Ann. Horizontal tangents, (3, 1), (3, 7);

vertical tangents, (7, 4), ( 1, 4).

18. jc = t~ - 2 /, //= r - 12 /. 20. x = sin 2 f, y = sin t.

19. s = 1i + r cos 0, .y A- + r siri 0. 21. .r = cos40, y = sin4 0.

In the following curves (figures in Chapter XXVI) find lengths of

(a) subtangent, (b) subnormal, (c) tangent, (d) normal, at any point.

rt _ m , l a = a (cos f -f / sin /),22. The curve

;

. .

'

t

?/ a (sin / / cos n.

4r?s. (a) 11 ctn /, (b) ?y tan /, (c) -^- , (d)--

sin / cos f

23. The hypocvcloid (astroid) /x =

fa c

.

OR''f

'

(Figure, p. 533)I//= 4 sin-*/.

.4?js. (a)-

// ctn t, ( ?/ tan t, (c)-~

(d)-^

sin / cos t

_, r ., fx~ rcos/,

24. The circle ^ . /[^/= r sin /.

nc rm. j- -j f x = </(- cos/ cos2 /), (Figure,25. The cardioid

{ \ . . . ./'h

..^ //= a (2 sin / sm J () P- 11

i "1+P26. The folium J 7., (Figure, p. 533)

I2/=

J^jT^-

r -^COS /,

27. The hyperbolic spiral ! (Figure, p. 534)i ?/=

ysin <.

L f

82. Parametric equations. Second derivative. Usin^ y/' as symbolfor the first derivative of y with respect to :r, then (A), Art. 81, will

give y' as a function of /,

(1) V'=

To find the second derivative y", use this formula (A) again, re-

placing y by y'. Then we have

dt

ifx=/(0,asin (1), Art. 81.

ILLUSTRATIVE EXAMPLE. Find y" for the cycloid (see Illustrative Example 2,

Art. 81)

fx = a(d -sin 6),

]^y a(l cos 0).

Solution. We found y' =l ^"^' and

^|= a(1 " cos ^

Also, differentiating,

c^_' __ (1- cos 6) cos sin 2

__ cos ,j_. __L__.^~ l-cos^) 15 (l-coc0J- (l-cos(9)

Substituting in (),

--- Ans.

Note that y" is negative, and the curve therefore is concave downward, as in the

figure for the cycloid, p. 117.

PROBLEMS

1. In each of the following examples find -~ and -7-^ in terms of t.

ax ax 2

(a)x = t-l,y = t* + l. Ans. = 2 /, ^ 2.

-. _ _._-/^ T _ o

, 7. __ <_!. (e) x = a cos /, |/

= b sin f.

^c; x - L i, y -3 (f) x = 2(1- sin 0, = 4 cos *.

/<ix T _ P>)J = (_

2.

(K) x = sin '> ?/= sin 2 <

v ;6

y2 (h) x = cos 2 /, y = sin ?.

2. Show that the curve r = sec ^, ?/= tan ^ has no point of inflection.

3. In each of the following examples plot the curve and find the maxi-

mum, minimum, and inflectional points :

(a) x = 2 a ctn 0, y = 2 a sin 2 6.

Arts. Max. (0, 2 a) ; points of inflection, ( =>"V"

(b) x = tan /, y siri t cos /.

Ans. Max. (1, -j); min. ( 1, \}\

p>oints of inflection, ( V3, - V (0, 0), /V3, V, , ,

4 / \ 4 /

83. Curvilineai motion. Velocity. When the parameter / in the

parametric equations (1), Art. 81, is the time, and the functions /(/)

and </>(/) are continuous, if /, varies continuously the point P(jr, ?/) will

trace the curve or path. We then have a curvilinear motion, and

(1) *=/(0, 0=0(0are called the equations of motion.

The velocity v of the moving point P(x, ?/) at any instant is de-

termined by its horizontal and vertical components.The horizontal component vx is equal to the velocity along OX

of the projection M of P, and is therefore the time rate of change of x.

Hence, from (C), Art. 51, when s is replaced by .r, we get

In the same way the vertical component vv,

or time rate of change of ?/, is

m *-J.'

Lay off the vectors vx and vv from P as in the figure, complete the

rectangle, and draw the diagonal from P. This is the required vector

velocity v. From the figure, its magnitude and direction are givenby the formulas .

vx dx

~dt

Comparing with (4), Art. 81, we see that tan r equals the slopeof the path at P. Therefore the direction of v lies along the tangentline at P. The magnitude of the vector velocity is called the speed.

84. Curvilinear motion. Component accelerations. In treatises onmechanics it is shown that in curvilinear motion the vector accelera-

tion a is not, like the vector velocity, directed along the tangent, buttoward the concave side of the path of motion. It may be resolvedinto a tangential component, a

t , and a normal component, an , where

_dv t _?/2at~dt'

a"~jT

(R is the radius of curvature. See Art. 105.)

The acceleration may also be resolved into components parallel tothe axes of coordinates. Following the same plan used in Art. 83for component velocities, we define the component accelerations paral-lel to OX and OY,

/m dvXm dv,,(F) , = ,

av- -

Also, if a rectangle is constructed with vertex P and sides ax and

avy then a. is the diagonal from P. Hence

(G) a

which gives the magnitude (always positive) of the vector accelera-

tion at any instant.

In Problem 1 below we make use of the equations of motion of a

projectile, which illustrate very well this and the preceding article.

PROBLEMS

1. Neglecting the resistance of the air, the equations of motion for a

projectile are

x = vi cos <t> t, y = 0i sin <t>t 16.1 1

2;

where v\ = initial velocity, <f> = angle of projection with

horizon, and t = time of flight in seconds, x and y being "of

measured in feet. Find the component velocities, com-

ponent accelerations, velocity, and acceleration (a) at any instant ; (b) at

the end of the first second, having given Vi = 100 ft. per second, </>= 30.

Find (c) the direction of motion at the end of the first second; (d) the

rectangular equation of the path.

Solution. From (C) and (Z>),

(a) TJ = PI cos </> ;vv v\ sin , sin </> -I- 1036.8 t2

.

From (F) and (G), a., = ;av 32.2 ;

a 32.2, direction downward.

(b) Substituting t 1, t'i 100, < = 30' in these results, we get

r, = 86.6 ft. per sec. ax = 0.

vy= 17.8 ft. per sec. a, = - 32.2 ft. per (sec.)

2.

v (speed; = 88.4 ft. per sec. a 32.2 ft. per (sec.;2

.

(c) r = arc tan - arc tan ^ 11" 37' = angle of direction of motion with

the horizontal.v* 8(j ' 6

(dj When v\ = 100, 4> 30, the equations of motion become

x - 50 /V3, y - 50 /- 16.1 /-.

Eliminating /, the result is y ^ -x~, a parabola.

\ 3 <5

2. Show that the rectangular equation of the path of the projectile

in Problem 1 is ir ^

y = .r tan -^ (1 + tan 20).r

2.

3. If a projectile be given an initial velocity of 160 ft. per second in a

direction inclined 45 with the horizontal, find (a) the component velocities

at the end of the second and fourth seconds; (b) the velocity and direc-

tion of motion at the same instants.

Ans. (a) When t 2, ?v 113.1 ft. per sec., vy = 48.7 ft. per sec.,

when / 4, rx = 113.1 ft. per sec., rv= 15.7 ft. per sec.

;

(b) when / = 2, r = 123.1 ft. per sec., r = 23 18',

when / = 4, r = 114.2 ft. per sec., r = 172 6'.

4. With the data as in Problem 3 find the greatest height reached bythe projectile. If the projectile strikes the ground at the same horizontal

level from which it started, find the time of flight and the angle of impact.

5. A projectile with an initial velocity of 160 ft. per second is hurled

at a vertical wall 480 ft. away. Show that the highest point on this wall

that can be hit is at a height above the o*-axis of 253 ft. What is </> for

this height? Ans. = 59.

6. If a point referred to rectangular coordinates moves so that

x a cos t + b and y = a sin t -f c,

show that its velocity has a constant magnitude.

7. If the path of a moving point is the sine curve

I

.r = at,

\ u b sin a/,

show (a) that the .r-component of the velocity is constant; (b) that theacceleration of the point at any instant is proportional to its distance fromthe /-axis.

8. Given the equations of motion .r = t\ //

-(t-

1)-'. ( a ) Find the

equation of the path in rectangular coordinates, (b) Draw the path withthe velocity and acceleration vectors for / -- A, / ~- 1, / 2. (c) For whatvalues of the time is the speed a minimum? (d) Where is the point whenits speed is 10 ft. per second ?

Ans. (a) Parabola, .r-1

+ i/~ = 1 ; (c) / = ; (d) (16, 9).

9. In uniform motion (speed constant) in a circle, show that the ac-

celeration at any point P is constant in magnitude and directed alongthe radius from P toward the center of the circle.

10. The equations of a curvilinear motion are j- ~ '2 cos 2 /, // 3 cos /.

(a) Show that the moving point oscillates on an arc of the parabola4 ?/-

- 9s - 18 0. Draw the path, (b) Draw the acceleration vectors

at the points where r 0. (c) Draw the velocity vector at the pointwhere the speed is a maximum.

Given the following equations of curvilinear motion, find at the

given instant rr ,v in r

;ax , <v,; , c\ ; position of point (coordinates) ;

direction

of motion. Also find the equation of the path in rectangular coordinates.

11. x = t2

, y = 2t; / = 2.

12. x = 2 /, ?/= /<; t= 1.

13. x = P, y = /-'; ? = 2.

14. x = 3 /, y = t'2 - 3

; / = 3.

15. x- 2 -t,y= 1 -f /-; / = 0.

16. x = a cos /, y = a sin /;

f =j

TT.

17. x = 4 sin /, y 2 cos /;

/ J TT.

18. x sin 2 t, y 2 cos /;

t = 1, w.

19. x = 2 sin t; y = cos 2 f

;/ = 1 T.

20. x = tan t; y = ctn /;

t = \ TV.

85. Polar coordinates. Angle between the radius vector and the tan-

gent line. Let the equation of a curve in polar coordinates p, 6, be

CD p


We proceed to prove the

Theorem. // \f/is the angle between the radius vector OP and the

tangent line at P, then

(H) tan $ =

where .

dd

Proof. Through P and a point

Q(p + Ap, + A0) on the curve near

P draw the secant line AB. Draw PRperpendicular to OQ.

Then (see figure) OQ = p+ Ap, angle POQ = A0, PR = p sin A0,

and OR = p cos A0. Also,

PR PR __ p sin A0"

RQ~OQ OR

~~

p + Ap p cos A0(2)

Denote by ^ the angle between the radius vector OP and the

tangent line PT. If we now let A0 approach zero as a limit, then

(a) the point Q will approach P ;

(b) the secant AB will turn about P and approach the tangent

line PT as a limiting position ;and

(c) the angle PQR will approach ^ as a limit.

Hence,rtx , . v p sin A0(3)v }

, . vtan ^ = hmY

o p + Ap p cos A0

To get this fraction in a form so that the theorems of Art. 16 will

apply, we transform it as shown in the following equations:

p sin A0 p sin A0~Pd

fSince from (5), p. 3, p - p cos A0 = p(l cos A0 )= 2 p sin 2 -1

sin A0p. A0

A0A0

[Dividing both numerator and denominator by A0 and factoring.]

PARAMETRIC AND POLAR EQUATIONS

When A0 - 0, then, by Art. 68,

125

sin

lim =1, and lim =L

Also,

\f\

lim sin = 0,

Hence the limits of numerator and denominator are, respectively,

p and p'. Thus (H) is proved.

To find the slope (tan r in the figure), proceed as follows. Take

rectangular axes OX, OY, as usual. Then for P(x, y) we have

(4) 3 p cos 0, ]1= p sin 0.

Using (1), these equations become parametric equations of the curve,

being the parameter. The slope is found by (4). Thus, from (4),

-? = p'

cos -p sin 0,

-& =p' sin + p cos 0,

du du

Slope of tangent = tan rp' sin 4 P cos

p' cos p sin 6

Formula (7) is easily verified for the figure* on page 124. For, from 1he triangle

SubstitutingOPT, T = 4- Then tan r = tan (0 f ^) =Uin

1 tan tan ^tan

""' *7

> tan \L = and reducing, we have (7).COS0 P

ILLUSTRATIVE EXAMPLE 1. Find tan ^ and the

slope for the cardioid p a(\ cos 0).

Solution. --- = p' a sin 6. Substituting in (77)

and (7),dd

i P (i(ltan \I/

t

= -^

P a si

tan r =

2 a sin \ cos J

= tan J 0. ((5), p. 3)

a sin 2 + "(1- cos 0) cos

a sin cos 6 a(l- cos i

cos cos 2

) sin

;

= tanl!0. ((5), (6), p. 3)sin 2 - sin 6

At P in the figure, ^ = angle OPT = =J angle XO7\ If the tangent

line PT is produced to cross the axis OX, forming with it the angle T, we have

angle XOP = 180 - angle OPT + r.

Therefore r = i- 180, and tan r = tan 0, as above ((3), p. 3).

NOTE. Formula (H) has been derived for the figure on page 124. In each prob-

lem, the relations between the angles i/s T, and 6 should be determined by examin-

ing the signs of their trigonometric functions and drawing a figure.

To find the angle of intersection (/>of two curves C and C" whose

equations are given in polar coordinates, we may proceed as follows :

Angle TPT' = angle OPT 1 - angle OPT,

or <j)=

\//' \[/.Hence

/ n * JLtan ^r'

- tant/r

(7) tan^ = z-p'1 + tan

t/r'tan

r/r

where tan \[/' and tan \{/are calculated by

(//) from the equations of the curves and

evaluated for the point of intersection.

ILLUSTRATIVE EXAMPLE 2. Find the angle of

intersection of the curves p = a sin 20, p a cos 2 0.

Solution. Solving the two equations simultaneously, we get, at the point of

intersection,tan 2 = 1, 2(9 = 45, = i>L

>

J.

From the first curve, using (H),

tan \[/

f =i tan 2 =

J, for = 22.^,

From the second curve,

tan ^ = -J cot 20 = -?,, for = 22\.

Substituting in (/),

tan </>= * -f =

$. /.</>= arc tan

The curves are shown in Chapter XXVI.

86. Lengths of polar subtangent and polar subnormal. Draw a line NTthrough the origin perpendicular to the radius vector of the point Pon the curve. If PT is the tangent and PN the normal

to the curve at P, then

OT length of polar subtangent,

and ON length of polar subnormal,

of the curve at P.

In the triangle OPT, tani// = ThereforeP

(1) OT = p tan^ - p2 ~ = length of polar subtangent.*

In the triangle OPN, tani/' = -- ThereforeON

(2) ON =tan

\[/

dp= - length of polar subnormal.

do* When increases with p,

- is positive and i/' is an acute angle, as in the above figure.tip

Then the subtangent OT is positive and is measured to the right of an observer placed at O

and looking along OP. When - is negative, th^ subtangent is negative and is measureddp

to the left of the observer.

The length of the polar tangent (= PT) and the length of the polarnormal ( PN) may be found from the figure, each being the hypotenuseof a right triangle.

ILLUSTRATIVE EXAMPLE. Find the lengths of the polar subtangont and polaisubnormal to the lemniscate p~ = a 2 cos 2 (figure in Chapter XX YJ).

Solution. Differentiating the equation of the curve, regarding p as an implicitfunction of 6, -,

dO dO

Substituting in (1) and (2), we get

Length of polar subtangent = -,

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 givenequation and substitute. Thus, in the above, p = i </ \ cos 20; therefore thelength of the polar subtangent = a ctn 2 Veos~2~0.

PROBLEMS

1. In the circle p a sin 0, find \p and r in terms of #.

A//*. t=0,T = 20.

2. In the parabola p = a sec- - show that r -f ^ -- TT.

3. Show thatt/'

is constant in the logarithmic spiral p e<l

. Since the

tangent makes a constant angle with the radius vector, this curve is also

called the equiangular spiral. (Figure, p. 534)

4. Show that tan \p= in the spiral of Archimedes, p = aO. Find

values of ty when = 2 TT and 4 TT. (Figure, p. 5.*J4)

-4 w.s'. ^ ^ 80 57' and 85 27'.

Find the slopes of the following curves at the points designated.

5. p = a(l- cos 0)',

= ~- Am. -1.

6. p = a sec 2; p 2 a. 3.

7. p = a sin 4 ; origin. 0, 1, oo, 1.

8. p2 = a 2 sin 4

; origin. 0, 1,oo

, 1.

9. p a sin 3; origin. 0, V3, V3.

10. p = a cos 3 ; origin.

11. p = acos20; origin. 14. p ad; 6 = ~-

13. p =


Find the angle of intersection between the following pairs of curves.

17. p cos 6 = 2 a, p 5 a sin 0. Am. arc tanif.

18. p = a sin 6, p = a sin 2 0. _j4w. At origin, 0; at two other points, arc tan 3V3.

n

19. p sin = 2 a, p = a sec- -. y4ws. 45 .

20. p = 4 cos 0, p = 4(1 - cos 0). 60.

21. p = (5 cos 0, p = 2(1 + cos 0). 30.

22. p = sin 0, p = cos 2 0. and arc tan ^

23. p'2 sin 2 = 4, p-'

= 16 sin 2 0. 60.

24. p = a(\ + cos 0), p = 6(1 - cos 0).

25. p = sin 2 0, p = cos 2 + 1.

26. p2 sin 2 = 8, p = 2 sec 0.

Show that the following pairs of curves intersect at right angles.

27. p = 2 sin 0, p 2 cos 0.

28. p - a0, p0 a.

29. p - a(l + cos 0), p = a(l - cos 0).

30. p2 sin 20 = a 2

, p- cos 20 = 6 2.

31. p = a sec- y p = /> esc'2 -

32. Find the lengths of the polar subtarigent, subnormal, tangent, and

normal of the spiral of Archimedes, p = aO.

Ann. Subtangent = > tangent = - Va 2 + p2

,

subnormal = a, normal = Va 2-f p

2.

The student should note the fact that the subnormal is constant.

33. Find the lengths of the polar subtangent, subnormal, tangent, and

normal in the logarithmic spiral p = a ft.

A NS. Subtangent = ~-> tangent = p^/l-ftj

subnormal = p In a, normal = pVl -f In 2a.

34. Show that the reciprocal spiral p0 = a has a constant polar sub-

tangent.

87. Real roots of equations. Graphical methods. A value of x which

satisfies the equation

(1) /(J)=0is called a root of the equation (or a root o//0)). A root of (1) maybe a real number or an imaginary (complex) number. Methods of

determining real roots approximately will now be developed.

Location and number of the roots.

FIRST METHOD. If the graph of /fr), that is, the locus of

(2) 0=/fr)

is constructed, following Art. 58, the intercepts or? the x-axis are the

roots. From the figure, therefore, we know at once the number of

roots and their approximate values.

ILLUSTRATIVE EXAMPLE. Locate all real roots of

(3) !'< - 9 j-2 f 24 JT- 7 = 0.

Solution. The graph has been con-

structed in Art. 58, Problem 1. It

crosses the axis of x between and 1.

Hence there is one real root betweenthese values, and then 4 are no other real roots.

The table gives the values of /(()) and /(I ), show-

ing a change of sign.

The table of values of x and // used in

plotting the graph may locate a root exactly,

namely, if y = for some value

of x. If not, the values of'//for

two successive values x = a,

x b may have opposite signs.

The corresponding points

X

a

XQ

b

y

Kb)

=

P(a,f(a)), Q(b,f(b)) are, therefore, on opposite sides of the r-axis, and

the graph of (2) joining these points will cross this axis. That is, a root

XQ will lie between a and b.

An exact statement of the principle involved here is as follows.

// a continuous function /fr) changes sign in an interval a < x < b

and if its derivative does not change sign, then the equation f(x) = has

one root, and only one, between a and b.

Location of a root by trial depends upon this principle. If a and 6

are not far apart, a further approximation can be found by interpo-

lation. This amounts to determining the intercept on the x-axis of the

chord PQ. That is, the portion of the graph joining P and Q is re-

placed, as an approximation, by the chord.

ILLUSTRATIVE EXAMPLE (CONTINUED). The root

between and 1 may be located by calculation more

closely between 0.3 and 0.4. See table. Let 0.3 + *

be this root. Then, by interpolation (proportion),

__ = -583 = Q32DiflF. 0.1 1.807

0.1 1.807'

Hence x 332 is a second approximation. This is the intercept on the x-axis

of the line that joins the points Q(0.4, 1.224) and P(0.3, - 0.583), which He

_z f(x) = y

0.4 E2240.3 + z(root)

0.3 - 0.583


on the graph of (3;. In the figure, MP = - 0.583,

NQ = 1.224, drawn to scale. The abscissas of M and

N are 0.3 and 0.4 respectively. Also, MC = z, and the

homologous sides of the similar triangles MPC and

PQIt give the above proportion.

For an algebraic equation, of which (3) is*"

an example, Homer's method is best adapted

to calculating a numerical root to any desired

degree of accuracy, as explained in textbooks on algebra.

88. Second method for locating real roots. The method of Art. 58 is

well adapted to constructing quickly the graph of f(jr). By this graph

the roots are located and their number determined. In many ex-

amples, however, the same result is attained more quickly by drawing

certain intersecting curves. The following example shows how this

is done.

10'

20" '10'"

Id 50 ItiO' 7U'

M)"( )

1 -Hc

Rad ion

I I'JO 200 'X.

ILLUSTRATIVE EXAMPLE. Determine the num-

ber of real roots U in radians) of the equation

(1) ctnj--j- = 0,

and locate the smallest root.

Solution. Transpose and write (1) thus :

(2) ctri x x.

If we draw the curves

(3) y ctn x and y x

on the same axes, the abscissas of Ihc points of in-

tersection will be roots of (1\ For, obviously,

eliminating y from (3) gives equation (1), from

which the values of x of the points of intersec-

tion are to be obtained.

In plotting it is well to lay off carefully both scales (degrees and radians ^ on OX.Number of solutions. The curve y

- ctn .r consists of an infinite number of branches

congruent to AQB of the figure i,see

Art. 70;. The line y = x will obviouslycross each branch. Hence the equation

(1) has an infinite number of solutions.

Using tables of natural cotangents

and radian equivalents of degrees, we

may locate the smallest root more closely

as shown in the table. By interpolation

we find x = 0.860. Ann.

The Second Method may be described as follows.

Transpose certain selected terms of f(.r] so that it becomes

(4) /iW-MO.Plot the curves

(5) y=fiW, 2/=/u(ar)

on the same axes, choosing suitable scales (wot necessarily the same on

both axes).

The number of points of intersection of these curres equate the num-

ber of real roots of f(x) = 0, and the abscissas of these points are the roots.

The terms selected in (4) can often be chosen so that one or both

of the curves in (5) are standard curves.

For example, to locate the real roots of

.r: < + 4 x - 5 = 0,

write the equation r{ = r>- 4 x.

The curves in (5) are now the standard curves

y = JC'A

, y ~ f> 4 x,

a cubical parabola and a straight line.

As a second example, consider

2 sin 2x4-1- x z = 0.

Write this in the form sin 2 x = I O 2 - 1 )

Then the curves in (5) are the standard curve

y = sin 2 x

and the parabola y = \ (** ~ l )

89. Newton's method. Having located a root, Newton's method

affords a procedure to calculate its approximate value.

The figure shows two points

on the graph of /(x) on opposite sides of the x-axis. Let PT be

the tangent line at P (Fig. a). The intercept a' of this line on the

x-o, x^a'\j

X[

/ = &' = & X

(,

FIG. a FIG. 6

x-axis is, obviously, an approximate value of the intercept of the

graph and hence of the corresponding root of f(x) == 0. Newton's

method determines the x-intercept of PT.

We find this intercept a' as follows. The coordinates of P are

xi = a, '//i /(a). The slope of PT is mi =f'(a). Hence the equation

of PT'is ((1), Art. 43)

0) y-f(a)^f'(a)(x-a).

Putting ?/= and solving for x(= a') gives Newton's formula for

approximation

,-a-M.

Having found a' by (K), we may substitute a' for a in the right-

hand member, obtaining f, ,,

<-<-$&as a second approximation. The process might be continued, giving

a sequence of values , ,

U, U-,

u/, Li )

* ' '

approaching the exact root.

Or the tangent may be drawn at Q (Fig. b). Then replacing a

in (K) by &, we obtain b', and from b' we obtain 6" etc., giving values

6', 6", 6'",-

approaching the exact root.

ILLUSTRATIVE EXAMPLE. Find the smallest root of

ctn x x =by Newton's method.

Solution. Here /(*) = ctn x j,

/'(x) = - csc'J x - 1 = - 2 - ctn 2 x.

By the illustrative example of Art. 88, we take a = 0.855. Then, by the table in

Art ' 88'

Afl>= 0.014.

Also, /'(a) = - 2 - ^0.869)* = - 2.76.

Hence, by (K), a' = 0.855 -f~~ = 0.860. Ans.

If we used 6 = 0.873 in (K), then

b'- 0.873 - ~^~ -

0.861.

By interpolation we found x = 0.860. The above results are valid to three

places of decimals.

From the figures on page 132 we observe that the graph crosses

the o>axis between the tangent PT and the chord PQ. Hence tlie exact

root lies between the value found &// Newton's method and that found by

interpolation. This statement is, however, subject to the reservation

that-f"(x) = has no root between a and b, that is, that there is no

point of inflection on the arc PQ.

PROBLEMS

Determine graphically the number and approximate location of the real

roots of each of the following equations. Calculate each root to two decimals.

1. JT* + 2 jr- 8 = 0. A //*. 1 .07.

2. .r'- 4 .r + 2 = 0. -

2.21, 0.54, 1.07.

3. .r3 - 8 x - 5 = 0. -

2.44,-

0.00, 3.10.

4. .ra - 3 .f

- 1 = 0. -1.53,

- 0.35, 1.88.

5. ^ - 3 jr* + 3 = 0. -0.88, 1.35, 2.53.

6. ^ + 3 x 2 - 10 = 0. 1.49.

7. .r3 - 3 x'

2 - 4 .r + 7 = 0. - 1.71, 1.14, 3.57.

8< j.3 + 9 j-2 _ 5 .r- 8 = 0. - 2.70, -

1 .30, 2.12.

9. 2 x* - 14 .r2-f 2 x -f 5 = 0. - 0.51, 0.71, 0.80.

10. x* -f 8 JT- 12 = 0.

- 2.30, 1.22.

11. x4 - 4 ^ - 6 x 2 + 20 .r -f 9 = 0. - 2.10, - 0.41, 2.41, 4.16.

12. r4 -I- 4 :r< - 6 x'2 - 20 x - 23 = 0. -4.00, 2.00.

Determine graphically the number of real roots of each of the following

equations. Calculate the smallest root (different from zero), using both

interpolation and Newton's formula.

13. cos x + x 0. Ans. One root ; x = - 0.739.

14. tan x x 0. Infinite number of roots.

15. cos 2 x - x = 0. One root ;x = 0.515.


16. 3 sin x - x = 0. Ans. Three roots ; x = 2.279.

17. 2 sin x - x 2 - 0. Two roots;x = 1.404.

18. cos x - 2 x 2 = 0. Two roots ; x = 0.635.

19. ctn x -f x 2 = 0. Infinite number of roots;x = 3.032.

20. 2 sin 2 x - x 0. Three roots ; x = 1.237.

21. sin x + x - 1 = 0. One root; x = 0.511.

22. cos x -f x 1 = 0. One root;f 0.

23. c'x cos j* = 0. Infinite number of roots

;x 1.29.

24. tan x log x 0. Infinite number of roots;x 3.65.

25. ('-f .r

- 3 = 0. One root;x = 0.792.

26. sin 3 x cos 2 .r = 0. Infinite number of roots; x = 0.314.

27. 2 sin J .r cos 2x^0. Infinite number of roots;

.r = 0.517.

28. tan x 2 c* 0. Infinite number of roots;x = 1.44.

29. The inner radius (r) and outer radius (R) in inches of the hollow

steel driving shaft of a steamer transmitting 77 horse power at a speed

of N revolutions per minute satisfy the relation 7i4 r4 = -^- If

77 = 2500, N = 160, r = 6, find R.Nir "

30. A cylindrical shell with a hemispherical end has a diameter d in.,

and contains V cu. in. The length of the cylindrical part is // in. Show

that (P + 3 hd'2 = ^-^ - Given // = 20, V = 800, find d. Ans. d = 6.777T

31. The quantity of water Q cu. ft. per second flowing over a weir of

width 7> ft. is given by Francis's formula

where 77 is the height of the water (the head) above the crest of the weir.

Given Q 12.5, H = 3, find 77. (Solve the formula for the factor 77^ andthen plot.) Ans. II = 1.23.

32. If V cu. ft. is the volume of 1 Ib. of superheated steam at a tem-

perature 7" F. and pressure P Ib. per square inch,

V = 0.6490 ^-~~4?.P PS *

Given V = 2,8, T = 420, find P. A

33. The chord r of an arc .s in a circle of radius r is

given approximately by the formula

-24 2

If r = 4 ft., c = 5.60 ft., find s. Am. s = 6.23 ft.


34. The area u of a circular segment whose arc s subtends the central

angle x (in radians) is u = % r'2(x sin x) Find the value of x if r = 8 in.

and u = 64 sq. in. Ans. x = 2.554 radians.

35. The volume V of a spherical segmentof one base of height CD = h is

V= ir(rh'2 -

i/?3).

Find h if r = 4 ft., V = 150 cu. ft.

Ans. h =4.32 ft.

36. The volume V of a spherical shell of

radius R and thickness / is

Derive this result. If R = 4 ft. and V is one

half the volume of a solid sphere of equal

radius, find t. Ans. t = 0.827 ft,

37. A solid wooden sphere of specific gravity S and diameter d sinks

in water to a depth h. Let x - and show that 2 .r:{

\\ x- -f tf 0.

(See Problem 35.) Find x for a maple ball for which S 0.786.

Ans. 0.702.

38. Find the smallest positive value of for which the curves p cos

and p = e~ e intersect. Find the angle of intersection at this point.

Ans. = 1.29 radians; 29(>

.

ADDITIONAL PROBLEMS

1. Find the angle of intersection of the curves p = 2 cos and p~

e

at the point of intersection farthest from the origin.

Ans. Point of intersection is = 0.54 radian ; 75 56'.

2. Show that the curve p = a sin4\ cuts itself at right angles.

3. Any radius vector of the cardioid p = a(l + cos 0) is OP. From the

center C of the circle p = a cos a radius of the circle CQ is drawn parallel

to OP and in the same direction. Prove that PQ is normal to the cardioid.

4. A square, one of whose diagonals lies along the polar axis, is cir-

cumscribed about the cardioid p = a(l- cos 0). Show that its area is

fi(2 + V3)o 2.

5. The path of a particle is the ellipse p = gp The particlei.

*~ C COS u

moves so that the radius vector p describes area at a constant time rate.

Find the ratio of the velocities of the particle at the ends of the major axis.

A 1 -Ans. -

1 + e

CHAPTER IX

DIFFERENTIALS

90. Introduction. Thus far we have represented the derivative of

y=f(x) by the notation

^-f'fcr)dx~ J(X) '

We have taken special pains to impress on the student that the

symbold/n

dx

was to be considered not as an ordinary fraction with dy as numeratorand dx as denominator, but as a single symbol denoting the limit of

the quotient

All

Ax

as Ax approaches zero as a limit.

Problems occur where it is important to give meanings to dx

and dy separately, and this is especially useful in applications of the

integral calculus. How this may be done is explained in what follows.

91. Definitions. If f'(x) is the derivative of f(x} for a particular

value of JT, and Ax is an arbitrarily chosen increment of x, then the

differential of J(x), denoted by the symbol df(s), is defined by the

equation

(A) d/(x)=

If now /(;r) = :r, then f'(x) = 1, and (4) reduces to

dx = Ax.

Thus, when x is the independent variable, the differential of x(= dx)is identical with Ax. Hence, if ?/ f(x), (A) may in general be written

in the form

(B) dy=f'(x)dx*=^-dx.dx

* On account of the position which the derivative f'(x) here occupies, it is sometimescalled the differential coefficient

136

DIFFERENTIALS 137

The differential of a function equals its derivative multiplied by the

differential of the independent variable.

Let us illustrate what this means geomet-

rically.

Draw the curve y =/(x).

Let/'(.r) be the value of the derivative at P. ~-4

Take dx = PQ, then

dy = f'(x)dx = tan r - PQ -^ PQ = QT.t y

Therefore dy, or df(x), is the increment (= QT) of the ordinale of the

tangent corresponding to dr.

This gives the following interpretation of the derivative as a

fraction.

// an arbitrarily chosen increment of the independent variable j/or

a point P (x, y) on the curve y = f('jc) be denoted by dx, then in the

derivative

dy denotes the corresponding increment of the ordinate of the tangent

line at P.

The student should note especially that the differential (= dy)

and the increment (= A?/) of the function corresponding to the same

value of d'jr(=A'jr) are not in general equal. For, in the figure,

dy = QT, but A'# = QP'.

92. Approximation of increments by means of differentials. From

Art. 91 it is clear that Ay(= QP' in the figure; and r///(- QT) are

approximately equal when dx(= PQ) is small. When only an ap-

proximate value of the increment of a function is desired, it is usually

easier to calculate the value of the corresponding differential and use

this value.

ILLUSTRATIVE EXAMPLE 1 . Find the volume approximately of a spherical shell

of outside diameter 10 in. and thickness ^ in.

Solution. The volume y of a sphere of diameter x is

(1, V=l*x\

Obviously, the exact volume of the shell is the difference AV between the

volumes of two solid spheres with diameters 10 in. and 9J in. respectively. Since

only an approximate value of AV is required, we find dV. From (1) and (),

dV = \ 7TX 2dx, since =

2 ^ 2-


Substituting x = 10, dx = -J, we obtain dV - 19.63 cu. in., approximately,

neglecting the sign, which merely means that V decreases as x decreases. The

exact value is AVr ~ 19.4 cu. in. Note that the approximation is close, for dx is

relatively Kmalltthat is, small a* compared with x (= 10*. The method would be

worthless otherwise.

ILLUSTRATIVE EXAMPLE 2. Calculate tan 4G', approximately, using differen-

tials, given tan 45 = 1, sec 45 = \7/2. 1 = 0.01745 radians.

Solution. Let y tan x. Then, by (J3),

When x changes to x + dx, y will change to y -f dy, approximately. In (1),

substitute x - \ IT (45"), dx = 0.0175. Then dy = 0.0350. Since y - tan 45 = 1,

y -f dy 1.0350 ~ tan 46 ", approximately. Anx.

(Four-place tables give tan 46r = 1.0355.)

93. Small errors. A second application of differentials is afforded

when small errors in calculation are to be determined.

ILLUSTRATIVE EXAMPLE 1. The diameter of a circle is found by measurement

to be 5.2 in., with a maximum error of 0.05 in. Find the approximate maximumerror in the area when calculated by the formula

(1) A\

TTX". (x diameter)

Solution. Obviously, the exact maximum error in A will be the change (AA)in its value found by (1) when x changes from 5.2 in. to 5.25 in. The approximateerror is the corresponding value of dA. Hence

dA -i irxdx - I TT x 5.2 x 0.05 - 0.41 sq. in. Ans.

Relative and percentage errors. If di( is the error in u, then the ratio

(2)=- the relative error ;v ' u

(3) 100 = the percentage error.

The relative error may be found directly by logarithmic differen-

tiation (Art. 66).

ILLUSTRATIVE EXAMPLE 2. Find the relative and percentage errors in the

preceding example.

Solution. Taking natural logarithms in (1),

log A log \ TT -f- 2 log x.

~.~ ,. ..1 dA 2 , dA 2dx

Differentiating, "TT"^"' an T~ =--A dx, x A x

Substituting x - 5.2, dx = 0.05, we find

Relative error in .4 = 0.0192 ; percentage error = l-fifa %. Ans.

The errors in calculation considered here are due to small errors

in the data upon which the calculation is based. The latter mayarise from lack of precision in the measurements or from other causes.

DIFFERENTIALS 139

PROBLEMS

1. If A is the area of a square of side x, find dA. Draw a figure show-

ing the square, dA, and A A. Aus. dA = 2 x dx.

2. Find an approximate formula for the area of a circular ring of

radius r and width dr. What is the exact formula ?

Ans. dA = 2 TIT dr; AA = ir(2 r + Ar)Ar.

3. What is the approximate error in the volume and surface of a cubeof edge 6 in. if an error of 0.02 in. is made in measuring the edge?

Ans. Volume, 2.16 cu. in. ; surface, 1.44 sq. in.

4. The formulas for the surface and volume of a sphere are S = 4 -rrr2

and VJ

Trr3

. If the radius is found to be 3 in. by measuring, (a) whatis the approximate maximum error in S and V if measurements are accu-

rate to 0.01 in.? (b) what is the maximum percentage error in each case?

Ans. (a) S, 0.24 TT sq. in. ; V, 0.36 TT cu. in.;

(b) s, I <;'< ; v, 1 ;,.

5. Show by means of differentials that

L(approximately).

x -f djc x x 2

6. Find an approximate formula for the volume of a thin cylindrical

shell with open ends if the radius is r, the length h, and the thickness L

Ans. 2 irrht.

7. A box is to be constructed in the form of a cube to hold 1000 cu. ft.

How accurately must the inner edge be made so that the volume will

be correct to within 3 cu. ft.? Ans. Error ^ 0.01 ft.

8. If ?/= a:

5 and the possible error in measuring x is 0.9 when x = 27,

what is the possible error in the value of ?/? Use this result to obtain ap-

proximate values of (27.9)^ and (26.1)". Am. 0.2; 9.2; 8.8.

Use differentials to find an approximate value of each of the following

expressions.

9. V66. 11. ^120. 13. T^. 15. \/35.

1 A __

10. V98. 12. \^1010. 14.

/

7=' - -

V51

17. If In 10 2.303, approximate In 10.2 by means of differentials.

Ans. 2.323.

18. If e 2 = 7.39, approximate e2 - 1 by means of differentials. Ans. 8.13.

19. Given sin 60 = 0.86603, cos 60 = 0.5, and 1 = 0.01745 radians,

use differentials to compute the values of each of the following functions

to four decimals : (a) sin 62; (b) cos 61 ; (c) sin 59 ; (d) cos 58.

Ans. (a) 0.8835; fb) 0.4849; fc) 0.8573; fd) 0.5302.


20. The time of one vibration of a pendulum is given by the formula

t* , where t is measured in seconds, g = 32.2, and /, the length of theV

pendulum, is measured in feet. Find (a) the length of a pendulum vibrat-

ing once a second ; (b) the change in / if the pendulum in (a) is lengthened

0.01 ft. ; (c) how much a clock with this error would lose or gain in a day.

An*, (a) 3.26 ft. ; (b) 0.00153 sec. ; (c)- 2 min. 12 sec.

21. How exactly must the diameter of a circle be measured in order

that the area shall be correct to within 1 per cent? Arts. Error ^ '

;.

22. Show that the relative error in the volume of a sphere, due to an error

in measuring the diameter, is three times the relative error in the radius.

23. Show that the relative error in the ?rth power of a number is

n times the relative error in the number. -

24. Show that the relative error in the nth root of a number is - times

the relative error in the number.

25. When a cubical block of metal is heated, each edge increases

r^ per cent per degree increase in temperature. Show that the surface

increases fa per cent per degree, and that the volume increases f\ per cent

per degree.

94. Formulas for finding" the differentials of functions. Since the

differential of a function is its derivative multiplied by the differen-

tial of the independent variable, it follows at once that the formulas

for finding differentials are the same as those for finding derivatives

given in Arts. 29 and 60, if we multiply each one by dx.

This gives

I d(c )- 0.

II d(j) - dy.

III d(u -f- v ?r) du + dv dw.

IV d(r/) = fd/>.

V d('ur) u dv + v du.

VI d(vn)= wn ~ } dv.

Via d(jr")= nj?

n - } dx.

vn

Vila rf(") = y-dpXv

XI d(a!

')= ar

In a dv.

XI a d(e*) = e*dv.

XII d(w.") = vu r ~ } du + In u ur dv.

XIII d(sin r) = cos vdv.

DIFFERENTIALS 141

XIV d(cos r) = sin vdv.

XV d(tan r) = sec 2 rdv. Etc.

XX d(arcsinr) = --^L==. Etc.

The term "differentiation" also includes the operation of findingdifferentials.

In finding differentials the easiest way is to find the derivative as

usual, and then multiply the result by dx.

ILLUSTRATIVE EXAMPLE 1. Find the differential of

Solution. dy = d(X + *

}- (

x * + :

\ara + 3/

(x'2 + 3 )dx (x + 3 )2 .r (te ( 3 - G j: - J-

1

)

. .

(x2 + 3)- (x- -f 3)-

ILLUSTRATIVE EXAMPLE 2. Find </// from

h'2x 2 ~

a~y~~

a'2 b'2 .

Solution. 2 6 2x dx - 2 a'2y dy ~ 0.

, b-jc j 4/. aij a/. Ans.

<*'2y

ILLUSTRATIVE EXAMPLE 3. P'ind dp from

p~ = a'2 cos 2 #.

Solution. 2 p dp = a'2 sin 20-2 <i#.

, a'2 sin 2 ,n.". op =--- (/. >inK.

P

ILLUSTRATIVE EXAMPLE 4. Find d[arc sin (3 /- 4 /')_).

Solution. d[arc sin (3 /- 4 /

:<

)

\ 1 - (3 /- 4 /')' \l-r-4

PROBLEMS

Find each of the following differentials.

1. y x* 3 x.

ax3. T/ Vax -f- 6.

-f 6

,- C/-I2 __ O r24/00 7 _ ' '* t-t A

. y = xv a 2 x 2. ay =

a 2

5. s = ae bt. ds abc ht

dt.

i j dv6. u = In rr. aw =

v

7. p sin ad. dp = a cos aO d6.

8. y = In sin x.

9. p cos 0.

10. = rf cos TT/.

,4ns. </?/ = ctn

f/p- (cos - 0sin 8)(W.

(In r'(cos TT/ Trsin

Find the differential of each of the following functions.

12. v = VV -f 1.

10,,, r==rlo. // /Va~ ~ x-

19. If jr2-f ?r = a-, show that f/?/

=

Find

20. 2 JT'~ -f 3 .r?/ -f 4 ?/2 = 20.

in terms of .r, //, and dx from each of the following equations.

_ _ (4 jr 4 3 ?/)dx

3x48//

21. x :{ + 6 x/r 4 2 //"= 10. 24. x :'

-f /' = a\

22. x -f 4 Vx// 4 12 //= a. 25. x - ?/

= rj4

^.

23. Vx 4- V?/ = Va. 26. sin (x //)- cos (x 4 ?/).

27. The legs of a right triangle are found by measurement to be

14.5 ft. and 21.4 ft. respectively. The maximum error in euch measure-

ment is 0.1 ft. Find the maximum error in degrees in calculating the

angle opposite the smaller side by using the formula for the tangent of

that angle.

95. Differential of the arc in rectangular coordinates. Let s be

the length of the arc AP measured from a fixed point A on the

curve. Denote the increment of s (= arc PQ) by As. The following

proof depends on the assumption that, as Q ap-

Ay

arc

From the figure, ^(1) (Chord PQ) 2 = (A*)

2 + (A?/)2

.

Multiplying and dhdding by (As)2 in the left-hand member and

dividing both members by (Ax)2

,we get

(2)/Chord PQV2

/ AsV_V As

\2 /A8\ 2_

/ VAx/~

DIFFERENTIALS 143

Now let Q approach P as a limiting position ; then Ax > andwe have

(3)\dx

Multiplying both members by dr% we get the result

(C) ds2 = dx2 + dy2

.

Or, if we extract the square root in (3) and multiply both members

by dr, i

From (C), we may readily show also thati

All these forms are useful.

From (>), since

I-T sec*"7%

we obtain r/,s = sec T djc, assuming the angle r to be acute. Hence

we may easily prove

(F)dx du= cos r,

-~ = sin r.ds ds

^ tan T COB r = sin r.

f/8 I

For later reference, we add the formulas, setting yf =

-J-

i y1

(G) COSTj>

sinr = 2p

(1 -f y'2)"2

(1 + yr2)*

If the angle r is obtuse (?/' < 0), a negative sign must be placed before

the denominators in (G) and before cos r in (F). y j

In the accompanying figure, I'Q = A:r = rfy,

FT is tangent at P, and r is acute. Angle

PQT is a right angle.

Therefore QT = tan r rfj = r/y/. By Art. 91

Then /> 71 = W/r 2 + rf?/-

- rfs. By (C)

The figure will help in memorizing the relations above.

The assumption made at the beginning of this article is proved

in Art. 99.


96. Differential of the arc in polar coordinates. From the relations

(1) j = p cos 0, y = p sin 6

between the rectangular and polar coordinates of a point, we obtain,

by V, XIII, XIV, of Art. 94,

(2) dx = cos dp p sin d6, dy = sin dp + p cos dO.

Substituting in (C), Art. 95, reducing, and extracting the square

root, we obtain the result

(H) ds = Vd/>2 + p

2d02.

This may be written

r /^*\2 o

dO.

The figure is drawn so that the angle \f/

between the radius vector OP and the tan-

gent line PT is acute (Art. 85). Also p, SO,and Sp (= OP' - op) are positive. Take

p for the independent variable. Then

Sp = dp. In the right triangle PQT, take

PQ = dp. Then QT = tan ^ r/p.

But

Therefore

hence

/tan ^ = p -7-P

= p-pdp

PT = Vrfp- + L> = rfs.

ILLUSTRATIVE EXAMPLE 1. Find the differential of the arc of the circle

x 2 + y* = ra .

Solution. Differentiating, ^ = _.aj: y

To find rfs in terms of a* we substitute in (/)), giving

To find ds in terms of y we substitute in (), giving

ILLUSTRATIVE EXAMPLE 2. Find the differential of the arc of the cycloidx = a (6

- sin 0), y = a(l - cos 0), in terms of and dd. (See Illustrative Ex-ample 2, Art. 81.)

Solution. Differentiating,

dx = a(l - cos 6)d6, dy = a sin dd,

DIFFERENTIALS 145

Substituting in (C)

ds'2 - a" (I- cos 6}~d6" -f a 2 sin- dO* - 2 a-(l - cos 0}dQ'

2.

From (5), Art. 2, 1 - cos 6 = 2 sin- .\ 0. Hence ds = 2 a sin J <J0. A?ks.

ILLUSTRATIVE EXAMPLE 3. Find the differential of the arc of the cardioid

p = a(l cos 0) in terms of 6.

Solution. Differentiating,(- = a sin 6.du

Substituting in (/) gives

cte = LaHl-cos0j- + a 2 sin- 0]adt? = a^

PROBLEMS

For each of the following curves find ds- in terms of jr and dr.

1. 2 y = .r2

. Ans. ds- = Vl + j* dr.

2. 7/2 = 2 px. (/.s - ^ dr.

3. & 2x 2-f a 2

//

2 - a 26 2. ds - \\

>, c 4 , o ;(^ -f-

4. 6 x?y = a*4

-f 3. d^ = ---

5. ?y~ In sec jr. ds- = sec jr djc.

6. a 2//= JT\ 9. 2 ?/

= cj-f

J.

7. a?/2

a*:?

. 10. //= sin jr.

8. V.r -f V/y = Va. 11. //= cos 2 x

For each of the following curves find ds- in terms of ?/ and dy.

,

:j?y

14. x^J

4- ?y^J = a*. <is =

^J~ d?/.

15. a 2?y= x3

.

16. ?y2 - 2 .r

- 3 ?/= 0. 17. 2 x//

2 -y2 -4 = 0.

For each of the following curves find ds, sin r, and cos r in terms of

t and dt.

18. x = 2 f -f 3, y t2 2. 20. x a sin /, y a cos J.

19. x = 3 /2

, ?y= 2 P. 21. x = 4 cos f, ?/ 3 sin /.

For each of the following curves find dx in terms of 6 and dO.

22. p a cos 0. A tin. dn a dO.

23. p ~ 5 cos 4- 12 sin 0. <& = 13 d8.

24. p = 1 - sin 0. ds = V2 - 2 sin e/0.

25. p 3 sin 4 cos 0. 30. p a cos 0.

26. p = 1 + cos 0. 31. p = 4 sin- {

f-

427. p^sec 2

^- 32. p =Z 1

'

28. p = 2 - cos 0. 33. p-

:

29. p = 2 4- 3 sin 0. 34. p =

1 + cos i

4

3 cos i

4

1- 3 cos 6

97. Velocity as the time-rate of change of arc. In the discussion of

curvilinear motion in Art. 83, the velocity, or, more correctly, the

speed v was given by (),

(1) fl2 = rx *+7y>.

By (C) and (D) in Art. 83, r, - > r, =d-%-

a/ a/

Substituting in (1), using differentials and (C), Art. 95, the result is

Extracting the square root, taking the positive sign, we have

dsV =

3T

Hence, in curvilinear motion the speed of the moving point is the

time-rate of change of the length of arc of the path.

This statement should be compared with the definition of velocityin rectilinear motion as the time-rate of change of distance (Art. 51).

98. Differentials as infinitesimals. In applied mathematics differ-

entials are often treated as infinitesimals (Art. 20), that is, as vari-

ables approaching zero as a limit. Conversely, relations betweeninfinitesimals are frequently established in which these are replaced

by differentials. The "principle of replacement"

involved here is

very useful.

DIFFERENTIALS 147

If x is the independent variable, we have seen that A.r = dr, andthus AJ may be replaced by djc in any equation. If A.r > 0, so will

also dy 0. On the other hand, A// and rf// are not in general equal.But, when x has a fixed value and A.r (= dx] is an infinitesimal, soalso is A?/, and, from (B), Art. 91, dy as well. Furthermore it is easyto prove the relation

(1) lim^-l.JLX-O W?/

Proof. Since \im=& =/'(?),AX ^ o A.r

we may write ^ =/'(.r) + /, if lim i = 0.

-AT Aj- .0

Clearing of fractions, and using (B),

A// = dy + i A.r.

Dividing both members by A//, and transposing, the result is

dy _ 1- AJ

i ^ _^

A?/ A//

Hence lim ^ = 1,

or also lim ^ = 1 Q. E. D.Ar -, A'// Ax-0"?/

We now state, without proof, the

Replacement Theorem. //? problems involving only the ratios of in-

finitesimals which simultaneously approach zero, an 'infinitesimal maybe replaced by a second infinitesimal so related lo it that the limit of their

ratio is unity.

From the above theorem, Ay/ may be replaced by dy, and, in

general, any increment by the corresponding differential.

In an equation which is hom.ogeneom in infinitesimals the above

theorem is simple in application.

ILLUSTRATIVE EXAMPLE 1. By (5), p. 3, if x - \ i, 1 - cos i = 2 sin 2J i. Let

{ be an infinitesimal. Then, by (B), Art. 68, sin i may be replaced by i, sin 2J i

by ^ i2

,and therefore 1 cos i by J i

2. Also tan i ( cos i sin i) may be replaced

by i.

ILLUSTRATIVE KxAMPLE 2. In (1), Art. 95, all quantities are infinitesimals

ultimately, since A.r -0. The equation is homogeneous, each term being of the

second degree. By the theorem, we may replace the infinitesimals as follows :

Chord PQ by arc PQ - As, and As by ds ; A// by ety ; and Ax by dx.

Then (1) becomes ds 2 - dx- + d?y2

, that is, (C).

99. Order of infinitesimals. Differentials of higher order. Let i and

; be infinitesimals which simultaneously approach zero, and let

lim 4 = L.i

If L is not zero, ? and j are said to be of the same order.

If L = 0, j is said to be of higher order than i.

If L becomes infinite, j is said to be of lower order than i.

Let L 1. Then j i is of higher order than i.

[hm(^)=Iim^-l)

=hmj-

1 =O.j

The converse also is true. In this case (L = 1), j is said to differ

from i by an infinitesimal of higher order.

For example, dy and AJ are of the same order if f'(jr] neither vanishes nor

becomes infinite. Then A// and AJ are of the same order, but A/y- dy is of higher

order than A.r. For this reason dij is called the "principal part of A?/." Obviously

powers of an infinitesimal i are of higher order than i.

ILLUSTRATIVE EXAMPLE. Prove the assumption of Art. 95,

Solution. In the figure we have, by geometry,

chord PQ <' arc PQ < PT + TQ.

Therefore, by division,

1 <chord PQ

"chord PQ chord PQ

Now

and hence

chord PQ = sec</> A.r, PT = sec r AJ, TQ = Ay -

dy,

PT A//- dy

Axchord PQ sec <' chord PQ

lim /_Z^_\ = lf ^limjTQ

^

\ = .

Jim/^rc

PQ \ = L

Differentiate of higher order. Let // =/(j). The equation

defines the second differential of //. If //" neither vanishes nor becomes

infinite, d-y is of the same order as Ax 2 and therefore of higher order

than dy. In a similar manner d*y,- -

,d n

y may be defined.

PROBLEM

In triangle ABC the sides a, b, c are infinitesimals which simultaneously

approach zero, and r is of higher order than b. Prove lim r = !

CHAPTER X

Let us find the

CURVATURE. RADIUS AND CIRCLE OF CURVATURE

100. Curvature. In Art. 55 the direction of bending of a curve

was discussed. The shape of a curve at a point (its ilatness or sharp-

ness) depends upon the rate of change of direction. This rate is called

the curvature at the point and is denoted by K.

mathematical expression for K.

In the figure, P f

is a second point on

a curve near P. When the point of con-

tact of the tangent line describes the

arc PP'(= As), the tangent line turns

through the angle AT. That is, AT is

the change in the inclination of the tan-

gent line. We now set down the follow-

ing definitions. ^r = average curvature of the arc PP'.

The curvature at P ( K) is the limiting value of the average curvature

when P' approaches P ax a limiting position, that is

(A) K = lim = = curvature at P.AS-. oAs ds

In formal terms the curvature is the rate of change of the inclination

with respect to the arc (compare Art. 50).

Since the angle AT is measured in radians and the length of arc As

in units of length, it follows that the unit of curvature at a point is one

radian per unit of length.

101. Curvature of a circle

Theorem. The curvature of a circle a,t any point equals the reciprocal

of the radius, and is therefore the same at

all points.

Proof. In the figure the angle AT be-

tween the tangent lines at P and P' equals

the central angle PCP' between the radii

As

_. = 1,As

~As ~As~ R

149

CP and CP'. Hence

AT angle

since the angle PCP' is measured in radians. That is, the averagecurvature of the arc PP' is equal to a constant. Letting As > 0, wehave the result stated in the theorem.

From the standpoint of curvature, the circle is the simplest curve,

since a circle bends at a uniform rate. Obviously, the curvature of a

straight line is everywhere zero.

102. Formulas for curvature; rectangular coordinates

Theorem. When the equation of a curve is given in rectangular co-

ordinates, then

(B) /; = __,,

where y' and //" are, w>/w/?/W//, the Jirst and second derivative* of ywith respect to jr.

Proof. Since r arc tun //, (//(

-^)\ dx'

differentiating, we have

(1 )

'yl _/>!_ . P>y XXII, Art. 60(Lc 1 -| //'"

Hut

(-) Y- (1 -h //''-'A l>.v (o>, Art. 95

Dividing (1) by (2) gives (B). Q.E.D.

EXKIUUSK. If ?/ is the indo])endent variable, show that

(O A' -"*"

,

(1 4- *'2)*

where jj and jc" are, respectively, the first and second derivatives

of .r with respect to //.

Formula (C) can be used as an alternative formula in cases wheredifferentiation with respect to y is simpler. Also, (B) fails when y' be-

comes infinite, that is, when the tangent at P is vertical. Then in (C)

.r' - and K - - .r".

Sign of K. Choosing the positive sign in the denominator of (B),

we see that A* and //" have like signs. That is, A' is positive or nega-tive according as the curve is concave upward or concave downward.

ILLUSTKATIVIO EXAMPLE 1. Find the curvature of the parabola y2 = 4 x

(a) at the point (1, '2); (Iri at the vertex.

> f1 /'>\ '7 -,/'

Solution. ii'= -< v

" = ^-(:) = -.rJK..

//

'

dx\ij/ y2

(a) When x = 1 and // ^ 2, then //'= !, y" = - L Substituting in

8

" =8 \2~ 0.177. Hence at (1, 2) the curve is concave downward and the

RADIUS AND CIRCLE OF CURVATURE 151

inclination of the tangent is changing at the rute 0.177 radian per unit arc. Since

0.177 radian = 10 7', the angle between the tangent lines at P( 1. 2) and at a point

Q such that arc PQ ~\ unit is approximately 10 .

(b) At the vertex (0, 0), //' becomes infinite. Hence use (C).

' I " I (J-ll

1 r 1

2'

2 dy 2 2

ILLUSTRATIVE EXAMPLE 2. Find K for the cycloid (see Art. 81)

x a(0 sin 0], // (!-- cos 0).

Solution. In Illustrative Example 2, Art. 81, we found

, _. sin

.1 cos 6

>)

Hence 1 +?/'-- T-

1 cos

Also, in the Illustrative Example, Art. 82, it was shown that

_ 1

Substituting in (B),

'

c(} ~(>(>s 0)

~

1 1

2 a\ 2 2 cos 4 a sin \

103. Special formula for parametric equations. From equation

Art. 81, we have, by differentiation,

djr d-y

,- dyf dtdf2 dtdl'2

(lj ~j"^^ '

(ft)

Whence, using (B), Art. 82, and substituting in (B), Art. 102, and

reducing, we obtain

where the accents indicate derivatives with respect to /, ;that is,

, dx ,, d2x , di/ ,, d-y'?' . x t II 11

~ ~w* Ti ** lj'

'*]i ' 7i'>

dt dt 2 di dt~

Formula (D) is convenient, but it is often better to proceed as in

Illustrative Example 2, Art. 102, finding //'as in Art. 81, '//" as in

Art. 82, and substituting directly in (B).

104. Formula for curvature ; polar coordinates

Theorem. When the equation of a curve is given in polar coordinates,

p2. 2 P'2 PP"

(E) K = P ^^*f-,

(P2 + P'

2)"2

where p' and p" are, respectively, the first and second derivatives of pwith respect to 0.

Proof. By (/), Art. 85, r = + ^.

Hence

<> s-'+S-Also, by (#), Art. 85, ^ = arc tan

Hence

Then, by (1),

Dividing (2) by (3) gives (). Q.E.D.

ILLUSTRATIVE EXAMPLE. Find the curvature of the logarithmic spiral p = e"9

at any point.

Solution.(-t~ -

p' = acat) = ap ; ^ = p" </->"" = a- p.

Substituting in (),p\ 1 -f a-

105. Radius of curvature. The radius of curvature R at a pointon a curve equals ( he reciprocal of the curvature at that point. Hence,from (),

(F) Jf = I == (l2l.y

ILLUSTRATIVE EXAMPLE. Find the radius of curvature at any point of the

/ _ i\

catenary y =- -\ra -f r / (figure in Chapter XXVI).

/- -\ / - r

\

a . ?/

' = I\ (, ._

,, uj.

y" = J_

\ ffl ^_ tf a) = JL.2 2 a a-

1 -fl/'2 ^ 1 +7\f"-f / =Mfa+f

2

Solution.

= - Ans.a

106. Railroad or transition curves. In laying out the curves on arailroad it will not do, on account of the high speed of trains, to pass

abruptly from a straight stretch of track to a circular curve. In

order to make the change of curvature gradual, engineers make useof transition curves to connect the straight part of a track with a


circular track. This curve should have zero curvature at its point

of junction with the straight track and the curvature of the circular

track where it joins the latter. Arcs of cubical parabolas are generally

employed as transition curves.

ILLUSTRATIVE EXAMPLE. The transition curve on a railway track has the shapeof an arc of the cubical parabola //

=J ,r

l

. At what rate is a car on this track

changing its direction (I mi. = unit of length) when it is passing through (a) the

point (3, 9)? (b) the point (2, J)? (c) the point (1, J)?

Solution. ^ = ^. ^ = 2x.dx dx 2

Substituting in (B), A" =(1 -f

(a) At (3, 9), K = radian per mile = 28' per mile. ATM.

(812)2

(b) At (2, 5), K = -~ radian per mile = 3 16' per mile. Ans.

(17)3

(c) At (1 , I \ K = -^ = U radian per mile - 40" 30' per mile. Am.(2)2 V2

107. Circle of curvature. Consider any point /' on the curve (\

The tangent line drawn to the curve at P has the same slope as the

curve itself at P (Art. 42). In an analogous /0

manner we may construct for each point of the

curve a tangent circle whose curvature is the /

same as the curvature of the curve itself at that',

point. To do this, proceed as follows. Draw \the normal to the curve at P on the concave

side of the curve. Lay off on this normal the dis-

tance PC = radius of curvature (= K) at P. With c as a center draw

the circle passing through P. The curvature of this circle is then

*-*which also equals the curvature of the curve itself at P. The circle

so constructed is called the circle of curvature for the point P on

the curve.

In general, the circle of curvature of a curve at a point will cross

the curve at that point. This is illustrated in the above figure.

(Compare with the tangent line at a point of inflection (Art. 57).)

Just as the tangent at P shows the direction of the curve at P, so

the circle of curvature at P aids us very materially in forming a geo-

metric concept of the curvature of the curve at P, the rate of changeof direction of the curve and of the circle being the same at P.


In a subsequent section (Art. 114) the circle of curvature will

be defined as the limiting position of a secant circle, a definition

analogous to that of the tangent

given in Art. 28.

ILLUSTRATIVE EXAMPLE. Find the ra-

dius of curvature at the point (U, 4 ) on the

equilateral hyperbola xy 12, and draw the

corresponding circle of curvature.

~ i * dySolution. - -

ax

d^y 2 v~~l - *.dx- jr*

For W. 4).ix .'J dx'2 9

. K Jl^lLi*j=5..vM

-4

The circle of curvature crosses the curve at two points.

iLLi'STHATivE K.VAMi'LE 2. Find /i at (2, 1 ) for the hyperbola

j j-f 4 ///

- 2 //-'- 10.

Solution. I )iiferent iiitui^, regarding // us an implicit function of x, we get

.r } 2// f 2 JTIJ'

- 2 ////'- ().

Differentiating again, regarding ij and //' as implicit functions of x, we get

1 4- 4 //'- 2

.//"-' -f 2(.r-

//)//'- 0.

Substituting the given values .r 2, // 1, we find ?/'-- 2, y"

-l--.

Hence, by (F), U -?|\/5. .\.s.

The method of this example (namely, regarding // and //' as implicit functions

of .r) can often be used to advantage when only the numerical values of ?/' and y"are required, and not general expressions for them in terms of x and y.

PROBLEMS

Find the radius of curvature for each of the following curves at the

point indicated. I )raw the curve and the corresponding circle of curvature.

1. 2 //-- .r

j; i(), 0). An*. 7i = 1.

2. 6 //= .r

;

; (2, .',). ft = g Vs.

3. ?/L> ~

.r;

; (1, 1).

4. // sin .r ; ^ TT, 1).

5. //= r

j; (0, 1).

6. .r2 - 4 //

2 = 9; (5, 2). 8. y = 2 sin 2 x ; ( J TT, 2).

7. //2 = ^-f8; (1,8). 9. y = tanx; (| TT, 1).


Calculate the radius of curvature at any point (.r\, y\] on each of

the following curves.

10.,, = ^. A,:,. K =(>

11. jr'= 2 JU-.

12. Wx* - aW - a*b*.*

13. b-jr- -f a-y- a'2b'2

.

14. ,* + ** = *. /, = ^' +/>'a*

15. j-3 + 2/

S = aI A* = 3 (an //,)*.

16. x = r arc vers - V2 rz/ z/~. /i 2\/2 n/i.

17. ?y In sec x. A*~ sec jv

18. If the point of contact of the tangent line at (2, 4) to the parabola

?/* 8 x moves along the curve a distance As ().l, through what angle,

approximately, will the tangent line turn? (Use differentials. )

19. The inclination of the curve 27 // .r{ at the point .\(IJ, 1) is 45.

Use differentials to find approximately the inclination of the curve at the

point H on the curve such that the distance along the curve from A to Bis A.s- = 0.2 units.

Calculate the radius of curvature at any point (pi, (M on each of

the following curves.

20. The circle p = a sin 0. An*. R \ a.3

21. The spiral of Archimedes p - ad. (Fig., p. 5:M) K =-^ _^ 2 ^

22. The cardioid p = a (I- cos 0). (Fig., p. 5153; # = 3^ a Pi-

a 2

23. The lemniscate p2 = a 2 cos 2 0. (Fig., p. 5.'52) A' =

^-^

24. The parabola p = a sec 2J 0. (Fig., p. f>:J7) R = 2a sec3 J 0j.

25. The curve p = a sin3J 0. ^ = 4 sin 2

J 0^

m r> _flf5-4cos0,)j

26. The tnsectnx p = 2 a cos a. ^(

fi

-7

27. The equilateral hyperbola p2 cos 2 = a 2

.

a(l - e 2 ) p aft'

.3

oo TU -

28. The conic p = -

1 e cos fl - -' cos L )

{

Find the radius of curvature for each of the following curves at the

point indicated. Draw the curve and the corresponding circle of curvature.

29. jr = 2 /, ?/= t

2 - 1;

/ = 1. A MS. ft = 4>/2.

30. jr = 3 t2

, //= 3 /

-/

: <

;t = 1. tf = .

31. .r = 2 r ', //= 2 f

'

;/ = 0. 72 = 2 V2.

32. -r = a cos /,, ?/= a sin /

;/ t\. ft a.

33. jr~ 2 /, ?/

= -;

/ - 1 . 36. jr = 2 sin /, ;//= cos 2 /

;t = TT.

34. .r - /- -f 1, //= t

' - 1;

/ = 1 . 37. x - tan /, //= ctn /

;/ - \ IT.

35. x 4 cos /, ?/= 2 sin /;// = !. 38. r / sin /, // 1 cos /

;/ TT.

39. Find the radius of curvature at any point (f = /i ) on the hypocy-cloid j = a cos- 1

/, // a sin :!

/. ,4//,s. ft ;i a sin /i cos /i.

40. Find the radius of curvature at any point (t= t\) on the involute

of the circle.r = aloe* / + / sin /;,

//~ a (sin / t cos /). An*, ft at\.

41. Find the point on the curve ?/= r' where the curvature is a

maximum. An$. s 0.347.

42. Find the points on the curve 3 y ~ .r! 2 x where the curvature

is a maximum. ,4^/.^. jr 0.931.

43. Show that the radius of curvature becomes infinito at a point of

inflection.

44. Given the curve // 3 jr .r\

(a) Find the radius of curvature at the maximum point of the curve

and draw the corresponding circle of curvature.

(b) Prove that the maximum point of the curve is not the point of

maximum curvature.

(c) Find to the nearest hundredth of a unit the abscissa of the point

of maximum curvature. An*, .r 1.01.

45. Find the radius of curvature at each maximum and minimumpoint on the curve //

~ x 4 2 .rL>

. Draw the curve and the circles of cur-

vature. Find the points on the curve where the radius of curvature is a

minimum.

46. Show that at a point of minimum radius of curvature on the

curve y = f(x] we have

'^Vl,dx) \

47. Show that the curvature of the cubical parabola 3 a 2?/= x3 in-

creases from zero to a maximum value when x increases from zero to4 _____

I aV125. Find the minimum value of the radius of curvature.

An*. 0.983 a.


108. Center of curvature. The tanpent line at P(r, y) has the

property that x, y, and yf have the same values at P for the tangent

line and the curve. The circle of curvature at P has a similar prop-

erty; namely, x, y, y', and //" have the same values at P for the

circle of curvature and the curve.

DEFINITION. The center of curvature (a, /j) for a point P(x, y) on

a curve is the center of the circle of curvature.

Theorem. The coordinates (a, /j) of the center of curvature for

P(x, y) are

y y

Proof. The equation of the circle of curvature is

(1) Cr-a^+fo-jCO-^/i'-,

\\here K is given by (F). Differentiating (1),

/ x a ,, R~(2) ?/"

= -y-P "

(y

From the second of these equations, after substituting the value

of R from (F), we obtain

- , ' y* I

From the first of equations (2), we get, using (3),

Solving in (3) for /3, in (4) for a, we

have (G). Q.E.D.

EXERCISE 1. Work out (G) di-

rectly from the accompanying figure,

using (G), Art. 95. (a = x R sin r,

/3= y -f- R cos r, etc.)

EXERCISE 2. If x' and x" are, re-

spectively, the first and second deriv-

atives of x with respect to y, derive

(G) in the form

a ~x" x"

Formulas (H) may be used when y' becomes infinite, or if dif-

ferentiation with respect to y is simpler.


ILLUSTRATIVE EXAMPLE. Find the coordinates of the center of curvature of

the parabola y'2 4 px corresponding (a) to any point on the curve; (b) to the

vertex.

1

"'

Solution. Use (//). Then x' =-j-

. x" =

y'2 + 4

Hence a = x = 3 x + 2 p,

Therefore (a) (3 x -f 2 p, ~~~ t } Ls the center of

V 4 p-/

curvature corresponding to any point on the curve.

(b) (2 p, 0) is the renter of curvature correspondingto the vertex (0, Oj.

From Art. 57 we know that at a point of inflection (as Q in the

next figure)

Therefore, by (B), Art. 102, the curvature K = 0; and from (F),

Art. 105, and (G), Art. 108, we see that in general a, 0, and R in-

crease without limit as the second derivative approaches zero, unless

the tangent line is vertical. That is, if we sup-

pose P with its tangent to move along the curve

to P', at the point of inflection Q the curvature

is zero, the rotation of the tangent is momen-tarily arrested, and as the direction of rotation

changes, the center of curvature moves out in-

definitely and the radius of curvature becomesinfinite.

109. Evolutes. The locus of the centers of

curvature of a given curve is called the evolnte

of that curve. Consider the circle of curva-

ture at a point P on a curve. If P moves along the curve, we maysuppose the corresponding circle of curvature to roll along the curvewith 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 bythe center of the circle is the evolute of PP7 .

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 equationof the evolute in terms of the parameter x.


To find the ordinary rectangular equation of the evolute we elimi-

nate x and y between the two expressions and the equation of the

given curve. No general process of elimination can be given that

will apply in all cases, the method to be adopted depending on the

form of the given equation. In a large number of cases, however,

the student can find the rectangular equation of the evolute by taking

the following steps.

General directions for finding the equation of the evolute in rec-

tangular coordinates.

FIRST STEP. Find a and /3from (G) or (H), Art. 108.

SECOND STEP. Solve the two resulting equations for y and y in terms

of a and /3.

THIRD STEP. Substitute these values of x and y in the given equation

and reduce. This gives a relation between the variables a and (3 which

is the equation of the evolute.

ILLUSTRATIVE EXAMPLE 1. Find the equation of the evolute of the parabola

V=4p*.-

,,

C,,l

dx y dx-

a = 3 ar + 2 p, = -

Solution.

First Step.

Second Step.

Third Step. (4

Remembering that a denotes the abscissa and p the

ordinate of a rectangular system of coordinates, we see

that the evolute of the parabola AOB is the semicubical parabola DC'E, the

centers of curvature for O, P, PI, P>2 being at C', C, d, C2 respectively.

ILLUSTRATIVE EXAMPLE 2. Find the equation of the evolute of the ellipse

^2X 24. n -iyi

= aw.

c . ..Solution. -- = -- .

dc a 2!/

T-Ta 2

?/3

p- / ei/" tm Mtfp. a =

Second Step, x ./1<* \*

"\a2 - b'

z)

(au; + (/^j = (a 2 - 6 2)% the

c?quation of tho evolute KHK'H' of the ellipse

A^/l'/^'. /, A1

', ^/', // are the centers of curvature corresponding to the points

A, A', K, It', on the curve, and (\ (v

, ("" correspond to the points P, P', P".

ILLUSTRATIVE EXAMPLE a. The paranietric equations of a curve are

(1)

/24- 1

:

4'

Find the equation of the evolute in parametric form, plot the curve and the

evolute, find the radius of curvature at the point where / = 1, and draw the cor-

responding circle of curvature.

X

Solution.

Substituting in (G) and reducing gives

(2) ^^^1

By (A), Art. 81

By (B), Art, 82

-f 3 t

calculate x, ^ from (1) and a, j8 from

(2), and tabulate the results.

Now plot the curve and its evolute.

The point (J, 0) is common to the

given curve and its evolute. The givencurve (semicubical parabola) lies en-

tirely to the right and the evolute

entirely to the left of x J.

The circle of curvature at A( Ji , Jl ) ,

where f = 1, will have its center at

A'( k, 1,} on the evolute and ra-

dius = AA'. To verify our work find

the radius of curvature at A. From

(F), Art. 105, we get

This should equal the distance

AA' = " By (1), Art. 3

ILLUSTRATIVE EXAMPLE 4. Find the parametric equations of the evolute of the

cycloid

V ;

r.r = </</

COS/).

Solution. As in the Illustrative Example of Art. cSl?, we get

d// _ sin / d'2y 1

dx 1 cos/ dx~ a(\ cos/)2

Substituting these results in formulas (G), Art. 108, we get

(4)

f a ait 4 sin /),

1 jg-

t/d - cos /). Ann.

NOTE. If we eliminate / between equations (4), there results the rectangular

equation of the evolute OO'Q* referred to the axes O'a and O'fi. The cocirdinates

of with respect to these axes are (~ mi, - 2 a). Let us transform equations (4) to

the new set of axes OX and OY. Then

a x ?ra, ft= y 2 a.

Also, let t t' TT.

Substituting in (4) and reducing, the ^

equations of the evolute become

(5)x = a(t'

- sin /'),

y = a(l cos /')

Since (5) and (3) are identical in o

form, we have:

The evolute of a cycloid is itself a cycloid whose generating circle equals that of the

given cycloid.


110. Properties of the evolute. The evolute has two interesting

properties.

Theorem 1. The normal at P(x, y) to the given curve is tangent to

the evolute at the center of curvature C(a, p) for P. (See figures in the

preceding article.)

Proof. From the figure,

(Ij a = x Rsin T,

ft= y + R COS T.

The line PC lies along the normal at

P, and the

(2) Slope of PC -x a tan r

slope of normal at P.

We show now that the slope of the evolute equals the slope of PC.

Note that j/j

Slope of evolute = -

da

since a arid fi are the rectangular coordinates of any point on the

evolute.

l^et us choose as independent variable the length of arc on the

given curve ;then x, //, R, T, a, are functions of 8. Differentiating

(1) with respect to s gives

(8)

(4)

da d'jr dr dR)-- # cos r __ sin T J

as ds as ds

rf/i dy . dr,

rf/2

j=

j

~~ ^ sm T T" H- cos r ~T~'ds ds ds ds

But(

~r = cos T, -^ = sin r, from Art. 95 : and - = .

rf.s ds ds R

Substituting in (3) and (4), and reducing, we obtain

(5)

Dividing the second equation in (5) by the first gives

da dR=- sin rrfs r/.s*

rf/J dR~ = COST-T--

(6)= - ctn T - --

da. tan r= slope of PC. Q.E.D.

Theorem 2. The length of an arc of the evolute is equal to the dif-

fererne between the radii of curvature of the given curve which are tangent

to this arc at its extremities, provided tfiat along the arc of the given curvt

R increases or decreases.

Proof. Squaring equations (5) and adding, we get

<" (DXiNff-But if s' = length of arc of the evolute,

ds'~ da- + dfi-,

by (C), Art. 95, if s = s', x = a, y = /i Hence (7) asserts that

dff(8)

= ^T ' or i =ds

Confining ourselves to an arc on the given curve for which the

right-hand member does not change sign, we may write

(9)= 4-1 or -,~ = -

dR~ '^ w

dtf

That is, Jfee rate of change of the arc of the evolute with respect to Ris _|_ i or _ i. Hence, by Art. 50, corresponding increments of s' and

R are numerically equal. That is,

(10) s' - s' = (R- flu),

or (first figure, p. 159) Arc CCi = (PiCi- PC).

Thus the theorem is proved.

In Illustrative Example 4, Art.109, we observe that at O', R = ;

at Pv,R = 4 a. Hence arc O'QQV = 4 a.

Tfce length of one arch of the cycloid (as OO'Q V) / eight times the

length of the radius of the general ing circle.

111. Involutes and their mechanical construction. Let a flexible

ruler be bent in the form of the curve r,r.,, the evolute of the curve

PI Po, and suppose a string of length

Ru, with one end fastened at Co, to

be stretched along the ruler (or

curve). It is clear from the results

of the last article that when the

string is unwound and kept taut, the

free end will describe the curve PjP<>.

Hence the name evolute.

The curve PiP9 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.

The involutes PiP*, /Y/V, /Y'/V are called parallel curves since

the distance between any two of them measured along their common

normals is constant.

The student should observe how the parabola and the ellipse on

pages 159, 160 may be constructed in this way from their evolutes.

PROBLEMS

Find the radius and center of curvature for each of the following

curves at the given point. Check your results by proving (a) that the

center of curvature lies on the normal to the curve at the given point,

and (b) that the distance from the given point to the center of curvature

is equal to the radius of curvature.

1. 2py = x'2; (0, 0). Am. (0, p).

2. x'2 + 4 y'

2 = 25 ; (3, 2). (iVb~ H)-

3. rj

-i/;'= 19 ; (3, 2). (W, 5?)-

4.x?/ = 6; (2, 3). (-?i )

5. //= <"; (0, 1). (- ~ 3 )-

6. y = cosr; (0, 1). (0, 0).

7. ?/= lnr; (1, 0). (3

~ 2) -

8. = 2 sin 2.r; (J TT, 2). (i TT, )-

9. (/ -f 6):i + ^?/

2 =; (-3, 3). (- 13, 8).

10. 2?/ = x 2 -4; (0, -2).

11. xy = x'2 + 2; (2, 3).

12. y = sin TT.T; (J, 1).

13. ?/=

-?, tan2o-; (-J TT, J).

Find the coordinates of the center of curvature at any point (x, y)

of each of the following curves.

A 3 <r -f 214. ?/

2 = 2'

a 4-f 15 ?/

4,. a4

?/- 9

15. ^ = a^. a=2

-. P= 4

16. 6 2x 2 - a 2?/2 = a 25 2

. a =

2 +

17. j.8 + y = a. a = x 4- 3 x?/,

18. Find the radii and centers of curvature for the curve xy 4 at the

points (1, 4) and (2, 2). Draw the arc of the evolute between these cen-

ters. What is its length ? __Ans. At (1, 4), A>, = V Vl7, a = V, = V ;

at (2, 2), A> 2 = 2 \/2, a = 4, = 4;

A>,- ft, = 5.938.

Find the parametric equations of the evolute of each of the following

curves in terms of the parameter t. Draw the curve and its evolute, and

draw at least one circle of curvature.

19. x = 3/ 2,

= 3f-? :i. Am. a = %(l + 2 t'

2 -/*), = - 4 <.

20. x = 3 t, # = f2 - 6. a = -

J f:l

,= 3 /-'

-jj.

21. x = 6 - *2

, y = 2 f. t* = 4 - 3 /2

,= - 2 P.

22. a- = 2 /, i/= J

2 - 2. a - - 2 /*,= 3 f

2.

23. x = 4 f, y = 3 4- J2

.= - /\ 0=11 + 3 /

2.

24. a- = 9 - f2

, ?/= 2 f. = 7 - 3 /

2f

= - 2 P.

3 12/ 4 + 9 . 27 + 4 /4

25.* = 2*, ?/=r

-4 t

.>

' ^=el

/^2 fo'2\

26. j* = a cos f, ?/= ?> sin L a - cos :i

/,

z

27^ x _ a COS3 ^ a a cos-* / + 3 a cos / sin 2,

?y= a sin :i

f.= '* " cos 2

/ sin / + a sin :{iL

28. x = a(cos t + / sin /), ot = a cos /,= a sin t.

y = a(sin / t cos /).

29. x = 4 - /2

, ?/= 2 /.

30. JT = 2 /, ?/= 16 - /

2.

31. x = /2

, ?y= i ^.

32. x = 1 cos f, y = t- sin /.

33. x = cos4/, ?/

= sin4/.

34. x = a sec , ?/= b tan f.

35. x = cos t, y = .

36. x = 6 sin t, y = 3 cos J.

37. x = 3 esc /, i/= 4 ctn J.

38. x = a(t + sin 0-

7/= a(l cos/)-

39. x = 2 cos t + cos 2 i.

y = 2 sin ? + sin 2 ^.


40. Show that in the parabola x* H- y2 = a 2 we have the relation

+ (3= X(x + y).

41. Given the equation of the equilateral hyperbola 2 xy = a 2, show

that - *

2

From this derive the equation of the evolute

(+ 0)1 - (a- 0)3 = 2 a*.

112. Transformation of derivatives. Some of the formulas derived

above independently can be deduced from others by formulas which

establish relations between derivatives. Two cases will be presentedhere.

Interchange of dependent and independent variables.

NOTATION. Let ?/'= 2/" = ! =

f, etc.,

, dx ,, dx' d2x .

x' = , x" = -, etc.dy dy dy

z

By IX, Art. 29,

XT // 'fNow y" -~~ = -7dx x'

Using (/), we Ret ^ - - ~>

(/// x' 2

dy"

Again, y'" = *f = *3L.dx x/

A,tt ,y.r~rrf Q ,

Using (/), ?- = - ~dy x'4

v'v'" O v"|

.-.y'" = -iX -*x '

Y'5

And so on for higher derivatives. By these formulas equations in

y'> y", '//'"> etc. can be transformed into equations in x 1

, x", x'", eta

ILLUSTRATIVE EXAMPLE. Transform (B), Art. 102, into (C) in that article.

Solution. Using (/) and (/) above,

1L

x" m '

Transformation from rectangular to polar coordinates. Given the

relations

(1) x = p cos 6, y = p sin 6

between the rectangular and polar coordinates of a point. If the

polar equation of a curve is p=f(0), then equations (1) are para-

metric equations for that curve, being the parameter.

NOTATION. The independent variable is 0, and x', .r", y', y", p', p"

denote successive derivatives of these variables with respect to 0.

Differentiating (1),

(2) x' = p sin + p' cos 0, y' = p cos + p' sin;

(3) z" = -2p'sin0 + (p"-p)cos0, <//"-2p'cos0 + (p"-p)sin0.

By formulas (1), (2), (3), equations in j, ?/, x', ?/, r", ?/" may be

transformed into equations in p, 0, p', p".

ILLUSTRATIVE EXAMPLE. Derive (), Art, 104, directly from (/)), Art. 10,'J.

Solution. Taking numerator and denominator in (D) separately, substituting

from (2), (3), and reducing, we obtain the results

X'y_

y>x" = p + 2 p'

2 - PP" ; ^ /1! + //"2 = P" + P

/2-

Putting these values in (D) gives ().

PROBLEMS

In Problems 1-5 interchange the dependent and independent variables,

4 d2* /dx\2

AAns. z -?/(-7-) =0.

rf V(]f

4

dy ~

6. Transform by assuming x = p cos 6, y = p sin 6.

A~<. P2

~ - - - - =7. Transform the equation y~- -

(-~-f

- - - = by assuming

8. Transform the equation s 2 ~? + 2 .c~^

-f ^7 y = by assuming .r = -

cLr2 dr .r-

'

/

A d'2 ij . v AAns. -7*7 4- a v = 0.~

ADDITIONAL PROBLEMS

1. Given the curve or 3 cos f 4- cos 3 t, y = 3 sin t sin 3 t. Find the

parametric equations of the evolute. Find the center of curvature for

/ = arid show that it coincides with the corresponding point on the given

curve. Ans. a = 6 cos i- 2 cos 3 t,

= 6 sin t 4- 2 sin 3 t.

2. If It is the radius of curvature at any point of the ellipse

b 2jc2-f a*//

2 = a'2b'2 and D the perpendicular distance from the origin to

the tangent drawn at this point, prove that RD* = a 2b 2.

3. Find the equations of the evolute of the parabola ?/2 = 4 xt using x

as a parameter. Find the points of the parabola for which the correspond-

ing centers of curvature are also points of the parabola. Hence find the

length of the part of the evolute inside the parabola.

Ans. (2, 2V2); 4(V7 -l).

4. (a) At every point (jc, y} of a certain curve, its slope is equal to

?/ and the curve passes through the point (2, 0). Verify that the

equation of the curve is log (1 -f- y) = 1 V5 jc2

.

(b) Find the curvature of the curve at this point and draw a small

portion of the curve near it. Ans. K = oVVH.

(c) Draw the circle of curvature at this point. Ans. cv = f , ft=

|.

5. The slope of the tangent to a certain curve C at any point P is

given by-jr-

= - where .s is the length of arc (measured from some fixed

point) and a is a constant. The center of curvature of C at P is P 1

'. De-note the radius of curvature of C at P by R and the radius of curvature of

the evolute of C at P' by 7i'. Prove

R =a 2

CHAPTER XI

THEOREM OF MEAN VALUE AND ITS APPLICATIONS

113. Rolle's Theorem. A theorem which lies at the foundation of

the theoretical development of the calculus will now be explained.

Let y=J(x) be a single-

valued function of x, con-

tinuous throughout the in-

terval [a, 6] (Art. 7) and

vanishing at its extremities

(/(a)= 0, f(b) = 0). Sup-

pose also that /(x) has a

derivative/' (2) at each in-

terior point (a < x < 6) of

the interval. The function will then be represented graphically by

a continuous curve as in the figure. Geometric intuition shows us at

once that for at least one value ofx between a and b the tangent is par-

allel to the x-axis (as at P) ; that is, the slope is zero. This illustrates

Rolle's Theorem, ///(x) is continuous throughout the interval [a, b]

and vanishes at its extremities, and if it has a derivative /'(x) at every

interior point of the interval, then f'(x) must vanish for at least one

value of x between a and b.

The proof is simple. For/(x) must be positive or negative in some

parts of the interval if it does not vanish at all points. But in this

special case the theorem is obviously true. Suppose, then, that /(x)

is positive in a part of the interval. Then /(x) will have a maxi-

mum value at some point within the interval. Similarly, if f(x)

is negative, it will have a minimum value. But if f(X) is a maxi-

mum or minimum (a < X < b), then f'(X) = 0. Otherwise, /(x)

would increase or decrease as x passes

through X (Art. 51).

The figure illustrates a case in which Rolle's

Theorem does not hold ; f(x) is continuous through-

out the interval [a, 6]. f'(x), however, does not

exist for x = c, but becomes infinite. At no point

of the graph is the tangent parallel to the x-axis.

169


We give first two applications of Rolle's Theorem to geometry.114. Osculating circle. If a circle be

drawn through three neighboring points

Po, Pi, P'2 on a curve, and if Pj and

P 2 be made to approach Pf , along the

curve as a limiting position, then this

circle will in general approach in mag- tx

nitude and position a limiting circle

caUed the osculating circle of the curve

at the point PO.

Theorem. The osculating circle is identical with the circle of curvature.

Proof. Let the equation of the curve be

U) ?/=/(*);

and let x ( , xi, x>> be the abscissas of the points Po, PI, Po respectively,

(a', fi

f

) the coordinates of the center, and R f

the radius of the circle

passing through the three points. Then the equation of the circle is

and since the coordinates of the points P, PI, P-2 must satisfy this

equation, we have

(2)

Now consider the function of x defined by

F(*) = (x~ a')2 + (y

-i3

f

)2 - R'2

,

in which y is defined by (1).

Then from equations (2) we get

)= 0, F(X!) = 0, F(x2 )

= 0.

Hence, by Rolle's Theorem (Art. 113), F'(x) must vanish for at

least two values of x, one lying between xo and x\, say xf

,and the

other lying between xi and x^ say x";that is,

F'(x') = 0, F'(x") = 0.

Again, for the same reason, F"(x] must vanish for some value of

x between x' and x", say x-^ ; hence

F"(z3 )= 0.

THEOREM OF MEAN VALUE AND ITS APPLICATIONS 171

Therefore the elements a', 0', R f

of the circle passing through the

points Po, PI, P2 must satisfy the three equations

F(zo) = 0, F'(x') = 0, F"(.r.i)= 0.

Now let the points PI and PL> approach P as a limiting position ;

then xi, ?2 , x', x", x3 will all approach :r () as a limit, and the elements

a, /3, R of the osculating circle are therefore determined by the three

equations F(* )= 0, F'(a- )

=f F"(x )=0;

or, dropping the subscripts, by

(3) (x- a)

2 + (y-

0)2 =

ff-',

(4) (x- a) + (y

-]8)'//'

- 0, differentiating (3).

(5) 1 + if2 + (y

-J3)y" = 0, differentiating (4).

Solving (4) and (5) for x a and ?/ 0, we get (?/" ^ 0),

(6) x a = >

7/ rf~ ^-

?/ , '//

Solving (6) for a and 0, the result, is identical with (G), Art. 108.

Substituting from (6) in (3), and solving for R, the result is (F),

Art. 105. Hence the osculating circle is identical with the circle of

curvature.

In Art. 28 the tangent line at P was defined as the limiting posi-

tion of a secant line drawn through P and a neighboring point Qon the curve. We now see that the circle of curvature at P may be

defined as the limiting position of a circle drawn through P and two

other points Q, R on the curve.

115. Limiting point of intersection of consecutive normals

Theorem. The center of curvature (^ for a point P on a curve is the

limiting position of the intersection of the normal to the curve at P with

a neighboring normal. C(a f /*j

Proof. Let the equation of a curve bec[<tt'^

(1) =/(*)

The equations of the normals to the

curve at two neighboring points Po andD ^ A'.^o'l/o)^ are

Oo - y) + ('//o-

?/)/'Oo) = 0,

If the normals intersect at C'(a', /3'), the coordinates of this point

must satisfy both equations, giving

Now consider the function of x defined by

<t>(x)= (x

-a') + (y- p')y' f

in which y is defined by (1).

Then equations (2) show that

</>(z )= 0, 0(xi) = 0.

But then, by Rolle's Theorem (Art. 113), <'(j) must vanish for

some value of x between XQ and x\, say x'. Therefore a' and /3' are

determined by the two equations

If now PI approaches P as a limiting position, then x' approachesx

, giving

and f"(a', 0') will approach as a limiting position a point C(a, ]8)

on the normal at /V Dropping the subscripts and accents, the last

two equations are

(x-a) + (y- &)y' = 0,

1 + y'2 + (y

- fry" = 0.

Solving for a and /3, the results are identical with (G), Art. 108. Q.E.D.

116. Theorems of Mean Value (Laws of the Mean). For later appli-

cations we need the

Theorem. ///Or) and F(x) and their first derivatives are continuous

throughout the interval [a, 6"|, and if, moreover, F'(x) does not vanish

within the interval, then for some value x = xi between a and 6,

--Proof. Form the function

Evidently </>() = </>(?>)= 0, and Rolle's Theorem, Art. 113, may

be applied. Differentiating,

This must vanish for a value x xi between a and 6.

~/(a) r*/

Dividing through by F'(XI) (remembering that F'(XI) does not

vanish), and transposing, the result is (4). Q.E.D.

If F(x) = x, (A) becomes

b - aJ v A>

In this form the theorem has a simple *

geometric interpretation. In the figure

the curve is the graph of /(x). Also,

OC = a, CA=/(a),OD = b, DB=f(b).

Hencer/_\

of chord AB.b a

Now f'(xi) in (B) is the slope of the curve at a point on the arc

AB, and (B) states that the slope at this point equals the slope of

AB. Hence there is at least one point on the arc AB at which the tan-

gent line is parallel to the chord A B.

The student should draw curves (as the first curve in Art. 113)

to show that there may be more than one such point in the interval,

and curves to illustrate, on the other hand, that the theorem maynot be true if f(x) becomes discontinuous for any value of x between

a and 6, or if f'(x) becomes discontinuous (as in the second figure of

Art. 113).

Clearing (B) of fractions, we may also write the theorem in the form

(C) /(&)=/(<*) + (b-

<0/'(*0-

Let 6 = a -r Aa ; then b a = Aa, and since x\ is a number lying

between a and 6, we may write

Xl =a + 6- Aa,

where 6 is a positive proper fraction. Substituting in (C), we get

another form of the Theorem of Mean Value,

(D) f(a + Aa) -/(a) = Aa/'(a + Aa). (0 < < 1)

PROBLEMS

1. Verify Rolle's Theorem by finding the values of x for which f(x)

and/'fa) vanish in each of the following cases.

(a) f(x) = x3 - 3 x. (e^ f(x) = sin irx - cos TTX.

(b) f(x) = 6 x2 - x3. (f) f(x) = tan x - x.

(c) f(x) = a + bx -f ex2. (g) f(x) = x In x.

(d) f(x) = sin x. (h) /(x) = xe*.

2. Given f(x) = tan x. Then /(O) = and /(ir) = 0. Does Rolle's

Theorem justify the conclusion that f'(jc) vanishes for some value of x

between and TT? Explain your answer.

3. Given (y -f l):i = x 2

. Then y = when x = 1 and //= when

x = + 1. Does Rolle's Theorem justify the conclusion that y' vanishes

for some value of x between - 1 and + 1 ? Explain your answer.

4. In each of the following cases find x\ such that

f ( b) = /() + (b-a)f'(jci).

(a) f(x) = x-, a 1, b 2. Ans. Xi = 1.5.

(b) /(x) = Vx, a = 1, b = 4. x, = 2.25.

(c) /(x) = 6", a = 0, b = 1. Xi = In (<?-

1) = 0.54.

(dj /(x) - -, r; = 1, 6 = 2.

(e) /(x) = In x, a 0.5, b 1.5.

5. Given f(x) = - a - 1,6 = 1. For what value of Xi, if any, willx

f(b) = /() + (b- )/'(x!)?

6. Given /(x) = x^, a = 1, /> 1. For what value of x,, if any, will

/(ft) =/(*;) -f (fc-aWi)?

117. Indeterminate forms. When, for a particular value of the

independent variable, a function takes on one of the forms

L\ 2?, X oo, oo - oo, 0, oc", I- t

it is said to be jndetcrvmuite, and the function is 'not defined for that

value of the independent variable by the given analytical expression.

For example, suppose we have

where for some value of the variable, as x = a,

/(a) - 0, F(a) = 0.

For this value of x our function is not defined and we may there-

fore assign to it any value we please. It is evident from what has

gone before (Case II, Art. 17) that it is desirable to assign to the

function a value that will make it continuous when x = a whenever

it is possible to do so.

118. Evaluation of a function taking on an indeterminate form. If

the function f(x) assumes an indeterminate form when x = a, then ii

exists and is finite, we assign this value to the function for x = a,which now becomes continuous for jc a (Art. 17).

The limiting value can sometimes be found after simple transfor-

mations, as the following examples show.

ILLUSTRATIVE EXAMPLE 1. Given /(j)^ ~

^. Prove lim f(x] =4.

Solution. /(2) is indeterminate. But, dividing numerator by denominatorfix} = x + 2. Then lim (x + 2) = 4.

z - 2

ILLUSTRATIVE EXAMPLE 2. Given /(x) = sec x tan j\ Prove lim f(x) = 0..r -

^ TT

Solution. /(*) ir> indeterminate (*>- oc). Transform as follows:

sec x - tan x = L^J^LZ =L^L . Lin* __ cor x^

fos j- cos .r 1 -f sin .r 1 4 sin xHence the limit is 0.

See also Art, 18. General methods for evaluating the indeter-

minate forms of Art. 117 depend upon the calculus.

119. Evaluation of the indeterminate form - (iiven a function of

ff.r)the form

j~~-such that /(a) and F(a) ~ 0. The function is

indeterminate when x a. It is then required to find

We shall prove the equation

Proof. Referring to (4), Art. 116, and setting 6 = -jr, rememberingthat /(a) = F(a) = 0, we have

m=m- <<-<'>If x > a, so also .TI

-> a. Hence, if the right-hand member of (1)

approaches a limit when r\ -> a, then the left-hand member will ap-

proach the same limit. Thus (E) is proved.

From (), if /'(a) and F'(a) are not both zero, we shall have

(2) i iWRule 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.


In case it happens that /'(a) = and F'(a) = 0, that is, the first

derivatives also vanish for x = a, then () can be applied to the ratio

F'(x)

and the rule will give us ijmx-aF(x) F"(a)

It may be necessary to repeat the process several times.

The student is warned against the very careless but common mistakeof differentiating the whole expression as a fraction by VII.

If a = oo, the substitution jc = - reduces the problem to the evaluationof the limit for z 0.

z

Thus lim = lim = lim

Therefore the rule holds in this case also.

ILLUSTKATIVK EXAMPLE 1. F^ove lim ^H!^ = n ,

x .0 JT

Solution. l^tf(x)=^Kinnx l F(x) = jr. Then/(0) =0, F(0) = 0. Therefore, by (),

Hm /(l = lim Hl = Hm ncoanr = w1 -b

ILLUSTRATIVE EXAMPLE 2. Prove lim*"* 3 ? + 2 _ 3

-' -jr^-x + 1 2

Solution. Let -/(.r) - ^ - 3 x + 2, F(.r) =** -x* - x + 1. Then /(I) ==0, Therefore, by (),

ILLUSTRATIVE EXAMPLE 3. Prove lim -t

_ = 2.

Solution. Let /(x) = e* - e~r - 2 x, F(x) = x - sin x. Then /(O) = 0, F(0) =Therefore, by (),


PROBLEMS

Evaluate each of the following indeterminate forms by differentiation.*

1. lim~ 2 ~ 1C

z-4

2. lim

_ o x sin

ev + sin y 1

x :4X 2-f x-20

jr-a 1

cM a" na"

,. In x n3. lim-- 1.

. ,. e e~ o4. lim :

-- 2.

x -. o sin .r

2.

. v In sin JT i

6. hm - -2

- ~4-

x- -

7. lim ^-I= ln r

,. ^ arc sin i8. hm- -- -

*

0-*o sin-1 6

15 limtan 6 + sec - 1

15. hm^ e _ ^ ^ + l

,/> v r sin ;

16. hm. o

x-,0 sin :i

18. Given a circle with center at O, radius r,

and a tangent line A T. In the figure, A M equals

arc AP, and B is the intersection of the line

through M and P and the line through A and O.

Find the limiting position of B as P approaches

A as a limiting position. An*. OB = 2 r.

* After differentiating, the student should in every case reduc the resulting expression

to its simplest possible form before substituting the value of the variable.

120. Evaluation of the indeterminate form ~. In order to find

limx-^a

when both f(x) and F(x) become infinite when x > a, we follow the

same rule as that given in Art. 119 for evaluating the indeterminate

form -. Hence

Rule for evaluating the indeterminate form ^. Differentiate the

numerator for a new 'numerator mid the denominator for a new denomi-

nator. The value of this new fraction for the assigned value of the mi /-

able will be the limiting value of the original fraction.

A rigorous proof of this rule is beyond the scope of this book.

ILLUSTRATIVE KXAMPLE. Prove lim --n/r- 0.

.r C-.C JT

Solution. I*t /(j) = In r, F(JT) = osc .r. Then /(O) - - o, F(Q) - oo . Hence,

by the rule,I

,. f(jr) ,. f'(jf) .. .r_ .. siriL'j

lim $-L-

) im -J-f~- = hm- - lim-

j , o f (jc) j ,<)/<'(>) j. , (,--

c'.-.c J ctn -i*

rni I /I^ 1-"

"I'

" '/I <-l T7 T^

Then, by (A), hm- - lim-:

- - 0. Q.E.D.j- o JT cos jr

, o cos j; J-

sin x

121. Evaluation of the indeterminate form oo. If a function

f(y)-

</>(**') takes on the indeterminate form oo for x a, we write

the Bi en function

so as to cause it to take on one of the forms -r or ^, thus bringing it

under Art. 119 or Art. 120.

As shown, the product f(jr)-

c/>(x) may be rewritten in either of

the two forms given. As a rule, one of these forms is better than

the other, and the choice will depend upon the example.

ILLUSTHATHK KXAMPLE. Prove lim (sec 3 J* cos 5 .r)=

.

X -+/, 7T

Solution. Since sec-J TT = oo

,cos J IT = 0, we write

o c 1 c C()S 5 Xsec 3 .r cos 5 J ~-7- cos 5 x =-

cos 3 .r cos 3 j"

Let/(j) = cos 5 or, F(x) - cos 3 jr. Then/(| TT)= 0, F(\ TT) = 0. Herice, by (),

-3sm3.r

122. Evaluation of the indetei ruinate form oo oo. It is possible in

general to transform the expression into a fraction which will assume

either the form - or ~-

ILLUSTRATIVE EXAMPLE. Prove lim (sec .r tan x] 0.

X -tl, JT

Solution. We have sec \ TT tan J T = x ~ x . .-. indeterminate.

^ y o *1 sni -v 1 sin *"

Bv (2), p. 2, sec jr tan .r = ~~

cos j* cos x cos r

= I - sin jc, F(rt = cos j-. Then/(\ TT) = 0, F(\ TT) - 0. Hence, by (E),

p

(.r) __ ^ 0;

X- '.TT ^H-

,.hm-~ s

'

inQ.E.D.

PRO3LEMS

Evaluate each of the following indeterminate forms.

2. limctn x

j , o ctn 2 r

,. tan 3 ^3. hm --

^ _ tan 6

4. lim

cs

5. lim :--

x - .r. In x

, .

6. hm

7. lim

ctn x

In sin 2 x

AAns. 0.

,o In sin .r

8. lim jc In sin jr.

JT-

rt T 7T . 7T09. lim --- tan - -

10. lim r sin -x -. oo ^

11. lim (TT 2 .r)tan x.

12. lim (1- tan 0) sec 2 0. Ann. 1.

-4.13. lim

22. lim (sec 5 <? tan 6.)

i-

i-


TT 7T 1 r* _" ' t\f* i *('

23. lim .

4 ^*

24. limf-4-

28.

123. Evaluation of the indeterminate forms 0, 1, 00. Given a func-

tion of the formf(x)+ (T)

In order that the function shall take on one of the above three

forms, we must have, for a certain value of x,

/(x)=0, 0Cr)=0, giving 0;or /(*) = !, 0tr) = oo, giving I

00

;

or /(j) = oo, (f)(x)= 0, giving 00.

Let y=f(x)+(f)

.

Taking the natural logarithm of both sides,

Invy = 000 In /(a-).

In any of the above cases the natural logarithm of y (the function)will take on the indeterminate form

-oo.

Evaluating this by the process illustrated in Art. 121 gives the

limit of the logarithm of the function. This being equal to the loga-rithm of the limit of the function, the limit of the function is known.For if limit In // a, then lim //

- e".

ILLUSTRATIVE EXAMPLE 1. Prove lim xz1.

j .

Solution. The function assumes the indeterminate form when x = 0.

Let y = x1;

then In y a* In j = oo, when x = 0.

r> * u. 1 ni i In x ooBy Art, 121, In ?/=--- , whenz = 0.

By Art, 120, lim = limr - 1 J- -

Therefore lim In y-

0, and lim y- lim x* = e = 1. O.E.D.' -* X () X -

i2

ILLUSTRATIVE EXAMPLE 2. Prove lim (2- j-)

tan 5 WJr = f*

jr-l

Solution. The function assumes the indeterminate form l when jr = 1.

Let y = (2-

.r)tttn * ^

;

Ini/= tan i 7r.r In (2

- j) - oo .

0, when j = 1 .

By Art. 121, ,n^!

By Art. 119, limln (2

~ ^ = hm L>~

'r

.T-.I ctn i TTJ .r-i - \ TT csc-.^Trr TT

o i 2

Therefore lim In y- -, and lim y ._ lim (2

-.r)

tan ^ ^ = r77

"

O E D3--1 7T x .1

'

j. .1

" v

ILLUSTRATIVE EXAMPLE 3. Prove lim (c-tn j-)sinj"

= i.j- *o

Solution. The function assumes the indeterminate form 00 when x ~0.

Let y=- (ctnx)sinj

;

then In y sin x In ctn j- = oc, when j = 0.

By Art. 121, In =^ r =, whon x = 0.

csc j: oo

- esc- j*

By Art. 120, lim l!L*!L = ]im_ilH _ ^ nm LL. =.

x -*o csc J j-^o csc x ctn .r T .<> cos'*' .r

Therefore lim In y 0, and lirn y = lim (ctn .r)Hinj" = c = 1. Q.E.D.

- -

PROBLEMS

Evaluate each of the following indeterminate forms.

1. lim (sinx)tanx

. Ans. 1. g i[m / COS -Vx -

I

'

z . r \ y/

2. lim/|+ lY. e*. 9- Hm

(cos^ .

S-limx^.J-

10. Hm(COB

?

4. lim fl -h -V. ca . 11. lim (r^ + 2 x)^i/--* x V y/ T-. o

5. lim (1 -f sin x)ctnT

. r. 12. lim (x -4- l)ctnx

.

x-O x-0- /I \sina;

6. lim (cx-f x)

x. e

2. 13. lim

( j

'.

1

7. lim (1 + ntY. c\ 14. lim (I -f r)lnr

.

/ - J- -

124. The Extended Theorem of Mean Value. Let the constant Rbe defined by the equation

(1) J(b) -f(o.) -(b- a)}' (a)-

(&- a^R = -

Let F(y) be a function formed by replacing b by x in the left-hand

member of (1) ;that is,

(2) F(x) = f(x) -J(a) - (x-

a)/' (a)-

\(x- a)*R.

From (1), F(b) = ; and from (2), F(a) = ; therefore, by Rolle's

Theorem (Art. 113), at least one value of x between a and b, say xhwill cause F'(x) to vanish. Hence, since

F'(x)=J'(x)-f'(<i)-(x-a)R,

we get F'(Xi) =f'(xi) ~J'(o) - (x l

- a)R = 0.

Since F f

(x\ ) and F'(a) = 0, it is evident that F'(r) also satisfies

the conditions of Eolle's Theorem, so that its derivative, namely F"(x),

must vanish for at least one value of x between a and x\, say x*, and

therefore x<2 also lies between a and b. But

F"(x) =f"(jr) - R\ tlierefore F"(jrJ =f /f

(^) - R = 0,

and R=--f"(xJ.

Substituting this result in (1 ), we get

' "a)

2/"(^). (a

By continuing this process we get the general result,

(G)

Equation (G) is called the Extended Theorem of Mean Value, or the

Extended Law of the Mean.

125. Maxima and minima treated analytically. By making use of

Art. 116 and the results of the last section we can now give a general

discussion of maxima and minima of functions of a single independent

variable.

Given the function /(.r). Let h be a positive number as small as

we please ; then the definitions given in Art. 46 may be stated as

follows.

If, for all values of :r different from a in the interval |a h, a + H(1) /(x) /(fl)

= a negative number,

then f(jc) is said to be a majriwinn -whew x = a.

If, on the other hand,

(2) /(.r) /(a) = a positive number,

then /(.r) is said to be a minimum when x == a.

We begin with an analytical proof of the criterion on page 51.

A function is increasing when the derivative is positive, and decreas-

ing when the derivative is negative.

For, let y=f(x). When A;/ is small, numerically, and theA.r

derivative /'(.r) will agree in sign (Art. 24). I jet /'(/) > 0. Then,when Ax is positive, so is A?/, and when A.r is negative, so is A?/.

Therefore f(jr) is increasing. A similar proof holds when the deriv-

ative is negative.

The truth of the following statement is now easily deduced.

///(a) is a maximum or minimum value of f(x), then f (a) = 0.

For, if /'(a) 5^ 0, f(x) would increase or decrease as x increases

through a. But then f(a) is neither a maximum nor a minimum.We seek general sufficient conditions for maxima and minima.

Consider the following cases.

I. Let /'(a) = 0, and/"(a)^ 0.

From (f), Art. 124, replacing 6 by n and transposing /(a),

(3) /(a-) -/(a) = (*~ a}2f"(.r<2). (a < x>> < x)

Since /"(a) ^ 0, and /"Or) is assumed as continuous, we maychoose our interval \a h, a + h\ so small that /"(/) will have the

same sign as /"(a). Also, (/ a)'2 does not change sign. Therefore

the second member of (3) will not change sign, and the difference

/Or) -/(a)

will have the same sign for all values of x in the interval [a h, a + h],

and, moreover, this sign mil be the same as the sign o//"(a). It there-

fore follows from our definitions (1) and (2) that

(4) /(a) is a maximum iff'(a) and f"(a) = a negative number;

(5) /(a) is a minimum if f'((i)= and j"(a) = a positive number.

These conditions are the same as those in Art. 56.

II. Let /'(a) =/"(a; - 0, and /'"(a) ^ 0.

From (G), Art. 124, putting n = 3, replacing b by r, and trans-

posing /(a),

(6) f(x)-

/(a) = ,4 te - a)3/"W (a < a* < a-)

12

As before, /'"(or) will have the same sign as f'"(a). But (x a):i

changes its sign from to -f as x increases through a. Therefore

the difference ., . fi-f(a)

must change sign, and /(a) is neither a maximum nor a minimum.

III. Let /'(a) =/"(a) - =/<"-'>(a) =0, and/(n) (a)^ 0.

By continuing the process as illustrated in I and II, it is seen that

if the first of the derivatives of f(x) which does not vanish for x = a

is of even order (= ri), then

(H) /(a) is a maximum if / (n)(a) = a negative number;

(/) /(a) is a minimum if f (n)(d) = a positive number.*

If the first derivative of f(x) which does not vanish for x = a is of

odd order, then /(a) will be neither a maximum nor a minimum.

ILLUSTRATIVE EXAMPLE 1. Examine x* - 9 x'2 -f 24 x - 7 for maximum andminimum values.

Solution. /(x) = z< { - 9 x'2 + 24 x - 7.

/'Or) = 3 a;2 - 18 z + 24.

Solving, 3 x 2 -18x4- 24 =

gives the critical values x = 2 and z = 4. /./'(2) = 0, and /'(4) = 0.

Differentiating again, /"(*)~ 6 x 18.

Since /"(2) = -6, we know, from (//), that/(2) = 13 is a maximum.

Since /"(4) = -f 6, we know, from (/), that /(4) = 9 is a minimum.

ILLUSTRATIVE EXAMPLE 2. Examine e* + 2 cos x 4- e~ x for maximum and mini-mum values.

Solution. /(r) = ex -f 2 cos .r -f f-*,

/'(j) = <?*- 2 sin x - e- 1 = 0, for x = 0,t

f"(x} = ex 2 cos x -f <?"* = 0, for x 0,

/"'(z) = g* -f 2 sin x - e~ x = 0, for x = 0,

/"( = e' -f 2 cos x -f- <r T = 4, for x = 0.

Hence, from (/), /(O) = 4 is a minimum.

* As in Art. 46, a critical value x = a is found by placing the first derivative equal to

zero and solving the resulting equation for real roots.

t x = is the only root of the equation e* 2 sin r e~x 0.


PROBLEMS

Examine each of the following functions for maximum and minimumvalues, using the method of the last section.

1. x4 4 x3 + 5. Arts, jc = 0, gives neither,

x = 3, gives min. = 22.

2. x3 + 3 x2 + 3 x, x = -1, gives neither.

3. x3(x-

2)2

. x = 0, gives neither,

x = L gives max. = 1.11,


4. x(x-l) 2(x+l) 3.

5. Investigate 4 x6 - 15 x4 + 20 x3 - 10 x2 at x = 1.

6. Show that if the first derivative of /(x) which does not vanish for

x = a is of odd order (= n), then/(x) is an increasing or decreasing function

when x = a, according as /W (a) is positive or negative.

ADDITIONAL PROBLEMS

1. If y = ex -f e~ x,find dx in terms of ?/ and dy. Ans. dx = ^ .

V;//2 - 4

2. Prove that

T- In (3 x + 2 + V9 x 2-f 12 x) = .

Vi) jf- -f 12

3. Prove that

Vx- -f 1i

= x 2Vx 2-f 1.

4. Show that the curve x = t* + 2 *2

, ?/= 3 /

2 + 4 f has no point of

inflection.

5. Prove that the points of intersection of the curves 2 ?/ x sin x and

y = cos x are points of inflection of the first curve. Sketch both curves on

the same axes.

6. Given the damped harmonic motion

s =r ae~ ht sin ct,

where a, 6, and c are positive constants ; prove that the successive values

of t for which v = form an arithmetic progression and that the correspond-

ing values of s form a decreasing geometric progression.

7. The abscissa of a point P moving on the parabola //- ax2 is increas-

ing at the rate of one unit per second. Let O denote the origin and let T

be the intersection of the x-axis with the tangent to the parabola at P.

Show that the rate of increase of the arc length OP is numerically equal

TPto the ratio


8. Let MP be the ordinate at any point P on the catenary (see page532). The line MA is drawn perpendicular to the tangent at P. Provethat the length of MA is constant and equal to a.

9. The curve x 2y -f 12 y = 144 has one maximum point and two pointsof inflection. F"ind the area of the triangle formed by the tangents to the

curve at these three points. Arts. 1.

10. Given In 6 = 1.792 and In 7 = 1.946. Calculate In 6.15 first byinterpolation and second by differentials. Show graphically that the true

value lies between the two approximations.

11. Given the ellipse b 2jr

2 + a 2?/2 = a'2 b'2

, find the length of the shortest

tangent intercepted between the coordinate axes. Ans. a 4- b.

12. Given the area in the first quadrant bounded by the curves y 2 = xand y'

2 = x3. 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 constructedin this way. Ans. 0.019.

13. Rectangles are drawn with one side along the ./--axis, a second side

along the line x = * and one vertex on the curve ?/ = c~ r\ Find the areaof the largest of these rectangles. A/?s. e~ - 2r> = 0.7788.

14. Find the maximum and minimum values of y, if

X X

y aea 3 x 2 ac .

Ans. Max. = a ; min. a(l 3 log 2).

15. Given x 2-f 3 xy -h 2 y 2 5 x 6 y -h 5 0, find the maximum and

minimum values of y. Ans. Max. 1 ; min. = 5.

INTEGRAL CALCULUS

CHAPTER XII

INTEGRATION ;RULES FOR INTEGRATING STANDARDELEMENTARY FORMS

126. Integration. The student is already familiar with the mutu-

ally inverse operations of addition and subtraction, multiplication

and division, raising to a power and extracting roots. In the ex-

amples which follow, the second members of one column are, respec-

tively, the inverse of the second members of the other column.

y = a*, x = log,, // ;

y = sin x, x = arc sin;//.

From the differential calculus we have learned how to calculate

the derivative f'(x) of a given function f(jr), an operation indicated by

-f A') =/'(*),dx

'

or, if we are using differentials, by

df(jr)=f'U)dx.

The problems of the integral calculus depend on the inverse

operation, namely :

To find a function f(x) whose derivative

(1) /'(*)- 4>W

is given.

Or, since it is customary to use differentials in the integral cal-

culus, we may write

(2) dj(x) = f'(x)dx = cf>(x)dx

and state the problem as follows :

Having given the differential of a function, to find the function itself.

187

188 INTEGRAL CALCULUS

The function /(x) thus found is called an integral of the givendifferential expression, the process of finding it is called integration,

and the operation is indicated by writing the integral sign* /in front

of the given differential expression ; thus ^

(3)ff'(x)dx=f(x

} ,

read the integral of f'(x)dx equals f (or). The differential dx indicates

that x is the variable of integration. For example,

(a) If/(x) = xa, then/'(x)dx = 3 x'2 dx, and

(b) If /(x) = sin x, then/'(x)dx = cos xdx, and

cosxcfx = sin x.

/r(c) If /(x) = arc tan x, then/'(x)rfx = .,, and

1 "T~ X"

- arc tan x.+ x~

Let us now emphasize what is apparent from the preceding ex-

planations, namely, that

Differentiation and integration are inverse operations.

Differentiating (3) gives

(4)

'

dff'(x)dx=J'(x)dx.

Substituting the value of f'(x)dx [= tf/(x)] from (2) in (3), we get

(5) fdf(x)=f(x).i /*

Therefore, considered as symbols of operation, 3- and /- dx are

dx Jinverse to each other

; or, if we are using differentials, d and / areinverse to each other. r

J

When d is followed by I they annul each other, as in (4), but

when I is followed by d, as in (5), that will not in general be the

case. The reason for this will appear at once from the definition of

the constant of integration given in the next section.

* Historically this sign is a distorted S, the initial letter of the word sum. See Art. 155.

INTEGRATION 189

127. Constant of integration. Indefinite integral. Prom the preced-ing section it follows that

snce rf(:r3)= 3 x 2

dx, we have f 3 .r~dr = x3;

since d(x* + 2) = 3 x~dx, we have f3 x-dx = x 3 + 2 ;

since d(x3 -

7) = 3 x-djr, we have / 3 x 2dx = x37.

In fact, since d(x* + c) = 3 .r-rfjr,

where C is any arbitrary constant, we have

3 a-2rfjp = a-

3 + C.f

A constant C arising in this way is called a constant of integration, anumber independent of the variable of integration. Since we can giveC as many values as we please, it follows that if a given differential

expression has one integral, it has infinitely many differing only byconstants. Hence r

Jf'(x)dx=f(x)+ C;

and since C is unknown and indefinite, the expression

/Or) + C

is called the indefinite integral of ff

(x)dx.

It is evident that if </>(#) is a function the derivative of which is

f(x), then <f>(x) + C, where C is any constant whatever, is likewise

a function the derivative of which is /(a;). Hence the

Theorem. // two functions differ by a constant, they have the samederivative.

It is, however, not obvious that if 0(#) is a function the derivative

of which is /(#), then all functions having the same derivative f(x)

are of the form_j_ ^

where C is any constant. In other words, there remains to be

proved the

Converse theorem. // two functions have the same derivative, their

difference is a constant.

Proof. Let <t>(x) and \l/(x) be two functions having the same deriva-

tive /(x). Place

F(x) = </>(x) \l/(x) ; then, by hypothesis,

(1) F*(x)=

But from the Theorem of Mean Value (Z>), Art. 116, we have

F(x + Ax) - F(x) = Ax F'(x + Ax). (0 < 6 < 1)

.-. F(x + Ax) - F(x) = 0,

[Since by (1) the derivative of F(x) is zero for all values of x.J

and F(x + Ax) = F(x).

This means that the function

F(x) =

does not change in value at all when x takes on the increment Ax,

that is, <t>(x) and \f/(x) differ only by a constant.

In any given case the value of C can be found when we know the

value of the integral for some value of the variable, and this will be

illustrated by numerous examples in the next chapter. For the pres-

ent we shall content ourselves with first learning how to find the

indefinite integrals of given differential expressions. In what fol-

lows we shall assume that every continuous function has an indefinite

integral, a statement the rigorous proof of which is beyond the scope

of this book. For all elementary functions, however, the truth of

the statement will appear in the chapters which follow.

In all cases of indefinite integration the test to be applied in veri-

fying the results is that the differential of the integral must be equal

to the given differential expression.

128. Rules for integrating standard elementary forms. The differen-

tial calculus furnished us with a General Rule for differentiation

(Art. 27). The integral calculus gives us no corresponding general

rule that can be readily applied in practice for performing the inverse

operation of integration.* Each case requires special treatment, andwe arrive at the integral of a given differential expression throughour previous knowledge of the known results of differentiation. That

is, we must be able to answer the question, What function, when dif-

ferentiated, will yield the given differential expression ?

Integration, then, is essentially a tentative process, and to expeditethe work, tables of known integrals are formed called standard forms.To effect any integration we compare the given differential expressionwith 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, westrive to reduce it to one of the standard forms by various methods,

many of which employ artifices which can be suggested by practice

* Even though the integral of a given differential expression may be known to exist, yetit may not be possible for us actually to find it in terms of known functions.

INTEGRATION 191

only. Accordingly a large portion of our text will be devoted to the

explanation of methods for integrating those functions which fre-

quently appear in the process of solving practical problems.From any result of differentiation may always be derived a formula

for integration.

The following two rules are useful in reducing differential expres-sions to standard forms.

(a) The integral of any algebraic sum of differential expressions

equals the same algebraic sum of the integrals of these expressionstaken separately.

Proof. Differentiating the expression

/ du + I dv I dw,

u, v, w being functions of a single variable, we get

du + dv - dw. By III, Art. 94

(1) /. / (du + dv- dw) = (du + Cdv - Cdw.

(b) A constant factor may be written either before or after the inte-

gral sign.

Proof. Differentiating the expression

gives adv. By IV, Art. 94

(2) .'. I adv= a I dv.

On account of their importance we shall write the above two rules

as formulas at the head of the following list ofMStandard Elementary

Forms."

STANDARD ELEMENTARY FORMS

(1) i(du + dv- dw) = idu + I dv - Cdw.

(2) I adv = a I dv.

(3) Cdx = x + C.

/vn + l

V"dV =^+l

+ C' (*-!)

Cdv(5) J T = lnv+C

= In v + In c = In cv.

(6)

(7) C

(8) / sin v </v = cos v + C.

(9) I cos v dv = sin v + C._/

(10) I sec2 i; c/v = tan v + C.

(11) fesc2

i; rfi; = - ctn v + C.

(12) I sec i; tan v dv = sec v+C.

(13) I esc i; ctn v dv = esc v+C.

(14) / tan v dv = In cos v In sec v + C.

(15) I ctn v dv In sin v + C.

(16) / sec v dv = ln (sec v + tan v) + C.

(17) I esc v dv = In (esc v ctn v) + C.

(18) f*"

2= iarc tan-+ C.

J v2 + a2 a G

fffvv

x= arc sin - + C.

Va2 - v2

INTEGRATION 193

129. Formulas (3), (4), (5). These are easily proved.

Proof of (3). Since d(x + C) = dr, II, Art. 94

we get I dx = x + C.

Proof of (4). Since

(

7,n + 1 \

j-j-j-+

Cj= tcdr, VI, Art. 94

/?,n-H

Vndv =-T + (7.

W+ 1

This holds true for all values of n except n = 1. For whenn 1, (4) involves division by zero.

The case when n = 1 comes under (5).

Proof of (5). Since,/.,

d(lnr+D =-f X, Art. 94v

we get f = In w + r.

J v

The results we get from (5) may be put in more compact form if

we denote the constant of integration by logc c. Thus

C^l = ]n v + in c = in CVf

J v

Formula (5) states that if the expression under the integral sign is a

fraction whose numerator is the differential of the denominator, then the

integral is the natural logarithm of the denominator.

ILLUSTRATIVE EXAMPLES*

Work out the following integrations :

7-6-6 + 1/ 7- r1. Cx^dx =

-j-r + C = *- + C, by (4), where v = x and n = 6.

J o + 1 /

2.JVxdz =fx*dx

= 2L+C = z*+C, by (4), where = a: and n = J.

* When learning to integrate, the student should have oral drill in integrating simple

functions.

3Cdx = r

x.3dx = ^ + c = _J_ + Ciby (4), where r = x and

'

J y* J X

4./ax*dx=a/Vdx

=<^+C.

By (2) and (4)

5./(2x3 - 5x2 -3x

x -/5 x 2 d(x -/3 xrfx +4 dx by (1)

= 2/x3 dx - 5/x

2 dr - 3/xdx + 4/dx by (2)

NOTE. Although each separate integration requires an arbitrary constant, we

write down only a single constant denoting their algebraic sum.

=f2 ax'^dx - fbx~'2 dx -f/3 rx^dx by (1)

= 2 a fx~*dx-bfx~

2 dx + 3 cfx$dx by (2)

= 2a .2-6. + 3r.^ + C by (4)

= 4 aVx -f- + 1 rx

5' 4- C.

./ 5

HINT. First expand.

Solution. This may be brought into the form (4). For Comparison with (4).

insert the factor 2 b'2after the integral sign before x dx, and v a'2 + b'2x'\ n = -

2 >

its reciprocal fo/orr the integral sign. Those operations dv = 2b*xdx.

balance each other by (2).

: =5L

/

%

(a + 6x)l(26*zdI)[=^5 /*i*=^5 + C, by

(4)]

( fl2 + &2X2)1

"^3 ^

NOTE. The student is warned against transferring any function of the variable

from one side of the integral sign to the other, since that would change the value

of the integral.

By <2)

INTEGRATION 195

This resembles (5). If we insert the factor 2 c 2

after the integral sign and its reciprocal before it,

Comparison with (5).

v = 6 2 + c 2* 2, dv = 2 c 2x dx.

the value of the expression will not be changed.

C x dx 3 a C2c 2xdx [ 3 a Cdv 3 a ,, , , /rvlHence 3 a I

-7 = ^7 I r^ ~ =

^r~, I"~ = ;r^ ln p + ^ by ^5>

J 6 2 + c 2x2 2 o-j 6^ + c-jc 2L 2 r -j v 2 c2

j

Solution. First divide the numerator by the denominator. Then

x -f 1 x + 1

Substituting in the integral, using (1), and integrating gives the answer.

11. f2/

~I dx = x - In (2 JT -f 3)

2-f T.

J 2 x + 39

3._ J ^

Solution. Dividing, ^-r = 1 --- Substitute and use (1) etc.-- --

The function to be integrated is called the integrand. Thus in

Illustrative Example 1, p. 193, the integrand is x (].

PROBLEMS

Verify the following integrations.

1. fx* dx = ^ + C. 6.J*3

a?/2d?/ = a?/

34- c:.

l* = _? +c .

t" t

*5

3. / .r5 dx = =- + C. /

- " ~8

= V2.r + r.

.r

INTEGRAL CALCULUS

= ~- 6x 4- 5 In x 4- C.

2Va by

x -h I)2 cfcr = ^r

4 4- ~ "^ 7 + c -

/* 4 x'2 dx _V Vx:i 4- 8

~

r 6 z dz 1

V(5-32r 2)2 S-Sz^"*"

r /2

rf/ _ _"J (a 4- 6/3 )

2

4- C\

4- C.

INTEGRATION 197

36'/ x*

=a"

3" +C '

/f* ^QlTl 1*^3 Rin*^ T"

sin 2 x cos x dx = I (sin x)2 cos x dx = l "

; -f C = ^-^ -f C,/ o o

HINT. Use (4), making v = sin x, dt> = cos x dx, n = 2.

oo r -j sin 2 aj^

, /,38. / sin ax cos ax ax = h C .

J 2 a

39.Jsin

2 x cos 2 2 x dx = - CQS

^

2 'r-f C.

40. ftan|

see-1dx = tan 2 ~

-f r.

4i/ cosaxdr = 2V6 4- sin a.r ^ ^"

J V6 + sin ax a

rf42.

tan

44 Cr'<i* = ln(2 + r)V 2 + a-

:! 3

_ In (a + M z) .~. f

f d< _5>J ^+1^

~26

= In (^ + 3

In (;/" + 4 ?/)> L-.

'lie

.

- In (a + be 9) .

+ bee~

b+

Ar. r sin x dx i /- N , /-t

49> i- In (1 cos x) + T .

j 1 cos x

50. fs

f^/(^ =lin (a + b tan ^ 4- C.

J a -h o tan ?/ o

51.. f(2 ^ +

^)rf:r = 2 x - In (x + 2) -f C.

J x + 2

62./Lt|* = f- + 3 In (X + 1) + C.

55. d = 2 In6

Work out each of the following integrals and verify your results by

differentiation.

r 2xdx56. 177===;.> V 6 - 5 x 2

Solution./6 - 5 x*

Verification, d [- ft (6- 5 x*)l + ('[

= --ft 3S(6

- 5 x 2)4(- 10 x)dx

2 x dx

57. / O'* -f 3 x 2)dx.

j

58.

66.

67,

68

/ X(/J

v Vi^^5

'

C' <lt

-J ITFT1

sin (iC (10

71. f_EL___J cos a^? + b

esc 2(&fr

72

130. Proofs of (6) and (7). These follow at once from the corre

spending formulas for differentiation, XI and XI a, Art. 94./^x *

ba'2T dx =

Solution. Cba*r dx = b fa 2' dx.

h

By (2)

This resembles (6). Let r = 2 x; then d?? = 2 dx. If we then insert the factor

2 before dx and the factor J before the integral sign, we have

= . + C. By (6)

INTEGRATION 199

PROBLEMS

Work out the following integrals.

1. f6 e** dx = 2 c*r + C. 4. f10' dx = r-^- + C.J J In 10

2.JV dr = nr " + C. 5.JV* <///= ^^ + C.

7. fU" + f ~)(ir = a(r- r" 'V -f C

r( - -"\2 n/ ^

8.JVf

a ~ rn) dx - %\c

(i - ca)

*-ffCxZ dx = I c

r2 + C.

10. f<' Bin x cos x dx = r sin x + T.

11. /Vanfl sec 2 ^ f/^ C i&nf) + C.

12. V? (// = 2VP -f T.

13. dx = + C.1 -f In a


differentiation.


131. Proofs of (8)-(17). Formulas (8) (13) follow at once from the

corresponding formulas for differentiation, XIII, etc., Art. 94.

Proof of (14). ftan v dv = f'sin r dv

cos v

C sin r dv

_ rJ

COS 7'

r)

COS V

In cos v + C by (5)

= In sec r + r.

[Since- In COM r In - In 1 -f In see r In sec r.

wc r J

Proof of (15). fctn v d, = f**^ = f^^J J sin v J sin v

- In sin v + C. By (5)

Proof of (16). Since sec v = sec v fsec v + tan v

sec v tan v + sec 2?>

sec v + tan ?;

/, Tsec r tan ?' + sec 2 v ,

sec r dr ~ I! dv

J sec* v + tan r

(sec r + tan r)

Proof of (17). Since esc v = esc v

sec ?> + tan r

=- In (sec v + tan r) + f. By (5)

esc r ctn r

esc r ctn r

esc v ctn 71 + esc 2 v

esc r ctn v

"2Mdv/, f esc r ctn ?' + esc 2 v

esc v ay = I!

J esc r ctn v

/(/(esc

?> ctn v)

esc r ctn v

= In (esc v ctn r) + C. By (5)

An alternative form of (17) is

esc v dv = In tan %v+C (see Problem 4, p. 201 )

INTEGRATION 201

ILLUSTRATIVE EXAMPLE 1. Prove the following integration.

Jsin2a*dj= - co*-

fl

flx+r.

Solution. This resembles (8). For let r = 2 ax ; then dr 2 a dr. If we now

insert the factor 2 o before ds and the" factor before the integral sign, we get

fsin 2 ax c?x - -i- fsin 2 aj- 2 a dr[

= fsi

J 2 aj L - <*Jsin r eft' -- - ~- cos r 4- T, by (8)

- -.

2 a 2 (i

iLLfSTRATH'E EXAMPLE 2. Prove the following integration.

f (tan 2 x - 1 )'-' dx = J tan 2 x + In cos 2 -f- ( \

Solution. (tan 2 x - 1 )* = tan-' 2 x - 2 tan 2 s + 1 .

tan 2 2 x = sec- 2 x - 1 .

Hence, substituting,

|(tan 2 .S

F

1 )- ds|

(sec'-' 2 *? 2 tan 2 ^-V/x / sec" 2 x dx 2|tan 2 d.

Now let r = 2 x. Then d?' 2 ds. Using (10) and (14), the steps are as follows.

By (2), Art. 2

I sec* 2 x ds = \ I sec- 2 x d(2 x)

ftan 2 x dx = J ftan 2 x d(2 x)

=I I sec 1'

r dr -- \ tan ? = ^ tan 2 x.

v dv .', 1 n cos r I In cos 2 x.

PROBLEMS


r ds sin rns -h C.m

2. / tan bs ds = - In sec bs -f ( \8./U/^sec as djc - In (sec as -f tan <ur) + C.

a

4. f esc v dv In tan J r -f ^Y

.

5. fsec 3 / tan / f// =?,sec 3 / -f C./Iesc ay ctn a?/ c///-- esc ay + C.

7. fcsc 2 3 j: ds = -J ctn 3 j- + C.

8. fctnIdr = 2 In sin

^+ C.

9.fs'

2 sec 2 J:i dx - J tan x:i

H- C.

10.sin- x

= - ctn x -f C.

11. f-^-- = tan x + r.J COS- X

12. f (tan -f ctn Or dO = tan - ctn 6 + C1

.

13. I (sec tan <f))2d$

~ 2 (tan sec 0) + C.

14 JLL = ctn jr + esc J -f C.

'J 1 -f OOB.T

HINT. Multiply both numerator and denominator by 1 cos x and reduce

before integrating.

15. f = tan r - sec jr + C.J 1 -f sin j'

16. ( .

Sin ** = - In (1 -f cos ) + C../ 1 -f cos ,s

. .

./ 1 -f tan jr

18. I j- cos ^'J djc I sin r L>

-f (\

19. f(.r 4~ sin 2 .r)(/.r i(.r2 cos 2 .r) -f C.

20. /" ^l^^L, = 2V4-COS.,- + C .

J >/4 _ cos j>

21. rn + cos^dj = }n(x + sin r) + r _

J jr -f sin j^

22./-J^L^L^ = Vl -f 2 tan fl -f r.

; V 1 -f 2 tan 6

Work out each of the following integrals and verify your results bydifferentiation.

23. /"sin ~ dr. 29.|sec

2 2 or (Lr.

24.Jcos (

b -f flj-V/x.30.

Jtan |dr.

r ^i f f//

25.Jesc* (a

- brtdjr.01

-J tan 5 ^

ftw r </>. a</> 7

, 33.27. I esc

y-ctn

-y

1(/0. J ctn 7 ?/

/> Q . /"sin V.T c/j*

28.J

rx ctn f rfj.

M -

J y^

35.sin 2 3 t

INTEGRATION 203

43. f (1- esc //)

2dy.

J

O.T f a rfjr37. I-

J cos" bj"

45. (V-^J 1 sin .r

"./ 3 -f cos !2 .r

47. f 0? / ( f.

' Vu -j- fo sin f

csc ctn (10rcsc c

'J 5 4

49 rcscj\rV -

csc

csc\r(/.r

ctn .r

38.|(sec 2 0- csc^

39. I (tan </> -f sec </>)-

40.f(tan

4 sctn^

41. f(ctn.r - I)2 dr.

42. I (sec / I) 2(//.

132. Proofs of (18)-(21). Formulas (18) and (20) follow easily

from the corresponding formulas for differentiation.

Proof of (18). Since ,

r\

(II-

1

d(- arc tan - + rW ~ ^^ - -r^-:, by XXII, Art. 60\a a / a

we getr rf'

1 i4

??i ^

-n -"= " arc ian " + r

./ ?'~ + "' a

Proofs of (19) and (19 a). We prove (19) first. By algebra, we have

1 1

Hence

Then

v a r -f a

1 1

a2

_J LI./) a r -f a

\

=7T- ln ( ??

- a)-7r- ln^1 d Ll (I

= _Lin

2a

To prove (19 a), by algebra,

1

i

+ C '

2 a

by(1)

By (2), Art. 1

a + v a v a

The rest of the proof proceeds as above.

NOTE. The integrals in (19) and (19 a) satisfy the relation

r dv _ _ C dv

J r 2 - '-'

~J a'2 - v'

2

'

Hence either formula can be applied in any given case. Later we shall see that

one form must be chosen in many numerical examples.

Proof of (20). Since

arc sin- + (j =a,

*(-\a

we get

>K )'

/dr. v . ,,-^ arc sin - + ( .

Va 2 - r'2 a

=> by XX, Art. 94

Proof of (21). Assume r = a tan z, where z is a new variable ;

differentiating, dv = a sec" z dz. Hence, by substitution,

dr C <J sec 2 z dz C sec 2 z dzr dr __ C a see- z dz __ C sec-sera

./ V?' 2 + a 2 J Va 2 tan 2 z + a~ J Vtan 2 z + 1

I sec z dz In (sec z + tan z) + r by (16)

- In (tan z + Vtan- a +1) + r. By (2), Art. 2

But tan z = -; hence,a

P*= =ln(*+

/ir

J V?'2 + a- \ ^ a ~

In (r + Vr 2 + a~) In a + c.

Placing r = In a + r, we get

I

= In (v + Vtr' + a 2) + C.

In the same manner, by assuming v a sec z, dv = a sec z tan z dz,

we get

/dvra sec a tan z dz r

i \ , \ sec z arV r 1 a 1 J va'-sec^z a L J= In (sec a + tan z) + c by (16)

= In (sec z + Vsec 2z - 1) + c by (2), Art. 2

INTEGRATION 205

ILLUSTRATIVE EXAMPLE. Work out the following integration.

f_*_ = l arc taii^ + r.J 4 x'2 + 9 6 3

Solution. This resembles (18). For, let f ' = 4 r* and a* = 9 ; then r = 2 x

dv = 2dt, and a = 3. Hence if we multiply the numerator by 2 and divide in front

of the integral sign by 2, we get

r_j_-ir 2d* r=!f_*l_:=-Larctan2+r. By (18)1

J 4 z~ + 9 2J (2 x)2 + (3)

2L 2j r 2 + a 2 a J

= - arc tan -- -f C.o <>

PROBLEMS


1 r-^_ = larctanf + C.J x 2

-f 9 3 3

f ,

dg = In ( + V.s^ - 16)2 - 6

6 fJ

dx

Vl6 - 9 x 2 ^= - arc sin '- + C.

9 r_lL = arc tan f + C.

_ ln EL + C.

or

f ax dx = aarc tan |! 4. C .

'

J x4 -f 64 2 & 2 ?>2

i f. ^ = - arc tan(

T-^)+ C.

J (- 2)

2-f 9 3 \ o

15. fdy = i In (ay + Vl + ay) 4- C.

J Vl + aV2 a

18. f <** = arc sin ("*\ + C .

J V4 - (u + 3j2 \ 2 y


The standard formulas (18)-(21) involve quadratic expressions

(7^'ia-, a- ?;L>

) with two terms only. If an integral involves a

quadratic expression containing three terms, the latter may be re-

duced to one with two terms by completing the square, as shown in

the following examples.

ILLUSTRATIVE EXAMPLE 1. Verify the following:

= 1 arc tan ^4_1 + (\+ 2 x -f 5

Solution. x'-' + 2 x + 5 = x- -f 2 j- -f 1 + 4 - (x -f I)2-f 4.

. r ^ = r ^"

J jr2

-f 2 j + 5 J (r + I)2 + 2 2

This is in the form (18). For let r = x + 1, and a = 2. Then dt; = dx.

Hence the above becomes

/-;-

;

= - arc tan --f C ~ - arc tan ~t y C.

r 2 + a* a a 2 2

/9 /fv. 9 y.1

^- 2 arc sin :1-:i + C.V2 +x - x'2 *

Solution. This is in the form (20), since the coefficient of x'2 is negative.

Now 2 -f a* x 2 = 2 (x'2 x -f J) 4- 1 = S (x J)

2.

Let v = x I, a ?. Then dt> = dr.

f_2dr = 2r dr = 2 f d; = 2 arcsin 2 + c> by (20)

^v2+x-x 2 ^ v| -(j-

J)-' JVa-'-r2 a

INTEGRATION

ILLUSTRATIVE EXAMPLE 3. f--~- = ~ In3 x ~~ 3 + C.

J 3 x 2 + 4 x 7 10 3x4-7

Solution. 3z 2 +4x-7 = 3(x2 + 1 x - j) = 3 (or

3 + J * + I- V)

= 3[(:r+{jV-Vl.

. f<*T - C dx = \ C dr

'

'J 3 x 2 + 4 x - 7 J 3[(x -f J)3 -

J 3j r 3 - a-'

by form (19), if r = x -f ,a = *j, since also dr = (ir. Then wo have

PROBLEMS


4. f .

dj< = = arc sin (2 s - 3) -f C.J Va x - jc* - 2

' o 2J 2 x 2 2 x

dx

/2 x - x 2arc sin (x 1) -f C.

/2 as + s'2= = ln

+ 3 y + 1

~V5

A"

\2 I/ -f 3 + V5

^ = 2arc tan

/2 + l\

l-fx-fx 2 V3 \ V3 /

207


16 . f Jfa= 1 arc tan (2^tl) + C.

J4x J 4-4x4-5 4 \ 2 /

= arc tan =Vll \ Vll

17. f .

* =|arc sin (*Z*\ + C.

J V2 - 3 x - 4 x 2 2\ V41 /


differentiation.

10 jc09' f

4(^V Vx 2 - 4 x -f 13

.,31r_a

. I .

J Vi> 2 -

Vl -t- 2 P

r* 4- 4x4-1*

2 w 2 4-2 w 4- l"

r x 2dfx

36-J 9 j.o _ 3 ^ _ i"

37.15 4- 4 t

-t2

S8./:V9 x 24- 12 x -f 8

28 f 39 - r y

dx =*'J r 2 _ 2 r - 3 ^ V4 x2 - 12 x + 7

When the integrand is a fraction of which the numerator is an

expression of the first degree while the denominator is an expression

of the second degree or the square root of such an expression, the

integral can be reduced to standard forms, as shown in the following

examples*

INTEGRATION 209

ILLUSTRATIVE EXAMPLE 1. Prove the following :

3x " 1r = V4^-4-9-lln (2 x 9) 4- C.

Solution. Multiply through by dr and apply (1).

C 3x-l ^ = r 3jdr__ r dJ

J V4 x 2-f 9 > V4 x a

-h 9 J ^4 jr2

Then by (4) and (21) we obtain the answer.

ILLUSTRATIVE EXAMPLE 2.

-f 9

Solution. 3x2 +4z-7=r 3[U -f SV -V'l, by Illustrative Example 3, p. 207.

Let v = x + 1. Then x = v - ?, and dr = dr. Substituting,

r 2 x - 3 ^ _ T2 ( r - "i )- 3 1 C<r^ ,,

j3x^4x~7dr

~J air^-V) J ^-^- 1 r 2 r (/r _ 12 r_^'~~aj f 5 - v 9 J r a - V'

Using (5) and (19), and substituting back t> = jr + 3, we have the above result.

PROBLEMS


-f 2

~*)<** = _ VF^T^ - arc sin j- -f C.

10 f(2x1U'Jx2 +

12.

13.

14.

!^JI

ln(1 _ 6 ,_ 9:r2)- 6x 9 r 2 6

4 \3 x -h 1 + V2+ C.

= Vx 2 + 2 a- -f 2 In (x + 1 + Vx 2 + 2 x) + C.

V4 x - x 2

arc sin -f r.

15.

/27 -f 6 x - x 2

.= - V27 -f 6 x - x 2 + 3 arc sisn

B .f fy-+ ~')</x _ ,.

'J Vl9 - 5 x + x 2

17 rjyizjjd'

J V4 x 2 - 4 x

-I- V In (r - -f Vl9 - 5 x -h .r2) + C.

-f 5

1*

-:|In (2 x - 1 + V4 x 2 - 4 x + 5) -f C.

.. ^^ J ~ ^ ^^'r - = 2Vl2 x ~ 4 x 2 5

'J Vl2x~4 x 2 - 5Q / 9 X 3\

+ ~arcsinr-r

)-f C.


19-/^T^'20

jCi^-4V/x.

21'jT-:ij- 2

'

2g-/yo-^'

23 r(4 -r - lw-r

.

^ V3 + 5r-

27.(3- 4

V3 x - a-2 - 2

28 . f(5^ + 2)^

.

J Vx- + 2 a- 4- 5

30 f(8- 3 x)<ir

'

J x a + x + 1

QI r (J" + 4)dr

24. r^4-J X- -

32.(2r +

25 r(i4J^.^ V3j'-x 2

2 x 2 + 2 x -h 1

(3o, f (3a- + )riJ-33'

J 9 x* - 3 x - 1

f.(6-

r)(fa34

V4x 2 - 12 x + 7

INTEGRATION

133. Proofs of (22) and (23). To prove (22), substitute

t' = a sin z.

dv = a cos z dz,

211

Then

and

Hence

Va2 v2 = Va2 a 2 sin 2 z = a cos z.

fVa2 -z;2<fo = a2 fcos2 z dz = f(cos 2z+ l)dz by (5), Art. 2

To obtain the result in terms of r, we have, from above,

z = arc sin - and sin 2 z 2 sin z cos z = 2 - a "~??

a a a

Substituting, we obtain (22).

Proof of (23). By substituting r = a tan z, we show (see Art. 132)

readily that

(1) I V?'2 + a2 dr = I a sec z a sec- z dz = a2I sec* r dz.

In a later section it is proved that

(2) fsec3 zdz = l sec z tan z + I In (sec z -f tan z) + r.

we derive, from (1) and (2),Since tan z = - and sec- z =a

(3) fx/V + a2 d -IV?- + a 2 + ~ In (r + v/r-' + a-) + C'

9

where C'=C-^- In a. Hence (23) is proved when the positive sign

holds.

By substituting v = a sec z, we obtain (see Art. 132)

(4)IvV2 a2 dv =

|a tan z a sec z tan z dz = a2

jtan 2 z sec z dz

= a2( sec 3 2 dz a2

Jsec z dz.

Comparing (4) with (2), we have

(5) fVfl2 a2dtf = y sec z tan z - ^ In (sec z + tan z) + C.

But sec z = - and hence tan z = Substituting in (5),

we obtain (23) for the negative sign before a2 .

ILLUSTRATIVE EXAMPLE 1. Prove the following:

f \/4 - 9 x 2 dx = 7 \/4 - 9 x24-

1arc sin

-^ +C.

Solution. Compare with (22), and let a 2 = 4, r = 3 x. Then dp = 3 dx. Hence

f \/4 - 9 x'2 dx = 7 fV4 - 9x 2 3 dx = - f v a* - t>2 d0.

,/ 3 J <> /

Using (22), and setting P = 3 x, a 2 = 4, we have the answer.

ILLUSTRATIVE EXAMPLE 2.

f \/3 x 2-f 4 x - 7 dx

= 1(3 x -f 2)\/3x* + 4z-7- 25-^2 In (3 x 4- 2 -f V9 x 12 + 12 x - 2l) -f C.1 18

Solution. By Illustrative Example 3, p. 207,

3 x 2-f 4 x - 7 = 3[(x -f I)

2 - V] = 3 ( r2- fl2 )

if v = x 4<j ,

a =/}

. Then d?? dr.

/. C\'\lx* + 4 x - 7dx = V^ f\ t'2 - a 2 dr.

Using (23), and setting r = x 4- ,=

.we obtain the answer.

PROBLEMS

Verify the following integrations :

1. JVl -4x*dx = ^ Vl - 4 .r2

4-1arc sin 2 x 4- C.

2. fVl 4- 9.r 2 dx = ? Vl -f- 9 J 24- 1 In (3 a- 4- Vl 4- 9 .r

2) 4- C.

/ li D

3. /\ /~ - 1 cij: = 7 Vx 2 -4 - In (a- 4- V.r L> - 4) 4- C.J \ 4 4

4. fV25 - 9 x 2 dx - 7 V25 - 9 x'2 + ~ arc sin ^

4- C./ 2t t> o

5. fV4 :r24- 9 dx = - V4 x 2

4- 9 4- 7 In (2 x 4- V4 x 2-f 9) 4- C.

./ 247. TV3 -2x- x 2 dx = 21--l V3 - 2 x - x 2

4- 2 arc sin ^--f C.

8. rV5- 2x4-x 2rf.r = ^-^ V5 - 2 x 4- x 2

x 2 dx =4- 2 In (x - 1 -f V5-2x4-x 2

) 4- C.

2 x - x 2 + arc sin (x-

1) 4- C.

10.JVlO-4x4-4x2 dx = 2J

4

~ 1 VlO - 4 x -f 4 x2

+ 5 In (2 x - 1 4- VlO-4x-4-4x2) 4- C.

INTEGRATION 213


11.JV16-9x 2 dx. 16. fV5 - 4 x - J a dr.

12.JV4 + 25 x 2 dx. 17. JVs 4 2 JT +7* dr.

13.JV9 r 2 - 1 rfr. 18. (Vr-- 8 .r + 7 dr.

14.JV8- 3 .r

2 dr. 19. fV4 - 2 .r- .,- dr.

is. fV5 -f 2 .r- dr. 20. rVJ^^TTTn dr.

134. Trigonometric differentials. We shall now consider some trigo-

nometric differentials of frequent occurrence which may be readily

integrated by transformation into standard forms by means of simple

trigonometric reductions.

Example I. To find \ sin" 1 u cos n u (hi.

When either m or n is a positive odd integer, no matter what the

other may be, this integration may be performed by means of simple

transformations and formula (4),

it- -T

For example, if m is odd, we write

sinm

?/ = sn m ] u sin u.

Then, since m 1 is even, the first term of the right-hand mem-ber will be a power of sin 2

w, and can be expressed in powers of cos2 u

by substituting sin->

u = 1 _ Cos2 M>

Then the integral takes the form

(1) I (sum of terms involving cos ?/) sin u du.

Since sin u du r/(cos M), each term to be integrated is of the

form vn dv if v = cos u.

Similarly, if n is odd, write cos" u = cos 7" 1 u cos u, and use the

substitution cos2?/ = 1 sin 2 u. Then the integral becomes

(2) I (sum of terms involving sin u) cos u du.

ILLUSTRATIVE EXAMPLE 1 . FindJ sin 2 x cos5 x dx.

Solution. I sin 2 x COB* x dx I sin 2 x cos4 x cos x dx

= fsin 2x(l - sin 2

x)2 cos x dx by (2), Art. 2

= I (sin2 x 2 sin 4 x -f sin" x) cos x dx

= f (sin x)'2 cos x dx 2 I (sin x) 4 cos x dx +

|(sin x)

fi cos x dx

sin'1 x 2 sin fl xBy(4)

5-

Here r = sin x, dr - cos x dx, and n ~ 2, 4, and 6 respectively.

ILLUSTRATIVE EXAMPLE 2. Prove I sin' J x dx = I cos :i

J x - 2 cos J x -f C.

Solution, l^t J x = ?/. Then x = 2u, dx = 2du. Substituting,

(3) /sin-' \ x dx - 2j

sin :! w du.

Now f sin-' w dujsin 2

?/ sin ?i dw - I (1- cos 2

w) sin ?* d?^

~ / sin u du I cos 2?/ sin w d?/ = cos u -f \ cos r<

?^ 4- C.

Using this result in the right-hand member of (3) and substituting back u x,

we have the answer.

PROBLEMS


1. fsin : <

j- dx =?,cos : '

JT cos JT -f C.

2. fsin 2 B cos 8 dO = J sin-^ ^ 4- r\

3.|cos 2 sin (/> d</> i cos :< $ + C.

4. Tsin-'^ 6 j* cos B .r f/.r = ^T sin'1 6 jr -f r\

5. fcos^ 2 8 sin 2 <f fl = -J cos4 2 fl + C.

6. r^~^ d.r - esc x - \ csc :}j- + C.

J sin4JT

7. f51^ rf0 = sec + cos + C.

J COS 2

8. fcos4 J sin3jc d.r = -

\ cos 5JT + \ cos7 x + C.

9. Tsin 5 x dx = cos x -f 1 cos3or i cos5 x -f- C.

INTEGRATION

10. jcos5 x dx = sin j f sin3

jc -f* sin 5

.r -f C.

11. fs)

n5 ?/

d?/ = - 2Vcos ?/ (1 - = cos- v -f

- cos4/A -f- C.

J V cos y*

\ o '9 *

/

12 .

r COBUL d, = 3sin s

, /!i

sin ,,

iBin ,

A r''N/riin

2 V 2 /


13. fsin3 2 6 dO. 18. /sin 1 ' m/ cos 2 wf <Jf.

/j ^

14. fcos3"d0. 19.

Jsin-'

1 r </j-.

15.J"sin

2 a- cos 2 a- dr. 20.J'coa

1

( + bt)dt.

[tan

16.J

sin :!/ cos:t

< <.Z1J V

17. cos^' sin*

Example II. To ^?irf I tan" u du or \ ctnn u (in.

These forms can be readily integrated, when n is an integer, on

somewhat the same plan as in the previous examples.

The method consists in writing, as the first step,

tan n u tan"~ 2 u tan 2 u tan n -

?v (?ec*L> u 1) ;

or ctn n u = ctnr'" a

?i ctn 1' w - ctn ri L>?/ (esc- w 1 ) . By (2), Art. 2

The examples illustrate the subsequent steps.

ILLUSTRATIVE EXAMPLE 1 . Find / tan 1 r dx.

Solution. I tan4 x dx =|

tan'-' x(sec?

j-- 1 )djc

= ftan 2a- sec 2 x (Lr - I tan 2 x do-

= f(tan x)- d(tan x)-

\ (sec2 J ~ l)dx

= ianiLH: _ tan x -f x -f- C. By (4) and (10)3

ILLUSTRATIVE EXAMPLE 2. Prove

fctn'! 2 x dx = -i ctn- 2 x - i In sin 2 x -f C.

Solution. Let 2 x = u. Then x = i w, dx = J dw. Substituting,

(4)fctn :i 2 x dx = U ctn :i u du.


Now / ctn a u du = I ctn u ctn 2 u du

I ctri w(csc2 u l)du

= / ctn u esc 2 u du I ctn u du

= -\ ctn 2 u - In sin u + C. By (4) and (15)

Using this result in the right-hand member of (4) and substituting back u = 2 z,

we have the answer.

Example III. To find I sec^udu or I cscn udu.

These can be easily integrated when n is a positive even integer.

The first step is to writen 2

secr'u = sec"" 2 u sec-w = (tan-w + 1)

2 sec2w- ;

n-2

or cscn w = csc"~ 2?/, csc 2 ^ = (ctn

2 w + 1)2 esc2 w. By (2), Art. 2

The example shows the subsequent steps.

ILLUSTRATIVE EXAMPLE ,'j. Prove fsec 4 \ x dx = *tan l

?, z + 2 tan 3 z + C.

Solution. Let \ x n. Then x 2 u, dx = 2 du. Substituting,

f- /

sec 4.j

j* r/j = 2 / sec 4 M dit.

//sec 4 w d?/~

I sec 2 w . sec 2 u du

=I*(tan

2 M -f 1 ) sec2 M d?z by (2), Art. 2

= I tan 2 w sec 2 u du -f I sec L> ?/dzt

-J tan ' u -f tan u + C. By (4) and (10)

Substituting back in the right-hand member of (5) and putting u = \ x gives

the answer.

EXERCISE. Set sec L w = 1 -f tan 2 w in the right-hand member of (5), square,

and follow Illustrative Example 1 above.

Example IV. To find I taw it secn u du or I ctnm u cscn u du.

When 7? is a positive even integer we proceed as in Example III.

ILLUSTRATIVE EXAMPLE 4. Find ftan x sec*xdx.

Solution. ftanx sec 4 z dx =|\an

fi

x(tan2 z -f 1) sec2 x dx by (2), Art. 2

=I(tan x)

8 sec 2xdx+ Ct&n*xsec2xdx

=^ +^ + C. By(4)

Here r = tan x, dv = sec2 xdx, etc.

INTEGRATION 217

When m is odd we may proceed as in the following example.

ILLUSTRATIVE EXAMPLE 5. Find / tan* x sec ' x dr.

Solution. I tan 6 x sec 3 x dx = I tan 4 j sec- .r sec j tan x dx

=j

(sec- j--

1 )* sec- .r sec x tan .r dr by (2), Art. 2

= / (secft j 2 sec 4 j 4 sec- .r) sec .r tan J fir

=^! _!** + **lr + Ci (4)< o o

Here r = sec x, dr = sec x tan x dr, etc.

The methods used in the above examples are obviously limited in

their application. For example, they fail in the following case.

I sec3 u du = I sec u sec- ?/ du

= I sec u tan 2 u (In + In (sec ?/ + t^n u).

For we cannot proceed further by the elementary standard forms.

Later other methods will be developed of more general use.

PROBLEMS


1.J tan3 xdxl tan 2 x + In cos jr + C.

2. fctn3| dx = - ~ ctn- 1

- U In sin ^ 4- C.Jo O '>

3. fctn3 2 x esc 2 x dr -^>esc 2 r -

J,csc :i 2 T -f- C.

4. fcsc4|dx = -

|ctn- {

|- 4 ctn

^-f r:.

5.Jtan5 30d6 = TV tan4 3 -

i tan 2 30+ J In sec 3 f? + C.

cos4

= tan 2 j + \ tan3 2 x - \ ctn 2 x + C.- -sin 2 2 x cos4 2 x 6 2

r A ,3 x mn 2 x

,

sin 4 x . n3. / cos4 x dx = -f 1 i" c -

5x sin_2jr sin* 2 x 3 sin 4 a:

16 4~~ 48 64

5 x sin 2 x sin3 2 x . 3 sin 4 x

~C.

5. /cos x dx =-:y

-f48

1

64/xsin 2 ax dx = -L*

r . ,, x o x 7

7./sin 2 - cos- - dx =

/ tL M

s sin or .,

h C .

4a

x sin 2 x

8~

16+

r . . . 3 x sin 2 ax . sin 4 ax ^8.J

S1n^arrL: = T + -|^- + C.

Z_ _i_8 ^ n:< 4 x

__sin 8 x

~~16 96 1289.

10./(2

12.

C.

sn

. fsin2xx cos 4

. n sin 2 .

4 cos h T.

cos 2 x cos 6 x

4 12

sin x sin 5 x, T o o j

sin :r sin t> x ^13. I sin 3 x sin 2 x dx = ~ h C.

r . o j sin x . sin 7 x . ^14. I cos 4 x cos 3 x dr =

7; 1 rr h C.


differentiation.

15.Jcos 2 ax dx.

16. / cos4 ax dx.

17. I sin 2 ax cos 2 ax dx.

18. fsin4 - cos 2 - dO.J ^ ij

19. (sin4 2 a cos4 2 a da.

20. / sin 2 x cos6 x dx.

-fcosx)3 dx.

22.f(Vsin2 6 - cos 2 0)

2d0.

23. r(VSoT5 - 2 sin 0)2d0.

24./(sin

2 x - sin 3 x)2 dx.

25.J (sin x -f cos 2 x)

2 dx.

26.J(cos

x + 2 cos 2 x)2 dr.

INTEGRATION 221

135. Integration of expressions containing Va2 - u2 or Vu 2fl2by

trigonometric substitution. In many cases the shortest method of in-

tegrating such expressions is to change the variable as follows.

When Va 2 u 2occurs, let u = a sin z.

When Va2 + w 2occurs, let u a tan z.

When V?-/ 2 a- occurs, let ?/ = a sec1

z.

These substitutions were used in Arts. 132-133. By them the radi-

cal sign is in each case eliminated. For

(1)

(2)

(3)

Va 2 a'2 sir\~z =

Va2 + a 2 tan 1':: = a Vl 4 tan- z a sec z ;

sin L> r = a cos z;

Va2 sec 2 z a 2 a-Vsec 2 ^

~

1 = a tan z.

ILLUSTRATIVE EXAMPLE 1. Find f ^ -.

J( fl

l! _ M '.')!!

Solution. Let u a sin z;then d</ cos c (/. and, using (1),

du _ r a cos z dz = j_ / ck _ J_ /'^..j.^,

a2_ w 2)fJ a :i cos^ fl-.' cos-r r/*.'

'

For, since sin z - draw a right triangle and mark thea

sides as in the figure. Then tan z

ILLUSTRATIVE EXAMPLE 2. Prove f =ii=r = - In -

'

j-\ 4 j- } 9 l{

Solution. Here v'4 x 2-f 9 \''u- -f </

Jif = - J*. <i ~ 3.

Hence let 2 r = w ; then j = ,\ M, dx = ,1 (///. Substituting,

^ '

/* (/J*__ /_jj_//^____ r JTJ^/

J y\ 4 j 2-f 9 ~/ J u \ 7/* -f

-' ^ wx'w 2-I- a 2

Let M = a tan c. Then dn - a sec-r (/r, and, using (lij,

/rf?// a sec- c Jc

__ 1 T sec z dz__ !_ r_dz_

uVu 2-f a 2 ~^ a tan -

' w se(> -

~""' tan c

~(/t ' sin

= -1- fcsc z dz - - In (esc z - ctn z) -f r.o ./

Since tan c = - draw a right triangle and mark the sides as in the figure. Then

. f = -u

/du 1

/ .. 9= ~

uV u2-\- a o

In

Substituting back in (4), and setting, as above, u = 2 x, /\a = 3, we have the answer.

PROBLEMS


dxr dx _ x_

C dx __ xi. I ...,= ~r i-

r f2 df _ _ I

r ir2 dx x

a C ?/2(/ " - 1Lo. I

"

.; ,

J(9_ W 2)&

v9 -v/

2

4 - /-' -f 2 arc sin ^-f r.

arc sin h C.

7.

9,

10.

.rVj' 2-f 4 2 \2 4. V.r 2

-f 4

./*

5 \5 -f V25 - .r2

r.

-f c.

.f^ ^^~ 7

+ r.J

//

2V/y

2 7 ^.'/

/2

/


differentiation.

13 ..

rJ

A/2 - 9 <///

l*.f j -;

9 fa

VlOO -?/

2 du

- 2 - 5

INTEGRATION 223

136. Integration by parts. If u and r are functions of a single inde-

pendent variable, we have, from the formula for the differentiation of

a product (V), Art. 94,F

or, transposing, 'u dr - d(uv) r du.

Integrating this, we get the inverse formula,

(A) I u dv - uv - I v duy

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 mustbe separated into two factors, namely, ?/ and dr. No general direc-

tions can be given for choosing these factors, except that

(a) dx is always a part of dv;

(b) it must be possible to integrate dv; and

(c) when the expression to be integrated is the product of two fwnc-

tions, it is usually best to choose the 'most complicated-looking one that it

is possible to integrate as part of dv.

The following examples will show in detail how the formula is

applied :

ILLUSTRATIVE EXAMPLE 1.FindJ

x cos xdx.

Solution. Let u = x and dv cos x dx ;

then du = dx and v \ cos xdx = sin x.

Substituting in (A),

u dv u v v du

f x cos x dx = x sin x ~ f *in xdx~ x sin x -f cos x -f C.

ILLUSTRATIVE EXAMPLE 2. Findj

x In x dx.

Solution. Let u = \n x and dr xdx;

then du=: ^ and v


; xdx

ILLUSTRATIVE EXAMPLE 3. Findj*xe**dx.

Solution. Let u = c aT and dv = xdx;

r x*then du = cai a dx and r =

|x dx =

J


Cxe tt'dx =- cax - ^ -

f~

<>ar a dx

But x 2caT dx is not as simple to integrate as xcar dx, which fact indicates that we

did not choose our factors suitably, instead, let

u = x and dv e"r dx ;

/xiozcai dx --

d


,. f,<iri'('

(lx

I xcar dx x - --I dx

J a J a

It may be necessary to apply the formula for integration by parts

more than once, as in the following example.

ILLUSTRATIVE EXAMPLE 4. Findf

x'2c"f dx.

Solution. Let u x~ and dv = c a*dx;

fcax

c (lj dx =

Substituting in (4),

The integral in the last term may be found by applying formula (4) again,

which gives

a

Substituting this result in (1), we get

a a 2


Csec*zdz - \ sec z tan z -f \ In (sec z + tan 2) + C.

Solution. Let u = sec z and dv = seQ 2 zdz;

then du sec z tan z dz and v = tan z.

INTEGRATION 225


j sec^zdz = sec z tan z - fsec z tan 2 rdz.

In the new integral, substitute tan-'c = sec*,: - 1. Then we get

Jsec :i zcfc = sec r tan c -

|see- rdc + In ^sec r -f tan z) + C.

Transposing the integral in the right-hand member and dividing by 2, we have therequired result.

ILLUSTRATIVE EXAMPLE (>. Prove

ff' sin ru-dr - l!lL<L!LH~ * < MJ)

r' (/' 4 n~

Solution. Let H = c ar and c/r = sin HJ- (/JT ;

then dw = w a'd.r and r = - ^Ji.

/i

Substituting in formula (.4), the result is

(2) fc< sin nx dr = - ^j^nx a r.

(.()s ^^

-' N ,/

Integrate the new integral by parts.

Let u eax and t/r = cos nx djr;

then dw = acar dx and r = ^LM.n

Hence, by (v4),

(3 ) />* cos nx dx = I^'JJIIL^ _ 1! />, sin nj f/x>/ n 11 J

Substituting in (2), we obtain/Cnf/ \ (]

^ /*

c (i:r sin nx djr a sin nx n cos no*) |

c"-1 sin nx dx.

n'2 \ / w- J

The two integrals in (4) are the same. Transposing the one in the right-handmember and solving, the result is as above.

Among the most important applications of the method of inte-

gration by parts is the integration of

(a) differentials involving products,

(b) differentials involving logarithms,

(c) differentials involving inverse circular Junctions.

PROBLEMS


1. I jr sin jr djr sin x x cos x -f C.

2.Jinx dx = x(ln x - 1) + C.

3. \x sin ^ dx 4 sin ~J

A C. I xJ

2 or cos ~ + C.

7cos nx dxcos nx--

n 2

sin nxn

. ,,h C.

5. / w sec 2 u du ~ u tan ?/ + In cos u + C.

6.J v sin 2 3 v dr = \ r'2 ^ r sin 6 r ^ cos f> r + C.

7. > Bin n7/ rf*COS 2 ?/ Rin cos

C.

8.

9.

a' (Y^-- =-4-1 + r -

[In a In 2 a

j

In x <Lr --f

In-j-

4- r.

10. f arc sin x dx ~ x arc sin x + "N/l x~ + r.

11.Jarc tan .r dx x arc tan r I In (1 + j~~) + C.

12. / arc cot ?/ d/y = ?/ arc cot y + J In (1 + y 2) + r\

13.|arc cos 2 x dx ~ x arc cos 2 x -~

JVl 4 .r2 + C.

14. / arc sec ?/ r/?/ ?/ arc sec ?/ In < ?y + V//- 1 ) + C.

15. fare esc - dl t arc esc ^ + 2 In (/ + V/ 2 4) + C./x 2 + 1 ^^ arc tan x dx - arc tan x - + c\

17. fare tan Vj" ^ (x + 1) arc tan W Vx + r.

18. \ JT-C x^/.r = c '(2 + 2 x + .r

L>

) + r.

19. Cc cos dO ~~ (sin + cos 0) + C'.

In x ~ In (x + 1) + C.

= 2V [In (x + 1) - 21 + C.

INTEGRATION 227


25. Cx sec 2^ dr. 36. fare tan V.r dr.

26.J JT cos 2 2 x <Lr. 37 f r

.r*

37.j

.r{ arc tan .r dr.

27. J jr~ cos JT dr. 38. f (<'

-f '2 .r}- dr.

28.J arc sin rw.r dr. 39. I (U1

-h .r*)-' dr.

29. fare ctn ^ dr. 40. /r '' cos ^ (/fl.

./ - 7 !L

30.Jarc

cos - dr. 41j",

'

sin 7r/ (//

31. fare sec -<///. 42. /<'' cos

"'

dr.J //

'

.^ >

32.Jarc esc nt dt. 43. / c - cos \1 f ill.

33. Tare sin ^1^ dr. 44 /^J (.()S nf (ff

34.J .r' arc sin r dr. 45.fr 4 sin <//.

/.r

arc sin .r dr r/^ _ 2 46. / csc' ; dO.

137. Comments. Integration is, on the whole, a more difficult op-

eration than differentiation. In fact, so simple an integral (in

appearance) as r\f'jr sin x dx

cannot be worked out ;that is, there is no clvmentary function whose

derivative is V? sin x. To assist in the technique of integration,

elaborate tables of integrals have been prepared. A short table is

given in Chapter XXVII in this book. The use of this table is ex-

plained below in Art. 176. At this point let it suffice to remark that

the methods thus far presented are adequate for many problems.

Other methods will be developed in later chapters.

MISCELLANEOUS PROBLEMS


12.JV2' ~ 2x) 2 dx.

IQ r ^13 '

J ^ - 4 r'

14.jsin 2 ax cos ax dx.

15.jsin 2 ax cos2 ax dx.

IG.Jln (1- Vx)dx.

17.J(2tan2^-ct

19 '

20

r tfdx.

'^ Vx 2 + l

32.j

(c* + sin x)2 dx.

33.J"(x

- ctn x)2 dx.

34.y (1 -f tan x)3 dx.

35. f(

J (1 cos 0)-'

(1 -f sin Q 3 dt36.r- cos

Vx 2 - 1

37.y^-'sin 2 t dt.

38.ysin2 6 cos 3 6 dd.

39.jsin sin 4 d</>.

40.Jcos a cos 2 a da.

CHAPTER XIII

CONSTANT OF INTEGRATION

138. Determination of the constant of integration by means of initial

conditions. As was pointed out on page 1 90, the constant of integration

may be found in any given case when we know the value of the

integral for some value of the variable. In fact, it is necessary, in

order to be able to determine the constant of integration, to havesome data given in addition to the differential expression to be

integrated. Let us illustrate this by means of an example.

ILLUSTRATIVE EXAMPLE. Find a function whoso first derivative is 3 x'2 2 x -f 6,

and which shall have the value 12 when .r -- 1.

Solution. (3 x'2 2 .r -f 5k/.r is the differential expression to be integrated.

Thus = .H - JT--

-f T,

where C is the constant of integration. From the conditions of our problem this

result must equal 12 when .r = 1;that is,

12 = 1-

1 -i 54- <\ or r = 7.

Hence x :$ x 2+- 5 j* -f 7 is the required function.

139. Geometrical signification of the constant of integration. Weshall illustrate this by means of examples.

ILLUSTRATIVE EXAMPLE 1. Determine the equation

of the curve at every point of which the tangent line has

the slope 2 x.

Solution. Since the slope of the tangent to a curve at

any point is -^, we have, by hypothesis,

Integrating,

cfc'

dy = 2xdx.

y = 2 Cx dx, or

(1) y = x'* + C,

where C is the constant of integration. Now if we give to

C a series of values, say 6, 0,-

3, (1 ) yields the equations

y x 2 4- 6, y x'2, y - x'2 - 3,

whose loci are parabolas with axes coinciding with the jy-axis and having 6, 0, 3

respectively as intercepts on the y-axis. ,

All the parabolas (1) have the same value of ^;that is, they have the same

direction (or slope) for the same value of x. It will also be noticed that the difference

229

in the lengths of their ordinates remains the same for all values of j. Hence all the

parabolas can be obtained by moving any one of them vertically up or down, the

value of r in this case not affecting the slope of the curve.

If in the above example we impose the additional condition that the curve shall

pass through the point (1, 4), then the coordinates of this point must satisfy (1),

fciving 4 ~1 -f r, or r ~ M.

Hence the particular curve required is the parabola y y'2 -4- 3.

ILLUSTRATIVE EXAMPLE 2. Determine the equation of a curve such that the

slope of the tangent line to the curve at any point is the ^ratio of the abscissa to the ordinate with sign changed. I

Solution. The condition of t he problem is expressed

by the equation ({,i /

or, separating the variables,

?/ </// dx.

Integrating,

or

This, we see, represents a family of con centric circles with their centers at the origin.

If, in addition, we impose the condition that the cur\<> must pas.^ through the

point (H, 4), then 9 4-1(5~ 2 r.

Hence the particular curve required is the circle .r-' -f //-'^ 25.

PROBLEMS

The following expressions have been obtained by differentiating certain

functions. Find the function in each case for the given values of the vari-

able and the function.

Junction

1. IT- 3

2. 3 4- r - f> S-

O n n\ 1\ '2 ii

4. sin 4~ cos

M-J-h6. sec 2

4- tan

*

:r24- (i

2

8. for34- a.r 4- 4

9. Vt 4- rVt

10. ctn 6 - esc 2 6

11. 3fe 2 '2

of

vnlnc of junction

9

-20

Answer

\j- - 3 JT+ 13.

304 4- 3 J- 4- i .r2 -

f; jr\

\ .V

1 -2 ^ L1

//2 + 2 6 2 - 4.

sin cos 04-1.

tan</> 4- In sec

</> 4- 5.

X 7T

a 4 a

CONSTANT OF INTEGRATION 231

Find the equation of the family of curves such that the slope of the

tangent at any point is as follows.

12. m. A us. Straight lines, // nur -f ('.

13. x. Parabolas, //

-k j-

2-f (\

14. -Parabolas, A

//-'.r -f (\

x 2

15. Semicubical parabolas, \ //'-' J,.r

l

-f C

16. -L^ Seinicubical parabolas, ;\ //*

-..J

.r~ -f (\

17. tt x 2. Cubical parabolas, \t

-- .H -f- r.

18. Cubical parabolas, \ //<-- .r + r.

?/"' '

JC

19. --Equilateral hyperbolas, \r

~~J"'

2 C.

20. - -- Equilateral hyperbolas, .r//= r.

21. - Hyperbolas, b-jc2 a 12

//1-' r.

a 2//

22. -^- Ellipses, //-'.r- 4 a'-'//- r.a'-y

23. i-i^- Circles, j- -f //" -f 'J .r-

i //= r.

In each of the following examples find the equation of the curve whose

slope at any point is the given function of the coordinates arid which

passes through the assigned particular point.

24. .r; (1,1). -4//.S-. '1 //----- jr* -f 1.

25. 4 //; (1, 1). In //- 4 jr

- 4.

26. 2 JT//; (8, 1). In u~ s~ - 9.

f''

27. -jry; (0, 2). y = 2 r ^.

28. ^4; (0,1). (?/+ 1;;! = U-f l)

2-h 3.

?/ + 1

29. -; (0, 0). ^ + y*

- 2 f<s - (

1 ky = 0.

y - k

30. 4,; (1, 1). x In //= r- 1.

x 2

31. ?yV7; (4, 1). 3 In //- 2(xV^ -

8).

32. 4^. (2fl) 4^-^-15.4 j*

2 15

33. ; (1,4). 37.2 y - 3

'

; (1, 9). 3g

35. ~~Q

*3C ____L__ /'I O \"

x 2-f 4

' '

40. x cos 2y ; (4, J TT).

41. Given dy (2 j* ~h l)<tr, ?/= 7 when y = 1. Find the value of y

when x 3. A??.s. 17.

42. Given dA "N/2 />./ r/.r, A ~- when x = ~ Find the value of Awhen .r = 2 7>.

' ",4r?*. g ;?

2.

43. Given dy j*Vl()0 ~j'

2r/^, ?y when .r

~0. Find the value of

?/ when x = 8.

44. Given dp cos 2 6 dO, p 6 when # J TT. Find the value of pwhen =

ijTT.

45. Given ^s = /V4 f -f 1 dt, K = when / = 0. Find the value of s

when t = 2.

46. At every point of a certain curve //" = .r. Find the equation of the

curve if it passes through the point (3,0) and has the slope ^ at that point.

.4w,s\ 6 y = j"3 - 6 jc 9.

147. At every point of a certain curve y" ~ Find the equation of

the curve if it passes through the point (1, 0) and is tangent to the line

6 x + y 6 at that point. Ans. xy + 6 x = 6.

48. Find the equation of the curve at every point of which ?/" = /

and which passes through the point (1, 1) with an inclination of 45.

49. Find the equation of the curve at every point of which y" - andwhich passes through the point (1, 0) with an inclination of 135'\

50. Find the equation of the curve whose subnormal is constant and

equal to 2 a. Aws. y'2 4 ax -f C, a parabola.

HINT. From (4), Art. 43, subnormal = y-^--ax

51. Find the curve whose subtangent is constant and equal to a

(see (3), Art. 43). Ans. a In y = x + C.

52. Find the curve whose subnormal equals the abscissa of the pointof contact. Ans. y

2 x 2 = 2 C, an equilateral hyperbola.

53. Find the curve whose normal is constant (= /?), assuming that

y = R when JT = 0. An*, a*2 + ?r = A>2

,a circle.

HINT. From Art. 43, length of normal - ij^/l -f f^\", or*

54. Determine the curves in \\hich the length of the subnormal is

proportional to the square of the ordinate. Ans. //= OK

55. Find the equation of the curve in which the angle between theradius 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 andthe tangent at any point is n times the vectorial angle. .4 ;/s. p" = c sin ?/0.

140. Physical signification of the constant of integration. The fol-

lowing examples will illustrate what is meant,

ILLUSTRATIVE EXAMPLE 1. Find the laws governing the motion of a pointwhich moves in a straight line with constant acceleration.

Solution. Since the acceleration from (A), Art. 59 is constant, nay /,

we have * -I

<* _ ,

or dv=.fdt. Integrating,

(1) *=// + ('.

To determine C, suppose that the initial velocity is ?><>; that is, let v - p, when1 = 0.

These values substituted in (1) give

p()=. -f C, or C = Vo.

Hence (1) becomes

(2) v=ft + v .

Since = ^ ^c> Art" 51) we ^ct from (2)

*=/ + *,

or dx = ft dt + vodl . Integrating,

(3) s = \ //2 4 ?'o/4 r\

To determine C, suppose that the initial distance is ; that is, let =o when /=0.

These values substituted in (3; give

s = -h -f C, or C = 8 .

Hence (3) becomes

(4) B = J/J2 + (,/ + *U.

By substituting the values / 0, vo = 0, 8 = 0, = h, in (2) and (4), we get thelaws of motion of a body falling from rest in a vacuum, namely,

?' = (//, and h ^ gl*.

Eliminating / between these equations gives r v t> <//?.

ILLUSTRATIVE EXAMPLE 2. Discuss the motion of a projectile having an initial

velocity ih\ inclined at an angle a. with the horizontal,

the resistance of the air being neglected.

Solution. Assume the A'OV-plane as the planeof motion, OX as horizontal, and ()Y as vertical, andlet the projectile be thrown from the origin.

Suppose the projectile to be acted upon by ^gravity alone. Then the acceleration in the hori-

zontal direction will be zero and in the vertical direction -g. Hence, from (F),

Art ' 84' dr dv_* = 0. and ^ = -0.

Integrating, vf = fi and ru=

yt -f C-2 .

But 0o cos a ~- initial velocity in the horizontal direction,

and o sin a = initial velocity in the vertical direction.

Hence (\ ?>< cos a and (\> ?>o sin , giving

(6) 7v = ?'o c'os <.* and ru~ gt + ?'o sin a.

But from (C) and (Z>), Art. 83, r, = ~ and ru~ ^ ;

therefore (5) gives

dx , </// .,~- = ?,'() cos a and -~ = yt -f- ?

(

o sin a,d/ d/

or dj' Vo cos a <:// and dy g\ dt + i\> sin a dL

Integrating, we get

((5) a: f) cos a / -f ( '.-i and y ~. ~ \ f//

1

-'

-f ?;,, sin cv / 4- ^4.

To determine r 3 and (\ we obser\e that when / = 0, x and ^/= 0.

Substituting these values in (G gives

( \ 4- arid C , 0.

Hence

(7) x = 7>o cos a -

/, and

(8) y \ f/'~ + ''o sin <x -t.

Eliminating t between (7) and (8), we obtain

(9) y = x tan a'2 r<>- cos 2 a

which is the equation of the trajectory and shows that the projectile will move in a

parabola.

PROBLEMS

In the following problems the relation between r and / is given. Find

the relation between s and Mf s = 2 when / = 1.

Ans. s = a(t-

1) -f b(t2 -

1) + 2.

3. r = ^ -f -

In the following problems the expression for the acceleration is given.Find the relation between r and / if r 2 when t = 3.

4. 4 - f2. An*, r = 4 f-

J /' - 1. 5. V/ -f 3.

In the following problems the expression for the acceleration is given.Find the relation between .s- and / if .x = 0, v 'JO when t 0.

7. - 32. An*, s = 20 t- 16 /-'. 8. 4 - /. 9. - 16 cos 2 t.

10. With what velocity will a stone strike the ground if dropped fromthe top of a building 120 ft. high? vl /

= ;?'J.) An*. 87.61 ft. per second.

11. With what velocity will the stone of Problem 10 strike the groundif thrown downward with a speed of 'JO ft. per second? if thrown upwardwith a speed of 20 ft. per second? Ans. 89.89 ft. per second.

12. A stone dropped from a balloon \\hich was rising at the rate of

15 ft. per second reached the ground in 8 sec. How high was the balloon

when the stone was dropped? Ans. 904 ft.

13. In Problem 12, if the balloon bad been falling at the rate of 15 ft.

per second, how long \\ould the stone have taken to reach the ground?.4//,s

%

. 7j' (

< sec.

14. A train leaving a railroad station has an acceleration of 0.5 -f- 0.02 / ft.

per second per second. Find how far it \vill move in 20 sec. AHS. 126.7 ft.

15. A particle sliding on a certain inclined plane is subject to an ac-

celeration downward of 4 ft. per second per second. If it is started up-ward from the bottom of the plane with a velocity of 6 ft. per second,

find the distance moved after / sec. How far will it go before sliding

backward? An*. 4.5ft.

16. If the inclined plane in Problem 15 is 20 ft. long, find the necessary

initial speed in order that the particle may just reach the^tpp of the plane.

Aits. 4VlO ft. per second.

17. A ball thrown upward from the ground reaches a height of 80 ft.

in 1 sec. Find how high the ball will go.

18. A projectile with an initial velocity of 160 ft. per second is fired at

a vertical wall 480 ft. from the point of projection.

(a) If a 45, find the height of the point struck on the wall.

An. 192 ft.

(b) Find a so that the projectile will strike the base, of the wall.

Anx. 18 or 72.

(c) Find a. so that the projectile will strike 80 ft. above the base.

AUK. 29 or 70.

(d) Find a for the maximum height on the wall and this height.

Ana. 59; 256 ft.


19. If the acceleration of a particle moving with a variable velocity r

is kv 2,where k is a constant, and if r is the velocity when / = 0, show

that - = - + kt.v r,,

20. The resistance of the air to an automobile, within certain limits

of speed, is proportional to the speed. Hence if F is the net force gen-

erated by the motor, we have M - = F AT. Express the velocity in

F *-terms of /, knowing that r when / 0. Ans. v (1 c

A/).

K

ADDITIONAL PROBLEMS

1. The temperature of a liquid in a room of temperature 20 is ob-

served to be 70", and after 5 min. to be 60 . Assuming the rate of cool-

ing to be proportional to the difference of the temperatures of the liquid

and the room, find the temperature of the liquid 30 min. after the first

observation. Ans. 33.1.

2. Find the equation of the curve whoso polar subtangent is n times

the length of the corresponding radius vector and which passes throughthe point (, 0). 1

Ans. p = ac n

3. Find the equation of the curve whose polar subnormal is n times

the length of the corresponding radius vector, and which passes throughthe point (a, 0). Ans. p = ac" e

.

4. A particle moves in the .r//-plane so that the components of velocity

parallel to the .r-axis and the //-axis are ky and kx, respectively. Prove that

the path is an equilateral hyperbola.

5. A particle projected from the top) of a tower at an angle of 45 abovethe horizontal plane strikes the ground in 5 sec. at a horizontal distance

from the foot, of the tower equal to its height. Find the height of the

tower (g= 32). AUK. 200 ft.

6. A particle starts from the origin of coordinates and in / sec. its

x-component of velocity is P 4 and its //-component is 4 t.

(a) Find the position of the particle after / sec.

Atis. .r- i/

:i - 4 /, i/= 2 /

2.

(b) Find the distance traversed along the path. Ans. s ~ J t* + 4 /.

(c) Find the equation of the path. Ans. 72 .r2 =

?/M 48 ?/

2 + 576 y.

7. Find an equation of a curve for which the length of the tangent

(Art. 43) is constant (= c).

SUGGESTION. Choose the minus sign in Problem 2 (a), p. 85, and assume y = c

when .r 0. / / AAns. jr = c\n

(

c + VfL ~?/"

)

- Vc 2 -y*.

\ V /

8. Find the equation of the curve for which (Art. 96) a 2 ds = p3 dO

t

and which passes through the point (a, 0). Ans. p2 = a 2 sec 2 6.

CHAPTER XIV

THE DEFINITE INTEGRAL

141. Differential of the area under a curve. Consider the con-

tinuous function <t>(x), and let

y = 0(x)

be the equation of the curve An. Let CD be a fixed and MP avariable ordinate, and let u be the measure of the area CMPD.When x takes on a small increment Ax, ?/ takes on an increment

Aw (= area MNQP). Completing the rectangles 1\1\KI> and MNQS,we see that

Area MNRP < are&MNQP < area MNQS,

or MP - Ax < Aw < NQ - Ax ;

and, dividing by Ax, if*-x

Now let Ax approach zero as a limit ; then since M P remains fixed

and NQ approaches MP as a limit (since y is a continuous function

of x), we get 1

or, using differentials, dn = y rfr.

Theorem. T/?e differential of the area bounded by any curve, the

x-axis, a fixed ordinate, and a variable ordtnate is equal to the product oj

the variable ordinate and the differential of the corresponding abscissa.

142. The definite integral. It follows from the theorem in the last

article that if the curve AB is the locus of

then du = ?/ dx, or

(1) dw = <f)(x)dx,

* In this figure MP is less than NQ ; if MP happens to be greater than NQ, simply

'erse the inequality signs,

237reverse

where du is the differential of the area between the curve, the x-axis,

and two ordinates. Integrating, we get

Denote ( </>(/) ds byf(x) + C.'

(2)

We determine r by observing that u =when x = a. Substituting these values in (2), we get

o=/(o)-f r,

and hence r = /(a;.

Then (2) becomes

(3) w -/(.r) -/(a).

The required area CKF1) is the value of u in (3) when x = 6,

Hence we have

(A) Area CEFD = f(b)-

f(a).

Theorem. The difference of the values of I yds for x = a and x = 6

gives the area bounded by the curve whose ordfnate is y, the x-axis, and

the ordinatex corresponding to x = a and, .r = b.

This difference is represented by the symbol*

(4)

S*h fhI

yd* or I 0(r)rf.r,/a ^'u

and is read "the integral from a to b of //rf.r." The operation is

called integration between limits, a being the lower and 6 the wpper

limit.f

Since (4) always has a definite value, it is called a definite integral.

For, if

I c/)(.r)dr-/(.r) + T,t/

then

or

r h

I 0(.rW.r-Ja

/(r) + T = [/(&) + T]-

[/(a) + C],

the constant of integration having disappeared.

* This notation is due to Joseph Fourier U 768- 1830).

f The word "limit"

in this connection means merely the value of the variable at one end

of its range (end value), and should not be confused with the meaning of the word in the

Theory of Limits.

THE DEFINITE INTEGRAL 239

We may accordingly define the symbol

r*> /*.

j <t>(jc)djr or I n df'a J

as the numerical measure of the area bounded by the curre //~

</>(>),*

the x-ajris, and the ordinates of the curre at j ~ a and JT h. This

definition presupposes that these lines hound an area; that is, the curve

does not rise or fall to infinity and does not cross the .r-uj/.s, ami both

a and b are finite.

143. Calculation of a definite integral. The process may be sum-marized as follows :

FIRST STEP. Integrate the (jircn differential expression.SECOND STEP. Substitute in this indefinite inlcynrt first the upper

limit and then the loirer limit for the rar/ahle, and subttacf the last

result from the first.

It is not necessary to brinpc in the constant of integration, since

it always disappears in subtracting.

ILLUSTRATIVE FAAMPLE 1. Findj

x-' dr.

Solution, f*

JT~ djc -I~

1

4 - - - -- -

1-

12 1 . \ ws.

J\ I 'Hi '* ^

/7T

ILLUSTRATIVE FXAMPLE 12. Find I MH .r d,r.

rn i I?1I 111

Solution. I sin j- djr ~ - cos / ( 1 ) '

"-- ^ nK -

Jo l J' T1 f)

~ ""' 134 -

]4 J -

.> 1

Solution. Comparing with (19) or (19 a), r- 2 r, ry

-:j

fr/?> ^- 2 c/j.

To decide between the use of (19) or (19 a), consider the limits. The values of

x increase from - 1 to 0.

Then v (= 2 x) increases from 2 to 0.

Hence v2 - 4. But o 2 = 9. Therefore ?'~'

-, and (19 a) must be used. Thus

r(} dr /*

Evaluating in (1) gives the answer. The result is negative because the curve and

the bounding ordinates lie below the j%axis.

* d)(j) is continuous und single-valued throughout the interval [a, b].

144. Change in limits corresponding to change in variable. Whenintegrating by the substitution of a new variable it is sometimes

rather troublesome to translate the result back into the original

variable. When integrating between limits, however, we may avoid

the process of restoring the original variable by changing the limits

to correspond with the new variable. This process will now be

illustrated by an example.

ILLUSTRATIVE KXAMPLK. Calculate \

x

Solution. Assume x z 1.

Then dx - 4 z '

dz, x^ = z'-', x* = z. Also, to change the limits we observe that when

JT = 0, z - 0,

and when JT 16, z 2.

r* dx _ r2 z 4 z'' dz

1 + X*

! -f 4 arc tan 2. Ans.

The relation between the old and the new variable should be such that to each

value of one within the limits of integration there is always one, and only one, finite

value of the other. When one is given as a many-valued function of the other, care

must be taken to choose the right values.

PROBLEMS

1. Prove that f V(.r)dr = - fa

f(x)dx.Ja

'

Jl>


ra 042. / (a'

2JC-

x*)dx = ~- 9. - = 5.6094.Jo 4 Jo x -f 1

0.3167.

-2x# 9

-

~ ln 2>

12. fW^/2 + 2 cos 6 d6 = 4.

g i*

.

i<-v _ ^ i^ o ^o

'() X + 1 3 7T

in3 x cos3 x dx =


Find the value of each of the following definite integrals.

20.I

l

jcc" dr.

241

21./'"cos-

Jv(3 J0.

22. I "sin- ~ cos|dO.

23.

145. Calculation of areas. On page 238 it was shown that the area

between a curve, the j-axis, and the ordinates

x = a and x == b is given by the formula

rb

(B) Area= I y dx,Ja

where the value of y in terms of x is substituted

from the equation of the given curve.

ILLUSTRATIVE EXAMPLE 1. Find the area bounded by

the parabola y = x-, the /-axis, an4 the ordinates x =-- 2

and x 4.

Solution. Substituting in the formula,

Area ABDC =

ILLUSTRATIVE EXAMPLE 2. Find the area bounded by the circle x'2 + y'2 =25,

the x-axis, and the ordinates x -3, x ~ 4.

Solution. Solving, y V'25 -r-. Hence

by (22)

= 6 4- arc sin | -f 6 - ^ arc sin (- g)= 31 .6. Am.

The answer should be compared with the area of the semicircle, which is

) =39.3.

ILLUSTRATIVE EXAMPLE 3. Area under a parabola wtwsc axis i parallel to ttw

y-axis. In the figure (p. 242) the point P' on the parabolic arc PP" is chosen so

that AO = OB. The ordinates of P, P', P" are, respectively, y, y', y" . Prove that

the area between the parabola, the z-axis, and the ordinates of P and P" equals

i h(y + 4 y' + y"} if 2 fe is the distance apart of the ordinates of P and P".

Solution. Take the ?/-axis along the ordinate of P', as in the figure. ThenAB = 2 k. The equation of a parabola with axis parallel to the #-axis is, by (7),

p. 4, (x h)'z 2 [>(y k). If this is solved for y,

the result takes the form

(1) //

~ as* + 2 bx + r.

The required area A /'/'"/? (= u) is, by (B),

(2) H- (" (ax- + 2 bx + r)ds = * ah' + 2 ch.J h

if x - 0, ?/'- O/'' r ;

if j - /,, y/

"-_- /{/>" - r///- -f 2 />// -f r.

Therefore ', //(// + 4 ?/' + ?/" )

-;/?

(

-f 2 r// = M.

P"

y'

o /^ ^\"

O.E.D.

146. Area when the equations of the curve are given in parametricform. 4 Let the equations of the curve be Driven in the parametric form

jr -/(/). y/=

(/>(/).

We then have //--

</>(/), and djr J'(t)dt. Hence

[!> fhArea--

/ // <1x = / </>(/)/'(/)</*,./a .//,

r;, and / = i^ when j = b.

CD

where when

ILLUSTKATIVK KXAMPLK. Find the area, of the ellipse whose parametricequations (Art. SI ) are

.r a cos (/>, //^ b sin c/>.

Solution. Here // b sin </^,

and ds = a sin </> c?</>.

When j- --= 0, </>=--

i IT ;

and when jr (7, </> 0.

Substituting these in (1\ above, we get

Area f" , f ,.

., , , , irabI // as I ah sin" 7r

4

Hence the entire area eciuals TT?). .4ns.

PROBLEMS

1. Find by integration the area of the triangle bounded by the line

y = "2 .r, the .r-axis, and the ordinate .r = 4. Verify your result by findingthe area as half the product of the base and altitude.

2. Find by integration the area of the trapezoid bounded by the line

jr -f //= 10, the .r-axis, and the ordinates jr = 1 and x 8. Verify your

result by finding the area as half the product of the sum of the parallel

sides and the altitude.

* For a rigorous proof of this substitution the student is referred to more advancedtreatises on the calculus.

Find the area bounded by the given curve, the j-axis, and the given

ordinates.

3. y .r3

;.r = 0, jr = 4. Aw. 64.

4. ?/= 9 -.r 2

;.r = 0, .r = 3.

'

18.

5. y = r< + 3 .r- -f 2 .r;

.r = -3, .r = 3. 54.

6. ?/= j-2-f j-f 1; .r = 2, j- = 3. 9-;.

7. .r//= A-

2; jr = a, j- = fc. F In

f~)-

9. //=

,

10;

.r = 0, j- - 5. 20.

Vj- 4-4

10. a// = .rV J -.r

1'

;a* = 0, .r

-a.

;\ -.

11. ?/- + 4 .r = ;JT = -

1, .* = 0. 14. //= 4 .r

-.r

L>

;jr -- 1, .r = 3.

12. /r = 4 .r + 16;

.r = -2, JT - 0. 15. //-'

- <> - .r ;.r-

0, jr = 8.

13. y = .r2 + 4 jr

;.r = -

4, j- = - 2. 16. 2 //- .r

{

;./ - 0, .r -= 2.

Find the area l>ounded by the given curve, the //-axis, and the given lines.

17. ?r = 4 jr; //

= 0, //- 4. AH*. 5.\.

18. ?/- 4 - j-- ; //

= 0, //= 3. 4;|.

19. jc 9 ?/-

/y

:{

; y 0, ?/ 3. 21. //

:i - rrV ; //

-0, //

-a.

20. jry 8; y 1, //

= 4. 22. <//r- .r^

; //

~0, //

-a.

Sketch each of the following curves and find the area of one arch.

23. = 2 cos.r. AUK. 4.

824. ?/

= 2 sin \ irx.-

25. ?/ cos 2 jr.1

26. ?/= sin \x. 4.

27. Find the area bounded by the coordinate axes and the parabola

Vx -f V?/ = Va.

28. Prove that the area of any segment of a parabola cut off by a chord

perpendicular to the axis of the parabola is two thirds of the circumscrib-

ing rectangle.

29. P and Q are any two points on an equilateral hyperbola xy = k.

Show that the area bounded by the arc PQ, the ordinates of P and Q,

and the x-axis is equal to the area bounded by PQ, the abscissas of P

and Q, and the 7/-axis. ^30. Find the area bounded by the catenary y = %a(c

n + c V, the

x-axis, and the lines x = a and x = a. ^ns a^( e _ 1\

\ e)


31. Find the area included between the two parabolas t/2 = 2 px and

x'2 = 2 py. Am. f p2

.

32. Find the area included between the two parabolas y2 = ax and

x 2 = &#. Ans. J a6.

33. Find the area inclosed by the loop of the curve whose equation is

4 y'2 = x'2 (4

-x). Ans. *f-

34. Find the area bounded by the curve whose equation is

y2 - x'2 (jr'

2 -1) and by the line x-2. Ans. 2\/3.


36. Find the area bounded by the curve whose equation is y2 = j*

3 x 2

and by the line JT = 2. Ans. ff .


7/2 x(x 2)'2

. Ans. ^|V2.


4 y~ r'(4 x). Ans. *.39. Find the area bounded by the hyperbola x 2 - y

2 a 2 and the line

x = 2a. Ans. a 2[2V3 - In (2 + V3)].

40. Find the area bounded by the hyperbola x 2 4 y2 4 and the

line x (>.

41. Find the area bounded by one arch of the cycloid x = a(0 sin 0),

y a(l cos 0), and the .r-axis. Ans. 3 ira 2.

42. Find the area of the cardioid

x ~ a (2 cos t cos 2 t),

y = a(2 sin / sin 2 /). Ans. 6 ?ra 2 .

43. The locus in the figure is called the "companion to the cycloid."

Its equations are

x = aO, Tl

y = a (l COS 0).

Find the area of one arch.

Am. 2 ?ra2 .

44. Find the area of the hypocycloid

x = a cos3 0,

y = a sin3 0,

c- 9

being the parameter. Ans. - *

that is, three eighths of the area

of the circumscribing circle.

147. Geometrical representation of an integral. In the precedingarticles the definite integral appeared as an area. This does not

necessarily mean that every definite integral is an area, for the


physical interpretation of the result depends on the nature of the

quantities represented by the abscissa and the ordinate. Thus, if x

and y are considered as simply the coordinates of a point, then the

integral in (B), Art. 145, is indeed an area. But suppose the ordinate

represents the speed of a moving point, and the corresponding ab-

scissa the time at which the point has that speed ;then the graph

is the speed curve of the motion, and the area under it and between

any two ordinates will represent the distance passed through in the

corresponding interval of time. That is, the number which denotes

the area equals the number which denotes the distance (or value of

the integral). Similarly, a definite integral standing for volume, sur-

face, mass, force, etc. may be represented geometrically by an area.

148. Approximate integration. Trapezoidal rule. We now prove two

rules for evaluating

Jo

approximately. These rules are useful when the integration in (1) is

difficult, or impossible in terms of elementary functions.

The exact numerical value of (1 ) is the measure of t he area bounded

by the curve

(2) y=f(x),

the x-axis, and the ordinates

x = a, x = b. This area may be

evaluated approximately by add-

ing together trapezoids, as follows.

Divide the segment b a on

OX into n equal parts, each of

length Ax. Let the successive

abscissas of the points of division

be x (= a), xi, x2 , -, xn (= b). At these points erect the correspond-

ing ordinates of the curve (2). Let these be

2/0 =f(xo), y\ =/(i) 2/2 =f('jr2),-

, yn =/Orn).

Join the extremities of consecutive ordinates by straight lines

(chords) forming trapezoids. Then, the area of a trapezoid being

one half the product of the sum of the parallel sides multiplied by

the altitude, we get

y + 2/1 )Ax = area of first trapezoid,

yi + 2/2)Ax = area of second trapezoid,

i(2/n_ 1 _j_ /n)Ax = area of nth trapezoid.


Adding, we get the trapezoidal rule,

(7) Area = (jy + JA + ?2 + - + yn - i -h |yn)Ax.

It is clear that the greater the number of intervals (that is, the

smaller Ax is), the closer will the sum of the areas of the trapezoids

approach the area under the curve.

c 12ILLUSTRATIVE KXAMPLK 1. Calculate \ x~dx by the trapezoidal rule, dividing

x = 1 to x = 12 into eleven intervals.l

^ _ a ]f > _ ]

Solution. Here -^ ,1 = Aa*. The area in question is under the

,12 in this equation, we

n 11

curve y ~- x~. Substituting the abscissas x 1, 2, 3,

get the ordinates // 1, 4, 9, -, 144. Hence, from (T),

Area = ( i + 4 + 9+16 + 25 + 36 +49+64 + 81 + 100 + 121 + J 144) 1 = 577?,.

By integration /":r- dx - " = 575^. Hence, in this example, the trape-

M I'> h

zoidal rule is in error by less than one

third of 1 per cent.

If we take n 10, we obtain / =4.826,a closer approximation.

O X

PROBLEMS

Compute the approximate values of the following integrals by the

trapezoidal rule, using the values of n indicated. Check your results byperforming the integrations.

io/ r1

rio/ r.1 ; w = 7.^3 X

2. f xV25 - x 2dx; n = 10.

/o

Compute the approximate values of the following integrals by the

trapezoidal rule, using the values of n indicated.

5.^

4

-^==; n = 4. Ans. 1.227. 10. {V^O + .r4dr; w = 5.

fcjjVl+ r'dx; n = 4. 3.283. 11.

jfWl<;~H dr ; w = 4.

7. f1

\/125-a' 2(ir; n = 5. 44.17. 12. T vT^'TiT^ cir ;

= 5./o Ji

8. r6

Vl26-o^dr; n = 4. 34.78. 13. (

4

--^L^ ; n = 6.

9. = 6. 9.47.

VIO ~f .r2

l\

149. Simpson's rule (parabolic rule). Instead of connecting the

extremities of successive ordinates by chords and forming trapezoids,we can get a still closer approxi-

mation to the area by joining

them with arcs of parabolas and

summing up the areas under these

arcs. A parabola with a vertical

axis may be passed through anythree points on a curve, and a

series of such arcs will fit the curve

more closely than the broken line

of chords. In fact, the equation ~<;" ^ ]u

1 ~M.,""My ~ji/

of such a parabola is of the form

(1) in Illustrative Example 3, Art. 14f>, and the values of the con-

stants a, b, c may be determined so that this parabola shall pass

through three given points. In the present investigation, however,this is not necessary.

We now divide the interval from x a = OA/o to x = b OMn

into an even number (= n) of parts, each equal to Ax. Througheach successive set of three points /'<>, ]\, P2 ; /^, /'n, /*4 ; etc., are

drawn arcs of parabolas with vertical axes. The ordinates of these

points are 2/0, y\, yz, , y*> as indicated in the figure. The area

MoP P2 PnM n is thus replaced by a set of "double parabolic

strips" such as MoPvP\P<2M<2 ,whose upper boundary is in every

case a parabolic arc (1) of Illustrative Example 3, Art. 145. Thearea 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


Area of first parabolic strip MoPGPiP2M2 = -(#o + 4 y\ +

Similarly, Second (double) strip = -rp (#2 + 4 2/3 + 1/4),o

AT-Third (double) strip = (z/4 + 4 y5 + o),

AxLast (double) strip = (y n-z + 4 yn~i + f/n).

Adding, we get Simpson's rule (n being even),

A*(S) Area=(y + 4^ + 2

As in the case of the trapezoidal rule, the greater the number of

parts into which MuM n is divided, the closer will the result be to the

area under the curve.rl()

ILLUSTRATIVE EXAMPLE 1. Calculate I jc:{ dx by Simpson's rule, taking ten

intervals.^

Solution. Here> = =

1 = Ax. The area in question is under then 10

curve y - .r\ Substituting the abscissas x 0, 1, 2, -, 10 in y = x'\ we get the

ordinates y = 0, 1, 8, 27, -, 1000. Hence, from (S),

Area = Jj(0 + 4 + 1 6 4 108 -f 128 + 500 + 432 + 1372 -f 1024 +2916 4- 1000) =2500.

By integration, f ^'{ dr = I

"2500, so that in this example Simpson's rule

gives an exact result.

ILLUSTRATIVE. EXAMPLE 2. Find the approximate value of

1 \ "\/4 -f a-* dxby (S), taking n = 4. Jo

Solution. The table of values is given in Illustrative Example 2 of the preced-

ing article. Hence

1 = (2.000 -f 8.124 + 4.472 4- 10.864 -f 3.464) x ^ = 4 .82 i.o

Compare this result with that given by (7) when n = 10, namely 4.826.

In this case formula (S) gives a better approximation than (T) when n = 4.

PROBLEMS

Compute the approximate values of the following integrals by Simp-son's rule, using the values of n indicated. Check your results by perform-

ing tbe integrations.

_ . n = 6 .

Ja 4 -f x

2. jV25-r2 dx; n = 4. . f Vl6 + x 2 dx;n = 6

*/4

Compute the approximate values of the following integrals by Simp-son's rule, using the values of n indicated.

* f4 dx

5. / ;n = 4.

Jo >/4 _j_ j-a

6. rVl + .r'drJo

= 4.

7. f5

Vl26 -.r'dr; n = 4.

Ans. 1.236.

3.239.

35.68.

9.49.

9. / V6 + X 2dx; n = 4.

r5 ~ri A*10.

'* VTT^'

n.

12.

,r*

jr (Lr ;n 4 .

-- 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.

13. f4

Vl6 - x'2dx\ n = 4. 18.

|"r - dr; n = 4.

Jz V3 -f r3 ^o

150. Interchange of limits. Since

f <fr(x)dx=f(b)-f(a),/a

and C4>(x)dx = /(a)-

/(&) = -[/(&)

-/(a)],

Xb

/

(/>(x)dx= - I c/)(x)cfx.

Jb

Theorem. Interchanging the limits is equivalent to changing the sign

of the definite integral.

151. Decomposition of the interval of integration of the definite

integral. Since

fTl

<l>(x)dx=f(xi) -/(a), (a < xl < 6)Jtt

r h

and I <t>(x)dx=f(b)-J(xi),Jx\

we get, by addition,

( "<KJK/j + f *<t>(x)dx =/(fc) -/(a).Ja Jxi

r h

But I <t>(x)dx=f(b)-f(a)',Ja

therefore, by comparing the last two expressions, we obtain

Interpreting this theorem geometrically, as in Art. 142, we see

that the integral on the left-hand side

represents the whole area CEFL), the

first integral on the right-hand side the

area CM PL), and the second integral on

the right-hand side the area MEFP. Thetruth of the theorem is therefore obvious.

Evidently the definite integral may be

decomposed into any number of separate definite integrals in this way.152. The definite integral a function of its limits

FromrI

'a

<p(jc}dx= /(?>)- /(a)

we see that the definite integral is a function of its limits. Thus

C b C b

I <f>(f)dz has precisely the same value as I (j)(x}dx.J Of J O.

Theorem. A definite integral is a function of its limits.

153. Improper integrals. Infinite limits. So far the limits of the

integral have been assumed as finite. Even in elementary work,however, it is sometimes desirable to remove this restriction and to

consider integrals with infinite limits. This is possible in certain

cases by making use of the following definitions.

When the upper limit is infinite,

= lim / <t>(x)dx,

and when the lower limit is infinite,

251

X

/t>

/- b

4>(jc)dx = lim I <t>(x)dx,oo <i -* - oc t / a

provided the limits exist.

ILLUSTRATIVE EXAMPLE 1. Find (

x

./, JP-

Solution. f = lim ffr^= lim |-1]"= lim \-}+l\=\.

J\ x2 b- -i xJi x 2b , f oc I

.r1

1 /, .i oc L &

.1

ILLUSTRATIVE EXAMPLE 2. Find /

J

Ans.

x'

r + 8a ;< Ar v r^ 8''f/j r \

i n - X ^TI^ =fc L X FTT^

=, l'

mK [

= lim 4 a'2 arc tan

r 1 f)

JU |o

4 a'2 "2 7R7". .Arts.

Let us interpret this result geometri-

cally. The graph of our function is the

witch, the locus of

y8a :{

x 2 + 4 a 2

. ,= 4 a ? arc tan -

Then as the ordinate bQ moves indefinitely to the right, the area OPQb ap-

proaches a finite limit 2 Tra1

-'.

/-f 00^3.

ILLUSTRATIVE EXAMPLE 3. Find I

'

Jj x

Solution. = Hm :c=: lim (ln

X b--foo

The limit of In b as b increases without limit does not exist ; hence the integral

has in this case no meaning.

154. Improper integrals. When y (/>(*) is discontinuous. Let us

now consider cases where the function to be integrated is discontinuous

for isolated values of the variable lying within the limits of integration.

Consider first the case where the function to be integrated is con-

tinuous for all values of x between the limits a and 6 except x = a.

If a (x)dx= \im I 4>(x}dx.

e->OJa+t

Likewise, when <fr(x) is continuous except at x = b, we definer b r*>-<

(2) I 0(z)az = lim I <t>(x)dx,Ja - Ja

dxprovided the limits exist.

ILLUSTRATIVE EXAMPLE 1. Find

Solution. Here

o2 x'2

becomes infinite for x = a. Therefore, by (2),

.

\ a 2 -

= lim arc sin /I -- -\ 1 = arc sin 1 = ? Ans-

* - o L \ a/ J 2

ILLUSTRATIVE EXAMPLE 2. Find f -

J x 2

Solution. Here becomes infinite for x = 0. Therefore, by (1),

In this case there is no limit, and therefore the integral does not exist.

If c lies between a and b, and 0(j) is continuous except at x = c,

then, e and e' being positive numbers, the integral between a and 6

isdcjirwd by

(3) f 0Cr)d:r = lim f' 0(^)dj + lim f 0(o;)dx,i/ -0./ a e'_0,L/r + e'

provided each separate limit exists.

ILLUSTRATIVE EXAMPLE 3. Find2 x rfx

Solution. Hefe the function to be integrated becomes discontinuous for x = a,

that is, for a value of x between the limits of integration and 3 a. Hence the abovedefinition (3) must be employed. Thus

=limf 3(z

2 - a 2)*l

a '

+ lim \S(x*- a 2

*-oL Jo '-*oL

= lim [3S^(a c)2 - a 2 + 3 a^J + lim

t - o *'-* o

= 3 o* -f 6 a* = 9 a'. Ans.

To interpret this geometrically, let us

plot the graph, that is, the locus, of

2xv =

(x*-a*rtand note that x = a is an asymptote.

--< 2xox

o + c')2 - a 2

]

Then as PE moves to the right toward the asymptote, that is, as c approaches

zero, the area OPE approaches 3 a^ as a limit.

Similarly,

2 J dx /

Area E'QRG - a 4- e')a - a a

approaches 6 a5 as a limit as QE' moves to the left toward the asymptote, that is,

as c' approaches zero. Adding these results, we get 9 <i*.

ILLUSTRATIVE EXAMPLE 4. Find\ f

( r

./o (.r- a^-

Solution. This function also becomes infinite between the limits of integration.

Hence, by (3), 8

t^ rf_ = lim fV^ + lim f"-^-o (z-a) J

-.oJo (x-a)- '-,o-'a+t'(-r-

")-

=Um /i_iU,im /_i + ie->0\6 a/ f'-0\ rt

In this case the limits do not exist and the inte-

gral has no meaning.If we plot the graph of this function the condi-

tion of things appears very much the same as in the

last example. We see, however, that the limits do not exist, and therein lies the

difference.

That it is important to note whether or not the given function becomes infinite

within the limits of integration will appear at once if we apply our integration

formula without any investigation. Thus

dx \ 12 _ 2

a'

r^i

J (x- a)'

2

result which is absurd in view of the above discussion.

PROBLEMS

Work out each of the following integrals.

C+ dx _ TT

Jo x 2 -hl"27T

r+ dx _ ^TT_

o a 24- 6 2x 2

~~2 a6

7.fJo

9.

10.

= V2.

1z

4ira

x 2 + 2x + 2

oo x dx _ 1

(1 +x 2)2 ~4*

CHAPTER XV

INTEGRATION A PROCESS OF SUMMATION

155. Introduction. Thus far we have defined integration as the

inverse of differentiation. In a great many of the applications of the

integral calculus, however, it is preferable to define integration as

a process of summation. In fact, the integral calculus was invented

in the attempt to calculate the area bounded by curves, by supposingthe given area to be divided into an "infinite number of infinitesimal

parts called elements, the sum of all these elements being the area

required." Historically, the integral sign is merely the long S, used

by early writers to indicate "sum."This new definition, as amplified in the next article, is of fun-

damental importance, and it is essential that the student should

thoroughly understand what is meant in order to be able to applythe integral calculus to practical problems.

156. The Fundamental Theorem of integral calculus. If </>O) is the

derivative of /(, then it has been shown in Art. 142 that the value

of the definite integral

(1) fh

<t>(x)dx=f(b)-f(a)

gives the area bounded by the curve

y = (/>(/), the j-axis, and the ordinates

erected at x a and .r = b.

Now let us make the following

construction in connection with this

area. Divide the interval from :r a to x = b into any numbern of equal subintervals, erect ordinates at these points of divi-

sion, and complete rectangles by drawing horizontal lines throughthe 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 whentheir number n is indefinitely increased will equal the area under

the curve.254

INTEGRATION A PROCESS OF SUMMATION 255

Let us now carry through the following more general construction.

Divide the interval into n subintervals, not necessarily equal, and

erect ordinates at the points of division. Choose a point within each

subdivision in any manner, erect ordi-

nates at these points, and through their

extremities draw horizontal lines to form

rectangles, as in the figure. Then, as

before, the sum of the areas of these

n rectangles (the shaded area) equals

approximately the area under the curve ;

and the limit of this SUM as n increases

without limit, and each subinterval ap-

proaches zero as a limit, is precisely the area under the curve. These

considerations show that the definite integral (1) may be regarded

as the limit of a sum. Let us now formulate this result.

(a) Denote the lengths of the successive subintervals by

(b) Denote the abscissas of the points chosen in the subintervals

*^ 'v_ v.v <->". . ...<"f 3

Then the ordinates of the curve at

these points are

(c) The areas of the successive rec-

tangles are obviously

Ail1

',;\

(d) The area under the curve is therefore equal to

lim [<Kxi)Azi + 0fejAx2 + 0fe)Ax;{ H-----h-

rBut from (1) the area under the curve - / ^(x

Therefore our discussion gives"

(A) f (f>(x)dx= lim [^(xi) Axi + ^(

Ja n ^

This equation has been derived by making use of the notion of area.

Intuition has aided us in establishing the result. Let us now regard

(A) simply as a theorem in analysis, which may then be stated as

follows.

FUNDAMENTAL THEOREM OF THE INTEGRAL CALCULUS

Let (f)(r

jr) be continuous for the interval x = a to x = b. Let this in-

terval be divided into n subintervals whose lengths are Aa*i, Aj2 , ,

Azn ,and points be chosen, one in each subinterval, their abscissas being

Xi, 2*2 1

*

>fjrn respectively. Consider the sum

(2) 0(j] )Aj, + </)(r2 )Ar2 H-----h </>(

Then the limiting value of this sum when n increases without limit,

and each subinterval approaches zero ax a limit, equals the value of the

definite integral

fJa <t>(x)dx.

Equation (A) may be abbreviated as follows.

C h ^(3) I (p(x)dx = lim

2,(/>0\)Aa%-.Ja n * oct_

!

The importance of this theorem results from the fact that we are

able to calculate by integration a magnitude which is the limit of a sumof the form (2).

It may be remarked that each term in the sum (2) is a differen-

tial expression, since the lengths AJI, A;r2 , ,Aj ri approach zero as

a limit. Each term is also called an element of the magnitude to becalculated.

The following rule will be of service in applying this theorem to

practical problems.

FUNDAMENTAL THEOREM. RULE

FIRST STEP. Divide the required magnitude into similar parts suchthat it is clear that the desired result will be found by taking the limit ofa sum of such parts.

SECOND STEP. Find expressions for the magnitudes of these partssuch that their sum will be of the form (2).

THIRD STEP. Having chosen the proper limits x = a and x = b f we

apply the Fundamental Theorem

it //>

lim ^j<t>(Xi)&Xt= I <t>(x)dx

-*i = 1 Ja

and integrate.

x


157. Analytical proof of the Fundamental Theorem. As in the last ar-

ticle, divide the interval from x = a to x = b into any number n of sub-intervals, not necessarily equal, and denotethe abscissas of these points of division byfeit fe2, , fen- if and the lengths of thesubintervals by Axi, Ax2 , , A.rn . Now,however, we let x'\, jr'2> , x'n denote ab-

scissas, one in each interval, determined bythe Theorem of Mean Value (Art. 116), erect

ordinates at these points, and through their

extremities draw horizontal lines to form

rectangles, as in the figure. Note that here

4>(x) takes the place of f'(x). Applying (5), Art. 116 to the first interval(a = a, fe = fei, and JT'I lies between a and feO, we have

or, since

Also, /(7> 2 ) /(fei) = </>(.r' 2 )A.r2 , for the second interval,

/(fea) ~/(fe2) = 0Cr'3)A.r:,, for the third interval,

/(fe) -/(fen-i) = 0(x'w)AxB> for the wth interval.

Adding these, we get

(1) /(fe) -/(a) = 0(x'i)Axi + 0(.r' 2 )A.r a -f -h (/>(:r'n)A:rn.

But 0(x'i) Axi = area of the first rectangle,

<j>(jc

f

2) Aj2 = area of the second rectangle, etc.

Hence the sum on the right-hand side of (1) equals the sum of theareas of the rectangles. But from (1), Art. 156, the left-hand side of (1)

equals the area between the curve y = c/>(j-), the ar-axis, and the ordinatesat x = a and x = b. Then the sum

(2)

fei a

fei a == A.TI,

equals this area. And while the corresponding sumtT

(3) ^P (/>(>)Axt (where xt is any abscissa of

1 = 1 the subinterval Ax,)

(formed as in last article) does not also give the area, nevertheless we

may show that the two sums (2) and (3) approach equality when n in-

creases without limit and each subinterval approaches zero as a limit.

For the difference </>(x\) 0(x t ) does not exceed in numerical value the

difference of the greatest and smallest ordinates in Ax,. And, furthermore,

it is always possible* to make all these differences less in numerical

* That such is the case is shown in advanced works on the calculus.

value than any assignable positive number c, however small, by continu-

ing the process of subdivision far enough, that is, by choosing n sufficiently

large. Hence for such a choice of n the difference of the sums (2) and (3)

is less in numerical value than t(h a), that is, less than any assignable

positive quantity, however small. Accordingly, as n increases without

limit, the sums (2) and (3) approach equality, and since (2) is always

equal to the area, the fundamental result follows that

/ f>n

I <t>(jr)djc= lim

'\ is subdivided in any manner whatever and x t

is any abscissa in the corresponding subinterval.

158. Areas of plane curves; rectangular coordinates. As already

explained, the area between a curve, the j-axis, and the ordinates at

x a and x ~ b is given by the formula

Area

the value of // in terms of x being substi-

tuted from the equation of the curve.

Equation (B) is readily memorized byobserving 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) betweenthe ordinates A I* and BQ.

Let us now apply the FundamentalTheorem (Art. 150) to the calculation of

the area bounded by the curve .r = </>(//)

(AB in figure), the //-axis, and the hori-

zontal lines ?/= c and //

= (/.

FIRST STEP. Construct the n rectan-

gles as in the figure. The required area

is clearly the limit of the sum of the

areas of these rectangles as their numberincreases without limit and the altitude of each one approacheszero as a limit.

SECOND STEP. Denote the altitudes by A?/i, A#L>, etc. Take a

point in each interval at the upper extremity and denote their

ordinates by y } , y, etc. Then the bases are 0(?/i\ 0C//2), etc., andthe sum of the areas of the rectangles is

THIRD STEP. Applying the Fundamental Theorem gives

259

* /(/

h'm 2>U/<>.A//,== / <f>G71 ' x ~

1 Jc

Hence the area between a curve, the //-axis, and the horizontallines y = c and y = d is given by the formula

(C)r=

J c

xdy,

X

the value of x in terms of// being substi-

tuted from the equation of the curve.

Formula (C) is remembered as indicatingthe limit of the sum of all horizontal strips

(rectangles) within the required area, jr and </// being, respectively,the base and altitude of any strip. The element of the area is oneof these rectangles.

Meaning of the negative sign before an area. In formula (B), a is

less than b. Since we now interpret the right-hand member as the

limit of the sum of ti terms resulting from //, A.r, by let ting i = 1 , 2, :*,

, ?/, then, if y is 'ticyatire, each term of this sum will be negative,and (B) will give the area with a negative sign prefixed. This meansthat the area lies below the .r-axis.

rILLUSTRATIVE EXAMPLE 1. Find the urea of

one arch of the sine curve y sin jr.

find

Solution. Placing y = and solving for .r, we

x 0, TT, 2 TT, etc.

Substituting in (B),

Also,

Area GAB = C 'ijdx - f \\nxdx = 2.>'u >'()

Area BCD -f 'yds

-|"**\nxdjr - - 2.

ILLUSTRATIVE EXAMPLE 2. Find the area bounded

by the semicubical parabola ay'2

x'', the //-axis, and

the lines y = a and y = 2 a.

Solution. By (C) above, and the figure, the ele-

ment of area = xdy a V" dy f substituting the value

of x from the equation of the curve MN. Hence

Area BMNC = f%V dyJa

-I

2(^2 - l)= 1.304 a 2. Ans.

Note that a 2 = area OLMB.

r

rT

7vN?(*,y)

In the area given by (B) one boundary is the x-axis. In (C) one

boundary is the $/-axis. Consider now the area bounded by two curves.

ILLUSTRATIVE EXAMPLE 3. Find the area bounded by the parabola y2 = 2 x

and the straight line x y = 4.

Solution. The curves intersect at A (2, 2), B(8, 4). Divide the area into

horizontal strips by a system of equidistant lines parallel to OA' drawn from the

parabola AOH to the line AH. Let

their common distance apart be dy.

Consider the strip in the figure whose

upper side has the extremities (xi, y),

(x<2, y}. From these points drop per-

pendiculars on the lower side. Thena rectangle is formed and its area is

given by

(1) dA = (x2 -xi)dy. (x-2 > x,)

This is the element of area. For the

required area is obviously the limit of

the sum of all such rectangles. That

is, by the Fundamental Theorem,

(2) Areard

=jc

to-*')

in which x^ and x\ are functions of y determined from the equations of the boundingcurves. Thus, in this example, from x y 4 we find x X* 4 + y ; from

y* 2 x we find x Ji = \ y'2

. Then we have, by (1),

(3) dA = (4 + y- I y'^dy.

This formula will apply to the rectangle formed from any strip. The limits are

r = - 2 (at A), d = 4 (at B}. Hence

Area = -i y'

2)dy = 18. Ans.

In this example the area can be divided also into strips by a system of equidis-

tant lines parallel to OK. Let A.r be their common distance apart. The upperend of each line will lie on the parabolaO7>. 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 upperextremity, and (x, y\) the lower, the

rectangle whose area is

(4) dA = (y,-

y,)dx (y, > y,}

is the element of area. But in this ex-

ample it is not possible to find by (4) a

single formula to represent the area of

every one of the rectangles. For while

y.2= V2 x, we have ?/i = V2 x or

?/!= x 4, according as the lower ver-

tex of dA is on the parabola or on AB. Thus from (4), we have two forms of dAand two integrations are necessary.

X


In any problem, therefore, the strips should be constructed so that only a

single formula for the element of area is necessary. Formula (4) is used when these

strips are constructed by drawing lines parallel to the y-axis.

In the Fundamental Theorem some or all of the elements 0(x,) Ax,

may be negative. Hence the limit of their sum (the definite integral)

may be zero or negative. For example, if 0(x) = sin .r, a 0, 6 = 2 TT,

the definite integral (3), Art. 156, is zero. The interpretation of this

result using areas appears from Illustrative Example 1 above.

PROBLEMS

1. Find the area bounded by the hyperbola x\i= a 2

, the .r-axis, and

the ordinates x = a and x 2 a. Ans. a'2 In 2.

2. Find the area bounded by the curve //= In x, the .r-axis, and the

line a- = 10. AUK. 14.026.

3. Find the area bounded by the curve //= xc r

,the .r-axis, and the

line x = 4. An*. 164.8.

4. Find the area bounded by the parabola V.r + V# - Va, and the

coordinate axes. A u*. { a 2.

5. Find the entire area of the hypocycloid x^ -f ?r'= a 5

. Ans. $ ira 2.

Find the areas bounded by the following curves. In each case draw the

figure, showing the element of area.

6. y2 - 6 .r, x'2 = 6 y. A HK. 12. 10. ir = 2 .r, .r- -f ?/

2 = 4 x.

7. y'2 = 4 x, x 2 - 6 y. 8. 11. //

= 6 .r-

x~, y-

x.

8. y'2 - 4 JT, 2 x - y = 4. 9. 12. //

= .r' -- 8 x, /y

= x.

9. y = 4 - x 2, y - 4 - 4 x. lOg. 13. y- = 4x,x = 12 + 2 ?/

-//

2.

14. Find the area bounded by the parabola y 6 -f 4 a- -r2 and the

chord joining (- 2,-

6) and (4, 6). An*. 36.

15. Find the area bounded by the semicubical parabola y* = x 2 and

the chord joining (- 1, 1) and (8, 4). ^4n. 2.7.

16. Find an expression for the area bounded by the equilateral hyper-

bola x2y

2 = a 2, the .r-axis, and a line drawn from the origin to any

point (x, y) on the curve. Am q^_ ^ ix + y\

"2 \ a )

17. Find the area bounded by the curve y - x(l Vx) and the line

r = 4. Aw. -4*.

18. Find the area bounded by the curve x'2y = x'2 1 and the lines

y = 1, x = 1, and x = 4. ^s. f .

19. Find the area bounded by the curve y = x3 - 9 x 2 + 24 x - 7, the

j/-axis, and the line y = 29. Ans. 108.

A square is formed by the coordinate axes and parallel lines throughthe point (1, 1). Calculate the ratio of the larger to the smaller of the twoareas into which it is divided by each of the following curves.

20. y = x 2.

21. y = x*.

22. y = x4 .

23. y* = x 2.

24. Vx +

25. x8 + ir*

ATMS. 2.

3.

4.

5.

32-3 TT

3 7T

26. y = sin ~- -

27. ?/= jrc

J

28. = tan '

29.

Ans.7T - 2

For each of the following curves calculate the area in the first quad-rant lying under the arc which extends from the //-axis to the first inter-

cept on the .r-axis.

30. jr -f // + y'2 = 2. An*. I}-.

31. ?/= ^-8r 2 + 15 j. 15J.

32. y = cr sin .r. 12.07.

33. 2 = (4- x)\

34. ?/= e- cos 2 j\

x

35. ?/ 4 r - cos JT

36. y sin (j- + 1).

159. Areas of plane curves; polar coordinates. Let it be required

to find the area bounded by a curve and two of its radii vectores.

Assume the equation of the curve

to bep=/(0), \D

and let OP\ and OI) be the two radii.

Denote by a and /j the angles which the

radii make with the polar axis. Applythe Fundamental Theorem, Art. 156.

FIRST STEP. The required area iso<

clearly the limit of the sum of circular

sectors constructed as in the figure.

SECOND STEP. Let the central angles of the successive sectors be

A0i, A0i>, etc., and their radii pi, p 2 ,etc. Then the sum of the areas

of the sectors isn

pr A0i + % p 22 A02 H h i Pn

2 A0 7t= V J Pl

-'

A0,.

For the area of a circular sector = J- radius X arc. Hence the area of

the first sector J PI p\ A0j = o pr' A0i, etc.

THIRD STEP. Applying the Fundamental Theorem,

lim 4

= f Ap2

rfff.

./a

Hence the area swept over by the radius vector of the curve in mov-

ing from the position OP\ to the position OD is given by the formula

(D)

the value of p in terms of 6 being substituted from the equation of

the curve.

The element of area for (D) is a circular sector with radius p and

central angle dO. Hence its area is i p" dO.

ILLUSTHATI\ E EXAMPLE, Find the entire area of the lemniscate p2 a 2 cos 2 8.

Solution. Since the figure is symmetrical

with respect to both OX and OY, the whole

area = 4 times the area of OAB.

Since p = when 9 =j.

we see that if

varies from to , the radius vector OP sweeps4

over the area OAB. Hence, substituting in (D),

Entire area = 4 x area OA B = 4 if" p- dO = 2 '-'

f* cos 2 <W = a'2 ;

*- v/<i Jo

that is, the area of both loops equals the area of a square constructed on OA as

one side.

PROBLEMS

1. Find the area bounded by the circle p = a cos and the lines

= and = 60. An*. 0.37 a 2.

2. Find the entire area of the curve p a sin 2 0. 4n,s. i Tra 2.

Calculate the area inclosed by each of the following curves.

3. p2 = 4 sin 2 0. An.s. 4. 10. p = 3 + cos 3 0.

4. p = a cos 3 0. i Tra 2.

1 1. p- a cos -f 6 sin ^.

$. p = a(l- cos 0). 1 Tra

2.

o ^12. p = 2 cos 2 -

6. p = 2 - cos 0. f TT. ^

7.p = sin4 ITT.13. p - a sin n0.

8. p - i + cos 2 0. J TT.14 - P = cos * ^ ~ cos fl -

9. p = 2 + sin 3 0. 1 TT. 15. p = cos 3 0-2 cos 0.

16. Find the area bounded by the parabola pfl + cos 0) = a and the

lines = and = 120. Am. 0.866 a2.

17. Find the area bounded by the hyperbola p2 cos 2 = a 2 and the

lines = and = 30. Ans. 0.33 a 2.

18. Prove that the area generated by the radius vector of the spiral

p = e eequals one fourth of the area of the square described on the radius

vector.a

19. Find the area of that part of the parabola p a sec 2 - which is

intercepted between the curve and the latus rectum.~

_

a 2.

20. Show that the area bounded by any two radii vectors of the hyper-bolic spiral pO a is proportional to the difference between the lengths

of these radii.

o.'2b'2

21. Find the area of the ellipse p'2 =-

:

-:

Ans. irab.a 2 sin 2 6 -f b'

2 cos 2 6

22. Find the entire area of the curve p = afsin 20 + cos 2 0).

Ans. ira'2

.

c\

23. Find the area below OA' within the curve p = a sin-^ -o

&(io 7r + 27V3)a 2.

24. Find the area bounded by p'2 a 2 sin 4 0. Ans. a2

.

Find the area bounded by the following curves and the given lines.

25. p = tan 0; = 0, =\ TT.

26. p = <*"; 0=1 TT, 6=1 TT.

27. p = sec + tan;

= 0, 6 =.]

TT.

28. p - a sin + ft cos ;(9 - 0, =

| ?r.

Calculate the area which the curves in each of the following pairs have

in common.

29. p = X cos 0, p 1 -f cos 0. Ans. TT.

30. p - 1 -f cos 0, p - 1. -.

31. p 1 cos 0, p sin 0. | TT 1.

32. p2 = 2 cos 2 0, p = 1. i TT + 2 - Vs.

33. p2 = cos 2 0, p

2 = sin 2 0. 1 - J V2.

34. p = V6 cos 0, p2 = 9 cos 2 0. i(?r + 9 - 3V).

35. p = V2 sin 0, p2 = cos 2 0. ( TT -f 3 - 3V).

36. p = V2 cos 0, p2 - V3 sin 2 0.

37. 3 p = V3 cos 0, p - cos 2 0.

38. 3 p = V6 sin 2 0, p2 = cos 2 0.

39. Find the area of the inside loop of the trisectrix p = a(l 2 cos 0).

For figure, see limayon, Chapter XXVI. Ans. \ a 2 (2 TT - sVs).


160. Volumes of solids of revolution. Let V denote the volume of

the solid generated by revolving the plane surface ABCD about the

x-axis, the equation of the planecurve DC being

FIRST STEP. Divide the seg-

ment AB into n parts of lengths

Axi, Ax2 , *, Ax u ,and pass a

plane perpendicular to the x-axis

through each point of division.

These planes will divide the solid

into n circular plates. If rectangles

are constructed with bases Axi,

AxL>, ,Ax n within the area

ABCD, then each rectangle gen-

erates a cylinder of revolution

when area ABCD is revolved. Thus a cylinder is formed correspond-

ing to each of the circular plates. (In the figure n = 4 and two cyl-

inders are shown.) The limit of the sum of these n cylinders (n > oo )

is the required volume.

SECOND STEP. Let the ordinates of the curve DC at the pointsof division on the x-axis be ?/i, ?/i>, , //. Then the volume of the

cylinder generated by the rectangle AEFD will be iry\~ Axi, and the

sum of the volumes of all such cylinders is

iry 2- Ax 2 ry n- Ax rt

= VV

THIRD STEP. Applying the Fundamental Theorem (using limits

OA = a and OB = b),

limpi>

,

-.IJa

Hence the volume generated by revolving about the x-axis the

area bounded by the curve, the x-axis, and the ordinates x a and

x b is given by the formula

(E) 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-


der of altitude dx with a base of area Try2 and hence of volume Try

2 dx.

This cylinder is the element of volume.

Similarly, when OY is the axis of revolution, we use the formula

(F) Vv - TT

jfx* dy,

where the value of x in terms of y must be substituted from the

equation of the given curve.

ILLUSTRATIVE EXAMPLE 1. Find the volume generated by revolving the ellipse

~4-

#_ i about the x-axis.a 2 b 2

Solution. Since y'2 = ^ (a

2 - x 2), and the

required volume is twice the volume gen-

erated by GAB, we get, substituting in (E),

V /vj /'! J,2

^ =7T/ y'

2 dx = TT ^(a'-L Jo Jo a-

2 irab*

To check this result, let b - a. Then 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

axis, the solid generated is called a prolate spheroid ;when about its minor axis, an

oblate spheroid.

ILLUSTRATIVE EXAMPLE 2. The area bounded by the semicubical parabola

(1) ay* = x\

the ?/-axis, and the line AB (y a) is revolved about AB.Find the volume of the solid of revolution generated.

Solution. In the figure, OPAB is the area revolved.

Divide the segment AB into n equal parts each of

length AJ. In the figure, NM is one of these parts. The

rectangle NMPQ when revolved about AB generates a

cylinder of revolution, whose volume is an element of

the required volume. Hence

Element of volume = irr-h TT(G i/)2 Ax,

since r = PM = RM - RP = a - y

and h = NM = Ax.

Then, by the Fundamental Theorem,

4(a,a)

(2)

ra paVolume of solid = V = TT / (a

-y)~ dx = ir I (a

2 - 2 ay +Jo Jo

for the limits are x = and x = AB = a. Substituting for y its value given by (l) f

the answer is V = 0.45 ira\ Ans.

This should be compared with the volume of the cone of revolution with

altitude AB (= a) and base of radius OB (= a). Volume of cone = J 7ra :?.

If the equations of the curve CD in the figure on page 265 are givenin parametric form . _ /,,\ , /A

y = </>(/),

then in () substitute y = <f>(t\ dx=f(t)dt, and change the limits

to /i and 2, if

t = /, when x = a, t = / L. when x = b.

Volume of a hollow solid of revolution. When a plane area is re-

volved about an axis not crossing the area, a hollow solid of revolu-

tion is formed. Consider the solid ob-

tained by revolving about the rr-axis the

area ACBDA of the figure. Pass throughthe solid a system of equidistant planes

perpendicular to the axis of revolution

OX. Let Ax be their common distance

apart. Then the solid is divided into

hollow circular plates each of thickness

Ax. If one of the planes dividing the solid passes through M, the

hollow circular plate with one base in this plane is approximately a

hollow circular cylinder whose inner and outer radii are respec-

tively MPi (=y\) and MP>2 (=2/2). Its altitude is A.r. Hence its

volume is ir(y>2- yr)Ax. Let there be w hollow cylinders, where

b a = n-Ax. The limit of the sum of these v hollow cylinders

when ft cc is the volume of the hollow solid of revolution. Hence

(3)r b

V; = TT

Ja

The element of volume in (3) is a hollow circular cylinder with

inner radius y it outer radius // L>, and altitude Ax. The radii y { and

y>> are functions of x (= OM) found from the equations of the curves

(or the equation of the curve) bounding the area revolved.

ILLUSTRATIVE EXAMPIK 3. Find the volume of the ring solid (anclwr ring or

torus) obtained by revolving a circle of radius n about an external axis in its planeb units from its center (b > a).

Solution. Let the equation of the circle be

x* + (y- &)2 = a\

and let the z-axis be the axis of revolution.

Solving for y, we have

.-. dV - ir(y22 -

2/12) Ax = 4 irbVa 2 - x2 Ax.

By (3), Vx = 4 irb x 2 dx = 2 7T2a 2

6. Aw.


A solid of revolution may be divided into cylindrical shells bypassing through it a system of circular cylinders whose common axis

is the axis of revolution. If the area

ACED of the figure be revolved about

the ?/-axis, it may be shown that

(4;r b

= 2ir \ (y>> -y } )xdx,Ja

where OM = x, MP\ y\, MP2 = 2/2-

The element dV is now a cylindrical shell

of radius r, altitude y? y\, and thickness Ax. Illustrative Ex-

ample 3 may be solved by (4).

PROBLEMS

1. Find the volume of the sphere generated by revolving the circle

x'2 -f y'2

r'2 about a diameter. Ans. f Trr*.

2. Find by integration the volume of the truncated cone generated byrevolving the area bounded by y = 6 x, y = 0, jr = 0, j* = 4 about OA~.

Verify geometrically.

3. Find the volume of the paraboloid of revolution whose surface is

generated by revolving about its axis the arc of the parabola y'2 = 2 px

between the origin and the point (j"i, ?/]).

Ans. irpxi'2

\ Tr\i\ix\ ; that is, one half of the volume

of the circumscribing cylinder.

4. Find the volume of the figure generated by revolving the arc in

Problem 3 about OY.Ans. I 7r.ri

2?/i ; that is, one fifth of the cylinder of

altitude y\ and radius of base x\.

Find the volume generated by revolving about OX the areas bounded

by the following loci.

5. ?/- jr\ u = 0, .r = 2.

6. ay'2

.r\ ?/= 0, .r = a.

7. The parabola Vj- -f V// = Vo, x = 0, y = 0.

8. The hypocycloid x$ -f y* = cA

9. One arch of y = sin x.

10. One arch of ?/= cos 2 x.

11. y = e-*, y = Q, x ^ 0, x = 5.

Ans.

TTO3

7T(1-

An$. 48 TT.12. 9 x 2 + 16 y2 = 144.

13. y = jrf, y = 0, x= 1.

14. The witch (jr2-f 4 a 2

)?/ = 8 a3, ?/= 0.

-1).

4 ?T2a3

.

16. ?/2(2 a - x) = JT*, y = 0, jr = a. 0.2115 wa\

17. y = a-2 - 6 x, //

= 0.

18. j/2 = (2

-.r)

3, ?/

= 0, jc = 0, j- = 1.

19. i/2(4 + .r

2)= 1, //

= 0, jc = 0, a* = oo.

20. Cr - 1 )y = 2, ?/= 0, j- = 2, r = 5.

Find the volume generated by revolving about OY the areas bounded

by the following loci.

21. y = .r:

', ?/= 0, x = 2. Aw.s. V TT.

22. 2 y2 = x:

\ ?/= 0, T = 2. TT-

23. ?/= <

>T, //= 0, x 0. 2 TT.

24. 9 JC2-f 16 y'

2 - 144. 64 TT.

25. (-

26. y2 = 9 - a% a- = 0.

27. x 2 = 16 - y, y = 0.

28. ?/

2 = GLT, ?/= 0, j- = a.

29. The equation of the curve OA in the figure is ?/2 = .r

3. Find the

volume generated when the area

(a) OAtt is revolved about OA'.

(b) GAB is revolved about AB.

(c) OAB is revolved about CA.

(d) O.4 # is revolved about OY.

(e) OAC is revolved about 07.

(f) OAC is revolved about CA.

(g) OAC is revolved about AB.

(h) OAC is revolved about OX.

y

Arts. 64 TT.

Z^J 7T.

192 TT.

(U.H)(4.8)

(4,0)

30. Find the volume of the oblate spheroid generated by revolving

the area bounded by the ellipse b 2x 2 + a 2?/2 = a z

b'2 about the t/-axis.

An.s. | ?ra 26.

31. A segment of one base of thickness h is cut from a sphere of radius r.

Show by integration that its volume is* l ' --

Calculate the volume generated by revolving about each of the follow-

ing lines the area which it cuts from the corresponding curve.

32. = 3; y = 4 x - x 2. Ann. jf TT.

33. x = 4; ?/

2 = y*. *%$* *

34. y - - 4; T/

= 4 + 6 x - 2 x2. -i^p TT.

35. y = r; y = j-a

.gig ;rV2.

36. = JT; # = 4 j - x 2

. 54 TTV2.

37. ?/= .r + 7

; ?/= 9 - x 2

. 4 TTVi38. s -f ?/

= 1; V7 + \/T/ = 1. ^ 7r\/2.

39. r -f ?/= 7

; .n/ = 6.

40. Find the volume generated by revolving one arch of the cycloid

x r arc vers ^ \/2 ry

about OX, its base.

r

HINT. Substitute dx - V:, and limits //

= 0, y - 2 r, in (), Art. 160.V2 r^ ~ ""' Aw. 5 7T

2 r :i

.

(IT \

c" 4- f "/

about the .r-axis from x = to r 6."

47T^'

5

( ? ?) _L^ 2^A w. V<'

" - f (/

/ -f-

42. Find the volume of the solid generated by revolving the cissoid

jTr<

y/

2 -- about its asymptote jr 2 (/. .4//,s. 2 TrV.L rt jc

43. Given the slope of the tangent to the tractrix - -- ?/find

the volume of the solid generated by revolving it about OX. An*. ?ra:

'.

44. Show that the volume of a conical cap of height a cut from the

solid generated by revolving the rectangular hyperbola .rj

//-= a 2 about

OX equals the volume of a sphere of radius a.

45. Using the parametric equations of the hypocycloid

f i* rr (j COS :i

0,

} y = a sin : *

6,

find the volume of the solid generated by revolving it about OX.Ans. -ffa Tra3 .

46. Find the volume generated by revolving one arch of the cycloid

x- a(0 - sin 0),

= ad cos 6}

about its base OA". Ans. 5 ?r2a3

.

Show that if the arch be revolved about ())', the volume generatedis 6 7r

3a3.


47. Find the volume generated if the area bounded by the curve

y = sec \ TTJC, the .r-axis, and the lines jr = } is revolved about the .r-axis.

Ans. 4.

48. The area under the curve //= r

1

sin .r from x to a* TT is re-

volved about the .r-axis. Find the volume generated.

49. Given the curve x /2

, ?/= 4 t

- K Find (a) the area of the loopand (b) the volume generated by the area inside the loop when revolved

about the x-axis. Ans. (a) vV ; (b) 67.02.

50. Revolve the area bounded by the two parabolas ir = 4 x and

y* = 5 x about each axis and calculate the respective volumes.

.4 wx. OX : 10 TT; OY : *%* IT.

51. Revolve about the polar axis the part of the cardioid p 4 -f 4 cos

between the lines 6 = and = ^ and compute the volume.

Ans. 160 TT.

161. Length of a curve. By the length of a straight line we com-

monly mean the number of times we can superpose upon it, another

straight line employed as a unit of length,

as when the carpenter measures the length

of a board by making end-to-end applica-

tions of his foot rule.

Since it is impossible to make a straight

line coincide with an arc* of a curve, wecannot measure curves in the same manner as we measure straight

lines. We proceed, then, as follows.

Divide the curve fas AB) into any number of parts in any mannerwhatever (as at r, I), E] and connect the adjacent points of division,

forming chords fas AC, r/>, ])K, EB}.

The length of the curve is defined as the limit of the sum of the chords

as the number of points of division increases without limit in such a waythat at the same time each chord separately approaches zero as a limit.

Since this limit will also be the measure of the length of some

straight line, the finding of the length of a curve is also called "the

rectification of the curve."

The student has already made use of this definition for the length

of a curve in his geometry. Thus the circumference of a circle is

defined as the limit of the perimeter of the inscribed for circum-

scribed) regular polygon when the number of sides increases without

limit.

The method of the next article for finding the length of a plane

curve is based on the above definition, and the student should note

very carefully how it is applied.

162. Lengths of plane curves; rectangular coordinates. We shall now

proceed to express, in analytical form, the definition of the last article,

making use of the Fundamental Theorem.

Given the curve

and the points P'(a, c), Q(b, d) on it; to find

the length of the arc P'Q.

FIRST STEP. Take any number n of points

on the curve between P' and Q and draw the

chords joining the adjacent points, as in the

// c

figure. The required length of arc P'Q is evidently the limit of the

sum of the lengths of such chords.

SECOND STEP. Consider any one of these chords, P'P" for ex-

ample, and let the coordinates of P' and P" be / /"

P'(xf

, yf

) and I>"(x' + Ax', //' + Ay').

Then, as in Art. 95,

or

P'P" = V(A:r')-

p'p" = 1 + a;j a:'* Ax'

[Dividing inside the radical by (A.r')z and multiplying outside by Ax'.]

But from the Theorem of Mean Value (Art. 116) (if Ay' is denoted

by /(ft) /(a) and A?' by b - a), we get

f^'=/W (x

f < x{ < x' + Ax')

Xi being the abscissa of a point Pi on the curve between P' and P" at

which the tangent is parallel to the chord.

Substituting, P'P" = [1 +f'(xi )2]*A:r' = length of first chord.

Similarly, p"p"' = [1 +f'(xz)2]*Ax" = length of second chord,

p()Q - [i +/'(jM )2]*Ajfw) = length of nth chord.

The length of the inscribed broken line joining P' and Q (sum of

the chords) is then the sum of these expressions, namely,

1=1


THIRD STEP. Applying the Fundamental Theorem,

lim V [1 +/'(j i )2)4A.ri" = f"[1 + f\r)*$ dx.

w-",^l Ja

Hence, denoting the length of arc P'Q by s, we have the

formula for the length of the arc

= fJa

(G)

where yf =

-^must be found in terms of x from the equation of the

given curve.

Sometimes it is more convenient to use y as the independent vari-

able. To derive a formula to cover this case, we know from Art. 39that

=-7- ; hence dx ~ x' dy.

dx or

dy

Substituting these values in (G), and noting that the correspond-

ing y limits are c and d, we get the formula for the length of the arc,

(H) *= rV+l]dy,J c

where x' = must be found in terms of y from the equation of thedy

given curve.

Formula (G) may be derived in another way. In Art. 95,

formula (D),

(1) ds = (I + y'2)* dx

gives the differential of the arc of a curve. If we proceed from (1)

as in Art. 142, we obtain (G). Also, (H) follows from (E) in Art. 95.

Finally, if the curve is defined by parametric equations

(2) *=/(*), =

it is convenient to use

(3) 8

since, from (2), dx=f'(t)dt, dy = 4>'(t}dt.

ILLUSTRATIVE EXAMPLE 1. Find the length of the circumference of the circle

Solution. Differentiating,*-& -(Is y

Substituting in (G),

Arc BA =

dx.Jo I y

[Substituting y'2 = r'2 jr'

2 from the equation of the i

Lcirele in order to get everything in tcrrm of jr. \

.-. arc BA = r f' -p^L^ =- ~ (See Illustrative Example 1, Art. 154.)./o vr' -

jr2 ~

Hence the total length equals 2 irr. Ans.

ILLUSTRATIVE EXAMPLE 2. P'ind the length of arc of one arch of the cycloid

y -_ a(0 sin 6), y a(\ - cos 0).

See Illustrative Example 2, Art. 81.

Solution. r/j = a(] cos #)(/#, (///= a sin f?r/^.

Then dx'2 -f <-///-' 2 -'( 1 cos ^ir/fl :'

4 <-/- sin jI 6 (W 2

. By (5), Art. 2

Using (,'}), ,s .

j2 (/ sin

^,^r/fy 8 . Ana.

'o

The limits are the values of at and 1) see figure, Illustrative Example 2,

Art. SI ), that is, = and = 2 TT.

ILLUSTRATIVE EXAMPLE ,'i. Find the length of the arc- of the curve 25 y2 - x 5

from x to x ~- 2.

Solution. Differentiating, y' = \ j2. Hence, by (G),

(4) =f

(1 fjjc*)*djr = \ I "(4 -f a-'M'^dr.

The integral in (4) was evaluated approximately in Illustrative Example 2,

Art. 148, by the trapezoidal rule, and also in Illustrative Example 2, Art. 149, bySimpson's rule. Taking the latter value, s ?, (4.821) - 2.41 linear units. .4/i.s.

163. Lengths of plane curves; polar coordinates. From (/), Art. 96,

by proceeding as in Art. 142, we get the formula Njc>

for the length of the arc,

where p and -^ in terms of must be substi-ctu

tuted from the equation of the given curve.

In case it is more convenient to use p as the independent variable,and the equation is in the form

thendp

Substituting this in [p2 d6- + dp

2]*

276

gives

Hence if pi and p2 are the corresponding limits of the independentvariable p, we get the formula for the length of the arc,

\dp.

where -r~ in terms of p must be substituted from the equation of thedp

given curve.

ILLUSTRATIVE EXAMPLE. Find the perimeter of the cardioid p u([ 4- cos 0).

Solution. Here -fl 2 dO = 2 a f "cos - </0 = 4 a.

,/o ,/(i li

.% = 8 a. Ans.

PROBLEMS

1. Find the length of the curve whose equation is ?/:{ = jr~ between the

points (0, 0) and (8, 4). AUK. 9.07.

2. Find the length of the arc of the semicubical parabola ay* x :i from

the origin to the ordinate x 5 a. A IW5 aA n .S .-

y3 1

3. Find the length of the curve whose equation is if -f--- from

6 2 x

'the point where x ~ 1 to the point where jr = 3. Ar/,s. -1,*-.

4. Find the length of the arc of the parabola //2 = 2 ;>.r from the vertex

to one extremity of the latus rectum. ^c. ?>\/2 , P i., M . x/o'i

/-I A/ ,S . I -p 111 \1 *7~ V ^j y

2 2

5. Find the length of arc of the curve y2 = x3 from the point where

x to the point where x \.Ans.

.]|.

6. Find the length of arc of the parabola 6 y = x2 from the origin to

the point (4, |).Ans. 4.98.

7. Approximate by Simpson's rule the length of arc of the curve

3 y = x3 from the origin to the point (1, -g).^s. 1.09,

8. Find the length of arc of the curve y log sec x from the origin to

the point (~ In 2\ Am. In (2 + Vs).

9. Find the length of arc of the hyperbola x2 -y

2 = 9 from (3, 0)

to (5, 4). Ans. 4.56.

10. Find the length of the arch of the parabola y 4 x x 2 which

lies above the x-axis. Ans. 9.29.

11. Find the entire length of the hypocycloid x* + y* = a*. Ans. 6 a.

12. Rectify the catenary y = -(c

u + ea

) from x = to the point (x, y).

'

2 ^

13. Find the length of one complete arch of the cycloid

x = r arc vers - V2 ry-

y'1

. Ans. 8 r.

HINT. Use (//), Art. 162. Here =9

V_= '

14. Rectify the curve 9 a?/2 = x(x 3 a)

2 from x = to x = 3 a.

Ans. 2 aVS.? 5J

15. Find the length in one quadrant of the curve(-J

-f (r) = !

a 2-f ab + b'2

Ans.a -f

c* 4- 116. Find the length between x = a and x = 6 of the curve fjy = -~-

x1

C 2h 1Ans. log + a b.H a

1

17. The equations of the involute of a circle are

(x= a (cos + 0sin 0),

\ y = a (sin cos B).

Find the length of the arc from 6 = to = ft. Ans.-J aft

2.

18. Find the length of arc of curve !*= c* sm

f from = to = -I // << cos 2

Ans. V2(e2 - 1).

Find the length of arc of each of the following curves.

19. y = In (1-

.r2) from .r = to x - \.

x 21

20. y = - In x from x = 1 to x ~ 2.

21. //= In esc x from x = ^ to x =

22. 3 x 2 =?/M from y = 1 to ?/

= 20.

23. One arch of the curve # = sin x.


24. Find the length of the spiral of Archimedes, p = a6, from theorigin to the end of the first revolution.

Am?. TraVl -f 4 TT- + ~ In (2 w + Vl -f 4 ?r2).

25. Rectify the spiral p ca9 from the origin to the point (p, 0).

HINT. Use (/).

n

26. Find the length of the curve p = a sec 2 -

from = to =

Aw*. [V2 + ln (V2 + i)]a .

27. Find the length of arc of the parabola

p = from = to =-7-

1 -f cos 2

V + ln (V2 + 1).

28. Find the length of the hyperbolic spiral

p0 = a from (p lf 0i) to (p 2 , 02).

- a In' (a +

.

4- Va'-' -f p,1

-)

29. Show that the entire length of the curve p = a sirr* ~ is -> Show

that OA, AB, BC (see figure) are in arithmetical progression.

30. Find the length of arc of the cissoid p = 2 a tan sin from =to e = f '

4

31. Approximate the perimeter of one leaf of the curve p = sin 2 0.

164. Areas of surfaces of revolution. A surface of revolution is

generated by revolving the arc CDof the curve

y=J(x)

about the axis of A'.

It is desired to measure the area

of this surface by making use of the

Fundamental Theorem.

FIRST STEP. As before, divide

the interval AB into subintervals

Aari, Ax2, etc., and erect ordinates

at the points of division. Drawthe chords CE, EF, etc. of the

curve. When the curve is revolved, each chord generates the lateral

surface of a frustum of a cone of revolution. The area of the required


surface of revolution is defined as the limit of the sum of the lateral

areas of these frustums.

SECOND STEP. For the sake of clearness let us draw the first

frustum on a larger scale. Let M be the middle point of the chord

CE. Then

(1) Lateral area = 2 ir\7M CE.*

In order to apply the Fundamental Theoremit is necessary to express this product as a

function of the abscissa of some point in the

interval AJI. As in Art. 162, we get, using the

Theorem of Mean Value, the length of the chord

(2) A/,,

I*- Axr

where Ji is the abscissa of the point P$r\, y$

on the arc CE where the tangent is parallel to

the chord CE. Let the horizontal line throughM intersect QP\ (the ordinate of 7'j) at /?, and denote RP\ byThen

(3) /V A/ = //!-,.

Substituting (2) and (3) in (1 ), we get

2 TT(//,-

6, )[1 + /'(*] )~J> A.ri = lateral area of first frustum.

Similarly,

2 irtite ej)[l + /'(A)-|*A.r 1> = lateral area of second frustum,

2 ir(y n c n )[l + /'u,,)~J*A.r,, = lateral area of last frustum.

Hence

rr

2 2 ir(y t c,)[l +f'(jr l Y2 \* AJ, = sum of lateral areas of frustums.

This may be written

(4) 2 2 iry^ - IT

1=1

* The lateral area of the frustum of a cone of revolution is equal to the circumference of

the middle section multiplied by the slant height.

t The student will observe that as A.TI approaches zero as a limit, ci also approaches the

limit zero.

THIRD STEP. Applying the Fundamental Theorem to the first

sum (using the limits OA = a and OB = 6), we get

lim V 2 7n/,[l +f'(x i r~] A.T, = iry[l + /W1* dx.

The limit of the second sum of (4) when n *oo is zero.* Hence the

area of the surface of revolution generated by revolving the arc CDabout OX is given by the formula

(TT\ Q 9 .

(A; OJT *

where Sz denotes the required area. Or we may write the formula

in the form>b

(I) S = 2 TT( y ds.Ja

Similarly, when OY is the axis of revolution we use the formula

(M) S - 2 TT f x ds.Jc

In (L) and (M) ds will have one of the three forms (C), (Z)),

of Art. 95, namely,

v~> i i

depending upon the choice of the independent variable. The last

form must be used when the given curve is defined by parametric

equations. In using (L) or (M), calculate ds first.

The formula (L) is easily remembered if we consider a narrow

band of the surface included between two planes perpendicular to the

axis of revolution, and regard it approximately as the convex surface

of a frustum of a cone of revolution of slant height r/s, with a middle

section whose circumference equals 2 Try, and hence of area 2 iry ds.

ILLUSTRATIVE EXAMPLE 1. The arc of the cubical parabola

(5) a'-y = * :<

between x = and x = a is revolved about OA', Find the area of the surface of

revolution generated.

* This is easily seen as follows. Denote the second sum by S n If equals the largest of

the positive numbersfd |, |

62 I, ,'!, then

7?

Sn ^ t^[l+f'(r1=1

The sum on the right is, by Art. 162, equal to the sum of the chords CE, EF, etc. Let

this sum be In Then Sn = dn Since lim c = 0, Sn is an infinitesimal, and therefore lim Sn = 0.

3 X 2

Solution. From (5), y' = - Hencea 2

= (1 + : = (a4

-f 9 dx.

O i

Then the element of area = 2 iryds = (a4-f 9 x4

)

2 x 3(

a*

Therefore, by (),

27-I)" 1' = 3.6 a 2

.

ILLUSTRATIVE EXAMPLE 2. Find the area of the ellipsoid of revolution gener-ated by revolving the ellipse whose parametric equations are (see (3), Art. 81)x '-- a cos 0, y = 6 sin about OX.

Solution. W7

e have

dx a sin d0, dy b cos d0,

and d.s-

(dx'2

4- dy'2)% (a'

2 sin 2-f b 2

co;,- 0)i d0.

Hence the element of area = 2 iry ds = 2 7r6(a2 sin 2

4- 6~ cos-' 0)i sin d0.

r(6) .'. i &, = 2 TT& / 2 (a

2 sin 2-f 6 2 cos- 0)i sin d0.

To integrate, let u cos 0. Then du = sin d0. Also,

o'J sin- 4 b'2 cos 2 = a'2 (I

- cos 20) 4- b 2 cos 2 = a'2 - (a

2 - 6 2)w

2.

Hence, using the new limits u = 1, w = 0, and interchanging the u limits (Art.150), the result is

-(a

2 - du. (a>6)

Working this out by (22), we get

i . ,

^Sj 2 TO- H2 irab . . .

arc sin f,where c eccentricity Ans.

ILLUSTRATIVE EXAMPLE 3. Find the area of the surface of revolution generated

by revolving the hypocycloid x<* 4- ?/s ^ about the x-axis.

Solution. Here -^ = ^-, w = (^Ti ja)3dx V A

1.= 1 4-

,

d.r =-..r 4- V fl-T~- -

X* / X*

-dx.

Substituting in (L), noting that the arc BAgenerates only one half of the surface, we get

f = 27ro^ rV^ Jo

This is an improper integral, since the func-tion to be integrated is discontinuous (becomesinfinite) when x = 0. Using the definition (1),

Art. 154, the result is _ 6 va<

2 12


PROBLEMS

1. Find by integration the area of the surface of the sphere generatedby revolving the circle x 2 + y

2 = r 2 about a diameter. Ans. 4 irr2

.

2. Find by integration the area of the surface of the cone generatedby revolving about OA~ the line joining the origin to the point (a, 6).

AUK. TrhVa 2-f b'

2.

3. Find by integration the area of the surface of the cone generatedby revolving the line y = 2x from x = to x = 2 (a) about OX

; (b) aboutO Y. Verify your results geometrically.

4. P'ind by integration the lateral area of the frustum of a cone gen-erated by revolving about OX the line 2 //

= .r- 4 from .r = to jr = 5.

Verify your result geometrically.

5. Find the area of the surface generated by revolving about OY thearc of the parabola //

= x 2 from ?/= to y 2. AUK. \? TT.

6. Find the area of the surface generated by revolving about O.V thearc of the parabola y = x 2 from (0, 0) to (2, 4).

7. Find the area of the surface generated by revolving about OX thearc of the parabola ?/

2 = 4 - jr which lies in the first quadrant, AUK. 3G.18.

8. Find the area of the surface generated by revolving about OX thearc of the parabola ?/

2 = 2 px from x = to .r 4 p. AUK. :v* irp2

.

9. Find the area of the surface generated by revolving about OY thearc of //

= x* from (0, 0) to (2, 8).

Find the area of the surface generated by revolving each of the follow-

ing curves about OX.

10. 9 y = x*, from y = to x = 2. Am.jj J TT.

11. y2 = 9 jr, from x to x 4. 49 TT.

12. //2 = 24 - 4 jr, from x = 3 to .r = 6. '$ w.

13. 6 //-

j-2, from .r = to .r = 4.f820 - 81 In 3)^

72

14. y = c~r

, from j- - to r = oo. 7r[V2 -f In ( 1 -f \/2 )].

15. The loop of 9 ay2

.r(3 a x)2

. 3 ira2

.

16. 6 a V?/ x4-f 3 a4

, from x = a to x 2 a.-} (

7;ira

2.

17. One loop of 8 a 2?/2 = a 2x 2 - x4

. j vrrz2

.

18. ?/2 + 4 jr = 2 log y t from ?y

= 1 to y - 2. -\? TT.

19. The cycloid \

X = aff

"" sin

J-, y = a(l cos 6).

20. The cardioid /x = a(2 cos

J~ cos 2 *>

i y = a(2sin 6 - sin 2 (9).


21. y'2 = 4 r, from a- = to x = 3.

22. x 2 + ?/2 = 4, from x = 1 to x = 2.

23. x 2 + 4 ?/2 = 36.

24. 9 x 2 + 4 ?/2 = 36.

Find the area of the surface generated by revolving each of the fol-

lowing curves about OY.

25. x = y'\ from y = to y = 3. Ans. ^ 7r[(730)*-

1].

26. y = x3,from //

= to ?/= 3.

27. 6 a 2.r// jr + 3 a 1

, from jr = a to x = 3 a. (20 + In 3)?ra2

.

28. 4 ?/= .r

2 - 2 In .r, from x = 1 to a- = 4. 24 TT.

29. 2 ?/- .rvV' - 1 + log (V - V.r 2 - 1 ), from x = 2 to j- - 5.

Am. 78 TT.

30. ?/2 = jr

{

,from x to x 8. 713.

31. 4 //= .r-, from //

= to y = 4. 33. 4 .r2-f /y

2 64.

32. jr* -f 4 if = 16. 34. 9 jr = //:j

,from //

= to //= 3.

Find the area of the surface generated by revolving each of the fol-

lowing curves.b A About OY

35. The ellipse ^ + f!= 1. 2 vra 2

-f^lnla- r>- r 1

HINT, c = eccentricity of ellipse

36. The catenary ?/= -

\r" -f r "/>^

>

TTfl

trom a- = tox = a.^j-(^

24-4-^~ 2

). 2 ?ra 2(l -^^ 1

(Figure, [). 532)

37. r4 -h3 = 6x//,fromx^ltoa*= 2. H w. 7r(V?

-hln2).

40. The slope of the tractrix at any point of the curve in the first

quadrant is given by -2 =~ y

. Show that the surface generatedOf. Vc 2 -

//-'

by revolving about OX the arc joining the points (xj, ?/i) and (x2t 1/2) onthe tractrix is 2 TJT(//I // L ). (Figure, p. 537)

41. The area in the first quadrant bounded by the curves whose equa-tions are y x* and y = 4 .r is revolved about OA'. Find the total surface

of the solid generated. Ar?s. 410.3.

42. The area bounded by the ?/-axis and the curves whose equationsare x 2 = 4 y and x 2 y -f 4 = is revolved about OY. Find the total

surface of the solid generated. Ans. 141.5.

43. Find the surface generated by revolving about OX the arc of the

X3 1curve whose equation is y = + from .r = 1 to .r = 8. Am.

*

IT

44. Find the entire surface of the solid generated by revolving aboutOX the area bounded by the two parabolas y'

2 = 4 x and //2

.r -f 3.

An*, i 7r(17Vl7 -f 32V2 - 17) = 51.53.

45. Find the area of the surface generated by revolving about OA' onearch of the curve y = sin jr. Ans. 14.42.

165. Solids with known parallel cross sections. In Art. 160 we dis-

cussed the volume of a solid of revolution, such as is shown in the

accompanying figure. All cross sec-

tions in planes perpendicular to the

.r-axis are circles. If OA1 = x, MC =y,

then

(1) Area cross section

ACBD=iry* =TT[</>(*) ]

2,

if 2/=

(/>(x) is the equation of the

generating curve OCG. Herice the area

of the cross section in any plane perpen-

dicular to OX is a function of its perpendicular distance (~ x) fromthe point O.

We shall now discuss the calculation of volumes of solids that are

not solids of revolution when it is possible to express the area of any

plane section of the solid which is

perpendicular to a fixed line (as OX)as a function of its distance from a

fixed point (as O).

Divide the solid into n slices by

equidistant sections perpendicular to

OX, each of thickness AJ.

Let FDE be one face of such a

slice, and let ON = x. Then, by hypothesis,

L)

(2) Area FDE = A(x).

The volume of this slice is equal, approximately, to

(3) Area FDE X A:r = A(x)Ax (base X altitude).

24 INTEGRAL CALCULUSn

Then ^ A(x l )Ax l

= sum of volumes of all such prisms. It is evi-i = i

dent that the required volume is the limit of this sum ; hence, by the

Fundamental Theorem,n

lim Y A(x l)Ax l= I A(x)dx,n-^ t:= ,J

and we have the formula

(N) V

were A(x) is defined in (2).

The element of volume is a prism (in some cases a cylinder)whose altitude is dx and whose base has the area A(x). That is,

civ = A(x)dx.

ILLUSTRATIVE EXAMPLE 1. The base of

a solid is a circle of radius r. All sections

perpendicular to a fixed diameter of the

base are squares. Find the volume of the

solid.

Solution. Take the circle x- + y~ r-

iri the AT-plane as base, and OA" as the

fixed diameter. Then the section PQRSperpendicular to OX is a square of area

4 y~, if 1>Q 2 //. (In the figure, the por-(ion of the solid on the right of the section

1'QRS is omitted.)

Hence /\ (.r) - 4 //'-'

-4(r

2 -.r--), and,

by (/V),

Volume 4 I (r-' j'~)djr = V r ;j. Am.

ILLUSTRATIVE EXAMPLE 2. Find the

volume of a right conoid with circular

base, the radius of the base being r and the altitude a.

Solution. Placing the conoid as shown in the figure, consider a section PQRperpendicular to OA". This section is an isosceles triangle ; and, since

RM = V2 rx - x~

(found by solving .r2 + ^ - 2 r.r, the equation of the

circle ORAQ, for //) and

MP = a,

the area of the section is a

Substituting in (N), a L.J.J4-E~^

This is one half the volume of the cylinder of the same base and altitude.


ILLUSTRATIVE EXAMPLE 3. Calculate the volume of the ellipsoid

by a single integration.

Solution. Consider a section of the ellipsoid perpendicular to OX, as ARCD,with semiaxes b' and c'. The equation of the ellipse HKJG in the YOV-plane is

Solving this for y (= b') in terms of

x (= OM) gives

Similarly, from the equation of the

ellipse EFGI in the A"OZ-plane we get

Hence the area of the ellipse (section) ARC!) is

, , , irbc , ., ,,. . , N-n-b'c' - (a

2 -j-')

-- A(x}.

Substituting in (W),

, r TTbc C*"

,,

\ -/ (a- jr~

a 2 J,)(/.r

- TTabc. Ana.o

PROBLEMS

1. A solid has a circular base of radius r. The line A H is a diameterof the base. Find the volume of the solid if every plane section perpen-dicular to AB is

(a) an equilateral triangle ; Am. % r:iV5.

(b) an isosceles right triangle with its hypotenuse in the plane of the

base ; AUK. ^ r:{

.

(c) an isosceles right triangle with one leg in the plane of the base;

An.s. $ r\

(d) an isosceles triangle with its altitude equal to 20 in. ; Ann. 10 ?rr2

.

(e) an isosceles triangle with its altitude equal to its base. Am. 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 everysection perpendicular to the major axis is

(a) a square ;

(b) an equilateral triangle ;

(c) an isosceles triangle with altitude 10 in.

An*. 1,333 cu. in.

577.3 cu. in.

785.4 cu. in.


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 distant8 in. from the vertex of the parabola. Find the volume of the solid if

every section perpendicular to the axis of the base is

(a) a square ; Arts. 1024 cu. in.

(b) an equilateral triangle; 443.4 cu. in.

(c) an isosceles triangle with altitude 10 in. 426. 7 cu. in.

4. A football is 16 in. long, and a plane section containing a seam is an

ellipse, the shorter diameter of which is 8 in. Find the volume (a) if theleather is so stiff that every cross section is a square; (b) if the cross

section is a circle. Anx. (a) 341:\ cu. in.

; (b) 535.9 cu. in.

5. A wedge is cut from a cylinder of radius 5 in. by two planes, one

perpendicular to the axis of the cylinder and the other passing through adiameter of the section made by the first plane and inclined to this planeat an angle of 45". Find the volume of the wedge. Ans. *%

Q cu. in.

6. Two cylinders of equal radius r have their axes meeting at right

angles. Find the volume of the common part. Ans. -^ 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 perpendicularto the plane of the given circle. Find the volume of the solid it will

generate. Ans. I a ri

(3 TT + 8).

8. A variable equilateral triangle moves with its plane perpendicularto the .r-axis and the ends of its base on the points on the curves y'

2 = 16 asand \r = 4 cue, respectively, above the .r-axis. Find the volume generatedby the triangle as it moves from the origin to the points whose abscissaifl- Ans. ^V5a3

.

9. A rectangle moves from a fixed point, one side being always equalto the distance from this point, and the other equal to the square of this

distance. What is the volume generated while the rectangle moves adistance of 2 ft.? Am. 4 cu. ft.

10. On the double ordinates of the ellipse + ^-=1, isosceles tri-a'~ b'

2

angles of vertical angle 90 are described in planes perpendicular to thatof the ellipse. Find the volume of the solid generated by supposing sucha variable triangle moving from one extremity to the other of the majoraxis of the ellipse. Am. $ a/A'.

Calculate the volumes bounded by the following quadric surfaces andthe given planes.

11. z = .r'2 + 4 y* ; z = 1. Ans. \ TT.

12. 4 x'2 + 9 c 2 + ?/=

; # + 1 = 0. & TT.

13. a-2 + 4 y

2 = 1 + * 2; z + 1 = ; z - I = 0. i

TT.

14. 25 ir + 9 z 2 = 1 + .r- ; .r =;

.r = 2. ft TT.


15. .r2-f 4 ?/

2 + 9 z 2 = 1. Ans. $ w.

16. z* = jr~ + 9 f/2; c -f 1 = 0.-i TT.

17. Given the parabola z - 4 -jr

2 in the A'Z-plane and the circle

0-2 4- 1/2 = 4 in the A' V-plane. From each point on the parabola lyingabove the circle two lines are drawn parallel to the r#-plane to meet the

circle. Calculate the volume of the wedge-shaped solid thus formed.

Ans. 6 TT.

18. Find the volume of the solid bounded by the hyperboloid of one" " T ~

11^

sheet ^ = -f- -f- 1 and the planes jr = 0, j- = a. AUK. $ Trafor.

r* a- b 2 tj

19. A solid is bounded by one nappe of the hyperboloid of two sheets

^ - ~ - ~ - 1 and the plane jr = 2 a. Find the volume. Ans. % irabr.

20. Find the volume of the solid bounded by the surfac

'~^ 4- TT; -f -7, 1 . Ann. > irahc.a* 0- r"

ADDITIONAL PROBLEMS

1. Find the area of the loop of the curve

y*-

(jr -f 4)(.rL' -

jr -f 2 y -4). AUK. vV'-

2. A point moves along a parabola in such a way that the radius

joining it to the focus generates area at a constant rate. If the point movesfrom the vertex to one end of the latus rectum in 1 sec., what will be its

position at the end of the next 8 sec. ?

Ans. Distance from focus latus rectum.

3. Find the perimeter of the figure bounded by the lino y =_l and the

curve 4 y = e'2r

-f c~'2r . Ans. V.3 -f In (2 -f V.'J ) 3.05.

4. The arc OP of the curve xy = x - y joins the origin to the point

P(jr\, 7/1 ), and bounds with the x-axis and the line .r = jr\ an area A. Thesame arc bounds with the ?/-axis and the line y = ?/, an area />'. Prove that

the volumes obtained by revolving A about the .r-axis and tt about the

?/-axis are equal.

5. The area bounded by the curve 16 y'2

(x -f 4):i and its tangent

at the point (12, 16) is revolved about the r-axis. Find the volumeinclosed. Am. 1J)#4 *.

6. The base of a solid is the area bounded by the parabola y2 = 2 px

and its latus rectum. Every section of the solid made by a plane at right

angles to the latus rectum is a rectangle whose altitude is equal to the

distance of the section from the axis of the parabola. Find the volume of

the solid. Ans. j p3.

7. Given the ellipse 9 s'2 + 25 y2 = 225. A solid is formed about this

curve in such a way that all plane sections perpendicular to the :r-axis are

ellipses whose foci are on the given ellipse. The major and minor axes of

each section are proportional to those of the given ellipse. Find thevolume of the solid. Ans. *-&

TT.

8. Let (r, y) be a point on the curve of Art. 159, O being the originand <)A the x-axis. Show that (D) may be written

Area = \j (x dy - y ds),

by using the transformation (5), p. 4. The limits are determined by thecoordinates of the extremities of the curve.

9. Derive the formula of the preceding problem directly from a fig-

ure, making use of (B) and (C), Art. 158.

The formula (1) of Problem 8 is useful for parametric equations. Findthe following areas by (1).

10. The area between the involute of a circle

.r r cos + rO sin 6, y r sin rO cos 6

and the .r-axis produced to the left, in the figure of Chapter XXVI,p. 537.

11. The entire area of the hypocycloidof three cusps

j

j' = 2 r cos -f r cos 2 6,

1?/= 2 r sin - r sin 2 6.

(See figure.) Ans. 2 ?rr2

.

12. A straight uniform wire attracts a

particle P according to the law of gravita-tion. The particle is in the line of the wire

but not in the wire. Prove that the wire

attracts the particle as if the mass of the

wire were concentrated at a point of the

wire whoso distance from P is the meanproportional of the distances from P to the ends of the wire.

13. Find the area of the loop of the folium of Descartes,r

HINT. Let y = tx; then .r-

:

1 +/<

* at\, and dx = \

~ 2t;\ 3 a dt.

(1 -f

The limits for / are and

CHAPTER XVI

FORMAL INTEGRATION BY VARIOUS DEVICES

166. Introduction. Formal integration depends ultimately uponthe use of a table of integrals. If, in a given case, no formula is

found in the table resembling the given integral, it is often possible

to transform the latter so as to make it depend upon formulas in the

tables. The devices which may be used are

(a) integration by parts (Art. 136),

(b) application of the theory of rational fractions,

(c) use of a suitable substitution.

We proceed to discuss (b) and (c).

167. Integration of rational fractions. A rational fraction is a frac-

tion the numerator and denominator of which are integral rational

functions, that is, the variable is not affected with negative or frac-

tional exponents. If the degree of the numerator is equal to or greater

than that of the denominator, the fraction may be reduced to a

mixed quantity by dividing the numerator by the denominator. For

example, 40* r* + e* J '

== j2 _|_ x __ 3 _|_

T + <*.

The last term is a fraction reduced to its lowest terms, havingthe degree of the numerator less than that of the denominator. It

readily appears that the other terms are at once integrable, and hence

we need consider only the fraction.

hi order to integrate a differential expression involving such a

fraction, it is often necessary to resolve it into simpler partial frac-

tions, that is, to replace it by the algebraic sum of fractions of forms

such that we can complete the integration. That this is always pos-

sible when the denominator can be broken up into its real primefactors is shown in algebra.*

Case I. When the factors of the denominators are all of the first

degree and none are repeated.

*See Chapter XX in Hawkes's "Advanced Algebra" (Ginn and Company, Bonton).

289


To each nonrepeated linear factor, such as x a, there corre-

sponds a partial fraction of the form

Ax a

where A is a constant. The given fraction can be expressed as a

sum of fractions of this form. The examples show the method.

ILLUSTRATIVE KXAMPLK. Find / -vX

..

' HfJ x ' 4 x 2 - 2 x

Solution. The factors of the denominator being x, x 1, x 4 2, we assume*

vx(x - 1)U 4 2) x x - 1 x42

where A, K, r are constants to be determined.

Clearing ( 1 ) of fractions, we get

(2) 2x 4 3 = A(JC - l,(x 4 2 i f- B(JC 4 2)x 4 C(x - l)x,

2 x -I 3 = 1/1 4 n 4 Ox- 4 (A 4 2 7? - Dor - 2 A.

Since this equation is an identity, we equate the coefficients of the like powersof x in the two members according to the method of Undetermined Coefficients,

and obtain three simultaneous equations

r A 4 7? + T = 0,

cn -;A 4 2 /? - r -2,

Solving equations (in, we get

Substituting. these values in U),

2x4 a 3 5 I

.r(,r-lHx42) 2x .S(x-l) 6(.r42)

_ i r_^L_l 6./X42",/ x(x- DU42) 2. x ;K

= - In x 4 rl In (.r- I) - J In (j 4 2) 4 In c

, r(x-l)>;

t= In -~-~ A HS.

.r'^Cr 42)"

A shorter method of finding the values of A, B, and f from (2) is the following:

Let factor x ~ ; then 3 = - 2 A, or A - -.

Let factor x --I = 0, or x = I ; then 5 = 3 #, or B = .

Let factor x 4 2 = 0, or x = - 2 ; then -I- 6 C, or C = -

/;.

In every example in rational fractions the number of constants to

be determined is equal to the degree of the denominator.

* In the process of decomposing the fractional part of the given differential neither

the integral sign nor dx enters,

FORMAL INTEGRATION BY VARIOUS DEVICES 291

Case II. When the factors of the denominator are all of the first

degree and some are repeated.

To every w-fold linear factor, such as (x a)n

t there will corre-

spond the sum of n partial fractions,

(x-a)" (x-a)"" 1 x-a

in which A, B, ,L are constants. These partial fractions are

readily integrated. For example,

C X 1* + 1

ILLUSTRATIVE EXAMPLE. FindJ _ 4

ctr.

Solution. Since x 1 occurs three times as a factor, we assume

4.cr-1) 2 ar-1

Clearing of fractions,

x< + 1 = A(x- 1);< + B;r + Or(j-- 1) + />j(j- I)

2.

x' + 1 = (A +/.*' + (-3^4 4-r-2/x 2 + (3.4 -f /? - T + /jr - A.

Equating the coefficients of like powers of j, we get the simultaneous equations

A + />= 1,

- 3 v4 4- C -L* /> - 0,

3 A -f B - T + D -- 0,

- A = 1.

Solving, A = -1, B = 2, C = 1, /> = 2, and

y(jr 1)" x (x

',. ln LH^L

a

+Cr-l)-'

PROBLEMS


JT* X2.

Q / (4 x -r o;ax _ J- i ^ i r*'

r3 -f 8 x 2-f 3 x

~~

2 (2 x + 1 N /0 " ' v

r(lr* + 2.r 2

J 4 rr1 -

l)n

(2 x + l)(2or- I)2

{ c2 r 2

.

-J ?/

J + 2 ?/< 2 ?/

4 3"

f -t i v o cJ] x* -f x^ 3 <:

10.

12 /" (a^ + 7 J)i._ = ln4 = 2877-

Ju (.' + l)(j- + 2)U + :i) 3

^ a -3)^ = in 1 _ = _ o.4K!9.

JLCllk = 5 in n _ 4 = 1.4930.


n5.r~ + 14 j- -h 10)dr

i f(5j' 2 " 9)(^ 2^lb

'J ^-9.r"

17 f<] - + 7v/ ~___ 24

'

10

n fy'./ (2(2 j- + 8)(4 .r^-

1 - 1)

/4 ~ 3 y*)dy

20.

01 r(^2 - J - 5)(ir OR f (2 /

4-f 3 /

- 20 /- 28)d/

J^x. I' "" * <wO. I

*

Case III. W r/w7i //^' denominator contains factors of the second degree

but none are repeated.

To every nonrepeated quadratic factor, such as x2 + px -f q t there

corresponds a partial fraction of the form

Ax + Bx2 + px + q

The method of integration of this term is explained on page 209

[Illustrative Example 2).

If p is not zero, we complete the square in the denominator,

Let x + \ p = u. Then x = u \ p, dx = du. Substituting these

values, the new integral in terms of the variable u is readily integrated.

ILLUSTRATIVE EXAMPLE 1. Find|/ X -f~ 4 X

4 A.Ex -f C

Solution. AssumeJ)=

*+ ~xMT

"

Clearing of fractions, 4 = ,4(x2

4- 4) + x(Bx 4 C1

)= M 4 ft)x* 4- Cx + 4 A.

Equating the coefficients of like powers of x, we get

A + B - 0, C = 0, 4 A = 4.

= 0,so that

/- 4 rfar _ rdx _ r xdx

"J x(x2

-f 4) ~J x J x 2-f 4

= In r - - In Or1

-' 4 4) 4- In c = InC*

Ans.2 Vx2 4 4


r_&E--in^^ 2)2

4-^-V^arctan^J x a + 8

~24 x2 - 2 x -f 4

^12 V3

Solution. Factoring, x <

-I- 8 = (x + 2) (x 2 - 2 x -f 4).

1 Ax -f B Cx :< + 8

~~x 2 - 2 x + 4 x 4- 2

'

1 = (Ax -f B)(x 4- 2) -f C(x 2 -2x4-4),

1 = (A 4- Ox 2 + (2 A + B - 2 C)x -f 2 B -f 4 C.

Then A = -^ B = J, C = ^7-

Now x 2 - 2 x + 4 = (x-

I)2

-f 3 = w24- 3, if x - 1 = w.

Then x = u + l,dx = du, and

Substituting back u = x - 1, using (4), and reducing, we have the answer.

Case IV. When the denominator contains factors of the second degree

some of which are repeated.

To every n-fold quadratic factor, such as (x2 + px + q)

n, there

will correspond the sum of n partial fractions,

A* + BI

Cx + DI . .

Lx+M _

, \n' (~2 J_ /n^ _L n \

n ~ 1 ^2(x

2 + px + q)n

(x'2 + px + q)

n ~ l x2 + px + q

To carry out the integration, the "reduction formula"

C du 1 I u(0)

J (u'2 + a 2r

proved in the next chapter, is necessary. If n > 2, repeated applica-

tions of (5) are necessary. If p is not zero, we complete the square,

jc2 + px + q = (x + p)

2 + \ (4 q- p

2)= u2 + a 2

, etc., as before.

ILLUSTRATIVE FA AMPLE. Prove

Solution. Since x 2 + 1 occurs twice as a factor, we assume

2 x-* + x 4- X __ Ax 4- 7*,O 4- />

(x- -f \)'z

~(^ + I)

2 x 2 + 1

*

Clearing of fractions,

L J :< 4x-f3 = >lxfB4- (Car + D)(x 2-f 1).

Kcjuuting the coefficients of like powers of x arid solving, we get

,4^-1, 7? = 3, r = L\ 7^ = 0.

The first of these two integrals is worked out by the power formula (4), the

second by (5) above, with u - x, a - 1, n = 2. Thus we obtain

Reducing, we have the answer.

Conclusion. Since a rational function may always be reduced to

the quotient of two integral rational functions, that is, to a rational

fraction, it follows from the above discussion that any rational

function whose denominator can be broken up into real quadratic

and linear factors may be expressed as the algebraic sum of integral

rational functions and partial fractions. The terms of this sum have

forms all of which we have shown how to integrate. Hence the


Theorem. The integral of every rational function whose denominator

can be broken up into real quadratic and linear factors may be found,

and is expressible in terms of algebraic, logarithmic, and inverse trigono-

metric functions, that is, in terms of the elementary functions.

PROBLEMS


J ^ + 3.r"'" N

" "'

L f (*"* + ^^ = In (j-- 1) -h arc tan j- + C.

2 f2 - 8 /

-8)rf/ 4>

, /2-f 4

,

_ .., . .^, - in -+ ( .

4r(^ +

^-iQ)^ = iln fii + arctan

J (2x- 3)(jT-H-4) 12 2J--3 12

5- = In + arc tan -f C.

4^-f9o- ^ L 6 :J

(2?/:

'? + ^+t2)% - ln v + >2) + arc lan

7 r__^_ - _ 1 __ arc tan z + C.J z4 -f z'

2 z

-h ^ tl

I)2

_

15 . fA^r = In 2-^|- 2 arc tan x + C.

J or1 - 1 x + 1

=2 , (z + 2 )_ arc tan (z + 1) + C.

* = arc tan (t + 2>~

rr-'

1 + 4 jr

19. f'

14^ = ln I +* = 0.667.

Jo U -f 2) Or 2-f 1) 9 4

u *> T^ ^ i j ;n.< i ^ j/i ___ rt $ 7 1

r

o U" -f 1)(^2-h 1 j 4

22.

23.


A f(6 -r2 + ?** + 4 )dr 9q f(4 Ja -h3j 2

4 -

J ^T^ 29 -

J ^I)

I 1 : t . 30. I' i

J (z + l)(z2-h 1) ^o (2 y -h 1)(4 i/

2-f

r(3 r'J

4- 3 a- + l)(Lr 01 T 1 (2 jr:t

4)dj'I ; o ^ 31-

I / v, IN/ r~r^*J .r

4 + 3 J-2

./o (x2

H- l)(j -f I)2

f(3^ -f jc'

2-h 3)<ijr 32

r3(x -f 10)dj

J j*4 -f3^

'

*Ji x3 -h2r 2-h 5x"

f(5 ^ 2-f 12 .r + 9)dr QQ fa

(2 x3-f 18

.I '"

^

5O. I

168. Integration by substitution of a new variable;

rationalization.

In the last article it was shown that all rational functions whose

denominators can be broken up into real quadratic and linear factors

may be integrated. Of algebraic functions which are not rational, that

is, such as contain radicals, only a small number, relatively speaking,

can be integrated in terms of elementary functions. By substituting

a new variable, however, these functions can in some cases be trans-

formed into equivalent functions that are either in the list of standard

forms (Art. 128) or else rational. The method of integrating a func-

tion that is not rational by substituting for the old variable such

a function of a new variable that the result is a rational function is

sometimes called integration by rationalization. This is a very im-

portant artifice in integration, and we shall now take up some of the

more important cases coming under this head.

Differentials containing fractional powers of x only. Suck an expres-

sion can be transformed into a rational form by m.eans of the substitution

x = zn

,

where n is the least common denominator of the fractional exponents ofx.

For x, dx, and each radical can then be expressed rationally in

terms of z.

ILLUSTRATIVE EXAMPLE 1 . Prove fJ *** =

| x* -1 In U + x% ) + C.

J1 + .r*

6 6

Solution. Here n - 4. Hence let x - ?.

Then x* = z2,

x* = z-<, da- = 4 z'A dz.

=!/(''- rfp)"=l''- 1

'-in-rt + r.

Substituting back 2 = x*", we have the answer.

The general form of the irrational expression here treated is then

where R denotes a rational function of x'1

.

Differentials containing fractional powers of a + bx only. Such an

expression can be transformed into a rational form by mea,m of the

substitution a + bx = z",

where n is the least common denominator of the fractional exponents of

the expression a + bx.

For x, dx, and each radical can then be expressed rationally in

terms of z.

fdx-_--.

-(1 -fx)'

2-f (1 +x) 2

Solution. Assume 1 4- x z2 .

Then dx = 2 z dz, (1 -f x)* = z\ and (1 -f xY1 = z.

2zdz _ 2

= 2 arc tan z + C = 2 arc tan (1 -I- x)2

-f C

when we substitute back the value of z in terms of x.

The general integral treated here has then the form

R[x, (a + bx)v]dx,

where R denotes a rational function.

PROBLEMS


2 sl

r <*

J

V7 - 1 V3

r3.

/ ijJ x x+ C.

1 -x

=I?

x " "B * 2 + c-

- ^ 6 r^ + 6 x + 1c

)

;;

12(4 j+1)'

= ~f- 4- 2 In ~-^l -f 4 arc tan x 8 + C.-v. f v. M ' *

do* 2(2 a -f

'/^(a -f ^'a -f />.r

8. /v'aH- // (///=

t

-~(4 ?/

- 3 )(a -f //)" -h C.

9. __/.( + 1

-1

*frrjk10. / ^ = ~ u -f )'! -

3(x -f )

: '

-f 3 In (1 + Vx 4- a) -f C.

~ 2V/ + 2 + \/2 arc tan

= = 2 arc tan 2 -

1 -f V .r -f a

(/ + 5k//

(/ +

12. f !

./() ( r 4-

13./-

4_^r 4 -21n;j.Jo l _j_ V.r

14 r4 y w = L^ V5.

''i V2 -f 4 //2 16 *

r ! ^^7r^4*

Jo a- -f- 1 2 3'

15.d/

3 9 arc tan1 r^r

l7 - 1 2V? + V/

- = 5.31.

(.r- 2) s +3



19. I -^~ 23. f - .

20-/^fe' 4.

(.r + 5)Vr + 4

22./

'-"t

. Mj(^-^rr+Tt:

27. Find the area bounded by the curve ?/ jr -f V.r -f 1, the .r-axis,

and the ordinates JT ~ 3 and .r = 8. AN*. 40 /;.

28. Find the volume generated by revolving about the .r-axis the area

of the preceding problem.

29. Find the volume generated by revolving about the .r-uxis the area

in the first quadrant bounded by the coordinate axes and each of the

following curves.

(a) y 2 Vx. (c) // a VJIr.

(b) y-2- \/jr. (d) ?/= 4 - A

30. Find the area bounded by the curves //- 2 .r -f V'J j -f 1 and

y JT v2 j* -f 1 and the ordinates .r 4 and j% ~ 12.

31. Find the area bounded by the curve

and the ordinates x 3 and x = 8. V5 4- In

4 - V2

169. Binomial differentials. A differential of the form

(1) xm (a + bx nr dx,

where a and b are any constants and the exponents m, n, p are rational

numbers, is called a binomial differential.

Let x = za;

then dx aza ~'

rfc,

and xm (a + bx n)p dx = azm " ' ff

-]

fa + bz"")f> dz.

If an integer a be chosen such that ma and w* are also integers,*

we see that the given differential is equivalent to another of the same

form where m and n have been replaced by integers. Also,

(2) x(a + bx n)p dx = xmJf ni

'(ax~n + b)

l) dx

* It is always possible to choose a so that ma and not are int^ors, for we can take a an

the L C.M. of the denominators of m and n.


transforms the given differential into another of the same form where

the exponent n of x has been replaced by n. Therefore, no matter

what the algebraic sign of n may be, in one of the two differentials

the exponent of x inside the parentheses will surely be positive.

When p is a positive integer the binomial may be expanded and

the differential integrated termwise. In what follows p is regarded

as a fraction ; hence we replace it by -> where r and s are integers.*s

We may then make the following statement.

Every binomial differential may be reduced to the formr

xm(a+ bxn)*dx,

where m, n, r, s are integers and n is positive.

In the next section we prove that (1) can be rationalized under the

following conditions.

CASE I. Whenm = an integer or zero, by assumingn

a + bxn == z*.

CASE II. When ---h - = an inteqer or zero, lyy assumingn s

a + bxn = z*xn .

ILLUSTRATIVE EXAMPLE 1. f -- = fx*(aJ

(a + bx'2 )lJ

_ 1 2 a 4- bx'2 c~~b *

\ a + fa-2

Solution. m = 3,n = 2,r = 3, 8 = 2; and herem "*"

2, an integer. Hencethis comes under Case I, and we assume n

a + bx 2 = z2; whence x = (-f- ' dx =

1

-. and (a -f

b

r x*dx _ rlz* - a\ zdz'

J(a -f bx*)'*

""' I &

=p f (1

~ az~ = (z

_ 1 2 a -f bx2

c~V \ a + bx*

ILLUSTRATIVE EXAMPLE 2. f f

x * x ~~

\-

Solution, m = -4, n = 2, ^

= -\ ;

and herem "^ + - = -

2, an integer.82 n s

* The case where p is an integer is not excluded, but appears as a special case, namely,

r = p, = 1.

Hence this comes under Case II, and we assume

(\

whence x 2 =

also x = -i

; and

u~ - D*

1

-

PROBLEMS


._2(3r<-2)(1. /VVl + x*dx=J

r r/o ,-

3.Jj-5(8 4- xj

(a _|_ f^)* 3 f>-va -

rfjr (1 4-

4 C.

4 r.

fi r djr - - (1 + 'r3)^

4- r6 -

1 r7~~

> j-J J :H1 4- x3)7 " ^

J -r2/1 4. r4 ^4 C.

rfr

(1 4-r")

1 4- 3 J3

(^-1^( ;*_ })

\

(2 -r'J - I'd 4 .r

2)

i__L_i__ 4 r.

loTT_, ,,h C.

-4 C.

4- C.



11.jV'Vl- x3 dx. IP f

xdx_ m 13

170. Conditions of rationalization of the binomial differentialr

(A) xm(a + bxn);dx.

CASE I. Assume a + bs 71 = z*.

1 r

Then (a + bx j=

2, and (a + fxr" )

; = sf ;

1

, /z* a\" Am /z* a\ Tl

also =(^

) and rm ~(-

);

j^j

hence dx = 7^- 2*~

J

(.

) dz.on \ o

'

Substituting in 04), we getm + 1

1f

*

hn~

V fe'

The second member of this expression is rational when

m + 1

nis an integer or zero.

CASE II. Assume a + fe.r" zxn.

Then xn = ^ r and a + bx n zx" ~ ^-r

-

^ fe s- b

r r _ r

Hence (a + fexnV = a 8

(^ fe) *zr

;

also x a"(z* 6) ", xm = a rl

(zs

fe)n

;

A J 8 ^ U M-- 1^and dx a^*" 1

(z8

fe)n dz.

n

Substituting in (A), we get

xm (a + bxn)*dx = - -a~ +

*(z*- fe)'^"^"

4"^ 1

)^^*-n


The second member of this expression is rational when 2?Lf--us

is an integer or zero.

Hence the binomial differentialr

xm (a + bx n)*dx

can be rationalized in the cases given in the preceding article.

171. Transformation of trigonometric differentials.

Theorem. A trigonometric differential involving sin u and cos u

rationally only can be transformed />// means of the substitution

(1) tan|=

z,

or, what is the same thing, by the substitutions

2 z 1 z2 2 dz(2) sin u = -> cos u ~> du ~ ~~-Z-

~ ~ ~

into another differential expression which is rational hi z.

Proof. From the formula for the tangent of half an angle in (5),

Art. 2, after squaring both members, we have

, o 1 1 cos ?/

tan- -u ^T2 1 + cos ?/

Substituting tan \ u = z, and solving for cos u,

-i ^2

(3) cos u- *.2

> i-z*

one of the formulas (2). The right triangle in the figure shows the

relation (3) and gives also sin u as in (2). Finally, from (1),

u 2 arc tan z,

2 dzand hence du .. :

2-

Thus the relations (2) are proved.

It is evident that if a trigonometric differential involves tan u,

ctn u, sec u, esc u rationally only, it will be included in the above

theorem, since these four functions can be expressed rationally in

terms of sin u, or cos ?/, or both. It follows, therefore, that any rational

trigonometric differential can be integrated, provided the transformed

differential in terms of z can be separated into partial fractions (see

Art. 167).

ILLUSTRATIVE EXAMPLE. Prove f-- - = - arc tan /5tanx + 4 \ + C.

J 5 -f 4 sin 2 x 3 \ 3 /

Solution. Let 2 x = M. Then .r = ^ ?<, djr = i c/?^ Substituting these values, andthen using (2), we have

/'_jfr_ = I f f/" = i/544sin^j- 2 ./

-f 4 sin M ~2 8 ; ~. 5 2- + 82 + 5

Substituting back z = tan i // = tan j- gives the above result.

PROBLEMS

Work out th< 4 following integrals.

,/ df) / n

l - = ln 1 + tan -

. __... 4 4- 5 cosr 3

tanf-3/

5. /^-7'->- - -

-: arctan (- tan ~) -f C.

*> 3 -f- cos <v v '3V Vu ~ '

6. r_--^- - arc tan(

1 4 t> tan -') -f- C.' J sin .r cos j* ~f 3 \ 2'

r cos ^ <1Q 0.5 . / . 0\ ,7. 7-----

7,- - -

4- ^ arc tan 2 tan --f C.

; o :? cos ^ 3 6 ^

2/

c.

'.(.4 -3 0080 V?' 11. f2 ^

:

.'o 2 + sin x

10 _ < - In - 191U'

~ n 12 '_

o 1'J 4- 13 cos~

r 2 ',o 3 -f 5 sin a-

4

Work out each of the following integrals

n.f ^ 16 r j

J 1 -f sm x cos x .'13 cos /- 5

14 C d t* r dx'J cot 4- esc 6 \l 2 cos x -f 1*

15. f ^ 18 /'_jrf<L_.J 13 - 5 cos <

*

J i> -f sin a

19. r f. . 21. /v-^ 23. r" ~J I -f 2 sm x J 5 sec f - 4 .' 3 -f 2 cos

0. fsin ^^

og r~ * djr

24 T 2rf<y

*/ 5 -f 4 sin Vo T> -f 3 cos x "'o 2 -f cos tv

172. Miscellaneous substitutions. So far the substitutions con-

sidered have rationalized the given differential expression. In a

great number of cases, however, integrations may be effected bymeans of substitutions which do not rationalize the given differential,

but no general rule can be given, and the experience gained in work-

ing out a large number of problems must be our guide.

A very useful substitution is

1 dzx ~~

z z'2

'

called the reciprocal substitution. Let us use this substitution in the

next example.

ILLUSTRATIVE EXAMPLE. Find f ,

dx.

Solution. Making the substitution x = -. dx = - ~-

, we get

/vQ a

I

(a 2 -^-j

, .,-_3 fl2

PROBLEMS


I./d

^= In (__^===). Let x = -

2- x -f 2 V2 \Vx 2 - T -f 2 -f x -f

Let j2- r-f 2 = 2 - x.

3 frfj = 2 arc tan (x + Vx^T 2 x - 1 )+ C.

' xVx 2 + 2 * - 1 T*t x/^TTTTLet Vx^ -f2x-l=z-x.

4 C dx - * /V2 4- 2 x - V2 - x\

"J xV2 4- x - x2~~

V2 W2 4- 2 x 4- V2 - x/C.

Let V2 4- x - x 2 = (x -f 1)2.

/

x- 6-

dx

Let V5 x - 6 - x2 = (x-

2)z.

xV3 x 2 - 2 x - 1

-dx'

^ xVl 4-4x4-5x 2

.r .

d-J xVx 2

4- 4 x - 4

8.fy

dx=-.

J x 2Vl 4- 2 x 4- 3 x2

= - arc sin 14- C.

-f 2 x 4- Vl -f 4 x -f 5 x2

C.rV2/

.-_ Vl -f 2 x -f 3 x 2

InVl -f 2 x -f 3 x 2

Let x =

Let x = -

Let x = -2

-f C.

Let x = -z

10.r^J -r2x/<>7 -r^

V27 x 2-f 6 x - 1 r

3 arc sin (

f

^i -f C.\ 6 x /

11. f1

<*-?**" =6.J X^

Let x = -z

12. f- = arc tan e./o rr -f r~ r 4

18. f/0

14.

ax x 2

/2 d/ = V3-^ In (2

Let x = a sin 2 ir.

Let / 4- 1 = z.


15>xVx 2 -2x4-3

4 x-dx"J

(x2 - 2 x 4- 3)*

2dx

V5 x - 6 - x2

2xdx

V5 x - 6 - x2

Let Vx 2 -2x4-3 z - x.

Let Vx 2 -2x4-3 = z - x.

Let V5 x - 6 - x 2 = (x-

2)2.

Let V5x - 6 - x2 =r (x- 2)2

CHAPTER XVII

REDUCTION FORMULAS. USE OF TABLE OF INTEGRALS

173. Introduction. In this chapter formal integration is completed.The aim is eventually to lay down directions for using a table of in-

tegrals. Methods of deriving certain general formulas, called reduc-

tion formula's, given in all tables are developed, since these methodsare typical in problems of this sort.

174. Reduction formulas for binomial differentials. When the bi-

nomial differential cannot be integral ed readily by any of the methods

shown so far, it is customary to employ reduction formulas deduced

by the method of integration by parts. By means of these reduction

formulas the given differential is expressed as the sum of two terms,

one of them not affected by the sign of integration, and the other

an integral of the same form as the original expression, but one

which is easier to integrate. The following are the four principal

reduction formulas.

'/ym n

xm (a + bxn)

f) dx =(np+m-i l)b

(m-n +

(np + m + l)b,

I xJ

n(a-f bxn

)t>dx.

np + m+

/x

+- - CXm (Q + fort

lj

b r,

B( fr

J

/xm+ i /a 4. bx")''

4 1

*~ (a + wy dx = - x

n [p

++1)

>

l rxm (fl + bxnyJn(/> -f l)a

307


While it is not desirable for the student to memorize these for-

'mulas, he should know what each one will do and when each one fails.

Thus :

Formula (A) diminisfos m by n. (A) fails when np + m -f 1 = 0.

Formula (B) diminishes pbyl. (B) fails when np + ra + 1 = 0.

Formula (C) increases w by n. (C) fails when w + 1 = 0.

Formula (D) increases p by 1 . (D) fails when p + 1 = 0.

I. To derive formula (A). The formula for integration by parts is

(1) (u dv - uv - fr du. (A), Art, 136

We may apply this formula in the integration of

rr

"(a + bs n)

J> dx

by placing u = zm~ n + ' * and dv = (a + ftr

n;

7>rn~

J da*;

then du - (m n + 1 )/m ~ ndx and v = '

, \?/.&(/> + l

y

Substituting in (1),/,rm n i \ ( n _[_ /, rn \ /' -f 1

~m(n j JLJ (tt -f- 0.

CJ

But Cxm' n(a + bxn)^ [ dx= r.rm

-n

= a I xm" n

(a + bxn)*dx

Substituting this in (2), we getm~ n f |

/ o

nb(p

Uar m-) J

Transposing the last term to the first member, combining, and

solving for I xm (a + bjrn)"dx, we obtain (A).

* In order to intoRrate r/r by the power formula it is necessary that x outside the

parenthesis shall have the exponent w - 1. Subtracting n 1 from m leaves m n + 1 for

the exponent of j in u,

REDUCTION FORMULAS 309

It is seen by formula (A) that the integration of rm (a + bxn )pdx is

made to depend upon the integration of another differential of the

same form in which m is replaced by m n. By repeated applicationsof formula (-4), m may be diminished by any multiple of n.

When np + m + 1 = 0, formula (A) evidently fails (the denomi-

nator vanishing). But in that case

hence we can apply the method of Art. 169, and the formula is not

needed.

II. To derive formula (B). Separating the factors, we may write

(3) Cxm (a + bxn )" dx =

jxm(a + bxn )

"~ '

(a + bx) dx

= a I xm (a + bx")p~ l dx

+ bjxm * n

(a + bxn)p~ l dx.

Now let us apply formula (^4) to the last term of (3) by substi-

tuting in the formula m + n for rn, and p I for p. This gives

/x^ np

Substituting this in (3), and combining like terms, we get (B).

Each application of formula (B) diminishes p by unity. Formula

(B) fails for the same case as (4).

III. To derive formula (C). Solving formula (A) for

and substituting m + n for m, we get (C).

Therefore each time we apply (C), m is replaced by m + n. Whenm + 1 0, formula (C) fails, but then the differential expression can

be rationalized by the method of Art. 169, and the formula is not

needed.

IV. To derive formula (D). Solving formula (B) for

(a+bx n)J'- l

dx,

and substituting p + 1 for p, we get (D).

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 = a2,6 = 1.

ILLUSTRATIVE EXAMPLE 1. f ?===== = - - CH -f 2)(1 -z*)* + CJ V 1 z* 3

Solution. Here ra = 3, n = 2, />=

J, a = 1, 6 = i.

We apply reduction formula (4) in this case because the integration of the dif-

ferential would then depend on the integration of Cx(\ - x'2 )~ *dx, which comes

under the power formula. Hence, substituting in (A), we obtain

(

2. f ^^ - - /I T ; + ij a^\ Va 2 -;

(aa -a- a 4 ^ 4 8 /

ILLUSTRATIVE EXAMPLE;

(aa -a- a

)4

+ -a 4 arc sin- -f C.cS a

HINT. Apply (A) twice.

ILLUSTRATIVE EXAMPLE 3. C (a2-f a-dz = \ a 1'

-f x*

-f ^ In (x -f V^THim1

. Here w = 0, n = 2, />=

.J,a = -', b = 1. Apply (B) once.

ILLUSTRATIVE EXAMPLE 4. fa-'

1

HINT. Apply (C) once.

PROBLEMS


/^3,] T 1

J_/ =

(a-2 - 2 a 2

)Va 2-f x 2

-f C.Va 2 + .r

2 *

/r5 r/r 1

t- - = -7r(3x4 + 4r 2 + 8)VT^ + r.V 1 .r

2 l^

. f^ 2Va 2 - x* df = ^ (2 .r2 - a 2

)Va 2 -.r

2-f ^ ar<* sin - + C.J o 8 a

*/ / o

,

J.no

=^ .,/ T . ^ 4- TT^T arc tan -

-f C.J(a- + Qc

2Y 2 a-(a2-f x 2

) 2 a-{ a

g - Va 2

j3Va 2 - x 2 2 a 2x 2 2 a ;<

x3 dx x 2-f 2 a 2

-f C.

(a2-f x 2

)

(a1

"

Q C dx _ x(3 a 2 - 2 x 2)

O-/ 5 3

"*- '

J(a

2 -x 2)* 3a4

(a2 -x 2

)

2

9. f(x2 4 a 2

)^ dx = J x(2 x 2 4 5 a 2)Vx- 4 a 2 4 3 a'1 In (x 4 Vx 2 4a 2

) 4 C.

10. fx 2Vx 2 4-a 2 dx = J x(2 x 2 4 a 2)Vx 2 4a 2 - a' In (x 4 Vx 2 4 a 2) 4- C.

C x 2 dxV ^

11(x 4- 3 a)V2 ax x 2

. 3 a 2 /- x\ . ,,1

arc cos 1 4- ( .

2 2 V a)

i twice.

V2 ax - x 2 2

HINT, f .

yl> rf>C=r fx

?(2 a - .r) idr. Apply

J V2 ax - x 2'

l2.f^M= = - r 0/L> + 5 ?/ + 30)V4//- ^ 4 20 arc cos

(l- 1A -f C.

r ds K 3_^ 3 - 4- r'^ (a

2-f s 2

):* 4 a 2

(a24- ^'

2)

2 8 a l (~ 4- s'") 8 a- 1 a

14

-/vfirp= ~

5-wV9 - 4 ?/a + S arp sin + c -

'J Vl 4- 4 /24

16. f?/2V4 - 9 //- d?/ = ?/(9 ?/

2 - 2)V4 - 9 //- 4 ~; arc sin ^ 4 C.J

' 3o - / ^

17

'^(1+ 9/ 2)

1 81Vl 49/ 2

18.2Vl 4- 4 /

2r// = T/2 /(I -h 8 /

2)Vl +4/ 2 - ^ In (2 / + Vl-f4/ 2) + C.


/x2 dx o r XK dx

7~^ ^TiT***

/ /A r'(a

2 x 2)2 J V4 xh

dx

25..s7 ds

20. fJx 2

(l +x 2)

2

28.

23. rVa'

2 + yig-

24fa-*'V x

29.

26

27.

l

d|/.

'/;

(a-f ?>,s

dx


175. Reduction formulas for trigonometric differentials. The methodof the last article, which makes the given integral depend on another

integral of the same form, is called successive reduction.

We shall now apply the same method to trigonometric differen-

tials by deriving and illustrating the use of the following trigonometricreduction formulas :

(E) I sinmx cosn x dx =sin"

m -f n

n - 1 C .

/ si

m-h njsinm xcosn ~ 2

;tcte.

f . _,sinm ~

(F) I sinm x cosnxdx =m -f ni /*

-f I sinm 2 xcosn xdx.m-j- n J

, _. . . , sinm f l x cosn j * x(G) I sinm jtcosnxdx=

/fi+ 1

sinm f * x cosn f 1 x

m-f 1

m+ 1

Here the student should note that

Formula (E) diminishes n by 2. (E) fails when m + n 0.

Formula (F) diminishes m by 2. (F) /a?"fe wtoi m + r^ = 0.

Formula (G) increases n by 2. (G) /a?k when n + 1 = 0.

Formula (H) increases m by 2. (H) fails when m + 1 = 0.

To derive these we apply, as before, the formula for integration

by parts, namely,

(1 ) f?/ dv = uv - Cv du. (A), Art. 136

Let u = cos"" 1

x, and dv sin"7 x cos x dx ;

,. , / 1 x > ; j sin7"^^

then dw = (n l)cosri ~-x sin jdx, and v =

m + 1

Substituting in (1), we get

^ i /*

+ T~T Ismm+2 xcosn ~-

?/i + 1 J


In the same way, if we let

u = sinm~

l

x, and dv = cosn x sin x dx,

we obtain

/ClTl"*1

'/ fO^" "^ ^ Xsin

m x cosn x dx = --~^

+^fSin-,oan+lj

But|sinm + -x cosn ~-x dx I sin"'/ (1 cos2

/) cos" 2 x dx

C CI sin'

J'/cosr'~~xdx I sin"'/cos"xdx.

Substituting this in (2), combining like terms, and solving for

I sinm x cos"xdx, we get ().**

Making a similar substitution in (3), we get (F).

Solving formula (E) for the integral on the right-hand side, and

increasing ?/ by 2, we get (G).

Jn the same way we get (H) from formula (F).

Formulas (E) and (F) fail when m + n = 0, formula (G) when

n + 1 = 0, and formula (H) when m + 1 = 0. But in such cases we

may integrate by methods which have been explained previously.

It is clear that when m and n are integers, the integral

sinm x cos7J xdx

may be made to depend, by using one of the above reduction for-

mulas, upon one of the following integrals :

fdx, fsin x dx, /cos x dx, / sin x cos x dx, I 4^ =j

esc x dx,

' C-d? = fsec x dx, f . ftan x dx, I ctn / dx,

J cos x J J cos x sin x J J

all of which we have learned how to integrate.


r?* cos' r dx = -** !'* + **SS* + (sin * cos x + x) + G

Solution. First applying formula (F), we get

/i sin .r cos r> x 1 r,,ria 4

sm 2 * cos 4 xc/x = +g J

cos

[Here w = 2, w = 4.]

Applying formula () to the integral in the second member of (4), we get

/c\ C 4 j &m x cos3 xi

3 C ., ,

(5) I cos 1 x dx ----1--

I cos- x dx.

[Here m = 0, n - 4.]

Applying formula (E) to the second member of (5) gives

,, C -j j(6) J

cos 2 x dx = sin -r cos x,x--- 4 -

Now substitute the result (6) in (5), and then this result in (4). This gives the

answer as above.


Aan^2jr djI

sef. 2 T tan 2 x - - In (sec 2 j + tan 2 jr) + C.J cos2x4 4

. .. tan j 2 x sin- 2 JP 1 sin" 2 xSolution. -- =-7---- -

cos ^J jc cos* Z x cos 2 x cos 1 ' J j*

Let 2 x = u. Then x \ u, djr \ du, arid

(7) / sin- 2 x cos ' 2 r dx J / sin 2 M cos *

?/ c?//.

Apply (G) to the new integral in (7), with rw = 2, ?? 3, rej)lacirig x by u.

(8) Ann' M cos-' M rfw = - A<< ^S l

+ Tsiri^' u cos ' M c/u.

Apply (F) to the new integral in (8), with m = 2, ?i -- 1.

(9) / sin 2/v cos J u du = sin M + / cos J u du = sin // + In (sec w -f tan ?0.

Substituting from (9) and (8) into (7), reducing, and sotting M = 2 x, we have

the answer.

PROBLEMS


1. I sin4jc cos 2 x djc sin jc cos jc\

- sin'1

jc- s in2 J* 7 -f-

- h C.J

[6 24 16 16

2. ftan-'5

f dr =4tan 2

f -f 3 In cos + r.%/ AJ O t>

/pf n ;{ /9

ctn 4 OdO=-^~ + ctnO + + C.

4. Tsec :{/ r// - \ sec / tan / -f J In (sec / + tan /) 4- C.

5. fcsc3 .r djc I esc x ctn x -f J In (esc .r ctn j-) -f C.

6.J*csc

6 OdO = - SC g

4

ctn ^

('csc-^ -f

j})+|

In (csc (9- ctn 0) + C.

7. Csin- cos- (t> d<t>= i sin cos (2 sin 2

1) -f J -f r.

8. fctn2 2 B

fe = -

7 ctn 2 esc 26-] In (esc 2 - ctn 2 (9) + C.J sin 2 4 4

9 r dj? - _ cosx - 2cos -r+ c

'.' sin4 * 3 sin 3 x 3 sin .r

10. (Wfldfl = [8 cos'fl + 10 cos 24-15] + - + C.

11. f2

sin<J o I o

12. f "cos4 jcdx = ^- 14.Jo o o 128

TT n

13. r'sin2 fld = |f- 15.^2^ = 5 _ 3_7r

Jo 32 Jn sm-x 4 84

Work out each of the following integrals :

16. fsin2 BdO. 18.J cos-v

17. fcsc< c/ff. 19. r-- 21.J 2 J sin jl cos J Jo

ZL ?

22. fasin 3 6 cos :<

c/0. 23. ('"(! + sin 6)4 dO.

Jo Jo

176. Use of a table of integrals. The methods of integration de-

veloped in Chapters XII, XVI, and XVII have been directed to re-

ducing a given integral to one or more of the Standard Elementary

Forms in Art. 128. Various devices have been elaborated to this

end, such as

integration by parts (Art. 136) ;

integration by partial fractions (Art. 167) ;

integration by substitution of a H.CW variable (Arts. 168-172) ;

use oj'reduction formulas (Arts. 174 175).

When, however, a more or less extensive table of integrals is

available, the first step in any problem in formal integration is to

search for a formula in the table by which the problem can be solved

without the use of any of these devices. Such a table is given in

Chapter XXVII. Some examples will now be given.

ILLUSTRATIVE EXAMPLE 1. Prove, by the Table of Integrals,

Solution. Use 14, with a = 2, 6 = 1, and u = x.

This example, without the table, would be worked out as in Case II, Art. 167.

ILLUSTRATIVE EXAMPLE 2. Verify, by the Table of Integrals,

dx__ = J_ , /_i_ v , r9 + 4:rO 18 \9+4:r7"

1"

"

Solution. Use 22, with a = 3, b = 2, and ?/ = /.

This example, without the table, is solved as in Case III, Art. 167.


r dx _ 1 V4 + 3 JT- 2 ,,

I /-- ^ in . ....... r-- -f ' .

JxV4-f3x 2 N/4+3J-+2

Solution. Use 31, with a = 4, 6 = 3, and a x.

This example, without the table, is worked out, by the substitution 4 + 3 x = z 2,

as shown in Art. 1 68.


f./

xdx V3 Jr" + 4x-7 2

V 3 x* -}- 4 j - 7 3 3V3

Solution. Use 113, with a 7, 6 = 4, c 3, and u = x.

Without the table the example would be solved by completing the square asin Illustrative Example 2, p. 206.


cos 2 x dx = g'^2 sin 2 J -f 3 cos 2 x) ^13

^

Solution. Use 154, with a 3, n 2, u x.

Without the table the example would be solved by integration by parts. SeeIllustrative Example 6, Art. 136.

In many problems the given integral cannot be identified withone in the table as easily as in the preceding examples. In such cases

we search for a formula in the table similar to the given integral,and such that the latter can be transformed into the former by a

simple change of variable. This method has been used constantlyin Chapter XII and in all integration problems hitherto.

ILLUSTRATIVE EXAMPLE 6. Verify, by the Table of Integrals,1 O r

;=iln- ==l= + r.zV4 x~ -f 9 3 3 + V4 or

2 + 9

Solution. Formula 47 is similar. Let u = 2 x. Then x - \ u, dx = J du, and,substituting the values in the given integral, we obtain

C <fr = r jdM = r du

J xV4^ 2 +9 J\ wVw 2

-f 9 J u^/u'2 + 9

Hence, applying 47, with a = 3, and substituting back v = 2 x, a = 3, we have

fJ x

Without tables we should proceed as in Illustrative Example 2, Art. 135.

r

J

\ 9 J - 4 . 2 (9 x - 4-Solution. Formula 84 is similar. Let u = 2 x. Then x = t n, dx = | dt*.

Substituting, we obtain

r\ '

J

'9ar-4 ^ rVjj u -" 2(/K

J J M 3 2du.

This is now 84 with a = j

;. Hence, applying 84, and substituting backu = 2 x, we get the required result.

If no formula from the table can be applied as in the precedingtwo cases, there remains the possibility that the use of one or moreof the devices mentioned at the beginning of this article will lead

to new integrals solvable by the table. No general directions can be

given other than the rules already developed in the text for the

employment of these devices.

The student should study the arrangement of the table. He will

find that the Standard Forms of Art. 1 28 appear in their proper places.

The reduction formulas of Art. 174 are given, with modifications, by96-104. Also, the reduction formulas of Art. 175, with additional

ones for various cases, are numbered 157-174. Increased powerin the technique of integration will come from familiarity with the

table and practice in using it.

PROBLEMS


-4) + C.

1. Vz 2 + 5 dx = A (3 x 1 - 10 )(x'2 + 5) + C.

= ^V9 x2 - 4 -f- In (3 x -f V9 x

7. Ce< sin 2 ^ dt = \ e<(2 - sin t- cos rt -f C.

J Z 4

g Csn26jd0 ^ 2 In (1 + cos 0)- 2 cos -f C.

J 1 4- cos

9-/2 + 2t + ^

= arCtan(j:+1) + C -

10. fx :< sin x 2 dx = \ sin x 2 -\ x 2 cos x 2

-f C.

dx

'J V(x- 1)(2 -x)= 2 arc sin Vx 1 -f (;.

C


4 -f sin 2

Evaluate each of the following definite integrals.

37.

38.

.

Jo (1 + J")2

dT35. /

36.

x 2V25 - 9 x 2 25

dx _ 2

(4x 2-h 9)

>:j

39. f l<L^>du=ira.J~a \ a -f w

0. rf

x2dX- = 1-3:

J\ V9 2 x'240. = 1.338. 42.

44.

43. / /V '

<// = 0.1605.

dx.jf

2 d"2

46.jf-^ ,- 48.

jf

3

r~ *i

45. f 2cos 2 sin4 6 dO. 47./

e i cos A TT/ dt. 49 f^ Cos A d> dd>.Jo Jo Vo

ADDITIONAL PROBLEMS

1. Verify the following results.

2. A parabola with its axis parallel to the //-axis passes through the

origin and the point (1, 2). Find its equation if the area between the

parabola and the x-axis is a maximum or a minimum.Anx. //

= 6 x 4 s'2gives a minimum.

3. Sketch the curve y V!r = In jr. Find the volume of the solid of

revolution formed by revolving about the .r-axis the area bounded by the

curve, the .r-axis, and two ordinates, one through the maximum point

and the other through the point of inflection. AUK. -^-^ TT.

4. A solid right circular cone of metal is formed so that the density at

any point P is 20(5 r) Ib. per cubic foot, where r is the distance in feet

of the point P from the axis of the cone. Find the weight of the cone if

its altitude and the radius of the base are each 3 ft. Am. 630 TT Ib.

NOTE. The weight of an element of uniform density is its volume times its

density.

5. A hollow metal sphere has an inside radius of 6 in. arid an outside

radius of 10 in. The density of the metal at any point varies inversely as

the distance of the point from the center of the sphere, and at the outside

surface the density is 2 oz. per cubic inch. Find the weight of the sphere.

Am. 2560 TT oz.

6. If n is an even integer, show that

7T TT

r 2 j r 27 (n - D(n -

3) (!) TT

I sm 71 x dx = I cos" x dx = ^^r TT

Jo JQ n (n 2) (Z)

7. If n is an odd integer, find the value of

/*2

/sin" x dx.

Jo

CHAPTER XVIII

CENTROIDS, FLUID PRESSURE, AND OTHER APPLICATIONS

177. Moment of area; centroids. The centroid of a plane area is

defined in the following manner.

A piece of stiff, flat cardboard will balance in a horizontal positionif supported at a point directly under its center of gravity. This

point of support is the centroid of the area of the flat surface of the

cardboard.

For certain areas considered in elementary geometry the cen-

troids are obvious. For a rectangle or a circle t he centroid coincides

with the geometrical center. In fact, if a plane figure possesses a

center of symmetry, that point is the centroid. Furthermore, if a

plane figure has an axis of symmetry, the centroid will lie on that axis.

The following considerations lead to the determination of the

centroid by mathematical means. It is beyond the purpose of

this book to justify the argument bymechanics.

Consider the area AMPNB of the

figure. Divide it into n rectangles, each

with base A.r, as heretofore. The figure

shows one of these rectangles. Let dA be

its area, and C(h, k) its centroid. Then

(l)dA = x, k

The moment of area of this elementary

rectangle about OA' (or O Y) is the product of its area by the per-

pendicular distance of its centroid from OX (or OY). If these

moments are, respectively, dMx and dMu , then

(A) dMx =kdA, dMv =hdA.

The moment of area for the figure AMPNB is obtained by applying

the Fundamental Theorem (Art. 156) to the sum of the moments of

area of the n elementary rectangles. Thus we obtain

1x = Ck dA, Mv= f

320

hdA.

CENTROIDS, FLUID PRESSURE 321

Finally, if (x, y) is the centroid of the area AMPNB, and A its

area, then the relations between the moments of area (B) and x

and y are given by

(C) Ax = MVt Ay = Mx .

To calculate (x, y), find the moments of area Mx and Mv . From(1) and (), these are, for the above figure,

(2)

ri> s*b

x = % y2dx, M v

= /

*Ja Juxydx,

in which the value of y in terms of jr must be substituted from the

equation of the curve MPN.If the area A is known, we have, from (C),

(3) x =

If A is not known, it may be found by integration, as in Art. 145.

ILLUSTRATIVE EXAMPLE 1. Find the centroid of the area under one arch of the

sine curve

(4) y = sm x.

Solution. Constructing an elementary rectan-

gle, we have

(5) dA = ydx sinx dx,

dMx = k dA = I y'2 dx = \ sin 2 x dx, dM v

= h dA = xy dx x sin x dx.

The limits are x 0, x = TT. Hence

(6) A = f^sinzdz = 2, Mx = \ f "sin 2 * do- = \ IT, MV =C zsinzdz = 7T.

J *' ^

Then, from (3), x = \ TT, y = I TT. Aws.

The value of x might have been anticipated, since the line x = } TT is an axis of

symmetry.

ILLUSTRATIVE EXAMPLE 2. In the figure the curve OPA is an arc of the parabola

y'2 2 px. Find the centroid of the area OPAB.

Solution. Draw an elementary rectangle, as in

the figure, and mark its centroid (h, k). Then

dA = x dy, h = \ x, k = y.

Using (4), dMz = kdAxy dy,

<?A (a, 6)

Finding x in terms of y from y2 = 2 px, and

integrating between the limits y = 0, y = 6, we find

Hence x = . y = 7 b. But x = a, y = 6 satisfy the equation i/= 2 px.

20 p 4

Hence 62 = 2 pa, and x = r3^ a. The centroid is therefore ( T

'

{

a, J b). Am.


PROBLEMS

Find the centroid of each of the areas bounded by the following curves.

1. y2 = 2 px, x = h. Ans. (f h, 0).

2. y = x'\ jr = 2, y = 0. (, -1

^).

3. y /*, y 4 x. (First quadrant.) (j|, f^).

4. .r 4 ?/ ?/*, ?/ j. (^, ij).

5. y2 = 4x, 2.r-0 = 4. (, 1).

8. j/= .f

2f i/

= 2.r + 3. (!,)7. ?/

= x 2 - 2 x - 3, //= 6 jr

- j 2 - 3. (2, 1).

8. y = /*, ?/= 8, .r = 0.

9. //= 6 x .r

2, // jr.

10. ?/= 4 jr

~ x'2, y = 2x- 3.

11. ?/ j-H 3 jr, y r. (First quadrant.)

12. y2 a- cur, jr = 0, y = 0. (First quadrant.)

13. =^- ^ = 1, //

= 0, x = 2 a. (First quadrant.)a- fr

14. Find the_centroid of the area bounded by the coordinate axes and

the parabola V.r -f V// = Va. /b?x. 7 ]/=

I a.

15. Find the centroid of the area bounded by the loop of the curve

y* 4 .r2 r*. 4/f.x. J = 1

/' 7/ ^-

16. Find the centroid of thejortion in the first quadrant of the ellipse

jr2

. y'2

1_ 4 a - 4 b

; + ~~l. Aws. x = -//=

oO" />

2t5 7T J 7T

17. Find the centroid of the area bounded by the parabola y2 2 ;xr

and the line ?/= rax. , - 4 />

- pAns. x = *-?/

5 m 2 m

18. Find the centroid of the area included by the parabolas y2 = ax

and .r- = ?>?/. 4 -R k -

__ 9 J/i/i AiS . J 2 oi

^ " // 20 ^ ^

19. Find the centroid of the area bounded by the cissoid //

2(2 a x) x*

and its asymptote x 2 a. .4 MS. 7 =!; a, //

= 0.

20. Find the centroid of the area bounded by the witch x 2y = 4 a 2

(2 a ?/)

and the .r-axis. Ans. x = 0, ]y ^ a.

21. Find the distance from the center of the circle to the centroid of

the area of a circular sector of angle 2 6. .^

2 r sin 6

3022. Find the distance from the center of the circle to the centroid of

the area of a circular segment the chord of which subtends a central

an le261 ' Am 2 r sin-^

3(6 - sin flcos 6)'


23. Find the centroid of the area bounded by the cardioid

p = a(l + cos 6). An$. jc = a, //= 0.

24. Find the centroid of the area bounded by one loop of the curve

p a cos 2 8.Ans. Distance from origin =

128aV2105 TT

25. Find the centroid of the area bounded by one loop of the curve

p = a cos 3 0.AHS. Distance from origin -

81

80 TT

178. Centroid of a solid of revolution. The center of gravity of

a homogeneous solid is identical with the centroid of that bodyconsidered as a geometrical solid.

The centroid will lie in any plane

of symmetry which the solid maypossess.

To achieve a mathematical def-

inition of the centroid of a solid

of revolution, it is necessary to

modify the discussion of the pre-

ceding article only in the details.

Let OX be the geometrical axis

of the solid. The centroid will

then lie on this axis. Let dV be an

element of volume, that is, a cylin-

der of revolution with altitude Axand radius y. Then dV iry- Ax.

The moment of volume of this cylinder with respect to the plane

through OY perpendicular to OX is

(1) dMv= x dV = 7TX//

2 Ax.

The moment of volume for the solid is then found by the Funda-

mental Theorem, and x is given from

(2) Vx = My I irxy- dx.

ILLUSTRATIVE EXAMPLE. Find the centroid

of a solid cone of revolution.

Solution. The equation of the element OB is

y = AB ^rx OA h'

or !,= --

Hence Mv= f TTX ^~ dx = - irr2h 2

.

Jo h* 4

Since V = 3 Trr'2h, x \ h. Ans.


Ans. jr jj a.

PROBLEMS

Find the centroid for each of the following solids.

1. Hemisphere. (See

figure.) Ans. x = I r.

2. Paraboloid of rev-

olution. (See figure.)

Ans. ~x%h.

The area bounded byOX and each curve given

below is revolved about

OX. Find the centroid of

the solid of revolution

generated.

3. x'2 y'2

a'2

,x 2 a.

4. 2 xy = a'2

, x - J a, x - 2 a.

5. ay = x'2

,x = a.

6. i/2 = 4 x, x = 1, x = 4.

7. x 24-

2 = 4, x = 0, x = 1. ? = fJ.

8. ?/= a sin x, x = ^ IT.

The area bounded by OY and each of the curves given below is re-

Tolved about OY. Find the centroid of the solid of revolution generated.

9. y'2 = 4 ax, y b. Ans. y = | 6.

10. jr2 - ir = 1, y = 0, ?/

= 1. y = A-

11. a//2 = /', ?/ a.

12. The radii of the upper and lower bases of a frustum of a cone of

revolution are, respectively, 3 in. and 6 in., and the altitude is 8 in.

Locate the centroid.

13. Find the centroid of the solid formed by revolving about the ?/-axis

the area in the first quadrant bounded by the lines ?/= 0, x a, and the

parabola y* = 4 ax. Ans. y = f a.

14. Find the centroid of the solid formed by revolving about the r-axis

that part of the area of the ellipse 2+

|r,= 1 which lies in the first

quadrant.a

Ans. x = a.

15. Find the centroid of the solid formed by revolving about the z-axis

the area in the first quadrant bounded by the lines y 0, x 2 a, andx'

2t/

2

the hyperbola ~ = 1 .


16. Find the centroid of the solid formed by revolving about the .r-axis

the area bounded by the lines x 0, JT = a, y 0, and the hyperbola

17. Find the centroid of the solid formed by revolving about the .r-axis

the area bounded by the lines //= 0, x = , and the curve // sin 2 x.

18. Find the centroid of the solid formed by revolving about the x-axis

the area bounded by the lines .r = 0, .r = a, //= 0, and the curve // e

r.

19. The area bounded by a parabola, its axis, and its iatus rectum is

revolved about the Iatus rectum. Find the centroid of the solid generated.Am. Distance from focus fa of Iatus rectum.

179. Fluid pressure. We will now take up the study of fluid pressureand learn how to calculate the pressureof a fluid on a vertical wall.

Let ABDC represent part of the

area of the vertical surface of one wall

of a reservoir. It is desired to deter-

mine the total fluid pressure on this area.

Draw the axes as in the figure, the

//-axis lying in the surface of the fluid.

Divide AR into n subintervals and

construct horizontal rectangles within

the area. Then the area of one rec-

tangle (as EP) is y Ax. If this rectangle

was horizontal at the depth x, the fluid pressure on it would be

Wxy Aj,

[The

pressure of a fluid on any given horizontal surface equals the weight!of a column of the fluid standing on that surface as a base and of heightequal to the distance of this surface below the surface of the fluid. J

where W = the weight of a unit volume of the fluid. Since fluid pres-sure'is the same in all directions, it follows that Wxy A.r will be ap-

proximately the pressure on the rectangle EP in its vertical position.

Hence the sum n

D

represents approximately the pressure on all the rectangles. The

pressure on the area ABDC is evidently the limit of this sum. Hence,by the Fundamental Theorem,

lim

n x>

im V Wxflt Ax, = / Wxy dx.-oo

i = 1 J

Hence the fluid pressure on a vertical submerged surface bounded

by a curve, the /-axis, and the two horizontal lines x = a and x = b

is given by the formula b

(D) Fluid pressure- W \ yx dx,

Ja

where the value of y in terms of x must be substituted from the equa-

tion of the given curve.

We shall assume 62 Ib. ( W) as the weight of a cubic foot of water.

ILLUSTRATIVE KXAMPLE 1. A circular water main

6 ft. in diameter is half full of water. Find the pres-

sure on the gate that closes the main.

Solution. The equation of the circle is x'2 + y'2 9.

Hence y= \ 9 - x~,

and the limits are from .r to x = II. Substituting

in (>), we get the pressure on the right of the

x-axis to be

Pressure = 62 C'\ 9 - x* x dx = [- <. (9 - x u;a]jj

= 558.'o

Hence Total pressure = 2 x 558 1 1 16 Ib. Ans.

The essential part of the above reasoning is that the pressure

( dP) on an elementary horizontal strip is equal (approximately)

to the product of the area of the strip (= dA) by its depth (= h)

and the weight (= W) of unit volume of the fluid. That is,

() dP = WhdA.

With this in mind, the axes of coordinates may be chosen in any

convenient- position.

JLLI 'STRATIVK KxAMPLE !2. A trapezoidal gate in a dam is shown in the figure.

Find the pressure on the gate when the vsurface of the water is 4 ft. above the

top of the gate.

Solution. Choosing axes OX and OYas shown, and drawing an elementaryhorizontal strip, we have, using (),

dA 2 .r (///,

WATER

The equation of AB is y = 2 x 8.

Solving this equation for .r, and substi-

tuting, the result is

dP W(% y)(y +

Integrating with limits y and y 4, we obtain

P = W f4

(64 - y~)dy = ^^ W = 14,549 Ib. Ans.t/O


PROBLEMS

In the following problems the //-axis is directed vertically upward, andthe .r-axis is at the surface level of a liquid. Denoting the weight of a

cubic unit of the liquid by ir, calculate the pressure on the areas formed

by joining with straight lines each set of points in the order given.

1. (0, 0), (3, 0), (0,-

6), (0, 0). Ans. 18 \V.

2. (0, 0), (3,-

6), (0,-

6), ^0, 0). 36 W.

3. (0, 0), (2,-

2), (0,-

4), i- 2,-

2), (0, 0). 16 W.

4. Calculate the pressure on the lower half of an ellipse whose semi-

axes are 2 and 3 units respectively, (a) when the major axis lies in the

surface of the liquid; (b) when the minor axis lies in the surface.

AH*, (a) 8 If; (b) 12 \V.

5. Each end of a horizontal oil tank is an ellipse of which the horizontal

axis is 12 ft. long and the vertical axis 6 ft. long. Calculate the pressureon one end when the tank is half full of oil weighing 60 lb. per cubic foot.

An^. 2160 lb.

6. The vertical end of a vat is a segment of a parabola (with vertex

at the bottom) 8 ft. across the top and 16 ft. deep. Calculate the pressureon this end when the vat is full of a liquid weighing 70 lb. per cubic foot.

/b/K. 38,22911).

7. The vertical end of a water trough is an isosceles

right triangle of which each leg is 8 ft. Calculate the

pressure on the end when the trough is full of water

(W = 62.5). AUK. 3771 lb.

8. The vertical end of a water trough is an isosceles triangle 5 ft.

across the top and 5 ft. deep). Calculate the pressure on the end when the

trough is full of water. A//,s. 1302 lb.

9. A horizontal cylindrical tank of diameter 8 ft. is half full of oil

weighing 60 lb. per cubic foot. Calculate the pressure on one end.

Ans. 2560 lb.

10. Calculate the pressure on one end if the tank of Problem 9 is full.

11. A rectangular gate in a vertical dam is 10 ft. wide and 6 ft. deep.

Find (a) the pressure when the level of the water ( W 62.5) is 8 ft. above

the top of the gate ; (b) how much higher the water must rise to double

the pressure found in (a). Am. (a) 41,250 lb. ; (b) 11 ft.

12. Show that the pressure on any vertical surface is the product of

the weight of a cubic unit of the liquid, the area of the surface, and the

depth of the centroid of the area.

13. A vertical cylindrical tank, of diameter 30 ft. and height 50 ft., is

full of water. Find the pressure on the curved surface. Ans. 3682 tons.


180. Work. In mechanics the work done by a constant force Fcausing a displacement d is the product Fd. When F is variable, this

definition leads to an integral. Twoexamples will be considered here.

Work done in pumping out a tank.

Let us now consider the problem of

finding the work done in emptyingreservoirs of the form of solids of rev-

olution with their axes vertical. It

is convenient to assume the r-axis of

the revolved curve as vertical, and the

?/-axis as on a level with the top of

the reservoir.

Consider a reservoir such as the one

shown ; we wish to calculate the workdone in emptying it of a fluid from the depth a to the depth b.

Divide AB into w subintervals, pass planes perpendicular to the

axis of revolution through these points of division, and construct

cylinders of revolution, as in Art. J60. The volume of any such

cylinder will be 7n/~A:r and its weight M'7r//2A.r, where W = weight

of a cubic unit of the fluid. The work done in lifting this cylinderof the fluid out of the reservoir (through the height jr) will be

Wiry2

fWork done in lifting equals the weight multiplied by the vertical height.]

The work done in lifting all such cylinders to the top is the sum

The work done in emptying that part of the reservoir will evi-

dently be the limit of this sum. Hence, by the Fundamental Theorem,

limr

cl= / Wiry2 xdx.J

Therefore the work done in emptying a reservoir in the form of asolid of revolution from the depth a to the depth 6 is given by the

formula

(F) Work= WIT f y2

xdx,Ja

where the value of y in terms of x must be substituted from the

equation of the revolved curve.

ILLUSTRATIVE EXAMPLE 1. Calculate the work done in pumping out the water

filling a hemispherical reservoir 10 ft. deep.

Solution. The equation of the circle is

& + 2/-= 100.

Hence y* = 100 - *-',

W = 62,

and the limits are from x = to r = 10.

Substituting in (F), we get

/> 10

Work = 62 TT I 000 - x'^rdr = 155,000 TT ft.-lb.Jo

The essential principle in the above reasoning is that the element

of work (= dw} done in lifting an elementary volume ( dV) through

a height (=A) isdw=KhdV,

where W = weight of unit volume of the fluid. With this in mind,

the axes of coordinates may be chosen in any convenient manner.

ILLUSTRATIVE EXAMPLE 2. A conical cistern is 20 ft. across the top and 15 ft.

deep. If the surface of the water is 5 ft. be-

low the top, find the work done in pump-

ing the water to the top of the cistern.

Solution. Take axes OX and OY as in

the figure. Then

dV = TTX 2dy,

h = 15-y,dw VT(15 y}irx~ dy.

1(10, 15)

The equation of the element OA is

x = y. Substituting,

dw = 7rW(15 - |/)J uz<ly=

i 7rHT

(15 y'2 -

y*)dy.

The limits are y = and y = 10, since the water is 10 ft. deep. Integrating,

w = $ irW f10

n5 it'

2 -y"')dy

= 216,421 ft.-lb. An*.'

Jo

Work done by an expanding gas. If a gas in a cylinder expands

against a piston head from volume v cu. ft. to vi cu. ft, the external

work done in foot-pounds is

(G) Workr*>i

= I pJvQ

where p = pressure in pounds per square foot.

Proof. Let the volume increase from v to v + dv.

Let c area of cross section of the cylinder.

Then = distance the piston moves.c


Since pc = force causing the expansion dv,

Element of work done = pc - = p dv.c

Then (G) follows by the Fundamental Theorem. To use (G), the

relation between /; and v during the expansion must be known. This

relation has the form

0) pi? = constant,

the exponent n being a constant.

Isothermal expansion occurs when the temperature remains con-

stant. Then n 1, and the pressure-volume relation is

(2) pv por ()= piVi.

If a graph of (1) is made (pressure-volume diagram), plotting

volumes as abscissas and pressures as ordinates, the area under this

curve gives, numerically, the work done, as calculated by (G). In

isothermal expansion the graph of (2) is a rectangular (equilateral)

hyperbola.

PROBLEMS

1. A vertical cylindrical cistern of diameter 16 ft. and depth 20 ft. is

full of water (VT 62.5). Calculate the work necessary to pump the

water to the top of the cistern. Am. 800,000 TT ft. Ib.

2. Jf the cistern of Problem 1 is half full, calculate the work necessary

to pump the water to the top.

3. A conical cistern 130 ft. across the top and 20 ft. deep is full of water

(IV rr 62.5). Calculate the work necessary to pump the water to a height

of 15 ft. above the top of the cistern. , . 2,500,000 TT ,,J\ / / 8 .

~1 1 . ID.

4. A hemispherical tank of diameter 10 ft. is full of oil weighing 60 Ib.

per cubic foot. Calculate the work necessary to pump the oil to the top

of the tank. Am. 9375 TT ft. Ib.

5. A hemispherical tank of diameter 20 ft. is full of oil weighing 60 Ib.

per cubic foot. The oil is pumped to a height of 10 ft. above the top of

the tank by an engine of } H.P. (that is, the engine can do work at the

rate of 16,500 ft. II). per minute). How long will it take the engine to

empty the tank ?

6. Find the work done in pumping out a semi-elliptical reservoir full

of water (W 62). The top is a circle of diameter 6 ft., and the depth

is 5ft. Ans. 3487 \ TT ft. Ib.

7. A conical reservoir 12 ft. deep is filled with a liquid weighing 80 Ib.

per cubic foot. The top of the reservoir is a circle 8 ft. in diameter. Cal-

culate the work necessary to pump the liquid to the top of the reservoir.

-4ns. 15,360 TT ft. Ib.

8. A water tank is in the form of a hemisphere, 124 ft. in diameter, sur-

mounted by a cylinder of the same diameter and 10 ft. high. Find the

work done in pumping it out when it is filled within 2 ft. of the top.

9. A bucket of weight M is to be lifted from the bottom of a shaft

h ft. deep. The weight of the rope used to hoist it is in Ib. per foot. Find

the work done.

10. A quantity of air with an initial volume of 200 cu. ft. and pres-

sure of 15 Ib. per square inch is compressed to 80 Ib. per square inch.

Determine the final volume and the \\ork done if the isothermal law holds,

that is, pr = C. -4//s. :i7.f> on. ft. ; 72:5,000 ft. Ib.

11. Determine the final volume and work done in Problem 10 if the

adiabatic law holds, that is pr7'

(\ assuming n~

1.4.

An*. 60 cu. ft.; 648,000 ft. Ib.

12. Air at pressure of 15 Ib. per square inch is compressed from

200 cu. ft. to 50 cu. ft. Determine the final pressure and the work done if

the law is pv C. A//s. 60 11). per square inch ; 599,000 ft. Ib.

13. Solve Problem 12 if the law is pr" - <\ assuming // 1.4.

AHS. 104.5 Ib. per square inch ; 801,000 ft. Ib.

14. A quantity of gas with an initial volume of 16 on. ft. and pressure

of 60 Ib. per square inch expands until the pressure is ;}() Ib. per square

inch. Determine the final volume and the \\ork done by the gas if the

law is pr = C. An*. :*2 on. ft. ; 95,800 ft. Ib.

15. Solve Problem 14 if the law is pr" (\ assuming n 1.2.

An*. 2K.f> cu. ft.; 75,600 ft. Ib.

16. A quantity of air with an initial volume of 200 cu. ft. and pressure

of 15 Ib. per square inch is compressed to 150 cu. ft. Determine the final

pressure and the work done if the law is pr ~-- (\

17. Solve Problem 16 if the law is pr7' = (\ assuming r? = 1.4.

18. A gas expands from an initial pressure of 80 11>. per square inch and

volume of 2.5 cu. ft. to a volume of 9 cu. ft. Find the work done if the

law-is prn

0, assuming n 1.0646.

19. Solve Problem 18 if n = 1.131.

20. Determine the amount of attraction exerted by a thin, straight,

homogeneous rod of uniform thickness, of length /, and of mass M upon

a material point P of mass m situated at a distance of a from one end of

the rod in its line of direction.

Solution. Suppose the rod to be divided into equal infinitesimal portions (ele-

ments) of length dx. M = mass of a unit length of rod ;

hence ~ dx = mass of any element.


Newton's law for measuring the attraction between any two masses is t

Force of attraction = Produrt of masses

'distance between them;*

therefore the force of attraction between the particle at P and an element of the rod is

which is then an clement of the force of attraction required. The total attractionbetween the particle at V and the rod being the limit of the sum of all such elementsbetween x = and j /, we have

/-./M ,

r'-r-mdx- I J- =^ C

l

-'o (j + aj-2

f JForce of attraction

l dx M

21. Determine the amount of attraction in the last example if P lies

in the perpendicular bisector of the rod at the distance a from it.

Am. 2mM2

-

22. A vessel in the form of a right circular cone is filled with water.If // is its height and r the radius of the base, what time will it requireto empty itself through an orifice of area a at the vertex?

Solution. Neglecting all hurtful resistances, it is known that the velocity of dis-

charge through an orifice is that acquired by a body falling freely from a heightequal to the depth of the water. If, then, x denotes the depthof the water, -

v = \ 2 gx.

Denote by dQ the volume of water discharged in timedt, and by dx the corresponding fall of surface. The volumeof water discharged through the orifice in a unit of time is

a \

being measured_as_ a right cylinder of area of base a andaltitude r (= \ 2 r/.r). Therefore in time dt

(1)

Denoting by S the area of the surface of the water when the depth is x, we have,from geometry, '' >

, or, ' VIh~

But the volume of water discharged in time dt may also be considered as thevolume of a cylinder AB of area of base S and altitude dx; hence

Equating (1) and (2) and solving for dt,

Therefore t = JL = ^. Ans.2 gx 5o\ 2 g

CENTROIDS, FLUID PRESSURE.

333

181. Mean value of a function,

value) of n numbers y\, y<2 , -,#

(i) y = \

The arithmetic mean (or averages

+ y<2 + - + //).

We proceed to establish the formula

r b

I 6(x)dx(H) Mean value of 4>(x) 1 = Ja

T

from x = a to x = bj

b a

The figure shows the graph of

(2) y = </>(/).

The mean value (= y) of the ordinates of the arc PQ is to be de-

fined. Divide AB into n equal parts each equal to A.r and let //i,

yz,' *

2/n, be the ordinates at the w points of division. Then (1 ) will

give an approximate value for the mean value required. Multiply

numerator and denominator of the right-hand member of (1) by Ax.

Then, since n Ax = b a, we get

(3) y (approximately) = ^ ,__

But the numerator in (3) is, approximately, the area APRQB.The average value of y (or (/>(j)) is defined as the limit of the right-

hand member in (3) when n > oo. This gives (//).

In the figure the mean value of <j>(x) equals CR if area rectangle

ABML = area ABQRP.Taking y as the function (dependent variable), then (H) becomes

fJa ydx

GivenILLUSTRATIVE EXAMPLE.the circle

(4) x2 + y2 = r2 .

Find the average value of the or-

dinates in the first quadrant

(a) when y is expressed as a func-

tion of the abscissa x ;

(b) when y is expressed as a func-

tion of the angle = Z MOP.

Solution, (a) Since y = \V2 - x2,

the numerator in (/) is

X

CJoVr2 - x2 dx = I Trr

2. Then y = \ irr = 0.785 r. Ans.


(b) Since y r sin 0, and the limits are 6 = = a, 8 \ IT b, the numeratorin (/) is

A *2 r

I r sin 6 d6 = r. Since b - a =\ TT, we have y = = 0.637 r. Ans.

JO 7T

Thus we have quite different values of y, depending upon the independentvariable with respect to which the mean value is taken.

As shown in the above example, the average value of a givenfunction y will depend upon the variable chosen as the independentvariable. For this reason, we write (/) in the form

C h

I ydxx

b a

in order to indicate explicitly the variable with respect to which yis averaged.

Thus, in the Illustrative Example, we have~yx 0.785 r, and y ft

= 0.637 r.

PROBLEMS

1. Find the average value of ?/= s~ from jr = to x 10. Ans. 33f3 .

2. Find the average value of the ordinates of y2 = 4 r from (0, 0) to

(4, 4) taken uniformly along the .r-axis. Ans. 2.3. Find the average value of the abscissas of y'

2 = 4 jc from (0, 0) to

(4, 4) when uniformly distributed along the ?/-axis. Ans. 1^.

4. Find the average value of sin .r between .r and .r = TT. Ans.IT

5. Find the average value of sin 2.r between jr = and *c = TT. (This

average value is frequently used in the theory of alternating currents.)

Ans. %.

6. If a particle in a vacuum were thrown downward with an initial

velocity of v(} ft. per second, the velocity after t sec. would be given by

/- N i /

The velocity after falling s ft. would be given by

(2) r - Vz\r + 2 gs. (Take g = 32)

Find the average value of v

(a) during the first 5 sec., starting from rest ; Ans. 80 ft. per second.

(b) during the first 5 sec., starting with an initial velocity of 36 ft.

per second; Am. 116 ft. per second.

(c) during the first 2-J sec., starting from rest; Ans. 40 ft. per second.

(d) during the first 100 ft., starting from rest; Ans. 53 J ft. per sec.

(e) during the first 100 ft., starting with an initial velocity of 60 ft.

per second. Ans. 81 ft. per second.


7. In simple harmonic motion s = a COP nt. Find the average valueof the velocity during one quarter of a period (a) as to the time

; (b) as

to the distance.

8. Show that in simple harmonic motion the average kinetic energywith respect to the time for any multiple of a quarter period is half the

maximum kinetic energy.

9. A point is taken at random on a straight line of length a. Prove

(a) that the average area of the rectangle whose sides are the two seg-ments is a 2

; (b) that the average value of the sum of the squares onthe two segments is fa 2

.

10. If a point moves with constant acceleration, the average as to the

time of the square of the velocity is J(r--f r

( ,rj + r,2), where r,> is the

initial and i\ the final velocity.

11. Show that the average horizontal range of a particle projectedwith a given velocity at an arbitrary elevation is 0.6366 of the maximumhorizontal range.

HINT. Take a = in the formula of Problem 35, p. 1 14.

The formulas ~ ~

I x ds I y ds

(6) *'

where (x, y) is any point on a curve for which ds is the element of arc,

define the centroid of the arc. They give, respectively, the average

values of the abscissas and ordinates of points on the curve when

distributed uniformly along it. (Compare Art. 177.)

12. Show that the area of the curved surface generated by revolving

an arc of a plane curve about a line in its plane not cutting the arc equals

the length of the arc times the circumference of the circle described byits centroid (6). (Theorem of Pappus. Compare Art. 250.)

HINT. Use (L), Art. 164.

13. Find the centroid of the arc of the parabola y2 4 x from (0, 0)

to (4, 4).Am. 7 = 1.64, y = 2.29.

14. Find the centroid of an arc of the circle p a between arid -f 6.

A ns.

15. Find the centroid of the perimeter of the cardioid p = a(l -f cos 0).

Ans. ~jc = I a, y == 0.

16. Find by the Theorem of Pappus the centroid of the arc of the circle

x'2 + 3/2-

r 2 which lies in the first quadrant, ^^ x = T =


17. Find by the Theorem of Pappus the surface of the torus generated

by revolving the circle (x b')2-f y

2 = a 2(b > a) about the |/-axis.

18. A rectangle is revolved about an axis which lies in its plane and is

perpendicular to a diagonal at its extremity. Find the area of the surface

generated.

ADDITIONAL PROBLEMS

1. An area is bounded by the lines y = x 2,x + y = 6, y 0, and x = 3.

Find its centroid. Ans. x = f , y = f .

2. The abscissa of the centroid of the area bounded by the curve

2 //= x'

2 and a certain line through the origin is 1. Find the ordinate of

the centroid. An.s. f .

3. Find the centroid of the area bounded by y = x n(n > 0), the x-axis,

and x 1. Discuss the locus of the centroid as n varies.

_ n-hl- n+ 1AUK. x =

n + 2y

2(2 n-f 1)

4. Find the equation of the locus of the centroid of the area bounded

by the .r-axis and the parabola y = ex x'2 when c varies. Ans. 5 y = 2 x'2 .

5. Given the parabola x'2 = 2 py and any oblique line y mx + b meet-

ing the parabola in the points A and B. Through C, the midpoint of

A /*, draw a line parallel to the axis of the curve meeting the parabola at

D. Prove that (a) the tangent to the parabola at /> is parallel to the line

.4 H ; (b) the centroid of the area ACIU) lies on the line CD.

6. Let P be a point on the parabola // x 2, and let C be the centroid

of the area bounded by the parabola, the .r-axis, and the ordinate throughP. Find the position of P so that the angle OPC is a maximum.

4ns. Ordinate = -f%.

7. A cistern has the form of a solid generated by revolving about its

vertical axis a parabolic segment cut off by a chord 8 ft. long, perpendicu-

lar to the axis and at a distance of 8 ft. from the vertex. The cistern is

filled with water weighing 62.5 Ib. per cubic foot. Find the amount of

work required to pump over the top of the cistern one half the volume of

water it contains. AnSm 16,000 (\/2-

1) f = 6937 ft.-lb.o

8. A hemispherical cistern of radius r is full of water. Two men, Aand B, are to pump it out, each doing half the work. If A starts first, whatwill be the depth d of the water when he has finished his share of the work ?


9. A tank in the shape of an inverted circular cone is full of water.

Two men are to pump the water to the top of the tank, each doing half

the work. When the first man has finished his share of the work, let s

denote the ratio of the depth of water left in the tank to the original depth.Show that z is determined by the equation 6 r4 8 r :{

4- 1 = 0. Calculate

the value of z to two decimals. An*. 0.61.

10. A well is 100 ft. deep. A bucket, weighing 3 lb., has a volume of

2 cu. ft. The bucket is filled with water at the bottom of the well and is

then raised at a constant rate of 5 ft. per second to the top. Neglectingthe weight of the rope, find the work done in raising the bucket if it is

discovered that the water is leaking out at a constant rate of 0.01 cu. ft.

per second. (A cubic foot of water weighs 62.4 lb.) An*. 12,156 ft.-lb.

11. The area OAB is divided into elements such as OPQ by lines from

O. Show that the area A and the moments of area

Mx and Mv are given by Y j

r\\ (xy

r

y)dx,

=3 I

^My

rM x = JJ y(jcy'

n/ y)djr.

(The centroid of a triangle is on any median at

two thirds of the distance from the vertex to the -}

opposite side.)

12. Find the centroid of the hyperbolic sector bounded by the equi-

lateral hyperbola x a sec 6, y a tan and radii from the origin to the

points (a, 0) and (x, y).

2 tan - _ 2 secAn*. x - a V

3 in (sec + tan 0)

CHAPTER XIX

SERIES

182. Definitions. A sequence is a succession of terms formed ac-

cording to some fixed rule or law.

For example, 1, 4, 9, 16, 25

,-,

X2 X* X* Xr'

and i,_ 3, _,__.,_,__are sequences.A scries is the indicated sum of the terms of a sequence. Thus

from the above sequences we obtain the series

1+4 + 9 + 16 + 25

, ! ,

X2 X* . X* X*and 1 _ :r + .__._ + T

-~.

When the number of terms is limited, the sequence or series is

said to be finite. When the number of terms is unlimited, the sequenceor series is called an infinite sequence or infinite series.

The ycH-cral term, or r/th term, is an expression which indicates

the law of formation of the terms.

ILLUSTRATIVE EXAMPLE 1. In the first example given above, the general term,or nth term, is n 2

. The first term is obtained by setting n = l, the tenth term bysetting u = 10, etc.

ILLUSTRATIVE EXAMPLE 2. In the second example given above, the nth term,

except for n = 1, is ^ : -

?il

If the sequence is infinite, this fact is indicated by the use of dots, as

1, 4, 9, --, r/-,

--.

Factorial numbers. An expression which occurs frequently in con-

nection with series is a product of successive integers, beginningwith 1. Thus, 1x2x3x4x5 is called 5 factorial and is indicated

by [5 or 5 !

In general, [w= lx2x3x---x(?i-l)xw

is called n factorial. It is understood that n is a positive integer. The

expression [n has no meaning if n is not a positive integer.

338

SERIES 339

183. The geometric series. For the geometric series of n terms,

(1) Sn = a + ar + ar 2 + + arn ~\it is shown in elementary algebra that

ororn _ ^ r)

_____

the first form being generally used if \r\ < 1, and the second form

if \r\> 1.

If \r\ < 1, then rn decreases in numerical value as // increases and

lim (rn

)= 0.

n - ao

From formula (2) we see, therefore, that (Art. 16)

(3) limS^T-^--n - oo l T

Hence if \r\< 1 the sum S n of a geometric series approaches a

limit as the number of terms is increased indefinitely. In this case

the series is said to be convergent.

If \r\> 1, then rn

will become infinite as // increases indefinitely

(Art. 18). Hence, from the second formula in (2), the sum Sn will

become infinite. In this case the series is said to be divergent.

A peculiar situation presents itself if r 1. The series then

becomes

(4) a a + r; a -f a a - -.

If t] is even the sum is zero. If n is odd the sum is a. As //, increases

indefinitely the sum does not increase indefinitely and it does not

approach a limit. Such a series is called an oscillating series.

ILLUSTRATIVE EXAMPLE. Consider the geometric series with

(5) Sn = 1

We find, by (2), that S n =

Then

(6) lim Sn = 'A which agrees with (3), when a = 1, r = \.n -*oo

It is interesting to dis- 1- +;*

cuss (5) geometrically. To 6l 'Sj! -5 ^do this, lay off successive values of S n on a straight line, as in the figure.

n 1 2 3

Sn 1 1-2- ij

Each point thus determined bisects the segment between the preceding point and

the point 2. Hence (6) is obvious.

340 DIFFERENTIAL AND INTEGRAL CALCULUS

PROBLEMS

In each of the following series (a) discover by inspection the law of

formation ; (b) write three more terms ; (c) find the nth, or general, term.

1. 2 + 4 + 8 + 16 + -. Arts, nth term = 2n.

3. - i + + J + jj+

-jj+

6 *

2~f2.4~

f2-4.f>~

f 2.4.6.8~f*"'

6 ^_l'j_^ 5L1

+ . . .y **'3579 2 n + 1

Write the first four terms of the series whose nth, or general, term is

given below.

7.^- Ans. l + -4- + 4- + -7= + "*-V2 V3 V4

8 JLA. 3 + 1 + ^6o. rt ., o-r-t--f-r.O O i

n ".

1 . ? ,3

, _4_ ,

^7i J-^Q^o^97^O O C/ i-j <

,fH 1 'y v2 /j3i o - 1 + -^ + - I- -1U.

/J- i^ / 1^ , i^ i I

V^ V^ V3 V4- - ( 1 )

' X X '

. X'* X,

llt . __}___{ ..

12.

n -M>

f/^"""^ r?y ~l~ n^^ w ^ *^2

[w

'

2"[n'

Vn~T2

184. Convergent and divergent series. In the series

Sn = Ui + U2 + Us + + 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 I. Sn approaches a limit, say u, indicated by

(1) lim Sn = u.

SERIES 341

The infinite series is now said to be convergent and to converge to the

value u, or to have the value u.

CASE II. Sn approaches no limit. The infinite series is now said

to be divergent.

Examples of divergent series are

1 + 2 + 3 + 4 + 5H---- ,

1-1 + 1-1 + ....

As stated above, in a convergent series the value of the series is the

number u (sometimes called the sum) defined by (1). No value is

assigned to a divergent series.

In the applications of infinite series, convergent series are of major

importance. Thus it is essential to have means of testing a given

series for convergence or divergence.

185. General theorems. Before developing special methods for

testing series, attention is called to the following theorems, of which

proofs are omitted.

Theorem I. // Sn is a variable that always increases as n increases

but never exceeds some definite jired nuwher A, then as n increases

without limit, Sn will approach a limit n which is not (jreater than A.

The figure illustrates the statement. The points determined by

the values Si, Sa, S, etc. approach the point w, where

and u is less than or equal to A.

ILLUSTRATIVE EXAMPLE. Show that the infinite series

a) 1 + 1+ rL2+ rnh) + --- +

[i+ ---

is convergent.

Solution. Neglect the first term, and write

(2) Sn = 1 +2+ JT^l + ' * * +

1 -2-3---n"

Consider the variable sn defined by

(3) *n = l+7,+~ + '-+~in which we have replaced all integers in the denominators of (2), except 1, by 2.

Obviously S B < Also, in (3j we have a geometric series with r = I and .<> < 2

no matter how large n may be (see Art. 1H3). Hence S w as defined by (2) is a vari-

able which always increases as n increases but remains less than 2. Hence S n ap-

proaches a limit as n becomes infinite, and this limit is less than 2. Therefore the

infinite series (1) is convergent, and its value is less than 3.

We shall see later that the value of (1; is the constant e = 2.71828 , the

natural base (Art. 61).


Theorem II. // S n is a variable that always decreases as n increases

but is never less than sonw definite fixed number B, then as n increases

without limit, S n will approach a limit which is not less than B.

Consider now a convergent series

for which lim S', t= u.

n oo

Ijet the points determined by the values S'i, .S\>, Sz, etc. be plotted

on a directed line. Then these points, as n increases, will approachthe point determined by ?/ (terms in S n all of the same sign; or

cluster about this point. Thus it is evident that

(A) 1imM M= 0.

it oo

That is, in a convergent series, the terms must approach zero as a limit.

On the other hand, if the general (or //th) term of a series does

not approach zero as // becomes infinite, we know at once that the

series is divergent.

(A) is not, however, a suflicient condition for convergence; that

is, even if the //th term does approach zero, we cannot state posi-

tively that t he series is convergent . For, consider the harmonic series

1 +2+

a+

4+ ' " ' +

w'

Hero lim (//) ~ lim (-) = 0;.* V W'

that is, condition (A) is fulfilled. Yet we shall show in Art. 186 that

the series is divergent.

We shall now proceed to deduce special tests which, as a rule, are

easier to apply than the above theorems.

186. Comparison tests. In many cases it is easy to determine

whether or not a given series is convergent by comparing it term

by term with another series whose character is known.

Test for convergence. Let

(1) Mi + Wj + U;\+ ' ' '

be a scries of posit ire terms irhich it is desired to test for convergence. If

a series of positive terms already kmnm to be convergent, namely,

(2) fli +fl- + :i + ,

raw be found whose terms are never less than the corresponding terms in

the series (1) to be tested, then (1) is a convergent series and its value

does not exceed that of (2).

SERIES 343

Proof. Let s n = MI + w 2 + MS + + u> n ,

and Sn = fli + <io -f a.< + -f a n ,

and suppose that lim Nfl== A.

- oo

Then, since S n < A and s n ^ tf llf

it follows that s n < A. Hence, by Theorem I, Art. 185, s n approaches

a limit and the series (1) is convergent and its value is not greater

than .4.

ILLUSTRATIVE EXAMPLE 1. Test the series

(3) 1 +-;

Solution. Compare with the geometric series

(4) 1 +|+

r*+

if-+ ? "'

which is known to be convergent. The terms of (4) are never less than the corre-

sponding terms of (3). Hence (3) also is convergent.

Following a line of reasoning similar to that applied to (1) and (2),

we may prove the

Test for divergence. Let

(5) '//i + W- -f W:i +be a series of positive terms to he tested, which are never less than the

corresponding terms of a series of positive terms, namely,

(6) fci +&i. + b:i-4 ,

known to be divergent. Then (5) is a divergent series.

ILLUSTRATIVE EXAMPLE 2. Show that the harmonic series

(7) 1 + i + .4 + 1 +

is divergent.

Solution. Rewrite (7) as below and compare with the series written under it.

The square brackets are introduced to aid in the comparison.

(8) 1 + ^ + [J + 11 + [' + i + 7 -4- Jl + [ J + + f',,1 + -.

(9) 4 + i -Kd + 1] + [ 1 + t + t + 11 + Li'e + ' ' ' + A] + -.

We observe the following facts. The terms in (8) are never less than the

corresponding terms in (9).

But (9) is divergent. For the sum of the terms in each square bracket is 4,

and S n will increase indefinitely as n becomes infinite.

Hence (8) is divergent.


1+ i +-L + -L + ...

\/2 V3 V4

Solution. This series is divergent, since its terms are greater than the cor-

responding terms of the harmonic series (7) which is divergent.

The "p series/'

(10; 1-

is useful in applying the comparison lest.

Theorem. The p series is convergent when p > 1, and divergent for

other values of p.

Proof. Rewrite (10) as below and compare with the series written

under it. The square brackets are used to aid in the comparison.

(11) M

(12) 1 +

J^

2"

!2"

i

If p > 1, the terms in (12) are never less than the correspondingterms in (11 ). Hut, in (12), the sums within the brackets are

1. 1 _ 2. _ . .]_

/>'

;) 2''

" ~

2 ;'

and so on. Hence, to test (12) for convergence, we may consider

the series

<" '+^+(rO'+(FWhen p> 1, series (13) is a geometric series with the common

ratio less than unity, and is therefore convergent. Therefore (10) is

also convergent. When p 1, series (10) is the harmonic series and

is divergent,. When p ^ 1, the terms of series (10) will, after the

first, be greater than the corresponding terms of the harmonic series ;

hence (10) is now divergent. Q.E.D.

ILLUSTRATIVE KXAMPLK 4. Show that the series

4(14) -f

2-3-4is convergent.

Solution. In (14),

, __3-4-54 5-6

2n(n

or -u n < ;

that is,.],u n is less than the general term of the /) series when p = 2. Hence the

series each of whose terms is half the corresponding term in (14) is convergentand therefore (14) also is convergent.

SERIES 345

PROBLEMS

Test each of the following series.

126

187. Cauchy's test-ratio test. In the infinite geometric series

a + ar + ar~ + + ar" + or 1 ' M + -,

the ratio of the consecutive general terms ar" and ar" f '

is the com-

mon ratio r. Moreover, we know that the series is convergent when

i

r\

< 1 and divergent for other values. We now explain a ratio test

which may be applied to any series.

Theorem. Let

(]) U\+ Wli + U-A + ' ' ' +Un + Mn + 1 + ' ' '

be an infinite scries of positive terms. Consider consecutive general

terms ?/ n and ?/ n i i >^^rf /orrw the test ratio.

Test ratio = ^2LLi .

/"/'A/// JAtf //w?'/ o/ /r/,s tes/ ratio when n becomes infinite. Let this be

l- W<n 4 1

p lim --n - cr W'n

I. H7/cw p-'

1, the scries is convergent.

II. M ;//r/? p > 1, ///r scries is divergent.

III. H/7/r// p = 1, ///r test foils.

Proof. I. H7/(v/ p 1. My the definition of a limit (Art. 14) we

can choose u so lar^e, say //~~

m, (hat when n ^ m the ratio ^-^ will

diller from p by as little as we please, and therefore be less than a

proper fraction r. Hence

?/, f i

'

?/,,,r ; //, ,

- '" //, , \r < n ,,,r'2

;w m , ;?

<-w,,,r* ;

and so on. Therefore, after the term ?/ each term of the series (1)

is less than the corresponding term of the geometric series

(2) u ,n r + y/,,,r- + //,,,r' +

J^ut since r ^ 1, the series (2), and therefore also the series (1),

is convergent (Art. 1S(>).

II. \\'hen p ^1 (or p --- oo). Following the same line of reasoning

as in I, the series (1 ) may be shown to be divergent.

III. When p = 1,

t he series may be eit her convergent or divergent ;

that is, the test fails. For, consider the p series, namely,

The test,st, ratio is^ = (-^-rY= (l- -^)" ;

n n \H. + 17 \ n + I/

and lim l'^ = lim (l--~-- (1)" - 1 (= p).

n * \ W H 7 w -. x \ W + 1 7

Hence p = 1, no matter what value p may have. But in Art. 186

we showed that

when p > 1, the series converges, and

when p ^ 1, the series diverges.

SERIES 347

Thus it appears that p can equal unity both for convergent and for

divergent series. There are other tests to apply in cases like this,

but the scope of our book does not admit of their consideration.

For convergence it is not enough that the test ratio is less than

unity for all values of n. This test requires that the limit of the test

ratio shall be less than unity. For instance, in the harmonic series

the test ratio is always less than unity. The limit, however, equals

unity.

The rejection of a group of terms at the beginning of a series

will affect the value but not- the existence of the limit.

188. Alternating series. This is the name given to a series whoseterms are alternately positive and negative.

Theorem. // MI M L> + M: < M

} ~f

is an alternating series in which each term is numerically less than the.

one which precedes it, and if]jm u ___ ^n * >

then the series is convergent.

Proof. When n is even, S n may be written in the two forms

(1) S n = (MI- M L.) + (M.I

-MI) -f -f (n n ,

-?/),

(2) S tt MI (M L>~ M.O (M, t 2 M,, i

) M,t .

Each expression in parentheses is positive. Hence when n in-

creases through even values, d ) shows that Stl increases and (2) shows

that S H is always less than ?/j ; therefore, by Theorem I, Art. 185,

S n approaches a limit /. But S' n t jalso approaches this limit /,

since S H + \~ X n + '"*} i and lirn u n n = 0. Hence when n increases

through all integral values, <S' n > / and the series is convergent.

ILLUSTRATIVE EXAMPLE. Test the alternating scries 1 -i -f i

-} +

Solution. Each term is less in numerical value than tl

Hfcnce lim (w n ) 0, and the series is convergent.

Solution. Each term is less in numerical value than the preceding one, and un -~ -

An important consequence of the above proof is the statement :

The error made by breaking a convergent alternating series off at anyterm does not exceed numerically the value of the first of the terms discarded.

Thus, the sum of ten terms in the above Illustrative Example is 0.646, and the

value of the series differs from this by less than one eleventh.

In the above statement it is assumed, however, that the series has

been carried far enough so that the terms are decreasing numerically.

189. Absolute convergence. A series is said to be absolutely or un-

conditionally convergent when the series formed from it by makingall its terms positive is convergent. Other convergent series are said

to be conditionally convergent.

For example, the series 19^ + 0^ 77 + FI

~~ ' ' '

is absolutely convergent, since the series (3), Art. 186, is convergent.

The alternating series

!1,1 1,1 ____

1-9+3-4+5is conditionally convergent, since the harmonic series is divergent.

A series with some positive and some negative terms is convergent if

Ike series deduced from it by waking all the signs positive is convergent.

The proof of this theorem is omitted.

190. Summary. Assuming that the test-ratio test of Art. 187 holds

without placing any restriction on the signs of the terms, we may sum-

marize our results in the following

General directions for testing the series

U\ 4" U2 + H3 + H4 + ' + l/n + Un + l + ' ''.

When it is an alternating series whose terms never increase in numer-

ical value, and 'if lim w w =- 0,Tl '00

then, the series is convergent.

In any series in which the above conditions are not satisfied, we deter-

mine the form of ?/ n and u n 1 i, Jorm the test ratio, and calculate

lim =p.

I. When| p |

< 1, the series is absolutely convergent.

II. When| p i

> 1, the series is divergent.

III. When\p = 1, the test faifo, and we compare the series with

some series which we know to be convergent, as

a + ar + ar~ + ar* + ; 0" < 1) (geometric series)

1 + ~~ + -~~ + ^ + J (P > 1) (P series)*j O ~r

or compare the given series with some series which is known to be

divergent, as1 +

l+ l + l+... ; (harmonic series)

1+ + r~ + + (p<l) (P series)

SERIES 349


Solution. Here Mn =,

-n+i=r

Mn+lsJ"-1

= 1

u n [n n

p = lim - = 0,n-oc K

and the series is convergent.


Ji + .yL + Ji +

Solution. Here

10" n + 1

T7=~io~

-

P= lim,

-"'n -* oo J U

and the series is divergent.


1}

1

j

1

,

...

Solution. Here =(2 n _* 1)2 w

*M~' =

J2"n + i,(2 n + 2)*

7i ni i _ (2 r?-

1 ) 2 n _4 r?

2 - 2 n^

w n

~(2ri -f lK2n-f 2)

""

4 n 2-f 6 n -f 2*

,. 4 n 2 2 n1

p lim - = 1,77 , oc 4 w- -h 6 w 4- 2

by the rule in Art. 18. Hence the test-ratio test fails.

But if we compare the given series with the p series, when p = 2, namely,

we See that it must be convergent, since its terms are less than the corresponding

terms of this ;; series, which was proved convergent.

PROBLEMS

Test each of the following series.

..... Ans. Convergent.

23 ,_l + A + .-H . Convergent.*'2 2 2 23 24

3 3 2 3 ;* 34

3. --f

~-~2+ ~~ + -~-^ + . An,s. Divergent.

212 1^13 2 :i

|4 -i 4 . h10. 1 -f -p- 4- -77- -f -rb1 + . 12. ^ -h

--f

~--f~

-f .

f> 10 \t ;> ;j- 5-! ;V

13.;;

-f^ + .7~ir~i7 -fjj.^.j^.jo

+

191. Power series. A series whose terms are monomials in ascend-

ing positive integral powers of a variable, say j, of the form

(1 ) flu + fli J + u L>.r2 + a;ir :{ + -,

where the coefficients uo, i, LS ure independent of a*, is called a

power scries in x. Such series are "of prime importance in the study of

calculus.

A power series in j may converge for all values of .r, or for no value

except x =;or it may converge for some values of x different from

and be divergent for other values.

We shall examine (1) only for the case when the coefficients are

such that . .

Mm (^A = L,

where L is a definite number. To see the reason for this, form the

test ratio (Art. 187) for (\], omitting the first term. Then we have

i<n a,txn a n

Hence for any fixed value of x,

i- A* * I \ ,. /dni 1

p = hm I-* i

j) =xhm \-JL-L

)=xL.

SERIES 351

We have two cases :

I. If L = 0, the series (P will converge for all values of j, since

p = 0.

II. If L is not zero, the series will converge when rL(= p) is nu-

merically less than 1, that is, when x lies in the interval

and will diverge for values of .r outside this interval.

The end points of the interval, called the interval of convergence,

must be examined separately. In any given series the test ratio

should be formed and the interval of convergence determined byArt. 187.

ILLUSTRATIVE EXAMPLE 1. Find the interval of convergence for the series

Solution. The test ratio here is

Hn .1 ft'' A i i- n'2_-L-_L = ~-jc. Also, lim = 1,

U n (/< + lr n . oc (n i- })'2

by Art. 18. Hence p j*. and the series converges when JT is numerically less

than 1 and diverges when .r is numerically greater than 1 .

Now examine the end points. Substituting jr~

1 in (2), we get

1-

r^+ -;

-^ + -,

which is an alternating series that converges.

Substituting x = 1 in (L! i, we get

which is convergent by comparison with the p series (p > 1).

The series in the above example has [ 1, 1] as the interval of convergence. This

may be written 1 - x r 1, or indicated graphically as follows:

ILLUSTRATIVE EXAMPLE 2. Determine the interval of convergence for the series

Solution. Omitting the first term, the test ratio is

1

Also lim r- = 0. Hence the series converges for all values of x.- (2w + l)(2n + 2)

PROBLEMS

For what values of the variable are the following Graphical representations

BCries convergent?of intervals of convergence*

1. 1 + x + z2 + x3 + . Ann. - 1 < x < 1.

3. x -f x44- x9 + x16 + -. Am. - 1< x < 1.

4. z + -V -f -~= + -. A?i. - 1 < x < 1.

V2 V3~

5. l-hx+|~-fr H . Arcs. All valuesof x. ^ Ll~_ UL o

02 04 ^r>-

6. 1 r~-+ r~ rrrH . Ans. All values of 0. "^ L

7. 0-+ ,~ ,

+ . Anx. All valuesof 0.

1 - 3 z51 3 5

- 4 6'

s. - 1 <z< 1.

8v _i_-1 ^

-1_ - -i-J *

I I. X -f- ~r T

~~~r ^ :;

:; "^ ~r *. . |

2 3 2-4 5 2 4 (> < ITJ 5

9. x -f 2 x 2 + 3 x3-f 4 x4 H . Ans. - 1 < x < 1.

10. l_:r + i!-^ + .... -1 gjr gl.

^. r 2 r.T r4

1i x -*

i _ ^i

. . . _ 9 <^ r <J>1 '-I O O . 02 '

J . 03 4 . >4~

' - \ .* - -.Jlfcrf 4-1 ^ O- T*rf

O r 2 'J Wi /I r4

12. x 4- + + + All values.

z 2.l-3x4

,l-3-5z

2 2 21 Q 1 4_

**_L i i

. . .

2 4 24 2 4 - 6 2'

* End points that are not included in the interval of convergence have circles drawnabout them.

SERIES 353

ld - l+l+ f^ + f^ + '"- (a>0) AM. -agjr<a.

92 ->. 32 r 2 42 r3

16. 1 +^ +^ -f i^ 4- - All values.

Lfi l Li

17 '

3+2^3^

+22 . 33

+ 03 . 34H - 6 < x < 6.

18. x + 4 J 2-f 9 x3

-f 16 x4 + -.

19*

,

2 * 2

,

3 J3 4 x*2 -<

1 2 2 2 3 2 2 3 4 2-< 4 5

.r |2 x2

13-

10 i|

100 j|

1000 j*; 23. r + ^' + ^ + ^

192. The binomial series. This important series is

/iv .. , , w(?w 1) o, in(m \}(m 2) ...

(1) 1 + mx + ^ 9J- + ^

l 9 ; 3- ^ H----

i

w (w 1 ) (w ~ 2 ) (?w n + 1 ) w ,

+ ^ +

where m is a constant.

If m is a positive integer, (1 ) is a finite series of m -f 1 terms, since

all terms following that containing xm have the factor m m in the

numerator and vanish. In this case (1) is the result obtained by

raising ! + / to the wth power. If m is not a positive integer, the

series is an infinite series.

Now test (1) for convergence. We have

- l)(m-2) - (m - n + 2) , .

1 -2-3- (w-1)'

, m(m l)(m 2) - - (m n + 2)(m ri + 1) nand ^^ J~

Hence ^ = m-n + 1x = tml

un n V n

Then, since lim (^---

1)= 1, we see that p = x,n -. oo \ tl /

and the series is convergent if x is numerically less than 1, and di-

vergent when x is numerically greater than 1.

Maclaurin's Series, Art. 194, implies the following statement.

Assuming that ra is not a positive integer and that \x\ < 1, the

value of the binomial series is precisely the value of (1 + x)m

. That is,

(2) (1 + xr = 1 + mx

If m is a positive integer, the series is finite and equals the value

of the left-hand member for all values of x.

Equation (2) expresses the special binomial theorem. We may also

write

(3) (a + b)m = a r

"(l + x)w

, if x = --a

Thus the left-hand member of (3) may also be expressed as a

power series.

Examples of approximate computation by the binomial series are

given below.

ILLUSTRATIVE EXAMPLE. Kind v (;w approximately, using the binomial series.

Solution. The perfect square nearest to G30 is 625. Hence we write

\ GIJO = \ (52~5~T5 = 25( 1 +]I -;)*

Now write out (2) with ?w J. The result is

(1 -f ar)i = 1 4- ?, .r-

jj,

.r-' f ,V -r'( -

T ? ftx 4 + %

In this example, j =1 J r>

~ 0.008. Hence

(1 -f1

.\f| )i

-1 -f ().()04 ~ O.OOOOOS -f ().0000()00:r2 H---- .

(4) 25(1 -fT ^2 = 25 -f O.i - 0.0002 4 0.0000008 =25.099S01 (to the nearest

figure in the sixth decimal place). Ans.

The series in (4) is an alternating series, and the error in the answer is less

than 0.0000008.

193. Another type of power series. We shall frequently use series

of the form

(1) 60 + MJT - a) + 62 (.r- a) 2

H-----h b,,(x- a)

nH----

in 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 test-ratio test to (1), as in Art. 191. Then, if

lim ^-i = M,

SERIES 355

we shall have, for any fixed value of j,

= lim ^-^ = (x a)M.

We have two cases :

I. If .17 = 0, series (1) is convergent for all values of x.

II. If M is not zero, series (1) will converge for the interval

A convergent power series in .r is adapt ed for computation when x

is near zero. Series (1), if convergent, is useful when x is near the

fixed value a, given in advance.

ILLUSTRATIVE EXAMPLE. Test the infinite series

1 /, ] , f( -r - 1>" (*~- 1"'

, . . .i-

tor i M^

-\-

for convergence.

Solution. Neglecting the first term, we have

n + 1

Also, lim(

- )= 1.

w - oc \ N -f 1 /

Hence |p|=

\x-

11

, and the seri(\s will converge when x lies between and 2.

The end point jr '2 may he included.

PROBLEMS

1. Using the binomial series, show that

1 + jr

'

Verify the answer by direct division.

Using the binomial series, find approximately the values of the fol-

lowing numbers.

2. V98. 5. Yii5. 8. -rp 11.

V990^ 1 1

4 --7" -x4.^/630. V412 V30 \16

For what values of the variable are the following series convergent ?

/~ t 1 \ 2 /-.. i i \a /,.- i 1 ^4

-. Ans. 2 < x < 0.

15. (jr- 1) +

16. 2(2.

23 4

(T- I)2

jfr-_l)

3

{

(x-l)X/2 V3 V4

^T < 2.

All values.

_

17. 1^(^-2)^ + ^

18. 1 -2(2jr- U) 4-3(2 jr - H) 2 ~ 4 (2 ^ - 3 )-{ H .

19.jr - 3

, (.r- 3)'

2

i - a; ''./.. i ..'.-

' 'i

2 - 3- 3 3 :t

(x - 3) 4

4 - 34

CHAPTER XX

EXPANSION OF FUNCTIONS

194. Maclaurin's series. In this chapter the question of represent-

ing a function by a power series, or, otherwise expressed, developing

(expanding) the function in a power series, will be discussed.

A convergent power series in jc is obviously a function of j for all

values in the interval of convergence. Thus we may write

(1) /(*")= 00 + J

J + a*r ~ + + a,,/" + --.

If, then, a function is represented by a power series, what must be

the form of the coefficients a , fli, -, (in, etc.? To answer this ques-

tion we proceed thus :

Set x = in (1). Then we must have

(2) /(())- oo.

Hence the first coefficient a in (1) is determined. Now assume that

the series in (1) may be differentiated term by term, and that this

differentiation may be continued. Then we shall have

(f(jr) = a, + 2 a 2r + 3 a ;<:r

2H h nanx

n ~ l

H

(3)J /"(j) = 2 a 2 + 6 flsr H h n(n - 1 )ar,j

n ~ 2H

[f'"(x)= 6 a.-H h n(n -l)(n- 2)anx

n -*-\

etc.

Letting x = 0, the results are

(4) /'(O) = ai, /"(O) = 12 02, /;//

(0) = [3 a3 , , /(n)(0) = lwon .

Solving (4) for a\ t a, -, On, etc., and substituting in (1 ), we obtain

(4) /(x) - /(O) + /'(o) i + /- (o)|+ - + / (w)

(o) g-f - -

-.

This formula expresses /(x) as a power series. We say, "the function

/Or) is developed (or expanded) in a power series in x." This is

Maclaurin's series (or formula).*

* Named after Colin Maclaurin (1698-1746), and first published in his"Treatise of

Fluxions" (Edinburgh, 1742). The series is really due to Stirling (1692-1770).

357

It is now necessary to examine (4) critically. For this purposerefer to (G), Art. 124, and rewrite it, letting a = 0, 6 = x. The result is

(5) f(x) =/(0; +

IT

where # ^/'r

"(^i) r (0 < jj < jj

The term R is called the remainder after n terms. The right-handmember of (5; agrees with Maclaurin's series (A) up to n terms. If

we denote this sum by N/,, then (5) is

Sn + K, or

Now assume that, for a fixed value x ~ xn, R approaches zero as a

limit when n. becomes infinite. Then Stl will approach /(.r ) as a limit

(Art. 14). That is, Maclaurin's series (4) will converge for x = x

and its value is/Oo). Thus we have the following result :

Theorem. /// order that the series (A) should converge arid represent

the function f(x) it is necessary and sufficient that

(G) lim 7? = 0.n oc

It is usually easier to determine the interval of convergence (as

in the preceding chapter) than that for which ((>) holds. Hut in

simple cases the two are identical.

To represent a function /(.n by the power series (A), it is obviously

'necessary that the function and its derivatives of all orders should be

finite. This is, however, not sufficient.

Examples of functions that cannot be represented by a Maclaurin's

series areIn x and ctn x,

since both become infinite when .r is zero.

The student should not fail to note the importance of such an

expansion as (A). In all practical computations results correct to a

certain number of decimal places are sought, and since the process

in question replaces a function perhaps difficult to calculate by an

ordinary polynomial with constant coefficients, it is very useful in sim-

plifying such computations. Of course we must use terms enough to

give the desired degree of accuracy.

In the case of an alternating series (Art. 188) the error made byStopping at any term is numerically less than that term.

EXPANSION OF FUNCTIONS 359

ILLUSTRATIVE EXAMPLE 1. Expand cos x into an infinite power series and de-

termine for what values of J it converges.

Solution. Differentiating first and then placing x - 0, we get

/(jrt= cos x. /(O) = 1,

f'(x) = - sinx, /'(( ~-0,

f"(x] = - cos x, f'\0) = -I,

/'"(j) = sin J, /'"(O) = 0,

/iv(j-)= COST, /1V((

rr \ f

fv(x )= _ sin jr, /v(0) = 0,

/Vl(j) zr: _ COS J, / vi (0) = -1,

etc., etc.


(7) C08T=l-|-f^

-jj

+ --"

Comparing with Problem G, Art. 191, we sec that the series converges for all

values of J*.

In the same way for sin j,

(8) sin,-,-?!

+ -

t

- ?1 -.

which converges for all values of JT (Problem 7, Art. 191 .

In (7) and (8) it is not difficult to show that the remainder /v approaches zero

as a limit as n becomes infinite, when jc has aw// fixed value. Consider (7). Here we

may write the nth derivative in the form

,,

Hence A = cos x

Now cos(ji 4-)never exceeds 1 in numerical value. Also, the second factor

of R is the nth term of the series

which is convergent for all values of x. Therefore it approaches zero as n become:-!

infinite (see (A), Art. 1S5). Hence (6) holds.

Jim - = 0.,i .* [W

From the above example we see that

r n

Jim -

,i .* [W

Also, on the page preceding, we had

K= fn)(

Hence lim /? = if f(n]

(x\) remains finite when n becomes infinite.

. (0< X, < X)

\n

ILLUSTRATIVE EXAMPLE 2. Using the series (8) found in the teat example, cal-

culate sin 1 correct to four decimal places.

Solution. Here x \ radian ; that is, the angle is expressed in circular measure.

Therefore, substituting x = 1 in (8) of the last example,

Summing up the positive and negative terms separately,

1 = 1.00000 -

-J-= 0.16667

~ - 0.008,'W ,-- 0.00020

y?_ y;_1.008U3 0.16687

Hence sin 1-

1 .008W - 0.16687 = 0.84146

which is correct to five decimal places, since the error made must be less than;

I?that is, less than 0.000003. Obviously the value of sin 1 may be calculated to anydesired degree of accuracy by simply including a sufficient number of additionalterms.

PROBLEMS

Verify the following expansions of functions by Maelaurin's series anddetermine for what values of the variable they are convergent.

1. rr - 1 + .r + ~ + !' + ...+ -1-1-

-|- .... Ans. All values.-

r-'l r-r>

(_ 1 \ " r

2. sin , = ,--+- ---- + 1_1J-:L- + . ... A11 values _

.r-' .r;t

,r4. In U -

j-) - - x - TT- ~ -

j-----

^j-

----.

- 1 ^ JT < 1.

:$' ^

5. arc sin jr = x

6. arc tan j- = r - -f ---- +

A n,s. All values.

8. In (a + J-) = In a -f-- - -^-

4-^-----

a 2 a- 3 aa

(n -

EXPANSION OF FUNCTIONS

Verify the following expansions.

361

(

-4

\412. tan

(

-4+ JT = I -f 2 jc + 2 x- +

13. arc tan - = ^ - x + - 4- -.

j* 2 3 5

or LS

16.1ncos J-=-^-^-g..,Find three terms of the expansion in powers of JT of each of the fol-

lowing functions.

17. cos j*~

18. sin (.r + 1).

19- e

20. J(r'- c

Compute the values of the following functions by substituting di-

rectly in the equivalent power series, taking terms enough to make the

results agree with those given below.

21. c = 2.7182 -.

Solution. Let x 1 in the series of Problem 1 ; then

(Dividing third term by 3.)

(Dividing fourth term by 4.)

(Dividing fifth term by 5.)

(Dividing sixth term by 6.)

(Dividing seventh term by 7.)

22. arc tan (I) 0.1973 -

; use series in Problem 6.

23. cos 1 = 0.5403 ;use series in (7), Illustrative Example 1.

24. cos 10 = 0.9848 -

;use series in (7), Illustrative Example 1.

25. sin 0.1 = 0.0998 ;use series in Problem 1>.

26. arc sin 1 1.5708 -;use series in Problem 5.

27 sin - ~ 0.7071 ;use series in Problem 2.

4

28. sin O.T> ~ 0.4794 ;use series in Problem 2.

29. <'-' = 1 + li f^'

4 + - - - 7.3890.

30. V? = 1 4- +7^7,

-foJTTj

-f . - l.b'487.

195. Operations with infinite series. One can carry out many of the

operations of algebra and the calculus with convergent series just as

one can with polynomials. The following statements are given with-

out proof.

],et f/o + fl,/4 <*#' + + s" + '

and fro f- b\jr 4 b^jr- 4 * 4 b ns" -[

be convergent power series. Then we obtain new convergent power

series from them as follows :

1. fry adding (or subtracting) term, hi/ term..

(a< ( i bu) 4 (MI i ^1 )*" 4 4 ( a n b n )jrn 4 *

2. #// multiplication, and (jroupnnj terms.

ILLUSTRATIVE K\ AMPLE 1. Computation of logarithm*. From the series (Prob-

lems 3 and 4, Art.. 194)

In (1 4 oO = J-- } J-'-f-

1

,-'" -i *' ^' '

'

In (1- J*)

- - J"-

1 -r'

J -a -r -

\ -r* - '

'

we obt ain, by subt raet ion of corresponding terms, and using (2) ,Art . 1

,the new series

(1) lnL, 2lJr+ .,. + UM!,. + -,

This series converges when',

j*' ^ 1.

To transform (1) into a form better adapted to computation, lot A be a positive

number. Then, if we set

1 , 1 + JT __ A* -f 1

/2) x = -:

whence;

r:

then|

j|< 1 for all values of .V. Substituting in (l\ we get the formula

x- o 1 1__1__4-1_ i_ 4- 1(3) In (A' + 1) = In .\ + ~

iTy"^+

3 (2 A' + 1 )' 5 (2 :V 4- 1> 5J

This series converges for all positive values of AT

, and is well adapted to com-

putation. For example, let A" = t. Then

In (A + 1) = In 2, r - =|

Substituting in (3), the result is In 2 = 0.69315.

Placing A* = 2 in (3\ we get

1,1 1

4- 1.09861

It is only necessary to compute the logarithms of prime numbers in this way,

the logarithms of composite numbers being then found by using formulas (2),

Art. 1. Thus, l n 8 = In 2 ' = 3 In 2 = 2.07944

In 6 = In 3 + In 2 - 1.791 76- -.

All the above are Xapicrian, or natural, logarithms, that is, the bust* is

r 2.71828 . If we wish to find /ir/f/f/s's, or common, logarithms, where the base

10 is employed, all we need to do is to change the base by means of the formula

, In 11

'<>^

In the actual computation of a table of logarithms only a few of the tabulated

values are calculated from series, all the rest being found by employing theorems

in the theory of logarithms and various ingenious devices designed for the purpose

of saving work.

ILLUSTRATIVE FAAMPLE 2. Find the power series for CT sin x.

Solution. From the series

sin x = x - ^ -f--^

---- Problem 2, Art. 1 94

and <*= * +* + ? + ? + +fSi

?robl"m ]'AH ' * 94

we obtain, by multiplication,

c* sin x = x~f-x 2 -f^-~-f- terms in x'5 etc. An.

3. By division. A special case is shown in the example below.

ILLUSTRATIVE EXAMPLE 3. Find the series for sec x from the series for cos x

(see (7), Art. 194).X"' X* X

(4) cos x = 1~n

+|jj

~-^+ ' '

Solution. From the formula sec x = , we see that we have to carry through

the division of 1 by the series (4). This is best done as follows.

Write (4) in the form cos x = 1 - z, where

r 2 r \ 7-,r<. __ __i _ __ ...lo; "

2 24 720

Then

(6) sec x =-^

= 1 + 2 + z'2 + 23 + ' ;

if!

2!< 1 (Problem 1, Art. 193).

364 DIFFERENTIAL AND INTEGRAL CALCULUSFrom (6), we have the series

T* x r>

z4"

~24

"*"" terrns f higher degree,

Substituting in (6), the result is

sec x = 1 -f - x - + x4-f rr + . .

.. .4ns.

PROBLEMS

Given In 2 - 0.69315, In 3 = 1.09861, calculate the following naturallogarithms by the method of the example above.

1. In 5 = 1.60944. 3. In 11 -2.397902. In 7 -1.94591. 4. In 13 - 2.56495.

Work out the following series.

5. c ' cos / = 1 / + ;{

/'<_ ii

/4 _j_. . ,

f

6.(

'X

=1 + 2^-4- j-2 +? ^ + 55^

7 JL55JL 1-1 _ 1 .2 __ J_ 3 jilL 4

10. r' tan .r = JT + .r2 + ~ j* -f

-.r

4-f . . .

-

1 1. f r sec j- 1 j- 4- j-2 - "

.r< -fI ^4 _j_

. . %

12. c 2 sin 2 /-

i> /-

/2 - ~

/.'* -f-5

/4 + . .B>

13. (1+ x) cos V.7 = 1 + ia- - 11 j.2 + ^i. ^ _ _11_

2 24 ^720 8064

14. (1 -f 2 j*) arc sin jc = x -f 2 r'2 -f-

j-a

-f-

.r4

-f-6 3

"

15. Vl -JT arc tan j- = j- - 1 ^ - 11 ^ -f A ^ + . . .

^4 48

16. Vl - tan x = l--x--x 2 -~^-^L.r^^X ^ ^

17. Vsecx = 1 + 7 J 24-~

J*4 + -

.

4 96

.-

1+smx 2 b 12

20. V4 + sin </>= 2

16 2;>b 6144

'

For the following functions find all terms of the aeries which involve

powrers of x less than j*

5.

-*-. 24. \/\\ -f c J

.

21. c r> smx.

22.<*cosjVI 25.^jy^,VI 4- xsin JT

23.x 26. V5 - cos x.

196. Differentiation and integration of power series. A convergent

power series

(1) oo + aiJ + a 2x2 + aa.r

:{

-f + a nxn +

may be differentiated term by term for any value of x within the

interval of convergence, and the resulting series is also convergent.

For example, from the series

we obtain, by differentiation, the new series

Both series converge for all values of x (see Problems 6 and 7,

Art. 191).

Again, the series (1) may be integrated term by term if the limits

lie within the interval of convergence, and the resulting series will

converge.

ILLUSTRATIVE EXAMPLE 1. Find the series for In (1 + x) by integration.

Solution. Since -r- In (1 -f x) -- we havedx 1 -f x

(2) ln(l -f x) = I r/() L ~T X

NOW T-T- = 1 ~ X + * 2 ~ X * + ^ -

when |x|<l (Art. 192). Substituting in (2) and integrating the right-handmember term by term, we obtain the result

In (1 -f ji = x - \ x'2 -f 3 x :i -I .T

4-f .

This series also converges when \x\ < I (see Problem 2, Art. 191).

ILLUSTRATIVE EXAMPLE 2. Find the power series for arc sin x by integration.

Solution. Since -- arc sin x.

we havedx \ i

_ X 2

dxr f dx

.'f, \l-j-l'

By the binomial series ((2j, Art. 192;, letting /// J, and replacing x by x2,

we have

1

2-4-6'

This series converges when \x\ < 1. Substituting in (3i and integrating term byterm, we get

an- sin , = T -, -,

This series converges also when \x\-

I (see Problem 8, Art. 191).

By this scries, the value of TT is readily computed. For, since the series con-

verges for values of / between 1 and + 1, we may let x = , giving

or TT = .'{.1415 -.

Evidently we might have used the series of Problem (>, Art. 194, instead. Both

of these series converge nit her slowly, but there are other series, found by more

elaborate methods, by means of which the correct value of TT to a large number of

decimal places may easily be calculated.

ILLUSTRATIVE EXAMPLE ,'{. Using series, find approximately the value of

Isin jr-dx.

Solution. Let r - x ?. Then

sin z- z -

rr +fp- * * ' Problem 2, Art. 194

jtt flOHence sin x 2 = x'2 - r~ + 'rr

' '

%lii 1A

r !, r 1 / r (i TU)

\ ,. ,

and / sin x-dx = I

(a*

2r^ -f V

)dx, approximately,

= I^1_ _

4. --I1

= 0.3333 - 0.0238 -f 0.0008

= 0.3103. Aws.

PROBLEMS

1. Find the series for arc tan x by integration.

2. Find the series for In (1 JT) by integration.

3. Find the series for sec 2.r by differentiating the series for tan .r.

4. Find the series for In cos .r by integrating the series for tan .r.

Using series, find approximately the values of the following integrals.

i~~~"' -4"-s'- 0.0295.

0.185. 10. f1

c ''dx.Jo

7. fVln (1 -f jr)djr. 0.0(528. 11. fin U 4 V7 ) (/.r.

/o A)

i

8. f(

',- 0.1815. 12. fVsin V7'</.r.Jo -\/i _ j.'J Jo

13. fVl - .H <Lr. 14. ( 'r J cos Vr f/.r. 15. / \/2 - sirTx dr.Jo ./o -A)

197. Approximate formulas derived from Maclaurin's series. Byusing a few terms of the power series by which a function is repre-

sented, we obtain for the function an approximate formula which

possesses some degree of accuracy. Such approximate formulas are

widely used in applied mathematics.

For example, taking the binomial series ((2), Art. 192), we maywrite down at once the following approximate formulas.

First <i}>j)}(>j'ini(ition Second dj>proriniatioH

(1 + x}m

1 + mx 1 + mx + \ m(w 1 )x- ;

_ = 1 mx = 1 mx + 2 'm(m l)x2.

In these \x\ is small and m is positive.

Again, consider the sine series> r

X X'*

Then

(2) sin x x,

' ''___sin x x ~

etc.

(3)

are approximate formulas. Let us examine the first.

In the series in (I), assume values of x such that the terms de-

crease. If the first term only is retained, the value of the remainingseries is numerically less than its first term j :{

(Art. 188). That is,

sin x = x, with|

error |< | x*\.

We may inquire, For what range of values of x will (2) hold to three

decimal places? Then "|

< 0.0005,

that is, LH < V0.003 < 0.1443 rad.

We then conclude that formula (2) is correct to three decimal placesfor values of x between 0.144.'* and +0.1443, or, in degrees, for

values between - 8.2 and + 8.2.

PROBLEMS

1. How accurate is the approximate formula sin x x when(a)jr = 30? (b).r-60 ? (c).r~90?

6

Am. (a) Error < 0.00033; (b) error < 0.01 ; (c) error < 0.08.

2. How accurate is the approximate formula cos jr = 1 ~ when(a) jc - 30? (h) JT = (JO ? (c) .r = 90? 2

Am. (a) Error < 0.0032; (}>) error < 0.05; (c) error < 0.25.

3. How accurate is the approximate formula e~ f = 1 jc when

(a) j- = 0.1? (b) .r = 0.,r>?

4. How accurate is the approximate formula arc tan x = x ~ when

(a)jr=0.1? (h)j^ = 0.5? (c) jc-

I ?3

Am. (a) Error < 0.000002 ; (b) error < 0.006; (c) error < 0.2.

5. How many terms of the series sin .r .rj^-

-f- must be

taken to give sin 45 correct to five decimals? LL L2- .477.8. Four.

jc~ x4

6. How many terms of the series cos j- = 1 ~irr + r

'** must be

taken to give cos 60 correct to five decimals? '- I

7. How many terms of the series In (1 -f x} = x ~-f- ^ - must

be taken to give log 1.2 correct to five decimals?~

Ans. Six.

Verify the following approximate formulas:

10. r~" cos = 1 - 6 + ~- 14. fare sin xdx = C -f -f ~o ' 2 24

11. cosVj(/j'=r+ x-~+- 15. (>sin0(/0= C -f -f/ 4 ^2 */ 23

EXPANSION OF FUNCTIONS 3(59

198. Taylor's series. A convergent power series in x is well adapted

for calculating the value of the function which it represents for small

values of x (near zero). We now derive an expansion in powers of

x a (see Art. 193), where o is a fixed number. The series thus

obtained is adapted for calculation of the function represented by it

for values of x near o.

Assume that

(1) f(x )= fr + bi(j

- a) + b>2(Jc- a) 2 + + bn (x

- a)n + ,

and that the series represents the function. The necessary form of

the coefficients &( >, 61, etc. is obtained as in Art. 194. That is, we dif-

ferentiate (1) with respect to r, assuming that this is possible, and

continue the process. Thus we have

J'(x) = bi+2 6L>U - a] H---- + nbn (jr- a)

n "' + ,

/"(*) = 2 6,> + - + n(n - 1 )M-r - a)"~- + ,

etc.

Substituting x a in these equations and in (1 ), and solving for

bo, bi, bi>, ,we obtain

6 =/(o), 6,=/'(d). k

Replacing these values in (1 ), the result is

(B) /(x) - /(a) + /'(a)^ + /"(a) - +

The series is called Taylor's series (or formula).*Let us now examine (B) critically. Referring to (G), Art. 124, and

letting b = x, the result may be written thus :

(2) /(or) =/(o)

where

The term R is called the remainder after n terms.

Now the series in the right-hand member of (2) agrees with

Taylor's series (B) up to n terms. Denote the sum of these terms

by Sn . Then, from (2), we have

f(x) = Sn + R, or

* Published by Dr. Brook Taylor (1685-1731) in his"Mpthoclus Incrementorum "

(London, 1715J.


Now assume that, for a fixed value x = XQ, the remainder R ap-

proaches zero as a limit when n becomes infinite. Then

(3) lim Sn =/0o),n - w

and (B) converges for x xo and its value is /(.TO).

Theorem. 77^ infinite series (B) represents th-e function for those

values of j, and those only, for which the remainder approaches zero as

tlie number of terms increases without limit.

If the series converges for values of x for which the remainder does

not approach zero as // increases without limit, then for such values

of x the series does not represent the function f(x).

It is usually easier to determine the interval of convergence of the

series than that for which the remainder approaches zero; but in

simple cases the two intervals arc identical.

When the values of a function and its successive derivatives are

known and finite for some fixed value of the variable, as jr a, then

(B) is used for finding t he value of the function for values of x near a,

and (B) is also called the expansion off(x) in the neighborhood of x a.

iLLrsTKATIVK EXAMPLE 1. Expand In x in powers of (x 1 ).

Solution. f(jr )= In jr, /( 1 )

- 0,

f'(x] =-, /'(D = l,x

/'V) =-^. /"(1) = -1,

/"V)=J;.

/"'(I) = 2,

etc., etc.

the ririnity

ec., ec.

Substituting in (B), In x - .r - 1-

J(r- 1 )

2 + ^(x-

1)''-

. Ans.

Tliis converges for values of x between and 2 and is the expansion of In x in

ririnity of x 1. See Illustrative Example, Art. 198.

ILLUSTKATIX E EXAMPLE 2. Expand cos .r in powers of / x j jto four terms.

Solution. Here f(x) cos x and a ~ - Then we have4

f(x) = cos .r, M~|

= =\4/ \ 2

.

etc., etc.

The series is, therefore,

v 2 N 2

The result may be written in the form

To check this result let us calculate cos 50 . Then x 5 expressed in radians,> ., 4

or x _ 1L - 0.08727, (x - -\- 0.00762, (x - ~Y~- (UHMMJG. With these values4 \ 4 / \ 4 /

the series ahove gives cos 50 = 0. (54278. Five-place tables give cos 50 0.64279.

199. Another form of Taylor's series. In (), Art. 19S, if we re-

place a by x() and let .r- a = h, thut is, .r = a f- // = .ru + //, the result is

In this second form the new value of /(.r) when .r changes from j\> to

Jo + h is expanded in a power series in //, the increment of .r.

ILLUSTRATIVE EXAMPLE. Expand sin r in a power series in h when .r changes

from Jo to /o 4- h.

Solution. Here f(.r>- sin x, and f(x (} 4/0 " sin (j,, 4 h). Differentiate and ar-

range the work as below.

f(x) sin j, /(J ( >)~ sin JTO ,

/'(jri= cos .r, f'(xu) cos .r () ,

/"(ji sin x, / "(J<>) = sin X,

etc., etc.

Substituting in (C), we obtain

sin (xo -f h i= sin x 4- cos x ~ sin j ( , cos j ( 4- *. Am.

PROBLEMS

Verify the following series hy Taylor's formula.

1.

2. sin x = sin a 4- (x a) cos a^ r;

sin a'r

i

qcos a 4-

3. cos a- = cos a (x a; sin a '

"-~ cos a 4- 7 sin a 4-

x2 x3

5. cos (a -f x) = cos a x sin a nr cos a + pr sin a + .

6. tan (x + h) = tan x -f- /* sec 2 x + /^ sec 2 x tan x + .

7. (x + A)" = ^ -f nxn -*h + n(n ~ 1} x"- 2A 2

8. Expand sin x in powers of f x \ to four terms.

(-!)'12

Ans. sin r = _

9. Expand tan x in powers of Lr - ~\ to three terms.

Am. tan .r = 1 + 2 a- - + 1> j - -f

10. Expand In r in powers of (r-

2) to four terms.

11. Expand <" in powers of (-r-

1) to five terms.

12. Expand sin f- -f jr \ in powers of x to four terms.

13. Expand ctn f~

-f .rjin powers of x to three terms.

200. Approximate formulas derived from Taylor's series. Such

formulas are provided by using some terms of series (B) or (C).

For example, if /(.r)= sin j, we have (see Problem 2, Art. 199)

(1) sin y ~ sin a + cos a(x a]

as a first approximation.A second approximation results by taking three terms of the series.

This is

fy _ Q \ 2

(2) sin x = sin a + cos a(x a) sin a -^

From (1), transposing sin a and dividing by x a, we get

rinar- Sina==COBOj

j a

Since cos a is constant, this means that (approximately)

The change in the value of the sine is proportional to the change in

the angle for vahics of the angle near a.

Formula (3) illustrates interpolation by proportional parts.

ILLUSTRATIVE EXAMPLE 1. For example, let a = 30 = 0.5236 radian, and sup-pose it is required to calculate the sines of 31 and 32 by the approximate formula

(1). Then, since x - a = 1 = 0.01745 radian,

sin 31 = sin 30 + cos 30 (0.01745)= 0.5000 + 0.8660 x 0.01745= 0.5000 + 0.0151 =0.5151.

Similarly, sin 32 = sin 30 + cos 30 (0.03490) = 0.5302.

These values by (1) are correct to three decimal places only. If greater ac-

curacy is desired, we may use (2).

Then sin 31 = sin 30 + cos 30 (0.01745) - smj*

'

^0.01745)2= 0.50000 + 0.0151 1

- 0.00008~

= 0.51503.

sin 32 = sin 30 + cos 30M

(0.03490) -sin

.j* (0.03490)'-= 0.50000 + 0.03022 - 0.00030

"

= 0.52992.

These results are correct to four decimal places.

From (C) we derive approximate formulas for the increment of

/(j) when x changes from jr(} to Jo -f h- For, transposing the iirst

term of the right-hand member, we get

(4) /(J + h) -/(Jo) = /'(ju )// +/"(:ro) j~ + --.

Ik

The left-hand member expresses the increment of /(j) as a powerseries in the increment of JT (=//).

From (4) we derive, as a first approximation,

(5)

This formula was used in Art. 92. For the left-hand member is the

value of the differential of f(x) for x = J and AJ h.

As a second approximation, we have

(6) /(Jo + h)- /(J )

-/' (J )// + /"(J )

~<u

ILLUSTRATIVE EXAMPLE 2. Calculate the increment of tan x, approximately,when x changes from 45

tJ

to 46, by (5) and by (6).

Solution. From Problem 6, Art. 199, if x = x (] ,we have

tan (x -f h} tan x 4- sec 2 x tt h -f sec 2x<, tan x (} h 2 H---- .

In this example x ()= 45, and tan x()

= 1,sec2 XG = 2.

Also, fe = 1 expressed in radians = 0.01745. Hence, by (5),

tan 46 - tan 45 = 2(0.01745) = 0.0349 ;

by (6), tan 46 - tan 45 = 0.0349 + 2(0.()1745)2 = 0.0349 4- 0.0006 = 0.0355.

From the second approximation we get tan 46 = 1.0355, which is correct to four

places of decimals.


PROBLEMS

1. Verify the approximate formula

In (10 + *) = 2.303 +

Calculate the value of the function from this formula and compareyour result with the tables, when (a) x = 0.5 ; (b) x = 1.

Am. (a) Formula, 2.253; tables, 2.251.

(b) Formula, 2.203 ; tables, 2.197.2. Verify the approximate formula

sin + x =0.5 + 0.8660 x.\b /

Use the formula to calculate sin 27, sin 33, sin 40, and compare yourresults with the tables.


tan K- + x ) = 1 + 2 x + 2 x 2.

Use the formula to calculate tan 46, tan 50, and compare yourresults with the tables.


cos x cos a (x a) sin a.

Given cos 30 = sin 60 = 0.8660,

cos 45 = sin 45 = 0.7071,

and cos 60 = sin 30 = 0.5,

use the formula to calculate cos 32, cos 47, cos 62, and compare yourresults with the tables.

ADDITIONAL PROBLEMS

r\1. Given the definite integral /

x5 In (1 + x)dx.*) o

(a) Obtain its value by series correct to four decimal places. Ans. 0.0009.

(b) Obtain its value by direct computation and compare with the ap-

proximate value derived in (a).

(c) Prove that if n terms of the series are used in the computation the

1error is less than

x2. Given f(jr) c - cos -

7)

(a) Show that f(

*\x) = -} f(x).

(b) Expand f(x) in a Maclaurin series to six terms.

(c) What is the coefficient of x 12 in this series? Ans.64 [12*

CHAPTER XXI

ORDINARY DIFFERENTIAL EQUATIONS*

201. Differential equations order and degree. A differential equa-

tion is an equation involving derivatives or differentials. Differential

equations have been frequently employed in this book. The illus-

trative examples of Art. 139 afford simple examples. Thus, from the

differential equation (Illustrative Example 1)

g-2i.we found, by integrating,

(2) y = x~ + C.

Again (Illustrative Example 2 ) , integration of the differential equation

(3) - ? ~v } dx y

led to the solution

(4) x2 + y~ - 2 C.

Equations (1 ) and (3) are examples of ordinary differential equations

of the first order, and (2) and (4) are, respectively, the complete

solutions.

Another example is

(

This is a differential equation of the second order, so named from the

order of the derivative.

The order of a differential equation is the same as that of the deriva-

tive of highest order appearing in it.

The derivative of highest order appearing in a differential equa-

tion may be affected with exponents. The largest exponent gives the

degree of the differential equation.

Thus, the differential equation

(6) y"2 = (1 + ?/'

2)3

,

* A few types only of differential equations are treated in this chapter, namely, such

types as the student is likely to encounter in elementary work in mechanics and physics.

375


where y1 and y" are, respectively, the first and second derivatives of

y with respect to x, is of the second degree and second order.

202. Solutions of differential equations. Constants of integration. Asolution or integral of a differential equation is a relation between the

variables involved by which the equation is satisfied. Thus

( 1 ) y = a sin x

is a solution of the differential equation

For, differentiating (1;,

(3) -7^ = a sin x.

Now, if we substitute from (1) and (3) in (2), we get

a sin x + a sin x = 0,

and (2) is satisfied. Here a is an arbitrary constant. In the samemanner

(4 ) y = b cos x

may be shown to be a solution of (2) for any value of b. The relation

(5) // fj sin r+c>2 cos x

is a still more general solution of (2). In fact, by giving particularvalues to c\ and 2 it is seen that the solution (5) includes the solu-

tions (1) and (4).

The arbitrary constants c\ and c>2 appearing in (5) are called con-

stants of integration. A solution such as (5), which contains a num-ber of arbitrary essential constants equal to the order of the equation(in this case two), is called the complete or general solution.* Solutions

obtained therefrom by giving particular values to the constants are

called particular solutions. In practice, a particular solution is ob-

tained from the complete solution by given conditions to be satisfied

by the particular solution.

ILLUSTRATIVE EXAMPLE. The complete solution of the differential equation

is y = C] cos x + C'2 sin x (see (5) above).

Find a particular solution such that

(2) y 2, y' = 1, when x = 0.

* It is shown in works on differential equations that the general solution has n arbi-

trary constants when the differential equation is of the nth order.

ORDINARY DIFFERENTIAL EQUATIONS 377

Solution. From the complete solution

(3) y = c i cos x + r2 sin x,

by differentiation, we obtain

(4) y' = c i sin x + c2 cos x.

Substituting in (3) and (4) from (12), we find c\ = 2, r2 1. Putting these

values in (3) gives the particular solution required, y = 2 cos r sin j. v4ns.

The solution of a differential equation is considered as having been

effected when it has been reduced to an expression involving integrals,

whether the actual integration can be effected or not.

203. Verification of the solutions of differential equations. Before

taking up the problem of solving differential equations, we show howto verify a given solution.

ILLUSTRATIVE EXAMPLE 1. Show that

(1 ) y = c\x cos In x -f c?x sin In x -f x In x

is a solution of the differential equation

(2) x^-4^2 W = x,nx.

Solution. Differentiating (1), we get

(3) ^ = (r2- d) sin In x -f (r^ -f o ) cos In .r -f In JT -f 1,

dx

Substituting from (1), (3), (4), in (2), we find that the equation is identically

satisfied.

ILLUSTRATIVE EXAMPLE 2. Show that

(5) y2 - 4 x =

is a particular solution of the differential equation

(6) *2/'2 -1 = 0.

Solution. Differentiating (5), the result is

2yy' 2 = 0, whence y' = -

Substituting this value of y' in (6) and reducing, we obtain 4 x - y2 = 0, which

is true by (5).

PROBLEMS

Verify the following solutions of the corresponding differential equa-tions.

Itifferentitil equations Solutions

(r jc

r <7r

'

4- r,

6 . f >l:Jl + ., 'Ht _j.,,

= o. ry = 2 f - 3 r '.

f/.TJ

(/J*

7. ---f 4 .S r_r (). rr r, COS (2 t -f <* 3 ).

8 -

(/

/X- r> '17 + 1:i J

" = ;5;) - *" = r" cos - * + 3 -

/* a/

9. // y^ (

10. .r?/ ^ 4-

11. ^-^'

<iv 1 -f r- 1- rr

12. ''---f 4 ,s = 8 t. * = 2 sin 2 t + cos 2 / + 2 /!.

<//-

13 .<!lll -z^- *

ll= r 2

'. //= r,r*' -f ce * -

J f 2'.

</.r* a.r

14.fll

_|_ <) j. 5 cos o/. j- = cos 2 / + 2 cos 3 / + 3 sin 3 /.

(//2

15> 1^ 4. 9 j. = 3 Cos 3 /. x = c, cos 3 / -f c 2 sin 3 / 4- \ t sin 3 /.

a/*

16. ^ -f .r// = J3//3

- A -J-'

24- 1 -f- re"',

(/.r*

/r

204. Differential equations of the first order and of the first degree.

Such an equation may be brought into the form

(A) Mdx


in which M and .V are functions of x and y. The more common dif-

ferential equations coming under this head may be divided into

four types.

Type I. Variables separable. When the terms of a differential

equation can be so arranged that it takes on the form

(1) /(J)rf.r+Fu/)rf// = 0,

where /(j) is a function of JT alone and F(y) is a function of \i alone,

the process is called separation of the variables, and the solution is

obtained by direct integration. Thus, integrating (1 ), we get the gen-

eral solution

where c is an arbitrary constant.

Equations which are not given in the simple form (1) may often

be brought into that form by means of the following rule for separat-

ing the variables.

FIRST STEP. Clear of fractions; and if the equation involves deriva-

tives, multiply through by the differential of the independent variable.

SECOND STEP. Collect all the terms containing the same differential

into a single term. If the equation, then takes on, the form

XYdx + -Y'y'(/i/ = 0,

where X, X' are functions ofx alone, and >', Y' are functions of y alone,

it may be brought to the form (1) by dividing through by X'Y.THIRD STEP. Integrate each part separately, as in (2).

ILLUSTRATIVE EXAMPLE 1. Solve the equation

dy __ 1 4- j/'

2

dx (1+ x*)jry

Solution. First Step. (1 4- x'2)xydy =(14- y'*)dx.

Second Step. (1 -\- y'2)dx x(l 4- x'2 )ydy ~ 0.

To separate the variables we now divide by o?(l 4- x 2j(l 4- y

2), giving

dx___ydy _x(\ 4-x*) 1 +y*

In x - J In (1 4- x 2)-

\ In (1 4- y2)= C,

In (1 4- x 2)(l 4- y*) = 2 In x - 2 C.

380 DIFFERENTIAL ANT) INTEGRAL CALCULUS

This result may be written in more compact form if we replace 2 C by In f,

that is, give a new form to the arbitrary constant. Our solution then becomes

In (1 + T*)(\ + y'2)= In j 2

4- In c,

In (1 -f r')fl -f y-) - In ex'2,

(1 4 Jr-)(l -f r') - rj 2. .4ns.

ILLI STKATIVE EXAMPLE 2. Solve the equation

Solution. Firat Step. as dy -f 2 a // djc sydy.

Second Step. 2 ay dx + jcla -y>dy = 0.

To separate the variables we divide by ry.

x y~

Third Step. 2 f -\-aC^~ Cdy = C,

a In j -f a In // ?/ C,

r/ In j'^/ =.- r -f ?/,

By passing from logarithms to exponentials this result may be written in theform

orc

Denoting the constant f hy r, we get our solution in the form

y

x'2y = ce. Ans.

Type II. Homogeneous equations. The differential equation

(A) M dx + N dy =

is said to be homogeneous when 717 and .V are homogeneous functions

of * and // of the same degree.* Such differential equations may be

solved by making the substitution

(3) y = rx.

This will give a differential equation in r and x in which the vari-

ables are separable, and hence we may follow the rule under Type I.

* A function of x and y is said to bo homogeneous in the variables if the result of replac-

ing .r and // by \jc and Xj/ (X being arbitrary) reduces to the original function multiplied bysome power of X. This power of X is called the degree of the original function.

In fact, from (A) we obtain

--Also, from (3),

dy dv .

<5) l= rdr*+

v-

The right-hand member of (4) will become a function of v only when

the substitution (3) is used. Hence, by using (5) and (3), we shall

obtain from (4)

(6) *g + t>=/(p).

and the variables x and v may be separated.

ILLUSTRATIVE EXAMPLE. Solve the equation

'+x'^ = ^^.dx dx

Solution. y2 dx -f (x-- xy }dy = 0.

Here M y2

,N = x'2 xy, and both are homogeneous and of the second degree

in or and y. Also we haveay _ y*

dx xy x 2

Substitute y = t?x. The result is

dx 1 v

or r dx -f x( 1 v)dr = 0.

To separate the variables, divide by rx. This givea

,

In x -f- In f - v T,

In rx = C -f ,

But v % Hence the complete solution is

X y

y = cex

.

PROBLEMS

Find the complete solution of each of the following differential equa-

tions.

1. (2 + y)dx-

(3- x)dy = 0. Ans. (2 -f y)C3

-x) = c.

2. xy dx - (1 -h ^ 2)d?/ = 0. c?/ = 1 + x 2

.

3. x(x -f 3)dy - i/(2 x + 3)dx = 0. t/= cx(x -f- 3).

4. Vl 4- f'2 dy Vl y'2 dx 0. Am. arc sin y = In c(x 4- Vl 4- x2

).

5. dp 4- p tan dO = 0. p = r cos 0.

6. (1 j-jd?/ y2 ds 0. j/ In rfl .rj

= 1.

7. (/- -f 2 y)(Lc + (2 jr- 3 ?/)</?/

= 0. x 24- 4 .n/

- 3 ?/2 - r.

8. (3 jr 4- 5 y)djr 4- (4 y 4- 6 ?/;r/?/= 0. (jr 4- ?/j

2(.r 4- 2 ?/)

= r.

9. 2O 4- y)ds 4- ?/ r/?/ 0.

Aim. \ In (2 r 2-f 2 r?/ 4- //*;

- arc tan (

J ?/

)

= r.

10. (8 // 4-10 jr )ds 4- (5 // 4- 7 .r )r///= 0. f.r 4- ?/ )

J(2 r 4- ?/)'*

= r.

11. (2 *r 4- y)djr 4- <V -f 3 y)(h{ = 0. 2 ./*L>

4- 2 ./*// 4- 3 //-'= r.

12. Vl - 4 /-' dx 4- 2Vl -,s~ f// = 0. .,-Vr^lT' 4- 2 / Vl - .s- - c.

13. 2cf3 c-f I)////' 4- Cl - 2 /r)r/-: - 0. (2 //-

1 )(1 4- 3 c) 3 r^.

14. 2 y r/c - 2 : ds ~ VT^TT^ <r. 1 4- 4 r: - r->- = 0.

15. 0' -f 4 //K/.r 4- 2 jr (///= 0. j-

{

4- (> jr~y = r.

16. (2 j'L>

-f //-' )(/.r 4- (2 sy 4- 3 /r )r///~

0. 2 .r{ + 3 .r//- 4- 3 //'<

= r.

</w 1 4:^L> r-hr

^-i/r'i r^' ^-rr^'18. (3 f 2 //)./ J.r 4- (./'

-2)(/// = 0. 22. (3 s -f //)</./ 4- (.r 4- //)<7//

~ 0.

19. 2(1 4- // )(/j- ( 1~

./)<///= 0. 23. jryi // 4- 2 )r/.r

-(// 4- 1 )r///

= 0.

20. ( 1 -f //).r r/.r-

( 1 -f .r )// (///= 0. 24. f 1 4- /-V// - (1

-//

L')r/.r - 0.

21. (a.r 4- />)^/// //- djr ~ 0. 25. O 2 //>/.r (2 ^* 4-- //)(///= 0.

26. (3 s 4- 2 //)</j- -f j-J// rr o.

27. 3(5 .r -f 3 i/)(ls 4- ( 1 1 ./* -f 5/y )(///

= 0.

28. (.r- 4- //

:>

U/.r -f (2 .r// 4- //L')<///

= 0.

29. 2 // </.r ~ (2 .r ?/)r/// 0.

In each of the following problems find the particular solution which is

determined by the given values of .r and //.

//* jr* .r.

1 + 40-4.^=: o.

31. (.r- 4- //

L'W/.r 2 .r// dy ;

.r = 1, //= 0.

32. .r dy - // J.r ~ Vr" 4- //- (Lr;

j- ~ ?,, // 0.

33. (1 + //-')r///=

// ( /.r ; .r = 2, //= 2.

34. Find the equation of the curve whose slope at any point is equal to

-(l4- -) and which passes through the point (2, 1). Anz. x 24- 2 nj 8.


j, j

;

and which passes through the point (1, 0). Am. y(\ 4- jr) = 1 .r.

Type III. Linear equations. The linear differential equation of the

first order in y is of the form

where P and Q are functions of .r alone, or constants.

Similarly, the equation

(C) ^ + Hx - J,<*y

where // and .7 are functions of// or constants, is a linear differential

equation.To integrate (#), let

(7) y = uz,

where z and u are functions of x to he determined. Differentiating (7),

( ,x,&rf= + ..!j.or a.r </.r

Substituting from (S) and (7) in (#), the result Is

We now determine u by integrating

do,iJ7+ /'" = '

in which the variables jr and ?/ are separable. Using the value of u

thus obtained, we find z by solving

(11) u^ = Q,df

in which j and z can be separated. Obviously, the values of u and z

thus found will satisfy (9), and the solution of (#) is then given by (1).

The following examples show the details.

ILLUSTRATIVE EXAMPLE 1. Solve the equation

Solution. This is evidently in the linear form (B), where

2 *

p = _ an(} Q=(x + l)*.x -f 1

Let y = uz; then . . .*a?y a2 aw

dx dx dx

Substituting in the given equation (12), we get

dz,

du 2 uz, ,

s

' i '\ i -tdx \ dx 1

To determine M we place the coefficient of a equal to zero. This gives

du 2 M

dx, 1 -f x

du 2 dx

u 1 -far

Integrating, we get In if = 2 In (1 -f x) = In (1 -f x)2.

(M) .-.u = (1 -f j) a.*

Equation (KJ) now becomes, since the term in z drops out,

Replacing M by its value from (14),

=<r + 1,dx

Integrating, we get dz = u -f 1 i

(15) s

'

,

Sut>stituting from (15) and (14 ) in y ~ uz, we get the complete solution

y =**i + C(r + l,*. An*.

ILLUSTRATIVE EXAMPLK 2. Derive a formula for the complete solution of (5).

Solution. Solving (10), we have

In u -f f/' dr = In k,

v\ here In ^r is the constant of integration ;

whence u ke 'Pd

*.

Substituting this value of // in (11 ), and separating the variables z and x, theresult is

fc = 2^ p ' e

'dr.

Integrating, and substituting in (7), we obtain

It should be observed that the constant k cancels out of the final result. Forthis reason it is customary to omit the constant of integration in solving (10).

* For the sake of simplicity wt hu\ assumed the particular value zero for the constantof integration. Otherwise we should have M = r(l -f j-)-. But in the work that followsr finally cancels out (See Illustrative Example 2 >


Type IV. Equations reducible to the linear form. Some equationsthat are not linear can be reduced to the linear form by means of a

suitable transformation. One type of such equations is

dx' *

where P and Q are functions of x alone or constants. Equation (D)

may be reduced to the linear form (B), Type 111, by means of the

substitution z = y~~w + 1

. Such a reduction, however, is not necessaryif we employ the same method for finding the solution as that givenunder Type III. Let us illustrate this by means of an example.

ILLUSTRATIVE EXAMPLE. Solve the equation

(16) ^ 4 & = fl In JT- y'2

.

dx x

Solution. This is evidently in the form (/>), where

P^i, Q = a In x, n = 2.j

Let y - uz-, then ^ - u ^ 4- z ^-dx dx dx

Substituting in (16), we get

dz,

du,uz

, ,,,

M T" + - -7- -f~- ~ In x u~z 2

,

dx dx x

(17)J" \r X

To determine u we place the coefficient of z equal to zero. This gives

,

dx x

du _ dx

u x

Integrating, we get In u In x In -,x

(18) u = i-

Since the term in z drops out, equation (17) now becomes

u^- =a In x u'*z*dx

= a \n x uz'z .

dx

Replacing u by its value from (18),

= a\nx .*-,dx x


1 a(m x)2

. rIntegrating, we get

2~adnx^-f 2C

Substituting from (19) and (18) in y = uz, we get the complete solution

_ 1 2*

x f/(lnj)~' + 2ror xy[a(\n t}- -f 2 f'J -f 2 = 0. An*.

PROBLEMS

Find the complete solution of each of the following differential equa-

tions.

I j.

(lli __ 2 // 2 r. An*. //

-- cs'2 - 2 j".

2. JT ^ - 2 /y~ -

J-. ?/= .' + ^' 1>

-

3.di( __ 2 //

-I- 2 r. ?/

= JT -f rr -".

4> j.

(J/ _;j y

- - 2 nj. y = us -f r.r{

.

5.(!ii -

;/rrr - 2 C

r. ?/

= Cr + C(" .

6. ~ - ctn / = 1-

(/ -f 2) ctn t. s = / + 2 + r sin /.

cx 2u 4- 2 j-?/ -1=0.

djr'

jr

'

8. 4 tan / = 2 / 4 /2 tan /, s rr /

L) 4 c cos /.

u/

(ili _ / _ 1 \ >r u c

*-j- ^ r

(f.r j*

12. nx -f 12 //= w n + l

. rj*2?/" 4- A'/

n -djr

13.-^+ i? = cos /

- sin /. 8 cos t -f cc~

14. ^ - x ctn / = fr

(l~ ctn /). 8 = ^'

( + c sin ?


15. x ^3L _ 2 y 4 3 x = 0. 20. ~* - * ctn / 4 csc / = 0.

16. ^ 4 ?/= 2 -f 2 .r. 21. 2

-^-f //

=(JT-

l)r'.

17. .r-~ 4 V = (1 4 -rU". 22. r ~ /y

= ^ cos .r sin x.ajc

*

ax

18.(-^ -

y = 1 - L> j-. 23. N ^ -// -f (.r

2-f 2 .r)?/"

+ 1 =

19. j- ^ -f ?/ -f j-2//-= 0. 24. ^ 4 * tan t = c

'

(tan /- 1 ).

In each of the following problems find the particular solution which is

determined by the given values of x and //.

25. 7^- ^ = /-V ;

x = 1 . ?/= 0. .4 //s. //

=. jr'2(c'

-e).

27. ~ -f ?/ tan jr sec ^;

JT~

0, // 1 . ?/ sin .r cos jr.

djr

28. ^ - -^- = (.r + 1 )' ;jc = 0, y/

= 1. 2 ?/= (r -fl )

4-f (jr -f I)

2,

dj* j" -f 1


, // 4- .r -f-an(j ^.^^.^ passes through the point (1, 0).

sAnn. 2 ?/

= tt.r2 - 2-r- 1.


y n f y ^^ which passes through the point (1 ,1 ). Anx. y(\ -f In x) = 1 .

205. Two special types of differential equations of higher order. Thedifferential equations discussed in this article occur frequently.

g-'-where X is a function of x alone or a constant.

To integrate, first multiply both members by dx. Then, inte-

grating, we have

Repeat the process (r? 1) times. Then the complete solution con-

taining n arbitrary constants will be obtained.

d*i/ILLUSTRATIVE EXAMPLE. Solve -rr = xe*.

dxA

Solution. Multiplying both members by dx and integrating,=* + <:,.

g = ^-e* + c,

Repeating the process,

^ =Jxe*dx -JV dx +Jd dx + C2 ,

or ^ = xr* - 2 <-* + T,x + 0V

j/= Cxe* dx - C r* dx -f fr,x dx -f fC2 dx + C3

= xe1 - c* -f ^TT- + r^x + c --

Hence y = xez - 3 r3"

-f r,x2

4- r2x -f f:t. ^4ns.

A second type of much importance is'

where }' is a function of y alone.

To integrate, proceed thus. Write the equation

d//'= Ydx,

and multiply both members by y'. The result is

y' dy''= r//'^.

But ^/' dx = d//, and the preceding equation becomes

y'dy' = Ydy.

The variables |/ and //' are now separated. Integrating, the result is

The right-hand member is a function of y. Extract the square root,

separate the variables, jr and //, and integrate again. The following

example illustrates the method.

ILLUSTRATIVE EXAMPLE. Solvec~~ f a*y 0.

Solution. HereC

~^- = ^ = a'2y, and hence the equation belongs to type (F).ax ox*

Multiplying both members by y'dx and proceeding as above, we get

y'

dy' = - a'2ydy.

Integrating, \ y"2 ~ -

\ a~y'2

-f C.

y' = \ 2 C - a'^2.

setting 2 C C\ and taking the positive sign of the radical. Separating the vari-

ables, we getd 'j = Ar

_

Integrating,- arc sin == = x -f i

\ r.

arc sin -~^^= = or -f uC2 .

This is the same as

Hence

-~= sin(j -f u("-)^ C,

= sin ax cos a( \, -f ros ax sin d(\., (4), Art. 2

# =-- oos </( \, sin a.r 4-! sin ( '- cos ax.a a

y c\ sin ax -f <'-j c-os ax. A us.

PROBLEMS

Find the complete solution of each of the following differential

equations.

3.-

4 sin 2 t. x = - sin 2 / + nf + r2 .

6.

Vasas

,72 Z-

9- TT + =

j= ar\ V? x -f r2 .

ro. Find /, having given that s = a, ^ = 0, when / rr Q.

at

Ans. t


206. Linear differential equations of the second order with constant

coefficients. Equations of the form

where p and q are constants, are important in applied mathematics.

To obtain a particular solution of (G), let us try to determine the

value of the constant r so that (G) will be satisfied by

(1 ) y = err

.

Differentiating (1 ), we obtain

Substituting from (1) and (2) in (G) and dividing out the factor

crr

, the result is

CJ) r2 + pr + q=-(),

a quadratic equation whose roots are the values of r required.

Equation (3) is called the auxiliary equation for (G). If (3) has dis-

tinct roots r\ and r^ y then

(4) y = cr ^ and y = e

7**

are distinct particular solutions of (G), and the complete solution is

(5) '//= c\e

r* r + fiK

>r -"r.

In fact, (5) contains two essential arbitrary constants, and (G) is

satisfied by this relation.

ILLUSTRATIVE EXAMPLE 1. Solve

(6 ) ^ - >>(lM _ a y = o.

dx~ dx

Solution. The auxiliary equation is

(7) r a -2r-3 = 0.

Solving (7), the roots are 8 and 1, and by (5) the complete solution is

y nrtx-f C'2e *. Ans.

Check. Substituting this value of y in (6), the equation is satisfied.

Roots of (8) imaginary. If the roots of the auxiliary equation (3)

are imaginary, the exponents in (5) will also be imaginary. A real

complete solution may be found, however, by choosing imaginaryvalues of c\ and r, in (5). In fact, let

(8) r, = a + 6\/^l, r2 = a - bV^Tbe the pair of conjugate imaginary roots of (3). Then

Substituting these values in (f>\ we obtain

(10) y = f"(rif'"x~+ c?c

''-rx~~0.

In the algebra of imaginary numbers it is shown that*

^-i Cos bx + V 1 sin bx, c'trx

' = cos hx V^~T sin bx.

When these values are substituted in (10), the complete solution

may be written

(11) y = c"J

(A cos bx + # sin M,if the new arbitrary constants .4 and B are determined from c\ and r-

by .4 = ci + CL>, B (o r L>)V 1. That i-vwe now take for c, nm^r-in ( 5 ) the imaginary values r

i= A ( .4 HV 1 ^

,r- ,\ ( .4 -f 7> \/ 1 ) .

By giving to A and B in (11) the values 1 and 0, and and 1, in

turn, we see that

( 12 ) //= (>

UJ cos bx and//

caf

sin bx

are real particular solutions of (G).


(18) g 4 A--'//= 0.

Solution. The auxiliary equation (3) is now

r' 4- /r-' = 0. /. r- i AV^l.

Comparing with (8), we see that a 0, b k. Hence, by (11), the complete

solution isy = ^4 C()S kjr + />> s j n A:J^

Check. When this value of y is substituted in (13), the equation is satisfied.

Compare this method with that used for the same example in Art. 205 (k a).

REMARK. A different form for the solution is obtained by setting A --- (' cos <*,

B = r sin o- in the above value of y. Then y - <' cos (A\r-

). (By (4), p. 3j.

* Let ? = V^~T, and assume that the scri(\s for c r in f'roblf-m 1, Art 191, rfprf.sonts

the function when J is replaced by ibs. Then, since ?- 1, /'*-

/,, *1 ~ I, etc.,

we have.

t ,

. ?/<>- .^J 1

,fcV

, ,

IW(ln flfc x =r ,4 lfcjr

._^l _4 +l

-^-.--.

Also, replacing j by bj in (7> and ''S), Art 104,

b^j-l- 5^4 . 63j.J fofijfi

cos 6x = 1 -^ +^ ----.

Bin ft* = fer - + ----.

Then, by Art. 195,

. . . , .. b'W bW . b*r* . b ;'sf>

(15) cos bx -f i sin bx - 1 -f- tkr - --i -f 4- i

-,

assuming that the series represents the function. The right-hand members of (14) and (15)

are identical. Therefore c t>>x - cos bx + i sin bx

Roots of (3) real and equal. The roots of the auxiliary equation (3)

will be equal if p2 = 4 q. Then (3) may be written, by substituting

q = \ P-,

04; r* + pr + lp2 =(r+>2 p)* = 0,

and r\ ~ r>2 = 3 p. In this case

(15) y~er>T and y - xertX

are distinct particular solutions, and

(16) //= ?'"(ri -f r,j)

is the complete solution.

To corroborate this statement, it is only necessary to prove that

the second equation in (15) gives a solution. But we have, by dif-

ferentiating,

(17) y - xer

>', ^ = <"'(! -f r,.r), ^ - <T"(2 r, -f rr'j).

Substituting from (17) into the left-hand member of (G), the

result is, after canceling e rijr,

(18) (rr + />r, -j- q}jr + 2 r} + p.

This vanishes, since r\ satisfies (3) and equals \ p.

ILLUSTRATIVE EXAMPLE ;*. Solve

+ ^., = o.

Find the particular solution such that

c/ss = 4 and ~ = 2 when / = 0.

cf/

Solution. The auxiliary ecjuation is

r-' -f L r -f 1 = 0, or (r -f 1 )* = 0.

Hence the roots are equal, r, = 1, and, hy (16),

(20) s = r-'(r, 4- r^/).

This is the complete solution.

To find the required particular solution, we must find values for the constants

d and C'2 such that the given conditions,

= 4 and -- = - 2 when t =are satisfied.

dt

Substituting in the complete solution (20) the given values s = 4, t = 0, wehave 4 = c\ t and hence

(21) s = e-'(

Now differentiate (21) with respect to t. We get

s'

By the given conditions, = 2 when / 0.a/

Substituting, the result is 2 c -2 4, and hence c% = 2. Then the particularsolution required is 8 = e ~'(4 + 2 f). Arts.

PROBLEMS

Find the complete solution of each of the following differential equa-tions.

1.(

^~-~-2x = Q. Atis. j- = r,r

r + r 2ra

'.

2. 7^ - 4 ^ + 3 ?/= 0. v = <v" -f r,r".

dr 2 dr

rj*4. -7^7 -f 16 -r = 0. j* c\ cos 4 / -f ('-2 sin 4 i.

d/ J

6. ~ - 4 = 0. = or 2 '

-f r.r -'.

(I/~

_ f/2?/ ,

. dij f. . Ar6. -7-^' -f 4 -f

- 0. ?/= r, 4 r 2r

4x.

(/u-2 djr

7. ^ -f 2 ~ + 2 .s-= 0. * =-- r

f

(n cos / -f r, sin /).dt 2 dt

Q d~s r/K, r _ A ^'

~<"'(o cos li / -f c* sin 2 0-

8 -~ ~ + ' * ~ -

In the following problems find the particular solution which satisfies

the given conditions.

17. ^!f + 3 ^f 4. 2 =;s = 0,

~ = 1, when f ^ 0. yin,s. = cdt 2 dt dt

18. ^ -h n 2x -;x = a,^ = 0, when / = 0. x - a cos

a< 2 ar

19. ~ - wV = (); r = 2,

~ = 0, when / = 0. An*. x = c"' + c '.

ar" /

20. ~f 4 2 ^- 8 //

=; y = 0, ^ = 24, when / = 0.

Arts. y = 4(>2 '

r~ 4

').

21. ~ - 8 ~ 4 16 K =;= 0, ^ = 1, when / = 0. 8 = tc

4t.

at- dt dt

22. -^~a~^(>; r = 0,^ = a,when/ = 0. a- = f-'-l.<//- dt dt

23. ~) -f 8 ~ 4- 25 =;

.s- = 4, ^ = -16, when / = 0.

Awx. ,s = 4 f 4f cos 3 /.

24. ^ - 6(

jf(-f 10 r = ; r = 1, ^ - 4, when / = 0.

Aw.s. J7 = e 3/ (cos/ -I- sin /).

25. 4 4 .s- =;= 0,

~ - 4, when t = 0.

26. - 4 j- -;

jr - 10, - 0, when t = 0.

no"

^ -r. ^ /v . ^J" i /x

28. ---;- 4 4- 4 r =

;.r 2, ;>, when / = 0.

dt- <11 dt

To solve the differential equation

where /> and r/ are constants and X is a function of the independentvariable .r or a constant., three steps are necessary.

FIRST STEP. Solre the equation (G). Let the complete solution be

(22) y = u.

Then u /,s called the complementary function for (H).

SKCOND STEP. Obtain a particular solution

(23) y = v

of (H) &// trial

THIRD STEP. 77/c complete solution of (H) is now

(24) = w + r .


In fact, when the value of // from (24) is substituted in (#), it is

seen that the equation is satisfied, and t,24) contains two essential

arbitrary constants.

To determine the particular solution (23), the following directions

are useful (see also Art. 208). In the formulas all letters except x,

the independent variable, are constants.

General case. If y = X is not a particular solution of (G), if

Form of X Form of v

X a + bx, assume //= r A f Hx ;

A" = ae t>jr

,assume //

= r ~ Ac'" ;

A' = a i cos b.r + a 2 sin bx, assume //= r = A

\cos bx -f A : sin />.r.

Special case. If ?/= X is a particular solution of (G), assume for /?

the above form multiplied by ;r (the independent variable).

The method consists in substituting // r, as given above, in (//),

and determining the constants A, R, A\, A-^ so that (//) is satisfied.

iLLt'STRATIVK EXAMPLK 4. Solve

(25) rffB_ 2 rfK_;, v = 2x.djr 2 djc

Solution. First Step. The complementary function u is found from the com-

plete solution of

S- 2S---By Illustrative Example 1, above, therefore

(27) y ~ u = rir'7

'

f c->cx

.

Second S/r;>. Since //= X 2 x is not a particular solution of (2(5), assume for

a particular solution of (25)

(28) y = r = A 4 Bx.

Substituting this value in (25) and collecting terms, the result is

(29 1

- 2 B - 3 A -:J Ilx = 2 x.

Equating coefficients of like powers of x, we get

-2B -XA=(), - U /* = 2.

Solving, A =|, B = -

|, and substituting in (28), we obtain the particular

solution

(30) y = v = $-Hx.

Third Step. Then, from (27) and (30), the complete solution is

y = u -f v = c^ 1-f c2e

*-f j

-!5

,396 DIFFERENTIAL AND INTEGRAL CALCULUS


dx*"dx

Solution. First Step. The complementary function is (27), or

(32 j y - u = cle'^ -f c2(''

x.

Second Step. Here y X = 2 e *is a particular solution of (26), for it is

obtained from the complete solution (32) by letting fi = 0, c2 = 2. Hence assumefor a particular solution v of (31 1,

(33; y = v = Axe x.

Differentiating (33), we obtain

(34;4r

= Ar '(1 ~ TI - S"'4 ''

'(X ~~>-

Substituting from (33) and (34) in (31), we obtain

(35; Ac r(x 2; 2 Ae r

(\ x) 3 Axe x = 2 c~ T

.

Simplifying, we get 4 Ac * 2 e *, and hence A ~-\. Substituting in (33)

we obtain

(30; y = r = -I xe *.

Third Step. The complete solution of (31 ) is, therefore,

y u -f r c\ru

-f c2eJ

\ xe f. Ana.

ILLUSTRATIVE EXAMPLE (5. Determine the particular solution of

(37) ~ + 4 * = 2 cos 2 /,

such that s - and ~ = 2 when t = 0.dt

Solution. Find the complete solution first.

First Step. Solving

(38)~

-f- 4 .s = 0,

we find the complementary function

(39) u = r, cos 2 / -f c2 sin 2 /.

Second Step. Considering the right-hand member in (37), we observe that

s = 2 "os 2 / is a particular solution of (38) resulting from (39) when = 2, c-2 0.

Hence for a particular solution s = v of (37) assume

(40) s - r = t(Ai cos 2 / -f A 2 sin 2 /).

Differentiating (40") , we obtain

f^' = 4! cos 2 / -f A - sin 2 /- 2 Mi sin 2 /

- -4 2 cos 2 /).

a/

(41)|'

^ = - 4 ^1, sin 2 / -f 4 .4, cos 2 /- 4 t( A } cos 2 / + .4 2 sin 2 /).

Substituting from (40) and (41 > in (37). and simplifying, the result is

(42) - 4 .4i sin 2 t + 4 A; cos 2 / = 2 cos 2 f.

This equation becomes an identity when A\ = 0, A a = J. Substituting in (40),

we get

(43) g = r = ? / sin 2 t.

Third Step. By (39) and (43) the complete solution of (37) is

(44) = r, cos 2 / + c-> sin 2 / 4 .\ / sin 2 t.

We must now determine d and r2 so that

(45 ) s = and ^ = 2 when / = 0.dt

Differentiating (44),

(46) = 2 Ci sin 2 t -f 2 r- cos 2 / -f -

1

, sin 2 / -f / cos 2 t.

at

Substituting the given conditions (45) in (44) and (16), the results are

= r,, 2 = 2 fa. .'.<, = 0, r, - 1.

Putting these values back in (44), the particular solution required is

(47) s = sin 2 / -f \ /sin 2 /. /I;i8.

PROBLEMS

Find the complete solution of each of the following diflVrenthil equa-

tions.

l. + jr = at + b. Ans. s ~ c\ cos / + c>2 sin / + at -I It.

dt 2

2.^~ -f x = ae ft/

. .f - O cos / -f r, sin / -f ^-j

3. ^ + f - 4 cos /,. s - r, cos / -f r- sin / f 2 / sin /.

dt-

4. + X = 4sln2t. JT = CI cos / -f c> sin /-

'\sin 2 /.

d/ 2

5<* _ 4 s = at + b. - r,^' -f r,r

^ -\(at -f 6j.

d/ 2

e^-4 s = *>^ = r,r-' -f r 2r2; -

1 r'.

d/ 2

7. ^! _ 4 s = e'2t

.R = o^' -f r 2f"

2 '

-f J ^'2t -

'

d^ 2

8 ^! 4 s = 2 cos 2 t. 8 = fie2 '

-h fzf'2 ' -

i cos 2 t.

dt 2

9^ 2y

_}_ 9 y 5 x 2. ?/

= d cos 3 x 4- c 2 sin 3 x 4- ^ x 2 -if.

*

dx2

10. - ~ - 2 s - 4 /. Anx. s - r,f' + r*--" -f 1 - 2 *.

11. 2'- -'~i '=- -r = r,<'' + r./r' + 8.

12. ^" 7- 1

''-{ :< = fj r~". * = r,r' -f r^r" - 6 c

2'.

///- <//

13.|-~

f 2j*

-f 2 - 8 H'. s = r 'fr, cos / -f c> sin /) -f | r 2'.

14.y/(

- 4 -^ -i ;{ .r ^ 4 r'. r = r,r' -f r ar*'-

li /r'.~

-f /!cos /

j

:{

sin /.

16.-,

--'2

~j--f 5

//-_- U sin 12 /. // r'(r, cos 1* / -f rj sin 12 /)

-f]s cos "2 f -f /V sin 2 /.

17. '-'; f 9 .<; - :i cos 12 /. 22. 4 ^ -f * - 5 cos

f-

21. f -.r - ^ - 2. 26.

|

+ f,

|

+ K, * = 5 sin o ^

In the following problems find the i>articular solution which satisfies

the tfiven conditions.

-

28.^ + 9,-9r^ ^=1, ^ = ^,when/ = 0.

J 3 / + r1

').

29. ~~ + 9 x = 5 cos 2 /;x = 1

,

~ = 3, when / = 0.

30.|~|

+ 9 * = X cos 3 / ;= 0,

~ = 6, when / = 0.

us. s = sin 3 t -f cos 2 f.

= 0.

An*, x = 2 sin 3 / -f it sin 3 t.

g-j..b ,. t-

Ans. or = i(e3 ' + e-'-6 J + l).

32.(

^~- 6 ^ + 13 JT - 39 ; jr = 4,

~ =- 3, when / = 0.(it- dt dt

, .., , ,

, ,t

.4 us. .r r'' cos 2 / -f 3.

33. ~ -f 9 * = 4 - 3 / ; * = 0, ^ = (), when / = 0.

34.^~

- 9 * = 6 /;

8 = 0,~ = 0, when / = ().

36. -f a- = Li cos 2 / ; jr = 0,^~

:----J, when / = 0.

37.^y + 4 ,s-

= 1> cos L> / ;.s- = 0, y =

"J, when / r- 0.

38. ^ - 2 ~ -f L> ^ - L> sin /; .r -- 0,

~- - 0, when / = ().dt 2

(It dt

39. ^ + 5 <& + 4 ,/= 12 ,- //

= 1, ? - 0, when .r = 0.a.r- c/.r

'

(/./

40. <%i + 4 ^ + 4 ,/= 4 ,- ;

= 0, ty^ 2. when , = 0.

dj'- as'

(/.r

207. Applications. Compound-interest law. A sim]>kk

application of

differential equations is afforded by problems in which the rate of

change of the function with respect to the variable (Art. f>0) for anyvalue of the variable is proportional to the corresponding value of

the function. That is, if // =/(.r),

-**

where k is a constant. In this differential equation the variables are

separable as in Type I, p. 379. Sohing, we get

(2) y-

c^ J

t

where c is an arbitrary constant. Thus the function y is an expo-nential function (Art. 62). Conversely, given (2), it is easily shown

by differentiation that y satisfies (1 ). The connection of (1 ) with the

name "compound-interest law" is shown as follows:

Let y = a sum of money in dollars arcumulatiri^ at compound interest;

i interest in dollars on one dollar for a year;

A/ = an interval of time measured in years ;

A?/ = the interest of y dollars for the interval of time At.

Then Ay = iy At. Therefore

Equation (3) states that the average rate of change of // (Art. 50) for the periodof time At is proportional to y itself. In business, interest is added to the principalat stated times only,

- -yearly, quarterly, etc. In other words, y changes dis-


continuously with /. But in nature, changes proceed on the whole in a continuous

manner. So that to adapt equation (3) to natural phenomena we must imaginethe sum // to accumulate continuously ; that is, assume the interval of time A/

to be infinitesimal. Then equation (3j becomes

dy _~dt

" ly >

and the rate of change of y is proportional to y, agreeing with (1) if k = i.

In ( 1 ) the function y is said to change according to the compound-interest law.

A second example is afforded by the complete solution of the

equation

(4) S =T/1'

tf + f '

where k and c are constants not equal to zero. For let c = ak. Then

(4) may be written

(r>)

~0/ + a) - k(y + a).

This equation states that the function y + a changes according to

the compound-interest law. The differential equation (4), or (5),

is solved as in Type I, p. 379. The solution is

(G) //= cc

kja.

ILLUSTKATIYIO EXAMPLE 1. The function // of x changes according to the

compound-interest law. When x 1, y-

4 ; when x - 2, y = 12. Find the law.

Solution. By (1) we have

(7) g = *,.

Separating the variables and integrating, we get

In //-- k.r 4 (\

We have to find the values of k and ('. Substitute the given values of x and y.

Then In 4 --- k + r, In 12 - 2 k + C.

Solving, k - In 12 - In 4 = In 3 - 1 .0980, (* = In 4 - In 3 = In J.

Therefore In //= 1.0986 JT -f In J, and y = J r 1 098Bj

. .4n,s.

ILLUSTRATIVE KXAMPLK 2. Washing down a solution. WTater is run into a

tank containing a saline (or acid) solution with the purpose of reducing its strength.The volume r of the mixture in the tank is kept constant. If s = quantity of salt

(or acid) in the tank at any time, and j amount of water which has run through,show that the rate of decrease of s with respect to x varies as s, and, in fact, that

Solution. Since s = quantity of salt in the mixture of total volume f, the

quantity of salt in any other volume u of the mixture is - u.

Suppose a volume Ax of the mixture is dipped out of the tank. The amount

of salt thusdipped

out will be^

A.r, and hence the change in the amount of salt

in the tank is given byv

(8) As = - *A.r.

r

Suppose now that a volume of water A.r is added to fill the tank to its originalvolume r. Then from (8) the ratio of the amount of salt removed to the volumeof water added is given by \ s

._ = _-.

When AJ- we obtain the instantaneous rule of change of s with respectto j, namely, rfs s.-~ - -

/b<s.dx r

Hence s changes according to the compound-interest law.

PROBLEMS

1. The rate of change of a function // with respect to .r equals -

{ //, and

y 4 when JT 1. Find the law connecting j- and //. A us. \i= 5.58 ci r

.

2. The rate of change of a function // with respect to .r equals "2 //,

and y 8 when jc = 0. Find the law*. An*. //--= f> c

'

-f 2.

3. In Illustrative Example li, if r 10,000 gal., how much water mustbe run through to wash down 50 per cent of the salt ? AUK. (>i),'U gal.

4. Neirton'K law of cooling. If the excess temperature of a body abovethe temperature of the surrounding air is .r degrees, the time rate of de-

crease of jc is proportional to .r. If this excess temperature was at first

80 degrees, and after 1 min. is 70 degrees, what will it be after 2 min.?

In how many minutes will it decrease 20 degrees?

5. Atmospheric pressure ]> at points above the earth's surface as a

function of the altitude // above sea level changes according to the com-

pound-interest law. Assuming ;>15 Ib. per sq. in. when //

-0, and 10 Ib.

when h = 10,000 ft., find p (a) when // = 5000 ft.; (b) when // = 15,000 ft.

AUK. (a) lU.'Jlb.; (b) 8.1511).

6. The velocity of a chemical reaction in which / is the amount trans-

formed in time t is the time-rate of change of T,

Reaction of the first order. Let a = concentration at the beginning of

the experiment. Then ~ k(a r), since the rate of change is propor-

tional to the concentration at that instant. (Note that a r, the concen-

tration, changes according to the compound-interest law.)

Prove that k, the velocity constant, is equal to - In/ a - f

7. In the inversion of raw sugar, the time-rate of change varies as the

amount of raw sugar remaining. If, after 10 hr., 1000 Ib. of raw sugarhave been reduced to 800 Ib., how much raw sugar will remain at the

expiration of 24 hr. ? Ans. 586 Ib.

8. In an electric circuit with given voltage K and current i (amperes),the voltage A1

is consumed in overcoming <]) the resistance ft (ohms) of

the circuit, and (2) the inductance L, the equation being

This process therefore comes under equation (4) above, h\ ft, L beingconstants. Given L ~ f>40, ft = 250, ft 500, and / = when / = 0, showthat the current will approach 2 amperes as / increases. Also find in howmany seconds / will reach W) per cent of its maximum value. Ann. 5.9 sec.

9. In a condenser discharging electricity, the time-rate of change of

(lie voltage c is proportional to r, and c decreases with the time. Given

/; -- 40, find / if r decreases to 10 per cent of its original value. A nx. 92 sec.

10. llniUlimj n /> a saline (or acid) XO/M//OJ/ by adding salt (or acid),

maintaining constant volume, leads to the equation - (r //), whereas r

^

r-- the constant volume, //

~salt (or acid) in the tank at any moment,

j' salt 'or acid) added from the beginning. Derive this result and com-

pare with Illustrative Example 2 above.

208. Applications to problems in mechanics. Many important prob-lems in mechanics and physics are solved by the methods explainedin this chapter. For example, problems in rectilinear motion often

lead to differential equations of the first or second order, and the solu-

tion of the problems depends upon solving these equations.

Before giving illustrative examples it is to be recalled (Arts. 51

and :><)) that

where /' and a are, respectively, the velocity and acceleration at anyinstant of time (- /), and N equals the distance of the moving pointat this time from a fixed origin on the linear path.

lu,rsTK-vn\ K !']\AMPLK 1. In u rectilinear motion the acceleration is inversely

proportional to the square of the distance x, and equals 1 when ,s' = 2. That is,

4(2) Acceleration a ---

Also, r 5, s- S, when /-

0.

(a) Find r when * 'J4.

Solution. From (2), using; the last form for a, we obtain

/o\ dr 4(3) r ^

a,^ 6'-

Multiplying both members by </s and integrating, the result is

(4) = - + r, or f-" = ? + r.

Substituting in (4) the above conditions r = ft, 8 =- 8, we find (" ~- 1M. Hence

(4> becomes

(5) r-1

,S"

From this equation, if s = lM. r =\\ 1M9 ~ 1.9;?. /I NX.

(hi Find the time which elapses when the point moves from s = S to ,s*= 24.

Solution. Solving (5j for r, we get

Separating the variables s and / and solving for / with limits as given, x -- 8,

s -4, we find, for the elapsed time,

(7) / =

NOTE. Using the first form in il for </, t~> is

^x _ 4

dl'"

.s'-

1

'

which ib of the form (F), Art. 'J05. The method of integration here is the same as

in Art. 205.

An important type of rectilinear motion is that in which tlie ac-

celeration and the distance are in a constant ratio and differ in si^n.

Then we may write

(8) a = - Ks- f

where k~ = magnitude of (i ut unit distance.

Remembering that, a force and the acceleration caused by it differ

in magnitude only, we see in the above cast 4 that the effective force

is directed always toward the point ,s -- and is, in magnitude, di-

rectly proportional to the distance ,s-. The motion is called MM pic

harmonic vibration.

From (8), we have, using (1 ),

a linear equation in s and / of the second order with constant, coeffi-

cients. Integrating (see Illustrative Example 2, Art. 20(lj, we obtain

the complete solution,

(10 ).s- = n cos ki + C'j sin kt.

From (10), by differentiation,

(ID r = k( C] sin /:/ + c- cos A-/).

It is easy to see that the motion defined by (10) is a periodic

oscillation between the extreme positions ,s-=

/>, and s = - fc, de-

termined by

(12)

In fact, we may replace the constants c\ and 2 in (10) by other

constants b and A, such that

(13j c\ = b sin Ay 02 = b cos A.

Substituting these values in (10), it reduces to

(14) s=bsm(kt + A), by (4), Art. 2

and now the truth of the above statement is manifest.

In the following examples we give cases when the simple harmonic

motion is disturbed by other forces. In all cases the problem de-

penis upon the solution of an equation of the form (G) and (H),

discussed above.

ILLUSTRATIVE KXAMPLK 2. In a rectilinear motion

(15) a - -'I

x - v.

Also, v = 2, K =- 0, when / 0.

(aj Find the equation of motion (x in terms of /).

Solution. Using (1), we have, from (15),

nr\ rf2 's'

_Ld$ ,5

df.) ^ + ^ +4

S ='

an equation of the form (G). The roots of the auxiliary equation r 2-f r + J

= are

r, = -J + \ ITT

,r2 = - - vC"I.

Hence the complete solution of (16) is

(17) s = r" 2'( r , cos / + r 2 sin /).

By the piven conditions, s when / = 0. Substituting these values in (17),

we find o = 0, and hence

(IS) s = c2e~

Diflerent.iuting to find r, we get

(19) r ~ r,f *'(- J sin / + cos /).

Substituting the given values r 2 when / = 0, we have 2 = 2.

With this value of r2 , (18) becomes

(b) For what values of r will r = 0?

Solution. When r = 0, the expression in the parenthesis of the right-handmember of (19) must vanish. Setting this equal to zero, we readily obtain

(21) tan < = 2.

For any value of / satisfying (21), r will vanish. These values are

(22) t = 1.10 + rnr. (n = an integer). Ans.


Successive values of t from (22) differ by the constant interval of time TT.

Discussion. This example illustrates damptd harmonic vibration. In fact, in

(15) the acceleration is the sum of two components

(23) fli = -J , ci 2 = - r.

The simple harmonic vibration corresponding to the component a\ is now dis-

turbed by a damping force with the acceleration a 2 ,that is, by a force proportional

to the velocity and opposite to the direction of motion. The etTects of this dumpingforce are twofold.

First, the interval of time between successive positions of the point where r =

is lengthened by the damping force. In fact, for the simple harmonic vibration

(24) i--- -

; *,

we have, by comparison with iS\ k - \\ f>-

1.12, and the half-period, by (12), is

0.89 TT. As we have seen above, for the damped harmonic vibration the correspond-

ing interval is TT.

Second, the values of * for the successive extreme positions where r ~ 0, instead

of being equal, now form a decreasing geometric progression. Proof is omitted.

ILLUSTRATIVE EXAMPLE IJ. In a rectilinear motion

(25) a - 4 s f 2 cos 2 t.

Also, s - 0, r = 2, when / = 0.

(a) Find the equation of motion.

Solution. By (1), we have, from (25),

(26) 44x^2 cos 2 t.

dt'

The particular solution required was found in Illustrative Example fi, Art. 206,

and is given by equation (47), p. 397. Hence

(27) R = sin 2 / + j / sin 2 /. Ans.

(b) For what values of / will r- 0?

Solution. Differentiating (27) to find r, and

setting the result equal to zero, we get

(28) (2 -f / icos 2 / -f \ sin 2 / - ;

or, dividing (28) through by cos 2 /,

(29) -Jtan 2 / 4- 2 4 / = 0.

The roots of this equation may be found as ex-

plained in Arts. 87-89. The figure shows the curves

(see Art. 88)

(30) y = ] tan 2 /, y = - 2 - /,

and the abscissas of the points of intersection are, approximately,

/ = O.SS, 2..'H>, etc. Anx.

Discussion. This example illustrates forced harmonic vibrut'ion. In fact, in (25)

the acceleration is the sum of two components

(31) a, = 4 s, a- = 2 cos 2 t.

The simple harmonic vibration corresponding to the component GJ with the

period TT is now disturbed by a force with the acceleration az ,that is, by a periodic


force whose period (= TT) is the mme as the period of the undisturbed simple

harmonic vibration. The effects of this disturbing force are twofold.

FirHt, the interval of time between successive positions of the point where

v~

is no longer constant, but decreases and approaches J TT. This is clear from

the above figure.

Second, the values of * for the successive extreme positions where r = now

increase arid eventually become, in numerical value, indefinitely great.

PROBLEMS

In each of the following problems the acceleration and initial conditions

are given. Find the equation of motion.

1. a = -A'

L'.s

;s =.(),v= r,,, when / - 0. A nx. * =

y'sin /;/.

2. a A--.S;

x ,s()> r 0, wrhen / = 0. s = *,, cos kt.

3. a A'^.s;

.s ,s,,, r r() ,\\ hen / 0.

4. a ft x;

.s- 0, r - 0, when / 0. = f> ( 1 cos /).

5. a - sin 2 /- s

;x - 0, r - 0, when / 0. s 5 sin /

-\ sin 2 t.

6. a 2 cos /- x

;x 2, r 0, when t 0. * - 2 cos / -f / sin /.

7. a = - 2 r - 2 x;

.s = .'5, r- -

ii, when / 0. ,s = 3 c' cos /.

8. a = -A-

L'.s- -f b

;x = 0, r = 0, when 1 = 0.

9. a - //r ;x 0, r ~ n, when / - 0.

10. a - 8 /- 4 x ;

x = 0, r = 4, \\ hen / - 0.

11. a 4 sin / 4 x; x = 0, r = 0, \\hen / = 0.

12. a = 2 sin 2 /- 4 s

;x = 0, r 0, when / -- 0.

13. (/- - 2 r 5 ,s-

;.s- 1 , r - 1 , when / 0.

14. Given a = 8 --4 x, and r 0, s - 0, when / 0. Show that the

motion is a simple harmonic vibration \\ith the center at s 2, with an

amplitude 2 and a period TT.

15. The acceleration of a particle is given by

a 5 cos 2 / 9 s.

(a) If the particle starts from rest at the origin, find its equation of

motion. A H*. * = cos 2 /- cos 3 /.

What is the greatest distance from the origin reached by the particle?

(b) If the particle starts from the origin with velocity v = f>, find its

equation of motion. AHS. s cos 2 t + 2 sin 3 t cos 3 /.

What is the greatest distance from the origin reached by the particle?

16. Answer the questions of the preceding problem if the acceleration

is given by a - 3 cos 37-9 *.

Ans. (a) x = \t sin 3 /; (b) s = \ t sin 3 / -f 2 sin 3 /.

17. A body falls from rest under the action of its weight and a small

resistance which varies as the velocity. Prove the following relations:

-.=

-^ (kt + <

; ' - 1).*"

18. A body falls from rest a distance of 80 ft. Assuming n = 32 r,

find the time. AH*. 3. -17 sec.

19. A boat moving in still \\ater is subject to a retardation propor-

tional to its velocity at any instant. Show that the velocity /sec. after

the po\ver is shut off is given by the formula r a '

', \\here c is the ve-

locity at the instant the power is shut off.

20. At a certain instant a boat drifting in still vater has a velocity of

4 mi. per hour. One minute later the velocity is 'J mi. per hour. Find

the distance moved.

21. Under certain conditions the equation defining the swing of a

galvanometer is

Show that it will not swing through the zero point if p. > k. Find the

complete solution iffj.< k.

209. Linear differential equations of the nth order with constant coef-

ficients. The solution of the linear differential equation

in which the coefficients p\, p>>, p,, are constants, will now be dis-

cussed.

The substitution of crx for y in the first member gives

(rn + pir

71"

1 + 7>>rn ~ 2 + + Pn)e".

This expression vanishes for all values of r which satisfy the

equation

(1 ) rn + p }

rn ~i + p->r

n ~'2 + - + pn = ;

and therefore for each of these values of r, crs

is a solution of (7).

Equation (1 ) is called the auxiliary equation of (/). We observe that

the coefficients are the same in both, the exponents in (1 ) correspond-

ing to the order of the derivatives in (7), and y being replaced by 1.

From the roots of the equation we may write down particular

solutions of the differential equation (/). The results are those of

Art. 2(K> extended to cases when the order exceeds two, and are

proved in more advanced textbooks.

Rule to solve the equation (/)

FIRST STEP. Write down the corresponding auxiliary equation

(1 ) r" + p.r"-

' + jw n ~'- + + Pn = 0.

SECOND STEP. Solve completely the auxiliary equation.

THIRD STOP. From the roots oj the auxiliary equation write down the

corresponding particular solutions of the differential equation as follows:

ArxiLiAHv EQUATION DIFFERENTIAL EQUATION

(a) Kach dlstin-ct real\ ,

,

f (jives a particular solution eT]X .

root r\ )

'

([}} Each distinct pair \ . ( two particular solutions c cos bx., . .

, i' <Jtves .

,

of nnaynuirtj roots a hi i

*

i e"-r sin b.r.

I

8 (or 2 s) particular solutions ob-

(c) A nndti}>h' root occur-\

.

>\

taiwd by multiplying the par-

ritiy s thne-s J*

'

ticular solutions (a) (or (b)) by

7, :r, .r1

-', -, j-- 1

.

FOURTH STI:P. Multiply each of the n *independent solutions thus

found hy an arbitrary constant and add the results. This result set equal

to y gives the complete solution.

ILLUSTRATIVE EXAMPLE 1. Solve(~-~

;i

(- - 4 ?/ 0.

CU'f

(U*'

Solution. Follow tho above rule.

First S/r/>. r' - ;i t- -f 4 0, auxiliary cxiuation.

Second N/r/>. Solving, the roots arc 1, -, !2.

Third Sh'p. (ii\ The root -1 gives the solution c ~*.

(r) The double root li givers the two solutions c-x, xe2x .

Fourth Step. The complete solution is

^ = dc z-I- c>>c~

J-f riiJY-*. A?is.

ILLUSTRATU-E EXAMPLE 2. Solve ^-4^-/

+ 10^~12^-h5|/ = 0.(/j

>l dr 1

c/j-J dr

Solution. Follow the above rule.

First Step, r 4 4 r' -f 10 r- 11! r -f 5 = 0, auxiliary equation.

Second Step, Sohdng, the roots are 1,

1,

1 2 i.

* A chock on the accuracy of the work is found in the fact that the first three stepsmust give n independent solutions.

Third Step, (bj The pair of imaginary roots 1 *J i* gives the two solutions

e* cos 2 x, <>J sin 2 x. (0 = 1,6 2)

(c) The double root 1 gives the two solutions r*. xe*.

Fourth Step. The complete solution is

y dc1-f ra-N'* -f c^c* cos 2 j -f <V'

T sin 2 x,

or I/=

(<*i -f r2x 4- fa cos 2 x + <'i sin 2 .rir1

. A?w.

The linear differential equation

in which pi, ;;_>, , pn are constants, and A' is a function of jc or a

constant, is solved by methods like those used in Art. 20(> for equa-tion (//). The three steps described on page 394 are to be followed

here also. That is, we solve first the equation (/), obtaining

(2) ?/= w,

the complete solution of (/). Then u is the complementary function

for (/).

Next we find in any manner a particular solution of (/),

(3) y = v.

Then the complete solution of (7) is

(4) y = ?/ + v.

In finding (3), methods of trial may be used analogous to those

given on page 395 for // = 2. The rules given there for the general

case apply also for any value of ?/. In any case we may follow the

Rule to find a particular solution of (7)

FIRST STEP. Differentiate successively tht (liven equation (7) and

obtain, either directly or by elimination, a differential equation of a

higher order of Type (/).

SECOND STEP. Solving this new equation by the rule on page 408,

we get its complete solution.

y ~ u + v,

where the part u is the complementary function of (/) already found

in the first step,* and v is the sum of the additional terms found.

THIRD STEP. To find ihv values of the constants of integration in

the particular solution v, substitute

* From the method of derivation it is obvious that every solution of the original

equation must also be a solution of the derived equation.

and its derivatives in the yiven equation (/). In the resulting identity

equate the coefficients of Like terms, solve for the constants of integration,

and substitute their values back in

?/= // + v,

(jiri)ty the complete solution of (J).

This method will now be illustrated by means of examples.

NOTK. The solution of the auxiliary equation of the new derived differential

equation is facilitated by observing that the left-hand member of that equation is

exactly divisible by the left-hand member of the auxiliary equation used in finding

the complementary function.

ILU'STRATIVH K\ AMPLE. Solve

Solution. We first find the complementary function u. Solving

I) y" ~ a \f 4 2 v = 0,

the result is

(7 ) y ?/ c\v~* 4 r-yff.

First ,SVr/>. We now dillerentiate (5), obtaining

(S) //'" \\ //" -f 2 //'= M* + ex .

Subtracting (5) from (S), the result is

(!) ) y'" - 4 //" -f 5 y'-

'2 y ez .

Dill'-reritiating (9>, we obtain

(10) ?/lv 4 //'"

-1-t5 //" 13 ?/'

= ex .

Subtracting (9) from (10), we get

(1 1 //'v - 5 //'" -f 9 y" - 1 ij' -f 2 //

= 0,

an equation of Type (/).

iSYrow/ N/r/>. Solve (11). The auxiliary equation is

( 12) r 1 - 5 r< + 9 r-' - 7 r -f 2 = ().

The left-hand member must be divisible by r- H r + 2, since the auxiliary

equation for (t>> is r'-1

8 r -f 2 = 0. In fact, we find that (12) may be written

(1M) (r-'-

;{ r -f 2>ir - 1 )- = 0.

'1 he roots are r = 1. 1, 1.2. Hence the complete solution of ( 11 ) is

(14) y = or 2 *4- c J

(c-2 -f r ;i j- -f c&*}.

Third Step. Comparing (7) and (14 \ we see that

(15) y ~ r r'(r.{j -f r 4J'J

>

will be a particular solution of (5) for suitable values of the constants c.\ and c^.

Differentiating (15), we obtain

(10) yr = r'co 4- (r.< 4- 2 r 4)x 4- r 4 .r

2),

I/" = p*(2(r 3 4- f 4) 4- (r 3 -f 4 r^or 4- r 4x2).

Substituting in (5) from (15) and (16\ dividing both members by cr and reduc-

ing, the result is

(17) 2<- 4- c - 2r,j = J-.

Equating coefficients of like powers of xt we obtain 2r~ 1, '-d ra = 0,

whence we find c\ = 1, r, = J. Substituting these values in (15), the particular

solution is

(18) y = r = r"\- j--

.1 J J),

and the complete solution is

PROBLEMS

Find tbe complete solution of each of the following differential equa-

tions.

IL-J. + }

_if o A H$ t // C] -f ^ , (os *j >r -|- c,\ sin 2 .r.

f/r{

djr

<2LJJ

_|_ 4 1 d o^ y r, -)_ ^.^j- -f r , c()S 2 j* -}- c \ sin 2 s.

3. - ^ 0. // r, -f C'2(>r

4- r. ( rJ4- r

icos .r 4- o, sin u*.

(/j--1 as

//4 ^ ^/L\-

4.- '

4- ;r 4 ,s = 0. ,s-- CiC* 4- r.r

'

4- r/, cos 2 / 4 risin 2 /.

,4 ?/,s. //= r, -I- cy" -f r. { cos JJ .r -f r

isin 3 .r.

-f r L./) cos4

J / -f (r :i -f r ,/) sin 2 /.

6 l + I L> + 8 ^ = 0. -4/**. a* - r L>r

(r, + c*t -f r :i /a).

J/ L>(//

;j/ t ^ WtSt .s- = nr'-r r-r' + r.j cos / -f r

i sin /- P - 3 ^.

9.i _ 4 = 2 ,-'. 7/

= r, + r,r-' + r.,r

rfr :l d.r

IQ.fHlL - (

l! = 4 .r . v = n + r,r + r:,f

d.r* (Lr~

1L g _ 3 g + 2 = ". y = c,c- + r,f-'-

2 r" + i se>.

18.^-9^+ 20 = /V. A,. . = .,.' + ^'+ f:"(7 +

V+ 2<2)

(^/L> a/ 4

13. ^j + 4 .s = / sin 2/.

4. .

,

i t cos 2 t t2 sin 2

Am. s = cj cos 2 ^ + c2 sin 2 / -f-- ---

"--- , 9 . + ,3 + 36, = 18-36.

,6.i;

- 1:1 + 30 , = (I. 20. - 2 + g = '.4 dt^ at 2 at

MISCELLANEOUS PROBLEMS

Find the complete solution of each of the following differential equations.

. Am. ?/=(/ + O\

4. (1 4- .!'-')<///=- Vl - y-dx.

// __ -r 4- r

Vi -//-'

5. (.r -f v/)r/.r ~ (.r //)(///. In (JT~ -f 'r ) 2 arc tan *- = c.x

7. -4-f li = 0. = o<-' + r 2^'.(//^ <//

8. ~ -1

(

^ -H 4 = 0. .s-= r,r-" 4- r,/r

2'.

(//^ <//

9-(^' _ 4 ^ + 8 s = 0. A//K. = r j/

(r, cos 2 f + c>2 sin 2 <).(//^ dt

10. ^ - 2 ^ - 8 //= r-'. //

- r,r" + ^^'r -

\ e-<.

<//* (//

11. ~ + A--.r = / -f b. j- - d cos A-/ -f r,> sin A-/ -f(^

dr A"*

.r = n cos A'/ -f r.. sin A'/ +2

-

!3^(l:-r _

/; i> - a Cos A-/. .r - c\ckt + cc~ u - -^ cos kt.

<//-' 2 A"^

14. ^4 -f A-2.r = a sin A*/. .r = c\ cos A'/ -f <" 2 sin kt -j t cos kt.

(if" n

15. (.r2 2 /rW.r -f 2 .r// (/// 0. //

2-f .r

2 In CJT = 0.

16. ^ -f 4ri- =.>l ivr ?vU'

L) + I)2 = arc tan x + c.

dr j- 4- 1 U'~ + 1)'

17. ~j - 5^ -f 4 * = 0. Ans. * = nc 1

4- c 2e<

4-W 4- <V 2'.

a/4 ar

Anx. s c\ cos / 4- r-j sin f 4- o cos 2 / 4- c* sin 2 f.

19. .n/2

(i?/=

(.r* 4- li'^djr. po [^ , * . __ /

20. (/// 4- j-z/d~

-rV-'W-r = 0.(//

"'

21. <-^ _ 8 -r4- 25 .r - 0. 24. '-

4- 4 .r = 6 cos 3 t.

22. 4- 4 jr = 8 / 4- 2. 25. -^ 4- 4 .r = 2 cos 2 /.

df2 (it-

Solve each of the following differential equations by making the trans-

formation suggested.

26. 1' ~ - 2 $t - *' = 0. Let ,s-

= ~- A ns. ~ 4-~ = r.

27. (/-' 4- 0(/.s- = (^'J

4- 2 x/ 4- .s1

)^/. Let .s-= r/. ^ws. * = r/(l + t)

-t.

28. C5 4- 2 K/)S r/^ = (3 - 2 .s7)/ (/.s\ Let x/ r.

29. (.r 4- ?/)'J -^ = 2 jr 4- 2 // 4- 5. Let .r 4- //

= r.

ADDITIONAL PROBLEMS

1. For a certain curve the area bounded by the curve, the .r-axis, and

any two ordinates is k times the length of arc intercepted between the

ordinates, arid the curve passes through the point (0, /;). Show that the

curve must be a catenary.

2. The acceleration of a man dropping in a parachute from a stationary

balloon is 32 i r-' ft. per second per second, where ?> is the velocity in

feet per second. If he reaches the ground in one minute, prove that the

height of the balloon is a little more than 950 ft.

'

3. A point moving on the .r-axis is subject to an acceleration directed

toward the origin and proportional to its distance from the origin and to

a retardation proportional to its velocity. Given that the differential

equation for jr is of the form

where m and n are positive, and given the initial conditions x=W,- 0, when / = 0, find in each of the following cases x and ~ and dis-

dt at

cuss the motion.

(a) m = 4, n = 5; (b) m - 4, n = 4 ; (c) m = 4, n = 3.

CHAPTER XXII

HYPERBOLIC FUNCTIONS

210. Hyperbolic sine and cosine. Certain simple expressions in-

volving exponential functions (Art. 02) occur frequently in appliedmathematics. They are called hyperbolic functions. The justifica-

tion for this name is brought out below in Art. 215. Two of these

functions, the hyperbolic sine and hyperbolic cosine of a variable r,

written, respectively, sinh v and cosh r, are defined by the equations

r C v C 1 '

~4~v

(4) sinh v = cosh v = !

where e is, as usual, the Napierian base (Art. (>1 ). These functions

are not, however, independent, for we have from (A)

(B) cosh2 v sinh2 v ^ 1.

r-'1

4 '2 f e'

,

f-' - 2 -f r -''

From (A), Hquarm^. cosh-' r :

~~~~ sum-' r = :

Hence, by suht ruction, cosh- r Mnh-' r - I.

From (A), by solving for the exponential functions, we get

(1) er cosh r -f sinh r, e

' = cosh r sinh r.

ILLUSTRATIVE EXAMPLE. Show that the complete 1 solution of the differential

equation/n\ rf'"'/ . f\

(2)Ji

-'//()

a.r-

may be written // A sinh as -4- />* cosh (/.r,

where .4 and /? are constants.

Solution. By Art. 2()(> the auxiliary equation for (2) is r~ a'2 = 0, whose roots

are a and a. Therefore the complete solution of (2) is

\l rif"-r

4- or "'.

The values of r"z and r

~ nj> are found from (1) by taking r ax. Hence

r" x~ cosh n.r -f- sinh u.r, r

" x = cosh a.r sinh a.r,

y = n(eosh (i.r + sinh a.r} -f- r-j(cosh ax sinh ajr]

=(c\ -f- r\'Vosh ax -j (ri ^Osinh a.r.

Letting ri r-j = .4, TI -f c ~/-?, we obtain the desired form.

The result should be compared with Illustrative Example 2, p. 391.

414

HYPERBOLIC FUNCTIONS 415

211. Other hyperbolic functions. The hyperbolic tangent, tanh r, is

defined bysinh v e 1 e v

C tanh v -^ r --r"lcosh v e 1 + e

The equations

define, respectively, the hyperbolic cotangent, hyperbolic secant, and

hyperbolic cosecant. The ratios used in (C) and (1) are the same as

those in (2), p. 2, for the corresponding trigonometric functions.

The following relations hold :

(2) 1 tanh 1'r = sech j

r, ctnh-' r 1 = csch- r,

analogous to formulas in (2), p. 2. The proof of the first formula is

given below.

The following statements hold for the values of the hyperbolic

functions. They should be verified. ^

sinh r can have any value ;cosh r, any positive value not less

than 1;sech r, any positive value not exceeding 1

;tanh r, any value

numerically less than 1 ;ctnh /', any value numerically greater than 1

;

csch r, any value except zero. Also, from the definitions, we have

sinh (- r) =- - sinh r, csch (- r) = - csch v,

(3) cosh (- n = cosh r, sech (- r) = sech v,

tanh (- r) = tanh r, ctnh (- r) = - ctnh v.

II.USTRATIVK EXAMPLE. Given 1anh JT- \. Find the values of the other

hyperbolic functions.

Solution. In () divide each term by cosh 2.r. Then we got

^"1

>r ^

cosh*' J* cosh* x

Therefore 1- tanh- .r - sech 2 x. By (C) and (1)

Since tanhjr= ?, this equation gives sech JT = J, the negative value being in-

admissible. Then

cosh jr = \ =~> bv (*)sech x 3

sinh j = cosh x tanh 3 = ? by (C)

ctnh x = J, arid csch jr = 4'. By (1)

212. Table of values ot the hyperbolic sine, cosine, and tangent.

Graphs. A table giving the values, to four significant figures, of

sinh r, cosh ?>, tanh r for values of r from to 5.9 is shown on p. 416.

For negative values of r, use the relations (3;, Art. 211.


When r -* + oo, sinh r and cosh r become infinite, while tanh v

approaches unity as a limit.

Graphs of sinh x, cosh x, and tanh a* (Figs. 1, 2, 3) are easily

drawn by making use of the table.

(O.-l)

(0,1)

ZlL

tanh x

FIG. 1 Fie. ,'J

213. Hyperbolic functions of v -f w. Formulas for hyperbolic func-

tions, corresponding to two of (4), p. 3, are

(D) sinh (i; + w) = sinh v cosh w + cosh v sinh w,

() cosh (v + w) ~ cosh i; cosh w + sinh i> sinh i#.

Proof of (/)). From the definition (A), replacing r by r + w, we have

(1 ) sinh ( + w }- - ~~ >

(2) cosh (v -f w)

The right-hand member of (1) is transformed as follows, makinguse of (1), Art. 210.

e e'e11

e (>

2 2

(cosh ??+ sinh ?>)(cosh w + sinhw) (cosh ?* sinh ?>j(eosh7^--sinh

Multiplying out and reducing, we get (D). Formula (E) is proved

in the same way.If we set w = v in (D) and (E), we have

(3) sinh 2 # = 2 sinh v cosh v,

(4) cosh 2 t? = cosh 2?; + sinh 2

v.

These are analogous to the formulas for sin 2 x and cos 2 x, re-

spectively, of (5), p. 3. From (B) and (4), we get results which

correspond to the formulas for sin- x and cosj x in (5), p. 3. These are

(5; sinh 2 v = | cosh 2 r \, cosh 1-'

r .V cosh 2 r + ?,.

Other relations for hyperbolic functions, which may be comparedwith those on page 3 for trigonometric functions, are given in the

problems.

ILLUSTRATIVE Ex AMPLE. Derive the formula

sinh 2 ?'

(6) tanhcosh 2 ?> -f 1

Solution. From (5), by division, we get

rush 2 r - 1

en;- h 2 r | 1

cosh 2 r 1 cosh "2 r i 1 cosh- 2 ?> 1

(7) tanh-' r -

Nowcosh 2 7' + 1 cosh 2 M 1 (cosh 2 /' -f 1 j-'

By (B), cosh- 2 v ~ ]~- sinh* 2 r. Hence ('7,< becomes

i ;.,i, - > .,1

(8) tanh-'r-

and therefore tanh r- i

(cosh 2 /' + 1 )-'

sinh 2 r

>sh 2 r 4 1

The sign before the right-hand member must now be examined. From (3)

we have . ,

, 2 sinh r ,, ,, . , , ,smh 2 r ~~

,cosli- r 1 tanh r cosh- r.

cosh r

Therefore sinh 2 f and tanh r will always a^ree in sign. Also cosh 2 r -f- 1 is

always positive. Hence the positive sign must be used, and we get (f>>.

If r is replaced by \ r, (T>) becomes

sinh r

(9) tanh =cosh r -f- 1

PROBLEMS

1. The value of one hyperbolic function is given. F"ind the values of

the others and check as far as possible by the table on page 416.

(a) cosh JT 1.25. (c) sinh .r = 10.

(b) csch .r 0.75. (d) ctnh s = 2.5.

Prove each of the formulas in Problems 27, and compare with the

corresponding formula (if any) in (2), (4)-(6), pp. 2, 3.

2. 1 ctnh 2 r = csch 2r.

3. sinh (r ?r ) = sinh r cosh w cosh r sinh ir,

cosh (r w] = cosh r cosh ir sinh r sinh ?r.

4. tanh (r w)


tanh r tanh w1 tanh r tanh w

, . i r . /cosh r 1 , r /cosh r -f 15. sinh - = A/ cosh - = -f A/

1>

6. sinh r -f sinh // 2 sinh \(r 4 /r) cosh \(r w),

cosh r -f cosh /r = 2 cosh -\(v -f w) cosh J(r /r).

sinh r sinh //

7. tanh J (r w) =cosh r -f cosh /r

8. Show that the equation of the catenary (figure, p. 532) may bei'

written ?/ a cosh -a

9. Solve the differential equation // in terms of hyperbolicu.r-

functions, given that //-= 3 when .r 0, and // when tanh x J.

/l>/,si

. //rr ;{ cosh .r 3.75 sinh jr.

10. Show that sech ( s) = sech s. Draw the ^rapli and provelim sech .r 0.

j- oc

11. Show that ctnh ( .r)-- ctnh .r. Draw the graph arid prove

lim ctnh .r 1.

X-* i QC

12. Show that csch ( .r)= csch .r. Draw the graj)h and prove

lim csch s 0.jr -> ex,

13. Prove (a) sinh 3 i< = 3 sinh // -f 4 sinh' 5 u;

(b) cosh X // 4 coshr // 15 cosh //.

14. Show that (sinh .r -f cosh .r)" sinh ns 4 cosh tijr. (n any positive

integer.)

15. Prove that sinh- ./ sinh jy sinh O -f //) sinh O //).

16 . simplifyC0

;

" + TOS,

h * " A *. ctnh (u + 2 v).sinh 2 // -f sinh 4 r

17. Parametric equations for the fmctrijr may be written

S = t a tanh -?/~ a sech -

a'

a

The parameter is /, and a is a constant. Plot the curve when a 4. (The

tractrix is the curve for which the length of the tangent (Art. 43) is con-

stant and equal to a. Figure in Chapter XXV7

!.)

18. Solve = n 2(y-

ax 2

2 tfi,

Ans. y = A cosh nx -f B sinh nx 4- rax 2H---


214. Derivatives. The formulas, in which v is a function of x, are as

follows.

XXVII 2- sinh v = cosh v ^ -

dx dx

XXVIII 4- cosh v = sinh v -

dx dx

XXIX 4- tanh z; - sech2 ydx dx

XXX ctnh v csch 2 fdx dx

XXXI - sech v = sech v tanh vdx dx

XXXII csch v csch v ctnh i;-^

Proof of XXVII. liy 01), sinh r -( =^

rru ' u1 hen -7- sinh ?'

~dx 2

lJL ''

~

i= COSH /' -;- >

dx

^1).

Formula XXVIII is proved in a similar manner. The proof of

XXIX is analogous to that given in Art. 72 foi the derivative of

tan r. To prove XXX XXXII, differentiate the forms as given in (1),

Art. 211. The details are left as exercises.

215. Relations to the equilateral hyperbola. The curve for which

(1) x = a cosh r, y = a sinh r

are parametric equations is the equilateral hyperbola x- y2 = a 2

.

For, eliminating the parameter r by squaring and subtracting, wehave X L> __ y

- = a 2( c.osh- v ~ sinh'-' r) = a~. By (B)

Fig. 2, p. 421, shows a hyperbolic sector OA P\ bounded by the

arc A PI of (1), the semitransverse axis 0.4, and the radius vector

or i. At PI, ? = ri.


Theorem. The area of the hyperbolic sector OAI*i equals A- ai.

Proof. Let (p, 6) be the polar coordinates of any point on the arc

A PL Then the element of area is (Art. 159) dA =J p- dQ.

Or ,,?/,)

FIG. 1

But p~ = x 2 + y'2 = a- (cosh- + sinh- r).

Also by (5), p. 4,

= arc tan = arc tan (tanh r).

Using (1)

By (1)

Therefore (j[d _ sechr v

dv 1 + tanh- v

Using (C) and (1), Art. 211, we getBy XXII, p. 87, and XXIX

cosh 2 v + sinh- v

and therefore dA i a- r/?;.

The theorem follows by integration, since /? = at ^4. Q.E.D.

The parametric equations of the circle in Fig. 1 are

x r cos /, y = r sin /. Art. 81

The parameter / equals t\ at PI, and /j is the measure of the central

angle AOP\ in radians. Hence the area of the circular sector AOP\is \ r-fi.

Let r = a = 1. Then in Fig. 1, for P(x, y),

x = cos t, y = sin, J t = area ^0/\

In Fig. 2, for P(x, ?/),

x = cosh v, ?/= sinh ?;, i ?; = area AOP.

Hyperbolic functions, therefore, have the same relations to the

equilateral hyperbola as the trigonometric functions do to the circle.

PROBLEMS

1. Show that the element of length of arc for the catenary z/= a cosh -

j.a

is given by J,x cosh - ds.

2. In the catenary of Problem 1 prove that the radius of curvature

i y2

equalstL--

Verify the following expansions of functions by Maclaurin's series, anddetermine for what values of the variable they are convergent.

3. sinh jr = jr 4-~

-f + -f ,./*"_', 4- . Aws. All values

4. cosh x = 1 -f --: -f V -f -f h -A NX. All values.\~ A

Verify the following expansions, using the series in Problems 15 and 4

and the methods explained in Art. 195.

5. sech .r = 1-

.1 .r-' -f .;:,jr*-

7

(>

,' / + .

6. tanh .r = jr-

.',

x'< + A r :> -.r,

7

:,^ 7 + -

7. Test the function 5 cosh .r -f 4 sinh .r for maximum or minimumvalues. .4 NX. Minimum value, .'5.

8. Test the function .4 sinh x -f /> cosh .r for maximum or minimumvalues. .4 NX. If /*- > A 2

,a maximum value \//>- ,4

Jif B < 0;

a minimum value -f \/ />*'-' .4* if B > 0.

9. Derive the series in Problems 15 and 4 from the series for cj and c

T

by subtraction and addition. (Use (^1) and Art. 195.)

10. Let (/x ~ length of the element of arc ; let p~ VVJ

-f ir~ radius

vector of /'(.r, //) for the circle or equilateral hyperbola of Art. 215, and

take limits of integration for the arc .4 l\ in Figs. 1 and 2, p. 421. Prove

(a) |

^ =/, for the circle ; (b) / ?', for the hyperbola.

J p J p

11. Prove lim (cosh .r sinh .r) = 0.j- f oc

12. Evaluate each of the following indeterminate forms.

f ^ v sinh x /u\ v tanh jr

(a) lim (b) hm

An*, -

13. Given tan = sinh x. Prove - = sech x.dx

14. Derive the expansion

arc tan (sinh x} .r J .r< -f ^4 x* Tj J| x

by integration as in Art. 196, using the result yj

in Problem 5.

15. Prove the following theorems for the

tractrix (see figure)

-f

jr = 1 a tanht

y = a sech -

(a) The parameter / equals the intercept of

the tangent on the .r-axis.

(b) The constant a equals the length of the

tangent (Art. 43).

(c) The evolute is the catenary fi= a cosh

/

a

(d ) The radius of curvature PC is a sinh -M X

216. Inverse hyperbolic functions. The relation

(1 ) y = sinh v

is also written

(2) v = sinh l

y,

and read"

v equals the inverse hyperbolic sine of y." Therefore

sinh v and sinhr l

y are inverse functions (Art. 39). The same nota-

tion and nomenclature are used for the other inverse hyperbolic

functions, cosh '

v (" inverse hyperbolic cosine of v "), etc.

The curves

(3) y = sinh x, y = cosh x, y = tanh x

are shown again on page 424. Assume now that y is given.

In Fig. 1, y may have any value, positive or negative, and then

the value of x is uniquely determined.

in Fig. 2, y may have any positive value not less than 1. When

y > 1, x has two values equal numerically and differing in sign.

In Fig. 3, y may have any value numerically less than 1, and then

the value of x is uniquely determined.

Summarized, the results are

The function sinh" 1 v is uniquely determined for any value of v.

Also sinh~ l

( v)= sinh~ !

v.

The function cosh" 1

v, when v > 1, has two values differing only

in sign. Also cosh" 1 1 = 0.

The function tanh" 1 v is uniquely determined when v-< 1. Also

tanh'K 0) = tann" 1v.

FIG. 1 FIG. 2 FIG. 3

Hyperbolic functions were defined in Art. 210 in terms of ex-

ponential functions. The inverse hyperbolic functions are expressiblein terms of logarithmic functions. The relations are

sinh l x --ln(jc l). (Any x)

(O cosh l x =-- In (x i N//2 l).

tanh 1 x = I In/ 1 y

Jl A

Proof of (F). Let v sinh ]

.r. Then

(4) sinh r By

To solve (4) for v, write it as follows :

<"'- 2 JT = 0, or

r

This is a quadratic equation in ev

. Solving, rr = x Vj>' + 1.

Since r1

is always positive, the negative sign before the radical

must be discarded. Hence, using Napierian logarithms, we have (F).

Proof of (G). Let v = cosh l x. Then

(5) jc = cosh /?= - ^ By

Clearing and reducing, we have c2 '' 2 arc

1' + 10.

Solving, cr = x Vx2 ~ 1.

Both values must be retained. Taking logarithms gives (G).


Proof of (//) Let r = tanh- l *. Then

(6) j = tanhr = -=^- By (C)

Clearing of fractions and simplifying, the result is

l+x(x \)e' + (jc + l)e

' = 0. Hence t'-r =

Taking logarithms, we have (//).

1-z

ILLUSTRATIVK EXAMPLE. Transform

(7) 5 cosh .r -f 4 sinh jc

into the form (" cosh (jr -f a), where r and u are constants, and find C and a.

Solution. By (), Art. 213, we have

(8) C cosh (.r + a) T cosh .r cosh a -f (' sinh j* sinh a.

Hence (7) will hav^e the desired form if T and a satisfy the equations

(9) C cosh a - 5, C sinh a 4.

Squaring, subtracting, and using (/?), Art. 210, we get ("' = 9. Then C = -4 3,

since cosh must he positive. Also, by division, tanh a/,.

Hence

H = tanh ' 0.8 - i In 9. By (//)

Therefore a = 1.099 and

(10) 5 cosh .r f 4 sinh JT~~ 3 cosh (JT -f 1.099).

The graph of the function 5 cosh .r j 4 sinh .r may be obtained from the graph

of 5 cosh x by translating the //-axis to the new origin (1.099, 0). (Compare with

Illustrative Example 2, p. 391.)

When x is given, the values of sinh l

.r, cosh ]

j, or Umh !

jc can

be determined by the table on page 41 (> to not more than three signifi-

cant figures. For example, sinh ' 0.25 = 0.247;cosh '

IJ = 1.76.

For greater accuracy (F), (G), or (//) may be used if tables of

Napierian logarithms are at hand.*

217. Derivatives (continued). The formulas, in which v is a function

of x, are as follows. ^v

XXXIII - sinh- 1 v - ,^ - - (Any v)/i;

2 + 1

dv

XXXIV - cosh- 1

i; = d*(v > 1)

dv/I rlY

XXXVcfx 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 correspond-

ing inverse functions to live significant figures may be found from these tables.

Proof of XXXIII. (Compare Art. 75.; Let

y = sintr l

v;

then r = sinh y.

Differentiating with respect to y, by XXVII,

dv ,

-= cosh y ;

dy

therefore ^ = ~ By (C), Art. 39dv cosh y

J \ "

Since v is a function of x, this may be substituted in (A), Art. 38,

djr cosh // dx vV' + 1 d.

[cosh //= Vsinh'-' // 4 1 vV' -f 1, by

The proofs of XXXIV and XXXV are similar. Other formulas are

the following.

(/) ctnh J x - In (x2 > 1)

(7) sech- 1 x = ln- --l- (0 < x g 1)

(K) csch 1 ^-!\A

_dvft rlv

XXXVI -,

_

XXXVII sech ii; =

*(0 < i; < 1)

XXXVIII csch~i t; =- =- (^2 > 0)

Details of the proofs are called for in Problems 58 on the next

page.

PROBLEMS

1. Show that the two values of cosh '

.r in (G) differ only in sign.

2. Draw the graph of ij- \ sinh >

jr. Check in the figure the values

of // and //' when s = 2. An*, tj= 0.72, //'

= 0.21236.

3. Prove XXXIII directly by differentiating (f).

4. Draw the graph of each of the following and check in the figure

the values of ?/ and //' for the given value of jr.

la) ?/= cosh l

JT ; ./ 2.

(bl //= tanh !

.r; .r = - O.Tfi.

5. Prove XXXIV and XXXV.

6. Derive (7) and XXXVI.

7. Derive (/) and XXXVII.

8. Derive (K) and XXXVIII.


tanh '

j .' } -: + -- +.{ ;>

by Art. 19 r>.

10. (riven sinh .r tan c/>.I 'rove

,</'

/

(a) .r = In (sec -f tan 0) ;(b)

^--

- sec 0.

11. Show that csch '

r =- sinh ' 1- Derive XXXVIII from XXXIII,

using this relation.

12. Evaluate lim j- ctnh ' f. An*- 1 -

J X

13. Evaluate lim s csch ] j\

J - X


sinh '' =.'--Ji

-rl + ^lf ""

15. Evaluate lim fsinh' f - In .0. ^w.s. In '2

4- X

16. Show that ctnh ' r = tanh '

y sech ' r ^ cosh J -. and verify

XXXVI and XXXVII from these relations.

d , . tanh a tan r _ s'nb a17. Prove - tanh ^ fl + se(

.

,.

~j + cosh a cos r

'

18. Draw the graphs off a) ?/= ctnhJ r; rb)?y= sech ]

j; (c) ?y=c8ch"1 A

using the theorem of Problem 28, p. 41.


218. Telegraph line. Assume in a telegraph line that a"steady

state"

of flow of electricity from A, the home end, to B, the receiving

end, has been established, with perfect A j ^ p f Binsulation and uniform linear leakage. Pis any intermediate point. We need to

consider :

the electromotive force (volts;, e.m.f., L\\ at A, En at B, E at P;the current strength (amperes), 7 A at A, IB at B, 7 at P

;

the characteristic constants a and ro, whose values depend uponthe linear resistance and leakage. They are positive numbers.

Let x = A P. Then it is shown in books on electrical engineeringthat E and 7 are functions of x such that

We wish to find the e.m.f. and current strength at P. They are

(8) E = E.\ cosh ax r() 7.i sinh ax,

(4) 7 7,1 cosh ax - sinh ax.

Proof. The complete solution of (1) is (Illustrative Example,Art. 210)

(5; E A cosh ax + B sinh ax.

Substituting in (2), the result is

(0) r ( )7 A sinh ax B cosh ax.

But E -~ E i, 7 = J A when x 0. Therefore A ~ EA , B nJA ,

and (5) and ((>) become (8) and (4) respectively.

For the solution in terms of the e.m.f. and current strength at

the receiving end, see Problem 2 below.

PROBLEMS

All refer to a telegraph line in a "steady state," and L AB.

1. Given EA = 200 volts, L = 500 kilometers, r- 4000 ohms, a = 0.0025,

In = 0. Find IA and En.

Ans. 1 1= 0.05 tanh 1.25 = 0.04238 ampere ;

EB = 200 sech 1.25 = 105.8 volts = 0.53 EA.

2. If y = PB = distance of P from the receiving end, show that

E = EB cosh ay + r//? sinh a//, / I fi cosh ay + sinh ay.ro

3. Given EA = 200 volts, I A = 0.04 ampere, r> = 4000 ohms, a = 0.0025.

Show that

E = 120 cosh (1.099 - 0.0025 jr}, I = 0.03 sinh (1.099 - 0.0025 x).

(See the Illustrative Example, Art. 216. Thus A1

tends towards a minimumvalue of 120 volts and / approaches as .r approaches 439.6.)

4. Given E.\ - 1GO volts, 7,i = 0.05 ampere, rn = 4000 ohms, cv = 0.0025,

Show that

E = 120 sinh (1.099 - 0.0025 x\ I = 0.03 cosh (1.099 - 0.0025 .r).

(See the Illustrative Example, Art. 216. Thus E approaches zero and 1

decreases to a minimum value of 0.03 ampere when .r approaches 439.6.)

d'2 I

5. Prove that -pr, a*l = 0. (Thus /; and / are solutions of the samedx 2

linear differential equation, which has the form //" a'2y 0.)

6. Given E,\ = ru/.4. Prove

(a) K= EA caf

, I = IAC*

;

(b) E = r 7 ;

(c) E * when .r becomes infinite.

(Tor example, if TO = 4000, and the impressed e.m.f. at the home end

of the line is 4000 times the current strength, then at mr// point of the line

the e.m.f. is 4000 times the current strength and diminishes towards zero

as the length of the line is indefinitely increased.)

7. In Problem 6 show that the decrease in E at P at unit distance alongthe line from P equals Ec~"

t where e is the Napierian base.

8. Prove the following.

(a) If IK - 0, then E\ - r,,/.i ctnh aL.

(b) If En = 0, then EA = r(J A tanh <*L.

ADDITIONAL PROBLEMS

Derive the following relations.

1. If EA > rnl A and T = tanh- 1 ^~A thenEA

E = EA sech r cosh (r ax), I IA csch T sinh (r ax).

2. If EA < rQIA and T = tanh^ 1 -^X thenro/A

E = EA csch T sinh (T ax), I = IA sech r cosh (r ax).


219. Integrals. A list of elementary integrals involving hyperbolicfunctions and supplementing Art. 128 is given here.

(24) I sinh v dv = cosh v + C.

(25) / cosh vdv= sinh v + C.

(26) jtanh v dv = In cosh v + C.

(27) I ctnh vdv=la sinh v+C.

(28) I sech2 vdv = tanh v+C.

(29) I csch2 vdv=^~ ctnh v+C.

(30) I sech v tanh v dv = sech v+C.

(31) I csch v ctnh v dv csch v+C.

The proofs follow immediately from XXVII-XXXH, except for

(26) and (27). To prove (26), we have

ftanh v dv - C^LH rfr by (c)J J cosh ?'

J \ /

f(/(cosh r) , , . _,i-- _]n eosh ? ,

_j_ ^J cosh /'

The proof of (27) is similar.

ILLUSTRATIVE FAAMPLE. Derive the formulas

(1) I sech v dv = arc tan (sinh r) -f C;

(2) fcsch P dr = In tanh + C.

Solution. Since sech r = -J- = -^LL = cosh r

cosh r cosh- r l-fsmh 2 rJ v 7

we have fsech r dr = f ^^df - f ^sinh

,f)

J J 1 -f sinh- r J 1 -f sinh 2z;

=J d[arc tan (sinh f)] = arc tan (sinh v) + C.

To derive (2) we have (compare Art. 131)

, , csch r ~\ ctnh rcsch r = csch ?

;

-csch r 4 ctnh r

csch- r 4- csch f ctnh /

/ csch r dv = I

=-f-

ctnh r 4- csch r

csch 2 r csch r ctnh r

ctnh r 4 csch r

'(/(ctnh r 4 csch M

dr

ctnh v 4- csch r

= In (ctnh r 4- csch r) 4- ('

= _ln /ilU + .__L_\ + r bv (n Art. 211\suin r sinh rl

In ( cosh r -f 1 ) 4 In sinh r f- (" In ^ f (

cosh r 4 1

= In tanh + r. By (9), Art. 213

PROBLEMS


1. fsinh" r dr ~ \ sinh 'J r \ r 4- r.

2. Tcosh L' r dr = 4 sinh4

J r 4- i r 4- C.

3.|tanh'J r dv = r tanh r 4- C.

4.|ctnh 2 r dr ~ r ctnh r 4- r.

5.|sinh :< r dr = cosh :i

71 cosh ?' 4- C.

6. (cosh :? r d? 1 =fjsinh :? r 4- sinh 7

1

4- r.

7.|tanh3 v dv In cosh r J tanh 2

?' 4- r.

8.Itanh 4 r dr r tanh v >

T tanh:< r 4- r.

9.Jcsch

Hr dr = I csch r ctnh ?'

-\ In tanh ~

4- C.

10.|x sinh x dx x cosh r sinh x 4- C.

1 1. fcos x sinh x dx = I (cos x cosh r 4- sin x sinh r) 4- C./Isinh (rax) sinh (ns) = r fw sinh (nx) cosh (rax)ra~ n z

n cosh (nx) sinh (rax)] + C.



13.J

sinh 4 x dx. 15. Ann x cosh x dx.

14.J^sec-h

4 2 x dx. 16-/-

r tanh x djr '

17.jx 2 cosh x dx. 18.JV sinh a- dx. 19. /V" cosh x dx.

Work out each of the following, using the hyperbolic substitution in-

dicated. (Compare Art. 135.)

20.JVx'2 - 4dx; / = 2 cosh r. Am. \ x^/x'2 - 4 - 2 cosh ' x + C.

/-

"/^TaTTs^" 2 " 1

23. The arc of the catenary //= a cosh - from (0, a) to (x, y) is revolved

about the //-axis. Find the area of the curved surface generated, usinghyperbolic functions.

24. Find the centroid of the hyperbolic sector OA P\ in Fig. 2, Art. 215.

(Compare Problem 12, p. 337.) . -__ 2 sinh r,

- 2 cosh r,- 1/r AtiS. .r (I

~y en

(j,.

'> ?'i

*

3 ?i

220. Integrals (continued). From XXXIII XXXVIII we mayderive integrals. Some of them we have already met in Art. 128.

Their values are now expressible in terms of inverse hyperbolicfunctions.

/______ 2

Vi>2 + a2 dv = V72 + a2 +|-

sinh- 1 - + C.

(39) fV> - a2 rfi; = Vi^~^ _ f*_ cosh-i + cj * <& a

In (33) and (39) the positive value of the inverse hyj>erbolic

cosine must be used, and in (36) the positive value of the inverse

hyperbolic secant.

Proofs of (32) and (33). Let x = - in (F). Then

In (r + vV + <7~) In a.a \a \ u,~ i

Hence

(1) In (v + Vr- + a~) = sinh ! --' + In a.

In the same way, from (G) we get

sinh- 1 - = In - + + 1

(2) In (v + Vr- - a 1') - cosh ' - 4 In a.

Using these results in the right-hand member of (21), p. 193, we

get (32) and (33).

Proofs of (34) and (35). Let x = - in (#). Then

/r^ 11

ft + /;i i ,

?'

(3;2

ln ^T7= t;inh

'a'

Then (34) follows from (3) and (19 a), p. 192.

In the same way, from (I) and (19), p. 192, we get (35).

[in (19), In^ = - In ^- I

I'

r + t> u J

Proof of (36). Since

d (-}

by XXXVIIv~

we have I /= - sech ] -

if the ijositive sign before theJ vV a 2 r- a a

radical is chosen. The proof of (37) is similar.

Formulas (38) and (39) follow from (23), p. 193, using (1) and (2).

REMARK. Since

,,,# 1,0 1,0 i,^ ,, v .,.

ctnh" 1 - = tanh" 1 - seen' 1 - = cosh" 1 - csch l - = smh^ 1 ->

a v a v a v

the integrals (35) -(37) may also be expressed in terms of the functions most con-

venient for use of the table of Art. 212.

ILLUSTRATIVE EXAMPLE. Derive (37) by means of the substitution v a csch z.

(Compare Art. 135.)

Solution. We have

By (2), Art. 211

Also

iv 2 + a'2 - Va~ csch- z + a 2 = a ctrih z.

dr = -a csch z ctnh z dz. By XXXII

t r dv C a csch z 01Therefore /

r- = I

J v vV -f a* J a (>sch z '

a csch 2 ctnh z dz 1,

~: z -f- C.

ctnh z a

Since 2 csch" 1 - we have (37).

PROBLEMS

1. In the figure the curve is the equilat-

eral hyperbola .r'2

y'2 = a-. Using Art. 142,

prove that

(a) area A MJ >

triangle OMP - a'2 cosh ' -;

(b) sector OA I> - \ a 2 cosh l - = a'2r,

if x a cosh ?'. (Thus we have an alternative

proof of the theorem of Art. 215.)

2. Derive each of the following powerseries by integration (Art. 19G).

(a) tanh

(b) sinh

= jr -f~

.r< -f | .rr>

4-<> .)

2 ;* 2-4 5


3. fsinh lJT djr JT sinh '

.r VI -f .r-' 4- C.

4. Mann '

.r dr. 5. (.rcosh" 1 ^

Work out, using hyperbolic functions.

'farrb-9. Find the length of arc for the parabola .r

2 = 4 y from (0, 0) to (4, 4),

using hyperbolic functions. Ans. V20 + sinh J 2 = 5.92.

10. Find the area bounded by the catenary y a cosh - and the line

y = 2a.

11. The downward acceleration a of a falling body is given bya = 32 - I r'

2,and r = 0, s = 0, when / = 0. Find r and ,s.

A nx. r = 8 tanh 4 /, ,s = 2 In cosh 4 /.

221. The gudermannian. The function arc tan (sinh v), which occurs

frequently in mathematics (for example, in (1), Illustrative Example,

Art. 219), is called the gudermannian* of i. The symbol used is

gd r (read"gudermannian of v "). Thus

(1) gd r = arc tan (sinh r).

The derivative is

XXXIX 4- gd. dv

sech v >

dx

assuming r to be a function of jr.

Proof. Differentiating (1), we get-

d

dv

cosh r

But

and

1 + sinh-' v

1 + sinh- r = cosh- v,

1

Then

cosh v

d

sech r.

dxgd v sech r

dr

By XXII and XXVII

by (B)

By (1), Art. 211

By (A), Art. 38

From the definition (1 ) and Art. 77, we have

(2) gd(0)=0;gd (-r)=-z<lt>; gd ( f oo )= .1 TT ; gd (-00)=-- \ TT.

When r increases, gd /' increases (since sech r> 0).

Its value lies between J TT and + i K-

values are given in the accompanying table.

By (1), Art. 219,

(40) I cech vdv = gdv+ C.

To find the inverre function (Art. 39) let

(3)= arc tan (sinh v), ( i TT ^ <t>

<

and solve for r. The result is

(4) v = sinh" l

(tan </>).

From (3) we have tan (p sinh ?/. Since cosh 2v = 1 + sinh'J w,

by (B), the trigonometric functions of <f>,when v > 0, can be read

off from the accompanying right triangle.

Thus sin </>= tanh v, cos = sech v, etc.

The inverse function j (4) may be written ^^^?; = In (sec

7T)

sinh v

(5) + tan 0).

* Named after the mathematician Gudermann. His papers were published in 1830.

t The symbol ^d '0 is used by some writers (r tfd '</>).

Proof. Replace x in (f), Art. 216, by tan <, and note that1 -f tan 2

4> is equal to sec 20, by (2), p. 2.

Conversely, given (5), then = gd ?;.

Proof. Changing to exponentials, (5) becomes

sec + tan (/>= e

r,

or tan cf> er = sec </>.

Squaring both members, substituting sec 2</>= ! + tan 2

0, and

reducing, the result is

- 2 tan </> ^' + e 2" = 1.

Solving for tan</>, we get

/>'' />~

'*

tan </>= - = sinh v.

Hence

2

(j> arc tan (sinh v) gd v.

By W)

Q.E.D.

ILLUSTRATIVE EXAMPLE. In the

tractrix let

a length of the tangent PT(constant by definition) ;

/ - intercept of the tangent onthe x-axis ;

= angle between the tangentline directed upward and the //-axis;

0-0 when / = 0.

Then #((), a) is on the curve.

To prove

(6) <f>=

f T'X

Proof. When / is given, </> is determined. Hence is a function of t. Let thevalues of t and for the tangent line at Q t a point near P, be, respectively,f 4- A/(=O7V

) and</> 4 A</>. Draw 77 r perpendicular to QT

1

'. Let the tangentlines at P and Q intersect at 8. Then in the right triangles UTT and STU wehave TV - !Tr cos LTr ; TV - 7\S sin TSU.Therefore TT' cos [

TT71; = TS sin 7VS7T

.

But angle ITT1 ' =</> + A</>, angle TSV = A0, rr = A/. Hence

A/ cos (0 4 A0) = rs sin A0.Let (? move along the curve towards r, and let Ac/> 0. Then A/ and A</> are in-finitesimals. Also S approaches P and TS . Hence, by the ReplacementTheorem, Art. 98, and (5), Art. 68, we have

dt cos</>= a d<f>, or cf

- = sec d> dd>a

^ v

Integrating, and remembering that 4>= when / = 0, we get

- In (sec </> -f tan </>).

Therefore, by (5), Q.E.D.

PROBLEMS

1. The figure shows the circle .r2-f //- 1 and equilateral hyperbola

x 2y'

1 = 1 in the first quadrant. From M, the foot of the ordinate MPof any point P on the hyperbola, draw AIT

tangent to the circle. Let v = area of the hyper- Y>

bolic sector OAP (Art. 215), and ^ = angle AOT.Prove </>

= gd v.

2. Prove

(a) gd v = 2 arc tan ev TT;

(b) I sinh r tanh v dr = sinh v gd r -f C.

3. Draw the graph of y = gd .r. Calcu-

late y and y' when x = 1. See figure.

Am. # = 0.8ti, ?/-0.65.

4. In the Illustrative Example, p. 43f>, if

P is (x, #), prove that

x = a sin 0, ?/= a cos </>.

From these and (6) derive the parametric equations

x = f a tanh -*?/= a sech -

a'

a

for the tractrix. Find the rectangular equation also.

5. Derive fsech3 v dv | sech v tanh v+Jgdr-fC.

6. If the length of the tangent of a curve (Art. 43) is constant ( a),

/ x dy(a) prove -r1 = -- .

dx Va 2 -y*

(b) Integrate by the hyperbolic substitution y = a sech - and the con-

dition x = when / = 0, and in this way derive the equations of the

tractrix in Problem 4.

- 7. Evaluate each of the following by differentiation.

gd :r xK

r->0(b) lim

X--.0

gd JT sin x

(a)-

i; (b) ^.

8. Using the expansion of Problem 14, Art. 215, we have

gd x = x - \ x3-f & x* - sU?) x 7

-f -.

Calculate the value of gd 0.5 to four places of decimals. Am. 0.4804.

9. The equation (5), p. 435, may be written

v = In tan (\ TT + J (/>).

Prove this statement, making use of (2), (4), and (5), pp. 2, 3,

Ar North Pole

222. Mercator's Chart. The figure shows a portion (one eighth)

of a sphere representing the earth. North Pole N, equator EF,longitude 61 and latitude <t>\ of the point P\ are indicated. Q, with

longitude 0j + A0, latitude

0j + A</>, is a second point,

near P\ on the curve P\QV.The meridians and parallels

through PI and Q are shown.

They form the quadrilateral

PiSQR. We seek expressions

for the circular arcs RQ and 7'j R.

Since O is the center of the

equal arcs RQ and 7'jtf, each

with central angle Ac/>, we have

(1 ) arc 7^Q = arc PiS = a Ac/>.

Also, r is the center of arc PiR, ^with central angle A6. Hencearc 7'i/t = CI'i A0. But, in the right triangle OPjT (right angle

at ('), ^P\ = a cos (/>,. Hence

arc 7'1 7? = a cos 0i A0.

The line 7Ve' is tangent at 7^ to the parallel PiK. The line l\Tis tangent* at V\ to the curve 7'iQV. The angle at P\ between the

curve and the parallel is the angle R'P\ T. Then

(2) tan iT = sec fa -^

the value of the derivative being found from the equation

(3) 0=/0satisfied by the latitude and longitude of each point of the curve

PiQV.

Proof of (2). By the Replacement Theorem, Art. 98, it can be

shown f that

(4) tan R'P 1 T = limarc KQ

A0-o arc P\R

Substituting the values from (1), we get (2).

* Defined as in Art. 28 as the limiting position of the secant through Pi and Q whenQ approaches P\ along the curve P\Q.

t The details are indicated in the Additional Problems (p. 143) Note that arc RQ andarc P\R are, respectively, opposite and adjacent to the angle at Pi in the curvilinear

triangle PiRQ.

On Mercator's* Chart of the earth's surface the point withlatitude <t>, longitude 6, is represented by the point (x, y) such that

(5) x = 0, y = In (sec </> + tan 0),

or, inversely,

(6) = art

= gdy. By Art. 221

In (5) and (6), 6 and 4> are expressed in radians. Meridians (0 = con-

stant) are represented on the chart by lines parallel to the ?/-axis,

parallels (<f>= constant) by lines parallel to the j-axis. The curve

(3) is given by the parametric equations

(7) x = A(/>), y = In (sec c/> -f tan c/>).

Theorem. The angle between a curve on the sphere and an intersect-

ing parallel is unchanged by mapping.

Proof. Let (x\, //i) be the point on (7) where c/>=

c/>i. The parallel

becomes the line y ~ y\ on the chart. Hence we have to prove that

the curve (7) is such that

From (7) and (3), we get

(\H , dx f , , . , (W

^sec</>,_=/(*):,_.

Then (8) follows from (4), Art, 81, and (C), Art, 39. Q.E.D.

Two important corollaries follow.

Cor. /. The angle at 7^ on the sphere formed by two curves,

P\QR and P\Q'Rr

,will equal the angle on the chart at (x\, y\)

formed by the corresponding curves. Hence angles remain un-

changed by mapping.Cor. 77. A straight line on the chart with slope tan a corresponds

to a curve on the sphere cutting all parallels under the same angle a.

This curve is called a rhumb line (or loxodrome).

Along a rhumb line

(9) < = gd(0tan + &).

This follows from (6) and y = x tan a + b. The course of a ship

proceeding always in the same direction lies along a rhumb line. In

the representation (5), 0, and therefore x, has values from -K

to+ TT, inclusive. On the other hand,?/ may have any value (Art. 221).

Hence the entire surface of the earth is mapped on the strip of the

r?/-plane determined by the lines x = TT and x = + IT.

* Grrardus M creator fl 512-1 594), a noted cartographer, published his Chart of tho

World in lf>69 His name is the Latinized form of Gerhard Kremer.

By the table of Art. 221 we may find the latitude in degrees of the parallels

which are given on the chart by the lines y = constant.

223. Relations between trigonometric and hyperbolic functions. Let

the exponent v of the exponential function e1

'

be a complex number

x -f /// (JT and // real numbers; / V l). Then we assume as a

definition

(1 ) cj * <" = (>*(>' = (''(cos y + i sin y).

If x = 0, we have (sec p. 391)

(2) e'" cos'// + i sin y.

Change y to y. Then (2) becomes

(3) (>-'" = cos // / sin ?/.

Solving (2) and (3) for sin y and cos y, the results are

cn< - c

'"

(4) sin y = cos yz /

"2

Thus the sine and cosine of a real variable are expressed in terms

of exponential functions with imaginary exponents.

Formulas (4) and (A) suggest definitions of the functions con-

cerned when the variable is any complex number z. These definitions

are(>''

**(>1Z -4_

W

(5) sin z = ' cos z =

sinh z = cosh z

* The lines .r = - TT and j- = -f TT re-present the same meridian (180 W. or 180 E.).

It is assumed that this meridian does not cross the curvilinear triangle. In the figure, A\

and B represent the same point on the earth, us do also B\ and C.

The other trigonometric and hyperbolic functions of z are defined

by the same ratios as are used when the variable is a real number.

From (5) we may prove the following :

(L) sinh iz = i sin 2, cosh iz cos z.

[fit_ e it 1

sinh iz = = i sin z, using (5) ; etc. I

From (L), by division, we get

(6) tanh iz = i tan z.

The similarity of many formulas in this chapter to others for

trigonometric functions is explained by the relations (L) and (6)

(see Illustrative Example 2). The right -hand members of (f>) are

expressible as complex numbers whose real parts involve only trigo-

nometric and hyperbolic functions of real variables. This appears

below in Illustrative Example 1.

ILLUSTRATIVE EXAMPLE 1. Derive the formula

(7) sinh (jc + iy) = sinh x cos // -f i cosh x sin y.

Solution. By (5), if z - x + iy, we have

c* * "/ _ c

r iv

(8) sinh (x + iy}=

TJ

(9)= ^(cos y + i sin ?y)

-f-*(ros y

- i sin y)^ By (1 )

By (1), Art. 210, if v = x, we have

cr cosh x + sinh x, c~T = cosh x - sinh x.

Substitute these values in (9), and reduce. The result is (7).

Changing i to ?, (7) becomes

sinh (x iy} sinh x cos //i cosh j- sin y.

'

The form of the right-hand member here and in (7) should be noticed.

ILLUSTRATIVE EXAMPLE 2. Prove directly by (5) the relations

sin 2 z + cos 2 2=1, cosh 2 z sinh-' 2=1.

, Solution. The details are the same as in the proof of (#), Art. 210.

The first relation may be derived from the second as follows:

Let z- iv. Then cosh 2 iv - sinh 2 iv = 1. But, by (L), cosh iv - cos r,

sinh iv i sin v. Hence cos 2 v -f- sin 2 v 1.

PROBLEMS

1. Using differentials, show that the distance apart on M creator's Chart

of the lines parallel to the z-axis which represent the parallels at latitudes

</>i and , respectively, varies as sec c/>i.

2. Along a rhumb line = gd (6 tan a + b). Prove by differentiation

j d<t>that tan a = sec cp -f-

du

442 DEFERENTIAL AND INTEGRAL CALCULUS

3. The altitude h of the zone on the sphere bounded by the parallels

</>= fa, c/>

= 0i (fa > <t>}) is " ( iri fasin fa) (see figure, p. 438). If

=?/2 ,'

=?/, are "the corresponding parallels on the map, prove the

following.

(a) h aCtanh 7/2 tanh ?/i ) ;

i

(b) (li/~~ sec 2

</>i d//, if 0;> 0i 4~ cz0.

4. Using (b), Problem U, show that equal zones of small altitude whose

lower bases are parallels at latitudes 0, 30", 45', 60, respectively, map

into rectangles whose areas are as 3:4:6: 12. (The area of a zone equals

its altitude times the circumference of a great circle.)

5. Describe the direction of a curve on the sphere (a) if -;

(V>) if becomes infinite.

dO

6. Derive each of the following formulas by the method used in Il-

lustrative Example 1.

(a) cosh (jr -f /'//)cosh jr cos y 4- ? sinh x sin y ;

(b) sin (x 4 ?'//)= sin x cosh ?/ -f / cos x sinh y ;

(c) cos (x -f /'//)= cos x cosh ?/

- / sin x sinh ?/.

From these write the values of cosh (x-

?'//), sin (x-

??/), cos (x-

iy).

7. Prove (a) sinh ( ? J .r

j

= ? cosh x;

(b) cosh ( i~ x

}= / sinh x,

8. Evaluate each of the following to two places of decimals.

(a) sinh (1.5 -f /) ; (b) cosh (1-

?);

(c) cos (0.8 -f 0.5 /') ; (d) sin (0.5 + 0.8 0-

ANK. (a) 1.15 -h 1.98 i; (c) 0.78 - 0.37 i.

ADDITIONAL PROBLEMS

1. In the figure of Art. 2l?2, P\M\ is drawn perpendicular to CR,

and therefore perpendicular to the plane of the meridian NQR. Then

triangle PiQMi is a right triangle (the chord PiQ is not shown), and

tan MiPiQ = -^^.- When M -> 0, the line Pi A/i (produced) approachesPi Mi

the tangent PiR', and angle MjPiQ approaches angle R'PiT. Therefore

(10) tan/

Pir = lim^^-v '

Compare with (4), Art. 222, and show that

(a) lim Pl**\ = I (see Fig. 1) ;

Afl-oarc P\It

(h) lim Afl(^ = 1 (see Fig. 2, which showsv . oarc KQ

the plane of the meridian In triangle .V

show that Af i K is an infinitesimal of

higher order than QR when A6 and JK/> are of

the same order (Art. 99;. Then see Problem,p. 148.

Using fa) and (h) and the Replacement Theo-

rem, Art. 98, (10) becomes (4), Art. 222.

2. If r/.s'i is the element of the length of arc for

a curve on the sphere of Art. 222, prove that

dx }

'

2 =a'-((l<f>'

2-4- cos-' <W2

). (In the figure of

Art. 222, (chord I\Q} 2 = Pi MI" + yj

andlirnc->nord/>l(?^l.)arc PiQ

3. If dx is the differential of the arc of a curve on M creator's Chart,show that r/.s- sec 2 0u70 L'

-f cos 2<10~). (Comparing with Problem 2,

we have dx\~ = a'2 cos 2(/.s-

2.)

4. Find the length of a rhumb line between points whose difference

of latitude is A0. Ann. a esc a A0. (a = radius of the earth.)

5. Prove that the first four formulas in (4), p. 3, and (>), (), Art. 213,

hold when JT, y, r, /r are replaced by complex numbers. (Use the defi-

nitions (5).)

6. Prove the formulas of Problem 6, p. 442, by using the results in

Additional Problem 5 and (L).

~ n <^u ^ 4. u / i \ smh 2 J' + ' sm 2 ?/

7. Prove that tanh (.r 4- /?/)=:

:~*- m

cosh 2 r -I- cos 2 ?/

8. I>erive the formula for tan (.r -f- it/) from the result in the preceding

problem.

CHAPTER XXIII

PARTIAL DIFFERENTIATION

224. Functions of several variables. Continuity. The preceding

chapters have been devoted to applications of the calculus to func-

lions of one variable. We now turn to functions of more than one

independent variable. Simple examples of such functions are af-

forded by formulas from elementary mathematics. Thus, in the

relation for the volume v of a right circular cylinder,

(1 ) v = irjr'-y,

v is a function of the two independent variables x ( radius) and

?/ ( altitude). Again, in the formula for the area u of an oblique

plane triangle,

("2) u =], yy sin otj

u is a function of the three independent variables r, ?/, and a, rep-

resenting, respectively, two sides and the included angle.

Obviously, in (1), as well as in (2), the values which can be as-

signed to the variables in the right-hand member are entirely in-

dependent of one another.

The relation

('*> z = f(f,y)

can be represented graphically by a surface, the locus of the equation

(I}) obtained by interpreting j, //, z as rectangular coordinates, as in

solid analytic geometry. This surface is the graph of the function

of two variables, f(jr, ?/).

A function /(.r, ;//) of two independent variables x and y is defined

as continuous for .r = a, y ~ b, when

04) lim/(x,y)=/(a, b),X -*

*- >>

no matter in what way x and y approach their respective limits a

and b.

This definition is sometimes roughly summed up in the statement

that o very small change in one or both of the variables produces a

very small change in the value of the function.*

* This will be hotter understood if the student again reads over Art. 17 on continuousfunctions of a single variable.

444

PARTIAL DIFFERENTIATION 445

We may illustrate this geometrically by considering the surface

represented by the equation

(3) z=f(x,y^.

Consider a fixed point P on the surface where :r = a and y b.

Denote by A.r and A// the increments of the variables x and

//, and by Ac the corresponding increment of the function z, the

coordinates of P' being

At P the value of the function is

z=f(a, b) = MP.

If the function is continuous at P, then how-

ever A.r and A?/ may approach zero as a limit , A::

will also approach zero as a limit. That is, 717 'P' will approach coin-

cidence with MP, the point P' approaching the point /' on the sur-

face from any direction whatever.

A similar definition holds for a continuous function of more than

two variables.

In what follows, only values of the variables are considered for

which a function is continuous.

225. Partial derivatives. In the relation

(1) z=/Cr, ?/),

we may hold y fast and let x alone vary. Then z becomes a function

of one variable x, and we may form its derivative in the usual manner.

The notation is

= partial derivative of z with respect to x (y remains constant).*-

c)x

Similarly,

= partial derivative of z with respect to y (x remains constant).*c)y

Corresponding symbols are used for partial derivatives of functions

of three or more variables.

In order to avoid confusion the round b \ has been generally

adopted to indicate partial differentiation.

* The constant values are substituted in the function before differentiating.

t Introduced by Jacobi (1804-1851).


ILLUSTRATIVE EXAMPLE 1 . Find the partial derivativas of z = ax 2 + 2 bxy + cy2.

Solution. Y1 = 2 ax -f 2 fr#, treating # as a constant,

i)"-r- - 2 ore -f 2 n/, treating x as a constant.

ILLUSTRATIVE EXAMPLE 2. Find the partial derivatives of u sin (ax + by + cz).

Solution. ~ ~ a cos (ax -f by 4- re), treating # and z as constants,

~fr cos (ax 4 by + cz), treating x and z as constants,

-f by -f r^), treating # and x as constants.~ _ c cos

Referring to (I), we have, in the notations commonly used,

~ =T^/(y,

y) = -^=/r (j, ?/) =/, = zx ;

Similar notations are used for functions of any number of variables.

Referring to Art. 24, we shall have

(2)

(3)

*) = MmAJ- .

/;,(.r(l , y) = lirn ^^-Aj;

226. Partial derivatives interpreted geometrically. Let the equationof the surface shown in the figure be

z= /Or, y).

Pass a plane EFGI1 through the

point /' (where x ~ a and y ft) on

the surface parallel to the XOZ-

plane. Since the equation of this

plane is ,

y^b*

the equation of the curve of inter-

section JPK with the surface is

2=/Cr, b\

if we consider EF as the axis of Z and EH as the axis of X. In this

U(] :

plane TT- means the same as -r1

' and we havefa dx

(1) |5 = tan MTP = slope of curve of intersection JX at P.ox

Similarly, if we pass the plane BCD through P parallel to the

YOZ-plane, its equation isj a

and for the curve of intersection DPI, means the same asj^-

Hence

(2)~ = tan MT'P slope of curve of intersection DI at P.

ILLUSTRATIVE EXAMPLE. Given the ellipsoid~- 4- ~-

4-~ 1 ; find the slope

LI4 1 !_ (>

of the curve of intersection of the ellipsoid made (a > by the plane // 1 at the point

where x = 4 and z is positive ; (b) by the plane j = 2 at the point where j/= H and

z is positive.

Solution. Considering y as constant,

24 6 to~~

'

to~

4 z

?/4> "*

()~ ft" yWhen J is constant, ~f + -rr

1 = 0, or - - --f-

12 () c^/ _ f).v 2 ^

(a) When ?/= 1 and r 4, 2; = x/h"

*

T.~

1 <*)' *J

(b) When x - 2 and y = ,'J,2 = = /. ~ - - - \ 2. A/ts.

PROBLEMS

Work out the following partial derivatives.

2. /(j, ?

a-f 2 to?/ -f

2-f 2 f:jr// -f

_ A JT + /?// x ^ - M/>- /^') ?y

3. /(JT, ?y)- AMS - ":

-

-- 2 '

4. 7 < = /?/ -f z/z 4- ZJT. An.s. M T-

// 4- z ;w v= s 4- 2

; n, ~jr + y.

5. /(j, ?/)= Cr 4- ?y) sin fx - ?/).

Ans. //JT, i/)- sin (r - ?/) + (r + ?/) cos (x

-?/) ;

/y (j- t T/)= sin (x - y)

~ (* 4- y) cos (r y).

6. p = sin 2 cos 3 </>.An.

|^= 2 cos 2 cos 3 ^ ;

^= 3 sin 2 6 sin 3 (/>.

7)- sin (fl-

</>)} ;

^P^ e 4c/>rcog ^_ 0) + sin (^_ 0^| e

Find the partial derivatives of the following functions.

8. f(s, y) = 3 jr4 - 4 s*y + 6 jr*tf. 11. f(s, y) = (x + 2y) tan (2 x + I/).

Q y + ^ ?/

-~0 + 2/ 12. p = tan 2 (9 ctn 4 0.

10. z = fMn 2. 13. p = < cos^-^"

14. If /(V, ?/)= 2 .r* - 3 jry -f 4 ?/*, show that /z (2, 3) = -

l,/v (2, 3) = 18.

15. If/O, //)=

r~j-show that/, (3, 1) = -i,/,,(3, lj = jj.

16. If/(r, //)= <' 'sin (JT + 2 ?/), show that /,((>, TW ~

1, ^0, -W 0.\ 4/ \ 4/

17. If w = ^1r4

-f 2 Rr 1'//

2-f ry , show that jr -f ?/

= 4 w.r.r

'

f'?/

18. If M = ~;^--

, show that jr -f ?/= 3 ?/.

^ 4- // r'x'

r'/y

19. If u - jr*u -f ifz -f- rV, show that -f + =(jr -f y + z)

2.

fj- cy (Z

20. If u = Ajrl+

'ty*' ^ow that .r -f ?/= (w - 2)w.TJ- -f />//- rx r -,;

21. The area of a triangle is given by the formula K o be sin /4.

Given b 10 in., r - 20 in., .4 = GO".

(a) Find the area.

(b) Find the rate of change of the area with respect to the side b if r

and A remain constant.

(c) Find the rate of change of the area with respect to the angle .4 if

/> and c remain constant.

(d) Using the rate found in (c), calculate approximately the change in

area if the angle is increased by one degree.

(e) Find the rate of change of c with respect to b if the area and the

angle remain constant.

22. The law of cosines for any triangle is a 2 = b 2-f c 2 - 2 be cos A.

Given b = 10 in., c = 15 in., A = GO .

(a) Find a.

(b) Find the rate of change of a with respect to b if c and A remainconstant.

(c) Using the rate found in (b), calculate approximately the change in

a if b is decreased by 1 in.

(d) Find the rate of change of a with respect to A if b and r remainconstant.

(e) Find the rate of change of c with respect to A if a and b remainconstant.

227. The total differential. We have already considered the dif-

ferential of a function of one variable in Art. 91. Thus, if

=/(*),we defined and proved

(1) cfy=

We shall next consider a function of two variables. Consider the

function

(2) w=/(j, //).

Let Ax and A?/ be the increments of x and // respectively, and let

Aw be the corresponding increment of u. Then

(3) Aw = /(x + Ax, // + A//)-

f(jr, //)

is called the total increment of u.

Adding and subtracting /U, y -f A-//) in the second member,

(4) Aw = f f(x + Ax, '// -f A-//) ~/{x, // + A//) ]

-H/Or, '// + A//) -/U, //)|.

Applying the Theorem of Mean Value (Z)), Art. 116, to each of

the two differences on the right-hand side of (4), we get, for the

first difference,

(5) /(x + Ax, y + A?/)-

J(x, y + Ay/) = /, (x + 0, Ax, y/ + Ay/) Ax.

|fj= x, Aa ~ A/, arid since* j varies while n f A// remains!

'constant, \vt' ^ct the partial derivative \vith resj)ecl toj.l

For the second difference,

(6) /(x, v + Ay/)-

/(x, //)- //y (x, // + 6. A//) A?/.

I

=//, A^/ = A?/, and since // varies while .r reinuins con-

J

stant, we got the partial derivative with respect to //-I

Substituting from (5) and (6) in (4) gives

(7) A^ - f,(r + 0i Ax, y + Ay/)Ax + /,,(x, y + 0- A?/) A?/,

where 6\ and 02 are positive proper fractions.

Since J,(x, y) and fu (x, y) are continuous functions of x and y, the

coefficients of Ax and A?/ in (7) will approach f r (x 9 y) and fu (x, y),

respectively, as limits when Ax and &y approach zero as common

limits. Hence if e andf

are infinitesimals such that

lim 6 = 0, lim e' = 0,AJ- - Aj -.

ly A?y-

we may write

(8) /,(z + 0i Ax, ?/ + Ay) - /x (x, |/) + ,

(9) fv (x, y + 6<2 A?/)= fy (x, y) + e',

and (1) will become

(10) &u=ff (x, y)kx+Sy(x, y)&y + *x + 'A?y.

We then define as the total differential (= du) of u

(11) <fa =/,(*, y)&x+fy (x, y)y.

The right-hand member in (11 ) is the "principal part" of the right-

hand member of (10), that is, du is a close approximate value of \u

for small values of \s and A// (compare Art. 92). Obviously, if

u = x, (11 ) becomes rfj = Ax. If w = //, (11 ) becomes dy = A//.

Substituting, then, in (11) for AJ and A// their corresponding dif-

ferentials, we obtain the important formula

(B) du - /,(*, y) dx + /(*, y) dy -j^

dx + ^ dy - ^ dx + ^ dy,

which should be compared with (1) at the beginning of this article.

If u is a function of three variables, its total differential is

(C) dv^dx + ^dy+'^dzi(x (y (i

and so on for any number of variables.

A geometric interpretation of (B) is given in Art. 238.

ILLUSTRATIVE EXAMPLE 1. Compute An and du for the function

(12) n = 2 J-- -4 3 /r,

when x= 10, /y~ 8, A>r = 0.2, Ji/y

= 0.3, and compare the results.

Solution. Substitute in (12) for j-, //, /<, respectively, x + AJ*, // + A/y, M + A?J,

and proceed as below (compare Art. 27).

u + A// - 2(.r -r- A.D- 4- 3(/y -f A?y)J

^ 2 jr* -f 3 /y-' 4- 4 J- A.r -f 6 yy A/y 4- 2(A.r)^' + 3(A?y)2

.

M = 2 .r- + 3 /y

j

(13) AM = 4 j- A.r 4 G // A/y -f 2( A.r)"' + 3(A#)2

.

Differentiating (12), we find

(111 . ()l(= 4 .r, =6 y.dx dy

Substituting in (B), the result is

(14) du = 4 x dx 4- 6 /y J/y.

Remembering that AJ = dr, A?y = <///, we see that the right-hand member in

(14) is the "principal part" of the right-hand member in (13), for the additional

terms are of the second degree in AJ* or A?y. This statement illustrates (10) and

(11) above (namely, e = 2 A.r,' = 3 A/y).

Substituting the given values in (13) and (14), we get

(15) AM = 84- 14.4 -f 0.08 4- 0.27 = 22.75 ;

(16) du = 8 + 14.4 =22.4.

Then AM - du = 0.35 = 1.6% of AM. Ans.

ILLUSTRATIVE EXAMPLE 2. Given u - arc tan ^, find du.x

e i * ^" V <"'* .rSolution.

(V .r* -f ir diji'

2 +

Substituting in (5), (/

^ ( ^~

^.(J

.4ns.

PROBLEMS

Find the total differential of each of the following functions.

1. z = 2 JL':{ - 4 .n/-' + 3 //

:l

. ^ N,S-. d; - (fi .r-'- 4 ij-)djr -f (9 //

2 - 8 .r?/ )<///.

((V-f- />//)'-'

3. u = jryW. dn = ir^ djc 4 2 .r//c-!

dy -f .S .r//'-',:

1

-'

(/c.

4. u .r- cos ^ //. 5. 6> = arc tan -6. n (.r

-//) In (.r -f //).

7. If ^ 4- >r + ^ - a^, show that d; - - J'

<*' +

8. Find ^ if 4 j-' - 9 //-'~ 16 c j = 100.

9. Compute A// and </// for the function u .r- 3 .r// -f 12 //

L> when.r = 2, y = -

3, A.r = -().:>>, A// =r 0.2. An*. A// = -

7.15, du = - 7.5.

10. Compute dn for the function // (j- -f //) V.r // when .r (J, //= 2,

J.r ~ \, dij-

A. /1//X. 1.

11. Compute A// and r/w for the function M j*/y -f 2 r - 4 ?/ whenx - 2, t/

= 3, A.r = 0.4, A?/ = - 0.2.

12. Compute <7p for the function p = c* sin (0 </>) when 0,

<t>=

J TT, cZO = 0.2, c/0 = - 0.2.

228. Approximation of the total increment. Small errors. For-

mulas (B) and (C) are used to calculate Au approximately. Also,

when the values of x and y are determined by measurement or ex-

periment, and hence subject to small errors AJ and A?/, a close ap-

proximation to the error in u can be found by (B). (CompareArts. 92, 93.)

ILLUSTRATIVE EXAMPLE 1 . Find, approximately, the volume of tin in a thin

cylindrical can without a top if the inside diameter and height are, respectively,

6 in. and 8 in., and the thickness is 1 in.

Solution. The volume v of a solid right circular cylinder with diameter x and

height y is

(1) v = \ vx*y.

Obviously, the exact volume of the can is the difference Ar between the volumes of

two solid cylinders for which x = 6J, y - 8, and x = 6, y = 8, respectively. Since

only an approximate value is required, we calculate dv instead of Av.

Differentiating (1), and using (B), we get

(2) df = \ Trxy dx + | irx'2 dy.

Substituting in (2) x = 6, y = 8, dx = \, dy = 1, the result is

dv 7i TT = 22.4 cu. in. Ans.

The exact value is At? - 23.1 cu. in.

ILLUSTRATIVE KXAMI'LE 2. Two sides of an oblique plane triangle measured,

respectively, 63 ft. and 78 ft., arid the included angle measured 60'. These meas-urements were subject to errors whose maximum values are 0.1 ft. in each lengthand 1 in the angle. Find the approximate maximum error and the percentageerror made in calculating the 1 third side from these measurements.

Solution. Using the law of cosines ((7), Art. 2),

(3) it- = x 2 + y'2 - 2 xy cos

,

where x, ij are the given sides, a the included angle, and u the third side. The givendata are

(4) r = 63, ?/-78, = 60 =, dx = d/y= 0.1, da = 0.01745 (radian).

o

Differentiating (3), we get

J ~~y (l()s c*

Of* y ~ x (><)s (y ^ u ~~ *y s^n adx H dy n Oa >'

Hence, using (C),

, (.r //'_cos (\')(1x -} (// .r (*os <v)dij -\- xij sin cy dcx

Substituting the values from (4), we find

/2.4 +4.65 + 74.25

, 10 . .

du ~ - = 1.13 ft. Aiis.< 1. 1

The percentage error = 100 =. 1.6'\ . An*.H

PROBLEMS

1. The legs of a right triangle measured 6 ft. and 8 ft. respectively,with maximum errors in each of 0.1 ft. Find approximately the maximumerror and percentage error in calculating (a) the area, (b) the hypotenuse,from these measurements. AHX. (a) 0.7 sq. ft., 2.9 (

\ ; (b) 0.14 ft., 1.4 <

;.

2. In the preceding problem find the approximate error in calculatingthe angle opposite the longer side from the given dimensions, and the

approximate maximum error in that angle in radians and degrees.

3. The radii of the bases of a frustum of a right circular cone measure5 in. and 11 in. respectively, and the slant height measures 12 in. Themaximum error in each measurement is 0.1 in. Find the approximateerror and percentage error in calculating from these measurements (a) the

altitude; (b) the volume (see (12), Art. 1).

-4?w. (a) 0.23 in., 2.2 <; ; (b) 32 TT cu. in., 4^ '~,


4. One side of a triangle measures 2000 ft., and the adjacent angles

measure 30 and 60 respectively, with a maximum error in each angle

of 30'. The maximum error in the measurement of the side is 1 ft.

Find the approximate maximum error and percentage error in calculating

from these measurements (a) the altitude on the given side; (b) the area

of the triangle. AN*, (a) 17.88 ft., 2.1 %.

5. The diameter and altitude of a right circular cylinder are found by

measurement to be 12 in. and 8 in. respectively. If there is a probable

error of 0.2 in. in each measurement, what is approximately the greatest

possible error in the computed volume? Ans. 16.8 TT cu. in.

6. The dimensions of a box are found by measurement to be 6 ft.,

8 ft., 12 ft. If there is a probable error of 0.05 ft., (a) what is approxi-

mately the greatest possible error in the computed volume? (b) What is

the percentage error? Aw,s. (a) 10.8 cu. ft. ; (b) *%.

7. Given the surface z - ^-^-If, at the point where x - 4, y = 2,

s + //

x and y are each increased by A, what is the approximate change in 2?

P8. The specific gravity of a solid is given by the formula s where

P is the weight in a vacuum and w is the weight of an equal volume of

water. How is the computed specific gravity affected by an error of

A in weighing P and ^ in weighing w, assuming P - 8 and w = 1

in the experiment, (a) if both errors are positive? (b) if one error is

negative? (c) What is approximately the largest, percentage error?

An*, (a) 0.3; (b) 0.5; (c) 6i%-

9. The diameter and slant height of a right circular cone are found by

measurement to be 10 in. and 20 in. respectively. If there is a probable

error of 0.2 in. in each measurement, what approximately is the greatest

possible error in the computed value_of (a) the volume? (b) the curved

surface? Af^ (a)37 7rVl5 = ^ cu Jn .

(h) 8 ^ =^ ^ m18

10. Two sides of a triangle are found by measurement to be 63 ft. and

78 ft. and the included angle to be 60. If there is a probable error of

0.5 ft. in measuring the sides and of 2 in measuring the angle, what is

approximately the greatest possible error in the computed value of the

area? (See (7), Art. 2.) Am. 73.6 sq. ft.

^11. If specific gravity is determined by the formula s = _ where

A is the weight in air and W the weight in water, what is (a) approxi-

mately the largest error in s if A can be read within 0.01 Ib. and W within

0.02 Ib., the actual readings being A = 9 Ib., W - 5 Ib.? (b) the largest

relative error? An. (a) 0.0144; (b) ?$fa.

454 DIFFERENTIAL AND INTEGRAL CALCULUSjrt

12. The resistance of a circuit was found by using the formula C =R

where C = current and / = electromotive force. If there is an error of

1*5 ampere in reading C and 3*5 volt in reading E, (a) what is the approxi-mate error in It if the readings are C = 15 amperes and E 110 volts?

(b) What is the percentage error? Am. (a) 0.0522 ohms; (b) ||f

;.

13. If the formula sin (x -f y) = sin jr cos // -f cos x sin y were used to

calculate sin (x + y), what approximate error would result if an error of

0.1 were made in measuring both x and ?/, the measurements of the twoacute angles giving sin x

j?and sin y -f3 ? Ans. 0.0018.

14. The acceleration of a particle down an inclined plane is given bya = g sin i. If g varies by 0.1 ft. per second per second, and ?, which is

measured as 30, may be in error 1, what is the approximate error in the

computed value of a ? Take the normal value of g to be 32 ft. per second

per second. Ans. 0.534 ft. per second per second.

15. The period of a pendulum is P 2 ir\-

(a) What is the greatest* g

approximate error in the period if there is an error of 0.1 ft. in measur-

ing a 10-foot suspension and (/, taken as 32 ft. per second per second, maybe in error by 0.05 ft. pe*r second per second? (b) What is the percentageerror? Ann. (a) 0.0204 sec. ; (b)gj

r;.

16. The dimensions of a cone are radius of base = 4 in., altitude = 6 in.

What is the approximate error in volume and in total surface if there is a

shortage of 0.01 in. per inch in the measure used?

Ans. dV - 3.0159 cu. in. ; dS = 2.818 sq. in.

17. The length I and the period P of a simple pendulum are connected

by the equation 4 ir'2l P'2g. If / is calculated assuming P = 1 sec. and

g = 32 ft. per second per second, what is approximately the error in / if

the true values are /' = 1.02 sec. and g~ 32.01 ft. per second per second?

What is the percentage error?

18. A solid is in the form of a cylinder capped at each end with a

hemisphere of the same radius as the cylinder. Its measured dimensions

are diameter = 8 in. and total length 20 in. What is approximately the

error in volume and surface if the tape used in measuring has stretched

uniformly r; beyond its proper length ?

19. Assuming the characteristic equation of a perfect gas to be rp Rt,

where r = volume, p = pressure, t = absolute temperature, and R a con-

stant, what is the relation between the differentials dr, dp, dt ?

Ans. v dp + p dv = R dt.

20. Using the result in the last example as applied to air, suppose that

in a given case we have found by actual experiment that t = 300 C.,

p = 2000 Ib. per square foot, v = 14.4 cu. ft. Find the change in p, assum-

ing it to be uniform, when / changes to 301 C., and v to 14.5 cu. ft.

R = %. Ans. 7.22 Ib. per square foot.

229. Total derivatives. Rates. Turn now to the case where x and y in

CD M=/(*,0)

are not independent variables. Assume, for example, that both are

functions of a third variable /, namely,

(2) r = <t>(U, ?/=M)-When these values are substituted in (1), M becomes a function of

one variable t, and its derivative may be found in the usual manner.We now have

(3) <f = f *. ^ = f *. dv^dt.Formula (B) was established with the assumption that x and y

were independent variables. We may easily show that it holds also

in the present case. To this end, return to (10), Art. 227, and divide

both members by A/. Changing the notation, this may be written

(A) r - i

4. 4./

M ox M ry A/ \ Af A/

Now when A? > 0, A:r and A// -> 0. Hence (see Art. 227)

lim e = 0, lim e' = 0.A/ - A/ -

Therefore, when A/ > 0, (4) becomes

rfu _ fiu c?x,

f^u dy( '

dt~~ ax dt ny dt

'

Multiplying both members by dt and using (3), we obtain (B).

That is, (B) holds also when x and y are Junctions of a third variable t.

In the same way, if u /(x, y, z),

and x, ?/, z are all functions of /, we get,

du __ (hddx_ , _ ^ ,

f

( 'dt~

f>x dt cJydt

'

<)z dt

and so on for any number of variables.

In (D) we may suppose t x;then y is a function of x, and u is

really a function of the one variable x, giving

djc f)x r>y dx

In the same way, from (E) we have, when y and 2 are functions

of x'

du __

f

du,

r;u jy . <r)udz

dx 'ox by dx f)z dx

The student should observe that and -^ have quite differentox dx

meanings. The partial derivative is formed on the supposition'Ox

that the particular variable x alone varies, all other variables being

held fast. But , . A(111 i / *_a 1JL \= limdx AX .0 \Ax/

where AM is the total increment of u due to changes in all the variables

caused by the change Ax in the independent variable. In contra-

distinction to partial derivatives,(

-j> ^ are called total derivatives

with respect to / and x respectively.

It should be observed that has a perfectly definite value for any

point U, y), while -, depends not only on the point (x, y) but also on

the particular direction chosen to reach that point.

ILLUSTRATIVE EXAMPLE 1. Given M sin -, JT - <', y = t'2

; find~y dt

, A . du 1 x du x x d.r, dy .

Solution. - cos -,- cos -

;= c',

- - 2 /.

dx // y dy // y dt dt

Substituting in (D), (/ 2) cos Am.at r i

2

iLLUSTKATivK'KxAMPLE 2. Given // = f"T

(y z), y = a sin x, z cos x ; find ~-ax

Solution. <u"'J(y

-

z}, =- r"r

,

-

r- = -

IM *

;

- a cos x,- sin x.

dx'

dy dz dx dx

Substituting in (G),

M __ acnr(y _-)-(- (ienr cos x -I- r

T sin x = f"T(a'

2 + 1) sin x. Ans.

NOTE. In examples like the above, n could, by substitution, be found explicitly

in terms of the independent variable and then differentiated directly ; but generally

this process would be longer and in many cases could not be used at all.

Formulas (D) and (E) are useful in all applications involving

time-rates of change of functions of two or more variables. The

process is practically the same as that outlined in the rule given in

Art. 52, except that, instead of differentiating with respect to t

(Third Step), we find the partial derivatives and substitute in (D)or (E). Let us illustrate by an example.

ILLUSTRATIVE EXAMPLE 3. The altitude of a circular cone is 100 in., anddecreases at the rate of 10 in. per second; and the radius of the base is 50 in.,

and increases at the rate of 5 in. per second. At what rate is the volume

changing ?

Solution. Let x = radius of base, y altitude; then

1 , , dn 'J dn Iu -

7r.r-?/ volume, -7- = :; Trrj/, T~~

flM".o ox o ()y o

Substituting in (>),-- =

^ irxy~ 4

\2

^f-

But a- = 50, y = 100, Y-

5, ^ = -10.

/. ^-r= S TT 5000 5 - TT 2500 -10 15.15 cu. ft. per second, increasing. /Ins.

at o o

230. Change of variables. If the variables in

(1) M=M 0)

are changed by the transformation

(2) r ~~~<p(r, s), //

= ^(r, 8),

the partial derivatives of u with respect to the new variables r and s

can be obtained by (Z>). For, if we hold s fast, then x and y in (2)

are functions of r only. Hence we have

(3)^M + ^4/,^r r-'wT ( r (

// fr

all derivatives with respect to r now being partial.

In the same way,(ju chi 'vx . 'cu (hi

(4)== r~ ~r 7*~'

r's r/x 6\s r''//rAs

In particular, let the transformation be

(5) x = z' + /A, ;-y=

?/' + &,

the new variables being x' and //', and h, k being constants. Then

rx - ex^ ()y __ /v ^?y i

r?~

'

(jyf

~'

vxf

~'

vy'

~~

Then we obtain, from (3) and (4),

(>IL !L

'

fix 'ox1

'

vy by'

Hence the transformation (5) leaves the value of the partial derivatives

unchanged.

If the values of x and y in (5) are substituted in (1), the result is


The results in (6) may now be written

(8) fx (x, y) = FA*', y'), /(*, y) = Mr', #')

In Art. 229 it was shown that (B) is true when x and y are func-

tions of a single independent variable t. We prove now that (B)

holds also when x and y are functions of two independent variables

r, s, as in (2). For by (B), when r and s are the independent variables,

we have ^ r/:r r ^dx dr + ds, rf?/

= -^rfr H

** ds.r^r fs rr r;s

Substitute these values in the expression

(9) dx + dyox vy

and reduce by (3) and (4). The result is

/in '(''U j i'(''M> j

(10; dr + ds.vr els

But, by (1 ) and (2), u becomes a function of the independent vari-

ables r and s. Then, by (B), (10) equals du. Hence (9) also equals du.

Therefore (B) holds when x and y are functions of one or of two

independent variables. In the same way, it can be shown that (C)

holds when x, y, z are functions of one, two, or three independentvariables.

231. Differentiation of implicit functions. The equation

(1) M ?/)= o

defines either x or y as an implicit function of the other. It repre-sents an equation containing x and y when all its terms have been

transposed to the first member. Let

(2) M=/Cr,0);

,u du W,

?/ dythen -7-^7- +^* by (F)dx ex f y dx J \ /

and y is an arbitrary function of x. Now let y be the function of x

satisfying (1). Then u = and du = 0, and hence

(3) 4^ + ^? = 0.f)x eydx

Solving, we get

(H) ^ =_. W^dx fl \dy

('y


Thus we have a formula for differentiating implicit functions.This formula in the form (3) is equivalent to the process employedin Art. 41 for differentiating implicit functions, and all the exampleson pages 40 and 41 may be solved by it.

When the equation of a curve is in the form (1), formula (H)affords an easy way of getting the slope.

ILLUSTRATIVE EXAMPLE 1. Given j-v + sin u = find ^-dx

Solution. Let f(x, y) = x 2y* + sin y.

V - O 2L A ,, j.-~ L xy\ -- 4 x-y* -f cos y.

dx dy

Therefore, from (//),~ - - -

,

"An*.

ILLUSTRATIVE EXAMPLE 2. If jr increases at the rate of 2 in. per second as it

passes through the value x = 3 in., at what rate must // change when //~

1 in., in

order that the function 2 xy'2 - 3 j-'y shall remain constant?

Solution. Let u = 2 xy- - 3 x'-y ; then, since u remains constant,-j:

0. Sub-

stituting this value in the loft -hand member of (/>), transposing, and solving for

dy' we Kpt du

___dt

~~

du dt

dy

Also, -y-- 2 /y

? - 6 j#, -T = 4 xy - 3 x2.

Now, substituting in (4), "77~ "

4*

-^- 17

But x = 3, //-I, ^"=2-

dyTherefore, -r- 2j\ in. per second. ATIS.

In like manner, the equation

(5) F(x, y,z) = Q

defines z as an implicit function of the two independent variables x

and y. To find the partial derivatives of z with respect to x and to y,

proceed as follows.

Let u=f(x,y,z).dF dF dF

Then du = - dx + T dy + -7 dz,ox tiy dz

by (#), and this holds no matter what the independent variables are

(Art. 230). Now let z be chosen as that function of the independent

variables x and y which satisfies (5). Then u = 0, du 0, and we have

oF dF,

cF(6) dx + dy + dz = 0.

rx <">y rz

P>ut now dz = -r^ dx + -^ rfy. Ey (5)

Substituting this value in (6) and simplifying gives

(vF (F f)z\ I'cF cF rz\( + }dx + ( +-- }dy

- 0.\r>x vz fx/ \f)y (Z ryj

Here dx (= Ax) and dy (= A?/) are independent increments. Wemay therefore set dy = 0, dx ^ 0, divide through by dx, and solve

()Z

for T The result is

( z

Proceeding in a similar manner, we may prove

(J) =_-( y f_l

( Z

Formulas (7) and (7) are to be interpreted as follows: In the

left-hand members z is the function of x and y satisfying (5). In

the right-hand members F is the function of three variables, x, y, z,

given in the left-hand member of (5).

The generalization of (//), (/), (J) to an implicit function u of

any number of variables is now obvious.

ILLUSTRATIVE EXAMPLE. By the (HI nation

S +12

+*?-

1 = -

z is defined as an implicit function of .r and ?/. Find the partial derivatives of this

function.

Solution. F = ^l

+ -\.

dF J dF y dF zTlPTIfP _ __ ._ _00

r'ir~12' fy~6' ~d~z

~3

Substituting in (7) and (7), we get

dz .r dz y .

a-= -T- ?-=-?r~- ^ws-

ar 4 z dy 2 z

(Compare with the Illustrative Example in Art. 226.)

PROBLEMS

In Problems 1-5 findat

1. u = x 2 3 xy -f 2 ?/2

; a* = cos /, y = sin f.

Aws. = sin 2 f- 3 cos 2 /.

at

2.u=x + 4 VJv - 3 // ;.r = f

3, u = \-

~ = 3 /2 + 4 4- 75-

/ (// /J

3. K = f* sin y + <v{/ sin j- ; .r = J f, ?/

= 2 /.

Ans. = rJ

'(> sin 2/4-2 cos 2 /) 4- f"(2 sin J / + j cos J /)

4. w = 2 .r2

j-?/ -f y'2

;j- = cos 2t,y sin /.

5. ?i = xy 4- ?/- -f ^ ; j- = -?/ = r', 2 = f "'.

In Problems 6-10 find(

-f-' by formula (/f).a.r

6. Ajr 2-f 2 7ij-?/ 4- Cy 2

-f 2 /Xr -f 2 /fy -h ^ = 0.

-f T// 4-

o r w i8. e* sin // f j/ cos x 1.

au' r" cos JT rj cos //

9. x4 - .ry - s'2 4- 2 ?/2 = 8.

10. Ax4-f 2 Rr 2

//2

-f r'//4 = (r

24- ?/

2)2

.

In Problems 11 15 verify that the given values of x and ?/ satisfy the

equation, and find the corresponding value of -

11. x 2-f 2 j-?/ 4- 2 /y

= 22;

jc = 2, //= 3. Am. ^ = -

|-

12. j'3 -

//

34- 4 ^/y

rr ();

j- = 2, y = - 2. - = 1.

13. Ax 4- By 4- O'" = C ;a- = 0, y = 0. ~

14. 2 x - V27^ 4- i/= 4

;.r = 2, y = 4.

15. ez cos ?y 4- ^y sin x 1

;x 0, ?/

= 0-

In Problems 16-20 find and~r^j <^?/

16. Ax>+BV> + Cz>=D. Ans. ^ = -*t;te = _*L

tfx Cz dy Cz

f)z Ay -i- Cz dz _ Ax 4- Bz17. Axy 4- Byz + Czx = D. - = - - - ~ "

. o . o /-

1 n A()Z V

18. x -f 2 // + z 2vxyz = 10. Ans. ='

()Z VZ ~ -rtJZ CZ XZ -.

Vxyz xy W Vxyz xy19. x* + ?/< -f z'

A - 3 axyz = 0.

20. ^^ -f /ty2-f Cz2

-f 2 Dxy + 2 7/2 -f 2 Fz.r = G.

21. A point is moving on the curve of intersection of the sphere

x'2 -f ir -f z 2 ~ 49 and the plane y = 2. When x is 6 and is increasing

4 units per second, find fa) the rate at which z is changing and (b) the

speed with which the point is moving.Ans. (a; 8 units per second

; (b) 4 V5 units per second.

22. A point is moving on the curve of intersection of the surface

x'2 -f xy -f ?/- z 1 and the plane x y + 2 = 0. When / is 3 and is

increasing 2 units per second, find (a) the rate at which // is changing,

(b) the rate at which z is changing, (c) the speed with which the point

is moving. Aws. (a) 2 units per second; (b) V units per second;

(c) 4.44 units per second.

23. The characteristic equation of a perfect gas is Rd pr, where 6

is the temperature, p the pressure, r the volume, and R a constant. At

a certain instant a given amount of gas has a volume of 15 cu. ft. and is

under a pressure of 25 Ib. per square inch. Assuming R 96, find the

temperature and the rate at which the temperature is changing if the

volume is increasing at the rate of \ cu. ft. per second and the pressure

is decreasing at the* rate of ,'Ib. per square inch per second.

Ans. Temperature is increasing at the rate of .^ degrees per second.

24. A triangle AHC is being transformed so that the angle A changes

at a uniform rate from O1

to 90L

in 10 sec., while side AC decreases 1 in.

per second and side A /> increases 1 in. per second. If at the time of ob-

servation A GO11

, AC = 1(1 in., and A /> - 10 in., (a) how fast is BC

changing? (b) how fast is the area of ABC1

changing?Atis. (a) 0.911 in. per second ; (b) 8.88 sq. in. per second.

232. Derivatives of higher order. If

(1) w =/(j, ?/ ),

then

/o\ f) u f / \ vn f f x

(2)- / r (r, ?/),

- fv (x, y)(

1X f'lf

are themselves functions of ;r and ?/, and can, in turn, be differentiated.

Thus, taking the first function and differentiating, we have,

(3) "/,,Cr, ?/), 7r =/i;,(:r,?/).

tfx- fy fx

In the same manner, from the second function in (2), we obtain

/A\ ^"U f f \

2U f / x

(4)- /,(*, y), = fvu (x, y}.)-

In (3) and (4) there are apparently four derivatives of the secondorder. It is shown below that

eycx exey

provided, merely, that the derivatives concerned are continuous.

That is, the order of differentiating vuccewirelH with respect to x and y is

immaterial. Thus /(x, //) has only three partial derivatives of the

second order, namely,

(5) fxjrfr, y\ &(*, '//)= /^(x, //), /,/?/ (x, y).

This may be easily extended to higher derivatives. For instance,since (K) is true,

_r*H _ (>__/ (~u \ C''H __ e_-

/( H\ e- /eu\ _ Ir^u

cx'2 ( }y rxVxr/// rxetjfx <(it\(jc) eyex\cx/ (]

y ex 2

Similar results hold for functions of three or more variables.

ILLUSTRATIVE EXAMPLE. Given // .r1

//-

,'J .r2

//-' ; verify~~~ 7^-.r')// oV d.r dy

Solution. -,-- *J f'"'y (> .r//'. T- :~ % J*" - IX -H/2

.

Hence the formula is verified.

Proof of (AT). Consider the expression

(6) F=f(jr+ Ax, // + A//)-

/(x + Ax, y)-

/(x, y + &y]

Introduce the function

(7) 0(w) =/(M, ?/ + A?/) /(M, ?/),

-where u is an auxiliary variable. Then

4>(x + Ax) =f(x + Ax, y + A?/) -/(x + Ax, ?/),

(8) c/>(x)= /(x, ?/ + Ay) - /(x, |/).

Hence (6) may be written

(9) F = <t>(x + Ax) -

Applying the Theorem of Mean Value (Z>), Art. 116,

(10) F = Az0'(x + 6li Ax). (

f /"(or) = 0(w), = a-, ^a = Ax.J

The value of 0'(z + 0i Ax) is obtained from (8) by taking the

partial derivative with respect to x and replacing x by x + 6\ Ax.

Thus (10) becomes

(11) F = Ax(/,(x + 61 Ax, y + A?/) -/,(* + 0, Ax, ?/)).

Now apply (/)), Art. 116, to /a (x + #j Ax, v), regarding v as the

independent variable. Then

(12) F = Ax &yfvf (x + 61 Ax, y + 0, A-//). (0 < 2 < 1)

If the second and third terms of the right-hand member of (6)

are interchanged, a similar procedure will give

(13) F = A?/ Ax/,,(x + ft. Ax, y + 4 Ay). (0< ft* < 1,

< 4 < 1 )

Hence from (12) and (13),

(14) fvf (x + Oi Ax, y + L, A?/) = /,(* + ft, Ax, y + 4 A/y).

Taking the limit of both sides as Ax and A?/ approach zero as

limits, we have

(15) fv,(x,y)=JrV (r,y),

since these functions are assumed to be continuous.

PROBLEMS

Find the second partial derivatives of each of the following functions.

1. /(x, //)= Ax 2

-f 2 lljry + O/ 2-

An*, /^(x, ?/)= 2 .4

; /, tf (x, ?y)= 2 It ; /^(x, 7y)

= 2 C.

2. /(x, /y)= Ax:< + Rr 2

?/ -f Tx//2-f /;//'.

. / (x, ?y)= 6 AJT + 2 Bij ; /^(x, ?/)

= 2 Rr -f 2 r\v ; / (x, ?y)= 2 t\r + 6 /)?/.

3. /(x, ?/)- .4.r -f- KU -f rr'".

-4w. f2f (x, y) = Cy'2ctv ; f*v =C(l

-

4-

5. /(x, ?/)= x 2 cos ?/ 4- ?y

2 sin x.

6. If /(x, |/)= a-

3-f 3 x 2

y 4- 6 x?y2 -

?y3

, show that

/xx (2, 3) = 30, /XJ,(2, 3) - 48, /yy (2, 3) = 6.

7. If /(x, //)= x 1 - 4 xa

?/ 4- 8 jrj/3 -

7/4

, show that

/(2, -1) = 96, .^(2,

-1) = - 24, /w (2,

-1) = - 108.

8. If /(x, ?/) = 2 x' - 3 x 2?/

2-f /y

4, find the values of

/(2, -2), /T1/ (2,

~2), /vy (2, - 2).

9. If u = Ax44- Rjr*y -f O* 2?r -f />.n/

a4- ?/

4, verify the following re-

sults. ,I3 p3- = 24 .4r 4- 6 #?/.- fc Rr -f 4 O/,

2!L = 4 fj + 6 /; ?/ ,

^1 = 6 Rr -f 24 //.

r'j ( }

\f <"//

10. If w = (ax2-f ft?/

2 + cz-}'\ show thatr'.r

= __!/_ - 2L If M = ___, Show that j-2

-f 2 xy-- 4- ?/2 = 0.

f>.r'2

"

r'j* r??/ <"/r

12. If M = In .H + /y2

, show that 4- = 0.2 *

ADDITIONAL PROBLEMS

1. A circular hill has a central vertical section in the form of the

curve whose equation is .r2 + 160 y - 1600 = 0, where the unit is 1 yd.

The top is being cut down in horizontal layers at the constant rate of

100 cu. yd. per day. How fast is the area of the horizontal cross section

increasing when the top has been cut down a vertical distance of 4 yd.?Am. 25 sq. yd. per day.

2 if w = -^~, show that + = Or -f ?/-

l)w.cf + c y fa ry

3 If w = , where r = Vj- 2 + y'2 + 2 2

,show that

7/ cr '()'*% 'f* ?/2

4. If z = x 2 arc tan ~ -y'

2 arc tan - show that -^2

_^_

'

5. If u z arc tan -> show that --f

-4-

- = 0.

y c)x2

()y2 (^

ffin

6. If u = In (cr

4- e" 4- r*), show that -:

~ - 2 ^

7. If u f(?, y) and x = r cos ^, y = r sin 0, show that

tfu n ()u sin 6 '(^u_____

8. Let M = (/i2 + x*'

2 + ----h * n2)'. What values of * will satisfy the

^ ---+ = 0? An*. * = l-(n>2).

J?

CHAPTER XXIV

APPLICATIONS OF PARTIAL DERIVATIVES

233. Envelope of a family of curves. The equation of a curve gen-erally involves, besides the variables x and //, certain constants uponwhich the size, shape, and jxxsition of that particular curve depend.For example, the locus of the equation

(x a)2 + y

2 = r2

is a circle whose center lies on the j-axis at a distance of a from the

origin, its size depending on the radius r. Suppose a to take on aseries of values while r is held fast ; then

we shall have a corresponding series of

circles of equal radius differing in their

distances from the origin, as shown in

the figure.

Any system of curves formed in this

way is called a family of curves, and the

quantity a, which is constant for any one curve, but changes in

passing from one curve to another, is called a variable parameter.To indicate that a enters as a variable parameter it is usual to

insert it in the functional symbol, thus:

/Or, y, a) = 0.

The curves of a family may be tangent to the same curve or groupof curves, as in the above figure. In that case the name envelope of

the family is applied to the curve or group of curves. We shall nowexplain a method for finding the equation of the envelope of a familyof curves. Suppose tha;, the curve whose parametric equations are

0) J- = </>(), y=\f,(a)

is tangent to each curve of the family

(2) /U, y, a) = 0,

the parameter a being the same in both cases. For any commonvalue of a equations (1) will satisfy (2). Hence, by (), Art. 229,since u =f(x, y, or), du = df = 0, and z is replaced by a, we have

(3) fx(x, y, a)(f>'(a} + fy (x, y, a)\f/'(a) +/(>, y, a) = 0.

466

APPLICATIONS OF PARTIAL DERIVATIVES 467

The slope of (1) at any point is

g-*$.and the slope of (2) at any point is

(5)= - - r ' ' <v

-(tf), Art. 231

rfj- ./(*, //, a)

Hence if the curves (1) and (2) are tangent, the slopes at a point

of tangency will be equal, giving

__ fj(x, ?/, nr)or

</> Uv) fu (x, //, a)

(6) /,(*, //, <v)</>'(aO + /(/, //, W() = 0.

Comparing (6) and (3) gives

(7) fa (x, y, a) = 0.

Therefore the coordinates of the point of tangency satisfy

(8) f(x, ?/, <v) - and /n (j, //, a) - ;

that is, the parametric equations of the envelope, in case an envelope

exists, may be found by solving these equations for x and y in terms

of the parameter a.

General rule for finding the parametric equations of the envelope

FIRST STEP. Write the equation of the family of curves in the form

/(x, y, a) and derive the equation fa (x, y, a) = 0.

SECOND STEP. Solve these two equations for x and. y in terms of the

parameter a.

The rectangular equation may be found either by eliminating a

-between the equations (8) or from the parametric equations (Art. 81).

ILLUSTRATIVE EXAMPLE 1. For the family of circles at the beginning of this

article,^ (j> ^ a) = (j

_ a) , + ^ _ r -z = .

Hence /(-r, 2/, J- (f

-)= 0.

Eliminating a, the result is /y2 - r 2 - 0, or y = r, y - -

r, and these are the equa-

tions of the lines .4K and CD in the figure.

ILLUSTRATIVE EXAMPLE 2. Find the envelope of the family of straight lines

x cos a -4- y sin a -;>, a being the variable parameter.

Solution. (9) f(x, y, a]- x cos a + y sin a - p - 0.


First Step. Differentiating with respect to a,

(10) fa (x, y, a) - - x sin a 4- y cos a = 0.

Second Step. Multiplying (9) by cos a and (10) by sin a and subtracting, we get

x p cos a.

Similarly, eliminating x between (9) and (10),

y p sin a.

The parametric equations of the envelope are therefore

(11) P^Iina'a being the parameter. Squaring equations (11) and

adding, we get ^ _^_ yiw^

the rectangular equation of the envelope, a circle.

ILLUSTRATIVE EXAMPLE 3. Find the envelope of a line of constant length a,

whose extremities move along two fixed rectangular axes.

Solution. Let AH = o in length, and let

(12) x cos a 4 y sin a p -

be its equation. Now as Aft moves, both a and p will vary. But p may be found

in terms of a. For AO Att cos a o cos a, and also

p = AO sin a = a sin cos a. Substituting in (12), we get

(13) x cos a -f y sin a a sin a cos a = 0,

where a is the variable parameter. This equation is in

the form f(x, y, a] 0. Differentiating with respect to

a, the equation fa(x, y, a) = is

(14) x sin a- -f y cos a -f a sin 2a. a cos- a: = 0.

Solving (13) and (14) for x and y in terms of a, the

result is

\

x = a sin- 4

a,

] y a cos-' a,(15)

the parametric equations of the envelope, a hypocycloid. The corresponding rec-

tangular equation is found from equations (15) by eliminating a: as follows:

Adding,

a*** = a d sin- a,

? 2

^3 a^ cos 2 a.

228r j + ?/

a = fl-\ the rectangular equation of the hypocycloid.

Many problems occur in which it is convenient to use two param-eters connected by an equation of condition. By using the latter,

one parameter may be eliminated from the equation of the familyof curves. It is, however, often better to proceed as in the following

xample.


ILLUSTRATIVE EXAMPLE 4. Find the envelope of the family of ellipses whoseaxes coincide and whose area is constant.

Solution. (16) + ^ = 1

o 2 b'2

is the equation of the ellipse, where a andb are the variable parameters connected

by the equation

(17) irab A",

irab being the area of an ellipse whose

somiaxes are a and b. Differentiating,

regarding a and b as variables and .r and

y as constants, we have, using differentials,

= 0, from (16),U" U"

and bda -f adb 0, from (17).

Transposing one term in each to the second member and dividing, the result is

Therefore, using (16),

whence a = j\ '2 and ft ~ //% 2.

Substituting these values in (17), we get the envelope xy ~, a pair of

conjugate rectangular hyperbolas (see figure).~ w

234. The evolute of a given curve considered as the envelope of its

normals. Since the normals to a curve are all tangent to the evolute

(Art. 110), it is evident that the evolute of a curve

may be defined as the envelope of its normals. It

is also interesting to notice that if we find the

parametric equations of the envelope by the

method of the previous article, we get the coor-

dinates x and y of the center of curvature ;so

that we have here a second method for fiivditm the

coordinates of the center of curvature. If we elimi-

nate the variable parameter, we obtain the rec-

tangular equation of the evolute.

ILLUSTRATIVE EXAMPLE. Find the evolute of the parabola y~ = 4 px considered

as the envelope of its normals.

Solution. The equation of the normal at any point (x\ t y\) is

from (2), Art. 43. As we are considering the normals all along the curve, both Ji

and 2/1 will vary. Eliminating x\ by means of y\2 4 px\, we find the equation of

the normal to be

(1) y-Vi^--j 1' or xyi +2py - 2 pj/i

- = 0.o p & p ^ /'

Setting the partial derivative of the left-hand member with respect to the

parameter y\ equal to zero, and solving for x, we find

4 p

Substituting this value of i in (1) and .solving for y,'

Equations (2) and (3) arc the coordinates of the center of curvature of the pa-

rabola. Taken together, they are the parametric equations of the evolute in terms

of the parameter y\. Eliminating ?/i, we obtain

27 pi/2=-4(.r ~2 p)\

the rectangular equation of tho evolute of the parabola. This is the same result we

obtained in Illustrative Example 1, Art. 109, by the first method.

PROBLEMS

Find the envelopes of the following systems of straight lines and draw

the figures.

1. y rns -f nr. A ns. jr2

-f- 4 y = 0.

2. //= -f nr. 27 x 2 = 4 ?y

3.

w3. //

= m'2j' 2 m ;<

. 27 ?/= ^ :i

.

4. ?/= 2 mjr -f- ?w

4. 16 ?/

:^

-f 27 u*1 = 0.

5. y tjr t'2

. 6. // f-s -h /. 7. ?/= m.r 2 m 2

.

Find tbe envelopes of the following systems of circles and draw the

figures.

8. (x c}'24- ir = 4 r. Am. y

2 4 jr -f 4.

9. a-2-h (?/

- O 2 = 2 /. 10. (T- O 2 + (?/ + O 2 = /

2-

Find the envelopes of the following systems of parabolas.

11. y* = c(x c). Ans. 2 y = jr.

12. cy'2 = 1 - c 2

jc.

13. Find the evolute of the ellipse b~x 2 + a 2y2 a 25 2

, taking the equa-

tion of tbe normal in the form by = ax tan (a2 6 2

) sin </>, the eccen-

tric angle c/> being tbe parameter.

Am. x = a2 ~ ?>2cos3

0, ?/= b2 ~ a "

sin3; or (ax)

?-f- (byY = (a

2 - 6 2)l'

2 2 2

14. Find the evolute of the hypocycloid .rs

-f ir = a 3, the equation of

whose normal is

r being the parameter.

r cos r x sin r = a cos 2 r,

Arts, (.r -f '< (JT- ?/)= 2 a.

15. Find the envelope of the circles which pass through the origin andhave their centers on the hyperbola x'

2?/- = r 2

.

Ans. The lemniscate (x'2-f ?/

2)2 = 4 r 2

(.r2

?/2).

16. Find the envelope of a line such that the sum of its intercepts onthe axes equals c. AnKf The parabola ,1 + y

\ = r*.

17. Find the envelope of the family of ellipses b 2x'

2-f a 2

//2 a 2

?)2 when

the sum of its semiaxes equals r. 4 ?/<s.

f^e hypocycloid x** -f ifi = c^.

18. Projectiles are nred from a gun with an initial velocity r>. Sup-posing the gun can be given any elevation and is kept always in the

same vertical plane, what is the envelope of all possible trajectories, the

resistance of the air being neglected?

HINT. The equation of any trajectory is

y = x tan ex - - ~,

2 r<r cos- a

(i being the variable parameter.

Ans. The parabola y = - ~^--2g 2 i,r

19. If the family

has an envelope, show that it is

g2(x, y) L

, ?/) + /'Or, ?/)=

235. Tangent line and normal plane to a skew curve. The student

is already familiar with the parametric representation of a planecurve (Art. 81). In order to extend

this notion to curves in space, let the

coordinates of any point P(x, y, z) on

a skew curve be given as functions of

some fourth variable which we shall

denote by t; thus,

(1) * = 2=x(0-

The elimination of the parametert between these equations two by two

will give us the equations of the

projecting cylinders of the curve on the coordinate planes.

Let the point P(.r, y, z) correspond to the value t of the param-eter, and the point P'(x + Az, y + Ay, z + Az) correspond to the

value t + A2 where Ax, A?/, Az are the increments of x, y, z due to the

increment AZ as found from equations (1). From analytic geometryof three dimensions, we know that the direction cosines of the secant

(diagonal; PI*' are proportional to

Ax, A#, Az;

or, dividing through by A/ and denoting the direction angles of the

secant by a', ft', y',cos a' __ cos (3

f

_ cos y'( } Ax

~Ag

"Az

AZ At A*

Now let P' approach P along the curve. Then AJ, and therefore

also AJ, Ay, Az, will approach zero as a limit, and the secant PP' will

approach the tangent line to the curve at P as a limiting position.

Now lim ^ = ^ =</>'(0, etc.

A* - o A at

Hence, for the tangent line,

cos a _ cos P _ cos y04) cfo (/y dz

dt dt dt

When the point of contact is P$x\, y\, z$, we use the notation

(3)= value of

c

-j-when x = .TI, y = y\, z = z\,

and similar notation for the other derivatives.

Hence, by (2) and (4), p. 5, we have the following result.

The equations of the tangent line to the curve whose equations are

(1) .r = <W), y = t(V, z

at the point PI(XI, y\, zO are

x xi_y y\ _ z z\

(B) \dx\dt \dt

I I

\dt\

The normal plane of a skew curve at a point P\(x\, y\, Zi) is the

plane which passes through PI and is perpendicular to the tangent

line at PI. The denominators in (B) are the direction numbers of

the tangent line at PI. Hence we have the following result.

The equation of thf normal plane to the curve (1) at Pi (x\, y\, z\) is

(x-\dt

(w-

ILLUSTRATIVE EXAMPLE. Find the equations of the tangent line and the equa-

tion of the normai plane to the circular helix (6 being the parameter)

( x = a cos 0,

-

y fl sin 6,(4)

(a) at any point (xi, 3/1, d) ; (b) when - 2 TT.

Solution. -^ = - a sin = -?/, ^ = a cos = .r, -f-

-b.

do du (<('

Substituting in (B) and (C), we get, at (.n, //,, z\ >,

(5)x ~ x- V

~V 1 = T"^' tangent line,

and

normal plane.

When '2 TT, the point on the curve is

(a, 0, 2 &TT), giving

g_--ji _ ?y _ z - 2 bir~a

~6

or a* = a, &# = a^ 2 O^TF,

the equations of the tangent line, and

ny -f ?w - 2 ?>27T = 0,

the equation of the normal plane.

REMARK. For the tangent line (5) we have by (2) and (4), p. 5,

fo 6cos 7 = : = a constant

-f yi2 + b'

2 Va~ -f b 2

That is, the helix cuts all elements of the cylinder x2 + y'

2 = a 2 under the same angle.

236. Length of arc of a skew curve. From the figure of the pre-

ceding article we have

/i x (Chord PP'Y _ /AxV-

Let arc PP' = As. Proceeding as in Art. 95, we easily prove

(2) (*)

2=

From this we obtain

(D) s =

where x 0(0, y = ^(0> z x(0 as in (1^ A*t. 23").

The direction cosines of the tangent line can now be given a simpleform. For, from (A) of the preceding article, and by the above

equation (2), using formulas in (2), p. 5, we have

(3 ) cos > ft ^ ~ds

"

ds"

ds

ILLUSTRATIVE EXAMPLE. Find the length of arc of the skew cubic

(4) x = t, y = } 1*, z = J f

between the points where / = and t = 4.

Solution. Differentiating (4), we obtain

dx = dt, dy = / (it, dz = t'2dt.

Substituting in (D), = f Vl + t'2 + t* dt = 23.92,

Jo

approximately, by Simpson's Rule, letting n = 8.

PROBLEMS

Find the equations of the tangent line and the equation of the normal

plane to each of the following skew curves at the point indicated.

1. x = at, y = bt2

,z = </; / = 1.

AUK. ^^ = ^-^ = ^p-f ; ax 4- 2 by -f 3 cz = a 24- 2 6 2

4- 3 c 2.

3. a- = /2 -

1, ?/= / + 1, r = /

:<

;/ = 2.

4. j = P -1, ?/

= r2 -f /, ^ = 4 /3 - 3 / -hi ; / = 1.

5. a- = 2 /-

S, ?/= 5 - /

2,c = -

;/ = 2.

6. x = a cos /, // b sin /, z ~ /;

/ = J TT.

7. x = /, ?/= r', ^ = r

'

;/ = 0.

8. x = cos /, z/ sin t, z = tan /;

f = 0.

9. Find the length of arc of the circular helix

x = a cos 0, y = a sin 6, z = b6

between the points where B = and 6 = 2 TT. Ans. 2 ?rVa 2-f 6-.

10. Find the length of arc of the curve

x = 3 cos 0, y = 3 6 sin 8, c = 4

between the points where 6 = and = 4. Ann. 26 +^ In 5 = 32.70.

11. Find the length of arc of the curve

x = 2 t, y = t'2 -2, ; = 1 - t

2

between the points where t = and / = 2.

12. Given the two curves

(5) x = t, y = 2t*, z =-y;

(6) x = 1 - 0, y = 2 cos 0, c = sin 6 - 1.

(a) Show that the two curves intersect at the point A(l, 2, 1).

(b) Find the direction cosines of the tangent line to (5) at A.

41 4 1

Ans. ;= p= 7=-Vl8 Vl8 Vl8

(c) Find the direction cosines of the tangent line to (6) at A.

(d) Find the angle of intersection of the curves at A . Ans. 90.

13. Given the two curves

a- = 2-/, i/= /

2 -4, 2 = f-8;:r = sin 0, y = 0, z = 1 cos 0.

(a) Show that the two curves intersect at the origin O.

(b) Find the direction cosines of the tangent line to each curve at O.

(c) Find the angle of intersection of the curves at O.

14. (a) If OF, OE, ON in the first figure of Art. 222 are chosen as axes

of coordinates OX, OY, OZ, respectively, and if P(JC, ?/, z) is a point on the

sphere, prove that r = a cos $ sin 0, ?/= a cos < cos 0, z = asin</>, if

</>and

are, respectively, the latitude and longitude of P.

(b) Using (3), and (3) on page 5, find the angle u at P between a curve

on the sphere for which =/(</>) and the parallel through P.

Ans. tan a sec </> -777 ,as in Art. 222.

au

237. Normal line and tangent plane to a surface. A straight line is

said to be tangent to a surface at a point P if it is the limiting position

of a secant line through P and a neighboring point P' on the surface

when P f

is made to approach P along a curve on the surface. Wenow proceed to establish a theorem of fundamental importance.

Theorem. All tangent lines to a surface at a given point lie in a plane.

Proof. Let

(1) F(x, y,z)=0be the equation of the given surface, and let P(x, y, z) be the given

Doint on the surface. If now P' be made to approach P along a curve

C lying on the surface and passing through P and P', then evidentlythe secant line PP' approaches the position of a tangent line to the

curve C at P. Now let the equations of the curve C be

Then the equation (1) must be satisfied identically by these

values. Hence, if u = F(x, y, z), then u = 0, du = 0, and by (), Art. 229,

tix dt 'dy dt cz dt~

This equation (see (3), Art. 4) shows that the tangent line to (2),

whose direction cosines are proportional to

dx d/n dz

dt dt dt

is perpendicular to a line whose direction cosines are proportional to

(4) > > By (3), p. 5(>x oy cz

Let PI(:TI, y\, z\) be a point on the surface and

/e\ <)F ?)F(O) TT" >

->

vx i vy \

the values of the partial derivatives in (4) when x = Xi, y = y\ 9

z z\. The line passing through P\ whose direction numbers are

given by (5) is called the normal line to the surface at PI. Hence wehave the following result :

The equations of the normal line to the surface

H 1 F(x y ~] =\i )r \A., y j

** i \/

at PI('X\, '//i, zO are

_. y y\ __ z z\x -

dF

dx

IdF

\9y

dF\

'dz\\

The preceding argument shows that all tangent lines to the sur-

face (1) at PI are perpendicular to the normal line at PI. Hence

they lie in a plane. Thus the theorem is proved.This plane is called the tangent plane at (Pi).

We may now state the following result.

The equation of the tangent plane, to the surface (1) at the point oj

contact PI(XI, y\, z\) i-s

<*> i?i(.- = 0.


REMARK. If all the denominators in (E) vanish, the normal line and tangentplane are indeterminate. Such points are called singular points and are excludedhere.

In case the equation of the surface is given in the formz = /(z, y), let

(6) F(x, ?/, z) =/(j, //)- z = 0.

rp^?>F ?j C'Z (

] F (f f}

Z (}F ,i nen -r ~T~ TT~ -r ~r~

~~ ~ == !ra OT ex dy r

}

y (}

y cz

Hence, by (), we have the following result.

The equations of the normal line to the surface z /(J, y) at (x\ , y\ , z\ )

are

(G)2-- 1

Also, from (F), we obtain

-xi) 4- (y

-yi)

-(z-

zi) =

which is then the formula for the equation, of a plane tangent at (x\ t y\, z\)

to a surface whose equation is given in the form z =f(x, y).

238. Geometric interpretation of the total differential. We are nowin a position to discuss formula (#), Art. 227, by geometry, in a man-ner entirely analogous to that in Art. 91.

Consider the surface

(1)

and the point

A'hen

(2)

i, y\, z\) on it. Then the total differential of (1) is,

x = x\ t y~yi,

Ayy,

using (5), Art. 227, and replacing dx and dy by their equivalents, Axand A^/, respectively. Let us find the z-coordinate of the point in

the tangent plane at P\ where

x = xi + Ax, y = yi + A?/.

Substituting these values in (H) of Art. 237, we find

(3)


Comparing (2) and (3), we get dz = z z\. Hence the

Theorem. The total dif-

ferential of afunction f(x, y)

corresponding to the incre-

ments Ax and Ay equals the

corresponding increment of

the z-coordinate of the tan-

gent plane to the surface

Thus, in the figure, PP'is the plane tangent to

surface PQ at P(x, y, z).

Let

and CD &y,

then dz = z - z\ = DP' - DE = EP'.

Notice also that Az = DQ - DE = EQ.

ILLUSTRATIVE EXAMPLE. Find the equation of the tangent plane and the equa-tions of the normal line to the sphere x'2 -f y

2-f z 2 14 at the point (1, 2, 3).

then

Solution. Let F(x, y, z) = x 2 + y* + z2 - 14 ;

dFdx

Therefore = 4,

vSubstituting in (F), 2(x - 1 ) + 4(y - 2) + 6(z - IJ) = 0,

or x + 2 y + 3 c = 14, the tangent plane.

y- 2 _ s - 3

4"

6'Substituting in (),

giving z = 3 x and 2 s = 3 y, equations of the normal line. Arts.

PROBLEMS

Find the equation of the tangent plane and the equations of the nor-

mal line to each of the following surfaces at the point indicated.

1. x'2 + y'2 + z2 = 49 ; (6, 2, 3).

An*. 6 x 4-

2. z = 8 -l; (2, 1,4).

x-6 = y-2 = z-3J

'

6~

2 3

- 1 2-4

3. r 2 + xy24- ?/

3-f z 4- 1 =

; (2,-

3, 4).

Ans. 13 .r -f 15 y + * + 15 =; ^-^ = i^J = L=-l.

A o 1 1

4. x~ 4- 2 jry 4- ?/2-f z - 7 =

; (1,-

2, 6).

5. j*2?/2-f .nr - 2 ?/<

- 10 =; (2 t 1, 4).

Ans. 4 .r + // -f z - 13 =;

^-=-? = 2-^ = ^-4116. .r

2 -y

2 -z'2 =: 1; (3, 2, 2).

7. ar2 + 2 - z 2 = 25

; (5, 5, 5).

8. 2 x 2 + 3 j/2 + 4z-' = 6; (1, 1, J).

9. y-f ?/- r 2 ^3; (3, 4, 2).

10. Find the equation of the tangent plane to the hyperboloid of two

sheets 1 - ^ - ! = i at (.r,, ?/ z,). .4**. l _M _l|= ! .

a- 6 2 r- 2?>2 r 2

11. Find the equation of the tangent plane at the point (.r,, ?/ ]t z\) onthe surface ax 2 + 6//

2-f re 2 + d ~ 0. ^^,s. a.ri.r -f ^?/i// -f rr,^ -f d = 0.

12. Show that the equation of the plane tangent to the sphere

x 2 + r -f z- -f 2 L.r -f 2 Af?/ -f 2 A^ + /> =at the point (xi, y it z\) is

^i^ 4- 3/1 1/ + Ziz + L(J- + j-i) 4- M(/y -f ?/]) -h A/T

(- 4- Ci) 4- D = 0.

13. Find the equation of the tangent plane at any point of the surface

j% -f y* -f 2 rr a",

and show that the sum of the squares of the intercepts on the axes made

by the tangent plane is constant.

14. Prove that the tetrahedron formed by the coordinate planes and

any tangent plane to the surface -jryz a :<

is of constant volume.

/2 415. The surface x 2 4 y

2 4 z = is cut by the curve x > y = -

/ _ 9 /2J t

2 = --- at the point (2, 2, 3). What is the angle of intersection?

Am. 90 - arc cos1

/!L-. = 32 37'.

3V1383

16. The surface x 2 + ?/2-f 3 z 2 ~ 25 and the curve x = 2t, y = ->

t

z = 2 f2 intersect at the point on the curve given by / = 1. What is

the angle of intersection ? A^ 9Qo _ arc CQS _J9 = 3QO 16/>

7V29

17. The ellipsoid x2-h 2 y

24- 3 z 2 = 20 and the skew curve x - \(P 4- 1 ),

i/= f

4-f 1, z P meet at the point (3, 2, 1). Show that the curve cuts

the surface orthogonally.

239. Another form of the equations of the tangent line and normal

plane to a skew curve. If the curve in

question be the curve of intersection ABof the two surfaces F (x, y, z) = and

G(x, y, z)= 0, the tangent line PT at

P(x\, y\ t Z}) is the intersection of the

tangent planes CD and CE at that,

point, for it is also tangent to both sur-

faces and hence must lie in both tangent

planes. The equations of the two tan-

gent planes at P are, from (F),

(1)

tix

f)G

IrJx

(x-

tiy

^ <*-">

vFc)z

(z Zi) = 0.

Taken simultaneously, these equations are the equations of the

tangent line PT to the skew curve AB.If A, B, C are direction numbers for the line of intersection of the

planes (1), then, by (6), Art. 4,

dG

(2)

c)JP

f)Z

C =

'dy

B = bF\ VvG_

'c)z i cjy

8Fdx dz

8y

c)F_

fry

c)G

Then the equations of the tangent line CPT are

(3) A_y y\ ^z

B C

The equation of the normal plane PHI is

(4) A(x - j,) + B(y - yi) + C(z - Z]) = 0.

ILLUSTRATIVE EXAMPLE 1. Find the equations of the tangent line and the

equation of the normal plane at (r, r, r N/2) to the curve of intersection of the sphere

and cylinder whose equations are, respectively, x2 + y

2H- z2 = 4 r2 ,

x2 + j/2 = 2 rx.

Solution. Let F = a-2 4 ?/

24- z2 - 4 r2 and G = x2 + y

2 - 2 nc.

aFax

= 0, dG]dy\


Substituting in (2), we find

A = -4r2\/2, B = 0, C = 4r2.

Hence, by (3), we have

x r y r z rV2

481

or

- V2

^ = r, z

the equations of the tangent PT at P to

the curve of intersection.

Substituting in (4), we get the equa-tion of the normal plane,

or

-r) 4- 0(y - r) + (z - r\ V =. 0.

\ 2 x - z 0.

ILLUSTRATIVE EXAMPLE 2. Find the angle of intersection of the surfaces in the

preceding example at the point given.

Solution. The angle of intersection equals the angle between the tangent planesor normal lines. We have found direction numbers for these lines above in Illus-

trative Example 1 (see (), Art. 237).

These are a = 2 r, b = 2 r, r - 2 r\/2.

a' = 0, 6' = 2 r, c' = 0.

Hence, by (6), Art. 4

= t| = . = 60. Am.

PROBLEMS

Find the equations of the tangent line and the equation of the normal

plane to each of the following curves at the point indicated.

1. x 2 + y2 + z 2 = 49, x* + y

2 13 ; (3, 2,-

G).

Aw*. rLzJ = -=, 2+6 = 0; 2 s - 3 # = 0.

2. 2 = a-2 + y

2 -1, 3 ? 2 + 2 //

2 + z 2 = 30; (2, 1, 4).

Am.5 -11

3. x* + y*_ Z 2 = 16. x - + 4 ?/* + 4 z 2 = 84

; (2, 4, 2).

x 2 _ y 4 _ 2 2

5,4ns.

16 - 5 6

4. x a + ?y2 + 3 z 2 = 32, 2 r 2 + ?/

2 - z 2 =; (2, 1, 3).

;16 j-- 5 ?y + 6,7 = 24.

5. x2 - y2 - z 2 = 1, x 2 - y

2 + z 2 = 9; (3, 2, 2).

6. x 2 + 4 #2 - 4 z2 - 0, 2 x + ?/ + 2 - 24 = ; (8, 3, 5).

7. The equations of a helix (spiral) are

X 2 + y'2_

f2>

iy= x tan -

r

Show that at the point O], ?/i, zj) the equations of the tangent line are

c(x - si) -f 1/1(2- ?])= 0,

r(?y-

?y } )- xi(z - Zj) = 0;

and the equation of the normal plane is

. 7/1 .r~ x\y r(z

~z\) = 0.

8. The surfaces x 2?/- + 2 j- -f z :< 16 and 3 jc

24- ?/

2 2 z = 9 intersect

in a curve which passes through the point (2, 1, 2). What are the equa-

tions of the respective tangent planes to the two surfaces at this point?

Am. 3a- + 4i/ + 6c = 2'2; 6 x -f y - z~

11.

9. Show that the ellipsoid x~ -f 3 y'2 + 2 z'

2 9 and the sphere

^2 _|_ ?y2 _|_~2 _ & r __ g ^ _ g _}_ 24 = are tangent to each other at the

point (2, 1, 1).

10. Show that the paraboloid 3 x~ -f 2 ?/

2 22 = 1 and the sphere

x 2-f y'

2 + z'2 - 4 y - 2 z -f 2 = cut orthogonally at the point (1, 1, 2).

240. Law of the Mean. The applications of partial derivatives to be

given now depend upon the Law of the Mean for functions of several

variables. The result to be derived is based upon the discussion in

Art. 116. We proceed to establish the formula

(1) /fa, + h, yo + k) = /(a-o, 2/0) + V,(ro + Oh, ?/o + 0k)

+ kftl(x*+0h,yv+0k). (0<0<1)To this end let

(2) F(t)=f(ro + ht,yo + kt).

Apply (Z)), Art. 116, to F(t), with a == 0, and Aa = 1. Then wehave

(3) F(l) = F(0) + F'(0). (0 < 6 < 1)

But from (2), by (D), Art. 229, since x = xn + hi, y = yo + kt,

(4) f'(0 = fc/,(:ro + fa

Then, from (2), we get

(5) F(l) = /(Jo

and, from (4),

(6) F'tf) = hfx (x + eh, yo + 6k) + kfv (x + Oh, yo + Ok).

When these results are substituted in (3), we obtain (1).

If we desire a formula analogous to (F), Art. 124, we must form

F"(t). Applying again (D), Art. 229, we get

x , kt) = hftf (xo + ht, y(} + kt) + kfvr (x + hi, y + kt) ;

at

v-/y(aro + kt, .2/0 + kt) = hf,(x + hi, // + kt) + kftw (x() + ht, y + kt).at

Hence from (4), we have by differentiating with respect to t,

(7) F"(t) = k*fxf (x<> + ht, ?/ + kt) + 2 hkfxu (xo + kt. //o + kt)

+ A--/,,, (Jo + kt., ?/o + kt).

From (F), Art. 124, letting 6 = 1, a = 0, ^2 = 0, we get

(8) ni) = no) + r(0) +r|r'(0).

We may easily prove now the extended Law of the Mean for a

function of two variables by substituting in (8) from (5), (4), and (7).

Thus we get

(9) /(jr + h, ?/o + k) =/(TO, v/o) + A/r(:r<>, ?/o) + A/i/to), ?/o)

+r|

t^2/xX^> + Oh, /A, + 0A-) + 2 Wn/ao + 0A, y() + Ok)

+ Ar2/^(y + 0//, ?/o + 0*)]- (0 < < 1).

There is no difficulty in establishing the corresponding formulas

for functions of more than two variables, nor in extending the laws

in a manner analogous to that at the end of Art. 124.

241. Maxima and minima of functions of several variables. In

Art. 46, and again in Art. 125, were derived necessary and sufficient

conditions for maximum and minimum values of a function of one

variable. We now take up this problem when several independent

variables are present.

The function /(.r, ?/) is said to be a maximum, at x = a, y = b

when /(a, b) is greater than f(x, y) for all values of x and y in the

neighborhood of a and b. Similarly, /(x, y) is said to be a minimum

at x a, y = b when /(a, 6) is less than f(x, y) for all values of x and y

in the neighborhood of a and 6.

These definitions may be stated in analytical form as follows:

If, for all values of h and k numerically less than some small

positive quantity,

Q) j(a + hfb + k) /(a, 6) = a negative number,

then /(a, b) is a maximum value of f(x, y). If

(2) /(a + A, 6 + k) /(a, 6) = a positive number,

then /(a, b) is a minimum value of /(x, #).


These statements may be interpreted geometrically as follows.

A point P on the surfacez = f(x,y)

l\ a maximum point when it is"higher" than all other points on the

surface in its neighborhood, the coordinate plane XOY being assumedhorizontal . Similarly,

P' is a minimum point

on the surface whenit is "lower" than all

other points on the

surface in its neigh-

borhood.

Hence if

21= /(a, 6)

is a maximum or mini-

mum, the tangent plane

at (a, b, zi) must be horizontal, that is, parallel to XOY. But the

tangent plane (H), Art. 237, is parallel to XOY when the coeffi-

cients of x and // are zero. Hence we have the following result.

A necessary condition that /(a, b) shall be a maximum or minimumvalue off(x, y) is that the equations

(3)c)x 8y'

shall be satisfied by x a, y b.

The conditions (3) may be obtained without use of the tangent

plane. For, when y ~ b, the function /(.r, b) can neither increase nor

decrease when x passes through a (see Art. 45). Hence follows the

first of equations (3). The same statement applies to the function

/(a, y}. Thus we have the second equation in (3).

The method just expounded applies to a function of three variables

f(x, y, z). That is, a necessary condition that /(a, 6, c) shall be a

maximum or a minimum value is that the equations

(4)ftp 8z

=

shall have the common solution x = a, y b, z c.

For necessary and sufficient conditions the problem is much more

difficult (see below). But in many applied problems the existence of

a maximum or minimum value is known in advance, and no test is

necessary.


ILLUSTRATIVE EXAMPLE 1. A long piece of tin 24 in. wide is to be madeinto a trough by bending up two sides. Find the widthand inclination of each side if the carrying capacity is a

maximum.

Solution. The area of the cross section shown in the

figure must be a maximum. The cross section is a _trapezoid of upper base 24 2 x -f 2 x cos a, lower r

* 24~2a? *i

base 24 - 2 x, and altitude x sin a. The area A is given by

(5) A 24 x sin a - 2 x2 sin a + x 2 sin a cos a.

By differentiation we have

-r = 24 sin a 4 x sin a + 2 jr sin a cos .

dx

dA-r = 24 x cos a 2 .r- cos a -f a*" (cos-' a sin- a).d

Setting the partial derivatives equal to zero, we have the two equations

2 sin (12- 2 x + x cos a) = 0.

[24 cos a 2 .r cos a 4- x(cos2

cv sin a)J= 0.

One solution of this system is a = 0, x = 0, which has no meaning in the

physical problem. Assuming a ^ 0, x & 0, and solving the e-quations, we get

cos a 2, x 8.

A consideration of the physical problem shows that there must exist a maxi-

mum value of the area. Hence this maximum value occurs when a = 60" andx = 8".

We now establish a sufficient condition. Assuming that equa-tions (3) hold, we obtain from (9), Art. 240, substituting x()

= a,

yo = b, and transposing,

(6) /(a +M + A') -/(a, b) = r \h^fxx (x y y}+2 kkffll (x, y)^ +k*fw (x,y)\,

where we have set x = a + Oh, y b + 6k. By (1) and (2), /(a, b) will

be a maximum (or a minimum) if the right-hand member is negative

(or positive) for all values of h and k sufficiently small in numerical

value (zero excluded). Set

(7) A =/(*, y), B=:fxv (x, y), r=fyy (x, y),

and consider the identity

(8) Ah2 + 2Bhk+ Ck* = [(Aft + Bt)2 + (AC - B*)k*].

The expression within the square brackets in the right-handmember in (8) is always positive if

(9) AC - B2 > 0,

and the left-hand member therefore has the same sign as A (or C,

since, by (9), A and C must agree in sign). The question now is,

therefore, to interpret the criterion (9; for the right-hand member in

(6), in which, as already stated, h and k are numerically small.

Assume that (9) holds when x = a, y = b. Then, the derivatives in

(1) being continuous, it will hold also for values of x, y near a, b.

Also, the sign of A (or C) will be the same as the sign of fxx (a, b)

(or fvv (a, b)). Thus we have established the following rule for find-

ing maximum and minimum values of a function / (x, y).

FIRST STEP. Solve the simultaneous equations

# = 0,# = 0.

c)x ( jy

SECOND STEP. Calculate for these values of x and y the value of

A^LM /J^LY\()jc~ '(

]

>\i~ \<rxvyj

THIRD STEP. The Junction will have

a maximum value if A > and ^ (or ~{} < ;

()x* \ f)y2/

a minimum value if A > and -^ (or ^-~

}> 0.

vx~ \ r)y-J

If A is negative, it is not difficult to see that /(or, y) will have

neither a maximum nor a minimum value.

The student should notice that this rule does not necessarily give

all maximum and minimum values. For a pair of values of x and y

determined by the First Step may cause A to vanish, and may lead

to a maximum or a minimum or neither. Further investigation is

therefore necessary for such values. The rule is, however, sufficient

for solving many important examples.

The question of maxima and minima of functions of three or more

independent variables must be left to more advanced treatises.

ILLUSTRATIVE EXAMPLE 2. Examine the function 3 axy x3 y5 for maxi-

mum and minimum values.

Solution. f(x, y] = 3 axy - x3 - y3

.

First Step. |= 3 ay - 3 z 2 - 0,

|= 3 ax. - 3 y* = 0.

Solving these two simultaneous equations, we get

x = 0, x a,

y = 0, y = a.

SecondStep. = - 6 x, = 3 a, = -6 V ;

\dxdyj-"

Third Step. When x = and{/=

(), A - - 9 a 2, and there can be neither a

maximum nor a minimum at (0, OK ^When x = a and y = a, A = 4 27 a 2

; and since ~ = -(5 a, we have the con-

ditions for a maximum value of the function fulfilled at (a, </K Substituting u* = a,

y a in the given function, we get its maximum value equal to a :{.

ILLUSTRATIVE EXAMPLE 3. Divide a into three parts such that their product

shall be a maximum.

Solution. Let x = first part, //= second part ; then a-(x + y)=a-x-y*=*

third part, and the function to be examined is

f(x t y} xy(a x ?/).

First Step. |= ay - 2 xy - y'

2 = 0, ^= -r

- 2 J?/- x 2 = 0.

(7 _ a

Solving simultaneously, we get as one pair of values .r -' y - -

Second Step. ^= - 2 ?/, ^ = a - 2 ar

- 2 ?/,

~^T,

= - 2 x ;

A = 4 xy - (a - 2 .r- 2 //)".

r>. When jr ="and y = ^ A = ^ ;

and since^ = - =~. it is seen

that our product is a maximum when x = -'y^r^'

Therefore the third .part is

also 7 , and the maximum value of the product is

PROBLEMS

Discuss for maxima and minima the following functions.

1. x 24- xy + y'

2 - 6 x 4- 2. Arts, r = 4, ?/= - 2 gives min.

2 . 4 j. + 2 ?/- r a 4 r//

-?/

2. ^ = J

.'?, ?/=

S gives max.

3. 2 x 2 - 2 A// 4- /y

2 + 5 j-- 3 ?/.

^ = - L //=

i gives min.

4 x>3 3 a^y 4- if-x = y a gives min.

5. sin x 4- sin y 4- sin (x 4- y). x = y = ~gives max.

x = y-- gives min.o

6. x 2 - xy 4- !/2 + oar 4- 60 4- c.

7 ^ , 2! + L' ^

x y2

8. Show that the maximum value of ^ . J2 . ^is a2 + fr2 "^ c2 '


9. Find the rectangular parallelepiped of maximum volume which

has three faces in the coordinate planes and one vertex in the plane

5 + 2+5 = 1. Ans. Volume =~a b c 27

10. Find the volume of the largest rectangular parallelepiped that can

be inscribed in the ellipsoid- + 2_ + = 1. Ans.

a 2 b 2 c 2

11. A pentagon is composed of a rectangle surmounted by an isosceles

triangle. If the perimeter of the pentagon has a given

value P, find the dimensions for maximum area.

PAUK. a 30, 2 JT = >

24-2 sec a tan a yp

y = .r(l + sec a).

12. Find the shortest distance between the lines x = ~ == - and, j Z o

13. A manufacturer produces two lines of candy at constant averagecosts of 50 cents and 60 cents per pound respectively. If the selling price

of the first line is jr cents per pound and of the second line is y cents per

pound, the number of pounds which can be sold each week is given by the

formulas N ^

_25Q(?/

_ ^ ^ = 3^()00 + 25Q(x __ 2 y).

Show that for maximum profit the selling prices should be fixed at 89 cents

and 94 cents per pound respectively.

14. A manufacturer produces razors and blades at a constant averagecost of 40 cents per razor and 20 cents per dozen blades. If the razors

are sold at x cents each and the blades at y cents per dozen, the demand

,., , . , , . 4,000,000 , 8,000,000 , ,, ,

of the market each week is ---- razors and ---- dozen blades..r?/ ry

Find the selling prices for maximum profit.

242. Taylor's theorem for functions of two or more variables. The

expansion of /(x, y) is found by using the methods and results of

Arts. 194 and 240. We consider

(1) F(t}=f(x + ht,y + kt),

and expand F(t) as in (5), Art. 194. The result is

' "(2)

We obtain the values of F(0), F'(Q), F"(0), by substituting t =in (2), (4), (7), Art. 240. By differentiating (7), and putting 1 = 0,

the expressions for F r/

'(0) etc. will result. These are omitted here.

Note, however, that F'"(0) is homogeneous and of the third degree


in h and k. A similar property holds for higher derivatives. If these

values are substituted in (2), and we set / = 1, the result is

(3) /(:r + h, y + k) =/(j, y) + hf,(x. y) +

R.

The expression for R is complicated and will be omitted from this

point on.

In (3) write x = a, # 6, and then replace A by (x a) and k by(?/

~~&) The result is Taylor's theorem for a function of two variables,

-a) +/y (a,

-a)(y

-

Finally, setting a = 6 = 0, we obtain an expansion correspondingto Maclaurin's series (4), Art. 194,

The right-hand member in (7) may be written as the infinite series

fA\ I

U lI

W iiI

(4) Mo +[T+

f|"+ '"'

where uo = f(0, 0),

w2 =/(0, 0)x2 + 2/,w (O f

etc.

These terms in (4) are homogeneous polynomials in (x, y). The

degree of each is equal to the subscript. That is, by (7) the function

is expanded into a sum of polynomials homogeneous in (x, y) and of

ascending degree. Similarly, in (/) the terms in the expansion are

polynomials homogeneous in (x a, y b).

Formula (/) is called the expansion off(x, y} at the point (a, b).

Reference must be made to more advanced treatises for proof of

the problem of determining those values of O, y) for which the

expansions (/) and (/) hold.

By breaking off series (4) at any term, an approximate formula for

f(x, y) is obtained for values near (a, b} or (0, 0). Compare Art. 200.


ILLUSTRATIVE EXAMPLE. Expand

xy 24- sin xy

at the point (1, \ IT) up to terms of the third degree.

Solution. Here a = 1, b = | IT,

and f(x, ?/)= *y* + si" *y

fy (x, y) =2xy + x cos xy,

= -y 2 smxy,= 2 # + cos x|/

- xy sin xy,

Substituting x = 1, y = J TT, the results are

/(I, JT) = iir'

/,(!,-

2 TT) - 1 7T2,

/(!, 2 T) = T.

=2

Substituting in (/), we get

xy2 + sin xy= l + -J

: 1) +7r(2/ ITT)

Formulas for expanding a function of three variables f(x t y, z) are

readily derived, and are left as problems.

PROBLEMS

1. From (1) above, show that

dr+ 3 + 3

2. Verify the following expansion.

cos x cos y = 1 TTT *

4- 6

J4

_ x6 + 15 J4?/2 + 15 j-V 4- y

6....

3. Expand sin .r sin y in powers of x and y.

4. Verify the following expansion.

a' log (1 + 10=0+ U2^1oga-t/2 + x 2i/log

2 a

5. Expand .r3-f x?/

2 at the point (1,2).

6. Verify the following expansion.- 3 x *y + 3

sin (x 4- !/)= x + y

.+

Verify the following approximate formulas for small values of x and y.

7. rr sin T/

=?/ -f .r//. (

8. PJ In (1 -h /)

=I' 4- ry

CHAPTER XXV

MULTIPLE INTEGRALS

243. Partial and successive integration. Corresponding to partial

differentiation in the differential calculus we have the inverse processof partial integration in the integral calculus. As may be inferred

from the connection, partial integration means that, having given a

differential expression involving two or more independent variables,

we integrate it, considering first a ///(//< on.e only as varying and all

the rest constant. Then we integrate the result, considering another

one as varying and the others constant, and so on. Such integrals

are called double, triple, etc., according to the number of variables,

and are known as multiple integrals.

In the solution of this problem the only new feature is that the

constant of integration has a new form. We shall illustrate this bymeans of examples. Thus, suppose we wish to find u, having given

Integrating this with respect to j, considering ?/ as constant, wehave

u - x 2 + .r// + 3 x + 0,

where c/> denotes the constant of integration. But since y was re-

garded as constant during this integration, c/> may involve y. Weshall then indicate this dependence of 4> on y by replacing (/> by the

symbol <t>(y). Hence the most general form of u is

where </>(?/) denotes an arbitrary function of y.

As another problem let us find

u^\ I (x2

This means that we wish to find u, having given

dxdy491

Integrating first with respect to y, regarding x as constant, we get

()U 0,2,

where \l/(x) is an arbitrary function of x.

Now integrating this result with respect to x, regarding y as con-

stant, we have

where <b(y) is an arbitrary function of y, and

244. Definite double integral. Geometric interpretation. Let/(j, y)

be a continuous and single-valued function of x and y. Geometrically,

(1) *=f(*,V)

is the equation of a surface, as KL. Take some area 5 in the A'OY-

plane and construct upon S as a base the right cylinder whoseelements are parallel to OZ. Let this cylinder inclose the area S'

on KL. Let us now find the volume V of the solid bounded byS, S', and the cylindrical surface. We proceed as follows :

At equal distances apart (= Ar) in the area 5 draw a set of lines

parallel to OY, and then a second set parallel to OX at equal distances

apart (= Ay). Through these lines pass planes parallel to YOZ andXOZ respectively. Thenwithin the areas S and S f

we have a network of

lines, as in the figure, that

in S being composed of

rectangles, each of area

AJ Ay. This construction

divides the cylinder into

a number of vertical col-

umns, such as MNPQ,whose upper and lower

bases are corre-

sponding portionsof the networks

in S' and 5 re-

spectively. As the upper bases of these columns are curvilinear, weof course cannot calculate the volume of the columns directly. Let

MULTIPLE INTEGRALS 493

us replace these columns by prisms whose upper bases are found thus :

each column is cut through by a plane parallel to XOY passed through

that vertex of the upper base for which x and y have the least

numerical values. Thus the column MNPQ is replaced by the right

prism MNPR, the upper base being in a plane through P parallel

to the X01T

-plane.

If the coordinates of P are (x, y, z), then MP = z = /(x, y), and

therefore

(2) Volume of M \PIt = /(x, ?/)A// Ax.

Calculating the volume of each of the other prisms formed in the

same way by replacing ;r and // in (2) by corresponding values, and

adding the results, we obtain a volume \rf

approximately equal to

V;that is,

(3) y'-vv j( .

where the double summation sign ^^ indicates that values of

two variables x, y must be taken account of in the quantity to be

summed up.

If now in the figure we increase the number of divisions of the

network in S indefinitely by letting Ax and A// diminish indefinitely,

and calculate in each case the double sum (3), then obviously V will

approach V as a limit, and hence we have the fundamental result

(4) V = HmA./ (I

A/y- (I

We show now that this limit, can be found by successive integration.

The required volume may be found as follows: Consider any

one of the slices into which the solid is divided by two successive

planes parallel to YOZ ;for example, the slice whose faces are FIHG

and JTL'K'. The thickness of this slice is Ax. Now the values of z

along the curve HI are found by writing x ~ OD in the equation

z = /(x, y) ;that is, along ///

Hence Area FIHG - f /(OD, y)dy.JltF

The volume of the slice under discussion is approximately equal

to that of a prism with base FIHG and altitude Ax ; that is, equal to

Ax - area FIHG = Ax f /(OD, y)dy.JDF

The required volume of the whole solid is evidently the limit of

the sum of all prisms constructed in like manner, as x (= OD) varies

from OA to OB ; that is,/ OB /^DG

(5) V =/

dx I J(x, y)dy.JOA JDF

Similarly, it may be shown that

/m r S*KV

(6) V=l dy f(x,y)dx.Joe JKW

The integrals (5) and (6) are also written 'in the more compact fonnno II nlHi nOV nEUI I /(*, y)dy dx and I I f(y, y)dx dy.JOA JDF Joe JEW

In (5) the limits DP and DO are functions of a% since they are

found by solving the equation of the boundary curve of the base of

the solid for y.

Similarly, in (6) the limits EW and EV are functions of y. Nowcomparison of (4), (5), and (6) gives the result

(A) V=limf(x,y)&y. Ax-A*-0Av-o

'i r fli

/ Hx.. _2 JV2

where v\ and r L> are, in general, functions of //, and ?/i and y/2 functions

of ;r. The second integral sign in each case applies to the first dif-

ferential.

Equation (A) is an extension of the Fundamental Theorem of

Art. 150 to double sums.

Our result may be stated in the following form.

The definite double integral

I

'

//Or, y)dy d2

Ja-2 J u-t

may be interpreted as that portion of the volume of a right cylinder which

is included between the plane, XOY and the surface

z^/0*, J/),

the base of the cylinder being the area in the XOY-plane bounded by the

/7,/r?v><?CM/IM.|/=

1*1, y = U2, x = ai, x a2 .

A similar statement holds for the second integral.

It is instructive to look upon the above process of finding the vol-

ume of the solid as follows.


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. Thevolume of the whole solid is then found by summing up all suchslices from x = OA to x = OB.

In successive integration involving two variables the order of

integration denotes that the limits on the second integral sign cor-

respond to the variable whose differential is written first, the differ-

entials of the variables and their corresponding limits being written

in the reverse order. Before attempting to apply successive integra-tion to practical problems it is best that the student should acquire

by practice some facility in evaluating definite multiple integrals.

ILLUSTRATIVE EXAMPLE 1. Find the value of the definite double integral

/a /"A/a-1

x~

I (.r -f y)dydx.Jo Jo

Solution,J ( O -f y)dy dxJo Jo

'

Jo I Jo

ir1

)^= I (x v a- - x'2 -f

-

Anx.

Interpreting this result geometrically, we have found the volume of the solid of

cylindrical shape* standing on OAB as base and bounded at the top by the surface

(plane) z = x + y.

The solid here stands on a base in the A'OV-plane bounded by

y (line OB)/ , , . . .

, I from y limits ;

y v a~ x~ (quadrant of circle AH] J

x- (line OA)^

x^a (line BE) _

, .. ..

from x limits.

ILLUSTRATIVE EXAMPLE 2. Verify

Solution,

(a y}x2 dydx

ra rZ^V ?v 2la r2h/i~

|(a - y}x'

2 dydx= I \ay-- \

x 2dx= ~Jo J b L ^ lo Jft -

/ a / ~\/a2 x2 9^j.'i

ILLUSTRATIVE EXAMPLE 3. Verify /I _xdydx -

JQ JVa^ x2 "

/* /*"

v/a " y2, , /*a f "|"v/a

2 - a; 2

Solution./

I _xdydx=l \xy\ _dxJQ J_Va ^ _ x2 J L J-Vaa-x^

In successive integration involving three variables the order of

integration is denoted in the same way as for two variables; that

is, the order of the limits on the integral signs, reading from the

inside to the left, is the same as the order of the corresponding

variables whose differentials are read from the inside to the right.

c 3 r 2 c r> *} *~\

ILLUSTRATIVE FA AMPLE 4. Verify I I/ ry* dz dy djr =

Ai -h J'2.**

nt>fr

/->;>r>

j nr^~

rjj /*2f ~15"I j-y~ dz dy dj- I I

"

/ .n/- dz dy dr = I

'

I xy2z dy dx

J''

-k '! \.'h'

*k J\ L J2

f\\r'2. r<\\ r2 ~\

/ x u* dyds^'3 I xy* dy \dx

-~ J\ ->2 Ml 1

35

In Problems 1 10 in the following list the solid whose volume

equals the value of the integral should be described.

PROBLEMS

Work out the following definite integrals.

(.r + L> )</// d.r = 5. 6.'

t -

^

/(/ -" '''' ^ l "*8. f f

"'

(.rt/ l./o

2/

4. r r " ^ <''/ = ,

7 9 -

('" f (j~" + " 2) ''-" ^

Ji /o'

()

2 rr ? ri r r '-' -

\ //f///(/.r= ]

-r'. 10. I

/f dij(Lr= I.

- ., ./o' ' '

- ;o At

11. f" PV' sin (JO dp-

\(a{ -

/r)(cos /J- cos a').

.//, .V

12. /

>7T

/"Mt(>sf/

p s jn ^ dp dO =,\a 1

'.

Jo . 'ii

/^TT /^a(1 ) t-(Y*(h

13. / / p2 sin ^ (/p ^/

A) .'o

15.~

//-'c (/c c/// (^ = i a^ia-* - 63 ).

16.

c " r^ c *~ x

17./

I I jrdzdjrdii = 3*5.Jo <Jtf 'o

is. r1

r1 " x

f1

"'zdzduds --

Jo Jo /o

/2/? rr^ I

r /Jo Jo

20.. f'

fT

fr +V ' " ' *d: <ln itr = i r 1 -

Jo Jo Jo

7

245. Value of a definite double integral taken over a region S. In the

last article the definite double integral appeared as a volume. This

(joes not necessarily mean that every definite double integral is a vol-

ume, for the physical interpretation of the result, depends on the

nature of the quantities represented by z, //, z. If x, y, z are the co-

ordinates of a point in space, then the result is indeed a volume. In

order to give the definite double integral

in question an interpretation not neces-

sarily involving the geometric concept of

volume, we observe that the variable z

does not occur explicitly in the integral,

and therefore we may confine ourselves

to the A'OV-plane. In fact, let us con-

sider simply a region S in the A'O V-plane,

and a given function /(z, y). Within this

region construct rectangular elements of

area by drawing a network of lines, as in Art. 244. Choose a point

(z, y) of the rectangular element, of area Az A?/, either with in the rec-

tangle or on its periwrtrr. Form the product

/(z, ?/)Az A//,

and similar products for all other rectangular elements. Sum up

these products. The result is

Finally let Ax - 0, and A?/ * 0.

We write the result

(1) lim yV/(x, ?y)Az A?y - ff/Yz, y}dx dy,Aa'-O*""'^ .''y->Q S

and call it the double integral of the function f(x, ?/) taken over the region S.

By (A) the value of the left-hand member in (1) was found by

successive integration when /(y, y) had no negative values for the

region S. The reasoning of Art. 244 will hold, however, if the portion


S' of the surface z =/(z, y) lies below the plane XOY. The limit of

the double sum will then be the volume with a negative sign. The

integrals in (A) will give the same negative number. Finally, if

/(x, y) is sometimes positive, sometimes negative for points of S, we

may divide S into subregions in which J(x, y) will be either always

positive or always negative. The reasoning will hold for each sub-

region and therefore for the combined region S. Hence the con-

clusion: the double integral in (1; may be evaluated in all cases by

successive integration .

It remains to explain the method of determining the limits of

integration. This is done in the next article.

246. Plane area as a definite double integral. Rectangular coordi-

nates. The problem of plane areas has been solved by single integra-

tion in Art-. 145. The discussion using double integration is useful

chiefly because the determination of limits for the general problemof Art. 245 is made clear. To set up the desired double integral,

proceed as follows.

Draw a network of rectangles as before. Then, in the figure,

(1) Element of area = Ax A//.

If A is the entire area of the region S, obviously, by (1), Art. 245,

Ay - ffdJ JOB) A = lim ]l;t Ay - dx dy.

Ax-0 ^"-f J JAy *0 S

Referring to the result stated in Art. 245, we may say :

The area of any region is the value of the double integral of the

function f(x, y) = 1 taken over that region.

Or, also : The area equals numerically the volume of a right cylinder

of unit height erected on the base S. (Art. 244.)

The examples show how the limits of integration are found.

ILLUSTRATIVE EXAMPLE 1. Calculate that portion of the area above OX whichis bounded by the semicubical parabola y'

2 x* and the straight line y = x.

Solution. The order of integration is indicated in the figure. Integrate first

with respect to jc. That is, sum up first the elements dr dy in a horizontal strip.Then we have

rAC pACI djc dy = dy I dx area of a horizontal strip of altitude dy.JAB JAB

Next, integrate this result with respect to y. This corresponds to summing upall horizontal strips. In this way we obtain

rOD rACA =

\ dxdy.Jo JAB

The limits AB and AC are found by solving each of the equations of the bound-ing curves for x. Thus from the equation of the line, jr = AB -

y, and from the

equation of the curve, .r = AC -?/-"

{

. To determine ()/>,solve the two equations simultaneously to find the pointof intersection E. This gives the point (1, 1); henceOD = 1. Therefore A

Am.

Or we may begin by summing up the elements dr dyin a vertical strip, and then sum up these strips. Weshall then have

'*~

/ /^ IJ d'r ~

I^"r~ jl>^r "

i~

In this example either order of integration may be chosen. This is not al-

ways the case, as the following example shows.

ILLUSTRATIVE EXAMPLE 2. Find the area in

the first quadrant bounded by the .r-axis arid thecurves j;2 + 2,2=10, tf = $ Xf

Solution. Here we first integrate with respectto x to cover a horizontal strip, that is, from the

parabola to the circle. We then have, for theentire area.

i<{

A = I I d.r (///,

Jo .'IK;

since the point of intersection is (1, ,3). To find

}JG, solve y~ = 9 x for .r. Then

To find ///, solve x~ -f y2 = 10 for x. We get

jc HI = + VlO ?/-'.

Hence

^ _ (' C yfly ,1*, \ li

i-|3-

^- i/M =6.75.27

|

Ans.

If we integrate first with respect to y, using ver-

tical strips, two integrals are necessary. Then

na'v'j

/^/

Io /

dydx+ /

Ji Jo

The order of integration should be such that the

area is given by one integral, if this is possible.

The examples above show that we set

or A=ffdydx

according to the nature of the curves bounding the area. The figures

below illustrate, in a general way, the difference in the summation

processes indicated by the two integrals.

PROBLEMS

1. Find by double integration the area between the two parabolas,'J \f

l 25 jr and 5 jr'' 9 //, (a) by integrating first with respect to y ;

(b) by integrating first with respect to .r.

,<\ ~^LL ^5 /-V ^AH*, (a) I I </// f/.r 5

; (b) I \

'

(i.r dy 5.

*> 25

Calculate by double integration the finite area bounded by each of the

following pairs of curves.

2. y =: 4 .r-

.r1', y ~ jr.

3. y'2 = 4 j-, 2 j-

-//= 4.

4. ?/

~.r

2, 2 .r

-// + H = 0.

5. r' = 2 .r, .r2 ^ (> //.

6. y'2 4 .r, jr = 12 + 2 // ?/~.

7. //2 2 .r, .r

2 + //'' 4 jc.

8. ?/2 = 9 + J, //

2 = 9 - :i s.

9. (j-2 + 4 2

)//= 8 \ 2 //

=.r, j- = 0.

10. jr^ + //* =a-', JT + //= a.

11. .r' + //

r< = a-\ jr + //= a.

12. ?/= j-

:i - 2 j-, //= 6 .r

-~ r :

.

9.

4.

48.

a 2(7r-l).

- 3 7r)a2

.

16.

13. j = 6 ?/-

?/

2, 2/

= .r. 16. .r2 + ?/

2 = 25, 27 ?/2 = 16 x?.

14. 4//2 = x'\ u

-x. 17. (2 a - .r)//

2 = .r3

, ?/ = or.

15. ?/2 = .r + 4, /y

2 = 4-2 .r. 18. ;r2 -

7/2 = 14, x- + ?/

2 = 36.

247. Volume under a surface. In Art 244 we discussed the volume

of a solid bounded by a surface

(l) ; =M 0),

the A'Oy-plane, and a cylinder. The elements of the cylinder were

parallel to OZ, and its base was a region S in the A'OV-plane. The

volume of this solid is, by 04),

(2) V =c, y)dx dy.

The order of integration and the limits are the same as for the

area of the region S. The volume of a solid of this type is the "volume

under the surface (1)." The analogous problem for the plane, "area

under a curve," has been treated in Chapter XIV. As a special case

the volume may be bounded by the surface and the A'OF-plane itself.

Note that the element of volume in (2 ) is

a right prism with base d.r dy and altitude z.

ILLUSTRATIVE EXAMPLE 1. Find the volume

bounded by the elliptic paraboloid

(3) 4 z= 16 ~-4.r 2 -\r

and the AW-plane.

Solution. Solving (3) for z, we get

(4) 2 = 4- a>J -1 //'.

Letting 2 = 0, we obtain

(5) 4 JT'J + //-

= 16,

which is the equation of the perimeter of the base

of the solid in the AW -plane. Hence by (2), using

the value of z in (4),

(6) V -- 4 1 I (4 - x- -\ ij'

2}dy dx = 16 TT. Am.

The limits are taken for the area OAB of the ellipse (5) lying in the first quadrant.

ILLUSTRATIVE K \AMPLE 2. Find the volume of the

solid bounded by the paraboloid of revolution

(7) -r* + ?r - nz,

the A'Oy-plane, and the cylinder

(8) .r- 4 !i2 = 2 ax.

Solution. Solving (7) for 2, and finding the limits

for the area of the base of the cylinder (8) in the XOY-

plane, we get, using (2;,

Jo Jody dx

l- 7ra :<. Ans.

For the area ONA (see figure), MN = v2 ax - x'2, (solving (8) for #), and OA = 2 a.

These are the limits.


PROBLEMS

1. Find the volume under z = 4 x'2, above z 0, and within y'

1 = 4 x.

Ann. V = 2f'

2

f2 X Z

(4- x 2

)dy dx = 17.24.Jo Jo

2. Find the volume under the plane x -f z = 2, above z 0, and within

x 2-f ?y

2 = 4. r2 /-A- 4

An*. F = r2 /-A- 4 x-1

2// (2

-.r;c/?/ dr = 8 TT.

J- 2 J()

3. Find the volume bounded by the plane --f

y-f

- 1 and the co-

ordinate planes.a ; r

^^ iabCu

4. Plnd the volume bounded above by j" -f z = 4, below by ^ = 0, and

laterally by //2 = 4 .r. Ans. \

l

.

5. Find the volume of the solid bounded above by //'-'= a 2

az, below

by 2 = 0, and within x'2-f //- a 2

. -4s. % ira*.

6. Find the volume under the elliptic paraboloid z 1]

.r2 J

//2

and above z 0. A/i,s\ 3 TV.

7. Find the volume under the plane x -f y -f z = 8, above ^ = 0, and

between the planes j: -f 2 // 8, x 2 //= 8. Ans. 170^.

8. Find the volume bounded by the cylindrical surface x 2-f ac a 2

and the planes x -f //= a, //

= 0, c = 0. yl/^s'. 4{a ;i

.

9. A solid is bounded by the surfaces ?/2-f

2 == 4 ar, x = 3 c/, and

lies within //2 = a.r. Find the volume. Ann. (6 TT 4- 9\/3)a :i

.

10. Plnd the volume below the cylindrical surface ?/2 = a 2

a^, above

2 = 0, and wit?iin the cylindrical surface .r2 + //

2 ax. Anx. % yra-5

.

11. Find the volume below ,: 2 x -f , al>ove ^ = 0, and within

x 2-f- //

2 = 2 ax. .4j/s. 3 ?ra :i.

12. Find the volume under //2 + ^ = 4, above r = 0, and within the

cylindrical surfaces ?/2 2 .r 0, //

2 8 2 .r. Ans. -\l

.

13. A solid is bounded by the paraboloid .r2 + ?/

2az, the cylindrical

surface ?/2 = a 2

a.r, and the planes .r 0, z = 0. Find the volume.

Ana. $ a3.

14. Find the volume under 4 z = 16 4 j-2

?/2

, above ^ = 0, and within

x* + y* = 2x.'

A?w?. ']! TT.

15. The axes of two circular cylindrical surfaces intersect at right angles

and their radii are equal (= r}. Find the common volume. .4??.s\ }^ r3 .

16. Find the volume of the closed surface jr:J + y

' + - y = a 3. (The trace

on each coordinate plane is the astroid, Chapter XXVI.) Ans. $ ?ra3 .

17. Find the volume common to ?/24- z 2 4 a.r and x 2

-f ?/2 = 2 a.r.

(2 TT + J#)a3

.

248. Directions for setting up a double integral. We shall now state

a rule for forming the double integral which will give a required

property. Applications are made in the following articles. For

single integration the corresponding rule is given in Art. 156.

FIRST STEP. Draw the curves 'which bound the region,, or area, con-

corned.

SECOND STOP. At any point I\.r, y) irithin the area construct the

rectangular element of area AJ A//.

THIRD STEP. Work out the function f(jr, //), which, 'when, multiplied

by AT A?/, gives the required property for the rectangular element of area.

FOURTH STEP. The required integral is

ffi/(.r, y)djr <

taken over the given region, or area. The order of integration and limits

are determined in the same wanner a,s in finding the area, itself.

249. Moment of area and centroids. This problem is treated in

Art. 177 by single integration. Double integration is often more

convenient.

We follow the rule of the preceding article. The moments of area

for the rectangular element of urea are, respectively,

.r AJ A?/, with respect to OV,

y AJ A'//, with respect to () V.

Hence for the entire area, using the notation of Art. 177, we have

(C) Mx

=j*J*ydxdy,My

=ffxdxdy.

*

The centroid of the area is given by

(D) x =area

"area

In (C) the integrals give the values of the .

double integrals of the functions

f(x, y) = y an(l fte, y) = x>

respectively, taken over the area. (Art. 245.)

For an area bounded by a curve, the x-axis, and two ordinates

(the "area under a curve"), w^ derive from (C)

(Dfft

rv rbI y dy dx - i I y'

2dx,

J() Jar\> rv r^>

My ==I I

x c dx= I xy dx.Ja Jo Ja


These agree with (2), Art. 177. Note that y in (!) is the ordinate

of a point on the curve, and its value in terms of x must be found

from the equation of the curve and substituted in the integrand be-

fore integration.

ILLUSTRATIVE EXAMPLE. Find the centroid of the area in the first quadrantbounded by the semicubical parabola y~ j ( and the

straight line y x.

Solution. The order and the limits of integration

were found in Illustrative Example 1, Art. 24(5. Ileuce,

using (C),

r1

c r1

?j ., i

1

~J() Jy*

'

Jo2

M v =11 y djc dy = i / (.va

y~)dy = fa.Jo J y ,/o

Since A areai

1

,,,

we have, from (D), ?=J-0.48, //= A> = 0.42. An*.

250, Theorem of Pappus. A useful relation between centroids and

volumes of solids of revolution is expressed in the following theorem.

// a plane area is reroltwl about an axis lying in its plane and not

crossing it, the volume of the solid of revolution, thus generated is equal

to the prod net of the plane area by the circumference described by its

centroid.

Proof. Let the area in the figure be

revolved about the x-axis. The rectangu-

lar element of area within the region X at

P will generate a hollow circular cylinder

whose volume A Y is given by

_\ y = TT(// + A//)L> AJ Try

2 A.r.

Factoring and simplifying, we get

AV-2 TTG/+ -2 A//) Ax A//.

Now, in (1), Art. 245, (x, y) in/(j, y) is a point"either within the

rectangle PQ or on its perimeter." But (x, y + \ A//) is a point on

the perimeter of PQ. Therefore let f(x, y >= 2 Try. Then A V has the

form fix, //) Ax A//, and, by (1), Art. 245, and (C),

ii) rx = 2

Finally, using (D), we get

(2) V, - 2 iry A,


where A is the area of the region 5. The right-hand member is the

product of the area by the circumference described by its centroid.

Hence the theorem is proved. We write the result

(3) V = 2 iry- A.

If two of the quantities V, y, A are known, the other can be found

by (3).

ILLUSTRATIVE EXAMPLE. Find the centroid of the trapezoid OMPB of the

figure by the Theorem of Pappus.

Solution. Area OMPR = J(3 + 5^,8 = 32. Revolving the figure about OX, the

solid formed is a frustum of a cone of revolution. Hence, by (12), Art. 1, since

a = 8, R = 5, r = 3,

Hence, by (3), "y= ~-~ = f^ = 2.04.

2 7T-4 19J

Revolving the figure about OY, the volume generatedis the difference of the volumes of the cylinder generated

by OrPM and the cone generated by the triangle HCP.

M

Hence

V u= 320 7T - 12S TT

3

Hence, by the theorem, x

The centroid is (4j, 12.04). Arts.

_3*

PROBLEMS

Find the centroid of the area bounded by each of the following curves:

1. y = .r3

, y 4 jr. (Area in first quadrant.) Am.([ Ji, !j \).

2. y = 6 x jr2

, ?/ .r.

3. ?/= 4 x - j'~, //

= 2 jr - 3.

4. x2 = 4 ?/, r - 2 ?/ -f 4 - 0.

5. ?/= ^ 2

,2 .r

-y 4-3 = 0.

6. y = x 2 - 2 .r-

3, ?/= 2 .r

- 3.

7. y* = jrf

jr 4- ?y= 2, ?y

= 0. (First quadrant.)

8. ?/2 =

.r, r -f ?/= 2, x = 0.

9. ?/3 = J 2

,2 //

= r.

10. 4 ?/- 3 j-

2,2 ?/

2 = 9 x.

11. ?/2 = 2 x, ?/

= r ~ x 2.

12. ?/ = 8 x, x + ?/= 6,

13. y2 = 4 .r, y2 = 5 - x.

14. y = 6 x - x2, x 4- ?/ = 6.

(. 5).

(1, I)-

0. g)-

0. -'e1)-

(2,-f).

(I!. A)-/ H 1 1 \

V2 T,' TOV'/J 4 ONV 3" dn)-/ 9 2 7 \

Vii)' J'o>

(U-- H

(-4).(.0).

Ans. (-V-, I).15. x = 4 y y2

, y = x.

16. ?/= 4 x x 2

, # 5 - 2 a;.

17. ?/2 = 4 x, 2 x - y = 4.

18. ?/= x 2 - 2 x - 3, y = 6 x - x2 - 8.

19. *2 + 2 = l, z + jy= l.

20. x 2 + ?/2 = 32, ?y

2 = 4 x.

21. i/2 = 4 x, 2 x + y = 4.

22. x 2 + ?y2 - 10 x = 0, x 2 = y.

23. x 2 =?y, 2 |/

= 6 x - x 2.

24. x 3 -f y 3 = a 3. (Area in first quadrant.)

25. x1

*-f ?y^

= a*, x = 0, y = 0.

(3, I).

(2, 1).

(0.58.5. 0.585).

/25B(7 256 a \

\31f> Tr'ttlo 7T/

(-' -V\5 5/

26. P'ind the controid of the area under one arch of the cycloid

x = a(9 sin 0), y = <j(l cos 0). *

^/ 5 o\

I71

*

'

6 /

27. Using the Theorem of Pappus, find the centroid of a semicircle.

Ans. Distance from diameter = -3 TT

28. Using the Theorem of Pappus, find the centroid of the area of the

ellipse \-~-~l which lies in the first quadrant. ..,,,,. , , ,

a 2ft- \3 TT 3 7T/

29. Using the Theorem of Pappus find the volume of the torus gen-

erated by revolving the circle (x ft)2

-f- ?y2 = a 2

(ft > a) about the ?y-axis.

A M. 2 ?T2a 2

ft.

30. A rectangle is revolved about an axis which lies in its plane and is

perpendicular to a diagonal at its extremity. Find the volume of the

solid generated.

251. Center of fluid pressure. The

problem of calculating the pressure of

a fluid on a vertical wall was discussed

in Art. 179.

The pressures on the rectangular ele-

ments of the figure constitute a systemof parallel forces, since they are per-

pendicular to the plane of the area

XOY. The resultant of this system of

forces is the total fluid pressure P, given

by (Z)), Art. 179.

D

(Dr b

' = W I yxdx.Ja

The point of application of /' is called the center of fluid pressure.

We wish to find the .r-coordinate (~ .ru ) 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, byArt. 179,

(2) dP = TT>// A:r.

The turning moment of this force about the axis OY is the productof dP by its lever arm OK (== .r), or, using (2),

(3) Turning moment of dP = x dP = Wx 2y Ax.

Hence we have, for the entire turning moment for the distributed

fluid pressure,r'

1

(4) Total turning moment / Wx~{/dx.

But the turning moment of the resultant fluid pressure P is x^P.

Hence

(5) x()P W I x'

2ydx.

Solving for /<> and using (1 ), we get the formula for the depth of the

center of pressure

f *x2dA

(6) a-o= -

/'Jii

xdA

where dA element of area = ydx.

The denominator in (6) is the moment of area of ABCD with

respect to OY (see Art. 177). The numerator is an integral not met

with hitherto. It is called the moment of inertia of the area ABCDabout OY.

The letter / is commonly used for moment of inertia about an

axis, and a subscript is attached to designate the axis. Thus (6)

becomes


The usual notation for moment of inertia about an axis I is

(8)

in which

(9) r = distance of the element dA from the axis I.

The problem of this article is one of many which lead to momentsof inertia. In the following section the calculation of moments of

inertia by double and single integration is explained. Applicationsare also given.

252. Moment of inertia of an area. In mechanics the moment of

inertia of an area about an axis is an important concept. The calcu-

lation of moments of inertia will now be explained. We follow the

rule of Art. 248. A

For the elementary rectangle PQ at

P(j, y) the moment of inertia about OA'

is defined as

(1) '//

2AxA*/,

and about the ?/-axis it is

(2)

Then, if Ix and Iy are the corresponding moments of inertia for the

entire area, we have (compare (8), Art. 251)

The radii of gyration rx and rv are given by

*U2 _ 2

area area

In (E) the functions whose integrals are taken over the area are,

respectively, /(.r, y) = y2

, and f(x, y) = .r2

.

Formulas (E) become simple for an area "under a curve," that is,

an area bounded by a curve, the x-axis and two ordinates. Thus weobtain r \, ry ~ b

I*=/

V2dydx=l I y*dx,

Ja J() J u.

(3) /*& rv rb,= \ x2dydx= / yzydx

J a Jo Ja

In these equations y is the ordinate of a point on the curve, andits value in terms of x must be found from the equation of this curve

and substituted in the integrand.


Formulas for moments of inertia / are written in the form

(G) I=Ar*,

where A = area and r radius of gyration. Solving (F) for Ix and Iv

will give this form.

Dimensions. If the linear unit is 1 in., the moment of inertia

has the dimensions in. 4. By (F), rx and rv are lengths, in inches.

ILLUSTRATIVE EXAMPLE 1. Find /,, /, and the corresponding radii of gyration

for the area of Illustrative Example 1, Art. 246.

Solution. Using the same order of integration and

the same limits as before, we have, by (E),

Since A - area = 1J

n ,we find, by (F),

rt = 0.48, rt = 0.53. Am.

ILLUSTRATIVE EXAMPLE 2. Find Is and Iv for the parabolic segment BOC in

the figure.

Solution. With the axes of coordinates as drawn, the equation of the bounding

parabola is

(4) y'2 = 2 px.

Since B(a, b} is a point on the curve, we get, by substitut-

ing x a, y = b in (4), b'2 = 2 pa. Solving this equation for

2 p and substituting its value in (4), we obtain

2 &*jr bx*y2 = , or 2/

~a ^7

(5)O

The moments of inertia for the area under the parab-

ola OPB in the first quadrant will be half the required

moments. Hence, using (3j, and substituting the value of

y from (5), we get

B(a,b)

15

For the area of the segment, we find

Hence, by (f), rS = = 6 s, and /, = A62

,

The results are in the form (G). Ans.


In the figure on page 325 the axis Y lies in the surface of the fluid.

If we denote this axis in any figure by s, then the depth of the center

of pressure, is by (7), Art. 251

if r = radius of gyration about the axis s,

and ha= depth of centroid below the axis s.

ILLUSTRATIVE EXAMPLE 3. Find the depth of the center of pressure on the

trapezoidal water gate of the figure. Compare Illustrative Example 2, Art. 179.

Solution. Choose axes OX and OY as shown, and draw an elementary hori-

zontal strip. Let the distance of this strip from the axis s at the water level be r.

r = 8 - y, dA = 2 x dy.

Hence, by (8), Art. 251, and by the

definition of moment of area (Art. 177),

we have

(7)

(8) M, =

The equation of AB is y = 2 x - 8.

Solving this for x, substituting in (7) and (8), and integrating with limits y = 0,

y = 4, we obtain4

/. = f (8-

i/)2(8 + y)dy = 1429 J,

A)

M. = f4

(64- y*)dy = 234.

Hence, by (7), Art. 251, x tt= 6.09. Ans.

253. Polar moment of inertia. The moment of inertia of the ele-

mentary rectangle PQ about the origin O is the product of the area

and OP 2, that is,

(1) (^ + \

Hence, by Art. 248, for the entire area

y

(2) ij2)dx dy.

We may, however, write the right-handmember as the sum of two integrals, for

(2) is clearly the same as

(3)lo-jy^dxdy+JJy*

Hence we have the following theorem.

The moment of inertia of an area about the origin equals the sum oj

its moments of inertia about the x-axis and the y-axis.

/.

PROBLEMS

Find /*, /, and / for each of the areas described below.

1. The semicircle which is to the right of the ?/-axis and which is

bounded by x2 + y2 = r 2

. ., __ . _ Ar 2

/ins. If iy -"~:

2. The isosceles triangle of height h and base a whose vertices are

/A m /i, a \ li a\ A jAa'2 T Ah'2

(0> 0)'r' 2/'\' ~2/

LS>

'^"iSr v=~'

3. The right triangle whose vertices are (0, 0), (b, a), (b, 0).

r - ^ a27 - Ab *

8. 1 x T7 ' - y Q*

4. The ellipse + | = 1. 4ns. /x = 4r~ ^ = '

a 2 b 2 445. The area in the first quadrant bounded by y'

2 4 .r, x = 4, ?/= 0.

j ~5

~7

J*2

772

6. The area included between the ellipse- + *- = 1 and the circle

y* = 2y. ;4r

19 A7

53 AAm. y^-^-,/^-^.j-2 ^i2 f<2

7. The area included between the ellipses -f TT 1 ano x 2 + 4- = 1.16 9 4

,r

5 Ar

19 AArts. /x ^ , 7V ^

|

8. The area included between the circle x 2-f 2/

2 16 and the circle

(y + 2)* = l. . _239 A _17ArAw*. /,-

6Q, /-

4

9. The area included between the circle x'2 + ?/2 = 36 and the circle

( + 3)* = 4.4ns j^ OA>

T 2772

, 10. The area betwee'n the circle x 2-f 7/

2 = 4 and the ellipse -f f- = 1.36 16

. r 23 A r 53 AAws. /, = , /

tf= -y

11. The entire area bounded by xf -h y* = a^. Ans. Ix = Iv = Z_4.64

12. Find the depth of the center of pressure on a triangular water

gate having its vertex below the base, which is horizontal and on a level

with the surface of the water.

13. Find the depth of the center of pressure on a rectangular water

gate 8 ft. wide and 4 ft. deep when the level of the water is 5 ft. abovethe top of the gate. Ans. 7.19 ft. below the surface of the water.


14. Find the depth of the center of pressure on the end of a horizontal

cylindrical oil tank of diameter 5 ft. when the depth of oil is (a) 2.5 ft.;

(b) 4ft.; (c) 6ft.(a) ==1 .47ft .. (b) approximately 2.4 ft.

;

= 3*95 ft.

254. Polar coordinates. Plane area. When the equations of the

curves bounding an area are given in polar coordinates, certain modi-

fications are necessary.

The area is now divided into elementary portions, as follows:

Draw arcs of circles with the common center with successive

radii differing by Ap. Thus, in Fig. 1, OP = p, OS = p + Ap. Then

-Ar U -x

FIG. 1

draw radial lines from O such that the angle between any two con-

secutive lines is the same and equal to A0. Thus, in Fig. 1, angle

FOR = A0.

The area will now contain a large number of rectangular portions,

such as PSQR in Fig. 1.

Let PSQR = AA. Now AA is the difference of the areas of the

circular sectors FOR and SOQ. Hence

(1) AA == i^ Ap 2 A0.

The function f(x, ?/) of Art. 245 is to be replaced by a function

using polar coordinates. Let this be F(p, 6). Then, proceeding as

in Art. 245, we choose a point (p, 6) of AA, form the product

F(p, 0)AA

for each AA within the region S, add these products, and finally let

Ap and A0 > 0. It is shown in Art. 258 that the limiting value

of this double sum may be found by successive integration. We nowwrite (compare (1), Art. 245)

(2) limAp -

'

AH -

AA =f (V(pJ J

, d)pdpd6,


and call it the double integral of the function F(p, 8} taken over the

region S.

Note in (2) that the value of AA in (1) has been replaced in the

integral by p dp dO.

The simplest case of (2) is that of finding the area of the region S.

We then have

(H) A= CCp dp dB =JTpdO dp.

These are easily remembered if we think of the elements (checks)

as being rectangles with dimensions p dd and dp, and hence of area

p dd dp.

The figures below illustrate, in a general way, the difference in

the processes indicated by the two integrals.

In the first, we integrate first with respect to p, since dp precedes

dd, keeping 6 constant. This process will cover the radial strip KGHLin Fig. 2, p. 512. The limits for p are p = OG and p = OH, found

by solving the equation (or equations) of the bounding curve (or

curves) for p in terms of 0. Then integrate varying 0, the limits

being = Z JOX and 6 = Z IOX.

The second integral in (2) is worked out by integrating with re-

spect to 0, p remaining constant. This step covers the circular strip

ABCD in Fig. 1, p. 512, between two consecutive circular arcs.

Then integrate varying p.

When the area is bounded by a curve and two of its radii vectores

(area swept over by the radius vector), we obtain from the first form

r ft r? + r= I pdpdd = %Jot t/0 Jci

which agrees with (D), Art. 159.

Double integrals in polar coordinates have one of the forms

(3) ffF(p,0)pdpd0 orJjF(p,0)pdOdp.


ILLUSTRATIVE EXAMPLE 1. Find the limits for the double integral giving somerequired property related to the ar^a inside the circle p = 2 r cos 6 and outside thecircle p ~ r.

Solution. The points of intersection are

A(T,~j.

and B(T, -|V Use the first form

in (3).

The limits for p are

p = O//-2r costf;

for 6 they are ~ and ^ - Ans.o o

ILLUSTRATIVE EXAMPLE 2. Find the area inside the circle p = 2 r cos 6 andoutside the circle p r.

Solution. From Illustrative Example 1 above, we have

7L

r(4 r2 cos 2 6 - r*)d6 = r2(i IT + J Vs) = 1 .91 r\ Ans.

255. Problems using polar coordinates. There should now be no

difficulty in establishing the following formulas:

(1) Mx = ffp 2 sin 9 dp d6.

(2) Mv= CCp*ca&8dpdO.

(3) /,= CCp*siY\26dpd8.

(4) /

(5)

The order of the differentials will have to be changed if integration

with respect to 6 is performed first.

ILLUSTRATIVE EXAMPLE 1. On account of important applications the momentsof inertia of a circle are now worked out.

Let a = radius. Then, by (5), the polar momentof inertia with respect to the center is

(6) 7 =

where A = area of the circle. ~TV\~\ V'

/ / / / / xAlso, since I, = Iv , by symmetry , we have, by (3),

Art. 253,

In words : The polar moment of inertia of a circle with respect to its center equal*the product of half the area and the square of the radius; the polar moment of inertia

with respect to any diameter equate the product of one fourth the area and the square

of the radius.

ILLUSTRATIVE EXAMPLE 2. Find the centroid of a loop of the lemniscate

p* = a'2 cos 2 6.

Solution. Since OA" is an axis of symmetry, wehave y = 0.

i r -V * r (i v f t>H 2 ft

\A^\< f pdpd.6Jo Jo

X

p" f" Vros2

V'cosJo '(}

33

f*

"

(1- 2 sin 2 0)* cos 9 dO

Jo

Hence z = = a v 2 = 0.55 a. Ana.

by (5), Art. 2

ILLUSTRATIVE EXAMPLE 3. Find 7 over the region bounded by the circle

p = 2r cos B.

Solution. Summing up for the elements in

the triangular-shaped strip OP, the p-limits

are zero and 2 r cos (found from the equationof the circle).

Summing up for all such strips, the 0-limits

are - ^ and - Hence, by (5),

. Ann.

Or, summing up first for the elements in a circular strip (as QR), we have

PROBLEMS

1 Find the area inside the circle p = $and to the right of the line

4 p cose = 3.

"

Arts .

3(4, -3Vi).ID

2. Find the area which is inside the circle p = 3 cos and outside the

circle p = f . A 718.3(2 7T + 3V3)


3. Find the area which is inside the circle p = 3 cos 6 and outside the

circle p = cos 6. Ans. 2 IT.

4. Find the area inside the cardioid p = 1 + cos 6 and to the right of

the line 4 p cos 6 = 3. ^ 9\/3Ans.

g+-16-

5. Find the area which is inside the cardioid p = 1 -f cos 6 and outside

the circle p = 1. AnK I + 2 .

8. Find the area which is inside the circle p = 1 and outside the cardi-

oid p=a + cos0. AnSt 2 __]r t

4

7. Find the area which is inside the circle p = 3 cos 6 and outside the

cardioid p = 1 + cos 9. Ans. IT.

8. Find the area which is inside the circle p 1 and outside the parab-

ola p(l + cos 0) = 1. 4ws. !_?.-in.s.2 3

9. Find the area which is inside the cardioid p = 1 -f cos 6 and outside

the parabola p(l + cos 0) = 1. x_ 3jr,4.4310. Find the area which is inside the circle p = cos -f sin and out-

side the circle p 1. Ans. %.

11. Find the area which is inside the circle p = sin 6 and outside the

cardioid p = 1 cos 0. Ans. 14

12. Find the area which is inside the lemniscate p2 2 a 2 cos 2 6 and

outside the circle p = a. Ans. 0.684 a 2.

13. Find th area which is inside the cardioid p = 4(1 -f cos 6) and

outside the parabola p(l- cos 9) = 3. Ans. 5.504.

14. Find the area which is inside the circle p = 2 a cos 9 and outside

the circle p a. Find the centroid of the area and If arid /.

-+ -

15. Find the centroid of the area bounded by the cardioid^ g ^

p = a(l + cos0). Ans. x =

16. Find the centroid of the area bounded by a loop of the curve

= acos20.105 ir

17. Find the centroid of the area bounded by a loop of the curve

= acos30. A - 81Via.Ans - x ~SO IT


!

18. Find / for the lemniscate p2 - a 2 cos 2 0.

19. Find Ir for the cardioid p = a(l -h cos 6}.

20. Find /, and Iv for one loop of the curve p = a cos 2 0.

21. Prove from (1), Art. 254, that

Am. ~ (3 TT + 8)a2

.

48

limmilo p Ap SO

-

and therefore SA "differs from p Ap SO by an infinitesimal of higher

order" (Art. 99). Then SA in the left-hand member of (2), Art. 254, maybe replaced by p Ap SO. (Proof omitted.)

256. General method for finding the areas of curved surfaces. Themethod given in Art. 164 applied only to the area of a surface of

revolution. We shall now give a more general method. Let

(1; z=f(jc,y)

be the equation of the surface KL in the figure, and suppose it is

required to calculate the area of the region S f

lying on the surface.

Denote by tf the region on the A'Oi'-plane which is the orthogonal

projection of Sf

on that plane. Now pass planes parallel to YOZ and

XOZ at common distances

Ax and Sy respectively. As

in Art. 244, these planes form

truncated prisms (as PB)bounded at the top by a

portion (as PQ) of the given

surface whose projection on

the A'O r-plane is a rectangle

of area Ax A// (as AB). This

rectangle also forms the lower

base of the prism. The coor-

dinates of P are (x, y, z).

Now consider the plane

tangent to the surface KLat P. Evidently the same rectangle AB is the projection on the

A'OY-plane of that portion of the tangent plane (PR) which is in-

tercepted by the prism PB. Assuming 7 as the angle the tangent

plane makes with the XOV-plane, we have

Area AB = area PR cos 7,

["The projection of a plane area upon a second plane is equal to the area of the!

or

[portion projected multiplied by the cosine of the angle between the planes

Sy Sx = area PR - cos 7

518 DEFERENTIAL AND INTEGRAL CALCULUS

Now 7 is equal to the angle between OZ and a line from per-

pendicular to the tangent plane. Hence from (H), Art. 237, and (2)

and (3), Art. 4, we have1

cos 7 =

L \f)x) WThen Area PR = 1 1 + (-Y+ (\~ |

J

Ay Ax.L \tixl \vy! J

This we take as the element of area of the region S'. We then define

the area of the region S' as

limAx^y -* o

the summation extending over the region S, as in Art. 245. Denoting

by A the area of the region N', we have

the limits of integration depending on the projection on the XOY-plane

of the region whose area we wish to calculate. Thus, for (/) we choose

our limits from the boundary curve or curves of the region S in the

.YOY-plane precisely as we have been doing in the previous sections.

Before integrating, the expression

must be reduced to a function of j and y only, by using the equation

of the curved surface on which the area lies.

If it is more convenient to project the required area on the XOZ-

plane, use the formula

(/) A-'

S

where the limits are found from the boundary of the region S, which

is now the projection of the required area on the A'OZ-plane.

Similarly, we may use

S

the limits being found from the projection of the required area on the

VOZ-plane.


In some problems it is required to find the area of a portion of onesurface intercepted by a second surface. In such cases the partialderivatives required for substitution in the formula should be foundfrom the equation of the surface whose partial area is wanted.

Since the limits are found by projecting the required area on oneof the coordinate planes, it should be remembered that

To find the projection of the area required on the XOY-plane, elimi-

nate z beticeen the equations of the surfaces whose intersections form the

boundary of the area.

Similarly, ire eliminateij to find the projection o)i, the XOZ-plane, and

x to find it on the YOZ-plane.

This area of a curved surface gives a further illustration of inte-

gration of a function over a given area. Thus in (/) we integrate the

function

over the projection of the required curved surface on the -YOV-plane.As remarked above, (J) and (AT) must, be reduced to

I lf(jr, z)dz dx and I

ff(y,z}dy dz,

respectively, by means of the equation of the surface on which the

required curved surface lies.

ILLUSTRATIVE EXAMPLE 1. Find the urea of the surface of the sphere

j>2 _|_ y'-' -|_~? _ r ?

by double integration.

Solution. Let ABC in the figure be one eighth of the surface of the sphere. Here

dx

and (fs

)'=\dy/

. The projection of the area required on the A'OV-plane is AOB, a region bounded

by x = (= OB) ; y = (= OA ) ;& + ^- r- (= 7M ).

Integrating first with respect to y, we sum upall the elements along a strip (as DEGF) which is

also projected on the A^OY-plane in a strip (as

MNGF) ; that is, our ^--limits are zero and MF(= Vr 2 x 2

). Then integrating with respect to x

sums up all such strips composing the surface ABC ;

that is, our x-limits are zero and OA ( r). Sub-

stituting in (7), we get

r dy dx _ irr2

Vr- x'2 y2 %

A = 4 Trr 2 . Ans. 5

A = r p-'-8 J Jo

JfU.


ILLUSTRATIVE EXAMPLE 2. The center of a sphere of radius r is on the surface

of a right cylinder, the radius of whose base is - Find the area of the surface of

the cylinder intercepted by the sphere.

Solution. Taking the origin at the center of the sphere, an element of the cylin-

der for the z-axis, and a diameter of a right section of the cylinder for the .r-axis, the

equation of the sphere is

x'2 + y'2 + z'

2 = r 2,

and of the cylinder x'2 + y'2 = rx.

ODATB is evidently one fourth of the cylindrical

surface required. Since this area projects into the

semicircular arc ODA on the ArOlr

-plane, there is

no region S from which to determine our limits

in this plane ;hence we shall project our area on,

say, the A'OZ-plane. Then the region N over

#hich we integrate is OACB, which is bounded

by z = (~ OA), x (- OB), and z 2 + rx = r'2

(= ACTt), the last equation being found by elim- y/inating y between the equations of the two sur-

faces. Integrating first with respect to z moans that we sum up all the elements in

a vertical strip (as 7V, the z-limits being zero and Vr- - rx. Then, on integrating

with respect to j, we sum up all such strips, the j-limits being zero and r.

Since the required surface lies on the cylinder, the partial derivatives required

for formula (J) must be found from the equation of the cylinder.

dy r 2 x dy _Hence -^ ^ ?/

'

TH


A = rr^^~4 Jo Jo

Substituting the value of y in terms of / from the equation of the cylinder,

PROBLEMS

1. In the preceding example find the surface of the sphere intercepted

by the cylinder. A frr r^'-** dy dx _ 9( 9 , 2

Am. 4rl/ -j=======--6(Tr ^)r.

Jo Jo Vr 2 a- y-

2. The axes of two equal right circular cylinders, r being the radius

of their bases, intersect at right angles. Find the surface of one inter-

cepted by the other.

HINT. Take x 2 + z2 = r 2 and x 2 + y2 = r 2 as the equations of the cylinders.

A o Cr f^ra -*2 dy dxAns. 8 r

/,

=Jo Jo v r 2 _ x 2

or-.


3. Find the area of that portion of the sphere .r2-f ?/

2-f z 2 = 2 a// cut

out by one nappe of the cone .r- 4- :'-' //-. An*. 2 Tra2

.

4. Find the surface of the cylinder .r2

-f //- r 2 included between the

plane z = nix and the A'OV-plane. An*. 4 r-m.

5. Find the area of that part of the plane-

-f--

-f- 1 which is in-

tercepted by the coordinate planes.a

^_An*. \Vlrc~ -f c-a 2

-f a 2b'

2.

6. Find the area of the portion of the sphere .r2-f ir 4- z'

2 2 ay whichlies within the paraboloid by f'

2-f r 2

. An*. 2 irab.

7. In the preceding example find the area of the portion of the pa-raboloid which lies within the sphere.

8. Find the area of the surface of the paraboloid //2-f z'

2 \ a.r in-

tercepted by the parabolic cylinder //'-'a.r and the plane JT = 3 a.

An*, -y Tra2

.

9. In the preceding problem find the arra of the surface* of the cylinder

intercepted by the paraboloid and plane. A , /lo^/Hi i\ a2/A )l*. ( 1 ) V 1 ) 1 ) ^r

Vii

10. Find the surface of the cylinder -:J

-f (.r cos c\ -f // sin cv)2 ~ r 2 which

is situated in the positive compartment of coordinates.

HINT. The axis of this cylinder is the line 2-

0, jr cos a -f // sin a ; and the

radius of the base is r. r 2

An*.sin (\ cos rv.

11. Find the area of that portion of the surface of the cylinder

y* -|~- !5

a'! bounded by a curve whose projection on the A' V-plane is

x% + yr* = a*. An*. V5 a 2

.

12. P"ind by integration the area of that portion of the surface of the

sphere jr2 + ?y

2-f z- - 100 which lies between the parallel planers jc = 8

and x 6.

257. Volumes found by triple integration. In many cases the vol-

ume of a solid bounded by surfaces whose equations are given maybe calculated by means of three successive integrations, the process

being merely an extension of the methods employed in the preceding

articles of this chapter (see also Art. 247).

Suppose the solid in question is divided by planes parallel to the

coordinate planes into rectangular parallelepipeds having the dimen-

sions Az, AT/, Ax. The volume of one of these parallelepipeds is

Az A?/ Ax,

and we choose it as the element of volume.

Now sum up all such elements within the region R bounded bythe given surfaces by first summing up all the elements in a column

parallel to one of the coordinate axes, then summing up all such


columns in a slice parallel to one of the coordinate planes containing

that axis, and finally summing up all such slices within the region in

question. The volume V of the solid will then be the limit of this

triple sum as Az, A?/, Ax each approach zero as a limit. That is,

(1) V= lim YVVA; A?/ A*,R

the summations being extended over the entire region H bounded bythe given surfaces. This limit is denoted by

By extension of the principle of Art. 245, we speak of (L) as the

triple integral of the Junction f(x, y, z)= I throughout the region R.

Many problems require the integration of a variable function of x, y,

and z throughout a given region. The notation is

(2) y> dx>

which is, of course, the limit of a triple sum analogous to the double

sums we have already discussed. In more advanced treatises it is

shown that the triple integral (2) is evaluated by successive integra-

tion. The limits are found in the same manner as for (L).

Simple examples of (2) are afforded by the formulas for thecentroid

&> y> z) (center of gravity) of a homogeneous solid, namely,

Vx = ff y dx dy dz, Vz = z dx dy dz.

They are obtained by reasoning as in Art. 249, using moments of

volume. In the integrands, (x, y, z) is an interior point. The centroid

will lie in any plane of symmetry.

ILLUSTRATIVE EXAMPLE 1. Find the volume of that portion of the ellipsoid

iI

1, )'

. i

a* o- r*-

which lies in the first octant.

Solution. Let O-ABC be that portion of the

ellipsoid whose volume is required, the equationsof the bounding surfaces being

(3)

(4)

(5)

(6)

= i (= ABC),c 2

2 = (= GAB),

y = (= OAC),x = (= OBC) .

PQ is an element, being one of the rectangular parallelepipeds with dimensions

Az, A?/, A.r into which the planes parallel to the coordinate planes have divided the

region.

Integrating first with respect to z, we sum up all such elements in a column

(as ftS), the z-limits being zero (from (4)) and TR = rWl - -f- (from (3) by

solving for z).V (/

~ 6 *

Integrating next with respect to //, we sum up all such columns in a slice (as

I>EM\GF\ the /y-limits being zero (from (5)) and M<1 =b^J

I- ~ (from the

e(iuation of the curve AGF>, namely -f^-

1, by solving for */).

a 2 b-

Lastly, integrating with respect to .r, we sum up all such slices within the entire

region O-ABC, the .r-limits being zero (from ((>)) and OA a.

Hence dz dy dx

Therefore the volume of the entire ellipsoid is

ILLUSTRATIVE K\ -\MPLE 2. Find the volume of t he solid bounded by the surfaces

(7) z~

4 - .r~ -', //-,

(8) 2 = 3 /-' + 1 //-.

Solution. The surfaces are the elliptic paraboloids of

the figure. Eliminating z between (7) and (8), we find

(9) 4 J a + 1 y- =4,

which is the equation of the cylinder AB(^I> (see figure)

that passes through the curve of intersection of (7) and

(8) and has its elements parallel to OZ.

We have

(10)r =

4/M

/'

2V2(1 ' ) f4

"V1

J{) Jo J'.ljr- \ } y-

The limits are found as follows :

Integrating with respect to z, we sum up the elements of volume dz dy dx in

a column of base dy dx from the surface (8) to the surface (7) (MP to MQ in figure).

The limits for z are, then, #iven by the right-hand members in these equations.

Thus we find

(1

/'2 V^ (] - s- )

\(4- 4 x~ - \ y*)dy dx.

., Jo

The limits on this double integral are those for the region OAB, the portion of

the area of the base of the cylinder (9) which lies in the first quadrant. Working

out (11), we find V ~ 4 ?rV2 = 17.77 cubic units. Ans.

The problem given may be such that the first integration should

be performed with respect to x or ?/, and not with respect to z, as

above. The limits must then be determined in accordance with the

preceding discussion.


258. Volumes, using cylindrical coordinates. In many problems in-

volving integration the work is much simplified by employing cy-lindrical coordinates (p, 0, z) as defined in (7), p. 6. The cylindrical

equation of any one of the bounding surfaces may often be written

down directly from its definition. In any case it may be found from

its rectangular equation by the substitution

(1) x p cos 6, y~

p sin 0.

Cylindrical coordinates are especially useful when a boundingsurface is a surface of revolution. For the equation of such a surface,

when the axis is OZ, will have the form z =f(p) ;that is, the coordi-

nate 6 will be absent.

Volume under a surface. Let

(2) z = F(p, 0)

be the cylindrical equation of a surface, as KL in the figure. Wewish to find the volume of the solid bounded above by this surface,

below by the plane XO >',

and laterally by the cylin-

drical surface whose right

section by the plane XOYis the region S. This cylin-

drical surface intercepts on

the surface (2) the region S'.

Divide the solid into ele-

ments of volume as follows :

Divide S into elements of

area AA by drawing radial

lines from and arcs of cir-

cles about 0, as in Art. 254.

Pass planes through the

radial lines and OZ. Pass

cylindrical surfaces of revo-

lution about OZ standing on the circular arcs within S. Then the

solid is divided into columns such as MNPQ, where area MN = AA,and MP = z. The element of volume is then a right prism with base

AA and altitude z. Hence

R

(3) V = z AA.

The volume V is found by summing up the prisms (3) whose bases

lie within S and finding the limit of this sum when the radial lines

and circular arcs within 5 increase in number so that Ap > andA0->0. That is,

AH-O

We now show that the double limit in (4) may be found by suc-

cessive integration. (Compare Art. 244.) This is done by finding

the volume, approximately, of a slice of the solid included between

two radial planes such as ROZ and XOZ, and then taking the limit

of the sum of these slices.

Let DEFG be the section of the solid in the plane ROZ. Thevalues of z along the curve GTF are given by (2) when (= angle XOR)is held fast. In the plane ROZ take OR and OZ as rectangular axes

and (p, z) as coordinates. Let (p, z) be the centroid of area DEFG.Then by (2) and (3), Art. 177,

p- area DEFG =

f'

pz dp =$* pf(p, 0)dp.

The integral will be a function of 0.

Now revolve area DEFG about OZ. By Art. 250, the volume of the

solid of revolution thus generated is 2 ir~p area DEFG. The planes

ROZ and SOZ cut out a wedge from 1 his solid of revolution whose vol-

ume is A0p area DEFG, since angle ROS A0 (radians). Therefore

(5)

is equal, approximately, to the volume of the slice of the solid in-

cluded between the planes ROZ and $OZ. The limit of the sum of

the wedges (5) when A0 > is the exact volume.

Hence

(6) V

where a - Z XOA, = Z A'O/*, pi- O/> =/,(), P2 = OE =/->(0),

values to be found from the polar equations of the curves bounding S.

The element of the integral in (6j, namely,

F(p, e)p<lpd6 = zpdp(W,

may be thought of as the volume of a right prism of altitude z and

base of area p dp d6. Thus AA in (3) is replaced by p Ap A0, as in

Art. 254.

We have now the formula*

(M) V =//*P dp d6=fff(p, fl)P ^P ^

\s s

*The orf/<T of integration is immaterial. Proof is omitted.


X

for the volume under the surface (2), and the limits are found as in

Art. 254 for the area of the region S.

From (M) and (4) we may derive (2), Art. 254.

ILLUSTRATIVE EXAMPLE 1. Show that the volume of the solid bounded bythe ellipsoid of revolution b-(x 2

-f y") -f a-z~ = Q"b~ and the cylindrical surface

x '2 + y*_ ax o js given by

/) r\"K f'U COH

(7) V =*7,f i

o-Jo Jo

Evaluate this integral.

Solution. By (1) the cylindrical equa-tion of the ellipsoid is b~p~ + a-z~ a-b-.

Hence

(8)

The polar equation of the circle

x -i

_j_ y-2_ ax 9 jn the AT-plane

bounding S is, by (1),

(9) p - a cos 6.

For the semicircle the limits for p are

zero and o cos 9, when is held fast, andfor 0, zero and \ TT. Substituting in (M) the value of z from (8) and the above limits,

we get (7). Integrating,

V = a 2&(3 TT - 4)

- 1.20G a'2b.

Volume by triple integration. The clement of volume A V will nowbe 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. Wenow have

(10) AV = AzA4.

By summation and taking the limit when Az--0, Ap 0,

A0 * 0, we have

(N)

for AA may be replaced by p ApM as before.

Formulas (3) in Art. 257 for the centroid become

Vx=fffp

2 cos 6 dz dp d6, Vy ==

fffp2 sin 6 dz dp dO,

Vz^CCCpzdzdpdO,

when cylindrical coordinates are used.


ILLUSTRATIVE EXAMPLE 2. Find the volume of the solid whose upper surface

is on the sphere

(in .r~-fr+2- = 8

and whose lower surface is on the

paraboloid of revolution

(12) .r-4 r' = --.

Solution. The figure showsthe sphere arid the paraboloid in

the first octant. The curve of in-

tersection All lies in the planez = 2. Its projection DK on the

AT-plane is the circle

(13) *-' + r' = 4.

The cylindrical equations

are, by (1) :

(14) p'2

-f z 2 = 8 (the sphere

(ID);

(15) p'2 - 2 2 (the paraboloid

(12));

(1G) p = 2 (the circle (1,T.

An element of area A/1 in the circle 1

(1(1) is drawn at A/(p, 0) in the figure. Anelement of volume AT is shown at /'(/>, 0, z).

We have, by (AO,

r>

/r r2 / v N -p-

I I p c/s (/p dO... Jo J

J, /(

"

The limits are found as follows: Integrating with respect, to z (holding p and

6 fast), we sum up the elements of volume (10) in a column from the surface (15)

to the surface (14) (M/Vt^J/7'i in the figure). From (15), z = MP-> = J p2

;

from (14), 2 = M/'i = V8 -p~, the 2-limits. The limits for p and are those

for the area of the circle (16). Integrating with respect to p gives the sum of the

columns in the slice included between the plane passing through ()Z and OM and

the plane passing through OZ and ON. The final integration sums up these slices.

Integrating in (17),

AUK.

In the following problems, formulas (M) and (AO are to be used

when the equations of bounding surfaces are in cylindrical coordi-

nates (cylindrical equations;. If the corresponding rectangular equa-

tions are needed for drawing a figure, they may be found by the

transformation

(18) p2 =

to which may be added

y'2 6 = arc tan ->

x

(19) sin c; = cos =

PROBLEMS

1. Find the volume of the solid below the cylindrical surface x 2-f z = 4,

above the plane jr -f z = 2, and included between the planes y = 0, y 3.

/!{ /~2 r4 - r 2

j. V / / / d2 dr dy = 13^ cubic units.JO J \J2.~ jr

2. Work out Illustrative Example 2, Art. 247, using cylindrical co-

ordinates. rjTT /*2rtros0p.'$ 3

An.s. I7 = 2 / I

i-~ dp d6 - Tra3 .

Jo Jo a 2

3. Find the volume of the solid bounded above by the cylinder

2 = 4 x 2 and below by the elliptic paraboloid z = 3 .r2

-f- 2/2

.

/"I /"2 x 1- J -' /*4 - J 2

Ans. V 4 // / dzdydx =

J() JO J3 j" -I y-1

4. Two planes forming an angle <* radians with each other meet alonga diameter of a sphere of radius a. Find the volume of the spherical wedgeincluded between the planes arid the spherical surface, using cylindrical

coordinates. Ans. - aa3.

5. Find the volume below the plane z = .r and above the elliptic

paraboloid z = .r24- y'

2. Ans. ^ TT.

6. Work Problem 5 using cylindrical coordinates.

/?, 7T /^COS fl //) COS ^?

= 2/- / /Jo Jo Jp^

7. Find the volume bounded by the sphere p2-f c 2 = a 2 within the

cylinder p = a cos ^. Aw.s-. g a^(w J).

8. Find the volume above c = 0, below the cone c- jc'2-f ?y

2, and

within the cylinder jr'2 + y'

2 2 aj-, using cylindrical coordinates.

A??,s. -^ a3.

9. Find the volume of the solid bounded by z ~ .r -f 1 and 2 z ~ x 2-f y

2.

Ans. | TT.

10. In Problem 3 show that integration with respect to z gives (with-

out further integration) V 4 A ~ 4 I v 1 M \vhere .4 is the area of the

ellipse 4 a*2

-f y'2 = 4, and 7 X and /

tfare moments of inertia for this ellipse

as given by (E), Art. 252.

11. Find the volume below the plane 2 z 4 -f p cos 0, above z 0,

and within the cylinder p = 2 cos 0. Ans. f TT.

12. A solid is bounded by the paraboloid of revolution az = p'2 and the

plane z c. Find the centroid. Ans. (0, 0, c).

13. A solid is bounded by the hyperboloid z 2 = a 2 + p2 and the upper

nappe of the cone z 2 = 2 p2

. Find the volume. Ans. 7ra3 (V2 1).

14. Find the centroid of the solid in Problem 13.

Ans. (0,0, ia(V2+l)).

15. Find the centroid of the solid in Problem 1. An*. (A, , V )

16. Find the centroid of the solid in Problem 2. Ans. ($ a, 0, V}a).

17. Find the centroid of the solid in Problem 8.

18. Find the volume of the solid bounded below by z 0, above by

the cone z = a p, and laterally by p = a cos 0. A NX. T<\. a ;<

(9 TT 16).

19. Find the centroid of the solid in the preceding problem.

20. Find the volume of the solid below the spherical surface pL>

-f z* = 25

and above the upper nappe of the conical surface z = p -f 1.

21. Compare Illustrative Example 3, Art, 165, and Illustrative Ex-

ample 1, Art. 257, and derive (N), Art. 165, from (L), Art. 257.

22. Derive formula (2), Art. 178, from the first formula in (;*), Art. 257.

ADDITIONAL PROBLEMS

1. Find the volume bounded above by the sphere p-' -f z-~

r~, below

by the cone z = p ctn 0, and included between the planes ft,

6 ft -f A#, </>and jtf being acute angles. (The solid is part of a spherical

wedge, like O-.SQA' in the figure of Art. 222 when OQ is drawn.)

An*.;\r !

Atf(l- cos 0).

2. Find (without integration) the volume bounded by the sphere

p^ _|_ Z 2 r i>

tthe cones z - p ctn 0, ,: p ctn (0 -f A0), and the planes

ft,=

ft + A^ f using the result in the preceding ])roblem. CTho solid

is like 0-Pi/iQS in the figure of Art. 222 when OR arid OQ are drawn.)

Xb/x.']

r> A/i sin (0 -f- -\ A0) sin \ A0.

3. Find (without integration) the volume bounded by ;' = p ctn 0,

2 p ctn (0 -f A0), 6 - rf, ft -f A/i and included l>etween the spheres

p'2 _|_ /:2 _ f

j> ^2 _j_ z

*>

(r -f Ar) 1

', using the answ( r in Problem 2.

.4??.s. 2 A/i Ar sin (0 4- A A0) sin \ A0(r- -f r Ar -f ', Ar*).

(The solid is obtained from the figure in Art. 222 by producing each of the

radii OPi, OR, OQ, OS a distance Ar to /',', /'', Q', Sf/ on the sphere

p* -f z 2-

(r 4- Ar)-. The cones intersect this sphere in the circular arcs

JYA" and Q'S', and the planes in the ar^s of great circles /Y,S", H'Q'- The

solid has the vertices PJtQS-Pi'H'Q'H'.)

4. The solid of Problem U is the element of volume AV7 when spherical

coordinates (8), p. 6, are used. Replace ft by 0. Then one vertex P of

AV has the spherical coorclinat.es (r, 0, 0). Prove, from Problem 3,

AVAr->o r

2 sin Ar A0v , o

A./. -

Therefore AT7differs from r2 sin </>

Ar A0 A0 by an infinitesimal of higher

order (Art, 99).

5. In the solid of the preceding problem prove that the edges of AVmeeting at any vertex are mutually perpendicular, and that the lengthsof those intersecting at (r, 0, 0) are, respectively, Ar, r A0, r sin A0.

6. Describe the three systems of surfaces (spheres, cones, planes)which must be drawn to divide a solid R into elements of volume AV(Problem 4) when spherical coordinates are used. Let (r, 0, 0) be anypoint of AV. Then we write

^ V^/^'fr, 0, 0;A V =fffF (r > 0J r ~ sin </> ^r ^0 f/ #-lim

In the left-hand member Al r

may be replaced by r 1 -' sin <^ Ar A< A0(see Problem 4), that is, by the product of the throe edges in Problem 5.

The right-hand member is calculated by successive integration. (Proof

omitted.)

7. Work out the integral in the preceding problem if F(r, </>, 0) = r,

and A' is the sphere r = "2 a cos (/;, that is, .r~ -f /r -f z- 'J a^.

2 7T /^ * TT /^2 <l COS

/'-'

/./o .At

8. Work out the integral in Problem 6 if /^tr, 0, 0) = r-' cos and A'

is the region r 12 a cos 0. /I//*. I;/, 7r<zr'.

CHAPTER XXVI

CURVES FOR REFERENCE

For the convenience of the student a number of the more commoncurves employed in the text are collected here.

CUBICAL PARABOLA SEMICUBICAL PARABOLA

y2 = ax3 .

THE WITCH OF AGNESI THE Cissoin OF DIOCLES

x2y = 4 a2

(2 a y).

2/2(2 a - x) = x3 .


THE LEMNISCATE OF BERNOULLI THE CONCHOID OF NICOMEDES

(x 2 -+ y/2

)

2

- = a- cos 2 0. p a csc + 6.

(In the figure, b

CYCLOID, ORDINARY CASIO

~ a arc vers ~~ V2 a//-

/r

:r = d(0 sn ^),

U ^ a(l cos 0).

CYCLOID, VERTEX AT ORIGIN

y

.r = a arc vers ^ + V2 a?/ y2

a

- sin 0),

- cos 6).

PARABOLA

a cosh - -

a

CURVES FOR REFERENCE 533

HYPOCYCLOID OF Fouu CUSPS ASTROII^ KVOLUTK OF ELLIPSE

r

= a siv .

+ (ft//)"- (a-- ft

2)

~~ a cos'5

0,

/) sin ;<f?.

CARDIOIP

yr + ^^* =p a(l cos 0).

3 axy = 0.

SINE CURVE COSINE CURVE

y

' = sin x. i = COS X.


STROPHOID

Y

p = fr a cos 0.

(in the figure, b - a.)

SPIRAL OF ARCHIMEDES

p = a0.

LOdARITHMIC OR EQUIANGULARSPIRAL

p = c'l(

*, or

log p ad.

HYPERBOLIC OR RECIPROCALSPIRAL

= a.

LITUUS

= a2 .


PARABOLIC SPIRAL LOGARITHMIC CURVE

r

(p- a)- = 4ac0. y = log x.

EXPONENTIAL CURVE PROBABILITY CURVE

v

SECANT CURVE

|37T II

y = sec

TANGENT CURVE

y = tan x.

53G DIFFERENTIAL AND INTEGRAL CALCULUS

THREE-LEAVED ROSE THREE-LEAVED ROSE

p -~ a sin 13 0. p =: a cos 3 6.

FOUR-LEANED ROSE FOUR-LEAVED ROSE

p = a sin 2 p = cos 2 0.

T\VO-LK\VEI> ROSE LEMNISCATE

r

EIGHT-LEAVED ROSE

= a 2 sin 2 5. p = a sin 4 0.


PARABOLA EQUILATERAL HYPERBOLA

x

.r//= a.

INVOLUTE OF A CIRCLE

r cos -f rO sin 6,

r sin rO cos 0.

CHAPTER XXVII

TABLE OF INTEGRALS

Some Elementary Forms

l.fdf(x) =ff'(s)dx=f(s) + C.

2. fa du = aCdu.

u dv dw )= Cdu Cdv Cdw

Rational Forms containing a + bu

See also the Binomial Reduction Formulas 96-104.

U ~ ln (a + ''" )] + C

- f. It v .= r:\-rT- + '"(J (a -f bw) 2

6^La -f ^>w

= a + bu ~ r- 2 a in (a + 6zo -f c.

3 -X

- T-T-T~

rJ (a -f ftw)2 b3

{_a -f

r rfM i r i a

J (a -h 6w)3

^>2L a -f 6w

^2(a -h 6w)

2

C .

IA r <*M _ 1,

^>i /a -f *>M\ , x,

14 -/ "~77 ,

r x= ---

1

--o In

(

-)-f C.

J u 2(a 4- OM) aw a^ \ u I

15 f du = 1 - 1 In ^a + ftM>

\ 4- r

538

TABLE OF INTEGRALS

Rational Forms containing a2

1 , bu= arc tan + r -

539

2 ab \a -

C d = JL in (bu ~ a

\ 4- rJ b'

2 u 2 -a'2 2ab \bu + a'

18. f *(a2

'

(a2

(a2

20(a

2b'2u'

2)'' b 2

(m 2 p -f l)(a'2

b'2u'

2}

l1~

l

b'~(m 2 /> -f 1 ).' (<r b2 u'2 )J>a

J(a'

2 b-u'2 )1

'

2 a'2(p 1 )(a~ b'

2 u 2)v l

_ _L1 L f _ i __,

22. f f ..*!".. .,=^ In

( ..";., .,)+ C.

'

u((i" />*//-) J a*- \a- n*M*/

23. f(/" !

u m (a'2

b'2u'

2)" a-(m ~ l)ir'-

l

(u'2 6 2 M 2

)""!

_ fe2(7w + 2 7)

- 3) / du

a 2 (m -/_ u__

./ v rn -<2

(a'2 b'W) p

'

24 f_-_ -_-

J ur

"(a'2

b'2

i(12

)" 2 a'2 (p- l)u

r"- [

(a'2

b*u*)"

m -f- 2 p 3-f- p r a?/

2 a 2(7>-

1) J u m (a'2

b'2^'2 )

1'- 1

'

Forms containing Va -f- bu

The integrand may be rationalized by setting a + bu = v*. See also

the Binomial Reduction Formulas 96-104.

25. fu V^T^rfK = - 2(2 a ~ 3 6<

jVfl +M*+ C.

*/ lo o 2

26.

27.

28. fMtfM = -

^ Va -f bu

28 a2 ~ 12

105 b 3

2

2(2 a - 6M)V(/ -f bu

3b'2

u'2 du __ 2(8 a 2 - 4 abu + 3 6 2M 2)Va + bu

29-/;

30. f->

31./'-

32.f~

/

oo. f~"

Jt

34-/

:

u r"du 2 K m Va -f 2am

'

+ C.

um~

l du

/a + bu b(2m+l) b(2m+ 1) J \^~+~l

- 4= In ( V "*" "" ~ v^ ) 4- T, for a > 0.

/a 4- bu Va VVa 4- btt + Va(/K

arc tan4-

~ a4- T, for < 0.

dw,

w \/a 4- />M ( ^ v 'w ~ 1 ^ ^" ' 2 a ( r/?

fu + budu tl ,/~~TT~ , /' d?/

a 4-

2 Va 4- ow 4- a f-

(a -f-

u Va 4- ?>M

2 rr/ a 4-

a(m 1 )//"' 1 )

Forms containing Vu2 a2

In this group of formulas we may replace

36.

\n(u 4- V?/ 24- a-) by sinh" 1 -

u 4- Vw 2 a 2) by cosh" 1

a 4- Vu 24- a 2)(j)>y sinh" 1 -

?^

= ~ Vw 2 a 2 ~ln (w. 4- Vw^ia 2) 4- C.

?/ -f 1 4-

38. /'//CM1'

a 2 Y~dn ~ (*r ^ fl2

\" h C,'

?/ 4- 2

// 4- m + 1

= ln (M 4- VM- a j) 4- C.

TABLE OF INTEGRALS 541

4- In (M -f Vw 2 a 2) + r.

i'. _i m n -f 1 J

n m~'2 du

49. fJ

-}/ 2 f >/ -u*(u*<i*)* a'

2 "

50*/~; 7"^ 77

=^aw'* + ^ ln

(

q * v^

+ QW

arc sec --f C.

c , r ^M V?/ 2 a 2 .151.

/

= -f-

r

Jii(w 2 - a 2

)*- M - a

52.C/M 1

a 2 (m --f n

2-1O 2 a 2

)2

53./-C/M

_ J^_-ZLzi:L f.a 2(m l)J

u

\

,

a 2(n 2)?/

m "!

(

m -f n - 3 r

a 2(n- 2)J

"

- 1

a 2)2

du

um (u2 a 2

)2

. A r(u2 + a^du A /~r^ ^ , /a -f Vu 2

-f a 2^ , ^

54. I = Vu 2-f a 2 a In I / -f C.

J M V u /

KK r(u 2 a 2)^du55, / ^ L

t/ 24

66 rluiIJ u*

58.

a'2 (w 1 )' n m~ 2

(u*a*p ,

fl2 w

(n~ TW -f 1 )"r" J w - w -h

Forms containing v a2 i/2

59. f(a2 - w^^r/w = ^ Vr/ 1' - M* -h if arc sin

^-f C.

Tl

60. (,^- *^ du = "(--' )5

+ (z -

61.

/ ., ., n j

62.J

<- -)-d =

-f

63 ^ = arc sin - + C.

(,/2 -,/^ w

r/// ^ -f C.

.>}

r>

C t /" // dn ((J1 M*) "

i r-bO. rr h ( .

'

IL N 2

66 arc sn" a

67. r?<2(///

,=" -

^fa 2 -

?/2

)3 V 2 - w 2

- arc sin - 4- C.

/ua 2 (m 1) rj

2-1 m n + lJ(a

2 - w 2)2

AQ r_irdu_- -um + l m-n + Zr <u^au

69J n- 2-1 a 2(n-2)J il -i

(a2- w 2

)2 a 2(n-2)(a 2 -

i/2)2 (a

2 - w2)2

70.

u(a2

u'2fi

a

, r (in V- u*71

-J uw-*^~~*^~72 . f & . = .

73-/

(In

1 ,,a . ^cosh

~' -

-f C.a-

{

1

a -(m 1 ) //'"

m 4- // '* /'

<r(m - 1 )-'

5 .

r(

.'

2

-f- <///

- 2) J.

!

w m (

a ~""/"

> _ a ln

va j?/

L' a cosh" J --f C.

w

76. arc slna

77.

70(a

2f/

2)

2,

(n m -f 1 jM"'""1 n m +

Forms containing V2 au i/2

The Binomial Reduction Formulas 96-104 may be applied by writing

\/2 au u~ - y}(2 a u}*.

-2i 2-f~ arc + C.

80. V2^3 a 2-f an - 2 w s

,a '3

/i w \ i /-+ arc cos(1 ~-)+C>

. ,,^ 3

8L/>V2au-w2 dw-- m -f-

m + 2

82. f M - M f M _ V2 aw - w 2-f a arc cos

(1 - -) -f C.

J u \ a/

/"\/

*) fj1

}! Ij'^filJ '^ \/'^ fl 11 It '* / 1J\-2

- - ~ - arc cos 1 - - + C.w 2 w \ a/

04 r^/2 aw K 2 dM _ __(2 aw w 2 )^ ,,

f'V2 a?/ w 2 dw (2 aw w 2

)

7 w J /^"v 2 a?/

w m~

a(2 ?w ,S)w'" a (2 w 3)J Mm"

!

86. f y

?/

=== = arc cos (1 -

J V2 aw - u'2 \ a

87. - = In (w -f a -f V2 a?^ -f u 2) -f r.

V2 aw -f- u 2

88. CF(U, V2 a?^ 4- ?/2) r/// = f^fc a, V^ 2

<i'2) dz, where 2 w 4- a.

89.?/(/?/

. f f^ V2 aw - w

QH f"*<*" =

J V2a?/-?/ 2

92./

93.

= - V^CIW-H* 4- a arc cos(1 - ^) + C.

2 v a/

94./

95. r^

V2 aw - w 2

dw.

wV2 aw w 2

dw

7/m\/2 aw w 2

(2 aw - w 2)*

udu

m

V2 au w^-f r.

V2 a?/ i/2 w 1 _ / a

a(2 m - 1 ).' M- ^V2

a 2V2 aw - w 2-f C.

(2 aw w 2)* aV2 aw w 2

-f C.

96. Cu m (aJ-f

Binomial Reduction Formulas

ll m-<]+ 1 (n i

7)7/';V'4 *

dM = --1.

b(pq+ m -f 1)

m

97./um (a + bu qYdu - ^ -~^-

J pq + m + 1

pq -f w -f 1'

8i r tj "

v ?/m

( 1 }um l (n -f btcn*' l

~(/ ~^ y )f

y""" ^

/

'

'^w,

100.rd"

== -Lin /-j^)+c .

c/ M( -f f>" <;> ar/ \a -f bu' 1 '

1A1 f (a 4- bit'1)

1' da __ ( -}- />/rn ;> M

1U1. I

J u m:

a ( m 1 ) //'" l

b(_m __~ </

~-j>(]

1 i ,'(a -f

102 .

r(a + bnWdu _ (// + bit")'1

;>r/ m -f 1

'

//"

u m du //'" (/'

'

103. f

104.

(a -f ^> M </)/' ?>(m - pq + 1 )(a 4- />//(/

)

;'~ '

_ (m (/ -f 1 ) r H"' '' dn

b(m pq -f 1).' (a -f bu v)*

u m dn u7" 4 l

m 4- g-

p(( H- 1

(iq(p 1)'

(a -f

Forms containing a -f bw ci/2(c > 0)

The expression a + &M -f rw 2 may be reduced to a binomial by writing

w = z AC =4 r 2

Then a + 6w -f r?^2 = r(z 2 -

A").i lien u f wo, -f r// - ( \z A;.

The expression a 4- bu cu 2 may be reduced to a binomial by writing

==2: "f

"27'=

4r 2

Then a -f bu - cu 2 - c(k - z 2).

106 .f -^ . = 2

arc tan(2

.

CU + 6

)-f C,J a -h 6w -f rw 2 V4 ac - b 2 \ V4 or - & 2/ wheiwhen 52 < 4 ac<

Cdu = 1

In (2cu + b -^ - 4 ac\ . c

J a + Jm -f cw 2 Vb 2 - 4 a \2 rw -f 6 -f Vfc 2 - 4 or/'


106.^

107. L

[

>2 4 4 ar 2 rw 4 b/

108. CLMJLJ^lll = Min (a + & M ra 2

} }; a -I- 6w cu l 2 r

when b 2 > 4 ac.

* + 4 r + 2rM-^ + c<

(/W

109. fVa -f- bu -f r?/^/?i = ^-i- Va -f bn -f rw 2

./ 4 r4 r

- ^"-iC] n (9 a -f- bu

110. ~cv*du = 2 ^?/~

/; Va4 r

/>2

4- 4 (ic . / 2 r?urc sln

'

- 6

'/;= -Vln (2rw 4-6-f 2

Va -f bu -h rw 2 Vr

112. r-7=J^L== - 4= urc sin( ~r^"~ b= }+ c -

tf Va 4- bu rw 2 Vr v Vh 2 + 4 ar/

// du113. fJ

-f />// 4-

4 ow 4 r?/2 <'

~ln (2 ru 4 /) 4 2 \/rVa 4 few 4 <

-j- ^ ^i r ^ " ?/ _ _ v 4- f>?/.4. I . _J Va 4 ww- rw 2 r

H ~ arc sin /-y >( j

4 C.

Other Algebraic Forms

117.'

118.

it

_-f (a b) log, (Va H- w -f V?> -f M)+ C._

+ M arc sin >- + C.

b) arc sin

f\ T 1/

du Vl M 2 4 arc sin ?^ 4- C.

= 2arcsin x /^-^4C.^w a)(b u)

TABLE OF INTEGRALS

Exponential and Logarithmic Forms

120. Ccan dw = -f T.J a

121. Cbau du = -^-r + C.J a In o

122. Cue*" du = ^ ut - 1) -f C.J a-

123. Cu neau

d?* = Cu" V" du.J a aJ

124. /V'fta " du = ^^ r-r f " !

ft'1 " dw + C.

J (Mn ft (i In ftJ

/"ft"d?/ ft"'

11 a In ft rb"" du

J u" ( H 1 )//"l n IJ u" l

547

105

126. An u du = ?/ In u - u 4- r.

10 - T N i ;127. I u" In ?/ d// M"

l 1)-'4- C.

128. I u"' In" // d?/ =-- In" //----

; I u m In" '

/( d//.J m -f 1 m -f IJ

r , i 7r""

i n ?/ i rr""

;

. I f"3 " In u du --- I dn.

J a aJ n

inn129

130.?/ In ?/

rrln (In M) -h r.

Trigonometric Forms

In forms involving tan //, ctn ?/, sec M, esc //, which do not appear

below, first use the relations

1sin u . cos u 1

tan // = ctn ?/ = -: sec ?/ =cos // sin ?/ cos z/

131. I sin // d// cos u -f- C.

132. I cos ?/ d// = sin // -h (\

133. I tan ?y d?/ = In cos M + C = In sec ?/ -h C.

134. f ctn ?/ d?/ In sin ?/ -f r.

//" d ?^

sec ?/ dw /= In fsec w -f tan w) + C

J cos ?/ 7

esc u =sin

136. /esc u du I .

( ?/ = In (esc w ctn w) -f CJ J sin w

= In tan --f C.

137. /see2 u du = tan u 4- C.

138.J esc 2 M dw = ctn u 4- C.

139. / sec w tan u dv = see w 4- r.

140. / esc u ctn w du esc ?/ 4- C.

141.jsin 2 u du ~ \ u \ sin 2 u + C.

142. / cos 2 u du = ^ M 4- ;Jsin 2 ?/ 4- C.

143. /cos" u sin r< d?,* 4- r.J w 4- 1

144. fsin" u cos w du = - -^4- c:.

/ n 4" 1

145. fsin mn sin WM rfw = - -" (w + " '"+

s

'^(wt ~ n)u + c.

146. fcos m?/ cos n JH = sin (m + w)l<+

sin (m ~"Jf + r .

/ 2(m 4 w) 2(?w w)

147./sin mw cos riw dw = ( QH w "^~ r/ ;/ _ cos ( m ?? ?/

_f_ (\-/ 2(m 4- w) 2(rw ?o

du2 esc a arc tan (tan ?, a tan J w) 4- r.+ cos a cos

149.1cos a 4- cos u \1- tan i a tan I ul

= 2 esc a tanh ]

(tan i a tan A ?/ ) -f r i tan- J u < ctn 2J

= 2 CS a arc tan (CSC fl tan i M + ctn fl) + r -

1. f-^__ = cscaln / tan a- tan.^i/

- secaxJ cos a 4- sin ?/ \tan a 4- tan J ?/ 4- sec a I

[ (ctn a tan // 4- esc a) 2 < 1]= 2 esc a tanh J

(ctn a tan -I u -f esc a } 4- r

[ (ctn a tan \ u -f esc a)2 < 1]

152. f rf ".

= 1Bre tan (Itanj'N a

./ a 2 cos 2?/ 4- h 2 sin 2

?/ a/> \ a /

163. ff sin rf = f" (a sin M " ~ M cos " M) + C.

?2

a cos

a 14- w 2

155.JM sin w dw = sin ?/ u cos M -f C.

156.Jw cos M du = cos w 4- u sin M 4- r.

154. /> cos d = '""

^

Trigonometric Reduction Formulas

/sin*1~

" u cos ?/ it "~ 1 /*smnudu h i sin"~'2 u du.

n n J

ito r j cos"" 1!/ sin M

,r? 1 / _ 9 ,

168. / COS"M du = h I cos"~ 2 w du.J n n J

159 C du = cos Mj_

>? 2 r dj/

V sin" M (n 1 ) sin"~

l u n 1 J sin"~ 2 M

160 . f-rfJL- = smj/ + !L-^ r__cluJ cos" w (n 1 ) cos" ' u n 1 J cos"

" 2?/

161. /cosm M sin"wdz/ = 1

: icos~'2 u sin^wdu.J m -f n m -f n J

162. fcosm u sin" M du = ^ c s-f^-^ fcos

m M sin"~ 2 w du.

J m + n m -f n/

V cosm w sinn M (m IJsin"" 1?^ cos"1

"1

!/

,

"*"

m + n 2 r dun r- 1 J Bm -2 ,m 1 ./ cos 7" z u sin" M

164. f - -

^cosm usinn w (n 1) sin"~ ' u cos*" 1 u

m + n 2 / dw

n 1 J cosm i< sin"~ 2 u

rcosm udu cos"1 ^ 1 u m r? -f- 2 rcoRmudu'J sin" u (n 1 ) sin"

~]

M, n 1 / sin"~ 2ii

IAA rcosm ?/ d/>< ___ cos*"" 1

?/ m 1 rcosm~ 2 u (

V sin"w~"

(m w) sin"" 1

it m nJ sinn u

- fl /sin n M d?i sin" 4 ] w n m + 2 /sinn im + rt* W "J cos'"?/ (m l)cos

m ~1 w m 1 ./ cosm

~ 2 w

16 r^Hllli-^ sin"~

]i/ n 1 rsinn ~ 2 u du

J cosm u (n m) cos"1"

1?/ n mJ cosm u

169. ftan"?/ dw = ^- ftan"~ 2 w d?>t../ n 1 J

170. fctnnw du = - ctnW

""lM _ rctr\ n

~ 2 u du.J n ~ 1 J

171. Ceau coBn udu = ^-^ ^ '

J a z + n z

. n(n 1) / _t n o ,+ ^T^/^ cos wrfM -

172./e-"8inM dM = -"n-'(asin - n COBU)

a + n

173. Cum coa au du = (au sin au 4- m cos au)v a

777(771 1 )/"*, o j, L I um ~ 2 cos au aw.a J

174./Mm sin aw dw = ~ (m sin a?v au cos a?/)

*/ a

in (m 1 ) r m . r,.

,___ / M w ^ sln aw C/ Wy-

Inverse Trigonometric Functions

175.J arc sin w dz/ = M arc sin u 4- Vl w 2 4 C.

176.Jarc cos u du ?/ arc cos w Vl u* 4 C.

177. farc tan ?/ r/// = ?> arc tan w In Vl -f u'2 4- r.

178. / arc ctn u du ?/ arc ctn ?/ 4 In Vl 4 u'z 4 r.

179. / arc sec */ dw -- u arc sec // In ( u 4 vV-' 1 ) 4 C.

= u arc sec // cosh ]// 4 (*.

180. I arc esc u du ~ n arc esc ?/ 4- In ( // 4- Vw J1 ) 4- C

u arc esc u 4- coh J?/ 4 r.

Hyperbolic Functions

181. / sinh ?/ d?^~ cosh // 4- C\

182. I cosh ?/ d?/ = sinh tt -\- ('.

183.|tanli // du In cosh w -}- C.

184.jctnh // d>'/ = In sinh u 4- (\

185. | sech M </// = arc tan (sinh ?/) -f C = gd M 4 C

186. /"cschM du = In tanh J M 4- C.

187.j"sechL> M du = tanh ?^ 4- C.

188. |csch'J

?f dw = ctnh ?/ 4- C.

189.|sech w tanh u du sech n 4- C.

190. I csch u ctnh u du = csch u -f C.

191.jsinh* u du \ sinh "2 n \ u -f C.

192. / cosh 2 u <ln = i sinh 2 M -f J u -f f.

193. / tanh- M <i?/ = M tanh u -f (\

194. I ctnh 2 u du u ctnh u -f C.

195.j

?/ sinh M du = u cosh //- sinh u -f f.

196. I n cosh M du u sinh >/ cosh // ~h r.

/>

___sinh ' u du = u sinh ' u VlH- n~ + (\

198.jcosh * u du ?/ cosh '

// vV- 1 -f r\

199. f tanh J

// c/// = M tanh '

// -f \ In (1- u :

) -f C.

200. fctnh '

i/ du u ctnh '

// -f J In (1 ?/-) -f r.

201. I sech ' M <^/^ = n sech '

// -f g \

203. / sinh m?-< sinh nu du ^~-- r-- ~r.

--;

---h < . (m^n\J Km + u) l(m-~ n) \ /

n ~ A ft, u ;sinh fw -f ?/)w . sinh (m u)u . ,, ( > \

204. / cosh mu cosh nu du = --- -- ----\- ( .

[m ^ n

Jll(m+ u) 'Km - n) \

^/

206

nn - r . , u 7 cosh (w-f ?/)M , f'osh (?w n ) u , ,, / > \205. / sinh ?^^?/ cosh nu du ~ -} i

-\-- --' f (\ m ^ n

J 'Km + n] 2(w-r/) \<

/

. f :-r = 2 csch a tanh

-' (tanh .', ?/ tanh i a) -f r:.

J cosh a -f cosh u"

207.. f- . - 2 esc a arc tan ftanh J ?/ tan i a) + r../ cos a -f cosh u

208. f- -- = 2 esc a tanh ]

(tanh J ?/ tan i a) 4 r:

Jl + cosacosh,(tanh*i W <ctn*Ja)

209. f- sinh m/ du =-

c<J a j n 2

rt , rt r , , pa"(a cosh TIU n sinh nw) , ~,

210. I ra '^ cosh n?( du =-------f C.

J a 2 n 2

INDEX

(The numbers refer to pages)

Absolute convergence, 348

Acceleration, curvilinear motion,121

;rectilinear motion, 83

Adiabatic law, 70

Agnesi, witch of, 531

Anchor ring, 267

Angle of intersection, of plane

curves, 43; polar form of, 126

;

of skew curves, 475;

of surfaces,

481

Approximate formulas, 367, 372,

490

Arc, centroid of, 335;

differential

of, 142, 144, 473; length of, plane

curve, 271;skew curve, 473

Archimedes, 127, 128, 155, 277, 534

Area, of a curved surface, 517;mo-

ment of, 320, 503, 514;moment

of inertia of, 508, 514; plane, 241,

258, 498 ; in polar coordinates,

262, 512 ;of a surface of revolu-

tion, 277

Astroid, 119, 533

Auxiliary equation, 390

Bending, direction of, 75

Bernoulli, lemniscate of, 532

Binomial differentials, 299, 307

Binomial theorem, 1, 353

Boyle's law, 70

Calculation, of e, 361 ; of loga-

rithms, 362;

of TT, 366

Cardioid, 117, 119, 125, 135, 145,

155, 244, 271, 275, 281, 323, 335,

516, 517, 533

Catenary, 152, 270, 276, 282, 423,

432, 434, 532

Cauchy, 345

Center of fluid pressure, 506

Centroid, of a homogeneous solid,

522, 526; of a plane area, 320,

337, 503;of a solid of revolution,

323

Change of variable, 166, 240, 457

Cissoid, 44, 46, 270, 277, 322, 531

Complementary function, 394

Complex number, 440

Compound-interest law, 399

Conchoid, 532

Conoid, 284

Constant, 7; absolute, 7; arbi-

trary, 7;of integration, 189, 229,

233, 376 ; numerical, 7

Continuity of functions, 12, 444

Convergence, 340

Coordinates, cylindrical, 6, 524-

527; polar, 123; spherical, 6,

529, 530

Cosine curve, 533

Critical values, 52

Cubic, skew, 474

Curvature, 149; center of, 157,

171; circle of, 153, 170; radius

of, 152

Curve-tracing, 81

Curvilinear motion, 120, 146

Cycloid, 116, 119, 144, 151, 161,

244, 270, 274, 276, 281, 532

Cylindrical coordinates, 6, 524-527

Derivative, definition, 21; inter-

pretation of, by geometry, 25,446

; partial, 445; as a rate, 64 ;

symbols for, 22, 445 ; total, 455 ;

transformation of, 166

Descartes, folium of, 46, 119, 288,533

553

554 INDEX

Differential, 136; of arc, 142, 144,

473; of area, 237; formulas for,

140; geometric interpretation,

137, 477; as an infinitesimal,

146; total, 449

Differential coefficient, 21, 136

Differential equations, applicationsto mechanics, 402

; definitions,

375 ; first order, 378; higher

order, 387; homogeneous, 380

;

linear, 383, 390, 407

Differentiation, 22;

formulas for,

28, 29, 86, 87, 115, 119, -120, 425,

426, 435; general rule for, 23

; of

implicit functions, -40, 73, 154,

458; logarithmic, 93; partial,

445, 462; successive, 73, 462

Diocles, cissoid of, 44, 46, 270, 277,

322, 531

Direction of a curve, 42

Ellipsoid, 280, 285, 522

Envelopes, 466

Equations, graphical solution, 128;

interpolation, 129 ;of motion,

120; Newton's method, 131

Errors, 138, 451; percentage, 138;

relative, 138

Evolute, 158, 469; of the cycloid,

161 ; of the ellipse, 160, 533; of

the parabola, 159, 470; proper-ties of, 1 62

Exponential curve, 89, 535

Exponential function, 89

Factorial number, 338

Family of curves, 230, 466

Fluid pressure, 325; center of, 506

Fluxions, 19, 357

Folium of Descartes, 46, 119, 288,

533

Formulas, approximate, 367, 372,

490;

for reference, 1-6

Fourier, 238

Function, complementary, 394 ;

continuity of, 12, 444 ; decreas-

ing, 50; definition of, 8, 444

;

derived, 21; differentiate, 21,

23 ; discontinuous, example of,

108; exponential, 89 ; of a func-

tion, 37; graph of, 10, 444; hy-

perbolic, 414, 415; implicit, 39,

73,458; increasing, 50 ; inverse,

38; inverse hyperbolic, 423

;in-

verse trigonometric, 105 ; loga-

rithmic, 89; mean value of, 333

;

periodic, 97;of several variables,

444; sine, 97 ;

table of hyper-bolics, 416 ; transcendental, 86

;

trigonometric, 99

Fundamental theorem of the in-

tegral calculus, 254-257

Graph of a function, 10, 444

Gravity, center of, 320, 323

Greek alphabet, 6

Gudermann, 435

Gudermannian, 435; inverse, 435

Gyration, radius of, 508

Harmonic vibration, 403, 405

Helix, circular, 473, 474, 482

Homer's method, 130

Hyperboloid, 528

Hypocycloid, 46, 119, 156, 244, 268,

270, 276, 280, 288, 468, 533

Increments, 19; approximation of,

137, 451

Indeterminate forms, 174

Inertia, moment of, 508, 514

Infinitesimals, 17; replacement

theorem, 147

Infinity, 13

Inflectional points, 79

Initial conditions, 229

Integrals, 188 ; change in limits,

240 ; decomposition of interval,

250 ; definite, 237 ; discontinu-

ous, 251 ; geometric representa-

tion, 244, 492; improper, 250

253 ; indefinite, 189 ; interchangeof limits, 249 ; multiple, 491 ;

table of, 538 ; use of table, 315

INDEX 555

Integrand, 195

Integration, 187; approximate, 245 ;

of binomial differentials, 299;formulas for, 191 193, 430, 432,

433, 435 ; fundamental theorem

of, 254 ; by mlsceJhmpotrs subsU-

tutions, 221, 305,J32 ; by parts,

223; by raTional^fracIions, 289 ;

by rationalization, 221, 296; by

reciprocal substitution, 305; byreduction formulas, 307, 312;

successive, 491; of trigonometric

forms, 213, 303, 312

Interpolation, 129, 372

Interval of a variable, 7

Involute, 163 ; of a circle, 156, 276,

288, 537

Isothermal expansion, 330

Jacobi, 445

Laplace, 19

Laws of the mean, 172, 182, 482

Leibnitz, 26

Lemniscate, 127, 155, 263, 515, 516,

517, 532

Length of arc, plane curves, 271 ;

in polar coordinates, 274;

of

skew curves, 473

Limac.on, 534

Limit of a variable, 10

Limits, change in, 240;

of an inte-

gral, 238; theorems on, 11, 17

Lituus, 534

Logarithmic curve, 89, 535

Logarithmic differentiation, 93

logarithmic function, 89

Logarithms, common, 88; natural,

87

Loxodrome, 439

Maclauriri's series, 357, 367

Maxima and minima, 47; analytic

treatment, 182; definitions, 52 ;

first method, 53; functions of two

variables, 483 ; second method,76

Mean value, extended theorem of,

182; of a function, 333 ; theo-

rems of, 172, 482

Mechanics, 402

Mercator, 439 ; chart, 439

Moment, of area, 320, 503; of in-

ertia, 507, 508, 514; polar, 510,

514

Motion, curvilinear, 120;

recti-

linear, 65, 83

Napierian logarithms, 88

Xewton, 19, 27, 131, 132, 133, 332,

401

Nicomedes, conchoid of, 532

Normal, to a plane curve, 43 ; planeto a skew curve, 471, 480; to a

surface, 475

Osculating circle, 170

Pappus, theorems of, 335, 504

Parabola, 532, 537; cubical, 156,

531; semicubical, 266, 498, 531

Parabolic rule, 247

Paraboloid of revolution, 268, 521,

528

Parameter, 1 15, 466

Parametric equations, 115; first

derivative, 115; second deriva-

tive, 119

Point of inflection, 79

Polar coordinates, 123;moment of

inertia, 514; subnormal, 126;

subtangent, 126

Power rule, 32

Pressure, fluid, 325;

center of,

506

Probability curve, 535

Projectile, 121, 234, 471

Quadratic equation, 1

Radius, of curvature, 152; of

gyration, 508

Railroad curves, 152

Rates, 64, 455

556 INDEX

Rectification, of plane curves, 271 ;

in polar coordinates, 274 ;of

skew curves, 473

Reduction formulas, 307, 312

Replacement theorem, 147

Rhumb line, 439

Rolle's theorem, 169

Roots of equations, 128

Rose-leaf curves, 536

Secant curve, 535

Sequence, 338

Series, 338 ; absolute convergence,

348; alternating, 347; approxi-mate formulas from, 367, 372

;

binomial, 353; Cauchy's test,

345; comparison tests, 342

;con-

vergent ,340

;differentiation and

integration of, 365 ; divergent,

340; geometric, 339 ; harmonic,

342; Maclaurin's, 357; opera-tions with, 362

; oscillating, 339;

p, 344; power, 350, 354;

Taylor's, 369, 488

Simpson's rule, 247

Sine curve, 533

Skew curves, 471, 480; length of,

473

Slope of a curve, 42 ; parametric

form, 115; polar form, 125

Solids of revolution, centroid of,

323 ; surface of, 277;volume of,

265, 267, 268

Speed, 121, 146

Spherical coordinates, 6, 529, 530

Spheroid, oblate, 266; prolate,

266

Spiral, of Archimedes, 127, 128, 155,

277, 534 ; hyperbolic or recip-

rocal, 119, 128, 264, 277, 534;

logarithmic or equiangular, 127,

128, 534 ; parabolic, 535

Stirling, 357

Strophoid, 534

Subnormal, 43; polar, 126

Subtangent, 43 ; polar, 126

Successive differentiation, 73, 462

Successive integration, 491

Table, of hyperbolic functions, 416;

of integrals, 538

Tangent, horizontal, 42, 117; to a

plane curve, 43;to a skew curve,

471, 480; plane to a surface, 475 ;

vertical, 42, 117

Tangent curve, 535

Taylor's theorem, 369, 488

Telegraph line, 428-429

Torus, 267

Tractrix, 85, 270, 282, 419, 423, 436,

537

Transformation of derivatives, 166

Transition curves, 152

Trapezoidal rule, 245

Triple integration, 521

Trisectrix, 155, 264

Variable, change of, 166, 457;

defi-

nition, 7; dependent, 8

;inde-

pendent, 8

Velocity, curvilinear motion, 120,

146; rectilinear motion, 65

Vibration, damped harmonic, 405;

forced harmonic, 405; simple

harmonic, 403

Volume, of a hollow solid of revo-

lution, 267;

of a solid with

known cross section, 283;

of a

solid of revolution, 265, 267;under a surface, 501, 524; bytriple integration, 521, 526

Witch, 62, 251, 269, 322, 531

Work, 328

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elements of the differential and integral calculus

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